Question

The speed of water flowing out from the small opening at a depth of \(h\) from the surface of water in a large tank is

• A. \(\sqrt{gh}\)
• B. \(\sqrt{4gh}\)
• C. \(gh\)
• D. \(2gh\)
• E. \(\sqrt{2gh}\)

Hint:

Torricelli's law shows that the speed of efflux from a small hole is the same as the speed a freely falling body would acquire after falling from rest through a distance \(h\).

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
This phenomenon is described by Torricelli's Law, which is a specific application of Bernoulli's principle to fluid flowing out of an orifice (a small opening). The law relates the speed of the exiting fluid (efflux speed) to the height of the fluid above the opening.
Step 2: Key Formula or Approach:
We can apply Bernoulli's equation between two points: point 1 at the free surface of the water in the tank and point 2 at the opening. Bernoulli's equation is: \[ P_1 + \rho g h_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho g h_2 + \frac{1}{2}\rho v_2^2 \] where P is pressure, \(\rho\) is fluid density, g is acceleration due to gravity, h is height, and v is speed.
Step 3: Detailed Explanation:
Let's set up the variables for our two points.
Point 1 (at the surface of the water):
The pressure \(P_1\) is the atmospheric pressure, \(P_{atm}\).
Let the height of the opening be our reference level, so the height of the surface is \(h_1 = h\).
Since the tank is large, the speed at which the water level at the surface drops is negligible, so we can approximate \(v_1 \approx 0\).

Point 2 (at the small opening):
The water flows out into the atmosphere, so the pressure \(P_2\) is also atmospheric pressure, \(P_{atm}\).
The height is at our reference level, so \(h_2 = 0\).
The speed of the water flowing out is \(v_2 = v\), which is what we need to find.

Now, substitute these into Bernoulli's equation: \[ P_{atm} + \rho g h + \frac{1}{2}\rho (0)^2 = P_{atm} + \rho g (0) + \frac{1}{2}\rho v^2 \] Simplifying the equation: \[ P_{atm} + \rho g h = P_{atm} + \frac{1}{2}\rho v^2 \] The \(P_{atm}\) terms cancel out: \[ \rho g h = \frac{1}{2}\rho v^2 \] The density \(\rho\) also cancels out: \[ g h = \frac{1}{2}v^2 \] Now, solve for the speed v: \[ v^2 = 2gh \] \[ v = \sqrt{2gh} \] This result is known as Torricelli's Law. It shows that the speed of efflux is the same as the speed an object would acquire by free-falling from a height h.
Step 4: Final Answer:
The speed of the water flowing out is \(\sqrt{2gh}\). This corresponds to option (E).