Question

A uniform cylindrical wire is subjected to a longitudinal tensile stress of $5 \times {10^7}\,N\,{m^{ - 2}}$ . Young's modulus of the material of the wire is $2 \times {10^{11}}\,N\,{m^{ - 2}}$ . The volume change in the wire is $0.02\,\% $ . The fractional change in the radius is
A. \[0.25 \times {10^{ - 4}}\]
B. \[0.5 \times {10^{ - 4}}\]
C. \[0.1 \times {10^{ - 4}}\]
D. \[1.5 \times {10^{ - 4}}\]

Answer 1

Hint: When a wire is under stress it obeys Hooke’s law which states that the stress is directly proportional to the strain. Mathematically it is expressed as $\sigma = Y\dfrac{{\Delta L}}{L}$ where $\dfrac{{\Delta L}}{L}$ is the longitudinal strain. In this question, we are given the value of stress and the young’s modulus through which we will calculate the fractional change in the length. Then we will use the relation $V = \pi {r^2}L$ since the volume change in the wire is given to be $0.02\,\% $ .

Complete step by step answer:
We know that $\sigma = Y\dfrac{{\Delta L}}{L}$ where $\dfrac{{\Delta L}}{L}$ is the fractional change in length, Y is the Young’s modulus and $\sigma $ is the stress.Given that $\sigma = 5 \times {10^7}\,N\,{m^{ - 2}}$ and $Y = 2 \times {10^{11}}\,N\,{m^{ - 2}}$ ,
Substituting in the equation we get,
$5 \times {10^7}\,N\,{m^{ - 2}} = 2 \times {10^{11}}\,N\,{m^{ - 2}} \times \dfrac{{\Delta L}}{L}$
$\Rightarrow \dfrac{{\Delta L}}{L} = 2.5 \times {10^{ - 4}}$

Now we know that $V = \pi {r^2}L$ where V is the volume and r is the radius of the cylindrical wire.
$\dfrac{{\Delta V}}{V} = \dfrac{{\pi \Delta ({r^2}L)}}{{\pi {r^2}L}}$
Further solving this, we get
$\dfrac{{\Delta V}}{V} = \dfrac{{2Lr\Delta r + {r^2}\Delta L}}{{{r^2}L}}$
$ \Rightarrow \dfrac{{\Delta V}}{V} = \dfrac{{\Delta L}}{L} + 2\dfrac{{\Delta r}}{r}$
We are given that $\dfrac{{\Delta V}}{V} = 0.02\,\% $
\[ \Rightarrow \dfrac{{\Delta V}}{V} = 2 \times {10^{ - 4}}\]

Now substituting in the equation, we get,
$ \Rightarrow 2 \times {10^{ - 4}} = 2.5 \times {10^{ - 4}} + 2\dfrac{{\Delta r}}{r}$
Further solving the equation, we get
$2\dfrac{{\Delta r}}{r} = - 0.5 \times {10^{ - 4}}$
$\therefore \dfrac{{\Delta r}}{r} = - 0.25 \times {10^{ - 4}}$

Hence, the correct option is A.

Note: The negative sign denotes that the radius of the wire is decreasing. As the wire is stretched by the tensile stress, its area of cross section reduces. Also, we must carefully note that here $L$ and $r$ both were variables. So, we applied the product rule of differentiation. Had one of these been a constant, we would have simply differentiated the quantity.