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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 _____ \(\times 10^{-3} \mathrm{~mm}\)

  1. A 30
  2. B 40
  3. C 35
  4. D 45
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Answer & Solution

Correct Answer

(C) 35

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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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