Answer (d) A bangle is in the form of a ring as shown in the adjacent diagram. The centre of mass lies at the centre, which is outside the body (boundary).

• (a) A pencil: The centre of mass of a pencil lies within its body, typically along its length at the midpoint, assuming it is uniform in density.

• (b) A shotput: The centre of mass of a shotput, which is a solid sphere, lies at its geometric centre, which is within the body.

• (c) A dice: The centre of mass of a dice, which is a solid cube, lies at its geometric centre, which is within the body.

Thinking Process

In a system of particles, the centre of mass of a body lies closer to heavier mass or masses.

Answer (c) Centre of mass of a system lies towards the part of the system, having bigger mass. In the above diagram, lower part is heavier, hence $CM$ of the system lies below the horizontal diameter.

• Option (a) $A$: This point is located above the horizontal diameter, which is closer to the lighter upper part of the system. Since the centre of mass lies towards the part with the bigger mass, $A$ cannot be the centre of mass.

• Option (b) $B$: This point is located at the horizontal diameter, which would imply that the masses are evenly distributed above and below this line. However, the lower part of the system is heavier, so the centre of mass must be below this point.

• Option (d) $D$: This point is located above the horizontal diameter, similar to point $A$. Since the lower part of the system is heavier, the centre of mass cannot be at $D$ as it is closer to the lighter upper part.

Thinking Process

In elastic collision, KE of the system remains conserved. Therefore, the ball will bounce back with the same speed $v$ but in opposite direction i.e., along -ve $y$-axis.

Answer (b) The initial velocity is $v_i$ = $v{\hat{e}}_y$ and after reflection from the wall, the final velocity is $v_f$ = $-v{\hat{e}}_y$. The trajectory is described as position vector r = $y{\hat{e}}_y+a{\hat{e}}_z$.

Hence, the change in angular momentum is $r\times m(v_f-v_j)=2mva{\hat{e}}_x$.

• Option (a) $mva{\hat{\mathbf{e}}}_x$:

  • This option is incorrect because it does not account for the factor of 2 that arises from the change in velocity due to the elastic collision. The correct change in angular momentum involves the difference between the initial and final velocities, which results in a factor of 2.

• Option (c) $ymv{\hat{\mathbf{e}}}_x$:

  • This option is incorrect because it incorrectly uses the variable $y$ instead of $a$. The change in angular momentum depends on the position where the trajectory intersects the $z$-axis, which is given as $z=a$.

• Option (d) $2ymv{\hat{\mathbf{e}}}_x$:

  • This option is incorrect because it incorrectly uses the variable $y$ instead of $a$. The correct expression for the change in angular momentum should involve the position $z=a$ where the trajectory intersects the $z$-axis.

Answer (d) We know that angular acceleration

$$\alpha =\frac{d\omega }{dt}\text{, given }\omega =\text{ constant }$$

where $\omega$ is angular velocity of the disc

$$\Rightarrow \alpha =\frac{d\omega }{dt}=\frac{0}{dt}=0$$

Hence, angular acceleration is zero.

• (a) The sense of rotation remains same: This statement is true because if the disc rotates with a uniform angular velocity, the direction of rotation does not change.

• (b) The orientation of the axis of rotation remains same: This statement is true because for a disc rotating with uniform angular velocity, the axis of rotation remains fixed in direction.

• (c) The speed of rotation is non-zero and remains same: This statement is true because uniform angular velocity implies that the rotational speed is constant and non-zero.

Thinking Process

For two bodies having same mass, the body having mass distributed at greater distance from an axis, will have more moment of inertia.

Answer (b) In the given diagrams, when the small piece $Q$ removed and glued to the centre of the plate, the mass comes closer to the $z$-axis, hence, moment of inertia decreases.

• Option (a) increased: This is incorrect because moving the mass closer to the axis of rotation (the $z$-axis) reduces the moment of inertia. The moment of inertia depends on the distribution of mass relative to the axis, and bringing mass closer to the axis decreases it.

