Moment of Inertia of Compound Bodies and Theorem of Moment of Inertia Questions in English

(A) The moment of inertia of the system about the axis $YY'$ passing through the center of the right sphere is the sum of the moments of inertia of both spheres about this axis.
For the right sphere,the axis passes through its center,so its moment of inertia is $I_1 = \frac{2}{5} M (R/2)^2 = \frac{2}{5} M (R^2/4) = \frac{1}{10} M R^2$.
For the left sphere,the axis is at a distance $d = 2R$ from its center. Using the parallel axis theorem,its moment of inertia is $I_2 = I_{cm} + Md^2 = \frac{2}{5} M (R/2)^2 + M(2R)^2 = \frac{1}{10} M R^2 + 4 M R^2 = \frac{41}{10} M R^2$.
The total moment of inertia is $I = I_1 + I_2 = \frac{1}{10} M R^2 + \frac{41}{10} M R^2 = \frac{42}{10} M R^2 = \frac{21}{5} M R^2$.

(D) Consider one rod of the square frame,say rod $AB$. The moment of inertia of a rod of mass $M$ and length $l$ about an axis passing through its center and perpendicular to its length is $I_{cm} = \frac{Ml^2}{12}$.
The distance $d$ from the center of the rod $AB$ to the center of the square frame is $d = \frac{l}{2}$.
Using the parallel axis theorem,the moment of inertia of one rod about the axis passing through the center of the square and perpendicular to its plane is:
$I_{rod} = I_{cm} + Md^2 = \frac{Ml^2}{12} + M\left(\frac{l}{2}\right)^2 = \frac{Ml^2}{12} + \frac{Ml^2}{4} = \frac{Ml^2 + 3Ml^2}{12} = \frac{4Ml^2}{12} = \frac{Ml^2}{3}$.
Since there are four identical rods forming the square frame,the total moment of inertia $I$ is:
$I = 4 \times I_{rod} = 4 \times \frac{Ml^2}{3} = \frac{4}{3}Ml^2$.

(B) According to the theorem of parallel axes,the moment of inertia of a thin rod of mass $M$ and length $L$ about an axis passing through one of its ends is given by:
$I = I_{CM} + Md^2$
Where $I_{CM}$ is the moment of inertia of the rod about an axis passing through its center of mass (midpoint) and perpendicular to its length,and $d$ is the perpendicular distance between the two parallel axes.
Here,$I_{CM} = I_0$ and the distance between the center and the end is $d = L/2$.
Substituting these values into the theorem:
$I = I_0 + M(L/2)^2$
$I = I_0 + ML^2/4$

(A) Mass per unit area of the disc is $\sigma = \frac{M}{\pi R^2}$.
The radius of the removed circular hole is $r = \frac{R}{2}$.
The mass of the removed portion is $M' = \sigma \times \pi r^2 = \frac{M}{\pi R^2} \times \pi \left(\frac{R}{2}\right)^2 = \frac{M}{4}$.
The moment of inertia of the removed portion about its own central axis (perpendicular to the plane) is $I_{cm} = \frac{1}{2} M' r^2 = \frac{1}{2} \left(\frac{M}{4}\right) \left(\frac{R}{2}\right)^2 = \frac{MR^2}{32}$.
Using the parallel axis theorem,the moment of inertia of the removed portion about the axis passing through the center $O$ of the original disc is $I'_{0} = I_{cm} + M' d^2$,where $d = \frac{R}{2}$ is the distance between the center of the disc and the center of the hole.
$I'_{0} = \frac{MR^2}{32} + \left(\frac{M}{4}\right) \left(\frac{R}{2}\right)^2 = \frac{MR^2}{32} + \frac{MR^2}{16} = \frac{3MR^2}{32}$.
The moment of inertia of the complete disc about the center $O$ is $I_{0} = \frac{1}{2} MR^2$.
The moment of inertia of the remaining part is $I = I_{0} - I'_{0} = \frac{1}{2} MR^2 - \frac{3MR^2}{32} = \frac{16MR^2 - 3MR^2}{32} = \frac{13MR^2}{32}$.

(B) Given: Mass of the rod $m = 0.6 \ kg$,Length $l = 1 \ m$.
The center of mass of the rod is at $l/2 = 0.5 \ m$ from one end.
The axis is at $20 \ cm = 0.2 \ m$ from one end.
The distance $x$ between the center of mass and the given axis is $x = |0.5 \ m - 0.2 \ m| = 0.3 \ m$.
Using the parallel axis theorem,$I = I_{cm} + mx^2$.
$I = \frac{ml^2}{12} + mx^2 = m \left( \frac{l^2}{12} + x^2 \right)$.
Substituting the values: $I = 0.6 \left( \frac{1^2}{12} + (0.3)^2 \right) = 0.6 \left( \frac{1}{12} + 0.09 \right)$.
$I = 0.6 \left( 0.0833 + 0.09 \right) = 0.6 \times 0.1733 = 0.104 \ kg \cdot m^2$.

(C) The moment of inertia of a solid sphere about its diameter is $I_{cm} = \frac{2}{5}MR^2$.
According to the parallel axis theorem,the moment of inertia about a tangent is $I = I_{cm} + MR^2$.
Substituting the values: $I = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2$.
Given $M = 10\; kg$ and $R = 0.5\; m$,we have:
$I = \frac{7}{5} \times 10 \times (0.5)^2$
$I = 14 \times 0.25 = 3.5\; kg\cdot m^2$.

