InterviewSolution
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InterviewSolution
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A body is rolling down an inclined plane. If kinetic energy of rotation is `40 %` of kinetic energy in translatory start then the body is a.A. ringB. cylinderC. hollow ballD. solid ball |
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Answer» Correct Answer - d Rotational kinetic energy is `K_(R) = 1/2Iomega^(2) = 1/2Mk^(2) (v/r)^(2)` `(therefore I=Mk^(2)` and `v=Romega)` `=1/2Mv^(2)(k^(2)/R^(2))` Translation kinetic energy is `K_(T) = 1/2Mv^(2)` As per question, `K_(R) = 40% K_(T)` `therefore 1/2Mv^(2)(k^(2)/R^(2)) = 40% " of " 1/2 Mv^(2)` or `k^(2)/R^(2) = 40/100 = 2/5` For solid sphere, `k^(2)/R^(2) = 2/5` Hence, the body is solid ball. |
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