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 √ρ (B) T propto (1/√A) (C) T propto (1/ρ) (D) T propto (1/√m)

Tardigrade Admin · Accepted Answer

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)

= - A x ρ g

Where

A

is area of cross-section of the block and

ρ

is its density. This must be equal to force

(= ma)

applied, where

m

is mass of the block and

a

is acceleration.

∴ ma = - A x ρ g

or

a = - \frac{A ρ g}{m} x = - ω^{2} x

This is the cquation of simple harmonic motion.
Time period of oscillation

T = \frac{2 π}{ω} = 2 π \frac{m}{A ρ g}

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

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:

$T \propto ρ$

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

$T \propto \frac{1}{ρ}$

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

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∝ρ​

T∝A​1​

T∝ρ1​

T∝m​1​

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:

$T \propto ρ$

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

$T \propto \frac{1}{ρ}$

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