Class 11

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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9

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Resolve the following rational expressions into partial fractions

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Equating nuemarator on bothsides we get

Question 6.

Solution:

Equating numerator on both sides
(x – 2)² = A(x² + 1) + (Bx + c)(x)
Put x = 0
1 = A
Equating co-eff of x²
1 = A + B
(i.e.,) 1 + B = 1 ⇒ B = 0
put x = 1
A(2) + B + C = 0 (i.e.,) 2A + B + C = 0
2 + 0 + C = 0 ⇒ C = -2

Question 7.

Solution:
Since numerator and denominator are of same degree
we have divide the numerator by the denominator

Substituting the value in ….(1)

Question 8.

Solution:
Numerator is of greater degree than the denominator
So dividing Numerator by the denominator

⇒ 21x + 31 = A(x + 3) + B(x + 2)
Put x = -3
-63 + 31 = B(-1)
B = 32
Put x = -2
-42 + 31 = A(1) + B(0)
A = -11

Question 9.

Solution:

Question 10.

Solution:

Equating Numerator on both sides we get
6x² – x + 1 = A(x² + 1) + (Bx + c)(x + 1)
6 + 1 + 1 = A(2) + 0 ⇒ 2A = 8 ⇒ A = 4
Equating co-eff of x²
6 = A + B
(i.e.,) 4 + B = 6 ⇒ B = 6 – 4 = 2
put x = 0
1 = A+ C
4 + C = 1 ⇒ C = 1 – 4 = -3

Question 11.

Solution:
Since Numerator and are of same degree divide Numerator by the denominator

equating Numerator on both sides we get
x – 5 = A(x + 3) + B(x – 1)
Put x = -3
-3 -5 = A(0) + B(-4)
-4B = -8 ⇒ B = 2
Put x = 1
1 – 5 = A(4) + B(0)
4A = -4 ⇒ A = -1

Question 12.

Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9 Additional Questions

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Equating nemerator on b/s
9 = A(x+2)² + B(x – 1)(x + 2) + C(x – 1)
Put x = -2
9 = A(0) + B(0) + C(-3)
-3C = 9 ⇒ C = -3
Put x = 1
9 = A (1 + 2)² + B (0) + C(0)
9A = 9
A = 1
Put x = 0
9 = 4A – 2B – C
9 = 4(1) – 2B + 3
9 – 7 = -2B
2 = -2B
B = -1

Question 4.

Solution:

Question 5.

Solution:

0 = 0 + B(1 + 2)
3B = 0 ⇒ B = 0
Put x = -2
(-2)³ – 1 = A(-2 – 1) + B(0)
-8 – 1 = -3A
-9 = -3A
A = 9/3 ⇒ A = 3

Students can Download Physics Chapter 5 Motion of System of Particles and Rigid Bodies Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Physics Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 5 Motion of System of Particles and Rigid Bodies

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Textual Questions Solved

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Multiple Choice Questions

Question 1.
The center of mass of a system of particles does not depend upon,
(a) position of particles
(b) relative distance between particles
(c) masses of particles
(d) force acting on particle
Answer:
(d) force acting on particle

Question 2.
A couple produces, [AIPMT 1997, AIEEE 2004]
(a) pure rotation
(b) pure translation
(c) rotation and translation
(d) no motion [AIPMT 1997]
Answer:
(a) pure rotation

Question 3.
A particle is moving with a constant velocity along a line parallel to positive X – axis. The magnitude of its angular momentum with respect to the origin is –
(a) zero
(b) increasing with x
(c) decreasing with x
(d) remaining constant [IIT 2002]
Answer:
(d) remaining constant

Question 4.
A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force 30 N?
(a) 0.25 rad s^-2
(b) 25 rad s^-2
(c) 5 m s^-2
(d) 25 m s^-2
[NEET 2017]
Answer:
(b) 25 rad s^-2

Question 5.
A closed cylindrical container is partially filled with water. As the container rotates in a horizontal plane about a perpendicular bisector, its moment of inertia,
(a) increases
(b) decreases
(c) remains constant
(d) depends on direction of rotation. [IIT 1998]
Answer:
(a) increases

Use this kinetic energy calculator to help you find out the energy of an object in motion. This KE calculator makes use of the kinetic energy formula

Question 6.
A rigid body rotates with an angular momentum L. If its kinetic energy is halved, the angular momentum becomes,
(a) L
(b) L / 2
(c) 2 L
(d) L / 2 [AFMC 1998, AIPMT 2015]
Answer:
(d) L / 2

Question 7.
A particle undergoes uniform circular motion. The angular momentum of the particle remain conserved about –
(a) the center point of the circle.
(b) the point on the circumference of the circle
(c) any point inside the circle.
(d) any point outside the circle. [IIT 2003]
Answer:
(a) the center point of the circle.

Question 8.
When a mass is rotating in a plane about a fixed point, its angular momentum is directed along –
(a) a line perpendicular to the plane of rotation
(b) the line making an angle of 45° to the plane of rotation
(c) the radius
(d) tangent to the path [AIPMT 2012]
Answer:
(a) a line perpendicular to the plane of rotation

Question 9.
Two discs of same moment of inertia rotating about their regular axis passing through center and perpendicular to the plane of disc with angular velocities ω₁ and ω₁. They are brought in to contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is-
(a) \(\frac {1}{4}\) I(ω₁ – ω₂)ω²
(b) I(ω₁ – ω₂)ω₂
(c) \(\frac {1}{8}\) I(ω₁ – ω₂)ω²
(d) \(\frac {1}{2}\) I(ω₁ – ω₂)ω²
Answer:
(a) \(\frac {1}{4}\) I(ω₁ – ω₂)ω²

Question 10.
A disc of moment of inertia I_a is rotating in a horizontal plane about its symmetry axis with a constant angular speed to. Another disc initially at rest of moment of inertia I_b is dropped coaxially on to the rotating disc. Then, both the discs rotate with same constant angular speed. The loss of kinetic energy due to friction in this process is-

Answer:

Question 11.
The ratio of the acceleration for a solid sphere (mass m and radius R) rolling down an incline of angle 0 without slipping and slipping down the incline without rolling is –
(a) 5 : 7
(b) 2 : 3
(c) 2 : 5
(d) 7 : 5
[AIPMT 2014]
Answer:
(a) 5 : 7

Question 12.
From a disc of radius R a mass M, a circular hole of diameter R, whose rim passes through the center is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis passing through it?
(a) 15MR²/32
(b) 13MR²/32
(c) 11MR²/32
(d) 9MR²/32 [NEET 2016]
Answer:
(b) 13MR²/32

Question 13.
The speed of a solid sphere after rolling down from rest without sliding on an inclined plane of vertical height h is,
(a) \(\sqrt{\frac{4}{3} g h}\)
(b) \(\sqrt{\frac{10}{7} g h}\)
(c) \(\sqrt{2gh}\)
(d) \(\sqrt{\frac{1}{2} g h}\)
Answer:
(a) \(\sqrt{\frac{4}{3} g h}\)

Question 14.
The speed of the center of a wheel rolling on a horizontal surface is vQ. A point on the rim in level with the center will be moving at a speed of speed of,
(a) zero
(b) v₀
(c) \(\sqrt{2}\)v₀
(d) 2 v₀
[PMT 1992, PMT 2003, IIT 2004]
Answer:
(c) \(\sqrt{2}\)v₀

Question 15.
A round object of mass m and radius r rolls down without slipping along an inclined plane. The fractional force,
(a) dissipates kinetic energy as heat.
(b) decreases the rotational motion.
(c) decreases the rotational and transnational motion ,
(d) converts transnational energy into rotational energy [PMT 2005]
Answer:
(d) converts transnational energy into rotational energy

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Short Answer Questions

Question 1.
Define center of mass.
Answer:
The center of mass of a body is defined as a point where the entire mass of the body appears to be concentrated.

Question 2.
Find out the center of mass for the given geometrical structures.
(a) Equilateral triangle
(b) Cylinder
(c) Square
Answer:

(a) For equilateral triangle, center of mass lies at its centro-id.
(b) For cylinder, center of mass lies at its geometrical center.
(c) For square, center of mass lies at the point where the diagonals meet.

Question 3.
Define torque and mention its unit.
Answer:
Torque is defined as the moment of the external applied force about a point or axis of rotation. The expression for torque is,
\(\vec{\tau}\) = \(\vec{r}\) x \(\vec{F}\)

Question 4.
What are the conditions in which force cannot produce torque?
Answer:
The forces intersect (or) passing through the axis of rotation cannot produce torque as the perpendicular distance between the forces is 0 i.e. r = 0.
∴ \(\vec{\tau}\) = \(\vec{r}\) x \(\vec{F}\) = 0

Question 5.
Give any two examples of torque in day – to – day life.
Answer:

• The opening and closing of a door about the hinges.
• Turning of a nut using a wrench.

Question 6.
What is the relation between torque and angular momentum?
Answer:
We have the expression for magnitude of angular momentum of a rigid body as, L = I ω. The expression for magnitude of torque on a rigid body is, τ = I α.
We can further write the expression for torque as,
τ = I\(\frac {dω}{dt}\) (∴ α = \(\frac {dω}{dt}\))
Where, ω is angular velocity and α is angular acceleration. We can also write equation,
τ = \(\frac {d(Iω)}{dt}\)
τ = \(\frac {dL}{dt}\)

Question 7.
What is equilibrium?
Answer:
A rigid body is said to be in mechanical equilibrium where both its linear momentum and angular momentum remain constant.

Question 8.
How do you distinguish between stable and unstable equilibrium?
Answer:
Stable Kquilibrium:

• The body tries to come back to equilibrium if slightly disturbed and released.
• The center of mass of the body shifts slightly higher if disturbed from equilibrium.
• Potential energy of the body is minimum and it increases if disturbed.

Unstable Equilibrium:

• The body cannot come back to equilibrium if slightly disturbed and released.
• The center of mass of the body shifts slightly lower if disturbed from equilibrium.
• Potential energy of the body is not minimum and it decreases if disturbed.

Question 9.
Define couple.
Answer:
A pair of forces which are equal in magnitude but opposite in direction and separated by a perpendicular distance so that their lines of action do not coincide that causes a turning effect is called a couple.

Question 10.
State principle of moments.
Answer:
Principle of moment states that when an object is in equilibrium the sum of the anticlockwise moments about a point is equal to the sum of the clockwise moments.

Question 11.
Define center of gravity.
Answer:
The center of gravity of a body is the point at which the entire weight of the body acts, irrespective of the position and orientation of the body.

Question 12.
Mention any two physical significance of moment of inertia
Answer:
Moment of inertia for point mass,
I = \(m_{i} r_{i}^{2}\)
Moment of inertia for bulk object,
I = ∑\(m_{i} r_{i}^{2}\)

Question 13.
What is radius of gyration?
Answer:
The radius of gyration of an object is the perpendicular distance from the axis of rotation to an equivalent point mass, which would have the same mass as well as the same moment of inertia of the object.

Question 14.
State conservation of angular momentum.
Answer:
The law of conservation of angular momentum states that when no external torque acts on the body the net angular momentum of a rotating rigid body remains constant.

Question 15.
What are the rotational equivalents for the physical quantities, (i) mass and (ii) force?
Answer:
The rotational equivalents for (i) mass and (ii) force are moment of inertia and torque respectively.

Question 16.
What is the condition for pure rolling?
Answer:
In pure rolling, there is no relative motion of the point of contact with the surface when the rolling object speeds up or shows down. It must accelerate or decelerate respectively.

Question 17.
What is the difference between sliding and slipping?
Sliding:

• Velocity of center of mass is greater than Rω i.e. V_CM > Rω.
• Velocity of transnational motion is greater than velocity of rotational motion.
• Resultant velocity acts in the forward direction.

Slipping:

• Velocity of center of mass is lesser than Rω. i.e. V_CM < Rω
• Velocity of translation motion is lesser than velocity of rotational motion.
• Resultant velocity acts in the backward direction.

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Long Answer Questions

Question 1.
Explain the types of equilibrium with suitable examples.
Answer:

• Transnational motion – A book resting on a table.
• Rotational equilibrium – A body moves in a circular path with constant velocity.
• Static equilibrium – A wall – hanging, hanging on the wall.
• Dynamic equilibrium – A ball decends down in a fluid with its terminal velocity.
• Stable equilibrium – A table on the floor
• Unstable equilibrium – A pencil standing on its tip.
• Neutral equilibrium – A dice rolling on a game board.

Question 2.
Explain the method to find the center of gravity of a irregularly shaped lamina.
Answer:
There is also another way to determine the center of gravity of an irregular lamina. If we suspend the lamina from different points like P, Q, R as shown in figure, the vertical lines I PP’, QQ’, RR’ all pass through the center of gravity. Here, reaction force acting at the point of suspension and the gravitational force acting at the center of gravity cancel each other and the torques caused by them also cancel each other.

Determination of center of gravity of plane lamina by suspending

Question 3.
Explain why a cyclist bends while negotiating a curve road? Arrive at the expression for angle of bending for a given velocity.
Answer:
Let us consider a cyclist negotiating a circular level road (not banked) of radius r with a speed v. The cycle and the cyclist are considered as one system with mass m. The center gravity of the system is C and it goes in a circle of radius r with center at O. Let us choose the line OC as X – axis and the vertical line through O as Z – axis as shown in Figure.

The system as a frame is rotating about Z – axis. The system is at rest in this rotating frame. To solve problems in rotating frame of reference, we have to apply a centrifugal force (pseudo force) on the system which will be \(\frac{m v^{2}}{r}\) This force will act through the center of gravity. The forces acting on the system are,

• gravitational force (mg)
• normal force (N)
• frictional force (f)
• centrifugal force (\(\frac{m v^{2}}{r}\)).

As the system is in equilibrium in the rotational frame of reference, the net external force and net external torque must be zero. Let us consider all torques about the point A in Figure.
For rotational equilibrium,
τ_net = 0
The torque due to the gravitational force about point A is (mg AB) which causes a clockwise turn that is taken as negative. The torque due to the centripetal force is I BC which causes an (\(\frac{m v^{2}}{r}\) BC) Which causes an anticlockwise turn that is taken as positive.


While negotiating a circular level road of radius r at velocity v, a cyclist has to bend by an angle 0 from vertical given by the above expression to stay in equilibrium (i.e. to avoid a fall).

Question 4.
Derive the expression for moment of inertia of a rod about its center and perpendicular to the rod.
Answer:
Let us consider a uniform rod of mass (M) and length (l) as shown in figure. Let us find an expression for moment of inertia of this rod about an axis that passes through the center of mass and perpendicular to the rod. First an origin is to be fixed for the coordinate system so that it coincides with the center of mass, which is also the geometric center of the rod. The rod is now along the x axis. We take an infinitesimally small mass (dm) at a distance (x) from the origin. The moment of inertia (dI) of this mass (dm) about the axis is, dI = (dm) x²

As the mass is uniformly distributed, the mass per unit length (λ) of the rod is, λ = \(\frac {M}{l}\)
The (dm) mass of the infinitesimally small length as, dm = λ dx = \(\frac {M}{l}\) dx
The moment of inertia (I) of the entire rod can be found by integrating dl,

As the mass is distributed on either side of the origin, the limits for integration are taken from to – l/2 to l/2.

Question 5.
Derive the expression for moment of inertia of a uniform ring about an axis passing through the center and perpendicular to the plane.
Answer:
Let us consider a uniform ring of mass M and radius R. To find the moment of inertia of the ring about an axis passing through its center and perpendicular to the plane, let us take an infinitesimally small mass (dm) of length (dx) of the ring. This (dm) is located at a distance R, which is the radius of the ring from the axis as shown in figure.
The moment of inertia (dl) of this small mass (dm) is,
dI = (dm)R²
The length of the ring is its circumference (2πR). As the mass is uniformly distributed, the mass per unit length (λ) is,
λ = \(\frac {mass}{lengh}\) = \(\frac {M}{2πR}\)
The mass (dm) of the infinitesimally small length is,
dm = λ dx = \(\frac {M}{2πR}\) dx
Now, the moment of inertia (I) of the entire ring is,
To cover the entire length of the ring, the limits of integration are taken from 0 to 2πR.

Question 6.
Derive the expression for moment of inertia of a uniform disc about an axis passing through the center and perpendicular to the plane.
Answer:
Consider a disc of mass M and radius R. This disc is made up of many infinitesimally small rings as shown in figure. Consider one such ring of mass (dm) and thickness (dr) and radius (r). The moment of inertia (dl) of this small ring is,
dI = (dm)R²
As the mass is uniformly distributed, the mass per unit area (σ) is σ = \(\frac {mass}{area}\) = \(\frac{M}{\pi R^{2}}\)
The mass of the infinitesimally small ring is,
dm = σ 2πr dr = \(\frac{\mathrm{M}}{\pi \mathrm{R}^{2}}\) 2πr dr
where, the term (2πr dr) is the area of this elemental ring (2πr is the length and dr is the thickness), dm = \(\frac{2 \mathrm{M}}{\mathrm{R}^{2}}\) r dr
dI = \(\frac{2 \mathrm{M}}{\mathrm{R}^{2}}\) r³ dr
The moment of inertia (I) of the entire disc is,

Question 7.
Discuss conservation of angular momentum with example.
Answer:
When no external torque acts on the body, the net angular momentum of a rotating rigid body remains constant. This is known as law of conservation of angular momentum.
τ = \(\frac {dL}{dt}\)
If τ = 0 then, L = constant.
As the angular momentum is L = Iω, the conservation of angular momentum could further be written for initial and final situations as,
I_iω_i = I_iω_i (or) Iω = constant
The above equations say that if I increases ω will decrease and vice – versa to keep the angular momentum constant.

There are several situations where the principle of conservation of angular momentum is applicable. One striking example is an ice dancer as shown in Figure A. The dancer spins slowly when the hands are stretched out and spins faster when the hands are brought close to the body.

Stretching of hands away from body increases moment of inertia, thus the angular velocity decreases resulting in slower spin. When the hands are brought close to the body, the moment of inertia decreases, and thus the angular velocity increases resulting in faster spin. A diver while in air as in Figure B curls the body close to decrease the moment of inertia, which in turn helps to increase the number of somersaults in air.

Question 8.
State and prove parallel axis theorem.
Answer:
Parallel axis theorem:
Parallel axis theorem states that the moment of inertia of a body about any axis is equal to the sum of its moment of inertia about a parallel axis through its center of mass and the product of the mass of the body and the square of the perpendicular distance between the two axes.

If I_C is the moment of inertia of the body of mass M about an axis passing through the center of mass, then the moment of inertia I about a parallel axis at a distance d from it is – given by the relation,
I = I_C + M d²
Let us consider a rigid body as shown in figure. Its moment of inertia about an axis AB passing through the center of mass is I_C. DE is another axis parallel to AB at a perpendicular distance d from AB. The moment of inertia of the body about DE is I. We attempt to get an expression for I in terms of I_C. For this, let us consider a point mass m on the body at position x from its center of mass.

The moment of inertia of the point mass about the axis DE is, m (x + d)². The moment of inertia I of the whole body about DE is the summation of the above expression.
I = ∑ m (x + d)²
This equation could further be written as,
I = ∑ m(x² + d² + 2xd)
1= ∑ (mx² + md² + 2 dmx)
l = ∑ mx² + md² + 2d ∑ mx
Here, ∑ mx² is the moment of inertia of the body about the center of mass. Hence,I_C = ∑ mx²
The term, ∑ mx = 0 because, x can take positive and negative values with respect to the axis AB. The summation (∑ mx) will be zero.
Thus, I = I_C + ∑ m d² = I_C + (∑m) d²
Here, ∑ m is the entire mass M of the object (∑ m = M).
I = I_C + Md²

Question 9.
State and prove perpendicular axis theorem.
Answer:
Perpendicular axis theorem:
This perpendicular axis theorem holds good only for plane laminar objects. The theorem states that the moment of inertia of a plane laminar body about an axis perpendicular to its plane is equal to the sum of moments of inertia about two perpendicular axes lying in the plane of the body such that all the three axes are mutually perpendicular and have a common point.

Let the X and Y – axes lie in the plane and Z – axis perpendicular to the plane of the laminar object. If the moments of inertia of the body about X and Y-axes are I_X and I_Y respectively – and I_Z is the moment of inertia about Z-axis, then the perpendicular axis theorem could be expressed as,
I_Z = I_X + I_Y

To prove this theorem, let us consider a plane laminar object of negligible thickness on which lies the origin (O). The X and Y – axes lie on the plane and Z – axis is perpendicular to it as shown in figure. The lamina is considered to be made up of a large number of particles of mass m. Let us choose one such particle at a point P which has coordinates (x, y) at a distance r from O.

The moment of inertia of the particle about Z – axis is, mr².
The summation of the above expression gives the moment of inertia of the entire lamina about Z – axis as, I_Z = ∑ mr²
Here, r² = x² + y²
Then, I_Z = ∑ m (x² + y²)
I_Z = ∑ m x² + ∑ m y²
In the above expression, the term ∑ m x² is the moment of inertia of the body about the Y-axis and similarly the term ∑ m y²is the moment of inertia about X- axis. Thus,
I_X = ∑ m y² and I_Y = ∑ m x²
Substituting in the equation for Iz gives,
I_Z = I_X + I_Y
Thus, the perpendicular axis theorem is proved.

Question 10.
Discuss rolling on inclined plane and arrive at the expression for the acceleration.
Answer:
Let us assume a round object of mass m and radius R is rolling down an inclined plane without slipping as shown in figure. There are two forces acting on the object along the inclined plane. One is the component of gravitational force (mg sin θ) and the other is the static frictional force (f). The other component of gravitation force (mg cos θ) is cancelled by the normal force (N) exerted by the plane. As the motion is happening along the incline, we shall write the equation for motion from the free body diagram (FBP) of the object.

For transnational motion, mg sin θ is the supporting force and f is the opposing force, mg sin θ f = ma
For rotational motion, let us take the torque with respect to the center of the object. Then mg sin 0 cannot cause torque as it passes through it but the frictional force f can set torque of Rf = Iα
By using the relation, a = rα, and moment of inertia I = mK² we get,
Rf = mK² \(\frac {a}{R}\); f = ma \(\left(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right)\)
Now equation becomes,
mg sin θ – ma \(\left(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right)\) = ma
mg sin θ = ma + ma \(\left(\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right)\)
a \(\left(1+\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right)\) = g sin θ
After rewriting it for acceleration, we get,
a = \(\frac{g \sin \theta}{\left(1+\frac{K^{2}}{R^{2}}\right)}\)
We can also find the expression for final velocity of the rolling object by using third equation of motion for the inclined plane.
v² = u² + 2as. If the body starts rolling from rest, u = 0. When h is the vertical height of the incline, the length of the incline s is, s = \(\frac {h}{sin θ}\)
By taking square root,

The time taken for rolling down the incline could also be written from first equation of motion as, v = u + at. For the object which starts rolling from rest, u = 0. Then,
t = \(\frac {v}{a}\)

The equation suggests that for a given incline, the object with the least value of radius of gyration K will reach the bottom of the incline first.

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Conceptual Questions

Question 1.
When a tree is cut, the cut is made on the side facing the direction in which the tree is required to fall. Why?
Answer:
A cut on the tree is made on the side facing the direction in which the tree is required to fall because that side will no longer be supported by the normal force from the bottom, therefore the gravitational force tries to rotate it. So the torque given by the gravity to the tree makes the tree fall on the side as anticipated.

Question 2.
Why does a porter bend forward while carrying a sack of rice on his back?
Answer:
When a porter carries a sack of rice, the line of action of his center of gravity will go away from the body. It affects the balance, to avoid this he bends. By which center of gravity will realign within the body again. So balance is maintained.

Question 3.
Why is it much easier to balance a meter scale on your finger tip than balancing on a match stick?
Answer:
A meter scale is larger then a match stick. So the center of gravity for meter scale is higher than a matchstick when we keep it vertically. It is easier to balance the object whose center of gravity is higher than the object whose centro of gravity is lower. So, it is hard to balance a match stick than a meter scale.

Question 4.
Two identical water bottles one empty and the other filled with water are allowed to roll down an inclined plane. Which one of them reaches the bottom first? Explain your answer.
Answer:
Mass of the empty water bottle mostly concentrated on its surface. So moment of inertia of empty water bottle is more than the bottle filled with water. As we know, moment of inertia is inversely proportional to angular velocity. Therefore, the bottle filled with water whirls with greater speed and reaches the ground first.

Question 5.
Write the relation between angular momentum and rotational kinetic energy. Draw a graph for the same. For two objects of same angular momentum, compare the moment of inertia using the graph.
Answer:
Let a rigid body of moment of inertia I rotate with angular velocity ω.
The angular momentum of a rigid body is, L = Iω
The rotational kinetic energy of the rigid body is, KE = \(\frac { 1 }{ 2 }\) Iω².
By multiplying the numerator and denominator of the above equation with I, we get a relation between L and KE as,

It resembles with y = Kx². If angular momentum is same for two objects, kinetic energy is inversely proportional to moment of inertia.
Moment of inertia of the object whose kinetic energy is lesser will have greater magnitude.

Question 6.
Three identical solid spheres move down through three inclined planes A, B and C all same dimensions. A is without friction, B is undergoing pure rolling and C is rolling with slipping. Compare the kinetic energies E_A, E_B and E_C at the bottom.
Answer:
Even though, the three identical solid spheres of same dimensions move down through three different inclined plane, according to the law of conservation of energy, the potential energy possessed by these three solid spheres will be converted into kinetic energies. So the kinetic energies E_A, E_B and E_C are equal at the bottom.

Question 7.
Give an example to show that the following statement is false. Any two forces acting on a body can be combined into single force that would have same effect.
Answer:
A single force i.e. resultant of two forces acting on a body depends upon the angle between them also. The simple example for this is if two forces 5 N and 5 N acting on the object in the opposite direction, the single resultant force acting on the body is zero. But, if two forces acting on the object along the same direction, then the resultant i.e. the single force is 5 + 5 = 10 N. Hence the given statement “any two forces acting on a body can be combined into single force that would leave same effect” is wrong.

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Numerical Problems

Question 1.
A uniform disc of mass 100 g has a diameter of 10 cm. Calculate the total energy of the disc when rolling along a horizontal table with a velocity of 20 cm s^-2.
Answer:
Given,
Mass of the disc = 100 g = 100 x 10^-3 kg = \(\frac { 1 }{ 10 }\)kg
Velocity of disc = 20 cm s^-1 = 20 x 10^-2 ms^-1 = 0.2 ms^-1

Question 2.
A particle of mass 5 units is moving with a uniform speed of v = \(3 \sqrt{2}\) units in the XOY plane along the line y = x + 4. Find the magnitude of angular momentum.
Answer:
Given,
Mass = 5 units
Speed = v = \(3 \sqrt{2}\) units
Y = X + 4
Angular momentum = L = m(\(\bar{r} \times \bar{v}\))
= m(x\(\hat{i}\) +y\(\hat{j}\))x(v\(\hat{i}\) + v\(\hat{j}\)) = m[xv\(\hat{k}\)-vy\(\hat{k}\)] = m[xv\(\hat{k}\)– v(x + 4)\(\hat{k}\)]
L = -mv\(\hat{k}\) = -4 x 5 x \(3 \sqrt{2}\)\(\hat{k}\) = – 60\(\sqrt{2}\)\(\hat{k}\)
L = 60\(\sqrt{2}\) units.

Question 3.
A fly wheel rotates with a uniform angular acceleration. If its angular velocity increases from 20π rad/s to 40π rad/s in 10 seconds, find the number of rotations in that period.
Answer:
Given,
Initial angular velocity ω₀ = 20 π rad/s
Final angular velocity ω = 40 π rad/s
Time t = 10 s
Solution:
Angular acceleration α = \(\frac{\omega-\omega_{0}}{t}\) = \(\frac {40π – 20π }{ 10 }\)
α = 2π rad/s²
According to equation of motion for rotational motion

The number of rotations = n = \(\frac {θ}{ 2π }\)
n = \(\frac {300π}{ 2π }\) = 150 rotations.

Question 4.
A uniform rod of mass m and length / makes a constant angle 0 with an axis of rotation which passes through one end of the rod. Find the moment of inertia about this gravity is.
Answer:
Moment of inertia of the rod about the axis which is passing through its center of gravity is


Moment of inertia of a uniform rod of mass m and length l about one axis which passes through one end of the rod

Question 5.
Two particles P and Q of mass 1 kg and 3 kg respectively start moving towards each other from rest under mutual attraction. What is the velocity of their center of mass?
Answer:
Given,
Mass of particle P = 1 kg Mass of particle Q = 3 kg
Solution:
Particles P and Q forms a system. Here no external force is acting on the system,

We know that M = \(\frac {d}{dt}\) (V_CM ) = f
It means that, C.M. of an isolated system remains at rest when no external force is acting and internal forces do not change its center of mass.

Question 6.
Find the moment of inertia of a hydrogen molecule about an axis passing through its center of mass and perpendicular to the inter-atomic axis.
Given: mass of hydrogen atom 1.7 x 10²⁷kg and inter atomic distance is equal to 4 x 10^-10m.
Answer:
Given,
Inter-atomic distance : 4 x 10^-10 m
Mass of H₂ atom : 1.7 x 10^-27 kg
Moment of inertia of H₂ =

Question 7.
On the edge of a wall, we build a brick tower that only holds because of the bricks’ own weight. Our goal is to build a stable tower whose overhang d is greater than the length l of a single brick. What is the minimum number of bricks you need?
(Hint: Find the center of mass for each brick and add.)
Answer:
Given:

Length of the brick = l
Length of the overhang = d
The mono of bricks can be decided only by using the concept of position of center of gravity. The first brick is in contact with the ground and it will not fall over.
Let one end of brick 2 is coinciding with the center of brick 1 i.e. x = 0.
∴ The position of n brick is
x_n = (n – 1) \(\frac {L}{4}\)
The center of gravity is in the midway between the center of brick 2 and the center of brick n.
position of G =

brick tower will fall when G >\(\frac {L}{4}\) it shows that n > 4.

Question 8.
The 747 boing plane is landing at a speed of 70 m s^-1. Before touching the ground, the wheels are not rotating. How long a skid mark do the wing wheels leave (assume their mass is 100 kg which is distributed uniformly, radius is 0.7 m, and the coefficient of friction with the ground is 0.5)?
Answer:
The types of the plane will leave a skid mark if the speed of the types in contact with ground is lesser than the velocity of the plane. The condition for this is –
v > ω
(When the type attained an angular velocity of V/R)
The types will stop the skidding and starts the rolling.
The forces acting on the wheel after the plane touches down are,
N – P Normal force W – weight
The wheel is not accelerating means
N = ω
The torque about the center of the wheel is
τ = RF = µωR
The angular acceleration is

According to equation of motion, time taken to stop the skidding by the wheel is,

The six mark will have a length of
l = vt = 70 x 0.03 = 2.1 m
Note:
The 747 is resting on the runway, supported by 16 wheels under the wing, and 2 under the nose total length is 68.63 m. The normal force experienced by plane through its 16 wheels is ω = 232 KN.

