In the system shown in figure ${\mathrm{m}}_{\mathrm{B}}$ = 4kg, and ${\mathrm{m}}_{\mathrm{A}}$ = 2 kg. The pulleys are massless and friction is absent everyrwhere. The acceleration of block A  is

For movable pulley

${\mathrm{a}}_{\mathrm{B}} = \frac{\mathrm{O}+{\mathrm{a}}_{\mathrm{A}}}{2}$

${\mathrm{a}}_{\mathrm{A}} = 2{\mathrm{a}}_{\mathrm{B}}$--------(i)

Free body diagrams of A and B:

Equation of motion of block A and B

For A : $\mathrm{T}-\frac{{\mathrm{m}}_{\mathrm{A}}\mathrm{g}}{2} = {\mathrm{m}}_{\mathrm{A}}{\mathrm{a}}_{\mathrm{A}} = {\mathrm{m}}_{\mathrm{A}}\left(2{\mathrm{a}}_{\mathrm{B}}\right)-----\left(\mathrm{ii}\right)$

For B : ${\mathrm{m}}_{\mathrm{B}}\mathrm{g}-2\mathrm{T} = {\mathrm{m}}_{\mathrm{B}}{\mathrm{a}}_{\mathrm{B}}$-------------(iii)

From (ii) and (iii) [2(ii) + (iii)]

${\mathrm{m}}_{\mathrm{B}}\mathrm{g}-{\mathrm{m}}_{\mathrm{A}}\mathrm{g} = [4{\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}]{\mathrm{a}}_{\mathrm{B}}$

$\Rightarrow {\mathrm{a}}_{\mathrm{B}} = \frac{[{\mathrm{m}}_{\mathrm{B}}-{\mathrm{m}}_{\mathrm{A}}]\mathrm{g}}{[4{\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}]} = \frac{[4-2]\times 10}{[4\times 2+4]}$

           = $\frac{20}{12} = \frac{5}{3}\mathrm{m}/{\mathrm{s}}^2$

Hence  ${\mathrm{a}}_{\mathrm{A}} = 2{\mathrm{a}}_{\mathrm{B}} = \frac{10}{3}\mathrm{m}/{\mathrm{s}}^2$