• Option (c) the same: This is incorrect because the redistribution of mass affects the moment of inertia. Removing a piece and then gluing it to the center changes the mass distribution, thus altering the moment of inertia.

• Option (d) changed in unpredicted manner: This is incorrect because the change in the moment of inertia can be predicted based on the principles of rotational dynamics. Specifically, moving mass closer to the axis of rotation decreases the moment of inertia.

Answer (c) Consider the adjacent diagram, there is a line shown in the figure drawn along the diagonal. First, centre of mass of the system was on the dotted line and was shifted towards $Q$ from the centre (Ist quadrant).

When mass is removed, it will be on the same line but shifted away from the centre and below (Illrd quadrant). Position of $CM$ is shown by $X$ in the diagram.

• Option (a) I: The center of mass (CM) of the plate cannot be in the first quadrant because the mass was removed from the first quadrant, causing the CM to shift away from this region.

• Option (b) II: The center of mass (CM) of the plate cannot be in the second quadrant because the mass removal and the resulting shift in the CM do not affect this quadrant. The shift occurs along the diagonal line, moving the CM to the third quadrant.

• Option (d) IV: The center of mass (CM) of the plate cannot be in the fourth quadrant because the mass removal causes the CM to shift along the diagonal line towards the third quadrant, not the fourth.

Answer (a) Density is given as

$$\rho (x)=a(1+b{x}^2)$$

where $a$ and $b$ are constants and $0\le x\le 1$.

Let $barrow0$, in this case

$$\rho (x)=a=\text{ constant }$$

Hence, centre of mass will be at $x=0.5m$. (middle of the rod)

Putting, $b=0$ in all the options, only (a) gives 0.5 .

Note We should not check options by putting $a=0$, because $\rho =0$ for $a=0$.

• Option (b) is incorrect because when ( b = 0 ), it simplifies to (\frac{4 \cdot 2}{3 \cdot 3} = \frac{8}{9} \neq 0.5).

• Option (c) is incorrect because when ( b = 0 ), it simplifies to (\frac{3 \cdot 3}{4 \cdot 2} = \frac{9}{8} \neq 0.5).

• Option (d) is incorrect because when ( b = 0 ), it simplifies to (\frac{4 \cdot 3}{3 \cdot 2} = 2 \neq 0.5).

Answer (a) As no external torque acts on the system, angular momentum should be conserved. Hence $I\omega =$ constant where, $I$ is moment of inertia of the system and $\omega$ is angular velocity of the system. From Eq. (i) $I_1{\omega }_1=I_2{\omega }_2$ (where ${\omega }_1$ and ${\omega }_2$ are angular velocities before and after jumping)

$$\begin{array}{cc}\Rightarrow & I\omega =\frac{I}{2}\times {\omega }_2\\ \Rightarrow & \text{ (as mass reduced to half, hence, moment of inertia also reduced to half) }\\ & {\omega }_2=2\omega \end{array}$$

• Option (b) $\omega$: This option is incorrect because it assumes that the angular velocity remains unchanged after the person jumps off. However, since the moment of inertia of the system changes when the person leaves, the angular velocity must adjust to conserve angular momentum. The system’s moment of inertia is halved, so the angular velocity must double to maintain the same angular momentum.

• Option (c) $\frac{\omega }{2}$: This option is incorrect because it suggests that the angular velocity decreases to half of its original value. This would imply that the angular momentum of the system decreases, which contradicts the principle of conservation of angular momentum. Since the moment of inertia is halved, the angular velocity should increase, not decrease.

• Option (d) 0: This option is incorrect because it implies that the merry-go-round stops rotating entirely after the person jumps off. This would mean that the angular momentum of the system is not conserved, which is not possible in the absence of external torques. The system must continue to rotate, and the angular velocity should increase to conserve angular momentum.

