A wedge Y with mass of 10 kg and all frictionless surfaces and the inclined surface making $37^{\circ}$ with horizontal. A block X with mass 2 kg is placed at the highest point of the wedge as shown in figure is at rest. At $\mathrm{t}=0$ wedge (Y) is pulled toward right with constant force $(f)$ of 24 N. Taking the block X at rest at $\mathrm{t}=0$, the time taken by it to slide down 8.8 m on the slope, while Y is on the move, is ______ s. (take $\tan(37^{\circ})=3/4$ and $\mathrm{g}=10$ m/s$^{2}$)

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Laws of Motion

A wedge Y with mass of 10 kg and all frictionless surfaces and the inclined surface making

3 7^{\circ}

with horizontal. A block X with mass 2 kg is placed at the highest point of the wedge as shown in figure is at rest. At

t = 0

wedge (Y) is pulled toward right with constant force

(f)

of 24 N. Taking the block X at rest at

t = 0

, the time taken by it to slide down 8.8 m on the slope, while Y is on the move, is ______ s. (take

tan (3 7^{\circ}) = 3/4

and

g = 10

m/s

^{2}

)

Answer: (A)

On the frictionless surfaces the horizontal pull is shared by the wedge and the block, so to the standard approximation the wedge accelerates as

a_{w} = \frac{f}{M + m} = \frac{24}{10 + 2} = 2 m/s^{2}

towards the right. Working in the (non-inertial) frame of the wedge, the block on the

3 7^{\circ}

face feels gravity pulling it down the slope and the pseudo-force pushing it up the slope, so its acceleration along the slope is

a_{r e l} = g sin 3 7^{\circ} - a_{w} cos 3 7^{\circ} = 10 \times \frac{3}{5} - 2 \times \frac{4}{5} = 6 - 1.6 = 4.4 m/s^{2}

. Starting from rest, the distance along the slope is

s = \frac{1}{2} a_{r e l} t^{2}

, so

8.8 = \frac{1}{2} \times 4.4 \times t^{2}

gives

t^{2} = 4

. The block takes

2

s to slide

8.8

m down the slope.

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Single correct option