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A wire of cross sectional area A, modulus of elasticity $2 \times 10^{11} \mathrm{Nm}^{-2}$ and length $2 \mathrm{~m}$ is stretched between two vertical rigid supports. When a mass of $2 \mathrm{~kg}$ is suspended at the middle it sags lower from its original position making angle $\theta=\frac{1}{100}$ radian on the points of support. The value of $\mathrm{A}$ is _______ $\times 10^{-4} \mathrm{~m}^2$ (consider $x< < \mathrm{L}$ ).
(given : $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )
(given : $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )
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The correct answer is:
1
In vertical direction
$2 \mathrm{~T} \sin \theta=20$
using small angle approximation $\sin \theta=\theta$
$\begin{aligned}
& \theta=\frac{1}{100} \\
& \therefore \mathrm{T}=\frac{10}{\theta} \\
& \mathrm{T}=1000 \mathrm{~N}
\end{aligned}$
$\begin{array}{ll}
\text {Change in length } \Delta \mathrm{L} =2 \sqrt{\mathrm{x}^2+\mathrm{L}^2}-2 \mathrm{~L} \\
=2 \mathrm{~L}\left[1+\frac{\mathrm{x}^2}{2 \mathrm{~L}^2}-1\right] \\
\Delta \mathrm{L} =\frac{\mathrm{x}^2}{\mathrm{~L}}
\end{array}$
$\therefore$ Modulus of elasticity $=\frac{\text { stress }}{\text { strain }}$
$\begin{aligned}
& 2 \times 10^{11}=\frac{10^3}{\mathrm{~A} \times \frac{\mathrm{x}^2}{\mathrm{~L}}} \times 2 \mathrm{~L} \\
\therefore & \mathrm{A}=1 \times 10^{-4} \mathrm{~m}^2
\end{aligned}$
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