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Students can Download Physics Chapter 1 Electrostatics Questions and Answers, Notes Pdf, Samacheer Kalvi 12th 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 12th Physics Solutions Chapter 1 Electrostatics

Samacheer Kalvi 12th Physics Electrostatics Textual Evaluation Solved

Samacheer Kalvi 12th Physics Electrostatics Multiple Choice Questions

Question 1.
Two identical point charges of magnitude -q are fixed as shown in the figure below. A third charge +q is placed midway between the two charges at the point P. Suppose this charge +q is displaced a small distance from the point P in the directions indicated by the arrows, in which direction(s) will +q be stable with respect to the displacement?

(a) A₁ and A₂
(b) B₁ and B₂
(c) both directions
(d) No stable
Answer:
(b) B₁ and B₂

Question 2.
Which charge configuration produces a uniform electric field?
(a) point charge
(b) the infinite uniform line charge
(c) uniformly charged infinite plane
(d) uniformly charged spherical shell
Answer:
(c) uniformly charged infinite plane

Question 3.
What is the ratio of the charges \(\left|\frac{q_{1}}{q_{2}}\right|\) for the following electric field line pattern?

(a) \(\frac { 1 }{ 5 }\)
(b) \(\frac { 25 }{ 11 }\)
(c) 5
(d) \(\frac { 12 }{ 25}\)
Answer:
(d) \(\frac { 12 }{ 25}\)

Question 4.
An electric dipole is placed at an alignment angle of 30° with an electric field of 2 x 10⁵ N C^-1. It experiences a torque equal to 8 N m. The charge on the dipole if the dipole length is 1 cm is-
(a) 4 mC
(b) 8 mC
(c) 5 mC
(d) 1 mC
Answer:
(b) 8 mC

Question 5.
Four Gaussian surfaces are given below with charges inside each Gaussian surface. Rank the electric flux through each Gaussian surface in increasing order-

(a) D < C < B < A
(b) A < B = C < D
(c) C < A = B < D
(d)D > C > B > A
Answer:
(a) D < C < B < A

Question 6.
The total electric flux for the following closed surface which is kept inside water-

(a) \(\frac { 80q }{{ ε }_{0}}\)
(b) \(\frac { q }{{ 40ε }_{0}}\)
(c) \(\frac { q }{{ 80ε }_{0}}\)
(d) \(\frac { q }{{ 40ε }_{0}}\)
Answer:
(b) \(\frac { q }{{ 40ε }_{0}}\)

Question 7.
Two identical conducting balls having positive charges q₁ and q₂ are separated by a center to center distance r. If they are made to touch each other and then separated to the same distance, the force between them will be- (NSEP 04-05)
(a) less than before
(b) same as before
(c) more than before
(d) zero
Answer:
(c) more than before

Question 8.
Rank the electrostatic potential energies for the given system of charges in increasing order

(a) 1 = 4 < 2 < 3
(b) 2 = 4 < 3 < 1
(c) 2 = 3 < 1 < 4
(d) 3 < 1 < 2 < 4
Answer:
(a) 1 = 4 < 2 < 3

Question 9.
An electric field \(\vec { E } \) = 10x\(\hat{i} \) exists in a certain region of space. Then the potential difference V = V₀ – V_A, Where V₀ is the potential at the origin and V_A is the potential at x = 2 m is-
(a) 10 J
(b) -20 J
(c) + 20 J
(d) – 10 J
Answer:
(a) 10 J

Question 10.
A thin conducting spherical shell of radius R has a charge Q which is uniformly distributed on its surface. The correct plot for electrostatic potential due to this spherical shell is-

Answer:

Question 11.
Two points A and B are maintained at a potential of 7 V and -4 V respectively. The work done in moving 50 electrons from A to B is-
(a) 8.80 x 10^-17 J
(b) -8.80 x 10^-17 J
(c) 4.40 x 10^-17 J
(d) 5.80 x 10^-17 J
Answer:
(a) 8.80 x 10^-17 J

Question 12.
If voltage applied on a capacitor is increased from V to 2V, choose the correct conclusion.
(a) Q remains the same, C is doubled
(b) Q is doubled, C doubled
(c) C remains same, Q doubled
(d) Both Q and C remain same
Answer:
(c) C remains same, Q doubled

Question 13.
A parallel plate capacitor stores a charge Q at a voltage V. Suppose the area of the parallel plate capacitor and the distance between the plates are each doubled then which is the quantity that will change?
(a) Capacitance
(b) Charge
(c) Voltage
(d) Energy density
Answer:
(d) Energy density

Question 14.
Three capacitors are connected in a triangle as shown in the figure. The equivalent capacitance between points A and C is

(a) 1 μF
(b) 2 μF
(c) 3 μF
(d) \(\frac { 1 }{ 4 }\) μF
Answer:
(b) 2 μF

Question 15.
Two metallic spheres of radii 1 cm and 3 cm are given charges of -1 x 10^-2 C and 5 x 10^-2 C respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is (AIIPMT 2012)
(a) 3 x 10^-2 C
(b) 4 x 10^-2 C
(c) 1 x 10^-2 C
(d) 2 x 10^-2 C
Answer:
(a) 3 x 10^-2 C

Samacheer Kalvi 12th Physics Electrostatics Short Answer Questions

Question 1.
What is meant by quantisation of charges?
Answer:
The charge q on any object is equal to an integral multiple of the fundamental unit of charge ‘e’.
q = ne
Where ‘n’ is an integer e
e = charge of an electron =1.6 × 10^-19 C.

Question 2.
Write down Coulomb’s law in vector form and mention what each term represents.
Answer:
The force on a charge q₁ exerted by a point charge q₁ is given by
\(\vec { F } \)₁₂ = \(\frac { 1 }{{ 4πε }{0}}\) \(\frac {{ q }_{1}{ q }_{2}}{{ r }^{2}}\) \(\hat{r} \)₂₁
Here \(\hat{r} \)₂₁ is the unit vector from charge q₁ to q₁.
But \(\hat{r} \)₂₁ = –\(\hat{r} \)₁₂,

Therefore, the electrostatic force obeys Newton’s third law.

Question 3.
What are the differences between the Coulomb force and the gravitational force?
Answer:

Coulomb force	Gravitational force
1. It can be attractive or repulsive depends on the nature of the charge	1. It is always attractive
2. The value of Proportionality constant K = 9 x 109 Nm2 C-2	2. The value of Gravitational constant G = 6.626 x 1011 Nm2 Kg-2
3. It depends on the medium which it exists.	3. It is independent of the medium which it exists.

Question 4.
Write a short note on the superposition principle.
Answer:
According to this superposition principle, the total force acting on a given charge is equal to the vector sum of forces exerted on it by all the other charges.
\({ \vec { F } }_{ 1 }^{ tot }\) = \(\vec { F } \)₁₂ + \(\vec { F } \)₁₃ + \(\vec { F } \)₁₄ + \(\vec { F } \)_1n

Question 5.
Define ‘Electric field’.
Answer:
It is defined as the force experience by a unit positive charge, kept at that point
It is a Vector quantity.
Unit: NC^-1

Question 6.
What is mean by ‘Electric field lines’?
Answer:
Electric field vectors are visualized by the concept of electric field lines. They form a set of continuous lines which are the visual representation of the electric field in some region of space.

Question 7.
The electric field lines never intersect. Justify.
Answer:
If two lines cross at a point, then there will be two different electric field vectors at the same point which is not possible. hence, they do not intersect.

Question 8.
Define ‘Electric dipole’
Answer:
Two equal and opposite charges separated by a small distance constitute an electric dipole.

Question 9.
What is the general definition of electric dipole moment?
Answer:
The electric dipole moment for a collection of ‘n point charges is given by
\(\overrightarrow{\mathrm{P}}=\sum_{i=1}^{\mathrm{n}} \mathrm{q}_{\mathrm{i}} \mathrm{r}_{\mathrm{i}}\)
where r̂_i is the position ofvector of change q_i from origin.

Question 10.
Define “electrostatic potential”.
Answer:
The electric potential at a point P is equal to the work done by an external force to bring a unit positive charge with constant velocity from infinity to the point P in the region of the external
electric field \(\vec { E } \).

Question 11.
What is an equipotential surface?
Answer:
An equipotential surface is a surface on which all the points are at the same electric potential.

Question 12.
What are the properties of an equipotential surface?
Answer:
Properties of equipotential surfaces
(i) The work is done to move a charge q between any two points A and B,
W = q (V_B – V_A). If the points A and B lie on the same equipotential surface, work done is zero because of V_A = V_B.

(ii) The electric field is normal to an equipotential surface. If it is not normal, then there is a component of the field parallel to the surface. Then work must be done to move a charge between two points on the same surface. This is a contradiction. Therefore the electric field must always be normal to the equipotential surface.

Question 13.
Give the relation between electric field and electric potential.
Answer:
The electric field is the negative gradient of the electric potential.
\(\mathrm{E}=-\frac{\mathrm{d} \mathrm{v}}{\mathrm{d} \mathrm{x}}\)

Question 14.
Define electrostatic potential energy?
Answer:
The potential energy of a system of point charges may be defined as the amount of work done in assembling the charges at their locations by bringing them in from infinity.

Question 15.
Define ‘electric flux’.
Answer:

1. The number of electric field lines crossing a given area kept normal to the electric field lines is called electric flux.
2. Scalar quantity.
3. Unit: Nm²C^-1

Question 16.
What is meant by electrostatic energy density?
Answer:
The energy stored per unit volume of space is defined as energy density u_E = \(\frac { U }{ Volume }\)
From equation u_E = \(\frac { 1 }{ 2 }\) \(\frac{\left(\varepsilon_{0} A\right)}{d}\) (Ed)² = \(\frac { 1 }{ 2 }\) ε₀ (Ad) E² or u_E = \(\frac { 1 }{ 2 }\) ε₀E²

Question 17.
Write a short note on ‘electrostatic shielding’.
Answer:

1. The process of isolating a certain region of space from the external field. It is based on the fact that the electric field inside a conductor is zero.
2. Whatever the charges at the surfaces and whatever the electrical disturbance outside, the electric field inside the cavity are zero.

Question 18.
What is Polarisation?
Answer:
Polarisation \(\vec { P } \) is defined as the total dipole moment per unit volume of the dielectric.
\(\vec { P } \) = X_e \(\vec { P } \)_ext

Question 19.
What is dielectric strength?
Answer:

1. The maximum electric field the dielectric can withstand before it breakdowns is called dielectric strength.
2. The dielectric strength of air 3 × 10⁶ Vm^-1

Question 20.
Define ‘capacitance’. Give its unit.
Answer:
The capacitance C of a capacitor is defined as the ratio of the magnitude of charge on either of the conductor plates to the potential difference existing between the conductors.
C = \(\frac { q }{ V }\) or Q ∝ V.
The SI unit of capacitance is coulomb per volt or farad (F).

Question 21.
What is corona discharge?
Answer:
The total charge of the conductor near the sharp edge gets reduces due to ionization of surrounding air. It is called corona discharge.

Samacheer Kalvi 12th Physics Electrostatics Long Answer Questions

Question 1.
Discuss the basic properties of electric charges.
Answer:
The electric charge is an inherent property of particles.
Conservation of electric charge:

1. Total electric charge in the universe is constant.
2. Charge can be neither created nor destroyed.
3. In any physical process, the net change in charge will always be zero.
4. The charge ‘q’ of any object is equal to an integral multiple of the fundamental unit of charge ‘e’.
   q = ne
5. n is any integer
6. e is charge of an electron = 1.6 × 10^-19C.

Question 2.
Explain in detail Coulomb’s law and its various aspects.
Answer:
Consider two point charges q₁ and q₂ at rest in vacuum, and separated by a distance of r. According to Coulomb, the force on the point charge q₂ exerted by another point charge q₁ is
\(\vec { F } \) ₂₁ = K\(\frac{q_{1} q_{2}}{r_{2}}\) \(\hat{r} \)₁₂,
where [/latex] \(\hat{r} \)₁₂ is the unit vector directed from charge q₁ to charge q₂ and k is the proportionality constant.

Important aspects of Coulomb’s law:
(i) Coulomb’s law states that the electrostatic force is directly proportional to the product of the magnitude of the two point charges and is inversely proportional to the square of the distance between the two point charges.

(ii) The force on the charge q₂exerted by the charge q₁ always lies along the line joining the two charges. \(\hat{r} \)₂₁is the unit vector pointing from charge q₁ to q₂ Likewise, the force on the charge q₁ exerted by q₂ is along – (i.e., in the direction opposite to \(\hat{r} \)₂₁).

(iii) In SI units, k = \(\frac { 1 }{{ 4πε }_{0}}\) and its value is 9 x 10⁹ Nm²C^-2. Here e0 is the permittivity of free space or vacuum and the value of ε₀ = \(\frac { 1 }{{ 4πε }_{0}}\) = 8.85 x 10^-12 C² N^-1 m^-2

(iv) The magnitude of the electrostatic force between two charges each of one coulomb and separated by a distance of 1 m is calculated as follows:
[F] = \(\frac{9 \times 10^{9} \times 1 \times 1}{1^{2}}\) = 9 x 10⁹N. This is a huge quantity, almost equivalent to the weight of one million ton. We never come across 1 coulomb of charge in practice. Most of the electrical phenomena in day-to-day life involve electrical charges of the order of pC (micro coulomb) or nC (nano coulomb).

(v) In SI units, Coulomb’s law in vacuum takes the form \(\vec { F } \) ₂₁ = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{q_{1} q_{2}}{r_{2}}\) \(\hat{r} \)₁₂. sin Since ε > ε₀, the force between two point charges in a medium other than vacuum is always less than that in vacuum. We define the relative permittivity for a given medium as ε = \(\frac { ε }{{ ε }_{0}}\) .For vacuum or air, ε_r = 1 and for all other media ε_r > 1

(vi) Coulomb’s law has same structure as Newton’s law of gravitation. Both are inversely proportional to the square of the distance between the particles. The electrostatic force is directly proportional to the product of the magnitude of two point charges and gravitational force is directly proportional to the product of two masses.

(vii) The force on a charge q₁ exerted by a point charge q₂ is given by \(\vec { F } \)₁₂ = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{q_{1} q_{2}}{r_{2}}\) \(\hat{r} \)₁₂ Here \(\hat{r} \)₂₁ is sthe unit vector from charge q₂ to q₁.

Therefore, the electrostatic force obeys Newton’s third law.

(viii) The expression for Coulomb force is true only for point charges. But the point charge is an ideal concept. However we can apply Coulomb’s law for two charged objects whose sizes are very much smaller than the distance between them. In fact, Coulomb discovered his law by considering the charged spheres in the torsion balance as point charges. The distance between the two charged spheres is much greater than the radii of the spheres.

Question 3.
Define ‘Electric field’ and discuss its various aspects.
Answer:
The electric field at the point P at a distance r from the point charge q is the force experienced by a unit charge and is given by

Here \(\hat{r} \) is the unit vector pointing from q to the point of interest P. The electric field is a vector quantity and its SI unit is Newton per Coulomb (NC^-1).

Important aspects of the Electric field:
(i) If the charge q is positive then the electric field points away from the source charge and if q is negative, the electric field points towards the source charge q.

(ii) If the electric field at a point P is \(\vec { E } \), then the force experienced by the test charge qo placed at the point P is \(\vec { F } \) = q₀ \(\vec { E } \). This is Coulomb’s law in terms of the electric field. This is shown below Figure.

(iii) The equation implies that the electric field is independent of the test charge q0 and it depends only on the source charge q.

(iv) Since the electric field is a vector quantity, at every point in space, this field has a unique direction and magnitude as shown in Figures (a) and (b). From the equation, we can infer that as distance increases, the electric field decreases in magnitude. Note that in Figures (a) and (b) the length of the electric field vector is shown for three different points. The strength or magnitude of the electric field at point P is stronger than at the point Q and R because the point P is closer to the source charge.

(v) In the definition of the electric field, it is assumed that the test charge q₀ is taken sufficiently small, so that bringing this test charge will not move the source charge. In other words, the test charge is made sufficiently small such that it will not modify the electric field of the source charge.

(vi) The expression is valid only for point charges. For continuous and finite-size charge distributions, integration techniques must be used. However, this expression can be used as an approximation for a finite-sized charge if the test point is very far away from the finite-sized source charge.

(vii) There are two kinds of electric field: uniform (constant) electric field and non-uniform electric field. A Uniform electric field will have the same direction and constant magnitude at all points in space. The non-uniform electric field will have different directions or different magnitudes or both at different points in space. The electric field created by a point charge is basically a non-uniform electric field. This non-uniformity arises, both in direction and magnitude, with the direction being radially outward (or inward), and the magnitude changes as distance increases.

Question 4.
How do we determine the electric field due to a continuous charge distribution? Explain. Electric field due to continuous charge distribution.
Answer:
The electric charge is quantized microscopically. The expressions of Coulomb’s Law, superposition principle force and electric field are applicable to only point charges. While dealing with the electric field due to a charged sphere or a charged wire etc., it is very difficult to look at individual charges in these charged bodies. Therefore, it is assumed that charge is distributed continuously on the charged bodies, and the discrete nature of charges is not considered here. The electric field due to such continuous charge distributions is found by invoking the method of calculus.

Consider the following charged object of irregular shape. The entire charged object is divided into a large number of charge elements ∆q₁, ∆q₂, ∆q₃ ……..∆q_n,…… and each charge element Δq is taken as a point charge.
The electric field at a point P due to a charged object is approximately given by the sum of the fields at P due to all such charge elements.

Here ∆ qi is the i^th charge element, r_ip is the distance of the point P frome the i^th charge element, r_ip is the unit vector from ith charge element to the pont P.
However the equation is only an approximation. To incorporate the continuous distribution of charge, we take the limit ∆q → 0(= dq). In this limit, the summation in the equation becomes an integration and takes the following form

Here r is the distance of the point P from the infinitesimal charge dq and \(\hat{r} \) is the unit vector from dq to point P. Even though the electric field for a continuous charge distribution is difficult to evaluate, the force experienced by some test charge q in this electric field is still given by \(\vec { F } \) = q\(\vec { E } \).

(a) Line charge distribution: If the charge Q is uniformly distributed along the wire of length L, then linear charge density (charge per unit length) is λ = \(\frac { Q }{ L }\). Its unit is colomb per meter (Cm^-1). The charge present in the infinitestimal length dl is dq = λdl.

The electric field due to the line of total charge Q is given by

(b) Surface charge distribution: If the charge Q is uniformly distributed on a surface of area A, then surface charge density (charge per unit area) is σ = \(\frac { Q }{ A }\). Its unit is coulomb per square meter (C m^-2). The charge present in the infinitesimal area dA is dq = σdA. The electric field due to a of total charge Q is given by

(c) Volume charge distribution: If the charge Q is uniformly distributed in a volume V, then volume charge density (charge per unit volume) is given by ρ = \(\frac { Q }{ V }\). Its unit is coulomb per cubic meter (Cm^-3) The charge present in the infinitesimal volume element dV is dq = ρdV. The electric field due to a volume of total charge Q is given by

Question 5.
Calculate the electric field due to a dipole on its axial line and the equatorial plane.
Answer:
Case (I) :
Electric field due to an electric dipole at points on the axial line. Consider an electric dipole placed on the x-ax is as shown in the figure. A point C is located at a distance of r from the midpoint O of the dipole along the axial line. Axial line

The electric field at a point C due to +q is

Since the electric dipole moment vector \(\vec { P } \) is from -q to +q and is directed along BC, the above equation is rewritten as

where \(\hat{p} \) is the electric dipole moment unit vector from -q to +q. The electric field at a point C due to -q is

Since +q is located closer to the point C than -q, \(\vec { E } \) _. \(\vec { E } \) + us stronger than \(\vec { E } \). Therefore, the length of the E + vector is drawn large than that of \(\vec { E } \) _vector.
The total electric field at point C is calculated using the superposition principle of the electric field.

Note that the total electric field is along \(\vec { E } \)₊, since +q is closer to C than -q.

The direction of \(\vec { E } \)_tot is shown in Figure
If the point C is very far away from the dipole then (r >> a). Under this limit the term(r² – a²)² ≈ r⁴ Substituting this into equation, we get

If point C is chosen on the left side of the dipole, the total electric field is still in the

Case (II) :
Electric field due to an electric dipole at a point on the equatorial plane
Consider a point C at a distance r from the midpoint O of the dipole on the equatorial plane as shown in Figure. Since point C is equidistant from +q and -q, the magnitude of the electric fields of +q and -q are the same. The direction of E+ is along with BC and the direction of E is along with CA. E+ and E_ are resolved into two components; one component parallel to the dipole axis and the other perpendicular to it.

The perpendicular components \(\left|\vec{E}_{+}\right|\) sin θ and \(\left|\vec{E}_{-}\right|\) sin θ are oppositely directed and cancel each other. The magnitude of the total electric field at point C is the sum of the paralle component of \(\vec { E } \)₊ and \(\vec { E } \)_– and its direction is along \(\hat{-p} \).


