JEE Advanced 2021 Full Test 4 Paper 2

Let length I of the cylinder is dipped in liquid, the weight of liquid displaced by the volume of l-part of cylinder πr2lρg=W ......(i)
where, ρ= density of the liquid when cylinder is further dipped through a distance x, then volume of displaced [liquid πr2(l+x)]
Therefore, vertically upward force, (i.e. force of buoyance) acting on mass equals πt2(1+x)ρg.
At situation of equilibrium, net force on the mass
F=W−πr2(1+x)ρg
=W−πr2lρg−πr2ρ‌xg .......(ii)
From equations (i) and (ii), we get
F=−πr2ρgx=−kx
R=πr2ρg
The time period, T=2π√

m

k

=2π√

m

πr2ρg

∴ Frequency of oscillations,
v=

1

T

=

1

2π

√

πr2gρ

m