Question

In a screw gauge, the fifth division of the circular scale coincides with the reference line, when the ratchet is closed. There are $50$ divisions on the circular scale, and the main scale moves by $0.5\text{mm}$ on a complete rotation. For a particular observation, the reading on the main scale is $5\text{mm}$ and the ${20}^{\text{th}}$ division of the circular scale coincides with the reference line. Calculate the true reading.

• A. $5.20\text{mm}$
• B. $5.25\text{mm}$
• C. $5.15\text{mm}$
• D. $5.00\text{mm}$

Answer 1

The correct option is C $5.15\text{mm}$

The distance covered by the main scale in one complete rotation is equal to pitch.
$\therefore \text{pitch}=0.5\text{mm}$

The least count of the screw gauge is,

$LC=\frac{\text{pitch of the screw}}{\text{total no. of div. on circular scale}}$

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

$\text{Zero error}=(CSR\times LC)=5\times 0.01=0.05\text{mm}$

Now, the fifth division of the circular scale coincides with the reference line, when the ratchet is closed.
$\Rightarrow$ The zero error should be subtracted from the reading obtained, to get the true reading.

$\therefore \text{True reading}=MSR+(CSR\times LC)-\text{zero error}$

$=5+(20\times 0.01)-0.05=5.15\text{mm}$

Hence, option $\left(\text{C}\right)$ is correct.