The magnitudes \(\vec { E } \)₊ and \(\vec { E } \)_– are the same and are given by

By substituting equation (1) into equation (2), we get

At very large distances (r >> a), the equation becomes
\(\vec { E } \)_tot \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { p }{{ r }^{3}}\) (r >>) …… (4)

Question 6.
Derive an expression for the torque experienced by a dipole due to a uniform electric field.
Answer:
Torque experienced by an electric dipole in the uniform electric field:
Consider an electric dipole of dipole moment \(\vec { p } \) placed in a uniform electric field E whose field lines are equally spaced and point in the same direction. The charge +q will experience a force q\(\vec { E } \) in the direction of the field and charge -q will experience a force -q\(\vec { E } \) in a direction opposite to the field.

Since the external field \(\vec { E } \) is uniform, the total force acting on the dipole is zero. These two forces acting at different points will constitute a couple and the dipole experience a torque. This torque tends to rotate the dipole. (Note that electric field lines of a uniform field are equally spaced and point in the same direction). The total torque on the dipole about the point O
\(\vec { τ } \) = \(\overrightarrow{\mathrm{OA}}\) × (-q\(\vec { E } \)) + \(\overrightarrow{\mathrm{OB}}\) × q\(\vec { E } \)
Using the right-hand corkscrew rule, it is found that total torque is perpendicular to the plane of the paper and is directed into it.

The magnitude of the total torque
\(\vec { τ } \) = \(|\overrightarrow{\mathrm{OA}}|\)(-q\(\vec { E } \)) sin θ + \(|\overrightarrow{\mathrm{OB}}|\) \(|q \overrightarrow{\mathrm{E}}|\) sin θ
where θ is the angle made by \(\vec { P } \) with \(\vec { E } \). Since p = 2aq, the torque is written in terms of the vector product as
\(\vec { τ } \) = \(\vec { p } \) x \(\vec { E } \)
The magnitude of this torque is τ = pE sin θ and is maximum Torque on dipole
when θ =90°.
This torque tends to rotate the dipole and align it with the electric field \(\vec { E } \). Once \(\vec { E } \) is aligned with \(\vec { E } \), the total torque on the dipole becomes zero.

Question 7.
Derive an expression for electrostatic potential due to a point charge.
Answer:
Electric potential due to a point charge:
Consider a positive charge q kept fixed at the origin. Let P be a point at distance r from the charge q.
The electric potential at point P is


Electric field due to positive point charge q is

The infinitesimal displacement vector, d\(\vec { r } \) = dr\(\hat{r} \) and using \(\hat{r} \) . \(\hat{r} \) = 1, we have

After the integration,

Hence the electric potential due to a point charge q at a distance r is
V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { q }{ r }\) …… (2)
Important points (If asked in the exam)
(i) If the source charge q is positive, V > 0. If q is negative, then V is negative and equal to
V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { q }{ r }\)
(ii) The description of the motion of objects using the concept of potential or potential energy is simpler than using the concept of field.
(iii) From expression (2), it is clear that the potential due to positive charge decreases as the distance increases, but for a negative charge, the potential increases as the distance are increased. At infinity (r = ∞) electrostatic potential is zero (V = 0).
(iv) The electric potential at a point P due to a collection of charges q₁,q₂,q₃… q_n is equal to the sum of the electric potentials due to individual charges.

Where r₁, r₂,r₃,…..r_n are the distances of q₁,q₂,q₃… q_n
respectively from P

Question 8.
Derive an expression for electrostatic potential due to an electric dipole.
Answer:
Electrostatic potential at a point due to an electric dipole:
Consider two equal and opposite charges separated by a small distance 2a. The point P is located at a distance r from the midpoint of the dipole. Let 0 be the angle between the line OP and dipole axis AB.
Let r₁ be the distance of point P from +q and r₁ be the distance of point P from -q.
Potential at P due to charge +q = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { q }{{ r }_{1}}\)
Potential at P due to charge -q = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { q }{{ r }_{2}}\)
Total Potential at the point P,
V = \(\frac { 1 }{{ 4πε }_{0}}\)q \(\left(\frac{1}{r_{1}}-\frac{1}{r_{2}}\right)\) ….. (1)
Suppose if the point P is far away from the dipole, such that r >> a, then equation can be expressed in terms of r. By the cosine law for triangle BOP,

\(r_{1}^{2}\) = r² + a² – 2ra cos θ = r² \(\left(1+\frac{a^{2}}{r^{2}}-\frac{2 a}{r} \cos \theta\right)\)
Since the point P is very far from dipole, then r >> a. As a result the term \(\frac {{ a }^{ 2 }}{{ r }^{ 2 }}\) is very small and can be neglected. Therefore

since \(\frac { a }{ r }\) << 1, we can use binominal theorem and retain the terms up to first order
\(\frac { 1 }{{ r_{1}} }\) = \(\left(1+\frac{a}{r} \cos \theta\right)\) ……. (2)
Similarly applying the cosine law for triangle AOP,
\(r_{2}^{2}\) = r² + a² – 2ra cos (180 – θ)
Since cos (180 – θ) = cos θ we get
\(r_{2}^{2}\) = r² + a² + 2ra cos θ
Neglecting the term \(\frac {{ a }^{ 2 }}{{ r }^{ 2 }}\) (because r >> a)
\(r_{2}^{2}\) = r² \(\left(1+\frac{2 a \cos \theta}{r}\right)\) (or) r₂ = r \(\left(1+\frac{2 a \cos \theta}{r}\right)^{\frac{1}{2}}\)
Using Binomial theorem, we get
\(\frac { 1 }{{ r_{2}} }\) = \(\frac { 1 }{ r }\) \(\left(1-a \frac{\cos \theta}{r}\right)\)
Substituting equations (3) and (2) in equation (1)

But the electric dipole moment p = 2qa and we get,
V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\left(\frac{p \cos \theta}{r^{2}}\right)\)
Now we can write p cos θ = \(\vec { P } \), \(\hat{r} \) where \(\hat{r} \) is the unit vector from the point O to point P. Hence the electric potential at a point P due to an electric dipole is given by
V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{\vec{p} \cdot \hat{r}}{r^{2}}\) (r >> a) ….. (4)
Equation (4) is valid for distances very large compared to the size of the dipole. But for a point dipole, the equation (4) is valid for any distance.
Special cases:

Case (I):
If the point P lies on the axial line of the dipole on the side of +q, then θ = 0. Then the electric potential becomes
V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { p }{{r}^{ 2 }}\)

Case (II):
If the point P lies on the axial line of the dipole on the side of -q, then θ = 180°, then
V = – \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { p }{{r}^{ 2 }}\)

Case (III):
If the point P lies on the equatorial line of the dipole, then θ = 90°. Hence, V = 0.

Question 9.
Obtain an expression for potential energy due to a collection of three-point charges which are separated by finite distances.
Answer:
Electrostatic potential energy for a collection of point charges:
The electric potential at a point at a distance r from point charge ql is given by
V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac {{ q }_{ 1 }}{r}\) …… (1)
This potential V is the work done to bring a unit positive charge from infinity to the point. Now if the charge q₂ is brought from infinity to that point at a distance r from qp the work done is the product of q₂ and the electric potential at that point. Thus we have W = q₂V …… (2)
This work done is stored as the electrostatic potential energy U of a system of charges q₁ and q₂ separated by a distance r. Thus we have
U = q₂ V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{q_{1} q_{2}}{r}\) …… (3)
The electrostatic potential energy depends only on the distance between the two point charges. In fact, the expression (3) is derived by assuming that q₁ is fixed and q₂ is brought from infinity. The equation (3) holds true when q₂ is fixed and q₁ is brought from infinity or both q₂and q₂ are simultaneously brought from infinity to a distance r between them.
Three charges are arranged in the following configuration as shown in Figure.

To calculate the total electrostatic potential energy, we use the following procedure. We bring all the charges one by one and arrange them according to the configuration.
(i) Bringing a charge q₁ from infinity to point A requires no work, because there are no other charges already present in the vicinity of charge q₁

(ii) To bring the second charge q₂ to point B, work must be done against the electric field created by the charge q₁ So the work done on the charge q₁ is W = q₂V_1B. Here V_1B is the electrostatic potential due to the charge q₁ at point B.
U = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{q_{1} q_{2}}{{r}_{12}}\) ….. (4)
Note that the expression is the same when q₂ is brought first and then q₁ later.

(iii) Similarly to bring the charge q3 to point C, work has to be done against the total electric field due to both charges q₁ and q₂. So the work done to bring the charge q₃ is = q₃ (V_1C + V_2C). Here V_1C is the electrostatic potential due to charge q₁ at point C and V_2C is the electrostatic potential due to charge q₂ at point C. The electrostatic potential is

(iv) Adding equations (4) and (5), the total electrostatic potential energy for the system of three charges q₁,q_2, and q₃ is

Note that this stored potential energy U is equal to the total external work done to assemble the three charges at the given locations. The expression (6) is the same if the charges are brought to their positions in any other order. Since the Coulomb force is a conservative force, the electrostatic potential energy is independent of the manner in which the configuration of charges is arrived at.

Question 10.
Derive an expression for the electrostatic potential energy of the dipole in a uniform electric field.
Answer:
The electrostatic potential energy of a dipole in a uniform electric field:
Consider a dipole placed a torque when kept in an uniform electric field \(\vec { E } \). A dipole experiences a torque when kept in an uniform electric field \(\vec { E } \). This torque rotates the dipole to align it with the direction of the electric field. To rotate the dipole (at constant angular velocity) from its initial angle θ’ to another angle θ against the torque exerted by the electric field, an equal and opposite external torque must be applied on the dipole.

The work done by the external torque to rotate the dipole from angle θ’ to θ at constant angular velocity is

Since τ_ext is equal and opposite to τ_E = \(\vec { P } \) x \(\vec { E } \), we have
\(\left|\overrightarrow{\mathrm{r}}_{\mathrm{ext}}\right|\) = \(\left|\overrightarrow{\mathrm{r}}_{\mathrm{E}}\right|\)= \(|\overrightarrow{\mathrm{P}} \times \overrightarrow{\mathrm{E}}|\) …. (2)
Substituting equation (2) in equation (1) We get,

This work done is equal to the potential energy difference between the angular positions θ and θ’.
U(θ) – (Uθ’) = AU = -pE cos θ +PE cos θ’.
If the initial angle is = θ’ = 90° and is taken as reference point, then U(θ’) + pE cos θ’ = θ.
The potential energy stored in the system of dipole kept in the uniform electric field is given by El = -pE cos θ = –\(\vec { P } \) . \(\vec { E } \) ….. (3)
In addition to p and E, the potential energy also depends on the orientation θ of the electric dipole with respect to the external electric field.
The potential energy is maximum when the dipole is aligned anti-parallel (θ = π) to the external electric field and minimum when the dipole is aligned parallel (θ = 0) to the external electric field.

Question 11.
Obtain Gauss law from Coulomb’s law.
Answer:
Gauss law: Gauss’s law states that if a charge Q is enclosed by an arbitrary closed surface, then the total electric flux Φ_E through the closed surface is
Φ_E = \(\oint { \vec { E } } \) .d \(\vec { A } \) = \(\frac{\mathrm{Q}_{\mathrm{end}}}{\varepsilon_{0}}\)
A positive point charge Q is surrounded by an imaginary sphere of radius r as shown in the figure. We can calculate the total electric flux through the closed surface of the sphere using the equation.

Φ_E = \(\oint { \vec { E } } \) .d \(\vec { A } \) = \(\oint { EdA } \) cos θ …… (1)
The electric field of the point charge is directed radially outward at all points on the surface of the sphere. Therefore, the direction of the area element d \(\vec { A } \) is along the electric field \(\vec { E } \) and θ = 0°.
Φ_E = \(\oint { EdA } \) since cos 0° = 1 ….. (2)
E is uniform on the surface of the sphere,
Φ_E = \(\oint { EdA } \) ….. (3)
Substituting for
\(\oint { dA } \) = 4π² and E = \(\frac { 1 }{{ 4πε }_{0}}\) Q in equation 3, we get
Φ_E = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { q }{{ r }^{2}}\) × 4π² = 4π \(\frac { 1 }{{ 4πε }_{0}}\) = \(\frac { q }{{ ε }_{0}}\) ……. (4)
The equation (4) is called as Gauss’s law. The remarkable point about this result is that the equation (4) is equally true for any arbitrary shaped surface which encloses the charge Q.

Question 12.
Obtain the expression for electric field due to an infinitely long charged wire.
Answer:
Electric field due to an infinitely long charged wire:
Consider an infinitely long straight wire having uniform linear charge density λ. Let P be a point located at a perpendicular distance r from the wire. The electric field at the point P can be found using Gauss law. We choose two small charge elements A₁ and A₁ on the wire which are at equal distances from the point P.

The resultant electric field due to these two charge elements points radially away from the charged wire and the magnitude of electric field is same at all points on the circle of radius r. From this property, we can infer that the charged wire possesses a cylindrical symmetry.

Let us choose a cylindrical Gaussian surface of radius r and length L. The total electric flux in this closed surface is

It is seen that for the curved surface, \(\vec { E } \) is parallel to \(\vec { A } \) and \(\vec { E } \).d \(\vec { A } \) = EdA. For the top and bottom surface, \(\vec { E } \) is perpendicular to \(\vec { A } \) and \(\vec { E } \).d\(\vec { A } \) = 0
Substituting these values in equation (2) and applying Gauss law


Since the magnitude of the electric field for the entire curved surface is constant, E is taken out of the integration, and Q_encl is given by Q_encl = λL.

Here,

dA = total area of the curved surface = 2πrL. Substituting this in
equation (4), We get

The electric field due to the infinite charged wire depends on \(\frac { 1 }{ r }\) rather than \(\frac { 1 }{{r}^{ 2 }}\) for a point charge.
Equation (6) indicates that the electric field is always along the perpendicular direction (\(\hat{r} \) ) to wire. In fact, if λ > 0 then E points perpendicular outward (\(\hat{r} \) ) from the wire and if λ < 0, then E points perpendicular inward (- \(\hat{r} \) ).

Question 13.
Obtain the expression for the electric field due to a charged infinite plane sheet.
Answer:
Electric field due to the charged infinite plane sheet: Consider an infinite plane sheet of charges with uniform surface charge density o. Let P be a point at a distance of r from the sheet. Since the plane is infinitely large, the electric field should be the same at all points equidistant from the plane and radially directed at all points. A cylindrical shaped Gaussian surface of length 2r and area A of the flat surfaces is chosen such that the infinite plane sheet passes perpendicularly through the middle part of the Gaussian surface.

Applying Gauss law for this cylindrical surface,

The electric field is perpendicular to the are element at all points on the curved surface and is parallel to the surface areas at P and P’. Then,

Since the magnitude of the electric field at these two equal surfaces is uniform, E is taken out of the integration, and Q_encl is given by Q_encl = σA, we get

The total area of surface either at P or P’

Hence 2EA = \(\frac { σA }{{ ε }_{0}}\) or E = \(\frac { σ }{{ 2ε }_{0}}\) …… (3)
In vector from, E = \(\frac { σ }{{ 2ε }_{0}}\) \(\hat{n} \) ….. (4)
Hence \(\hat{n} \) is the outward unit vector normal to the plane. Note that the electric field due to an infinite plane sheet of the charge depends on the surface charge density and is independent of the distance r.

The electric field will be the same at any point farther away from the charged plane. Equation (4) implies that if o > 0 the electric field at any point P is outward perpendicular n to the plane and if σ < 0 the electric field points inward perpendicularly (\(\hat{n} \) ) to the plane. For a finite charged plane sheet, equation (4) is approximately true only in the middle region of the plane and at points far away from both ends.

Question 14.
Obtain the expression for the electric field due to a uniformly charged spherical shell.
Answer:
Electric field due to a uniformly charged spherical shell:
Consider a uniformly charged spherical shell of radius R and total charge Q. The electric field at points outside and inside the sphere is found using Gauss law.

Case (a):
At a point outside the shell (r > R): Let us choose a point P outside the shell at a distance r from the center as shown in figure (a). The charge is uniformly distributed on the surface of the sphere (spherical symmetry). Hence the electric field must point radially outward if Q > 0 and point radially inward if Q < 0. So we choose a spherical Gaussian surface of radius r and the total charge enclosed by this Gaussian surface is Q. Applying Gauss law,
\(\oint { \vec { E } } .d\vec { A } \) = \(\frac { Q }{{ ε }_{0}}\) …….(1)
The electric field \(\vec { E } \) and d\(\vec { A } \) point in the same direction (outward normal) at all the points on the Gaussian surface. The magnitude of \(\vec { E } \) is also the same at all points due to the spherical symmetry of the charge distribution.

But

dA = total area of Gaussian surface = 4πr². Substituting this value in equation (2).

The electric field is radially outward if Q > 0 and radially inward if Q < 0. From equation (3), we infer that the electric field at a point outside the shell will be same as if the entire charge Q is concentrated at the center of the spherical shell. (A similar result is observed in gravitation, for gravitational force due to a spherical shell with mass M)

Case (b):
At a point on the surface of the spherical shell (r = R): The electrical field at points on the spherical shell (r = R) is given by
\(\vec { E } \) = \(\frac{\mathrm{Q}}{4 \pi \varepsilon_{0} \mathrm{R}^{2}}\) \(\hat{r} \) …… (4)

Case (c):
At a point inside the spherical shell (r < R): Consider a point P inside the shell at a distance r from the center. A Gaussian sphere of radius r is constructed as shown in the figure (b). Applying Gauss law

Since Gaussian surface encloses no charge, So Q = 0. The equation (5) becomes E = 0 (r < R) …(6)
The electric field due to the uniformly charged spherical shell is zero at all points inside the shell.

Question 15.
Discuss the various properties of conductors in electrostatic equilibrium.
Answer:
Properties of conductors in electrostatic equilibrium:
(i) The electric field is zero everywhere inside the conductor. This is true regardless of whether the conductor is solid or hollow. This is an experimental fact. Suppose the electric field is not zero inside the metal, then there will be a force on the mobile charge carriers due to this electric field.

As a result, there will be a net motion of the mobile charges, which contradicts the conductors being in electrostatic equilibrium. Thus the electric field is zero everywhere inside – the conductor. We can also understand this fact by applying an external uniform electric field on the conductor.

Before applying the external electric field, the free electrons in the conductor are uniformly distributed in the conductor. When an electric field is applied, the free electrons accelerate to the left causing the left plate to be negatively charged and the right plate to be positively charged.

Due to this realignment of free electrons, there will be an internal electric field created inside the conductor which increases until it nullifies the external electric field. Once the external electric field is nullified the conductor is said to be in electrostatic equilibrium. The time taken by a conductor to reach electrostatic equilibrium is in the order of 10^-6s, which can be taken as almost instantaneous.

(ii) There is no net charge inside the conductors. The charges must reside only on the surface of the conductors. We can prove this property using Gauss law. Consider an arbitrarily shaped conductor. A Gaussian surface is drawn inside the conductor such that it is very close to the surface of the conductor.

Since the electric field is zero everywhere inside the conductor, the net electric flux is also zero over this Gaussian surface. From Gauss’s law, this implies that there is no net charge inside the conductor. Even if some charge is introduced inside the conductor, it immediately reaches the surface of the conductor.

(iii) The electric field outside the conductor is perpendicular to the surface of the conductor and has a magnitude of \(\frac { σ }{{ ε }{0}}\) where a is the surface charge density at that point. If the electric field has components parallel to the surface of the conductor, then free electrons on the surface of the conductor would experience acceleration. This means that the conductor is not in equilibrium. Therefore at electrostatic equilibrium, the electric field must be perpendicular to the surface of the conductor.

We now prove that the electric field has magnitude \(\frac { σ }{{ ε }{0}}\) just outside the conductor’s surface. Consider a small cylindrical Gaussian surface. One half of this cylinder is embedded inside the conductor. Since electric field is normal to the surface of the conductor, the curved part of the cylinder has zero electric flux. Also inside the conductor, the electric field is zero. Hence the bottom flat part of the Gaussian surface has no electric flux. Therefore the top flat surface alone contributes to the electric flux. The electric field is parallel to the area vector and the total charge inside the surface is σA. By applying Gauss’s law,
EA = \(\frac { σA }{{ ε }{0}}\)
In vector from, \(\vec { E } \) = \(\frac { σ }{{ ε }{0}}\) \(\hat{n} \)
Here n represents the unit vector outward normal to the surface of the conductor. Suppose σ < 0, then electric field points inward perpendicular to the surface.

(iv) The electrostatic potential has the same value on the surface and inside of the conductor. We know that the conductor has no parallel electric component on the surface which means that charges can be moved on the surface without doing any work. This is possible only if the electrostatic potential is constant at all points on the surface and there is no potential difference between any two points on the surface. Since the electric field is zero inside the conductor, the potential is the same as the surface of the conductor. Thus at electrostatic equilibrium, the conductor is always at equipotential.

Question 16.
Explain the process of electrostatic induction.
Answer:
Whenever a charged rod is touched by another conductor, charges start to flow from the charged rod to the conductor. This type of charging without actual contact is called electrostatic induction:

(i) Consider an uncharged (neutral) conducting sphere at rest on an insulating stand. Suppose a negatively charged rod is brought near the conductor without touching it, as shown in figure (a). The negative charge of the rod repels the electrons in the conductor to the opposite side.

Various steps in electrostatic induction
As a result, positive charges are induced near the region of the charged rod while negative charges on the farther side. Before introducing the charged rod, the free electrons were distributed uniformly on the surface of the conductor and the net charge is zero. Once the charged rod is brought near the conductor, the distribution is no longer uniform with more electrons located on the farther side of the rod and positive charges are located closer to the rod. But the total charge is zero.

