Question

A thick uniform rubber rope of density \[1.5\,{\text{gc}}{{\text{m}}^{ - 3}}\] and Young’s modulus \[5 \times {10^6}\,{\text{N}}{{\text{m}}^{ - 2}}\] has a length of \[8\,{\text{m}}\]. When hung from the ceiling of a room, the increase in the length of the rope due to its own weight will be:
A. \[9.6 \times {10^{ - 2}}\]
B. \[19.2 \times {10^{ - 2}}\]
C. \[9.6 \times {10^{ - 3}}\]
D. \[9.6\]

Answer 1

Hint: First of all, we will find the expression of Young’s modulus. Since, the force applied on the rope is not supplied, we will make necessary substitution for the applied force. We know, force is the product of mass and acceleration due to gravity. We will substitute the required values and obtain the result accordingly.

Complete step by step answer:
In the given question, we are supplied with the following details:
The rubber rope has a density of \[1.5\,{\text{gc}}{{\text{m}}^{ - 3}}\] .
The rubber rope has Young’s modulus of \[5 \times {10^6}\,{\text{N}}{{\text{m}}^{ - 2}}\] .
The total length of the rubber rope is \[8\,{\text{m}}\] .
We are asked to find the increase in the length of the rope due to its own weight.

We will use the formula of Young’s modulus, which is given by:
\[Y = \dfrac{{FL}}{{A\Delta l}}\] …… (1)
Where,
\[Y\] indicates Young’s modulus.
\[F\] indicates applied force on the rope.
\[L\] indicates the original length of the rope.
\[A\] indicates the cross-sectional area of the rubber rope.

Again, we have,
\[F = mg\] …… (2)
Where,
\[m\] indicates the mass of the rope.
\[g\] indicates acceleration due to gravity.

Again, we have,
\[m = \rho V\] …… (3)
Where,
\[\rho \] indicates density of the material of the rubber rope.
\[V\] indicates volume of the rope.

Again,
\[V = A \times L\]
Where,
\[V\] indicates volume of the rope.
\[A\] indicates the cross-sectional area of the rope.
\[L\] indicates the original length of the rope.

Using equation (3) in equation (2), we get:
\[
F = m \times g \\
\Rightarrow F = \rho \times V \times g \\
\Rightarrow F = \rho \times A \times L \times g \\
\]
But the weight of the rope acts at the midpoint of the rope. So, we will replace \[L\] by \[\dfrac{L}{2}\] in the equation (1).

Now, from equation (1), we get:
\[
Y = \dfrac{{F \times \dfrac{L}{2}}}{{A\Delta l}} \\
\Rightarrow Y = \dfrac{{\rho ALg \times \dfrac{L}{2}}}{{A\Delta l}} \\
\Rightarrow Y = \dfrac{{g{L^2}\rho }}{{2\Delta l}} \\
\]
\[\Rightarrow \Delta l = \dfrac{{g{L^2}\rho }}{Y}\] …… (4)
Now, substituting the required values in the equation (4), we get:
\[
\Delta l = \dfrac{{g{L^2}\rho }}{Y} \\
\Rightarrow \Delta l = \dfrac{{9.8\,{\text{m}}{{\text{s}}^{ - 2}} \times {8^2} \times 1500\,{\text{kg}}{{\text{m}}^{ - 3}}}}{{5 \times {{10}^6}\,{\text{N}}{{\text{m}}^{ - 2}}}} \\
\Rightarrow \Delta l = 9.6 \times {10^{ - 2}}\,{\text{m}} \\
\]
Hence, the increase in length of the rubber rope is \[9.6 \times {10^{ - 2}}\,{\text{m}}\] .
The correct option is A.

Note: While solving this problem, many students tend to make mistakes in taking the length into account. The weight of the rope acts at the centre of the rope, hence the length of the rope should be taken as half the original length while solving the problem. All the units should be in the S.I system.