Answer is 6.04Nm. How do i get this answer please ?

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Answer:-Given that,Number of tecth of the input egear, N_{1}=45No. of teeth of the output ejear, N_{2}=80Gear efficiency =95 \%=0.95=\eta_{G}.The output shaft is to deliver torque T_{0}=10.2 \mathrm{Nm}.Therefove,The power deliver from Motor input eyear, N_{1}=45 input to output shaft via ejear.so, By using formulae for power flow equation is\begin{aligned}& P_{\text {out }}+\eta_{G} \times P_{i n}=0 . \\\because P_{\text {out }}= & Z_{\text {orit }} \times \omega_{0} . \\P_{\text {in }}= & T_{\text {ipit }} \times \omega_{\text {in }} .\end{aligned}outpat year, N_{2}=80^{\circ}substituat\begin{array}{l}T_{0} \times \omega_{0}+\eta_{G}\left[T_{i} x \omega_{i}\right]=0 . \\T_{0}=10.2 \mathrm{Nm}, \eta_{G}=0.95 \\\end{array}substituted valus. in eq(2), we eget\begin{array}{l}(10.2) \omega_{0}+\left(\eta_{G}=0.95\right)\left(T_{i} \times \omega_{i}\right)=0 . \\10.2 \omega_{0}=-0.95 T_{i} \omega_{i} \\\frac{10.2}{0.95}=-T_{i} \frac{\omega_{\text {input }}}{\omega_{\text {output }}}\end{array}\therefore rejer of lar is.(2)\frac{N_{2}}{N_{1}}=\frac{\omega_{1}}{\omega_{0}}=\frac{80}{45}substituted\begin{array}{c}\frac{10.2}{0.95}=-T_{i}\left[\frac{80}{45}\right] \\T_{i}=-\left[\frac{10.2 \times 45}{80 \times 0.95}\right] \\T_{i}=-6.03947 \mathrm{Nm} \\\therefore T_{i} \simeq-6.04 \mathrm{Nm}\end{array}Here-ve sign shows reverse rotation asTherefove, Given Answer.The torque required on the input shaft, \tau_{i}=6.04 \mathrm{Nm} ...