JEE Main 2026 Apr 6 Shift 1, Physics Q7: Work Energy Theorem And Its Applications

JEE Main2026Apr 6, Shift 1Physics

A smooth inclined plane ends in a vertical circular loop, as shown in the figure. A small body is released from height $h$ as shown. If the body exerts a force of three times its weight on the plane at the highest point of circle then the height $h = α R$ . The value of $α$ is

A
$2$
B
$4$
C
$3$
D
$6$

Solution

At the highest point of the loop, both the weight $m g$ and the normal contact force $N$ point toward the centre and provide centripetal force:

$m g + N = \frac{m v ^{2}}{R}$ .

Given the body presses on the track with $3 m g$ , by Newton's third law $N = 3 m g$ . So $4 m g = \frac{m v ^{2}}{R} \Rightarrow v^{2} = 4 g R$ .

Energy conservation between release height $h$ and the top of the loop (at height $2 R$ ):

$m g (h - 2 R) = \frac{1}{2} m v^{2} = 2 m g R \Rightarrow h - 2 R = 2 R \Rightarrow h = 4 R$ .

$∴ α = 4$ .

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JEE Main 2026 Apr 6 Shift 1, Physics Q7: Work Energy Theorem And Its Applications

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