JEE Main 2026 — Jan 28, Shift 2 — Physics solution

JEE Main2026Jan 28, Shift 2Physics

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
$\frac{2 L}{g sin 2 θ - a _{0} ( 1 + cos 2 θ )}$
B
$\frac{4 L}{g sin 2 θ - a _{0} ( 1 + cos 2 θ )}$
C
$\frac{4 L}{g cos ^{2} θ - a _{0} sin θ cos θ}$
D
$\frac{2 L}{g sin θ - a _{0} cos θ}$

Solution

Work in the non-inertial frame of the wedge. A pseudo-force $m a_{0}$ acts on the block to the right. Resolve along the incline (taking down-the-slope as positive):

Net force along incline $= m g sin θ - m a_{0} cos θ$ . Acceleration $a = g sin θ - a_{0} cos θ$ .

Length along the incline $= \frac{L}{cos θ}$ . Using $S = \frac{1}{2} a t^{2}$ :

$\frac{L}{cos θ} = \frac{1}{2} (g sin θ - a_{0} cos θ) t^{2} \Rightarrow t^{2} = \frac{2 L}{cos θ ( g sin θ - a _{0} cos θ )} = \frac{2 L}{g sin θ cos θ - a _{0} cos ^{2} θ}$ .

Multiply numerator and denominator by 2: $t^{2} = \frac{4 L}{2 g sin θ cos θ - 2 a _{0} cos ^{2} θ} = \frac{4 L}{g sin 2 θ - a _{0} ( 1 + cos 2 θ )}$ .

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

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