Least count, accuracy and precision — practice questions

Question 1 · difficulty L2 · understanding

In a Vernier calipers, when both jaws touch each other, zero of the Vernier scale is shifted to the right of zero of the main scale and $7^{\text {th }}$ Vernier division coincides with a main scale reading. If the value of 1 main scale division is 1 mm and there are 10 Vernier scale divisions, then the Vernier caliper has

• A. 0.07 cm negative zero error
• B. 0.7 cm negative zero error
• C. 0.07 cm positive zero error
• D. 0.7 cm positive zero error

Question 2 · difficulty L2 · understanding

In a screw gauge when the circular scale is given five complete rotations it moves linearly by 2.5 mm . If the circular scale has 100 divisions, the least count of screw gauge is $\_\_\_\_$ mm.

• A. $1 \times 10^{-2}$
• B. $1 \times 10^{-3}$
• C. $5 \times 10^{-2}$
• D. $5 \times 10^{-3}$

Question 3 · difficulty L2 · understanding

In an experiment, a set of reading are obtained as follows - 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. 0.05
• C. 0.001
• D. 0.1

Question 4 · difficulty L2 · understanding

When both jaws of vernier callipers touch each other, zero mark of the vernier scale is right to zero mark of main scale, $4{ }^{\text {th }}$ mark on vernier scale coincides with certain mark on the main scale. While measuring the length of a cylinder, observer observes 15 divisions on main scale and $5^{\text {th }}$ division of vernier scale coincides with a main scale division. Measured length of cylinder is $\_\_\_\_$ mm. (Least count of Vernier calliper $=0.1 \mathrm{~mm}$ )

• A. 15.5
• B. 15.4
• C. 15.9
• D. 15.1

Question 5 · difficulty L2 · understanding

In a vernier callipers, 50 vernier scale divisions are equal to 48 main scale divisions. If one main scale division $=0.05 \mathrm{~mm}$, then the least count of the vernier callipers is $\_\_\_\_$ mm.

• A. 0.002
• B. 0.02
• C. 0.05
• D. 0.005

Question 6 · difficulty L2 · understanding

In a screw gauge, the zero of the circular scale lies 3 divisions above the horizontal pitch line when their metallic studs are brought in contact. Using this instrument thickness of a sheet is measured. If pitch scale reading is 1 mm and the circular scale reading is 51 then the correct thickness of the sheet is $\_\_\_\_$ mm. [Assume least count is 0.01 mm ]

• A. 1.54
• B. 1.50
• C. 1.51
• D. 1.48

Question 7 · difficulty L2 · understanding

For the determination of refractive index of glass slab, a travelling microscope is used whose main scale contains 300 equal divisions equals to 15 cm . The vernier scale attached to the microscope has 25 divisions equals to 24 divisions of main scale. The least count (LC) of the travelling microscope is (in cm ) :

• A. 0.002
• B. 0.0025
• C. 0.0005
• D. 0.001

Question 8 · difficulty L2 · understanding

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. Both Statement I and Statement II are false
• B. Statement I is true but Statement II is false
• C. Both Statement I and Statement II are true
• D. Statement I is false but Statement II is true

Question 9 · difficulty L2 · understanding

One main scale division of a vernier caliper is equal to $$\mathrm{m}$$ units. If $$\mathrm{n}^{\text {th }}$$ division of main scale coincides with $$(n+1)^{\text {th }}$$ division of vernier scale, the least count of the vernier caliper is :

• A. $$\frac{1}{(\mathrm{n}+1)}$$
• B. $$\frac{m}{(n+1)}$$
• C. $$\frac{n}{(n+1)}$$
• D. $$\frac{\mathrm{m}}{\mathrm{n}(\mathrm{n}+1)}$$

Question 10 · difficulty L2 · understanding

While measuring diameter of wire using screw gauge the following readings were noted. Main scale reading is $$1 \mathrm{~mm}$$ and circular scale reading is equal to 42 divisions. Pitch of screw gauge is $$1 \mathrm{~mm}$$ and it has 100 divisions on circular scale. The diameter of the wire is $$\frac{x}{50} \mathrm{~mm}$$. The value of $$x$$ is :

• A. 42
• B. 71
• C. 21
• D. 142