Question

A uniform steel rod of length $1\,{\text{m}}$ and area of cross section $20\,{\text{c}}{{\text{m}}^{\text{2}}}$ is hanging from a fixed support. Find the increase in the length of the rod ( ${Y_{{\text{steel}}}} = 2.0 \times {10^{11}}\,{\text{N}}{{\text{m}}^{ - 2}}$ , ${\rho _{{\text{steel}}}} = 7.85 \times {10^3}\,{\text{kg}}\,{{\text{m}}^{ - 3}}$ )
(A) $1.923 \times {10^{ - 5}}\,{\text{cm}}$
(B) $2.923 \times {10^{ - 5}}\,{\text{cm}}$
(C) $1.123 \times {10^{ - 5}}\,{\text{cm}}$
(D) $3.123 \times {10^{ - 5}}\,{\text{cm}}$

Answer 1

A rod elongates due to its own weight, when hung from a particular point or a fixed support. We use Hooke’s law in order to determine the elongation of the rod.

Complete step by step solution:
In the given question, we are supplied the following data:
The length of the uniform steel rod is $1\,{\text{m}}$ .
The area of the cross section is $20\,{\text{c}}{{\text{m}}^{\text{2}}}$ .
Young’s modulus of steel is given as ${Y_{{\text{steel}}}} = 2.0 \times {10^{11}}\,{\text{N}}{{\text{m}}^{ - 2}}$ .
Density of steel is given as ${\rho _{{\text{steel}}}} = 7.85 \times {10^3}\,{\text{kg}}\,{{\text{m}}^{ - 3}}$ .
We are asked to find the increase in length of the rod.
To begin with, we will need to find the weight of the rod, which is acting in the downward direction. The elongation shown by the rod is due to its own weight as far as concerned. The weight of the rod is distributed all along its length. The rod is attached to the ceiling at its one end while the other end experiences the force.
We can write the expression for the weight of the rod, which is given by:
$W = mg$ …… (1)
Where,
$W$ indicates the weight of the rod.
$m$ indicates the mass of the rod.
$g$ indicates the acceleration due to gravity.
Again, we can write for density as:
$\rho = \dfrac{m}{V}$ …… (2)
Where,
$\rho$ indicates the density of the rod.
$V$ indicates the volume of the rod.
We now, rearrange the equation (2), and we get:
$\rho = \dfrac{m}{V} \\\ \Rightarrow \rho = \dfrac{m}{{AL}} \\\ \Rightarrow m = \rho AL$
Where,
$A$ indicates the area of the cross section of the rod.
$L$ indicates the length of the rod.
Now, we rearrange the equation (1) as follows:
$W = mg \\\ \Rightarrow W = \rho ALg$
The formula which gives the elongation of the rod is given below:
\Delta L = \dfrac{{WL}}{{2AY}} \\\ \Rightarrow \Delta L = \dfrac{{\rho ALg \times L}}{{2AY}} \\\
$\Rightarrow \Delta L = \dfrac{{\rho {L^2}g}}{{2Y}}$ ……. (3)
Now, we substitute the required values in the equation (3) and we get:
$\Delta L = \dfrac{{\rho {L^2}g}}{{2Y}} \\\ \Rightarrow \Delta L = \dfrac{{7.5 \times {{10}^3} \times {1^2} \times 9.8}}{{2 \times 2 \times {{10}^{11}}}} \\\ \Rightarrow \Delta L = \dfrac{{76.93 \times {{10}^3}}}{{4 \times {{10}^{11}}}} \\\ \Rightarrow \Delta L = 19.23 \times {10^{ - 8}}\,{\text{m}}$
Therefore, the elongation of the rod is found to be $19.23 \times {10^{ - 8}}\,{\text{m}}$ .
Now, we convert into centimetres:
We know,
$1\,{\text{m}} = 1000\,{\text{cm}}$
So,
$19.23 \times {10^{ - 8}}\,{\text{m}} \\\ \Rightarrow 19.23 \times {10^{ - 8}} \times 1000 \\\ \therefore 1.923 \times {10^{ - 5}}\,{\text{cm}}$
Hence, the required answer is $1.923 \times {10^{ - 5}}\,{\text{cm}}$ .

The correct option is (A).

Note: A rod should be hung in a vertical position so as to minimize the elongation because of its own weight along with a load at its lower end. Hooke’s law comes handy so as to determine the elongation shown by the rod. It is important to note that we should use the S.I units for the calculations.