A body of mass 100 g moves in a circular path of radius 2 m on a vertical plane. The velocity at point A (bottom) is 10 text m/s . The ratio of kinetic energies at point B (at 30° from bottom) and C (at top, 90° from A) is: (Take g = 10 text m/s ^2):

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JEE MAIN Physics QUESTION #993

Question 1

A body of mass 100 g moves in a circular path of radius 2 m on a vertical plane. The velocity at point \(A\) (bottom) is \(10\ \text{m/s}\). The ratio of kinetic energies at point \(B\) (at \(30°\) from bottom) and \(C\) (at top, \(90°\) from \(A\)) is: (Take \(g = 10\ \text{m/s}^2\)):

\(\dfrac{2+\sqrt{2}}{3}\)
\(\dfrac{2+\sqrt{3}}{3}\)✔️
\(\dfrac{3+\sqrt{3}}{2}\)
\(\dfrac{3-\sqrt{2}}{2}\)

Correct Answer Explanation

Using energy conservation: \(KE_B = KE_A - mgh_B\) and \(KE_C = KE_A - mgh_C\). Height of B: \(h_B = r - r\cos30° = 2(1-\frac{\sqrt{3}}{2})\). Height of C: \(h_C = r = 2\) m. Ratio \(\frac{KE_B}{KE_C} = \frac{5-mg(2-\sqrt{3})}{5-mg\cdot2} = \frac{2+\sqrt{3}}{3}\).

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