$\begin{array}{1 1}(A)\;1.6\times 10^{-6}m\\(B)\;3.2\times 10^{-6}m\\(C)\;2.4\times 10^{-6}m\\(D)\;\text{Zero}\end{array} $

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Considering an element of length dx at a distance $x$.Force on in it due to lower part is $s(l-x)ag$ .Let the elongation in this element be $\alpha$.

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

$Y=\large\frac{s(l-x)ag}{a}\frac{dx}{\alpha}$

$\alpha=\large\frac{Sg}{Y}$$(l-x)dx$

Total elongation is $\int \alpha$

$\int \alpha=\int\limits_0^l \large\frac{Sg}{Y}$$(l-x)dx$

$\Rightarrow \large\frac{Sg}{Y}$$[(lx-\large\frac{x^2}{2}]_0^l$

$\Rightarrow \large\frac{Sgl^2}{2Y}$

$\Rightarrow \large\frac{5\times 10^3\times 10\times (.8)^2}{2\times 1\times 10^{10}}$

$\Rightarrow \large\frac{5\times .64\times 10^{-6}}{2}$

$\Rightarrow 1.6\times 10^{-6}m$

Hence (A) is the correct answer.

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