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

Options

A. 6
B. 9
C. 11
D. 13

Answer

b

Explanation

Let initial velocity of bullet be u.
After penetrating plank of thickness x
Final velocity of bullet `v=u-(u)/(20)=(19)/(20)u`
Let us assume the bullet crossed n number of planks `v^(2)=u^(2)+2ax`
`((19)/(20)u)^(2)=u^(2)+2axrArr2ax=(-39)/(400)u^(2)`
After penetrating n plank, bullet is stopped
`therefore v^(2)=u^(2)+2a(nx)`
`0=u^(2)+(2ax)n`
`0=u^(2)+((-39u^(2))/(400))n`
`0=u^(2)-(39u^(2))/(400)n`
`ucancel=0, n=(400)/(39)=10.26`
So, the bullet will stop at `11^(th)` plank.