(A) The moment of inertia of a single rod $AB$ about an axis passing through its centre $P$ and perpendicular to its length is $I_{cm} = \frac{1}{12}Ml^2$.
To find the moment of inertia of this rod about an axis passing through the centre $O$ of the square and perpendicular to the plane,we use the parallel axis theorem: $I = I_{cm} + Md^2$,where $d = l/2$ is the distance between the parallel axes.
$I_{rod} = \frac{1}{12}Ml^2 + M(l/2)^2 = \frac{1}{12}Ml^2 + \frac{1}{4}Ml^2 = \frac{1+3}{12}Ml^2 = \frac{4}{12}Ml^2 = \frac{1}{3}Ml^2$.
Since the system consists of $4$ identical rods,the total moment of inertia of the system is $I_{system} = 4 \times I_{rod} = 4 \times \frac{1}{3}Ml^2 = \frac{4}{3}Ml^2$.

(B) The moment of inertia of a circular disc about its diameter is given by $I = \frac{1}{4}MR^2$,which implies $MR^2 = 4I$.
To find the moment of inertia about an axis perpendicular to the plane and passing through a point on the rim,we use the parallel axis theorem.
The moment of inertia about an axis passing through the center of mass and perpendicular to the plane is $I_{cm} = \frac{1}{2}MR^2$.
The distance between the center of mass and the point on the rim is $R$.
By the parallel axis theorem,$I_{rim} = I_{cm} + MR^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.
Substituting $MR^2 = 4I$ into the expression,we get $I_{rim} = \frac{3}{2}(4I) = 6I$.

(B) The moment of inertia of a circular disc about an axis passing through its center and perpendicular to its plane is $I_{cm} = \frac{1}{2}MR^2$.
Given mass $M = 2 \ kg$ and diameter $D = 0.2 \ m$,the radius is $R = \frac{D}{2} = 0.1 \ m$.
According to the parallel axis theorem,the moment of inertia about an axis passing through the edge and perpendicular to the plane is $I = I_{cm} + MR^2$.
Substituting the values: $I = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.
$I = \frac{3}{2} \times 2 \times (0.1)^2 = 3 \times 0.01 = 0.03 \ kg \cdot m^2$.

(B) The moment of inertia of a uniform ring about an axis passing through its center and perpendicular to its plane is $I_{cm} = Mr^2$.
According to the parallel axis theorem,$I = I_{cm} + Md^2$,where $d$ is the distance between the parallel axes.
For a tangent in the plane of the ring,the distance $d$ between the center of mass axis and the tangent axis is equal to the radius $r$.
Substituting the values,we get $I = Mr^2 + M(r)^2 = Mr^2 + Mr^2 = 2Mr^2$.
Wait,the question asks for a tangent in its own plane. The moment of inertia about a diameter is $I_{diameter} = \frac{1}{2}Mr^2$.
Using the parallel axis theorem for the tangent in the plane: $I = I_{diameter} + Md^2 = \frac{1}{2}Mr^2 + Mr^2 = \frac{3}{2}Mr^2$.

(A) The moment of inertia of a disc about its central axis is given by $I = \frac{1}{2}MR^2$,which implies $MR^2 = 2I$.
According to the theorem of parallel axes,the moment of inertia about a tangent in its plane is given by $I_{tangent} = I_{diameter} + MR^2$.
The moment of inertia about the diameter is $I_{diameter} = \frac{1}{4}MR^2 = \frac{1}{2}I$.
Substituting these values,we get $I_{tangent} = \frac{1}{2}I + 2I = \frac{5}{2}I$.

(C) The moment of inertia of a ring about its natural axis (passing through the center and perpendicular to the plane) is $I_{cm} = mr^2$.
According to the parallel axis theorem,the moment of inertia about an axis passing through the edge and parallel to the natural axis is given by $I = I_{cm} + mr^2$.
Substituting the values,we get $I = mr^2 + mr^2 = 2mr^2$.
Given $m = 3 \ gm$ and $r = 1 \ cm$,we have $I = 2 \times 3 \times (1)^2 = 6 \ gm \cdot cm^2$.

(D) The axis passing through the surface of the cylinder and parallel to its central axis is called the generator axis.
According to the parallel axis theorem,the moment of inertia $I$ about this axis is given by:
$I = I_{cm} + Md^2$
Here,$I_{cm} = \frac{1}{2}MR^2$ is the moment of inertia about the central axis of the cylinder,and $d = R$ is the perpendicular distance between the two parallel axes.
Therefore,$I = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.

(A) The moment of inertia of a disc about its diameter is $I_{diameter} = \frac{1}{4}MR^2$.
According to the parallel axis theorem,the moment of inertia about an axis parallel to the diameter and passing through the rim (tangent in the plane) is given by $I = I_{cm} + Md^2$.
Here,the distance $d$ between the diameter and the tangent is $R$.
Substituting the values,we get $I = \frac{1}{4}MR^2 + MR^2 = \frac{5}{4}MR^2$.

(A) According to the perpendicular axis theorem,the moment of inertia of a planar body about an axis perpendicular to its plane $(I_z)$ is the sum of the moments of inertia about two mutually perpendicular axes in its plane ($I_x$ and $I_y$) passing through the same point.
For a rectangular plate of mass $M$,length $l$,and breadth $b$:
The moment of inertia about an axis passing through the center and parallel to the breadth is $I_x = \frac{Ml^2}{12}$.
The moment of inertia about an axis passing through the center and parallel to the length is $I_y = \frac{Mb^2}{12}$.
By the perpendicular axis theorem,$I_z = I_x + I_y$.
Therefore,$I_z = \frac{Ml^2}{12} + \frac{Mb^2}{12} = \frac{M}{12}(l^2 + b^2)$.

