Physics - Gravitation Question with Solution | TestHub

Loss of $\mathrm{KE}=$ Gain in $\mathrm{PE}$

$\Rightarrow \frac{1}{2}{\mathrm{mv}}^2=mgh\left(\frac{\mathrm{R}}{\mathrm{h}+\mathrm{R}}\right)$

CASE I.

$\Rightarrow \frac{1}{2}\mathrm{m}\left(\frac{2\mathrm{gR}}{3}\right)={\mathrm{mgh}}_1\left(\frac{\mathrm{R}}{{\mathrm{h}}_1+\mathrm{R}}\right)$     ...(i)

$\Rightarrow \frac{1}{2}m\left(\frac{4gR}{3}\right)={mgh}_2\left(\frac{R}{h_2+R}\right)$      ...(ii)

Solving (i)

$\Rightarrow \frac{1}{3}=\frac{h_1}{h_1+R}$

$\Rightarrow {\mathrm{h}}_1=\frac{\mathrm{R}}{2}$        ...(iii)

Solving (ii)

$\Rightarrow \frac{2}{3}=\frac{h_2}{h_2+R}$        ...(iv)

$\Rightarrow {\mathrm{h}}_2=2\mathrm{R}$

From (iii) and (iv)

$\Rightarrow \frac{{\mathrm{h}}_1}{{\mathrm{h}}_2}=\frac{\mathrm{R}/2}{2\mathrm{R}}$

$\therefore \frac{{\mathrm{h}}_1}{{\mathrm{h}}_2}=\frac{1}{4}=0.25$