One end of a string of length L is tied to the celling of lift accelerating upwards with an accelerating 2g The other end of the string is free. The linear mass density of the string varies linearly from 0 of `lamda` from bottom to top :-

Options

A. The velocity of the wave in the string will be 0
B. The acceleration of the wave in the string will be `3g//4` every where
C. The time taken by a pulse to reach from bottom to top will be `sqrt(8 L//3g)`
D. The time taken by a pulse to reach from bottom to top will be `sqrt(4 L//3g)`

Answer

B, C

Explanation

`mu = (lamda)/(L)x`
(at a distance 'x' from free end)
`:. T= underset(0)overset(x)(int)mudx(g+2g)=(3lamdagx^(2))/(2L)`
`:. v_("wave")=sqrt((T)/(mu))=sqrt((3 lamdagx^(2))/(2L((lamdax)/(L))))=sqrt((3xg)/(2))`
`rArrv^(2)=(3xg)/(2)rArr 2v(dv)/(dx)=(3g)/(2)rArr a=3g//4`
Now `S=ut+(1)/(2)at^(2)rArr L=0+(3g)/(8)t^(2)`
`rArr t=sqrt(8L//3g)`