JEE Main 2024 — 9 April, Shift 2 — Question 40

Total energy $=\frac{-\mathrm{GMm}}{2(2 \mathrm{R})}$

if energy $=\frac{10^{4} R}{6}$ is added then

$\frac{-\mathrm{GMm}}{4 \mathrm{R}}+\frac{10^{4} \mathrm{R}}{6}=\frac{-\mathrm{GMm}}{2 \mathrm{r}}$

where $r$ is new radius of revolving and $g=\frac{G M}{R^{2}}$

$-\frac{\mathrm{mgR}}{4}+\frac{10^{4} \mathrm{R}}{6}=-\frac{\mathrm{mgR}^{2}}{2 \mathrm{r}}\left(\mathrm{m}=10^{3} \mathrm{~kg}\right)$

$-\frac{10^{3} \times 10 \times \mathrm{R}}{4}+\frac{10^{4} \mathrm{R}}{6}=-\frac{10^{3} \times 10 \times \mathrm{R}^{2}}{2 \mathrm{r}}$

$-\frac{1}{4}+\frac{1}{6}=-\frac{\mathrm{R}}{2 \mathrm{r}}$

$r=6 R$