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 __________.
Pick an option to check your answer.
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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