A rubber pipe of density 1.5 times 10^3 , kg/ | vedclass

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(D) The change in length $\Delta L$ of a rod or pipe of length $L$,density $d$,and Young's modulus $Y$ due to its own weight is given by the formula:

Vedclass · Accepted Answer

$9.6 	imes 10^{-2} \, m$

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(D) The change in length $\Delta L$ of a rod or pipe of length $L$,density $d$,and Young's modulus $Y$ due to its own weight is given by the formula: $\Delta L = \frac{L^2 dg}{2Y}$.Given values are: $L = 8 \, m$,$d = 1.5 	imes 10^3 \, kg/m^3$,$Y = 5 	imes 10^6 \, N/m^2$,and taking $g = 10 \, m/s^2$.Substituting these values into the formula:$\Delta L = \frac{(8)^2 	imes (1.5 	imes 10^3) 	imes 10}{2 	imes (5 	imes 10^6)}$$\Delta L = \frac{64 	imes 1.5 	imes 10^4}{10 	imes 10^6} = \frac{96 	imes 10^4}{10^7} = 9.6 	imes 10^{-2} \, m$.

$A$ rubber pipe of density $1.5 \times 10^3 \, kg/m^3$ and Young's modulus $5 \times 10^6 \, N/m^2$ is suspended from the roof. The length of the pipe is $8 \, m$. What will be the change in length due to its own weight?

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