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A solid sphere is projected up along an inclined plane of inclination $30^{\circ}$ with the horizontal with a speed of $4 \mathrm{~ms}^{-1}$. If it rolls without slipping, the maximum distance traversed by it is ........$\left(g=10 \mathrm{~ms}^{-2}\right)$
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The correct answer is:
2.24 m
Given, $u=4 \mathrm{~ms}^{-1}$
$$
a=\frac{g \sin \theta}{1+\frac{I}{m R^2}}=\frac{10 \times \sin 30}{1+\frac{2}{5}}=\frac{5}{\frac{7}{5}}=\frac{25}{7} \mathrm{~m} / \mathrm{s}^2
$$
As $v^2=u^2+2 a S$
$$
\Rightarrow|S|=\frac{u^2}{2 a}=\frac{4^2}{2 \times \frac{25}{7}}=\frac{16 \times 7}{50}=\frac{112}{50}=2.24 \mathrm{~m}
$$
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