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Question: A rectangular block of mass m and area of cross-section a floats in a liquid of density r. If it is ...

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

A

T∝1 m\mathrm { T } \propto \frac { 1 } { \sqrt { \mathrm {~m} } }

B

C

T∝1A\mathrm { T } \propto \frac { 1 } { \sqrt { \mathrm { A } } }

D

T∝1ρ\mathrm { T } \propto \frac { 1 } { \rho }

Answer

T∝1A\mathrm { T } \propto \frac { 1 } { \sqrt { \mathrm { A } } }

Explanation

Solution

Refer figure,

Let h be the height of block immersed in liquid, when the block is floating.

……. (i)

If the block is given a vertical displacement y, then the effective restoring force is

F=−[A(h+y)ρg−mg]=−[A(h+y)ρg−Ahρg]\mathrm { F } = - [ \mathrm { A } ( \mathrm { h } + \mathrm { y } ) \rho \mathrm { g } - \mathrm { mg } ] = - [ \mathrm { A } ( \mathrm { h } + \mathrm { y } ) \rho \mathrm { g } - \mathrm { Ah } \rho \mathrm { g } ] ….(ii)

i.e. F∝yF \propto y and –ve sign shows that F is directed towards its equilibrium position of block. So if the block is left free. It will executes SHM

For SHM, F=−mω2yF = - m \omega ^ { 2 } y …… (iii)

Comparing (ii) and (iii), we get

Aρg=mω2A \rho g = m \omega ^ { 2 }

ω=Aρgm\omega = \sqrt { \frac { \mathrm { A } \rho \mathrm { g } } { \mathrm { m } } }

∴\therefore time period