Check the correctness of the relation `pi = I alpha` whare `pi` is torque acting on the body, I is moment of inertia and `alpha` is angular acceleration.

Explanation

To check the correctness of the relation \( \tau = I \alpha \) where \( \tau \) is the torque acting on a body, \( I \) is the moment of inertia, and \( \alpha \) is the angular acceleration, we will analyze the dimensions of each term involved in the equation.

### Step 1: Determine the dimensions of torque (\( \tau \))
Torque is defined as the product of force and the distance from the pivot point (radius). The formula for torque is:
\[ \tau = F \times r \]
Where:
- \( F \) is the force
- \( r \) is the radius (distance)

The dimension of force (\( F \)) is given by:
\[ F = m \cdot a \]
Where \( a \) (acceleration) has dimensions of \( \frac{L}{T^2} \). Thus, the dimension of force is:
\[ [F] = [M^1 L^1 T^{-2}] \]

Now substituting this into the torque formula:
\[ [\tau] = [F] \times [r] = [M^1 L^1 T^{-2}] \times [L^1] = [M^1 L^2 T^{-2}] \]

### Step 2: Determine the dimensions of moment of inertia (\( I \))
The moment of inertia (\( I \)) is defined as:
\[ I = m \cdot r^2 \]
Where \( m \) is mass and \( r \) is the distance. Therefore, the dimensions of moment of inertia are:
\[ [I] = [M^1] \times [L^2] = [M^1 L^2] \]

### Step 3: Determine the dimensions of angular acceleration (\( \alpha \))
Angular acceleration (\( \alpha \)) is defined as the change in angular velocity per unit time. The dimension of angular acceleration can be expressed as:
\[ [\alpha] = \frac{\text{angular velocity}}{T} = \frac{\text{radians/second}}{T} \]
Since radians are dimensionless, we have:
\[ [\alpha] = [T^{-2}] \]

### Step 4: Combine the dimensions on the right-hand side of the equation
Now we can analyze the right-hand side of the equation \( I \alpha \):
\[ I \alpha = [I] \times [\alpha] = [M^1 L^2] \times [T^{-2}] = [M^1 L^2 T^{-2}] \]

### Step 5: Compare the dimensions of both sides
Now we compare the dimensions of both sides of the equation:
- Left-hand side (\( \tau \)): \( [M^1 L^2 T^{-2}] \)
- Right-hand side (\( I \alpha \)): \( [M^1 L^2 T^{-2}] \)

Since both sides have the same dimensions, we conclude that:
\[ \tau = I \alpha \text{ is dimensionally correct.} \]

### Conclusion
The relation \( \tau = I \alpha \) is dimensionally correct.