**Given:**
- Torque applied, 𝑇 = 60 Nm
- Mass moment of inertia of the flywheel, 𝐼 = 12 kgm²
- Time, 𝑡 = 20 s

**To find:**
1) Angular velocity of the flywheel after 20s
2) Number of revolutions the flywheel will make after 20s

**Solution:**

1) **Angular Velocity:**
The torque applied to a flywheel is given by the equation:

𝑇 = 𝐼 𝛼

Where 𝛼 is the angular acceleration.

To find the angular velocity, 𝜔, after a given time, we need to find the angular acceleration first.

The moment of inertia, 𝐼, is given by the equation:

𝐼 = 𝑚𝑎𝑠𝑠 × 𝑟²

Where 𝑚𝑎𝑠𝑠 is the mass of the flywheel and 𝑟 is the radius of the flywheel.

Given that 𝑚𝑎𝑠𝑠 = 12 kgm², we can rearrange the equation to solve for 𝑟:

12 kgm² = 𝑚𝑎𝑠𝑠 × 𝑟²
𝑟² = 12 kgm² / 𝑚𝑎𝑠𝑠
𝑟² = 12 kgm² / 12 kg
𝑟² = 1 m²
𝑟 = 1 m

Now, we can substitute the values of 𝑡 = 20 s, 𝐼 = 12 kgm², and 𝑟 = 1 m into the torque equation:

60 Nm = (12 kgm²) × 𝛼
𝛼 = 60 Nm / 12 kgm²
𝛼 = 5 rad/s²

The angular acceleration, 𝛼, is 5 rad/s².

Using the equation:

𝜔 = 𝜔₀ + 𝛼𝑡

Where 𝜔₀ is the initial angular velocity (which is zero in this case), 𝜔 is the final angular velocity, 𝛼 is the angular acceleration, and 𝑡 is the time.

Substituting the values into the equation:

𝜔 = 0 + (5 rad/s²) × 20 s
𝜔 = 100 rad/s

Therefore, the angular velocity of the flywheel after 20s is 100 rad/s.

2) **Number of Revolutions:**
The number of revolutions, 𝑁, can be calculated using the equation:

𝑁 = 𝜔 / (2𝜋)

Substituting the value of 𝜔 = 100 rad/s into the equation:

𝑁 = 100 rad/s / (2𝜋)
𝑁 ≈ 15.92 revolutions

Therefore, the flywheel will make approximately 15.92 revolutions after 20s.

In summary,
- The angular velocity of the flywheel after 20s is 100 rad/s.
- The flywheel will make approximately 15.92 revolutions after 20s.