Diameter of a steel ball is measured using a Vernier callipers which has divisions of 0. 1,cm on its

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1.Units, Dimensions and Measurement

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Diameter of a steel ball is measured using a Vernier callipers which has divisions of $0. 1\,cm$ on its main scale $(MS)$ and $10$ divisions of its vernier scale $(VS)$ match $9$ divisions on the main scale. Three such measurements for a ball are given as

S.No. $MS\;(cm)$ $VS$ divisions

$(1)$ $0.5$ $8$

$(2)$ $0.5$ $4$

$(3)$ $0.5$ $6$

If the zero error is $- 0.03\,cm,$ then mean corrected diameter is ........... $cm$

A

$0.52$

B

$0.59$

C

$0.56$

D

$0.53$

(JEE MAIN-2015)

Solution

$\begin{array}{l}
Lets\,count\, = \frac{{0.1}}{{10}} = 0.01\,\,cm\\
{d_1} = 0.5 + 8 \times 0.01 + 0.03 = 0.61\,cm\\
{d_2} = 0.5 + 4 \times 0.01 + 0.03 = 0.57\,cm\\
{d_3} = 0.5 + 6 \times 0.01 + 0.03 = 0.59\,cm\\
Mean\,diameter\, = \frac{{0.61 + 0.57 + 0.59}}{3}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.59\,cm
\end{array}$

Standard 11

Physics

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