Question

The pitch of the screw gauge is $1\text{mm}$ and there are $100$ divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies $8$ divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while ${72}^{\text{nd}}$ division on circular scale coincides with the reference line. The radius of the wire is :

• A. $1.64\text{mm}$
• B. $1.80\text{mm}$
• C. $0.82\text{mm}$
• D. $0.90\text{mm}$

Answer 1

The correct option is C $0.82\text{mm}$

Least count, $\text{LC}=\frac{\text{Pitch}}{\text{Number of divisions on circular scale}}$

$\Rightarrow \text{LC}=\frac{1}{100}=0.01\text{mm}$

Also, positive zero error $e=+8\times \text{LC}=+8\times 0.01=+0.08\text{mm}$

Now,
True reading
$=$ Measured reading $-e$
$=1+72\times 0.01-0.08=1.64\text{mm}$

Hence, radius $=\frac{1.64}{2}=0.82\text{mm}$