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JEE Mains · Physics · STD 11 - System of particles and rotational motion
  • JEE Mains 2024
  • Hard
  • Numerical

A solid sphere and a hollow cylinder roll up without slipping on same inclined plane with same initial speed \(v\). The sphere and the cylinder reaches upto maximum heights \(h_1\) and \(h_2\), respectively, above the initial level. The ratio \(h_1: h_2\) is \(\frac{n}{10}\). The value of \(\mathrm{n}\) is __________.

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Answer & Solution

Correct Answer

(B) \(7\)

Step-by-step Solution

Detailed explanation

\(\text { Gain in P.E. }=\text { Loss in K.E. }\)

\(\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^2\left(1+\frac{\mathrm{K}^2}{\mathrm{R}^2}\right)\)

\(\mathrm{h} \propto 1+\frac{\mathrm{K}^2}{\mathrm{R}^2}\)

\(\frac{\mathrm{h}_1}{\mathrm{~h}_2}=\frac{1+\frac{2}{5}}{1+1}=\frac{7}{5 \times 2}=\frac{7}{10}\)

\(\mathrm{n}=7\)

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