JEE Main 2026 — Physics Mechanics

Let us analyze the forces acting on blocks $P$P and $Q$Q in the non-inertial frame of reference of the cube $S$S, which is moving to the right with an acceleration $a = \dfrac{g}{2} = 5 \text{ m/s}^2$a=2g​=5 m/s2.

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

For block $P$P (mass $m_P = 2 \text{ kg}$mP​=2 kg):

The forces in the horizontal direction are the tension $T$T in the string (acting towards the right) and the pseudo force $F_{pP}$FpP​ (acting towards the left). Since block $P$P is stationary relative to the cube and there is no friction on the top surface:

$$T = m_P a = 2 \times 5 = 10 \text{ N}$$T=mP​a=2×5=10 N

For block $Q$Q (mass $m_Q = 1.5 \text{ kg}$mQ​=1.5 kg):

The horizontal forces are the normal reaction $N$N from the side surface of the cube (acting towards the right) and the pseudo force $F_{pQ}$FpQ​ (acting towards the left). Since block $Q$Q is stationary horizontally relative to the cube:

$$N = m_Q a = 1.5 \times 5 = 7.5 \text{ N}$$N=mQ​a=1.5×5=7.5 N

The vertical forces acting on block $Q$Q are the downward gravitational force $m_Q g$mQ​g, the upward tension $T$T, and the frictional force $f$f.

The downward force is $m_Q g = 1.5 \times 10 = 15 \text{ N}$mQ​g=1.5×10=15 N.

Since $m_Q g > T$mQ​g>T ($15 \text{ N} > 10 \text{ N}$15 N>10 N), the block has a tendency to slide downwards, so the static friction $f$f must act upwards.

For vertical equilibrium of block $Q$Q:

$$T + f = m_Q g$$T+f=mQ​g

$$10 + f = 15 \Rightarrow f = 5 \text{ N}$$10+f=15⇒f=5 N

To keep the block stationary, the required friction must be less than or equal to the maximum static friction ($f \le \mu N$f≤μN). Assuming the limiting case to find the value of $\mu$μ:

$$\mu N = f$$μN=f

$$\mu (7.5) = 5$$μ(7.5)=5

$$\mu = \dfrac{5}{7.5} = \dfrac{2}{3} \approx 0.67$$μ=7.55​=32​≈0.67

Answer: $0.67$0.67