Question

The cylinder having the hoop and solid cylinder having the same mass and radius made of permanent magnetic material with their magnetic moment parallel with their axes respectively. It is given that the magnetic moment of the hoop is twice that of the solid cylinder, they are placed in uniform magnetic moments to make a small angle with the field. The oscillation periods of hoop and cylinder are ${T_h}$ and ${T_c}$ respectively, so:
A. ${T_h} = 0.5{T_c}$
B. ${T_h} = 2{T_c}$
C. ${T_h} = 1.5{T_c}$
D. ${T_h} = {T_c}$

Answer 1

Hint: Find the moment of inertia for hoop and moment of inertia for cylinder and in the presence of magnetic field, time period is directly proportional to the square root of inertia and inversely proportional to the square root of magnetic moment and square root of magnetic field.

Complete step by step answer:
We know that, during the presence of magnetic field, time period is directly proportional to the square root of inertia and inversely proportional to the square root of magnetic moment and square root of magnetic field. So, let $T$ be the time period.
$
  \therefore T \propto \sqrt I \\
   \Rightarrow T \propto \dfrac{1}{{\sqrt \mu }} \\
   \Rightarrow T \propto \dfrac{1}{{\sqrt B }} \\
 $
So, we can write it as –
$ \Rightarrow T \propto \sqrt {\dfrac{I}{{\mu B}}} $
Now, using the above formula of time period for ${T_h}$ and ${T_c}$ -
$\therefore {T_c} = k\sqrt {\dfrac{{{I_c}}}{{\mu B}}} \cdots \left( 1 \right)$
where, $k$ is the proportionality constant
$\therefore {T_h} = k\sqrt {\dfrac{{{I_h}}}{{\mu 'B}}} \cdots \left( 2 \right)$
Now, we know that –
Moment of inertia for hoop, ${I_h} = M{R^2}$
Moment of inertia for cylinder, ${I_c} = \dfrac{{M{R^2}}}{2}$
According to the question, it is given that magnetic moment of hoop is twice of the solid cylinder –
$\therefore \mu ' = 2\mu $
Putting all these values in their respective places in the equation $\left( 1 \right)$ and equation $\left( 2 \right)$ and dividing them, we get –
$
  \therefore \dfrac{{{T_c}}}{{{T_h}}} = \dfrac{{\sqrt {\dfrac{{M{R^2}}}{{2\mu B}}} }}{{\sqrt {\dfrac{{M{R^2}}}{{2\mu B}}} }} \\
   \Rightarrow \dfrac{{{T_c}}}{{{T_h}}} = 1 \\
  \therefore {T_c} = {T_h} \\
 $
Hence, the correct option is (D).

Note:Here we can commit an error in taking the wrong value of $I$. We need to make sure that the value of I is taken about the center.
Also, Time period is proportional to the square root of the quantities. Although, this will not lead to the wrong answer but it may hurt us in other questions.