Answer $(a,c)$

For a general rotational motion, where axis of rotation is not symmetric. Angular momentum $L$ and angular velocity $\omega$ need not be parallel. For a general translational motion momentum $\mathbf{p}=m\mathbf{v}$, hence, $\mathbf{p}$ and $\mathbf{v}$ are always parallel.

• For option (b): For a rotational motion about a fixed axis, angular momentum $\mathbf{L}$ and angular velocity $\omega$ are not always parallel. This is because the distribution of mass around the axis can cause the angular momentum to have components that are not aligned with the angular velocity.

• For option (d): For a general translational motion, acceleration $\mathbf{a}$ and velocity $\mathbf{v}$ are not always parallel. This is because acceleration can have components that change the direction of the velocity, such as in the case of circular motion where the acceleration is perpendicular to the velocity.

Answer $(a,b)$

The angular momentum $\mathbf{L}$ of a particle with respect to origin is defined to be $\mathbf{L}=\mathbf{r}\times \mathbf{p}$ where, $\mathbf{r}$ is the position vector of the particle and $\mathbf{p}$ is the linear momentum. The direction of $L$ is perpendicular to both $d\mathbf{r}$ and $\mathbf{p}$ by right hand rule.

For particle $1,I_1=r_1\times m\mathbf{v}$, is out of plane of the paper and perpendicular to $r_1$ and $\mathbf{p}(m\mathbf{v})$ Similarly $I_2=r_2\times m(-\mathbf{v})$ is into the plane of the paper and perpendicular to $r_2$ and $-\mathbf{p}$.

Hence, total angular momentum

$$l=l_1+l_2=r_1\times m\mathbf{v}+(-r_2\times m\mathbf{v})$$

$|l|=mv{d}_1-mv{d}_2$ as $d_2>d_1$, total angular momentum will be inward

Hence, $I=m\mathbf{v}(d_2-d_1)\otimes$

Note In the expression of angular momentum $I=\mathbf{r}\times \mathbf{p}$ the direction of $l$ is taken by right hand rule.

• Option (c): The total angular momentum of the system about point ( A ) is not ( I = m v (r_1 + r_2) ). This is incorrect because the angular momentum contributions from the two particles are in opposite directions due to their opposite velocities. Therefore, the correct expression should account for the subtraction of the angular momenta, not their addition.

• Option (d): The total angular momentum of the system about point ( A ) is not ( I = m v (d_2 - d_1) \otimes ). This is incorrect because the symbol ( \otimes ) represents a unit vector coming out of the page, but the correct direction for the total angular momentum, given the problem’s conditions, is into the page. The correct notation should use ( \odot ) to represent a unit vector going into the page.

Answer $(a,b,c,d)$

We know that torque on a system of particles $\tau =\mathbf{r}\times \mathbf{F}=\mathbf{F}\sin \theta \hat{\mathbf{n}}$

where, $\theta$ is angle between $\mathbf{r}$ and $\mathbf{F}$, and $\hat{\mathbf{n}}$ is a unit vector perpendicular to both $\mathbf{r}$ and $\mathbf{F}$.

(a) When forces act radially, $\theta =0$ hence $\tau =0$

[from Eq. (i)]

(b) When forces are acting on the axis of rotation, $r=0,\tau =0$ [from Eq. (i)]

(c) When forces acting parallel to the axis of rotation $\theta =0^{\circ },\tau =0$

(d) When torque by forces are equal and opposite, ${\tau }_{\text{net }}={\tau }_1-{\tau }_2=0$ [from Eq. (i)]

• There are no incorrect options in the given answer. All options (a, b, c, d) are compatible with the condition that the net external torque on a system of particles about an axis is zero.

Answer ( $b,c$ )

(a) Consider the adjacent diagram, where $r>r^{\prime }$

Torque $\tau$ about $z$-axis $\tau =\mathbf{r}\times \mathbf{F}$ which is along $\hat{\mathbf{k}}$

(b) ${\tau }^{\prime }=r^{\prime }\times F$ which is along $-\hat{k}$

(c) ${\tau }_z=F{r}_{\perp }=$ magnitude of torque about $z$-axis where $r_{\perp }$ is perpendicular distance between $\overline{F}$ and z-axis.

