JEE Physics 2025 Question with Solution

Given: Original least count is $0.01 \, \text{mm}$. The pitch is increased by $75\%$, so the new pitch becomes $1.75P$. The number of divisions on the circular scale is reduced by $50\%$, so the new number of divisions becomes $\frac{N}{2}$.

Find: The new least count of the screw gauge.

The least count of a screw gauge is given by

$$LC = \frac{\text{Pitch}}{\text{Number of divisions on the circular scale}}$$

Using the changed values,

$$LC_{\text{new}} = \frac{1.75P}{N/2}$$ $$LC_{\text{new}} = \frac{1.75P \times 2}{N} = 3.5\frac{P}{N}$$

Since

$$LC_{\text{original}} = \frac{P}{N} = 0.01 \, \text{mm}$$

therefore,

$$LC_{\text{new}} = 3.5 \times 0.01 = 0.035 \, \text{mm}$$

Therefore, the new least count is $0.035 \, \text{mm}$.

Given: Least count varies directly with pitch and inversely with the number of circular scale divisions.

Find: The multiplication factor for the new least count.

Because

$$LC \propto \frac{P}{N}$$

if pitch becomes $1.75$ times and divisions become $\frac{1}{2}$ times, then the least count changes by the factor

$$\frac{LC_{\text{new}}}{LC_{\text{old}}} = \frac{1.75}{1/2} = 3.5$$

So,

$$LC_{\text{new}} = 3.5 \times 0.01 = 0.035 \, \text{mm}$$

Hence, the required numerical value is 0.035.