91Ó°ÊÓ

Diameter of hole above the ground = 50.00 cm

Acceleration due to gravity$=9.80m/s^2$

Liquid is non viscous.

Hole diameter is small compared to tank’s diameter.

When the speed of a liquid increases, pressure decreases, and when its speed decreases, pressure increases, according to Bernoulli's principle.

As a result of the Bernoulli equation, After the liquid has exited the tank, we get free-fall projectile motion. Atmospheric pressure p0 is the pressure at the opening through which the liquid departs. In a liquid, the top pressure is equal to the sum of the gauge pressure and the zero-point pressure.

$p_1+\frac{1}{2}\mathrm{ÒÏ}{v}_1^2+\mathrm{ÒÏ}g{y}_1=p_2+\frac{1}{2}\mathrm{ÒÏ}{v}^2+\mathrm{ÒÏ}g{y}_2$

Applying Bernoulli’s equation between top and bottom of liquid tank,

$\begin{array}{r}{p}_0+p_g+\mathrm{ÒÏ}gh=\frac{1}{2}\mathrm{ÒÏ}{v}^2+p_0\\ p_g+\mathrm{ÒÏ}gh=\frac{1}{2}\mathrm{ÒÏ}{v}^2\end{array}$

As it is free fall motion after it leaves tank, therefore,

$\begin{array}{l}vt=R\\ y=\frac{1}{2}g{t}^2\end{array}$

Here, y = 50.0 cm.

Simplifying and substitute in Bernoulli’s equation get,

$\begin{array}{r}{v}^2=\left(\frac{\mathrm{ÒÏ}g}{4y}\right)×R^2\\ R^2=\left(\frac{4y}{\mathrm{ÒÏ}g}\right)×p_g+4yh\end{array}$

Slope of straight line$=\frac{4y}{pg}$

Which gives,

$\mathrm{ÒÏ}=\frac{4y}{\left[g\left(slope\right)\right]}$…………………….. (i)

y-intercept = 4yh

$R^2=(25.679m^2/atm){\mathrm{ÒÏ}}_g+16.385m^2$

From y-intercept,

$h=\left(\frac{y-intercept}{4y}\right)=\left(\frac{16.385m^2}{\left[4\right(0.500\left)\right]}\right)=8.2m$

Here, h is height of liquid of liquid in tank.

Substitute value in equation (i) we get density,

$\mathrm{ÒÏ}=\frac{4y}{\left[g\right(slope\left)\right]}=\frac{4(0.500m)}{\left[\right(9.80m/s^2\left)\right(25.679m^2/atm\left)\right(1atm/1.01×{10}^{5}Pa\left)\right]}=\frac{803kg}{m^3}$

Thus, liquid is 80% dense as water. Height of fluid in tank is 8.2m and density of liquid is $803\mathrm{kg}I{m}^3$.