Samacheer Kalvi 9th Science Solutions Chapter 1 Measurement

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Tamilnadu Samacheer Kalvi 9th Science Solutions Chapter 1 Measurement

Samacheer Kalvi 9th Science Measurement Textbook Exercises

I. Choose the correct answer.

Question 1.
Choose the correct one
(a) mm < cm < m < km
(b) mm > cm > m > km
(c) km < m < cm < mm
(d) mm > m > cm > km
Answer:
(a) mm < cm < m < km

Question 2.
Rulers, measuring tapes and metre scales are used to measure
(a) Mass
(b) Weight
(c) Time
(d) Length
Answer:
(d) Length

Samacheer Kalvi 9th Science Solutions Chapter 1 Measurement

Question 3.
1 metric ton is equal to
(a) 100 quintals
(b) 10 quintals
(c) 1/10 quintals
(d) 1/100 quintals
Answer:
(b) 10 quintals

Question 4.
Which among the following is not a device to measure mass?
(a) Spring balance
(b) Beam balance
(c) Physical balance
(d) Digital balance
Answer:
(a) Spring balance

II. Fill in the blanks.

  1. Metre is the unit of …………..
  2. 1 kg of rice is weighed by …………
  3. The thickness of a cricket ball is measured by ………….
  4. The radius of a thin wire is measured by ………….
  5. A physical balance measures small differences in mass up to …………….

Answer:

  1. Length
  2. Beam balance
  3. Vernier Caliper
  4. Screw Gauge
  5. 1 mg

III. True or False.

Question 1.
The SI unit of electric current is kilogram.
Answer:
False
Correct Statement: The SI unit of electric current is ampere. Kilogram is the unit of mass.

Question 2.
Kilometre is one of the SI units of measurement.
Answer:
False
Correct.Statement: Metre only SI unit. Kilometre is multiple of metre.

Samacheer Kalvi 9th Science Solutions Chapter 1 Measurement

Question 3.
In everyday life, we use the term weight instead of mass.
Answer:
True

Question 4.
A physical balance is more sensitive than a beam balance.
Answer:
True

Question 5.
One Celsius degree is an interval of IK and zero degree Celsius is 273.15 K.
Answer:
False
Correct Statement: One Celsius degree is an interval 1K is true, but zero degree Celsius is equal to -273.15K.

Question 6.
With the help of vernier caliper we can have an accuracy of 0.1 mm and with screw gauge we can have an accuracy of 0.01 mm.
Answer:
False
Correct Statement: With the help of vernier caliper we can have an accuracy of 0.01 cm and with screw gauge we can have an accuracy of 0.01 mm.

IV. Match the following.

1. 

Column – I Column – II
(a) Length (i) Kelvin
(b) Mass (ii) meter
(c) Time (iii) Kilogram
(d) Tempature (iv) second

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

2.

Column – I Column – II
(a) Screw gauge  (i) Vegetables
(b) Vernier Caliper  (ii) Coins
(c) Beam balance  (iii) Gold ornaments
(d) Digital balance  (iv) Cricket ball

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

V. Assertion and Reason Type.

In the following questions, statement is given, followed by reason. Answer the questions as below.
(a) Both A and R are true but R is not the correct reason.
(b) Both A and R are true and R is the correct reason.
(c) A is true but R is false.
(d) A is false but R is true.

Samacheer Kalvi 9th Science Solutions Chapter 1 Measurement

Question 1.
Assertion(A): The scientifically correct expression is “ The mass of the bag is 10 kg”
Reason (R): In everyday life, we use the term weight instead of mass.
Answer:
(a) Both A and R are true but R is not the correct reason.

Question 2.
Assertion (A): 0°C = 273.16 K. For our convenience, we take it as 273 K after rounding off the decimal.
Reason (R): To convert a temperature on the Celsius scale we have to add 273 to the given temperature.
Answer:
(b) Both A and R are true and R is the correct reason.

Question 3.
Assertion (A): The distance between two celestial bodies is measured in terms of a light-year.
Reason (R): The distance traveled by the light in one year is one light year.
Answer:
(b) Both A and R are true and R is the correct reason.

