Here, $F = 600\,\text{N};$ $m_1 = 10\,\text{kg}; m_2 = 20$ kg. Let T be tension in the string and a be the acceleration of the system, in the direction of force applied.(a) When force is applied on lighter block A, as shown in figure (a) The equation of motion of masses $m_1 \text{ and } m_2$ are given by$F - T = m_1 a$ ...(i)$T = m_2 a$ ...(ii)Divide (i) by (ii), we get$\frac{F-T}{T} = \frac{m_1 a}{m_2 a};$$\frac{F}{T} - 1 = \frac{m_1}{m_2} = \frac{10}{20} = \frac{1}{2}$$\Rightarrow \frac{F}{T} = \frac{1}{2} + 1 = \frac{3}{2};$$T = \frac{2}{3} F$$= \frac{2}{3} \times 600 = 400\,\text{N}$From equation (ii), $a = \frac{T}{m_2} = \frac{400}{20} = 20\,\text{m}\,\text{s}^{-2}$(b) When force is applied on heavier block B as shown in figure (c) $T = m_1 a$ ...(iii)$F - T = m_2 a$ ...(iv)Divide (iv) by (iii), we get$\frac{F-T}{T} = \frac{m_2 a}{m_1 a} = \frac{20}{10} = 2;$$\frac{F}{T} - 1 = 2; \quad \frac{F}{T} = 3,$$T = \frac{F}{3} = \frac{600}{3} = 200\,\text{N}$From equation (iii) $a = \frac{T}{m_1} = \frac{200}{10} = 20\,\text{m}\,\text{s}^{-2}$