Q80 / 90 · JEE Main 2021 (22 Jul Shift 1) · Physics

A body is projected vertically upwards from the surface of earth with a velocity sufficient enough to carry it to infinity. The time taken by it to reach height $h$ is $S .$

A
$\sqrt{\frac{R_{e}}{2 g}} [{(1 + \frac{h}{R_{e}})}^{\frac{3}{2}} - 1]$
B
$\sqrt{\frac{2 R_{e}}{g}} [{(1 + \frac{h}{R_{e}})}^{\frac{3}{2}} - 1]$
C
$\frac{1}{3} \sqrt{\frac{R_{e}}{2 g}} [{(1 + \frac{h}{R_{c}})}^{\frac{3}{2}} - 1]$
Correct answer
D
$\frac{1}{3} \sqrt{\frac{2 R_{c}}{g}} [{(1 + \frac{h}{R_{e}})}^{\frac{3}{2}} - 1]$

Solution

Applying energy conservation from ( $(1) to (2)$ ,
$\frac{1}{2} m \cdot (\frac{2 GM}{R_{e}}) - \frac{G M m}{R_{e}} = \frac{1}{2} m v^{2} - \frac{G M m}{R + r}$

$\Rightarrow \frac{1}{2} m v^{2} = \frac{G M m}{R + r}$

$\Rightarrow v = \sqrt{\frac{2 G M}{R + r}} = \frac{d r}{d t}$

$\Rightarrow \sqrt{2 G M} \int_{0}^{t} d t = \int_{R_{e}}^{R_{e} + h} (\sqrt{R + r}) d r$

$\sqrt{2 G M} \cdot t = \frac{2}{3} {[(R + r)^{\frac{3}{2}}]}_{R_{e}}^{R_{e} + h}$

$t = \frac{2}{3} \sqrt{\frac{R_{e}^{3}}{2 GM}} [{(1 + \frac{h}{R_{e}})}^{\frac{3}{2}} - 1]$

$\frac{G M}{R_{e}^{2}} = g$

$t = \frac{1}{3} \sqrt{\frac{2 R_{e}}{g}} [{(1 + \frac{h}{R_{e}})}^{\frac{3}{2}} - 1]$

Hence, the correct answer is option: 'D'.

A body is projected vertically upwards from the surface of earth with a velocity sufficient enough to carry it to infinity. The time taken by it to reach height h is S.

Solution

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A body is projected vertically upwards from the surface of earth with a velocity sufficient enough to carry it to infinity. The time taken by it to reach height $h$ is $S .$