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Additional Questions Solved

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Multiple Choice Questions

Question 1.
The changes produced by the deforming forces in a rigid body are –
(a) very large
(b) infinity
(c) negligibly small
(d) small
Answer:
(c) negligibly small

Question 2.
When a rigid body moves all particles that constitute the body follows-
(a) same path
(b) different paths
(c) either same or different path
(d) circular path
Answer:
(b) different path

Question 3.
For bodies of regular shape and uniform mass distribution, the center of mass is at –
(a) the comers
(b) inside the objects
(c) the point where the diagonals meet
(d) the geometric center
Answer:
(d) the geometric center

Question 4.
For square and rectangular objects center of mass lies at –
(a) the point where the diagonals meet
(b) at the comers
(c) on the center surface
(d) any point
Answer:
(a) the point where the diagonals meet

Question 5.
Center of mass may lie –
(a) within the body
(b) outside the body
(c) both (a) and (b)
(d) only at the center
Answer:
(c) both (a) and (b)

Question 6.
The dimension of point mass is –
(a) positive
(b) negative
(c) zero
(d) infinity
Answer:
(c) zero

Question 7.
The motion of center of mass of a system of two particles is unaffected by their internal forces –
(a) irrespective of the actual directions of the internal forces
(b) only if they are along the line joining the particles
(c) only if acts perpendicular to each other
(d) only if acting opposite
Answer:
(a) irrespective of the actual directions of the internal forces

Question 8.
A circular plate of diameter 10 cm is kept in contact with a square plate of side 10 cm. The density of the material and the thickness are same everywhere. The center of mass of the system will be
(a) inside the circular plate
(b) inside the square plate
(c) At the point of contact
(d) outside the system
Answer:
(6) inside the square plate

Question 9.
The center of mass of a system of particles does not depend on
(a) masses of particles
(b) position of the particles
(c) distribution of masses
(d) forces acting on the particles
Answer:
(d) forces acting on the particles

Question 10.
The center of mass of a solid cone along the line from the center of the base to the vertex is at –
(a) \(\frac { 1 }{ 2 }\) th of its height
(b) \(\frac { 1 }{ 3 }\) of its height
(c) \(\frac { 1 }{ 4 }\) th of its height
(d) \(\frac { 1 }{ 5 }\) th of its height
Answer:
(d) \(\frac { 1 }{ 5 }\) th of its height

Question 11.
All the particles of a body are situated at a distance of X from origin. The distance of the center of mass from the origin is –
(a) ≥ r
(b) ≤ r
(c) = r
(d) > r

Question
A free falling body breaks into three parts of unequal masses. The center of mass of the three parts taken together shifts horizontally towards –
(a) heavier piece
(b) lighter piece
(c) does not shift horizontally
(d) depends on vertical velocity
Answer:
(c) does not shift horizontally

Question 13.
The distance between the center of carbon and oxygen atoms in the gas molecule is 1.13 A. The center of mass of the molecule relative to oxygen atom is –
(a) 0.602 Å
(b) 0.527 Å
(c) 1.13 Å
(d) 0.565 Å
Answer:
(b) 0.527 Å
Given,
Inter atomic distance = 1.13 Å
Mass of carbon atom = 14
Mass of oxygen atom = 16
Let C.M. of molecule lies at a distance of X from oxygen atom-
i.e. m₁r₁ = m₂r₂
16 X = 14(1.13 – X)
30 X = 15.82
X = 0.527 Å

Question 14.
The unit of position vector of center of mass is-
(a) kg
(b) kg m²
(c) m
(d) m²
Answer:
(c) m

Question 15.
The sum of moments of masses of all the particles in a system about the center of mass is-
(a) minimum
(b) maximum
(c) zero
(d) infinity
Answer:
(c) zero

Question 16.
The motion of center of mass depends on-
(a) external forces acting on it
(b) internal forces acting within it
(c) both (a) and (b)
(d) neither (a) nor (b)
Answer:
(a) external forces acting on it

Question 17.
Two particles P and Q move towards with each other from rest with the velocities of 10 ms^-1 and 20 ms^-1 under the mutual force of attraction. The velocity of center of mass is-
(a) 15 ms^-1
(b) 20 ms^-1
(c) 30 ms^-1
(d) zero
Answer:
(d) zero

Question 18.
The reduced mass of the system of two particles of masses 2 m and 4 m will be –
(a) 2 m
(b) \(\frac {2 }{ 3 }\)y m
(c) \(\frac {3}{ 2 }\)y m
(d) \(\frac { 4 }{ 3 }\)m
Answer:
(d) \(\frac { 4 }{ 3 }\)m

Question 19.
The motion of the center of mass of a system consists of many particles describes its –
(a) rotational motion
(b) vibratory motion
(c) oscillatory motion
(d) translator y motion
Answer:
(c) oscillatory motion

Question 20.
The position of center of mass can be written in the vector form as –

Answer:

Question 21.
The positions of two masses m₁ and m₂ are x₁ and x₂. The position of center of mass is –

Answer:

Question 22.
In a two particle system, one particle lies at origin another one lies at a distance of X. Then the position of center of mass of these particles of equal mass is –

Answer:
(a) \(\frac {X}{2}\)

Question 23.
Principle of moments is –

Answer:

Question 24.
Infinitesimal quantity means –
(a) collective particles
(b) extremely small
(c) nothing
(d) extremely larger
Answer:
(b) extremely small

Question 25.
In the absence of external forces the center of mass will be in a state of –
(a) rest
(b) uniform motion
(c) may be at rest or in uniform motion
(d) vibration
Answer:
(c) may be at rest or in uniform motion

Question 26.
The activity of the force to produce rotational motion in a body is called as –
(a) angular momentum
(b) torque
(c) spinning
(d) drive force
Answer:
(b) torque

Question 27.
The moment of the external applied force about a point or axis of rotation is known as –
(a) angular momentum
(b) torque
(c) spinning
(d) drive force
Answer:
(b) torque

Question 28.
Torque is given as –
(a) \(\vec{r}\) . \(\vec{F}\)
(b) \(\vec{r}\) x \(\vec{F}\)
(c) \(\vec{F}\) x \(\vec{r}\)
(d) r F cos θ
Answer:
(b) \(\vec{r}\) x \(\vec{F}\)

Question 29.
The magnitude of torque is –
(a) rF sin θ
(b) rF cos θ
(c) rF tan θ
(d) rF
Answer:
(a) rF sin θ

Question 30.
The direction of torque ácts –
(a) along \(\vec{F}\)
(b) along \(\vec{r}\) & \(\vec{F}\)
(c) Perpendicular to \(\vec{r}\)
(d) Perpendicular to both \(\vec{r}\) & \(\vec{F}\)
Answer:
(d) Perpendicular to both \(\vec{r}\) & \(\vec{F}\)

Question 31.
The unit of torque is –
(a) is
(b) Nm^-2
(c) Nm
(d) Js^-1
Answer:
(c) Nm

Question 32.
The direction of torque is found using –
(a) left hand rule
(b) right hand rule
(c) palm rule
(d) screw rule
Answer:
(b) right hand rule

Question 33.
if the direction of torque is out of the paper then the rotation produced by the torque is –
(a) clockwise
(b) anticlockwise
(c) straight line
(d) random direction
Answer:
(a) clockwise

Question 34.
If the direction of the torque is inward the paper then the rotation is –
(a) clockwise
(b) anticlockwise
(c) straight line
(d) random direction
Answer:
(a) clockwise

Question 35.
if \(\vec{r}\) and \(\vec{F}\) are parallel or anti parallel, then the torque is –
(a) zero
(b) minimum
(c) maximum
(d) infinity
Answer:
(a) zero

Question 36.
The maximum possible value of torque is –
(a) zero
(b) infinity
(c) \(\vec{r}\) + \(\vec{F}\)
(d) rF
Answer:
(d) rF

Question 37.
The relation between torque and angular acceleration is –
(a) \(\vec{τ}\) = \(\frac{1}{\alpha}\)
(b) \(\vec{α}\) = \(\frac{\vec{\tau}}{\mathrm{I}}\)
(c) \(\vec{α}\) = I \(\vec{τ}\)
(d) \(\vec{τ}\) = \(\frac{\vec{\alpha}}{\mathrm{I}}\)
Answer:
(b) \(\vec{α}\) = \(\frac{\vec{\tau}}{\mathrm{I}}\)

Question 38.
Angular momentum is –
(a) \(\vec{P}\) x \(\vec{r}\)
(b) \(\vec{r}\) x \(\vec{P}\)
(c) \(\overrightarrow{\frac{r}{\vec{p}}}\)
(d) \(\vec{r}\) . \(\vec{P}\)
Answer:
(b) \(\vec{r}\) x \(\vec{P}\)

Question 39.
The magnitude of angular momentum is given by –
(a) rp
(b) rp sin θ
(c) rp cos θ
(d) rp tan θ
Answer:
(b) rp sin θ

Question 40.
Angular momentum is associated with –
(a) rotational motion
(b) linear motion
(c) both (a) and (b)
(d) circular motion only
Answer:
(c) both (a) and (b)

Question 41.
Angular momentum acts perpendicular to –
(a) \(\vec{r}\)
(b) \(\vec{P}\)
(c) both \(\vec{r}\) and \(\vec{P}\)
(d) plane of the paper
Answer:
(c) both \(\vec{r}\) and \(\vec{P}\)

Question 42.
Angular momentum is given by –
(a) \(\frac {I}{ω}\)
(b) τω
(c) Iω
(d) \(\frac {ωI}{2}\)
Answer:
(c) Iω

Question 43.
The rate of change of angular momentum is –
(a) Torque
(b) angular velocity
(c) centripetal force
(d) centrifugal force
Answer:
(a) Torque

Question 44.
The forces acting on a body when it is at rest –
(a) is gravitational force
(b) Normal force
(c) both gravitational as well as normal force
(d) No force is acting
Answer:
(c) both gravitational as well as normal force

Question 45.
The net force acting on a body when it is at rest is –
(a) gravitational force
(b) Normal force
(c) Sum of gravitational and normal force
(d) zero
Answer:
(d) zero

Question 46.
If net force acting on a body is zero, then the body is in –
(a) transnational equilibrium
(b) rotational equilibrium
(c) both (a) and (b)
(d) none
Answer:
(a) transnational equilibrium

Question 47.
If the net torque acting on the body is zero, then the body is in –
(a) transnational equilibrium
(b) rotational equilibrium
(c) mechanical equilibrium
(d) none
Answer:
(b) rotational equilibrium

Question 48.
when the net force and net torque acts on the body is zero then the body is in –
(a) transnational equilibrium
(b) rotational equilibrium
(c) mechanical equilibrium
(d) none
Answer:
(d) none

Question 49.
When the net force and net torque acts on the body is zero then the body is in –
(a) static equilibrium
(b) Dynamic equilibrium
(c) both (a) and (b)
(d) transnational equilibrium
Answer:
(c) both (a) and (b)

Question 50.
When two equal and opposite forces acting on the body at two different points, it may give –
(a) net force
(b) torque
(c) stable equilibrium
(d) none
Answer:
(b) torque

Question 51.
The torque in rotational motion is analogous to in transnational motion –
(a) linear momentum
(b) mass
(c) couple
(d) force
Answer:
(d) force

Question 52.
Which of the following example does not constitute a couple?
(a) steering a car
(b) turning a pen cap
(c) ball rolls on the floor
(d) closing the door
Answer:
(c) ball rolls on the floor

Questioner 53.
If the linear momentum and angular momentum are zero, then the object is said to be in –
(a) stable equilibrium
(b) unstable equilibrium
(c) neutral equilibrium
(d) all the above
Answer:
(d) all the above

Question 54.
When the body is disturbed, the potential energy remains same, then the body is in –
(a) stable equilibrium
(b) unstable equilibrium
(c) neutral equilibrium
(d) all the above
Answer:
(c) neutral equilibrium

Question 55
The point where the entire weight of the body acts is called as –
(a) center of mass
(b) center of gravity
(c) both (a) and (b)
(d) pivot
Answer:
(b) center of gravity

Question 56.
The forces acting on a cyclist negotiating a circular Level road is /are –
(a) gravitational force
(b) centrifugal force
(c) frictional force
(d) all the above
Answer:
(d) all the above

Question 57.
While negotiating a circular level road a cyclist has to bend by an angle θ from vertical to stay in an equilibrium is-
(a) \(\tan \theta=\frac{r g}{r^{2}}\)
(b) θ = \(\tan ^{-1}\left(\frac{v^{2}}{r g}\right)\)
(c) θ = \(\sin ^{-1}\left(\frac{r g}{r^{2}}\right)\)
(d) zero
Answer:
(b) θ = \(\tan ^{-1}\left(\frac{v^{2}}{r g}\right)\)

Question 58.
Moment of inertia for point masses –
(a) m²r
(b) rw²
(c) mr²
(d) zero
Answer:
(c) mr²

Question 59.
Moment of inertia for bulk object –
(a) rm²
(b) rw²
(c) \(m_{i} r_{i}^{2}\)
(d) \(\Sigma m_{i} r_{i}^{2}\)
Answer:
(d) \(\Sigma m_{i} r_{i}^{2}\)

Question 60.
For rotational motion, moment of inertia is a measure of –
(a) transnational inertia
(b) mass
(c) rotational inertia
(d) invariable quantity
Answer:
(c) rotational inertia

Question 61.
Unit of moment of inertia –
(a) kgm
(b) mkg^-2
(c) kgm²
(d) kgm^-1
Answer:
(c) kgm²

Question 62.
Dimensional formula for moment of inertia is –
(a) [ML^-2]
(b) [M²L^-1]
(c) [M^-2]
(d) [ML²]
Answer:
(d) [ML²]

Question 63.
Moment of inertia of a body is a –
(a) variable quantity
(b) invariable quantity
(c) constant quantity
(d) measure of torque
Answer:
(a) variable quantity

Question 64.
Moment of inertia of a thin uniform rod about an axis passing through the center of mass and perpendicular to the length is –
(a) \(\frac { 1 }{ 3 }\)Ml²
(b) \(\frac { 1 }{ 12 }\)Ml²
(c) \(\frac { 1 }{ 2 }\)M(l² + b² )
(d) Ml²
Answer:
(b) \(\frac { 1 }{ 12 }\)Ml²

Question 65.
Moment of inertia ofa thin uniform rod about an axis passing through one end and perpendicular to the length is-
(a) \(\frac { 1 }{ 3 }\)Ml²
(b) \(\frac { 1 }{ 12 }\)Ml²
(c) \(\frac { 1 }{ 2 }\)M(l² + b² )
(d) Ml²
Answer:
(a) \(\frac { 1 }{ 3 }\)Ml²

Question 66.
Moment of inertia of a thin uniform rectangular sheet about an axis passing through the center of mass and perpendicular to the plane of the sheet is-
(a) \(\frac { 1 }{ 3 }\)Ml²
(b) \(\frac { 1 }{ 12 }\)Ml²
(c) \(\frac { 1 }{ 2 }\)M(l² + b² )
(d) Ml²
Answer:
(c) \(\frac { 1 }{ 2 }\)M(l² + b² )

Question 67.
Moment of inertia of a thin uniform ring about an axis passing through the center of gravity and perpendicular to the plane is –
(a) MR²
(b) 2 MR²
(c) \(\frac { 1 }{ 2 }\)MR²
(d) \(\frac { 3 }{ 2 }\)MR²
Answer:
(a) MR²

Question 68.
Moment of inertia of a thin uniform ring about an axis passing through the center and lying on the plane (along diameter) is –
(a) MR²
(b) 2 MR²
(c) \(\frac { 1 }{ 2 }\) MR²
(d) \(\frac { 2 }{ 3 }\)MR²
Answer:
(c) \(\frac { 1 }{ 2 }\) MR²

Question 69.
Moment of inertia of a thin uniform disc about an axis passing through the center and perpendicular to the plane is –
(a) MR²
(b) 2 MR²
(c) \(\frac { 1 }{ 2 }\) MR²
(d) \(\frac { 2 }{ 3 }\)MR²
Answer:
(c) \(\frac { 1 }{ 2 }\) MR²

Question 70.
Moment of inertia of a thin uniform disc about an axis passing through the center lying on the plane (along diameter is)
(a) MR²
(b) \(\frac { 1 }{ 2 }\) MR²
(c) \(\frac { 3 }{ 2 }\) MR²
(d) \(\frac { 1 }{ 4 }\) MR²
Answer:
(d) \(\frac { 1 }{ 4 }\) MR²

Question 71.
Moment of inertia of a thin uniform hollow cylinder about an axis of the cylinder is –
(a) MR²
(b) \(\frac { 1 }{ 2 }\) MR²
(c) \(\frac { 3 }{ 2 }\) MR²
(d) \(\frac { 1 }{ 4 }\) MR²
Answer:
(a) MR²

Question 72.
Moment of inertia of a thin uniform hollow cylinder about an axis of the cylinder is –
(a) MR²
(b) M\(\left(\frac{\mathrm{R}^{2}}{2}+\frac{l^{2}}{12}\right)\)
(c) \(\frac { 1 }{ 2 }\) MR²
(d) M\(\left(\frac{\mathrm{R}^{2}}{4}+\frac{l^{2}}{12}\right)\)
Answer:
(b) M\(\left(\frac{\mathrm{R}^{2}}{2}+\frac{l^{2}}{12}\right)\)

Question 73.
Moment of inertia of a uniform solid cylinder about an axis passing through the center and along the axis of the cylinder is –
(a) MR²
(b) M\(\left(\frac{\mathrm{R}^{2}}{2}+\frac{l^{2}}{12}\right)\)
(c) \(\frac { 1 }{ 2 }\) MR²
(d) M\(\left(\frac{\mathrm{R}^{2}}{4}+\frac{l^{2}}{12}\right)\)
Answer:
(c) \(\frac { 1 }{ 2 }\) MR²

Question 74.
Moment of inertia of a uniform solid cylinder about as axis passing perpendicular to the length and passing through the center is –
(a) MR²
(b) M\(\left(\frac{\mathrm{R}^{2}}{2}+\frac{l^{2}}{12}\right)\)
(c) \(\frac { 1 }{ 2 }\) MR²
(d) M\(\left(\frac{\mathrm{R}^{2}}{4}+\frac{l^{2}}{12}\right)\)
Answer:
(d) M\(\left(\frac{\mathrm{R}^{2}}{4}+\frac{l^{2}}{12}\right)\)

Question 75.
Moment of inertia of a thin hollow sphere about an axis passing through the center along its diameter is
(a) \(\frac { 2 }{ 3 }\)MR²
(b) \(\frac { 5 }{ 3 }\)MR²
(c) \(\frac { 7 }{ 5 }\)MR²
(d) \(\frac { 2 }{ 5 }\)MR²
Answer:
(a) \(\frac { 2 }{ 3 }\)MR²

Question 76.
Moment of inertia of a thin hollow sphere about an axis passing through the edge along its tangent is –
(a) \(\frac { 2 }{ 3 }\)MR²
(b) \(\frac { 5 }{ 3 }\)MR²
(c) \(\frac { 7 }{ 5 }\)MR²
(d) \(\frac { 2 }{ 5 }\)MR²
Answer:
(b) \(\frac { 5 }{ 3 }\)MR²

Question 77.
torment of inertia of a uniform solid sphere about an axis passing through the center along its diameter is –
(a) \(\frac { 2 }{ 3 }\)MR²
(b) \(\frac { 5 }{ 3 }\)MR²
(c) \(\frac { 7 }{ 5 }\)MR²
(d) \(\frac { 2 }{ 5 }\)MR²
Answer:
(d) \(\frac { 2 }{ 5 }\)MR²

Question 78.
Moment of inertia of a uniform solid sphere about an axis passing through the edge along its tangent is –
(a) \(\frac { 2 }{ 3 }\)MR²
(b) \(\frac { 5 }{ 3 }\)MR²
(c) \(\frac { 7 }{ 5 }\)MR²
(d) \(\frac { 2 }{ 5 }\)MR²
Answer:
(c) \(\frac { 7 }{ 5 }\)MR²

Question 79.
The ratio of K²/R² of a thin uniform ring about an axis passing through the center and perpendicular to the plane is-
(a) 1
(b) 2
(c) \(\frac { 7 }{ 5 }\)
(d) \(\frac { 3 }{ 2 }\)
Answer:
(a) 1

Question 80.
The ratio of K²/ R² of a thin uniform disc about an axis passing through the center and perpendicular to the plane is –
(a) 1
(b) 2
(c) \(\frac {1}{ 2 }\)
(d) \(\frac { 3 }{ 2 }\)
Answer:
(c) \(\frac {1}{ 2 }\)

Question 81.
When no external torque acts on the body, the net angular momentum of a rotating body.
(a) increases
(b) decreases
(c) increases or decreases
(d) remains constant
Answer:
(d) remains constant

Question 82.
Moment of inertia of a body is proportional to –
(a) ω
(b) \(\frac { 1 }{ ω }\)
(c) ω²
(d) \(\frac{1}{\omega^{2}}\)
Answer:
(b) \(\frac { 1 }{ ω }\)

Question 83.
When the hands are brought closer to the body, the angular velocity of the ice dancer –
(a) decreases
(b) increases
(c) constant
(d) may decrease or increase
Answer:
(b) increases

Question 84.
When the hands are stretched out from the body, the moment of inertia of the ice dancer –
(a) decreases
(b) increases
(c) constant
(d) may decrease or increase
Answer:
(b) increases

Question 85.
The work done by the torque is –
(a) F. ds
(b) F. dθ
(c) τ dθ
(d) r.dθ
Answer:
(c) τ dθ

Question 86.
Rotational Kinetic energy of a body is –
(a) \(\frac { 1 }{ 2 }\)mr
(b) \(\frac { 1 }{ 2 }\) Iω²
(c) \(\frac { 1 }{ 2 }\)Iv²
(d) \(\frac { 1 }{ 2 }\)mω²
Answer:
(b) \(\frac { 1 }{ 2 }\) Iω²

Question 87.
Rotational kinetic energy is given by –
(a) \(\frac { 1 }{ 2 }\)mr
(b) \(\frac { 1 }{ 2 }\)Iv²
(c) \(\frac{\mathrm{L}^{2}}{2 \mathrm{I}}\)
(d) \(\frac{2 \mathrm{I}}{\mathrm{L}^{2}}\)
Answer:
(c) \(\frac{\mathrm{L}^{2}}{2 \mathrm{I}}\)

Question 88.
If E is a rotational kinetic energy then angular momentum is-
(a) \(\sqrt{2 \mathrm{IE}}\)
(b) \(\frac{\mathrm{E}^{2}}{2 \mathrm{I}}\)
(c) \(\frac{2 \mathrm{I}}{\mathrm{E}^{2}}\)
(d) \(\frac{E}{I^{2} \omega^{2}}\)
Answer:
(a) \(\sqrt{2 \mathrm{IE}}\)

Question 89.
The product of torque acting on a body and angular velocity is –
(a) Energy
(b) power
(c) work done
(d) kinetic energy
Answer:
(b) power

Question 90.
The work done per unit time in rotational motion is given by –
(a) \(\vec{F}\) .v
(b) \(\frac {dθ}{dt}\)
(c) τ ω
(d) I ω
Answer:
(c) τ ω

Question 91.
While rolling, the path of center of mass of an object is –
(a) straight line
(b) parabola
(c) hyperbola
(d) circle
Answer:
(a) straight line

Question 92.
In pure rolling, the velocity of the point of the rolling object which comes in contact with the surface is –
(a) maximum
(b) minimum
(c) zero
(d) 2 V_CM
Answer:
(c) zero

Question 93.
In pure rolling velocity of center of mass is equal to –
(a) zero
(b) Rω
(c) \(\frac { ω }{ R }\)
(d) \(\frac { R }{ ω }\)
Answer:
(b) Rω

Question 94.
In pure rolling, rotational velocity of points at its edges is equal to-
(a) Rω
(b) velocity of center of mass
(c) transnational velocity
(d) all the above
Answer:
(a) Rω

Question 95.
Sliding of the object occurs when –
(a) V_trans < V_rot
(b) V_trans = V_rot
(c) V_trans > V_rot
(d) V_trans = 0
Answer:
(c) V_trans > V_rot

Question 96.
Sliding of the object occurs while –
(a) V_trans = V_rot
(b) V_CM = Rω
(c) V_CM < Rω
(d) V_CM > Rω
Answer:
(d) V_CM > Rω

Question 97.
Slipping of the object occurs when –
(a) V_trans < V_rot
(b) V_trans = V_rot
(c) V_trans > V_rot
(d) V_trans = 0
Answer:
(a) V_trans < V_rot

Question 98.
Slipping of the object occurs when –
(a) V_trans = V_rot
(b) V_CM = Rω
(c) V_CM < Rω
(d) V_CM > Rω
Answer:
(c) V_CM < Rω

Question 99.
In sliding, the resultant velocity of a point of contact acts along –
(a) forward direction
(b) backward direction
(c) either (a) or (b)
(d) tangential direction
Answer:
(a) forward direction

Question 100.
In slipping, the resultant velocity of a point of contact acts along –
(a) forward direction
(b) backward direction
(c) either (a) or (b)
(d) tangential direction
Answer:
(b) backward direction

Question 101.
When a solid sphere is undergoing pure rolling, the ratio of transnational kinetic energy to rotational kinetic – energy is –
(a) 2 : 5
(b) 5 : 2
(c) 1 : 5
(d) 5 : 1
Answer:
(b) 5 : 2

Question 102.
Time taken by the rolling object in inclined plane to reach its bottom is –

Answer:

Question 103.
The velocity of the rolling object on inclined plane at the bottom of inclined plane is –

Answer:

Question 104.
Moment of inertia of an annular disc about an axis passing through the centre and perpendicular to the plane of disc is –

Answer:

Question 105.
Moment of inertia of a cube about an axis passing through the center of mass and perpendicular to face is –
(a) \(\frac{\mathrm{Ma}^{2}}{6}\)
(b) \(\frac {1}{3}\) Ma²
(c) \(\frac {Ma}{6}\)
(d) \(\frac{\mathrm{Ma}^{2}}{12}\)
Answer:
(a) \(\frac{\mathrm{Ma}^{2}}{6}\)

Question 106.
Moment of inertia of a rectangular plane sheet about an axis passing through center of mass and perpendicular to side b in its plane is –
(a) \(\frac{\mathrm{Ml}^{2}}{12}\)
(b) \(\frac{\mathrm{Ma}^{2}}{12}\)
(c) \(\frac{\mathrm{Mb}^{2}}{12}\)
(d) \(\frac{\mathrm{Ml}^{2}}{6}\)
Answer:
(c) \(\frac{\mathrm{Mb}^{2}}{12}\)

Question 107.
Rotational kinetic energy can be calculated by using –
(a) \(\frac{1}{2}\) I ω²
(b) \(\frac{\mathrm{L}^{2}}{2I}\)
(c) \(\frac{1}{2}\) Lω
(d) all the above
Answer:
(b) \(\frac{\mathrm{L}^{2}}{2I}\) )

Question 108.
The radius of gyration of a solid sphere of radius r about a certain axis is r. The distance of that axis from the center of the sphere is –
(a) \(\frac{2}{5}\)r
(b) \(\sqrt{\frac{2}{5}}\)r
(c) \(\sqrt{0.6r}\)
(d) \(\sqrt{\frac{5}{3}}\)
Answer:
(c) \(\sqrt{0.6r}\)
From parallel axis theorem
I = I_G + Md²
mr² = \(\frac{2}{5}\) mr² + md²
d = \(\sqrt{\frac{3}{5}}\)r = \(\sqrt{0.6r}\)

Question 109.
A wheel is rotating with angular velocity 2 rad/s. It is subjected to a uniform angular acceleration 2 rad/s² then the angular velocity after 10 s is
(a) 12 rad/s
(b) 20 rad/s
(c) 22 rad/s
(d) 120 rad/s
Answer:
(c) 22 rad/s
ω = ω₀ + αt
Here ω₀ = 2 rad/s,
α = 2 rad/s2
ω = 10 s
ω = 2 + 2 x 10 = 22 rad/s

Question 110.
Two rotating bodies A and B of masses m and 2m with moments of inertia I_A and I_B (I_b > I_A) have equal kinetic energy of rotation. If L_A and L_B be their angular momenta respectively,
then,
(a) L_B > L_A
(b) L_A > L_B
(c) L_A = \(\frac{L_{B}}{2}\)
(d) L_A = 2L_B
Answer:
(a) L_B > L_A

Question 111.
Three identical particles lie in x, y plane. The (x, y) coordinates of their positions are (3, 2), (1, 1), (5, 3) respectively. The (x, y) coordinates of the center of mass are –
(a) (a, b)
(b) (1, 2)
(c) (3, 2)
(d) (2, 1)
Answer:
(c) The X and Y coordinates of the center of mass are

Question 112.
A solid cylinder of mass 3 kg and radius 10 cm is rotating about its axis with a frequency of 20/π. The rotational kinetic energy of the cylinder
(a) 10 π J
(b) 12 J
(c) \(\frac{6 \times 10^{2}}{\pi}\) J
(d) 3 J
Answer:
(b) 12 J
Given,
M = 3 kg
R = 0.1 m
v = 20 / π
Angular frequency ω = 2πv = \(\frac{2π x 20}{π}\) = 40 rad/s^-1
Moment of inertia of the cylinder about its axis = I = \(\frac{1}{2}\) mR² = \(\frac{1}{2}\) x 3 x (0.1)² = 0.015 kg m²
K.E. = \(\frac{1}{2}\) Iω² = \(\frac{1}{2}\) x 0.015 x (40)² = 12 J

Question 113.
A circular disc is rolling down in an inclined plane without slipping. The percentage of rotational energy in its total energy is
(a) 66.61%
(b) 33.33%
(c) 22.22%
(d) 50%
Answer:
(b) 33.33%
Rotational K.E. = \(\frac{1}{2}\)Iω² \(\frac{1}{2}\)(\(\frac{1}{2}\)MR²)ω² = \(\frac{1}{4}\) MR²ω²
Transnational K.E. = \(\frac{1}{2}\)MV² = \(\frac{1}{2}\)M(Rω)² = \(\frac{1}{2}\) MR²ω²
Total kinetic energy = E_rot + E_trans = \(\frac{1}{4}\) MR²ω²\(\frac{1}{2}\)M(Rω)² = \(\frac{3}{4}\) MR²ω²
% of E_rot = \(\frac{E_{\text {rot }}}{E_{\text {Tot }}}\) x 100% = 33.33%

Question 114.
A sphere rolls down in an inclined plane without slipping. The percentage of transnational energy in its total energy is
(a) 29.6%
(b) 33.4%
(c) 71.4%
(d) 50%
Answer:
(c) 71.4%
Rotational K.e. E_rot =

Question 115.
Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of the center of mass is –
(a) 30 m/s
(b) 20 m/s
(c) 10 m/s
(d) 5 m/s
Answer:
(c) According to law of conservation of linear momentum
MV = (M + M) V_CM
V_CM = \(\frac{MV}{M + M}\) = \(\frac{10 × 10}{10 + 4}\) = 10 ms^-1

Question 116.
A mass is whirled in a circular path with constant angular velocity and its angular momentum is L. If the string is now halved keeping the angular velocity the same, the angular momentum is –
(a) \(\frac{L}{4}\)
(b) \(\frac{L}{2}\)
(c) L
(d) 2L
Answer:
(a) \(\frac{L}{4}\)
We know that
angular momentum L = Mr²
Here, m and co are constants L α r²
If r becomes \(\frac{r}{2}\) angular momentum becomes \(\frac{1}{4}\) th of its initial value.

Question 117.
The moment of inertia of a thin uniform ring of mass 1 kg and radius 20 cm rotating about the axis passing through the center and perpendicular to the plane of the ring is –
(a) 4 x 10^-2 kg m²
(b) 1 x 10^-2 kg m²
(c) 20 x 10^-2 kg m²
(d) 10 x 10^-2 kg m²
Answer:
(b) Moment of inertia I = MR² = 1 x (10 x 10^-2)² = 1 x 10^-2 kg m².

Question 118.
A solid sphere is rolling down in the inclined plane, from rest without slipping. The angle of inclination with horizontal is 30°. The linear acceleration of the sphere is –
(a) 28 ms^-2
(b) 3.9 ms^-2
(c) \(\frac{25}{7}\)ms^-2
(d) \(\frac{1}{20}\)ms^-2
Answer:
(c) \(\frac{25}{7}\)ms^-2
We know that,a =

Question 119.
An electron is revolving in an orbit of radius 2 A with a speed of 4 x 10⁵ m /s. The angular momentum of the electron is [Me = 9 x 10^-31 kg]
(a) 2 x 10^-35 kg m² s^-1
(b) 72 x 10^-36 kg m² s^-1
(c) 7.2 x 10^-34 kg m² s^-1
(d) 0.72 x 10^-37 kg m² s^-1
Answer:
(b) Angular momentum L = mV x r = 9 x 10^-31 x 4 x 10⁵ x 2 x 10^-10 = 72 x 10^-36kg m² s^-1

Question 120.
A raw egg and hard boiled egg are made to spin on a table with the same angular speed about the same axis. The ratio of the time taken by the eggs to stop is –
(a) =1
(b) < 1
(c) > 1
(d) none of these
Answer:
(d) When a raw egg spins, the fluid inside comes towards its side.
∴ “1” will increase in – turn it decreases ω. Therefore it takes lesser time than boiled egg.
∴\(\frac {time fìr raw egg}{time for boiled egg}\) < 1

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Short Answer Questions (1 Mark)

Question 1.
What is a rigid body?
Answer:
A rigid body is the one which maintains its definite and fixed shape even when an external force acts on it.

Question 2.
When an object will have procession? Give one example.
Answer:
The torque about the axis will rotate the object about it and the torque perpendicular to the axis will turn the axis of rotation when both exist simultaneously on a rigid body the body will have a procession.
Example:
The spinning top when it is about to come to rest.

Question 3.
Define angular momentum. Give an expression for it.
Answer:
The angular momentum of a point mass is defined as the moment of its linear momentum.
\(\vec{L}\) = \(\vec{r}\) x \(\vec{p}\) or L = rp sin θ

Question 4.
When an angular momentum of the object will be zero?
Answer:
If the straight path of the particle passes through the origin, then the angular momentum is zero, which is also a constant.

Question 5.
When an object be in-mechanical equilibrium?
Answer:
A rigid body is said to be in mechanical equilibrium when both its linear momentum and angular momentum remain constant.

Question 6.
Derive an expression for the power delivered by torque.
Answer:
Power delivered is the work done per unit time. IF we differentiate the expression for work done with respect to time, we get the instantaneous
power (P).
p = \(\frac{dw}{dt}\) = τ \(\frac{dθ}{dt}\)
p = τ dω

Question 7.
A boy sits near the edge of revolving circular disc

1. What will be the change in the motion of a disc?
2. If the boy starts moving from edge to the center of the disc, what will happen?

Answer:

1. As we know L = Iω = constant if the boy sits on the edge of revolving disc, its I will be increased in turn it reduces angular velocity.
2. If the boy starts moving towards the center of the disc, its I will decrease in turn that increases its angular velocity.

Question 8.
Are moment of inertia and radius of gyration of a body constant quantities?
Answer:
No, moment of inertia and radius of gyration depends on axis of rotation and also on the distribution of mass of the body about its axis.

.

Question 9.
A cat is able to land on its feet after a fall. Which principle of physics is being used? Explain.
Answer:
A cat is able to land on its feet after a fall. This is based on law of conservation of angular ~ momentum. When the cat is about to fall, it curls its body to decrease the moment of inertia and increase its angular velocity. When it lands it stretches out its limbs. By which it increases its moment of inertia and inturn it decreases its angular velocity. Hence, the cat lands safety.

Question 10.
About which axis a uniform cube will have minimum moment of inertia ?
Answer:
It will be about an axis passing through the center of the cube and connecting the opposite comers.

Question 11.
State the principle of moments of rotational equilibrium.
Answer:
∑ =\(\bar{\tau}\) = 0

Question 12.
Write down the moment of inertia of a disc of radius R and mass m about an axis in its plane at a distance R / 2 from its center.
Answer:
\(\frac { 1 }{ 2 }\) MR²

Question 13.
Can the couple acting on a rigid body produce translator motion ?
Answer:
No. It can produce only rotatory motion.

Question 14.
Which component of linear momentum does not contribute to angular momentum?
Answer:
Radial Component.

Question 15.
A system is in stable equilibrium. What can we say about its potential energy ?
Answer:
PE. is minimum.

Question 16.
Is radius of gyration a constant quantity ?
Answer:
No, it changes with the position of axis of rotation.

Question 17.
Two solid spheres of the same mass are made of metals of different densities. Which of them has a large moment of inertia about the diameter?
Answer:
Sphere of smaller density will have larger moment of inertia.

Question 18.
The moment of inertia of two rotating bodies A and B are I_A and I_B (I_A > I_B) and their angular momenta are equal. Which one has a greater kinetic energy ?
Answer:
K = \(\frac{\mathrm{L}^{2}}{2 \mathrm{I}}\) ⇒ K_A > K_A

Question 19.
A particle moves on a circular path with decreasing speed. What happens to its angular momentum?
Answer:
As \(\vec{L}\) = \(\vec{r}\) x m\(\vec{v}\) i.e., \(\vec{L}\) magnitude decreases but direction remains constant.

Question 20.
What is the value of instantaneous speed of the point of contact during pure rolling ?
Answer:
Zero.

Question 21.
Which physical quantity is conserved when a planet revolves around the sun ?
Answer:
Angular momentum of planet.

Question 22.
What is the value of torque on the planet due to the gravitational force of sun ?
Answer:
Zero.

Question 23.
If no external torque acts on a body, will its angular velocity be constant ?
Answer:
No.

Question 24.
Why there are two propellers in a helicopter ?
Answer:
Due to conservation of angular momentum.

Question 25.
A child sits stationary at one end of a long trolley moving uniformly with speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, then what is the effect of the speed of the centre of mass of the (trolley + child) system ?
Answer:
No change in speed of system as no external force is working.

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Short Answer Questions (2 Marks)

Question 26.
State the factors on which the moment of inertia of a body depends.
Answer:

• Mass of body
• Size and shape of body
• Mass distribution w.r.t. axis of rotation
• Position and orientation of rotational axis

Question 27.
On what factors does radius of gyration of body depend?
Answer:
Mass distribution.

Question 28.
Why the speed of whirl wind in a Tornado is alarmingly high?
Answer:
In this, air from nearly regions get concentrated in a small space, so I decreases considerably. Since Iω = constant so ω increases so high.

Question 29.
Can a body be in equilibrium while in motion? If yes, give an example.
Answer:
Yes, if body has no linear and angular acceleration then a body in uniform straight line of motion will be in equilibrium.

Question 30.
There is a stick half of which is wooden and half is of steel, (i) it is pivoted at the wooden end and a force is applied at the steel end at right angle to its length (ii) it is pivoted at the steel end and the same force is applied at the wooden end. In which case is the angular acceleration more and why?
Answer:
I (first case) > 1 (Second case)
∴ τ r = l α
⇒ α (first case) < α (second case)

Question 31.
If earth contracts to half of its present radius what would be the length of the day at equator?
Answer:

Question 32.
An internal force cannot change the state of motion of center of mass of a body. Flow does the internal force of the brakes bring a vehicle to rest?
Answer:
In this case the force which bring the vehicle to rest is friction, and it is an external force.

Question 33.
When does a rigid body said to be in equilibrium? State the necessary condition for a body to be in equilibrium.
Answer:
For translation equilibrium
∑ F_ext  = 0
For rotational equilibrium
∑ \(\overline{\mathrm{τ}}\)_ext  = 0

Question 34.
How will you distinguish between a hard boiled egg and a raw egg by spinning it on a table top’
Answer:
For same external torque, angular acceleration of raw egg will be small than that of Hard boiled egg.

Question 35.
Equal torques are applied on a cylinder and a sphere. Both have same mass and radius. Cylinder rotates about its axis and sphere rotates about one of its diameter. Which will acquire greater speed and why?
Answer:
τ = I α α = \(\frac { τ }{ I }\)
α in cylinder, α_C = \(\frac{\tau}{I_{C}}\)
α in sphere, α_S = \(\frac{\tau}{I_{S}}\)

Question 36.
In which condition a body lying in gravitational field is in stable equilibrium?
Answer:
When vertical line through center of gravity passes through the base of the body.

Question 37.
Give the physical significance of moment of inertia. Explain the need of fly wheel in Engine.
Answer:
It plays the same role in rotatory motion as the mass does in translator y motion.

