Physics - NLM Question with Solution | TestHub

From free body diagram of body $A$,$m_{1}a=T_1-m_{1}g$....(i) For body $B$, $m_2\mathrm{a}=m_{2}g+T_2-T_1$...(ii) For body C, $m_{3}a=m_{3}g-T_2$....(iii) On solving Eqs. (i), (ii) and (iii), we get $a=\frac{\left[\left(m_2+m_3\right)-m_1\right]g}{m_1+m_2+m_3}$ As, $\begin{array}{rl}{m}_1& =8\mathrm{kg},m_2=m_3=7\mathrm{kg}\\ a& =\frac{6}{22}\mathrm{g}=0.27\mathrm{g}\end{array}$