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(B) The least count $(LC)$ of the Vernier callipers is given by:$LC = \frac{	ext{Value of 1 MS division}}{	ext{Total number of VS divisions}} = \fra

Vedclass · Accepted Answer

$0.59$

Vedclass · Answer

(B) The least count $(LC)$ of the Vernier callipers is given by:$LC = \frac{	ext{Value of 1 MS division}}{	ext{Total number of VS divisions}} = \frac{0.1\,cm}{10} = 0.01\,cm$.The observed diameter $d$ is calculated as: $d = 	ext{Main Scale Reading} + (VS 	ext{ division} 	imes LC)$.Given zero error $= -0.03\,cm$,the correction is $-(	ext{zero error}) = +0.03\,cm$.Corrected diameter $d' = d + 0.03\,cm$.For measurement $(1)$: $d_1 = 0.5 + (8 	imes 0.01) + 0.03 = 0.5 + 0.08 + 0.03 = 0.61\,cm$.For measurement $(2)$: $d_2 = 0.5 + (4 	imes 0.01) + 0.03 = 0.5 + 0.04 + 0.03 = 0.57\,cm$.For measurement $(3)$: $d_3 = 0.5 + (6 	imes 0.01) + 0.03 = 0.5 + 0.06 + 0.03 = 0.59\,cm$.Mean corrected diameter $= \frac{0.61 + 0.57 + 0.59}{3} = \frac{1.77}{3} = 0.59\,cm$.

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

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