Question

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 _____.

Answer 1

Correct Answer: 15

To determine the extension of the wire when a transverse wave passes through, we first find the tension (T) in the wire due to the wave's speed (v). 

The wave speed formula is: v = √(T/μ), where μ is the linear mass density (mass per unit length).
Linear mass density, μ = (mass of string)/(length of string) = (10 g)/(50 cm) = (10 × 10⁻³ kg)/(0.50 m) = 0.02 kg/m.
Substitute v and μ into the wave speed formula:

60 m/s = √(T/0.02 kg/m).
Squaring both sides:
(60)² = T/0.02.
T = (60)² × 0.02 = 72 N.
The tension in the wire is 72 N.

Next, we determine the extension (ΔL) using Young's modulus (Y). The formula is: ΔL = (FL)/(AY), where F is the force (tension), L is the original length, A is the cross-sectional area, and Y is Young's modulus.
Substituting values:
ΔL = (72 N × 0.50 m)/(2 × 10⁻⁶ m² × 1.2 × 10¹¹ N/m²).
ΔL = 72 × 0.50/(2.4 × 10⁵)
ΔL = 36/240000.
ΔL = 1.5 × 10⁻⁴ m.

Hence, x is given as:
ΔL = x × 10⁻⁵ m
1.5 × 10⁻⁴ m = x × 10⁻⁵ m.
x = 15.

The value of x is 15, which falls within the expected range of (15, 15).

Answer 2

\(v=\sqrt{\frac{T}{μ}}\)
So,\( T=60^2×\frac{10×10^{−3}}{0.5}\)
=72 N
\(Δℓ=\frac{Tℓ}{YA}=\frac{72×0.5}{1.2×10^{−11}×2×10^{−6}}\)
=15×10⁻⁵ m
Given that, the extension of the wire over its natural length due to its tension is x × 10^–5 m.
On comparing, x = 15
So, the answer is 15.