Question

Four identical thin rods each of mass M and Length l, form a square frame. Moment of inertia of this frame about an axis through the center of the square and perpendicular to its plane is:
$
  A.\dfrac{1}{3}M{l^2} \\
  B.\dfrac{4}{3}m{l^2} \\
  C.\dfrac{2}{3}M{l^2} \\
  D.\dfrac{{13}}{3}M{l^2} \\
 $

Answer 1

Hint: concept of moment of inertia through the center and axis perpendicular to the ends of a rod and parallel axis theorem. As the moment of inertia about the axis passing through the centre of square and perpendicular to its plane is due to combined effect of moment of inertia of four rods at the centre of square.

Formula used:
(i) Moment of inertia from center and perpendicular to rod
$I = \dfrac{{M{L^2}}}{{12}}$
(ii) Parallel axis theorem
$I = I\,cm + M{d^2}$

Complete step by step answer:
Now, we know that the moment of inertia of a rod of length L about a perpendicular axis passing through its center O, as shown in figure given by
$ I = \dfrac{{M{L^2}}}{{12}} ...(i) $
Where M $ = $ Mass of rod
L $ = $ total length of rod
Here, in the question we have to find moment of inertia about
an axis through the center and perpendicular to plane as shown in figure.
Now,
L $ = $ length of each rod
M $ = $ mass of each rod
Now,
Moment of inertia of each rod about a perpendicular axis passing through its centre $ = \dfrac{{M{L^2}}}{{12}}$
For finding at centre, we have to apply the parallel axis theorem.
According to parallel axis theorem, the moment of inertia of a body about any axis is equal to moment of inertia about a parallel axis through the centre of mass plus the product of the mass of the body and the square of the perpendicular distance between two parallel axis.
i.e.
$I = I\, + M{d^2}$
Here, for each rod, moment of inertia at centre of square
$
  I = \dfrac{{M{L^2}}}{{12}} + M{\left( {\dfrac{L}{2}} \right)^2} ...as,\,\left( {d = \dfrac{L}{2}} \right) \\
  I = \dfrac{{M{L^2}}}{{12}} + \dfrac{{M{L^2}}}{4} \\
 \implies I = M{L^2}\left( {\dfrac{1}{{12}} + \dfrac{1}{4}} \right) \\
\implies I = M{L^2}\left( {\dfrac{{1 + 3}}{{12}}} \right) \\
 \implies I = M{L^2} \times \dfrac{4}{{12}} \\
\implies I = \dfrac{{M{L^2}}}{{30}} \\
   \\
$
This is the moment of inertia due to each rod at centre of square
Total moment of inertia at centre
$ = $ Number of rods $ \times $ I
$
   = 4 \times \dfrac{{M{L^2}}}{3} \\
   = \dfrac{4}{3}M{L^2} \\
 $

So, the correct answer is “Option B”.

Note:
The distance between the rod and centre is $\dfrac{L}{2}$not L. So, when applying the theorem of parallel axis to find the moment of inertia due to a single rod at the centre of square the distance is taken to be L/2.