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QJEEM24Ph612e: JEE Main 2024 Physics - Class 11 JEEM24Ph612e Solution: One main scale division of a vernier caliper is equal to units. …

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

Multiple Choice Options

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

Step-by-Step Verified Solution

The least count of a vernier caliper is defined as the smallest distance that it can measure and is calculated by the difference in length between one main scale division and one vernier scale division. It can be represented as:

Least Count=Main scale division−Vernier scale division\text{Least Count} = \text{Main scale division} - \text{Vernier scale division}

Given that one main scale division is equal to m\mathrm{m} units and the nth \mathrm{n}^{\text {th }} division of main scale coincides with the (n+1)th (n+1)^{\text {th }} division of the vernier scale, this means that nn divisions on the main scale is equal to (n+1)(n+1) divisions on the vernier scale.

Since one main scale division is m\mathrm{m} units, nn divisions on the main scale would be n×mn \times \mathrm{m} units. If nn divisions on the main scale are equal to (n+1)(n+1) divisions on the vernier scale, we can determine the length of one vernier scale division as

Length of one vernier scale division=n×mn+1\text{Length of one vernier scale division} = \frac{n \times \mathrm{m}}{n+1}

Thus, the least count, which is the difference between one main scale division and one vernier scale division, is:

Least Count=m−n×mn+1=m(1−nn+1)=m(n+1−nn+1)=m(1n+1)\text{Least Count} = \mathrm{m} - \frac{n \times \mathrm{m}}{n+1} = \mathrm{m} \left(1 - \frac{n}{n+1} \right) = \mathrm{m} \left( \frac{n+1-n}{n+1} \right) = \mathrm{m} \left( \frac{1}{n+1} \right)

This simplifies to:

Least Count=mn+1\text{Least Count} = \frac{\mathrm{m}}{\mathrm{n + 1}}

Therefore, the correct option is:

Option B: mn+1\frac{\mathrm{m}}{\mathrm{n+1}}

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