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 option is 2000.

Analysis of the Gear Train:
Based on the provided diagram, we can identify a compound gear train configuration consisting of three shafts:
1. Input Shaft (Left): Carries a gear with 12 teeth, subjected to an input torque of $T=100\text{N-m}$.
2. Intermediate Shaft (Middle): Carries a compound gear system consisting of a larger gear with 48 teeth (which meshes with the input gear) and a smaller gear with 12 teeth sharing the same shaft.
3. Output Shaft (Right): Carries a gear with 60 teeth, which meshes with the 12-teeth intermediate gear.

Step 1: Calculate the Speed Ratios
For any pair of meshing gears, the speed of rotation is inversely proportional to the number of teeth. Let $N$ represent the rotational speed (rpm) and $Z$ represent the number of teeth.

For the first gear mesh (input gear with 12 teeth to the intermediate gear with 48 teeth):

$\frac{N_1}{N_2}=\frac{Z_2}{Z_1}=\frac{48}{12}=4$

Since the 48-teeth gear and the second 12-teeth gear are mounted on the same intermediate shaft, they rotate at the same speed:

$N_2=N_3$

For the second gear mesh (intermediate gear with 12 teeth to the output gear with 60 teeth):

$\frac{N_3}{N_4}=\frac{Z_4}{Z_3}=\frac{60}{12}=5$

Step 2: Determine the Overall Gear Train Ratio
The overall speed ratio (velocity ratio) of the compound gear train is the product of the individual stage ratios:

$\frac{N_{\mathrm{in}}}{N_{\mathrm{out}}}=\frac{N_1}{N_4}=\left(\frac{N_1}{N_2}\right)\times \left(\frac{N_3}{N_4}\right)=4\times 5=20$

Step 3: Calculate the Output Torque
Since the gear train is frictionless, mechanical power is conserved during transmission (${\mathrm{Power}}_{\mathrm{in}}={\mathrm{Power}}_{\mathrm{out}}$):

$T_{\mathrm{in}}\times {\omega }_{\mathrm{in}}=T_{\mathrm{out}}\times {\omega }_{\mathrm{out}}$

where $\omega$ represents angular velocity. This relationship allows us to solve for the output torque as follows:

$T_{\mathrm{out}}=T_{\mathrm{in}}\times \left(\frac{N_{\mathrm{in}}}{N_{\mathrm{out}}}\right)$

Substituting the given values:

$T_{\mathrm{out}}=100\text{N-m}\times 20=2000\text{N-m}$

Therefore, the output torque from the 60-teeth gear is 2000 N-m.