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JIPMER 2019
Exam
JIPMER
Year
2019
Subject
Physics
Chapter
Mechanical Properties of Solids

JIPMER 2019 · Physics · Mechanical Properties of Solids

Two wire of same material having radius in ratio 2: 1 and lengths...

Two wire of same material having radius in ratio 2:12: 1 and lengths in ratio 1:21: 2. If same force is applied on them, then ratio of their change in length will be

Options

  1. A1:11: 1
  2. B1:21: 2
  3. C1:41: 4
  4. D1:81: 8

Correct answer

D. 1:81: 8

Step-by-step solution

Given, ratio of radius of two wires r1:r2=2:1⇒r1r2=21r_1: r_2=2: 1 \Rightarrow \quad \frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{2}{1} and ratio in their lengths, 11:12=1:2⇒l112=121_1: 1_2=1: 2 \Rightarrow \quad \frac{l_1}{1_2}=\frac{1}{2} When same force is applied on them, then ratio r change in their length Δl1Δl2=\frac{\Delta \mathrm{l}_1}{\Delta \mathrm{l}_2}= ? We know that, Young's modulus Y=FlAΔl⇒Δl=FlAYΔl∝1 A∴ΔI1ΔI2=l1 A1/l2 A2=l1 A2l2 A1⇒Δl1ΔI2=l1l2⋅r22r12=12⋅(12)2=18\begin{aligned} & \mathrm{Y}=\frac{\mathrm{Fl}}{\mathrm{A} \Delta \mathrm{l}} \Rightarrow \Delta \mathrm{l}=\frac{\mathrm{Fl}}{\mathrm{AY}} \\ & \Delta \mathrm{l} \propto \frac{1}{\mathrm{~A}} \\ \therefore \quad & \frac{\Delta \mathrm{I}_1}{\Delta \mathrm{I}_2}=\frac{\mathrm{l}_1}{\mathrm{~A}_1} / \frac{\mathrm{l}_2}{\mathrm{~A}_2}=\frac{\mathrm{l}_1 \mathrm{~A}_2}{\mathrm{l}_2 \mathrm{~A}_1} \\ \Rightarrow \quad & \frac{\Delta \mathrm{l}_1}{\Delta \mathrm{I}_2}=\frac{\mathrm{l}_1}{\mathrm{l}_2} \cdot \frac{\mathrm{r}_2^2}{\mathrm{r}_1^2}=\frac{1}{2} \cdot\left(\frac{1}{2}\right)^2=\frac{1}{8} \end{aligned}
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