Work Energy Power

JEE Main level, single correct, medium

60 of 183

Question 60 of 183

A bob of mass mm is suspended by a light string of length LL. It is imparted a minimum horizontal velocity at the lowest point AA such that it just completes half circle reaching the top most position BB. The ratio of kinetic energies (K.E.)A(K.E.)B\dfrac{(K.E.{)}_{A}}{(K.E.{)}_{B}} is :

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Answer: (B) 5:15:1

The velocity given is minimum, just enough to complete verticle circle. At the top most point, tension of the string will be zero and gravitational force will provide the required centripetal force.
Therefore,
mg=m(VH)2L⇒12m(VH)2=12gLmg=\dfrac{m{\left({V}_{H}\right)}^{2}}{L} \Rightarrow \dfrac{1}{2}m{\left({V}_{H}\right)}^{2}=\dfrac{1}{2}gL
Apply energy conservation between point AA and BB, we get
12mVL2=12mVH2+mg(2L)\dfrac{1}{2}m{V}_{L}^{2}=\dfrac{1}{2}m{V}_{H}^{2}+mg(2L)
⇒VL=5gL\Rightarrow {V}_{L}=\sqrt{5gL}
Also, VH=gL{V}_{H}=\sqrt{gL}
Required ratio, (K.E)A(K.E)B=12m(5gL)212m(gL)2=51\dfrac{(K.E{)}_{A}}{(K.E{)}_{B}}=\dfrac{\dfrac{1}{2}m(\sqrt{5gL}{)}^{2}}{\dfrac{1}{2}m(\sqrt{gL}{)}^{2}}=\dfrac{5}{1}

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