JEE Main 2026 — Jan 24, Shift 2 — Physics solution

JEE Main2026Jan 24, Shift 2Physics

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
$5 g r$
B
$\frac{7}{2} g r$
C
$4 g r$
D
$\frac{5}{2} g r$

Solution

At the position shown (radius makes 60° with the downward vertical, equivalently 30° from the top), $T = 0$ and gravity provides the centripetal force component along the string:

$m g cos 6 0^{\circ} = \frac{m V ^{2}}{r} \Rightarrow V^{2} = \frac{g r}{2}$ .

Height of this point above A: $h = r + r cos 6 0^{\circ} = r + \frac{r}{2} = \frac{3 r}{2}$ .

Energy conservation from A (speed $u$ ) to the upper point:

$\frac{1}{2} m u^{2} = m g \cdot \frac{3 r}{2} + \frac{1}{2} m V^{2}$

$u^{2} = 3 g r + V^{2} = 3 g r + \frac{g r}{2} = \frac{7 g r}{2}$ .

$∴ u = \frac{7 g r}{2}$ .

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