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 a U-tube, pressure is same. On pouring water on left side, mercury rises x cm (say) from its previous level in the right limb of 17-tube creating a difference of levels of mercury by 2x cm. Equating pressures at A and B, we get                 \[{{P}_{A}}={{P}_{B}}\] \[\therefore \]     \[11.2\times {{10}^{-2}}\times {{\rho }_{water}}\times g=2x\times {{\rho }_{mercury}}\times g\] \[\Rightarrow \] \[11.2\times {{10}^{-2}}\times 1000\,\,kg/{{m}^{3}}\]                 \[=2x\times 13600\times kg/{{m}^{3}}\] \[\Rightarrow \] \[x=\frac{11.2\times {{10}^{-2}}\times 1000\,\,m}{2\times 13600}\] = 0.41 cm