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NVAEasyJEE 2024Vernier Calipers & Screw Gauge

JEE Physics 2024 Question with Solution

If 5050 Vernier divisions are equal to 4949 main scale divisions of a traveling microscope and one smallest reading of the main scale is 0.5 mm0.5 \, \text{mm}, the Vernier constant of the traveling microscope is:

Answer

Correct answer:0.01

Step-by-step solution

Standard Method

Given:

  • 5050 Vernier scale divisions are equal to 4949 main scale divisions.
  • 1 MSD=0.5 mm1 \, \text{MSD} = 0.5 \, \text{mm}

Find: The Vernier constant of the traveling microscope.

The Vernier constant is calculated as

VC=Value of 1 MSD−Value of 1 VSD\text{VC} = \text{Value of 1 MSD} - \text{Value of 1 VSD}

Given that 5050 Vernier divisions correspond to 4949 main scale divisions,

50 VSD=49 MSD50 \, \text{VSD} = 49 \, \text{MSD}

So,

1 VSD=4950 MSD1 \, \text{VSD} = \frac{49}{50} \, \text{MSD}

Substituting 1 MSD=0.5 mm1 \, \text{MSD} = 0.5 \, \text{mm},

1 VSD=4950×0.5 mm=0.49 mm1 \, \text{VSD} = \frac{49}{50} \times 0.5 \, \text{mm} = 0.49 \, \text{mm}

Now,

VC=0.5 mm−0.49 mm=0.01 mm\text{VC} = 0.5 \, \text{mm} - 0.49 \, \text{mm} = 0.01 \, \text{mm}

Therefore, the Vernier constant of the traveling microscope is 0.01 mm0.01 \, \text{mm}.

Step-by-Step Method

Given:

  • The value of 11 main scale division is 0.5 mm0.5 \, \text{mm}.
  • 5050 Vernier scale divisions coincide with 4949 main scale divisions.

Find: The least count or Vernier constant.

Concept Used: The Vernier constant or least count is the difference between the value of one main scale division and one Vernier scale division.

LC=1 MSD−1 VSD\text{LC} = 1 \, \text{MSD} - 1 \, \text{VSD}

First find the value of 1 VSD1 \, \text{VSD} from

50 VSD=49 MSD50 \, \text{VSD} = 49 \, \text{MSD}

Dividing by 5050,

1 VSD=4950 MSD1 \, \text{VSD} = \frac{49}{50} \, \text{MSD}

Now substitute 1 MSD=0.5 mm1 \, \text{MSD} = 0.5 \, \text{mm},

1 VSD=4950×0.5 mm1 \, \text{VSD} = \frac{49}{50} \times 0.5 \, \text{mm} 1 VSD=0.49 mm1 \, \text{VSD} = 0.49 \, \text{mm}

Hence,

LC=0.5 mm−0.49 mm\text{LC} = 0.5 \, \text{mm} - 0.49 \, \text{mm} LC=0.01 mm\text{LC} = 0.01 \, \text{mm}

Therefore, the correct numerical answer is 0.010.01.

Common mistakes

  • Using 5049\frac{50}{49} instead of 4950\frac{49}{50} for 1 VSD1 \, \text{VSD}. This is wrong because 5050 VSD equals 4949 MSD, so one VSD must be slightly smaller than one MSD. Write

    1 VSD=4950 MSD1 \, \text{VSD} = \frac{49}{50} \, \text{MSD}

    not the reverse.

  • Taking the Vernier constant as the value of 1 VSD1 \, \text{VSD} itself. This is wrong because the Vernier constant is the difference between 1 MSD1 \, \text{MSD} and 1 VSD1 \, \text{VSD}. After finding 1 VSD=0.49 mm1 \, \text{VSD} = 0.49 \, \text{mm}, subtract it from 0.5 mm0.5 \, \text{mm}.

  • Missing the unit conversion logic and writing the answer as 0.010.01 without recognizing it is in millimetres. The numerical answer field should contain only the number, but during working the quantity must be treated as 0.01 mm0.01 \, \text{mm}.

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