JEE Physics 2026 Question with Solution

Given: A block of mass $m$ slides on a frictionless incline. The inclined plane accelerates left with constant acceleration $a_0$. The incline makes angle $\theta$ with the ground and has base length $L$.

Find: The time taken by the block to reach the lowest point of the incline.

Analyze the motion in the non-inertial frame of the inclined plane. In this frame, a pseudo force $ma_0$ acts on the block opposite to the acceleration of the plane, that is, horizontally to the right.

Along the plane:

$$mg\sin\theta$$

acts downward along the incline, while

$$ma_0\cos\theta$$

acts opposite to the downward motion.

Therefore, the effective acceleration along the incline is

$$a_{\text{eff}} = g\sin\theta - a_0\cos\theta$$

The length of the incline is

$$s = \frac{L}{\cos\theta}$$

Since the block starts from rest, use

$$s = \frac{1}{2} a_{\text{eff}} t^2$$

So,

$$\frac{L}{\cos\theta} = \frac{1}{2}\left(g\sin\theta - a_0\cos\theta\right)t^2$$

Solving,

$$t^2 = \frac{2L}{\cos\theta\left(g\sin\theta - a_0\cos\theta\right)}$$

which simplifies to

$$t = \sqrt{\frac{2L}{g\sin\theta - a_0\cos\theta}}$$

Therefore, the correct option is B.

Given: The wedge accelerates left with acceleration $a_0$, so in the wedge frame the block experiences a pseudo force to the right.

Find: Time of descent of the block.

The two relevant force components along the incline are:

• Gravity component along incline: $mg\sin\theta$
• Pseudo-force component along incline: $ma_0\cos\theta$

Hence net force along the incline is

$$F_{\parallel} = m\left(g\sin\theta - a_0\cos\theta\right)$$

Thus,

$$a_{\parallel} = g\sin\theta - a_0\cos\theta$$

Now the geometry gives the incline length as

$$\text{incline length} = \frac{L}{\cos\theta}$$

Starting from rest,

$$\frac{L}{\cos\theta} = \frac{1}{2}\left(g\sin\theta - a_0\cos\theta\right)t^2$$

Rearranging gives the same result reported in the solution:

$$t = \sqrt{\frac{2L}{g\sin\theta - a_0\cos\theta}}$$

Hence the answer is B.