Thank you for reporting, we will resolve it shortly
Q.
A bullet looses $\left(\frac{1}{n}\right)^{th}$ of its velocity passing through one plank. The number of such planks that are required to stop the bullet can be :
Let the bullet be traveling at a speed v before entering the first plank.
If it loses $\left(\frac{1}{ n }\right)^{\text {th }}$ velocity its new velocity will become $v-\frac{v}{n}$
Kinetic energy lost $\Delta E =\frac{1}{2} m v^{2}-\frac{1}{2} m \left( v -\frac{ v }{ n }\right)^{2}=\frac{1}{2} m v^{2}\left(1-\left(\frac{ n -1}{ n }\right)^{2}\right)$
Now, let $N$ be the number of planks after which the velocity of the bullet becomes 0 .
Thus we get
$N \times($ Energy lost in one plank $)=$ Total Energy or
$ N \times \Delta E =\frac{1}{2} mv ^{2} $
$ \frac{m v^{2}}{2 \Delta E }= N =\frac{1}{1-\left(\frac{ n -1}{ n }\right)^{2}} $
$ N=\frac{n^{2}}{2 n-1} $
The velocity will become $o$ after passing through $\frac{n^{2}}{2 n-1}$ planks