A uniform string of length `20m` is suspended from a rigid support. A short 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. `2pisqrt(2)s`
B. `2 s`
C. `2sqrt(2)s`
D. `sqrt(2)s`

Answer

C

Explanation

As `mu` is mass per unit length of the rop, then `mu=m//L`
As `upsilon=sqrt((T)/(mu))`
`:. (dx)/(dt)=sqrt((mgx//L)/(m//L))=sqrt(g)sqrt(x)` or `(dx)/(sqrt(x))=sqrt(g)dt`
Integrating both sides, we get
`int_(0)^(L)x^(-1//2)dx=sqrt(g)int_(0)^(L)dt`
`[(x^(1//2))/(1//2)]_(0)^(L)=sqrt(g)t` or` (sqrt(L)-0)/(1//2)=sqrt(g)t`
`t=(2sqrt(L))/(sqrt(g))=2xxsqrt((20)/(10))=2sqrt(2)s`