JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Work, Power & Energy

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

(A)2
(B)4✓ Correct
(C)3
(D)6

Answer: (B)

Solution

At the highest point the track pushes on the body with

N = 3 m g

, and both this force and gravity point toward the centre:

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

, giving

4 m g = \frac{m v ^{2}}{R}

and

v^{2} = 4 g R

. Energy conservation on the smooth track from height

h

to the top of the loop at height

2 R

gives

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

. Solving,

h = 4 R

, so

α = 4

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Single correct option

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