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(D) The least count $(LC)$ of the screw gauge is $0.005 \ mm$. The number of divisions on the circular scale is $N = 200$. The pitch of the screw is $

Vedclass · Accepted Answer

$2.05$

Vedclass · Answer

(D) The least count $(LC)$ of the screw gauge is $0.005 \ mm$. The number of divisions on the circular scale is $N = 200$. The pitch of the screw is $P = LC 	imes N = 0.005 \ mm 	imes 200 = 1 \ mm$. Zero error $(ZE)$ = $5 	imes LC = 5 	imes 0.005 \ mm = 0.025 \ mm$. When the sphere is placed,the main scale reading $(MSR)$ is $4 	imes P = 4 	imes 1 \ mm = 4 \ mm$. The circular scale reading $(CSR)$ is $5 	imes 5 = 25$ divisions. The observed reading $(OR)$ = $MSR + (CSR 	imes LC) = 4 \ mm + (25 	imes 0.005 \ mm) = 4 \ mm + 0.125 \ mm = 4.125 \ mm$. The true diameter $(D)$ = $OR - ZE = 4.125 \ mm - 0.025 \ mm = 4.1 \ mm$. The radius of the sphere is $r = D / 2 = 4.1 \ mm / 2 = 2.05 \ mm$.

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