A student measured the diameter of a small steel ball using a screw gauge of least count $0.001 c m .$ The main scale reading is $5 m m$ and zero of circular scale division coincides with $25$ divisions above the reference level. If screw gauge has a zero error of - $0.004 c m,$ the correct diameter of the ball is

0.521 c m

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0.525 c m

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0.529 c m

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0.053 c m

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Solution

The correct option is A $0.529 c m$
$L \cdot C = 0.001 c m$

$M \cdot S \cdot R = 5 m m = 0.5 c m$

$C \cdot S \cdot R = coinciding division$

$\begin{matrix} = 25 \times 0.001 = 0.025 c m zero error = - 0.004 c m \end{matrix}$

$\begin{matrix} Final reading = m s R + C S R - z ero error = 0.5 + 0.025 - (- 0.004) = 0.525 + 0.004 = 0.529 c m \end{matrix}$

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