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BITSAT 2020
Exam
BITSAT
Year
2020
Subject
Physics
Chapter
Mechanical Properties of Solids

BITSAT 2020 · Physics · Mechanical Properties of Solids

There are two wire of same material and same length while the...

There are two wire of same material and same length while the diameter of second wire is two times the diameter of first wire, then the ratio of extension produced in the wires by applying same load will be

Options

  1. A1: 1
  2. B2: 1
  3. C1: 2
  4. D4: 1

Correct answer

D. 4: 1

Step-by-step solution

∵\because Both wires are same materials so both will have same Young's modulus, and let it be Y. Y= stress  strain =F A. (ΔL/L),F=Y=\frac{\text{ stress }}{\text{ strain }}=\frac{F}{\text{ A. }(\Delta L / L)}, F= applied force A=\mathrm{A}= area of cross-section of wire Now, Y1=Y2⇒FL(A1)(ΔL1)=FL(A2)(ΔL2)\mathrm{Y}_{1}=\mathrm{Y}_{2} \Rightarrow \frac{\mathrm{FL}}{\left(\mathrm{A}_{1}\right)\left(\Delta \mathrm{L}_{1}\right)}=\frac{\mathrm{FL}}{\left(\mathrm{A}_{2}\right)\left(\Delta \mathrm{L}_{2}\right)} Since load and length are same for both ⇒r12ΔL1=r22ΔL2,(ΔL1ΔL2)=(r2r1)2=4ΔL1:ΔL2=4:1\begin{array}{l} \Rightarrow \mathrm{r}_{1}^{2} \Delta \mathrm{L}_{1}=\mathrm{r}_{2}^{2} \Delta \mathrm{L}_{2},\left(\frac{\Delta \mathrm{L}_{1}}{\Delta \mathrm{L}_{2}}\right)=\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{2}=4 \\ \Delta \mathrm{L}_{1}: \Delta \mathrm{L}_{2}=4: 1 \end{array}
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