Physics: Units and Measurement

A comprehensive guide for ICSE Class 9 Physics – Chapter 1: Units and Measurement. This resource covers important definitions, solved numerical questions, detailed explanations of measuring instruments like vernier callipers and screw gauge, and the working of a simple pendulum. Ideal for quick revision and concept clarity with neatly organized Q&A format.

A. UNITS AND MEASUREMENT

1. What are the three requirements for selecting a unit of a physical quantity?
The three requirements are:
  1. The unit should be well-defined
  2. It should be reproducible
  3. It should be invariable
2. Name the three systems of units and state the various fundamental units in them.
The three systems of units are:
  • CGS (Centimeter-Gram-Second): centimeter, gram, second
  • FPS (Foot-Pound-Second): foot, pound, second
  • MKS (Meter-Kilogram-Second): meter, kilogram, second
3. What are the fundamental units in the S.I. system? Name them along with their symbols.
The fundamental units in the S.I. system are:
  • meter (m)
  • kilogram (kg)
  • second (s)
  • ampere (A)
  • kelvin (K)
  • mole (mol)
  • candela (cd)
4. Name the units which are used to measure very large astronomical distances.
The units used are:
  • Light year
  • Parsec
5. The unit of force divided by the unit of area is the unit of which physical quantity?

Force is measured in Newtons (N) and area is measured in square meters (m²).

The unit of force divided by the unit of area is N/m², which is the unit of pressure.

6. It takes 5 years for light to reach the earth from a star. Express the distance of the star from the earth in: (i) light years (ii) kilometres

Given: Speed of light = 3 × 10⁸ m/s, time = 5 years = 5 × 3.1 × 10⁷ s

(i) Distance in light years:

Distance = speed × time = 1 light year × 5 = 5 light years

(ii) Distance in kilometers:

Distance = 3 × 10⁸ m/s × 5 × 3.1 × 10⁷ s = 4.65 × 10¹⁶ m = 4.65 × 10¹³ km

Final answer: (i) 5 light years, (ii) 4.65 × 10¹³ km

7. The wavelength of light is 589 nm. What is its wavelength in Å?

Given: Wavelength = 589 nm

1 nm = 10 Å, so:

589 nm = 589 × 10 Å = 5890 Å

Final answer: 5890 Å

8. It takes 8 min for light to reach the earth from the sun. Find the distance from the sun to the earth in km.

Given: Time = 8 min = 8 × 60 s, speed of light = 3 × 10⁸ m/s

Distance = speed × time:

3 × 10⁸ m/s × 8 × 60 s = 1.44 × 10¹¹ m = 1.44 × 10⁸ km

Final answer: 1.44 × 10⁸ km

9. The heart beats once in 4/5 s. How many times does the heart beat in 60 years?

Given: Time period of heartbeat = 4/5 s

Number of heartbeats per second = 5/4

Number of heartbeats in 60 years:

(5/4) × 60 × 365 × 24 × 60 × 60 = 2.365 × 10⁹

Final answer: 2.365 × 10⁹

10. The size of a bacteria is 1μ. Find the number of bacteria in 1 metre length.

Given: Size of bacteria = 1 μm = 1 × 10⁻⁶ m

Number of bacteria in 1 meter:

1 m / (1 × 10⁻⁶ m) = 10⁶

Final answer: 10⁶

11. The earth is approximately a sphere of radius 6.37 × 10⁶ m. What is its circumference in kilometres?

Given: Radius of earth = 6.37 × 10⁶ m

Circumference = 2πr:

2 × 3.14 × 6.37 × 10⁶ m = 4 × 10⁷ m = 4 × 10⁴ km

Final answer: 4 × 10⁴ km

12. Rina uses the length of her school desk as a unit to measure the length of a room. Why is this method not suitable for scientific measurements?

The length of Rina’s desk may vary from person to person and is not a standard unit. Scientific measurements require standard units that are:

  1. Well-defined
  2. Reproducible
  3. Invariable
13. Tom claims that the length of a pencil can be used as a standard unit of length. Justify whether this is a valid choice.

No, it’s not a valid choice because:

  • The length of a pencil is not fixed (varies between pencils)
  • It cannot be precisely reproduced
  • It changes with wear and tear

Scientific measurements require standardized, invariable units.

14. Priya wrote the speed of a car as “50 Km/hr”. The teacher asked her to correct the unit. What mistake did she make?

The correct unit should be km/h (kilometers per hour). The mistakes were:

  • “Km” should be “km” (SI unit convention)
  • “hr” should be “h” (standard abbreviation)
15. Rita wrote the unit of pressure as “Newtons per square meter”. Her friend wrote it as “N/m²”. Who is correct and why?

Both Rita and her friend are correct because:

  • “Newtons per square meter” is the verbal description of the unit
  • “N/m²” is the standard symbolic notation

Both represent the same physical quantity (pressure).

