Question

When the gap is closed without placing any object in the screw gauge whose least count is $0.005\text{mm}$, the $5^{th}$ division on its circular scale coincides with the reference line on main scale. When a small sphere is placed, the reading on the main scale advances by 4 divisions, whereas the ${25}^{th}$ division on its circular scale coincides with the reference line on the main scale. There are 200 divisions on the circular scale. The radius of the sphere measured is

• A. $4.10\text{mm}$
• B. $4.05\text{mm}$
• C. $2.10\text{mm}$
• D. $2.05\text{mm}$

Answer 1

The correct option is D $2.05\text{mm}$

Pitch of the screw gauge $=L.C\times$ No. of circular scale divisions $\left(CSD\right)$
$=200\times 0.005=1\text{mm}$
& Zero error $(Z.E)=0.005\times 5$

Reading of the diameter of the sphere
$d=MSD+(CSD\times L.C)-Z.E$
$d=4\text{mm}+25\times 0.005-0.005\times 5$
$=4.10\text{mm}$
$\therefore r=\frac{d}{2}=2.05\text{mm}$ is the radius of the sphere.