Similarly,

Clearly

$$|\tau {|}_{z^{\prime }}=F{r}_{\perp }^{\prime }$$ $$r_{\perp }>r_{\perp }^{\prime }\Rightarrow |\tau {|}_{z>}|\tau {|}_{z^{\prime }}$$

(d) We are always calculating resultant torque about a common axis.

Hence, total torque $\tau \ne \tau +{\tau }^{\prime }$, because $\tau$ and ${\tau }^{\prime }$ are not about common axis.

• (a) Incorrect Reason: The torque (\tau) caused by (\mathbf{F}) about the (z)-axis is given by (\mathbf{r} \times \mathbf{F}). According to the right-hand rule, the direction of the torque vector is along (\hat{\mathbf{k}}), not (-\hat{\mathbf{k}}). Therefore, option (a) is incorrect.

• (d) Incorrect Reason: The total torque is not simply the sum of (\tau) and (\tau^{\prime}) because they are torques about different axes ((z) and (z^{\prime}) respectively). To find the resultant torque, we need to consider torques about a common axis. Hence, (\tau \neq \tau + \tau^{\prime}), making option (d) incorrect.

Thinking Process

Moment of inertia about two symmetrical axes are same. To calculate net moment of inertia we can apply the concept of symmetry.

Answer $(a,b,d)$

(a) Theorem of perpendicular axes is applicable only for laminar (plane) objects. Thus, option (a) is false.

(b) As $z^{\prime }|z$ and distance between them $=a\frac{\sqrt{2}}{2}=\frac{a}{\sqrt{2}}$

Now, by theorem of parallel axes

$$I_{z^{\prime }}=I_z+m(\frac{a}{\sqrt{2}}{)}^2=I_z+\frac{m{a}^2}{2}$$

Hence, choice (b) is true.

(c) $z^{\prime \prime }$ is not parallel to $z$ hence, theorem of parallel axis cannot be applied.

Thus, option (c) is false.

(d) As $x$ and $y$-axes are symmetrical.

Hence,

$$I_x=I_y$$

Thus, option (d) is true.

• Theorem of perpendicular axes is applicable only for laminar (plane) objects. Thus, option (a) is false.

• $z^{\prime \prime }$ is not parallel to $z$ hence, theorem of parallel axis cannot be applied. Thus, option (c) is false.

Thinking Process

Centre of gravity is centre of a given structure but centre of mass is a point where whole mass of the body can be assumed to be concentrated.

Answer When the vertical height of the object is very small as compared to the earth’s radius, we call the object small, otherwise it is extended.

(i) Building and pond are small objects.

(ii) A deep lake and a mountain are examples of extended objects.

Answer The moment of inertia of a body is given by $I=\mathrm{\Sigma }{m}_i{r}_i^2$ [sum of moment of inertia of each constituent particles]

All the mass in a cylinder lies at distance $R$ from the axis of symmetry but most of the mass of a solid sphere lies at a smaller distance than $R$.

Answer Consider the below diagram

Moment of the force $F$ about point $A,{\tau }_1=F\times a$ (anti-clockwise)

Moment of weight $mg$ of the cube about point $A$,

Cube will not exhibit motion, if ${\tau }_1={\tau }_2$

$(\because$ In this case, both the torque will cancel the effect of each other)

$\therefore F\times a=mg\times \frac{a}{2}\Rightarrow F=\frac{mg}{2}$

Cube will rotate only when, ${\tau }_1>{\tau }_2$

$$\Rightarrow F\times a>mg\times \frac{a}{2}\Rightarrow F>\frac{mg}{2}$$

Let normal reaction is acting at $\frac{a}{3}$ from point $A$, then

$$mg\times \frac{a}{3}=F\times a\text{ or }F=\frac{mg}{3}$$

(For no motion)

When $F=\frac{mg}{4}$ which is less than $\frac{mg}{3}$,.

$$(F<\frac{mg}{3})$$

there will be no motion.