(A) The moment of inertia of a disc about its diameter is $I_d = \frac{1}{4}MR^2$.
Using the parallel axis theorem,the moment of inertia about a tangent parallel to its plane is $I = I_d + MR^2 = \frac{1}{4}MR^2 + MR^2 = \frac{5}{4}MR^2$.
From this,we find $MR^2 = \frac{4}{5}I$.
Now,the moment of inertia of a disc about its central axis perpendicular to its plane is $I_c = \frac{1}{2}MR^2$.
Using the parallel axis theorem,the moment of inertia about a tangent perpendicular to its plane is $I' = I_c + MR^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.
Substituting $MR^2 = \frac{4}{5}I$ into the expression for $I'$,we get $I' = \frac{3}{2} \times (\frac{4}{5}I) = \frac{6}{5}I$.

(C) The moment of inertia of a solid cylinder about its central axis is given by $I_{cm} = \frac{1}{2}MR^2$.
According to the parallel axis theorem,the moment of inertia $I$ about an axis parallel to the central axis at a distance $d$ is given by $I = I_{cm} + Md^2$.
Here,the axis is on the surface of the cylinder,so the distance $d$ from the central axis is equal to the radius $R$.
Substituting the values,we get $I = \frac{1}{2}MR^2 + M(R)^2$.
$I = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.
Therefore,the correct option is $C$.

(A) The moment of inertia of a uniform circular ring about an axis passing through its center and perpendicular to its plane is $I_{cm} = MR^2$.
Using the parallel axis theorem,$I = I_{cm} + Md^2$,where $d$ is the distance between the center of mass and the tangential axis.
Here,the distance $d$ is equal to the radius $R$.
Substituting the values,we get $I = MR^2 + M(R)^2$.
Therefore,$I = 2MR^2$.

(D) The moment of inertia of a solid sphere of mass $M$ and radius $R$ about its diameter is given by:
$I_{\text{diameter}} = \frac{2}{5} M R^2$
According to the parallel axis theorem,the moment of inertia about a tangent is:
$I_{\text{tangent}} = I_{\text{CM}} + M d^2$
Here,the distance $d$ between the center of mass and the tangent is $R$.
So,$I_{\text{tangent}} = \frac{2}{5} M R^2 + M R^2$
$I_{\text{tangent}} = \frac{7}{5} M R^2$

(D) According to the perpendicular axis theorem,for a planar body,the moment of inertia about an axis perpendicular to its plane $(I_z)$ is equal to the sum of the moments of inertia about two mutually perpendicular axes in its plane ($I_x$ and $I_y$):
$I_z = I_x + I_y$
For a circular ring,by symmetry,the moment of inertia about any diameter is the same,so $I_x = I_y = I_D$.
Given $I_z = 200 \, gm \cdot cm^2$.
Substituting these into the theorem:
$200 = I_D + I_D = 2I_D$
$I_D = \frac{200}{2} = 100 \, gm \cdot cm^2$.
Thus,the correct option is $D$.

(A) The moment of inertia of the full disc about point $O$ is:
$I_{full} = \frac{1}{2}(9M)R^2 = \frac{9}{2}MR^2$ . . . $(i)$
The radius of the removed disc is $r = R/3$. Since the mass is proportional to the area $(M \propto R^2)$,the mass of the removed disc is $m = 9M \times (r/R)^2 = 9M \times (1/3)^2 = M$.
The moment of inertia of the removed disc about its own center of mass is:
$I_{cm} = \frac{1}{2}m r^2 = \frac{1}{2}M(R/3)^2 = \frac{1}{18}MR^2$.
Using the parallel axis theorem,the moment of inertia of the removed disc about point $O$ is:
$I_{removed} = I_{cm} + m d^2$,where $d = 2R/3$ is the distance from $O$ to the center of the removed disc.
$I_{removed} = \frac{1}{18}MR^2 + M(2R/3)^2 = \frac{1}{18}MR^2 + \frac{4}{9}MR^2 = \frac{1+8}{18}MR^2 = \frac{9}{18}MR^2 = \frac{1}{2}MR^2$.
The moment of inertia of the remaining disc is:
$I_{remaining} = I_{full} - I_{removed} = \frac{9}{2}MR^2 - \frac{1}{2}MR^2 = \frac{8}{2}MR^2 = 4MR^2$.

(D) To find the moment of inertia about the given axis,we use the parallel axis theorem: $I = I_{\text{cm}} + M d^2$.
Here,$I_{\text{cm}}$ is the moment of inertia about the center of mass,which is $\frac{ML^2}{12}$.
The distance $d$ between the center of mass and the axis is the difference between the distance from the end to the center of mass $(L/2)$ and the distance from the end to the axis $(L/4)$:
$d = \frac{L}{2} - \frac{L}{4} = \frac{L}{4}$.
Substituting these values into the theorem:
$I = \frac{ML^2}{12} + M \left(\frac{L}{4}\right)^2$
$I = \frac{ML^2}{12} + \frac{ML^2}{16}$
Taking the least common multiple of $12$ and $16$,which is $48$:
$I = \frac{4ML^2 + 3ML^2}{48} = \frac{7ML^2}{48}$.