(ii) Now the conducting sphere is connected to the ground through a conducting wire. This is called grounding. Since the ground can always receive any amount of electrons, grounding removes the electron from the conducting sphere. Note that positive charges will not flow to the ground because they are attracted by the negative charges of the rod (figure (b)).

(iii) When the grounding wire is removed from the conductor, the positive charges remain near the charged rod (figure (c)).

(iv) Now the charged rod is taken away from the conductor. As soon as the charged rod is removed, the positive charge gets distributed uniformly on the surface of the conductor (figure (d)). By this process, the neutral conducting sphere becomes positively charged.

Question 17.
Explain dielectrics in detail and how an electric field is induced inside a dielectric.
Answer:
Induced Electric field inside the dielectric:
When an external electric field is applied to a conductor, the charges are aligned in such a way that an internal electric field is created which cancels the external electric field. But in the case of a dielectric, which has no free electrons, the external electric field only realigns the charges so that an internal electric field is produced.

The magnitude of the internal electric field is smaller than that of the external electric field. Therefore the net electric field inside the dielectric is not zero but is parallel to an external electric field with a magnitude less than that of the external electric field. For example, let us consider a rectangular dielectric slab placed between two oppositely charged plates (capacitor) as shown in the figure.

The uniform electric field between the plates Induced electric field lines inside the dielectric acts as an external electric field \(\vec { E } \)_ext which polarizes the dielectric placed between plates. The positive charges are induced on one side surface and negative charges are induced on the other side of the surface But inside the dielectric, the net charge is zero even in a small volume. So the dielectric in the external field is equivalent to two oppositely charged sheets with the surface charge densities +σb and -σb. These charges are called bound charges. They are not free to move like free electrons in conductors. This is shown in the figure.

(a) Balloon sticks to the wall
(b) Polarisation of the wall due to the electric field created by the balloon
For example, the charged balloon after rubbing sticks onto a wall. The reason is that the negatively charged balloon is brought near the wall, it polarizes opposite charges on the surface of the wall, which attracts the balloon.

Question 18.
Obtain the expression for capacitance for a parallel plate capacitor.
Answer:
The capacitance of a parallel plate capacitor:
Consider a capacitor with two parallel plates each of cross-sectional area A and separated by a distance d. The electric field between two infinite parallel plates is uniform and is given by E = \(\frac { σ }{{ ε }{0}}\) where σ is the surface charge density on the plates σ = \(\frac { Q }{ A }\) .If the separation distance d is very much smaller than the size of the plate (d² << A), then the above result is used even for finite-sized
parallel plate capacitor.

The capacitance of a parallel plate capacitor
The electric field between the plates is
E = \(\frac { Q }{{ Aε }{0}}\) ….. (1)
Since the electric field is uniform, the electric potential between the plates having separation d is given by
V = Ed = \(\frac { Qd }{{ Aε }{0}}\) ….. (2)
Therefore the capacitance of the capacitor is given by
C = \(\frac { Q }{ V }\) = \(\frac{\mathrm{Q}}{\left(\frac{\mathrm{Q} d}{\mathrm{A} \varepsilon_{0}}\right)}\) = \(\frac{\varepsilon_{0} \mathrm{A}}{d}\) ….. (3)
From equation (3), it is evident that capacitance is directly proportional to the area of cross-section and is inversely proportional to the distance between the plates. This can be understood from the following.

• If the area of cross-section of the capacitor plates is increased, more charges can be distributed for the same potential difference. As a result, the capacitance is increased.
• If the distance d between the two plates is reduced, the potential difference between the plates (V = Ed) decreases with the E constant.

Question 19.
Obtain the expression for energy stored in the parallel plate capacitor.
Answer:
Energy stored in the capacitor:
Capacitor not only stores the charge but also it stores energy. When a battery is connected to the capacitor, electrons of total charge -Q are transferred from one plate to the other plate. To transfer the charge, work is done by the battery. This work done is stored as electrostatic potential energy in the capacitor. To transfer an infinitesimal charge dQ for a potential difference V, the work done is given by
dW = VdQ ….. (1)
Where V = \(\frac { Q }{ C }\)
The total work done to charge a capacitor is

This work done is stored as electrostatic potential energy (U_E) in the capacitor.
U_E = \(\frac {{ Q }^{2}}{ 2C }\) = \(\frac { 1 }{ 2 }\) CV² (∴ Q = CV) ….. (3)
where Q = CV is used. This stored energy is thus directly proportional to the capacitance of the capacitor and the square of the voltage between the plates of the capacitor. But where is this energy stored in the capacitor? To understand this question, the equation (3) is rewritten as follows using the results
C = \(\frac{\varepsilon_{0} \mathrm{A}}{d}\) and V = Ed
U_E = \(\frac { 1 }{ 2 }\) \(\left(\frac{\varepsilon_{0} \mathrm{A}}{d}\right)\) (Ed)² = \(\frac { 1 }{ 2 }\) ε₀(Ad)² …… (4)
where Ad = volume of the space between the capacitor plates. The energy stored per unit volume of space is defined as energy density \(\overline { Volume } \). Frome equation (4) we get
u_E = \(\frac { 1 }{ 2 }\) ε₀E²
From equation (5), we infer that the energy is stored in the electric field existing between the plates of the capacitor. Once the capacitor is allowed to discharge, the energy is retrieved.

Question 20.
Explain in detail the effect of a dielectric placed in a parallel plate capacitor.
Answer:
(i) When the capacitor is disconnected from the battery:
Consider a capacitor with two parallel plates each of cross-sectional area A and are separated by a distance d. The capacitor is charged by a battery of voltage V₀ and the charge stored is Q₀. The capacitance of the capacitor without the dielectric is
C₀ = \(\frac {{ Q }_{0}}{{ V }_{0}}\) ….. (1)
The battery is then disconnected from the capacitor and the dielectric is inserted between the plates. The introduction of dielectric between the plates will decrease the electric field. Experimentally it is found that the modified electric field is given by

(a) Capacitor is charged with a battery
(b) Dielectric is inserted after the battery is disconnected
E = \(\frac {{ E }_{0}}{{ ε }_{r}}\) …… (2)
Here E₀ is the electric field inside the capacitors when there is no dielectric and ε_r is the relative permeability of the dielectric or simply known as the dielectric constant. Since ε_r > 1, the electric field E < E₀. As a result, the electrostatic potential difference between the plates (V = Ed) is also reduced. But at the same time, the charge Q₀ will remain constant once the battery is disconnected. Hence the new potential difference is
V = Ed = \(\frac {{ E }_{0}}{{ ε }_{r}}\)d = \(\frac {{ V }_{0}}{{ ε }_{r}}\) ….. (3)
We know that capacitance is inversely proportional to the potential difference. Therefore as V decreases, C increases. Thus new capacitance in the presence of a dielectric is
C = \(\frac {{ Q }_{0}}{ V }\) = ε_r \(\frac {{ Q }_{0}}{{ V }_{0}}\) = ε_r C₀ …… (4)
Since ε_r > 1, we have C > C₀. Thus insertion of the dielectric constant ε_r increases the capacitance. Using equation,
C = \(\frac { { \varepsilon }_{ 0 }A }{ d } \)
C = \(\frac{\varepsilon_{r} \varepsilon_{o} A}{d}\) = \(\frac { εA }{ d }\) …… (5)
where ε = ε_rε₀ is the permittivity of the dielectric medium. The energy stored in the capacitor before the insertion of a dielectric is given by U₀ = \(\frac { 1 }{ 2 }\) \(\frac{\mathrm{Q}_{0}^{2}}{\mathrm{C}_{0}}\) ….. (6)
After the dielectric is inserted, the charge Q₀ remains constant but the capacitance is increased. As a result, the stored energy is decreased.

Since ε_r> 1 we get U < U₀. There is a decrease in energy because, when the dielectric is inserted, the capacitor spends some energy in pulling the dielectric inside.

(ii) When the battery remains connected to the capacitor: Let us now consider what happens when the battery of voltage V₀ remains connected to the capacitor when the dielectric is inserted into the capacitor.
The potential difference V₀ across the plates remains constant. But it is found experimentally (first shown by Faraday) that when the dielectric is inserted, the charge stored in the capacitor is increased by a factor ε_r.

(a) Capacitor is charged through a battery
(b) Dielectric is inserted when the battery is connected.
Q = ε_rQ₀ ….. (1)
Due to this increased charge, the capacitance is also increased. The new capacitance is
C = \(\frac {{ Q }_{0}}{ V }\) = ε_r \(\frac {{ Q }_{0}}{{ V }_{0}}\) = ε_r C₀ …… (2)
However, the reason for the increase in capacitance in this case when the battery remains connected is different from the case when the battery is disconnected before introducing the dielectric.
Now, C₀ = \(\frac {{ ε }{_0}A}{ d }\) and, C = \(\frac { εA }{ d }\) …… (3)
U₀ = \(\frac { 1 }{ 2 }\) C₀ \({ V }_{ 0 }^{ 2 }\) ….. (4)
Note that here we have not used the expression
U₀ = \(\frac { 1 }{ 2 }\)\({{ V }_{ 0 }^{ 2 }}{{C}_{0}}\)
because here, both charge and capacitance are changed, whereas in equation 4, V₀ remains constant. After the dielectric is inserted, the capacitance is increased; hence the stored energy is also increased.
U = \(\frac { 1 }{ 2 }\) \({ CV }_{ 0 }^{ 2 }\) = \(\frac { 1 }{ 2 }\) ε_r \({ CV }_{ 0 }^{ 2 }\) = ε_r U₀
Since e_r > 1 we have U > U₀
It may be noted here that since voltage between the capacitor V₀ is constant, the electric field between the plates also remains constant.

Question 21.
Derive the expression for resultant capacitance, when capacitors are connected in series and in parallel.
Answer:
capacitors in series and parallel:
(i) Capacitors in series:
Consider three capacitors of capacitance C₁, C₂ and C₃ connected in series with a battery of voltage V as shown in figure (a).
As soon as the battery is connected to the capacitors in series, the electrons of charge -Q are transferred from the negative terminal to the right plate of C₃which pushes the electrons of the same amount -Q from left plate of C₃ to the right plate of C₂ due to electrostatic induction. Similarly, the left plate of C₂ pushes the charges of Q to the right plate of which induces the positive charge +Q on the left plate of C₁ At the same time, electrons of charge -Q are transferred from the left plate of C₁ to the positive terminal of the battery.

By these processes, each capacitor stores the same amount of charge Q. The capacitances of the capacitors are in general different so that the voltage across each capacitor is also different and are denoted as V₁, V₂ and V₃ respectively.
The total voltage across each capacitor must be equal to the voltage of the battery.
V = V₁ + V₂ + V₃ ….. (1)
Since Q = CV, we have V = \(\frac { Q }{{ C }_{1}}\) + \(\frac { Q }{{ C }_{2}}\) + \(\frac { Q }{{ C }_{3}}\)
Q = \(\left( \frac { 1 }{ { C }_{ 1 } } +\frac { 1 }{ { C }_{ 2 } } +\frac { 1 }{ { C }_{ 3 } } \right) \) ….. (2)
If three capacitors in series are considered to form an equivalent single capacitor Cs shown in figure (b), then we have V = \(\frac { Q }{{ C }_{s}}\)
Substituting this expression into equation (2) we get
V = \(\frac { Q }{{ C }_{s}}\) = Q\(\left( \frac { 1 }{ { C }_{ 1 } } +\frac { 1 }{ { C }_{ 2 } } +\frac { 1 }{ { C }_{ 3 } } \right) \)
\(\frac { 1 }{{ C }_{s}}\) = \(\frac { 1 }{{ C }_{1}}\) + \(\frac { 1 }{{ C }_{2}}\) + \(\frac { 1 }{{ C }_{3}}\) ….. (3)
Thus, the inverse of the equivalent capacitance C_s of three capacitors connected in series is equal to the sum of the inverses of each capacitance. This equivalent capacitance C_s is always less than the smallest individual capacitance in the series.

(ii) Capacitance in parallel:
Consider three capacitors of capacitance C₁,C₂ and C₃ connected in parallel with a battery of voltage V as shown in figure (a).

Since corresponding sides of the capacitors are connected to the same positive and negative terminals of the battery, the voltage across each capacitor is equal to the battery’s voltage. Since the capacitance of the capacitors is different, the charge stored in each capacitor is not the same. Let the charge stored in the three capacitors be Q₁,Q₂, and Q₂ respectively. According to the law of conservation of total charge, the sum of these three charges is equal to the charge Q transferred by the battery,
Q = Q₁ + Q₂ + Q₃ ….. (1)
Now, since Q = CV, we have
Q = C₁V + C₂ V + C₃ V ….. (2)
If these three capacitors are considered to form a single capacitance C_P which stores the total charge Q as shown in figure (b), then we can write Q = CPV. Substituting this in equation (2), we get
C_p V = C₁ V + C₂ V + C₃ V
C_p = C₁ + C₂ + C₃
Thus, the equivalent capacitance of capacitors connected in parallel is equal to the sum of the individual capacitance. The equivalent capacitance C_p in a parallel connection is always greater than the largest individual capacitance. In a parallel connection, it is equivalent as an area of each capacitance adds to give a more effective area such that total capacitance increases.

Question 22.
Explain in detail how charges are distributed in a conductor, and the principle behind the lightning conductor.
Answer:
Distribution of charges in a conductor: Consider two conducting spheres A and B of radii r₁ and r₂ respectively connected to each other by a thin conducting wire as shown in the figure. The distance between the spheres is much greater than the radii of either sphere.

If a charge Q is introduced into any one of the spheres, this charge Q is redistributed into both the spheres such that the electrostatic potential is same in both the spheres. They are now uniformly charged and attain electrostatic equilibrium. Let q₁ be the charge residing on the surface of sphere A and q₂ is the charge residing on the surface of sphere B such that Q = q₁ + q₂ The charges are distributed only on the surface and there is no net charge inside the conductor. The electrostatic potential at the surface of sphere A is given by
V_A = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac {{ q }_{ 2 }}{{ r }_{ 2 }}\) …. (1)
The electrostatic potential at the surface of sphere B is given by
V_B = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac {{ q }_{ 2 }}{{ r }_{ 2 }}\) ….. 2)
The surface of the conductor is an equipotential. Since the spheres are connected by the conducting wire, the surfaces of both the spheres together form an equipotential surface. This implies that
V_A = V_B or \(\frac {{ q }_{ 1 }}{{ r }_{ 1 }}\) = \(\frac {{ q }_{ 2 }}{{ r }_{ 2 }}\) ….. (3)
Let us take the charge density on the surface of sphere A is σ₁ and charge density on the surface of sphere B is σ₁. This implies that q₁ = \({ 4\pi r }_{ 1 }^{ 2 }\)σ₁ and q₁ = \({ 4\pi r }_{ 1 }^{ 2 }\)σ₂. Substituting these values into equation (3), we get
σ₁ r₁ = σ₂r₂ ….. (4)
from which we conclude that
σr = constant …. (5)
Thus the surface charge density o is inversely proportional to the radius of the sphere. For a smaller radius, the charge density will be larger and vice versa.

Lightning arrester or lightning conductor:
This is a device used to protect tall buildings from lightning strikes. It works on the principle of action at points or corona discharge. The device consists of a long thick copper rod passing from top of the building to the ground. The upper end of the rod has a sharp spike or a sharp needle.

The lower end of the rod is connected to the copper plate which is buried deep into the ground. When a negatively charged cloud is passing above the building, it induces a positive charge on the spike. Since the induced charge density on thin sharp spike is large, it results in a corona discharge.

This positive charge ionizes the surrounding air which in turn neutralizes the negative charge in the cloud. The negative charge pushed to the spikes passes through the copper rod and is safely diverted to the Earth. The lightning arrester does not stop the lightning; rather it divers the lightning to the ground safely.

Question 23.
Explain in detail the construction and working of a Van de Graaff generator.
Answer:
Principle: Electrostatic induction and action at points.
Construction:
A large hollow spherical conductor is fixed on the insulating stand. A pulley B is mounted at the center of the hollow sphere and another pulley C is fixed at the bottom. A belt made up of insulating materials as silk or rubber runs over both pulleys. The pulley C is driven continuously by the electric motor.

Two comb-shaped metallic conductors E and D are fixed near the pulleys. The comb D is maintained at a positive potential of 104 V by a power supply. The upper comb E is connected to the inner side of the hollow metal sphere.

Working:
Due to the high electric field near comb D, the air between the belt and comb D gets ionized. The positive charges are pushed towards the belt and negative charges are attracted towards the comb D. The positive charges stick to the belt and move up. When the positive charges reach the comb E, a large amount of negative and positive charges are induced on either side of comb E due to electrostatic induction. As a result, the positive charges are pushed away from the comb E and they reach the outer surface of the sphere. Since the sphere is a conductor, the positive charges are distributed uniformly on the outer surface of the hollow sphere. At the same time, the negative charges nullify the positive charges in the belt due to corona discharge before it passes over the pulley.

When the belt descends, it has almost no net charge. At the bottom, it again gains a large positive charge. The belt goes up and delivers positive charges to the outer surface of the sphere. This process continues until the outer surface produces the potential difference of the order of 10⁷ which is the limiting value. We cannot store charges beyond this limit since the extra charge starts leaking to the surroundings due to the ionization of air. The leakage of charges can be reduced by enclosing the machine in a gas-filled steel chamber at very high pressure. Uses: The high voltage produced in this Van de Graaff generator is used to accelerate positive ions (protons and deuterons) for nuclear disintegrations and other applications.

Samacheer Kalvi 12th Physics Electrostatics Numerical Problems

Question 1.
When two objects are rubbed with each other, approximately a charge of 50 nC can be produced in each object. Calculate the number of electrons that must be transferred to produce this charge.
Solution:
Charge produced in each object q = 50 nC
q = 50 x 10^-9 C
Charge of electron (e) = 1.6 x 10^-9 C
Number of electron transferred, n = \(\frac { q }{ e }\) = \(\frac {{ 50 × 10 }^{-9}}{{ 1.6 × 10 }^{-19}}\)
=31. 25 × 10^-9 × 10¹⁹
n = 31.25 x 10¹⁰ electrons
Ans. n = 31.25 x 10¹⁰ electrons

Question 2.
The total number of electrons in the human body is typically in the order of 10²⁸. Suppose, due to some reason, you and your friend lost 1% of this number of electrons. Calculate the electrostatic force between you and your friend separated at a distance of 1 m. Compare this with your weight. Assume the mass of each person is 60 kg and use point charge approximation.
Solution:
Number of electrons in the human body = 10²⁸
Number of electrons in me and my friend after lost of 1% = 10²⁸ x 1%
= 10²⁸ x \(\frac { 1 }{ 100 }\)
n = 10²⁶ electrons
Separation distance d = 1 m
Charge of each person q = 10²⁶ x 1.6 x 10^-19
q = 1.6 x 10⁷ C
Electrostatic force, F = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{q_{1} q_{2}}{r^{2}}\) = \(\frac{9 \times 10^{9} \times 1.6 \times 10^{7} \times 1.6 \times 10^{7}}{1^{2}}\)
F = 2.304 x 10²⁴N
Mass of the person, M = 60 kg
Acceleration due to gravity, g = 9.8 ms^-2
Weight (W) = mg
= 60 x 9.8
W = 588 N
Comparison: Electrostatic force is equal to 3.92 x 10²¹ times of weight of the person.

Question 3.
Five identical charges Q are placed equidistant on a semicircle as shown in the figure. Another point charge q is kept at the center of the circle of radius R. Calculate the electrostatic force experienced by the charge q.

Solution:
Force acting on q due to Q₁ and Q₅ are opposite direction, so cancel to each other.
Force acting on q due to Q₃ is F₃ = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac {{ qQ }_{3}}{{ R }^{2}}\)
Force acting on q due to Q₂ and Q₄

Resolving in two-component method:
(i) Vertical Component:
Q₂ Sin θ and Q₄ Sinθ are equal and opposite directions, so they cancel to each other.

(ii) Horizontal Component:
Q₂ Sin θ and Q₄ cos θ are equal and same direction, so they can get added.
F₂₄ = F_2q + F_4q = F₂ cos 55° + F₄ cos 45°
F₂₄ = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac {{ qQ }_{2}}{{ R }^{2}}\) cos 45° + \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { qQ 4}{{ R }^{2}}\) cos 45°
Resultant net force F

Question 4.
Suppose a charge +q on Earth’s surface and another +q charge is placed on the surface of the Moon, (a) Calculate the value of q required to balance the gravitational attraction between Earth and Moon (b) Suppose the distance between the Moon and Earth is halved, would the charge q change? (Take m_E = 5.9 x 10²⁴ kg, m_M = 7.348 x 10²² kg)
Solution:
Mass of the Earth, M_E = 5.9 x 10²⁴ kg
Mass of the Moon, M_M = 7.348 x 10²² kg
Charge placed on the surface of Earth and Moon = q
(a) Required charge to balance the F_G between Earth and Moon
F_C = F_G (or) \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac {{ q }^{2}}{{ r }^{2}}\) = \(\frac{\mathrm{G} \mathrm{M}_{\mathrm{E}} \times \mathrm{M}_{\mathrm{M}}}{r^{2}}\)
q² = G × M_E × M_M × 4πε₀ = 320.97 × 10²⁵
q = \(\sqrt { 320.97\times { 10 }^{ 25 } } \) = 5.665 x 10¹³ = 5.67 x 10¹³ C

(b) The distance between the Moon and Earth is

so q = 5.67 x 10¹³ C
There is no change.

