QJEEM24Phb66c: JEE Main 2024 Physics - Class 11 JEEM24Phb66c Solution: Least count of a vernier caliper is . The value of one division …

In a vernier caliper, the least count is the smallest distance measurable by the instrument. It can be defined using the difference between one main scale division and one vernier scale division. Given the least count, we can relate the number of divisions on the main scale to the divisions on the vernier scale.

The given least count of the vernier caliper is:

$\frac{1}{20 \mathrm{~N}} \mathrm{~cm}$

We know that the value of one division on the main scale is:

$1 \mathrm{~mm}$

To find the number of divisions on the main scale that coincide with N divisions of the vernier scale, let’s denote:

The number of divisions on the main scale = M

The number of divisions on the vernier scale = N

The least count formula for a vernier caliper is given by:

$\text{Least Count} = \text{Value of one main scale division} - \text{Value of one vernier scale division}$

The value of one main scale division is:

$1 \mathrm{~mm}$

The value of one vernier scale division can be expressed in terms of the number of divisions M and N:

$\text{Value of one vernier scale division} = \frac{M}{N} \mathrm{~mm}$

Given the least count:

$\frac{1}{20 \mathrm{~N}} \mathrm{~cm} = \frac{1}{20 \mathrm{~N}} \cdot 10 \mathrm{~mm} = \frac{1}{2 \mathrm{~N}} \mathrm{~mm}$

Using the least count formula, we have:

$\frac{1}{2 \mathrm{~N}} = 1 - \frac{M}{N}$

Rearranging the equation to solve for the number of main scale divisions (M), we get:

$1 - \frac{1}{2 \mathrm{~N}} = \frac{M}{N}$

Simplifying further:

$\frac{2 \mathrm{~N} - 1}{2 \mathrm{~N}} = \frac{M}{N}$

Multiplying both sides by N, we get the value of M:

$M = \frac{2 \mathrm{~N} - 1}{2}$

Therefore, the correct answer is option D:

$\left( \frac{2 \mathrm{~N} - 1}{2} \right)$