Let $m_1$ and $m_2$ be the masses suspended at the ends of a light inextensible string passing over the pulley.$m_1 = 8\text{ kg}, m_2 = 12\text{ kg},$Let T be the tension in the string and a be the common acceleration with which $m_1$ moves upwards and $m_2$ moves downward = ?The equation of motion of $m_1$ and $m_2$ are given by$T - m_1g = m_1a$ ...(i)and $m_2g - T = m_2a$ ...(ii)Adding (i) and (ii), we get$(m_2 - m_1) g = (m_1 + m_2)a$$\therefore a = \frac{(m_2 - m_1)g}{m_1 + m_2}$...(iii)From (i) and (iii), we get$T = m_1 g + m_1 \frac{(m_2 - m_1)g}{m_1 + m_2}$$= \frac{m_1 g}{m_1 + m_2} (m_1 + m_2 + m_2 - m_1)$or $T = \frac{2 m_1 m_2}{m_1 + m_2} g$...(iv)Putting $m_1 = 8\text{ kg}$ and $m_2 = 12\text{ kg}$ and $g = 10\text{ m s}^{-2},$ in equations (iii), (iv), we get$a = \left( \frac{12-8}{8+12} \right) \times 10$$= \frac{4}{20} \times 10 = 2\text{ m s}^{-2}$and $T = \frac{2 \times 8 \times 12}{8+12} \times 10$$= \frac{2 \times 96 \times 10}{20} = 96\text{ N}$$\therefore a = 2\text{ m s}^{-2}, T = 96\text{ N}$