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. If 15.0 cm of water and spirit 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).

Solution

The mercury column in two arms are in level with 10cm of water in one arm and 12.5cm of spirit in the second arm.

Thus the density of spirit is,

ρ_{s} = \frac{10}{12.5} \times 1 = 0.8 gm / cc

Let the difference in the level of mercury in two arms be x, on adding 15.0cm of water and 15.0cm of spirit in respective limbs.

Now,

Pressure due to 25 cm of water = Pressure due to x cm of mercury + Pressure due to 27.5 cm of spirit

ρ_{w} h_{w} g = ρ_{hg} xg + ρ_{s} h_{s} g \Rightarrow ρ_{w} h_{w} = ρ_{hg} x + ρ_{s} h_{s} \Rightarrow 25 \times 1 = (13.6) x + (0.8) (27.5) \Rightarrow x = \frac{25 - (0.8) (27.5)}{13.6} = 0.221 c m

Some More Questions From Mechanical Properties of Fluids Chapter

For Daily Free Study Material Join wiredfaculty