In case of vertical circular motion of a particle by a thread of length r if the tension in the thread is zero at an angle 30° shown in figure, the velocity at the bottom point (A) of the circular path is (g = gravitational acceleration) (a) √(7/2gr) (b) √(4gr) (c) √(5/2gr) (d) √(5gr)

Answer: (a). At the point 30° above the horizontal, the radial component of gravity is mgsin 30°. Tension zero: mgsin 30° = mv²/r ⇒ v² = gr/2. Height of this point above A: r + rsin 30° = 3r/2. Energy: vA² = v² + 2g · 3r/2 = gr/2 + 3gr = 7/2gr So the answer is (a) √(7/2gr).

In case of vertical circular motion of a particle by a thread of… | JEE Main 2026 (24 Jan, Shift 2)

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Critical Speed & Looping

In case of vertical circular motion of a particle by a thread of length $r$ if the tension in the thread is zero at an angle $30^\circ$ shown in figure, the velocity at the bottom point $(A)$ of the circular path is (g = gravitational acceleration)

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Answer: (a)

At the point $30^\circ$ above the horizontal, the radial component of gravity is $mg\sin 30^\circ$ .
Tension zero: $mg\sin 30^\circ = \frac{mv^2}{r} \Rightarrow v^2 = \frac{gr}{2}$ .
Height of this point above A: $r + r\sin 30^\circ = \frac{3r}{2}$ .
Energy: $v_A^2 = v^2 + 2g \cdot \frac{3r}{2} = \frac{gr}{2} + 3gr = \frac{7}{2}gr$
So the answer is (a) $\sqrt{\frac{7}{2}gr}$ .

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