Minimum velocity required to complete verticle circle of radius $R$ at lowest, horizontal and top most point respectively.

\sqrt{5 g R}, \sqrt{3 g R}, \sqrt{g R}

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\sqrt{5 g R}, \sqrt{2 g R}, \sqrt{g R}

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\sqrt{5 g R}, \sqrt{g R}, \sqrt{3 g R}

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\sqrt{g R}, \sqrt{3 g R}, \sqrt{5 g R}

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Solution

The correct option is A $\sqrt{5 g R}, \sqrt{3 g R}, \sqrt{g R}$
Equation of motion:
$T - m g cos θ = \frac{m V^{2}}{R}$
At $C,$ $T_{C} = 0$ (minimum case)
$⟹ \frac{m V_{c}^{2}}{R} = - m g cos (180^{o})$
$⟹ V_{C}^{2} = R g$
$⟹ V_{C} = \sqrt{R g}$
Using the law of conservation of energy:
$E_{A} = E_{C}$
$\frac{1}{2} m V_{A}^{2} = \frac{1}{2} m V_{C}^{2} + 2 m g R$
$⟹ \frac{V_{A}^{2}}{2} = \frac{g R}{2} + 2 g R$
$⟹ V_{A}^{2} = 5 g R$
$⟹ V_{A} = \sqrt{5 g R}$
Again,
$E_{A} = E_{B}$
From energy conservation:
$\frac{1}{2} m V_{A}^{2} = \frac{1}{2} m V_{B}^{2} + m g R$
$⟹ \frac{5}{2} g R = \frac{1}{2} V_{B}^{2} + g R$
$⟹ V_{B} = \sqrt{3 g R}$

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Minimum velocity required to complete verticle circle of radius R at lowest, horizontal and top most point respectively.

Minimum velocity required to complete verticle circle of radius $R$ at lowest, horizontal and top most point respectively.