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The M.I. of a thin rod of length L about the perpendicular axis through its centre is I. The M.I. of the square structure made by four such rods about a perpendicular axis to the plane and through the centre will be:
\[
  A.\;\;\;\;\;4I \\
  B.\;\;\;\;\;8I \\
  C.\;\;\;\;\;12I \\
  D.\;\;\;\;\;16I \\
\]

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Last updated date: 13th Jun 2024
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Answer
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Hint: The moment of inertia (M.I.) of a rectangular rod is given by
$I = \dfrac{1}{{12}}m{L^2}$
Where, I = Moment of inertia of the rod.
     m = Mass of the rod.
     L = Length of the rod.
Next use the parallel axis theorem to get the required result.

Complete step by step solution:
We know, the moment of inertia (M.I.) of a rectangular rod is given by
$I = \dfrac{1}{{12}}m{L^2}$……(i)
Where, I = Moment of inertia of the rod.
     m = Mass of the rod.
     L = Length of the rod.
From equation (i) we get
$12I = m{L^2}$……(ii)
Additional Information: Parallel axis theorem states that M.I. of any body about any axis is equal to the sum of the M.I. of the body about a axis parallel passing through the centre of gravity and the product of the mass of the body and square of the distance between the two axis.
Now, using parallel axis theorem, we get
$I = {I_{CM}} + m{h^2}$
Where, $I_{CM}$ = Moment of inertia about the centre of mass of the body.
     h = Distance between the two axes.
$\therefore $ Net moment of inertia = $4\left( {\dfrac{1}{{12}}m{L^2} + m\left( {\dfrac{L}{2}} \right)} \right)$
The RHS of the above equation is multiplied by 4 as there are 4 rods given in the question.
Putting the value of $mL^2$ from equation (ii) in above equation we get
Or, Net M.I. = $4\times\left( {\dfrac{{12I}}{{12}} + \dfrac{{12I}}{4}} \right)$
Or, Net M.I. = $16I$.

Therefore, the required answer is 16I (Option D)

Note: Try and remember all the formulas of moment of inertia of different bodies and both the theorems i.e. perpendicular axis theorem and parallel axis theorem. Equation (i) represents the expression for moment of inertia of rectangular lamina.