Student A and Student B used two screw gauges | vedclass

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(A) The least count $(LC)$ of the screw gauge is given by:$LC = \frac{	ext{Pitch}}{	ext{Total circular divisions}} = \frac{0.1 \, 	ext{cm}}{100} =

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(A) The least count $(LC)$ of the screw gauge is given by:$LC = \frac{	ext{Pitch}}{	ext{Total circular divisions}} = \frac{0.1 \, 	ext{cm}}{100} = 0.001 \, 	ext{cm}$.For Student $A$:Zero error $= +5 	imes LC = +0.005 \, 	ext{cm}$.Observed reading $= 	ext{MSR} + (	ext{CSR} 	imes LC) = 0.322 \, 	ext{cm}$.Assuming the main scale reading $(MSR)$ is $0.300 \, 	ext{cm}$,then:$0.300 + (	ext{CSR}_A 	imes 0.001) - 0.005 = 0.322 \implies 	ext{CSR}_A 	imes 0.001 = 0.027 \implies 	ext{CSR}_A = 27$.For Student $B$:Zero error $= -8 	imes LC = -0.008 \, 	ext{cm}$ (since $92$ is aligned,error is $92-100 = -8$).Assuming the main scale reading $(MSR)$ is $0.300 \, 	ext{cm}$,then:$0.300 + (	ext{CSR}_B 	imes 0.001) - (-0.008) = 0.322 \implies 	ext{CSR}_B 	imes 0.001 = 0.014 \implies 	ext{CSR}_B = 14$.Difference in circular scale readings $= |27 - 14| = 13$.

$S$.No.	$MS\;(cm)$	$VS$ divisions
$(1)$	$0.5$	$8$
$(2)$	$0.5$	$4$
$(3)$	$0.5$	$6$

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