Three blocks A, B and C are pulled on a horizontal smooth surface by a force of 80 N as shown in figure
The tensions T₁ and T₂ in the string are respectively :

Solution

Physics sir · 4 steps

Find the common acceleration first. Then each tension only has to pull the blocks behind it.

1. Common acceleration. The strings stay taut, so all three blocks move together. Treat them as one body of mass 5 + 3 + 2 = 10 kg. The surface is smooth, so the only horizontal force on this body is F.

   a = F/(m_A + m_B + m_C) = 80/10 = 8 m s⁻²

2. Tension T1. T₁ is the only horizontal force on A, so it alone gives A its acceleration: T₁ = m_A a = 5 × 8 = 40 N.
3. Tension T2. T₂ pulls A and B together, a total mass of 8 kg: T₂ = (5 + 3) × 8 = 64 N. As a check on C, 80 - T₂ = 2 × 8 = 16, which gives T₂ = 64 N again.
4. So. T₁ = 40 N and T₂ = 64 N, which is option (a).

The idea to keep. In a pulled train of blocks, the tension in any string equals (mass of everything behind it) × a. So tension is largest near the applied force and drops toward the last block.

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