
A vibration magnetometer consists of two identical bar magnets placed one over the other, such that they are mutually perpendicular and bisect each other. The period of oscillation of combination in a horizontal magnetic field is\[4s\] . if one of the magnets is removed, then the period of oscillations of the other in the same field is
A. ${2^{\dfrac{1}{4}}}\sec $
B. ${2^{\dfrac{7}{4}}}\sec $
C. ${2^{\dfrac{5}{4}}}\sec $
D. ${2^{\dfrac{3}{4}}}\sec $
Answer
483.3k+ views
Hint: Moment of inertia angular mass or rotational inertia can be defined concerning the rotation axis. A quantity defines the amount of torque needed for angular acceleration. The heavier particle has more inertia.
Complete step by step solution:
Given that the period of oscillation is 4s
We know that the period of oscillation is proportional to the moment of inertia
And the effective magnetic moment of the magnet is inversely proportional to the period of oscillation.
We can write this expression as follows,
$T \propto \sqrt {\dfrac{I}{L}} $
Now the initial moment of inertia and the magnetic moment of the combination are \[I\] and \[M\] respectively
When we remove one of the magnets then the result will be as follows,
$\dfrac{M}{{\sqrt 2 }}$
Similarly, the moment of inertia will be after removing the one magnet is as follows,
$\dfrac{I}{2}$ hence we directly added two magnets when a moment of inertia passing through the midpoint of the magnets along the perpendicular plane.
Now,
\[\dfrac{{{T_1}}}{{{T_2}}} = \sqrt {\dfrac{{{I_1}{M_2}}}{{{I_2}{M_1}}}} \]
Now the equation will be,
\[ = \sqrt {\dfrac{{I\dfrac{M}{{\sqrt 2 }}}}{{M\dfrac{I}{2}}}} = {2^{\dfrac{1}{4}}}\]
\[{T_2} = {2^{\dfrac{7}{4}}}\sec \]
hence,
\[{T_2} = \dfrac{{{T_1}}}{{{2^{\dfrac{1}{4}}}}}\sec \]
after substituting the values we get,
\[{T_2} = \dfrac{4}{{{2^{\dfrac{1}{4}}}}}\sec \]
Now simplify the above equation we get,
\[{T_2} = {2^{\dfrac{7}{4}}}\sec \]
So, the correct answer is option (B).
Note:
Due to its mass, inertia occurs.
if the mass of a body is high then the inertia is high
For example, we can throw a small stone farther than a heavier one. Because the heavier one has more mass, hence it resists change more, that is, it has more inertia.
Complete step by step solution:
Given that the period of oscillation is 4s
We know that the period of oscillation is proportional to the moment of inertia
And the effective magnetic moment of the magnet is inversely proportional to the period of oscillation.
We can write this expression as follows,
$T \propto \sqrt {\dfrac{I}{L}} $
Now the initial moment of inertia and the magnetic moment of the combination are \[I\] and \[M\] respectively
When we remove one of the magnets then the result will be as follows,
$\dfrac{M}{{\sqrt 2 }}$
Similarly, the moment of inertia will be after removing the one magnet is as follows,
$\dfrac{I}{2}$ hence we directly added two magnets when a moment of inertia passing through the midpoint of the magnets along the perpendicular plane.
Now,
\[\dfrac{{{T_1}}}{{{T_2}}} = \sqrt {\dfrac{{{I_1}{M_2}}}{{{I_2}{M_1}}}} \]
Now the equation will be,
\[ = \sqrt {\dfrac{{I\dfrac{M}{{\sqrt 2 }}}}{{M\dfrac{I}{2}}}} = {2^{\dfrac{1}{4}}}\]
\[{T_2} = {2^{\dfrac{7}{4}}}\sec \]
hence,
\[{T_2} = \dfrac{{{T_1}}}{{{2^{\dfrac{1}{4}}}}}\sec \]
after substituting the values we get,
\[{T_2} = \dfrac{4}{{{2^{\dfrac{1}{4}}}}}\sec \]
Now simplify the above equation we get,
\[{T_2} = {2^{\dfrac{7}{4}}}\sec \]
So, the correct answer is option (B).
Note:
Due to its mass, inertia occurs.
if the mass of a body is high then the inertia is high
For example, we can throw a small stone farther than a heavier one. Because the heavier one has more mass, hence it resists change more, that is, it has more inertia.
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