question_answer

A rectangular block of mass m and area of cross-section A floats in a liquid of density\[\rho \]. If it is given a small vertical displacement from equilibrium it undergoes oscillation with a time period T. Then

A) \[T\propto \frac{1}{\sqrt{m}}\]   

B) \[T\propto \sqrt{\rho }\]

C) \[T\propto \frac{1}{\sqrt{A}}\]   

D) \[T\propto \frac{1}{\rho }\]

Correct Answer: C

Solution :

[c]:   Let \[l\] be the length of block immersed in liquid, when the block is floating. \[\therefore \]\[mg=Al\rho g\]                                                  ...(i) If the block is given a vertical displacement y, then the effective restoring force is \[F=-[A(l+y)\rho g-mg]\] \[=-[A(l+y)\rho g-Al\rho g]=-A\rho gy\]            (Using (i)) i.e.\[F\propto y.-ve\] -ve sign shows that F is directed towards its equilibrium position. Therefore, if the block is left free, it will execute SHM. Acceleration, \[a=\frac{F}{m}=\frac{A\rho gy}{m}=-{{\omega }^{2}}y\] \[\therefore \]\[\omega =\sqrt{\frac{A\rho g}{m}}\] \[\therefore \]Time period, \[T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{A\rho g}}\] i.e.,\[T\propto \frac{1}{\sqrt{A}}.\]