question_answer

A test tube of uniform cross-section A and length \[\ell \] is immersed inverted into a liquid of density \[\rho \]. The liquid rises to a height \[x(<\ell )\] in the tube. The excess pressure at the interface between trapped air and the liquid in the inverted tube is [Given : Atmospheric pressure is h column of mercury \[\left( {{\rho }_{m}} \right)\]]

A)  \[h{{\rho }_{m}}g\]                       

B)  \[\frac{h({{\rho }_{m+\rho }})gx}{(\ell -x)}\]

C)  \[h({{\rho }_{m}}-\rho )g\]                        

D)  \[\frac{h{{\rho }_{m}}gx}{\ell -x}\]

Correct Answer: D

Solution :

Conserving PV before and after the tube is immersed in the liquid, we get  \[{{p}_{0}}A\ell =({{p}_{0}}+p)A(\ell -x)\] \[h{{\rho }_{m}}gA\ell =(h{{\rho }_{m}}g+p)A(\ell -x)\] \[p=\frac{h{{\rho }_{m}}gx}{\ell -x}\]