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

Answer

To find the dimensional formula for \( KM \) given the equation for torque \( \tau = KL + \frac{MI}{\omega} \), we will follow these steps:

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

### 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 (\( \omega \)):
\[ L = I \cdot \omega \]
The dimensional formula for moment of inertia (\( I \)) is:
\[ I = M \cdot L^2 \]
The dimensional formula for angular velocity (\( \omega \)) is:
\[ \omega = T^{-1} \]
Thus, the dimensional formula for angular momentum (\( L \)) becomes:
\[ L = (M \cdot L^2) \cdot (T^{-1}) = M \cdot L^2 \cdot T^{-1} \]

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

### Step 4: Identify the dimensions of angular speed (\( \omega \))
The angular speed (\( \omega \)) has the dimensional formula:
\[ \omega = T^{-1} \]

### Step 5: Substitute dimensions into the equation for \( KM \)
From the equation \( \tau = KL + \frac{MI}{\omega} \), both terms on the right must have the same dimensions as torque (\( \tau \)).

1. For the term \( KL \):
\[ [K] \cdot [L] = [K] \cdot (M \cdot L^2 \cdot T^{-1}) = M \cdot L^2 \cdot T^{-2} \]
Therefore, the dimensional formula for \( K \) can be derived as:
\[ [K] = \frac{[M \cdot L^2 \cdot T^{-2}]}{[M \cdot L^2 \cdot T^{-1}]} = T^{-1} \]

2. For the term \( \frac{MI}{\omega} \):
\[ \frac{[M] \cdot [I]}{[\omega]} = \frac{M \cdot (M \cdot L^2)}{T^{-1}} = \frac{M^2 \cdot L^2}{T^{-1}} = M^2 \cdot L^2 \cdot T \]
This must also equal \( M \cdot L^2 \cdot T^{-2} \), leading to:
\[ [M] = \frac{[M \cdot L^2 \cdot T^{-2}] \cdot [T]}{[M \cdot L^2]} = M \cdot T \]

### Step 6: Combine the results for \( KM \)
Now, we need to find the dimensional formula for \( KM \):
\[ [KM] = [K] \cdot [M] = (T^{-1}) \cdot (M) = M \cdot T^{-1} \]

### Final Answer:
The dimensional formula for \( KM \) is:
\[ \boxed{M \cdot T^{-1}} \]