A horizontal pipe of non-uniform cross-section allows water to flow through it with a velocity $1…
- $50 \mathrm{kPa}$
- $100 \mathrm{kPa}$
- $48.5 \mathrm{kPa}$
- $24.25 \mathrm{kPa}$
Solution

$\begin{aligned} & P_1+\frac{1}{2} \rho v_1^2=P_2+\frac{1}{2} \rho v_2^2 \\ & \Rightarrow \quad P_1+\frac{1}{2} \rho\left(v_1^2-v_2^2\right)=P_2 \\ & \Rightarrow \quad P_2=50 \times 10^3+\frac{1}{2} \times 10^3 \times\left(1^2-2^2\right) \\ & \Rightarrow \quad P_2=50 \times 10^3-1.5 \times 10^3 \\ & =48.5 \mathrm{kPa} \\ & \end{aligned}$
Asked in: AP EAMCET 2007
Practice more Mechanical Properties of Fluids questions on Aicharya