Question

An Indian rubber cord $L$ metre long and the area of cross-section A $m^2$ is suspended vertically. Density of rubber is $\rho \,{\text{kg/m}^3}$ and Young’s modulus of rubber is $Y\,{\text{N/m}^2}$. If cord extends by $l$ metre under its own weight, then extension is:

A. $\dfrac{{{L^2}\rho g}}{Y}$

B. $\dfrac{{{L^2}\rho g}}{{2Y}}$

C. $\dfrac{{{L^2}\rho g}}{{4Y}}$

D. $\dfrac{Y}{{{L^2}\rho g}}$

Answer 1

Use the formula for density of an object. Also use the formula for the stress, strain and Young’s modulus for the material of the object. Using all these formulas derive the expression for the stress,, strain and the Young’s modulus for the material of the rubber cord. From all these formulae, derive the relation for the extension in the rubber cord.

Formulae used:

The density $\rho$ of an object is given by

$\rho = \dfrac{m}{V}$ …… (1)

Here, $m$ is the mass of the object and $V$ is the volume of the object.

The stress on an object is given by

${\text{Stress}} = \dfrac{F}{A}$ …… (2)

Here, $F$ is the force acting on the object and $A$ is the area on which the force is applied.

The strain acting on a wire given by

${\text{Strain}} = \dfrac{{\Delta L}}{L}$ …… (3)

Here, $\Delta L$ is the change in length of the wire and $L$ is the original length of the wire.

The Young’s modulus $Y$ of the material of the object is

$Y = \dfrac{{{\text{Stress}}}}{{{\text{Strain}}}}$ …… (4)

Complete step by step solution:

We have given that length of the rubber cord is $L$, cross-sectional area of the cord is A $m^2$, density of the rubber is $\rho kg/m^3$ and Young’s modulus for the material of the rubber cord is $Y N/m^2$.

The cord extends by the length $l$ because of its own weight. The weight of the rubber cord is given by

$F = mg$

According to equation (1), substitute $\rho V$ for $m$ in the above equation.

$F = \rho Vg$

Substitute $AL$ for $V$ in the above equation.

$F = \rho ALg$

The stress acting on the rubber cord is given by equation (2).

Substitute $\rho ALg$ for $F$ in equation (2).

${\text{Stress}} = \dfrac{{\rho ALg}}{A}$

$\Rightarrow {\text{Stress}} = \rho Lg$

Hence, the stress acting on the rubber cord is $\rho Lg$.

Let us now calculate the strain acting on the rubber cord. The extension in the length of the cord is due to its own weight and the weight of the cord is acting at the centre of the rubber cord. Hence, the original length of the rubber cord is half of its original length. Substitute $l$ for $\Delta L$ and $L/2$ for $L$ in equation (3).

${\text{Strain}} = \dfrac{l}{{L/2}}$

$\Rightarrow {\text{Strain}} = \dfrac{{2l}}{L}$

Hence, the strain on the rubber cord is $\dfrac{{2l}}{L}$.

Let us now calculate the Young’s modulus of the material of the rubber cord. Substitute $\rho Lg$ for stress and $\dfrac{{2l}}{L}$ for strain in equation (4).

$Y = \dfrac{{\rho Lg}}{{\dfrac{{2l}}{L}}}$

$\therefore l = \dfrac{{{L^2}\rho g}}{{2Y}}$

Therefore, the extension in the length of the rubber cord is $\dfrac{{{L^2}\rho g}}{{2Y}}$.

Hence, the correct option is (B).

Note: The student should keep in mind that the weight of the rubber cord acts at its centre of gravity which is at the centre of the rubber cord. Hence, the original length of the rubber cord while determining the strain acting on the wire must be taken half of the whole length of the rubber cord. Otherwise, the derivation for the extension rubber cord will be incorrect.