Statement $I$ : When speed of liquid is zero everywhere, pressure difference at any two points depends on equation $\mathrm{P}_1-\mathrm{P}_2=\rho \mathrm{g}\left(\mathrm{h}_2-\mathrm{h}_1\right)$
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.
$\mathrm{P}_1+\rho \mathrm{gh}_1+\frac{1}{2} \rho v_1^2=\mathrm{P}_2+\rho \mathrm{gh}_2+\frac{1}{2} \rho v_2^2$
$\left[h_1 \& h_2\right.$ are height of point from any reference level]
$\text { Given } \mathrm{V}_1=\mathrm{V}_2=0 \text { (for statement-1) }$
$\therefore \mathrm{P}_1-\mathrm{P}_2=\rho \mathrm{g}\left(\mathrm{h}_2-\mathrm{h}_2\right)$
For statement-2
$\mathrm{P}_1+\frac{1}{2} \rho v_1^2=\mathrm{P}_2+\frac{1}{2} \rho v_2^2$
$\mathrm{P}_1-\mathrm{P}_2=\rho \mathrm{gh}$
$\mathrm{P}_1-\mathrm{P}_2=\frac{1}{2} \rho v_2^2-\frac{1}{2} \rho v_1^2$
$\rho \mathrm{gh}=\frac{1}{2} \rho v_2^2-\frac{1}{2} \rho v_1^2$
$2 \mathrm{gh}=\mathrm{v}_2^2-\mathrm{v}_1^2$
Hence answer $(4)$

