Two blocks A and B are placed on an inclined plane of inclination 30° as shown in figure. If A and B each have mass $m$ and $μ_{A} = 0$ and $μ_{B} = \frac{2}{3}$ , where $μ_{A}$ and $μ_{B}$ are the coefficients of friction between plane and the bodies $A & B$ respectively,

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Let the acceleration of the system be ' $a$ ' and tension in the string be ' $T$ '.

Free Body Diagram (F.B.D.) of A:

$m g sin θ - T = ma$

$T = m g sin θ - ma$ ...........(1)

Free Body Diagram (F.B.D.) of B:

$T + m g sin θ - f = ma$ .........(2)

$N = m g cos θ$ ...........(3)

$f = μ N$ ............(4)

Substituting $T$ and $f$ :

$m g sin θ - ma + m g sin θ - f = ma$

$\Rightarrow 2 m g sin θ - μ m g cos θ = 2 ma$

$\Rightarrow 2 m g \times \frac{1}{2} - \frac{2}{3} m g \frac{3}{2} = 2 ma$

$\Rightarrow a = \frac{g}{2} [1 - \frac{1}{2}]$ ............(A)

On putting the value of ' $a$ ' in equation (1), we have:

$T = m g \times (\frac{1}{2}) - m \frac{g}{2} [1 - \frac{1}{2}]$

$\Rightarrow T = \frac{m g}{2 2}$ ................(B)

Options (A) and (B) are correct.

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