Moment of Inertia and Radius of gyration Questions in English

(A) In translational motion,the property of a body to resist a change in its state of rest or uniform motion is called mass $(m)$.
In rotational motion,the property of a body to resist a change in its state of rotational motion is called the moment of inertia $(I)$.
The rotational analogue of Newton's second law $(F = ma)$ is $\tau = I\alpha$,where $\tau$ is torque,$I$ is the moment of inertia,and $\alpha$ is angular acceleration.
Thus,the moment of inertia plays the same role in rotational motion as mass does in translational motion.
Therefore,the correct option is $A$.

(D) Let the mass per unit length of the wire be $\lambda$.
The circumference of loop $P$ is $2\pi r$,so its mass is $M_P = \lambda(2\pi r)$.
The moment of inertia of loop $P$ about an axis passing through its center and perpendicular to its plane is $I_P = M_P r^2 = \lambda(2\pi r)r^2 = 2\pi\lambda r^3$.
The circumference of loop $Q$ is $2\pi(nr)$,so its mass is $M_Q = \lambda(2\pi nr) = n M_P$.
The moment of inertia of loop $Q$ about a similar axis is $I_Q = M_Q (nr)^2 = (n M_P)(n^2 r^2) = n^3 M_P r^2 = n^3 I_P$.
Given that $I_Q = 8 I_P$,we have $n^3 I_P = 8 I_P$.
Therefore,$n^3 = 8$,which gives $n = 2$.

(C) The moment of inertia of a sphere about an axis passing through its centre is $I_{cm} = \frac{2}{5}MR^2$.
Using the parallel axis theorem,the moment of inertia $I$ about a tangent axis is $I = I_{cm} + MR^2$.
Substituting the values,$I = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2$.
The radius of gyration $K$ is defined by $I = MK^2$.
Therefore,$MK^2 = \frac{7}{5}MR^2$.
Solving for $K$,we get $K = \sqrt{\frac{7}{5}}R$.

(A) The distance of each particle from the center $P$ of the square is $r = \frac{l}{\sqrt{2}}$.
The moment of inertia $I$ of the system about an axis passing through the center $P$ and perpendicular to the plane is given by:
$I = \sum m_i r_i^2 = 4 \times m \times r^2 = 4m \left( \frac{l}{\sqrt{2}} \right)^2 = 4m \left( \frac{l^2}{2} \right) = 2ml^2$.
The radius of gyration $K$ is defined by the relation $I = MK^2$,where $M$ is the total mass of the system.
Here,$M = 4m$,so:
$4mK^2 = 2ml^2$
$K^2 = \frac{2ml^2}{4m} = \frac{l^2}{2}$
$K = \frac{l}{\sqrt{2}}$.

(B) The radius of gyration $K$ of a body is defined by the relation $I = MK^2$,where $I$ is the moment of inertia and $M$ is the total mass of the body.
From this,$K = \sqrt{I/M}$.
The moment of inertia $I$ of a body depends on the distribution of mass relative to the axis of rotation.
Since $K$ is derived from $I$ and $M$,and $I$ itself accounts for the mass distribution relative to the axis,the radius of gyration $K$ depends on the mass distribution and the chosen axis of rotation.
Therefore,the correct option is $B$.

(B) The moment of inertia $(I)$ of a disc about an axis passing through its center of gravity and perpendicular to its plane is given by $I = \frac{1}{2}MR^2$.
By the definition of radius of gyration $(K)$,the moment of inertia is also expressed as $I = MK^2$.
Equating the two expressions: $MK^2 = \frac{1}{2}MR^2$.
Simplifying,we get $K^2 = \frac{R^2}{2}$,which implies $K = \frac{R}{\sqrt{2}}$.
Given $R = 2.5 \, cm$,we calculate $K = \frac{2.5}{\sqrt{2}} \approx \frac{2.5}{1.414} \approx 1.767 \, cm$.
Rounding to the nearest provided option,we get $1.76 \, cm$.

(A) The moment of inertia $(I)$ of a disc about an axis passing through its centre of gravity and perpendicular to its plane is given by $I = \frac{1}{2}MR^2$.
By definition,the radius of gyration $(K)$ is related to the moment of inertia by $I = MK^2$.
Equating the two expressions: $MK^2 = \frac{1}{2}MR^2$.
Canceling mass $(M)$ from both sides: $K^2 = \frac{R^2}{2}$.
Taking the square root: $K = \frac{R}{\sqrt{2}}$.
Given the radius $R = 5\,cm$,we have $K = \frac{5}{\sqrt{2}} \approx \frac{5}{1.414} \approx 3.535\,cm$.
Rounding to two decimal places,we get $K = 3.54\,cm$.