Question 5.
Draw the free body diagram for the following charges as shown in figure (a), (b) and (c).

Solution:

Question 6.
Consider an electron travelling with a speed V_Φ and entering into a uniform electric field \(\vec { E } \) which is perpendicular to \(\overrightarrow{\mathrm{V}_{0}}\) as shown in the Figure. Ignoring gravity, obtain the electron’s acceleration, velocity and position as functions of time.
Solution:

Speed of an electron = V₀
Uniform electric field = \(\vec { E } \)
(а) Electron’s acceleration:
Force on electron due to uniform electric field, F = Ee
Downward acceleration of electron due to electric field, a = \(\frac { F }{ m }\) = – \(\frac { eE }{ M }\)
Vector from, \(\vec { a } \) = – \(\frac { eE }{ M }\) \(\hat{j} \)

(b) Electron’s velocity:
Speed of electron in horizontal direction, u = V₀ From the equation of motion, V = u + at
V = V₀ \(\frac { eE }{ M }\) t
Vector from \(\vec { V } \) = V₀ \(\hat{j} \) – \(\frac { eE }{ M }\) t \(\hat{j} \)

(c) Electron’s position:
Position of electron, s = r
From equation of motion, r = V₀ t + \(\frac { 1 }{ 2 }\) \(\left(-\frac{e \mathrm{E}}{\mathrm{M}}\right)\) t²
r = V₀ t + \(\frac { 1 }{ 2 }\) \(\frac { eE }{ M }\) t² \(\hat{j} \)
Vector from,
\(\vec { r } \) = V₀ t \(\hat{j} \) \(\frac { 1 }{ 2 }\) \(\frac { eE }{ M }\) t² \(\hat{j} \)

Question 7.
A closed triangular box is kept in an electric field of magnitude E = 2 × 10³ N C^-1 as shown in the figure.

Calculate the electric flux through the
(a) vertical rectangular surface
(b) slanted surface and
(c) entire surface.
Answer:
Electric field of magnitude E = 2 × 10³ NC^-1
(a) Vertical rectangular surface:
Rectangular area A= 5 × 10^-2 × 15 × 10^-2
A = 75 × 10^-24 m²
θ = 180°
⇒ cos 180° = -1

Electric flux, Φ_v.s = EA cos θ
= 2 × 10³ × 75 × 10^-4 × cos 180°
= -150 × 10^-1
Φ_v.s = -15 Nm² C^-1

(b) Slanted surface:
cos θ = cos 60° = 0.5
sin θ = sin 30° = \(\frac { Opposite }{ hyp }\)

hyp = \(\frac {{ 5 × 10 }^{2}}{ 0.5 }\)
hyp = 0.1m
Area of slanted surface A₂ = (0.1 × 15 × 10^-2)
A₂ = 0.015 M²
Electric flux, Φ_v.s = EA = cos θ
= 2 × 10³ × 0.015 × cos 60°
= 2 × 10³ × 0.015 × 10³
= 0.015 × 10³
Φ_v.s = 15 Nm² C^-1
Horizontal surface
θ = 90° ; cos 90° = 0
Electric flux, Φ_H.S = E. A₃ Cos 90° = 0

(c) Entire surface:
Φ_Total = Φ_V.S + Φ_S.S + Φ_H.S = -15 + 15 + 0
Φ_Total = 0

Question 8.
The electrostatic potential is given as a function of x in figure (a) and (b). Calculate the corresponding electric fields in regions A, B, C and D. Plot the electric field as a function of x for the figure (b).

Answer:
The relation between electric field and potential
E = – \(\frac { dv }{ dx }\)

(a) Region A :
dv = -3V ; dx = 0.2 m
Electric field, E_A = \(\frac { (-3) }{ 0.2 }\) = 15 V m^-1
Region B:
dv = 0V ; dx = 0.2 m
Electric field, E_B = \(\frac { 0 }{ 0.2 }\) = 0
Region C:
dv = 2V ; dx = 0.2 m
Electric field, E_C = \(\frac { -2 }{ 0.2 }\) = 10 V m^-1
Region D:
dv = -6V ; dx = 0.2 m
Electric field, E_D = \(-\left(\frac{-6}{0.2}\right)\) = 10 V m^-1 = 30 V m^-1

Electric field, E_A = 15 V m^-1
Electric field, E_B = 0
Electric field, E_C = \(\frac { (-3) }{ 0.2 }\) = 10 V m^-1
Electric field, E_D = 30 V m^-1

(b)

Question 9.
A spark plug in a bike or a car is used to ignite the air-fuel mixture in the engine. It consists of two electrodes separated by a gap of around 0.6 mm gap as shown in the figure.

To create the spark, an electric field of magnitude 3 x 10⁶Vm^-1 is required, (a) What potential difference must be applied to produce the spark? (b) If the gap is increased, does the potential difference increase, decrease or remains the same? (c) find the potential difference if the gap is 1 mm.
Answer:
Separation gap between two electrodes, d = 0.6 mm
d = 0.6 × 10^-3 m
Magnetude of electric field Electric field = E = 3 × 10⁶ V m^-1
Electric field E = \(\frac { V }{ d }\)
(a) Applied potential difference, V = E . d
= 3 × 10⁶ × 0.6 10^-13 = 1.8 × 10³
V = 1800 V

(b) From equation, V = E . d
If the gap (distance) between the electrodes increased, the potential difference also increases.

(c) Gap between the electrodes, d = 1mm = 1 x 10^-3 m
Potential difference, V = E.d
= 3 × 10⁶ × 1 × 10^-3 = 3 × 103
V = 3000 V

Question 10.
A point charge of +10 μC is placed at a distance of 20 cm from another identical point charge of +10 μC. A point charge of -2 μC is moved from point a to b as shown in the figure. Calculate the
change in potential energy of the system? Interpret your result.

q₁ = 10μC = 10 x 10^-6 C
q₂ = 2μC = -2 x 10^-6 C
distance, r = 5cm = 5 x 10^-2 m
Answer:
Change in potential energy,

= -36 × 1 × 10⁹ × 10^-12 × 10² = -36 × 10^-1
∆ U = -3.6 J

Negative sign implies that to move the charge -2pC no external work is required. System spends its stored energy to move the charge from point a to point b.
Ans:
∆ U = -3.6 J, negative sign implies that to move the charge -2μC no external work is required. System spends its stored energy to move the charge from point a to point b.

Question 11.
Calculate the resultant capacitances for each of the following combinations of capacitors.

Answer:

Parallel combination of capacitor 1 and 2
C_p = C₀ + C₀ = 2C₀
Series combination of capacitor C_p and 3
\(\frac { 1 }{{ C }_{S}}\) = \(\frac { 1 }{{ C }_{p}}\) + \(\frac { 1 }{{ C }_{3}}\) = \(\frac { 1 }{{ 2C }_{0}}\) + \(\frac { 1 }{{ C }_{0}}\) = (or) \(\frac { 1 }{{ C }_{S}}\) = \(\frac { 3 }{ 2 }\) C₀  (or)C_S = \(\frac { 2 }{ 3 }\) C₀ 

\(\frac { 1 }{ { C }_{ { S }_{ 1 } } } \) = \(\frac { 1 }{{ C }_{1}}\) + \(\frac { 1 }{{ C }_{2}}\) = \(\frac { 1 }{{ C }_{0}}\) + \(\frac { 1 }{{ C }_{0}}\) = \(\frac { 1 }{{ C }_{0}}\) (or)
\(\frac { 1 }{ { C }_{ { S }_{ 1 } } } \) = \(\frac { 2 }{{ C }_{0}}\) (or) \({ C }_{ { S }_{ 1 } }\) = \(\frac {{ C }_{0}}{ 2 }\)
Similarly 3 and 4 are series combination
\(\frac { 1 }{ { C }_{ { S }_{ 2 } } } \) = \(\frac { 1 }{{ C }_{3}}\) + \(\frac { 1 }{{ C }_{4}}\) = \(\frac { 1 }{{ C }_{0}}\) + \(\frac { 1 }{{ C }_{0}}\) = \(\frac { 2 }{{ C }_{0}}\) (or) \({ C }_{ { S }_{ 2 } }\) = \(\frac {{ C }{0}}{ 2 }\)

\({ C }_{ { S }_{ 1 } }\) and \({ C }_{ { S }_{ 2 } }\) are in parallel combination
C_p = \({ C }_{ { S }_{ 1 } }\) + \({ C }_{ { S }_{ 2 } }\) = \(\frac {{ C }_{0}}{ 2 }\) + \(\frac {{ C }_{0}}{ 2 }\) (or) C_p = \(\frac {{ 2C }_{0}}{ 2 }\) C_p = C₀

(c) Capacitor 1, 2 and 3 are in parallel combination
C_p = C₀ + C₀ + C₀ = 3C₀
C_p = 3C₀

(d) Capacitar C₁ and C₂ are in combination

Similarly C₃ and C₄ are in series combination


\({ C }_{ { S }_{ 1 } }\) and \({ C }_{ { S }_{ 2 } }\) are in parallel combination across RS:

(e) Capacitor 1 and 2 are series combination

Similarly 3 and 4 are series combination
\(\frac { 1 }{ { C }_{ { S }_{ 2 } } } \) = \(\frac { 2 }{{ C }_{0}}\) (or) \({ C }_{ { S }_{ 2 } }\) = \(\frac {{ C }_{0}}{ 2 }\)
Three capacitors are in parallel combination

Question 12.
An electron and a proton are allowed to fall through the separation between the plates of a parallel plate capacitor of voltage 5 V and separation distance h = 1 mm as shown in the figure.

(a) Calculate the time of flight for both electron and proton
(b) Suppose if a neutron is allowed to fall, what is the time of flight?
(c) Among the three, which one will reach the bottom first?
(Take m_p = 1.6 x 10^-27 kg, m_e= 9.1 x 10^-31 kg and g = 10 m s^-2)
Answer:
Potential difference between the parallel plates V = 5 V
Separation distance, h = 1 mm =1 x 10^-3 m
Mass of proton, mp = 1.6 x 10^-27 kg
Mass of proton, m =9.1 x 10^-31 kg
Charge of an a proton (or) electron, e— 1.6 x 10^-19 C
[u = 0; s = h]
From equation of motion, S = ut + \(\frac { 1 }{ 2 }\) at²
From equation of motion, h = \(\frac { 1 }{ 2 }\) at²
t = \(\sqrt { \frac { 2h }{ a } } \)
Acceleration of an electron due to electric field, a = \(\frac { F }{ m }\) = \(\frac { eE }{ m }\)
[E = \(\frac { V }{ d }\)]

(a) Time of flight for both electron and proton,

t_p = 63 ns……. (2)

(b) time of flight of neutron t_n = \(\sqrt { \frac { 2h }{ g } } \) = \(\sqrt{\frac{2 \times 1 \times 10^{3}}{10}}\) = \(\sqrt{0.2 \times 10^{-3}}\)
t_n = 0.0141 s = 14.1 x 10^-3 s
t_n = 14.1 x 10^-3 ms ……. (3)
(c) Compairision of values 1,2 and 3. The electron will reach the bottom first.

Question 13.
During a thunderstorm, the movement of water molecules within the clouds creates friction, partially causing the bottom part of the clouds to become negatively charged. This implies that the bottom of the cloud and the ground act as a parallel plate capacitor. If the electric field between the cloud and ground exceeds the dielectric breakdown of the air (3 x 10⁶ Vm^-1 ), lightning will occur.

(a) If the bottom part of the cloud is 1000 m above the ground, determine the electric potential difference that exists between the cloud and ground.
(b) In a typical lightning phenomenon, around 25C of electrons are transferred from cloud to ground. How much electrostatic potential energy is transferred to the ground?
Answer:
(a) Electric field between the cloud and ground,
V = E.d
V= 3 x 10⁶ x 1000 = 3 x 10⁹V
(a) Electrons transfered from cloud to ground,
q = 25 C
Electron static potential energy,
U = \(\frac { 1 }{ 2 }\) CV²
[C = \(\frac { q }{ V }\)]
= \(\frac { 1 }{ 2 }\) qV = \(\frac { 1 }{ 2 }\) x 25 x 3 x 10⁹
U = 37.5 x 10⁹ J

Question 14.
For the given capacitor configuration
(a) Find the charges on each capacitor
(b) potential difference across them
(c) energy stored in each capacitor.

Answer:
Capacitor b and c in parallel combination
C_p = C_b + C_c = (6 + 2) μF = 8 μF
Capacitor a, c_p and d are in series combination, so the resulatant copacitance
\(\frac { 1 }{{ C }_{s}}\) = \(\frac { 1 }{{ C }_{a}}\) + \(\frac { 1 }{{ C }_{cp}}\) + \(\frac { 1 }{{ C }_{d}}\) = \(\frac { 1 }{ 8 }\) + \(\frac { 1 }{ 8 }\) + \(\frac { 1 }{ 8 }\) = \(\frac { 3 }{ 8 }\)
C_s  = \(\frac { 8 }{ 3 }\) μF

(a) Charge on each capacitor,
Charge on capacitor a, Q_a = C_s V = \(\frac { 8 }{ 3 }\) x 9
Q_a = 24 μC
Charge on capacitor, d, Q_d = C_s V = \(\frac { 8 }{ 3 }\) x 9
Q_d = 24 μC
Capacitor b and c in parallel
Charge on capacitor, b, Q_b = \(\frac { 6 }{ 3 }\) x 9 = 18
Q_b = 18 μC
Charge on capacitor, c, Q_c = \(\frac { 2 }{ 3 }\) x 9 = 6
Q_c = 6 μC

(b) Potential difference across each capacitor, V = \(\frac { q }{ C }\)
Capacitor C_a, V_a = \(\frac{ { q }_{a}}{{ C }_{a}}\) = \(\frac {{ 24 × 10 }^{6}}{{ 8 × 10 }^{6}}\) = 3 V
Capacitor C_b, V_b = \(\frac{ { q }_{b}}{{ C }_{b}}\) = \(\frac {{ 18 × 10 }^{6}}{{ 6 × 10 }^{6}}\) = 3 V
Capacitor C_c, V_c = \(\frac{ { q }_{c}}{{ C }_{c}}\) = \(\frac {{ 6 × 10 }^{6}}{{ 2 × 10 }^{6}}\) = 3 V
Capacitor C_d, V_d = \(\frac{ { q }_{d}}{{ C }_{d}}\) = \(\frac {{ 24 × 10 }^{6}}{{ 8 × 10 }^{6}}\) = 3 V

(c) Energy stores in a capacitor, U = \(\frac { 1 }{ 2 }\) CV²
Energy in capacitor C_a, U_a = \(\frac { 1 }{ 2 }\) C_a \({ V }_{ a }^{ 2 }\) = \(\frac { 1 }{ 2 }\) x 8 x 10^-6 x (3)²
U_a = 36 μJ
Capacitor C_b, U_b = \(\frac { 1 }{ 2 }\) C_b \({ V }_{ b }^{ 2 }\) = \(\frac { 1 }{ 2 }\) x 6 x 10^-6 x (3)²
U_a = 27 μJ
C_c, U_c = \(\frac { 1 }{ 2 }\) C_c \({ V }_{ c }^{ 2 }\) = \(\frac { 1 }{ 2 }\) x 2 x 10^-6 x (3)²
U_a = 9 μJ
C_d, U_d = \(\frac { 1 }{ 2 }\) C_d \({ V }_{ d }^{ 2 }\) = \(\frac { 1 }{ 2 }\) x 8 x 10^-6 x (3)²
U_a = 36 μJ

Question 15.
Capacitors P and Q have identical cross-sectional areas A and separation d. The space between the capacitors is filled with a dielectric of dielectric constant as shown in the figure. Calculate the capacitance of capacitors P and Q.

Answer:
Cross-sectional area of parallel plate capacitor = A
Each area of different medium between parallel plate capacitor = \(\frac { A }{ 2 }\)
Separation distance = d
Capacitance of parallel plate capacitor, C = \(\frac { εA }{ d }\)
Air medium of dielectric constant, ε_r = 1
dielectric medium of dielectric constant = ε_r

Case 1:
Capacitance of air filled capacitor

Capacitance of dielectric-filled capacitor

Capacitance of parallel plate capacitor

Case 2:
Each distance of different medium between the parallel plate capacitor = \(\frac { d }{ 2 }\)
Capacitance of dielectric-filled capacitor

Capacitance of air filled capacitor,

Capacitance of parallel plate capacitor,

Samacheer Kalvi 12th Physics Electrostatics Additional Questions Solved

I. Multiple Choice Questions

Question 1.
When a solid body is negatively charged by friction, it means that the body has
(a) acquired excess of electrons
(b) lost some, problems
(c) acquired some electrons and lost a lesser number of protons
(d) lost some positive ions
Answer:
(a) acquired excess of electrons

Question 2.
A force of 0.01 N is exerted on a charge of 1.2 x 10^-5 G at a certain point. The electric field at that point is
(a) 5.3 x 10⁴ NC^-1
(b) 8.3 x 10^-4 NC^-1
(c) 5.3 x 10² NC^-1
(d) 8.3 x 10⁴ NC^-1
Answer:
(d) 8.3 x 10⁴ NC^-1
Hint:
E = \(\frac { F }{ q }\) = \(\frac { 0.01 }{{ 1.2 × 10 }^{-5}}\) = 8.3 x 10² NC^-1

Question 3.
The electric field intensity at a point 20 cm away from a charge of 2 x 10 5 C is
(a) 4.5 x 10⁶ NC^-1
(b) 3.5 x 10⁵ NC^-1
(c) 3.5 x 10⁶ NC^-1
(d) 4.5 x 10⁵ NC^-1
Answer:
(a) 4.5 x 10⁶ NC^-1
Hint:
E = \(\frac{q}{4 \pi \varepsilon_{0} r^{2}}\) = \(\frac{9 \times 10^{9} \times 2 \times 10^{-5}}{(0.2)^{2}}\) = 4.5 x 10⁶ NC^-1

Question 4.
How many electrons will have a charge of one coulomb?
(a) 6.25 x 10¹⁸
(b) 6.25 x 10¹⁹
(c) 1.6 x 10¹⁸
(d) 1.6 x 10¹⁹
Answer:
(a) 6.25 x 10¹⁸
Hint:
Number of electron, n = \(\frac { q }{ e }\) = \(\frac { 1 }{{ 1.6 × 10 }^{-19}}\) = 6.25 × 10¹⁸

Question 5.
The ratio of the force between two charges in air and that in a medium of dielectric constant K is
(a) K : 1
(b) 1 : K
(c) K² : 1
(d) 1 : K²
Answer:
(a) K : 1

Question 6.
The work done in moving a positive charge on an equipotential surface is
(a) finite and positive
(b) infinite
(c) finite and negative
(d) zero
Answer:
(d) zero

Question 7.
If a charge is moved against the Coulomb force of an electric field.
(a) work is done by the electric field
(b) energy is used from some outside source
(c) the strength of the field is decreased
(d) the energy of the system is decreased
Answer:
(b) energy is used from some outside source

Question 8.
No current flows between two charged bodies when connected
(a) if they have the same capacitance
(b) if they have the same quantity of charge
(c) if they have the same potential
(d) if they have the same charge density
Answer:
(c) if they have the same potential

Question 9.
Electric field lines about a negative point charge are
(a) circular, anticlockwise
(b) circular, clockwise
(c) radial, inwards
(d) radial, outwards
Answer:
(c) radial, inwards

Question 10.
Two plates are 1 cm apart and the potential difference between them is 10 V. The electric field between the plates is
(a) 10 NC^-1
(b) 250 NC^-1
(c) 500 N^-1
(d) 1000 NC^-1
Answer:
(d) 1000 NC^-1
Hint:
E = \(\frac { V }{ d }\) = \(\frac { 10 }{{ 1 × 10 }^{-2}}\) = 8.3 x 10² NC^-1

Question 11.
At a large distance (r), the electric field due to a dipole varies as
(a) \(\frac { 1 }{ r }\)
(b) \(\frac { 1 }{{ r }^{2}}\)
(c) \(\frac { 1 }{{ r }^{3}}\)
(d) \(\frac { 1 }{{ r }^{4}}\)
Answer:
(c) \(\frac { 1 }{{ r }^{3}}\)

Question 12.
Two thin infinite parallel plates have uniform charge densities +c and -σ. The electric field in the space between then is
(a) \(\frac { σ }{{ 2ε }_{0}}\)
(b) \(\frac { σ }{{ ε }_{0}}\)
(c) \(\frac { 2σ }{{ 2ε }_{0}}\)
(d) Zero
Answer:
(b) \(\frac { σ }{{ ε }_{0}}\)

