b
(b)Let the thickness of each plank is \(s\). If the initial speed of bullet is \(100 m/s\) then it stops by covering a distance \(2s\) By applying \({v^2} = {u^2} - 2as\)

\(⇒\) \(0 = {u^2} - 2as\) \(s = \frac{{{u^2}}}{{2a}}\)

\(s \propto {u^2}\) [If retardation is constant]

If the speed of the bullet is double then bullet will cover four times distance before coming to rest

i.e. \({s_2} = 4({s_1}) = 4(2s)\)

\(⇒\) \({s_2} = 8s\)

So number of planks required \(= 8\)