A thin rod of length `1 m` is fixed in a vertical position inside a train, which is moving horizentally with constant acceleration `4 ms^(-2)`.A bead can slide on the rod and friction coefficient between them is `0.5`. If the head is released from rest at the top of the rod , it will reach the bottom the in

Options

A. `sqrt2 s`
B. `1 s`
C. `2 s`
D. `0.5 s`

Answer

D

Explanation

`N = ma`
`f_(L) = mu N = 0.5ma`
Vertical acceleration,
`a_nu = (mg - f_(L))/(m) = g - 0.5 a`
`= 10 - 0.5 xx 4`
`= 8 m//s^(2)`
Applying `s = (1)/(2) at^(2)`
or `t = sqrt((2 s)/(a)` in vertical direction ,we have
`t = sqrt((2 xx 1)/(8)) = 0.5 s`