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

Question

A rectangular block of mass m and area of cross section . A floats in a liquid of density ρ. If it is given a small vertical displacement from equilibrium it undergoes oscillation with a time period T, then : (A) T propto √ m  (B)  T propto √ ρ  (C) T propto (1/√ A ) (D)  T propto (1/ρ)

Tardigrade Admin · Accepted Answer

Time period of SHM of small vertical oscillations in a liquid is given
by

T = 2 π \frac{1}{g}

, where

l

is the length of
cube/cylinder/block dipped in the liquid.
So according to law of floatation,
weight of the block

=

weight of the liquid displaced

m g = A l gg

\Rightarrow I = \frac{m}{A p}

\Rightarrow T = 2 π \frac{m}{A p g}

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

$T \propto m$

$T \propto ρ$

$T \propto \frac{1}{A}$

$T \propto \frac{1}{ρ}$

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

T∝m​

T∝ρ​

T∝A​1​

T∝ρ1​

Time period of SHM of small vertical oscillations in a liquid is given by T=2πg1​​, where l is the length of cube/cylinder/block dipped in the liquid. So according to law of floatation, weight of the block = weight of the liquid displaced mg=Algg ⇒I=Apm​ ⇒T=2πApgm​​

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

$T \propto m$

$T \propto ρ$

$T \propto \frac{1}{A}$

$T \propto \frac{1}{ρ}$