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

Answer 1

Step 1: Find the value of least count.
Formula used:
$L.C.=1MSD-1VSD$

Given that the zero of the Vernier scale lies between $5.10cm$ and $5.15cm$ of the main scale.

So, one main scale division $\left(MSD\right)$
$1MSD=5.15cm–5.10cm$
$1MSD=0.05cm\dots \left(i\right)$

and Vernier scale has $50$ divisions so one Vernier scale division $\left(VSD\right)$

$1VSD=\frac{2.45}{50}cm=0.049cm\dots \left(ii\right)$

As the Least count is,
$L.C.=1MSD-1VSD$

Substituting the values from equation $\left(i\right)$ and $\left(ii\right)$, we get
$L.C.=0.05-0.049=.001cm$

Step 2: Find the diameter
Formula used:
$D=MSR+VSR\times LC$

The diameter of the cylinder is
$D=MSR+VSR\times LC$

Substituting the values, we get
$D=5.10+24\times 0.001=5.124cm$

Final Answer: Option (b)