Question

A rubber rope of length $8m$ is hung from the ceiling of a room. What is the increase in length of rope due to its own weight? (Given: Young’s Modulus of elasticity of rubber $=5\times {{10}^{6}}\dfrac{N}{m}$ and density of rubber $=1.5\times {{10}^{3}}\dfrac{kg}{{{m}^{3}}}$ . Take $g=10\dfrac{m}{{{s}^{2}}}$ ).
$\text{A}\text{. }1.5mm$
$\text{B}\text{. }6mm$
$\text{C}\text{. }24mm$
$\text{D}\text{. }96mm$

Answer 1

Hint: When a rope is hanged, the weight of the rope causes it to elongate. Hooke's law is used to determine the elongation in rope and the calculus of variations can be used to find the taper which can minimize the elongation.

Formula used:
$\Delta L=\dfrac{MgL}{2AY}$

Complete step-by-step answer:
When a rope is hung through the ceiling, extension in rope length will be due to self-weight which is distributed all along the length. The extension is half of the extension produced when the same wire is connected to the ceiling and force $F$ is applied to the other end. We use Hooke's law to find the extension in the rope length.

Weight of rubber rope $W=mg$
Using $\text{density = }\dfrac{\text{mass}}{\text{volume}}$
$\rho =\dfrac{m}{AL}$
where $A$ is the cross sectional area of rope and $L$ is the length of rope
$W=mg=\rho ALg$
Elongation in the length of rope due to its own weight would be half of the elongation due to point load
Let the elongation in length of rope be$\Delta L=\dfrac{WL}{2AY}=\dfrac{8\times 1.5\times {{10}^{3}}\times A\times L\times g}{2A\times 5\times {{10}^{6}}}=96\times {{10}^{-3}}m$
$\Delta L=96mm$
Elongation in length of rubber rope is $96mm$
Hence, the correct option is D.

Note: In order to taper a heavy rope, it should be hanged vertically, to minimise the elongation due to its own weight plus a load at its lower end. Hooke's law can be used to determine the elongation in the rope.
While doing calculations, prefer to take all the terms in SI units to avoid calculation error.