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 $5.10\text{cm}$ and $5.15\text{cm}$ of the main scale. The Vernier scale has $50$ divisions equivalent to $2.45\text{cm}$. The ${24}^{\text{th}}$ division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is

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

Answer 1

The correct option is D

$5.124\text{cm}$

We can see that $1MSD=0.05\text{cm}(\because$given "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."$)$, vernier scale has $50$ divisions equivalent $2.45\text{cm}$.

No. of $MSD$ in $2.45\text{cm}=\frac{2.45}{0.05}=49$

$\therefore 50VSD=49MSD$

$\Rightarrow 1VSD=\frac{49}{50}MSD$

Least count,

$L.C.=1MSD-1VSD=1MSD-\frac{49}{50}MSD$

$L.C.=\frac{1}{50}MSD$

$L.C.=\frac{1}{50}\times 0.05\text{cm}$

$L.C.=0.001\text{cm}$

We know that,

$\text{Final reading}\left(FR\right)=MSR+L.C\times VSR$

$\therefore FR=5.10+0.001\times 24=5.124\text{cm}$

Hence, option $\left(\text{B}\right)$ is the correct answer.