A uniform string oflength \(20\ m\) 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 supports is (take \(g= 10 \) \(ms^{-2}\) )

\(\mathrm{v}=\sqrt{\frac{\mathrm{T}}{\mu}}\)              \(...(I)\)

where \(\mu=\frac{\mathrm{m}}{1}=\frac{\text { mass of string }}{\text { length of string }}\)

The tension \(\mathrm{T}=\frac{\mathrm{m}}{\ell} \times \mathrm{x} \times \mathrm{g}\)                   \(...(II)\)

From \((a)\) and \((b)\)

\(\frac{d x}{d t}=\sqrt{g x}\)

\(\mathrm{x}^{-1 / 2} \mathrm{dx}=\sqrt{\mathrm{g}} \mathrm{dt} \quad \therefore \int_{0}^{\ell} \mathrm{x}^{-1 / 2} \mathrm{dx}-\sqrt{\mathrm{g}} \int_{0}^{\ell} \mathrm{dt}\)

\(2 \sqrt{l}=\sqrt{g} \times t \quad \therefore t=2 \sqrt{\frac{\ell}{g}}=2 \sqrt{\frac{20}{10}}=2 \sqrt{2}\)