Question

The diameter of a sphere is measured using a vernier caliper whose 9 divisions of main scale are equal to 10 divisions of vernier scale. The shortest division on the main scale is equal to \(1 \mathrm{~mm}\). The main scale reading is \(2 \mathrm{~cm}\) and second division of vernier scale coincides with a division on main scale. If mass of the sphere is \(8.635 \mathrm{~g}\), the density of the sphere is:

(1) \( 2.5 \mathrm{~g} / \mathrm{cm}^{3}\)

(2) \(1.7 \mathrm{~g} / \mathrm{cm}^{3}\)

(3) \(2.2 \mathrm{~g} / \mathrm{cm}^{3}\)

(4) \( 2.0 \mathrm{~g} / \mathrm{cm}^{3}\)  

Answer 1

Correct option is : (4) \(2.0 \mathrm{~g} / \mathrm{cm}^{3}\) 

Given  \(9 \mathrm{MSD}=10 \mathrm{VSD}\)

mass \(=8.635 \mathrm{~g}\)

\(\mathrm{LC}=1 \ \mathrm{MSD}-1 \ \mathrm{VSD}\)

\(\mathrm{LC}=1 \ \mathrm{MSD}-\frac{9}{10} \mathrm{MSD}\)

\(\mathrm{LC}=\frac{1}{10} \mathrm{MSD}\)

\(\mathrm{LC}=0.01 \mathrm{~cm}\)

Reading of diameter \(=\mathrm{MSR}+\mathrm{LC} \times \mathrm{VSR}\)

\( \begin{aligned} & =2 \mathrm{~cm}+(0.01) \times(2) \end{aligned} \) 

\( \begin{aligned} =2.02 \mathrm{~cm} \end{aligned} \)

Volume of sphere \(=\frac{4}{3} \pi\left(\frac{\mathrm{d}}{2}\right)^{3}=\frac{4}{3} \pi\left(\frac{2.02}{2}\right)^{3}\)

\( =4.32 \mathrm{~cm}^{3} \)

Density \(=\frac{\text { mass }}{\text { volume }}=\frac{8.635}{4.32}=1.998 \sim 2.00 \mathrm{~g}\)