Q.

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 _____ ×1033 mm.

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answer is 35.

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Detailed Solution

Given least count of Screw Gauge = 0.01 mm
L.C=(pitch)No. of circular turn=PN=0.01mm
New pitch=P(1+0.75)N(10.5)=PN[1.750.5]
 =(0.01)3.5 =0.035mm =35×103mm Ans. is35

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