(A) The moment of inertia $(I)$ of a solid cylinder about its central longitudinal axis is given by the formula: $I = \frac{1}{2}MR^2$.
Given:
Mass $(M)$ = $500 \ g = 0.5 \ kg$.
Radius $(R)$ = $10 \ cm = 0.1 \ m$.
Substituting the values into the formula:
$I = \frac{1}{2} \times 0.5 \times (0.1)^2$
$I = 0.25 \times 0.01$
$I = 0.0025 \ kg \cdot m^2$
$I = 2.5 \times 10^{-3} \ kg \cdot m^2$.

(A) The moment of inertia of a thin circular disc about an axis passing through its center and perpendicular to its plane is $I_z = \frac{1}{2}MR^2$.
According to the perpendicular axis theorem,$I_z = I_x + I_y$.
Since the disc is symmetric,the moment of inertia about any diameter is the same,so $I_x = I_y = I_d$.
Therefore,$I_z = 2I_d$.
Substituting the value of $I_z$,we get $\frac{1}{2}MR^2 = 2I_d$.
Solving for $I_d$,we get $I_d = \frac{1}{4}MR^2$.

(A) The moment of inertia $I$ of a uniform thin rod of mass $M$ and length $L$ about an axis passing through its centre of mass and perpendicular to its length is given by $I = \frac{ML^2}{12}$.
By definition,the radius of gyration $K$ is related to the moment of inertia by the equation $I = MK^2$.
Equating the two expressions for $I$,we get $MK^2 = \frac{ML^2}{12}$.
Canceling $M$ from both sides,we have $K^2 = \frac{L^2}{12}$.
Taking the square root of both sides,we find $K = \frac{L}{\sqrt{12}}$.

(B) The moment of inertia $I$ of a ring about its central axis is given by $I = mR^2$,where $m$ is the mass and $R$ is the radius.
Since the diameter $D = 2R$,we can write $R = D/2$.
Substituting this into the formula,we get $I = m(D/2)^2 = m(D^2/4)$.
This implies $I \propto mD^2$,or $m \propto I/D^2$.
Given the ratios $I_1/I_2 = 2/1$ and $D_1/D_2 = 2/1$,we find the ratio of masses as:
$m_1/m_2 = (I_1/I_2) \times (D_2/D_1)^2$
$m_1/m_2 = (2/1) \times (1/2)^2 = 2 \times (1/4) = 2/4 = 1/2$.
Therefore,the ratio of their masses is $1:2$.

(B) The moment of inertia of a hollow sphere (thin spherical shell) about an axis passing through its center (diameter) is given by the standard formula $I = \frac{2}{3}MR^2$.
Here,$M$ is the mass of the hollow sphere and $R$ is its radius.
Therefore,the correct option is $B$.

(A) The moment of inertia $I$ of a system of point masses about an axis is given by $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the $i$-th mass from the axis.
For the given axis passing through $m_1$ and bisecting the side $m_2m_3$:
$1$. The distance of $m_1$ from the axis is $r_1 = 0$.
$2$. The distance of $m_2$ from the axis is $r_2 = a/2$.
$3$. The distance of $m_3$ from the axis is $r_3 = a/2$.
Therefore,the total moment of inertia is:
$I = m_1(0)^2 + m_2(a/2)^2 + m_3(a/2)^2$
$I = 0 + m_2(a^2/4) + m_3(a^2/4)$
$I = (m_2 + m_3) \frac{a^2}{4}$

(C) Let the vertices of the equilateral triangle be $A$,$B$,and $C$. Let the axis of rotation be the side $AB$.
The particles at $A$ and $B$ lie on the axis of rotation,so their perpendicular distances from the axis are $0$. Thus,their contribution to the moment of inertia is $0$.
The particle at $C$ is at a perpendicular distance $x$ from the side $AB$. In an equilateral triangle of side $a$,the altitude $x$ is given by $x = a \sin(60^\circ) = \frac{\sqrt{3}}{2} a$.
The moment of inertia of the system about side $AB$ is $I = \sum m_i r_i^2 = m(0)^2 + m(0)^2 + m(x)^2$.
$I = m x^2 = m \left( \frac{\sqrt{3}}{2} a \right)^2 = m \left( \frac{3}{4} a^2 \right) = \frac{3}{4} m a^2$.

