Question

The top of a water tank is open to air and its water level is maintained. It is giving out 0.74 m³ water per minute through a circular opening of 2 cm radius in its wall. The depth of the centre of the opening from the level of water in the tank is close to:
1. 6.0 m
2. 4.8 m
3. 9.6 m
4. 2.9 m

Answer 1

Correct Answer - Option 2 : 4.8 m

Concept:

The law states that the speed v of efflux of a fluid through a sharp-edged hole at the bottom of a tank filled to a depth h is the same as the speed that a body (in this case a drop of water) would acquire in falling freely from a height h.

\({\rm{v}} = \sqrt {2{\rm{gh}}} \) 

Calculation:

From question, the flow rate (Q) is:

Q = 0.74 m³/min

\({\rm{Q}} = \frac{{0.74}}{{60}}\;{{\rm{m}}^3}/{\rm{s}}\) 

Now, the flow rate and the radius are related by the formula:

Q = (Area of opening) × (Velocity of water)

Q = (πr²) × v

Where,

r = 2 × 10^-2 m

v = Velocity of efflux

\(\Rightarrow \frac{{0.74}}{{60}} = {\rm{\pi }} \times {\left( {2 \times {{10}^{ - 2}}} \right)^2} \times {\rm{v}}\) 

Here, Torricelli's law is taking place.

On substituting velocity of efflux in flow rate relation,

\(\Rightarrow \frac{{0.74}}{{60}} = {\rm{\pi }} \times {\left( {2 \times {{10}^{ - 2}}} \right)^2} \times \sqrt {2{\rm{gh}}} \) 

\(\Rightarrow 9.81 = \sqrt {2{\rm{gh}}} \) 

Squaring on both sides,

⇒ 96.3 = 2gh

We know that, g = 9.8 m/s²

⇒ 96.3 = 2(9.8)h

\(\Rightarrow {\rm{h}} = \frac{{96.3}}{{2\left( {9.8} \right)}}\) 

∴ h = 4.91 m ≈ 4.8 m