JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Laws of Motion

A small block of mass

m

slides down from the top of a frictionless inclined surface, while the inclined plane is moving towards left with constant acceleration

a_{0}

. The angle between the inclined plane and ground is

θ

and its base length is L . Assuming that initially the small block is at the top of the inclined plane, the time it takes to reach the lowest point of the inclined plane is ______ .

(A) $\frac{2 L}{g s i n 2 θ - a _{0} ( 1 + c o s 2 θ )}$
(B) $\frac{4 L}{g s i n 2 θ - a _{0} ( 1 + c o s 2 θ )}$ ✓ Correct
(C) $\frac{4 L}{g c o s ^{2} θ - a _{0} s i n θ c o s θ}$
(D) $\frac{2 L}{g s i n θ - a _{0} c o s θ}$

Answer: (B)

Solution

Work in the frame of the incline, which accelerates to the left at

a_{0}

; a pseudo-force

m a_{0}

then acts on the block towards the right. Along the incline (down the slope) gravity contributes

g sin θ

while the pseudo-force opposes the motion with

a_{0} cos θ

, giving a net acceleration

a = g sin θ - a_{0} cos θ

. With base length

L

the distance along the incline is

s = \frac{L}{cos θ}

. From

s = \frac{1}{2} a t^{2}

t = \frac{2 L}{cos θ ( g sin θ - a _{0} cos θ )} = \frac{2 L}{g sin θ cos θ - a _{0} cos ^{2} θ} .

Using

2 sin θ cos θ = sin 2 θ

and

2 cos^{2} θ = 1 + cos 2 θ

, this is

t = \frac{4 L}{g sin 2 θ - a _{0} ( 1 + cos 2 θ )}

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Single correct option

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JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Laws of Motion

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