$\begin{array}{1 1}(A)\;7.48\times 10^5N\\(B)\;8.53\times 10^5N\\(C)\;7.53\times 10^5N\\(D)\;7.63\times 10^5N\end{array} $

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$l$=15m

$r=$1mm

Area(a)=$\pi(1\times 10^{-3})^2m^2$

$\qquad\;\;=10^{-6}\pi m^2$

$Y=2.4\times 10^{11}N/m^2$

$\Delta l=5\times 10^{-2}$mm

$\quad\;=5\times 10^{-5}$m

As we know

Y=$\large\frac{\text{Stress}}{\text{Strain}}$

$\Rightarrow $Stress=Y.Strain

$\Rightarrow F=\large\frac{Y\Delta lA}{l}$

$\Rightarrow \large\frac{2.4\times 10^{11}\times 5\times 10^{-5}\times 10^{-6}\pi}{5\times 10^{-5}}$

$\Rightarrow 7.53\times 10^5N$

Hence (C) is the correct answer.

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