JEE Main 2026 Apr 6 Shift 1, Physics Q4: Constraint Relations

JEE Main2026Apr 6, Shift 1Physics

Two blocks (P and Q) with respectively masses $2$ kg and $1.5$ kg are joined by a massless thread. These blocks are mounted on a frictionless pulley which is fixed on the edge of a cube (S), as shown in the figure below. Block P is positioned on the top surface which has no friction and block Q is in contact with side-surface, having coefficient friction $μ$ . The cube (S) moves towards the right with acceleration of $\frac{g}{2}$ , where $g$ is gravitational acceleration. During this movement the block P and Q remain stationary. The value of $μ$ is _____. (take $g = 10$ m/s $^{2}$ )

A
$0.33$
B
$0.67$
C
$1$
D
$0.5$

Solution

Work in the non-inertial frame of the cube; a pseudo-force $- ma$ acts on each block (leftward, since cube accelerates rightward with $a = g /2$ ).

Block Q (vertical side): The pseudo-force $1.5 \cdot g /2$ pushes Q against the wall, giving normal reaction $N = 1.5 \cdot g /2 = 0.75 g$ . Vertical balance: weight $1.5 g$ downward is supported by the upward tension $T$ plus upward friction $f = μ N$ . So $T + μ (0.75 g) = 1.5 g$ .

Block P (top surface, frictionless): horizontal equilibrium in the cube frame requires $T =$ pseudo-force on P $= 2 \cdot g /2 = g$ .

Substitute: $g + 0.75 g μ = 1.5 g \Rightarrow 0.75 μ = 0.5 \Rightarrow μ = \frac{2}{3} \approx 0.67$ .

Concept behind this question

Constraint RelationsNotes, formulas and examples →

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2026 Apr 6 Shift 1, Physics Q4: Constraint Relations

Concept behind this question

More previous year questions on Constraint Relations

Practice more Physics