Question

A rubber string $10\,m$ long is suspended from a rigid support at its one end.Calculate the extension in the string due to its own weight. The density of rubber is $1.5 \times {10^3}\,kg\,{m^{ - 3}}$ and Young’s modulus for the rubber is $5 \times {10^6}\,N{m^{ - 2}}$. Take $g = 10\,N\,k{g^{ - 1}}$.

Answer 1

Hint: Let know about Young’s modulus. Young's modulus is a term used to describe the ability of a The Young modulus, also known as the modulus of elasticity in tension, is a mechanical property that assesses a solid material's tensile stiffness. In the linear elastic region of a material, it quantifies the relationship between tensile stress $\sigma $ (force per unit area) and axial strain \[\varepsilon \] (proportional deformation).

Formula used:
Young’s Modulus$ = \dfrac{{\sigma (\varepsilon )}}{\varepsilon } = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta L}}{L}}} = \dfrac{{FL}}{{A\Delta L}}$
Where, $F$ is the force exerted on an object under tension, $A$ is the real cross-sectional area, which is the cross-sectional area perpendicular to the applied force, $\Delta L$ is the amount by which the object's length varies and $L$ is the original length of the object.

Complete step by step answer:
Given: Length $L = 10\,m$, density of the rubber $\rho = 1.5 \times {10^3}\,kg\,{m^{ - 3}}$ and Young’s modulus Y$ = 5 \times {10^6}\,N{m^{ - 2}}$.
We have to find extend length,
$\Delta L = \dfrac{{FL}}{{AY}}$
Let us know about density. The amount of mass per unit of volume is referred to as "density."An object constructed of a relatively dense material (such as iron) will have less volume than one made of a less dense substance of comparable mass.
$\rho = \dfrac{\text{Mass}}{\text{Volume}}$
$\Rightarrow \rho = \dfrac{m}{{AL}}$
$\Rightarrow m = \rho AL$

Now Force $F = mg$
Now using value of m:
$F = \rho ALg$
$\Rightarrow \Delta L = \dfrac{{FL}}{{AY}}$
Now using the value:
$\Delta L = \dfrac{{\rho ALg \times L}}{{A \times Y}}$
$\Rightarrow \Delta L = \dfrac{{\rho g{L^2}}}{Y}$
$\Rightarrow \Delta L = \dfrac{{1.5 \times {{10}^3} \times 10 \times 10 \times 10}}{{5 \times {{10}^6}}}$
$\therefore \Delta L = 0.3\,m = 30\,cm$

Hence, the extension in the string due to its own weight is 30 cm.

Note: The Young's modulus of a material is not always the same in all orientations.The mechanical characteristics of most metals and ceramics, as well as many other materials, are the same in all orientations. Metals and ceramics, on the other hand, can be treated with particular impurities, and metals can be mechanically manipulated to create directed grain structures. These materials become anisotropic, and Young's modulus varies depending on the force vector's direction.