Question

The height of water level in a tank of uniform cross-section is 5 m. The volume of the water leaked in 5 s through a hole of area \(2.4~mm^{2}\) made at the bottom of the tank is (Assume the level of the water in the tank remains constant and acceleration due to gravity \(=10~ms^{-2}\))

• A. \(90\times10^{-6}m^{3}\)
• B. \(120\times10^{-6}m^{3}\)
• C. \(80\times10^{-6}m^{3}\)
• D. \(40\times10^{-6}m^{3}\)

Hint:

Torricelli's theorem is a special case of Bernoulli's principle. It states that the speed of efflux of a fluid through a sharp-edged orifice at the bottom of a tank filled to a depth \(h\) is the same as the speed that a body would acquire in falling freely from a height \(h\).

Answer 1

The Correct Option is B

The velocity of water coming out of the hole is given by Torricelli's theorem: \[ v = \sqrt{2gh} \] where \( g \) is the acceleration due to gravity and \( h \) is the height of the water level. Given: \begin{itemize} \item \( h = 5 \) m \item \( g = 10 \) ms\(^{-2}\) \end{itemize} Therefore, \[ v = \sqrt{2 \times 10 \times 5} = \sqrt{100} = 10 \text{ m/s} \] The area of the hole is given as \( A = 2.4 \text{ mm}^2 = 2.4 \times 10^{-6} \text{ m}^2 \). The volume flow rate \( Q \) is given by: \[ Q = Av \] \[ Q = 2.4 \times 10^{-6} \text{ m}^2 \times 10 \text{ m/s} = 2.4 \times 10^{-5} \text{ m}^3/\text{s} \] The volume of water leaked in 5 seconds is: \[ V = Q \times t \] \[ V = 2.4 \times 10^{-5} \text{ m}^3/\text{s} \times 5 \text{ s} = 12 \times 10^{-5} \text{ m}^3 = 120 \times 10^{-6} \text{ m}^3 \] Therefore, the volume of water leaked in 5 seconds is \( 120 \times 10^{-6} \text{ m}^3 \).