JEE Physics 2025 Question with Solution

Given: Satellite mass is $\frac{M}{2}$ and height above Earth's surface is $\frac{R}{3}$.

Find: The value of $x$ in $L = M \sqrt{\frac{GMR}{x}}$.

First, the orbital radius from Earth's center is

$$r = R + \frac{R}{3} = \frac{4R}{3}$$

For a circular orbit, gravitational force provides the centripetal force, so

$$\frac{GMm}{r^2} = \frac{mv^2}{r}$$

Hence,

$$v = \sqrt{\frac{GM}{r}}$$

The angular momentum of the satellite is

$$L = mvr$$

Substituting $m = \frac{M}{2}$ and $r = \frac{4R}{3}$,

$$L = \frac{M}{2} \cdot \frac{4R}{3} \cdot \sqrt{\frac{GM}{\frac{4R}{3}}}$$ $$L = \frac{M}{2} \cdot \frac{4R}{3} \cdot \sqrt{\frac{3GM}{4R}}$$ $$L = \frac{2MR}{3} \cdot \frac{\sqrt{3}}{2} \sqrt{\frac{GM}{R}}$$ $$L = \frac{M\sqrt{3}}{3} \sqrt{GMR} = \frac{M}{\sqrt{3}} \sqrt{GMR}$$

Comparing with the given form

$$L = M \sqrt{\frac{GMR}{x}}$$

we get

$$\frac{1}{\sqrt{3}} = \frac{1}{\sqrt{x}}$$

Squaring both sides,

$$\frac{1}{3} = \frac{1}{x}$$

So,

$$x = 3$$

Therefore, the correct option is B.

Given: $m = \frac{M}{2}$ and orbital radius $r = \frac{4R}{3}$.

Find: The value of $x$.

Using

$$L = mvr$$

and

$$v = \sqrt{\frac{GM}{r}}$$

we write

$$L = mr\sqrt{\frac{GM}{r}} = m\sqrt{GMr}$$

Now substitute the given values:

$$L = \frac{M}{2} \sqrt{GM \cdot \frac{4R}{3}}$$ $$L = \frac{M}{2} \cdot 2 \sqrt{\frac{GMR}{3}}$$ $$L = M \sqrt{\frac{GMR}{3}}$$

Comparing with

$$L = M \sqrt{\frac{GMR}{x}}$$

we obtain

$$x = 3$$

Therefore, the correct option is B. The answer key points to option $4$, but the extracted solution clearly concludes $x = 3$, so the solution is taken as authoritative.