The least count of a screw gauge is 0.01 mm. If the pitch is increased by 75% and the number of divisions on the circular scale is reduced by 50%, the new least count will be × 10⁻³ mm.

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Accepted Answer

35. L.C. =(pitch)/N. New pitch =1.75, new N=0.5 N. New L.C. =(1.75 pitch)/(0.5 N)=3.5×(pitch)/N=3.5×0.01=0.035 mm=35×10⁻³ mm.

The least count of a screw gauge is 0.01 mm. If the pitch is increased by 75% and the number of divisions on the circular scale is reduced by 50%, the new least count will be ______ × 10⁻³ mm.

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