To solve the problem, we need to find the velocity of water in the narrower section of a horizontal pipe when the cross-sectional area reduces to half.
1. Principle of Continuity:
According to the continuity equation for incompressible fluids:
$ A_1 v_1 = A_2 v_2 $
where $A$ is the cross-sectional area and $v$ is the velocity of the fluid.
2. Given Data:
Initial velocity, $v_1 = 2\, m/s$
Cross-sectional area reduces to half, so $A_2 = \frac{1}{2} A_1$
3. Calculate Velocity in Narrower Section:
Using continuity:
$ A_1 v_1 = A_2 v_2 \Rightarrow A_1 \times 2 = \frac{1}{2} A_1 \times v_2 $
Simplify:
$ 2 = \frac{1}{2} v_2 \Rightarrow v_2 = 4\, m/s $
Final Answer:
The velocity of water in the narrower section is $ {4\, m/s} $.
Consider two identical tanks with a bottom hole of diameter \( d \). One tank is filled with water and the other tank is filled with engine oil. The height of the fluid column \( h \) is the same in both cases. The fluid exit velocity in the two tanks are \( V_1 \) and \( V_2 \). Neglecting all losses, which one of the following options is correct?