A) \[T\propto \frac{1}{\sqrt{A}}\]
B) \[T\propto \frac{1}{\rho }\]
C) \[T\propto \frac{1}{\sqrt{m}}\]
D) \[T\propto \sqrt{\rho }\]
Correct Answer: A
Solution :
Let the body be depressed by distance x from its equilibrium position. The extra up thrust created is x \[\text{ }\!\!\rho\!\!\text{ Ag}\] which applies to whole body. If a be acceleration created then, \[\text{x }\!\!\rho\!\!\text{ Ag=mga}\Rightarrow \text{a=}\frac{\text{ }\!\!\rho\!\!\text{ A}}{\text{m}}x\] Since, acceleration ax. So it is equation of S.H.M. So, \[{{\omega }^{\text{2}}}\text{=}\frac{\text{ }\!\!\rho\!\!\text{ A}}{\text{m}}\Rightarrow T=2\pi \sqrt{\frac{m}{\rho A}}\] \[T\alpha \frac{1}{\sqrt{A}}\]
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