Physics - Fluid Question with Solution | TestHub

Suppose the radius at A is $R$ and it decreases uniformly to $r$ at B where

$S=\pi R^2$ and $s=\pi r^2$. Assume also that

$$\frac{R}{L_2}=\frac{r}{L_1}=\frac{y}{R-r}$$

So $y=r+\frac{r}{L_2-L_1}(x-L_1)$

where $y$ is the radius at the point P distant $x$ from the vertex O. Suppose the velocity with which the liquid flows out is $V$ at A, $v$ at B and $u$ at P.

Then by the equation of continuity

$$\pi R^{2}V=\pi y^{2}u=\pi r^{2}v$$

The velocity of efflux is given by

$$v=\sqrt{2gh}$$

and Bernoulli's theorem gives

$$p_p+\frac{1}{2}\rho u^2=p_0+\frac{1}{2}\rho v^2$$

where $p_p$ is the pressure at P and $p_0$ is atmospheric pressure.

The force on the nozzle tending to pull it out is then

$$F=\int (p_p-p_0)\sin \theta \cdot 2\pi yds$$

We subtract $p_0$ since it acts externally. The factor $\sin \theta$ gives horizontal component.

$$\tan \theta =\frac{R-r}{L_2-L_1}$$

$$ds=dx\sec \theta$$

$$\Rightarrow F={\int }_{L_1}^{L_2}\frac{1}{2}(v^2-u^2)\rho \cdot 2\pi y\cdot \frac{R-r}{L_2-L_1}dx$$

$$=\pi \rho \int v^2\left(1-\frac{r^4}{y^4}\right)ydy$$

$$=\pi \rho \frac{v^2}{2}\left(R^2-r^2+\frac{r^4}{R^2}-r^2\right)$$

$$=\rho gh\left(\frac{\pi (R^2-r^2{)}^2}{R^2}\right)$$

$$=\rho gh\frac{(S-s{)}^2}{S}=6.02N$$