Question

The young's modulus of a rubber string 8cm long and density 1.5kg/m^3 is 5x10^8N/m^2, is suspended on the ceiling in a room. The increase in length due to its own weight will be???

Answer 1

Answer: $\Delta L = 18.8 \times10^{-11}\ m$

Explanation:

Given that,

Young's modulus $Y = 5\times10^{8}\ N/m^{2}$

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

Length $l = 8\times10^{-2}\ m$

We know that,

The young modulus

$Y = \dfrac{\dfrac{F}{A}}{\dfrac{\Delta L}{L}}$

$Y = \dfrac{\dfrac{F\times d}{A\times d}}{\dfrac{\Delta L}{L}}$

$Y = \dfrac{F\times d\times L}{V\times \Delta L}$

$Y = \dfrac{mg\times A}{V\times \Delta L}$

$Y = \dfrac{\rho \times g\times A}{\Delta L}$

$5\times10^{8}\ N/m^{2} = \dfrac{1.5\ kg/m^{3}\times 9.8\ m/s^{2}\times64\times10^{-4}\ m}{\Delta L}$

$\Delta L = 18.8 \times10^{-11}\ m$

Hence, the increases in length due to its own weight will be $\Delta L = 18.8 \times10^{-11}\ m$

Answer 2

Answer:

The increase in length due to its own weight will be $9.4\times10^{-11}\ m$

Explanation:

Given that,

Length $l=8\times10^{-2}\ m$

Young's modulus $Y=5\times10^{8}\ N/m^{2}$

Density $\rho=1.5\ kg/m^3$

The young's modulus is defined as,

$Y=\dfrac{stess}{strain}$

$Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta l}{l}}$....(I)

Multiply  and divided by d in equation (I)

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

$Y=\dfrac{F\times d\times l}{V\times\Delta l}$

$Y=\dfrac{FA}{V\times\Delta l}$

Using newton's second law

$Y=\dfrac{maA}{V\time\Delta l}$

$Y=\dfrac{\rho a\times A}{\Delta l}$

$\Delta l=\dfrac{\rho a\times A}{Y}$....(II)

Here, $a =\dfrac{g}{2}$  = acceleration due to gravity

Put the value in the equation (II)

$\Delta l=\dfrac{1.5\times4.9\times64\times10^{-4}}{5\times10^{8}}$

$\Delta l =9.4\times10^{-11}\ m$

Hence, The increase in length due to its own weight will be $9.4\times10^{-11}\ m$