A simple pendulum oscillates in simple harmonic motion in water with a period t, while the period of oscillation of the bob is t₀ in air. Neglecting frictional force of water and given that the density of the bob is $\left.4/3\times 1000\mathrm{kg}/{\mathrm{m}}^3\right)$.  Which of the given relationship between t and t₀ is true?

The time period of simple pendulum in air

$T=t_0=2\pi \sqrt{\left(\frac{l}{\mathrm{g}}\right)}\dots$ (i)

l, being the length of simple pendulum. 
In water, effective weight of bob

$w^{\mathrm{\prime }}=$weight of bob in air - upthrust

$\begin{array}{l}\Rightarrow \rho V{\mathrm{g}}_{\text{eff }}=m\mathrm{g}-m/\mathrm{g}\\ =\rho V\mathrm{g}-{\rho }^{\mathrm{\prime }}V\mathrm{g}=\left(\rho -{\rho }^{\mathrm{\prime }}\right)V\mathrm{g}\end{array}$

where ${\rho }^{\mathrm{\prime }}=$ density of bob.

$\rho =$density of water

$\begin{array}{l}\therefore {\mathrm{g}}_{\mathrm{eff}}=\left(\frac{\rho -{\rho }^{\mathrm{\prime }}}{\rho }\right)\mathrm{g}\\ =\left(1-\frac{{\rho }^{\mathrm{\prime }}}{\rho }\right)\mathrm{g}\end{array}$

$\therefore$   $t=2\pi \sqrt{\left[\frac{l}{\left(1-\frac{{\rho }^{\mathrm{\prime }}}{\rho }\right)\mathrm{g}}\right]}\cdots$

Thus,    $\frac{t}{t_0}=\sqrt{\left[\frac{1}{\left(1-\frac{{\rho }^{\mathrm{\prime }}}{\rho }\right)}\right]}$

$=\sqrt{\left(\frac{1}{1-\frac{1000}{(4/3\times 1000)}}\right)}=\sqrt{\left(\frac{4}{4-3}\right)}=2$

$\Rightarrow t=2t_0$