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Short Answer Questions (3 Marks)

Question 38.
Three mass point m₁, m₂, m₃ are located at the vertices of equilateral A of side ‘a’. What is the moment of inertia of system about an axis along the altitude of A passing through mi?
Answer:

Question 39.
A disc rotating about its axis with angular speed ω₀ is placed lightly (without any linear push) on a perfectly friction less table. The radius of the disc is R. What are the linear velocities of the points A, B and C on the disc shown in figure. Will the disc roll?
Answer:

For A V_A = R ω₀ in forward direction
For B = V_B = R ω₀ in backward direction R
For C, V_C = \(\frac {R}{2}\) ω₀ in forward direction disc will not roll.

Question 40.
Find the torque of a force 7\(\hat{i}\) – 3\(\hat{j}\) – 5\(\hat{k}\) about the origin which acts on a particle whose position vector is \(\hat{j}\) +\(\hat{j}\) – \(\hat{j}\)
Answer:

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Numericals

Question 41.
Three masses 3 kg, 4 kg and 5 kg are located at the comers of an equilateral triangle of side 1 m. Locate the center of mass of the system.
Answer:
(x,y) = (0.54 m, 0.36 m)

Question 42.
Two particles mass 100 g and 300 g at a given time have velocities 10\(\hat{j}\) – 7\(\hat{j}\) – 3\(\hat{j}\) and 7\(\hat{i}\) – 9 \(\hat{j}\) + 6\(\hat{k}\) ms^-1 respectively. Determine velocity of center of mass.
Answer:
Velocity of center of mass = \(\frac{31 \hat{i}-34 \hat{j}+15 \hat{k}}{2}\) ms^-1

Question 43.
From a uniform disc of radius R, a circular disc of radius R / 2 is cut out. The center of the hole is at R / 2 from the center of original disc. Locate the center of gravity of the resultant flat body.
Answer:
Center of mass of resulting portion lies at R/6 from the center of the original disc in a direction opposite to the center of the cut out portion.

Question 44.
The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 16 seconds,

1. What is its angular acceleration (assume the acceleration to be uniform)
2. How many revolutions does the wheel make during this time ?

Answer:
a = 4π rad s^-2
n = 576

Question 45.
A meter stick is balanced on a knife edge at its center. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm, what is the mass of the meter stick?
Answer:
m = 66.0 gm.

Question 46.
A solid sphere is rolling op a friction less plane surface about its axis of symmetry. Find ratio of its rotational energy to its total energy.
Answer:

Question 47.
Calculate the ratio of radii of gyration of a circular ring and a disc of the same radius with respect to the axis passing through their centers and perpendicular to their planes.
Answer:
2 : 1

Question 48.
Two discs of moments of inertia I₁ and I₂ about their respective axes (normal to the disc and passing through the center), and rotating with angular speed col and ω₂ are brought into contact face to face with their axes of rotation coincident,

1. What is the angular speed of the two – disc system ?
2. Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy ? Take ω₁ ≠ ω₂.

Answer:

1. Let co be the angular speed of the two-disc system. Then by conservation of angular momentum
   
2. Initial K.E. of the two discs.
   

Hence there is a loss of rotational K.E. which appears as heat.
When the two discs are brought together, work is done against friction between the two discs.

Question 49.
In the HCL molecule, the separating between the nuclei of the two atoms is about 1.27 Å (1Å = 10^-10m). Find the approximate location of the CM of the molecule, given that the chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in all its nucleus.
Answer:
As shown in Fig. suppose the H nucleus is located at the origin. Then,
x₁ = 0, x₂ = 1.27 Å, m₁ = 1, m₂ = 35.5
The position of the CM of HCl molecule is


Thus the CM of HCl is located on the line joining H and Cl nuclei at a distance of 1.235 Å from the H nucleus.

Question 50.
A child stands at the center of turn table with his two arms out stretched. The turn table is set rotating with an angular speed of 40 rpm. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/3 times the initial value?

1. Assume that the turn table rotates without friction
2. Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation.

How do you account for this increase in kinetic energy ?
Answer:
Here ω = 40 rpm, I₂ = \(\frac { 1 }{ 2 }\) I₁
By the principle of conservation of angular momentum,
I₁ω₁ = I_2ω2 or I₁ x 4o = \(\frac {2}{5}\) I₁ ω₁ or ω₂ = 100 rpm
(ii) Initial kinetic energy of rotation –

new kinetic energy of rotation –

Thus the child’s new kinetic energy of rotation is 2.5 times its initial kinetic energy of rotation. This increase in kinetic energy is due to the internal energy of the child which he uses in folding his hands back from the out stretched position.

Question 51.
To maintain a rotor at a uniform angular speed of 200 rad s^-1 an engine needs to transmit a torque of 180 N m. What is the power required by the engine? Assume that the engine is 100% efficient.
Here ω = 200 rad s^-1, τ = 180 N m
Power, P = τω = 180 x 200 = 36,000 W = 36 kW.

Question 52.
A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its center of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front and back wheel.
Answer:

For transnational equilibrium of car
N_F + N_B = W = 1800 x 9.8 = 17640 N
For rotational equilibrium of car
1.05 N_F = 0.75 N_B
1.05 N_F = 0.75(17640 – N_F )
1.8 N_F = 13230
N_F = 13230 / 1.8 = 7350 N
N_B = 17640 – 7350 = 10290 N
Force on each front wheel = \(\frac {7350}{ 2 }\) = 3675 N
Force on each back wheel = \(\frac {10290}{ 2 }\) = 5145 N

Samacheer Kalvi 11th Physics Motion of System of Particles and Rigid Bodies Long Answer Questions (5 Marks)

Question 1.
Derive an expression for center of mass for distributed point masses.
Answer:
A point mass is a hypothetical point particle which has nonzero mass and no size or shape. To find the center of mass for a collection of n point masses, say,m₁ , m₂, m₃ ….. m_n  we have to first choose an origin and an appropriate coordinate system as shown in Figure. Let, x₁, x₂, x₃ …….. x_n be the X – coordinates of the positions of these point masses in the X direction from the origin.
The equation for the X coordinate of the center of mass is,

where, ∑ mi ¡s the total mass M of all the particles. ( ∑ mi = M).Hence,

Similarly, we can also find y and z coordinates of the center of mass for these distributed point masses as indicated in figure.

Hence, the position of center of mass of these point masses in a Cartesian coordinate system is (x_CM, y_CM z_CM). in general, the position of center of mass can be written in a vector form as,

where, is the position vector of the center of mass and \(\vec{r}_{i}\) = x_i \(\hat{j}\) + y_i\(\hat{j}\) + z_i\(\hat{k}\) is the position vector of the distributed point mass; where, \(\hat{i}\), \(\hat{j}\), and \(\hat{j}\) are the unit vectors along X, Y and Z-axis respectively.

Question 2.
Discuss the center of mass of two point masses with pictorial representation.
Answer:
With the equations for center of mass, let us find the center of mass of two point masses m₁ and m₂, which are at positions x₁ and x₂ respectively on the X – axis. For this case, we can express the position of center of mass in the following three ways based on the choice of the coordinate system.

(1) When the masses are on positive X-axis:
The origin is taken arbitrarily so that the masses m₁ and m₂ are at positions x₁ and x₂ on the positive X-axis as shown in figure (a). The center of mass will also be on the positive X- axis at x_CM as given by the equation,
\(x_{\mathrm{CM}}=\frac{m_{1} x_{1}+m_{2} x_{2}}{m_{1}+m_{2}}\)

(2) When the origin coincides with any one of the masses:
The calculation could be minimized if the origin of the coordinate system is made to coincide with any one of the masses as shown in figure (b). When the origin coincides with the point
mass m₁, its position x₁ is zero, (i.e. x₁ = 0). Then,
\(x_{\mathrm{CM}}=\frac{m_{1}(0)+m_{2} x_{2}}{m_{1}+m_{2}}\)
The equation further simplifies as,
x_CM = \(\frac{m_{2} x_{2}}{m_{1}+m_{2}}\)

(3) When the origin coincides with the center of mass itself:
If the origin of the coordinate system is made to coincide with the center of mass, then, x_CM = O and the mass rn1 is found to be on the negative X- axis as shown in figure (c). Hence, its position x1 is negative, (i.e. – x₁).

The equation given above is known as principle of moments.

Question 3.
Derive an expression for kinetic energy in rotation and establish the relation between rotational kinetic energy and angular momentum.
Answer:
Let us consider a rigid body rotating with angular velocity ω about an axis as shown in figure. Every particle of the body will have the same angular velocity ω and different tangential velocities v based on its positions from the axis of rotation. Let us choose a particle of mass m_i situated at distance r_i from the axis of rotation. It has a tangential velocity v_i given by the relation, v_i = r_i ω. The kinetic energy KE_i. of the particle is,
KE_i = \(\frac{1}{2} m_{i} v_{i}^{2}\)
Writing the expression with the angular velocity,
KE = \(\frac{1}{2}\) m_i(r_iω)² = \(\frac{1}{2} m_{i} r_{i}^{2}\)ω²

For the kinetic energy of the whole body, which is made up of large number of such particles, the equation is written with summation as,
KE = \(\frac{1}{2}\left(\sum m_{i} r_{i}^{2}\right)\)ω²
where, the term ∑ m_ir_i^r is the moment of inertia I of the whole body. ∑ m_ir_i^r
Hence, the expression for KE of the rigid body in rotational motion is –
KE = \(\frac{1}{2}\) Iω²
This is analogous to the expression for kinetic energy in transnational motion.
KE = \(\frac{1}{2}\) Mv²

Relation between rotational kinetic energy and angular momentum
Let a rigid body of moment of inertia I rotate with angular velocity ω.
The angular momentum of a rigid body is, L = Iω
The rotational kinetic energy of the rigid body is, KE = \(\frac{1}{2}\) Iω²
By multiplying the numerator and denominator of the above equation with I, we get a relation between L and KE as,

Question 4.
Discuss how the rolling is the combination of transnational and rotational and also be possibilities of velocity of different points in pure rolling.
Answer:
The rolling motion is the most commonly observed motion in daily life. The motion of wheel is an example of rolling motion. Round objects like ring, disc, sphere etc. are most suitable for rolling. Let us study the rolling of a disc on a horizontal surface. Consider a point P on the edge of the disc. While rolling, the point undergoes transnational motion along with its center of mass and rotational motion with respect to its center of mass.

Combination of Translation and Rotation:
We will now see how these transnational and rotational motions arc related in rolling. If the radius of the rolling object is R, in one full rotation, the center of mass is displaced by 2πR (its circumference). One would agree that not only the center of mass. but all the points Of l the disc are displaced by the same 2πR after one full rotation. The only difference is that the center of mass takes a straight path; but, all the other points undergo a path which has a combination of the transnational and rotational motion. Especially the point on the edge undergoes a path of a cyclonic as shown in the figure.

As the center of mass takes only a straight line path. its velocity v_CM is only transnational velocity v_TRANS (v_CM = v_TRANS). All the other points have two velocities. One is the transnational velocity v_TRANS (which is also the velocity of center of mass) and the other is the rotational velocity v_ROT (v_ROT = rω). Here, r ¡s the distance of the point from the center of mass and o is the angular velocity. The rotational velocity v_ROT is perpendicular to the instantaneous position vector from the center of mass as shown in figure (a). The resultant of these two velocities is v. This resultant velocity y is perpendicular to the position vector from the point of contact of the rolling object with the surface on which it is rolling as shown in figure (b).

We shall now give importance to the point of contact. In pure rolling, the point of the rolling object which comes in contact with the surface is at momentary rest. This is the case with every point that is on the edge of the rolling object. As the rolling proceeds, all’the points on the edge, one by one come in contact with the surface; remain at momentary rest at the time of contact and then take the path of the cycloid as already mentioned.
Hence, we can consider the pure rolling in two different ways.
(i) The combination of transnational motion and rotational motion about the center of mass.
(or)
(ii) The momentary rotational motion about the point of contact.
As the point of contact is at momentary rest in pure rolling, its resultant velocity v is zero (v = o). For example, in figure, at the point of contact, v_TRANS is forward (to right) and v_ROT is backwards (to the left).

That implies that, v_TRANS and v_ROT are equal in magnitude and opposite in direction (v = v_TRANS – v_ROT = 0). Hence, we conclude that in pure rolling, for all the points on the edge, the magnitudes of v_TRANS and v_ROT are equal (v_TRANS = v_ROT) As v_TRANS = v_CM and v_ROT = Rω, in pure rolling we have,
v_CM = Rω

We should remember the special feature of the above equation. In rotational motion, as per the relation v = rω, the center point will not have any velocity as r is zero. But in rolling motion, it suggests that the center point has a velocity v_CM given by above equation v_CM – Rω. For the topmost point, the two velocities v_TRANS and v_ROT are equal in magnitude and in the same direction (to the right). Thus, the resultant velocity v is the sum of these two velocities, v = v_TRANS + v_ROT In other form, v = 2 v_CM as shown in figure below.

Question 5.
Derive an expression for kinetic energy in pure rolling.
Answer:
As pure is the combination of transnational and rotational motion, we can write the total kinetic energy (KE) as the sum of kinetic energy due to transnational motion (KE_TRANS) and kinetic energy due to rotational motion (KE_ROT).
KE = KE_TRANS + KE_ROT ………(i)
If the mass of the rolling object is M, the velocity of center of mass is v_CM, its moment of inertia about center of mass is I_CM and angular velocity is ω, then

With center of mass as reference:
The moment of inertia (I_CM) of a rolling object about the center of mass is, I_CM = MK² and v_CM = Rω. Here, K is radius of gyration.

With point of contact as reference:
We can also arrive at the same expression by taking the momentary rotation happening with respect to the point of contact (another approach to rolling). If we take the point of contact as o, then,
KE = \(\frac {1}{2}\) I₀ω²

Here, I₀ is the moment of inertia of the object about the point of contact. By parallel axis
theorem, I₀ = I_CM + MK² Further we can write, I₀ MK² + MR². With v_CM = Rω or ω = \(\frac{v_{\mathrm{CM}}}{\mathrm{R}}\)

As the two equations (v) and (vi) are the same, it ¡s once again confirmed that the pure tolling problems could be solved by considering the motion as any one of the following two cases.
(i) The combination of transnational motion and rotational motion about the center of mass.
(or)
(ii) The momentary rotational motion about the point of contact.

Question 6.

1. Can a body in translator y motion have angular momentum? Explain.
2. Why is it more difficult to revolve a stone by tying it to a longer string than by tying it to a shorter string?

Answer:
(1) Yes, a body in translatory motion shall have angular momentum unless fixed point about which angular momentum is taken lies on the line of motion of body
\(|\overrightarrow{\mathrm{L}}|\) = rp sin θ
= 0 only when θ = O° or 180°

(2) MI of stone I = ml² (l – length of string) l is large, a is very small
τ = Iα
α = \(\frac {τ}{I}\) = \(\frac{\tau}{m l^{2}}\)
if l is large a is very small.
∴ more difficult to revolve.

Students can Download Bio Botany Chapter 7 Cell Cycle Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Bio Botany Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 7 Cell Cycle

Samacheer Kalvi 11th Bio Botany Cell Cycle Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the correct answer
Question 1.
The correct sequence in cell cycle is …………… .
(a) S-M-G1-G2
(b) S-G1-G2-M
(c) G1-S-G2-M
(d) M-G-G2-S
Answer:
(c) G1-S-G2-M

Question 2.
If mitotic division is restricted in G1 phase of the cell cycle then the condition is known as …………… .
(a) S Phase
(b) G2 Phase
(c) M Phase
(d) G₀ Phase
Answer:
(d) G₀ Phase

Question 3.
Anaphase promoting complex APC is a protein degradation machinery necessary for proper mitosis of animal cells. If APC is defective in human cell, which of the following is expected to occur?
(a) Chromosomes will be fragmented
(b) Chromosomes will not condense
(c) Chromosomes will not segregate
(d) Recombination of chromosomes will occur
Answer:
(b) Chromosomes will not condense

Question 4.
In S phase of the cell cycle …………… .
(a) Amount of DNA doubles in each cell
(b) Amount of DNA remains same in each cell
(c) Chromosome number is increased
(d) Amount of DNA is reduced to half in each cell
Answer:
(a) Amount of DNA doubles in each cell

Question 5.
Centromere is required for …………… .
(a) Transcription
(b) Crossing over
(c) Cytoplasmic cleavage
(d) Movement of chromosome towards pole
Answer:
(d) Movement of chromosome towards pole

Question 6.
Synapsis occur between …………… .
(a) mRNA and ribosomes
(b) Spindle fibres and centromeres
(c) Two homologous chromosomes
(d) A male and a female gamete
Answer:
(c) Two homologous chromosomes

Question 7.
In meiosis crossing over is initiated at …………… .
(a) Diplotene
(b) Pachytene
(c) Leptotene
(d) Zygotene
Answer:
(b) Pachytene

Question 8.
Colchicine prevents the mitosis of the cells at which of the following stage …………… .
(a) Anaphase
(b) Metaphase
(c) Prophase
(d) Interphase
Answer:
(b) Metaphase

Question 9.
The paring of homologous chromosomes on meiosis is known as …………… .
(a) Bivalent
(b) Synapsis
(c) Disjunction
(d) Synergids
Answer:
(b) Synapsis

Question 10.
Anastral mitosis is the characteristic feature of …………… .
(a) Lower animals
(b) Higher animals
(c) Higher plants
(d) All living organisms
Answer:
(c) Higher plants

Question 11.
Write any three significance of mitosis.
Answer:
Exact copy of the parent cell is produced by mitosis (genetically identical).

1. Genetic stability – daughter cells are genetically identical to parent cells.
2. Repair of tissues – damaged cells must be replaced by identical new cells by mitosis.
3. Regeneration – Arms of star fish.

Question 12.
Differentiate between Mitosis and Meiosis.
Answer:
Difference Between Mitosis and Meiosis:

Difference Between Mitosis and Meiosis
Mitosis	Meiosis
1. One division	1. Two divisions
2. Number of chromosomes remains the same	2. Number of chromosomes is halved
3. Homologous chromosomes line up separately on the metaphase plate	3. Homologous chromosomes line up in pairs at the metaphase plate
4. Homologous chromosome do not pair up	4. Homologous chromosome pairup to form bivalent
5. Chiasmata do not form and crossing over never occurs	5. Chiasmata form and crossingover occurs
6. Daughter cells are genetically identical	6. Daughter cells are genetically different from the parent cells
7. Two daughter cells are formed	7. Four daughter cells are formed

Question 13.
Given an account of G₀ phase.
Answer:
Some cells exit G₁ and enters a quiescent stage called G₀, where the cells remain metabolically active without proliferation. Cells can exist for long periods in G₀ phase. In G₀ cells cease growth with reduced rate of RNA and protein synthesis. The G₀ phase is not permanent. Mature neuron and skeletal muscle cell remain permanently in G₀ .Many cells in animals remains in G₀ unless called onto proliferate by appropriate growth factors or other extracellular signals. G₀ cells are not dormant.

Question 14.
Differentiate Cytokinesis in plant cells and animal cells.
Answer:
1. Cytokinesis in Plant Cells:
Division of the cytoplasm often starts during telophase. In plants, cytokinesis cell plate grows from centre towards lateral walls centrifugal manner of cell plate formation. Phragmoplast contains microtubules, actin filaments and vesicles from golgi apparatus and ER. The golgi vesicles contains carbohydrates such as pectin, hemicellulose which move along the microtubule of the pharagmoplast to the equator fuse, forming a new plasma membrane and the materials which are placed their becomes new cell wall.

The first stage of cell wall construction is a line dividing the newly forming cells called a cell plate. The cell plate eventually stretches right across the cell forming the middle lamella. Cellulose builds up on each side of the middle lamella to form the cell walls of two new plant cells.

2. Cytokinesis in Animal Cells:
It is a contractile process. The contractile mechanism contained in contractile ring located inside the plasma membrane. The ring consists of a bundle of microfilaments assembled from actin and myosin. This fibril helps for the generation of a contractile force. This force draws the contractile ring inward forming a cleavage furrow in the cell surface dividing the cell into two.

Question 15.
Write about Pachytene and Diplotene of Prophase I.
Answer:
1. Pachytene: At this stage bivalent chromosomes are clearly visible as tetrads. Bivalent of meiosis I consists of 4 chromatids and 2 centromeres. Synapsis is completed and recombination nodules appear at a site where crossing over takes place between non – sister chromatids of homologous chromosome. Recombination of homologous chromosomes is completed by the end of the stage but the chromosomes are linked at the sites of crossing over. This is mediated by the enzyme recombinase.

2. Diplotene: Synaptonemal complex disassembled and dissolves. The homologous chromosomes remain attached at one or more points where crossing over has taken place. These points of attachment where ‘X’ shaped structures occur at the sites of crossing over is called Chiasmata. Chiasmata are chromatin structures at sites where recombination has been taken place. They are specialised chromosomal structures that hold the homologous chromosomes together.

Sister chromatids remain closely associated whereas the homologous chromosomes tend to separate from each other but are held together by chiasmata. This substage may last for days or years depending on the sex and organism. The chromosomes are very actively transcribed in females as the egg stores up materials for use during embryonic development. In animals, the chromosomes have prominent loops called lampbrush chromosome.

Samacheer Kalvi 11th Bio Botany Cell Cycle Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer:
Question 1.
Most of the neurons in the brain are in …………… stage.
(a) G₁
(b) S
(c) G₂
(d) G₀
Answer:
(d) G₀

Question 2.
Short, constricted region in the chromosome is …………… .
(a) Kinetochore
(b) Centromere
(c) Satellite
(d) Telomere
Answer:
(b) Centromere

Question 3.
Robert Brown discovered the nucleus in the cells of …………… roots.
(a) Mirabilas
(b) Orchid
(c) Moringa
(d) Oryza
Answer:
(b) Orchid

Question 4.
Scientist who described chromosomes for the first time is …………… .
(a) Robert Brown
(b) Anton van Leeuwenhoek
(c) Boveri
(d) Anton Schneider
Answer:
(d) Anton Schneider

Question 5.
Number of chromosomes in onion cell is …………… .
(a) 8
(b) 16
(c) 32
(d) 64
Answer:
(a) 16

Question 6.
Longest part of the cell cycle is …………… .
(a) Prophase
(b) G₁ Phase
(c) Interphase
(d) Sphase
Answer:
(c) Interphase

Question 7.
Eukaryotic cells divides every …………… .
(a) 12
(b) 24
(c) 1
(d) 6
Answer:
(b) 24

Question 8.
Cell cycle was discovered by …………… .
(a) Singer & Nicolson
(b) Prevost & Dumans
(c) Schleider & Schwann
(d) Boveri
Answer:
(b) Prevost & Dumans

Question 9.
G₀ stage is called as …………… stage.
(a) Quiescent
(b) Metabolically active
(c) Synthesis of DNA
(d) Replication
Answer:
(a) Quiescent

Question 10.
…………… protein acts as major check point in phase.
(a) Porins
(b) Kinases
(c) Cyclins
(d) Ligases
Answer:
(c) Cyclins

Question 11.
Replication of DNA occurs at …………… phase.
(a) G₀
(b) G₁
(c) S
(d) G₂
Answer:
(c) S

Question 12.
Condensation of interphase chromosomes into mitotic forms is done by …………… proteins.
(a) MPF
(b) APF
(c) AMF
(d) MAF
Answer:
(a) MPF

Question 13.
Which of the following is also called as direct division?
(a) Amitosis
(b) Meiosis
(c) Mitosis
(d) Reduction division
Answer:
(a) Amitosis

Question 14.
Cells of mammalian cartilage undergoes …………… .
(a) Amitosis
(b) Meiosis
(c) Mitosis
(d) Equational division
Answer:
(a) Amitosis

Question 15.
Yeast cells undergo …………… .
(a) Open mitosis
(b) Closed mitosis
(c) Amitosis
(d) Meiosis
Answer:
(b) Closed mitosis

Question 16.
…………… is the longest phase in mitosis.
(a) Anaphase
(b) Telophase
(c) Prophase
(d) Interphase
Answer:
(c) Prophase

Question 17.
The DNA protein complex present in the centromere is …………… .
(a) Cyclin
(b) Kinesis
(c) MPF
(d) Kinetochore
Answer:
(d) Kinetochore

Question 18.
…………… protein induces the break down of cohesion proteins leading to chromatid separation during mitosis.
(a) APC
(b) MPF
(c) Cyclin
(d) Kinetochore
Answer:
(a) APC

Question 19.
Regeneration of arms of star fish is due to …………… .
(a) Meiosis
(b) Amitosis
(c) Mitosis
(d) Budding
Answer:
(c) Mitosis

Question 20
…………… is called as reduction division.
(a) Meiosis
(b) Mitosis
(c) Amitosis
(d) Budding
Answer:
(a) Meiosis

Question 21.
Bivalents occur at …………… stage.
(a) Zygotene
(b) Leptotene
(c) Pachytene
(d) Diplotene
Answer:
(a) Zygotene

Question 22.
Recombination of chromosomes occur at …………… .
(a) Zygotene
(b) Leptotene
(c) Pachytene
(d) Diplotene
Answer:
(c) Pachytene

Question 23.
Terminalisation of chiasmata occurs at …………… .
(a) Zygotene
(b) Leptotene
(c) Diakinesis
(d) Pachytene
Answer:
(c) Diakinesis

Question 24.
Number of daughter cells formed at the end of Meiosis I is …………… .
(a) 2
(b) 4
(c) 1
(d) 0
Answer:
(a) 2

Question 25.
…………… division leads to genetic variability.
(a) Mitotic
(b) Amitotic
(c) Meiotic
(d) Equational
Answer:
(c) Meiotic

Question 26.
Crossing over occurs at …………… stage.
(a) Leptotene
(b) Zygotene
(c) Pachytene
(d) Diplotene
Answer:
(c) Pachytene

Question 27.
Which of the following is not a mitogen?
(a) Giberellin
(b) Ethylene
(c) Kinetin
(d) Colchicine
Answer:
(d) Colchicine

Question 28.
In plants mitosis occurs at …………… cells.
(a) Sclerenchyma
(b) Meristem
(c) Xylem
(d) Parenchyma
Answer:
(b) Meristem

Question 29.
Which of the following alone is formed in the division of plant cells?
(a) Aster
(b) Centrioles
(c) Spindle
(d) Microtubules
Answer:
(c) Spindle

Question 30.
Amphiastral type cell division is seen in …………… cells.
(a) Fungal
(b) Algal
(c) Plant cells
(d) Animal
Answer:
(d) Animal

II. Very Short Answer Type Questions (2 Marks)

Question 1.
Name the two types of nuclear division.
Answer:
The two types of nuclear division:

1. Mitosis and
2. Meiosis.

Question 2.
Define Cell Cycle.
Answer:
A series of events leading to the formation of new cell is known as cell cycle.

Question 3.
Who discovered the Cell Cycle?
Answer:
Prevost & Dumans in 1824.

Question 4.
Draw a tabular column showing the duration of various phase in the cell cycle of human cell.
Answer:
A tabular column showing the duration of various phase in the cell cycle of human cell:

Cell cycle of a proliferating human cell
Phase	Time Duration (in hrs)
1. G2	1. 11
2. S	2. 8
3. G2	3. 4
4. M	4. 1

Question 5.
Define C – Value.
Answer:
C – Value is the amount in picograms of DNA contained within a haploid nucleus.

Question 6.
Which is the longest phase of cell cycle? What happens during that phase?
Answer:
Interphase is the longest phase. Cells are metabolically active and involved in protein synthesis and growth.

Question 7.
Name the phases which comprises the Interphase.
Answer:
The phases which comprises the Interphase:

1. G₁ Phase
2. S Phase and
3. G₂ Phase.

Question 8.
Name the proteins involved in the activation of genes & their proteins to perform cell division.
Answer:
Kinases & Cyclins.

Question 9.
What do you mean by G₀ stage?
Answer:
G₀ stage is called as quiescent stage, where the cells remain metabolically active without proliferation.

Question 10.
What is the role of MPF in Cell cycle?
Answer:
Maturation Promoting Factor (MPF) brings about condensation of interphase chromosomes into the mitotic form.

Question 11.
Distinguish between Karyokinesis & Cytokinesis.
Answer:
Between Karyokinesis & Cytokinesis:

• Karyokinesis: Karyokinesis refers to the nuclear division.
• Cytokinesis: Cytokinesis refers to the cytoplasmic division.

Question 12.
Point out any two cell – types which remain G₀ phase.
Answer:
Mature neurons and Skeletal muscle cells.

Question 13.
Why amitosis is called as incipient cell division?
Answer:
Amitosis is also called incipient cell division. Since there is no spindle formation and chromatin material does not condense.

Question 14.
List out the disadvantages of Amitosis.
Answer:
The disadvantages of Amitosis:

• Causes unequal distribution of chromosomes.
• Can lead to abnormalities in metabolism and reproduction.

Question 15.
Mitosis also called as equational division – Justify.
Answer:
At the end of mitosis the number of chromosomes in the parent and the daughter (Progeny) cells remain the same so it is also called as equational division.

Question 16.
Enumerate the stages of mitosis.
Answer:
Mitosis is divided into four stages prophase, metaphase, anaphase and telophase.

Question 17.
Define an aster.
Answer:
In animal cell the centrioles extend a radial array of microtubules towards the plasma membrane when they reach the poles of the cell. This arrangement of microtubules is called an aster. Plant cells do not form asters.

Question 18.
What is metaphase plate?
Answer:
The alignment of chromosome into compact group at the equator of the cell is known as metaphase plate.

Question 19.
What is Kinetochore?
Answer:
Kinetochore is a DNA – Protein complex present in the centromere DNA, where the microtubules are attached. It is a trilaminar disc like plate.

Question 20.
How will you calculate the length of the S period.
Answer:
Length of the S period = Fraction of cells in DNA replication × generation time.

Question 21.
Which type of cell division occurs in reproductive cells? What will be the result?
Answer:
Meiosis takes place in the reproductive organs. It results in the formation of gametes with half the normal chromosome number.

Question 22.
Define Synapsis.
Answer:
In Zygotene, pairing of homologous chromosomes takes place and it is known as synapsis.

Question 23.
What do you understand by independent assortment?
Answer:
The random distribution of homologous chromosomes in a cell in Metaphase I is called independent assortment.

Question 24.
Define Mitogen. Give an example.
Answer:
The factors which promote cell cycle proliferation is called mitogen.
Example: gibberellin. These increase mitotic rate.

Question 25.
What are mitotic poisons.
Answer:
Certain chemical components act as inhibitors of the mitotic cell division and they are called mitotic poisons.

Question 26.
Distinguish between Anastral & Amphiastral.
Answer:
Between Anastral & Amphiastral:
Anastral:

1. This is present only in plant cells.
2. No asters or centrioles are formed only spindle fibres are formed during cell division.

Amphiastral:

1. This is found in animal cells.
2. Aster and centrioles are formed at each pole of the spindle during cell division.

Question 27.
Draw a simple diagram to show the Amitosis.
Answer:
The Amitosis:

III. Short Answer Type Questions (3 Marks)

Question 1.
What is the role of nucleus in the cell?
Answer:
The role of nucleus in the cell:

• Control activities of the cell.
• Genetic information copied from cell to cell while the cell divides.
• Hereditary characters are passed onto new individuals when gametic cells fuse together in sexual reproduction.

Question 2.
What are restriction points? Mention its role in Cell cycle.
Answer:
The checkpoint called the restriction point at the end of G₁ it determines a cells fate whether it will continue in the cell cycle and divide or enter a stage called G₀ as a quiescent stage and probably as specified cell or die.

Question 3.
Point out the reasons responsible for the arresting of the cell in G₁ phase?
Answer:
Cells are arrested in G₁ due to:

• Nutrient deprivation
• Lack of growth factors or density dependant inhibition
• Undergo metabolic changes and enter into G₀ state.

Question 4.
Write a note on G₀ phase.
Answer:
Some cells exit G₁ and enters a quiescent stage called G₀, where the cells remain metabolically active without proliferation. Cells can exist for long periods in G₀ phase. In G₀ cells cease growth with reduced rate of RN A and protein synthesis. The G₀ phase is not permanent. Mature neuron and skeletal muscle cell remain permanently in G₀. Many cells in animals remains in G₀ unless called onto proliferate by appropriate growth factors or other extracellular signals. G₀ cells are not dormant.

Question 5.
List out the events taking place in S – Phase.
Answer:
S Phase – Synthesis phase – cells with intermediate amounts of DNA Growth of the cell continues as replication of DNA occur, protein molecules called histones are synthesised and attach to the DNA. The centrioles duplicate in the cytoplasm. DNA content increases from 2C to 4C.

Question 6.
Distinguish between Karyokinesis & Cytokinesis.
Answer:
Karyokinesis:

1. Involves division of nucleus.
2. Nucleus develops a constriction at the center and becomes dumbellshaped.
3. Constriction deepens and divides the nucleus into two.

Cytokinesis:

1. Involves division of cytoplasm.
2. Plasma membrane develops a constriction along nuclear constriction.
3. It deepens centripetally and finally divides the cell into two cells.

Question 7.
Explain the differences between closed and open mitosis.
Answer:
Between closed and open mitosis:

1. In closed mitosis, the nuclear envelope remains intact and chromosomes migrate to opposite poles of a spindle within the nucleus. Example: Yeast and slime molds.
2. In open mitosis, the nuclear envelope breaks down and then reforms around the 2 sets of separated chromosome. Example: Most plants and animals cells.

Question 8.
What happens to plant cells at the end of Telophase in Mitosis?
Answer:
In plants, phragmoplast are formed between the daughter cells. Cell plate is formed between the two daughter cells, reconstruction of cell wall takes place. Finally the cells are separated by the distribution of organelles, macromolecules into two newly formed daughter cells.

Question 9.
Bring out the significance of Meiosis.
Answer:
The significance of Meiosis:

• Meiosis maintains a definite constant number of chromosomes in organisms.
• Crossing over takes place and exchange of genetic material leads to variations among species. These variations are the raw materials to evolution. Meiosis leads to genetic variability by partitioning different combinations of genes into gametes through independent assortment.
• Adaptation of organisms to various environmental stress.

Question 10.
Differentiate between the mitosis of Plant Cell & Animal Cell.
Answer:
Plants:

1. Centrioles are absent
2. Asters are not formed
3. Cell division involves formation of a cell plate
4. Occurs mainly at meristem.

Animals:

1. Centrioles are present
2. Asters are formed
3. Cell division involves furrowing and cleavage of cytoplasm
4. Occurs in tissues throughout the body.

Question 11.
Explain briefly about Endomitosis.
Answer:
The replication of chromosomes in the absence of nuclear division and cytoplasmic division resulting in numerous copies within each cell is called endomitosis. Chromonema do not separate to form chromosomes, but remain closely associated with each other. Nuclear membrane does not rupture. So no spindle formation. It occurs notably in the salivary glands of Drosophila and other flies. Cells in these tissues contain giant chromosomes (polyteny), each consisting of over thousands of intimately associated, or synapsed, chromatids. Example: Polytene chromosome.

Question 12.
How G₀ cells help in Closing Technology?
Answer:
Since the DNA of cells in G₀, do not replicate. The researcher are able to fuse the donor cells from a sheep’s mammary glands into G₀, state by culturing in the nutrient free state. The G₀, donor nucleus synchronised with cytoplasm of the recipient egg, which developed into the clone Dolly.

IV. Long Answer Type Questions (5 Marks)

Question 1.
Draw and label the various stages of Prophase I.
Answer:
Label the various stages of Prophase I:

Question 2.
Explain in detail about the various stages of Prophase I.
Answer:
The various stages of Prophase I:
1. Prophase I – Prophase I is of longer duration and it is divided into 5 substages – Leptotene, Zygotene, Pachytene, Diplotene and Diakinesis.

2. Leptotene – Chromosomes are visible under light microscope. Condensation of chromosomes takes place. Paired sister chromatids begin to condense.

3. Zygotene – Pairing of homologous chromosomes takes place and it is known as synapsis. Chromosome synapsis is made by the formation of synaptonemal complex. The complex formed by the homologous chromosomes are called as bivalent (tetrads).

