Question

A wire 3m in length and 1 mm in diameter at \[30^\circ C\], kept in a low temperature at \[170^\circ \] and is stretched by hanging a weight of 10 kg at one end. Calculate the change in length of the wire. Given \[\alpha = 1.2 \times {10^{ - 5}}/^\circ C\], \[Y = 2 \times {10^{11}}\,N\,{m^2}\], \[g = 10\,m/{s^2}\].
(A) \[2 \times {10^{ - 3}}\,m\]
(B) \[7.04 \times {10^{ - 3}}\,m\]
(C) \[95.6 \times {10^{ - 5}}\,m\]
(D) \[5.2 \times {10^{ - 3}}\,m\]

Answer 1

Hint:Calculate the change in length of wire due to hanging load at its one end using formula for Young’s modulus. Also, determine the change in length due to change in temperature using the formula for linear expansion of the rod due to change in length.

Formula used:
1) \[Y = \dfrac{{F \times l}}{{A \times \Delta l}}\]
Here, \[\Delta l\] is the change in length, F is the force, l is the length of wire and A is the cross-sectional area of the wire.
2) \[\Delta L = L\alpha \,\Delta T\]
Here, L is the original length of wire, \[\alpha \]is the linear expansion coefficient of wire and \[\Delta T\] is the change in temperature.

Complete step by step answer:
 We can observe that the change in the length of wire is due to two factors: change in length due to change in temperature and change in length due to hanging load at its one end.
We can calculate the change in the length of the wire due to hanging load at its one end using the expression for Young’s modulus as follows,
\[Y = \dfrac{{F \times l}}{{A \times \Delta l}}\]
\[ \Rightarrow \Delta l = \dfrac{{F \times l}}{{A \times Y}}\]
Here, \[\Delta l\] is the change in length, F is the force, l is the length of wire and A is the cross-sectional area of the wire.
 The force on the wire is due to the weight of the load hanging from its one end. Therefore, we can write the above equation as follows,
\[\Delta l = \dfrac{{mgl}}{{\pi {r^2}Y}}\]
Here, r is the radius of the wire.
Substituting 10 kg for m, \[10\,m/{s^2}\] for g, 3 m for l, \[5 \times {10^{ - 4}}\,m\] for r and \[2 \times {10^{11}}\,N\,{m^2}\] for Y in the above equation, we get,
\[\Delta l = \dfrac{{\left( {10} \right)\left( {10} \right)\left( 3 \right)}}{{\left( {3.14} \right){{\left( {5 \times {{10}^{ - 4}}} \right)}^2}\left( {2 \times {{10}^{11}}} \right)}}\]
\[ \Rightarrow \Delta l = \dfrac{{300}}{{157000}}\]
\[ \Rightarrow \Delta l = 2 \times {10^{ - 3}}\,m\]
Now, the change in length due to increase in temperature can be expressed as,
\[\Delta L = L\alpha \,\Delta T\]
\[ \Rightarrow \Delta L = L\alpha \,\left( {{T_f} - {T_i}} \right)\]
Here, L is the original length of wire, \[\alpha \]is the linear expansion coefficient of wire and \[\Delta T\] is the change in temperature.
Substituting 3 m for L, \[1.2 \times {10^{ - 5}}/^\circ C\] for , \[170^\circ \] for \[{T_f}\]and \[30^\circ C\] for \[{T_i}\] in the above equation.
\[\Delta L = \left( 3 \right)\left( {1.2 \times {{10}^{ - 5}}} \right)\,\left( {170 - 30} \right)\]
\[ \Rightarrow \Delta L = 5.04 \times {10^{ - 3}}\,m\]
Therefore, the total increase in the length of wire is,
\[\Delta L' = \Delta l + \Delta L\]
\[ \Rightarrow \Delta L' = 2 \times {10^{ - 3}}\,m + 5.04 \times {10^{ - 3}}\,m\]
\[ \therefore \Delta L' = 7.04 \times {10^{ - 3}}\,m\]

So, the correct answer is option (B).

Note: If the change in length of wire \[\Delta L\] is positive quantity, then we can say that the length of the wire is increased and if it is negative, then the length of the wire is decreased. The diameter of the wire is given in mm. Therefore, you should convert it into meters. Remember, if the length of the wire changes on hanging it from the other end, the force responsible for its elongation is its weight.