VI. Very Short Answer Type.

Question 1.
Define measurement.
Answer:
Measurement is the process of comparison of the given physical quantity with the known standard quantity of the same nature.

Question 2.
Define standard unit.
Answer:
Unit is the quantity of a constant magnitude which is used to measure the magnitudes of other quantities of the same nature.

Question 3.
What is the full form of SI system?
Answer:
The full form of S.I. system is International System of Units.

Question 4.
Define least count of any device.
Answer:
Least count is the least distance measured in a given device by it.

Samacheer Kalvi 9th Science Solutions Chapter 1 Measurement

Question 5.
What do you know about pitch of screw gauge?
Answer:
Pitch of the screw gauge is the distance between two successive screw threads. It is measured by the ratio of distance travelled on the pitch scale to the number of rotations of the head scale.

Question 6.
Can you find the diameter of a thin wire of length 2 m using the ruler from your instrument box?
Answer:
Yes, first you have to wound the wire around the scale for 10 cm and count the number of turns in it. Then if you divide 10 cm by number of turns which gives the thickness of the wire.

VII. Short Answer Type.

Question 1.
Write the rules that are followed in writing the symbols of units in SI system.
Answer:

  1.  The units named after scientists are not written with a capital initial letter.
    E.g. newton, henry, ampere, and watt.
  2. The symbols of the units named after scientists should be written by the initial capital letter.
    E.g. N for newton, H for henry, A for ampere and W for watt.
  3. Small letters are used as symbols for units not derived from a proper noun.
    E.g. m for metre, kg for kilogram.
  4.  No lull stop or other punctuation marks should be used within or at the end of symbols.
    E.g. 50 m and not as 50 m.
  5.  The symbols of the units are not expressed in plural form.
    E.g. 10 kg not as kgs.

Question 2.
Write the need of a standard unit.
Answer:
Many of the ancient systems of measurement were based on the dimensions of human body. As a result, unit of measurement varied from person to person and also from location to location. In earlier time, different unit systems were used by people from different countries.
But, at the end of the Second World War there was a necessity to use worldwide system of measurement. Hence, SI (International System of Units) system of units was developed and recommended by General Conference on Weights and Measures in 1960 for international usage.

Question 3.
Differentiate mass and weight.
Answer:

S.No. Mass Weight
1. Fundamental quantity Derived quantity
2. Has magnitude alone – scalar quantity Has magnitude and direction – vector quantity
3. It is the amount of matter contained in a body It is the normal force exerted by the surface on the object against gravitational pull
4. Remains the same Varies from place to place
5. It is measured using physical balance It is measured using spring balance
6. Its unit is kilogram Its unit is newton

Samacheer Kalvi 9th Science Solutions Chapter 1 Measurement

Question 4.
How will you measure the least count of Vernier Caliper?
Answer:
Least count of Vernier Caliper is the ratio of value of one smallest main scale division to total
number of Vernier scale division.
i.e., L.C. = 0.1mm = 0.01cm
(or) L.C. = 1MD – 1VSD = 1.0 mm – 0.9 mm
= 0. 1mm = 0.01 cm

VIII. Long Answer Type.

Question 1.
Explain a method to find the thickness of a hollow tea cup.
Answer:
To find the thickness of a hollow teacup,

(i)Determine the pitch, of the least count and zero error of the screw gauge.

  • Pitch of the screw = \(\frac{\text { Distance moved by the pitch }}{\text { No. of rotations by Head scale }}\)
  • Least count (LC) = 0.01 mm
  • Zero error:
    Positive zero error (ZE) = + (n × LC)mm = + (n × 0.01) mm
    ∴ Zero correction (ZC) = – (n × 0.01) mm
    Negative zero error (ZE) = – (100 – n) × LC mm
    ∴ Zero correction (ZC) = (100 – n) × LC mm

(ii) Place the teacup between the two studs.

(iii) Rotate the head until the tea cup is held firmly but not tightly, with the help of ratchat.