$\therefore$

(a) $\to$ (ii)

(b) $\to$ (iii)

(c) $\to$ (i)

(d) $\to$ (iv)

Answer Consider the diagram where a sphere of $m$ and radius $R$, struck horizontally at height $h$ above the floor

The sphere will roll without slipping when $\omega =\frac{v}{r}$, where, $v$ is linear velocity and $\omega$ is angular velocity of the sphere.

Now, angular momentum of sphere, about centre of mass

[We are applying conservation of angular momentum just before and after struck]

$$\begin{array}{rl}mv(h-R)& =I\omega =(\frac{2}{5}m{R}^2)(\frac{v}{R})\\ \Rightarrow mv(h-R)& =\frac{2}{5}mvR\\ h-R& =\frac{2}{5}R\Rightarrow h=\frac{7}{5}R\end{array}$$

Therefore, the sphere will roll without slipping with a constant velocity and hence, no loss of energy, so

(d) $\to$ (i)

Torque due to applied force, $F$ about centre of mass

$$\tau =F(h-R)$$

For $\tau =0,h=R$, sphere will have only translational motion. It would lose energy by friction.

Hence,

$$\text{ (b) }\to \text{ (iv) }$$

The sphere will spin clockwise when $\tau >0\Rightarrow h>R$

Therefore,

(c) $\to$ (ii)

The sphere will spin anti-clockwise when $\tau <0\Rightarrow h<R$, (a) $\to$ (iii)

Answer No, not necessarily. Given,

$$\sum _i{F}_i\ne 0$$

The sum of torques about a certain point $O,\sum _i{\mathbf{r}}_i\times {\mathbf{F}}_i=0$

The sum of torques about any other point $O^{\prime }$

Here, the second term need not vanish.

$$\sum _i(r_i-a)\times F_i=\sum _i{r}_i\times F_i-a\times \sum _i{F}_i$$

Therefore, sum of all the torques about any arbitrary point need not be zero necessarily.

Answer Wheel is a rigid body. The particles that constitute the wheel do experience a centripetal acceleration directed towards the centre. This acceleration arises due to internal elastic forces, which cancel out in pairs.

In a half wheel, the distribution of mass about its centre of mass (through which axis of rotation passes) is not symmetrical. Therefore, the direction of angular momentum of the wheel does not coincide with the direction of its angular velocity. Hence, an external torque is required to maintain the motion of the half wheel.

Answer Consider the diagram, where weight of the door acts along negative $y$-axis.

A force can produce torque only along a direction normal to itself as $\tau =\mathbf{r}\times \mathbf{F}$. So, when the door is in the $xy$-plane, the torque produced by gravity can only be along $\pm z$ direction, never about an axis passing through $y$-direction.

Hence, the weight will not produce any torque about y-axis.

Thinking Process

The centre of mass of a regular n-polygon lies at its geometrical centre.

Answer Let $\mathbf{b}$ be the position vector of the centre of mass of a regular $n$-polygon.

$(n-1)$ equal point masses are placed at $(n-1)$ vertices of the regular $n$-polygon, therefore, for its centre of mass

$r_{CM}=\frac{(n-1)mb+ma}{(n-1)m+m}=0(\because \text{ Centre of mass lies at centre })$

$\Rightarrow (n-1)mb+ma=0$

$\Rightarrow b=-\frac{a}{(n-1)}$

Answer Let $M$ and $R$ be the mass and radius of the half-disc, mass per unit area of the half-disc

$$m=\frac{M}{\frac{1}{2}\pi R^2}=\frac{2M}{\pi R^2}$$

(a) The half-disc can be supposed to be consists of a large number of semicircular rings of mass $dm$ and thickness $dr$ and radii ranging from $r=0$ to $r=R$.

