Question

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

Answer 1

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 ${T_2} = 60^\circ {\rm{C}}$.
The speed of transverse waves in the given wire is $V = 10x{\rm{ m}}{{\rm{s}}^{ - 1}}$.
The temperature at the speed of transverse waves v is ${T_1} = 60^\circ {\rm{C}}$.
The value of Young's modulus of copper wire is ${Y_{Cu}} = 1.6 \times {10^{11}}{\rm{ Pa}}$.
The coefficient of linear expansion of copper wire is ${\alpha _{Cu}} = 1.8 \times {10^{ - 6}}{\rm{ }}^\circ {{\rm{C}}^{ - 1}}$.
The density of copper wire is ${\rho _{Cu}} = 9000{\rm{ kg }}{{\rm{m}}^{ - 3}}$.
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 ${T_1}$.
$\Delta l = l{\alpha _{Cu}}\Delta T$
Here l is the length of the given wire and $\Delta T$ 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:
$F = YA{\alpha _{Cu}}\Delta T$
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.
$V = \sqrt {\dfrac{F}{{A\rho }}}$
Substitute $YA\alpha \Delta T$ for F in the above expression.

V = \sqrt {\dfrac{{{Y_{Cu}}A{\alpha _{Cu}}\Delta T}}{{A\rho }}} \\\ = \sqrt {\dfrac{{{Y_{Cu}}{\alpha _{Cu}}\left( {{T_2} - {T_1}} \right)}}{\rho }} \end{array}$$ Substitute $$10x{\rm{ m}}{{\rm{s}}^{ - 1}}$$ for V, $$1.8 \times {10^{ - 6}}{\rm{ }}^\circ {{\rm{C}}^{ - 1}}$$ for $${\alpha _{Cu}}$$, $$1.6 \times {10^{11}}{\rm{ Pa}}$$ for $${Y_{Cu}}$$, $$60^\circ {\rm{C}}$$ for $${T_2}$$, and $$10^\circ {\rm{C}}$$ for $${T_1}$$ and $$9000{\rm{ kg }}{{\rm{m}}^{ - 3}}$$ for $${\rho _{Cu}}$$ in the above expression. $$\begin{array}{l} \Rightarrow 10x{\rm{ m}}{{\rm{s}}^{ - 1}} = \sqrt {\dfrac{{\left( {1.6 \times {{10}^{11}}{\rm{ Pa}}} \right)\left( {1.8 \times {{10}^{ - 6}}{\rm{ }}^\circ {{\rm{C}}^{ - 1}}} \right)\left( {60^\circ {\rm{C}} - 10^\circ {\rm{C}}} \right)}}{{9000{\rm{ kg }}{{\rm{m}}^{ - 3}}}}} \\\ \Rightarrow 10x{\rm{ m}}{{\rm{s}}^{ - 1}} = \sqrt {\dfrac{{\left( {1.6 \times {{10}^{11}}{\rm{ Pa}} \times \dfrac{{{{\rm{N}} {\left/ {\vphantom {{\rm{N}} {{{\rm{m}}^2}}}} \right. } {{{\rm{m}}^2}}}}}{{{\rm{Pa}}}} \times \dfrac{{{{{\rm{kg}} \cdot {\rm{m}}} {\left/ {\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {{{\rm{s}}^2}}}} \right. } {{{\rm{s}}^2}}}}}{{\rm{N}}}} \right)\left( {1.8 \times {{10}^{ - 6}}{\rm{ }}^\circ {{\rm{C}}^{ - 1}}} \right)\left( {60^\circ {\rm{C}} - 10^\circ {\rm{C}}} \right)}}{{9000{\rm{ kg }}{{\rm{m}}^{ - 3}}}}} \\\ \Rightarrow 10x{\rm{ m}}{{\rm{s}}^{ - 1}} = \sqrt {1600{\rm{ }}{{\rm{m}}^2}{{\rm{s}}^{ - 2}}} \\\ \Rightarrow 10x{\rm{ m}}{{\rm{s}}^{ - 1}} = 40{\rm{m}}{{\rm{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.