Suppose, the torque acting on a body, is given by `tau = KL+(MI)/(omega)`
  Where L = angular momentum, I = moment of inertia & `omega`= angular speed What is the dimensional formula for K & M?

Answer

To find the dimensional formulas for \( K \) and \( M \) in the equation \( \tau = KL + \frac{MI}{\omega} \), we will follow these steps:

### Step 1: Determine the dimensional formula for torque (\( \tau \))
Torque is defined as the product of force and distance. The dimensional formula for force (\( F \)) is given by:
\[ F = m \cdot a = m \cdot \frac{L}{T^2} = M L T^{-2} \]
where \( M \) is mass, \( L \) is length, and \( T \) is time.

Thus, the dimensional formula for torque (\( \tau \)) is:
\[ \tau = F \cdot r = (M L T^{-2}) \cdot L = M L^2 T^{-2} \]

### Step 2: Analyze the term \( KL \)
The term \( KL \) must have the same dimensional formula as torque since they are being added. The dimensional formula for angular momentum (\( L \)) is:
\[ L = M L^2 T^{-1} \]
Thus, the dimensional formula for \( KL \) can be expressed as:
\[ KL = K \cdot (M L^2 T^{-1}) \]
Setting this equal to the dimensional formula for torque, we have:
\[ K \cdot (M L^2 T^{-1}) = M L^2 T^{-2} \]

### Step 3: Solve for \( K \)
To find \( K \), we can rearrange the equation:
\[ K = \frac{M L^2 T^{-2}}{M L^2 T^{-1}} = T^{-1} \]
Thus, the dimensional formula for \( K \) is:
\[ K = M^0 L^0 T^{-1} = T^{-1} \]

### Step 4: Analyze the term \( \frac{MI}{\omega} \)
Next, we need to analyze the term \( \frac{MI}{\omega} \). The dimensional formula for angular speed (\( \omega \)) is:
\[ \omega = \frac{\Delta \theta}{\Delta t} \quad \text{(dimensionless angle over time)} \]
Thus, the dimensional formula for \( \omega \) is:
\[ \omega = T^{-1} \]

### Step 5: Set up the equation for \( \frac{MI}{\omega} \)
The term \( \frac{MI}{\omega} \) must also equal the torque:
\[ \frac{MI}{\omega} = \frac{M \cdot I}{T^{-1}} = M \cdot I \cdot T \]
Setting this equal to the torque gives:
\[ M \cdot I \cdot T = M L^2 T^{-2} \]

### Step 6: Solve for \( M \)
Rearranging gives:
\[ M \cdot I = M L^2 T^{-2} \cdot T^{-1} = M L^2 T^{-3} \]
Thus, we can conclude that:
\[ I = L^2 T^{-2} \]
This means the dimensional formula for \( M \) is:
\[ M = M^1 L^2 T^{-3} \]
Therefore, the dimensional formula for \( M \) is:
\[ M = M^1 L^2 T^{-3} \]

### Final Results
- The dimensional formula for \( K \) is \( T^{-1} \).
- The dimensional formula for \( M \) is \( M^1 L^2 T^{-3} \).