A $10 m e t e r$ long thick rubber pipe is suspended from one of its ends. The extension produced in the pipe under its own weight will be : $(Y = 5 \times 10^{6} N / m^{2}$ and density of rubber = $1500 k g / m^{3}$ )

1.5 m

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0.15 m

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0.015 m

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0.0015 m

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Solution

The correct option is A $0.15 m$
$Δ l = \frac{F L}{A Y} = \frac{\frac{m g}{2} L}{A Y} (a v e r a g e f o r c e d u e t o w e i g h t a t c e n t r e o f m a s s = \frac{m g}{2}) = \frac{(1500 \times A \times L \times g) . L}{2 A Y} = \frac{1500 \times 10 \times 10 \times 10}{2 \times 5 \times 1 0^{6}} = 0.15 m$

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A 10 meter long thick rubber pipe is suspended from one of its ends. The extension produced in the pipe under its own weight will be :(Y=5×106N/m2 and density of rubber =1500 kg/m3)

The correct option is A 0.15 mΔl=FLAY=mg2LAY(averageforceduetoweightatcentreofmass=mg2)=(1500×A×L×g).L2AY=1500×10×10×102×5×106=0.15m

A $10 m e t e r$ long thick rubber pipe is suspended from one of its ends. The extension produced in the pipe under its own weight will be : $(Y = 5 \times 10^{6} N / m^{2}$ and density of rubber = $1500 k g / m^{3}$ )