JEE Main 2024 — 8 April, Shift 1 — Question 48

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 $=\mathrm{MSR}+\mathrm{LC} \times \mathrm{VSR}$

$$\begin{aligned} & =2 \mathrm{~cm}+(0.01) \times(2) & =2.02 \mathrm{~cm} \end{aligned}$$

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

$$=4.32 \mathrm{~cm}^{3}$$

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