Question

In a U-tube, the diameter of the left-hand limb is half of that of the right one. First, mercury is poured in the U-tube so that the level of the mercury in the narrow limb is a distance l= 20 cm from the rim of the tube. Then, the left-hand limb is filled to the top with water. a) How much does the mercury level rise in the right-hand limb from the initial level, when there was just mercury? b) What is the distance between the levels of the mercury in the limbs of the tube? a) 0.3125 cm b) 1.5625 cm

Answer 1

Solution

The diamet ere off the lleft limb = d_l.l

Diameter of right limb = dr .r

d_l . = 0.5 d_r . ( given)

=> Area of right limb = 4 * area of left limb

Let rise of level = x

Level of mercury in the narrow limb = 20 cm + 4x cm

(20+4x)(1)(g) = (4x+x) (13.6)(g)

=> 20+4x = 5x * 13.6 => x = 20/(68-4) =0.3125

=> Rise in the Mercury level in the right hand limb = x= 0.3125 cm

b) The rise in the levels of mercuryin the limbs of the tube =4x + x =

= 5* x = 5 *0.3125

= 1.5625 cm