question_answer

An open\[U-\]tube contains mercury. When 11.2 cm of water is poured into one of the arms of the tube, how high does the mercury rise in the other arm form its initial level?

A)  0.56 cm                               

B)  1.35 cm

C)   0.41 cm                              

D)  2.32 cm

Correct Answer: C

Solution :

                 Key Idea: At the same level in the two limbs of the U-tube, pressure is same. On pouring water on the left side, mercury rises \[x\]cm (say) from its previous level in the right limb of \[U-\]tube creating a difference of levels of mercury by\[2xcm\]. Equating pressures at AB, we get \[{{P}_{A}}={{P}_{B}}\] \[\therefore \]\[11.2\times {{10}^{-2}}\times {{\rho }_{water}}\times g\]                 \[=2x\times {{\rho }_{mercury}}\times g\] \[11.2\times {{10}^{-2}}\times 1000\,kg/{{m}^{3}}\]                 \[=2x\times 13600\,kg/{{m}^{3}}\] \[\therefore \]\[x=\frac{11.2\times {{10}^{-2}}\times 1000}{2\times 13600}m=0.41\,cm\]