Question 13.
Two isolated, charged conducting spheres of radii R₁, and R₂ produce the same electric field near their surfaces. The ratio of electric potentials on their surfaces is-
(a) \(\frac {{ R }_{1}}{{ R }_{2}}\)
(b) \(\frac {{ R }_{2}}{{ R }_{1}}\)
(c) \(\frac { { R }_{ 1 }^{ 2 } }{ { R }_{ 2 }^{ 2 } } \)
(d) \(\frac { { R }_{ 2 }^{ 2 } }{ { R }_{ 1 }^{ 2 } } \)
Answer:
(b) \(\frac {{ R }_{2}}{{ R }_{1}}\)

Question 14.
A 100 μF capacitor is to have an energy content of 50 J in order to operator a flash lamp. The voltage required to charge the capacitor is
(a) 500 V
(b) 1000 V
(c) 1500 V
(d) 2000 V
Answer:
(b) 1000 V
Hint:

Question 15.
A 1 μF capacitor is placed in parallel with a 2 μF capacitor across a 100 V supply. The total charge on the system is
(a) \(\frac { 100 }{ 3 }\) μC
(b) 100 μC
(c) 150 μC
(d) 300 μC
Answer:
(d) 300 μC
Hint:
Equivalent capacitor = 1 + 2 = 3 μF
Total charge, q = CV = 3 x 100 = 300 μF

Question 16.
A parallel plate capacitor of capacitance 100 μF is charged to 500 V. The plate separation is then reduced to half its original value. Then the potential on the capacitor becomes
(a) 250 V
(b) 500 V
(c) 1000V
(d) 2000 V
Answer:
(a) 250 V
Hint:
Here, C’ = 2C, since the charge remains the same.
q = C’V’ = CV ⇒ V = \(\frac { CV }{ 2C }\) = \(\frac { 500 }{ 2 }\) = 250 V

Question 17.
A point charge q is placed at the midpoint of a cube of side L. The electric flux emerging from the cube is ‘
(a) \(\frac { q }{{ ε }_{0}}\)
(b) \(\frac { q }{{ 6Lε }_{0}}\)
(c) \(\frac { 6Lq }{{ ε }_{0}}\)
(d) zero
Answer:
(a) \(\frac { q }{{ ε }_{0}}\)

Question 18.
The capacitor C of a spherical conductor of radius R is proportional to
(a) R²
(b) R
(c) R^-1
(d) R⁰
Answer:
(b) R

Question 19.
Energy of a capacitor of capacitance C, when subjected to a potential V, is given by
(a) \(\frac { 1 }{ 2 }\) CV²
(b) \(\frac { 1 }{ 2 }\) C²V
(c) \(\frac { 1 }{ 2 }\) CV
(d) \(\frac { 1 }{ 2 }\) \(\frac { C }{ V }\)
Answer:
(a) \(\frac { 1 }{ 2 }\) CV²

Question 20.
The electric field due to a dipole at a distance r from its centre is proportional to
(a) \(\frac { 1 }{{ r }^{3/2}}\)
(b) \(\frac { 1 }{{ r }^{3}}\)
(c) \(\frac { 1 }{ r }\)
(d) \(\frac { 1 }{{ r }^{3}}\)
Answer:
(b) \(\frac { 1 }{{ r }^{3}}\)

Question 21.
A point charge q is rotating around a charge Q in a circle of radius r. The work done on it by the Coulomb force is
(a) 2πrq
(b) 2πQq
(c) \(\frac { Q }{{ 2ε }^{0}r}\)
(d) zero
Answer:
(d) zero

Question 22.
The workdone in rotating an electric dipole of moment P in an electric field E through an angle 0 from the direction of the field is
(a) pE (1 – cos θ)
(b) 2pE
(c) zero
(d) -pE cos θ
Answer:
(a) pE (1 – cos θ)
Hint:
W = pE(cos θ₀ – cos θ)
[θ₀ = cos 0, cos 0 = 1]
W = pE(1 – cos θ)

Question 23.
The capacitance of a parallel plate capacitor can be increased by
(a) increasing the distance between the plates
(b) increasing the thickness of the plates
(c) decreasing the thickness of the plates
(d) decreasing the distance between the plates
Answer:
(d) decreasing the distance between the plates

Question 24.
Two charges are placed in vacuum at a distance d apart. The force between them is F. If a medium of dielectric constant 2 is introduced between them, the force will now be
(a) 4F
(b) 2F
(c) F/2
(d) F/4
Answer:
(d) F/4

Question 25.
An electric charge is placed at the centre of a cube of side a. The electric flux through one of its faces will be
(a) \(\frac { q }{{ 6ε }^{0}}\)
(b) \(\frac { q }{ { ε }_{ 0 }{ a }^{ 2 } } \)
(c) \(\frac { q }{ { 4πε }_{ 0 }{ a }^{ 2 } } \)
(a) \(\frac { q }{{ ε }^{0}}\)
Answer:
(a) \(\frac { q }{{ 6ε }^{0}}\)
Hint:
According to Gauss’s law, the electric flux through the cube is \(\frac { q }{{ ε }^{0}}\). Since there are six faces, the flux through one face is \(\frac { q }{{ 6ε }^{0}}\).

Question 26.
The electric field in the region between two concentric charged spherical shells-
(a) is zero
(b) increases with distance from centre
(c) is constant
(d) decreases with distance from centre
Answer:
(d) decreases with distance from centre

Question 27.
A hollow metal sphere of radius 10 cm is charged such that the potential on its surface is 80V. The potential at the centre of the sphere is-
(a) 800 V
(b) zero
(c) 8 V
(d) 80 V
Answer:
(d) 80 V

Question 28.
A 4 μF capacitor is charged to 400 V and then its plates are joined through a resistance of 1 K Ω. The heat produced in the resistance is-
(a) 0.16 J
(b) 0.32 J
(c) 0.64 J
(d) 1.28 J
Answer:
(b) 0.32 J
Hint:
The energy stored in capacitor is converted into heat
U = H = \(\frac { 1 }{ 2 }\) CV² = \(\frac { 1 }{ 2 }\) x 4 x 10^-6 x (400)² = 0.32 J

Question 29.
The work done in carrying a charge Q, once round a circle of radius R with a charge Q₂ at the centre is-
(a) \(\frac{\mathrm{Q}_{1} \mathrm{Q}_{2}}{4 \pi \varepsilon_{0} \mathrm{R}^{2}}\)
(b) zero
(c) \(\frac{\mathrm{Q}_{1} \mathrm{Q}_{2}}{4 \pi \varepsilon_{0} \mathrm{R}}\)
(d) infinite
Answer:
(b) zero
Hint:
The electric field is conservative. Therefore, no work is done in moving a charge around a closed path in an electric field.

Question 30.
Two plates are 2 cm apart. If a potential difference of 10 V is applied between them. The electric field between the plates will be
(a) 20 NC^-1
(b) 500 NC^-1
(c) 5 NC^-1
(d) 250 NC^-1
Answer:
(b) 500 NC^-1
Hint:
\(\frac { V }{ d }\) = \(\frac { 10 }{{ 2 ×10 }^{-2}}\) 500 NC^-1

Question 31.
The capacitance of a parallel plate capacitor does not depend on
(a) area of the plates
(b) metal of the plates
(c) medium between the plates
(d) distance between the plates
Answer:
(b) metal of the plates

Question 32.
A capacitor of 50 μF is charged to 10 volts. Its energy in joules is
(a) 2.5 x 10^-3
(b) 5 x 10^-3
(c) 10 x 10^-4
(d) 2.5 x 10^-4
Answer:
(a) 2.5 x 10^-3
Hint:
U = \(\frac { 1 }{ 2 }\) CV² = \(\frac { 1 }{ 2 }\) x 50 x 10^-6 x (10)² = 2.5 x 10^-3 J

Question 33.
A cube of side b has a charge q at each of its vertices. The electric field due to this charge distribution at the centre of the cube is
(a) \(\frac { q }{{b}^{ 2 }}\)
(b) \(\frac { q }{{2b}^{ 2 }}\)
(c) \(\frac { 32q }{{b}^{ 2 }}\)
(d) zero
Answer:(d) zero
Hint:
There
is an equal charge at diagonally opposite comer. The fields due the these at the centre cancel out. Therefore, the net field at the centre is zero.

Question 34.
Total electric fulx coming out of a unit positive charge put in air is
(a) ε₀
(b) \({ \varepsilon }_{ 0 }^{ -1 }\)
(c) (4πε₀)^-1
(d) 4πε₀
Answer:
(b) \({ \varepsilon }_{ 0 }^{ -1 }\)

Question 35.
Electron volt (eV) is a unit of
(a) energy
(b) potential
(c) current
(d) charge
Answer:
(a) energy

Question 36.
A point Q lies on the perpendicular bisector of an electric dipole of dipole moment P. If the distance of Q from the dipole is r, then the electric field at Q is proportional to-
(a) p^-1 and r^-2
(b) p and r^-2
(c) p and r^-3
(d) p² and r^-3
Answer:
(c) p and r^-3

Question 37.
A hollow insulated conducting sphere is given a positive charge of 10 μC. What will be the electric field at the centre of the sphere is its radius is 2 metres?
(a) zero
(b) 8 μCm^-2
(c) 20 μCm^-2
(d) 5 μCm^-2
Answer:
(d) zero

Question 38.
A particle of charge q is placed at rest in a uniform electric field E and then released. The kinetic energy attained by the particle after moving a distance y is-
(a) qE²y
(b) q²Ey
(c) qEy²
(d) qEy
Answer:
(d) qEy
Hint:
Force on the particle = qE
KE = Work done by the force = F.y = qEy

Question 39.
Dielectric constant of metals is-
(a) 1
(b) greater then 1
(c) zero
(d) infinite
Answer:
(d) infinite

Question 40.
When a positively charged conductor is earth connected
(a) protons flow from the conductor to the earth
(b) electrons flow from the earth to the conductor
(c) electrons flow from the conductor to the earth
(d) no charge flow occurs
Answer:
(b) electrons flow from the earth to the conductor

Question 41.
The SI unit of electric flux is
(a) volt metre²
(b) newton per coulomb
(c) volt metre
(d) joule per coulomb
Answer:
(c) volt metre

Question 42.
Twenty seven water drops of the same size are charged to the same potential. If they are combined to form a big drop, the ratio of the potential of the big drop to that of a small drop is-
(a) 3
(b) 6
(c) 9
(d) 27
Answer:
(c) 9
Hint:
V’ = n^2/3 V
⇒ \(\frac { V’ }{ V }\) = (27)^2/3 = 9

Question 43.
A point charge +q is placed at the midpoint of a cube of side l. The electric flux emerging ’ from the cube is-
(a) \(\frac { q }{{ ε }^{0}}\)
(b) \(\frac {{ 6ql }^{2}}{{ ε }^{0}}\)
(c) \(\frac { q }{ { 6l }^{ 2 }{ { ε }^{ 0 } } } \)
(d) \(\frac { { C }^{ 2 }{ V }^{ 2 } }{ 2 } \)
Answer:
(a) \(\frac { q }{{ ε }^{0}}\)

Question 44.
The energy stored in a capacitor of capacitance C, having a potential difference V between the plates, is-
Answer:
(c)

Question 45.
The electric potential at the centre of a charged conductor is-
(a) zero
(b) twice that on the surface
(c) half that on the surface
(d) same as that on the surface
Answer:
(d) same as that on the surface

Question 46.
The energy stored in a capacitor is given by
(a) qV
(b) \(\frac { 1 }{ 2 }\)qV
(c) \(\frac { 1 }{ 2 }\) CV
(d) \(\frac { q }{ 2C }\)
Answer:
(b) \(\frac { 1 }{ 2 }\)qV

Question 47.
The unit of permitivity of free space so is
(a) coulomb/newton-metre
(b) newton-metre²/coulomb²
(c) coulomb²/newton-metre²
(d) coulomb/(newton-metre)²
Answer:
(c) coulomb²/newton-metre²

Question 48.
An electric dipole has the magnitude of its charge as q and its dipole moment is p. It is placed in a uniform electric field E. If its dipole moment is along the direction of the field, the force on it and its potential energy are, respectively.
(a) 2qE and minimum
(b) qE and pE
(c) zero and minimum
(d) qE and maximum
Answer:
(c) zero and minimum
Hint:
Potential energy, U = -pE cos θ
For q = 0°; U = -pE, which is minimum.

Question 49.
An electric dipole of moment \(\vec { P } \) is lying along a uniform electric field \(\vec { E } \) . The workdone in rotating the dipole by 90° is
(a) \(\frac { pE }{ 2 }\)
(b) 2pE
(c) pE
(d) √2pE
Answer:
(c) pE

Question 50.
A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates
(a) does not charge
(b) becomes zero
(c) increases
(d) decreases
Answer:
(c) increases

Question 51.
When air is replaced by a dielectric medium of constant K, the maximum force of attraction between two charges separated by a distance
(a) increases K times
(b) increases K^-1 times
(c) decreases K times
(d) remains constant
Answer:
(c) decreases K times

Question 52.
A comb runs through one’s dry hair attracts small bits of paper. This is due to the fact that
(a) comb is a good conductor
(b) paper is a good conductor
(c) the atoms in the paper get polarised by the charged comb
(d) the comb posseses magnetic properties
Answer:
(c) the atoms in the paper get polarised by the charged comb

Question 53.
Which of the following is not a property of equipotential surfaces?
(a) they do not cross each other
(b) they are concentric spheres for uniform electric field
(c) the rate of change of potential with distance on them is zero
(d) they can be imaginary spheres.
Answer:
(b) they are concentric spheres for uniform electric field

Question 54.
A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will be
(a) reduced to half
(b) doubled
(c) becomes 4 times
(d) remains the same
Answer:
(d) remains the same

Question 55.
If the electric field in a region is given by \(\vec { E } \) = 5\(\hat{j} \) + 4\(\hat{j} \) + 9\(\hat{k} \) , then the electric flux through a surface of area 20 units lying in the y-z plane will be-
(a) 20 units
(b) 80 units
(c) 100 units
(d) 180 units
Answer:
(c) 100 units
Hints:
The area vector \(\vec { A } \) = 20\(\hat{j} \); \(\vec { E } \) = (5\(\hat{j} \) + 4\(\hat{j} \) + 9\(\hat{k} \))
Flux (Φ) = \(\vec { E } \) – \(\vec { A } \) = 5 x 20 =100 units

Question 56.
A, B and C are three points in a uniform electric field. The electric potential is-

(a) maximum at A
(b) maximum at B
(c) maximum at B
(d) same at all the three points A, B, and C
Answer:
(b) maximum at B
Hint:
The potential decreases in the direction of the field. Therefore V_B > V_C>C_A.

Question 57.
A conducting sphere of radius R is give a charge Q. The electric potential and the electric field at the centre of the sphere are, respectively-
(a) zero, \(\frac { Q }{ { 4\pi ε }_{ 0 }{ R }^{ 2 } } \)
(b) \(\frac { Q }{ { 4\pi ε }_{ 0 }{ R } } \)
(c) \(\frac { Q }{ { 4\pi ε }_{ 0 }{ R } } \), zero
(d) zero,zero
Answer:
(c) \(\frac { Q }{ { 4\pi ε }_{ 0 }{ R } } \), zero.

II. Fill in the blanks

Question 1.
A dipole is placed in a uniform electric field with its axis parallel to the field. It experiences …………………
Answer:
neither a net force nor a torque

Question 2.
The unit of permittivity is…………………
Answer:
C²N^-1m^-2

Question 3.
The branch of physics which deals with static electric charges or charges at rest is …………………
Answer:
electrostatics

Question 4.
The charges in an electrostatic field are analogous to ………………… in a gravitational field.
Answer:
mass

Question 5.
The substances which acquire charges on rubbing are said to be …………………
Answer:
electrified

Question 6.
Electron means …………………
Answer:
amber

Question 7.
A glass rod rubbed with a silk cloth. Glass rod and silk cloth acquire…………………
Answer:
positive and negative charge respectively

Question 8.
When ebonite rod is rubbed with fur, ebonite rod and fur acquires …………………
Answer:
negative and positive charge respectively

Question 9.
………………… termed the classification of positive and negative charges.
Answer:
Franklin

Question 10.
Applications such as electrostatic point spraying and powder coating, are based on the property of ………………… between charged bodies.
Answer:
attraction and repulsion

Question 11.
Bodies which allow the charge to pass through them are called …………………
Answer:
conductor

Question 12.
Bodies which do not allow the charge to pass through them are called …………………
Answer:
insulators

Question 13.
The unit of electric charge is …………………
Answer:
coulomb

Question 14.
Total charge in an isolated system …………………
Answer:
remains a constant

Question 15.
The force between two charged bodies was studied by …………………

Answer:
coulomb

Question 16.
The unit of permittivity in free space (s0) is …………………
Answer:
C²N^-1m^-2

Question 17.
The value of s, for air or vacuum is …………………
Answer:1

Question 18.
Charges can neither be created nor be destroyed is the statement of the law of conservation of …………………
Answer:
charge

Question 19.
The space around the test charge, in which it experiences a force is known as field …………………
Answer:
electric

Question 20.
Electric field at a point is measured in terms of …………………
Answer:
electric field intensity

Question 21.
The unit of electric field intensity is …………………
Answer:
NC^-1.

Question 22.
The lines of force are far apart, when electric field E is …………………
Answer:
small

Question 23.
The lines of force are close together when electric field E is …………………
Answer:
large

Question 24.
Electric dipole moment …………………
Answer:
P = 2qd

Question 25.
Torque experienced by electric dipole is …………………
Answer:
x = PE sin θ

Question 26.
An electric dipole placed in a non-uniform electric field at an angle θ experiences …………………
Answer:
both torque and force

Question 27.
When thee dipole is aligned parallel to the field, its electric potential energy is …………………
Answer:
u = -PE

Question 28.
Change of potential with distance is known as …………………
Answer:
potential distance

Question 29.
The number of electric lines of force crossing through the given area is …………………
Answer:
electric flux

Question 30.
The process of isolating a certain region of space from the external field is called …………………
Answer:
electrostatic shielding

Question 31.
A capacitor is a device to store …………………
Answer:
charge

Question 32.
The charge density in maximum at …………………
Answer:
pointed

Question 33.
The principle made use of lightning arrestor is …………………
Answer:
action of points

Question 34.
Van de Graaff generator producers large electrostatic potential difference of the order of …………………
Answer:
10⁷ V

III. Match the following

Question 1.

Answer:
(i) → (d)
(ii) → (a)
(iii) → (b)
(iv) → (c)

Question 2.

Answer:
(i) → (c)
(ii) → (d)
(iii) → (a)
(iv) → (b)

Question 3.

Answer:
(i) → (b)
(ii) → (d)
(iii) → (a)
(iv) → (c)

Question 4.

Answer:
(i) → (b)
(ii) → (d)
(iii) → (a)
(iv) → (c)

IV. Assertion and reason type

(a) If both assertion and reason are true and the reason is the correct explanation of the assertion.
(b) If both assertion and reason are true but the reason is not correct explanation of the assertion.
(c) If the assertion is true but the reason is false.
(d) If the assertion and reason both are false.
(e) If the assertion is false but the reason is true.

Question 1.
Assertion: Electric lines of force cross each other.
Reason: Electric field at a point superimposed to give one resultant electric field.
Answer:
(e) Both assertion and reason are true but the reason is not the correct explanation of the assertion.
Explanation: If electric lines of forces cross each other, then the electric field at the point of intersection will have two directions simultaneously which is not possible physically.

Question 2.
Assertion: Charge is quantized.
Reason: Charge, which is less than 1 C is not possible.
Answer:
(c) If assertion is true but reason is false.
Explanation: Q = ±ne and charge lesser than 1 C is possible.

Question 3.
Assertion:
A point charge is brought in an electric field. The field at a nearby point will increase, whatever be the nature of the charge.
Reason: The electric field is independent of the nature of the charge.
(d) If the assertion and reason both are false.
Explanation: Electric field at the nearby-point will be resultant of the existing field and field due to the charge brought. It may increase or decrease if the charge is positive or negative depending on the position of the point with respect to the charge brought.

Question 4.
Assertion: The tyre’s of aircraft are slightly conducting.
Reason: If a conductor is connected to the ground, the extra charge induced on the conductor will flow to the ground.
Answer:
(b) Both assertion and reason are true but the reason is not the correct explanation of the assertion.
Explanation: During take-off and landing, the friction between treys and the runway may cause electrification of treys. Due to conducting to a ground and election sparking is avoided.

Question 5.
Assertion: The lightning conductor at the top of a high building has sharp ends.
Reason: The surface density of charge at sharp points is very high, resulting in the setting up of electric wind.
Answer:
(a) Both assertion and reason are true and the reason is the correct explanation of the assertion.

Samacheer Kalvi 12th Physics Electrostatics Short Answer Questions

Question 1.
What is meant by triboelectric charging?
Answer:
Charging the objects through rubbing is called triboelectric charging.

Question 2.
What is meant by the conservation of total charges?
Answer:
The total electric charge in the universe is constant and the charge can neither be created nor be destroyed. In any physical process, the net change in charge will always be zero.

Question 3.
State Gauss’s Law?
Answer:
Definition:
Gauss’s law states that if a charge Q is enclosed by an arbitrary closed surface, then the total electric flux OE through the closed surface is
Φ_E = \(\oint { \vec { E } } \) .d\(\vec { A } \) = \(\frac {{ q }_{encl}}{{ ε }_{0}}\)

Question 4.
What is meant by electrostatic shielding?
During lightning accompanied by a thunderstorm, it is always safer to sit inside a bus than in open ground or under a tree. The metal body of the bus provides electrostatic shielding, since the electric field inside is zero. During lightning, the charges flow through the body of the conductor to the ground with no effect on the person inside that bus.

