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JEE Main
Exam
JEE Main
Subject
Physics
Chapter
Mechanical Properties of Solids

JEE Main · Physics · Mechanical Properties of Solids

Four wires of the same material are stretched by the same load....

Four wires of the same material are stretched by the same load. Which one of them will elongate most if their dimensions are as follows

Options

  1. AL=100 cm,r=1 mmL=100 \mathrm{~cm}, r=1 \mathrm{~mm}
  2. BL=200 cm,r=3 mmL=200 \mathrm{~cm}, r=3 \mathrm{~mm}
  3. CL=300 cm,r=3 mmL=300 \mathrm{~cm}, r=3 \mathrm{~mm}
  4. DL=400 cm,r=4 mmL=400 \mathrm{~cm}, r=4 \mathrm{~mm}

Correct answer

A. L=100 cm,r=1 mmL=100 \mathrm{~cm}, r=1 \mathrm{~mm}

Step-by-step solution

\Delta L=\frac{F L}{A Y}Because,wiresofthesamematerialarestretchedbythesameload.So,Because, wires of the same material are stretched by the same load. So,FandandYwillbeconstant.will be constant.\DeltaL∝Lπr2\DeltaL1=100π×(1×10−3)2=100π×10−6=100π×10−6\DeltaL2=200π×(3×10−3)2=200π×9×10−6=22.2π×106\DeltaL3=300π×(3×10−3)2=300π×9×10−6=33.3π×106\DeltaL4=400π×(4×10−3)2=400π×16×10−6=25π×106\begin{aligned}\DeltaL & \propto \frac{L}{\pi r^{2}}\\ \DeltaL_{1} &=\frac{100}{\pi \times\left(1 \times 10^{-3}\right)^{2}}\\ &=\frac{100}{\pi \times 10^{-6}}=\frac{100}{\pi} \times10^{-6} \\ \DeltaL_{2} &=\frac{200}{\pi \times\left(3 \times 10^{-3}\right)^{2}}\\ &=\frac{200}{\pi \times 9 \times 10^{-6}}=\frac{22.2}{\pi} \times10^{6} \\ \DeltaL_{3} &=\frac{300}{\pi \times\left(3 \times 10^{-3}\right)^{2}}\\ &=\frac{300}{\pi \times 9 \times 10^{-6}}=\frac{33.3}{\pi} \times10^{6} \\ \DeltaL_{4} &=\frac{400}{\pi \times\left(4 \times 10^{-3}\right)^{2}}\\ &=\frac{400}{\pi \times 16 \times 10^{-6}}=\frac{25}{\pi} \times10^{6} \end{aligned}Wecanseethat,We can see that,L=100 \mathrm{~cm}andandr=1 \mathrm{~mm}$ will elongate most.
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