JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Work, Power & Energy

Let the string length be $\ell$. Point A is at the bottom, D at the top (height $2\ell$ above A), and B, C sit symmetrically at 60° steps up from A: B is 60° up from A, and C is a further 60° up (equivalently 60° short of D). The height of a point at angle $\theta$ measured from the bottom is $\ell (1-\cos \theta )$, giving


The string goes slack at D, so gravity alone supplies the centripetal force there: $mg=\frac{m{v}_D^2}{\ell }\Rightarrow v_D^2=g\ell$.

Energy conservation from A to D (height gain $2\ell$):


Kinetic energy at B, from energy conservation with A (height $h_B=\ell /2$):


Kinetic energy at C, from energy conservation with D (height drop $\ell /2$ from D down to C):


The ratio is