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A solid cylinder rolls down an inclined plane of height 3 m and reaches the bottom of plane with angular velocity of $2 \sqrt{2} rad \cdot s^{- 1}$ . The radius of cylinder must be(Take $g = 10 {ms}^{- 2}$ )

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detailed solution

Correct option is D

v=2gh1+Imr2=2×10×31+mr22×mr2=2×10×332=40⇒v=rω⇒r=vω=4022=408=5m

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A solid cylinder rolls down an inclined plane of height 3 m and reaches the bottom of plane with angular velocity of 22rad·s-1. The radius of cylinder must be(Take g=10 ms-2)

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A solid cylinder rolls down an inclined plane of height 3 m and reaches the bottom of plane with angular velocity of $2 \sqrt{2} rad \cdot s^{- 1}$ . The radius of cylinder must be(Take $g = 10 {ms}^{- 2}$ )