Unit 1: Units and Measurements — Online MCQ Test
PHYSICS · JEE MAINS · JEE
Practice Unit 1: Units and Measurements with a free chapter-wise online MCQ test.
This chapter covers: Units of measurements System of units SI Units fundamental and derived units least count significant figures Errors in measurements. Dimensions of Physics quantities dimensional an....
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Unit 1: Units and Measurements — Important Questions & Answers
Given below are two statements: Statement I: In a vernier callipers, one vernier scale division is always smaller than one main scale division. Statement II: The vernier constant is given by one main scale division multiplied by the number of vernier scale divisions. In the light of the above statements, choose the correct answer from the options given below.
- A. Statement I is true but Statement II is false
- B. Statement I is false but Statement II is true
- C. Both Statement I and Statement II are false
- D. Both Statement I and Statement II are true
Answer: C. Both Statement I and Statement II are false
Vernier constant is (1 MSD - 1 VSD). Statement II defines it incorrectly; it is the difference, not product.
Vernier constant is (1 MSD - 1 VSD). Statement II defines it incorrectly; it is the difference, not product.
Keeping the significant figures in view, the sum of the physical quantities 52.01 m, 153.2 m and 0.123 m is:
- A. 205 m
- B. 205.333 m
- C. 205.33 m
- D. 205.3 m
Answer: D. 205.3 m
When adding numbers, the result should have the same number of decimal places as the number with the fewest decimal places. 153.2 has one decimal place, so the sum must be rounded to one decimal place: 52.01 + 153.2 + 0.123 = 205.333, rounded to 205.3 m.
When adding numbers, the result should have the same number of decimal places as the number with the fewest decimal places. 153.2 has one decimal place, so the sum must be rounded to one decimal place: 52.01 + 153.2 + 0.123 = 205.333, rounded to 205.3 m.
In a vernier callipers, 50 vernier scale divisions are equal to 48 main scale divisions. If one main scale division = 0.05 mm, then the least count of the vernier callipers is ____ mm.
- A. 0.002
- B. 0.05
- C. 0.02
- D. 0.005
Answer: A. 0.002
Least Count (LC) = 1 MSD - 1 VSD. Given 50 VSD = 48 MSD, so 1 VSD = 48/50 MSD. LC = 1 MSD - (48/50) MSD = (2/50) MSD = 0.04 * 0.05 mm = 0.002 mm.
Least Count (LC) = 1 MSD - 1 VSD. Given 50 VSD = 48 MSD, so 1 VSD = 48/50 MSD. LC = 1 MSD - (48/50) MSD = (2/50) MSD = 0.04 * 0.05 mm = 0.002 mm.
In an experiment, a set of readings are obtained – 1.24 mm, 1.25 mm, 1.23 mm, 1.21 mm. The expected least count of the instrument used in recording these readings is ____ mm.
- A. 0.01
- B. 0.001
- C. 0.1
- D. 0.05
Answer: A. 0.01
The precision of the measurements (two decimal places in mm) indicates the least count is 0.01 mm.
The precision of the measurements (two decimal places in mm) indicates the least count is 0.01 mm.
The time period of a simple harmonic oscillator is T = 2π*sqrt(m/k). The measured value of mass (m) of the object is 10 g with an accuracy of 10 mg, and time for 50 oscillations of the spring is found to be 60 s using a watch of 2 s resolution. Percentage error in determination of spring constant (k) is ____ %.
- A. 3.43
- B. 3.35
- C. 7.60
- D. 6.76
Answer: D. 6.76
Error propagation formula for k = 4*pi^2*m / T^2 is (Δk/k) = (Δm/m) + 2*(ΔT/T). With T = 60/50 = 1.2s and ΔT = 2/50 = 0.04s, (Δk/k) = (0.01/10) + 2*(2/60) = 0.001 + 0.0666 = 0.0676 or 6.76%.
Error propagation formula for k = 4*pi^2*m / T^2 is (Δk/k) = (Δm/m) + 2*(ΔT/T). With T = 60/50 = 1.2s and ΔT = 2/50 = 0.04s, (Δk/k) = (0.01/10) + 2*(2/60) = 0.001 + 0.0666 = 0.0676 or 6.76%.