B. MEASURING INSTRUMENTS

16. Meera took the reading on a vernier callipers and noted it as 1.20 cm. The teacher accepted it. But when she wrote 1.2 cm in the next experiment, she lost marks. Why?

The difference is in the precision:

  • 1.20 cm implies precision to 0.01 cm (hundredths place)
  • 1.2 cm implies precision to only 0.1 cm (tenths place)

Vernier callipers can measure to hundredths of a centimeter, so the first reading properly represents the instrument’s precision.

17. When Rohit wrote the reading of a screw gauge as 2.3052 cm, the teacher said it was incorrect. Why?

A standard screw gauge typically has:

  • Least count of 0.001 cm (micrometer)
  • Cannot measure to 0.0001 cm precision

The reading 2.3052 cm implies precision beyond the instrument’s capability.

18. A boy measures the length of a piece of pencil by metre rule, vernier callipers, and screw gauge to be 1.2 cm, 1.24 cm, and 1.243 cm respectively. (a) State the least count of each measuring instrument. (b) Which instrument gives the most accurate result?

(a) Least count of each instrument:

  • Metre rule: 0.1 cm
  • Vernier callipers: 0.01 cm
  • Screw gauge: 0.001 cm

(b) The screw gauge gives the most accurate result as it has the smallest least count.

19. In an instrument, there are 25 divisions on the vernier scale which have the length of 24 divisions of the main scale. 1 cm on the main scale is divided into 20 equal parts. Find the least count.

Given:

25 vernier scale divisions = 24 main scale divisions

1 cm = 20 main scale divisions

Calculation:

1 main scale division = 1/20 cm = 0.05 cm

Least count = 1 main scale division / 25 = 0.05 cm / 25 = 0.002 cm

Final answer: 0.002 cm

20. Name the part of the vernier callipers which is used to measure the following: (a) External diameter of a tube (b) Internal diameter of a mug (c) Depth of a small bottle (d) Thickness of a pencil

The parts used are:

  • (a) External jaws
  • (b) Internal jaws
  • (c) Depth rod
  • (d) External jaws
21. What do you mean by zero error of a vernier calliper? Show with a labelled diagram the positions of the main scale and vernier scale in case of two types of zero error.

Zero error occurs when the vernier scale does not coincide with the main scale at zero when the jaws are closed.

Two types:

  1. Positive zero error: When the zero of the vernier scale is to the right of the main scale zero
  2. Negative zero error: When the zero of the vernier scale is to the left of the main scale zero

[Diagram would show main scale and vernier scale alignment for both cases]

22. When a vernier callipers (Least Count 0.01 cm) was checked for zero error, the zero mark on the vernier scale was found towards the right and its 3rd mark was in line with a main scale mark. Calculate the zero error.

Given:

Least count = 0.01 cm

Zero mark is towards the right, and the 3rd mark is in line with a main scale mark

Calculation:

Zero error = +3 × least count = +3 × 0.01 cm = +0.03 cm

Final answer: +0.03 cm

23. The circular head of a screw gauge is divided into 50 divisions and the screw moves 1 mm ahead in two revolutions of the circular head. Find its: (a) Pitch (b) Least Count

Given:

Number of divisions = 50

Distance moved in 2 revolutions = 1 mm

(a) Pitch:

Pitch = distance moved / number of revolutions = 1 mm / 2 = 0.5 mm

(b) Least Count:

Least count = pitch / number of divisions = 0.5 mm / 50 = 0.01 mm

Final answer: (a) 0.5 mm, (b) 0.01 mm

24. What do you mean by zero error of a screw gauge? Show with a labelled diagram the relative positions of the main scale and the circular scale in case of two types of zero error.

Zero error occurs when the circular scale does not coincide with the reference line on the main scale when the screw is fully closed.

Two types:

  1. Positive zero error: When the zero of the circular scale is below the reference line
  2. Negative zero error: When the zero of the circular scale is above the reference line

[Diagram would show main scale and circular scale alignment for both cases]

25. The screw of a screw gauge moves 1 mm ahead in two revolutions of the circular head. The circular head has 50 divisions. What is the least count of the screw gauge?

Given:

Distance moved in 2 revolutions = 1 mm

Number of divisions = 50

Calculation:

Pitch = 1 mm / 2 = 0.5 mm

Least count = 0.5 mm / 50 = 0.01 mm

Final answer: 0.01 mm

26. The dimensions of a pocketbook are measured as 33.2 mm, 54.6 mm, 13.8 mm. Which measuring tool could have been used to obtain these readings?

The measurements have precision to 0.1 mm, which suggests the instrument used was a vernier calliper (least count 0.1 mm or 0.01 cm).

