A frictionless gear train is shown in the figure. The leftmost 12-teeth gear is given a torque of 100 N-m. The output torque from the 60-teeth gear on the right in N-m is

The correct answer is 2000.

Based on the provided gear train diagram, we can observe the following labels and configurations for the gears:
- Input gear (leftmost shaft): has a teeth count of $N_1=12$ and is subjected to an input torque of $T=100\text{N-m}$.
- Intermediate compound shaft (middle): consists of a larger driven gear with $N_2=48$ teeth and a smaller driving gear with $N_3=12$ teeth mounted on the same shaft.
- Output gear (rightmost shaft): has a teeth count of $N_4=60$ teeth.

For a frictionless gear train, mechanical power is fully conserved. Therefore, the input power equals the output power:
$P_{\text{in}}=P_{\text{out}}\Rightarrow T_{\text{in}}{\omega }_{\text{in}}=T_{\text{out}}{\omega }_{\text{out}}$
This relationship allows us to determine the output torque as:
$T_{\text{out}}=T_{\text{in}}\times \frac{{\omega }_{\text{in}}}{{\omega }_{\text{out}}}$

Next, we determine the overall angular velocity ratio of the gear train step-by-step:
1. For the first mesh between the input gear and the intermediate driven gear:
$\frac{{\omega }_1}{{\omega }_2}=\frac{N_2}{N_1}=\frac{48}{12}=4$
2. Since Gear 2 and Gear 3 reside on the same physical intermediate shaft, they rotate at identical angular velocities:
${\omega }_2={\omega }_3$
3. For the second mesh between the intermediate driving gear and the final output gear:
$\frac{{\omega }_3}{{\omega }_4}=\frac{N_4}{N_3}=\frac{60}{12}=5$

By multiplying the individual stage ratios, we find the overall velocity ratio:
$\frac{{\omega }_{\text{in}}}{{\omega }_{\text{out}}}=\frac{{\omega }_1}{{\omega }_4}=\left(\frac{{\omega }_1}{{\omega }_2}\right)\times \left(\frac{{\omega }_3}{{\omega }_4}\right)=4\times 5=20$

Finally, we calculate the output torque at the 60-teeth gear:
$T_{\text{out}}=100\text{N-m}\times 20=2000\text{N-m}$