At first, machine $M_1$ will be associated with 2 object of job $P (30\times 2=60$ minutes$)$ and simultaneously, machine $M_2$ will do $1$ object of job $R$. So, in $1$ hour jobs $P$ and $R$ are completed.
Now, machine $M_1$ will do $3$ objects of job $Q$ for next hour $(20\times 3=60$ minutes$)$ while, machine $M_2$ will simultaneously complete $4$ objects of job $S(15\times 4= 60$ minutes$).$
Thus, minimum time needed to complete all the jobs $= 1 + 1 = 2$ hours.
Correct Answer: $A$