27. Name the instrument you would use to measure the following: (i) Diameter of a pin (ii) External diameter of a ball pen refill (iii) External/internal diameter of a calorimeter (iv) Thickness of a thin glass plate (v) Thickness of a paper (vi) Diameter of a sphere (vii) Internal diameter of a test tube (viii) Width of a table

Appropriate instruments:

  1. (i) Screw gauge
  2. (ii) Screw gauge or vernier calliper
  3. (iii) Vernier calliper
  4. (iv) Screw gauge
  5. (v) Screw gauge
  6. (vi) Vernier calliper or screw gauge
  7. (vii) Vernier calliper
  8. (viii) Metre rule or vernier calliper

C. SIMPLE PENDULUM AND TIME PERIOD

28. What do you mean by a second’s pendulum?

A second’s pendulum is a pendulum that has a time period of exactly 2 seconds (1 second for each swing in one direction).

29. A simple pendulum is set up in a lab at sea-level. How would its time period change if it is shifted to: (i) The moon (ii) A deep mine (iii) A place at Mount Everest (iv) An artificial satellite of Earth

The time period (T) of a simple pendulum depends on acceleration due to gravity (g): T ∝ 1/√g

Changes:

  1. (i) On the moon: g decreases ⇒ T increases
  2. (ii) In a deep mine: g decreases slightly ⇒ T increases slightly
  3. (iii) At Mount Everest: g decreases slightly ⇒ T increases slightly
  4. (iv) In an artificial satellite: g ≈ 0 ⇒ T becomes infinite (pendulum doesn’t oscillate)
30. A second’s pendulum is taken to a planet where g = 4 times that on Earth. What would be its time period there?

Given:

gplanet = 4 × gEarth

Time period on Earth = 2 seconds

Calculation:

Since T ∝ 1/√g:

Tplanet = TEarth × √(gEarth/gplanet) = 2 × √(1/4) = 2 × (1/2) = 1 second

Final answer: 1 second

31. Time periods of two pendulums are in the ratio 1.28 : 0.32. Calculate the ratio of lengths.

Given: Time period ratio = 1.28 : 0.32

Since T ∝ √L:

L ∝ T²

L1/L2 = (T1/T2)² = (1.28/0.32)² = (4)² = 16

Final answer: 16 : 1

32. Time periods of two pendulums are in the ratio 5.6 : 0.14. Calculate the simplest ratio of lengths.

Given: Time period ratio = 5.6 : 0.14

Since L ∝ T²:

L1/L2 = (T1/T2)² = (5.6/0.14)² = (40)² = 1600

Final answer: 1600 : 1

33. Find the length of a pendulum which completes 6 oscillations in 15 seconds. (Take g = 9.8 m/s², π = 3.14)

Given:

Number of oscillations = 6

Time taken = 15 seconds

g = 9.8 m/s², π = 3.14

Calculation:

Time period T = time/number of oscillations = 15/6 = 2.5 s

Using T = 2π√(L/g):

L = (T² × g)/(4π²) = (2.5² × 9.8)/(4 × 3.14²) = (6.25 × 9.8)/(4 × 9.8596) ≈ 1.55 m

Final answer: 1.55 m

34. If a simple pendulum is situated on the Moon, identical to that on Earth, what will be the ratio of its time period on Moon to Earth? (Given g on Earth = 10 m/s², g on Moon = 1.6 m/s²)

Given:

gEarth = 10 m/s²

gMoon = 1.6 m/s²

Calculation:

Since T ∝ 1/√g:

TMoon/TEarth = √(gEarth/gMoon) = √(10/1.6) = √6.25 = 2.5

Final answer: 2.5

35. Two pendulums P and Q of lengths 4 m and 8 m are made to oscillate. Which one makes more oscillations and why?

Given:

Length of P = 4 m

Length of Q = 8 m

Since T ∝ √L, pendulum P (shorter length) has a smaller time period and thus makes more oscillations in the same time period.

Final answer: Pendulum P makes more oscillations.

36. It takes 0.2 s for a pendulum bob to move from mean position to one end. What is the time period of the pendulum?

Given: Time to move from mean position to one end = 0.2 s

The time period is the time for one complete oscillation (mean → end → mean → other end → mean):

T = 4 × time to move from mean to end = 4 × 0.2 = 0.8 s

Final answer: 0.8 s

37. How much time does the bob of a second’s pendulum take to move from one extreme to the other?

A second’s pendulum has a time period of 2 seconds.

Time to move from one extreme to the other extreme (through mean position) = half the time period = 1 second.

Final answer: 1 second

38. In a simple pendulum experiment, draw a graph showing variation of square of time period with length of pendulum. How can you use this graph to determine the value of acceleration due to gravity (g)?

The graph between T² and L is a straight line.

From the relation T² = (4π²/g)L:

The slope of the graph = 4π²/g

Therefore, g = 4π²/slope

39. If the graph between the square of time period (T²) and length (L) of a simple pendulum is a straight line, what does the slope of this line represent? Write its relation with acceleration due to gravity.

The slope of the T² vs L graph represents 4π²/g.

The relation is:

slope = 4π²/g ⇒ g = 4π²/slope

Subscribe
Subscribe