JEE MAIN - Physics (2022 - 29th July Evening Shift - No. 11)
An object of mass $$1 \mathrm{~kg}$$ is taken to a height from the surface of earth which is equal to three times the radius of earth. The gain in potential energy of the object will be [If, $$\mathrm{g}=10 \mathrm{~ms}^{-2}$$ and radius of earth $$=6400 \mathrm{~km}$$ ]
48 MJ
24 MJ
36 MJ
12 MJ
Explanation
$\Delta U=U_{f}-U_{i}$
$ \begin{aligned} &=-\frac{G M m}{4 R}+\frac{G M m}{R} \\\\ &=\frac{3 G M m}{4 R}=\frac{3}{4} m g R \\\\ &=48 \mathrm{MJ} \end{aligned} $
$ \begin{aligned} &=-\frac{G M m}{4 R}+\frac{G M m}{R} \\\\ &=\frac{3 G M m}{4 R}=\frac{3}{4} m g R \\\\ &=48 \mathrm{MJ} \end{aligned} $
Comments (0)
