#### A bullet looses $\dpi{100} \left ( \frac{1}{n} \right )^{th}$of its velocity passing through one plank. The number of such planks that are required to stop the bullet can be :Option: 1 $\frac{n^{2}}{2n-1}$Option: 2 $\frac{2n^{2}}{n-1}$Option: 3 InfiniteOption: 4 $n$

Let the bullet be traveling at a speed $v$ before entering the first plank.

If it loses $(\frac{1}{n})^{th}$ velocity its new velocity will become $v-\frac{v}{n}$

Kinetic energy lost

$\Delta \mathrm{E}=\frac{1}{2} \mathrm{m} v^2-\frac{1}{2} \mathrm{~m}\left(v-\frac{v}{\mathrm{n}}\right)^2=\frac{1}{2} m v^2\left(1-\left(\frac{\mathrm{n}-1}{\mathrm{n}}\right)^2\right)$

Now, let N be the number of planks after which the velocity of the bullet becomes 0.
Thus we get
$\mathrm{N} \times(\text{ Energy lost in one plank} )= \text{Total Energy}$

or

$\mathrm{N} \times \Delta \mathrm{E}=\frac{1}{2} \mathrm{mv}^2$

\begin{aligned} &\frac{\mathrm{mv}}{2 \Delta \mathrm{E}}=\mathrm{N}=\frac{1}{1-\left(\frac{\mathrm{n}-1}{\mathrm{n}}\right)^2}\\ &\Rightarrow \mathrm{N}=\frac{\mathrm{n}^2}{2 \mathrm{n}-1} \end{aligned}