QJEEM25Ph2e70: JEE Main 2025 Physics - Class 11 JEEM25Ph2e70 Solution: The least count of a screw guage is 0.01 mm . If the pitch is…

The least count (L.C.) of a screw gauge is given by:

$\text{L.C.} = \frac{\text{Pitch}}{\text{Number of Divisions on Circular Scale}}$

Let the original pitch be $p$ and the original number of divisions be $n$ . Then, the original least count is:

$\frac{p}{n} = 0.01 \, \text{mm}$

After modifications:

The pitch is increased by 75%, so the new pitch is:

$p_{\text{new}} = p + 0.75p = 1.75p$

The number of divisions is reduced by 50%, so the new number of divisions is:

$n_{\text{new}} = 0.5n$

The new least count is then:

$\text{L.C.}_{\text{new}} = \frac{p_{\text{new}}}{n_{\text{new}}} = \frac{1.75p}{0.5n} = \frac{1.75}{0.5} \times \frac{p}{n} = 3.5 \times \frac{p}{n}$

Substitute the original least count:

$\text{L.C.}_{\text{new}} = 3.5 \times 0.01 \, \text{mm} = 0.035 \, \text{mm}$

Expressed in the form $\times 10^{-3} \, \text{mm}$ , this becomes:

$0.035 \, \text{mm} = 35 \times 10^{-3} \, \text{mm}$

Thus, the new least count is:

$35 \times 10^{-3} \, \text{mm}$