A bullet loses ` 1//20th` of its velocity is passing through a plank. What is the least number of planks required to stop the bullet ?

Options

A. 5
B. 10
C. 11
D. 20

Answer

C

Explanation

Let the thickness of one plank is s
if bullet enters with velocity u then it leaves with velocity
`v=(u-(u)/(20))=(19)/(20)u`
from `v^(2)=u^(2)-2as`

`rArr ((19)/(20)u)^(2)=u^(2)-2as rArr (400)/(39)=(u^(2))/(2as)`
Now if the n planks are arranged just to stop the bullet then again from `v^(2)=u^(2)-2as`
`0=u^(2)=2` ans
`rArr n = (u^(2))/(2as)=(500)/(39)`
`rArr n = 10.25`
As the planks are more than 10 so we can consider n = II