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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)`)A. `2pisqrt(2)s`B. `2 s`C. `2sqrt(2)s`D. `sqrt(2)s` |
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Answer» Correct Answer - C 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` |
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