A deflection magnetometer is adjusted and a magnet of magnetic moment M is placed on it in the usual manner and the observed deflection is $\theta $. The period of oscillation of the needle before settling to the deflection is T. When the magnet is removed, the period of oscillation of the needle is ${T_0}$ before settling to 0 degrees. If the earth’s magnetic field is ${B_H}$, the relation between T and ${T_0}$ is-
A. ${T^2} = T_0^2\cos \theta $
B. ${T^2} = \dfrac{{T_0^2}}{{\cos \theta }}$
C. $T = {T_0}\cos \theta $
D. $T = \dfrac{{{T_0}}}{{\cos \theta }}$

Hint: We will first understand that there will be two types of magnetics field working i.e. magnetic field due to magnet and magnetic field due to the earth. Then we will find out the resultant magnetic field and then solve it further by using the formula for time period. Refer to the solution below.

Formula used: $T = 2\pi \sqrt {\dfrac{I}{{MB}}} $.
The formula for time period is given as-
$ \Rightarrow T = 2\pi \sqrt {\dfrac{I}{{MB}}} $
Where, B is the horizontal magnetic field.

Step-By-Step answer:
As we know, a deflection magnetometer works in a horizontal magnetic field.
Now, it is said that the device was fit in a usual way. Now, we know that the magnetometer has a magnetic field of its own. And where the device is applied, earth has its own magnetic field there.
Let F be the magnetic field of the magnetometer. And ${B_H}$ be the magnetic field of the earth.
So, the resultant magnetic field (B) will be considered as-
$ \Rightarrow B = \sqrt {{F^2} + {B_H}^2} $
Thus, the formula for time period will be-
$ \Rightarrow T = 2\pi \sqrt {\dfrac{I}{{M\sqrt {{F^2}{B_H}^2} }}} $
Let this equation be equation 1.
$ \Rightarrow T = 2\pi \sqrt {\dfrac{I}{{M\sqrt {{F^2}{B_H}^2} }}} $ (equation 1)
This time period will be the initial time period.
After we remove the magnet, the time period becomes ${T_0}$. In this case, the magnetic field due to magnet (F) will become 0 since the magnet was removed-
$
   \Rightarrow {T_0} = 2\pi \sqrt {\dfrac{I}{{M\sqrt {{B_H}^2} }}} \\
    \\
   \Rightarrow {T_0} = 2\pi \sqrt {\dfrac{I}{{M{B_H}}}} \\
$
Let this equation be equation 2.
$ \Rightarrow {T_0} = 2\pi \sqrt {\dfrac{I}{{M{B_H}}}} $ (equation 2)
Now, dividing equation 1 and equation 2, we get-
$ \Rightarrow \dfrac{T}{{{T_0}}} = \sqrt {\dfrac{H}{{\sqrt {{B_H}^2 + {F^2}} }}} $
The relation between magnetic field due to magnet (F) and magnetic field due to the earth (${B_H}$) is-
$ \Rightarrow \dfrac{F}{{{B_H}}} = \tan \theta $
Now, the angle between magnetic field due to magnet (F) and magnetic field due to the earth (${B_H}$) is always 90 degrees.
$ \Rightarrow F = {B_H}\tan \theta $
Putting this value of F into the above equation, we get-
$
   \Rightarrow \dfrac{T}{{{T_0}}} = \sqrt {\dfrac{{{B_H}}}{{\sqrt {{B_H}^2 + {B_H}^2{{\tan }^2}\theta } }}} \\
    \\
   \Rightarrow \dfrac{T}{{{T_0}}} = \sqrt {\dfrac{{{B_H}}}{{{B_H}\sqrt {1 + {{\tan }^2}\theta } }}} \\
$
Since we know that $1 + {\tan ^2}\theta = {\sec ^2}\theta $. So-
$
   \Rightarrow \dfrac{T}{{{T_0}}} = \sqrt {\dfrac{1}{{\sqrt {{{\sec }^2}\theta } }}} \\
    \\
   \Rightarrow \dfrac{T}{{{T_0}}} = \sqrt {\dfrac{1}{{\sec \theta }}} \\
$
Since we know that $\dfrac{1}{{\sec \theta }} = \cos \theta $. So-
$
   \Rightarrow \dfrac{T}{{{T_0}}} = \sqrt {\cos \theta } \\
    \\
   \Rightarrow \dfrac{{{T^2}}}{{T_0^2}} = \cos \theta \\
    \\
   \Rightarrow {T^2} = T_0^2\cos \theta \\
$
Hence, it is clear that option A is the correct option.

Note: Magnetic field, a neighbourhood vector field of a magnet, electric current or electric field changing, where magnetic forces are observed. Magnetic fields like Earth cause magnetic compasses and other permanent magnets to line up in the field direction.