Question

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 $\mathrm{5.1.0}cm$ and $5.10cm$ of the main scale. The Vernier scale has $50$ divisions equivalent to $2.45cm$. The $24th$ division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is

• A. $5.112cm$
• B. $5.124cm$
• C. $5.136cm$
• D. $5.148cm$

Answer 1

Answer: (B)

$5.124cm$

$\begin{array}{cc}MSR& =5.10cm\\ 1MSD& =5.15-5.10cm\\ & =0.05m\\ \text{50 VSD}=& 2.45cm\\ \text{1 VSD}& =0.049cm\end{array}$

$\begin{array}{cc}L\cdot C\cdot & =1MSD-1VSD\\ & =0.05-0.045\\ & =0.001cm\\ VSR& =24\times L\cdot C.\\ & =24\times 0.001\\ & =0.024cm\end{array}$

$\begin{array}{cc}\text{Final reading}& =MSR+VSR\\ & =5\cdot 10+0.024\\ & =5.124cm\end{array}$