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 \(2.0\,mm ^{2}\) and its Young's modulus is \(1.2 \times 10^{11}\,Nm ^{-2}\). The extension of the wire over its natural length due to its tension will be \(x \times 10^{-5}\; m\). The value of \(x\) is \(...\)

Explanation:
Let us calculate the extension of the wire step-by-step by combining wave mechanics and elasticity principles:

1. Calculate the Linear Mass Density (\( \mu \)):
Linear mass density represents the mass per unit length of the string:
\( \text{Mass } (m) = 10\text{ g} = 10 \times 10^{-3}\text{ kg} = 0.01\text{ kg} \)
\( \text{Length } (L) = 50\text{ cm} = 0.5\text{ m} \)
\( \mu = \frac{m}{L} = \frac{10 \times 10^{-3}\text{ kg}}{0.5\text{ m}} = 0.02\text{ kg/m} \)

2. Determine the Tension (\( T \)) using Wave Speed:
The velocity of a transverse wave traveling along a stretched string is governed by the equation:
\( v = \sqrt{\frac{T}{\mu}} \)
Squaring both sides and solving for tension \( T \) gives:
\( v^2 = \frac{T}{\mu} \implies T = v^2 \times \mu \)
Substituting the given speed \( v = 60\text{ ms}^{-1} \):
\( T = (60)^2 \times 0.02 = 3600 \times 0.02 = 72\text{ N} \)

3. Calculate the Longitudinal Extension (\( \Delta l \)):
From Hooke's Law and the definition of Young's Modulus (\( Y \)):
\( Y = \frac{\text{Stress}}{\text{Strain}} = \frac{T / A}{\Delta l / L} \implies \Delta l = \frac{T L}{A Y} \)
Convert the cross-sectional area to standard SI units:
\( A = 2.0\text{ mm}^2 = 2.0 \times 10^{-6}\text{ m}^2 \)
Substituting the parameters into the elongation formula yields:
\( \Delta l = \frac{72 \times 0.5}{2.0 \times 10^{-6} \times 1.2 \times 10^{11}} \)
\( \Delta l = \frac{36}{2.4 \times 10^5} = 15 \times 10^{-5}\text{ m} \)

Comparing this with the format \( x \times 10^{-5}\text{ m} \) given in the problem statement confirms that \( x = 15 \).

Hence, Option B is the correct choice.