(A) For a uniform square plate,the moment of inertia about any axis passing through the centre and lying in the plane of the plate is the same,provided the axis is symmetric with respect to the square's geometry.
Let $I_Z$ be the moment of inertia about an axis passing through the centre and perpendicular to the plane of the square.
According to the perpendicular axis theorem,$I_Z = I_{AB} + I_{A'B'}$,where $AB$ and $A'B'$ are two mutually perpendicular axes in the plane of the plate.
Since the square is symmetric,$I_{AB} = I_{A'B'} = I$. Thus,$I_Z = 2I$.
Similarly,for any two mutually perpendicular axes $CD$ and $C'D'$ in the plane of the plate passing through the centre,$I_Z = I_{CD} + I_{C'D'}$.
Due to the rotational symmetry of the square,the moment of inertia about any axis in the plane passing through the centre is constant and equal to $I$. Therefore,$I_{CD} = I$.

(D) According to the theorem of perpendicular axes,the moment of inertia $I_z$ of a planar body about an axis perpendicular to its plane and passing through a point $O$ is equal to the sum of the moments of inertia about two mutually perpendicular axes in the plane of the body,both passing through $O$.
In the given figure,axes $1$ and $2$ are the diagonals of the square,which are mutually perpendicular. Thus,$I_0 = I_1 + I_2$.
Axes $3$ and $4$ are lines passing through the centre $O$ and parallel to the sides of the square. These are also mutually perpendicular. Thus,$I_0 = I_3 + I_4$.
Furthermore,due to the symmetry of the square,the moment of inertia about any axis passing through the centre and lying in the plane of the plate is the same if the angle of the axis is the same relative to the sides. Specifically,for a square plate,$I_1 = I_2 = I_3 = I_4 = \frac{ML^2}{12}$.
Therefore,$I_0 = I_1 + I_2 = I_3 + I_4 = I_1 + I_3$ (since $I_1 = I_3$).
Thus,all the given options are correct.

(C) The moment of inertia of the body about the axis passing through the center of mass $G$ is given by $I_{cm} = mK^2$.
According to the parallel axis theorem,the moment of inertia $I_{new}$ about the new parallel axis at a distance '$a$' from the center of mass is:
$I_{new} = I_{cm} + ma^2 = mK^2 + ma^2 = m(K^2 + a^2)$.
The rotational kinetic energy $K_R$ about the new axis is given by:
$K_R = \frac{1}{2} I_{new} \omega^2 = \frac{1}{2} m(K^2 + a^2) \omega^2$.

(B) The moment of inertia $(I_1)$ of a uniform cylinder of mass $M$,radius $R$,and length $L$ about its central longitudinal axis (passing through the centre and normal to the circular face) is given by $I_1 = \frac{1}{2} M R^2$.
The moment of inertia $(I_2)$ of the same cylinder about an axis passing through its centre and perpendicular to its length (transverse axis) is given by $I_2 = M \left( \frac{L^2}{12} + \frac{R^2}{4} \right)$.
According to the problem,$I_1 = I_2$,so:
$\frac{1}{2} M R^2 = M \left( \frac{L^2}{12} + \frac{R^2}{4} \right)$
Dividing both sides by $M$:
$\frac{R^2}{2} = \frac{L^2}{12} + \frac{R^2}{4}$
Subtracting $\frac{R^2}{4}$ from both sides:
$\frac{R^2}{4} = \frac{L^2}{12}$
Multiplying by $12$:
$3 R^2 = L^2$
Taking the square root of both sides:
$L = \sqrt{3} R$.

(B) The moment of inertia of a rectangular block of mass $m$,length $l$,breadth $b$,and thickness $t$ about the axes passing through its center of mass is given by:
$I_x = \frac{m}{12}(b^2 + t^2)$
$I_y = \frac{m}{12}(l^2 + t^2)$
$I_z = \frac{m}{12}(l^2 + b^2)$
Since $l > b > t$,it follows that $(l^2 + b^2) > (l^2 + t^2) > (b^2 + t^2)$.
Therefore,$I_z > I_y > I_x$.
Thus,the moment of inertia is maximum about the $z-z$ axis.

(B) Consider a right-angled triangle $ABC$ with mass $M$ and sides $AC = BC = a$. The hypotenuse $AB = a\sqrt{2}$.
If we join another identical triangle of mass $M$ along the hypotenuse $AB$,we form a square plate of side $a$ and total mass $2M$.
The axis passing through the midpoint $O$ of $AB$ and perpendicular to the plane of the square is the axis passing through the center of the square.
The moment of inertia of a square plate of side $L$ and mass $M_{total}$ about an axis passing through its center and perpendicular to its plane is $I = \frac{1}{6} M_{total} L^2$.
Here,$M_{total} = 2M$ and $L = a$.
Thus,the moment of inertia of the square plate is $I_{square} = \frac{1}{6} (2M) a^2 = \frac{Ma^2}{3}$.
Since the square is composed of two identical triangles,by the principle of superposition,the moment of inertia of the square is the sum of the moments of inertia of the two triangles about the same axis.
$I_{square} = I_{triangle1} + I_{triangle2} = 2 I_{triangle}$.
Therefore,$I_{triangle} = \frac{1}{2} I_{square} = \frac{1}{2} \left( \frac{Ma^2}{3} \right) = \frac{Ma^2}{6}$.

