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JEE Advanced 2018Paper 1
Exam
JEE Advanced
Year
2018
Paper
Paper 1
Subject
Physics
Chapter
Rotational Motion

JEE Advanced 2018 · Paper 1 · Physics · Rotational Motion

A ring and a disc are initially at rest, side by side, at the top...

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 ° 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\left(2 - \sqrt{3}\right) / \sqrt{10}s, then the height of the top of the inclined plane (in meters) is __________. Take g=10ms−2g = 10m{s}^{- 2}.A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle 60 ° with

Correct answer

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Step-by-step solution

A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle 60 ° with
ac=gsin\theta1+IcMR2{a}_{c} = \dfrac{gsin\text{\theta}}{1 + \dfrac{{I}_{c}}{M{R}^{2}}}
aring=gsin\theta2{a}_{ring} = \dfrac{gsin\text{\theta}}{2}
aring=2gsin\theta3{a}_{ring} = \dfrac{2gsin\text{\theta}}{3}
hsin\theta=12(gsin\theta2)t12⇒t1=4hgsin⁡2\theta=16h3g\dfrac{h}{sin\text{\theta}} = \dfrac{1}{2}\left(\dfrac{gsin\text{\theta}}{2}\right){t}_{1}^{2} ⇒ {t}_{1} = \sqrt{\dfrac{4h}{g\sin^{2}\text{\theta}}} = \sqrt{\dfrac{16h}{3g}}
⇒hsin\theta=12(2gsin\theta3)t22⇒t2=3hgsin⁡2\theta=4hg⇒ \dfrac{h}{sin\text{\theta}} = \dfrac{1}{2}\left(\dfrac{2gsin\text{\theta}}{3}\right){t}_{2}^{2} ⇒ {t}_{2} = \sqrt{\dfrac{3h}{g\sin^{2}\text{\theta}}} = \sqrt{\dfrac{4h}{g}}
⇒16h3g−4hg=2−310⇒ \sqrt{\dfrac{16h}{3g}} - \sqrt{\dfrac{4h}{g}} = \dfrac{2 - \sqrt{3}}{\sqrt{10}}
h[43−2]=2−3\sqrt{h}\left[\dfrac{4}{\sqrt{3}} - 2\right] = 2 - \sqrt{3}
h=(2−3)3(4−23)=32⇒h=34=0.75m\sqrt{h} = \dfrac{\left(2 - \sqrt{3}\right)\sqrt{3}}{\left(4 - 2\sqrt{3}\right)} = \dfrac{\sqrt{3}}{2} ⇒ h = \dfrac{3}{4} = 0.75m
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