A uniform string of length `20m` is suspended from a rigid support. A shirt wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is :
(take `g = 10ms^(-2)`)

Options

A. (a) `2sqrt(2s)`
B. (b) `sqrt(2s)`
C. ( c ) `2pisqrt(2s)`
D. (d) `2s`

Answer

A

Explanation

(a) We know that velocity in string is given by
`v = sqrt((T)/(mu))` ….(i)
where `mu = (m)/(1) = (mass of string)/(length of string)`
The tension `T = (m)/(l) xx x xx g` …(II)
From (a) and (b)
`(dx)/(dt) = sqrt(gx)`

`x^(-1//2)dx = sqrt(g)dt` :. `int_(0)^(l)x^(-1//2)dx - sqrt(g)int_(0)^(l)dt`
`2sqrt(l) = sqrt(g) xx t` :. `t = 2sqrt((l)/(g)) = 2sqrt((20)/(10)) = 2sqrt(2)`