Question

A screw gauge with a pitch of $0.5\text{mm}$ and a circular scale with $50$ divisions is used to measure the thickness of a thin sheet of aluminium. What is the thickness of the sheet if the main scale reading is $0.4\text{mm}$ and $17th$ division on the circular scale coincides with the main scale line?

Answer 1

We firstly need to calculate the least count of the screw gauge.

The Least count $LC$ of the screw gauge is given by

$LC=\frac{\text{Pitch}}{\text{No. of divisions on circular scale}}$

$LC=\frac{0.5\text{mm}}{50}=0.01\text{mm}$

We know that

Measured value $\left(MV\right)$= Main scale reading $\left(MSR\right)$ + Circular scale reading $\left(CSR\right)$

Here, main scale reading, $MSR=0.4\text{mm}$

And we know that
Circular scale reading = circular scale division $\times$ the least count

$\Rightarrow CSR=17\times 0.01\text{mm}=0.17\text{mm}$

So, Measured value $=(0.4+0.17)\text{mm}=0.57\text{mm}$

So, option $\left(c\right)$ is the correct choice.