question_answer

A U-tube is of non uniform cross-section. The area of cross-sections of two sides of tube are A and 2A (see fig.). It contains non-viscous liquid of mass m. The liquid is displaced slightly and free to oscillate. Its time period of oscillations is

A) \[T=2\pi \sqrt{\frac{m}{3\rho gA}}\]

B) \[T=2\pi \sqrt{\frac{m}{2\rho gA}}\]

C) \[T=2\pi \sqrt{\frac{m}{\rho gA}}\]

D) None of these

Correct Answer: A

Solution :

[a] Suppose the liquid in left side limb is displaced slightly by y, the liquid in right limb will increase by \[y/2\]. The restoring force \[F=-PA=-\rho g\left( \frac{3y}{2} \right)\times 2A=3\rho gA(-y)\]. \[a=\frac{F}{m}=3\rho gA(-y)/m\] On comparing with, \[a=-\,{{\omega }^{2}}y\], we get \[\omega =\sqrt{\frac{3\rho gA}{m}}\]   and  \[T=2\pi \sqrt{\frac{m}{3\rho gA}}\]