The Environmental Protection Agency is investigating an abandoned chemical plant. A large, closed cylindrical tank contains an unknown liquid. You must determine the liquid’s density and the height of the liquid in the tank (the vertical distance from the surface of the liquid to the bottom of the tank). To maintain various values of the gauge pressure in the air that is above the liquid in the tank, you can use compressed air. You make a small hole at the bottom of the side of the tank, which is on a concrete platform—so the hole is 50.0 cm above the ground. The table gives your measurements of the horizontal distance Rthat the initially horizontal stream of liquid pouringout of the tank travels before it strikes the ground and the gauge pressure pg of the air in the tank.(a) Graph R2 as a function of pg.[l1] Explain why the data points fall close to a straight line. Find the slope and intercept of that line. (b) Use the slope and intercept found in part (a) to calculate the height h(in meters) of the liquid in the tank and the density of the liquid (in kg/m³). Use g= 9.80 m/s². Assume that the liquid is no viscous and that the hole is small enough compared to the tank’s diameter so that the change in hduring the measurements

is very small.

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.

$p_1+\frac{1}{2}\rho v_1^2+\rho g{y}_1=p_2+\frac{1}{2}\rho v^2+\rho g{y}_2$

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

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

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,

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

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

Which gives,

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

y-intercept = 4yh

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

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,

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

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