First slide

Stress and strain

Question

An Indian rubber cord L metre long and area of cross-section A metre² is suspended vertically. Density of rubber is p kg/metre³ and Young's modulus of rubber is Y newton/metre². If the cord extends by l metre under its own weight, then extension I is

Moderate

A

$L^{2} ρg / Y$
B

$L^{2} ρg / 2 Y$
C

$L^{2} ρg / 4 Y$
D

$Y / L^{2} ρg$

Solution

Weight of the cord
$\begin{array}{l} W = Vρg = ALρg \\ ∴ Stress = \frac{W}{A} = Lρg \end{array}$
For the purpose of extension, the weight acts at the centre of gravity of the bar. Hence the upper half length, i.e., [L / 2) will be stretched.
$∴ Strain = \frac{l}{(L / 2)} = \frac{2 l}{L}$
Now, $Y = \frac{Stress}{Strain} = \frac{Lρg \times L}{2 l} = \frac{L^{2} ρg}{2 l}$
or $l = L^{2} ρg / 2 Y$

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