Question

A cylindrical tank having large diameter is filled with water to a height \(H\). A hole of cross-sectional area \(5\ \text{cm}^2\) in the tank allows water to drain out. If the water drains out at the rate of \[ 2\times10^{-3}\ \text{m}^3\text{s}^{-1}, \] then the value of \(H\) is
\[ (\text{acceleration due to gravity }=10\ \text{m s}^{-2}) \]

• A. \(80\ \text{cm}\)
• B. \(120\ \text{cm}\)
• C. \(60\ \text{cm}\)
• D. \(90\ \text{cm}\)

Hint:

For a large tank with a small outlet, \[ v=\sqrt{2gH} \] (Torricelli's theorem). Also, \[ Q=Av, \] where \(Q\) is the volume flow rate, \(A\) is the area of the hole, and \(v\) is the speed of efflux.

Answer 1

The Correct Option is A

Step 1: Use Torricelli's theorem for flow through hole.
Velocity of efflux: \( v = \sqrt{2gH} \). Flow rate \( Q = av \).

Step 2: Solve for H.
\( v = \frac{Q}{a} = \frac{2\times10^{-3}}{5\times10^{-4}} = 4\text{ m/s} \).
\( H = \frac{v^2}{2g} = \frac{16}{20} = 0.8\text{ m} = 80\text{ cm} \)

\[ \boxed{H = 80\text{ cm}} \]