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AIIMS 201827 May
Exam
AIIMS
Year
2018
Date
27 May
Subject
Physics
Chapter
Mechanical Properties of Solids

AIIMS 2018 · 27 May · Physics · Mechanical Properties of Solids

The Young's modulus of a rope of 10 ~m length and having diameter...

The Young's modulus of a rope of 10 m10 \mathrm{~m} length and having diameter of 2 cm2 \mathrm{~cm} is 20.0×1011dyne/cm220.0 \times 10^{11} \mathrm{dyne} / \mathrm{cm}^2. If the elongation produced in the rope is 1 cm1 \mathrm{~cm}, the force applied on the rope is

Options

  1. A6.28×105 N6.28 \times 10^5 \mathrm{~N}
  2. B6.28×104 N6.28 \times 10^4 \mathrm{~N}
  3. C6.28×1046.28 \times 10^4 dyne
  4. D6.28×1056.28 \times 10^5 dyne

Correct answer

B. 6.28×104 N6.28 \times 10^4 \mathrm{~N}

Step-by-step solution

Here, Length of the rope, L=10 mL=10 \mathrm{~m} Diameter of the rope, D=2 cm∴D=2 \mathrm{~cm}\therefore \quad Radius of the rope, r=D2=1 cm=10−2 mr=\frac{D}{2}=1 \mathrm{~cm}=10^{-2} \mathrm{~m} Young's modulus of the rope Y=20×1011 dyne cm2=20×1011×10−5(10−2)2Nm2=20×1011×10−5 N10−4 m2=20×1010Nm2\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, ΔL=1 cm=10−2 m\Delta L=1 \mathrm{~cm}=10^{-2} \mathrm{~m} As Y=F/AΔL/L or  F=YAΔLL=Yπr2ΔLL=20×1010×3.14×(10−2)2×10−210=6.28×104 N\begin{aligned} Y & =\frac{F / A}{\Delta L / L} \text{\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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