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

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 \[{T_2} = 60^\circ {\rm{C}}\].
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.
\[\begin{array}{l}
V = \sqrt {\dfrac{{{Y_{Cu}}A{\alpha _{Cu}}\Delta T}}{{A\rho }}} \\
= \sqrt {\dfrac{{{Y_{Cu}}{\alpha _{Cu}}\left( {{T_2} - {T_1}} \right)}}{\rho }}
\end{array}\]
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.