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JEE Main · Physics · Experimental Physics

A Vernier caliper has 1 mm marks on the main scale. It has 20 equal...

A Vernier caliper has 1 mm1\text{ mm} marks on the main scale. It has 2020 equal divisions on the Vernier scale which match with 1919 main scale divisions. When the measuring jaws are in contact, the zero of the Vernier scale is to the left of the zero of the main scale and the 12th12^{\text{th}} Vernier division coincides with a main scale mark. When this caliper is used to measure the diameter of a cylinder, the main scale reading is 2.5 cm2.5\text{ cm} and the 15th15^{\text{th}} Vernier division coincides with a main scale mark. The true diameter of the cylinder is

Options

  1. A2.515 cm2.515\text{ cm}
  2. B2.535 cm2.535\text{ cm}
  3. C2.635 cm2.635\text{ cm}
  4. D2.615 cm2.615\text{ cm}

Correct answer

D. 2.615 cm2.615\text{ cm}

Step-by-step solution

First, calculate the least count (LC) of the Vernier caliper.

Value of 1 MSD=1 mm1\text{ MSD} = 1\text{ mm}.

Since 20 VSD=19 MSD20\text{ VSD} = 19\text{ MSD}, we have 1 VSD=1920 mm1\text{ VSD} = \dfrac{19}{20}\text{ mm}.

LC=1 MSD−1 VSD=1 mm−1920 mm=120 mm=0.05 mm=0.005 cm\text{LC} = 1\text{ MSD} - 1\text{ VSD} = 1\text{ mm} - \dfrac{19}{20}\text{ mm} = \dfrac{1}{20}\text{ mm} = 0.05\text{ mm} = 0.005\text{ cm}.

Next, determine the zero error. Since the Vernier zero is to the left of the main scale zero, the zero error is negative.

Zero error =−(Total VSD−n)×LC= - (\text{Total VSD} - n) \times \text{LC}

Zero error =−(20−12)×0.005 cm=−8×0.005 cm=−0.040 cm= - (20 - 12) \times 0.005\text{ cm} = - 8 \times 0.005\text{ cm} = -0.040\text{ cm}.

Now, calculate the observed reading for the cylinder's diameter.

Observed reading =Main scale reading+(Coinciding VSD×LC)= \text{Main scale reading} + (\text{Coinciding VSD} \times \text{LC})

Observed reading =2.5 cm+(15×0.005 cm)=2.5 cm+0.075 cm=2.575 cm= 2.5\text{ cm} + (15 \times 0.005\text{ cm}) = 2.5\text{ cm} + 0.075\text{ cm} = 2.575\text{ cm}.

Finally, apply the zero correction to find the true reading.

True reading =Observed reading−Zero error= \text{Observed reading} - \text{Zero error}

True reading =2.575 cm−(−0.040 cm)=2.575 cm+0.040 cm=2.615 cm= 2.575\text{ cm} - (-0.040\text{ cm}) = 2.575\text{ cm} + 0.040\text{ cm} = 2.615\text{ cm}.

Answer: 2.615 cm2.615\text{ cm}
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