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JEE Mains · Physics · STD 11 - 5. work,energy,power and collision

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 _______.

  1. A \(3:2\)
  2. B \(5:1\)
  3. C \(2:5\)
  4. D \(1:5\)
Verified Solution

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