The speed of a transverse wave passing through a string of length 50 cm and mass 10 g is 60 ms⁻¹. The area of cross-section of the wire is 2.0 mm² and its Young's modulus is 1.2 × 10¹¹ Nm⁻². The extension of the wire over its natural length due to its tension will be x × 10⁻⁵ m. The value of x is __________.

Solution

Physics sir · 4 steps

The wave speed hands you the tension, and the tension then stretches the wire by TL/AY.

1. Mass per length. The linear mass density is μ = m/L = (0.01 kg)/(0.5 m) = 0.02 kg m⁻¹.
2. Tension from speed. For a transverse wave on a string, v = √(T/μ), so T = μ v² = 0.02 × 60² = 72 N.

   Δ L = TL/AY = (72 × 0.5)/((2.0 × 10⁻⁶) × (1.2 × 10¹¹))

3. Put in numbers. The area is 2.0 mm² = 2.0 × 10⁻⁶ m². The numerator is 36 and the denominator is 2.4 × 10⁵, so Δ L = 1.5 × 10⁻⁴ m.
4. So. Δ L = 1.5 × 10⁻⁴ m = 15 × 10⁻⁵ m, so x = 15.

The common slip: 1.5. Copying the 1.5 from Δ L = 1.5 × 10⁻⁴ m gives 1.5. The question wants the answer as x × 10⁻⁵, so x is 15.

The idea to keep. When a string question also gives you Young's modulus, the bridge between them is tension: T = μ v² from the wave, then Δ L = TL/AY. Convert every quantity to SI first.

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