The bob of a simple pendulum executes simple harmonic motion in water with a period t, while the period of oscillation of the bob is ${t_o}$ in air. Neglect frictional force of water and given that the density of the bob is $\left( {4/3} \right) \times 1000\;{\rm{kg/}}{{\rm{m}}^3}$. What relationship between t and ${t_0}$ is true?
A. $t = {t_0}$
B. $t = \dfrac{{{t_0}}}{2}$
C. $t = 2{t_0}$
D. $t = 4{t_0}$

Hint: To obtain the relation between $t$ and ${t_o}$, first we have to build the relation between acceleration due to gravity in water and acceleration due to gravity in air. After this, use the relation of accelerations in the expression of time period, so that we can obtain the correct answer.

Complete step by step solution:
Given:
The period of the oscillation of the bob in air is ${t_0}$.
The period of oscillation of the bob in water is $t$.
The density of the bob is $\rho = \dfrac{4}{3} \times 1000\;{\rm{kg/}}{{\rm{m}}^3}$.

From the expression of density, the mass of the bob is
$\begin{array}{l}
\rho = \dfrac{M}{V}\\
M = \rho V
\end{array}$

Here, $M$ is the mass of the bob and is the volume of the bob.

Substitute the values in the above equation
$M = \dfrac{4}{3} \times 1000\;{\rm{kg/}}{{\rm{m}}^3} \times V$ (1)

The expression of the effective weight of the bob in water is,
$\begin{array}{l}
w' = w - B\\
Mg' = Mg - B
\end{array}$ (2)

Here $w'$ is the effective weight of the bob, $w$ is the weight of the bob in air, $g'$ is the acceleration due to the gravity in water, $g$ is the acceleration due to gravity in air and $B$ is the buoyant force.

The expression of the buoyant force is,
$B = V{\rho _{water}}g$ (3)

Here, ${\rho _{water}} = 1000\;{\rm{kg/}}{{\rm{m}}^3}$ is the density of the water.

From equation (1), (2) and (3), the acceleration due to the gravity in water becomes,

\[Mg' = Mg - V{\rho _{water}}g\]

Substitute the values in the above equation
\[\begin{array}{l}
\left( {\dfrac{4}{3} \times 1000\;{\rm{kg/}}{{\rm{m}}^3} \times V} \right)g' = \left( {\dfrac{4}{3} \times 1000\;{\rm{kg/}}{{\rm{m}}^3} \times V} \right)g - \left( {V \times 1000\;{\rm{kg/}}{{\rm{m}}^3}} \right)g\\
\dfrac{4}{3} \times 1000\;{\rm{kg/}}{{\rm{m}}^3} \times g' = g\left( {\dfrac{{1000\;{\rm{kg/}}{{\rm{m}}^3}}}{3}} \right)\\
g' = \dfrac{g}{4}
\end{array}\] (4)

The expression of the time period of the bob in air is,
${t_o} = 2\pi \sqrt {\dfrac{l}{g}} $ (5)

Here, $l$ is the length of the pendulum.

The expression of the time period of the bob in water is,
$t = 2\pi \sqrt {\dfrac{l}{{g'}}} $ (6)

Now from equation (4) ,(5) and (6), the relation between t and ${t_0}$ becomes,
$\begin{array}{l}
t = 2\pi \sqrt {\dfrac{l}{{g'}}} \\
t = 2\pi \sqrt {\dfrac{{4l}}{g}} \\
t = 2 \times 2\pi \sqrt {\dfrac{l}{g}} \\
t = 2{t_0}
\end{array}$

Therefore, option (C) is correct, that is $t = 2{t_0}$.

Note: The expression of buoyant force is used to obtain the relation between acceleration due to gravity and air. So, remember the value of density of water because it is used in the expression of buoyant force