(B) raw egg contains liquid inside,which does not rotate with the shell. Thus,it behaves like a spherical shell with mass concentrated near the surface. The moment of inertia of a spherical shell is $I_{raw} = \frac{2}{3}MR^2$.
$A$ half-boiled egg is more viscous and behaves more like a solid sphere as the contents rotate with the shell. The moment of inertia of a solid sphere is $I_{boiled} = \frac{2}{5}MR^2$.
Therefore,the ratio of the moment of inertia of the raw egg to the half-boiled egg is:
$\frac{I_{raw}}{I_{boiled}} = \frac{\frac{2}{3}MR^2}{\frac{2}{5}MR^2} = \frac{5}{3} = 1.67$.
Since $1.67 > 1$,the ratio is more than one.

(A) The moment of inertia $(I)$ of a uniform rod of mass $M$ and length $L$ about an axis passing through its center and perpendicular to its length is given by the formula:
$I = \frac{ML^2}{12}$
Given:
Mass $(M)$ = $0.12 \ kg$
Length $(L)$ = $1 \ m$
Substituting the values into the formula:
$I = \frac{0.12 \times (1)^2}{12}$
$I = \frac{0.12}{12}$
$I = 0.01 \ kg \cdot m^2$
Therefore,the correct option is $A$.

(C) Let the two rings be $1$ and $2$.
For ring $1$,the axis of rotation is perpendicular to its plane and passes through its center. The moment of inertia of ring $1$ about this axis is $I_1 = mr^2$.
For ring $2$,the axis of rotation lies in its plane and passes through its center (i.e.,it is a diameter of ring $2$). The moment of inertia of ring $2$ about this diameter is $I_2 = \frac{1}{2}mr^2$.
The total moment of inertia of the system about the given axis is $I = I_1 + I_2 = mr^2 + \frac{1}{2}mr^2 = \frac{3}{2}mr^2$.

(D) For a uniform rectangular plate of mass $M$,length $l$,and width $b$,the moment of inertia about an axis passing through its center and parallel to its length $l$ is calculated by treating the plate as a series of rods of length $l$ stacked along the width $b$.
The moment of inertia of a thin rod of mass $m$ and length $l$ about its center is $I = \frac{ml^2}{12}$.
For the rectangular plate,we consider an elemental strip of width $dy$ at a distance $y$ from the central axis parallel to the length. The mass of this strip is $dm = \frac{M}{b} dy$.
The moment of inertia of this strip about the central axis is $dI = (dm) y^2 = \left( \frac{M}{b} dy \right) y^2$.
Integrating this from $y = -b/2$ to $y = b/2$:
$I = \int_{-b/2}^{b/2} \frac{M}{b} y^2 dy = \frac{M}{b} \left[ \frac{y^3}{3} \right]_{-b/2}^{b/2} = \frac{M}{b} \left( \frac{b^3}{24} - (-\frac{b^3}{24}) \right) = \frac{M}{b} \left( \frac{2b^3}{24} \right) = \frac{Mb^2}{12}$.
Thus,the correct option is $(D)$.

(D) The moment of inertia of a circular disc is given by $I = \frac{1}{2}MR^2$.
Since mass $M = \text{density} (\rho) \times \text{volume} (V) = \rho \times (\pi R^2 t)$,where $t$ is the thickness.
Substituting $M$ into the formula: $I = \frac{1}{2} (\pi R^2 t \rho) R^2 = \frac{1}{2} \pi t \rho R^4$.
Given that the discs are made of the same material (same $\rho$) and have the same thickness $(t)$,we have $I \propto R^4$.
Since diameter $D = 2R$,it follows that $I \propto D^4$.
Therefore,$\frac{I_A}{I_B} = \left( \frac{D_A}{D_B} \right)^4$.
Given $D_A = 2 D_B$,we get $\frac{I_A}{I_B} = (2)^4 = 16$.
Thus,the moment of inertia of $A$ is $16$ times that of $B$.

(B) The moment of inertia $I$ of a circular ring about its central axis is given by $I = MR^2$,where $M$ is the mass and $R$ is the radius.
Since the ratio of diameters is $2:1$,the ratio of radii $R_1/R_2$ is also $2:1$.
Given $M_1/M_2 = 1/2$ and $R_1/R_2 = 2/1$.
The ratio of moments of inertia is $\frac{I_1}{I_2} = \frac{M_1 R_1^2}{M_2 R_2^2} = \left( \frac{M_1}{M_2} \right) \left( \frac{R_1}{R_2} \right)^2$.
Substituting the values: $\frac{I_1}{I_2} = \left( \frac{1}{2} \right) \times \left( \frac{2}{1} \right)^2 = \frac{1}{2} \times 4 = 2$.
Thus,the ratio is $2:1$.

