
$p_{1}+\frac{1}{2} \rho v_{1}^{2}=p_{2}+\frac{1}{2} \rho v_{2}^{2}$
Here, $\quad p_{1}=p_{ atm }+\frac{F}{A}$
and $\quad p_{2}=p_{\text {atm }}$ (as area is too small)
$\therefore \quad p_{ atm }+\frac{F}{A}=p_{ atm }+\frac{1}{2} \rho v_{2}^{2} \quad\left(\because v_{1}=0\right)$
$\Rightarrow \quad v_{2}^{2}=\frac{2 F}{\rho A}$ ...........$(i)$
Range of efflux, $R=v \sqrt{\frac{2 h}{g}}$
$\Rightarrow \quad v^{2}=\frac{R^{2} g}{2 h}$ ...........$(ii)$
From Eqs. $(i)$ and $(ii)$, we get
$\frac{2 F}{\rho A}=\frac{R^{2} g}{2 h}$
$\therefore \quad F =\frac{R^{2} g \times \rho A}{4 h}$
$=\frac{(2)^{2} \times 10 \times 1000 \times 10 \times 10^{4}}{4 \times 1} =10 \,N$
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.


