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 _________.

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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.

Updated On: Mar 19, 2025
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Correct Answer: 2

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Step 1: Apply the Principle of Continuity

According to the principle of continuity, the volume flow rate is constant:

\[ A_1V_1 = A_2V_2 \]

where \(A_1\) and \(V_1\) are the area and velocity at the top surface of the water, and \(A_2\) and \(V_2\) are the area and velocity at the tap.

Step 2: Convert Units and Substitute Values

Given \(A_1 = 750 \, \text{cm}^2 = 750 \times 10^{-4} \, \text{m}^2\), \(A_2 = 500 \, \text{mm}^2 = 500 \times 10^{-6} \, \text{m}^2\), and \(V_2 = 30 \, \text{cm/s} = 0.3 \, \text{m/s}\):

\[ 750 \times 10^{-4} V_1 = 500 \times 10^{-6} \times 0.3 \]

\[ V_1 = \frac{500 \times 0.3 \times 10^{-6}}{750 \times 10^{-4}} = \frac{150 \times 10^{-6}}{750 \times 10^{-4}} = 2 \times 10^{-4} \, \text{m/s}. \]

Step 3: Relate \(V_1\) to \(\frac{dh}{dt}\)

The rate of change of height (\(\frac{dh}{dt}\)) of the water in the tank is equal to the velocity of the water at the top surface but with a negative sign, because the height is decreasing:

\[ \frac{dh}{dt} = -V_1 = -2 \times 10^{-3} \, \text{m/s}. \]

Step 4: Find the Value of \(x\)

Given \(\frac{dh}{dt} = x \times 10^{-3} \, \text{m/s}\), and we have found \(\frac{dh}{dt} = -2 \times 10^{-3} \, \text{m/s}\). Therefore, \(x = -2\). Since the magnitude is asked, take the absolute value:

\[ |x| = 2. \]

Conclusion: The value of \(x\) is 2.

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Approach Solution -2

The correct answer is 2.




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