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Physics · Rotational Dynamics

JEE Advanced 2018 — Paper 1 — Question 9

A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle 60∘60^{\circ} with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is (2−3)/10s(2-\sqrt{3}) / \sqrt{10} s, then the height of the top of the inclined plane, in metres, is _____\_\_\_\_\_ . Take g=10 ms−2g=10 \mathrm{~ms}^{-2}.

Answer: 0.75

Numerical answer — enter this value.

Step-by-step solution

hsin⁡θ=12gsin⁡θ1+ImR2t2\frac{h}{\sin \theta}=\frac{1}{2} \frac{g \sin \theta}{1+\frac{I}{m R^{2}}} t^{2}

∴t=1sin⁡θ2hg(1+ImR2)\therefore t=\frac{1}{\sin \theta} \sqrt{\frac{2 h}{g}\left(1+\frac{I}{m R^{2}}\right)}

∴2−310=232h10(2−32)\therefore \frac{2-\sqrt{3}}{\sqrt{10}}=\frac{2}{\sqrt{3}} \sqrt{\frac{2 h}{10}}\left(\sqrt{2}-\sqrt{\frac{3}{2}}\right)

∴h=0.75m\therefore h=0.75 m

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2018
Paper
Paper 1
Subject
Physics
Chapter
Rotational Dynamics
Topic
Rolling Motion

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