A simple pendulum oscillating in air has a period of 3 second. If...

The time period of a simple pendulum in air is given by:

$T_1 = 2\pi \sqrt{\dfrac{L}{g}} = \sqrt{3}$

When immersed in a liquid, the effective acceleration due to gravity becomes:

$g_{eff} = g \left(1 - \dfrac{\rho}{\sigma}\right)<br><br>The new time period is:<br><br>$ T_2 = 2$\pi \sqrt{\dfrac{L}{g_{eff}}} =2$

Taking the ratio of the two time periods:

$\dfrac{T_2}{T_1} =\sqrt{\dfrac{g}{g_{eff}}} = \dfrac{1}{\sqrt{1 - \dfrac{\rho}{\sigma}}}$

Substituting the given values $T_1 = \sqrt{3},$ T_2 = 2$, and$\rho = $\dfrac{\sigma}{x}:<br><br>\dfrac{2}{\sqrt{3}} = \dfrac{1}{\sqrt{1 - \dfrac{1}{x}}}<br><br>Squaring both sides:<br><br>\dfrac{4}{3} = \dfrac{1}{1 - \dfrac{1}{x}}<br><br>1 - \dfrac{1}{x} = \dfrac{3}{4}<br><br>\dfrac{1}{x} =1 - \dfrac{3}{4} = \dfrac{1}{4}<br><br>$ x = 4$<br><br>Answer: 4$