(D) Let the mass of the square plate be $M$ and the side length be $L$.
$1$. The axis $AOC$ passes through the center of the square and lies in the plane of the plate along the diagonal. The moment of inertia of a square plate about its diagonal is $I_1 = \frac{1}{12} M L^2$.
$2$. The axes $xDx'$ and $yBy'$ are parallel to the diagonal $AOC$ and pass through the corners $D$ and $B$ respectively.
$3$. The perpendicular distance between the diagonal $AOC$ and the parallel axes $xDx'$ or $yBy'$ is $d = \frac{L}{\sqrt{2}}$.
$4$. Using the parallel axis theorem,$I = I_{cm} + M d^2$,where $I_{cm} = I_1 = \frac{1}{12} M L^2$.
$5$. Thus,$I_2 = I_3 = \frac{1}{12} M L^2 + M \left( \frac{L}{\sqrt{2}} \right)^2 = \frac{1}{12} M L^2 + \frac{1}{2} M L^2 = \left( \frac{1 + 6}{12} \right) M L^2 = \frac{7}{12} M L^2$.
$6$. The ratio $I_1 : I_2 : I_3 = \frac{1}{12} M L^2 : \frac{7}{12} M L^2 : \frac{7}{12} M L^2 = 1 : 7 : 7$.

(B) The triangular sheet is a right-angled isosceles triangle with legs of length $l$. The area of the triangle is $A = \frac{1}{2} l^2$.
Let the mass per unit area be $\sigma = \frac{M}{A} = \frac{2M}{l^2}$.
The moment of inertia of a thin triangular plate about an axis passing through one of its sides of length $b$ and height $h$ is given by $I = \frac{Mh^2}{6}$.
In this case,for the axis $PR$,the base is the hypotenuse $PR = \sqrt{l^2 + l^2} = l\sqrt{2}$.
The height $h$ of the triangle from vertex $Q$ to the hypotenuse $PR$ is $h = \frac{l}{\sqrt{2}}$.
Substituting these values into the formula $I = \frac{Mh^2}{6}$:
$I = \frac{M (l/\sqrt{2})^2}{6} = \frac{M (l^2/2)}{6} = \frac{Ml^2}{12}$.

(D) Let the center of the semicircular ring be $O$. The moment of inertia of a complete ring of mass $2M$ and radius $R$ about an axis passing through its center and perpendicular to its plane is $I_{cm} = (2M)R^2$.
By symmetry,the moment of inertia of the semicircular ring of mass $M$ about the axis passing through $O$ and perpendicular to the plane is $I_O = MR^2$.
Now,we use the parallel axis theorem to find the moment of inertia about an axis passing through point $A$ and perpendicular to the plane. The distance between the center $O$ and point $A$ is $R$.
The parallel axis theorem states $I_A = I_O + Md^2$,where $d = R$.
Substituting the values,we get $I_A = MR^2 + M(R)^2 = 2MR^2$.

(B) The moment of inertia of a square plate about an axis passing through its center of mass and parallel to an edge is $I_{CM} = \frac{M L^{2}}{12}$.
Using the parallel axis theorem,the moment of inertia about an axis passing through the center of mass but rotated by an angle $\theta$ in the plane of the plate is $I_{\theta} = I_{CM} = \frac{M L^{2}}{12}$ (since for a square,the moment of inertia about any axis in the plane passing through the center of mass is the same).
Now,we use the parallel axis theorem: $I = I_{CM} + M d^{2}$,where $d$ is the perpendicular distance from the center of mass to the given axis.
The distance $d$ from the center of the square to the axis passing through a vertex at an angle of $15^{\circ}$ with the horizontal is $d = \frac{L}{\sqrt{2}} \sin(45^{\circ} + 15^{\circ}) = \frac{L}{\sqrt{2}} \sin(60^{\circ}) = \frac{L}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3} L}{2 \sqrt{2}}$.
Thus,$I = \frac{M L^{2}}{12} + M \left( \frac{\sqrt{3} L}{2 \sqrt{2}} \right)^{2} = \frac{M L^{2}}{12} + M \left( \frac{3 L^{2}}{8} \right) = \frac{M L^{2} + 4.5 M L^{2}}{12} = \frac{M L^{2}}{12} + \frac{9 M L^{2}}{24} = \frac{2 M L^{2} + 9 M L^{2}}{24} = \frac{11 M L^{2}}{24}$.

(C) Let the height of the triangular plate be $h$. Since the apex angle is $90^o$ and the triangle is isosceles,the height $h$ is half the base $L$,so $h = L/2$.
The area density of the plate is $\sigma = \frac{M}{\text{Area}} = \frac{M}{\frac{1}{2} \times L \times h} = \frac{M}{\frac{1}{2} \times L \times (L/2)} = \frac{4M}{L^2}$.
Consider a thin strip of thickness $dy$ at a distance $y$ from the apex. The length of the strip $l(y)$ at height $y$ is given by $l(y) = 2y$ (since the apex angle is $90^o$,the sides make $45^o$ with the vertical).
The mass of this strip is $dm = \sigma \times l(y) \times dy = \frac{4M}{L^2} \times 2y \times dy = \frac{8M}{L^2} y dy$.
The moment of inertia of this strip about the base (which is at $y = h = L/2$) is $dI = dm \times (h - y)^2$.
Integrating from $y = 0$ to $y = L/2$:
$I = \int_0^{L/2} \frac{8M}{L^2} y (L/2 - y)^2 dy = \frac{8M}{L^2} \int_0^{L/2} (y \frac{L^2}{4} - Ly^2 + y^3) dy$.
$I = \frac{8M}{L^2} [\frac{L^2}{4} \frac{y^2}{2} - L \frac{y^3}{3} + \frac{y^4}{4}]_0^{L/2} = \frac{8M}{L^2} [\frac{L^2}{4} \frac{L^2}{8} - L \frac{L^3}{24} + \frac{L^4}{64}] = \frac{8M}{L^2} [\frac{L^4}{32} - \frac{L^4}{24} + \frac{L^4}{64}] = \frac{8M}{L^2} [\frac{6L^4 - 8L^4 + 3L^4}{192}] = \frac{8M}{L^2} \frac{L^4}{192} = \frac{ML^2}{24}$.

