Question:
Speed of a transverse wave on a straight wire (mass 6.0 g, length 60 cm and area of cross-section 1.0 mm2) is 90 ms-1. If the Young's modulus of wire is 16 \( \times \) 1011 Nm-2, the extension of wire over its natural length is :
Speed of a transverse wave on a straight wire (mass 6.0 g, length 60 cm and area of cross-section 1.0 mm2) is 90 ms-1. If the Young's modulus of wire is 16 \( \times \) 1011 Nm-2, the extension of wire over its natural length is :
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Wave speed on a string is \(v=\sqrt{T/\mu}\); get the tension, then use Young's modulus \(Y=\dfrac{T/A}{\Delta L/L}\) to find the extension.
Updated On: Oct 31, 2026
- 0.03 mm
- 0.04 mm
- 0.02 mm
- 0.01 mm
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The Correct Option is A
Solution and Explanation
Step 1: Linear mass density \(\mu=\dfrac{6.0\times10^{-3}\text{ kg}}{0.60\text{ m}}=1.0\times10^{-2}\) kg/m.
Step 2: \(v=\sqrt{T/\mu}\Rightarrow T=\mu v^2=10^{-2}\times(90)^2=81\) N.
Step 3: Stress \(=\dfrac{T}{A}=\dfrac{81}{10^{-6}}=8.1\times10^{7}\) N/m². Strain \(=\dfrac{8.1\times10^{7}}{16\times10^{11}}=5.06\times10^{-5}\).
Step 4: Extension \(\Delta L=\text{strain}\times L=5.06\times10^{-5}\times0.60=3.04\times10^{-5}\) m \(\approx0.03\) mm.
Answer: (A) 0.03 mm
Step 2: \(v=\sqrt{T/\mu}\Rightarrow T=\mu v^2=10^{-2}\times(90)^2=81\) N.
Step 3: Stress \(=\dfrac{T}{A}=\dfrac{81}{10^{-6}}=8.1\times10^{7}\) N/m². Strain \(=\dfrac{8.1\times10^{7}}{16\times10^{11}}=5.06\times10^{-5}\).
Step 4: Extension \(\Delta L=\text{strain}\times L=5.06\times10^{-5}\times0.60=3.04\times10^{-5}\) m \(\approx0.03\) mm.
Answer: (A) 0.03 mm
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