Question

A screwgauage has pitch $1.5mm$ and there is no zero error. Linear scale has marking at $MSD=1mm$ and there are $100$ equal division of circular scale. When diameter of a sphere is measured with instrument, main scale is having $2mm$ mark visible on linear scale, but $3mm$ mark is not visible, $76th$ division of circular scale is in line with linear scale. What is the diameter of sphere.

• A. $2.64mm$
• B. $3.14mm$
• C. $1.14mm$
• D. $2.76mm$

Answer 1

The correct option is B $3.14mm$

$\text{pitch}=1.5mm$

$L\cdot C$

$\begin{array}{c}=\frac{\text{pitch}}{\text{no of div on circular scale}}\\ =\frac{1.5}{100}=0.015mm\end{array}$

$\begin{array}{c}\text{MSR}=2mm\\ \text{Final reading}=2+\mathrm{CSR}\times L\cdot C\\ =2+76\times 0.015\\ =2+1.140\\ =3.14mm\end{array}$