(D) Let the mass of the rectangle be $M$. Let $AB = l$ and $BC = 2l$.
The moment of inertia of a rectangle about an axis passing through its center and parallel to its sides is given by $I = \frac{M d^2}{12}$,where $d$ is the side perpendicular to the axis.
$1$. For axis $HF$ (parallel to $AB$): The perpendicular side is $BC = 2l$. So,$I_{HF} = \frac{M(2l)^2}{12} = \frac{4Ml^2}{12} = \frac{Ml^2}{3}$.
$2$. For axis $EG$ (parallel to $BC$): The perpendicular side is $AB = l$. So,$I_{EG} = \frac{Ml^2}{12}$.
Comparing the two,$I_{EG} < I_{HF}$.
Since $EG$ passes through the center and is parallel to the longer side,it results in the minimum moment of inertia among the given options.

(A) The moment of inertia $(I)$ depends on the distribution of mass relative to the axis of rotation.
For a circular ring of mass $m$ and radius $R$ about an axis perpendicular to its plane,$I = mR^2$.
For a disc of mass $m$ and radius $R$ about an axis perpendicular to its plane,$I = \frac{1}{2}mR^2$.
For a solid sphere of mass $m$ and radius $R$ about its diameter,$I = \frac{2}{5}mR^2$.
For a rod of mass $m$ and length $L$ about its center,$I = \frac{1}{12}mL^2$.
Comparing these,the ring has the largest moment of inertia because all its mass is concentrated at the maximum distance $R$ from the axis.

(B) The moment of inertia of a rigid body is defined as $I = \sum m_i r_i^2$.
For a ring of mass $M$ and radius $R$,all mass elements are at a constant distance $R$ from the axis passing through the center and perpendicular to the plane.
Therefore,$I = \int r^2 dm = \int R^2 dm = R^2 \int dm$.
Since $\int dm = M$,the moment of inertia is $I = M R^2$.

(C) The moment of inertia is a physical quantity that represents the rotational inertia of a rigid body. It is the rotational analogue of mass in linear motion. It measures the resistance of an object to changes in its rotational motion about a specific axis. Therefore,it is effective during rotational motion.

(A) The moment of inertia $I$ of a flywheel (disk) about an axis passing through its center is given by $I = \frac{1}{2}MR^2$,where $M$ is the mass and $R$ is the radius.
Since the diameter $D = 2R$,we have $R = D/2$. Substituting this into the formula,$I = \frac{1}{2}M(D/2)^2 = \frac{1}{8}MD^2$.
This shows that $I \propto D^2$.
Using the concept of percentage error,if $D$ increases by $1\%$,the percentage change in $I$ is given by $\frac{\Delta I}{I} \times 100 = 2 \times (\frac{\Delta D}{D} \times 100)$.
Substituting the given value,$\frac{\Delta I}{I} \times 100 = 2 \times 1\% = 2\%$.
Therefore,the percentage increase in the moment of inertia is $2\%$.

(B) Given: Mass $m = 1\,kg$,Diameter $D = 0.2\,m$.
Radius $r = D/2 = 0.1\,m = 10^{-1}\,m$.
The moment of inertia of a circular plate about its diameter is given by the formula $I = \frac{1}{4}mr^2$.
Substituting the values:
$I = \frac{1}{4} \times 1\,kg \times (0.1\,m)^2$
$I = \frac{1}{4} \times 1 \times 0.01\,kg\cdot m^2$
$I = 0.0025\,kg\cdot m^2 = 2.5 \times 10^{-3}\,kg\cdot m^2$.
Thus,the correct option is $B$.

(A) The moment of inertia $(I)$ of a circular disc of mass $M$ and radius $R$ about an axis passing through its center and perpendicular to its plane is given by $I = \frac{1}{2}MR^2$.
By definition,the radius of gyration $(k)$ is related to the moment of inertia by the formula $I = Mk^2$.
Equating the two expressions for $I$:
$Mk^2 = \frac{1}{2}MR^2$
$k^2 = \frac{R^2}{2}$
$k = \frac{R}{\sqrt{2}}$
Therefore,the correct option is $A$.

(C) The moment of inertia of the system about the axis $AX$ is given by the sum of the moments of inertia of individual particles:
$I = I_A + I_B + I_C$
Since particle $A$ lies on the axis $AX$,its perpendicular distance $r_A = 0$. Thus,$I_A = m(0)^2 = 0$.
Particle $B$ is at a distance $l$ from $A$ along the line $AB$. Since $AX$ is perpendicular to $AB$,the perpendicular distance of $B$ from $AX$ is $r_B = l$. Thus,$I_B = m(l)^2 = m l^2$.
Particle $C$ forms an equilateral triangle with $A$ and $B$. The perpendicular distance of $C$ from the axis $AX$ is the horizontal projection of side $AC$ onto the line perpendicular to $AX$ (which is parallel to $AB$). This distance is $r_C = l \cos(60^{\circ}) = l \cdot \frac{1}{2} = \frac{l}{2}$.
Thus,$I_C = m(r_C)^2 = m(\frac{l}{2})^2 = \frac{m l^2}{4}$.
Total moment of inertia $I = 0 + m l^2 + \frac{m l^2}{4} = \frac{5}{4} m l^2$.

