JEE Main · 2016 · 03 Apr 2016 · Offline · Waves & Sound

Because the string hangs freely, the tension at height $x$ above the lowest point only supports the weight of string below it, so $T=\frac{Mgx}{L}$ and the wave speed there is $v=\sqrt{T/\mu }=\sqrt{gx}$, growing as the pulse climbs. Writing $v=\frac{dx}{dt}=\sqrt{gx}$ and separating variables gives $\frac{dx}{\sqrt{x}}=\sqrt{g}dt$. Integrating from the bottom $x=0$ to the top $x=20$ m, $2\sqrt{x}{|}_0^{20}=\sqrt{g}t$, so $2\sqrt{20}=\sqrt{10}t$, which gives $t=2\sqrt{2}$ s.