JEE Physics 2025 Question with Solution

Given: A particle is released from height $S$ above the earth with initial speed $0$. At height $h$ above the surface, its kinetic energy is three times its potential energy.

Find: The height $h$ and the speed $v$ at that instant.

Using the given condition,

$$\frac{1}{2}mv^2 = 3mgh$$

So,

$$v^2 = 6gh$$

Now apply conservation of mechanical energy between the release point and height $h$:

$$mgS = mgh + \frac{1}{2}mv^2$$

Since $\frac{1}{2}mv^2 = 3mgh$,

$$mgS = mgh + 3mgh = 4mgh$$

Therefore,

$$h = \frac{S}{4}$$

Substitute this into $v^2 = 6gh$:

$$v^2 = 6g\left(\frac{S}{4}\right) = \frac{3gS}{2}$$

Hence,

$$v = \sqrt{\frac{3gS}{2}}$$

Therefore, the height is $\frac{S}{4}$ and the speed is $\sqrt{\frac{3gS}{2}}$. The correct option is C.

Given: The particle falls from height $S$ to height $x$ above the surface.

Find: $x$ and $v$ when $KE = 3PE$.

The distance fallen is $S-x$. From kinematics,

$$v^2 = 2g(S-x)$$

At height $x$,

$$PE = mgx, \qquad KE = \frac{1}{2}mv^2$$

Given,

$$\frac{1}{2}mv^2 = 3mgx$$

So,

$$v^2 = 6gx$$

Now equate the two expressions for $v^2$:

$$2g(S-x) = 6gx$$ $$2S - 2x = 6x$$ $$2S = 8x$$ $$x = \frac{S}{4}$$

Then,

$$v^2 = 2g\left(S-\frac{S}{4}\right) = 2g\cdot \frac{3S}{4} = \frac{3gS}{2}$$

Thus,

$$v = \sqrt{\frac{3gS}{2}}$$

Therefore, the correct option is C.