(C) For a circular disc of mass $M$ and radius $R$,the moment of inertia about a tangential axis in its plane is given by $I_{\text{disk}} = I_{\text{cm}} + MR^2 = \frac{1}{4}MR^2 + MR^2 = \frac{5}{4}MR^2$.
Since $I = MK^2$,the radius of gyration $K_{\text{disk}} = \sqrt{\frac{5}{4}}R = \frac{\sqrt{5}}{2}R$.
For a circular ring of mass $M$ and radius $R$,the moment of inertia about a tangential axis in its plane is given by $I_{\text{ring}} = I_{\text{cm}} + MR^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.
Since $I = MK^2$,the radius of gyration $K_{\text{ring}} = \sqrt{\frac{3}{2}}R = \frac{\sqrt{3}}{\sqrt{2}}R$.
The ratio of the radii of gyration is $\frac{K_{\text{disk}}}{K_{\text{ring}}} = \frac{\sqrt{5}/2}{\sqrt{3}/\sqrt{2}} = \frac{\sqrt{5}}{2} \times \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{5}}{\sqrt{2} \times \sqrt{3}} = \sqrt{\frac{5}{6}}$.

(C) The moment of inertia of a solid sphere of mass $M$ and radius $R$ about its diameter is given by $I_A = \frac{2}{5}MR^2 = 0.4 MR^2$.
The moment of inertia of a hollow sphere of mass $M$ and radius $R$ about its diameter is given by $I_B = \frac{2}{3}MR^2 \approx 0.67 MR^2$.
Since both spheres have the same mass $M$ and the same radius $R$,comparing the coefficients shows that $0.4 MR^2 < 0.67 MR^2$.
Therefore,$I_A < I_B$.

(A) The moment of inertia $I$ of a system of particles about an axis is given by $I = \sum m_i r_i^2$, where $r_i$ is the perpendicular distance of the $i$-th particle from the axis of rotation.
For the $x$-axis, the perpendicular distance of a point $(x, y, z)$ is $r = \sqrt{y^2 + z^2}$.
$1$. For mass $m_1 = 1 \text{ kg}$ at $(0,0,0)$: $r_1 = \sqrt{0^2 + 0^2} = 0 \text{ m}$. So, $I_1 = 1 \times 0^2 = 0 \text{ kg} \cdot \text{m}^2$.
$2$. For mass $m_2 = 2 \text{ kg}$ at $(2,0,0)$: $r_2 = \sqrt{0^2 + 0^2} = 0 \text{ m}$. So, $I_2 = 2 \times 0^2 = 0 \text{ kg} \cdot \text{m}^2$.
$3$. For mass $m_3 = 3 \text{ kg}$ at $(0,3,0)$: $r_3 = \sqrt{3^2 + 0^2} = 3 \text{ m}$. So, $I_3 = 3 \times 3^2 = 27 \text{ kg} \cdot \text{m}^2$.
$4$. For mass $m_4 = 4 \text{ kg}$ at $(-2,-2,0)$: $r_4 = \sqrt{(-2)^2 + 0^2} = 2 \text{ m}$. So, $I_4 = 4 \times 2^2 = 16 \text{ kg} \cdot \text{m}^2$.
The total moment of inertia $I = I_1 + I_2 + I_3 + I_4 = 0 + 0 + 27 + 16 = 43 \text{ kg} \cdot \text{m}^2$.

(B) The moment of inertia of a ring about its central axis is $I_{ring} = mr^2$.
When the ring is melted and recast into a sphere of the same mass $m$,let the radius of the sphere be $r'$.
Since the volume remains constant,the density is uniform. However,for a sphere of mass $m$ and radius $r'$,the moment of inertia about its diameter is $I_{sphere} = \frac{2}{5}mr'^2$.
Since the sphere is formed from the material of the ring,the radius of the sphere $r'$ will be significantly smaller than the radius of the ring $r$ (i.e.,$r' \ll r$).
Therefore,$I_{sphere} = \frac{2}{5}mr'^2 < mr^2 = I_{ring}$.
Thus,the moment of inertia of the sphere is less than that of the ring.

