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QJEEM23Ph41a9: JEE Main 2023 Physics - Class 11 JEEM23Ph41a9 Solution: A solid cylinder is released from rest from the top of an…

A solid cylinder is released from rest from the top of an inclined plane of inclination 30∘30^{\circ} and length 60 cm60 \mathrm{~cm} . If the cylinder rolls without slipping, its speed upon reaching the bottom of the inclined plane is __________ ms−1\mathrm{ms}^{-1} . (Given g=10 ms−2\mathrm{g}=10 \mathrm{~ms}^{-2} )

JEE Main 2023 (Online) 1st February Morning Shift Physics - Rotational Motion Question 56 English

Step-by-Step Verified Solution

Loss in potential energy =mgh=mg[60sin⁡30∘cm]=m g h=m g\left[60 \sin 30^{\circ} \mathrm{cm}\right]

⇒mg[30100]=12mv2+12mv22\Rightarrow m g\left[\frac{30}{100}\right]=\frac{1}{2} m v^{2}+\frac{1}{2} \frac{m v^{2}}{2}

⇒0.3×10=34v2\Rightarrow 0.3 \times 10=\frac{3}{4} v^{2}

⇒v2=4\Rightarrow v^{2}=4

⇒v=2 m/s\Rightarrow v=2 \mathrm{~m} / \mathrm{s}

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