The bob of a simple pendulum performs S.H.M with a period $T$in air and with a period $'{T_1}'$ in water. The relation between $T$and ${T_1}$ is (neglect friction due to water, the density of the material of the bob is $\dfrac{9}{8} \times {10^3}kg/{m^3},$ the density of water=$1g/cc$)
(A) ${T_1} = 3T$
(B) ${T_1} = 2T$
(C) ${T_1} = T$
(D) ${T_1} = \dfrac{T}{2}$

Hint: A simple pendulum consists of a mass suspended by an inextensible massless string of a length $l$. Here we have to compare the period of oscillation of the simple pendulum in air and the same simple pendulum if it is oscillating in water. We have to find the relation between the two time periods.
Formula used
The period of a simple pendulum can be written as,
$T = 2\pi \sqrt {\dfrac{l}{g}} $
Where $T$ stands for the period of oscillation of the simple pendulum, $l$stands for the length of the simple pendulum, and $g$stands for the acceleration due to gravity.

Complete step by step solution:
The period of oscillation of a simple pendulum can be written as,
$T = 2\pi \sqrt {\dfrac{l}{g}} $…………………………………………………………………………………………..(1)
When the simple pendulum is kept under water, there will be a buoyant force in a direction opposite to the direction of gravitational force.
Therefore, we can write the net force as,
${F_{net}} = {F_{gravity}} - {F_{buoyant}}$ ………………………………………………………………………….(2)
We know that force can be written as,
$F = ma$
The density can be written as,
$\rho = \dfrac{m}{V}$
Where $m$is the mass of the object and $v$stands for the volume of the object.
From this, we can write the mass as,
$m = \rho V$
Substituting this value of mass in the equation of force, we get
$F = \rho Va$
Now we can write equation (2) as,
${\rho _{bob}} \times {V_{bob}} \times {a_{net}} = {\rho _{bob}} \times {V_{bob}} \times g - {\rho _{water}} \times {V_{bob}} \times g$
Canceling the common terms and rearranging we can write the acceleration as
${a_{net}} = \dfrac{{{\rho _{bob}} - {\rho _{water}}}}{{{\rho _{bob}}}} \times g$
This equation can be written as,
${a_{net}} = \left( {1 - \dfrac{{{\rho _{water}}}}{{{\rho _{bob}}}}} \right)g$
This will be the acceleration of the bob under water.
Now we can write the period of oscillation of the simple pendulum under water as
${T_1} = 2\pi \sqrt {\dfrac{l}{{{a_{net}}}}} $
Substituting the value of ${a_{net}}$we get
$T = 2\pi \sqrt {\dfrac{l}{{\left( {1 - \dfrac{{{\rho _{water}}}}{{{\rho _{bob}}}}} \right)g}}} $……………………………………………………………………(3)
Dividing equation (1) with equation (3)
We get
\[\dfrac{T}{{{T_1}}} = \dfrac{{2\pi \sqrt {\dfrac{l}{g}} }}{{2\pi \sqrt {\dfrac{l}{{\left( {1 - \dfrac{{{\rho _{water}}}}{{{\rho _{bob}}}}} \right)}}} }}\]
From this, we can write
${T_1} = \dfrac{T}{{\sqrt {\left( {1 - \dfrac{{{\rho _{water}}}}{{{\rho _{bob}}}}} \right)} }}$
The density of water is given as, ${\rho _{water}} = 1g/cc = {10^3}kg/{m^3}$
The density of the bob can be written as, ${\rho _{bob}} = \dfrac{9}{8} \times {10^3}kg/{m^3}$
Putting these values in the above equation,
${T_1} = \dfrac{T}{{\sqrt {1 - \dfrac{{{{10}^3}}}{{\dfrac{9}{8} \times {{10}^3}}}} }}$
This will become,
${T_1} = \dfrac{T}{{\sqrt {1 - \dfrac{8}{9}} }}$
${T_1} = \dfrac{T}{{\sqrt {\dfrac{{9 - 8}}{9}} }}$
Finally,
${T_1} = \dfrac{T}{{\sqrt {\dfrac{1}{9}} }} = \dfrac{T}{{\dfrac{1}{3}}} = 3T$
Therefore,

The answer is:
Option (A): ${T_1} = 3T$

Note:
The time period of the oscillation of a pendulum does not depend on the mass of the bob. When a body is allowed to oscillate freely it will oscillate with a particular frequency. Such oscillations are called free oscillations. The frequency of free oscillations is called natural frequency. A pendulum that is adjusted in a way that it has a period of two seconds is called a second’s pendulum.