4. Pachytene – At this stage bivalent chromosomes are clearly visible as tetrads. Bivalent of meiosis I consists of 4 chromatids and 2 centromeres. Synapsis is completed and recombination nodules appear at a site where crossing over takes place between non – sister chromatids of homologous chromosome. Recombination of homologous chromosomes is completed by the end of the stage but the chromosomes are linked at the sites of crossing over. This is mediated by the enzyme recombinase.

5. Diplotene – Synaptonemal complex disassembled and dissolves. The homologous chromosomes remain attached at one or more points where crossing over has taken place. These points of attachment where ‘X’ shaped structures occur at the sites of crossing over is called.

6. Chiasmata: Chiasmata are chromatin structures at sites where recombination has been taken place. They are specialised chromosomal structures that hold the homologous chromosomes together. Sister chromatids remain closely associated whereas the homologous chromosomes tend to separate from each other but are held together by chiasmata. This substage may last for days or years depending on the sex and organism. The chromosomes are very actively transcribed in females as the egg stores up materials for use during embryonic development. In animals, the chromosomes have prominent loops called lampbrush chromosome.

7. Diakinesis – Terminalisation of chiasmata. Spindle fibres assemble. Nuclear envelope breaks down. Homologous chromosomes become short and condensed. Nucleolus disappears.

Question 3.
Describe the process of Cytokinesis in Plant cell & Animal Cell.
Answer:
1. Cytokinesis in Plant Cell: Division of the cytoplasm often starts during telophase. In plants, cytokinesis cell plate grows from centre towards lateral walls – centrifugal manner of cell plate formation. Phragmoplast contains microtubules, actin filaments and vesicles from golgi apparatus and ER. The golgi vesicles contains carbohydrates such as pectin, hemicellulose which move along the microtubule of the pharagmoplast to the equator fuse, forming a new plasma membrane and the materials which are placed there becomes new cell wall.

The first stage of cell wall construction is a line dividing the newly forming cells called a cell plate. The cell plate eventually stretches right across the cell forming the middle lamella. Cellulose builds up on each side of the middle lamella to form the cell walls of two new plant cells.

2. Cytokinesis in Animal Cells:
It is a contractile process. The contractile mechanism contained in contractile ring located inside the plasma membrane. The ring consists of a bundle of microfilaments assembled from actin and myosin. This fibril helps for the generation of a contractile force. This force draws the contractile ring inward forming a cleavage furrow in the cell surface dividing the cell into two.

Question 4.
What are the significances of Mitosis.
Answer:
Exact copy of the parent cell is produced by mitosis (genetically identical).

1. Genetic stability – Daughter cells are genetically identical to parent cells.
2. Growth – As multicellular organisms grow, the number of cells making up their tissue increases. The new cells must be identical to the existing ones.
3. Repair of tissues – Damaged cells must be replaced by identical new cells by mitosis.
4. Asexual reproduction – Asexual reproduction results in offspring that are identical to the parent. Example Yeast and Amoeba.
5. In flowering plants, structure such as bulbs, corms, tubers, rhizomes and runners are produced by mitotic division. When they separate from the parent, they form a new individual. The production of large numbers of offsprings in a short period of time, is possible only by mitosis. In genetic engineering and biotechnology, tissues are grown by mitosis (i.e. in tissue culture).
6. Regeneration – Arms of star fish

Question 5.
Explain the various phases in Cell Cycle.
Answer:
The different phases of cell cycle are as follows:
1. Interphase: Longest part of the cell cycle, but it is of extremely variable length. At first glance the nucleus appears to be resting but this is not the case at all. The chromosomes previously visible as thread like structure, have dispersed. Now they are actively involved in protein synthesis, at least for most of the interphase. C – Value is the amount in picograms of DNA contained within a haploid nucleus.

2. G₁ Phase: The first gap phase – 2C amount of DNA in cells of G₁ The cells become metabolically active and grows by producing proteins, lipids, carbohydrates and cell organelles including mitochondria and endoplasmic reticulum. Many checkpoints control the cell cycle. The checkpoint called the restriction point at the end of G₁ it determines a cells fate whether it will continue in the cell cycle and divide or enter a stage called G₀ as a quiescent stage and probably as specified cell or die. Cells are arrested in G₁ due to:

3. Nutrient deprivation: Lack of growth factors or density dependant inhibition. Undergo metabolic changes and enter into G₀ state. Biochemicals inside cells activates the cell division. The proteins called kinases and cyclins activate genes and their proteins to perform cell division. Cyclins act as major checkpoint which operates in G₁ to determine whether or not a cell divides.

4. G₀ Phase: Some cells exit G₁ and enters a quiescent stage called G₀, where the cells remain metabolically active without proliferation. Cells can exist for long periods in G₀ phase. In G₀ cells cease growth with reduced rate of RNA and protein synthesis. The G₀ phase is not permanent. Mature neuron and skeletal muscle cell remain permanently in G₀. Many cells in animals remains in G₀ unless called on to proliferate by appropriate growth factors or other extracellular signals. G₀ cells are not dormant.

5. S phase – Synthesis phase – cells with intermediate amounts of DNA. Growth of the cell continues as replication of DNA occur, protein molecules called histones are synthesised and attached to the DNA. The centrioles duplicate in the cytoplasm. DNA content increases from 2C to 4C.

6. G₂ – The second Gap phase – 4C amount of DNA in cells of G₂ and mitosis. Cell growth continues by protein and cell organelle synthesis, mitochondria and chloroplasts divide. DNA content remains as 4C. Tubulin is synthesised and microtubules are formed. Microtubles organise to form spindle fibre. The spindle begins to form and nuclear division follows.

One of the proteins synthesized only in the G₂ period is known as Maturation Promoting Factor (MPF). It brings about condensation of interphase chromosomes into the mitotic form. DNA damage checkpoints operates in G₁ S and G₂ phases of the cell cycle.

Question 6.
List out the important features of Chromosomes.
Answer:
The four important features of the chromosome are:
1. The shape of the chromosome is specific: The long, thin, lengthy structured chromosome contains a short, constricted region called centromere. A centromere may occur any where along the chromosome, but it is always in the same position on any given chromosome. The number of chromosomes per species is fixed: For example the mouse has 40 chromosomes, the onion has 16 and humans have 46.

2. Chromosomes occur in pairs: The chromosomes of a cell occur in pairs, called homologous pairs. One of each pair come originally from each parent. Example, human has 46 chromosomes, 23 coming originally from each parent in the process of sexual reproduction. Chromosomes are copied: Between nuclear divisions, whilst the chromosomes are uncoiled and cannot be seen, each chromosome is copied. The two identical structures formed are called chromatids.

V. Higher Order Thinking Skills (HOTs)

Question 1.
Given that the average duplication time of E.coli is 20 minutes. How much time will two E.coli cells takes to become 32 cells?
Answer:

One cells takes 80 minutes to form 16 cells. If 2 cells undergoes division simultaneously, it take 160 minutes (2 hours 40 minutes) to form 32 cells.

Question 2.
Complete the cell cycle by filling the gaps with respective phases.
Answer:
X= S phase or Synthesis phase
Y= M phase or Mitosis phase
Z= G₀ phase

Question 3.
Telophase is reverse of prophase – Comment
Answer:
Events in Prophase:

1. Nuclear membrane disappears
2. Nucleolus disappear
3. Spindle fibre begins to form
4. Chromosomes threads condeme to form chromosomes

Events in Telophase:

1. Nuclear membrane reappears
2. Nucleolus reappears
3. Spindle fibre disappears
4. Chromosomes decondeme to form chromosomes

Question 4.
Name the pathological condition when uncontrolled cell division occurs.
Answer:
Uncontrolled cell division & abnormal growth of cells leads to the pathological condition called tumor or cancer.

Question 5.
Microspores are produced in the multiples of four, why?
Answer:
Microspores are haploid spores produced from diploid microspores mother cells. Each microspores mother cell (2n) undergoes meiosis producing four Microspores (n). Because a complete meiotic division yields 4 cells. Thus microspores are produced in multiples of four.

Question 6.
Between Prokaryotes & Eukaryotes, which cell has a shorter cell division time.
Answer:
Prokaryotes like bacteria undergo simple form of cell division called binary fission which will get completed with in a hour, whereas Eukaryotic cell division (mitosis) takes nearly 24 hours to get completed. Hence Prokaryotes have shorter cell division time.

Question 7.
Though Prokaryotic cell division differs from Eukaryotic cell division, both show certain common aspects during cell division. Explain.
Answer:
Whether a cell is Prokaryote or Eukaryote, while undergoing division, the following events must occur in common.

1. Replication of DNA.
2. Cytokinesis at the end of cell division.

Question 8.
An anther has 1204 pollen grains. How many Pollen mother cells must have been there to produce them?Explain.
Answer:
301 – Pollen mother cells: 301 Pollen mother cells undergo meiosis producing 1204 pollen grains. Because at the end of meiosis, each pollen mother cells produces 4 pollen grains.

Question 9.
A cell has 32 chromosomes. It undergoes mitosis. What will be the chromosome number during metaphase?
Answer:
During S phase of interphase, the genetic material of the cell is duplicated. So during metaphase, the chromosome number(chromatid number) will be doubled thus 64 chromosomes (chromatids) will be present.

Question 10.
Why sibilings show disimilarities?
Answer:
Though born to same parents, siblings show dissimilarities and variation due to the crossing over and recombination of chromosomes during meiosis.

Question 11.
Ramu’s met with an accident while riding cycle and got wounded in his leg. After few days, the wound was healed and the skin becomes normal. How?
Answer:
Ramu’s wound was healed because of the mitotic division. As a result of mitosis, new cells are produced and damaged tissues were repaired resulting the damaged skin to become normal.

Question 12.
A flower of tomato plant following the process of sexual reproduction produces 240 viable seeds. What is the minimum number of microspore mother cells involved in this process?
Answer:
60 microspore mother cells are involved in providing 240 pollen grains. Because each microspore mother cell undergoes meiosis producing four pollen grains (i.e. 60 × 4 = 240). Each pollen grain produces two male gametes of which one undergoes true fertilization of ovule producing seeds. Other male gamete participate in double fertilization.

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5

Question 1.
Solve 2x² + x – 15 ≤ 0.
Solution:
To find the solution of the inequality
ax² + bx + c ≥ 0 or ax² + bx +c ≤ 0 (for a > 0)
First we have to solve the quadratic equation ax² + bx + c = 0
Let the roots be a and P (where a < P)
So for the inequality ax² + bx + c ≥ 0 the roots lie outside α and β
(i.e.,) x ≤ α and x ≥ β
So for the inequality ax² + bx + c ≤ 0. The roots lie between α and β
(i.e.,) x > α and x < β (i.e.) a ≤ x ≤ β

The inequality solver will then show you the steps to help you learn how to solve it on your own.

Question 2.
Solve -x² + 3x – 2 ≥ 0
Solution:
-x² + 3x – 2 ≥ 0 ⇒ x² – 3x + 2 ≤ 0
(x – 1) (x – 2) ≤ 0
[(x – 1) (x – 2) = 0
⇒ x = 1 or 2.
Here α = 1 and β = 2. Note that α < β]
So for the inequality (x – 1) (x – 2) ≤ 2
x lies between 1 and 2
(i.e.) x ≥ 1 and x ≤ 2 or x ∈ [1, 2] or 1 ≤ x ≤ 2

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5 Additional Questions

Question 1.
Solve for x.

Solution:

Select the intervals in which (3x +1) (3x – 2) is positive
(3x + 1) > 0 and (3x – 2) > 0 or
3x +1 < 0 and 3x – 2 < 0

Question 2.

Solution:

Question 3.

Solution:

Students can Download Bio Botany Chapter 9 Tissue and Tissue System Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Bio Botany Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 9 Tissue and Tissue System

Samacheer Kalvi 11th Bio Botany Tissue and Tissue System Text Book Back Questions and Answers

Question 1.
Refer to the given figure and select the correct statement:

(i) A, B, and C are histogen of shoot apex
(ii) A Gives rise to medullary rays
(iii) B Gives rise to cortex
(iv) C Gives rise to epidermis
(a) (i) and (ii) only
(b) (ii) and (in) only
(c) (i) and (iii) only
(d) (iii) and (iv) only
Answer:
(c) (i) and (iii) only

Question 2.
Read the following sentences and identify the correctly matched sentences.
(i) In exarch condition, the protoxylem lies outside of metaxylem.
(ii) In endarch condition, the protoxylem lie towards the centre.
(iii) In centarch condition, metaxylem lies in the middle of the protoxylem.
(iv) In mesarch condition, protoxylem lies in the middle of the metaxylem.
(a) (i), (ii) and (iii) only
(b) (ii), (iii) and (iv) only
(c) (i), (ii) and (iv) only
(d) All of these
Answer:
(c) (i), (ii) and (iv) only

Question 3.
In Gymnosperms, the activity of sieve tubes are controlled by:
(a) Nearby sieve tube members.
(b) Phloem parenchyma cells.
(c) Nucleus of companion cells.
(d) Nucleus of albuminous cells.
Answer:
(c) Nucleus of companion cells.

Question 4.
When a leaf trace extends from a vascular bundle in a dicot stem, what would be the arrangement of vascular tissues in the veins of the leaf?
(a) Xylem would be on top and the phloem on the bottom
(b) Phloem would be on top and the xylem on the bottom
(c) Xylem would encircle the phloem
(d) Phloem would encircle the xylem
Answer:
(a) Xylem would be on top and the phloem on the bottom

Question 5.
Grafting is successful in dicots but not in monocots because the dicots have:
(a) vascular bundles arranged in a ring
(b) cambium for secondary growth
(c) vessels with elements arranged end to end
(d) cork cambium
Answer:
(b) cambium for secondary growth

Question 6.
Why the cells of sclerenchyma and tracheids become dead?
Answer:
The cells of sclerenchyma and tracheids become dead because they lack protoplasm.

Question 7.
Explain sclereids with their types.
Answer:
Sclereids are dead cells, usually these are isodiametric but some are elongated too. The cell wall is very thick due to lignification. Lumen is very much reduced. The pits may simple or branched. Sclereids are mechanical in function. They give hard texture to the seed coats, endosperms etc., Sclereids are classified into the following types.

1. Branchysclereids or Stone cells: Isodiametric sclereids, with hard cell wall. It is found in bark, pith cortex, hard endosperm and fleshy portion of some fruits. eg: Pulp of Pyrus.
2. Macrosclereids: Elongated and rod shaped cells, found in the outer seed coat of leguminous plants. eg: Crotalaria and Pisum sativum.
3. Osteosclereids (Bone cells): Rod shaped with dilated ends. They occur in leaves and seed coats. eg: seed coat of Pisum and Hakea.
4. Astrosclereids: Star cells with lobes or arms diverging form a central body. They occur in petioles and leaves. eg: Tea, Nymphae and Trochodendron.
5. Trichosclereids: Hair like thin walled sclereids. Numerous small angular crystals are embedded in the wall of these sclereids, present in stems and leaves of hydrophytes. eg: Nymphaea leaf and Aerial roots of Monstera

Question 8.
What are sieve tubes? Explain.
Answer:
Sieve tubes are long tube like conducting elements in the phloem. These are formed from a series of cells called sieve tube elements. The sieve tube elements are arranged one above the other and form vertical sieve tube. The end wall contains a number of pores and it looks like a sieve. So it is called as sieve plate. The sieve elements show nacreous thickenings on their lateral walls. They may possess simple or compound sieve plates.

The function of sieve tubes are believed to be controlled by campanion cells In mature sieve tube, Nucleus is absent. It contains a lining layer of cytoplasm. A special protein (P. Protein = Phloem Protein) called slime body is seen in it. In mature sieve tubes, the pores in the sieve plate are blocked by a substance called callose (callose plug). The conduction of food material takes lace through cytoplasmic strands. Sieve tubes occur only in Angiosperms.

Question 9.
Distinguish the anatomy of dicot root from monocot root.
Answer:
The anatomy of dicot root from monocot root:

Question 10.
Distinguish the anatomy of dicot stem from monocot stem.
Answer:
The anatomy of dicot stem from monocot stem:

Samacheer Kalvi 11th Bio Botany Tissue and Tissue System Other Important Questions & Answers

I. Choose the correct answer. (1 Mark)
Question 1.
Who is the father of plant anatomy?
(a) David Muller
(b) Katherine Esau
(c) Nehemiah Grew
(d) Hofmeister
Answer:
(c) Nehemiah Grew

Question 2.
The study of internal structure and organisation of plant is called:
(a) plant taxonomy
(b) plant anatomy
(c) plant physiology
(d) plant ecology
Answer:
(b) plant anatomy

Question 3.
The book “Anatomy of seed plants” is written by:
(a) Hanstein
(b) Schmidt
(c) Nicholsen
(d) Katherine Esau
Answer:
(d) Katherine Esau

Question 4.
The term meristem is coined by:
(a) Nageli
(b) Robert
(c) Stevers
(d) Clowes
Answer:
(a) Nageli

Question 5.
Which of the statement is not correct?
(a) Meristematic cells are self perpetuating
(b) Meristematic cells are most actively dividing cells
(c) Meristematic cells have large vacuoles
(d) Meristematic cells have dense cytoplasm with prominent nucleus
Answer:
(c) Meristematic cells have large vacuoles

Question 6.
Apical cell theory is proposed by:
(a) David brown
(b) Hofmeister
(c) Land mark
(d) Clowes
Answer:
(b) Hofmeister

Question 7.
The tunica is:
(a) the peripheral zone of shoot apex, that forms cortex
(b) the inner zone of shoot apex, that forms stele
(c) the peripheral zone of shoot apex, that forms epidermis
(d) the inner zone of shoot apex, that forms cortex and stele
Answer:
(c) the peripheral zone of shoot apex, that forms epidermis

Question 8.
Which of the histogens gives rise to root cap?
(a) Plerome
(b) Periblem
(c) Dermatogen
(d) Calyptrogen
Answer:
(d) Calyptrogen

Question 9.
Quiescent centre concept was proposed by:
(a) Lindall
(b) Clowes
(c) Holstein
(d) Sanio
Answer:
(b) Clowes

Question 10.
Parenchyma cells which stores resin, tannins, calcium carbonate and calcium oxalate are termed as:
(a) critoblast
(b) chromoblasts
(c) idioblasts
(d) astroblasts
Answer:
(c) idioblasts

Question 11.
Petioles of banana is composed of:
(a) storage parenchyma
(b) stellate parenchyma
(c) angular collenchyma
(d) prosenchyma
Answer:
(b) stellate parenchyma

Question 12.
Which of the following statement is not correct?
(a) Sclerenchyma is a dead cell
(b) It lacks protoplasm
(c) The cell walls of these cells are uniformly thickened
(d) Sclerenchyma are actively dividing cells
Answer:
(d) Sclerenchyma are actively dividing cells

Question 13.
The seed coat of ground nut is made up of:
(a) stone cells
(b) osteosclereids
(c) macrosclereids
(d) parenchyma cells
Answer:
(b) osteosclereids

Question 14.
Plant fibers are modified:
(a) sclerenchyma cells
(b) collenchyma cells
(c) parenchyma cells
(d) none of the above
Answer:
(a) sclerenchyma cells

Question 15.
The term xylem was introduced by:
(a) Alexander
(b) Nageli
(c) Holstein
(d) Schemidt
Answer:
(b) Nageli

Question 16.
What type of xylem arrangement is seen in Selaginella sp?
(a) Endarch
(b) Exarch
(c) Centrarch
(d) Mesarch
Answer:
(c) Centrarch

Question 17.
In cross section, the tracheids are:
(a) hexagonal in shape
(b) rectangular in shape
(c) triangular in shape
(d) polygonal in shape
Answer:
(d) polygonal in shape

Question 18.
In grasses the guard cells in stoma are:
(a) bean shaped
(b) irregular shaped
(c) dumbbell shaped
(d) bell shaped
Answer:
(c) dumbbell shaped

Question 19.
Bulliform cells are present in:
(a) mango
(b) grasses
(c) ground nut
(d) potato
Answer:
(b) grasses

Question 20.
The sunken stomata:
(a) reduce water loss by transpiration
(b) increase water loss by transpiration
(c) increase heat loss by evaporation
(d) neither reduce nor increase water loss by transpiration
Answer:
(a) reduce water loss by transpiration

Question 21.
In Ocimum the trichomes are:
(a) non – glandular
(b) fibrous
(c) glandular
(d) none of these
Answer:
(c) glandular

Question 22.
In dicot stem, the hypodermis is generally:
(a) parenchymatous
(b) sclerenchymatous
(c) collenchymatous
(d) none of these
Answer:
(c) collenchymatous

Question 23.
Casparian strips contain thickenings of:
(a) calcium carbonate and calcium oxalate
(b) carbohydrate, protein and lignin
(c) crystal of calcium oxalate
(d) lignin, suberin and some other carbohydrates
Answer:
(d) lignin, suberin and some other carbohydrates

Question 24.
Indicate the correct statement:
(a) Albuminous cells in gymnosperms are a nucleated parenchyma cells.
(b) Albuminous cells in gymnosperms are nucleated collenchyma cells.
(c) Albuminous cells in gymnosperms are nucleated, thin walled parenchyma cells.
(d) Albuminous cells in gymnosperms are a nucleated sclerenchyma cells.
Answer:
(c) Albuminous cells in gymnosperms are nucleated, thin walled parenchyma cells.

Question 25.
Secondary phloem is derived from:
(a) apical meritesm
(b) vascular cambium
(c) primary phloem
(d) none of the above
Answer:
(b) vascular cambium

Question 26.
Which of the following statement is not correct?
(a) The outer most layer of the root is called piliferous layer.
(b) The chief function of piliferous layer is protection.
(c) Piliferous layer is made up of parenchyma cells with intracellular space.
(d) Piliferous layer is made up of parenchyma cells without intracellular space.
Answer:
(d) Piliferous layer is made up of parenchyma cells without intracellular space.

Question 27.
In beans, the metaxylem vessels are generally:
(a) polygonal in shape
(b) circular in shape
(d) rectangular in shape
(d) triangular in shape
Answer:
(a) polygonal in shape

Question 28.
Who discovered the Annular collenchyma?
(a) Clowes
(b) Sanio
(c) Nageli
(d) Duchaigne
Answer:
(d) Duchaigne

Question 29.
The main function of xylem is:
(a) to conduct the minerals to various parts of plants
(b) to conduct oxygen to various parts of plant body
(c) to conduct water and minerals from root to the other parts of the plant body
(d) to conduct stored food to various parts of plant body
Answer:
(c) to conduct water and minerals from root to the other parts of the plant body

Question 30.
In maize the vascular bundles are:
(a) scattered
(b) concentric
(c) excentric
(d) radial
Answer:
(a) scattered

Question 31.
stomata in leaves of a plant are used for:
(a) transpiration
(b) transpiration and gas exchange
(c) gas exchange
(d) none of the above
Answer:
(b) transpiration and gas exchange

Question 32.
Which of the statement is not correct?
(a) Palisade parenchyma cells are seen beneath the upper epidermis
(b) Palisade parenchyma cells contain more chloroplasts
(c) Palisade parenchyma cells are irregularly shaped
(d) The function of palisade parenchyma is photosynthesis
Answer:
(c) Palisade parenchyma cells are irregularly shaped

Question 33.
Spongy parenchyma cells are:
(a) irregularly shaped
(b) elongated cylindrical cells
(c) very lightly arranged cells
(d) with more number of chloroplasts than palisade parenchyma
Answer:
(a) irregularly shaped

Question 34.
The main function of spongy parenchyma is:
(a) photosynthesis
(b) exchange of gases
(c) exchange of minerals
(d) water transport
Answer:
(b) exchange of gases

Question 35.
All mesophyll cells in monocot leaf are nearly:
(a) isodiametric and thick walled
(b) irregular and thick walled
(c) isodiametric and thin walled
(d) irregular and thin willed
Answer:
(c) isodiametric and thin walled

Question 36.
Structurally, hydathodes are modified:
(a) cambium tissue
(b) parenchyma
(c) pith
(d) stomata
Answer:
(d) stomata

Question 37.
Hydathodes occurs in the leaves of:
(a) desert plants
(b) submerged aquatic plants
(c) floating aquatic weeds
(d) forest trees
Answer:
(b) submerged aquatic plants

Question 38.
The process of guttation is seen in:
(a) grasses
(b) dicot plants
(c) desert plants
(d) Nerium
Answer:
(a) grasses

Question 39.
Salt glands are present in:
(a) xerophytes
(b) hydrophytes
(c) halophytes
(d) merophytes
Answer:
(c) halophytes

Question 40.
The term sieve tubes is coined by:
(a) Schleiden
(b) Hanstein
(c) Tsehireh
(d) Hartig
Answer:
(d) Hartig

II. Answer the following. (2 Marks)

Question 1.
Define the tissue?
Answer:
A tissue is a group of cells that are alike in origin, structure and function. The study of tissue is called Histology.

Question 2. What are the different types of plant tissue?
Answer:
The two types of principal groups are:

1. Meristematic tissues
2. Permanent tissues.

Question 3.
Mention any, two characters of meriste – matic tissue.
Answer:
Two characters of meriste – matic tissue:

1. The meristematic cells are isodiametric and they may be, oval, spherical or polygonal in shape.
2. They have generally dense cytoplasm with prominent nucleus.

Question 4.
Mention the function of apical meristem.
Answer:
Present in apices of root and shoot. It is responsible for increase in the length of the plant, it is called as primary growth.

Question 5.
What is mean by carpus?
Answer:
It is the inner zone of shoot apex,that forms cortex and stele of shoot.

Question 6.
Explain apical cell theory?
Answer:
Apical cell theory is proposed by Nageli. The single apical cell or apical initial composes the root meristem. The apical initial is tetrahedral in shape and produces root cap from one side. The remaining three sides produce epidermis, cortex and vascular tissues. It is found in vascular cryptogams.

Question 7.
What is meant by angular collenchyma?
Answer:
It is the most common type of collenchyma with irregular arrangement and thickening at the angles where cells meet., eg: Hypodermis of Datum and Nicotiana.

Question 8.
Explain briefly Branchysciereids or Stone cells.
Answer:
Isodiametric sclereids, with hard cell wall. It is found in bark, pith cortex, hard endosperm and fleshy portion of some fruits. eg: Pulp of Pyrus.

Question 9.
What is Filiform Sclereids?
Answer:
The sclereids are present in the leaf lamina of Olea europaea. They are very much elongated fibre like and about 1mm length.

Question 10.
Distinguish between Libriform fibres and Fibre tracheids.
Answer:
Between Libriform fibres and Fibre tracheids:

Libriform fibres	Fibre tracheids
1. These fibres have slightly lignified secondary walls with simple pits. These fibres are long and narrow.	1. These are shorter than the libriform fibres with moderate secondary thickenings in the cell walls. Pits are simple or bordered.

Question 11.
What are bast fibres?
Answer:
These fibres are present in the phloem. Natural Bast fibres are strong and cellulosic. Fibres obtaining from the phloem or outer bark of jute, kenaf, flax and hemp plants. The so called pericyclic fibres are actually phloem fibres.

Question 12.
What is meant by endarch type of xylem arrangements?
Answer:
Protoxylem lies towards the .centre and meta xylem towards the periphery this condition is called Endarch. It is seen in stems.

Question 13.
What are the types of cells present in phloem?
Answer:
Phloem consists of four types of Cells:

1. Sieve elements
2. Companion cells
3. Phloem parenchyma
4. Phloem fibres.

Question 14.
Define epiblema?
Answer:
It is made up of single layer of parenchyma cells which are arranged compactly without intercellular spaces. It is devoid of epidermal pores and cuticle. Root hair is always single celled, it absorbs water and mineral salts from the soil. The another important function of piliferous layer is protection.

Question 15.
Explain bulliform cells in grasses.
Answer:
Some cells of upper epidermis (eg: Grasses) are larger and thin walled. They are called bulliform cells or motor cells. These cells are helpful for the rolling and unrolling of the leaf according to the weather change.

Question 16.
What is meant by Sunken Stomata?
Answer:
In some Xerophytic plants (eg: Cycas, Nerium), stomata is sunken beneath the abaxial leaf surface within stomatal crypts. The sunken stomata reduce water loss by transpiration.

Question 1 7.
Mention any two functions of epidermal tissue system in plants.
Answer:
Two functions of epidermal tissue system in plants:

1. This system in the shoot checks excessive loss of water due to the presence of cuticle.
2. Epidermis protects the underlying tissues.

Question 18.
Define chlorenchymo?
Answer:
Sometimes in young stem, chloroplasts develop in peripheral cortical cells, which is Called chlorenchyma.

Question 19.
What is meant by casparian strips?
Answer:
In true root endodermis, radial and inner tangential walls of endodermal cells possess thickenings of lignin, suberin and some other carbohydrates in the form of strips they are called casparian strips.

Question 20.
What are albuminous cells?
Answer:
The cytoplasmic nucleated parenchyma, is associated with the sieve cells of Gymnosperms. Albuminous cells in Conifers are analogous to companion cells of Angiosperms. It also called as strasburger cells.

Question 21.
Describe briefly radial types of vascular Bundles.
Answer:
Xylem and phloem are present on different radii alternating with each other. The bundles are separated by parenchymatous tissue. (Monocot and Dicot roots).

Question 22.
Define stele?
Answer:
All the tissues inside the endodermis comprise the stele. This includes pericycle, vascular system and pith.

Question 23.
What is meant by cambium?
Answer:
Cambium consists of brick shaped and thin walled meristematic cells. It is one to four layers in thickness. These cells are capable of forming new cells during secondary growth.

Question 24.
Define silica Cells?
Answer:
Some of the epidermal cells of the grass are filled with silica. They are called silica cells.

Question 25.
Define, hydathode?
Answer:
A hydathode is a type of epidermal pore, commonly found in higher plants. Structurally, hydathodes are modified stomata, usually located at leaf tips or margins, especially at the teeth. Hydathodes occur in the leaves of submerged aquatic plants such as ranunculus fluitans as well as in many herbaceous land plants.

III. Answer the following. (3 Marks)

Question 1.
Explain apical cell theory.
Answer:
Apical cell theory is proposed by Hofmeister (1852) and supported by Nageli (1859). A single apical cell is the structural and functional unit. This apical cell governs the growth and development of whole plant body. It is applicable in Algae, Bryophytes and in some Pteridophytes.

Question 2.
Explain histogen theory.
Answer:
Histogen theory is proposed by Hanstein (1868) and supported by Strassburgur. The shoot apex comprises three distinct zones.

1. Dermatogen: It is a outermost layer. It gives rise to epidermis.
2. Periblem: It is a middle layer. It gives rise to cortex.
3. Plerome: It is innermost layer. It gives rise to stele.

Question 3.
What is meant by quiescent centre concept?
Answer:
Quiescent centre concept was proposed by Clowes (1961) to explain root apical meristem activity. These centre is located between root cap and differentiating cells of the roots. The apparently inactive region of cells in root promeristem is called quiescent centre. It is the site of hormone synthesis and also the ultimate source of all meristematic cells of the meristem.

Question 4.
Explain the term “sclereids”.
Answer:
Sclereids are dead cells, usually these are isodiametric but some are elongated too. The cell wall is very thick due to lignification. Lumen is very much reduced. The pits may simple or branched. Sclereids are mechanical in function. They give hard texture to the seed coats, endosperms etc.

Question 5.
Explain briefly about plant fibres.
Answer:
Fibres are very much elongated sclerenchyma cells with pointed tips. Fibres are dead cells and have lignified walls with narrow lumen. They have simple pits. They provide mechanical strength and protect them from the strong wind. It is also called supporting tissues. Fibres have a great commercial value in cottage and textile industries.

Question 6.
Write briefly about xylem fibres.
Answer:
The fibres of sclerenchyma associated with the xylem are known as xylem fibres. Xylem fibres are dead cells and have lignified walls with narrow lumen. They cannot conduct water but being stronger provide mechanical strength. They are present in both primary and secondary xylem. Xylem fibres are also called libriform fibres.
The fibres are abundantly found in many plants. They occur in patches, in continuous bands and sometimes singly among other cells. Between fibres and normal tracheids, there are many transitional forms which are neither typical fibres nor typical tracheids. The transitional types are designated as fibre – tracheids. The pits of fibre – tracheids are smaller than those of vessels and typical tracheids.

Question 7.
Explain companion cells.
Answer:
The thin walled, elongated, specialized parenchyma cells, which are associated with die sieve elements, are called companion cells. These cells are living and they have cytoplasm and a prominent nucleus. They are connected to the sieve tubes through pits found in the lateral walls. Through these pits cytoplasmic connections are maintained between these elements. These cells are helpful in maintaining the pressure gradient in the sieve tubes. Usually the nuclei of the companion cells serve for the nuclei of sieve tubes as they lack them. The companion cells are present only in Angiosperms and absent in Gymnosperms and Pteridophytes. They assist the sieve tubes in the conduction of food materials.

Question 8.
Distinguish between meristematic tissue and permanent tissue.
Answer:
Between meristematic tissue and permanent tissue:

Meristematic tissue	Permanent tissue
1. Cells divide repeatedly	1. Do not divide
2. Cells are undifferentiated	2. Cells are fully differentiated
3. Cells are small and Isodiametric	3. Cells are variable in shape and size
4. Intercellular spaces are absent	4. Intercellular spaces are present
5. Vacuoles are absent	5. Vacuoles are present
6. Cell walls are thin	6. Cell walls maybe thick or thin
7. Inorganic inclusions are absent	7. Inorganic inclusions  are present

Question 9.
Write down the differences between tracheids and fibres.
Answer:
The differences between tracheids and fibres:

Tracheids	Fibres
1. Not much elongated	1. Very long cells
2. Posses oblique end walls	2. Posses tapering end walls
3. Cell walls are not as thick as fibtres	3. Cell wall are thick and lignified
4. Possess various types of thickenings	4. Possess only pitted thickenings
5. Responsible for the conduction and also mechanical support	5. Provide only mechanical support

Question 10.
Give a brief answer on subsidiary cells in plant leaves.
Answer:
Stomata are minute pores surrounded by two guard cells. The stomata occur mainly in the epidermis of leaves. In some plants addition to guard cells, specialised epidermal cells are present which are distinct from other epidermal cells. They are called Subsidiary cells. Based on the number and arrangement of subsidiary cells around the guard cells, the various types of stomata are recognised, The guard cells and subsidiary cells help in opening and closing of stomata during gaseous exchange and transpiration.

Question 11.
Explain the term trichomes.
Answer:
There are many types of epidermal outgrowths in stems. The unicellular or multicellular appendages that originate from the epidermal cells are called trichomes. Trichomes may be branched or unbranched and one or more one celled thick. They assume many shapes and sizes. They may also be glandular (eg: Rose, Ocimum) or non – glandular.

Question 12.
What do you understand about hypodermis in plant tissue system.
Answer:
One or two layers of continuous or discontinuous tissue present below the epidermis, is called hypodermis. It is protective in function. In dicot stem, hypodermis is generally collenchymatous, whereas in monocot stem, it is generally sclerenchymatous. In many plants collenchyma form the hypodermis.

Question 13.
What is meant by pith?’
Answer:
The central part of the ground tissue is known as pith or medulla. Generally this is made up of thin walled parenchyma cells with intercellular spaces. The cells in the pith generally stores starch, fatty substances, tannins, phenols, calcium oxalate crystals, etc.

Question 14.
Explain the piliferous layer as epiblema.
Answer:
The outermost layer of the root is known as piliferous layer. It consists of single row of thin – walled parenchymatous cells without any intercellular space. Epidermal pores and cuticle are absent in the piliferous layer. Root hairs that are found in the piliferous layer are always unicellular. They absorb waer and mineral salt from the soil. Root hairs are generally short lived. The main function of piliferous layer is protection of the inner tissues.

Question 15.
What is meant by stele in plant stem?
Answer:
The central part of the stem inner to the endodermis is known as stele. It consists of pericyle, vascular bundles and pith. In dicot stem, vascular bundles are arranged in a ring around the pith. This type of stele is called eustele.

