JEE Main 2026 — Laws of Motion Question with Solution

Let us analyze the forces acting on blocks $P$ and $Q$ in the non-inertial frame of reference of the cube $S$, which is moving to the right with an acceleration $a=\frac{g}{2}=5{\text{ m/s}}^2$.

In this frame, a pseudo force acts on both blocks towards the left.

For block $P$ (mass $m_P=2\text{ kg}$):
The forces in the horizontal direction are the tension $T$ in the string (acting towards the right) and the pseudo force $F_{pP}$ (acting towards the left). Since block $P$ is stationary relative to the cube and there is no friction on the top surface:
$T=m_{P}a=2\times 5=10\text{ N}$

For block $Q$ (mass $m_Q=1.5\text{ kg}$):
The horizontal forces are the normal reaction $N$ from the side surface of the cube (acting towards the right) and the pseudo force $F_{pQ}$ (acting towards the left). Since block $Q$ is stationary horizontally relative to the cube:
$N=m_{Q}a=1.5\times 5=7.5\text{ N}$

The vertical forces acting on block $Q$ are the downward gravitational force $m_{Q}g$, the upward tension $T$, and the frictional force $f$.
The downward force is $m_{Q}g=1.5\times 10=15\text{ N}$.
Since $m_{Q}g>T$ ($15\text{ N}>10\text{ N}$), the block has a tendency to slide downwards, so the static friction $f$ must act upwards.
For vertical equilibrium of block $Q$:
$T+f=m_{Q}g$
$10+f=15\Rightarrow f=5\text{ N}$

To keep the block stationary, the required friction must be less than or equal to the maximum static friction ($f\le \mu N$). Assuming the limiting case to find the value of $\mu$:
$\mu N=f$
$\mu (7.5)=5$
$\mu =\frac{5}{7.5}=\frac{2}{3}\approx 0.67$

Answer: $0.67$