d
\(\begin{array}{l}
Acceleration\,produced\,in\,upward\,direction\\
a = \frac{F}{{{M_1} + {M_2} + Mass\,of\,metal\,rod}}\\
= \frac{{480}}{{20 + 12 + 8}} = 12\,m{s^{ - 2}}\\
Tension\,at\,the\,mid\,{\rm{point}}\,\\
T = \left( {{M_2} + \frac{{Mass\,of\,rod}}{2}} \right)a\\
\,\,\,\, = \left( {12 + 4} \right) \times 12 = 192N
\end{array}\)