Q 11-01-081JEE MainJEE Main 2025 (24 Jan, Shift 1)Easy
The least count of a screw gauge is $0.01\ \text{mm}$. If the pitch is increased by $75\%$ and the number of divisions on the circular scale is reduced by $50\%$, the new least count will be ______ $\times10^{-3}\ \text{mm}$.
Numerical value type. Enter your answer.
Answer: 35
Least count $= \dfrac{\text{pitch}}{N}$.
$$LC' = \frac{1.75\,p}{0.5\,N} = 3.5\times\frac{p}{N} = 3.5\times0.01 = 0.035\ \text{mm} = 35\times10^{-3}\ \text{mm}$$
Solution by Sreeraj P, M.Sc Physics