(B) The moment of inertia of a single rod of mass $m$ and length $l$ about an axis passing through its center and perpendicular to its length is given by $I = \frac{ml^2}{12}$.
Since the cross is formed by two such rods joined at their centers,the total moment of inertia about the axis passing through their common center and perpendicular to the plane of the rods is the sum of the moments of inertia of the two individual rods.
$I_{total} = I_{rod1} + I_{rod2} = \frac{ml^2}{12} + \frac{ml^2}{12} = \frac{2ml^2}{12} = \frac{ml^2}{6}$.

(A) The moment of inertia of a ring of mass $M$ and radius $r$ about an axis passing through its center and perpendicular to its plane is $I = Mr^2$.
Since the rings are made of the same material and have the same cross-sectional area,their mass is proportional to their circumference ($M = \lambda \cdot 2\pi r$,where $\lambda$ is the linear mass density).
For the first ring: $M_1 = \lambda \cdot 2\pi R$ and $I_1 = M_1 R^2 = (\lambda \cdot 2\pi R) R^2 = 2\pi \lambda R^3$.
For the second ring: $M_2 = \lambda \cdot 2\pi (nR)$ and $I_2 = M_2 (nR)^2 = (\lambda \cdot 2\pi nR) (nR)^2 = 2\pi \lambda n^3 R^3$.
The ratio of the moments of inertia is given as $\frac{I_1}{I_2} = \frac{1}{8}$.
Substituting the expressions: $\frac{2\pi \lambda R^3}{2\pi \lambda n^3 R^3} = \frac{1}{n^3} = \frac{1}{8}$.
Therefore,$n^3 = 8$,which gives $n = 2$.

(B) The moment of inertia $I$ of a solid cylinder about its geometrical axis (longitudinal axis) is given by the formula:
$I = \frac{1}{2} M R^2$
Given:
Mass $M = 20 \ kg$
Radius $R = 0.2 \ m$
Substituting the values into the formula:
$I = \frac{1}{2} \times 20 \times (0.2)^2$
$I = 10 \times 0.04$
$I = 0.4 \ kg \cdot m^2$

(B) When a swimmer curls their body,the mass distribution shifts closer to the axis of rotation.
According to the formula $I = \sum mr^2$,this reduction in the distance $r$ of the mass from the axis leads to a decrease in the moment of inertia $(I)$.
Since the angular momentum $(L = I\omega)$ remains conserved in the absence of external torque,a decrease in the moment of inertia $(I)$ results in an increase in the angular velocity $(\omega)$.
Therefore,the correct option is $B$.

(C) When the person pulls their arms inward,the mass distribution shifts closer to the axis of rotation.
Since the moment of inertia $I = \sum m_i r_i^2$,decreasing the distance $r$ of the mass from the axis leads to a decrease in the moment of inertia.
According to the law of conservation of angular momentum,$L = I\omega = \text{constant}$.
Since $I$ decreases,the angular velocity $\omega$ must increase to keep $L$ constant.
Therefore,the correct observation is that the moment of inertia decreases.

(A) The moment of inertia of a complete uniform circular disc of mass $M_{total}$ and radius $R$ about an axis passing through its center and perpendicular to its plane is given by $I = \frac{1}{2}M_{total}R^2$.
Since the disc is uniform,the mass is distributed proportionally to the area. $A$ quarter sector has $\frac{1}{4}$ of the total area,so its mass $M$ is $\frac{1}{4}M_{total}$,which means $M_{total} = 4M$.
The moment of inertia of the quarter sector about the same axis is $\frac{1}{4}$ of the moment of inertia of the full disc.
Therefore,$I_{sector} = \frac{1}{4} \times (\frac{1}{2} M_{total} R^2) = \frac{1}{4} \times (\frac{1}{2} \times 4M \times R^2) = \frac{1}{2}MR^2$.

(B) The moment of inertia $I$ of a disc about an axis passing through its center and perpendicular to its plane is given by $I = \frac{1}{2}MR^2$.
Since the mass $M$ is related to density $\rho$,thickness $t$,and radius $R$ by $M = \rho \cdot V = \rho \cdot (\pi R^2 t)$,we can express the radius as $R^2 = \frac{M}{\pi t \rho}$.
Substituting this into the moment of inertia formula: $I = \frac{1}{2}M \left( \frac{M}{\pi t \rho} \right) = \frac{M^2}{2 \pi t \rho}$.
Given that the mass $M$ and thickness $t$ are constant for both discs,we have $I \propto \frac{1}{\rho}$.
Therefore,the ratio of the moments of inertia is $\frac{I_1}{I_2} = \frac{\rho_2}{\rho_1}$.
Given the ratio of densities $\rho_1 : \rho_2 = 1 : 3$,we have $\frac{\rho_2}{\rho_1} = \frac{3}{1}$.
Thus,$\frac{I_1}{I_2} = 3:1$.