Question 16.
Explain the nature of phloem in dicot stem.
Answer:
Primary phloem lies towards the periphery. It consists of protpphloem and metaphloem. Phloem consists of sieve tubes, companion cells and phloem parenchyma. Phloem fibres are absent in primary phloem. Phloem conduct organic foods material from the leaves to other parts of the plant body.

Question 17.
Explain the mesophyll layer of leaf.
Answer:
The ground tissue that is present between the upper and lower epidermis of the leaf is called mesophyll. Here, the mesophyll is not differentiated into palisade and spongy parenchyma. All the mesophyll cells are nearly isodiametric and thin walled. These cells are compactly arranged with limited intercellular spaces. They contain numerous chloroplasts.

Question 18.
Mention any three differences between stomata and hydathodes.
Answer:
Stomata:

1. Occur in epidermis of leaves, young stems.
2. Stomatal aperture is guarded by two guard cells.
3. The two guard cells are generally surrounded by subsidiary cell.

Hydathodes:

1. Occur at the tip or margin of leaves that are grown in moist shady place.
2. Aperture of hydathodes are surrounded by a ring of cuticularized cells.
3. Subsidiary cells are absent.,

Question 19.
What are halophiles? Explain briefly.
Answer:
Halophiles:

1. Plants that grow in salty environment are called Halophiles.
2. Plant growth in saline habitat developed numerous adaptations to salt stress. The secretion of ions by salt glands is the best known mechanism for regulating the salt content of plant shoots.
3. Salt glands typically are found in halophytes. (Plants that grow in saline environments)

IV. Answer in detail.

Question 1.
Explain Histogen theory, Korper Kappe Theory and Quiescent Centre Concept with diagrams.
Answer:
Histogen Theory: Histogen theory is proposed by Hanstein (1868) and supported by Strassburgur. The histogen theory as appilied to the root apical meristem speaks of four histogen in the meristem. They are respectively

1. Dermatogen: It is a outermost layer. It gives rise to root epidermis.
2. Periblem: it is a middle layer. It gives rise to cortex.
3. Plerome: It is innermost layer. It gives rise to stele.
4. Calyptrogen: it gives rise to root cap.
   

Korper Kappe Theory: Korper kappe theory is proposed by Schuepp. There are two zones in root apex – Korper and Kappe.

1. Korper zone forms the body.
2. Kappe zone forms The cap.

This theory is equivalent to tunica corpus theory of shoot apex.The two divisions are distinguished by the type of T (also called Y divisions). Korper is characterised by inverted T division and kappe by straight T divisions.

Quiescent Centre Concept: Quiescent centre concept was proposed by Clowes (1961) to explain root apical meristem activity. These centre is located between root cap and differentiating cells of the roots. The apparently inactive region of cells in root promeristem is called quiescent centre. It is the site of hormone synthesis and also the ultimate source of all meristematic cells of the meristem.

Question 2.
Describe the structure and function of different kinds of parenchyma tissues?
Answer:

Parenchyma is generally present in all organs of the plant. It forms the ground tissue in a plant. Parenchyma is a living tissue and made up of thin walled cells. The cell wall is made up of cellulose. Parenchyma cells may be oval, polyhedral, cylindrical, irregular, elongated or armed. Parenchyma tissue normally has prominent intercellular spaces. Parenchyma may store various types of materials like, water, air, ergastic substances. It is usually colourless. The turgid parenchyma cells help in giving rigidity to the plant body. Partial conduction of water is also maintained through parenchymatous cells. Occsionaliy Parenchyma cells which store resin, tannins, crystals of calcium carbonate, calcium oxalate are called idioblasts. Parenchyma is of different types and some of them are discussed as follows. Types of paranchyma:
(i) Aerenchyma: Parenchyma which contains air in its intercellular spaces. It helps in aeration and buoyancy. eg: Nymphae and Hydrilia.

(ii) Storage Parenchyma: parenchyma stores food materials. eg: Root and stem tubers.

(iii) Stellate Parenchyma: Star shaped parenchyma. eg: Petioles of Banana and Canna.

(iv) Chlorenchyma: Parenchyma cells with chlorophyll. Function is photosynthesis, eg: Mesophyll of leaves.

(v) Prosenchyma: parenchyma cells became elongated, pointed and slightly thick walled. It provides mechanical support.

Question 3.
Describe the types of tracheids with diagram.
Answer:

Types of secondary wall thickenings in tracheids and vessels:
Tracheids are dead, lignified and elongated cells with tapering ends. Its lumen is broader than that of fibres. In cross section, the tracheids are polygonal. There are different types of cell wall thickenings due to the deposition of secondary wall substances. They are annular (ring like), spiral (spring like), scalariform (ladder like) reticulate (net like) and pitted (uniformly thick except at pits). Tracheids are imperforated cells with bordered pits on their side walls. Only through this conduction takes place in Gymnosperms. They are arranged one above the other. Tracheids are chief water conducting elements in Gymnosperms and Pteridophytes. They also offer mechanical support to the plants.

Question 4.
Compare the different types of plant tissues.
Answer:
The different types of plant tissues:

Question 5.
Compare the vascular tissues of plant.
Answer:
The vascular tissues of plant:

Question 6.
Draw and label the various parts of T.S. of dicot root.
Answer:
Draw and label the various parts of T.S. of dicot root:

Question 7.
Explain in detail about the vascular bundles of monocot stem.
Answer:
1. Vascular bundles: Vascular bundles are scattered (atactostele) in the parenchyma ground tissue. Each vascular bundle is surrounded by a sheath of sclerenehymatous fibres called bundle sheath. The vascular bundles are conjoint, collateral, endarch and closed. Vascular bundles are numerous, small and closely arranged in the peripheral portion. Towards the centre, the bundles are comparatively large in size and loosely arranged. Vascular bundles are skull or oval shaped.

2. Phloem: The phloem in the monocot stem consists of sieve tubes and companion cells. Phloem parenchyma and phloem fibres are absent. It can be distinguished into an outer crushed protophloem and an inner metaphloem.

3. Xylem: Xylem vessels are arranged in the form of ‘Y’ the two metaxylem vessels at the base. In a mature bundle, the lowest protowylem disintegrates and forms a cavity known as protoxylem lacuna.

Question 8.
Draw and label the various parts of monocot stem.
Answer:
The various parts of monocot stem:

Question 9.
Explain the various parts of sunflower leaf with neat diagram.
Answer:
1. Anatomy of a Dicot Leaf – sunflower Leaf: Internal structure of dicotyledonous leaves reveal epidermis, mesophyll and vascular tissues.

2. Epidermis: This leaf is generally dorsiventral. It has upper and lower epidermis. The epidermis is usually made up of a single layer of cells that are closely packed. The cuticle on the upper epidermis is thicker than that of lower epidermis. The minute opening found on the epidermis are called stomata. Stomata are more in number on the lower epidermis than on the upper epidermis.

A stomata is surrounded by a pair of bean shaped cells are called guard cells. Each stoma internally opens into an air chamber. These guard cells contain chlotroplasts. The main function of epidermis is to give protection to the inner tissue called mesospyll. The cuticle helps to check transpiration. Stomata are used for transpiration and gas exchange.

3. Mesophyll: The entire tissue between the upper and lower epidermis is called mesophyll (GK meso = in the middle, phyllome = leaf). There are two regions in the mesophyll. They are palisade parenchyma and spongy parenchyma. Palisade parenchyma cells are seen beneath the upper epidermis. It consists of vertically elongated cylindrical cells in one or more layers. These are compactly arranged and are generally without intercellular spaces. Palisade parenchyma cells contain more chloroplasts than the spongy parenchyma cells.

The function of palisade parenchyma is photosynthesis. Spongy parenchyma lies below the palsied parenchyma. Spongy cells are irregularly shaped. These cells are very loosly arranged with numerous airspaces. As compared to palisade cells, the spongy cells contain number of chloroplasts. Spongy cells facilitate the exchange of gases with the help of air spaces. The air space that is found next to the stomata is called respiratory cavity or substomatal cavity. Å

4. Vascular tissue: Vascular tissue are present in the veins of leaf. Vascular bundles are conjoint collateral and closed. Xylem is present towards the upper epidermis, while the phloem towards the lower epidermis. Vascular bundles are surrounded by a compact layer by a parenchymatous cells called bundle sheath or border parenchyma.

Xylem consists of metaxylem and protoxylem elements. Protoxylem is present towards the upper epidermis, while the phloem consists of sieve tubes, companion cells and phloem parenchyma. Phloem fibres are absent. Xylem sonsists of vessels and xylem parenchyma. Tracheids and xylem fibres are absent.

Solution To Activity
Text Book Page No. 10
Question 1.
Cell lab: Students prepare the slide and identify the different types tissues.
Answer:
Preparing a slide of plant tissue.
Objective:

1. Using hand cutting method to make thin slice of dicot root.
2. To make slide and stain of plant sample.
3. To observe the plant sample under microscope.

Materials:

1. A young dicot root
2. Compound microscope
3. Slide
4. Cover slip
5. Eosin stain

Method:

1. Place 2 cm of young dicot root on a glass slide or plate.
2. Cut thin slices of the root through the region of maturation.
3. Stain it with Eosin.
4. Fix one or two of the sections in a slide and put a cover slip.
5. To observe the sample under a compound microscope and record the parts of the sample.

Students can Download Bio Botany Chapter 4 Reproductive Morphology Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Bio Botany Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 4 Reproductive Morphology

Samacheer Kalvi 11th Bio Botany Reproductive Morphology Text Book Back Questions and Answers

Choose the correct answer
Question 1.
Vexillary aestivation is characteristic of the family …………… .
(a) Fabaceae
(b) Asteraceae
(c) Solanaceae
(d) Brassicaceae
Answer:
(a) Fabaceae

Question 2.
Gynoecium with united carples is termed as …………… .
(a) apocarpous
(b) multicarpellary
(c) syncarpous
(d) none of the above
Answer:
(c) syncarpous

Question 3.
Aggregate fruit develops from …………… .
(a) multicarpellary, apocarpous ovary
(b) multicarpellary, syncarpous ovary
(c) multicarpellary ovary
(d) whole inflorescence
Answer:
(a) multicarpellary, apocarpous ovary

Question 4.
In an inflorescence where flowers are borne laterally in an acropetal succession the position of the youngest floral bud shall be …………… .
(a) proximal
(b) distal
(c) intercalary
(d) anywhere
Answer:
(b) distal

Question 5.
A true fruit is the one where …………… .
(a) only ovary of the flower develops into fruit
(b) ovary and calyx of the flower develops into fruit
(c) ovary, calyx and thalamus of the flower develops into fruit
(d) all floral whorls of the flower develops into fruit
Answer:
(a) only ovary of the flower develops into fruit

Question 6.
Find out the floral formula for a bisexual flower with bract, regular, pentamerous, distinct calyx and corolla, superior ovary without bracteole?
Answer:

Question 7.
Give the technical terms for the following:
(a) A sterile stamen
(b) Stamens are united in one bunch
(c) Stamens are attached to the petals
Answer:
(a) A sterile stamen – Staminode
(b) Stamens are united in one bunch – Monadelphous
(c) Stamens are attached to the petals – Epipetalous (petalostemonous)

Question 8.
Explain the different types of placentation with example.
Answer:
The different types of placentation with example:

1. Marginal: It is with the placentae along the margin of a unicarpellate ovary. Example: Fabaceae.
2. Axile: The placentae arises from the column in a compound ovary with septa. Example: Hibiscus, tomato and lemon.
3. Superficial: Ovules arise from the surface of the septa. Example: Nymphaeceae.
4. Parietal: It is the placentae on the ovary walls or upon intruding partitions of a unilocular, compound ovary. Example: Mustard, argemone and cucumber.
5. Free – central: It is with the placentae along the column in a compound ovary without septa. Example: Caryophyllaceae, Dianthus and primrose.
6. Basal: It is the placenta at the base of the ovary. Example: Sunflower (Asteraceae) Marigold.

Question 9.
Differentiate between aggregate fruit with multiple fruit.
Answer:
1. Aggregate fruit:
Aggregate fruits develop from a single flower having an apocarpous pistil. Each of the free carpel is developed into a simple fruitlet. A collection of simple fruitlets makes an aggregate fruit. An individual ovary develops into a drupe, achene, follicle or berry. An aggregate of these fruits borne by a single flower is known as an etaerio. Example: Magnolia, Raspberry, Annona and Polyalthia.

2. Multiple or Composite fruit: A multiple or composite fruit develops from the whole inflorescence along with its peduncle on which they are borne.

• Sorosis: A fleshy multiple fruit which develops from a spike or spadix. The flowers fused together by their succulent perianth and at the same time the axis bearing them become fleshy or juicy and the whole inflorescence forms a compact mass. Example: Pineapple, Jack fruit and Mulberry.
• Syconus: A multiple fruit which develops from hypanthodium inflorescence. The receptacle develops further and converts into fleshy fruit which encloses a number of true fruit or achenes which develops from female flower of hypanthodium inflorescence. Example: Ficus.

Question 10.
Explain the different types of fleshy fruit with suitable example?
Answer:
The fleshy fruits are derived from single pistil, where the pericarp is fleshy, succulent and differentiated into epicarp, mesocarp and endocarp. It is subdivided into the following:

1. Berry: Fruit develops from bicarpellary or multicarpellary, syncarpous ovary. Here the epicarp is thin, the mesocarp and endocarp remain undifferentiated. They form a pulp in which the seeds are embedded. Example: Tomato, date palm, grapes and brinjal.
2. Drupe: Fruit develops from monocarpellary, superior ovary. It is usually one seeded. Pericarp is differentiated into outer skinny epicarp, fleshy and pulpy mesocarp and hard and stony endocarp around the seed. Example: Mango and coconut.
3. Pepo: Fruit develops from tricarpellary inferior ovary. Pericarp terns leathery or woody which encloses, fleshy mesocarp and smooth endocarp. Example: Cucumber, watermelon, bottle gourd and pumpkin.
4. Hesperidium: Fruit develops from multicarpellary, multilocular, syncarpous, superior ovary. The fruit wall is differentiated into leathery epicarp with oil glands, a middle fibrous mesocarp. The endocarp forms distinct chambers, containing juicy hairs. Example: Orange and lemon.
5. Pome: It develops from multicarpellary, syncarpous, inferior ovary. The receptacle also develops along with the ovary and becomes fleshy, enclosing the true fruit. In pome the epicarp is thin skin like and endocarp is cartilagenous. Example: Apple and pear.
6. Balausta: A fleshy indehiscent fruit developing from multicarpellary, multilocular inferior ovary whose pericarp is tough and leathery. Seeds are attached irregularly with testa being the edible portion. Example: Pomegranate.

Textbook Activity Solved

Prepare a diet chart to provide balanced diet to an adolescent (a school going child) which includes food items (fruits, vegetable and seeds) which are non – expensive and are commonly available.
Diet Chart for an Adolescent:

Samacheer Kalvi 11th Bio Botany Reproductive Morphology Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer:
Question 1.
Placentation in tomato and lemon is …………… .
(a) parietal
(b) marginal
(c) free – central
(d) axile
Answer:
(d) axile

Question 2.
The coconut water and the edible part of coconut are equivalent to …………… .
(a) endosperm
(b) endocarp
(c) mesocarp
(d) embryo
Answer:
(a) endosperm

Question 3.
Geocarpic fruits are seen in …………… .
(a) carrot
(b) groundnut
(c) radish
(d) turnip
Answer:
(b) groundnut

Question 4.
Keel is characteristic petal of the flowers of …………… .
(a) Gulmohar
(b) Cassia
(c) Calotropis
(d) Bean
Answer:
(d) Bean

Question 5.
When the calyx is coloured and showy, it is called …………… .
(a) petaloid
(b) sepaloid
(c) bract
(d) spathe
Answer:
(a) petaloid

Question 6.
Bracts are modified leaves which bear flowers in their axils. Identify the plant which has a large showy brightly coloured bract …………… .
(a) Jasmine
(b) Euphorbia
(c) Hibiscus
(d) Bougainvillea
Answer:
(d) Bougainvillea

Question 7.
A flower which can be divided into equal vertical halves, by more than one plane of division is …………… .
(a) zygomorphic
(b) cyclic
(c) actinomorphic
(d) heteromorphic
Answer:
(c) actinomorphic

Question 8.
In Theobroma cocoa, the inflorescence arise from …………… .
(a) terminal shoot
(b) axillary part
(c) trunk of plant
(d) leaf node
Answer:
(c) trunk of plant

Question 9.
The type of inflorescence seen in Caesalpinia is …………… .
(a) corymb
(b) compound corymb
(c) capitulum
(d) umbel
Answer:
(a) corymb

Question 10.
Head is the characteristic of …………… family.
(a) Fabaceae
(b) Malvaceae
(c) Asteraceae
(d) Solanaceae
Answer:
(c) Asteraceae

Question 11.
Thyrsus is a type of …………… inflorescence.
(a) raceme
(b) cyme
(c) mixed
(d) special
Answer:
(c) mixed

Question 12.
Number of whorls in a complete flower is …………… .
(a) one
(b) two
(c) three
(d) four
Answer:
(d) four

Question 13.
Monoclinous flower will have …………… .
(a) androecium
(b) gynoecium
(c) both androecium & gynoecium
(d) none
Answer:
(c) both androecium & gynoecium

Question 14.
If unisexual and bisexual flowers are seen in same plant then the plant is said to be …………… .
(a) polyphyllous
(b) polygamous
(c) hermaphroditic
(d) dioecious
Answer:
(b) polygamous

Question 15.
…………… is a raceme of cymes.
(a) Verticil
(b) Cyathium
(c) Umbel
(d) Thyrsus
Answer:
(d) Thyrsus

Question 16.
Unit of perianth is …………… .
(a) petal
(b) sepal
(c) tepal
(d) stamen
Answer:
(c) tepal

Question 17.
Number of floral parts per whorl is called …………… .
(a) curosity
(b) atrocity
(c) merosity
(d) porosity
Answer:
(c) merosity

Question 18.
What is the green cap – like part of brinjal fruit?
(a) Corolla
(b) Perianth
(c) Calyx
(d) Pistil
Answer:
(c) Calyx

Question 19.
Butterfly shaped corolla is seen in …………… type.
(a) rosaceous
(b) caryophyllaceous
(c) cruciform
(d) papilionaceous
Answer:
(d) papilionaceous

Question 20.
Inflorescence seen in Daucas carota is …………… .
(a) umbel
(b) corymb
(c) compound umbel
(d) compound corymb
Answer:
(c) compound umbel

Question 21.
Arrangement of sepals and petals in flower bud is called …………… .
(a) adhesion
(b) aestivation
(c) placentation
(d) cohesion
Answer:
(b) aestivation

Question 22.
Which is not a part of pistil?
(a) Style
(b) Stigma
(c) Connective tissue
(d) carpel
Answer:
(c) Connective tissue

Question 23.
The type of calyx in brinjal is …………… .
(a) caducous
(b) deciduous
(c) persistent
(d) fugacious
Answer:
(c) persistent

Question 24.
Sterile stamen is called …………… .
(a) pistillode
(b) sessile
(c) staminode
(d) apostamen
Answer:
(c) staminode

Question 25.
Other name for gynoecium is …………… .
(a) carpel
(b) pistil
(c) style
(d) overy
Answer:
(b) pistil

Question 26.
Cavity found inside the ovary is called …………… .
(a) lobule
(b) locule
(c) lacuna
(d) labium
Answer:
(b) locule

Question 27.
Which part of saffron flower is used as flavouring agent?
(a) Carpel
(b) Anther
(c) Style
(d) Stigma
Answer:
(d) Stigma

Question 28.
If the ovary is inferior, then the flower is …………… .
(a) hypogynous
(b) epigynous
(c) perigynous
(d) epihypogynous
Answer:
(b) epigynous

Question 29.
Fabaceae members show …………… placentation.
(a) basal
(b) parietal
(c) superficial
(d) marginal
Answer:
(d) marginal

Question 30.
The side of the flower facing mother axis is called as …………… side.
(a) anterior
(b) posterior
(c) lateral
(d) dorsi – ventral
Answer:
(b) posterior

Question 31.
Which of the following option represents calyx?
(a) C
(b) Ca
(c) K
(d) Ka
Answer:
(c) K

Question 32.
…………… are the products of pollination & fertilization.
(a) Seeds
(b) Ovules
(c) Fruits
(d) Vegetables
Answer:
(c) Fruits

Question 33.
Study of fruits is called as …………… .
(a) honology
(b) pomology
(c) horticulture
(d) apology
Answer:
(b) pomology

Question 34.
Fruit wall can also be called as …………… .
(a) endocarp
(b) epicarp
(c) pericarp
(d) mericorp
Answer:
(c) pericarp

Question 35.
Which part of the apple fruit does we eat?
(a) Perianth
(b) Involucre
(c) Thalamus
(d) Bracteole
Answer:
(c) Thalamus

Question 36.
The false septum seen in siliqua fruits is …………… .
(a) frenulum
(b) micropyle
(c) raphae
(d) replum
Answer:
(d) replum

Question 37.
The type of fruit in Ricinus in …………… .
(a) lomentum
(b) cremocarp
(c) regma
(d) nut
Answer:
(c) regma

Question 38.
Jack fruit is an example for …………… .
(a) syconus
(b) siliqua
(c) sorosis
(d) nut
Answer:
(c) sorosis

Question 39.
Which of the following is not a schizocarpic fruit?
(a) Cremocarp
(b) Regma
(c) Samara
(d) Carcerulus
Answer:
(c) Samara

Question 40.
After fertilization …………… modifies into seed.
(a) ovary
(b) ovule
(c) carpel
(d) stigma
Answer:
(b) ovule

Question 41.
In groundnuts, which part nourishes the embryo?
(a) Endosperm
(b) Albumin
(c) Cotyledons
(d) Carpel
Answer:
(c) Cotyledons

Question 42.
…………… are the means for perpetuation of species.
(a) Fruits
(b) Seeds
(c) Corolla
(d) Flowers
Answer:
(b) Seeds

Question 43.
…………… is the second whorl of the flower.
(a) Calyx
(b) Corolla
(c) Gynoecium
(d) Perianth
Answer:
(b) Corolla

Question 44.
…………… is a ripened ovule.
(a) Carpel
(b) Pistil
(c) Seed
(d) Fruit
Answer:
(c) Seed

Question 45.
Imperfect flowers will have …………… essential whorl(s).
(a) only 1
(b) 2
(c) none
(d) 4
Answer:
(a) only 1

II. Very Short Answer Type Questions (2 Marks)

Question 1.
How will you define inflorescence?
Answer:
An inflorescence is a group of flowers arising from a branched or unbranched axis with a definite pattern.

Question 2.
Where does the inflorescence axis arise in cauliflorous type of inflorescence?
Answer:
In cauliflorous type, inflorescence developed directly from a woody trunk. Example: Theobroma cocoa.

Question 3.
Name any two mixed inflorescences.
Answer:
Two mixed inflorescences:

1. Thyrsus and
2. Verticillaster.

Question 4.
When a flower is said to be complete?
Answer:
A flower is said to be complete when it has all the four whorls (calyx, corolla, androecium & gynoecium).

Question 5.
What is a sessile flower?
Answer:
A flower without a pedicel or stalk is said to be sessile flower.

Question 6.
Define merosity.
Answer:
Number of floral parts per whorl is called merosity.

Question 7.
Write the units of (a) Perianth and (b) Calyx.
Answer:
The units of (a) Perianth and (b) Calyx:

1. (a) Perianth – tepals and
2. (b) Calyx – sepals

Question 8.
Name the three types of petals in papilionoceous corolla.
Answer:
The three types of petals in papilionoceous corolla:

1. Vexillum (standard)
2. wings (alae)
3. petals (carina)

Question 9.
What is a staminode? Give example.
Answer:
Sterile stamen is called staminode. e.g. Cassia

Question 10.
Define Pollinium.
Answer:
When the pollen grains are fused together as a single main, it is said to be pollinium.

Question 11.
List out the parts of a pistil.
Answer:
Ovary, style and stigma.

Question 12.
Define aestivation.
Answer:
Arrangement of sepals and petals in a floral bud.

Question 13.
What is mother axis?
Answer:
The branch that bears the flower is called mother axis.

Question 14.
What do you understand by the term “Pomology”?
Answer:
The branch of horticulture that deals with the study of fruits and their cultivation is called pomology.

Question 15.
How the seeds are classified based on endosperm?
Answer:
(a) Albuminous seed or Endospermous seed.
(b) Ex – Albuminous seed or non – Endospermous seed.

Question 16.
What is Spathe?
Answer:
In spadix, entire inflorescence is covered by a brightly coloured or hard bract called a spathe.

Question 17.
Differentiate Apocarpous and Syncarpous ovary.
Answer:
Apocarpous and Syncarpous ovary:

1. Apocarpous: A pistil contains two or more distinct carpels. Example: Annona
2. Syncarpous: A pistil contains two or more carpels which are connate. Example: Citrus and Tomato

Question 19.
Give examples for following fruit types: (a) Berry (b) Hesperidium
Answer:
(a) Berry: Tomato
(b) Hesperidium: Orange

III. Short Answer Type Questions (3 Marks)

Question 1.
Distinguish between Monoecious & Dioecious.
Answer:
Between Monoecious & Dioecious:

1. Monoecious: Both male and female flowers are present in the same plant, e.g., Coconut
2. Dioecious:  Male and female flowers are present on separate plants, e.g., Papaya

Question 2.
Explain Bilateral symmetry.
Answer:
In bilateral symmetry the flower can be divided into equal halves in only one plane. Zygomorphic flower can efficiently transfer pollen grains to visiting pollinators. Example: Pisum.

Question 3.
Differentiate Apopetalous from Sympetalous.
Answer:
Apopetalous from Sympetalous:

1. Apopetalous (or) Polypetalous: Petals are distinct, e.g., Hibiscus.
2. Sympetalous (or) Gamopetalous: Petals are fused, e.g., Datura.

Question 4.
Define Placentation & mention their types.
Answer:
The mode of distribution of placenta inside the ovary is called placentation. Types of placentation: Marginal, Axile, Superficial, Parietal, Free – central and Basal.

Question 5.
Write the floral formula for the Hibiscus rosasinensis.
Answer:

Question 6.
What are Parthenocarpic fruit?
Answer:
Development of fruits without fertilization are called Parthenocarpic fruit. They are seedless fruits. Example: Banana.

Question 7.
From which type of flowers does the aggregate fruit develops?
Answer:
Aggregate fruits develop from a single flower having an apocarpous pistil. Each of the free carpel is develops into a simple fruitlet. A collection of simple fruitlets makes an aggregate fruit.

Question 8.
Classify seeds based on their cotyledons.
Answer:
Based on the number of cotyledons present, two types of seeds are recognized.

1. Dicotyledonous seed: Seed with two cotyledons.
2. Monocotyledonous seed: Seed with one cotyledon.

Question 9.
List out any 3 significances of seed.
Answer:
3 significances of seed:

1. The seed encloses and protects the embryo for next generation.
2. Seeds of various plants are used as food, both for animals and human.
3. Seeds are the products of sexual reproduction so they provide genetic variations and recombination in a plant.

Question 10.
What is the importance of inflorescence.
Answer:
Function of inflorescence is to display the flowers for effective pollination and facilitate seed dispersal. The grouping of flowers in one place gives a better attraction to the visiting pollinators and maximize the energy of the plant.

Question 11.
Draw the line diagram for the following inflorescence.
Answer:
(a) Simple Dichasium:

(b) Compound Dichasium:

IV. Long Answer Type Questions (5 Marks)

Question 1.
Explain the various types of Schizocarpic fruit.
Answer:
This fruit type of intermediate between dehiscent and indehiscent fruit. The fruit instead of dehiscing rather splits into number of segments, each containing one or more seeds. They are of following types:

1. Cremocarp: Fruit develops from bicarpellary, syncarpous, inferior ovary and splitting into two one seeded segments known as mericarps. e.g., Coriander and Carrot.
2. Carcerulus: Fruit develops from bicarpellary, syncarpous, superior ovary and splitting into four one seeded segments known as nutlets, e.g., Leucas, Ocimum and Abutilon.
3. Lomentum: The fruit is derived from monocarpellary, unilocular ovary. A leguminous fruit, constricted between the seeds to form a number of one seeded compartments that separate at maturity, e.g., Desmodium, Arachis and Mimosa.
4. Regma: They develop from tricarpellary, syncarpous, superior, trilocular ovary and splits into one – seeded cocci which remain attached to carpophore, e.g., Ricinus and Geranium.

Question 2.
Explain the different types of flowers based on the position of ovary.
Answer:
Based on the position of ovary, a flower can be classified as:

1. Hypogynous: The term is used for sepals, petals and stamens attached at the base of a superior ovary, e.g. Malvaceae.
2. Epihypogynous: The term is used for sepals, petals and stamens attached at the middle of the ovary (half – inferior), e.g. Fabaceae and Rosaceae.
3. Epigynous: The term is used for sepals, petals and stamens attached at the tip of an inferior ovary, e.g. Cucumber, apple and Asteraceae.
4. Perigynous: The term is used for a hypanthium attached at the base of a superior ovary.
5. Epiperigynous: The term is used for hypanthium attached at the apex of an inferior ovary.

Question 3.
Classify the anthers based on their mode of attachment.
Answer:
The anthers based on their mode of attachment:

1. Basifixed: (Innate) Base of anther is attached to the tip of filament, e.g., Brassica, Datura
2. Dorsifixed: Apex of filament is attached to the dorsal side of the anther, e.g. Citrus, Hibiscus
3. Versatile: Filament is attached to the anther at midpoint, e.g., Grasses
4. Adnate: Filament is continued from the base to the apex of anther, e.g. Verbena, Ranunculus, Nelumbo.

Question 4.
Define aestivation. Explain its types with example.
Answer:

Question 5.
Distinguish between racemose and cymose inflorescence.
Answer:
Racemose inflorescence

1. Main axis of unlimited growth
2. Flowers arranged in an acropetal succession
3. Opening of flowers is centripetal
4. Usually the oldest flower at the base of the inflorescence axis.

Cymose inflorescence:

1. Main axis of limited growth
2. Flowers arranged in a basipetal succession
3. Opening of flowers is centrifugal
4. Usually the oldest flower at the top of the inflorescence axis.

Question 6.
Write in detail about head inflorescence.
Answer:
Head: A head is a characteristic inflorescence of Asteraceae and is also found in some members of Rubiaceae. Example: Neolamarkia cadamba and Mitragyna parvifolia; and in some members of Fabaceae – Mimosoideae, example: Acacia nilotica, Albizia lebbeck, Mimosa pudica (sensitive plant). Torus contains two types of florets:

1. Disc floret or tubular floret.
2. Ray floret or ligulate floret.

Heads are classified into two types:

1. Homogamous head: This type of inflorescence exhibits single kind of florets. Inflorescence has disc florets alone. Example: Vernonia, Ageratum or Ray florets alone. example: Launaea, Sonchus.
2. Heterogamous head: The inflorescence possesses both types of florets. Example: Helianthus, Tridax.

Disc florets at the centre of the head are tubular and bisexual whereas the ray florets found at the margin of the head which are ligulate pistilate (unisexual).

Question 7.
List out the significance of fruits.
Answer:
The significance of fruits:

1. Edible part of the fruit is a source of food, energy for animals.
2. They are source of many chemicals like sugar, pectin, organic acids, vitamins and minerals.
3. The fruit protects the seeds from unfavourable climatic conditions and animals.
4. Both fleshy and dry fruits help in the dispersal of seeds to distant places.
5. In certain cases, fruit may provide nutrition to the developing seedling.
6. Fruits provide source of medicine to humans.

Question 8.
Draw a flow chart depicting the classification of fruits.
Answer:

Question 9.
Explain in detail about any two special inflorescence.
Answer:
The inflorescences do not show any of the development pattern types are classified under special type of inflorescence.
1. Cyathium: Cyathium inflorescence consists of small unisexual flowers enclosed by a common involucre which mimics a single flower. Male flowers are organized in a scorpioid manner. Female flower is solitary and centrally located on a long pedicel. Male flower is represented only by stamens and female flower is represented only by pistil. Cyathium may be actinomorphic (Example: Euphorbia) or zygomorphic (Example: Pedilanthus) Nectar is present in involucre.

2. Hypanthodium: Receptacle is a hollow, globose structure consisting unisexual flowers present on the inner wall of the receptacle. Receptacle is closed except a small opening called ostiole which is covered by a series of bracts. Male flowers are present nearer to the ostiole, female and neutral flowers are found in a mixed manner from middle below. Example: Ficus sp. (Banyan and Pipal).

V. Higher Order Thinking Skills (HOTs)

Question 1.
Brinjal fruit has persistent calyx. Have you ever noticed the same in any other fruits? Name them.
Answer:
Tomato, Lady’s finger, guava and chilli also have persistent calyx.

Question 2.
Whether parthenocarpic fruits develop endosperm? Why?
Answer:
No, parthenocarpic fruits are developed without fertilization, but endosperm will form only after fertilization. So parthenocarpic fruits do not have endosperm.

Question 3.
Ovary develops into fruit after fertilization. While eating an Apple which part do you eat? Explain.
Answer:
Apple belongs to false fruit. In false fruits, apart from the ovary, non – carpellary parts also develop into fruit. In apple, the thalamus develops into fleshy edible part.

Question 4.
Cremocarp and Carcerulus both are schizocarpic fruits yet they differ. How?
Answer:
Cremocarp:

1. Fruit develops from syncarpous inferior ovary.
2. Ripened fruit split into two, one – seeded segments called mericarps.

Carcerulus:

1. Fruit develops from syncarpous superior ovary.
2. Ripened fruit split into four, one – seeded segments called nutlets.

Question 5.
Mango and coconut are ‘drupe’ type of fruits. In Mango, the edible part is fleshy mesocarp. What does the milk of tender coconut represent?
Answer:
Endosperm is the liquid (milk) potable part of tender coconut, which is rich in nutrients and is formed as a result of triple fusion.

Question 6.
Pick out correct ratio of the male flower to female flower in cyathium inflorescence and explain it?
(a) one : one
(b) one : many
(c) many : many and
(d) many : one.
Answer:
(a) One : one – Cyathium has single female flower represented by pistil and male flower represented by stamen.

Question 7.
Pollen differs from pollinium. How?
Answer:
Pollen are microspores which produces male gametes, whereas pollinium refers to the single mass of fused pollen grains.

Question 8.
Sunflower is not a flower – Justify your answer.
Answer:
Sunflower is actually an inflorescence not a single flower. The inflorescence of sunflower is capitulum composed of disc florets and ray florets.

Question 9.
Flower is a modified shoot for reproduction – Give possible evidence.
Answer:
Modified shoot for reproduction:

• Floral leaves (sepals and petals) are modified leaves.
• Floral and vegetative buds both emerge either in terminal or axillary position.
• Foliage leaves and floral leaves have identical arrangement on stem.

Question 10.
Is tomato a fruit or vegetable? Explain.
Answer:
Yes, tomato is a fruit. Because it develops from the ripened ovary and bears seeds, whereas vegetable refers to all other plant parts like root, stem and leaves.

Question 11.
What is caruncle? Where it is seen? How it helps the plant?
Answer:
Caruncle is the fleshy outgrowth at the base of seed. Usually caruncle helps in the seed dispersal, particularly by ants (Myrmecophily).

Question 12.
Both the prefixes (Uni – and Mono -) have the same meaning i.e. one in number. Does it mean that unisexual and monoecious are the same?
Answer:
That unisexual and monoecious:

1. Unisexual refers to the sex of flower (i.e. whether it has anther or carpel).
2. Monoecious refers to the plant bearing both the sexes in their flowers.