Question 5.
What is meant by dielectric?
Answer:
A dielectric is a non-conducting material and has no free electrons. The electrons in a dielectric are bound within the atoms. Ebonite, glass and mica are some examples of dielectrics.

Question 6.
What are non-polar molecules? Give examples.
A non-polar molecule is one in which centers of positive and negative charges coincide. As a result, it has no permanent dipole moment. Examples of non-polar molecules are hydrogen (H₂), oxygen (O₂), and carbon dioxide (CO₂) etc.

Question 7.
What are polar molecules? Give examples.
Answer:
In polar molecules, the centers of the positive and negative charges are separated even in the absence of an external electric field. They have a permanent dipole moment.
The net dipole moment is zero in the absence of an external electric field. Examples of polar molecules are H₂O, N₂O, HCl, NH₃.

Question 8.
What is a capacitor?
Answer:
A capacitor is a device used to store electric charge and electrical energy. Capacitors are widely used in many electronic circuits and have applications in many areas of science and technology.

Samacheer Kalvi 12th Physics Electrostatics Long Answer Questions

Question 1.
Derive an expression for the electric field due to the system of point charges?
Answer:
Electric field due to the system of point charges:
Suppose a number of point charges are distributed in space. To find the electric field at some point P due to this collection of point charges, the superposition principle is used. The electric field at an arbitrary point due to a collection of point charges is simply equal to the vector sum of the electric fields created by the individual point charges. This is called the superposition of electric fields.
Consider a collection of point charges q₁, q₂, q₃,…., q_n located at various points in space. The ‘ total electric field at some point P due to all these n charges is given by

Here r_1p, r_2p, r_3p,…., r_np, are the distance of the charges ₁, q₂, q₃,…., q_n from the point respectively. Also \(\hat{r} \)_1p + \(\hat{r} \)_2p + \(\hat{r} \)_3p,…., \(\hat{r} \)_np are the corresponding unit vectors directed from q₁, q₂, q₃,…., q_n tpo P.

Equation (2) can be re-written as,

For example in figure, the resultant electric field due to three point charges q₁, q₂, q₃ at point P is shown. Note that the relative lengths of the electric field vectors for the charges depend on relative distantes of the charges to the point P.

Question 2.
Derive an expression for the electric flux of rectangular area placed in a uniform electric field.
Answer:
(i) Electric flux for uniform Electric field:
Consider a uniform electric field in a region of space. Let us choose an area A normal to the electric field lines as shown in figure (a). The electric flux for this case is
Φ_E = EA ….. (1)
Suppose the same area A is kept parallel to the uniform electric field, then no electric field lines pierce through the area A, as shown in figure (b). The electric flux for this case is zero.
Φ_E = 0 ….. (2)
If the area is inclined at an angle θ with the field, then the component of the electric field perpendicular to the area alone contributes to the electric flux. The electric field component parallel to the surface area will not contribute to the electric flux. This is shown in figure (c). For this case, the electric flux
ΦE = (E cosθ) A …(3)
Further, θ is also the angle between the electric field and the direction normal to the area. Hence in general, for uniform electric field, the electric flux is defined as
Φ_E= \(\vec { E } \).\(\vec { A } \) = EA cos θ …(4)

Here, note that \(\vec { A } \) is the area vector \(\vec { A } \) = A\(\hat{n} \). Its magnitude is simply the area A and the direction is along the unit vector h perpendicular to the area. Using this definition for flux, Φ_E= \(\vec { E } \).\(\vec { A } \), equations (2) and (3) can be obtained as special cases.
In figure (a), θ = 0° so Φ_E= \(\vec { E } \).\(\vec { A } \) = EA
In figure (b), θ = 90° so Φ_E= \(\vec { E } \).\(\vec { A } \) = 0

(ii) Electric flux in a non uniform electric field and an arbitrarily shaped area: Suppose the electric field’is not uniform and the area A is not flat, then the entire area is divided
into n small area segments ∆\(\vec { A } \)₁ ∆\(\vec { A } \)₂, ∆\(\vec { A } \)₃,…..∆\(\vec { A } \)_n, such that each area element is almost flat and the electric field over each area element is considered to be uniform.
The electric flux for the entire area A is approximately written as

By taking the limit ∆\(\vec { A } \)₁ → 0 (for all i) the summation in equation (5) becomes integration. The total electric flux for the entire area is given by
Φ_E = ∫\(\vec { E } \).d\(\vec { A } \) ….. (6)
From Equation (6), it is clear that the electric flux for a given surface depends on both the electric field pattern on the surface area and the orientation of the surface with respect to the electric field.

(iii) Electric flux for closed surfaces: In the previous section, the electric flux for any arbitrary curved surface is discussed. Suppose a closed surface is present in the region of the non-uniform electric field as shown in figure (a).

The total electric flux over this closed surface is written as
Φ_E = \(\oint { \vec { E } } \).d\(\vec { A } \) …… (7)
Note the difference between equations (6) and (7). The integration in equation (7) is a closed surface integration and for each areal element, the outward normal is the direction of d\(\vec { A } \) as shown in figure (b).
The total electric flux over a closed surface can be negative,
positive or zero. In figure (b), it is shown that in one area element, the angle between d\(\vec { A } \) and \(\vec { E } \) is less than 90°, then the electric flux is positive and in another areal element, the angle between dA and E is greater than 90°, then the electric flux is negative. In general, the electric flux is negative if the electric field lines enter the closed surface and positive if the electric field lines leave the closed surface.

Samacheer Kalvi 12th Physics Electrostatics Numerical Problems

Question 1.
Electrons are caused to fall through a potential difference of 1500 volts. If they were initially at rest. Then calculate their final speed.
Solution:
The electrical potential energy is converted into kinetic energy. If v is the final speed then

Question 2.
Small mercury drops of the same size are charged to the same potential V. If n such drops coalesce to form a single large drop, then calculate its potential.
Solution:
Let r be the radius of a small drop and R that of the large drop. Then, since the volume remains conserved,
\(\frac { 1 }{ 2 }\) πR² = \(\frac { 4 }{ 3 }\) πR³n
⇒ R³ = r³n
R = r³(n)^1/3
Further, since the total charge remains conserved, we have, using Q = CV
C_large V = n C_small v
Where V is the potential of the large drop.
4πε₀ RV = n (4πε₀r)v
V = \(\frac { nrv }{ R }\) = \(\frac { nrv }{{ r(n) }^{1/3}}\)
V = vn^2/3

Question 3.
Two particles having charges Q₁ and Q₂ when kept at a certain distance, exert a force F on each other. If the distance between the two particles is reduced to half and the charge on each particle is doubled. Find the force between the particles.
Solution:
F = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{\mathrm{Q}_{1} \mathrm{Q}_{2}}{r^{2}}\)
If the distance is educed by half and two particles of charges are doubled.

Question 4.
Two charged spheres, separated by a distance d, exert a force F on each other. If they are immersed in a liquid of dielectric constant 2, then what is the force.
Solution:
Force between the charges (vacuum)

Force between the charges (medium)

Question 5.
Find the force of attraction between the plates of a parallel plate capacitor.
Solution:
Let d be the distance between the plates. Then the capacitor is
C = \(\frac { { \varepsilon }_{ 0 }A }{ d } \)
Energy stored in a capacitor,

Energy magnitude of the force is,

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Tamilnadu Samacheer Kalvi 12th Commerce Solutions Chapter 17 Consumer Protection

Samacheer Kalvi 12th Commerce Consumer Protection Textbook Exercise Questions and Answers

I. Choose the Correct Answer

Question 1.
The final aim of modem marketing is _________
(a) Maximum profit
(b) Minimum profit
(c) Consumer satisfaction
(d) Service to the society
Answer:
(c) Consumer satisfaction

Question 2
_________  is the king of modem marketing.
(a) Consumer
(b) Wholesaler
(c) Producer
(d) Retailer
Answer:
(a) Consumer

Question 3.
As the consumer is having the rights, they are also having _________
(a) Measures
(b) Promotion
(c) Responsibilities
(d) Duties
Answer:
(c) Responsibilities

Question 4.
Which of the following is not a consumer right summed up by John F. Kennedy?
(a) Right to safety
(b) Right to choose
(c) Right to consume
(d) Right to be informed
Answer:
(a) Right to safety

Question 5.
It is the responsibility of a consumer that he must obtain _________  as a proof for the purchase of goods.
(a) Cash receipt
(b) Warranty card
(c) Invoice
(d) All of these
Answer:
(c) Invoice

II. Very Short Answer Questions

Question 1.
Write short notes on “Right to be informed.”
Answer:
Consumers should be given all the relevant facts about the product. This implies that manufacturer and the dealer are expected to disclose all the material facts relevant and relating to the product.

Question 2.
What do you understand about “Right to Safety”?
Answer:
There may be few products that are more likely to cause physical danger to consumers health, lives and property. The health hazards which are likely to arise have to be eradicated or reduced altogether. In case of food items and drugs both life saving and life sustaining safety is to be guaranteed.

Question 3.
What are the rights of consumer according to John F. Kennedy?
Answer:
The former president of U.S.A Mr. John F. Kennedy defined the basic consumer rights as “The Right of Safety“ the Right to be informed, the Right to choose and the Right to be heard.”

Question 4.
Which is the supreme objective of business?
Answer:
The modem marketing concept recognises that the consumer is the pivotal point around which the business moves. Satisfaction of consumer needs / requirements is stated to be supreme objective of a business.

Question 5.
What are the important aspects to be kept in mind by consumer while purchasing goods related to the quality of goods?
Answer:
The consumer has to have the knowledge about the quality from his own experiences or from the experiences of other persons who used the product or by browsing the website.

III. Short Answer Questions

Question 1.
What do you understand by “Right to redressal”?
Answer:
The complaints and protests are not just to be heard: but the aggrieved party is to be granted compensation within a reasonable time period . There should be prompt settlement of complaints and claims lodged by the aggrieved customers.

Question 2.
Define “Consumer Rights”.
Answer:
Consumer Right is interpreted as “the right to have information about the quality, potency, quantity, purity, price, and standard of goods or services”.

Question 3.
What do you understand about “Right to protection of health and safety”?
Answer:
A few products may contain potentially harmful substances which are dangerous from the consumer welfare point of view. The best examples of this kind are food additives, colours, emulsifiers, preservatives. In case of food items and drugs both life saving and life sustaining safety is to be guaranteed.

IV. Long Answer Questions

Question 1.
What are the rights of consumers?
Answer:
As a consumer, everyone should know the basic rights as well as about the counts and procedures to be followed.
The rights of consumers as per Consumer Protection Act are given below:

1. Right to Protection of Health and Right of Safety: There may be products that cause physical danger to consumers health, lives and property. The health hazards which are likely to arise have to be eradicated or reduced altogether.
2. Right to be Informed: Consumers should be given all the relevant facts about the products. The manufacturer and the dealer should disclose all the material facts relating to the product.
3. Right to choose: Consumer satisfaction can be increased by giving the consumer the widest choice. From the widest range of products in quality and brand as well as price, the consumer can choose the goods.
4. Right to be Heard: Consumers have every right to ventilate and register the dissatisfaction, disagreements and get the complaint heard and aired.
5. Right to Seek Redressal: The aggrieved party is to be granted compensation within a reasonable time.
6. Right to Consumer Education: The consumer has a right to acquire knowledge and stay well-informed all through his life.

Question 2.
Explain the duties of consumers.
Answer:
Apart from rights, there are certain duties imposed on the consumer. The following are the duties of consumers

1. Buying Quality Products at Reasonable Price: It is the duty of a consumer to purchase a product after gaining a thorough knowledge of its price, quality and other terms and conditions.
2. Ensure the Weights and Measurement before Purchase: The consumer should ensure that he/she is getting the product of exact weight and measure.
3. Reading the Label Carefully: It is the duty of the consumer to read the label of the product thoroughly.
4. Beware of False and Attractive Advertisements: It is the prime duty of the consumer about the genuineness of the advertisement, before purchasing the product.
5. Ensuring the Receipt of Cash Bill: It is a legitimate duty of consumers to get the cash receipt and warranty card supplied along with the bill.

Question 3.
What are the responsibilities of consumers?
Answer:
Rights and responsibilities are two sides of the same coin.
The responsibilities of consumers are listed below:

1. The consumer must pay the price of the goods according to the terms and conditions.
2. The consumer has the responsibility to make the seller to deliver the goods in time.
3. The consumer has to bear any loss, which may arise to the seller, due to delay in taking delivery.
4. The consumer has to follow the instructions and precautions while using the product.
5. The consumer must collect the cash receipt as a proof of goods purchased from the seller.
6. The consumer must file a complaint with the seller about the defects in products or deficiency in service.

Samacheer Kalvi 12th Commerce Consumer Protection Additional Questions and Answers

I. Choose the Correct Answer

Question 1.
The consumer is to be protected against any __________
(a) unfair practices of trade
(b) family functions
(c) profit making firm
(d) loss in business
Answer:
(a) unfair practices of trade

Question 2.
The consumer is the __________ of the modem marketing.
(a) Manager
(b) Director
(c) King
(d) None of these
Answer:
(c) King

II. Very Short Answer Questions

Question 1.
Write a note on Right to consumer education.
Answer:
The consumer has a right to acquire knowledge and stay well-informed all through his life. He should be aware of the availability of the products.

Question 2.
How the consumer got the basic needs, as per right of the consumer?
Answer:
Every consumer has a right to get basic necessities of life such as food, clothing and water, and right to pine and healthy environment. It is the latest addition to consumer bill of rights.

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Tamilnadu Samacheer Kalvi 12th Commerce Solutions Chapter 16 Consumerism

Samacheer Kalvi 12th Commerce Consumerism Textbook Exercise Questions and Answers

I. Choose the Correct Answer

Question 1.
The term ‘consumerism’ came into existence in the year ________
(a) 1960
(b) 1957
(c) 1954
(d) 1958
Answer:
(a) 1960

Question 2.
Who is the father of Consumer Movement?
(a) Mahatma Gandhi
(b) Mr. John F. Kennedy
(c) Ralph Nader
(d) Jawaharlal Nehru
Answer:
(c) Ralph Nader

Question 3.
Sale of Goods Act was passed in the year?
(a) 1962
(b) 1972
(c) 1982
(d) 1985
Answer:
(c) 1982

Question 4.
The main objective of all business enterprises is ________
(a) Providing service
(b) Providing better standard of life
(c) Providing necessities to the society
(d) Earn profit
Answer:
(d) Earn profit

Question 5.
The Consumer Protection Act came into force with effect from ________
(a) 1.1.1986
(6) 1.4.1986
(c) 15.4.1987
(d) 15.4.1990
Answer:
(c) 15.4.1987

Question 6.
________ of every year is declared as a Consumer Protection Day to educate the public about their rights and responsibilities.
(a) August 15
(b) April 15
(c) March 15
(d) September 15
Answer:
(c) March 15

Question 7.
Any person who buys any goods or avails services for personal use, for a consideration is called as ________
(a) Customer
(b) Consumer
(c) Buyer
(d) User
Answer:
(b) Consumer

Question 8.
The General Assembly of United Nations passed resolution of consumer protection guide lines on ________
(a) 1985
(b) 1958
(c) 1986
(d) 1988
Answer:
(a) 1985

II. Very Short Answer Questions

Question 1.
Who is a consumer?
Answer:
A consumer is one who consumes goods manufactured and sold by others or created (air, water, natural resources) by nature and sold by others.

Question 2.
Define Consumerism.
Answer:
“Consumerism is an attempt to enhance the rights and powers by buyers in relation to sellers”

Question 3.
Give two examples of adulteration.
Answer:

1. Powdered rice / wheat is adulterated with starch.
2. Coffee powder is adulterated with tamarind seed.

Question 4.
What is Caveat Emptor?
Answer:
‘Caveat emptor’ is a Latin term that means “let the buyer beware.” Similar to the phrase”sold as is” this term means that the buyer assumes the risk that a product fails to meet expectations or have defects.

Question 5.
What is Caveat Venditor?
Answer:
Today, most sales in the U.S. fall under the principle of caveat venditor, which means “let the seller beware” by which goods are covered by an implied warranty of merchantability.

Question 6.
Write a short notes on Consumer Protection Act, 1986.
Answer:
The Central Government enacted a comprehensive law called the Consumer Protection Act in 1986. This Act came into force with effect from 15.04.1987. This Act was further amended in 1993. The Act is referred in short as ‘COPRA’.

III. Short Answer Questions

Question 1.
Which are the three constituent elements of business?
Answer:
The producer, the consumer and the government are the three constituent elements of business. The consumer is the most exploited constituent in the business world.

Question 2.
What are the important legislations related to consumerism in India?
Answer:
Consumer Legislation:

1. The Indian Contract Act, 1982
2. The Sale of Goods Act, 1982
3. The Essential Commodities Act, 1955
4. The Agricultural Products Grading and Marketing Act, 1937
5. The Prevention of Food Adulteration Act, 1954
6. Weights and Measures Act, 1958
7. The Trademark Act, 1999

Question 3.
What is meant by artificial scarcity?
Answer:
There are certain situations where the shop-keepers put up the board “No Stock” in front of their shops, even though there is plenty of stock in the store. In such situations consumers who are desperate to buy such goods have to pay hefty price to buy those goods and thus earning more profit unconscientiously.

Question 4.
Write the importance of consumerism.
Answer:
Importance of consumerism lies in:

1. Awakening and uniting consumers
2. Discouraging unfair trade practices
3. Protecting against exploitation
4. Awakening the government
5. Effective implementation of consumer protection laws
6. Providing complete and latest information
7. Discouraging anti-social activities

Question 5.
What is the role of Government in consumer protection?
Answer:
Since most of consumers including academically educated are illiterate about their rights and hence passive. Government should assure an active role in safeguarding the consumers. Government both the central and the state have brought out a number of legislations to protect the interest of consumers across the country. Law enforcement authorities should see that penal clause is not mere paper jaws-they should sting the offenders mercilessly.

IV. Long Answer Questions

Question 1.
How consumers are exploited?
Consumer is one who consumes the goods manufactured or created. Consumers are exploited in many’ways in the business.
They are as follows:

1. Selling at Higher Price: The price charged by the seller for a product or service may not be higher compared to the quality.
2. Adulteration: It refers to mixing or substituting undesirable material in food. This leads to heavy loss to the consumer, (e.g.) Mixing of stones with grains.
3. Duplicate or Spurious goods: Duplicate products of popular products are illegally produced and sold
4. Artificial Scarcity: There are certain situations where the shop-keepers put up the board “No Stock” in front of their shops, even though there is plenty of stock in the store.
5. Sub-standard: On opening a packet or sealed container one may find the content to be of poor quality.

Question 2.
Explain the role of business in consumer protection.
Answer:
Business enterprises should do the following towards protecting consumers.

1. Avoidance of Price Hike: Business enterprises should desist from hiking the price in the context of acute shortage of goods.
2. Avoidance of Hoarding: Business enterprises should not indulge in hoarding and black marketing to earn maximum profit.
3. Guarantees for Good Quality: Business enterprises should not give false warranty for the products.
4. Product Information: Business enterprises should disclose correct, complete and accurate information about the product viz. size, quality, quantity, weight etc.
5. The in advertising: Business enterprises should not convey false, untrue, bogus information relating to the product through the advertisement.
6. Money Refund Guarantee: Where the product becomes defective, business enterprises should replace it with new one or refund the purchase price.

Question 3.
What are the needs for consumer protection?
Answer:
Consumer is to be protected from the cheating business people. Though the consumer is said to be the king, his interests are neglected. Consumer protection is applicable to public sector, financial and co-operative enterprises. Recently even medical services have been brought under consumer-movement. Satisfaction and well being of the consumer should be the main objective of business. But in real practice consumer is not protected and safeguarded. Thus there is a need for consumer protection or movement.

Question 4.
Explain the role of consumers in Consumer Protection.
Answer:
Consumers have to be vigilant and organize themselves into a movement for concerned action.
Activation of Consumer Action Councils:

1. Consumer action councils established at village levels should educate consumers.
2. Consumer protection agencies should take necessary steps to investigate consumer complaints and grievances.
3. Voluntary consumer groups should provide information so as to educate consumers.
4. Consumer cooperatives need to be strengthened.
5. Consumer groups should contact the legislators to raise the consumer issue in Assembly and Parliament.
6. There should be testing laboratories at each district to test the purity of goods.

Question 5.
What are the objectives of Consumer Protection Act, 1986?
Answer:
The Central Government enacted a comprehensive law called the Consumer Protection Act in 1986. This Act came into force with effect from 15.04.1987. It is in short, called as ‘COPRA’.
Objectives:

1. Consumer protection Act protects the interests of the consumers.
2. This Act provides safeguards against defective goods and deficient services, untrade practices.
3. It also gives settlement of consumer disputes.
4. It is applicable to public sector, financial and co-operative enterprises.