(D) Let the square lamina have mass $M$ and side length $a$. The moment of inertia of a square lamina about an axis passing through its centre $O$ and perpendicular to its plane $(I_z)$ is given by $I_z = I_x + I_y$,where $I_x$ and $I_y$ are moments of inertia about the axes passing through the centre and parallel to the sides. By symmetry,$I_x = I_y = I_{EF} = I_{GH}$ (where $EF$ and $GH$ are axes passing through the midpoints of opposite sides). Thus,$I_z = 2 I_{EF}$.
Similarly,the moment of inertia about the diagonal axes $AC$ and $BD$ are equal by symmetry,so $I_{AC} = I_{BD}$. According to the perpendicular axes theorem,$I_z = I_{AC} + I_{BD} = 2 I_{AC}$.
Equating the two expressions for $I_z$,we get $2 I_{EF} = 2 I_{AC}$,which simplifies to $I_{AC} = I_{EF}$.

(A) The moment of inertia of a square plate of side '$a$' and mass '$m$' about an axis passing through its center of mass and perpendicular to its plane is given by:
$I_{cm} = \frac{1}{12}m(a^2 + a^2) = \frac{ma^2}{6}$
According to the parallel axis theorem,the moment of inertia about an axis passing through a corner and perpendicular to the plane is:
$I = I_{cm} + md^2$
Here,'$d$' is the distance from the center of the square to the corner. For a square of side '$a$',the diagonal is '$a\sqrt{2}$',so the distance from the center to a corner is:
$d = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$
Substituting the values:
$I = \frac{ma^2}{6} + m\left(\frac{a}{\sqrt{2}}\right)^2$
$I = \frac{ma^2}{6} + \frac{ma^2}{2}$
$I = \frac{ma^2 + 3ma^2}{6} = \frac{4ma^2}{6} = \frac{2}{3}ma^2$

(A) The moment of inertia of a uniform solid cylinder of mass $m$,length $l$,and radius $R$ about its perpendicular bisector is given by:
$I = \frac{mR^2}{4} + \frac{ml^2}{12}$
Assuming the density $\rho$ is constant,the mass $m = \rho V = \rho (\pi R^2 l)$.
Substituting $m$ in the expression for $I$:
$I = \frac{\rho \pi R^2 l R^2}{4} + \frac{\rho \pi R^2 l^3}{12} = \frac{\rho \pi R^4 l}{4} + \frac{\rho \pi R^2 l^3}{12}$
Since the volume $V = \pi R^2 l$ is constant,we can write $R^2 = \frac{V}{\pi l}$. Substituting this into the expression for $I$:
$I = \frac{m}{4} \left( \frac{V}{\pi l} + \frac{l^2}{3} \right)$
To find the minimum moment of inertia,we differentiate $I$ with respect to $l$ and set it to zero:
$\frac{dI}{dl} = \frac{m}{4} \left( -\frac{V}{\pi l^2} + \frac{2l}{3} \right) = 0$
$\frac{V}{\pi l^2} = \frac{2l}{3} \implies V = \frac{2\pi l^3}{3}$
Since $V = \pi R^2 l$,we have:
$\pi R^2 l = \frac{2\pi l^3}{3} \implies R^2 = \frac{2l^2}{3} \implies \frac{l^2}{R^2} = \frac{3}{2}$
Therefore,the ratio is $\frac{l}{R} = \sqrt{\frac{3}{2}}$.

(D) Let $\sigma$ be the mass per unit area of the disc.
The total mass of the original disc is $M_{total} = 9M$.
The radius of the original disc is $R$.
The mass of the removed small disc of radius $r = \frac{R}{3}$ is:
$m = \sigma \times \pi r^2 = \sigma \times \pi \left(\frac{R}{3}\right)^2 = \frac{\sigma \pi R^2}{9} = \frac{M_{total}}{9} = \frac{9M}{9} = M$.
The moment of inertia of the complete disc of mass $9M$ about an axis passing through its centre $O$ and perpendicular to its plane is:
$I_1 = \frac{1}{2}(9M)R^2 = \frac{9}{2}MR^2$.
The moment of inertia of the removed disc of mass $M$ about its own centre $O'$ is:
$I_{O'} = \frac{1}{2}M\left(\frac{R}{3}\right)^2 = \frac{1}{18}MR^2$.
Using the parallel axis theorem,the moment of inertia of the removed disc about the axis passing through $O$ is:
$I_2 = I_{O'} + M d^2$,where $d = \frac{2R}{3}$ is the distance between $O$ and $O'$.
$I_2 = \frac{1}{18}MR^2 + M\left(\frac{2R}{3}\right)^2 = \frac{1}{18}MR^2 + \frac{4}{9}MR^2 = \left(\frac{1+8}{18}\right)MR^2 = \frac{9}{18}MR^2 = \frac{1}{2}MR^2$.
The moment of inertia of the remaining disc is:
$I = I_1 - I_2 = \frac{9}{2}MR^2 - \frac{1}{2}MR^2 = \frac{8}{2}MR^2 = 4MR^2$.

