As per given figure, a weightless pulley P is attached on a double inclined frictionless surfaces. The tension in the string (massless) will be (if g = 10 m/s²)

Solution

Physics sir · 4 steps

The 4 kg block on the steeper side wins: get the acceleration first, then use the 1 kg block for T.

1. Pulling forces. Only the component of weight along each incline drives the motion. For the 4 kg block it is 4 × 10 × sin 60° = 20√3 N. For the 1 kg block it is 1 × 10 × sin 30° = 5 N.
2. Find acceleration. Since 20√3 > 5, the 4 kg block slides down and the 1 kg block slides up. Treat both blocks and the string as one system of mass 5 kg.

   a = (20√3 - 5)/(4 + 1) = (4√3 - 1) m s⁻²

3. Tension from lighter block. Along its incline, T - 5 = 1 × a, so T = 5 + 4√3 - 1 = 4√3 + 4. The 4 kg block confirms it: 20√3 - T = 4a = 16√3 - 4 also gives T = 4√3 + 4.
4. So. T = 4(√3 + 1) N ≈ 10.9 N, which is option (d).

The idea to keep. For blocks on a double incline, compare the mg sin θ components to find the direction, get a from the whole system, then apply F = ma to one block for the tension.

Answer from the NTA final answer key. Solution by PrepWiser.