JEE Main 2025 Jan 24 Shift 1, Physics Q22: Measurements

Least count of a screw gauge: $\text{LC}=\frac{\text{pitch}}{N}$, where $N$ is the number of circular-scale divisions. So

$\frac{{\text{LC}}_{\text{new}}}{{\text{LC}}_{\text{old}}}=\frac{{\text{pitch}}_{\text{new}}/N_{\text{new}}}{{\text{pitch}}_{\text{old}}/N_{\text{old}}}=\frac{1.75}{0.5}=3.5$

Therefore ${\text{LC}}_{\text{new}}=3.5\times 0.01\text{mm}=0.035\text{mm}=35\times 10^{-3}\text{mm}$.

$\therefore$ Answer: $35$.