Question

You
An inverted u-tube has one of its arms dipped into a pool of mercury and the other into water. Air is gradually pumped out of the tube through valve v. Calculate the heights to which the liquid will rise in the arms when the pressure drops to 95% of atmospheric.

Answer 1

1)

$$\rho_hgh_h=0.05p_a$$

$$(13600)(9.8)h_h=0.05(101325)$$

$$h_h=0.038\ m=38\ mm$$

2)

$$\rho_wgh_w=0.05p_a$$

$$(1000)(9.8)h_w=0.05(101325)$$

$$h_w=0.517\ m=517\ mm$$