In the below given problem, if 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms ? (Specific gravity of mercury = 13.6)
Problem:[A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other ]

Height of the water column, $h_1=10+15=25cm$
Height of the spirit column, $h_2=12.5+15=27.5cm$
Density of water, ${\rho }_1=1g/c{m}^3$
Density of spirit, ${\rho }_2=0.8g/c{m}^3$
Density of mercury $\rho =13.6g/c{m}^3$
Let h be the difference between the levels of mercury in the two arms.
Pressure exerted by height h, of the mercury column:
= $\rho hg$
= $h\times 13.6\times g$ (i)
Difference between the pressures exerted by water and spirit:
= ${\rho }_1{h}_{1}g-{\rho }_2{h}_{2}g$
= $g(25\times 1-27.5\times 0.8)$
= $3g$ (ii)
Equating equations (i) and (ii), we get:
13.6 hg = 3g
$h=0.221cm$
Hence, the difference between the levels of mercury in the two arms is 0.221 cm.