Question

A pressure meter attached to a closed water tap reads $1.5\times {10}^5\text{Pa}$. When the tap is opened, the velocity of flow of water is $10{\text{ms}}^{-1}$ and the reading of the pressure meter is

• A. $1.5\times {10}^5\text{Pa}$
• B. $3\times {10}^5\text{Pa}$
• C. $0.5\times {10}^5\text{Pa}$
• D. ${10}^5\text{Pa}$

Answer 1

Answer: (D)

${10}^5\text{Pa}$

Applying Bernoulli's principle on a horizontal pipe, we have
$P_1+\frac{1}{2}\rho {v_1}^2=P_2+\frac{1}{2}\rho {v_2}^2$
here $P_1=1.5\times {10}^5\text{pascal},v_1=0,v_2=10\text{m/s}$, $\rho =1000{\text{kg/m}}^3$
$P_1+\frac{1}{2}\rho {v_1}^2=P_2+\frac{1}{2}\rho {v_2}^2\Rightarrow 1.5\times {10}^5=P_2+\frac{1}{2}\times 1000\times {\left(10\right)}^2$
solving the above equation we have $P_2={10}^5\text{pascal}$