JEE Main 2026 — Apr 5, Shift 1 — Physics solution

JEE Main2026Apr 5, Shift 1Physics

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 $^{2}$ )

A
2
B
4
C
$2$
D
$22$

Solution

Let $a_{W}$ be the horizontal acceleration of the wedge (to the right) and $a_{r}$ be the acceleration of the block relative to the wedge along the incline (down the slope).

For the block along the slope (taking down-slope positive):

$m g sin 3 7^{\circ} - m a_{W} cos 3 7^{\circ} = m a_{r}$ .

Normal to the slope: $N = m g cos 3 7^{\circ} + m a_{W} sin 3 7^{\circ}$ .

For the wedge horizontally: $f - N sin 3 7^{\circ} = M a_{W}$ .

With $m = 2$ , $M = 10$ , $sin 3 7^{\circ} = 3/5$ , $cos 3 7^{\circ} = 4/5$ , $f = 24$ , $g = 10$ :

$N = 16 + \frac{6 a _{W}}{5}$ , and $24 - \frac{3}{5} (16 + \frac{6 a _{W}}{5}) = 10 a_{W}$ .

Simplifying: $24 - \frac{48}{5} - \frac{18 a _{W}}{25} = 10 a_{W} \Rightarrow \frac{72}{5} = \frac{268 a _{W}}{25} \Rightarrow a_{W} = \frac{45}{29}$ m/s $^{2}$ .

Substituting back: $a_{r} = g sin 3 7^{\circ} - a_{W} cos 3 7^{\circ} = 6 - \frac{45}{29} \cdot \frac{4}{5} = \frac{138}{29}$ m/s $^{2}$ .

Using $s = \frac{1}{2} a_{r} t^{2}$ with $s = 8.8$ m: $t = \frac{2 \cdot 8.8 \cdot 29}{138} \approx 3.69 \approx 1.92$ s, which rounds to $2$ s.

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