question_answer

A U tube of uniform bore of cross-sectional area A has been set up vertically with open ends facing up. Now m gm of a liquid of density d is poured into it. The column of liquid in this tube will oscillate with a period T such that

A)            \[T=2\pi \sqrt{\frac{M}{g}}\]

B)            \[T=2\pi \sqrt{\frac{MA}{gd}}\]

C)            \[T=2\pi \sqrt{\frac{M}{gdA}}\]                                      

D)            \[T=2\pi \sqrt{\frac{M}{2Adg}}\]

Correct Answer: D

Solution :

                   If the level of liquid is depressed by y cm on one side, then the level of liquid in column P is 2y cm higher than B as shown.                     The weight of extra liquid on the side \[P=2Aydg\].                     This becomes the restoring force on mass M.                     \[\therefore \] Restoring acceleration \[=\frac{-\,2Aydg}{M}\]                     This relation satisfies the condition of SHM i.e. \[a\propto -y\].                     Hence time period \[T=2\pi \sqrt{\frac{\text{Displacement}}{\text{ }\!\!|\!\!\text{ Acceleration }\!\!|\!\!\text{ }}}\]                    \[=2\pi \sqrt{\frac{y}{\frac{2Aydg}{M}}}\] \[\Rightarrow T=2\pi \sqrt{\frac{M}{2Adg}}\]