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Question: Answered & Verified by Expert
The Young's modulus of a rope of $10 \mathrm{~m}$ length and having diameter of $2 \mathrm{~cm}$ is $20.0 \times 10^{11} \mathrm{dyne} / \mathrm{cm}^2$. If the elongation produced in the rope is $1 \mathrm{~cm}$, the force applied on the rope is
PhysicsMechanical Properties of SolidsAIIMSAIIMS 2018 (27 May)
Options:
  • A $6.28 \times 10^5 \mathrm{~N}$
  • B $6.28 \times 10^4 \mathrm{~N}$
  • C $6.28 \times 10^4$ dyne
  • D $6.28 \times 10^5$ dyne
Solution:
2426 Upvotes Verified Answer
The correct answer is: $6.28 \times 10^4 \mathrm{~N}$
Here, Length of the rope, $L=10 \mathrm{~m}$
Diameter of the rope, $D=2 \mathrm{~cm}$
$\therefore \quad$ Radius of the rope, $r=\frac{D}{2}=1 \mathrm{~cm}=10^{-2} \mathrm{~m}$
Young's modulus of the rope
$\begin{aligned} Y & =20 \times 10^{11} \frac{\text { dyne }}{\mathrm{cm}^2}=20 \times 10^{11} \times \frac{10^{-5}}{\left(10^{-2}\right)^2} \frac{\mathrm{N}}{\mathrm{m}^2} \\ & =20 \times 10^{11} \times \frac{10^{-5} \mathrm{~N}}{10^{-4} \mathrm{~m}^2}=20 \times 10^{10} \frac{\mathrm{N}}{\mathrm{m}^2}\end{aligned}$
Elongation of rope, $\Delta L=1 \mathrm{~cm}=10^{-2} \mathrm{~m}$
As $\begin{aligned} Y & =\frac{F / A}{\Delta L / L} \text { or } \quad F=Y \frac{A \Delta L}{L}=\frac{Y \pi r^2 \Delta L}{L} \\ & =\frac{20 \times 10^{10} \times 3.14 \times\left(10^{-2}\right)^2 \times 10^{-2}}{10} \\ & =6.28 \times 10^4 \mathrm{~N}\end{aligned}$

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