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(B) To complete a full vertical circle,the minimum velocity at the lowest point $A$ must be $V_A = \sqrt{5gL}$.At the topmost point $B$,the minimum ve

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$5:1$

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(B) To complete a full vertical circle,the minimum velocity at the lowest point $A$ must be $V_A = \sqrt{5gL}$.At the topmost point $B$,the minimum velocity required to maintain tension in the string is $V_B = \sqrt{gL}$.The kinetic energy at point $A$ is $(K.E.)_A = \frac{1}{2} m V_A^2 = \frac{1}{2} m (\sqrt{5gL})^2 = \frac{5}{2} mgL$.The kinetic energy at point $B$ is $(K.E.)_B = \frac{1}{2} m V_B^2 = \frac{1}{2} m (\sqrt{gL})^2 = \frac{1}{2} mgL$.Therefore,the ratio of kinetic energies is $\frac{(K.E.)_A}{(K.E.)_B} = \frac{\frac{5}{2} mgL}{\frac{1}{2} mgL} = \frac{5}{1}$.

$A$ bob of mass $m$ is suspended by a light string of length $L$. It is imparted a minimum horizontal velocity at the lowest point $A$ such that it just completes a full vertical circle,reaching the topmost position $B$. The ratio of kinetic energies $\frac{(\text{K.E.})_A}{(\text{K.E.})_B}$ is:

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