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Experimental Physics

JEE Main · Physics · Experimental Physics

In a Vernier calliper, 1 cm on the main scale is divided into 10...

In a Vernier calliper, 1 cm1\text{ cm} on the main scale is divided into 1010 equal parts. It is given that 2020 divisions of the Vernier scale coincide with 1919 divisions of the main scale. When the two jaws of the instrument are brought into contact, the zero of the Vernier scale lies to the left of the zero of the main scale and the 14th14^{\text{th}} Vernier scale division coincides with a main scale division. When a small block is held between the jaws, the zero of the Vernier scale lies between 23 mm23\text{ mm} and 24 mm24\text{ mm}, and the 11th11^{\text{th}} Vernier scale division coincides with a main scale division. The true thickness of the block is:

Options

  1. A2.325 cm2.325\text{ cm}
  2. B2.425 cm2.425\text{ cm}
  3. C2.385 cm2.385\text{ cm}
  4. D2.355 cm2.355\text{ cm}

Correct answer

C. 2.385 cm2.385\text{ cm}

Step-by-step solution

First, find the value of one main scale division (MSD):
1 MSD=1 cm10=1 mm1\text{ MSD} = \dfrac{1\text{ cm}}{10} = 1\text{ mm}

Given that 20 VSD=19 MSD20\text{ VSD} = 19\text{ MSD}, the value of one Vernier scale division is:
1 VSD=1920 MSD=0.95 mm1\text{ VSD} = \dfrac{19}{20}\text{ MSD} = 0.95\text{ mm}

The least count (LC) is:
LC=1 MSD−1 VSD=1 mm−0.95 mm=0.05 mm\text{LC} = 1\text{ MSD} - 1\text{ VSD} = 1\text{ mm} - 0.95\text{ mm} = 0.05\text{ mm}

When the jaws are closed, the Vernier zero is to the left of the main scale zero, indicating a negative zero error. The 14th14^{\text{th}} division coincides, so the zero error is:
e=−(N−n)×LCe = -(N - n) \times \text{LC}
e=−(20−14)×0.05 mm=−6×0.05 mm=−0.30 mme = -(20 - 14) \times 0.05\text{ mm} = -6 \times 0.05\text{ mm} = -0.30\text{ mm}

For the measurement of the block, the main scale reading (MSR) is 23 mm23\text{ mm} and the Vernier scale reading (VSR) is 1111.
Measured reading =MSR+VSR×LC= \text{MSR} + \text{VSR} \times \text{LC}
Measured reading =23 mm+11×0.05 mm=23.55 mm= 23\text{ mm} + 11 \times 0.05\text{ mm} = 23.55\text{ mm}

The true reading is obtained by subtracting the zero error from the measured reading:
True reading=Measured reading−Zero error\text{True reading} = \text{Measured reading} - \text{Zero error}
True reading=23.55 mm−(−0.30 mm)=23.85 mm\text{True reading} = 23.55\text{ mm} - (-0.30\text{ mm}) = 23.85\text{ mm}

Converting to cm:
True thickness=2.385 cm\text{True thickness} = 2.385\text{ cm}

Answer: 2.385 cm2.385\text{ cm}
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