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 _______.

- A \(3:2\)
- B \(5:1\)
- C \(2:5\)
- D \(1:5\)
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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