Question

The diameter of brass of rod is 4mm and Young’s modulus of brass is $9\times {{10}^{10}}N/{{m}^{2}}$ . The force required to stretch by 0.1% of its length is :
$\begin{align}
  & \text{a)360 }\!\!\pi\!\!\text{ N} \\
 & \text{b)36 N} \\
 & \text{c)144 }\!\!\pi\!\!\text{ }\!\!\times\!\!\text{ 10}{{\text{ }}^{\text{3}}}\text{N} \\
 & \text{d)36 }\!\!\pi\!\!\text{ }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{5}}}\text{ N} \\
\end{align}$

Answer 1

Hint: In the above question the Young’s modulus is given to us. As per the definition of Young’s modulus it comprises the length of the rod change in the length of the rod on application of stress. The information given in the question is sufficient enough to calculate the force by using the definition of Young’s Modulus.

Complete step by step answer:

In the above figure we can see that on application of stress how the length of rod changes. The Young’s modulus can be mathematically written as,
$Y=\dfrac{\text{Longitudinal stress}}{\text{Longitudinal strain}}=\dfrac{{}^{F}/{}_{A}}{{}^{\Delta L}/{}_{L}}=\dfrac{F}{A}\centerdot \dfrac{L}{\Delta L}...(1)$ where F is the force applied along the length, A is the area of cross section, L is the original length of the rod before extension and $\Delta L$is the change in length or in the above case the increase in the length.
The diameter of the rod is given as 4mm, hence its area of cross section is given by $\begin{align}
  & A=\pi {{\left( \dfrac{d}{2} \right)}^{2}}\text{where d is the diameter of the rod}\text{. d=4mm, hence its area is,} \\
 & A=\pi {{\left( \dfrac{4\times {{10}^{-3}}m}{2} \right)}^{2}}=4\times {{10}^{-6}}\pi {{m}^{2}} \\
\end{align}$
The increase in length of the rod is 0.1% which is equal to 0.001L.
Now we are ready to calculate the force responsible for the extension of the rod. Using equation 1 we get,
$\begin{align}
  & Y=\dfrac{F}{A}\centerdot \dfrac{L}{\Delta L} \\
 & 9\times {{10}^{10}}=\dfrac{F}{4\times {{10}^{-6}}\pi }\centerdot \dfrac{L}{0.01L} \\
 & 9\times {{10}^{10}}=\dfrac{F}{4\times {{10}^{-6}}\pi }\centerdot \dfrac{1}{0.001} \\
 & F=9\times {{10}^{10}}\times 4\times {{10}^{-6}}\times 0.001\pi =360\pi N \\
\end{align}$

So, the correct answer is “Option A”.

Note:
Young’s modulus describes how the length of the object changes with the application of stress. The units of the Young’s modulus is Newton per meter square. Similarly there exists two more modulus of elasticity i.e. bulk modulus and shear modulus. Bulk modulus is the ratio of volumetric stress to its volumetric strain and shear modulus is the ratio of tangential stress on a body to its shear strain.