w of fluid, the product of cross-sectional area (A) and the velocity of flow (V) at any point is constant throughout the flow. We can pair this with the equation that determines the velocity of a freely falling object to address the change in velocity as the water falls.

First, let's denote:

Initial cross-sectional area as $$A_1 = 1 \mathrm{~cm}^2$$
Initial velocity as $$V_1 = 2 \mathrm{~m/s}$$
Final cross-sectional area as $$A_2$$, which we need to find.
Final velocity as $$V_2$$, which we can find using the formula for the velocity of a freely falling object.

The velocity of a freely falling object can be determined by the formula:

$$V_2 = \sqrt{V_1^2 + 2gh}$$

where

$$g = 10 \mathrm{~m/s^2}$$ is the acceleration due to gravity,
$$h = 20 \mathrm{~cm} = 0.2 \mathrm{~m}$$ is the height the water has fallen,

Let's calculate $$V_2$$:

$$V_2 = \sqrt{(2)^2 + 2 \cdot 10 \cdot 0.2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \mathrm{~m/s}$$

Now, using the equation of continuity $$A_1V_1 = A_2V_2$$, we can find $$A_2$$:

$$1 \cdot 2 = A_2 \cdot 2\sqrt{2}$$

$$A_2 = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \mathrm{~cm^2}$$

Since $$\sqrt{2} \approx 1.414$$, substituting this value gives us:

$$A_2 \approx \frac{1.414}{2} \mathrm{~cm^2} \approx 0.707 \mathrm{~cm^2}$$

Therefore, the cross-sectional area of the stream $$20 \mathrm{~cm}$$ below the tap is approximately $$0.707 \mathrm{~cm^2}$$, making the correct option C) 0.707 cm$$^2$$.

Fluid Mechanics

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