Two blocks A and B of mass m and 2m are placed on a fixed triangular wedge by  massless inextensible string as shown. The pulley is massless and frictionless. The coefficient of friction between blocks A and the wedge is 2/3 and that between B and the wedge is 1/3  $\left(\text{Take m=1kg,\hspace{0.17em}g\hspace{0.17em}=10}\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.\right)$

	List – I		List – II
A	Friction force between block A and wedge	P	$\frac{5}{\sqrt{2}}$ units
B	Friction force between block B and wedge	Q	$\frac{10}{3\sqrt{2}}$units
C	Tension in the string	R	$\frac{10\sqrt{2}}{3}$ units
D	Maximum friction force between block B and wedge	S	$30\sqrt{2}$ units
		T	$\frac{20\sqrt{2}}{3}$ units

Equivalent diagram:
  
 Given  ${\mu }_1=\frac{2}{3}and{\mu }_2=\frac{1}{3}$
$f_1\left(\max \right)={\mu }_1{N}_1=\frac{2}{3}mg\cos \theta =\frac{2}{3}\times 1\times 10\times \frac{1}{\sqrt{2}}=\frac{10\sqrt{2}}{3}$

$\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{f}_2\max ={\mu }_2{N}_2=\frac{1}{3}\left(2mg\cos \theta \right)=\frac{10\sqrt{2}}{3}$ 
 Maximum friction between B and the wedge is  $f_2\left(\max \right)=\frac{\text{10}\sqrt{\text{2}}}{\text{3}}\text{units}$
 $\therefore \left(D\to r\right)$
Total maximum friction on the system  is $f\max =\frac{10\sqrt{2}}{3}+\frac{10\sqrt{2}}{3}=\frac{20\sqrt{2}}{3}$  
 Net pulling force =  $F=mg\sin \theta =1\times 10\times \frac{1}{\sqrt{2}}=5\sqrt{2}$
 Since $\text{F < f max}$ system will be at rest since friction is not sufficient to keep A and B at rest individually , tension must   not be zero for tension to appear at least  one of the blocks must be in limiting  equilibrium.  Then $f_2=f_2\left(max\right)=\frac{10\sqrt{2}}{3}$  
$\therefore$  Tension in the string  $T=2mg\sin \theta -T_2$
$T=10\sqrt{2}-\frac{10\sqrt{2}}{3}=\frac{20\sqrt{2}}{3}\therefore \left(C\to t\right)$   
Force  of friction between A and the wedge  is $f_1=T-mg\sin \theta =\frac{21\sqrt{2}}{3}-5\sqrt{2}$

$\Rightarrow f_1=\frac{5\sqrt{2}}{3}=\frac{10}{3\sqrt{2}}\therefore \left(A\to q\right)$