Explanation

(d) Let the thickness of each plank is $d.$
From equation of motion
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v^{2} = u^{2} + 2as\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
Ist case
$\ \ \ \ \ \ \ \ \ \ s = 2d,\ u = 100\ {ms}^{- 1},\ v = 0$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 = (100)^{2} + 2a \times 2d$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ 4ad = - 100 \times 100$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a = - \frac{100\ \times 100}{4d}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a = - \frac{2500}{d}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
IInd case Let the bullet with double the previous speed will penetrate $n$ planks of equal thickness $d.$
Now, $v = 0,\ u = 200\ {ms}^{- 1},\ a = - \frac{2500}{d},\ s = nd$
$\ \ \ \ \ \ \ \ \ \ 0 = (200)^{2} - 2 \times \frac{2500}{d} \times nd$
Or $\ \ n = \frac{200\ \times \ 200}{2\ \times 2500} = 8$