(B) The moment of inertia of a uniform disc of mass $M$ and radius $R$ about the $z$-axis (passing through the centre and perpendicular to the plane) is $I_z = \frac{1}{2}MR^2$.
The line is given by $x - y + c = 0$. The perpendicular distance $d$ from the origin $(0,0)$ to this line is $d = \frac{|0 - 0 + c|}{\sqrt{1^2 + (-1)^2}} = \frac{|c|}{\sqrt{2}}$.
According to the parallel axis theorem,the moment of inertia about any axis parallel to a diameter is $I = I_{cm} + Md^2$,where $I_{cm}$ is the moment of inertia about the diameter,which is $I_{cm} = \frac{1}{4}MR^2$.
Thus,the moment of inertia about the line $y = x + c$ is $I = \frac{1}{4}MR^2 + M\left(\frac{c}{\sqrt{2}}\right)^2 = \frac{1}{4}MR^2 + \frac{Mc^2}{2}$.
Equating the two moments of inertia:
$\frac{1}{2}MR^2 = \frac{1}{4}MR^2 + \frac{Mc^2}{2}$
$\frac{1}{4}MR^2 = \frac{Mc^2}{2}$
$c^2 = \frac{R^2}{2}$
$c = \pm \frac{R}{\sqrt{2}}$.
Since the options provide $R/\sqrt{2}$,the correct value is $R/\sqrt{2}$.

(D) The given triangle $ABC$ is an isosceles right-angled triangle with base $BC = a$ and height $h = a/2$.
Consider a square of side $a$ and mass $4M$ formed by four such identical triangular plates.
The moment of inertia of this square about an axis passing through its center and perpendicular to its plane is $I_{square} = \frac{1}{6}(4M)a^2 = \frac{2}{3}Ma^2$.
By the parallel axis theorem,the moment of inertia of the square about an axis passing through one of its vertices (say $A$) and perpendicular to its plane is $I_{vertex} = I_{CM} + (4M)d^2$,where $d$ is the distance from the center to the vertex,$d^2 = (a/2)^2 + (a/2)^2 = a^2/2$.
So,$I_{vertex} = \frac{1}{6}(4M)a^2 + (4M)(a^2/2) = \frac{2}{3}Ma^2 + 2Ma^2 = \frac{8}{3}Ma^2$.
Since the square consists of four identical triangles,the moment of inertia of one triangle about the same axis passing through $A$ is $I_{triangle} = \frac{1}{4} I_{vertex} = \frac{1}{4} \times \frac{8}{3}Ma^2 = \frac{2}{3}Ma^2$.
However,checking the standard result for a triangle of base $a$ and height $h$ about an axis through the apex perpendicular to the plane,$I = M(\frac{h^2}{6} + \frac{a^2}{24})$. With $h = a/2$,$I = M(\frac{a^2}{24} + \frac{a^2}{24}) = M(\frac{2a^2}{24}) = \frac{Ma^2}{12}$.

$(a)$ The perpendicular axis theorem $I_z = I_x + I_y$ is only applicable to planar (two-dimensional) objects. A cube is a three-dimensional object, so this is incorrect.
$(b)$ The $A$-axis is parallel to the $z$-axis passing through the center $O$. The distance between the $z$-axis and the $A$-axis is $d = \sqrt{(\frac{a}{2})^2 + (\frac{a}{2})^2} = \frac{a}{\sqrt{2}}$. By the parallel axis theorem, $I_A = I_{cm} + md^2 = I_z + m(\frac{a}{\sqrt{2}})^2 = I_z + \frac{ma^2}{2}$. This is correct.
$(c)$ The $B$-axis is not parallel to the $z$-axis, so the parallel axis theorem cannot be applied in this manner. This is incorrect.
$(d)$ Due to the symmetry of the cube about the center $O$, the moment of inertia about the $x, y,$ and $z$ axes passing through the center are equal, i.e., $I_x = I_y = I_z$. Thus, $I_x = I_z$ is correct.

(B) The center of mass $(COM)$ of a uniform hollow hemisphere of radius $R$ lies at a distance of $R/2$ from the center of its base along the axis of symmetry.
Both axes $I_A$ and $I_B$ are parallel to the diameter of the base passing through the $COM$.
Axis $I_A$ is at a distance of $R/2$ from the $COM$.
Axis $I_B$ is at a distance of $R/2$ from the $COM$.
According to the parallel axis theorem,$I = I_{COM} + Md^2$.
Since both axes are at the same distance $d = R/2$ from the $COM$,the moment of inertia about both axes is equal.
Therefore,$I_A = I_B = I_{COM} + M(R/2)^2$.

