The bob of simple pendulum performs S.H.M. with a frequency 'n' in...

The frequency of a simple pendulum in air is given by $n = \dfrac{1}{2\pi} \sqrt{\dfrac{g}{l}}$.

When the bob is immersed in water, the effective acceleration due to gravity becomes $g' = g \left(1 - \dfrac{\rho_{water}}{\rho_{bob}} \right).<br><br>Given$\rho_{water} = 1$g/cc and$\rho_{bob} = x$g/cc, we have$g' = g \left( 1 - \dfrac{1}{x} \right) = g \left( \dfrac{x - 1}{x} \right).

The frequency$of the pendulum in water is n_1 =$\dfrac{1}{2\pi} \sqrt{\dfrac{g'}{l}}.

The ratio of the frequencies is \dfrac{n}{n_1} = \sqrt{\dfrac{g}{g'}} = \sqrt{\dfrac{x}{x - 1}}.

Given \dfrac{n}{n_1} =3$, we get \sqrt{\dfrac{x}{x - 1}}$=3$.<br><br>Squaring both sides,$\dfrac{x}{x - 1} =9$.<br><br>$x = 9x - 9$<br><br>8x = 9 \Rightarrow x =$\dfrac{9}{8}$g/cc.<br><br>Answer:$\dfrac{9}{8}