JEE MAIN - Physics (2023 - 8th April Evening Shift - No. 23)
A hollow spherical ball of uniform density rolls up a curved surface with an initial velocity $$3 \mathrm{~m} / \mathrm{s}$$ (as shown in figure). Maximum height with respect to the initial position covered by it will be __________ cm.
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Answer
75
Explanation
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Total initial kinetic energy
$$ =\frac{1}{2} m \mathrm{v}^2+\frac{1}{2} \mathrm{I} \omega^2 $$
$\mathrm{v}=\mathrm{R} \omega$ (for pure rolling)
$$ \mathrm{K} . E C=\frac{1}{2} m \mathrm{v}^2+\frac{1}{2} \times \frac{2}{3} m \mathrm{R}^2 \times \frac{\mathrm{v}^2}{\mathrm{R}^2}=\frac{5}{6} m \mathrm{v}^2 $$
Energy remains conserve during whole journey.
$\mathrm{K}_{\cdot} \mathrm{E}_i+\mathrm{P.E}_{\cdot i}=\mathrm{K}_{\cdot} \mathrm{E}_f+\mathrm{P.E}_{\cdot f}$
$$ \begin{aligned} & \Rightarrow \frac{5}{2} m \mathrm{v}^2=m g \mathrm{H} ~~~~~~~(\because {K.E.}_f=0)\\\\ & \Rightarrow \mathrm{H}=\frac{5}{6} \times \frac{\mathrm{v}^2}{g} \\\\ & =\frac{5 \times(3)^2}{6 \times 10} \\\\ & =\frac{15}{20} \mathrm{~m}=0.75 \mathrm{~m}=75 \mathrm{~cm} \end{aligned} $$
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