Given,
$\mu_2=\frac{2}{3} \text { and } \mu_0=\frac{1}{3}$
Let the acceleration of the whole system be a then,
As per the diagram,
For block $B$,
$2 mg \cos 45^{\circ}+2 ma = T +\mu_0 N _{ b }$
$N _{ b }=2 mg \sin 45^{\circ}$
And, For block $A$,
$T =m g \cos 45^{\circ}+\mu_g N_a$
$N_{ a }=m g \sin 45^{\circ}$
So,
$T=m g \cos 45^{\circ}+\mu_2 m g \sin 45^{\circ} \ldots \ldots$
From eq $(1)$ and $(2)$
$2 m g \cos 45^{\circ}+2 ma =m g \cos 45^{\circ}+\mu_2 m g \sin 45^{\circ}+\mu_0 2 m g \sin 45^{\circ}$
After solving, We get $a =0$
So the acceleration of the block $A$ will be zero.
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The correct option ($s$) is (are)
$(A)$ $q_{A C}=\Delta U_{B C}$ and $W_{A B}=P_2\left(V_2-V_1\right)$ $(B)$ $W _{ BC }= P _2\left( V _2- V _1\right)$ and $q _{ BC }= H _{ AC }$ $(C)$ $\Delta H _{ CA }<\Delta U _{ CA }$ and $q _{ AC }=\Delta U _{ BC }$ $(D)$ $q_{B C}=\Delta H_{A C}$ and $\Delta H_{C A}>\Delta U_{C A}$
