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  • A thick uniform rubber rope of density <img alt="\mathrm{1.5 \mathrm{~g} \mathrm{~cm}^{-3}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B1.5%20%5Cmathrm%7B%7Eg%7D%20%5Cmathrm%7

A thick uniform rubber rope of density \mathrm{1.5 \mathrm{~g} \mathrm{~cm}^{-3}}and Young's modulus \mathrm{5 \times 10^6 \mathrm{Nm}^{-2}} has a length of \mathrm{8 \mathrm{~m}.} When hung from the ceiling of a room, the increase in length of the rope due to its own weight will be

Option: 1

\mathrm{9.6 \times 10^{-2} \mathrm{~m}}


Option: 2

\mathrm{19.2 \times 10^{-3} \mathrm{~m}}


Option: 3

\mathrm{9.6 \times 10^{-3} \mathrm{~m}}


Option: 4

\mathrm{9.6 \mathrm{~m}}


Answers (1)

best_answer

The weight of the rope can be assumed to act at its mid-point. Now, the extension l is proportional to original length L. If the weight of the rope acts at its mid-point, the extension will be that produced by half the rope. So, replacing L by \mathrm{L / 2} in the expression for Young's modulus, we have

                     \mathrm{ \begin{gathered} Y=\frac{F L / 2}{A l}=\frac{F L}{2 A l} \\\\ \therefore \quad l=\frac{F L}{2 A Y}=\frac{g L^2 \rho}{2 Y} \\\\ (\because F=m g \text { and } m=\rho V=\rho A L) \end{gathered} }

Now \mathrm{\rho=1.5 \mathrm{~g} \mathrm{~cm}^{-3}=1500 \mathrm{~kg} \mathrm{~m}^{-3},} therefore

                       \mathrm{ l=\frac{10 \times(8)^2 \times 1500}{2 \times 5 \times 10^6}=9.6 \times 10^{-2} \mathrm{~m} }

Posted by

Kuldeep Maurya

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