Torricelli law describes the speed of efflux of a fluid through an orifice under the force of gravity.it states that the speed of fluid flowing out of an orifice under the influence of gravity is given by X, the height of the liquid inside a tank Mark is measuring the exit velocity of water from a tank at two different heights: x=9m 'and x=4m what is the ratio of exit velocities V9 and V4 at these heights

Torricelli's Law describes the speed of efflux \(v\) of a fluid under the force of gravity from an orifice at the bottom of a tank as:

\[
v = \sqrt{2gh}
\]

Where:
- \(v\) is the speed of the fluid exiting the orifice,
- \(g\) is the acceleration due to gravity (approximately \(9.8 \, \text{m/s}^2\)),
- \(h\) is the height of the fluid above the orifice.

Given the heights:
- \(h_1 = 9 \, \text{m}\)
- \(h_2 = 4 \, \text{m}\)

The exit velocities at these heights are:
- \(v_1 = \sqrt{2gh_1} = \sqrt{2g \times 9}\)
- \(v_2 = \sqrt{2gh_2} = \sqrt{2g \times 4}\)

To find the ratio of the exit velocities \( \frac{v_1}{v_2} \), we can divide the expressions for \(v_1\) and \(v_2\):

\[
\frac{v_1}{v_2} = \frac{\sqrt{2g \times 9}}{\sqrt{2g \times 4}} = \frac{\sqrt{9}}{\sqrt{4}} = \frac{3}{2} = 1.5
\]

So, the ratio of the exit velocities \( \frac{v_9}{v_4} \) is **1.5**.

Would you like further details or explanations?

Here are 5 related questions you could explore:

1. How does the shape and size of the orifice affect the exit velocity according to Torricelli's law?
2. What happens to the exit velocity if the height of the fluid is doubled?
3. How does Torricelli’s law apply to fluids with different densities?
4. Can Torricelli’s law be used to measure the flow rate from an orifice?
5. What are some practical applications of Torricelli's law in engineering and fluid mechanics?

**Tip:** When dealing with fluid mechanics problems, always check if energy conservation principles, such as Bernoulli's equation, can simplify your calculations.

Explore how Torricelli's Law defines the speed of fluid efflux from an orifice under gravity's influence. Calculate the exit velocities at heights of 9m and 4m using the formula v = sqrt(2gh). Discover the ratio of exit velocities V9 and V4 and its practical applications in fluid mechanics.

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