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 \[T=2\pi \sqrt{\frac{\text{Displacement}}{\text{ }\!\!|\!\!\text{ Acceleration }\!\!|\!\!\text{ }}}\] \[=2\pi \sqrt{\frac{y}{\frac{2Aydg}{M}}}\] \[\Rightarrow T=2\pi \sqrt{\frac{M}{2Adg}}\]You need to login to perform this action.
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