Question

A thick rope of rubber of density 1.5×10^3 kgm^-3 and
Young's modulus 5x10^6 Nm2,8 m in length is hung
from the ceiling of a room, the increase in its length due to
its own weight is
(a) 9.6 x 10-2 m
(b) 19.2 x 10-2 m
(c) 9.6 x 10-3 m
(d) 9.6 m​

Answer 1

Answer:

A thick rope of rubber of density 1.5×10^3 kgm^-3 and  Young's modulus 5x10^6 Nm2,8 m in length is hung

from the ceiling of a room, the increase in its length due to its own weight is:

9.6 x 10-2 m i.e Option (a) is correct.

The problem will be solved by using the following formula:

K= (F/A)/(Δl/l)

Answer 2

Answer:

The increase in its length due to  its own weight is $9.6\times10^{-2}\ m$

(a) is correct option.

Explanation:

Given that,

Density of rubber$\rho=1.5\times10^{3}\ kg/m^3$

Young's modulus $Y=5\times10^{6}\ Nm^2$

Length = 8 m

We need to calculate the increase length

Using formula of young's modulus

$Y=\dfrac{F\times l}{A\times\Delta l}$

$\Delta l=\dfrac{F\times l}{A\timesY}$

$\Delta I=\dfrac{mgl}{AY}$

$\Delta l=\dfrac{V\times\rho\times g\times l}{2AY}$

$\Delta l=\dfrac{A\times l\times\rho\times g\times l}{2AY}$

$\Delta l=\dfrac{\rho\times g\times l^2}{2Y}$

Put the value into the formula

$\Delta l=\dfrac{1.5\times10^{3}\times10\times8\times8}{2\times5\times10^{6}}$

$\Delta l=0.096=9.6\times10^{-2}\ m$

Hence, The increase in its length due to  its own weight is $9.6\times10^{-2}\ m$