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JEE Main 4-April-2025 Paper - Shift 1 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 4-April-2025 Paper - Shift 14 Marks
Question
A circular ring and a solid sphere having same radius roll down on an inclined plane from rest without slipping. The ratio of their velocities when reached at the bottom of the plane is $\sqrt{\frac{x}{5}}$ where $\mathrm{x}=$ _____________.
✓

Answer

Applying Mechanical Energy conservation :
$\mathrm{k}_{\mathrm{i}}+\mathrm{U}_{\mathrm{i}}=\mathrm{k}_{\mathrm{f}}+\mathrm{U}_{\mathrm{f}}$
$\Rightarrow 0+\mathrm{Mgh}=\frac{1}{2} \mathrm{mv}^{2}\left(1+\frac{\mathrm{k}^{2}}{\mathrm{R}^{2}}\right)+0$
$\Rightarrow \mathrm{V}=\sqrt{\frac{2 \mathrm{gh}}{1+\frac{\mathrm{k}^{2}}{\mathrm{R}^{2}}}}$
So Ratio of velocities
$\frac{\mathrm{V}_{\text {Ring }}}{\mathrm{V}_{\text {solids sphere }}}=\sqrt{\frac{1+\frac{2}{5}}{1+1}}=\sqrt{\frac{7}{10}}$
$x=3.5$ Rounding off $x=4$

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