A ball of mass m is dropped from a height H. At height H/3, the ratio of its potential energy (PE) to kinetic energy (KE) is equal to:

Correct option is C

​​​$$\text{Initially, the ball is at height } H,\text{ so its total mechanical energy is:}\\[6pt]E = mgH\\[10pt]\text{When the ball reaches a height of } \frac{H}{3} \text{ above the ground:}\\[10pt]\text{Potential Energy}\\[6pt]PE = mg\left(\frac{H}{3}\right) = \frac{mgH}{3}\\[10pt]\text{Kinetic Energy}\\[6pt]\text{By conservation of mechanical energy:}\\[6pt]KE = mgH - \frac{mgH}{3}\\KE = \frac{2mgH}{3}\\[10pt]\text{Ratio of PE to KE}\\[6pt]\frac{PE}{KE}=\frac{\frac{mgH}{3}}{\frac{2mgH}{3}}=\frac{1}{2}\\[10pt]\therefore \quad \frac{PE}{KE} = \frac{1}{2}$$​​