JEE MAIN - Physics (2023 - 1st February Morning Shift - No. 26)
A solid cylinder is released from rest from the top of an inclined plane of inclination $$30^{\circ}$$ and length $$60 \mathrm{~cm}$$. If the cylinder rolls without slipping, its speed upon reaching the bottom of the inclined plane is __________ $$\mathrm{ms}^{-1}$$. (Given $$\mathrm{g}=10 \mathrm{~ms}^{-2}$$)
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Answer
2
Explanation
Loss in potential energy $=m g h=m g\left[60 \sin 30^{\circ} \mathrm{cm}\right]$
$\Rightarrow m g\left[\frac{30}{100}\right]=\frac{1}{2} m v^{2}+\frac{1}{2} \frac{m v^{2}}{2}$
$\Rightarrow 0.3 \times 10=\frac{3}{4} v^{2}$
$\Rightarrow v^{2}=4$
$\Rightarrow v=2 \mathrm{~m} / \mathrm{s}$
$\Rightarrow m g\left[\frac{30}{100}\right]=\frac{1}{2} m v^{2}+\frac{1}{2} \frac{m v^{2}}{2}$
$\Rightarrow 0.3 \times 10=\frac{3}{4} v^{2}$
$\Rightarrow v^{2}=4$
$\Rightarrow v=2 \mathrm{~m} / \mathrm{s}$
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