Each of three blocks P, Q and R shown in figure has a mass of 3 kg. Each of the wires A and B has cross-sectional area 0.005 cm² and Young's modulus 2 × 10¹¹ N m⁻². Neglecting friction, the longitudinal strain on wire B is ________ × 10⁻⁴. (Take g=10 m / s²)

Solution

Physics sir · 4 steps

Find the common acceleration first. Wire B has to pull both P and Q, so its tension is 6a.

1. Common acceleration. With no friction, only the weight of R drives all three blocks together. The total mass moved is 3 + 3 + 3 = 9 kg, so a = (3 × 10)/9 = 10/3 m s⁻².
2. Tension in B. Wire B is the only horizontal force on the pair P and Q, which together have mass 6 kg. So T_B = 6 × 10/3 = 20 N. This is less than R's weight of 30 N because R is accelerating downward.
3. Convert the area. 1 cm² = 10⁻⁴ m², so A = 0.005 × 10⁻⁴ = 5 × 10⁻⁷ m².

   strain = T_B/AY = 20/(5 × 10⁻⁷)(2 × 10¹¹) = 20/(1 × 10⁵) = 2 × 10⁻⁴

4. So. The longitudinal strain on wire B is 2 × 10⁻⁴, so the answer is 2.

The common slip: 3. Taking the tension in B as the full weight of R, 30 N, gives a strain of 3. That ignores R's own acceleration.

The idea to keep. In a connected-block system, the tension in a string equals the total mass behind it times the common acceleration. Get the tension first, then strain is T/(AY).

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