Students can Download Bio Botany Chapter 14 Respiration Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Bio Botany Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 14 Respiration

Samacheer Kalvi 11th Bio Botany Respiration Text Book Back Questions and Answers

Question 1.
The number of ATP molecules formed by complete oxidation of one molecule of pyruvic acid is:
(a) 12
(b) 13
(c) 14
(d) 15
Answer:
(a) 12

Question 2.
During oxidation of two molecules of cytosolic NADH + H⁺, number of ATP molecules produced in plants are:
(a) 3
(b) 4
(c) 6
(d) 8
Answer:
(c) 6

Question 3.
The compound which links glycolysis and Krebs cycle is:
(a) succinic acid
(b) pyruvic acid
(c) acetyl CoA
(d) citric acid
Answer:
(c) acetyl CoA

Question 4.
Assertion (A): Oxidative phosphorylation takes place during the electron transport chain in mitochondria.
Reason (R): Succinyl CoA is phosphorylated into succinic acid by substrate phosphorylation.
(a) A and R is correct. R is correct explanation of A
(b) A and R is correct but R is not the correct explanation of A
(c) A is correct but R is wrong
(d) A and R is wrong.
Answer:
(a) A and R is correct. R is correct explanation of A

Question 5.
Which of the following reaction is not . involved in Krebs cycle.
(a) Shifting of phosphate from 3C to 2C
(b) Splitting of Fructose 1,6 bisphosphate of into two molecules 3C compounds.
(c) Dephosphorylation from the substrates
(d) All of these
Answer:
(d) All of these

Question 6.
What are enzymes involved in phosphorylation and dephosphorylation reactions in EMP pathway?
Answer:
(i) Enzymes involved phosphorylation in EMP pathway:

• Hexokinase
• Phospho – fructokinase
• Glyceraldehyde – 3 – phosphate dehydrogenase

(ii) Enzymes involved in dephosphorylation in EMP pathway:

• Phospho glycerate kinase,
• Pyruvate kinase

Question 7.
Respiratory quotient is zero in succulent plants. Why?
Answer:
In some succulent plants like Opuntia, Bryophyllum carbohydrates are partially oxidised to organic acid, particularly malic acid without corresponding release of CO₂ but O₂ is consumed hence the RQ value will be zero.

Question 8.
Explain the reactions taking place in mitochondrial inner membrane.
Answer:
In plants, an additional NADH dehydrogenase (External) complex is present on the outer surface of inner membrane of mitochondria which can oxidise cytosolic NADH + H⁺. Ubiquinone (UQ) or Coenzyme Quinone (Co Q) is a small, lipid soluble electron, proton carrier located within the inner membrane of mitochondria.

Question 9.
What is the name of alternate way of glucose breakdown? Explain the process involved in it?
Answer:
During respiration breakdown of glucose in cytosol occurs both by glycolysis (about 2 / 3) as well as by oxidative pentose phosphate pathway (about 1 / 3). Pentose phosphate pathway was described by Warburg, Dickens and Lipmann (1938). Hence, it is also called Warburg – Dickens – Lipmann pathway. It takes place in cytoplasm of mature plant cells. It is an alternate way for breakdown of glucose.

It is also known as Hexose monophosphate shunt (HMP Shunt) or Direct Oxidative Pathway. It consists of two phases, oxidative phase and non – oxidative phase. The oxidative events convert six molecules of six carbon Glucose – 6 – phosphate to 6 molecules of five carbon sugar Ribulose – 5 phosphate with loss of 6CO₂ molecules and generation of 12 NADPH + H⁺ (not NADH). The remaining reactions known as non – oxidative pathway, convert Ribulose – 5 – phosphate molecules to various intermediates such as Ribose – 5 – phosphate(5C), Xylulose – 5 – phosphate(5C), Glyceraldehyde – 3 – phosphate(3C), Sedoheptulose – 7 – Phosphate (7C), and Erythrose – 4 – phosphate (4C). Finally, five molecules of glucose – 6 – phosphate is regenerated. The overall reaction is:

The net result of complete oxidation of one glucose – 6 – phosphate yield 6CO₂ and 12 NADPH + H⁺. The oxidative pentose phosphate pathway is controlled by glucose – 6 – phosphate dehydrogenase enzyme which is inhibited by high ratio of NADPH to NADP⁺.

Question 10.
How will you calculate net products of one sucrose molecule upon complete oxidation during aerobic respiration as per recent view?
Answer:
When the cost of transport of ATPs from matrix into the cytosol is considered, the number will be 2.5 ATPs for each NADH + H⁺ and 1.5 ATPs for each FADH₂ oxidised during electron transport system. Therefore, in plant cells net yield of 30 ATP molecules for complete aerobic oxidation of one molecule of glucose. But in those animal cells (showing malate shuttle mechanism) net yield will be 32 ATP molecules. Since sucrose molecule gives, two molecules of glucose and net ATP in plant cell will be 30 × 2 = 60. In animal cell it will be 32 × 2 = 64.

Samacheer Kalvi 11th Bio Botany Respiration Additional Questions & Answers

I. Choose the correct answer (1 Mark)
Question 1.
The term respiration was coined by:
(a) Lamark
(b) Kerb
(c) Pepys
(d) Blackman
Answer:
(c) Pepys

Question 2.
In floating respiration the substrates are:
(a) carbohydrate or protein
(b) carbohydrate or fat
(c) protein or fat
(d) none of the above
Answer:
(b) carbohydrate or fat

Question 3.
The discovery of ATP was made by:
(a) Lipman
(b) Hans Adolt
(c) Warburg
(d) Karl Lohman
Answer:
(d) Karl Lohman

Question 4.
The end product of glycolysis is:
(a) pyruvate
(b) ethanol
(c) malate
(d) succinate
Answer:
(a) pyruvate

Question 5.
On hydrolysis, one molecule of ATP releases energy of:
(a) 8.2 K cal
(b) 32.3 kJ
(c) 7.3 K cal
(d) 7.8 K cal
Answer:
(c) 7.3 K cal

Question 6.
Which of the following is known as terminal oxidation:
(a) glycolysis
(b) electron transport chain
(c) Kreb’s cycle
(d) pyruvate oxidation
Answer:
(b) electron transport chain

Question 7.
Identify the link reaction:
(a) conversion of glucose into pyruvic acid
(b) conversion of glucose into ethanol
(c) conversion of acetyl CoA into CO₂ and water
(d) conversion of pyruvic acid into acetyl coenzyme – A
Answer:
(d) conversion of pyruvic acid into acetyl coenzyme – A

Question 8.
Who was awarded Nobel prize in 1953 for the discovery of TCA cycle?
(a) Lipmann
(b) Hans Adolf Kreb
(c) Petermitchell
(d) Dickens
Answer:
(b) Hans Adolf Kreb

Question 9.
Kreb’s cycle is a:
(a) catabolic pathway
(b) anabolic pathway
(c) amphibolic pathway
(d) hydrolytic pathway
Answer:
(c) amphibolic pathway

Question 10.
Electron transport system during aerobic respiration takes place in:
(a) cytoplasm
(b) mitochondria
(c) chloroplast
(d) golgi apparatus
Answer:
(b) mitochondria

Question 11.
The oxidation of one molecule of NADH + H⁺ gives rise to:
(a) 2 ATP
(b) 3 ATP
(c) 4 ATP
(d) 2.5 ATP
Answer:
(b) 3 ATP

Question 12.
In aerobic prokaryotes each molecule of glucose produces:
(a) 36 ATP
(b) 32 ATP
(c) 34 ATP
(d) 38 ATP
Answer:
(d) 38 ATP

Question 13.
Cyanide acts as electron transport chain inhibitor by preventing:
(a) synthesis of ATP from ADP
(b) flow of electrons from NADH + H⁺
(c) flow of electrons from cytochrome a₃ to O₂
(d) oxidative phosphorylation
Answer:
(c) flow of electrons from cytochrome a₃ to O₂

Question 14.
Respiratory quotient for oleic acid is:
(a) 0.69
(b) 0.71
(c) 0.80
(d) 0.36
Answer:
(b) 0.71

Question 15.
End products of fermentation in yeast is:
(a) pyruvic acid and CO₂
(b) lactic acid qnd CO₂
(c) ethyl alcohol and CO₂
(d) mixed acid and CO₂
Answer:
(c) ethyl alcohol and CO₂

Question 16.
The end products of mixed acid fermentation in enterobacteriaceae are:
(a) lactic acid, ethanol, formic acid, CO₂ and H₂
(b) lactic acid, formic acid and CO₂
(c) lactic acid, ethanol, CO₂ and O₂
(d) ethanol, formic acid, CO₂ and H₂
Answer:
(a) lactic acid, ethanol, formic acid, CO₂ and H₂

Question 17.
The external factors that affect the respiration are:
(a) temperature, insufficient O₂ and amount of protoplasm
(b) temperature, insufficient O₂ and high concentration of CO₂
(c) temperature, high concentration of CO₂ and respiratory substrate
(d) temperature, high concentration of CO₂ and amount of protoplasm
Answer:
(b) temperature, insufficient O₂ and high concentration of CO₂

Question 18.
Pentose phosphate pathway was described by:
(a) Pepys and Black man
(b) Kreb and Embden
(c) Warburg, Dickens and Lipmann
(d) Warburg and Pamas
Answer:
(c) Warburg, Dickens and Lipmann

Question 19.
The oxidative pentose phosphate pathway is controlled by the enzyme:
(a) glucose, 1, 6 diphosphate dehydrogenase
(b) glucose 6 phosphate dehydrogenase
(c) fructose – 6 – phosphate dehydrogenase
(d) none of the above
Answer:
(b) glucose 6 phosphate dehydrogenase

Question 20.
In pentose phosphate pathway the glucose – 6 – phosphate dehydrogenase enzyme is inhibited by high ratio of:
(a) FADH to FAD
(b) glucose to glucose – 6 – phosphate
(c) NADPH to NADP
(d) GTPH to GTP
Answer:
(c) NADPH to NADP

Question 21.
In plant tissue erythrose is used for the synthesis of:
(a) Erythromycin
(b) Xanthophill
(c) Erythrocin
(d) Arithocyanin
Answer:
(d) Arithocyanin

Question 22.
As per the recent view, when a glucose molecule is completely aerobically oxidised, the net yield of ATP in plant cell is:
(a) 38
(b) 36
(c) 30
(d) 32
Answer:
(c) 30

Question 23.
Identify the electron transport inhibitor:
(a) phosphophenol
(b) dinitrophenol
(c) xylene
(d) indol acetic acid
Answer:
(b) dinitrophenol

Question 24.
The phenomenon of climacteric is present in:
(a) banana
(b) coconut
(c) cauli flower
(d) brinjal
Answer:
(a) banana

Question 25.
Cyanide resistant respiration is known to generate heat in thermogenic tissues as high as:
(a) 35° C
(b) 38° C
(c) 40° C
(d) 51° C
Answer:
(d) 51° C

Question 26.
Match the following:

Substrate	RQ
A. Palmitic acid	(i) 1.6
B. Oleic acid	(ii) 4.0
C. Tartaric acid	(iii) 0.36
D. Oxalic acid	(iv) 0.71

(a) A – (ii), B – (iii); C – (i); D – (iv)
(b) A – (iii), B – (iv); C – (i); D – (ii)
(c) A – (ii); B – (iv); C – (i); D – (iii)
(d) A – (iii); B – (i); C – (iv); D – (ii)
Answer:
(b) A – (iii), B – (iv); C – (i); D – (ii)

Question 27.
Indicate the correct statement:
(a) In Bryophyllum, carbohydrates are partially oxidised to organic acid
(b) In opuntia, the Respiratory Quotient value is 0.5
(c) Alcoholic fermentation takes place in enterobacteriaceae
(d) Muscles of vertebrate does not have lactate dehydrogenase enzyme
Answer:
(a) In Bryophyllum, carbohydrates are partially oxidised to organic acid

Question 28.
The order of aerobic respiration in plant cell is:
(a) glycolysis, Kreb’s cycle, pyruvate oxidation and electron transport chain
(b) glycolysis, pyruvate oxidate, Kreb’s cycle, electron transport chain
(c) pyruvate oxidation, glycolysis, Kreb’s cycle, electron transport chain
(d) none of the above order
Answer:
(b) glycolysis, pyruvate oxidate, Kreb’s cycle, electron transport chain

Question 29.
The complete reactions of glycolysis take place in:
(a) mitochondria
(b) cristae
(c) cytoplasm
(d) outer membrane of mitochondria
Answer:
(c) cytoplasm

Question 30.
The Co – enzyme quinone is a proton carrier located within:
(a) outer membrane of mitochondria
(b) cytoplasm
(c) inner membrane of mitochondria
(d) matrix of mitochondria
Answer:
(c) inner membrane of mitochondria

Question 31.
How many molecules of CO₂ are produced during link reaction:
(a) 1
(b) 6
(c) 4
(d) 2
Answer:
(d) 2

Question 32.
In the case of ground nut, during seed germination they use:
(a) carbohydrate as respiratory substrate
(b) fat alone as respiratory substrate
(c) fat and protein as respiratory substrate
(d) protein alone as respiratory substrate
Answer:
(c) fat and protein as respiratory substrate

Question 33.
Lactic acid fermentation takes place in:
(a) yeast
(b) bacillus
(c) enterobacteriaceae
(d) none of the above
Answer:
(b) bacillus

Question 34.
The net result of complete oxidation of one glucose-6-phosphate in pentose phosphate pathway yield:
(a) 6 CO₂ and 12 NADPH + H⁺
(b) 6 CO₂ and 10 NADPH + H⁺
(c) 8 CO₂ and 16 NADPH + H⁺
(d) 8 CO₂ and 14 NADPH + H
Answer:
(a) 6 CO₂ and 12 NADPH + H+

Question 35.
Ribose – 5 – phosphate and its derivatives are used in the synthesis of:
(a) lignin
(b) coenzyme A
(c) anthocyanin
(d) xanthophyll
Answer:
(b) coenzyme A

II. Answer the following (2 Marks)

Question 1.
Define respiration?
Answer:
Respiration is a biological process in which oxidation of various food substances like carbohydrates, proteins and fats take place and as a result of this, energy is produced where O₂ is taken in and CO₂ is liberated.

Question 2.
What is meant by protoplasmic respiration?
Answer:
Respiration utilizing protein as a respiratory substrate, it is called protoplasmic respiration. Protoplasmic respiration is rare and it depletes structural and functional proteins of protoplasm and liberates toxic ammonia.

Question 3.
What do you understand by compensation of point?
Answer:
The point at which CO₂ released in respiration is exactly compensated by CO₂ fixed in photosynthesis that means no net gaseous exchange takes place, it is called compensation point.

Question 4.
Explain briefly about aerobic respiration.
Answer:
Respiration occurring in the presence of oxygen is called aerobic respiration. During aerobic respiration, food materials like carbohydrates, fats and proteins are completely oxidised into CO₂, H₂O and energy is released.

Question 5.
What is anaerobic respiration?
Answer:
In the absence of molecular oxygen glucose is incompletely degraded into either ethyl alcohol or lactic acid. It includes two steps:

1. Glycolysis
2. Fermentation

Question 6.
What do you know about transition reaction?
Answer:
In aerobic respiration the pyruvate with coenzyme A is oxidatively decarboxylated into acetyl CoA by pyruvate dehydrogenase complex. This reaction is irreversible and produces two molecules of NADH + H⁺ and 2CO₂. It is also called transition reaction or Link reaction.

Question 7.
Who is Sir Hans Adolf Krebs?
Answer:
Sir Hans Adolf Krebs was born in Germany on 25th August 1900. He was awarded Nobel Prize for his discovery of Citric acid cycle in Physiology in 1953.

Question 8.
Explain briefly about amphibolic pathway.
Answer:
Krebs cycle is primarily a catabolic pathway, but it provides precursors for various biosynthetic pathways thereby an anabolic pathway too. Hence, it is called amphibolic pathway.

Question 9.
Mention the role of NADH dehydrogenase enzyme in electron transport system.
Answer:
NADH dehydrogenase contains a flavoprotein (FMN) and associated with non – heme iron Sulphur protein (Fe – S). This complex is responsible for passing electrons and protons from mitochondrial NADH (Internal) to Ubiquinone (UQ).

Question 10.
What is oxidative phosphorylation?
Answer:
The transfer of electrons from reduced coenzyme NADH to oxygen via complexes I to IV is coupled to the synthesis of ATP from ADP and inorganic phosphate (Pi) which is called Oxidative phosphorylation.

Question 11.
Mention any two electron transport chain inhibitors.
Answer:
Two electron transport chain inhibitors:

1. 2, 4 DNP (Dinitrophenol) – It prevents synthesis of ATP from ADP, as it directs electrons from Co Q to O₂.
2. Cyanide – It prevents flow of electrons from Cytochrome a₃ to O₂.

Question 12.
Define respiratory quotient.
Answer:
The ratio of volume of carbon dioxide given out and volume of oxygen taken in during respiration is called Respiratory Quotient.

Question 13.
What are the significances of Respiratory Quotient?
Answer:
The significances of Respiratory Quotient:

1. RQ value indicates which type of respiration occurs in living cells, either aerobic or anaerobic.
2. It also helps to know which type of respiratory substrate is involved.

Question 14.
Explain the term alcoholic fermentation.
Answer:
The cells of roots in water logged soil respire by alcoholic fermentation because of lack of oxygen by converting pyruvic acid into ethyl alcohol and CO₂. Many species of yeast (Saccharomyces) also respire anaerobically. This process takes place in two steps:

Question 15.
Mention any two industrial uses of alcoholic fermentation.
Answer:
Two industrial uses of alcoholic fermentation:

1. In bakeries, it is used for preparing bread, cakes, biscuits.
2. In beverage industries for preparing wine and alcoholic drinks.

Question 16.
What do you understand by the term mixed acid fermentation?
Answer:
This type of fermentation is a characteristic feature of Enterobacteriaceae and results in the formation of lactic acid, ethanol, formic acid and gases like CO₂ and H₂.

Question 17.
Mention any two internal factors, that affect the rate of respiration in plants.
Answer:
Two internal factors, that affect the rate of respiration in plants:

1. The amount of protoplasm and its state of activity influence the rate of respiration.
2. Concentration of respiratory substrate is proportional to the rate of respiration.

Question 18.
What is the control mechanism of pentose phosphate pathway?
Answer:
The oxidative pentose phosphate pathway is controlled by glucose-6-phosphate dehydrogenase enzyme which is inhibited by high ratio of NADPH to NADP⁺.

Question 19.
Write down any two significance of pentose phosphate pathway.
Answer:
Two significance of pentose phosphate pathway:

1. HMP shunt is associated with the generation of two important products,
2. Coenzyme NADPH generated is used for reductive biosynthesis and counter damaging the effects of oxygen free radicals.

III. Answer the following (3 Marks)

Question 1.
In biosphere how do plants and animals are complementary systems, which are integrated to sustain life?
Answer:
In plants, oxygen enters through the stomata and it is transported to cells, where oxygen is utilized for energy production. Plants require carbon dioxide to survive, to produce carbohydrates and to release oxygen through photosynthesis, these oxygen molecules are inhaled by human through the nose, which reaches the lungs where oxygen is transported through the blood and it reaches cells. Cellular respiration takes place inside or the cell for obtaining energy.

Question 2.
What will happen, when you sleep under a tree during night time?
Answer:
If you are sleeping under a tree during night time you will feel difficulty in breathing. During night, plants take up oxygen and release carbon dioxide and as a result carbon dioxide will be abundant around the tree

Question 3.
What are the factors associated with compensation point in respiration?
Answer:
The two common factors associated with compensation point are CO₂ and light. Based on this there are two types of compensation point. They are CO₂ compensation point and light compensation point. C₃ plants have compensation points ranging from 40 – 60 ppm (parts per million) CO₂ while those of C₄ plants ranges from 1 – 5 ppm CO₂.

Question 4.
Why do you call ATP as universal energy currency of cell?
Answer:
ATP is a nucleotide consisting of a base- adenine, a pentose sugar – ribose and three phosphate groups. Out of three phosphate groups the last two are attached by high energy rich bonds. On hydrolysis, it releases energy (7.3 K cal or 30.6 KJ / ATP) and it is found in all living cells and hence it is called universal energy currency of the cell.

Question 5.
What is a redox reaction?
Answer:
NAD⁺ + 2e^– + 2H⁺ → NADH + H⁺
FAD + 2e^– + 2H⁺ → FADH₂
When NAD⁺ (Nicotinamide Adenine Dinucleotide – oxidised form) and FAD (Flavin Adenine Dinucleotide) pick up electrons and one or two hydrogen ions (protons), they get reduced to NADH + H⁺ and FADH₂ respectively. When they drop electrons and hydrogen off they go back to their original form. The reaction in which NAD⁺ and FAD gain (reduction) or f lose (oxidation) electrons are called redox reaction (Oxidation reduction reaction). These reactions are important in cellular respiration.

Question 6.
Write down any three differences between aerobic and anaerobic respiration.
Answer:
Aerobic respiration:

• It occurs in all living cells of higher organisms.
• It requires oxygen for breaking the respiratory substrate.
• The end products are CO₂ and H₂O.

Anaerobic Respiration:

• It occurs yeast and some bacteria.
• Oxygen is not required for breaking the respiratory substrate.
• The end products are alcohol, and CO₂ (or) lactic acid.

Question 7.
Mention the significance of Kreb’s cycle.
Answer:
The significance of Kreb’s cycle:

1. TCA cycle is to provide energy in the form of ATP for metabolism in plants.
2. It provides carbon skeleton or raw material for various anabolic processes.
3. Many intermediates of TCA cycle are further metabolised to produce amino acids, proteins and nucleic acids.
4. Succinyl CoA is raw material for formation of chlorophylls, cytochrome, phytochrome and other pyrrole substances.
5. α – ketoglutarate and oxaloacetate undergo reductive amination and produce amino acids.
6. It acts as metabolic sink which plays a central role in intermediary metabolism.

Question 8.
Derive the respiratory quotient for carbohydrate as substrate in oxidative metabolism.
Answer:
The respiratory substrate is a carbohydrate, it will be completely oxidised in aerobic respiration and the value of the RQ will be equal to unity.

Question 9.
Write down the characteristic of Anaerobic respiration.
Answer:
The characteristic of Anaerobic respiration:

1. Anaerobic respiration is less efficient than the aerobic respiration.
2. Limited number of ATP molecules is generated per glucose molecule.
3. It is characterized by the production of CO₂ and it is used for Carbon fixation in photosynthesis.

Question 10.
Distinguish between glycolysis and fermentation.
Answer:
Glycolysis:

1. Glucose is converted into pyruvic acid.
2. It takes place in the presence or absence of oxygen.
3. Net gain is 2ATR
4. 2 NADH + H⁺ molecules are produced.

Fermentation:

1. Starts from pyruvic acid and is converted into alcohol or lactic acid.
2. It takes place in the absence of oxygen.
3. No net gain of ATP molecules.
4. 2 NADH + H⁺ molecules are utilised.

Question 11.
Write down any three external factors, that affect respiration in plants.
Answer:
Three external factors, that affect respiration in plants:

1. Optimum temperature for respiration is 30°C. At low temperatures and very high temperatures rate of respiration decreases.
2. When sufficient amount of O₂ is available the rate of aerobic respiration will be optimum and anaerobic respiration is completely stopped. This is called Extinction point.
3. High concentration of CO₂ reduces the rate of respiration

Question 12.
How alcoholic beverages like beer and wine is made?
Answer:
The conversion of pyruvate to ethanol takes place in malted barley and grapes through fermentation. Yeasts carryout this process under anaerobic conditions and this Conversion increases ethanol concentration. If the concentration increases, it’s toxic effect kills yeast cells .and the left out is called beer and wine respectively.

IV. Answer the following (5 Marks)

Question 1.
Give the schematic representation of glycolysis or EMP pathway.
Answer:
The schematic representation of glycolysis or EMP pathway:

Question 2.
Write down the biochemical events in Kreb’s cycle.
Answer:
The biochemical events in Kreb’s cycle:

Question 3.
Mention the schematic diagram of the various steps involved in pentose phosphate pathway.
Answer:
The schematic diagram of the various steps involved in pentose phosphate pathway:

Question 4.
Describe the events in electron transport chain in plant cell.
Answer:
During glycolysis, link reaction and Krebs cycle the respiratory substrates are oxidised at several steps and as a result many reduced coenzymes NADH + H⁺ and FADH₂ are produced. These reduced coenzymes are transported to inner membrane of mitochondria and are converted back to their oxidised forms produce electrons and protons. In mitochondria, the inner membrane is folded in the form of finger projections towards the matrix called cristae.

In cristae many oxysomes (F₁ particles) are present which have election transport carriers are present. According to Peter Mitchell’s Chemiosmotic theory this electron transport is coupled to ATP synthesis. Electron and hydrogen (proton) transport takes place across four multiprotein complexes (I – IV). They are

(i) Complex – I (NADH dehydrogenase: It contains a flavoprotein (FMN) and associated with non – heme iron Sulphur protein (Fe – S). This complex is responsible for passing electrons and protons from mitochondrial NADH (Internal) to Ubiquinone (UQ).
NADH + H⁺ + UQ ⇌ NAD⁺ + UQH₂
In plants, an additional NADH dehydrogenase (External) complex is present on the outer surface of inner membrane of mitochondria which can oxidise cytosolic NADH + H⁺ Ubiquinone (UQ) or Coenzyme Quinone (Co Q) is a small, lipid soluble electron, proton carrier located within the inner membrane of mitochondria.

(ii) Complex – II (Succinic dehydrogenase): It contains FAD flavoprotein is associated with non – heme iron Sulphur (Fe – S) protein. This complex receives electrons and protons from succinate in Krebs cycle and is converted into fumarate and passes to ubiquinone.
Succinate + UQ → Fumarate + UQH₂

(iii) Complex – III (Cytochrome bc₁ complex): This complex oxidises reduced ubiquinone (ubiquinol) and transfers the electrons through Cytochrome bc₁ Complex (Iron Sulphur center bc₁ complex) to cytochrome c. Cytochrome c is a small protein attached to the outer surface of inner membrane and act as a. mobile carrier to transfer electrons between complex III to complex IV.
UQH₂ + 2Cyt c_oxidised  ⇌  UQ + 2Cyt c_reduced  + 2H⁺

(iv) Complex IV (Cytochrome c oxidase): This complex contains two copper centers (A and B) and cytochromes a and as. Complex IV is the terminal oxidase and brings about the reduction of 1/2 O₂ to H₂O. Two protons are needed to form a molecule of H₂O (terminal oxidation).
2Cyt c_oxidised + 2H⁺ + 1/2 O₂  ⇌  2Cyt c_reduced + H₂O

The transfer of electrons from reduced coenzyme NADH to oxygen via complexes I to IV is coupled to the synthesis of ATP from ADP and inorganic phosphate (Pi) which is called Oxidative phosphorylation. The F₀F₁ – ATP synthase (also called complex V) consists of F₀ and F₁. F₁ converts ADP and Pi to ATP and is attached to the matrix side of the inner membrane. F₀ is present in inner membrane and acts as a channel through which protons come into matrix.

Oxidation of one molecule of NADH + H⁺ gives rise to 3 molecules of ATP and oxidation of one molecule FADH₂ produces 2 molecules of ATP within a mitochondrion. But cytoplasmic NADH + H⁺ yields only two ATPs through external NADH dehydrogenase. Therefore, two reduced coenzyme (NADH + H⁺) molecules from glycolysis being extra mitochondrial will yield 2 × 2 = 4 ATP molecules instead of 6 ATPs. The Mechanism of mitochondrial ATP synthesis is based on Chemiosmotic hypothesis.

According to this theory electron carriers present in the inner mitochondrial membrane allow for the transfer of protons (H⁺). For the production of single ATP, 3 protons (H⁺) are needed. The terminal oxidation of external NADH bypasses the first phosphorylation site and hence only two ATP molecules are produced per external NADH oxidised through However, in those animal tissues in which malate shuttle mechanism is present, the oxidation of external NADH will yield almost 3 ATP molecules.

Complete oxidation of a glucose molecule in aerobic respiration results in the net gain of 36 ATP molecules in plants. Since huge amount of energy is generated in mitochondria in the form of ATP molecules they are called ‘power house of the cell’. In the case of aerobic prokaryotes due to lack of mitochondria each molecule of glucose produces 38 ATP molecules.

Question 5.
Define respiratory quotient. Explain the derivation of respiratory quotient for various substrates oxidised :
Answer:
The ratio of volume of carbon dioxide given out and volume of oxygen taken in during respiration is called Respiratory Quotient or Respiratory ratio. RQ value depends, upon respiratory substrates and their oxidation.

(i) The respiratory substrate is a carbohydrate, it will be completely oxidised in aerobic respiration and the value of the RQ will be equal to unity.

(ii) If the respiratory substrate is . a carbohydrate it will be incompletely oxidised when it goes through anaerobic respiration and the RQ value will be infinity.

(iii) In some succulent plants like Opuntia, Bryophyllum carbohydrates are partially oxidised to organic acid, particularly malic acid without corresponding release of CO₂ but O₂ is consumed hence the RQ value will be zero.

(iv) When respiratory substrate is protein or fat, then RQ will be less than unity.

(v) When respiratory substrate is an organic acid the value of RQ will be more than unity.

Question 6.
Describe an experiment to demonstrate the production of CO₂ in aerobic respiration.
Answer:
Take small quantity of any seed (groundnut or bean seeds) and allow them to germinate by imbibing them. While they are germinating place them in a conical flask. A small glass tube containing 4 ml of freshly prepared Potassium hydroxide (KOH) solution is hung into the conical flask with the help of a thread and tightly close the one holed cork. Take a bent glass tube, the shorter end of which is inserted into the conical flask through the hole in the cork, while the longer end is dipped in a beaker containing water.

Observe the position of initial water level in bent glass tube. This experimental setup is kept for two hours and the seeds were allowed to germinate. After two hours, the level of water rises in the glass tube. It is because, the CO₂ evolved during aerobic respiration by germinating seeds will be absorbed by KOH solution and the level of water will rise in the glass tube.
CO₂ + 2KOH → K₂CO₃ + H₂O
In the case of groundnut or bean seeds, the rise of water is relatively lesser because these seeds use fat and proteins as respiratory substrate and release a very small amount of CO₂. But in the case of wheat grains, the rise in water level is greater because they use carbohydrate as respiratory substrate. When carbohydrates are used as substrate, equal amounts of CO₂ and O₂ are evolved and consumed.

Textbook Page No. 145

Question 1.
How many ATP molecules are produced from one sucrose molecule?
Answer:
One sucrose molecules gives rise to two glucose molecules. The net production of ATP during complete oxidation of one glucose molecule in plant cell is 36 ATP. Therefore one sucrose molecule yields 36 x 2 = 72 ATP molecules.
As per recent view in plants cells, one molecules of glucose, after complete aerobic oxidation yields only 30 ATP molecules and hence one sucrose molecule yield only 30 x 2 = 60 ATP molecules.

Textbook Page No. 156

Question 1.
Why Microorganisms respire anaerobically?
Answer:
Some of the microorganism live in environments devoid of oxygen and they have to adopt themselves in anoxic condition. Hence they respire anaerobically and they are called anaerobic microbes.

Question 2.
Does anaerobic respiration take place in higher plants?
Answer:
Anaerobic respiration some time occur in the root of some water – logged plants.

Students can Download Bio Botany Chapter 15 Plant Growth and Development Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Bio Botany Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 15 Plant Growth and Development

Samacheer Kalvi 11th Bio Botany Plant Growth and Development Text Book Back Questions and Answers

Question 1.
Select the wrong statement from the following:
(a) Formative phase of the cells retain the capability of cell division.
(b) In elongation phase development of central vacuole takes place.
(c) In maturation phase thickening and differentiation takes place.
(d) In maturation phase, the cells grow further.
Answer:
(d) In maturation phase, the cells grow further.

Question 2.
If the diameter of the pulley is 6 inches, length of pointer is 10 inches and distance travelled by pointer is 5 inches. Calculate the actual growth in length of plant:
(a) 3 inches
(b) 6 inches
(c) 12 inches
(d) 30 inches
Answer:
(a) 3 inches

Question 3.
In uni sexual plants, sex can be changed by the application of:
(a) ethanol
(b) cytokinins
(c) ABA
(d) auxin
Answer:
(c) ABA

Question 4.
Select the correctly matched one:

(a) A – (iii), B – (iv), C – (v), D – (vi), E – (i), F – (ii)
(b) A – (v), C – (ii), D – (iv), E – (vi), F – (iii)
(c) A – (iii), B – (v), C – (vi),D – (i), E – (ii), F – (iv)
(d) A – (ii), B – (iii), C – (v), D – (vi), E – (iv), F – (i)
Answer:
(b) A – (v), C – (ii), D – (iv), E – (vi), F – (iii)

Question 5.
Seed dormancy allows the plants to:
(a) overcome un favorable climatic conditions
(b) develop healthy seeds
(c) reduce viability
(d) prevent deterioration of seeds
Answer:
(a) overcome unfavorable climatic conditions

Question 6.
What are the parameters used to measure growth of plants?
Answer:

• Increase in length or girth (roots and stems)
• Increase in fresh or dry weight
• Increase in area or volume (fruits and leaves)
• Increase in number of cells produced.

Question 7.
What is plasticity?
Answer:
Plants follow different pathways in response to environment or phases of life to form different kinds of structures. This ability is called plasticity,
eg : Heterophylly in cotton and coriander. In such plants, the leaves of the juvenile plant are different in shape from those in mature plants.

On the other hand, the difference in shapes of leaves produced in air and those produced in water in buttercup also represent he heterophyllous development due to the environment. This phenomenon of heterophylly is an example of plasticity.

Question 8.
Write the physiological effects of Cytokinins.
Answer:

1. Cytokinin promotes cell division in the presence of auxin (IAA).
2. Induces cell enlargement associated with IAA and gibberellins
3. Cytokinin can break the dormancy of certain light-sensitive seeds like tobacco and induces seed germination.
4. Cytokinin promotes the growth of lateral bud in the presence of apical bud.
5. Application of cytokinin delays the process of aging by nutrient mobilization. It is known as Richmond Lang effect.
6. Cytokinin:
   • increases rate protein synthesis
   • induces the formation of inter-fascicular cambium
   • overcomes apical dominance
   • induces formation of new leaves, chloroplast and lateral shoots.
7. Plants accumulate solutes very actively with the help of cytokinins.

Question 9.
Describe the mechanism of photoperiodic induction of flowering.
Answer:

Photoperiodic stimulus is perceived by the leaves. Floral hormone is synthesised in leaves and translocated to the apical tip to promote flowering. This can be explained by a simple experiment on Cocklebur (Xanthium pensylvanicum), a short day plant. Usually Xanthium will flower under short day conditions. If the plant is defoliated and kept under short day conditions it will not flower.

Flowering will occur even when all the leaves are removed except one leaf. If a cocklebur plant is defoliated and kept under long day conditions, it will not flower. If one of its leaves is exposed to short day condition and rest are in long day condition, flowering will occur.

Question 10.
Give a brief account on Pr grammed Cell Death (PCD).
Answer:
Senescence is controlled by plants own genetic program and death of the plant or plant part consequent to senescence is called Programmed Cell Death. In short senescence of an individual cell is called PCD. The proteolytic enzymes involving PCD in plants are phytases and in animals are caspases. The nutrients and other substrates from senescing cells and tissues are remobilized and reallocated to other parts of the plant that survives.

The protoplasts of developing xylem vessels and tracheids die and disappear at maturity to make them functionally efficient to conduct water for transport. In aquatic plants, aerenchyma is normally formed in different parts of the plant such as roots and stems which encloses large air spaces that are created through PCD. In the development of unisexual flowers, male and female flowers are present in earlier stages, but only one of these two completes its development while other aborts through PCD.

