JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A solid sphere and solid cylinder of identical radii approach an incline with the same linear velocity (see figure). Both roll without slipping all throughout. The two climb maximum heights \(h_{sph}\) and \(h_{cyl}\) on the incline. The radio \(\frac{{{h_{sph}}}}{{{h_{cyl}}}}\) is given by
- A \(1\)
- B \(\frac{4}{5}\)
- C \(\frac{2}{{\sqrt 5 }}\)
- D \(\frac{14}{15}\)
Answer & Solution
Correct Answer
(D) \(\frac{14}{15}\)
Step-by-step Solution
Detailed explanation
For solid sphere \(\frac{1}{2}m{v^2} + \frac{1}{2} \cdot \frac{1}{2}m{R^2} \cdot \frac{{{V^2}}}{{{R^2}}} = mg{h_{sph}}\) For solid cylinder \(\frac{1}{2}m{V^2} + \frac{1}{2} \cdot \frac{1}{2}m{R^2} \cdot \frac{{{V^2}}}{{{R^2}}} = mg{h_{cyl}}\)…
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