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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?

Option: 1

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


Option: 2

L=300 \mathrm{~cm}, r=3 \mathrm{~mm}


Option: 3

L=200 \mathrm{~cm}, r=3 \mathrm{~mm}


Option: 4

L=400 \mathrm{~cm}, r=4 \mathrm{~mm}


Answers (1)

best_answer

A=, \quad \Delta L=\frac{F L}{A Y}
Because, wires of the same material are stretched by the same load. So, F and Y wile be constant.

\begin{aligned} & \therefore \quad \Delta L \propto \frac{L}{\pi r^2} \\ & \Delta L_1=\frac{100}{\pi \times\left(1 \times 10^{-3}\right)^2}=\frac{100}{\pi \times 10^{-6}}=\frac{100}{\pi} \times 10^6 \\ & \therefore \quad \Delta L_2=\frac{200}{\pi \times\left(3 \times 10^{-3}\right)^2}=\frac{200}{\pi \times 9 \times 10^{-6}}=\frac{22.2}{\pi} \times 10^6 \\ & \therefore \Delta L_3=\frac{300}{\pi \times\left(3 \times 10^{-3}\right)^2}=\frac{300}{\pi \times 9 \times 10^{-6}}=\frac{33.3}{\pi} \times 10^6 \\ & \therefore \quad \Delta L_4=\frac{400}{\pi \times\left(4 \times 10^{-3}\right)^2}=\frac{400}{\pi \times 16 \times 10^{-6}}=\frac{25}{\pi} \times 10^6 \\ & \end{aligned}
We can see that, L=100 \mathrm{~cm} and r=1 \mathrm{~mm} will elongate most.

Posted by

Devendra Khairwa

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