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 need to analyze the dimensions of each term in the equation.

### Step 1: Identify the dimensions of torque (\( \tau \))
Torque (\( \tau \)) is defined as the product of force and distance. The dimensional formula for force is given by:
\[ [F] = [M][L][T^{-2}] \]
Thus, the dimensional formula for torque is:
\[ [\tau] = [F][L] = [M][L][T^{-2}][L] = [M][L^2][T^{-2}] \]

### Step 2: Identify the dimensions of angular momentum (\( L \))
Angular momentum (\( L \)) is defined as the product of moment of inertia (\( I \)) and angular velocity (\( \omega \)). The dimensional formula for moment of inertia is:
\[ [I] = [M][L^2] \]
And the dimensional formula for angular velocity (\( \omega \)) is:
\[ [\omega] = [T^{-1}] \]
Thus, the dimensional formula for angular momentum is:
\[ [L] = [I][\omega] = [M][L^2][T^{-1}] = [M][L^2][T^{-1}] \]

### Step 3: Identify the dimensions of moment of inertia (\( I \))
As stated above, the dimensional formula for moment of inertia is:
\[ [I] = [M][L^2] \]

### Step 4: Identify the dimensions of angular speed (\( \omega \))
As stated above, the dimensional formula for angular speed is:
\[ [\omega] = [T^{-1}] \]

### Step 5: Determine the dimensions of \( K \)
From the equation \( \tau = KL \), we can express \( K \) as:
\[ K = \frac{\tau}{L} \]
Substituting the dimensions we found:
\[ [K] = \frac{[M][L^2][T^{-2}]}{[M][L^2][T^{-1}]} = \frac{[M][L^2][T^{-2}]}{[M][L^2][T^{-1}]} = [T^{-1}] \]
Thus, the dimensional formula for \( K \) is:
\[ [K] = [M^0][L^0][T^{-1}] \]

### Step 6: Determine the dimensions of \( M \)
From the equation \( \tau = \frac{MI}{\omega} \), we can express \( M \) as:
\[ M = \frac{\tau \cdot \omega}{I} \]
Substituting the dimensions we found:
\[ [M] = \frac{[M][L^2][T^{-2}][T^{-1}]}{[M][L^2]} = \frac{[M][L^2][T^{-3}]}{[M][L^2]} = [T^{-3}] \]
Thus, the dimensional formula for \( M \) is:
\[ [M] = [M^0][L^0][T^{-3}] \]

### Final Answer:
- The dimensional formula for \( K \) is \( [T^{-1}] \).
- The dimensional formula for \( M \) is \( [T^{-3}] \).