(iv) Note the reading of the pitch scale crossed by the head scale (PSR) and the head scale
the division that coincides with the pitch scale axis (HSC).

(v) The thickness of the teacup is given by PSR + CHSR (Corrected HSR). Repeat the experiment for different positions of the teacup.

(vi) Tabulate the readings.

(vii) The average of the last column reading gives the thickness of the tea cup.

S.No. P.S.R

(mm)

H.S.C

(division)

CHSC = HSC ± ZC (Division) CHSR = CHSC x LC (mm) Total reading = PSR + CHSR (mm)
1.

2.

mean = mm

Thickness of the teacup = ……….. mm

Question 2.
How will you find the thickness of a one rupee coin?

  1. Determine the pitch, the least count and the zero error of the screw gauge
  2.  Place the coin between the two studs
  3.  Rotate the head until the coin is held firmly but not tightly, with the help of the ratchat
  4.  Note the reading of the pitch scale crossed by the head scale (PSR) and the head scale division that coincides with the pitch scale axis (HSC)
  5.  The width of the coin is given by PSR + CHSR (Corrected HSR). Repeat the experiment for different positions of the coin
  6.  Tabulate the readings
  7.  The average of the last column readings gives the width of the coin
S.No. P.S.R

(mm)

H.S.C

(division)

CHSC = HSC ± ZC (Division) CHSR = CHSC x LC (mm) Total reading = PSR + CHSR (mm)
1.

2.

mean = mm

Thickness of the coin = …….. mm

IX. Numerical problem.

Question 1.
Inian and Ezhilan argue about the light year. Inian tells that it is 9.46 × 10<sup>15</sup>? m and Ezhilan argues that it is 9.46 × 10<sup>12</sup> km. Who is right? Justify your answer.
Answer:
The magnitude of light year = 9.46 × 10<sup>15</sup> m. So Inian gave a correct answer.

Samacheer Kalvi 9th Science Solutions Chapter 1 Measurement

Question 2.
The main scale reading while measuring the thickness of a rubber ball using Vernier Caliper is 7 cm and the Vernier scale coincidence is 6. Find the radius of the ball.
Answer:
Given: The main scale reading = 7 cm
Vernier scale coincidence = 6
we know that least count of vernier = 0.01 cm
The radius of the ball = MSR + VC × LC
= 7 cm + 6 × 0.01 cm
= 7 cm + 0.06 cm = 7.06 cm

Question 3.
Find the thickness of a five rupee coin with the screw gauge, if the pitch scale reading is 1 mm and its head scale coincidence is 68.
Answer:
Given Pitch scale reading = 1 mm
Head scale coincidence = 68
The thickness of a fire rupee coin = PSR + HSC × L.C ± ZE
= 1 mm + 68 × 0.01 mm
= 1 mm + 0.68 mm
= 1.68 mm

ACTIVITY

Question 1.
Using Vernier caliper find the outer diameter of your pen cap.
Samacheer Kalvi 9th Science Solutions Chapter 1 Measurement 1
Answer:

S.No. P.S.R. H.S.C. C.H.S.C × L.C Total reading
1. 1mm 68 68 × 0.01 mm  P.S.R. + (H.S.C. × L.C) ± ZE
= 1 mm + (68 × 0.01 mm)
= 1 mm + 0.68 mm = 1.68 mm

Question 2.
Determine the thickness of a single sheet of your science textbook with the help of a Screw gauge.
Answer:
Pitch scale reading = 0.05 mm L.C. = 0.1 mm
Head scale coincidence = 02
The thickness of a single sheet of science text book = PSR + HSC × L.C. + ZE
= 0.05 mm+ (02 × 0.1)
= 0.05 mm + 0.2 mm
= 0.07 mm

Samacheer Kalvi 9th Science Solutions Chapter 1 Measurement

Question 3.
With the resources such as paper plates, teacups, thread, and sticks available at home make a model of an ordinary balance. Using standard masses find the mass of some objects.
Answer:

Resource avail Device mass of objects
Paper Plates Common balance 10g
Tea Cups Common balance 5g
Thread Physical balance 10 ms
sticks two par balance 15g

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