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Question: Answered & Verified by Expert
A solid cylinder of mass 'M' and radius 'R' rolls down a smooth inclined plane about its own axis and reaches the bottom with velocity ' $\mathrm{v}^{\prime}$. The height of the inclined plane
is $(\mathrm{g}=$ acceleration due to gravity $)$
PhysicsRotational MotionMHT CETMHT CET 2020 (12 Oct Shift 1)
Options:
  • A $\frac{3 v^{2}}{4 g}$
  • B $\frac{4 v^{2}}{5 g}$
  • C $\frac{7 v^{2}}{9 g}$
  • D $\frac{2 v^{2}}{3 g}$
Solution:
1070 Upvotes Verified Answer
The correct answer is: $\frac{3 v^{2}}{4 g}$
No Solution

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