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NEET
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Physics
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Experimental Physics

NEET · Physics · Experimental Physics

In a certain vernier callipers, 10 divisions of the vernier scale...

In a certain vernier callipers, 1010 divisions of the vernier scale coincide with 88 divisions of the main scale. The length of one main scale division is 11 mm. When measuring the diameter of a sphere, the zero of the vernier scale lies just after the 3.23.2 cm mark on the main scale, and the 4th4^{\text{th}} division of the vernier scale exactly coincides with a main scale division. The measured diameter of the sphere is:

Options

  1. A3.243.24 cm
  2. B3.603.60 cm
  3. C3.2083.208 cm
  4. D3.283.28 cm

Correct answer

D. 3.283.28 cm

Step-by-step solution

First, we determine the least count (LC) of the vernier callipers.

Given 1010 VSD =8= 8 MSD, we have:

11 VSD =810= \dfrac{8}{10} MSD =0.8= 0.8 MSD

Since 11 MSD =1= 1 mm, 11 VSD =0.8= 0.8 mm.

The least count is:

LC =1= 1 MSD −1- 1 VSD =1= 1 mm −0.8- 0.8 mm =0.2= 0.2 mm

Converting the least count to cm:

LC =0.02= 0.02 cm

Now, we calculate the total reading for the diameter of the sphere.

Main Scale Reading (MSR) =3.2= 3.2 cm

Vernier Scale Reading (VSR) =4= 4

Total Reading =MSR+(VSR×LC)= \text{MSR} + (\text{VSR} \times \text{LC})

Total Reading =3.2 cm+(4×0.02 cm)=3.2 cm+0.08 cm=3.28 cm= 3.2 \text{ cm} + (4 \times 0.02 \text{ cm}) = 3.2 \text{ cm} + 0.08 \text{ cm} = 3.28 \text{ cm}

Answer: 3.283.28 cm
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