$\begin{array}{1 1}(A)\;2\times 10^6\\(B)\;3\times 10^6\\(C)\;4\times 10^6\\(D)\;5\times 10^6\end{array} $

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Force applied by mass =$m\omega^2R$

$\Rightarrow 1\times (\large\frac{50\times 2\pi}{360})^2$$\times .28$

$\Rightarrow \large\frac{25\times \pi^2\times .28}{9}$

Extension $=\Delta l=R-l$

$\Rightarrow (28-25)$cm

$\Rightarrow 3$cm

Area $=\pi r^2$

$\Rightarrow \pi(2\times 10^{-3})^2$

$\Rightarrow 4\pi \times 10^{-6}m^2$

As we know

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

$\;\;\;=\large\frac{25\pi^2\times .28}{9\times 4\pi \times 10^{-6}}\times \frac{25}{3}$

$\;\;\;=\large\frac{625\pi\times .07\times 10^6}{27}$

$\;\;\;=5.09\times 10^6N/m^2\approx 5\times 10^6 N/m^2$

Hence (D) is the correct answer.

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