JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Work, Power & Energy

Let A be the lowest point of the vertical circle, with O the centre and radius R = 2 m. From the figure, B lies at 30° from A (measured at the centre) and C lies at 120° from A.

The height of a point at angle $\theta$ from the bottom of the circle, measured above A, is $h=R(1-\cos \theta )$.

For B ($\theta =30°$): $h_B=R\left(1-\frac{\sqrt{3}}{2}\right)$
For C ($\theta =120°$): $h_C=R\left(1+\frac{1}{2}\right)=\frac{3R}{2}$

By conservation of mechanical energy, taking A as reference, with $v_A=10$ m/s, R = 2 m, g = 10 m/s²:

$v_B^2=v_A^2-2g{h}_B=100-2(10)(2)\left(1-\frac{\sqrt{3}}{2}\right)=100-40+20\sqrt{3}=60+20\sqrt{3}$

$v_C^2=v_A^2-2g{h}_C=100-2(10)(2)\left(\frac{3}{2}\right)=100-60=40$

Since kinetic energy is proportional to $v^2$ for the same mass:
$\frac{K{E}_B}{K{E}_C}=\frac{v_B^2}{v_C^2}=\frac{60+20\sqrt{3}}{40}=\frac{3+\sqrt{3}}{2}$