Q61 / 90 · JEE Main 2014 (19 Apr Online) · Physics

A bullet loses ${(\frac{1}{n})}^{th}$ of its velocity passing through one plank. Considering uniform retardation, the number of such planks that are required to stop the bullet can be:

Solution

Let $t$ be the thickness of the plank

${\{υ_{0} (1 - \frac{1}{n})\}}^{2} = υ_{0}^{2} - 2 a t$

$∴ 2 a t = υ_{0}^{2} - υ_{0}^{2} {(1 - \frac{1}{n})}^{2}$

If $m$ number of planks are required to stop a bullet, then

$0 = υ_{0}^{2} - (2 a t) m$

$∴ m = \frac{υ_{0}^{2}}{2 a t}$

$= \frac{υ_{0}^{2}}{υ_{0}^{2} \{1 - {(1 - \frac{1}{n})}^{2}\}}$

$= \frac{n^{2}}{n^{2} - {(n - 1)}^{2}} = \frac{n^{2}}{(2 n - 1)}$

Hence, the correct answer is option: 'C'.