A body of mass 100 g is moving in circular path of radius 2 m on vertical plane as shown in figure. The velocity of the body at point A is 10 m/s. The ratio of its kinetic energies at point B and C is: (Take acceleration due to gravity as 10 m/s²) (a) (3 - √(2))/(2) (b) (2 + √(3))/(3) (c) (2 + √(2))/(3) (d) (3 + √(3))/(2)

Answer: (d). ABCO30°h_B = 2−√3 mh_C = 3 m A is the lowest point; B is 30° and C is 120° from OA. Heights above A: hB = R(1 - cos 30°) = 2 - √(3) m, hC = R(1 - cos 120°) = 3 m vB² = 100 - 20(2 - √(3)) = 60 + 20√(3), vC² = 100 - 60 = 40 (KB)/(KC) = (vB²)/(vC²) = (60 + 20√(3))/(40) = (3 + √(3))/(2) So the answer is (d).

A body of mass 100 g is moving in circular path of radius 2 m on… | JEE Main 2025 (22 Jan, Shift 2)

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Vertical Circle: Speed & Tension

A body of mass 100 g is moving in circular path of radius 2 m on vertical plane as shown in figure. The velocity of the body at point $A$ is $10\text{ m/s}$ . The ratio of its kinetic energies at point B and C is:

(Take acceleration due to gravity as $10\text{ m/s}^2$ )

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Answer: (d)

A is the lowest point; B is $30^\circ$ and C is $120^\circ$ from OA.
Heights above A: $h_B = R(1 - \cos 30^\circ) = 2 - \sqrt{3}$ m, $h_C = R(1 - \cos 120^\circ) = 3$ m
$v_B^2 = 100 - 20(2 - \sqrt{3}) = 60 + 20\sqrt{3}$ , $v_C^2 = 100 - 60 = 40$
$\frac{K_B}{K_C} = \frac{v_B^2}{v_C^2} = \frac{60 + 20\sqrt{3}}{40} = \frac{3 + \sqrt{3}}{2}$
So the answer is (d).

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