Initial speed of an `alpha` particle inside a tube of length 4m is `1 kms^(-1)`, if it is accelerated in the tube and comes out with a speed of `9 kms^(-1)` , then the time for which the particle remains inside the tube is

Options

A. `8xx10^(-3)s`
B. `8xx10(-4)s`
C. `80xx10^(-3)s`
D. `800xx10^(-3)s`

Answer

B

Explanation

Initial speed, `u=1 kms^(-1)=1000 ms^(-1)`
Final speed, `upsilon=9 kms^(-1)=9000 ms^(-1)`
By using the relation,
`upsilon^(2)=u^(2)+2as`
`(9000)^(2)=(1000)^(2)+2xx a xx4 rArr a=10^(7)ms^(-2)`
`therefore` The time for which the particle remains in the tube
`upsilon=u+at`
`rArr t=(upsilon-u)/(a)=(9000-1000)/(10^(7))=8xx10^(-4)s`