(C) The moment of inertia $I$ of a body is given by $I = \sum m_i r_i^2$. By concentrating the mass $M$ at the rim (at a distance $R$ from the axis of rotation),the radius of gyration is maximized. Since $I = MR^2$ for a ring-like structure,placing the mass as far as possible from the axis of rotation significantly increases the moment of inertia. $A$ higher moment of inertia allows the flywheel to store more rotational kinetic energy and resist changes in its rotational speed,which is the primary function of a flywheel.

(B) The moment of inertia $I$ of a rim (or thin hoop) about its central axis is given by the formula $I = mr^2$.
Given:
Mass $m = 6 \ kg$
Radius $r = 40 \ cm = 0.4 \ m$
Substituting the values into the formula:
$I = 6 \times (0.4)^2$
$I = 6 \times 0.16$
$I = 0.96 \ kg \cdot m^2$.
Thus,the moment of inertia is $0.96 \ kg \cdot m^2$.

(D) The moment of inertia of a disc is $I = \frac{1}{2} M r^2$.
Since mass $M = V \rho = (\pi r^2 t) \rho$,we can write $r^2 = \frac{M}{\pi t \rho}$.
Substituting $r^2$ into the moment of inertia formula: $I = \frac{1}{2} M \left( \frac{M}{\pi t \rho} \right) = \frac{M^2}{2 \pi t \rho}$.
Given that both discs have the same mass $M$ and same thickness $t$,the moment of inertia is inversely proportional to the density: $I \propto \frac{1}{\rho}$.
Therefore,the ratio of their moments of inertia is $\frac{I_1}{I_2} = \frac{\rho_2}{\rho_1}$.

(C) The moment of inertia of a solid sphere about its diameter is $I_S = \frac{2}{5}MR_S^2$.
The moment of inertia of a hollow sphere about its diameter is $I_H = \frac{2}{3}MR_H^2$.
Given that the moments of inertia are equal,$I_H = I_S$.
Substituting the formulas: $\frac{2}{3}MR_H^2 = \frac{2}{5}MR_S^2$.
Canceling common terms: $\frac{R_H^2}{3} = \frac{R_S^2}{5}$.
Rearranging for the ratio of radii: $\frac{R_H^2}{R_S^2} = \frac{3}{5}$.
Taking the square root on both sides: $\frac{R_H}{R_S} = \sqrt{\frac{3}{5}} = \frac{\sqrt{3}}{\sqrt{5}}$.
Thus,the ratio is $\sqrt{3}:\sqrt{5}$.

(B) The moment of inertia for each body about an axis passing through its center of mass is as follows:
For body $A$ (solid cylinder of radius $R$): $I_{A} = \frac{1}{2} mR^{2} = 0.5 mR^{2}$.
For body $B$ (square lamina of side $R$): The moment of inertia about an axis passing through the center and perpendicular to the plane is $I_{B} = \frac{m}{12}(R^{2} + R^{2}) = \frac{2mR^{2}}{12} = \frac{1}{6} mR^{2} \approx 0.167 mR^{2}$.
For body $C$ (solid sphere of radius $R$): $I_{C} = \frac{2}{5} mR^{2} = 0.4 mR^{2}$.
Comparing the values: $0.5 mR^{2} > 0.4 mR^{2} > 0.167 mR^{2}$.
Thus,$I_{A} > I_{C} > I_{B}$.
Therefore,body $B$ has the smallest moment of inertia.

(D) Let the total mass of each body be $m$. The moment of inertia $(I)$ about an axis passing through the centre of mass and perpendicular to the plane is:
$1$. For a disc of radius $a$: $I_{\text{disc}} = \frac{1}{2} m a^2 = 0.5 m a^2$.
$2$. For a ring of radius $a$: $I_{\text{ring}} = m a^2 = 1.0 m a^2$.
$3$. For a square lamina of side $L = 2a$: Using the perpendicular axis theorem,$I_z = I_x + I_y$. Since $I_x = I_y = \frac{m L^2}{12}$,we have $I_{\text{sq}} = \frac{m(2a)^2}{12} + \frac{m(2a)^2}{12} = \frac{4 m a^2}{12} + \frac{4 m a^2}{12} = \frac{8 m a^2}{12} = \frac{2}{3} m a^2 \approx 0.67 m a^2$.
$4$. For four rods of length $L = 2a$ forming a square: Each rod has mass $m/4$. The moment of inertia of one rod about the centre of the square is found using the parallel axis theorem: $I_{\text{rod}} = I_{\text{cm}} + (m/4)d^2 = \frac{(m/4)(2a)^2}{12} + (m/4)a^2 = \frac{m a^2}{3} + \frac{m a^2}{4} = \frac{7}{12} m a^2$. For four rods,$I_{\text{total}} = 4 \times \frac{7}{12} m a^2 = \frac{7}{3} m a^2 \approx 2.33 m a^2$.
Comparing the values,the four rods forming a square have the largest moment of inertia.

