Question

The diameter of a cylinder is measured using a Vernier calipers with no zero error. It is found that the zero of the Vernier scale lies between $5.10\text{cm}$ and $5.15\text{cm}$ of the main scale. The Vernier scale has $50$ divisions equivalent to $2.45\text{cm}$. The ${24}^{th}$ division of the Vernier scale excatly coincides with one of the main scale divisions. The diameter of the cylinder is :-

• A. $5.112\text{cm}$
• B. $5.124\text{cm}$
• C. $5.136\text{cm}$
• D. $5.148\text{cm}$

Answer 1

The correct option is B $5.124\text{cm}$

We can see that,

Given that ("It is found that the zero of the Vernier scale lies between $5.10\text{cm}$ and $5.15\text{cm}$ of the main scale.")
$1MSD=0.05\text{cm}$

Vernier scale has $50$ divisions equivalent $2.45\text{cm}$.

$50$ divisions $=2.45\text{cm}$
$\Rightarrow$ $1$ division $=\frac{2.45}{50}=0.049\text{cm}$
$\Rightarrow$ least count $1MSD-1VSD=0.05-0.049=0.001\text{cm}$

Final reading $=MSR+L.C\times VSR$

$=5.10+0.001\times 24$

$=5.124\text{cm}$