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$P_1,P_2$ momentum before and after let orginal K.E be 100 J

K.E increased by $300 \%$

new $KE=400 J$

$\large\frac{{P_2}^2}{{P_1}^2}=\frac{E_2}{E_1}=\frac{400}{100}$$=4$

$\large\frac{P_2}{P_1}$$=2=>P_2=2P_1$

Increase % $=\large\frac{P_2-P_1}{P_1} $$\times 100$

$\qquad=\large\frac{2P_1 -P_1}{P_1} $$ \times 100$

$\qquad=100 \;\%$

Hence b is the correct answer.

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