The diameter of a cylinder is measured using a Vernier callipers with no zero error. It is found that the zero of the Vernier scale lies between $$5.10 \,cm$$ and $$5.15 \,cm$$ of the main scale. The Vernier scale has $$50$$ divisions equivalent to $$2.45 \,cm.$$ The $${24^{th}}$$ division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is-
A.
$$5.112 \,cm$$
B.
$$5.124 \,cm$$
C.
$$5.136 \,cm$$
D.
$$5.148 \,cm$$
Answer :
$$5.124 \,cm$$
Solution :
Reading $$=$$ M.S.R. $$+$$ Number of division of V.S. matching the main scale division (1 M.S.D. $$-$$ 1 V.S.D.)
$$\eqalign{
& = 5.10 + 24\left( {0.05 - \frac{{2.45}}{{50}}} \right) \cr
& = 5.124\,cm \cr} $$
Option (B) is correct
Releted MCQ Question on Basic Physics >> Unit and Measurement
Releted Question 1
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