Question

The surface of water in a water tank of cross section area \(750 \, \text{cm}^2\) on the top of a house is \(h \, \text{m}\) above the tap level. The speed of water coming out through the tap of cross section area \(500 \, \text{mm}^2\) is \(30 \, \text{cm/s}\). At that instant, \(\frac{dh}{dt}\) is \(x \times 10^{-3} \, \text{m/s}\). The value of \(x\) will be _________.

Hint:

Remember the principle of continuity for fluid flow. Pay close attention to units and signs. The rate of decrease in height is equal in magnitude to the velocity at the top surface.

Answer 1

Correct Answer: 2

To tackle this problem, we will apply the principle of conservation of mass and Torricelli's law. With the given data and performing the required calculations:

Step 1: Apply Torricelli's Law for Fluid Flow Speed
According to Torricelli's law, the speed \(v\) of water flowing out of a hole at height \(h\) is given by:
\(v = \sqrt{2gh}\),
where \(g\) is the acceleration due to gravity (approx. \(9.81 \, \text{m/s}^2\)). However, we know the speed \(v\) is \(30 \, \text{cm/s}\) (or \(0.3 \, \text{m/s}\)).

Step 2: Use the Continuity Equation
The volume flow rate of water through the tap must equal the rate at which the water level decreases in the tank, hence:
\(A_{\text{tap}} v = A_{\text{tank}} \frac{dh}{dt}\),
where:
\(A_{\text{tap}} = 500 \, \text{mm}^2 = 500 \times 10^{-6} \, \text{m}^2\),
\(A_{\text{tank}} = 750 \, \text{cm}^2 = 750 \times 10^{-4} \, \text{m}^2\).

Step 3: Substitute and Solve for \(\frac{dh}{dt}\)
Solving the equation for \(\frac{dh}{dt}\):
\(\frac{dh}{dt} = \frac{A_{\text{tap}} \cdot v}{A_{\text{tank}}}\).
Substitute the known values:
\(\frac{dh}{dt} = \frac{(500 \times 10^{-6}) \times 0.3}{750 \times 10^{-4}}\).

Step 4: Calculate the Result
Solving the numerical expression:
\(\frac{dh}{dt} = \frac{150 \times 10^{-6}}{750 \times 10^{-4}} = 0.2 \times 10^{-3} \, \text{m/s}\).
Thus, \(x = 2\).

Step 5: Verify Within the Range
The computed value \(x = 2\) falls within the expected range of \(2, 2\). Therefore, the solution meets the specified requirement.