Question

A thick rope of rubber of density $1.5 \times {10^3}kg{m^{ - 3}}$ and Young’s modulus$5 \times {10^6}N{m^{ - 2}}$ . 8m length is hung from the ceiling of a room, the increase in its length due to its own weight is:
A. $g = 10m{s^{ - 2}}$
B. $9.6 \times {10^{ - 2}}m$
C. $19.2 \times {10^{ - 7}}m$
D. $9.6 \times {10^{ - 7}}m$
E. $9.6m$

Answer 1

Hint: We know that when an external force is applied to a body there is change in its configuration, and so after the removal force if it regains its original position then it is called elasticity of the material. For a more elastic body Young’s modulus is also large.

Formula used:
Elongation of the wire is given by,
$\Delta L = \dfrac{{\rho g{l^2}}}{{2Y}}$
Where, $\rho $ is the density of the rod, g is acceleration due to gravity, l is the length of the wire, y is the young’s modulus of the rod.

Complete step by step answer:
Given, density $\rho = 1.5 \times {10^3}kg{m^{ - 3}}$ young’s modulus$Y = 5 \times {10^6}N{m^{ - 2}}$, $g = 10m{s^{ - 2}}$ and l=8m
The centre of mass of the given rod is at its geometric centre. Therefore, the length of the wire from the centre of mass we can measure as $\dfrac{l}{2}$.
Let us Consider a wire of length l and area of cross section A, suspended from a rigid support. Let $\Delta l$ be the increase in length of the wire due to a load F applied at the other end of the wire.
We know that restoring force is developed when an object is stretched. That restoring force is called as stress and it is given as:
Longitudinal stress$\sigma = \dfrac{F}{A}$
The longitudinal strain is defined as the change in length divided by the original length. It is mathematically given as:
Longitudinal strain$\varepsilon = \dfrac{{\Delta l}}{l}$
The ratio of stress and strain is called Young's modulus. The dimensions of the young’s modulus are given as$M{L^{^{ - 1}}}{T^{ - 2}}$.
Young’ modulus,$Y = \dfrac{\sigma }{\varepsilon }$
The SI unit of the young’s modulus is $Pa$
$Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta l}}{l}}}$……...(1)
 $ = \dfrac{{Fl}}{{A\Delta l}}$
Now let's consider the density. We all know that mass by volume is density. That is,
Density, $\rho = \dfrac{M}{V}$
$ = \dfrac{M}{{A \times l}}$
Where A is the area of the cross section of the rod.
Therefore, $M = \rho lA$ and also $F = Mg = \rho lAg$
Now substituting all these values in equation (1)
We get,
$Y = \dfrac{{\rho gA{l^2}}}{{A\Delta l}}$
Therefore, elongation produced, $\Delta l = \dfrac{{\rho g{l^2}}}{{2Y}}$
Now substitute the given values in problem.
That is,
We get,
$\Delta L = \dfrac{{1.5 \times {{10}^3} \times 10 \times {8^2}}}{{5 \times {{10}^6} \times 2}}$
$ = 9.6 \times {10^{ - 2}}m$
$ = 9.6cm$
Thus the correct option is (a).

Additional information:
When a body is subjected to deforming force, a restoring force is developed in the body. In equilibrium position, this internal restoring force is equal in magnitude and opposite in direction to the applied force. The internal restoring force developed per unit area of a deformed body is called stress.

Under the action of deforming force, the configuration of a body changes, and the body is said to be strained by the force. When a body is deformed the ratio of change in configuration to the original configuration is called strain. A Strain is a dimensionless quantity and has no units.

The ratio of stress by strain is called the modulus of elasticity. Depending upon the type of stress applied and the resulting strain, there are three types of moduli of elasticity.

Note:
Elasticity is a property of a material that regains its original length, volume, and shape after the removal of deforming force.
When a solid is deformed by the application of external force, the atoms or molecules are displaced from its initial position.