Question

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 timeT.
a) $T\alpha \sqrt m$
b) $T\alpha \sqrt \rho$
c) $T\alpha \dfrac{1}{{\sqrt A }}$
d) $T\alpha \dfrac{1}{{\sqrt \rho }}$

Answer 1

Mass of the object is equal to the mass of liquid displaced. Thus, when the rectangular block is submerged in liquid it displaces a particular amount of the liquid which is equal to its weight.

Complete step by step answer:
Given, a rectangular block of mass floats in a liquid having a small vertical displacement from the initial equilibrium.
We know that , time period of Simple Harmonic Motion or SHM of small vertical oscillations in a liquid is given by $T = 2\pi \sqrt {\dfrac{l}{g}}$--(i)
Where l = length of the particle dipped in the liquid.
By the law of floatation,
Weight of the particle = weight of the water displaced

$$\Rightarrow mg = Al\rho g \\\ \Rightarrow l = \dfrac{m}{{A\rho }} - - - (ii) \\\$$

Substituting the value of l in (i) we get,
$T = 2\pi \sqrt {\dfrac{m}{{gA\rho }}}$
Thus, from the above equation it can be said
$T\alpha \sqrt m$ , $T\alpha \dfrac{1}{{\sqrt A }}$and $T\alpha \dfrac{1}{{\sqrt \rho }}$
Hence , for the given question the required options are a) $T\alpha \sqrt m$ c)$T\alpha \dfrac{1}{{\sqrt A }}$ and d) $T\alpha \dfrac{1}{{\sqrt \rho }}$
Additional information:
By the law of floatation, a particle when dipped in a liquid displaces the amount of water which is equal to its own weight.

Note: Before solving this sum, the student must be well acquainted with the concept of floatation to choose the correct formula. The student must also analyze the sum properly and choose the correct formula and way of solving the sum.Students also need to know the formula for simple harmonic motion with respect to the time taken by the object.