Question

A copper wire is held at the two ends by rigid supports. At $${60}^{\circ }\mathrm{C}$$, the wire is just taut with negligible tension. The speed of transverse waves in this wire at $${10}^{\circ }\mathrm{C}$$ is $$10x\mathrm{m}{\mathrm{s}}^{-1}$$. Then the value of 'x' is. ( Take Young's modulus, $$Y_{Cu}=1.6\times {10}^{11}\mathrm{P}\mathrm{a}$$, coefficient of linear expansion, $${\alpha }_{Cu}=1.8\times {10}^{-6}{}^{\circ }{\mathrm{C}}^{-1}$$ and density, $${\rho }_{Cu}=9000\mathrm{k}\mathrm{g}{\mathrm{m}}^{-3}$$)
(1) $$4\mathrm{m}/\phantom{\mathrm{m}\mathrm{s}}\mathrm{s}$$
(2)$$3\mathrm{m}/\phantom{\mathrm{m}\mathrm{s}}\mathrm{s}$$
(3) $$14\mathrm{m}/\phantom{\mathrm{m}\mathrm{s}}\mathrm{s}$$
(4) $$2\mathrm{m}/\phantom{\mathrm{m}\mathrm{s}}\mathrm{s}$$

Answer 1

Hint:We will utilize the concept of thermal expansion or compression of given copper wire. The expression for force developed in the wire provides us with the relationship between Young's modulus, cross-sectional area, thermal expansion coefficient, and change in the wire temperature.

Complete step by step answer:
Given:
The temperature at which wire is taut with negligible tension is $$T_2={60}^{\circ }\mathrm{C}$$.
The speed of transverse waves in the given wire is $$V=10x\mathrm{m}{\mathrm{s}}^{-1}$$.
The temperature at the speed of transverse waves v is $$T_1={60}^{\circ }\mathrm{C}$$.
The value of Young's modulus of copper wire is $$Y_{Cu}=1.6\times {10}^{11}\mathrm{P}\mathrm{a}$$.
The coefficient of linear expansion of copper wire is $${\alpha }_{Cu}=1.8\times {10}^{-6}{}^{\circ }{\mathrm{C}}^{-1}$$.
The density of copper wire is $${\rho }_{Cu}=9000\mathrm{k}\mathrm{g}{\mathrm{m}}^{-3}$$.
We have to evaluate the value of 'x'
Let us write the expression for the change in copper wire length when its temperature changed from $$T_2$$ to $$T_1$$.
$$\mathrm{\Delta }l=l{\alpha }_{Cu}\mathrm{\Delta }T$$
Here l is the length of the given wire and $$\mathrm{\Delta }T$$ is the temperature change.
We know that the expression for force of thermal expansion or compression of the given copper wire can be written as:
$$F=YA{\alpha }_{Cu}\mathrm{\Delta }T$$
Here A is the cross-sectional area of the wire.
Let us write the expression for the speed of transverse of the given copper wire.
$$V=\sqrt{{\displaystyle \frac{F}{A\rho }}}$$
Substitute $$YA\alpha \mathrm{\Delta }T$$ for F in the above expression.
$$\begin{array}{l}V=\sqrt{{\displaystyle \frac{Y_{Cu}A{\alpha }_{Cu}\mathrm{\Delta }T}{A\rho }}}\\ =\sqrt{{\displaystyle \frac{Y_{Cu}{\alpha }_{Cu}\left(T_2-T_1\right)}{\rho }}}\end{array}$$
Substitute $$10x\mathrm{m}{\mathrm{s}}^{-1}$$ for V, $$1.8\times {10}^{-6}{}^{\circ }{\mathrm{C}}^{-1}$$ for $${\alpha }_{Cu}$$, $$1.6\times {10}^{11}\mathrm{P}\mathrm{a}$$ for $$Y_{Cu}$$, $${60}^{\circ }\mathrm{C}$$ for $$T_2$$, and $${10}^{\circ }\mathrm{C}$$ for $$T_1$$ and $$9000\mathrm{k}\mathrm{g}{\mathrm{m}}^{-3}$$ for $${\rho }_{Cu}$$ in the above expression.
$$\begin{array}{l}\Rightarrow 10x\mathrm{m}{\mathrm{s}}^{-1}=\sqrt{{\displaystyle \frac{\left(1.6\times {10}^{11}\mathrm{P}\mathrm{a}\right)\left(1.8\times {10}^{-6}{}^{\circ }{\mathrm{C}}^{-1}\right)\left({60}^{\circ }\mathrm{C}-{10}^{\circ }\mathrm{C}\right)}{9000\mathrm{k}\mathrm{g}{\mathrm{m}}^{-3}}}}\\ \Rightarrow 10x\mathrm{m}{\mathrm{s}}^{-1}=\sqrt{{\displaystyle \frac{\left(1.6\times {10}^{11}\mathrm{P}\mathrm{a}\times {\displaystyle \frac{\mathrm{N}/\phantom{\mathrm{N}{\mathrm{m}}^2}{\mathrm{m}}^2}{\mathrm{P}\mathrm{a}}}\times {\displaystyle \frac{\mathrm{k}\mathrm{g}\cdot \mathrm{m}/\phantom{\mathrm{k}\mathrm{g}\cdot \mathrm{m}{\mathrm{s}}^2}{\mathrm{s}}^2}{\mathrm{N}}}\right)\left(1.8\times {10}^{-6}{}^{\circ }{\mathrm{C}}^{-1}\right)\left({60}^{\circ }\mathrm{C}-{10}^{\circ }\mathrm{C}\right)}{9000\mathrm{k}\mathrm{g}{\mathrm{m}}^{-3}}}}\\ \Rightarrow 10x\mathrm{m}{\mathrm{s}}^{-1}=\sqrt{1600{\mathrm{m}}^2{\mathrm{s}}^{-2}}\\ \Rightarrow 10x\mathrm{m}{\mathrm{s}}^{-1}=40\mathrm{m}{\mathrm{s}}^{-1}\end{array}$$
From the above expression, we get:
$$x=4$$
Therefore, we can say that the value of x is 4, and option (1) is correct.

Note:We can remember Pascal's conversion to Newton-meter and converting Newton into its base units (kilogram, meter, and second). We also have to be extra careful while rearranging the final expression.