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The rubber chord has a force constant k given by $k=\large\frac{yA}{L}$

$\qquad= \large\frac{5.0 \times 10^8 \times 1.0 \times 10^{-6}}{0.1}$

$\qquad= 5.0 \times 10^3 N/m$

Now from conservation of energy theorem.

The elastic potential energy of string is equal to the kinetic energy of the projectile

$\large\frac{1}{2} $$k ( \Delta l)^2=\large\frac{1}{2}$$mv^2$

Where $\Delta L$ is the extension in the string

$\therefore v=\sqrt {\large\frac{k}{m}}$$\Delta l$

$\qquad= \sqrt {\frac{5.0 \times 10^3}{5.0 \times 10^{-3}}}$$ 2 \times 10^{-2}$

$\qquad=20 m/s$

Hence b is the correct answer.

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