Physics · 2025
JEE Main · 24 January 2025, Shift 1 · Q46
The least count of a screw guage is 0.01 mm. If the pitch is increased by 75% and number of divisions on the circular scale is reduced by 50 %, the…
The least count of a screw guage is $\displaystyle 0.01$ mm. If the pitch is increased by $\displaystyle 75$% and number of divisions on the circular scale is reduced by $\displaystyle 50 \%$, the new least count will be $\displaystyle \_\_\_\_$ $\displaystyle \times 10^{-3} \mathrm{~mm}$
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35
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JEE Main 2025 Physics question, with the answer from NTA’s final answer key. Where our answers come from.