Correct Answer - Option 2 : 5 × 10
-5 m
2
Concept:
Bernoulli's Equation
The Bernoulli equation states that, \(p + \frac{1}{2}\rho {V^2} + \rho gh = constant\)
Where p is the pressure,
ρ is the density,
V is the velocity,
h is elevation
and g is the gravitational acceleration.
Where the points 1 and 2 lie on a streamline, the fluid has constant density, the flow is steady, and there is no friction.
Calculation:
To find the velocity of the water at 0.15 m from tap, we need to use the Bernoulli’s equation.
To find the area of the streamline at 0.15 m from tap, we need to use the continuity equation:
The continuity theorem states that the rate at which mass enters a system is equal to the rate at which mass leaves the system plus the accumulation of mass within the system.
By continuity equation,
A1 V2 = A2 V2
Where, A1 (10-4 m2) and V1 (1.0 m/s) are area and velocity of the water from tap.
A2 and V2 are area and velocity of the water at 0.15 m from tap
10-4(1) = A2 V2
⇒ A2 V2 = 10-4
The Bernoulli’s equation for this problem is:
\(\frac{1}{2}{\rm{\rho }}\left( {{\rm{V}}_2^2 - {\rm{V}}_1^2} \right) + {\rm{\rho gh}} = 0\)
\(\Rightarrow \frac{1}{2}{\rm{\rho }}\left( {{\rm{V}}_2^2 - {\rm{V}}_1^2} \right) = {\rm{\rho gh}}\)
\(\Rightarrow {\rm{\rho }}\left( {{\rm{V}}_2^2 - {\rm{V}}_1^2} \right) = 2{\rm{\rho gh}}\)
\(\Rightarrow {\rm{V}}_2^2 - {\rm{V}}_1^2 = 2{\rm{gh}}\)
\(\Rightarrow {\rm{V}}_2^2 = 2{\rm{gh}} + {\rm{V}}_1^2\)
\(\Rightarrow {\rm{V}}_2^2 = \left( {2 \times 10 \times 0.15} \right) + {\left( 1 \right)^2}\)
\(\Rightarrow {\rm{V}}_2^2 = 4\)
∴ V2 = 2 m/s
Now, substituting velocity in equation (1),
⇒ A2 (2) = 10-4
\(\Rightarrow {{\rm{A}}_2} = \frac{{{{10}^{ - 4}}}}{2}\)
∴ A
2 = 5 × 10
-5 m
2
