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a
$\mathrm{L}\left(1+\frac{2}{9}\frac{\mathrm{Mg}}{{\mathrm{\pi \gamma R}}^{2}}\right)$
b
$\mathrm{L}\left(1+\frac{1}{9}\frac{\mathrm{Mg}}{{\mathrm{\pi \gamma R}}^{2}}\right)$
c
$\mathrm{L}\left(1+\frac{1}{3}\frac{\mathrm{Mg}}{{\mathrm{\pi \gamma R}}^{2}}\right)$
d
$\mathrm{L}\left(1+\frac{2}{3}\frac{\mathrm{Mg}}{{\mathrm{\pi \gamma R}}^{2}}\right)$

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detailed solution

Correct option is A

Take a cross-section taken at a point x from the bottom; hence, at this point,

$\mathrm{r}=3\mathrm{R}-\frac{\mathrm{x}}{\mathrm{L}}2\mathrm{R}$

$\frac{\mathrm{Mg}}{{\mathrm{\pi r}}^{2}}=\mathrm{Y}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{Mg}}{\mathrm{\pi }{\left(3\mathrm{R}-\frac{2\mathrm{Rx}}{\mathrm{L}}\right)}^{2}}$

where dy is the elongation of the portion dx.

As we solve for x in the range of 0 to L, we obtain,

$\mathrm{Y}=\mathrm{L}\left(1+\frac{1}{3}\frac{\mathrm{Mg}}{{\mathrm{\pi \gamma R}}^{2}}\right)$

Hence the correct answer is $\mathrm{L}\left(1+\frac{1}{3}\frac{\mathrm{Mg}}{{\mathrm{\pi \gamma R}}^{2}}\right).$

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