Surface area of semicircular ring of radius $r$ and of thickness $dr=\frac{1}{2}2\pi r\times dr=\pi rdr$

$\therefore$ Mass of this elementary ring, $dm=\pi rdr\times \frac{2M}{\pi R^2}$

$$dm=\frac{2M}{R^2}rdr$$

If $(x,y)$ are coordinates of centre of mass of this element, then,

$$(x,y)=(0,\frac{2r}{\pi })$$

Therefore,

$$x=0\text{ and }y=\frac{2r}{\pi }$$

Let $x_{CM}$ and $y_{CM}$ be the coordinates of the centre of mass of the semicircular disc.

Then

$$\begin{array}{rl}{x}_{CM}& =\frac{1}{M}{\int }_0^{R}xdm=\frac{1}{M}{\int }_0^{R}0dm=0\\ y_{CM}& =\frac{1}{M}{\int }_0^{R}ydm=\frac{1}{M}{\int }_0^R\frac{2r}{\pi }\times (\frac{2M}{R^2}rdr)\\ & =\frac{4}{\pi R^2}{\int }_0^R{r}^{2}dr=\frac{4}{\pi R^2}[\frac{r^3}{3}{]}_0^R\\ & =\frac{4}{\pi R^2}\times (\frac{R^3}{3}-0)=\frac{4R}{3\pi }\end{array}$$

$\therefore$ Centre of mass of the semicircular disc $=(0,\frac{4R}{3\pi })$

(b) Centre of mass of a uniform quarter disc.

Mass per unit area of the quarter disc $=\frac{M}{\frac{\pi R^2}{4}}=\frac{4M}{\pi R^2}$

Using symmetry

For a half-disc along $y$-axis centre of mass will be at $x=\frac{4R}{3\pi }$

For a half-disc along $x$-axis centre of mass will be at $x=\frac{4R}{3\pi }$

Hence, for the quarter disc centre of mass $=(\frac{4R}{3\pi },\frac{4R}{3\pi })$

Thinking Process

Due to friction between the two discs, the system will acquire common angular speed after sometime.

Answer Consider the diagram below

Let the common angular velocity of the system is $\omega$.

(a) Yes, the law of conservation of angular momentum can be applied. Because, there is no net external torque on the system of the two discs.

External forces, gravitation and normal reaction, act through the axis of rotation, hence, produce no torque.

(b) By conservation of angular momentum

$$\begin{array}{rlr}& L_f& =L_i\\ \Rightarrow & I\omega & =I_1{\omega }_1+I_2{\omega }_2\\ \Rightarrow & \omega & =\frac{I_1{\omega }_1+I_2{\omega }_2}{I}=\frac{I_1{\omega }_1+I_2{\omega }_2}{I_1+I_2}(\because I=I_1+I_2)\end{array}$$

(c) $K_f=\frac{1}{2}(I_1+I_2)\frac{(I_1{\omega }_1+I_2{\omega }_2{)}^2}{(I_1+I_2{)}^2}=\frac{1}{2}\frac{(I_1{\omega }_1+I_2{\omega }_2{)}^2}{(I_1+I_2)}$

$$\begin{array}{rl}& K_i=\frac{1}{2}(I_1{\omega }_1^2+I_2{\omega }_2^2)\\ & \mathrm{\Delta }K=K_f-K_i=-\frac{I_1{I}_2}{2(I_1+I_2)}({\omega }_1-{\omega }_2{)}^2<0\end{array}$$

(d) Hence, there is loss in KE of the system. The loss in kinetic energy is mainly due to the work against the friction between the two discs.

Answer (a) Before being brought in contact with the table the disc was in pure rotational motion hence, $v_{CM}=0$.

(b) When the disc is placed in contact with the table due to friction velocity of a point on the rim decreases.

(c) When the rotating disc is placed in contact with the table due to friction centre of mass acquires some linear velocity.

(d) Friction is responsible for the effects in (b) and (c) .

