Experimental Physics

JEE Main level, numerical answer, medium

9 of 70

Question 9 of 70

The least count of a screw guage is 0.01 mm . If the pitch is increased by 75%75 \% and number of divisions on the circular scale is reduced by 50%50 \%, the new least count will be _____ ×10−3 mm\times 10^{-3} \mathrm{~mm}

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Answer: 35

 Given least count of Screw Gauge =0.01 mm L.C =( pitch ) No. of circular turn =PN=0.01 mm New pitch =P(1+0.75)N(1−0.5)=PN[1.750.5]=(0.01)3.5=0.035 mm=35×10−3 mm\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 }=\frac{\mathrm{P}(1+0.75)}{\mathrm{N}(1-0.5)}=\frac{\mathrm{P}}{\mathrm{N}}\left[\frac{1.75}{0.5}\right] \\ & =(0.01) 3.5 \\ & =0.035 \mathrm{~mm} \\ & =35 \times 10^{-3} \mathrm{~mm}\end{aligned}

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