A three$-$phase, Y$-$connected, $220-$V $($line$-$to$-$line$),7\mathrm{.}5-$kW$,60-$Hz$,6-$pole induction motor has the following parameter values in $/$phase referred to the stator: R$1=0\mathrm{.}294,$ R$2=0\mathrm{.}144,$ Rc $=415,$ X$1=0\mathrm{.}503,$ X$2=0\mathrm{.}209,$ Xm $=13\mathrm{.}25$ Two loads are available and can be modelled by the below equations: $1=60+0\mathrm{.}3752=80+0\mathrm{.}4872(1)$ Which load can be started by this motor? Justify your answer mathematically. $(2)$ For the chosen load pf Part $(1)$ calculate: a$.$ The rotor slip and rotor speed b$.$ The output torque and power c$.$ The stator current $(3)$ Calculate the value of resistance that can be inserted in series with the rotor that will be just sufficient to start the other load.