Answer

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**Hint:**The square of the velocity of transverse wave is the ratio of force and linear mass density. The thermal stress in the wire corresponds to change in temperature.

Formula Used: The formulae used in the solution are given here.

The speed of the transverse wave is given by, $ v = \sqrt {\dfrac{F}{m}} $ where $ m $ is linear mass density and $ F $ is force.

From thermal expansion, $ \Delta l = l\alpha \Delta \theta $ .

Hence, force $ F $ is given by, $ F = Y\alpha A\Delta \theta $ .

**Complete Step by Step Solution**

It has been given that, Young modulus of copper $ Y = 1.2 \times {10^{11}}N/{m^2} $ , Coefficient of linear expansion of copper $ \alpha = 1.6 \times {10^{ - 5}}/^\circ C $ , Density of copper $ \rho = 9.2 \times {10^3}kg/{m^3} $ .

Young’s modulus is also known as modulus of elasticity and is defined as, the mechanical property of a material to withstand the compression or the elongation with respect to its length.

It is denoted as $ E $ or $ Y $ . Young’s Modulus (also referred to as the Elastic Modulus or Tensile Modulus), is a measure of mechanical properties of linear elastic solids like rods, wires, and such.

Coefficient of Linear Expansion is the rate of change of unit length per unit degree change in temperature.

We know, the speed of the transverse wave is given by, $ v = \sqrt {\dfrac{F}{m}} $ where $ m $ is linear mass density and $ F $ is force.

From thermal expansion, $ \Delta l = l\alpha \Delta \theta $ .

The thermal stress in the wire corresponds to change in temperature.

Hence, $ \Delta T $ is $ F = Y\alpha \Delta T $

If A is the cross-sectional area of the wire, then tension produced in the wire, $ F = fA = Y\alpha \Delta TA $

Hence, force $ F $ is given by, $ F = Y\alpha A\Delta \theta $ .

$ {{Linear\, mass\, density = mass/length}} $

= $ {{volume \times density/length}} $

= $ {{area \times length \times density/length}} $

= $ {{area \times density = A}}\rho $

Putting equation (2) and (3) in equation (1),

$ v = \sqrt {\dfrac{F}{\mu } = } \sqrt {\dfrac{{Y\alpha A\Delta T}}{{\left( {A \times l} \right)\rho }}} = \sqrt {\dfrac{{Y\alpha \Delta T}}{\rho }} $ .

Here, $ Y = 1.2 \times {10^{11}}N/{m^2} $ , $ \alpha = 1.6 \times {10^{ - 5}}/^\circ C $ , $ \rho = 9.2 \times {10^3}kg/{m^3} $ .

The change in angle is $ \Delta \theta = {50^ \circ }C - {30^ \circ }C = {20^ \circ }C $ .

Here, the velocity of the transverse waves,

$ v = \sqrt {\dfrac{{1.2 \times {{10}^{11}} \times 1.6 \times {{10}^{ - 5}} \times 20}}{{9.2 \times {{10}^3}}}} = 64.60m/s $ .

**Hence the correct answer is option A.**

**Note**

Transverse wave, motion in which all points on a wave oscillate along paths at right angles to the direction of the wave’s advance. Surface ripples on water, seismic S (secondary) waves, and electromagnetic (e.g., radio and light) waves are examples of transverse waves. A simple transverse wave can be represented by a sine or cosine curve, so called because the amplitude of any point on the curve—i.e., its distance from the axis—is proportional to the sine (or cosine) of an angle.

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