Question

A cylindrical cork of uniform density \(\rho_1\) floats in a liquid of density \(\rho_1\). If the cork is depressed slightly and released, it oscillates harmonically with time period \(T\). If the same cork floats in another liquid of density \(\rho_2\), then the similar oscillation has time period \(2T\). The value of \(\dfrac{\rho_2}{\rho_1}\) is:

• A. \(\dfrac{1}{4}\)
• B. \(4\)
• C. \(2\)
• D. \(\dfrac{1}{2}\)

Hint:

For oscillations of a floating body, \[ T\propto \frac{1}{\sqrt{\rho}} \] A denser liquid provides a stronger restoring force and hence a smaller time period.

Answer 1

The Correct Option is A

Step 1: Set up the floating oscillator.
When a floating cork is pushed down by $x$, the extra buoyant force $A\rho g x$ acts as a restoring force, producing SHM. The period is $T = 2\pi\sqrt{\dfrac{m}{A\rho g}}$.
Step 2: Note the dependence.
For the same cork (same $m$, same $A$), only the liquid density $\rho$ changes, so $T \propto \dfrac{1}{\sqrt{\rho}}$.
Step 3: Write the ratio of periods.
$\dfrac{T_2}{T_1} = \sqrt{\dfrac{\rho_1}{\rho_2}}$.
Step 4: Plug in the data.
Given $T_1 = T$ and $T_2 = 2T$, so $\dfrac{2T}{T} = \sqrt{\dfrac{\rho_1}{\rho_2}}$, that is $2 = \sqrt{\dfrac{\rho_1}{\rho_2}}$.
Step 5: Square both sides.
$4 = \dfrac{\rho_1}{\rho_2}$, hence $\dfrac{\rho_2}{\rho_1} = \dfrac14$.
Step 6: Choose the option.
This matches option A.
\[ \boxed{ \frac{\rho_2}{\rho_1} = \frac14 } \]