JEE Main 2026 · 5 Apr · Shift 1 · Physics Q30

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PhysicsSingle correct+4 / -1Question 30 of 75

A wedge $Y$ with mass of 10 kg and all frictionless surfaces and the inclined surface making 37° 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 $\tan(37^\circ)=3/4$ and $g=10$ m/s²)

Right-angled wedge Y on a smooth floor, vertical face on the left and the 37° incline sloping down to the right; block X sits at the top of the incline; a force f = 24 N pulls the wedge to the right.

\sqrt2

2\sqrt2

Answer and worked solution

Official answer: A

The key's working. Treat the wedge's acceleration as $A=\dfrac{24}{10+2}=2$ m/s² to the right.

Block, in the wedge's frame. The incline falls to the right. Gravity gives $g\sin37^\circ=6$ m/s² down the slope, and the pseudo-force $mA$ (to the left) gives $-A\cos37^\circ=-1.6$ m/s². So the block slides down with $a_r=4.4$ m/s².

Time.
$t=\sqrt{\frac{2(8.8)}{4.4}}=2\text{ s}.$

Exact check. Solving the wedge and block together gives a slightly different value.

Normal force: $N=m(g\cos37^\circ+A\sin37^\circ)=16+1.2A$ .
Wedge: $10A=24-0.6N$ , which gives $A\approx1.34$ m/s².
Then $a_r\approx4.93$ m/s² and $t\approx1.9$ s.

The nearest option, and the official key, is 2 s.

(A)

Topics:Pseudo forces and non-inertial frames Constraint motion and pulleys

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JEE Main 2026 · 5 Apr · Shift 1, Physics question 30