A uniform string of length 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 support is (Take, $$g = 10$$ m s$$^{-2}$$)

We start with the standard formula for the speed of a transverse wave on a stretched string:

$$v \;=\; \sqrt{\dfrac{T}{\mu}}$$

where $$T$$ is the tension at the point under consideration and $$\mu$$ is the uniform linear mass density of the string. For a string that hangs vertically, the tension is created solely by the weight of the portion of string lying below the point.

Let us choose the origin $$x = 0$$ at the lowest end of the string and measure the coordinate $$x$$ upward. At a distance $$x$$ from the lower end, the length of string below is exactly $$x$$, so its mass equals $$\mu x$$. Therefore the tension at that point is simply its weight:

$$T(x) \;=\; (\mu x)\,g \;=\; \mu g x.$$

Substituting this expression for $$T(x)$$ into the velocity formula, we obtain the position-dependent wave speed:

$$v(x) \;=\; \sqrt{\dfrac{T(x)}{\mu}} \;=\; \sqrt{\dfrac{\mu g x}{\mu}} \;=\; \sqrt{g\,x}.$$

Because the speed changes continuously with height, the pulse takes different times to traverse different small segments. Consider a differential element of length $$dx$$ located at coordinate $$x$$. The small time $$dt$$ required to pass through this element is

$$dt \;=\; \dfrac{dx}{v(x)} \;=\; \dfrac{dx}{\sqrt{g\,x}} \;=\; \dfrac{dx}{\sqrt{g}\,\sqrt{x}}.$$

The total time $$t$$ for the pulse to travel from the lower end ($$x = 0$$) to the rigid support at the upper end ($$x = L$$) is obtained by integrating $$dt$$ over the entire length:

$$t \;=\; \int_{0}^{L} dt \;=\; \int_{0}^{L} \dfrac{dx}{\sqrt{g}\,\sqrt{x}} \;=\; \dfrac{1}{\sqrt{g}} \int_{0}^{L} x^{-\frac{1}{2}}\,dx.$$

We now perform the integral. The integral of $$x^{-\frac{1}{2}}$$ is $$2\,x^{\frac{1}{2}}$$, so

$$\int_{0}^{L} x^{-\frac{1}{2}}\,dx \;=\; 2\,x^{\frac{1}{2}}\Big|_{0}^{L} \;=\; 2\left(\sqrt{L} - \sqrt{0}\right) \;=\; 2\sqrt{L}.$$

Substituting this result back into the expression for $$t$$ gives

$$t \;=\; \dfrac{1}{\sqrt{g}} \,\big(2\sqrt{L}\big) \;=\; \dfrac{2\sqrt{L}}{\sqrt{g}}.$$

For the present problem the string length is $$L = 20 \,\text{m}$$ and the acceleration due to gravity is $$g = 10 \,\text{m s}^{-2}$$. Substituting these values, we find

$$t \;=\; \dfrac{2\sqrt{20}}{\sqrt{10}}.$$

To simplify, first evaluate the square roots:

$$\sqrt{20} \;=\; \sqrt{4\times5} \;=\; 2\sqrt{5},$$

and

$$\sqrt{10} \;=\; \sqrt{2\times5} \;=\; \sqrt{2}\,\sqrt{5}.$$

Substituting these into the time expression:

$$t \;=\; \dfrac{2 \times 2\sqrt{5}}{\sqrt{2}\,\sqrt{5}} \;=\; \dfrac{4\sqrt{5}}{\sqrt{2}\,\sqrt{5}}.$$

The factor $$\sqrt{5}$$ cancels out, leaving

$$t \;=\; \dfrac{4}{\sqrt{2}} \;=\; 4 \times \dfrac{1}{\sqrt{2}} \;=\; 4 \times \dfrac{\sqrt{2}}{2} \;=\; 2\sqrt{2}\; \text{s}.$$

Hence, the correct answer is Option A.