JEE Main 2026 — Apr 2, Shift 2 — Physics solution

JEE Main2026Apr 2, Shift 2Physics

If a body of mass $1$ kg falls on the earth from infinity, it attains velocity ( $v$ ) and kinetic energy ( $k$ ) on reaching the surface of earth. The values of $v$ and $k$ respectively are______. (Take radius of earth to be $6400$ km and $g = 9.8$ m/s $^{2}$ )

A
$11.2$ km/s; $6.27 \times 1 0^{7}$ J
B
$11.2$ km/s; $12.54 \times 1 0^{7}$ J
C
$8.8$ km/s; $6.27 \times 1 0^{7}$ J
D
$8.8$ km/s; $12.54 \times 1 0^{7}$ J

Solution

By conservation of mechanical energy from infinity to surface: $0 = \frac{1}{2} m v^{2} - \frac{GM m}{R _{e}}$ .

Using $g = \frac{GM}{R _{e}^{2}}$ , we get $v = 2 g R_{e} = 2 \cdot 9.8 \cdot 6.4 \times 1 0^{6} \approx 1.12 \times 1 0^{4}$ m/s $= 11.2$ km/s.

Kinetic energy: $K = \frac{1}{2} m v^{2} = \frac{1}{2} (1) (11200)^{2} \approx 6.27 \times 1 0^{7}$ J.

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