Consider a completely full cylindrical water tank of height 1.6 m and cross-sectional area 0.5 $ m^2 $. It has a small hole in its side at a height 90 cm from the bottom. Assume, the cross-sectional area of the hole to be negligibly small as compared to that of the water tank. If a load 50 kg is applied at the top surface of the water in the tank then the velocity of the water coming out at the instant when the hole is opened is : (g = 10 $ m/s^2 $)
Show Hint
- 3 m/s
- 5 m/s
- 2 m/s
- 4 m/s
The Correct Option is D
Approach Solution - 1
To find the velocity of the water coming out of the hole, we can use Torricelli's theorem, which is an application of Bernoulli's principle for fluid dynamics. The velocity of efflux, \(v\), from a hole at the side of a tank can be given by:
\(v = \sqrt{2gh}\)
where:
- \(g\) is the acceleration due to gravity (10 \(m/s^2\)).
- \(h\) is the effective height of the water column above the hole.
In addition to the gravitational force of the water column's height, we have an external load of 50 kg exerting additional pressure. The pressure due to this load can be calculated and converted to an equivalent height of water using the formula:
\(P = \rho gh + \frac{F_{\text{load}}}{A_{\text{tank}}}\)
Here, the additional effective pressure head can be given by:
\(h_{\text{extra}} = \frac{F_{\text{load}}}{\rho g A_{\text{tank}}}\)
- \( F_{\text{load}} = m \cdot g = 50 \cdot 10 = 500 \, \text{N} \)
- \( \rho = 1000 \, \text{kg/m}^3 \) (density of water)
- \( A_{\text{tank}} = 0.5 \, \text{m}^2 \)
Now calculate the height:
\(h_{\text{extra}} = \frac{500}{1000 \times 10 \times 0.5} = 0.1 \, \text{m}\)
The initial height of the water column above the hole is \(1.6 \, \text{m} - 0.9 \, \text{m} = 0.7 \, \text{m}\).
So the total effective height above the hole \( h = 0.7 \, \text{m} + 0.1 \, \text{m} = 0.8 \, \text{m} \).
Using Torricelli's theorem with the total height,
\(v = \sqrt{2 \times 10 \times 0.8} = \sqrt{16} = 4 \, \text{m/s}\)
Thus, the velocity of the water coming out of the hole is \(4 \, \text{m/s}\).
Approach Solution -2
Apply Bernoulli equation between points 1 and 2.
\( P_1 + \frac{1}{2} \rho v_1^2 + \rho g h = P_2 + \frac{1}{2} \rho v_2^2 + 0 \)
\( P_0 + \frac{mg}{A} + \rho g \frac{70}{100} = P_0 + \frac{1}{2} \rho v_2^2 \) \( \frac{5000}{0.5} + 10^3 \times 10 \times \frac{70}{100} = \frac{1}{2} \times 10^3 v_2^2 \) \( 10^4 + 10^3 \times 7 = \frac{10^3}{2} v_2^2 \) \( v_2^2 = 16 \) \( v_2 = 4 m/s \)
As the tank area is large \( v_1 \) is negligible compared to \( v_2 \).
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