JEE Main 2025 — Circular Motion Question with Solution

Step 1: Use Energy Conservation at Points A and B

At point A, the kinetic energy is from velocity and the potential energy is set at reference (usually zero). At point B, both kinetic and potential energy are present.

The conservation of energy formula between A and B is: $$\frac{1}{2}m\times 100+0=\frac{1}{2}m{V}_B^2+mg\left(R-\frac{R\sqrt{3}}{2}\right)$$

Step 2: Use Given Values

Given: $R=2$ m, $g=10m/{\mathrm{s}}^2$, and initial velocity at A is $10m/s$, so $(10{)}^2=100$.

Substitute these into the equation: $$100=V_B^2+2gR\left(1-\frac{\sqrt{3}}{2}\right)$$ Here, $m$ cancels out because it is on both sides.

Step 3: Simplify for $V_B^2$

Calculate $2gR=2\times 10\times 2=40$. So, $$100=V_B^2+40\left(1-\frac{\sqrt{3}}{2}\right)$$

$40(1-\frac{\sqrt{3}}{2})=40-20\sqrt{3}$

Rearrange to solve for $V_B^2$: $$V_B^2=100-40+20\sqrt{3}$$ $$V_B^2=60+20\sqrt{3}$$

So, kinetic energy at point B: $$K.E{.}_B=\frac{1}{2}m{V}_B^2=\frac{m}{2}(60+20\sqrt{3})$$

Step 4: Use Energy Conservation at Points A and C

At point C, use conservation of energy again: $$\frac{1}{2}m(10{)}^2=\frac{1}{2}m{V}_C^2+mg\left(\frac{3R}{2}\right)$$

The potential energy term is $mg\times \frac{3R}{2}$. Substitute values: $$100=V_C^2+2gR\times \frac{3}{2}$$ $2gR=40$, so $40\times \frac{3}{2}=60$

So: $$100=V_C^2+60$$ $$V_C^2=100-60=40$$

So, kinetic energy at point C: $$K.E{.}_C=\frac{1}{2}m{V}_C^2=\frac{1}{2}m(40)$$

Step 5: Find the Ratio

The ratio of kinetic energies at B and C is: $$\frac{K.E{.}_B}{K.E{.}_C}=\frac{\frac{1}{2}m(60+20\sqrt{3})}{\frac{1}{2}m(40)}=\frac{60+20\sqrt{3}}{40}=\frac{3}{2}+\frac{\sqrt{3}}{2}=\frac{3+\sqrt{3}}{2}$$