JEE Mains · Physics · STD 11 - 1. units,dimensions and measurement
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 new least count will be _____ \(\times 10^{-3} \mathrm{~mm}\)
- A 30
- B 40
- C 35
- D 45
Answer & Solution
Correct Answer
(C) 35
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { Given least count of Screw Gauge }=0.01 \mathrm{~mm} \\ & \text { L.C }=\frac{(\text { pitch })}{\text { No. of circular turn }}=\frac{\mathrm{P}}{\mathrm{N}}=0.01 \mathrm{~mm} \\ & \text { New pitch…
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