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 10^-3 mm

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Correct answer: 35. (35) Given least count of Screw Gauge =0.01 mm L .C= ( pitch ) No . of circular turn= P N=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

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

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