Answer to Question #218412 in Classical Mechanics for Buba

Question

a U-Tube manometer has a mercury of relative density of 13.6 as it monumetric liquid. kerosene of relative density of 0.6 was poured into one of the arms and so a drop of 2.0 cm result in that arm. Also, water with a density of 1 g per cm cube was poured into another arm and so the two arms become the same again. calculate the height of the kerosene

Answer 1

$\text {Let A be the point of the upper level of mercury}$

$\text{shoulders in the pipe where kerosene is injected}$

$\text{Let B be the point of the upper level of mercury}\newline \text{ of the other arm of the tube}$

$P_A = P_B$

$P_A =\rho_kgh_k$

$\text{where }$

$\rho_k- \text{density of kerosene}$

$h_k-\text{kerosene column height}$

$P_B =\rho_mg\Delta h$

$\text{where}$

$\rho_m - \text {mercury density }$

$\Delta h- \text{difference in mercury levels at points A and B}$

$\Delta h = 2cm \text{ by the condition of the problem}$

$P_A = P_B$

$\rho_kgh_k = \rho_mg\Delta h$

$h_k =\frac{ \rho_m\Delta h}{\rho_k}=\frac{ 13.6*2}{0.6}=45.3cm$

$\text{Answer: }h_k =45.3cm$