Question

three hundred milliliters of water is poured into a u-tube that has a cross-sectional area of 1 cm 2. then 120 milliliters of oil, with a density 80% that of water, is poured down one side of the u-tube so that the oil floats on the water. find the difference in height of the liquid surfaces on the two sides of the u-tube.

Answer 1

The difference in height of the liquid surface on two sides of the U-tube is 24 cm.

Oscillating U-tube is a method for calculating fluid densities. It has a measuring scale and a U-shaped glass tube. The principle behind this method is that when pressure is applied to one side of the tube, it causes the liquid to move.

Let h be the difference in the height of the liquid surface and H is the height of the oil. We know that the pressure at all horizontal points in a line is the same.

Therefore, pressure at point A = pressure at point B. Then, equation 1 is $(H+h)\rho_wg=H\rho_{\text{oil}}g$

Here, ρ(w) is the density of water, and ρ(oil) is the density of oil. Then, the density of oil is,

$\begin{aligned}\rho_{\text{oil}}&=80\%\times \rho_w\\&=0.8\rho_w\end{aligned}$

Substitute the above equation in equation 1.

Then,

$\begin{aligned}(H+h)\rho_{w}g&=H\times0.8\rho_{w}g\\H+h&=0.8\times H\\h&=0.8H-H\\&=-0.2H\end{aligned}$

The negative sign in the above equation indicates that the oil surface is at a higher height.

Given, the volume of oil is V(oil)=120mL=120×10⁻⁶ m³. Area is A=1 cm² = 10⁻⁴ m². Also, the volume of oil = volume of cylinder = AH

Then,

$\begin{aligned}120\times10^{-6} &=AH\\H&=\mathrm{120\times10^{-2}\;m}\\&=\mathrm{1.2\;m}\\|h|&=0.2\times\mathrm{1.2\;m}\\&=\mathrm{0.24\;m}\\&=\mathrm{24\;cm}\end{aligned}$

The answer is 24 cm.

To know more about the U-tube technique:

https://brainly.com/question/14244484

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