(e) When rolling starts $v_{CM}=\omega R$.

where $\omega$ is angular speed of the disc when rolling just starts. (f) Acceleration produced in centre of mass due to friction

Angular retardation produced by the torque due to friction.

$\alpha =\frac{\tau }{I}=\frac{{\mu }_{k}mgR}{I}[\because \tau =({\mu }_{k}N)R={\mu }_{k}mgR]$

$\therefore v_{CM}=u_{CM}+a_{CM}t$

$\Rightarrow v_{CM}={\mu }_{k}gt$

$\text{ and }\omega ={\omega }_0+\alpha t$

$\Rightarrow \omega ={\omega }_0-\frac{{\mu }_{k}mgR}{I}t$

For rolling without slipping, $\frac{V_{CM}}{R}=\omega$

$$\Rightarrow \begin{array}{rl}\frac{v_{CM}}{R}& ={\omega }_0-\frac{{\mu }_{k}mgR}{l}t\\ \frac{{\mu }_{k}gt}{R}& ={\omega }_0-\frac{{\mu }_{k}mgR}{l}t\\ t& =\frac{R{\omega }_0}{{\mu }_{k}g(1+\frac{m{R}^2}{l})}\end{array}$$

Note In this problem, frictional force helps in setting pure rolling motion.

Answer (a) Consider the situation shown below, we have shown the frictional forces.

Velocities at the point of contact Frictional

$F\uparrow$ force on left drum (upward)

Frictional

$F\downarrow$ force on right drum (downward) (b) $F^{\prime }=F=F^{\prime \prime }$ where $F^{\prime }$ and $F^{\prime \prime }$ are external forces through support.

$\Rightarrow F_{\text{net }}=0$

(one each cylinder)

External torque $=F\times 3R$, (anti-clockwise)

(c) Let ${\omega }_1$ and ${\omega }_2$ be final angular velocities of smaller and bigger drum respectively (anti-clockwise and clockwise respectively)

Finally, there will be no friction.

Hence, $R{\omega }_1=2R{\omega }_2\Rightarrow \frac{{\omega }_1}{{\omega }_2}=2$

Note We should be very careful while indicating direction offrictional forces.

Answer By given question

Area of square $=$ Area of rectangular plate

$\Rightarrow c^2=a\times b\Rightarrow c^2=ab$

Now by definition

(a)

$$\frac{I_{xR}}{I_{xS}}=\frac{b^2}{c^2}$$

$[\because I\propto (area{)}^2]$

From the diagram $b<c$

$\Rightarrow \frac{I_{xR}}{I_{xS}}=(\frac{b}{c}{)}^2<1\Rightarrow I_{xR}<I_{xS}$

(b)

$$\frac{I_{y_R}}{I_{y_S}}=\frac{a^2}{c^2}$$

as

$$a>c$$

$$\frac{I_{y_R}}{I_{y_S}}=(\frac{a}{c}{)}^2>1$$

(c) $I_{zR}-I_{zS}\propto (a^2+b^2-2c^2)=a^2+b^2-2ab=(a-b{)}^2[\because c^2=ab]$

$$\Rightarrow (I_{zR}-I_{zS})>0\Rightarrow \frac{I_{zR}}{I_{zS}}>1$$

Thinking Process

Frictional force on the disc will be in opposite direction of $F$ and supports the rotation of the disc in clockwise direction.

Answer Consider the diagram below

Frictional force $(f)$ is acting in the opposite direction of $F$.

Let the acceleration of centre of mass of disc be a then

$$F-f=Ma$$

where $M$ is mass of the disc

The angular acceleration of the disc is

$\alpha =a/R$

$\tau =/\alpha$

From Eqs. (i) and (ii), we get

$f=F/3[\because N=Mg]$

$\because f\le \mu N=\mu Mg$

$\Rightarrow \frac{F}{3}\le \mu Mg\Rightarrow F\le 3\mu Mg$ $\Rightarrow F_{max}=3\mu Mg$