Question

The pitch of a screw gauge is $1\text{mm}$ and there are $100$ divisions on circular scale. When the faces $A$ and $B$ are just touching each without putting anything between the studs, the ${32}^{\text{nd}}$ division of the circular scale coincides with the reference line. The error in measurement due to screw gauge will be (zero of linear scale is not hidden from circular scale when $A$ and $B$ touches)

• A. $-0.04\text{mm}$
• B. $+0.32\text{mm}$
• C. $+0.16\text{mm}$
• D. $+0.04\text{mm}$

Answer 1

The correct option is B $+0.32\text{mm}$

Least count $\left(LC\right)=$ pitch$/$ number of divisions on circular scale
$\Rightarrow LC=\frac{1}{100}\text{mm}=0.01\text{mm}$
When $A$ and $B$ touches each other, zero is not hidden from the circular scale. Thus, the screw gauge is having a positive error.
$\Rightarrow e=n\times LC$
$\Rightarrow e=+32\times 0.01=+0.32\text{mm}$
Thus correct reading will be = Measured - Positive error