Samacheer Kalvi 11th Bio Botany Plant Growth and Development Additional Questions & Answers

I. Choose T he correct answer (1 Mark)
Question 1.
Open form of the growth occurs in:
(a) leaves and flowers
(b) stem and root
(c) leaves and stem
(d) stem and flowers
Answer:
(b) stem and root

Question 2.
Bamboo is classified under:
(a) monocarpi c annual plants
(b) polycarpic perennials
(c) monocarpic perennials
(d) polycarpic annual plants
Answer:
(c) monocarpic perennials

Question 3.
Primary growth of the plant is due to the activity of:
(a) phloem parenchyma
(b) phloem meristem
(c) vascular cambium
(d) apical meristem
Answer:
(d) apical meristem

Question 4.
One single maize root apical meristem can give rise to more than:
(a) 17,500 hew cells per hour
(b) 18,500 new cells per hour
(c) 19,000 new cells per hour
(d) 500 new cells per hour
Answer:
(a) 17,500 hew cells per hour

Question 5.
Thickening and differentiation of cells take place during:
(a) elongation phase
(b) formative phase
(c) maturation phase
(d) flowering phase
Answer:
(c) maturation phase

Question 6.
When die total growth of a plant is plotted against time, the shape of the curve obtained is:
(a) hyperbolic curve
(b) ‘S’ shaped sigmoid curve
(c) linear curve
(d) none of the above
Answer:
(b) ‘S’ shaped sigmoid curve

Question 7.
The total growth of the plant consists of four phases in the following order.
(a) Log phase, lag phase, decelerating phase and maturation phase
(b) Log phase, lag phase, maturation phase and decelerating phase
(c) Lag phase, log phase, maturation phase and decelerating phase
(d) Lag phase, log phase, decelerating phase and maturation phase
Answer:
(d) Lag phase, log phase, decelerating phase and maturation phase

Question 8.
Internal factors, that influences the growth of the plant is:
(a) nutrition
(b) light
(c) C / N ratio
(d) oxygen
Answer:
(c) C / N ratio

Question 9.
Absence of light may lead to yellowish in color in plants and this is called:
(a) venation
(b) etiolation
(c) estivation
(d) vernation
Answer:
(b) etiolation

Question 10.
Differentiated cells, after multiplication again lose the ability to divide and mature to perform specific functions. This is called:
(a) plasticity
(b) differentiation
(c) dedifferentiation
(d) redifferentiation
Answer:
(d) redifferentiation

Question 11.
Indicate a plant growth regulator from the following:
(a) cytocin
(b) cytokinins
(c) acetic acid
(d) methylene
Answer:
(b) cytokinins

Question 12.
Some of the polyamines are known to behave like:
(a) growth inhibitors
(b) plant hormones
(c) flowering inhibitors
(d) fruit ripening agent
Answer:
(b) plant hormones

Question 13.
The activity of synergistic effect involves the activity of:
(a) auxin and gibberellins
(b) auxin and ethylene
(c) ABA and gibberellins
(d) none of the above
Answer:
(a) auxin and gibberellins

Question 14.
Phytohormones are usually produced to in tips of:
(a) root alone
(b) stem alone
(c) leaves alone
(d) root, stem and leaves
Answer:
(d) root, stem and leaves

Question 15.
The term auxin was first coined by:
(a) Charles Darwin
(b) Kogl
(c) F.W. Went
(d) Smith
Answer:
(c) F.W. Went

Question 16.
Indole Acetic Acid (IAA) is a:
(a) growth inhibitor
(b) hetero auxin
(c) root inhibitor
(d) synthetic auxin
Answer:
(b) hetero auxin

Question 17.
Indicate a synthetic auxin.
(a) Indole Acetic Acid
(b) Phenyl Acetic Acid
(c) Indole Butyric Acid
(d) Napthalene Acetic Acid
Answer:
(d) Napthalene Acetic Acid

Question 18.
Auxin has a similar chemical structure of:
(a) Indole acetic acid
(b) Napthalene acetic acid
(c) Phenyl acetic acid
(d) 2,4 – Dichloro phenoxy
Answer:
(a) Indole acetic acid

Question 19.
Auxin stimulates:
(a) transpiration
(b) respiration
(c) flowering
(d) none of the above
Answer:
(b) respiration

Question 20.
The term gibberllin was named by:
(a) Brain
(b) Yabuta
(c) Sumiki
(d) kurosawa
Answer:
(b) Yabuta

Question 21.
Who established the structure of gibberellic acid?
(a) Brain etal
(b) Kurosawa
(c) Cross et al
(d) Yabuta and Sumiki
Answer:
(c) Cross et al

Question 22.
Formation of seedless fruits without fertilization is induced by:
(a) auxin
(b) cytokinin
(c) ethylene
(d) gibberellin
Answer:
(d) gibberellin

Question 23.
Cytokinins inducing cell division was first demonstrated by:
(a) Haberlandt
(b) Charles Darwin
(c) Clarke
(d) Hubert
Answer:
(a) Haberlandt

Question 24.
Zeatin is first isolated from unripe grains of:
(a) paddy
(b) wheat
(c) maize
(d) com
Answer:
(c) maize

Question 25.
Indicate correct statements.
(i) Genes are intracellular factors for growth.
(ii) Temperature has no role in the growth of plant.
(iii) Oxygen has a vital role in the growth of plants.
(iv) CIN ratio of soil does not affect the growth of plant.
(a) (i) and (iv)
(b) (ii) and (iv)
(c) (i) and (iii)
(d) (ii) and (iii)
Answer:
(c) (i) and (iii)

Question 26.
Aspartic acid is classified under:
(a) freeauxin
(b) precursor of auxin
(c) chemical structure of auxin
(d) bound auxin
Answer:
(d) bound auxin

Question 27.
The stress phytohormones (Abscisic acid) was first isolated by:
(a) Linn et al
(b) Addicott et al
(c) Edward et al
(d) Stone and Black
Answer:
(b) Addicott et al

Question 28.
The chemical structure of abscisic acid resembles the structure of:
(a) indole Acetic Acid
(b) malanic acid
(c) carotenoid
(d) xanthophyll
Answer:
(c) carotenoid

Question 29.
Pick out the correct statement from the following:
(i) Abscisic acid is found abundantly inside the chloroplast of green cells.
(ii) ABA is a powerful growth promotor.
(iii) ABA is formed from pentose phosphate pathway.
(iv) ABA has anti-auxih and anti-gibberellin property.
(a) (i) and (iv)
(b) (i) and (ii)
(c) (ii) and (iii)
(d) (ii) and (iv)
Answer:
(a) (i) and (iv)

Question 30.
Abscisic acid induces male flower formation on female plants of:
(a) potato
(b) Cannabis sativa
(c) Vinca rosea
(d) Delomix regia
Answer:
(b) Cannabis sativa

Question 31.
Pea and barley are classified under:
(a) short day plants
(b) short long day plants
(c) long day plants
(d) long short day plants
Answer:
(c) long day plants

Question 32.
The term ‘photoperiodism’ was coined by:
(a) Miller and Amald
(b) Gamer and Allard
(c) Michael and Edward
(d) Darwin and Lamark
Answer:
(b) Gamer and Allard

Question 33.
Usually Xanthiumpensylvanicum will flower under:
(a) long day condition
(b) short long day condition
(c) photoneutral condition
(d) short day condition
Answer:
(d) short day condition

Question 34.
Phytochrome is a:
(a) reddish xanthophyll pigment
(b) bluish biliprotein pigment
(c) rodopsin pigment
(d) none of the above
Answer:
(b) bluish biliprotein pigment

Question 35.
Who found out the phytochrome in plants?
(a) Butler et al
(b) Michell et al
(c) Boumick et al
(d) Gamers and Allard
Answer:
(a) Butler et al

Question 36.
The term “vernalization” was first used by:
(a) Gamer
(b) Michell
(c) Lysenko
(d) Kawasacki
Answer:
(c) Lysenko

Question 37.
Pick out the wrong statement from the following:
(a) Vernalization increases the cold resistance of plants
(b) It increase the resistance of plants to fungal disease
(c) Vemalizatiqn increase the vegetative period of the plant
(d) It accelerates the plant breeding
Answer:
(c) Vemalizatiqn increase the vegetative period of the plant

Question 38.
In Oxalis, the seed viability ranges from:
(a) 10 to 15 years
(b) a few days
(c) more than 100 years
(d) upto 100 years
Answer:
(b) a few days

Question 39.
In apple and plum, the method of breaking seed dormancy involves the process of:
(a) impaction
(b) Scarification
(c) exposing to red light
(d) Stratification
Answer:
(d) Stratification

Question 40.
The proteolytic enzymes involved in – programmed cell death in plants are:
(a) phytochrome
(b) caspases
(c) phytaspases
(d) protolysis
Answer:
(c) phytaspases

II. Answer the following (2 Marks)

Question 1.
Define closed form of growth in plants.
Answer:
Leaves, flowers and fruits are limited in growth or of determinate or closed form growth.

Question 2.
What is meant by grand period of growth in plants?
Answer:
The total period from initial to the final stage of growth is called the grand period of growth. The total growth is plotted against time and ‘S’ shaped sigmoid curve (Grand period curve) is obtained.

Question 3.
Name the phases of growth in ‘S’ shaped growth curve.
Answer:

• Lag phase
• Log phase
• Decelerating phase
• Maturation phase

Question 4.
Define arithmetic growth rate in plant organ.
Answer:
If the length of a plant organ is plotted against time, it shows a linear curve and this growth is called arithmetic growth.

Question 5.
Distinguish between absolute growth rate and relative growth, rate.
Answer:
Absolute growth rate:
Increase in total growth of two organs measured and compared per unit time is called absolute growth rate.

Relative growth rate:
The growth of the given system per unit time expressed per unit initial parameter is called relative growth rate.

Question 6.
Define the term etiolation
Answer:
Light has its own contribution in the growth of the plant. Light is important for growth and photosynthesis. Light stimulates healthy growth. Absence of light may lead to yellowish in colour. This is called etiolation.

Question 7.
What is meant by redifferentiation of plant cells?
Answer:
Differentiated cells, after multiplication again lose the ability to divide and mature to perform specific functions. This is called redifferentiation, eg: Secondary xylem and Secondary phloem.

Question 9.
Mention any two synthetic auxins.
Answer:

• 2,4 – Dichloro Phenoxy Acetic Acid (2,4 – D)
• 2, 4, 5 – Trichloro Phenoxy Acetic Acid (2,4,5 – T)

Question 10.
Explain the synergistic effect of phytochromes.
Answer:
The effect of one or more substance in such a way that both promote each others activity, eg: Activity of auxin and gibberellins or cytokinins.

Question 11.
Name the natural auxins present in plants.
Answer:

• Indole Acetic Acid (IAA)
• Indole Propionic Acid (IPA)
• Indole Butyric Acid (IBA)
• Phenyl Acetic Acid (PAA)

Question 12.
Mention any two physiological effect of auxins in plant.
Answer:

• They promote cell elongation in stem and coleoptile.
• At higher concentrations auxins inhibit the elongation of roots but induce more lateral roots. Promotes growth of root only at extremely low concentrations.

Question 13.
Match the following.

(i) Indole acetic acid	(a) bolting
(ii) Napthalene acetic acid	(b) anti-auxin
(iii) Gibberellins	(c) synthetic auxin
(iv) Abscisic acid	(d) Natural auxin

Answer:
(i) – (d) Natural auxin
(ii) – (c) synthetic auxin
(iii) – (a) bolting
(iv) – (b) anti-auxin

Question 14.
Where do you find cytokinin hormone in plants?
Answer:
The distribution of cytokinin in plants is not as wide as those of auxin and gibberellins but found mostly in roots. Cytokinins appear to be translocated through xylem.

Question 15.
What is Richmond Lang effect?
Answer:
Application of cytokinin delays the process of aging by nutrient mobilization. It is known as Richmond Lang effect.

Question 16.
What is meant by non-climacteric fruits?
Answer:
All fruits cannot be ripened by exposure to ethylene. Such fruits are called nonclimacteric fruits and are insensitive to ethylene, eg: Grapes, Watermelon, Orange.

Question 17.
Why do you call Abscisic acid (ABA) as stress hormone?
Answer:
It inhibits the shoot growth and promotes growth of root system. This character protect the plants from water stress. Hence, ABA is called as stress hormone.

Question 18.
What is meant by short day plants?
Answer:
The plants that require a short critical day length for flowering are called short day plants or long night plants, eg: Tobacco, Cocklebur, Soybean, Rice and Chrysanthemum.

Question 19.
Write down the importance of photoperiodism in plants.
Answer:

• The knowledge of photoperiodism plays an important role in hybridisation experiments.
• Photoperiodism is an excellent example of physiological pre-conditioning that is using an external factor to induce physiological changes in the plant.

Question 20.
Define the term vernalization.
Answer:
Besides photoperiod certain plants require a low temperature exposure in their earlier stages for flowering. Many species of biennials and perennials are induced to flower by low temperature exposure (0°C to 5°C). This process is called Vernalization.

Question 21.
What is meant by Epigeal germination?
Answer:
During epigeal germination cotyledons are pushed out of the soil. This happens due to the elongation of the hypocotyl. eg: Castor and Bean.

Question 22.
Explain the term seed dormancy.
Answer:
The condition of a seed when it fails to germinate even in suitable environmental condition is called seed dormancy.

Question 23.
Define the term phytogerontology.
Answer:
The branch of botany which deals with ageing, abscission and senescence is called Phytogerontology.

Question 24.
Explain the term programmed cell death.
Answer:
Senescence is controlled by plants own genetic programme and death of the plant or plant part consequent to senescence is called Programmed Cell Death. In short senescence of an individual cell is called PCD.

Question 25.
Define the term “Abscission”.
Answer:
Abscission is a physiological process of shedding of organs like leaves, flowers, fruits and seeds from the parent plant body.

III. Answer the following (3 Marks)

Question 1.
Explain the different phases of growth in plants.
Answer:
There are three phases of growth.
1. Formative phase:
Growth in this phase occurs in. meristematic cells of shoot and root tips. These cells are small in size, have dense protoplasm, large nucleus and small vacuoles. Cells divide continuously by mitotic cell division. Some cells retain capability of cell division while other cells enter the next phase of growth.

2. Elongation Phase:
Newly formed daughter cells are pushed out of the meristematic zone and increases the volume. It requires auxin and food supply, deposition of new cell wall materials (intussusception), addition of protoplasm and development of central vacuole take place.

3. Maturation Phase:
During this stage cells attain mature form and size. Thickening and differentiation takes place. After differentiation, the cells do not grow further.

Question 2.
Draw the ‘S’ shaped growth curve and mark the different phases of growth
Answer:

Question 3.
Mention the internal factors, that affect the growth of plant.
Answer:

• Genes are intracellular factors for growth.
• Phytohormones are intracellular factors for growth, eg: auxin, gibberellin, cytokinin.
• C/N ratio.

Question 4.
What are the characteristic features of phytohormones?
Answer:

• Usually produced in tips of roots, stems and leaves.
• Transfer of hormones “from one place to another takes part through conductive systems.
• They are required in trace quantities.
• AH hormones are organic in nature.
• There are no specialized cells or organs for their secretion.
• They are capable of influencing physiological activities leading to promotion, inhibition and modification of growth.

Question 5.
List out the agricultural applications of auxins.
Answer:

• It is used to eradicate weeds, eg: 2,4 – D and 2,4,5 – T.
• Synthetic auxins are used in the formation of seedless fruits (Parthenocarpic fruit).
• It is used to break the dormancy in seeds.
• Induce flowering in Pineapple by NAA & 2,4 – D.
• Increase the number of female flowers and fruits in cucurbits.

Question 6.
Mention any three physiological effects of gibberellins.
Answer:

• It produces extraordinary elongation of stem caused by cell division and cell elongation.
• Rosette plants (genetic dwarfism) plants exhibit excessive inter modal growth when they are treated with gibberellins. This sudden elongation of stem followed by flowering is called bolting.
• Gibberellin breaks dormancy in potato tubers.

Question 7.
What are the uses of ethylene in agriculture?
Answer:

• Ethylene normally reduces flowering in plants except in Pine apple and Mango.
• It increases the number of female flowers and decreases the number of male flowers.
• Ethylene spray in cucumber crop produces female flowers and increases the yield.

Question 8.
What is meant by climacteric fruits?
Answer:
In most of the plants, there is sharp rise in respiration rate near the end of the development of fruit, called climacteric  rise. Such fruits are called climacteric fruits. The ripening on demand can be induced in these fruits by exposing them to normal air containing about 1 ppm of ethylene. A liquid called ethephon is being used in fruit ripening as it continuously releases ethylene, eg: Tomato, Apples, Banana, Mango.

Question 9.
Give the classification of plants based on photoperiodism.
Answer:

Question 10.
Explain the term photoperiodic induction.
Answer:
An appropriate photoperiod in 24 hours cycle constitutes one inductive cycle. Plants , may require one or more inductive cycles for flowering. The phenomenon of conversion of leaf primordia into flower primordia under the influence of suitable inductive cycles is called photoperiodic induction.
eg: Xanthium (SDP) – 1 inductive cycle and Plantago (LDP) – 25 inductive cycles.

Question 11.
What are the practical applications of vernalization in plants?
Answer:

• Vernalization shortens’ the vegetative period and induces the plant to flower earlier.
•  It increases the cold resistance of the plants.
• It increases the resistance of plants to fungal disease.
• Plant breeding can be accelerated.

Question 12.
Write down the internal factors, that affect seed germination.
Answer:
1. Maturity of embryo:
The seeds of some plants, when shed will contain immature embryo. Such seeds germinate only after maturation of embryo.

2. Viability:
Usually seeds remain viable or living only for a particular period. Viability of seeds range from a few days (eg: Oxalis) to more than hundred years. Maximum viability (1000 years) has been recorded in lotus seeds. Seeds germinate only within the period of viability.

3. Dormancy:
Seeds of many plants are dormant at the time of shedding.

Question 13.
Mention the factors causing dormancy of seeds.
Answer:

• Hard, tough seed coat causes barrier effect as impermeability of water, gas and restriction of the expansion of embryo prevents seed germination.
• Many species of seeds produce imperfectly developed embryos called rudimentary embryos which promotes dormancy.
• Lack of specific light requirement leads to seed dormancy.
• A range of temperatures either higher or lower cause dormancy.
• The presence of inhibitors like phenolic compounds which inhibits seed germination cause dormancy.

Question 14.
What are the significances of abscission?
Answer:

• Abscission separates dead parts of the plant, like old leaves and ripe fruits.
• It helps in dispersal of fruits and continuing the life cycle of the plant.
• Abscission of leaves in deciduous plants helps in water conservation during summer.
• In lower plants, shedding of vegetative parts like gemmae or plantlets help in vegetative reproduction.

IV. Answer the following (5 Marks)

Question 1.
Describe the geometric growth rate in plants with suitable diagram.
Answer:
Geometric growth rate:
This growth occurs in many higher plants and plant organs and is measured in size or weight. In plant growth, geometric cell division results if all cells of an organism or tissue are active mitotically. eg: Round three in the given figure, produces 8 cells as 2³ – 8 and after round 20 there are 2²⁰ = 1,048,576 cells. The large plant or animal parts are produced this way. In fact, it is common in animals but rare in plants except when they are young and small. Exponential growth curve can be expressed as,

W₁ = W₀e^rt
W₁ = Final size (weight, height and number)
W₀ = Initial size at the beginning of the period
r = Growth rate
t = Tittle of growth
e = Base of the natural logarithms

Here V is the relative growth rate and also a measure of the ability of the plant to produce new plant material, referred to as efficiency index. Hence, the final size of W₁ depends on the initial size W₀.

Question 2.
Describe the experiment to measure the increase in length of the stem tip using an arc auxanometer.
Answer:
The increase in the length of the stem tip can easily be measured by an arc auxanometer which consists of a small pulley to the axis of which is attached a long pointer sliding over a graduated arc. A thread one end of which is tied to the stem tip and another end to a weight passes over the pulley tightly.

As soon as the stem tip increases in length, the pulley moves and the pointer slide over the graduated arc (Figure). The reading is taken. The actual increase in the length of the stem is then calculated by knowing the length of the pointer and the radius of the pulley. If the radius of the pulley is 4 inches and the length of pointer 20 inches the actual growth is measured as follows:

Question 3.
Write an essay on the phytochrome, Gibberelf ns in plants.
Answer:
1. Discovery:
The effect of gibberellins had been known in Japan since early 1800 where certain rice plants were found to suffer from ‘Bakanae’ or foolish seedling disease. This disease was found . by Kurosawa (1926) to be caused by a fungus Gibberella fujikuroi. The active substance was separated from fungus and named as gibberellin by Yabuta (1935). These are more than 100 gibberellins reported from both fungi and higher plants. They are noted as GA₁, GA₂, GA₃ and so on. GA₃ is the first discovered gibberellin. In 1938, Yabuta and Sumiki isolated gibberellin in crystalline form. In 1955, Brain et al., gave the name gibberellic acid. In 1961, Cross et al., established its structure.

2. Occurrence:
The major site of gibberellin production in plants is parts like embryo, roots and young leaves near the tip. Immature seeds are rich in gibberellins.

3. Precursors:
The gibberellins are chemically related to terpenoids (natural rubber, carotenoids and steroids) formed by 5 – C precursor, an Isoprenoid unit called Iso Pentenyl Pyrophosphate (IPP) through a number of intermediates. The primary precursor is acetate.

4. Chemical structure:
All gibberellins have gibbane ring structure.

5. Transport in plants:
The transport of gibberellins in plants is non-polar. Gibberellins are translocated through phloem and also occur in xylem due to lateral movement between vascular bundles.

6. Bioassay (Dwarf Pea. assay):
Seeds of dwarf pea are allowed to germinate till the formation of the coleoptile. GA solution is applied to some seedlings. Others are kept under control. Epicotyle length is measured and as such, GA stimulating epicotyle growth can be seen.

7. Physiological Effects:
It produces extraordinary elongation of stem caused by cell division and cell elongation.

• Rosette plants (genetic dwarfism) plants exhibit excessive intermodal growth when they are treated with gibberellins. This sudden elongation of stem followed by flowering is called bolting.
  
• Gibberellin breaks dormancy in potato tubers.
• Many biennials usually flower during second year of their growth. For flowering to take place, these plants should be exposed to cold season. Such plants could be made to flower without exposure to cold season in the first year itself, when they are treated with gibberellins.

8. Agricultural role:

• Formation of seedless fruits without fertilization is induced by gibberellins eg: Seedless tomato, apple and cucumber.
• It promotes the formation of male flowers in cucurbitaceae. It helps in crop improvement.
  Uniform bolting and increased uniform seed production.
• Improves number and size of fruits in grapes. It increase yield.
• Promotes elongation of inter-node in sugarcane without decreasing sugar content.
• Promotion of flowering in long day plants even under short day conditions. .
• It stimulates the seed germination

Question 4.
What are their physiological effects of Abscisic acid in – plants and its role in agriculture?
Answer:
Physiological effects:

• It helps in reducing transpiration rate by closing stomata. It inhibits K+ uptake by guard cells and promotes the leakage of malic acid. It results in closure of stomata;
• It spoils chlorophylls, proteins and nucleic acids of leaves making them yellow.
•  Inhibition of cell division and cell elongation.
•  ABA is a powerful growth inhibitor. It causes 50% inhibition of growth in Oat coleoptile.
• It induces bud and seed dormancy.
• If promotes, the abscission of leaves, flowers and fruits by forming abscission layers.
• ABA plays an important role in plants dtiring water stress -and during drought conditions. It results in loss of turgor and closure of stomata,
• It has anti-auxin and anti-gibberellin property.
• Abscisic acid promotes senescence in leaves’ by causing loss of chlorophyll pigment decreasing the rate of photosynthesis and changing the rate of proteins and nucleic acid synthesis

Agricultural Role:

• In Cannabis sativa, induces male flower formation on female plants.
• Induction of flowers in short day plants.
• It promotes sprouting in storage organs like Potato.
• ABA plays an important role in plants during water stress drought conditions.
• It inhibits the shoot growth and promotes growth of root system. This character protect the plants from water stress. Hence, ABA is called as stress hormone.

Question 5.
Write an essay on the role of ethylene on plant physiology and agriculture.
Answer:
Almost all plant tissues produce ethylene gas in minute quantities.
1. Discovery:
In 1924, Denny found that ethylene stimulates the ripening of lemons. In 1934, R. Gane found that ripe bananas contain abundant ethylene. In 1935, Cocken et al., identified ethylene as a natural plant hormone.

2. Occurrence:
Maximum synthesis occurs during climacteric ripening of fruits and tissues undergoing senescence. It is formed in almost all plant parts like roots, leaves, flowers, fruits and seeds.

3. Transport in plants:
Ethylene can easily diffuse inside the plant through intercellular spaces.

4. Precursor:
It is a derivative of amino acid methionine, linolenic acid and fumaric acid.

5. Bioassay (Gas Chromatography):
Ethylene can be measured by gas chromatography. This technique helps in the detection of exact amount of ethylene from different plant tissues like lemon and orange.

6. Physiological Effects:

• Ethylene stimulates respiration and ripening in fruits.
• It stimulates radial growth in sterft and roof and inhibits linear growth.
• It breaks the dormancy of buds, seeds and storage organs.
• It stimulates formation of abscission zone in leaves, flowers and fruits. This makes the leaves to shed prematurely.
• Inhibition of stem elongation (shortening the intemode).
• In low concentration, ethylene helps in root initiation.
• Growth of lateral roots and root hairs. This increases the absorption surface of the plant roots.
• The growth of fruits is stimulated by ethylene in some plants. It is more marked in climacteric fruits.
• Ethylene causes epinasty.

7. Agricultural role:

• Ethylene normally reduces flowering in plants except in Pine apple and Mango.
• It increases the number of female flowers and decreases the number of male flowers.
• Ethylene spray in cucumber crop produces female flowers and increases the yield.

Question 6.
Explain the two hypothesis of explaining the mechanism of vernalization.
Answer:
Two main theories to explain the mechanism of vernalization are:

1. Hypothesis of phasic development:
According to Lysenko, development of an annual seed plant consists of two phases. First phase is thermostage, which is vegetative phase requiring low temperature and suitable moisture. Next phase is photo stage which requires high temperature for synthesis of florigen (flowering hormone).

2. Hypothesis of hormonal involvement:
According to Purvis (1961), formation of a substance A from its precursor is converted into B after chilling. The substance B is unstable. At suitable temperature B is converted into stable compound D called Vemalin. Vernalin is converted to F (Florigen). Florigen induces flower formation. At high temperature B is converted to C and devemalization occurs.

Question 7.
Write an essay on the types of senescence, its physiology and the factors affecting senescence.
Answer:
1. Types of Senescence:
Leopold (1961) has recognised four types of senescence.

(a) Overall senescence:
This kind of senescence occurs in annual plants when entire plant gets affected and dies, eg: Wheat and Soybean. It also occurs in few perennials also, eg: Agave and Bamboo.

(b) Top senescence:
It occurs in aerial parts of plants. It is common in perennials, underground and root system remains viable, eg: Banana and Gladiolus.

(c) Deciduous senescence:
It is common in deciduous plants and occurs only in leaves of plants, bulk of the stem and root system remains alive, eg: Elm and Maple.

(d) Progressive senescence:
This kind of senescence is gradual. First it occurs in old leaves followed by new leaves f then stem and finally root system. It is common in annuals.

2. Physiology of Senescence:

• Cells undergo changes in structure.
• Vacuole of the cell acts as lysosome and secretes hydrolytic enzymes.
• The starch content is decreased in the cells.
• Photosynthesis is reduced due to loss of chlorophyll accompanied by synthesis and accumulation of anthocyanin pigments, therefore the leaf becomes red.
• There is a marked decrease in protein content in the senescing organ.
• RNA content of the leaf particularly rRNA level is decreased in the cells due to increased activity of the enzyme RNAase.
• DNA molecules in senescencing leaves degenerate by the increased activity of enzyme DNAase.

3. Factors affecting Senescence:

•  ABA and ethylene accelerate senescence while auxin and cytokinin retard senescence.
• Nitrogen deficiency increases senescence whereas nitrogen supply retards, senescence. High temperature senescence but low retards senescence.
• Senescence is rapid in dark than in light.
• Water stress leads to accumulation of ABA leading to senescence.

Question 8.
Describe the methods of breaking dormancy of seeds in plants.
Answer:
The dormancy of seeds can be broken by different methods. These are:
1. Scarification:
Mechanical and chemical treatments like cutting or chipping of hard tough seed coat and use of organic solvents to remove waxy or fatty compounds are called as Scarification.

2. imp-action:
in some seeds water and oxygen are unable to penetrate micropyle due to blockage by cork cells. These seeds are shaken vigorously to remove the plug which is called Imp-action.

3. Stratification:
Seeds of rosaceous plants (Apple, Plum, Peach and Cherry) will not germinate until they have been exposed to well aerated, moist condition under low temperature (0°C to 10°C) for weeks to months. Such treatment is called Stratification.

4. Alternating temperatures: Germination of some seeds is strongly promoted by alternating daily temperatures. An alternation of low and high temperature improves the germination of seeds.

5. Light:
The dormancy of photoblastic seeds can be broken by exposing them to red light.

Solution To Activity

Textbook Page No : 164

Question 1.
Demonstration of phases of growth.
Answer:
To demonstrate and study the phases of growth, germinate a few seeds of bean on a circular filter paper soaked with water in a petridish. After two days of growth, select a few seedlings with straight radical of 2 to 3 cm length. Dry the surface of radical with a blotting paper and mark the radical from tip to base with at least 2 mm gap using water proof ink. Replace the seedlings in filter paper and observe further growth.

Observation:
The marked area in the radical will grow and increase in length and hence the marked area of 2mm is found to be grow beyond 2mm size due to the growth in the radical.

Textbook Page No: 169

Question 2.
Measurement of growth by direct method.
Answer:
Step 1: Take ordinary scale.
Step 2: Measure ground stem up to the growing point of the plant.
Step 3: Use Indian ink and mark at regular intervals to measure the length of root, stem, and girth of the trunk.

Observation:
At regular intervals measure the increase in length and girth of the trunk and it can be observed that the length of the root, stem and girth of the trunk increased with the increase in the period of growth.

 

Students can Download Bio Botany Chapter 8 Biomolecules Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Bio Botany Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 8 Biomolecules

Samacheer Kalvi 11th Bio Botany Biomolecules Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the correct answer
Question 1.
The most basic amino acid is …………… .
(a) Arginine
(b) Histidine
(c) Glycine
(d) Glutamine
Answer:
(a) Arginine

Question 2.
An example of feedback inhibition is  …………… .
(a) Cyanide action on cytochrome
(b) Sulpha drug on folic acid synthesiser bacteria
(c) Allosteric inhibition of hexokinase by glucose – 6 – phosphate
(d) The inhibition of succinic dehydrogenase by malonate
Answer:
(c) Allosteric inhibition of hexokinase by glucose – 6 – phosphate

Question 3.
Enzymes that catalyse interconversion of optical, geometrical or positional isomers are …………… .
(a) Ligases
(b) Lyases
(c) Hydrolases
(d) Isomerases
Answer:
(d) Isomerases

Question 4.
Proteins perform many physiological functions. For example some functions as enzymes. One of the following represents an additional function that some proteins discharge …………… .
(a) Antibiotics
(b) Pigment conferring colour to skin
(c) Pigments making colours of flowers
(d) Hormones
Answer:
(d) Hormones

Question 5.
Given below is the diagrammatic representation of one of the categories of small molecular weight organic compounds in the living tissues. Identify the category shown & one blank component “X” in it …………… .

Answer:
(a) Nucleoside
(b) Uracil.

Question 6.
Distinguish between nitrogenous base and a base found in inorganic chemistry.
Answer:
Nitrogenous Base:

1. Nitrogenous bases are organic molecules containing the element nitrogen & acts as a base in chemical reaction.
2. e.g. Adenine, Thymine

Base:

1. Bases are the substance that release hydroxide (OH^– ) ions in aqueous solution.
2. e.g. NaOH and Ca(OH)₂

Question 7.
What are the factors affecting the rate of enzyme reaction?
Answer:
(a) Temperature: Heating increases molecular motion. Thus the molecules of the substrate and enzyme move more quickly resulting in a greater probability of occurrence of the reaction. The temperature that promotes maximum activity is referred to as optimum temperature.

(b) pH: The optimum pH is that at which the maximum rate of reaction occurs. Thus the pH change leads to an alteration of enzyme shape, including the active site. If extremes of pH are encountered by an enzyme, then it will be denatured.

(c) Substrate Concentration: For a given enzyme concentration, the rate of an enzyme reaction increases with increasing substrate concentration.

(d) Enzyme Concentration: The rate of reaction is directly proportional to the enzyme concentration.

The Michaelis – Menton Constant (Km) and Its Significance:
When the initial rate of reaction of an enzyme is measured over a range of substrate concentrations (with a fixed amount of enzyme) and the results plotted on a graph. With increasing substrate concentration, the velocity increases – rapidly at lower substrate concentration. However the rate increases progressively, above a certain concentration of the substrate the curve flattened out. No further increase in rate occurs. This shows that the enzyme is working at maximum velocity at this point. On the graph, this point of maximum velocity is shown as V_Max.

Question 8.
Briefly outline the classification of enzymes.
Answer:
Enzymes are classified into six groups based on their mode of action.

Question 9.
Write the characteristic feature of DNA.
Answer:
The characteristic feature of DNA.

1. If one strand runs in the 5′ – 3′ direction, the other runs in 3′ – 5′ direction and thus are antiparallel (they run in opposite direction). The 5′ end has the phosphate group and 3’end has the OH group.
2. The angle at which the two sugars protrude from the base pairs is about 120°, for the narrow angle and 240° for the wide angle. The narrow angle between the sugars generates a minor groove and the large angle on the other edge generates major groove.
3. Each base is 0.34 nm apart and a complete turn of the helix comprises 3.4 nm or 10 base pairs per turn in the predominant B form of DNA.
4. DNA helical structure has a diameter of 20 Å and a pitch of about 3 Å. X – ray crystal study of DNA takes a stack of about 10 bp to go completely around the helix (360°).
5. Thermodynamic stability of the helix and specificity of base pairing includes
   • (a) The hydrogen bonds between the complementary bases of the double helix
   • (b) stacking interaction between bases tend to stack about each other perpendicular to the direction of helical axis. Electron cloud interactions (\({ \Pi -{ \Pi } }\)) between the bases in the helical stacks contribute to the stability of the double helix.
6. The phosphodiester linkages gives an inherent polarity to the DNA helix. They form strong covalent bonds, gives the strength and stability to the polynucleotide chain.
7. Plectonemic coiling – the two strands of the DNA are wrapped around each other in a helix, making it impossible to simply move them apart without breaking the entire structure. Whereas in paranemic coiling the two strands simply lie alongside one another, making them easier to pull apart.
8. Based on the helix and the distance between each turns, the DNA is of three forms – A DNA, B DNA and Z DNA.

Question 10.
Explain the structure and function of different types of RNA.
Answer:
1. mRNA (messenger RNA): Single stranded, carries a copy of instructions for assembling amino acids into proteins. It is very unstable and comprises 5% of total RNA polymer. Prokaryotic mRNA (Polycistronic) carry coding sequences for many polypeptides. Eukaryotic mRNA (Monocistronic) contains information for only one polypeptide.

2. tRNA (transfer RNA): Translates the code from mRNA and transfers amino acids to the ribosome to build proteins. It is highly folded into an elaborate 3D structure and comprises about 15% of total RNA. It is also called as soluble RNA.

3. rRNA (ribosomal RNA): Single stranded, metabolically stable, makeup the two subunits of ribosomes. It constitutes 80% of the total RNA. It is a polymer with varied length from 120 – 3000 nucleotides and gives ribosomes their shape. Genes for rRNA are highly conserved and employed for phylogenetic studies.

