Question

The pitch of a screw gauge is $2\text{mm}$ and there are $100$ divisions on the circular scale. In measuring the diameter of a sphere, there are $6$ divisions on the linear scale and $40$ divisions on circular scale coinciding with the reference line. The diameter of the sphere is:

• A. $6.4\text{mm}$
• B. $12.8\text{mm}$
• C. $3.2\text{mm}$
• D. $20\text{mm}$

Answer 1

The correct option is B $12.8\text{mm}$

Least count $\left(\text{LC}\right)$ of screw gauge
= $\frac{\text{pitch}}{\text{no. of division on circular scale}}=\frac{2}{100}=0.02\text{mm}$

From the data given in the question,
Circular scale reading
$=\text{(LC)}\times \text{no. of division on circular scale coinciding}=0.02\times 40=0.8\text{mm}$
Linear scale reading $=\text{no. of division on linear scale coinciding}\times \text{pitch}=6\times 2=12\text{mm}$
Total reading = Linear scale reading + Circular scale reading
= 12 + 0.8 = 12.8 mm
$\therefore$ Diameter $=12.8\text{mm}$.
Hence, option (b) is the correct answer.