A thick rope of rubber of density 1.5 $ imes 10^{3}\ kgm^{- 3}$ and Young’s modulus $5\ imes 10^{6}Nm^{- 2}$, 8m in length is hung from the ceiling of a room, the increase in $\phi$ its length due to its own weight is

Question

A thick rope of rubber of density 1.5 $	imes 10^{3}\ kgm^{- 3}$ and Young’s modulus $5\ 	imes 10^{6}Nm^{- 2}$, 8m in length is hung from the ceiling of a room, the increase in $\phi$ its length due to its own weight is

Dr. Abubacker · Accepted Answer

(a) If $(A)$ is the area of cross-section and $l$ is the length of rope, the mass of rope, $m = \frac{Al}{ho}$. As the weight of the rope acts at the mid-point of the rope.
So, $Y = \frac{mg}{A} 	imes \frac{(I/2)}{\mathrm{\Delta}l}$
$\mathrm{\Delta}l = \frac{mgl}{2AY} = \frac{Alho gl}{2AY} = \frac{gho l^{2}}{2AY}$
Or $\mathrm{\Delta}l = \frac{9.8 	imes 1.5 	imes 10^{3} 	imes 8^{2}}{2 	imes 5 	imes 10^{6}} = 9.6 	imes 10^{2}$m

Q. A thick rope of rubber of density 1.5 $\times 10^{3}\ kgm^{- 3}$ and Young’s modulus $5\ \times 10^{6}Nm^{- 2}$, 8m in length is hung from the ceiling of a room, the increase in $\phi$ its length due to its own weight is

Explanation

Reviewed by Dr. Abubacker

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