Entrance Examination Questions Solved
Choose the correct answer:
Question 1.
Who invented electron microscope? (2010 AIIMS, 2008 JIPMER)
(a) Janssen
(b) Edison
(c) Knoll and Ruska
(d) Landsteiner
Answer:
(c) Knoll and Ruska

Question 2.
Specific proteins responsible for the flow of materials and information into the cellare called …………… . (2009 AIIMS)
(a) Membrane receptors
(b) carrier proteins
(c) integral proteins
(d) none of these
Answer:
(b) carrier proteins

Question 3.
Omnis – cellula – e – cellula was given by …………… . (2007 AIIMS)
(a) Virchow
(b) Hooke
(c) Leeuwenhoek
(d) Robert Brown
Answer:
(a) Virchow

Question 4.
Which of the following is responsible for the mechanical support, protein synthesis and enzyme transport? (2007 AIIMS)
(a) cell membrane
(b) mitochondria
(c) dictyosomes
(d) endoplasmic reticulum
Answer:
(d) endoplasmic reticulum

Question 5.
Genes present in the cytoplasm of eukaryotic cells are found in …………… . (2006 AIIMS)
(a) mitochondria and inherited via egg cytoplasm
(b) lysosomes and peroxisomes
(c) Golgi bodies and smooth endoplasmic reticulum
(d) Plastids inherited via male gametes
Answer:
(a) mitochondria and inherited via egg cytoplasm

Question 6.
In which one the following would you expect to find glyoxysomes? (2005 AIIMS)
(a) Endosperm of wheat
(b) Endosperm of castor
(c) Palisade cells in leaf
(d) Root hairs
Answer:
(b) Endosperm of castor

Question 7.
A quantosome is present in …………… . (JIPMER 2012)
(a) Mitochondria
(b) Chloroplast
(c) Golgi bodies
(d) ER
Answer:
(b) Chloroplast

Question 8.
In mitochondria the enzyme cytochrome oxidase is present in …………… . (2012 JIPMER)
(a) Outer mitochondrial membrane
(b) inner mitochondrial membrane
(c) Stroma
(d) Grana
Answer:
(b) inner mitochondrial membrane

Question 9.
Which organelle is present in higher number in secretory cell? (2008 JIPMER)
(a) Mitochondria
(b) Chloroplast
(c) Nucleus
(d) Dictyosomes
Answer:
(d) Dictyosomes

Question 10.
Major site for the synthesis of lipids …………… . (2013 NEET)
(a) Rough ER
(b) smooth ER
(c) Centriole
(d) Lysosome
Answer:
(b) smooth ER

Question 11.
Golgi complex plays a major role in …………… . (2013 NEET)
(a) post translational modification of proteins and glycosidation of lipids
(b) translation of proteins
(c) Transcription of proteins
(d) Synthesis of lipid
Answer:
(a) post translational modification of proteins and glycosidation of lipids

Question 12.
Main arena of various types of activities of a cell is …………… . (2010 AIPMT)
(a) Nucleus
(b) Mitochondria
(c) Cytoplasm
(d) Chloroplast
Answer:
(c) Cytoplasm

Question 13.
The thylakoids in chloroplast are arranged in …………… . (2005 JIPMER)
(a) regular rings
(b) linear array
(c) diagonal direction
(d) stacked discs
Answer:
(d) stacked discs

Question 14.
Sequences of which of the following is used to know the phylogeny rRNA? (20022JIPMER)
(a) mRNA
(b) rRNA
(c) tRNA
(d) Hn RNA
Answer:
(b) rRNA

Question 15.
Structures between two adjacent cells which is an effective transport pathway? (2010 AIPMT)
(a) Plasmodesmata
(b) Middle lamella
(c) Secondary wall layer
(d) Primary wall layer
Answer:
(a) Plasmodesmata

Question 16.
In active transport carrier proteins are used, which use energy in the form of ATP to …………… .
(a) transport molecules against concentration gradient of cell wall
(b) transport molecules along concentration gradient of cell membrane
(c) transport molecules against concentration gradient of cell membrane
(d) transport molecules along concentration gradient of cell wall
Answer:
(c) transport molecules against concentration gradient of cell membrane

Question 17.
The main organelle involved in modification and routing of newly synthesised protein to their destinations is …………… . (AIPMT 2005)
(a) Mitochondria
(b) Glyoxysomes
(c) Spherosomes
(d) Endoplasmic reticulum
Answer:
(d) Endoplasmic reticulum

Question 18.
Algae have cell wall made up of …………… . (AIPMT 2010)
(a) Cellulose, galactans and mannans
(b) Cellulose, chitin and glucan
(c) Cellulose, Mannan and peptidoglycan
Answer:
(a) Cellulose, galactans and mannans

Samacheer Kalvi 11th Bio Botany Biomolecules Additional Questions and Answers

Question 1.
The percentage of water in the total cellular mass is …………… .
(a) 50%
(b) 60%
(c) 70%
(d) 80%
Answer:
(c) 70%

Question 2.
The metabolites which does not show any direct function in growth is called …………… metabolite.
(a) Primary
(b) Secondary
(c) Tertiary
(d) Quartemary
Answer:
(b) Secondary

Question 3.
Molecular formula for carbohydrates is …………… .
(a) (CH₂O)₂
(b) (CH₆O)
(C) (C₂H₂O)_n
(d) (CH₆O)_n
Answer:
(a) (CH₂O)₂

Question 4.
Number of carbon molecule in glucose is …………… .
(a) 4
(b) 6
(c) 8
(d) 12
Answer:
(b) 6

Question 5.
Number of sugar units in oligo saccharides are …………… .
(a) 6 to 10
(b) 1 to 10
(c) 2 to 8
(d) 2 to 10
Answer:
(d) 2 to 10

Question 6.
Which of the following is a trisaccharide?
(a) Maltose
(b) Stachyose
(c) Ramnose
(d) Aldose
Answer:
(c) Ramnose

Question 7.
…………… are also called as Glycan.
(a) Monosaccharides
(b) Disaccharides
(c) Polysaccharides
(d) Multisaccharides
Answer:
(c) Polysaccharides

Question 8.
Sucrose is a combination of …………… and fructose.
(a) α – glucose
(b) β – glucose
(c) Ketoses
(d) Maltose
Answer:
(a) α – glucose

Question 9.
…………… is also called as animal starch.
(a) Amylose
(b) Glycogen
(c) Glucose
(d) Glycerol
Answer:
(b) Glycogen

Question 10.
…………… reagent is used in starch test.
(a) Potassium permanganate
(b) Potassium iodide
(c) Calcium chloride
(d) Calcium iodide
Answer:
(b) Potassium iodide

Question 11.
Glycogen is not seen in …………… cells.
(a) liver
(b) skeletal
(c) muscle
(d) brain
Answer:
(d) Brain

Question 12.
Benedicts solution is nothing but …………… .
(a) Copper II sulphate
(b) Cuprous sulphate
(c) Cupric sulphate
(d) Copper I sulphate
Answer:
(a) Copper II sulphate

Question 13.
…………… is not a reducing sugar.
(a) Glucose
(b) Fructose
(c) Sucrose
(d) Ketose
Answer:
(c) Sucrose

Question 14.
…………… form the exoskeleton of insects & arthropods.
(a) N – acetyl glucosamine
(b) N – butyl glucosamine
(c) N – phenyl glucosamine
(d) N – methyl glucosamine
Answer:
(a) N – acetyl glucosamine

Question 15.
Number of fatty acids in triglyceride is …………… .
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3

Question 16.
The major structural component of cell membrane is …………… .
(a) glucolipids
(b) phospholipids
(c) proteolipids
(d) triglycerides
Answer:
(b) phospholipids

Question 17.
There are …………… different amino acids existing naturally.
(a) about 20
(b) about 10
(c) about 25
(d) about 22
Answer:
(a) about 20

Question 18.
A zwitterion also called as …………… ion.
(a) dipolar
(b) monopolar
(c) tripolar
(d) nonpolar
Answer:
(a) dipolar

Question 19.
…………… test is used as an indicator of the presence of protein.
(a) Biuret test
(b) Iodine test
(c) Benedict’s test
(d) Starch test
Answer:
(a) Biuret test

Question 20.
The competitive inhibitor is …………… for succinic dehydrogenase.
(a) malonate
(b) succinate
(c) oxalate
(d) citrate
Answer:
(a) malonate

Question 21.
…………… is the abundant protein in whole biosphere.
(a) RUBP
(b) NAD⁺
(c) NADPH
(d) RUBISCO
Answer:
(d) RUBISCO

Question 22.
…………… is an active enzyme with its non – protein component.
(a) Apoenzyme
(b) Holoenzyme
(c) Coenzymes
(d) Enzymes
Answer:
(b) Holoenzyme

Question 23.
Flavin adenine dinucleotide contains …………… which helps to accept hydrogen.
(a) ascolac acid
(b) cyanocobalamin
(c) riboflavin
(d) keratinine
Answer:
(c) riboflavin

Question 24.
…………… is a catalytic RNA.
(a) mRNA
(b) Ribozyme
(c) Ribonuclease
(d) rRNA
Answer:
(b) Ribozyme

Question 25.
…………… protects the end of the chromosomes from damage.
(a) Satellite
(b) Kinetochore
(c) Primary constriction
(d) Telomere
Answer:
(d) Telomere

Question 26.
Which is not a pyrimidine base?
(a) Cytosine
(b) Uracil
(c) Guanine
(d) Thymine
Answer:
(c) Guanine

Question 27.
Which type of DNA was described by Watson & Crick?
(a) Z – DNA
(b) α – DNA
(c) B – DNA
(d) A – DNA
Answer:
(c) B – DNA

Question 28.
According to Chargaff’s rule, the hydrogen bonding between Adenine and Thymine is …………… .
(a) 2
(b) 3
(c) 4
(d) Nil
Answer:
(a) 2

Question 29.
The first clear crystallographic evidence for helical structure of DNA was produced by …………… .
(a) Maurice Wilkins
(b) Rosalind Franklin
(c) Francis Crick
(d) Chargaff
Answer:
(b) Rosalind Franklin

Question 30.
According to Cargaff’s rule, A : T = G : C = …………… .
(a) 0
(b) 1
(c) >1
(d) <1
Answer:
(b) 1

Question 31.
A complete turn of the helix comprises …………… .
(a) 34 nm
(b) 3.4 nm
(c) 20 nm
(d) 2nm
Answer:
(b) 3.4 nm

Question 32.
Diameter of DNA helix is …………… .
(a) 34 Å
(b) 20 nm
(c) 34 nm
(d) 20 Å
Answer:
(d) 20 Å

Question 33.
RNA is …………… .
(a) Single stranded and stable
(b) Single stranded and unstable
(c) Double stranded and stable
(d) Double stranded and unstable
Answer:
(b) Single stranded and unstable

Question 34.
rRNA constitutes …………… of total RNA.
(a) 20%
(b) 70%
(c) 80%
(d) 15%
Answer:
(c) 80%

Question 35.
Shape to the ribosomes is provided by …………… .
(a) rRNA
(b) tRNA
(c) mRNA
(d) DNA
Answer:
(a) rRNA

Question 36.
Which RNA is also called as soluble RNA?
(a) rRNA
(b) tRNA
(c) mRNA
(d) ssRNA
Answer:
(b) tRNA

Question 37.
Which is the left – handed DNA?
(a) B – DNA
(b) A – DNA
(c) Z – DNA
(d) dsDNA
Answer:
(c) Z – DNA

Question 38.
Which of the following does not contain cell wall?
(a) Fungi
(b) Bacteria
(c) Mycoplasma
(d) Algae
Answer:
(c) Mycoplasma

Question 39.
The amino acid which is both an acid and a base is called …………… .
(a) Amphibolic
(b) Amphoteric
(c) Amphipathetic
(d) Anabolic
Answer:
(b) Amphoteric

Question 40.
…………… leads to the loss of 3D structure of protein.
(a) Annealing
(b) Extension
(c) Denaturation
(d) Polymerisation
Answer:
(c) Denaturation

Question 41.
Which of the following polysaccharides is used as solidifying agent in culture medium?
(a) Inulin
(b) Heparin
(c) Agar
(d) Keratan sulphate
Answer:
(c) Agar

Question 42.
Which is an anticoagulant?
(a) Inulin
(b) Heparin
(c) Agar
(d) Keratan sulphate
Answer:
(b) Heparin

Question 43.
Insulin is a polymer of …………… .
(a) sucrose
(b) fructose
(c) glucose
(d) maltose
Answer:
(b) fructose

II. Very Short Answer Type Questions (2 Marks)

Question 1.
Define cell pool and mention its constituents.
Answer:
The cell components are made of collection of molecules called as cellular pool, which consists of both inorganic and organic compounds.

Question 2.
Draw the molecular structure of water.
Answer:
the molecular structure of water:

Question 3.
Point out the percentage of water in human cell & a plant cell.
Answer:
Water makes upto 70% of human cell and upto 95% of mass of a plant cell.

Question 4.
What are metabolites?
Answer:
Metabolites are the organic compounds synthesized by plants, fungi and various microbes. They are the intermediates & products of metabolism.

Question 5.
Write the molecular formula for carbohydrates?
Answer:
(CH₂O)_n

Question 6.
Give an example for simple sugar with its formula.
Answer:
Glucose – C₆H₁₂O₆

Question 7.
Which type of sugar does sucrose belongs to? Write its monomer units.
Answer:
Sucrose is a disaccharides composed of α – glucose & fructose.

Question 8.
Classify polysaccharides based on function.
Answer:
Depending on the function, polysaccharides are of two types:

1. storage polysaccharide and
2. structural polysaccharide.

Question 9.
What are Glycans?
Answer:
Polysaccharides are also called as Glycans. They are made of hundreds of monosaccharide units.

Question 10.
Which is a common storage polysaccharide? Mention its monomer units.
Answer:
Starch is a storage polysaccharide made up of repeated units of amylose and amylopectin.

Question 11.
Which is an animal starch? Where can we see it in our body?
Answer:
Glycogen. It is found in liver cells & skeletal muscles.

Question 12.
Why oil does not get mixed if added with water?
Answer:
Oil is a lipid. Lipids are long hydrocarbon chains that are non-polar & thus hydrophobic, which avoids the oil to dissolve in water.

Question 13.
How saturated fatty acids differ from unsaturated fatty acids?
Answer:
Saturated factty acids have the hydrocarbon chain with single bond, whereas in unsaturated fatty acids the hydrocarbon chain will have double bonds.

Question 14.
How waxes are formed?
Answer:
Waxes are esters formed between a long chain alcohol and saturated fatty acids.

Question 15.
Why amino acids are amphoteric?
Answer:
The amino acid is both an acid and a base and is called amphoteric.

Question 16.
Name the various groups attached to the 4 valencies of carbon in an amino acid.
Answer:
The 4 valencies of carbon in an amino acid:

1. (NH₂)
   
2. an acidic carboxylic group (COOH) and
   
3. a hydrogen atom (H)
   
4. and side chain or variable R group.

Question 17.
Where the peptide bond is formed?
Answer:
A peptide bond is formed when the amino group of one amino acid reacts with carbonyl group of another amino acid.

Question 18.
Which was the first sequenced protein? Who had done it?
Answer:
First protein is insulin and it was sequenced by Fred Sanger.

Question 19.
Why proteins undergo conformational changes after its synthesis?
Answer:
After synthesis, the protein attains conformational change into a specific 3D form for proper functioning.

Question 20.
Mention the levels of protein organisation based on folding.
Answer:
According to the mode of folding, four levels of protein organisation have been recognised namely primary, secondary, tertiary and quaternary.

Question 21.
Define enzymes.
Answer:
Enzymes are globular proteins that catalyse the thousands of metabolic reactions taking place within cells and organism.

Question 22.
Name any four factors that affect enzyme reactions.
Answer:
Four factors that affect enzyme reactions:

1. pH
2. temperature
3. enzyme concentration and
4. substrate concentration.

Question 23.
What are inhibitors? Mention its types.
Answer:
Certain substances present in the cells may react with the enzyme and lower the rate of reaction. These substances are called inhibitors. It is of two types:

1. Competitive and
2. Non – competitive.

Question 24.
Differentiate Apoenzyme from Holoenzyme.
Answer:
Differ Apoenzyme from Holoenzyme:

Apoenzyme	Holoenzyme
1. Active enzyme with its non – protein component	1. Inactive enzyme without its non – protein component

Question 25.
What are Prosthetic groups? Give example.
Answer:
Prosthetic groups are organic molecules that assist in catalytic function of an enzyme. Example: Flavin adenine dinucleotide (FAD).

Question 26.
Draw a diagram showing the various components of enzymes.
Answer:
Catalytic site, Cofactor and Holoenzyme:

Question 27.
Write a note on Ribozyme.
Answer:
Ribozyme – Non – Protein Enzyme. A Ribozyme, also called as catalytic RNA; is a ribonucleic acid that acts as enzyme. It is found in ribosomes.

Question 28.
Give an example for following enzyme groups.
Answer:
An example for following enzyme groups:

1. Transferase – Ex: Transaminase
2. Isomerase – Ex: Isomerase
3. Oxidoreductase – Ex: Dehydrogenase
4. Lyase – Ex: Decarboxylase

Question 29.
Write the composition of DNA & RNA.
Answer:
Nitrogen base, pentose sugar and phosphate.

Question 30.
What is a nucleoside?
Answer:
A purine or a pyrimidine and a ribose or deoxyribose sugar is called nucleoside. A nitrogenous base is linked to pentose sugar through n-glycosidic linkage and forms a nucleoside.

Question 31.
What is a nucleotide?
Answer:
When a phosphate group is attached to a nucleoside it is called a nucleotide.

Question 32.
Name the two types of Purines and Pyrimidines.
Answer:
The two types of Purines and Pyrimidines:

1. Purines: Adenine and guanine
2. Pyrimidines: Cytosine and thymine (Uracil)

Question 33.
How DNA differs from RNA?
Answer:
DNA has thymine base, whereas RNA has uracil base. DNA has deoxyribose sugar, whereas RNA has ribose sugar.

Question 34.
Draw a simple diagram showing basic components of DNA.
Answer:
Deoxyribose sugar:

Question 35.
Which is the secondary structure of DNA? Who discovered it?
Answer:
B – DNA is the secondary structure of DNA. Watson & Crick discovered B – DNA.

Question 36.
State Chargaff’s rule.
Answer:
Chargaff’s Rule:
A = T; G = C
A + G = T + C,
A : T = G : C = 1.

Question 37.
Name the three forms of DNA.
Answer:
The three forms of DNA:

1. A – DNA
2. B – DNA and
3. Z – DNA.

Question 38.
Which is the soluble forms of RNA. Write its percentage composition of total RNA.
Answer:
tRNA is the soluble RNA which is about 15% of total RNA.

Question 39.
Name the types of RNA?
Answer:
The types of RNA:

1. mRNA
2. tRNA and
3. rRNA.

Question 40.
Draw the structure of transfer RNA.
Answer:
Transfer RNA (tRNA):

III. Short Answer Type Questions (3 Marks)

Question 1.
Distinguish between Macronutrients & Micronutrients.
Answer:
Macronutrients:

• Nutrients required in larger quantities for plant growth are called Macronutrients.
• e.g. Potassium and Calcium

Micronutrients:

• Nutrients required in trace amount for plant growth are called Micronutrients
• e.g. Zinc and Bora

Question 2.
Tabulate the various cellular components with their percentage.
Answer:
The various cellular components with their percentage:

Component	% of the total cellular mass
1. Water	1. 70
2. Proteins	2. 15
3. Carbohydrates	3. 3
4. Lipids	4. 2
5. Nucleic acids	5. 6
6. Ions	6. 4

Question 3.
List out the properties of Water.
Answer:
Properties of Water:

1. Adhesion and cohesion property
2. High latent heat of vaporisation
3. High melting and boiling point
4. Universal solvent
5. Specific heat capacity

Question 4.
How lattice formation occurs in water molecule?
Answer:
Two electro negative atoms of oxygen share a hydrogen bonds of two water molecule. Thus, they can stick together by cohesion and results in lattice formation.

Question 5.
Distinguish between Primary metabolite & Secondary metabolite.
Answer:
Between Primary metabolite & Secondary metabolite:

• Primary metabolites are those that are required for the basic metabolic processes like photosynthesis, respiration, etc Example: Lipase, a protein.
• Secondary metabolites does not show any direct function in growth and development of organisms. Example: Ricin, gums.

Question 6.
Define Polymerization.
Answer:
Polymerization, is a process in which repeating subunits termed monomers are bound into chains of different lengths. These chains of monomers are called polymers.

Question 7.
Explain the bond formation in sucrose molecule.
Answer:
Sucrose is formed from a molecule of α – glucose and a molecule of fructose. This is a condensation reaction releasing water. The bond formed between the glucose and fructose molecule by removal of water is called glycosidic bond. This is another example of strong, covalent bond.

Question 8.
How will you identify the presence of starch in a food sample.
Answer:
The presence of starch is identified by adding a solution of iodine in potassium iodide. Iodine molecules fit nearly into the starch helix, creating a blue – black colour.

Question 9.
Write a note on steroids.
Answer:
Steroids are complex compounds commonly found in cell membrane and animal hormones. e.g. Cholesterol which reinforces the structure of the cell membrane in animal cells and in an unusual group of cell wall deficient bacteria – Mycoplasma.

Question 10.
Draw the structure of basic amino acid.
Answer:
The structure of basic amino acid:

Question 11.
What is a Zwitterion? or What is an isoelectric point?
Answer:
A zwitterion also called as dipolar ion, is a molecule with two or more functional groups, of which at least one has a positive and other has a negative electrical charge and the net charge of the entire molecule is zero. The pH at which this happens is known as the isoelectric point.

Question 12.
Write briefly about protein denaturation.
Answer:
Denaturation is the loss of 3D structure of protein. Exposure to heat causes atoms to vibrate violently, and this disrupts the hydrogen and ionic bonds. Under these conditions, protein molecules become elongated, disorganised strands. Agents such as soap, detergents, acid, alcohol and some disinfectants disrupt the interchain bond and cause the molecule to be non – functional.

Question 13.
Why do some people have curly hair?
Answer:
Human hair is made of protein. The more the distance between the sulphur atoms, the more the proteins bend; the more the hair curls.

Question 14.
Write a note on RUBISCO.
Answer:
Ribulose biphosphate carboxylase oxygenase (RUBISCO) is an enzyme that catalyses the reaction between CO₂ and the CO₂ acceptor molecule in photosynthesis. It is the most abundant protein in the whole biosphere.

Question 15.
Differentiate Anabolic reaction and Catabolic reaction.
Answer:
Anabolic reaction:

• Anabolic reaction involves the building up of organic molecules.
• Ex: Synthesis of protein from amino acids.

Catabolic reaction:

• Catabolic reaction involves the breaking down of larger molecules.
• Ex: Breaking down of sugar in respiration.

Question 16.
What are Allosteric inhibitors?
Answer:
Compounds which modify enzyme activity by causing a reversible change in the structure of the enzyme active site. This in turn affects the ability of the substrate to bind to the enzyme. Such compounds are called allosteric inhibitors, e.g. The enzyme hexokinase which catalysis glucose to glucose – 6 phosphate in glycolysis is inhibited by glucose – 6 phosphate. This is an example for feedback allosteric inhibitor.

Question 17.
Explain in brief about End – product inhibitor. (Negative Feedback Inhibition)
Answer:
When the end product of a metabolic pathway begins to accumulate, it may act as an allosteric inhibitor of the enzyme controlling the first step of the pathway. Thus the product starts to switch off its own production as it builds up. The process is self – regulatory. As the product is used up, its production is switched on once again. This is called end – product inhibition.

Question 18.
Draw the structure of Purine & Pyrimidine.
Answer:
The structure of Pyrimidine & Purine:

Question 19.
Why the sugar in DNA is a deoxyribose?
Answer:
The sugar in DNA molecule is called 2’ – deoxyribose because there is no hydroxyl group at 2’ position.

Question 20.
How dinucleotide & polynucleotides are formed?
Answer:
Two nucleotides join to form dinucleotide that are linked through 3′ – 5′ phosphodiester linkage by condensation between phosphate groups of one with sugar of other. This is repeated many times to make polynucleotide.

Question 21.
Compare Plectonemic & Paranemic Coiling.
Answer:
Plectonemic coiling – the two strands of the DNA are wrapped around each other in a helix, making it impossible to simply move them apart without breaking the entire structure. Whereas in Paranemic coiling the two strands simply lie alongside one another, making them easier to pull apart.

Question 22.
Differentiate between Polycistronic & Monocistronic mRNA.
Answer:
Between Polycistronic & Monocistronic mRNA:

Polycistronic mRNA	Monocistronic mRNA
1. Polycistronic mRNA carry coding sequences for many Polypeptides	1. Monocistronic mRNA carry coding sequences for only one Polypeptide
2. Prokaryotic mRNA are polycistronic	2. Eukaryotic mRNA are monocistronic

Question 23.
What are Proteins?
Answer:
Proteins are polymers of 20 different amino acids, each of which has a distinct side chain with specific chemical properties. Each protein has a unique amino acid sequence which determines its 3D structure.

Question 24.
Herbivores can digest cellulose rich food, Why can’t human beings?
Answer:
Human cannot digest cellulose but herbivores can digest them with the help of bacteria present in the gut which produces enzymes cellulase. This is an example of mutualism.

Question 25.
How will you identify the presence of protein in food samples?
Answer:
The biuret test is used as an indicator of the presence of protein because it gives a purple colour in the presence of peptide bonds (-C- N-). To a protein solution an equal quantity of sodium hydroxide solution is added and mixed. Then a few drops of 0.5% copper (II) sulphate is added with gentle mixing. A distinct purple colour develops without heating.

Question 26.
Write a note on peptide bonds between amino acids.
Answer:
The amino group of one amino acid reacts with carboxyl group of other amino acid, forming a peptide bond. Two amino acids can react together with the loss of water to form a dipeptide. Long strings of amino acids linked by peptide bonds are called polypeptides. In 1953, Fred Sanger first sequenced the Insulin protein.

Question 27.
Which was the first alkaloid discovered? Mentions its uses.
Answer:
Morphine is the first alkaloid to be found. It comes from the plant Opium poppy (Papaver somniferum). It is used as a pain reliever in patients with severe pain levels and cough suppressant.

IV. Long Answer Type Questions (5 Marks)

Question 1.
How will you identify the presence of glucose in a given food sample?
Answer:
Aldoses and ketoses are reducing sugars. This means that, when heated with an alkaline solution of copper (II) sulphate (a blue solution called Benedict’s solution), the aldehyde or ketone group reduces Cu²⁺ ions to Cu⁺ ions forming brick red precipitate of copper (I) oxide. In the process, the aldehyde or ketone group is oxidised to a carboxyl group (-COOH).

This reaction is used as test for reducing sugar and is known as Benedict’s test. The results of Benedict’s test depends on concentration of the sugar. If there is no reducing sugar it remains blue. Sucrose is not a reducing sugar The greater the concentration of reducing sugar, the more is the precipitate formed and greater is the colour change.

Question 2.
Write a note on various levels of protein organisation.
Answer:
The primary structure is linear arrangement of amino acids in a polypeptide chain. Secondary structure arises when various functional groups are exposed on outer surface of the molecular interaction by forming hydrogen bonds. This causes the amino acid chain to twist into coiled configuration called α – helix or to fold into a flat β – pleated sheets.

Tertiary protein structure arises when the secondary level proteins fold into globular structure called domains. Quaternary protein structure may be assumed by some complex proteins in which more than one polypeptide forms a large multiunit protein. The individual polypeptide chains of the protein are called subunits and the active protein itself is called a multimer.

Question 3.
Enumerate the properties of Enzyme.
Answer:
The properties of Enzyme:

• Enzymes are globular proteins.
• They act as catalysts and effective even in small quantity.
• They remain unchanged at the end of the reaction.
• They are highly specific.
• They have an active site where the reaction takes place.
• Enzymes lower activation energy of the reaction they catalyse.

Question 4.
Draw a Flow Chart depicting the Carbohydrate Classification
Answer:
Flow Chart depicting the Carbohydrate Classification:

Question 5.
Explain the various types of chemical bonding in proteins.
Answer:
1. Hydrogen Bond: It is formed between some hydrogen atoms of oxygen and nitrogen in polypeptide chain. The hydrogen atoms have a small positive charge and oxygen and nitrogen have small negative charge. Opposite charges attract to form hydrogen bonds. Though these bonds are weak, large number of them maintains the molecule in 3D shape.

2. Ionic Bond: It is formed between any charged groups that are not joined together by peptide bond. It is stronger than hydrogen bond and can be broken by changes in pH and temperature.

3. Disulfide Bond: Some amino acids like cysteine and methionine have sulphur. These form disulphide bridge between sulphur atoms and amino acids.

4. Hydrophobic Bond: This bond helps some protein to maintain structure. When globular proteins are in solution, their hydrophobic groups point inwards away from water.

Question 6.
Explain Lock & Key Mechanism of Enzymatic reaction.
Answer:
Lock and Key Mechanism of Enzyme: In a enzyme catalysed reaction, the starting substance is the substrate. It is converted to the product. The substrate binds to the specially formed pocket in the enzyme – the active site, this is called lock and key mechanism of enzyme action. As the enzyme and substrate form a ES complex, the substrate is raised in energy to a transition state and then breaks down into products plus unchanged enzyme.

Question 7.
Describe the Competitive & Non – Competitive Inhibitors of enzyme.
Answer:
1. Competitive Inhibitor: Molecules that resemble the shape of the substrate and may compete to occupy the active site of enzyme are known as competitive inhibitors. For Example: the enzyme that catalyses the reaction between carbon dioxide and the CO₂ acceptor molecule in photosynthesis, known as ribulose biphosphate carboxylase oxygenase (RUBISCO) is competitively inhibited by oxygen / carbon – di – oxide in the chloroplast. The competitive inhibitor is malonate for succinic dehydrogenase.

2. Non – competitive Inhibitors: There are certain inhibitors which may be unlike the substrate molecule but still combines with the enzyme. This either blocks the attachment of the substrate to active site or change the shape so that it is unable to accept the substrate. For example the effect of the amino acids alanine on the enzyme pyruvate kinase in the final step of glycolysis.

Certain non – reversible / irreversible inhibitors bind tightly and permanently to an enzyme and destroy its catalytic properties entirely. These could also be termed as poisons. Example – cyanide ions which blocks cytochrome oxidase in terminal oxidation in cell aerobic respiration, the nerve gas sarin blocks a neurotransmitter in synapse transmission.

Question 8.
Give a detailed account on Enzyme Co – factors.
Answer:
Many enzymes require non – protein components called co – factors for their efficient activity. Co – factors may vary from simple inorganic ions to complex organic molecules.
They are of three types:

1. Inorganic ions, prosthetic groups and coenzymes.
2. Holoenzyme – active enzyme with its non – protein component.
3. Apoenzyme – the inactive enzyme without its non – protein component.

Inorganic ions help to increase the rate of reaction catalysed by enzymes. Example: Salivary amylase activity is increased in the presence of chloride ions. Prosthetic groups are organic molecules that assist in catalytic function of an enzyme. Flavin adenine dinucleotide (FAD) contains riboflavin(vit B2), the function of which is to accept hydrogen. ‘Haem’ is an iron – containing prosthetic group with an iron atom at its centre. Coenzymes are organic compounds which act as cofactors but do not remain attached to the enzyme. The essential chemical components of many coenzymes are vitamins. Eg. NAD, NADP, Coenzyme A, ATP.

Question 9.
Tabulate the various features of different forms of DNA.
Answer:
The various features of different forms of DNA:

Question 10.
Compare DNA with RNA?
Answer:
Compare DNA with RNA:

DNA	RNA
1. Deoxyribose sugar is present	1. Ribose sugar is present
2. Thymine is present	2. Uracil is present
3. More stable	3. Less stable
4. Double stranded	4. Single stranded
5. Types: A – DNA, B – DNA, Z – DNA	5. Type: mRNA, tRNA, rRNA
6. Genetic material for most of living organism except few viruses	6. Genetic material for few viruses only

V. Higher Order Thinking Skills (HOTs)

Question 1.
In which form does the glucose is stored in animal cells? Specify the cells?
Answer:
Glucose is stored in the form of glycogen. Glycogen is stored in liver cells and skeletal muscles, etc.

Question 2.
State the key differences between DNA & RNA.
Answer:
The key differences between DNA & RNA:

DNA	RNA
1. Double stranded	1. Single stranded
2. Thymine is the pyrimidine base	2. Uracil is the pyrimidine base

 

Question 3.
Aminoacids are the monomers of proteins. Similarly mention the monomers of nucleic acids along with its composition.
Answer:
The monomer unit of nucleic acids are nucleotides, which are composed of nitrogen base, pentose sugar and phosphoric acid.

Question 4.
Complete the equations.
(a) Nitrogen base + …………… . = Nucleoside.
(b) …………… + nucleoside = Nucleotide.
(c) Glucose + fructose = …………… .
Answer:
(a) sugar
(b) phosphoric acid and
(c) sucrose.

Question 5.
What happens if the sucrose is hydrolysed?
Answer:
On hydrolysis, the glycosidic bonds in sucrose gets splitted yielding glucose and fructose.

Question 6.
Name the types of bonds.
(a) Between amino acids of protein
(b) Between carboxyl group and glycerol of fatty acids and
(c) Between glucose units of cellulose.
Answer:
(a) Peptide bond
(b) Ester bond
(c) Glycosidic bond

Question 7.
Study the following equation and name the reaction A and B.

Answer:
Reaction A is glycogenolysis. Reaction B is glycogenesis.

Question 8.
Whether waxes are found in living organisms?
Answer:
Yes. Fur, feathers, fruits, leaves, skin, and exoskeleton of insects are naturally water – proofed with a coating of wax.

Question 9.
If dsDNA has 40% Guanine. Calculate the percentage of Adenine.
Answer:
According to Chargaff’s rule:
Guanine pairs with cytosine. It Guanine is 40%, then cytosine will also be 40%. Similarly, Adenine pairs with thymine, if guanine is 40%, the remaining 60% will be Adenine. So Thymine will also be 60%.
Thus, A : T = G : C = 1
and 60 : 60 = 40 : 40 = 1.

Question 10.
In an Eukaryotic cell, totally there are 10000 RNA molecules. Calculate the number of mRNA’s and tRNA’s if the count of rRNA is 8000.
Answer:
In a cell, rRNA contributes 80%, tRNA constitutes 15% and mRNA constitutes 5%. If rRNA is 8000 (80%), then tRNA count is 1500 (15%) and mRNA is 500 (5%).

Question 11.
Despite made of two different monomers amylose and amylopectin, starch is a homopolysaccharide – Comment.
Answer:
Starch is made up of amylose and amylopectin. Both are glucose polymers, hence starch is considered as homopolysaccharides.

Question 12.
How do you call a fatty acid as saturated or unsaturated?
Answer:
If the hydrocarbon chain is single bonded, then the fatty acid is said to be saturated. In unsaturated fatty acids, the hydrocarbon chain is double bonded.

Question 13.
Enzymes are biocatalysts – Justify.
Answer:
Enzymes are globular proteins that catalyze thousands of metabolic reactions taking place within cells and organisms. Hence enzymes are called as biological catalysts.

Question 14.
Starch, cellulose, glycogen and chitin are polysaccharides found among the following. Choose the one appropriate and write against each.
(a) Cotton fibre – …………… .
(b) Exoskeleton of ant – …………… .
(c) Liver –  …………… .
(d) Peeled potato – …………… .
Answer:
(a) Cellulose
(b) Chitin
(c) Glycogen and
(d) Starch.

Question 15.
Sucrose is not a reducing sugar. Why?
Answer:
Sucrose is a non – reducing sugar since it does not possess aldehyde or ketone group, which is responsible for reducing the alkaline solutions like copper (II) sulphate.

Question 16.
A DNA segment has a total of 1000 nucleotides, out of which 240 are adenine containing nucleotides. How many pyrimidine bases this DNA segment possess?
Answer:
Pyrimidine = 500.
According to Chargaff’s rule,
A = T,
A = 240, hence T = 240.
A + T = 240 + 240 = 480.
So, G + C = 1000 – 480 = 520.
G = C, Therefore, C = \(\frac {520}{2}\) = 260.
Thus, pyrimidine = C + T = 260 + 240 = 500.