Question

The extension produced in a wire by the application of a load is \[3.0mm\] . The extension produced in a wire of the same material and length but half the radius by the same load is:
(A) \[5mm\]
(B) \[9mm\]
(C) \[12mm\]
(D) \[13mm\]

Answer 1

Hint: We have been given the extension in the wire. We can assume the length and the radius of the wire to be of certain values. Further, we can use Hooke’s Law which states that the ratio of stress to strain in a material, when subjected to lengthwise tension, is a constant which is given by Young’s modulus of elasticity.

Formula Used:
\[\dfrac{FL}{A\Delta L}=Y\]

Complete step by step answer:
According to Hooke’s Law, stress to strain ratio is constant.
Mathematically, we can say that
\[\begin{align}
  & \dfrac{stress}{strain}=Y \\
 & \dfrac{\left( \dfrac{F}{A} \right)}{\left( \dfrac{\Delta L}{L} \right)}=Y \\
 & \Rightarrow \dfrac{FL}{A\Delta L}=Y \\
\end{align}\]
Where \[F\] is the tension applied on the body, \[A\] is the area of cross-section on which the tension is acting, \[L\] is the original length, \[\Delta L\] is the change in length and \[Y\] is the young’s modulus of the material of the body
Since the force (load applied) and the length and the material of the object remain constant, we can say that \[A\times \Delta L=\dfrac{FL}{Y}\] or \[A\times \Delta L=\] constant.
Further expanding it, we can write \[{{r}^{2}}\times \Delta L\] is constant since \[A=\pi {{r}^{2}}\] .
Hence \[{{r}_{1}}^{2}\times \Delta {{L}_{1}}={{r}_{2}}^{2}\times \Delta {{L}_{2}}\]
Substituting the values in the above equation,
\[\begin{align}
 \Rightarrow & {{r}_{1}}^{2}\times \Delta {{L}_{1}}={{r}_{2}}^{2}\times \Delta {{L}_{2}} \\
 \Rightarrow & {{r}^{2}}\times \Delta {{L}_{1}}={{\left( \dfrac{r}{2} \right)}^{2}}\times \Delta {{L}_{2}} \\
 & \Rightarrow 4\times \Delta {{L}_{1}}=\Delta {{L}_{2}} \\
\end{align}\]
Now the extension in the first case is known to us, so we’ll substitute its value to find the extension in the second case, that is,
\[\begin{align}
  & \Delta {{L}_{2}}=4\times 3.0mm \\
 & \Rightarrow \Delta {{L}_{2}}=12mm \\
\end{align}\]

Hence option (C) is the correct answer.

Additional Information: The applications of Young’s modulus of elasticity are:
1) It measures the stiffness of the material.
2) It gives us a relation between the stress and strain for a material
A flexible material has a low value of Young’s modulus and changes its shape considerably for a certain change in the stress.

Note: Although we have assumed the young’s modulus to be constant in this case, it is possible only in certain cases. Young’s modulus is significantly affected by changes in temperature and material. Hence read the question properly to double-check which quantities are changing and which quantities are not as your answer will depend on it.