The periodic time of a simple pendulum, (the simple pendulum with...

The time period of a simple pendulum in air is given by $T = 2\pi \sqrt{\dfrac{L}{g}}$.

When the bob is immersed in a liquid, the effective acceleration due to gravity $g'$ becomes:

$g' = g \left(1 - \dfrac{\rho}{\sigma}\right)<br><br>where$\rho is the $density of the liquid and \sigma is the density of the material of the bob.<br><br>Given that$\rho = $\dfrac{\sigma}{20}, we have:<br><br>$ g' = g \left(1 - \dfrac{1}{20}\right) = \dfrac{19g}{20}

The new time period $T'$ is:

$T' = 2$\pi \sqrt{\dfrac{L}{g'}} =2\pi \sqrt{\dfrac{L}{\dfrac{19g}{20}}} =2\pi \sqrt{\dfrac{L}{g}} \sqrt{\dfrac{20}{19}}

$T' =$\sqrt{\dfrac{20}{19}}$T$

Answer: \sqrt{\dfrac{20}{19}}$ T