Q. The area of cross-section of one limb of an $ U-tube $ is twice that of other. Both the limbs contain mercury at the same level. Water is poured in the wider tube so that mercury level in it goes down by 1 cm. The height of water column is (density of water $ ={{10}^{3}}\,kg\,{{m}^{-3}}, $ density of mercury $ =13.6\times {{10}^{3}} $ $ kg{{m}^{-3}} $ )
KEAMKEAM 2008
Solution:
The depression of level in the left limb is 1 cm, so the rise of level of mercury in right limb be 2 cm, because the area of cross-section of left limb is twice as that of right limb.
Now, equating pressure at interface of mercury and water (at AB)
$ (x+1)\times {{10}^{3}}\times g=(1+2)\times 13.6\times {{10}^{3}}g $
or height of water column $ x+1=40.8\text{ }m $
