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v-velocity with which the body reaches the sand

$v^2=u^2+2as$

$\quad=2 \times 9.8 \times 5$

$\quad=98$

It penetrates send 0.05 m and comes to stop

$a=\large\frac{-98}{2 \times 0.05}$

$\quad= -980 m/s^2$

retarding force$= m \times a$

$\qquad=0.02 \times -980$

$\qquad=-19.6 N$

Hence b is the correct answer.

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