Question 6.
Write about five important consumer legislations.
Answer:
To protect the consumers from the unfair traders, the government passed various legislative Acts.
They are follows:

1. The Indian Contract Act, 1982 was passed to bind the people on the promise made in the contract.
2. The Essential Commodities Act, 1955 protects the consumers against artificial shortages created by the sellers by hoarding the goods.
3. The Prevention of Food Adulteration Act, 1954 checks the adulteration of food articles and ensures purity of goods supplied.
4. Weights and Measures Act, 1958 protects the consumer against malpractices of underweight or under measurement.

Question 7.
What are the salient features of the Consumer Protection Act, 1986?
Answer:
Salient Features of The Indian Consumer Protection Act, 1986:

1. Protecting consumers against products and services which are harmful to the health of the consumers.
2. Ensuring consumers, with supply of goods at fair quality.
3. Ensuring the availability of goods in correct quantity and right size.
4. Protecting the consumers against pollution of various kinds.
5. Ensuring that consumers are charged fair price.
6. Protecting the consumers against unfair trade practices of unscrupulous trader.

Question 8.
What are the objectives of United Nations guidelines for consumer protection?
Answer:
The General Assembly of the United Nations passed a Resolution on April 9,1985 adopting a set of guidelines for consumer protection to persuade the member countries.

Objectives of United Nations Guidelines for Consumer Protection:

1. To assist countries in achieving or maintaining protection to consumers.
2. To facilitate production and distribution patterns responsive to the needs and desires of consumers.
3. To encourage high levels of ethical conduct for production and distribution of goods and services to consumers.
4. To facilitate the developing of independent consumer groups.
5. To encourage the development of market conditions which provide consumers with greater choice at lower prices.

Samacheer Kalvi 12th Commerce Consumerism Additional Questions and Answers

I. Choosy the Correct Answer

Question 1.
Match List I with List II and select the correct answer using the codes given below:

(a)	Prevention of Food Adulteration Act	–	(i)	1958
(b)	Weight and Measurement Act	–	(ii)	1982
(c)	Essential commodities Act	–	(iii)	1954
(d)	Sale of goods Act	–	(iv)	1955

Codes:

Answer:
(c) (i) 3, (ii) 1, (iii) 4, (iv) 2.

Question 2.
The term ‘Caveat emptor’ is a Latin term, which means _________
(a) Let the seller beware.
(b) Let the buyer beware
(c) Consumer
(d) Marketer
Answer:
(b) Let the buyer beware

II. Very Short Answer Questions

Question 1.
What is meant by sub-standard according to the Consumer Protection Act?
Answer:
On opening a packet or sealed container one may find the content to be of poor quality. If defective or damaged items are found in a pack, a consumer finds it difficult to exchange the defective one for good one.

Question 2.
What is meant by consumer protection?
Answer:
Consumer protection is a form of social action which is designed to attain the wellbeing of the society namely consumers. A consumer is said to be a king in a free market economy.

III. Short Answer Questions

Question 1.
Write a short note on False Advertisements.
Answer:
The main motive of advertisements is to educate the consumers regarding various aspects of the products and services. Sometimes it makes false representation about the quality, price, grade, composition and utility of products. Often the products are not as attractive as shown in the advertisement by the sellers. Consumers who buy the products on the faith of claims made in advertisements are cheated.

Case Study

Mr. Narasimachary bought a refrigerator of a familiar brand with a warranty for seven years. He uses the fridge as per the guidelines given by the manufacturer. After the completion of two years the fridge went out of order. He was shocked, and approached the dealer. But the dealer refused to service the fridge at free of cost.

Question 1.
What is your suggestion to Mr. Narasimachary to this grievance?
Answer:
Each and every consumer should know the rights and duties of consumers, according to the consumer protection Act – 1986. In this case, the manufacturer refused to repair and service the fridge, though there is a warranty period. So my suggestion is that the consumer Mr. Narasimachary can contact the manufacturer again to repair the fridge. If he refuses, Mr. Narasimachary can file a case in the consumer court according to the Act 1986. For proceeding to the case, the consumer has to make ready the cash bill, invoice and warranty card of the fridge.

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Tamilnadu Samacheer Kalvi 12th English Grammar Semantic Fields

Identify each of the following sentences with the fields given below:

Question 1.
(Sports; Religion; Environmental; Politics; Election)
(a) “Centre is cheating” says Mehta.
(b) Rajasthan passes law banning conversions.
(c) Vijayakanth vows to build a brand new city.
(d) EC goes in for a better ink.
(e) Lehmann gets world cup nod over Khan
Answer:
(a) Politics
(b) Religion
(c) Environmental
(d) Election
(e) Sports

Question 2.
(Paralympics; Philosophy; Railway; Media; Arts)
(a) In 1988, the first Shatabdi Express was introduced between New Delhi and Jhansi.
(b) Those days, the only entertainment and infotainment was the newspaper and radio.
(c) Sports was started for the differently-abled children.
(d) Vegetable carving is displayed at the entrance.
(e) This branch explores the foundations, methods, history’, implications and purpose of science.
Answer:
(a) Railway
(b) Media
(c) Paralympics
(d) Arts
(e) Philosophy

Question 3.
(Environment; Medicine; Crime; Urbanisation; Transportation)
(a) Taxis are mobile spaces that enable strangers to be physically proximate for a short time.
(b) The police traced his belongings to the taxi and retrieved them within two days.
(c) A baby less than a month old has received a pacemaker.
(d) Tirunelveli city has less vegetation and more buildings all just in one decade.
(e) With plastics out, old newspapers fetch more.
Answer:
(a) Transportation
(b) Investigation
(c) Medicine
(d) Urbanisation
(e) Environment

Question 4.
(Legal; Science; Education; Music; Tourism)
(a) Dropouts who come for mid-day meal are returning to class.
(b) The court extended the interim bail.
(c) At the Sarangkheda horse fair in Maharashtra, local breeds attract buyers from as far as Saudi Arabia.
(d) It is hard to draw the line between some early 1980s hard rock and heavy metal.
(e) Visitors were interacting with a security robot at a showroom.
Answer:
(a) Education
(b) Legal
(c) Tourism
(d) Music
(e) Science

Question 5.
(Sports; Nutrition and Dietetics; Travel; Literature; Agriculture)
(a) Shakespeare’s plays are read by many people.
(b) The yield of wheat has increased.
(c) Fast food is a growing health hazard.
(d) My brother is planning to go to the US.
(e) Dhoni was declared the Man of the Series.
Answer:
(a) Literature
(b) Agriculture
(c) Nutrition and Dietics
(d) Travel
(e) Sports

Question 6.
(Astronomy; Law; Flora; Geology; Sports)
(a) Sometimes A-class facilities are extended to even those imprisoned.
(b) Did an asteroid hit south India milllions of years ago?
(c) The investigations included megascopic and microscopic studies of rocks.
(d) Many foreigners come to watch Jallikatu at Madurai.
(e) The rapid depletion in indigenous aquatic plants would prove detrimental to wetlands.
Answer:
(a) Law
(b) Astronomy
(c) Geology
(d) Sports
(e) Flora

Question 7.
(Movie;Entertainment; Cuisine; Robotics; Electrodynamics)
(a) French actress Catherine’s clothing was designed by her friend Yas for her next project.
(b) In the world’s largest open-air theatre, stories of Krishna and Kansa area magically retold,
(c) Technology develops machines that can substitute for humans and replicate human actions.
(d) A normal bulb uses almost 80% energy to create heat and only 20% for production of light.
(e) A tropical diet is based on fruits and vegetables, while a polar diet rely on meat and fish.
Answer:
(a) Movie
(b) Entertainment
(c) Robotics
(d) Electrodynamics
(e) Cuisine

Question 8.
(Astronomy; Medicine; Politics; Literature; Agriculture)
(a) The deliberate suppression of texts by organisations shrouds the subject.
(b) The study of stars and stellar evolution is fundamental to our understanding of the universe.
(c) This shift to farming may have occurred because of climate change.
(d) Sovereign power may be vested on an individual or it may be vested on a group.
(e) Some medicines may not be safe during pregnancy.
Answer:
(a) Literature
(b) Astronomy
(c) Agriculture
(d) Politics
(e) Medicine

Question 9.
(Medical; Entertainment; Architecture; Education; Politics)
(a) DD to focus on Tamil novels, folk lore.
(b) PMK backs 49% quota in schools.
(c) Advani’s Yatra.
(d) Jaipur Royal glory restored in UK Town.
(e) A P Doctor gave Saudis Kingsize health system.
Answer:
(a) Entertainment
(b) Education
(c) Politics
(d) Architecture
(e) Medical

Question 10.
(Sports, Personality, Appointment, Politics, Commerce)
(a) Indira – Giant of her times.
(b) Benazir linked to oil – for food scam.
(c) Need a job – Sapco lab.
(d) Corporate governance must be exercised in spirit.
(e) Viswanathan Anand wins.
Answer:
(a) Personality
(b) Commerce
(c) Appointment
(d) Politics
(e) Sports

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Tamilnadu Samacheer Kalvi 12th English Grammar Proverbs

Fill in with a suitable word from the given choices to complete the Proverb.

Question 1.
Fortune favours the ………………………….
(a) bold
(b) rich
(c) poor
(d) friendly
Answer:
(a) bold

Question 2.
The squeaky …………………………. gets the grease.
(a) squirrel
(b) wheel
(c) machine
(d) screwdriver
Answer:
(b) wheel

Question 3.
A bad workman always blames his ………………………….
(a) co-workers
(b) time
(c) tools
(d) fortune
Answer:
(c) tools

Question 4.
…………………………. can’t be choosers.
(a) Poor
(b) Elite
(c) Elderly
(d) Beggars
Answer:
(d) Beggars

Question 5.
Cowards die many times before their ………………………….
(a) birth
(b) death
(c) feat
(d) chances
Answer:
(b) death

Question 6.
Cross the …………………………. where it is shallowest.
(a) river
(b) stream
(c) pond
(d) sea
Answer:
(b) stream

Question 7.
Cut your coat according to your ………………………….
(a) cloth
(b) linen
(c) court
(d) need
Answer:
(a) cloth

Question 8.
Don’t count you’re …………………………. before they hatch.
(a) ducks
(b) chickens
(c) eggs
(d) dumplings
Answer:
(b) chickens

Question 9.
Don’t have too many irons in the ………………………….
(a) fire
(b) building
(c) stock
(d) backyard
Answer:
(a) fire

Question 10.
Don’t make a …………………………. out of a molehill.
(a) mule
(b) mountain
(c) scene
(d) fountain
Answer:
(b) mountain

Question 11.
Don’t put all your eggs in one ………………………….
(a) plate
(b) oven
(c) basket
(d) egg case
Answer:
(c) basket

Question 12.
Don’t put the cart before the ………………………….
(a) horse
(a) horse
(b) cow
(c) dog
Answer:
(d) buggy

Question 13.
Every …………………………. has a wherefore.
(a) where
(c) why
(b) when
(c) why
Answer:
(d) who

Question 14.
Fine feathers make fine ………………………….
(a) birds
(b) peacocks
(c) fans
(d) weathers
Answer:
(a) birds

Question 15.
Fine words …………………………. no parsnips.
(a) cheese
(b) butter
(c) prepare
(d) cut
Answer:
(d) cut

Question 16.
Fools rush in where ………………………… fear to tread.
(a) devils
(b) angels
(c) ghosts
Answer:
(d) wise men

Question 17.
Half a loaf is better than ………………………….
(a) none
(b) one
(c) two
(d) many
Answer:
(a) none

Question 18.
It’s a long ………………………… that has no turning.
(a) love
(b) life
(c) lane
Answer:
(d) lap

Question 19.
It takes two to make a ………………………….
(a) handshake
(b) quarrel
(c) gossip
(d) conversation
Answer:
(b) quarrel

Question 20.
Let sleeping dogs ………………………….
(a) sleep
(b) bark
(c) whine
(d) lie
Answer:
(d) lie

Question 21.
………………………… is the mother of invention.
(a) Creation
(b) Necessity
(c) Father
(d) Husband
Answer:
(b) Necessity

Question 22.
One ………………………… doesn’t make a summer.
(a) swallow
(b) sparrow
(c) shower
(d) spring
Answer:
(a) swallow

Question 23.
The road to ………………………… is paved with good intentions.
(a) heaven
(b) hell
(c) highway
(d) hotel
Answer:
(b) hell

Question 24.
You can’t ………………………… with the hare and hunt with the hounds.
(a) race
(b) rent
(c) run
(d) read
Answer:
(c) run

Question 25.
You may lead a horse to but you cannot make him drink ………………………….
(a) water
(b) food
(c) porridge
(d) oatmeal
Answer:
(a) water

Question 26.
When the going gets tough, the tough get ………………………….
(a) coming
(b) going
(c) carried
(d) safer
Answer:
(b) going

Question 27.
I hope for the best, but ………………………… for the worst.
(a) prepare
(b) deal
(c) think
(d) act
Answer:
(a) prepare

Question 28.
Don’t bite the hand that ………………………… you.
(a) binds
(c) feeds
(b) beats
(d) hurts
Answer:
(c) feeds

Question 29.
One ………………………… turn deserves another.
(a) good
(b) bad
(c) wrong
(d) new
Answer:
(a) good

Question 30.
While there’s life, there’s ………………………….
(a) death
(b) disappointment
(c) hope
(d) an end
Answer:
(c) hope

Match the proverbs with their meanings 

Question 1.

Proverbs	Meanings
Manners maketh a Man	Try to read as much as possible.
Reading makes a man perfect	Never rush up.
Haste makes waste	Be perfect and respect others.

Ans.
(1) (c) Be pertìct and respect others.
(ii) (a) Try to read as much ãs possible.
(iii) (h) Never rush up.

Question 2.

Proverb	Meaning
Brevity is the soul of wit	Most appearances are deceptive
Make hay while the sun shines	Be short in speech and writing
All that glitters is not gold	Make optimum use of the opportunity

Answer:
(a) (ii) Be short in speech and writing
(b) (iii) Make optimum use of the opportunity
(c) (i) Most appearances are deceptive

Question 3.

Proverbs	Meanings
Actions speak louder than words	One person alone cannot be powerful
A single flower can’t make a garland	Great people also make mistakes
Even Homer nods	Actions are a better reflection of one’s character

Answer:
(i) (c) Actions are a better reflection of one’s character
(ii) (a) One person alone cannot be powerful
(iii) (b) Great people also make mistakes

Question 4.

Proverbs	Meanings
A rolling stone gathers no moss	Taking revenge.
Tit for tat	Be steady to achieve.
When one door shuts, another opens	Life offers many chances.

Answer:
(i) (b) Be steady to achieve.
(ii) (a) Taking revenge.
(iii) (c) Life offers many chances.

Question 5.

Proverbs	Meanings
Speech is silver, silence is gold	Prefer to live in groups.
No man is an island	Silence is better than speech.
Practise what you preach	Example is better than precept.

Answer:
(i) (b) Silence is better than speech.
(ii) (a) Prefer to live in groups.
(iii) (c) Example is better than precept.

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Tamilnadu Samacheer Kalvi 12th Commerce Solutions Chapter 22 The Negotiable Instruments Act 1881

Samacheer Kalvi 12th Commerce The Negotiable Instruments Act 1881 Textbook Exercise Questions and Answers

I. Choose the Correct Answer

Question 1.
Negotiable Instrument Act was passed in the year ________
(a) 1981
(b) 1881
(c) 1994
(d) 1818
Answer:
(b) 1881

Question 2.
Negotiable Instrument is freely transferable by delivery if it is a ________ instrument.
(a) Order
(b) Bearer
(c) Both a and b
(d) None of the above
Answer:
(b) Bearer

Question 3.
The transferee of a Negotiable Instrument is the one ________
(a) Who transfer the instrument
(b) On whose name it is transferred
(c) Who enchases it
(d) None of the above
Answer:
(b) On whose name it is transferred

Question 4.
Number of parties in a bill of exchange are ________
(a) 2
(b) 6
(c) 3
(d) 4
Answer:
(c) 3

Question 5.
Section 6 of Negotiable Instruments Act 1881 deals with ________
(a) Promissory Note
(b) Bills of exchange
(c) Cheque
(d) None of the above
Answer:
(c) Cheque

Question 6.
________ cannot be a bearer instrument.
(a) Cheque
(b) Promissory Note
(c) Bills of exchange
(d) None of the above
Answer:
(a) Cheque

Question 7.
When crossing restrict further negotiation ________
(a) Not negotiable crossing
(b) General Crossing
(c) A/c payee crossing
(d) Special crossing
Answer:
(a) Not negotiable crossing

Question 8.
Which endorsement relieves the endorser from incurring liability in the event of dishonour?
(a) Restrictive
(b) Facultative
(c) Sans recourse
(d) Conditional
Answer:
(b) Facultative

Question 9.
A cheque will become stale after ________ months of its date.
(a) 3
(b) 4
(c) 5
(d) 1
Answer:
(a) 3

Question 10.
Document of title to the goods exclude ________
(a) Lorry receipt
(b) Railway receipt
(c) Airway bill
(d) Invoice
Answer:
(d) Invoice

II. Very Short Answer Questions

Question 1.
What is meant by Negotiable Instrument?
Answer:
A negotiable instrument is a document which entitles a person to a certain sum of money and which is transferable from one person to another by mere delivery or by endorsement and delivery.

Question 2.
Define Bill of Exchange.
Answer:
According to section 5 of the Negotiable Instruments Act, “a bill of exchange is an instrument in writing containing an unconditional order, signed by the maker, directing a certain person to pay a certain sum of money only to, or to the order of a certain person or to the bearer of the – instrument”.

Question 3.
List three characteristics of a Promissory Note.
Answer:
Characteristics of a Promissory Note:

1. A promissory note must be in writing.
2. The promise to pay must be unconditional.
3. It must be signed by the maker.

Question 4.
What is meant by a cheque?
Answer:
According to section 6 of the Negotiable Instruments Act, 1881 defines a cheque as “a bill of exchange drawn on a specified banker and not expressed to be payable otherwise than on demand”.

Question 5.
Define Endorsement.
Answer:
“When the maker or holder of a negotiable instrument signs the name, otherwise that as such maker for the purpose of negotiation, on the back or face thereof, or on a slip of paper annexed thereto.”

III. Short Answer Questions

Question 1.
Explain the nature of a Negotiable Instrument.
Answer:
A negotiable instrument is transferable from one person to another without any formality, such as affixing stamp, registration, etc. When the instrument is held by holder in due course in the process of negotiation, it is cured of all defects in the instrument with respect to ownership. Though a bill, a promissory note or a cheque represents a debt, the transferee is entitled to sue on the instrument in his own name in case of dishonour, without giving notice to the debtor that he has become its holder.

Question 2.
Distinguish between Negotiability and Assignability.
Answer:

Question 3.
What are the characteristics of a bill of exchange?
Answer:
Characteristics of a Bill of Exchange:

1. A bill of exchange is a document in writing.
2. The document must contain an order to pay.
3. The order must be unconditional.
4. The instrument must be signed by the person who draws it.
5. The name of the person on whom the bill is drawn must .be specified in the bill itself.

Question 4.
Distinguish between Bill of Exchange and Promissory Note.
Answer:

IV. Long Answer Questions

Question 1.
Mention the presumptions of Negotiable Instruments.
Answer:
Presumptions of Negotiable Instrument:

1. Every negotiable instrument is presumed to have been drawn and accepted for consideration.
2. Every negotiable instrument bearing, a date is presumed to have been made or drawn on such a date.
3. It is presumed to have been accepted within a reasonable time after the date and before its maturity.
4. The transfer of a negotiable instrument is presumed to have been made before maturity.
5. When a negotiable instrument has been lost, it is presumed to have been duly stamped.
6. The holder of a negotiable instrument is presumed to be a holder in due course.

Question 2.
Distinguish a cheque and a bill of exchange.
Answer:

Question 3.
Discuss in detail the features of a cheque.
Answer:
A cheque is a negotiable instrument drawn on a particular banker.
Features:
(i) Instrument in Writings:
A cheque or a bill or a promissory note must be an instrument in writing. Though the law does not prohibit a cheque being written in pencil, bankers never accept it because of risks involved. Alternation is quite easy but detection is impossible in such cases.

(ii) Unconditional Orders:
The instrument must contain an order to pay money. It is not necessary that the word ‘order’ or its equivalent must be used to make the document a cheque. It does not cease to be a cheque just because the world ‘please’ is used before the word pay. Further the order must be unconditional.

(iii) Drawn on a Specified Banker Only:
The cheque is always drawn on a specified banker. A cheque vitally differs from a bill in this respect as latter can be drawn on any person including a banker. The customer of a banker can draw the cheque only on the particular branch of the bank where he has an account.

(v) A Certain Sum of Money Only:
The order must be for payment of only money. If the banker is asked to deliver securities, the document cannot be called a cheque. Further, the sum of money must be certain.

(v) Payee to be Certain:
The cheque must be made payable to a certain person or to the order of a certain person or to the bearer of the instrument. The word, person includes corporate bodies, local authorities, associations, holders of office of an institution etc.

(vi) Signed by the Drawer:
The cheque is to be signed by the drawer. Further, it should tally with specimen signature furnished to the bank at the time of opening the account.

Question 4.
What are the requisites for a valid endorsement?
Answer:
If an endorsement is to be valid, it must possess the following requisites:

1. Endorsement is to be made on the face of the instrument or on its back.
2. When there is no space for making further endorsements a piece of paper can be attached
3. Endorsement for only a part of the amount of the instrument is invalid.
4. Endorsement is complete only when delivery of the instrument is made.
5. Signing in block letters does not constitute regular endorsement.
6. If the payee is an illiterate person, he can endorse it by affixing his thumb impression on the instrument.