(B) The moment of inertia of a thin rod of mass $m$ and length $l$ about an axis passing through one of its ends and perpendicular to its length is given by $I = \frac{ml^2}{3}$.
$1$. For the rod along the $x$-axis: The axis of rotation is the $z$-axis. Since the rod lies in the $xy$-plane and is perpendicular to the $z$-axis at the origin,its moment of inertia about the $z$-axis is $I_x = \frac{ml^2}{3}$.
$2$. For the rod along the $y$-axis: The axis of rotation is the $z$-axis. Since the rod lies in the $yz$-plane and is perpendicular to the $z$-axis at the origin,its moment of inertia about the $z$-axis is $I_y = \frac{ml^2}{3}$.
$3$. For the rod along the $z$-axis: The rod lies along the axis of rotation itself. Therefore,the distance of every mass element of this rod from the $z$-axis is zero. Thus,$I_z = 0$.
Total moment of inertia $I_{\text{total}} = I_x + I_y + I_z = \frac{ml^2}{3} + \frac{ml^2}{3} + 0 = \frac{2ml^2}{3}$.

(C) The moment of inertia of a rod of mass $m$ and length $l$ about an axis passing through its center and making an angle $\theta$ with the rod is given by $I = \frac{ml^2}{12} \sin^2 \theta$.
For the rod along the $x$-axis,the angle with the line $x=y$ is $\theta = 45^\circ$. Thus,$I_1 = \frac{ml^2}{12} \sin^2 45^\circ = \frac{ml^2}{12} \times \frac{1}{2} = \frac{ml^2}{24}$.
For the rod along the $y$-axis,the angle with the line $x=y$ is also $\theta = 45^\circ$. Thus,$I_2 = \frac{ml^2}{12} \sin^2 45^\circ = \frac{ml^2}{24}$.
The total moment of inertia is $I = I_1 + I_2 = \frac{ml^2}{24} + \frac{ml^2}{24} = \frac{2ml^2}{24} = \frac{ml^2}{12}$.

(D) For a semicircular plate of mass $M$ and radius $R$,the moment of inertia about an axis passing through the centre and perpendicular to the diameter (the axis of symmetry) is $I = \frac{MR^2}{4}$.
Since the semicircular plate is symmetric about this axis,any axis passing through the centre in the plane of the plate will have the same moment of inertia due to the distribution of mass being identical relative to the axis of symmetry.
Thus,the moment of inertia about the axis $AA'$ is $I = \frac{MR^2}{4}$.

(B) The moment of inertia of a complete circular disc of mass $M'$ and radius $R$ about an axis passing through its center and perpendicular to its plane is $I_{total} = \frac{1}{2} M' R^2$.
For a semicircular disc of mass $M$,we can consider it as half of a complete circular disc of mass $2M$ and radius $R$.
The moment of inertia of this complete circular disc (mass $2M$) about the axis through its center $O$ is $I_{total} = \frac{1}{2} (2M) R^2 = MR^2$.
Since the semicircular disc is exactly half of this complete disc,its moment of inertia about the same axis is half of the total moment of inertia.
Therefore,$I = \frac{1}{2} I_{total} = \frac{1}{2} (MR^2) = \frac{1}{2} MR^2$.

(B) The square sheet has side length $L$. When it is rolled into a hollow cylinder,the side length $L$ becomes the circumference of the cylinder's base.
Therefore,$L = 2\pi R$,which gives the radius $R = \frac{L}{2\pi}$.
The total mass $M$ of the sheet is given by $M = \sigma \times \text{Area} = \sigma L^2$.
The moment of inertia $I$ of a hollow cylinder of mass $M$ and radius $R$ about its central axis is $I = MR^2$.
Substituting the values of $M$ and $R$:
$I = (\sigma L^2) \times \left(\frac{L}{2\pi}\right)^2$
$I = \sigma L^2 \times \frac{L^2}{4\pi^2}$
$I = \frac{\sigma L^4}{4\pi^2}$.