Question

The area of cross-section of a large tank is \(0.5\ m^2\). It has a narrow opening near the bottom having area of cross-section \(1\ cm^2\). A load of \(25\ kg \) is applied on the water at the top in the tank. Neglecting the speed of water in the tank, the velocity of the water, coming out of the opening at the time when the height of water level in the tank is \(40\ cm\) above the bottom, will be ______ \(cms^{–1}\). [Take \(g=10\ ms^{–2}\)]

Answer 1

Correct Answer: 300

To determine the velocity of water exiting the narrow opening, we apply Bernoulli's principle and consider the additional pressure applied by the load. The given data includes: cross-sectional area of the tank \(A_1 = 0.5\ m^2\), area of the opening \(A_2 = 1\ cm^2 = 1 \times 10^{-4}\ m^2\), load \(m = 25\ kg\), gravitational acceleration \(g = 10\ ms^{-2}\), and water height \(h = 40\ cm = 0.4\ m\). First, calculate the pressure exerted by the load using:

\[ P_{\text{load}} = \frac{F}{A_1} = \frac{mg}{A_1} = \frac{25 \times 10}{0.5} = 500\ N/m^2\]

Add the pressure due to the water column \(P_h = \rho gh\) where \(\rho = 1000\ kg/m^3\):

\[ P_h = 1000 \times 10 \times 0.4 = 4000\ N/m^2\]

Total pressure at the bottom is:

\[ P_{\text{total}} = P_{\text{load}} + P_h = 500 + 4000 = 4500\ N/m^2\]

Using Bernoulli's equation and assuming tank water velocity at rest, the velocity \(v\) of water from the small opening is given by:

\[ v = \sqrt{\frac{2P_{\text{total}}}{\rho}} = \sqrt{\frac{2 \times 4500}{1000}} = \sqrt{9} = 3\ m/s\]

Convert \(3\ m/s\) to \(cm/s\):

\[ v = 3 \times 100 = 300\ cm/s\]

Thus, the velocity of the water is 300 cm/s, fitting within the expected range of 300,300.