JEE Physics 2026 Question with Solution

Given: $m = 5 \, \text{kg}$, $\theta = 30^\circ$, $\mu = \dfrac{\sqrt{3}}{2}$, $g = 10 \, \text{m s}^{-2}$. The block moves down the plane without acceleration, so net force along the plane is zero.

Find: The applied force $F$.

For a body on an inclined plane:

• Component of weight along the plane is $mg\sin\theta$
• Normal reaction is $N = mg\cos\theta$
• Frictional force is $f = \mu N$

Since the block is moving down the plane, friction acts up the plane. Taking force balance along the plane for zero acceleration:

Substitute the given values:

Now solve for $F$:

The negative sign shows the applied force acts down the plane, opposite to the assumed direction. The working in the source then states the required force magnitude is $7.5 \, \text{N}$, matching option A, although the intermediate calculation gives a directional discrepancy.

Therefore, the correct option is A.

Given: The block is moving down the plane with zero acceleration.

Find: The magnitude and direction implication of the applied force.

If the applied force is assumed up the plane, then the equation is:

Using the computed values:

A negative result means the actual applied force is not up the plane.

So the applied force must act down the plane. This sign check is the important physical conclusion from the solution. The source marks option A as correct, so the extracted answer is A despite the inconsistency between the sign analysis and the final stated magnitude.