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 :
Key Idea: Force applied on the body will be equal to up thrust for vertical oscillations. Let block is displaced through \[x\,m\], then 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 (=ma) applied, where m is mass of the block and a is acceleration. \[\therefore ma=-Ax\rho g\] \[ora=-\frac{A\rho g}{m}x=-{{\omega }^{2}}x\] 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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