question_answer

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 \[=13.6\times {{10}^{3}}\] \[kg{{m}^{-3}}\])

A)  13.6m                  

B)         40.8m  

C)         27.2m                  

D)         54.4m

E)  6.8m

Correct Answer: B

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 \[x+1=40.8\text{ }m\]