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QJEEM19Phe65c: JEE Main 2019 Physics - Class 11 JEEM19Phe65c Solution: The following bodies are made to roll up (without slipping) the …

The following bodies are made to roll up (without slipping) the same inclined plane from a horizontal plane. : (i) a ring of radius R, (ii) a solid cylinder of radius R/2 and (iii) a solid sphere of radius R/4 . If in each case, the speed of the centre of mass at the bottom of the incline is same, the ratio of the maximum heights they climb is :

Multiple Choice Options

  • A 20:15:14
  • B 4:03:02
  • C 2:03:04
  • D 10:15:07

Step-by-Step Verified Solution

Total kinetic energy of a rolling body is given as

Etotal =12mv2[1+K2R2]E_{\text {total }}=\frac{1}{2} m v^2\left[1+\frac{K^2}{R^2}\right]

where, KK is the radius of gyration.

Using conservation law of energy,

12mv2[1+K2R2]=mgh or h=v22g[1+K2R2]\begin{array}{rlrl}\frac{1}{2} m v^2\left[1+\frac{K^2}{R^2}\right] =m g h \\\\\text { or } h =\frac{v^2}{2 g}\left[1+\frac{K^2}{R^2}\right]\end{array}

For ring, K2R2=1\frac{K^2}{R^2}=1

⇒h1=v22g[1+1]=2v22g=v2g\Rightarrow h_1=\frac{v^2}{2 g}[1+1]=\frac{2 v^2}{2 g}=\frac{v^2}{g}

For solid cylinder, K2R2=(R/22)2(R/2)2\frac{K^2}{R^2}=\frac{(R / 2 \sqrt{2})^2}{(R / 2)^2}

amp;=R28×4R2=12⇒h2amp;=v22g[1+12]=3v24g\begin{aligned}& =\frac{R^2}{8} \times \frac{4}{R^2}=\frac{1}{2} \\\\\Rightarrow h_2 & =\frac{v^2}{2 g}\left[1+\frac{1}{2}\right]=\frac{3 v^2}{4 g}\end{aligned}

For solid sphere, K2R2=25\frac{K^2}{R^2}=\frac{2}{5}

⇒h3=v22g[1+25]=7v210g\Rightarrow h_3=\frac{v^2}{2 g}\left[1+\frac{2}{5}\right]=\frac{7 v^2}{10 g}

So,the ratio of h1,h2h_1, h_2 and h3h_3 is

h1:h2:h3amp;=v2g:3v24g:710v2gamp;=1:34:710=20:15:14\begin{aligned}h_1: h_2: h_3 & =\frac{v^2}{g}: \frac{3 v^2}{4 g}: \frac{7}{10} \frac{v^2}{g} \\\\& =1: \frac{3}{4}: \frac{7}{10}=20: 15: 14\end{aligned}

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