Answer to Question #217616 in Classical Mechanics for Buba

Question

A U-tube manometer has Mercury of relative density off 13.68 as its manometric liquid. Kerosene of relative density of 0.6 was poured into one of the arms and so a drop of 2.0 cm results in that arm. Also, water with a relative density of 1g/cm3 was poured into the other arm and so the two arms became the same again. Calculate 1 the height of the kerosene to the height of the water column in details.

Answer 1

As per the given question,

Density of mercury $(\rho)=13.68$

Relative density of kerosene $(\rho_k)=0.6$

Drop in height $(\Delta H)= 2 cm$

Relative density of water $(\rho_w)=1g/cm^3$

Height of kerosene =?

Height of water = ?

$\rho_1 h_1=\rho_2h_2$

Now, substituting the values,

$h_1=\frac{\rho_2 h_2}{\rho_1}$

$=\frac{0.6\times 2}{13.68}cm$

$=0.0877cm$