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JEE Mains · Physics · STD 11 - Work,energy,power and collision
  • JEE Mains 2024
  • Hard
  • MCQ

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 half circle reaching the top most position B. The ratio of kinetic energies \(\frac{(\text { K.E. })_A}{(\text { K.E. })_B}\) is _______.

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Answer & Solution

Correct Answer

(B) \(5:1\)

Step-by-step Solution

Detailed explanation

Apply energy conservation between \(A\) & \(B\)

\( \frac{1}{2} \mathrm{mV}_{\mathrm{L}}^2=\frac{1}{2} \mathrm{mV}_{\mathrm{H}}^2+\mathrm{mg}(2 \mathrm{~L}) \)

\( \because \mathrm{V}_{\mathrm{L}}=\sqrt{5 \mathrm{gL}}\)

So, \(\mathrm{V}_{\mathrm{H}}=\sqrt{\mathrm{gL}}\)

\(\frac{(\mathrm{K} . \mathrm{E})_{\mathrm{A}}}{(\mathrm{K} . \mathrm{E})_{\mathrm{B}}}=\frac{\frac{1}{2} \mathrm{~m}(\sqrt{5 \mathrm{gL}})^2}{\frac{1}{2} \mathrm{~m}(\sqrt{\mathrm{gL}})^2}=\frac{5}{1}\)

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