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Question

A thick rope of rubber of density $$1.5 \times {10^3}{\text{kg/}}{{\text{m}}^{\text{3}}}$$ and Young's modulus $$5 \times {10^6}{\text{N/}}{{\text{m}}^2}$$ , $$8 m$$ length is hung from the ceiling of a room , the increases in its length due to its own weight is :

 $$\left( {g = 10{\text{m/}}{{\text{s}}^{\text{2}}}} \right)$$



A
9.6×10 - 2m
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B
19.2×10 - 7m
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C
9.6×10 - 7m
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D
9.6m
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Solution

The correct option is A $$9.6 \times {10^{{\text{ - 2}}}}{\text{m}}$$
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
Mg acts at centre of gravity which is at $$\dfrac{L}{2}$$
so we take length of rope$$=\dfrac{L}{2}$$

we know $$Y=\dfrac{stress}{strain}=\dfrac{F/A}{\Delta l/l}=\dfrac{Fl}{\Delta l.A}$$

$$\Rightarrow \Delta l=\dfrac{Fl}{AY}=\dfrac{mg L}{2AY}=\dfrac{Al\rho g l}{2AY}$$

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

$$=\dfrac{1.5\times 10^3\times 10\times 8^2}{2\times 5\times 10^6}$$

$$\Rightarrow \Delta l=9.6\times 10^{-2}m$$.
So (A) is correct option.

1446121_1163043_ans_1ed419d3ef2d4b4da913f03385721fc3.png

Physics

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