A) \[T\propto \sqrt{\rho }\]
B) \[T\propto \frac{1}{\sqrt{A}}\]
C) \[T\propto \frac{1}{\rho }\]
D) \[T\propto \frac{1}{\sqrt{m}}\]
Correct Answer: B
Solution :
Force applied on the body will be equal to upthrust for vertical oscillations. Let block is displaced through\[x\]m, then the weight of displaced water or upthrust (upwards) \[=-Ax\rho g\] Where A is area of cross-section of the block and \[\rho \]is its density. This must be equal to force (= md) applied, where m is mass of the block and a is acceleration. \[\therefore \] \[ma=-Ax\rho g\] or \[a=-\frac{A\rho g}{m}x=-{{\omega }^{2}}x\] This is the equation of simple harmonic motion. Time period of oscillation \[T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{A\rho g}}\Rightarrow T\propto \frac{1}{\sqrt{A}}\]
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