1.

An impulse `I` is applied at the end of a uniform rod if mass `m`. Then: A. `KE` of translation of the rod is `(I^(2))/(2m)`B. `KE` of rotation of the rod is `(I^(2))/(6m)`C. `KE` of rotation of the rod is `(3I^(2))/(2m)`D. `KE` of the rod is `(2I^(2))/m`

Answer» Correct Answer - A::C::D
`KE_(T)=1/2mv^(2).....(1) & I=mv....(2)`
so `KE_(T)=(I^(2))/(2m),` again `KE_(R)=1/2Iomega^(2)...(3)`
we have `Ixxl/2=1/12 ml^(2)xxomegarArromega=(6I)/(ml)`
Thus `KE_(R)=1/2xx1/12 ml^(2)xx(36I^(2))/(m^(2)l^(2))=(3I^(2))/(2m)`
and `KE=KE_(T)+KE_(R)=(2I^(2))/m`


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