A cylindrical vessel of cross-section A contains water to a height h. There is a hole in the bottom of radius 'a'. The time in which it will be emptied is:

We are given a cylindrical vessel with cross-sectional area $$A$$ filled with water to a height $$h$$. There is a hole at the bottom with radius $$a$$, so the area of the hole is $$\pi a^2$$. We need to find the time taken for the vessel to empty completely.

According to Torricelli's theorem, the speed of efflux $$v$$ of water through the hole is given by $$v = \sqrt{2gh}$$, where $$g$$ is the acceleration due to gravity and $$h$$ is the height of water above the hole at any instant.

The volume flow rate of water through the hole is the product of the area of the hole and the speed of efflux. Since the volume is decreasing, we write:

$$\frac{dV}{dt} = - (\pi a^2) \sqrt{2gh}$$

Here, $$V$$ is the volume of water in the vessel. For a cylinder, $$V = A h$$. Differentiating both sides with respect to time $$t$$:

$$\frac{dV}{dt} = A \frac{dh}{dt}$$

Setting the two expressions for $$\frac{dV}{dt}$$ equal:

$$A \frac{dh}{dt} = - \pi a^2 \sqrt{2gh}$$

We can rearrange this equation to separate the variables $$h$$ and $$t$$. First, express $$\sqrt{2gh}$$ as $$\sqrt{2g} \sqrt{h}$$:

$$A \frac{dh}{dt} = - \pi a^2 \sqrt{2g} \sqrt{h}$$

Bring terms involving $$h$$ to one side and $$t$$ to the other:

$$A dh = - \pi a^2 \sqrt{2g} \sqrt{h} dt$$

Divide both sides by $$\sqrt{h}$$ and rearrange:

$$dt = - \frac{A}{\pi a^2 \sqrt{2g}} \frac{dh}{\sqrt{h}}$$

Simplify the right side:

$$dt = - \frac{A}{\pi a^2 \sqrt{2g}} h^{-1/2} dh$$

To find the total time $$T$$ for the vessel to empty, integrate both sides. At $$t = 0$$, $$h = h$$ (initial height). At $$t = T$$, $$h = 0$$ (vessel empty). So:

$$\int_{0}^{T} dt = - \frac{A}{\pi a^2 \sqrt{2g}} \int_{h}^{0} h^{-1/2} dh$$

The left side integrates to $$T$$. For the right side, reversing the limits of integration removes the negative sign:

$$T = \frac{A}{\pi a^2 \sqrt{2g}} \int_{0}^{h} h^{-1/2} dh$$

Now, integrate $$h^{-1/2}$$:

$$\int h^{-1/2} dh = \frac{h^{1/2}}{1/2} = 2 h^{1/2}$$

Evaluate the definite integral:

$$\int_{0}^{h} h^{-1/2} dh = \left[ 2 \sqrt{h} \right]_{0}^{h} = 2\sqrt{h} - 2\sqrt{0} = 2\sqrt{h}$$

Substitute back:

$$T = \frac{A}{\pi a^2 \sqrt{2g}} \times 2 \sqrt{h}$$

Simplify:

$$T = \frac{2A \sqrt{h}}{\pi a^2 \sqrt{2g}}$$

Express $$\sqrt{h}$$ and $$\sqrt{2g}$$ in terms of $$\sqrt{\frac{h}{g}}$$:

$$T = \frac{2A}{\pi a^2} \cdot \frac{\sqrt{h}}{\sqrt{2g}} = \frac{2A}{\pi a^2} \sqrt{\frac{h}{2g}}$$

Now, $$\sqrt{\frac{h}{2g}} = \frac{1}{\sqrt{2}} \sqrt{\frac{h}{g}}$$, so:

$$T = \frac{2A}{\pi a^2} \cdot \frac{1}{\sqrt{2}} \sqrt{\frac{h}{g}} = \frac{2A}{\sqrt{2} \pi a^2} \sqrt{\frac{h}{g}}$$

Simplify $$\frac{2}{\sqrt{2}}$$:

$$\frac{2}{\sqrt{2}} = \frac{2}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{2\sqrt{2}}{2} = \sqrt{2}$$

Thus:

$$T = \frac{\sqrt{2} A}{\pi a^2} \sqrt{\frac{h}{g}}$$

Comparing with the options, this matches option B.

Hence, the correct answer is Option B.