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(D) Since the rod is about to slip,both friction forces will be at their limiting values:$f_1 = \mu_1 N_1$$f_2 = \mu_2 N_2$For option $(A)$ and $(D)$,

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$(C,D)$

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(D) Since the rod is about to slip,both friction forces will be at their limiting values:$f_1 = \mu_1 N_1$$f_2 = \mu_2 N_2$For option $(A)$ and $(D)$,$\mu_1 = 0$. The net torque about the floor contact point $A$ must be zero for equilibrium:$mg \cos 	heta \left(\frac{\ell}{2}ight) = N_1 \sin 	heta (\ell)$$\Rightarrow N_1 = \frac{mg \cot 	heta}{2}$$\Rightarrow N_1 	an 	heta = \frac{mg}{2}$This matches the condition in option $(D)$.For option $(B)$,$\mu_2 = 0$. There is no horizontal force to balance $N_1$,so the rod cannot remain in equilibrium.For option $(C)$,$\mu_1 
eq 0$ and $\mu_2 
eq 0$. Balancing forces:Horizontal: $N_1 = f_2 = \mu_2 N_2$Vertical: $N_2 + f_1 = mg \Rightarrow N_2 + \mu_1 N_1 = mg$Substituting $N_1 = \mu_2 N_2$ into the vertical equation:$N_2 + \mu_1 (\mu_2 N_2) = mg$$N_2 (1 + \mu_1 \mu_2) = mg$$N_2 = \frac{mg}{1 + \mu_1 \mu_2}$Thus,options $(C)$ and $(D)$ are correct.

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