Question 5.
Explain the different kinds of endorsements. .
Answer:
When the person signs on the back of the instrument to transfer his interest, it is known as endorsement. The endorsement are of various types:
(i) Blank or general endorsement:
When the endorser puts his mere signature on the back of an instrument without mentioning the name of the person to whom the endorsement is made, it is called Blank Endorsement

(ii) Endorsement in full or special endorsement:
If the endorser, in addition to his signature, mentions the name of the person to whom it is endorsed, is known as endorsement in full or special endorsement.

(iii) Conditional endorsement:
When the endorser of a negotiable instrument makes his liability dependent upon the happening of an event which may or may not happen, it is called conditional endorsement,

(iv) Restrictive endorsement:
When an endorsement restricts or prohibits further negotiability of the instrument, it is called Restrictive Endorsement.

(v) Partial Endorsement:
Where the endorsement seeks to transfer only a part of the amount payable under the instrument, the endorsement is called Partial Endorsement.

Samacheer Kalvi 12th Commerce The Negotiable Instruments Act 1881 Additional Questions and Answers

I. Choose the Correct Answer

Question 1.
A bill of exchange drawn on a specified banker is
(a) promissory note
(b) cheque
(c) hundi
(d) share
Answer:
(b) cheque

Question 2.
Grace days allowed to a Bill of exchange for calculation of due date is
(a) 4
(b) 10
(c) 3
(d) 5
Answer:
(c) 3

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Tamilnadu Samacheer Kalvi 12th English Grammar Tenses

Fill in the blanks with the correct tense forms of the verbs given in brackets

Question 1.
We (wait) for you for more than an hour.
Answer:
have been waiting

Question 2.
When Seetha reached the station, the train (leave).
Answer:
had already left

Question 3.
The burglar (attack) the tenant.
Answer:
attacked

Question 4.
Mohan’s company is greatly (seek) after.
Answer:
sought

Question 5.
The terrified people (flee) to the mountains.
Answer:
had fled

Question 6.
My school (hold) a food-and-fun fair next month to raise money for the school building fund.
Answer:
is holding

Question 7.
She had (look) very worried for the past few days; but when I (ask) what the matter was, she said that it was nothing.
Answer:
had been looking, asked

Question 8.
I thought that the grass (need) cutting, but the lawnmower (be) out of order.
Answer:
needed, was

Question 9.
Madhu told him what (happen) to his dad, so he (run) home to see how he was.
Answer:
had happened, ran

Question 10.
She says that she (send) the letter a month ago; but, so far, she (not receive) any reply.
Answer:
had sent, has not received

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Tamilnadu Samacheer Kalvi 12th English Different Grammatical Functions

Use the given word as an adjective/a noun/a verb/ a preposition/ an adverb in sentences of your own.

Question 1.
Use the word ‘well’ as a noun, verb and an adjective in your own sentences.
Answer:

• There is an unused well behind my house, (noun)
• Tears welled up in her eyes on hearing her father’s death, (verb)
• Sangeetha is well qualified for the Professor post, (adj)

Question 2.
Use the word ‘tear’ as a noun, verb, adjective and adverb in your own sentences.
Answer:

• Don’t shed tears over loss of money, (noun)
• Don’t tear the paper, (verb)
• He was given a tearful farewell, (adj)
• Sheetal tearfully left the house, (adverb)

Question 3.
Use the word ‘like’ as a noun, verb, adjective and preposition in your own sentences.
Answer:

• Great gurus are above likes and dislikes, (noun)
• You need to carry out your duties whether you like it or not. (verb)
• You must befriend like-minded persons, (adj)
• Rohit looks like his uncle, (preposition)
• She looked like she was about to scream. (Conj)

Question 4.
Use the word “equal” as a noun, adjective, verb and adverb in your own sentences.
Answer:

• Sara treated Tarun and Kavin as equals, (noun)
• Father gave equal share of his property to his sons and daughters, (adj)
• One plus one equals two. (verb)
• We decided to divide the amount equally, (adverb)

Question 5.
Use the word “fine” as an adjective, a noun and a verb.
Answer:

• She wore a fine dress which caught everyone’s eyes, (adj)
• Rohan was asked to pay fine for skipping the red signal, (noun)
• I was fined for over-speeding, (verb)

Question 6.
Use the word “after” as an adverb, a preposition and conjunction.
Answer:

• Jill came tumbling after, (adv)
• He returned after the accident, (prep)
• After Margeret finished her studies, she returned to India, (conjunction)

Question 7.
Use the word “fast” as a noun, an adjective and adverb.
Answer:

• He didn’t take anything during the fast, (noun)
• I missed the fast train, (adj )
• She speaks fast, (adv)

Question 8.
Use the word “each” as an adjective and a pronoun.
Answer:

• Each day brings its opportunity, (adj)
• I received a rupee from each, (pronoun)

Question 9.
Use the word “as” as an adverb and a conjunction.
Answer:

• We ran as fast as we could, (adv)
• As he was late, we went without him. (conj)

Question 10.
Use the word “wrong” as an adjective, adverb and a noun.
Answer:

• You have taken the wrong road, (adj)
• Hari was wrongly arrested, (adv)
• We must differentiate the right and wrong, (noun)

Question 11.
Use the word ‘love’ as a verb and as a noun in sentences of your own.
Answer:

• I love my neighbours. (Verb)
• God is love. (Noun)

Question 12.
Use the word ‘ act’ as a verb and as a noun in sentences of your own.
Answer:

• He seemed to be sincere but it could be an act. (Noun)
• He acted wisely. (Verb)

Question 13.
Use the word ‘ hope’ as a verb and as a noun in sentences of your own.
Answer:

• I just hope 1 reach on time. (Verb)
• There was still a hope for the friendship to be understood. (Noun)

Question 14.
Use the word ‘angry’ as an adjective and as a noun in sentences of your own.
Answer:

• He was angry. (Adj) The angry bird is rather a game.
• His anger needs to be controlled. (Noun)

Question 15.
Use the word ‘religious’ as an adjective and as a noun in sentences of your own.
Answer:

• I am a religious person. (Adj)
• My religion is Christianity. (Noun)

Question 16.
Use the word ‘doctor’ as a verb and as a noun in sentences of your own.
Answer:

• Experts suspected that the company’s report had been doctored. (Verb)
• The doctor can be seen in his clinic in the evenings. (Noun)

Question 17.
Use the word ‘water’ as a verb and as a noun in sentences of your own.
Answer:

• Water the plants. (Verb)
• I should drink juice now. (Noun)

Question 18.
Use the word ‘up’ as a preposition and as an adverb in sentences of your own.
Answer:

• They are travelling up the elevator. (Prep)
• I stood up. (Adverb)

Question 19.
Use the word ‘dare’ as a verb and as an adjective in sentences of your own.
Answer:

• Some snakes are not poisonous but I dare you to hold it in your hand. (Verb)
• He was always known as a dare devil. (Adj)

Question 20.
Use the word ‘file’ as a noun and as a verb in sentences of your own.
Answer:

• The file is missing. (Noun)
• I need to file all the papers. (Verb)

Question 21.
Use the word ‘long’ as an adjectiveand as a noun in sentences of your ow n.
Answer:

• It was a long speech. (Adj)
• It didn’t take long to reach there. (Noun) .

Question 22.
Use the word ‘care’ as a verb and as a noun in sentences of your own.
Answer:

• We don’t care what happens. (Verb)
• The care of my children is my first priority. (Noun)

Question 23.
Use the word ‘walk’ as a verb and as a noun in sentences of your own.
Answer:

• I go for a walk in the evenings. (Noun)
• We walk every Sunday on beach. (Verb)

Question 24.
Use the word ‘beauty’ as an adjective and as a noun in sentences of your own.
Answer:

• Sunitha is a beauty queen. (Adj)
• Her beauty is stunning. (Noun)

Question 25.
Use the word ‘money’ as an adjective and as a noun in sentences of your own.
Answer:

• Ram is a moneyed person. (Adj)
• He has a lot of money. (Noun)

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Tamilnadu Samacheer Kalvi 12th English Grammar Concord

Rule 1 – A verb agrees with its subject in number.
Singular subjects take singular verbs:
1. The car stays in the garage.
2. The flower smells good.

Rule 2 – The number of the subject (singular or plural) is not changed by words that come between the subject and the verb.
One of the eggs is broken. Of the eggs is a prepositional phrase.
The subject one and the verb is are both singular.

Fill in the blanks with concord.
1. Your friend ……………………… (talk-talks) too much.
2. The man with the roses ……………………… (look-looks) like your brother.
3. The women in the pool ……………………… (swim-swims) well.
4. Bala ……………………… (drive-drives) a cab.
5. The football players ……………………… (run-runs) five miles every day.
6. That red-haired lady in the fur hat ……………………… (live-lives) across the street.
7. He ……………………… (cook-cooks) dinner for his family.
8. The boys ……………………… (walk-walks) to school every day.
9. The weather on the coast ……………………… (appear-appears) to be good this weekend.
10. The center on the basketball team ……………………… (bounce-bounces) the ball too high.
11. Either Priya or Prema ……………………… (was-were) here.
12. Either the cups or the glasses ……………………… (are-is) in the dishwasher.
13. Mohan and Ravi ……………………… (need-needs) a ride to work.
14. There ……………………… (is-are) a dog, a cat, and a bird in the garage.
15. Neither Murali nor his brothers ……………………… (was-were) at the party.
16′. Here into the main ring of the circus ……………………… (come-comes) the trained elephants.
17. Either the workers or the boss ……………………… (deliver-delivers) the merchandise.
18. The committee ……………………… (work-works) hard for better schools.
19. There ……………………… (is-are) many things to do before the holidays.
20. The jury ……………………… (was-were) polled for their verdicts.
Answer:
1. talks
2. looks
3. swim
4. drives
5. run
6. lives
7. cooks
8. walk
9. appears
10. bounces
11. was
12. are
13. need
14. are
15. were
16. comes
17. delivers
18. works
19. are
20. were

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Tamilnadu Samacheer Kalvi 12th Commerce Solutions Chapter 21 The Sale Of Goods Act 1930

Samacheer Kalvi 12th Commerce The Sale Of Goods Act 1930 Textbook Exercise Questions and Answers

I. Choose the correct answer

Question 1.
Sale of Goods Act was passed in the year _______
(a) 1940
(b) 1997
(c) 1930
(d) 1960
Answer:
(c) 1930

Question 2.
Which of the below constitutes the essential element of contract of sale?
(a) Two parties
(b) Transfer of property
(c) Price
(d) All of the above
Answer:
(d) All of the above

Question 3.
Which of the below is not a good?
(a) Stocks
(b) Dividend due
(c) Crops
(d) Water
Answer:
(b) Dividend due

Question 4.
In case of the sale, the _______ has the right to sell.
(a) Buyer
(b) Seller
(c) Hirer
(d) Consignee
Answer:
(b) Seller

Question 5.
The property in the goods means the _______
(a) Possession of goods
(b) Custody of goods
(c) Ownership of goods
(d) Both (a) and (b)
Answer:
(c) Ownership of goods

Question 6.
Specific goods denote goods identified upon the time of of sale.
(a) Agreement
(b) Contract
(c) Order
(d) Obligation
Answer:
(b) Contract

Question 7.
In which of the following types, the ownership is immediately transferred to buyer?
(a) When goods are ascertained
(b) When goods are appropriate
(c) Delivery to the carrier
(d) Sale or return basis
Answer:
(c) Delivery to the carrier

Question 8.
_______ is a stipulation which is collateral to main purpose of contract.
(a) Warranty
(b) Condition
(c) Right
(d) Agreement
Answer:
(a) Warranty

Question 9.
Unpaid seller can exercise his right of lien over the goods, where he is in possession of the goods as _______
(a) Owner of goods
(b) Agent of buyer
(c) Bailee for buyer
(d) All of these
Answer:
(d) All of these

Question 10.
The unpaid seller can exercise his right of stoppage of goods in transit where the buyer _______
(a) Becomes insolvent
(b) Refuses to pay price
(c) Payment of price
(d) Both (b) and (c)
Answer:
(a) Becomes insolvent

II. Very Short Answer Questions

Question 1.
What is a contract of sale of goods?
Answer:
Contract of sale of goods is a contract whereby the seller transfers or agrees to transfer the property (ownership) of the goods to the buyer for a price.

Question 2.
List down the essential elements of a contract of sale.
Answer:
Following essential elements are necessary for a contract of sale:

1. Two Parties
2. Transfer of Property
3. Goods
4. Price
5. Includes both ‘Sale’ and ‘Agreement to Sell’

Question 3.
What is meant by goods?
Answer:
The term goods mean every kind of movable property other than actionable claim and money.

Question 4.
What is a Contingent Goods?
Answer:
Contingent goods are the goods, the acquisition of which by the seller depends upon a contingency (an event which may or may not happen). Contingent goods are a part of future goods.

Question 5.
What do you understand by warranty?
Answer:
Warranty represents a stipulation which is collateral to the main purpose of the contract. It is of secondary importance to the contract.

III. Short Answer Questions

Question 1.
Explain the meaning of Agreement to sell.
Answer:
The property (ownership or title) in the goods has to pass at a future time or after the fulfilment of certain conditions specified in the contract.

Question 2.
Discuss in detail about existing goods.
Answer:
Existing goods are those owned or possessed by the seller at the time of contract of sale. Goods , possessed even refer to sale by agents or by pledgers. Existing goods may be either:

1. Specific Goods
2. Ascertained Goods
3. Generic or Unascertained Goods

Question 3.
Discuss the implied conditions and warranties in sale of goods contract.
Answer:
In every contract of sale, there are certain expressed and implied conditions and warranties. The term implied conditions means conditions which can be inferred from or guessed from the context of the contract.
Following are the implied conditions:

1. Conditions as to Title
2. Conditions as to Description
3. Sale by Sample
4. Conditions as to Quality or Fitness
5. Conditions as to Merchantability
6. Condition as to Wholesomeness
7. Condition Implied by Trade Usage

Following are the implied warranties:

1. Quiet Possession
2. Free from Any Encumbrances
3. Warranty in the case of Dangerous Goods

Question 4.
Discuss in detail the rights of an unpaid seller against the buyer personally.
Answer:
Where the Property in the Goods does not pass to the Buyer. Right of an Unpaid Seller against the Buyer Personally:

1. Suit for price
2. Suit for Damages for Non-acceptance
3. Suit for Cancellation of the Contract before the Due Date- Where the buyer cancels the contract before the date of delivery, the seller may either treat the contract as continuing or wait till the due date.
4. Suit for Interest

IV. Long Answer Questions

Question 1.
Explain in detail the elements of Contract of sale.
Answer:
Sale means selling the ownership of the goods to the buyer for a price. Similarly purchase means buying the ownership of the goods from the seller for a price. Following essential elements are necessary for a contract of sale:

1. Two Parties: A contract of sale involves two parties – the seller and the buyer. The buyer and the seller should be two different persons.
2. Transfer of Property: To constitute sale, the seller must transfer or agree to transfer the ownership in the goods to the buyer.
3. Goods: The subject matter of contract of sale must be goods. It excludes money, actionable claims and immovable property.
4. Price: The monetary consideration for the goods sold is called price

Question 2.
Distinguish between sale and agreement to sell.
Answer:

Basis for Comparison	Sale	Agreement to Sell
1. Ownership	The property (ownership or title) in the goods passes from the seller to the buyer immediately.	The property (ownership or title) in the goods has to pass at a future time or after the fulfilment of certain conditions.
2. Risk of Loss	Where the goods sold under the contract of sale are destroyed, the loss falls fully on the buyer as the ownership has already passed.	Where the goods under the agreement to sell are destroyed, the loss falls fully on the seller as the ownership is still vested with seller.
3. Consequences of violating the contract	Where the buyer fails to pay the price, the seller cannot seize the goods.	Where the buyer violates the contract, the seller can repossess the goods from the former. .
4. Nature of contract	It is an executed contract.	It is an executory contract, i.e., contract yet to be performed.
5. Insolvency of the Buyer	In a sale, if a buyer becomes insolvent before he pays for the goods even though the goods sold are under the possession of the seller, the latter has to return them to the Official Receiver.	If the buyer becomes insolvent before the payment of the price, the seller can retain the goods if they are under his possession.

Question 3.
Classify goods under the Sale of Goods Act.
Answer:
The term goods mean every kind of movable property other than actionable claim and money. The goods are classified as follows:

1. Existing Goods- These goods are owned or possessed by the seller at the time of contract of sale. Existing goods may again be divided as:
   • Specific Goods- It denotes goods identified and agreed upon at the time of contract of sale.
   • Ascertained Goods- The term ‘ascertained goods’ is also used as similar in meaning to specific goods.
   • Unascertained Goods- These are goods which are not identified and agreed upon at the time of contract of sale.
2. Future Goods- These are goods which a seller does not possess at the time of contract of sale, but which will be manufactured or produced or acquired by him after entering into the contract.
3. Contingent Goods- These are the goods, the acquisition of which by the seller depends upon a contingency (an event which may or may not happen).

Question 4.
Distinguish between Condition and Warranty.
Answer:

Basis for Difference	Condition	Warranty
1. Meaning	It is a stipulation which is essential to the main purpose of the contract of sale.	It is a stipulation which is collateral to the main purpose of contract.
2. Significance	Condition is necessary to the contract that the breaking of which cancels out the contract.	The violation of warranty will not revoke the contract.
3. Transfer of Ownership	Ownership on goods cannot be transferred without fulfilling the conditions.	Ownership on goods can be transferred on the buyer without fulfilling the warranty.
4. Remedy	In case of breach of contract, the affected party can cancel the contract and claim damages.	In the ease of breach of warranty, the affected party cannot cancel the contract but can claim damages only.
5. Treatment	Breach of condition may be treated as breach of warranty.	Breach of warranty cannot be treated as breach of condition.

Question 5.
Discuss in detail the rights of an unpaid seller against the goods.
Answer:
A seller is deemed to be an unpaid seller when:
(a) the whole of the price has not been paid
(b) a bill of exchange or other negotiable instrument given to him has been dishonoured.

Rights of an Unpaid Seller against the Goods:
(I) Where the Property in the Goods has Passed to the Buyer:

A. Right of Lien: An unpaid seller has a right to retain the goods till he receives the price. But to exercise this lien-

1. He must be in possession of goods
2. The goods must have been sold without any condition.

B. Right of Stoppage in Transit: Where the seller has delivered the goods to a carrier or other bailee for the purpose of transmission to the buyer, but the buyer has not acquired them, then the seller can stop the goods.

C. Right of Resale: The unpaid seller can resell the goods-

1. Where they are of a perishable nature or
2. Where the seller has expressly reserved the right of resale in the contract itself.

Case Study

Question 1.
X purchased a hot water bottle from Y, retail chemist. X asked Y if it would stand boiling water. The chemist told him that the bottle was meant to hold hot water. The bottle burst when water was poured into it and injured his wife. State whether seller is liable for the injury suffered by the buyer and the consequent compensation, give your reasons.
Answer:
The seller is not liable for the injury because he already told that the bottle is hot water bottle. X’s wife poured boiling water in the bottle, so she was injured. But there is no need for compensation.

Question 2.
X asked a car dealer to suggest him car suitable for touring purposes. The dealer suggested a ‘Buggati Car’. Accordingly, X purchased it but found it unsuitable for touring purpose. State whether the car dealer is liable for breach of condition?
Answer:
Yes, the car dealer is liable for breach of condition, because he told ‘Buggati Car’ was suitable for touring purpose. But in usage, it became unsuitable for touring purpose.

Question 3.
X, a dealer sold a plastic catapult to B. While using the catapult in the usual manner, it broke due to the fact that the materials used in its manufacture were unsuitable. As a result, the boy who was using it, blinded in one of his eyes. State whether the seller is liable or not.
Answer:
The seller is liable, because the materials used to manufacture catapult were not of a good quality. That is why the boy was blinded in one of his eyes.

Question 4.
X bought from Y a heap of wheat the weight of which is 1000 kg at the rate of Rs. 8 per kg. and agrees to pay the price on the first day of the next month and the wheat is to be delivered at X’s godown on the following day. A fire broke out and the entire quantity of wheat was destroyed. State whether X is liable to pay the price or not. Why?
Answer:
X is not liable, because before delivery, the goods were destroyed in Y’s godown. So X need not pay the price.

Question 5.
X bought from Y a heap of wheat (weight 100 kg) at the rate of Rs. 8 per kg. and Y had to put the wheat in bags to deliver it to X. Y filled some bags in X’s presence, but before the remainder could be filled, a fire broke out and the entire quantity of wheat was destroyed. State whether X is liable to pay the price or not. Why?
Answer:
X is not liable, due to non-delivery of the goods by Y. Before the wheat was filled in the bags, the godown caught fire. So X need not pay the price of wheat.

Question 6.
X bought from Y a heap of wheat at a rate of Rs. 8 per kg and Y had to weigh the wheat. Before weighing was completed, the wheat was destroyed by fire. State whether X is liable to pay the price or not. Why?
Answer:
X is not liable, because the wheat was destroyed before weighing. So X need not pay the price.