Answer to Question #244994 in Physics for Diane

Question

Mercury is poured into a u-tube with two cylinders arms of different radii. Both arms are cylindrical tubes. The left arm of the tube has a radius of 8.4cm, and the right arm has a radius of 4.0cm. 160g of water are then poured into the right arm. Given that the density of mercury is 13.6 g/cm3, and density of water is 1g/cm3 to what height h above the mercury level in the right arm, does the mercury rise in the left arm? State your answer in cm to the nearest hundredth cm. Use g=9.8 m/s2

Answer 1

The pressure of the displaced mercury must be equal to the pressure of the water added in the right arm:

$$\frac{m_mg}{A_l}=\frac{m_wg}{A_r},\\\space\\ \frac{m_m}{A_l}=\frac{m_w}{A_r},\\\space\\ \frac{\rho_mA_lh_{m}}{A_l}=\frac{m_w}{A_r},\\\space\\ h_m=\frac{m_w}{A_r\rho_m}=\frac{m_w}{\pi r_r^2\rho_m}=0.23\text{ cm}.$$

This is how high the mercury in the left arm rise compared to its initial level. Find the final level of mercury in the right arm (let us denote the decrease as x). Treating mercury as an incompressible fluid, the volume that corresponds to 0.23 cm is

$$V_l=\pi r_l^2h_m,\\ V_r=\pi r_r^2x.\\ V_l=V_r,\\ \pi r_l^2h_m=\pi r_r^2x,\\\space\\ x=h_m\frac{r_l^2}{r_r^2}=1.01\text{ cm},\\\space\\ \Delta h=h_m+x=1.24\text{ cm}.$$