$\begin{array}{1 1}(A)\;1mm\\(B)\;2mm\\(C)\;3mm\\(D)\;4mm\end{array} $

Force on 1 leg=$\large\frac{50\times 10^3}{4}$

Area(a)=$\pi r^2$

$\Rightarrow \pi(5\times 10^{-2})^2$

$\Rightarrow 25\pi\times 10^{-4}$

$l=2m$

As $Y=(\large\frac{f}{a})\frac{l}{\Delta l}$

$\Rightarrow \Delta l=(\large\frac{f}{a})\frac{l}{Y}$

$\Rightarrow \large\frac{50\times 10^3}{4\times 25\pi\times 10^{-4}}\times \frac{2}{5\times 10^{10}}$

$\Rightarrow 100\times 10^{-5}m$

$\Rightarrow$ 1mm

Hence (A) is the correct answer.

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