By given question

Area of square $$=$$ Area of rectangular plate

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

Now by definition

(a) $$\frac{I_{x R}}{I_{x S}}=\frac{b^2}{c^2} \quad\left[\because I \propto(\text { area })^2\right]$$

From the diagram $$b< c$$

$$\Rightarrow \quad \frac{I_{x R}}{I_{x S}}=\left(\frac{b}{c}\right)^2 < 1 \Rightarrow I_{x R} < I_{x S}$$

$$\begin{aligned} &\text { (b) }\\ &\frac{I_{y_R}}{I_{y_S}}=\frac{a^2}{c^2} \end{aligned}$$

$$\begin{aligned} &\text { as }\\ &\begin{aligned} & a>c \\ & \frac{I_{y_R}}{I_{y_S}}=\left(\frac{a}{c}\right)^2>1 \end{aligned} \end{aligned}$$

$$\begin{aligned} & \text { (c) } I_{z R}-I_{z S} \propto\left(a^2+b^2-2 c^2\right)=a^2+b^2-2 a b=(a-b)^2 \quad\left[\because c^2=a b\right] \\ & \Rightarrow \quad\left(I_{z R}-I_{z S}\right)>0 \Rightarrow \frac{I_{z R}}{I_{z S}}>1 \end{aligned}$$

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

$$F-f=M a\quad \text{.... (i)}$$

where $$M$$ is mass of the disc

The angular acceleration of the disc is

$$\alpha=a / R\quad \text{(for pure rolling)}$$

from

$$\tau=l \alpha$$

$$f R=\left(\frac{1}{2} M R^2\right) \alpha \Rightarrow f R=\left(\frac{1}{2} M R^2\right)\left(\frac{a}{R}\right)$$

$$M a=2 f\quad \text{.... (ii)}$$

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

$$\begin{array}{cc} & f=F / 3[\because N=M g] \\ \because & f \leq \mu N=\mu M g \\ \Rightarrow & \frac{F}{3} \leq \mu M g \Rightarrow F \leq 3 \mu M g \\ \Rightarrow & F_{\max }=3 \mu M g \end{array}$$