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A rod of length L , density $\rho$ and young modulus Y is hung from the ceiling of a room . Increase in its length due to its own weigth is
Option: 1
$\frac{\rho gL^2}{2Y}$

Option: 2
$\frac{2Y}{\rho gL^2}$

Option: 3
$\frac{2\rho gL}{Y}$

Option: 4
$\frac{\rho^2 g^2L^2}{Y^2}$

Answers (1)

best_answer

As we have learned

Elongation in terms of density -

$\Delta l= \frac{MgL}{AY}$

$M=volume (AL)\times density (\rho )$

$\Delta l= \frac{\left ( AL \right )g\rho L}{AY}$

$\Delta l= \frac{\rho g L^{2}}{AY}$

- wherein

$\rho = density$

$Y =Young\: modulus$

$A =area\:$

$L =length\:$

Its centre of mass is at L/2 . Lets apply weight of rod at its centre of mass then elongation in its length is

$\Delta l = \frac{Mg(L/2)}{YA}= \frac{(\rho /AL)(g\frac{L}{2})}{AY}\\\\ \Delta l = \frac{\rho gL^2}{2Y}$

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jitender.kumar

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