1. Tension at a point of the hanging string

Let the string have length $L=20\,\text{m}$ and linear mass density $\mu$.

Take a point at distance $x$ from the lower end of the string. Then the tension at that point is due to the weight of the part of string below it:

$T(x)=\mu x g$

2. Wave speed at that point

The speed of a transverse wave on a string is

$v=\sqrt{\frac{T}{\mu}}$

So here,

$v(x)=\sqrt{\frac{\mu x g}{\mu}}=\sqrt{xg}$

3. Time taken by the pulse to travel upward

For a small element $dx$, the time taken is

$dt=\frac{dx}{v(x)}=\frac{dx}{\sqrt{xg}}$

Hence total time from bottom $(x=0)$ to top $(x=L)$ is

$t=\int_0^L \frac{dx}{\sqrt{xg}}$

$t=\frac{1}{\sqrt g}\int_0^L x^{-1/2}\,dx$

$t=\frac{1}{\sqrt g}\left[2\sqrt x\right]_0^L$

$t=\frac{2\sqrt L}{\sqrt g}=2\sqrt{\frac{L}{g}}$

4. Substitute values

$t=2\sqrt{\frac{20}{10}}=2\sqrt2\,\text{s}$

5. Check options

• A: $2\sqrt2\,\text{s}$ ✅
• B: $2\pi\sqrt2\,\text{s}$ ❌
• C: $2\pi\sqrt2\,\text{s}$ ❌
• D: $2\,\text{s}$ ❌

Therefore, the correct answer is A.