Question 53 of 192

Given below are two statements:
Statement I : When speed of liquid is zero everywhere, pressure difference at any two points depends on equation $P_1-P_2=\rho g\left(h_2-h_1\right)$.
Statement II : In ventury tube shown $2 \mathrm{gh}=v_1^2-v_2^2$

In the light of the above statements, choose the most appropriate answer from the options given below.

ABoth Statement I and Statement II are correct.BStatement I is correct but Statement II is incorrect.CStatement I is incorrect but Statement II is correct.DBoth Statement I and Statement II are incorrect.

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Answer: (B) Statement I is correct but Statement II is incorrect.

Applying Bernoulli's equation
$\mathrm{P}_1+\rho \mathrm{gh}_1+\frac{1}{2} \rho \mathrm{v}_1^2=\mathrm{P}_2+\rho \mathrm{gh}_2+\frac{1}{2} \rho \mathrm{v}_2^2$
$\left[h_1 \& h_2\right.$ are height of point from any reference level]
Given $\mathrm{V}_1=\mathrm{V}_2=0$ (for statement-1)
$\therefore \mathrm{P}_1-\mathrm{P}_2=\mathrm{g}\left(\mathrm{h}_2-\mathrm{h}_2\right)$
For statement-2
$\mathrm{P}_1+\frac{1}{2} \rho \mathrm{v}_1^2=\mathrm{P}_2+\frac{1}{2} \rho \mathrm{v}_2^2$
$\begin{aligned}
& P_1-P_2=\rho g h \\
& P_1-P_2=\frac{1}{2} \rho v_2^2-\frac{1}{2} \rho v_1^2 \\
& \rho g h=\frac{1}{2} \rho v_2^2-\frac{1}{2} \rho v_1^2 \\
& 2 g h=v_2^2-v_1^2
\end{aligned}$