Step 1: Understand the gear system.
Two mating spur gears have 36 teeth (pinion) and 108 teeth (gear). The pinion rotates at 1200 rpm and transmits a torque of 20 Nm. We need to find the torque transmitted by the gear.
Step 2: Determine the gear ratio.
The gear ratio is the ratio of the number of teeth on the gear to the number of teeth on the pinion:
\[
\text{Gear ratio} = \frac{\text{Number of teeth on gear}}{\text{Number of teeth on pinion}} = \frac{108}{36} = 3.
\]
Step 3: Relate the torque using the gear ratio.
In a gear system, the power transmitted is the same (assuming no losses):
\[
\text{Power} = \text{Torque} \times \text{Angular velocity}.
\]
Torque on pinion \( T_p = 20 \, \text{Nm} \),
Speed of pinion \( N_p = 1200 \, \text{rpm} \).
The angular velocity \( \omega \) is:
\[
\omega_p = \frac{2\pi N_p}{60} = \frac{2\pi \times 1200}{60} = 40\pi \, \text{rad/s}.
\]
Power transmitted by the pinion:
\[
P = T_p \omega_p = 20 \times 40\pi = 800\pi \, \text{W}.
\]
The gear ratio also gives the speed ratio (inverse of gear ratio):
\[
\frac{N_p}{N_g} = \frac{T_g}{T_p} = 3,
\]
\[
N_g = \frac{N_p}{3} = \frac{1200}{3} = 400 \, \text{rpm},
\]
\[
\omega_g = \frac{2\pi N_g}{60} = \frac{2\pi \times 400}{60} = \frac{40\pi}{3} \, \text{rad/s}.
\]
Power transmitted by the gear:
\[
P = T_g \omega_g,
\]
\[
800\pi = T_g \times \frac{40\pi}{3},
\]
\[
T_g = \frac{800\pi}{\frac{40\pi}{3}} = 800 \times \frac{3}{40} = 60 \, \text{Nm}.
\]
Alternatively, use the torque ratio directly:
\[
\frac{T_g}{T_p} = \text{Gear ratio},
\]
\[
T_g = T_p \times \text{Gear ratio} = 20 \times 3 = 60 \, \text{Nm}.
\]
Step 4: Select the correct answer.
The torque transmitted by the gear is 60 Nm, matching option (4).