(B) For a semicircular disc of mass $M$ and radius $R$,the moment of inertia about the diameter (axis $XX'$) is $I_x = \frac{1}{4} M R^2$.
Since the disc is symmetric about the axis perpendicular to the diameter in its plane (axis $AA'$),the moment of inertia about this axis is also $I_y = \frac{1}{4} M R^2$.
The product of inertia $I_{xy}$ for a semicircular disc about these axes is zero due to symmetry.
The moment of inertia $I$ about an axis passing through the center and making an angle $\theta$ with the diameter is given by the formula:
$I = I_x \sin^2 \theta + I_y \cos^2 \theta - 2 I_{xy} \sin \theta \cos \theta$
Substituting $I_x = \frac{1}{4} M R^2$,$I_y = \frac{1}{4} M R^2$,and $I_{xy} = 0$:
$I = \frac{1}{4} M R^2 \sin^2 \theta + \frac{1}{4} M R^2 \cos^2 \theta$
$I = \frac{1}{4} M R^2 (\sin^2 \theta + \cos^2 \theta)$
$I = \frac{1}{4} M R^2$

(D) The moment of inertia of a ring of mass $M$ and radius $R$ about its diameter is given by $I = \frac{1}{2}MR^2$.
According to the parallel axis theorem,the moment of inertia $I_{tangent}$ about a tangent parallel to the diameter is $I_{tangent} = I_{cm} + Md^2$,where $I_{cm}$ is the moment of inertia about the center of mass (diameter) and $d = R$ is the distance between the parallel axes.
Substituting the values,we get $I_{tangent} = I + MR^2$.
Since $I = \frac{1}{2}MR^2$,we have $MR^2 = 2I$.
Therefore,$I_{tangent} = I + 2I = 3I$.

(A) The moment of inertia of a uniform disc of mass $M$ and radius $R$ about the $z$-axis (passing through the center and perpendicular to the plane) is $I_z = \frac{1}{2}MR^2$.
The line $y = x + c$ can be rewritten as $x - y + c = 0$. The perpendicular distance $d$ of this line from the origin $(0,0)$ is $d = \frac{|0 - 0 + c|}{\sqrt{1^2 + (-1)^2}} = \frac{|c|}{\sqrt{2}}$.
According to the parallel axis theorem,the moment of inertia about any line in the $x-y$ plane parallel to a diameter is $I = I_{cm} + Md^2$,where $I_{cm}$ is the moment of inertia about a diameter,which is $\frac{1}{4}MR^2$.
Thus,$I_{line} = \frac{1}{4}MR^2 + M\left(\frac{c}{\sqrt{2}}\right)^2 = \frac{1}{4}MR^2 + \frac{Mc^2}{2}$.
Given that $I_{line} = I_z$,we have $\frac{1}{4}MR^2 + \frac{Mc^2}{2} = \frac{1}{2}MR^2$.
Subtracting $\frac{1}{4}MR^2$ from both sides,we get $\frac{Mc^2}{2} = \frac{1}{4}MR^2$.
Simplifying,$c^2 = \frac{R^2}{2}$,which gives $c = \pm \frac{R}{\sqrt{2}}$.

(A) The moment of inertia of a uniform circular disc about an axis passing through its centre and perpendicular to its plane is $I_{G} = \frac{1}{2}MR^2$.
By the theorem of parallel axes,the moment of inertia about an axis passing through the edge and normal to the disc is given by $I = I_{G} + Md^2$,where $d = R$ is the distance between the parallel axes.
Substituting the values,we get $I = \frac{1}{2}MR^2 + M(R)^2 = \frac{3}{2}MR^2$.

(B) For a planar object,the perpendicular axes theorem states that the moment of inertia about an axis perpendicular to the plane $(z-z')$ is the sum of the moments of inertia about two mutually perpendicular axes in the plane ($x-x'$ and $y-y'$).
Thus,$I_{zz'} = I_{xx'} + I_{yy'}$.
Since $I_{xx'}$ and $I_{yy'}$ are both positive quantities,it follows that $I_{zz'} > I_{xx'}$ and $I_{zz'} > I_{yy'}$.
Therefore,the moment of inertia is maximum about the $z-z'$ axis.

(C) The moment of inertia of a circular disc of mass $m$ and radius $R$ about an axis passing through its center of mass $(C.M.)$ and perpendicular to its plane is given by:
$I_z = \frac{1}{2} mR^2$
The axis $z'$ is a tangential axis parallel to the $z$-axis. According to the parallel axis theorem,the moment of inertia about an axis parallel to the central axis is given by $I = I_{cm} + md^2$,where $d$ is the distance between the axes. Here,$d = R$.
$I_{z'} = I_z + mR^2 = \frac{1}{2} mR^2 + mR^2 = \frac{3}{2} mR^2$
The ratio of the moment of inertia about $z$ and $z'$ axes is:
$\frac{I_z}{I_{z'}} = \frac{\frac{1}{2} mR^2}{\frac{3}{2} mR^2} = \frac{1}{3}$
Thus,the ratio is $1:3$.

(D) For a thin uniform square sheet of side $a$ and mass $m$,the moment of inertia about an axis passing through the center and perpendicular to the plane is $I_z = \frac{ma^2}{6}$.
By the perpendicular axis theorem,$I_z = I_x + I_y$,where $I_x$ and $I_y$ are the moments of inertia about the two perpendicular axes passing through the center in the plane of the sheet.
Due to symmetry,$I_x = I_y = \frac{ma^2}{12}$.
Let $I_d$ be the moment of inertia about a diagonal. By the perpendicular axis theorem applied to the two diagonals (which are also perpendicular to each other and lie in the plane),$I_z = I_{d1} + I_{d2}$.
Since the square is symmetric,$I_{d1} = I_{d2} = I$.
Thus,$I_z = 2I$,which implies $I = \frac{I_z}{2} = \frac{ma^2/6}{2} = \frac{ma^2}{12}$.
Wait,let us re-evaluate: The moment of inertia of a square plate about an axis passing through the center and parallel to a side is $I_{side} = \frac{ma^2}{12}$.
The moment of inertia about a diagonal is $I_{diag} = \frac{ma^2}{12}$.
Therefore,$I = \frac{ma^2}{12}$.