The time period of a compound pendulum is given by
`T=2pisqrt((I)/(mgl))`
`{:(,"where","I=moment of inertia about the centre of the suspension,"),(,,"g=acceleration due to gravity , m=mass of the pendulam"),(,,"l=distance of the centre of the gravity from the centre of the suspension."):}`

Answer

To verify the dimensional correctness of the formula for the time period of a compound pendulum, we will follow these steps:

### Step 1: Identify the formula
The formula for the time period \( T \) of a compound pendulum is given by:
\[ T = 2\pi \sqrt{\frac{I}{mgl}} \]
where:
- \( I \) = moment of inertia about the center of suspension
- \( m \) = mass of the pendulum
- \( g \) = acceleration due to gravity
- \( l \) = distance from the center of gravity to the center of suspension

### Step 2: Determine the dimensions of the left-hand side (LHS)
The left-hand side of the equation is the time period \( T \). The dimension of time \( T \) is:
\[ [T] = [M^0 L^0 T^1] \]

### Step 3: Determine the dimensions of the right-hand side (RHS)
The right-hand side involves the expression \( \sqrt{\frac{I}{mgl}} \). We need to find the dimensions of \( I \), \( m \), \( g \), and \( l \).

1. **Moment of Inertia \( I \)**:
The moment of inertia for a point mass is given by \( I = mr^2 \). Thus, its dimension is:
\[ [I] = [M^1 L^2 T^0] \]

2. **Mass \( m \)**:
The dimension of mass is:
\[ [m] = [M^1 L^0 T^0] \]

3. **Acceleration due to gravity \( g \)**:
The dimension of acceleration is:
\[ [g] = [M^0 L^1 T^{-2}] \]

4. **Distance \( l \)**:
The dimension of distance is:
\[ [l] = [M^0 L^1 T^0] \]

### Step 4: Combine the dimensions in the RHS expression
Now we can substitute these dimensions into the RHS:
\[ \frac{I}{mgl} = \frac{[M^1 L^2 T^0]}{[M^1][M^0 L^1 T^{-2}][M^0 L^1]} \]
This simplifies to:
\[ \frac{[M^1 L^2 T^0]}{[M^1 L^2 T^{-2}]} = [M^0 L^0 T^2] \]

### Step 5: Take the square root
Now, we take the square root of the result:
\[ \sqrt{[M^0 L^0 T^2]} = [M^0 L^0 T^1] \]

### Step 6: Multiply by \( 2\pi \)
Since \( 2\pi \) is a dimensionless constant, it does not affect the dimensions:
\[ T = 2\pi \sqrt{\frac{I}{mgl}} \implies [T] = [M^0 L^0 T^1] \]

### Step 7: Compare LHS and RHS
Now we compare the dimensions of LHS and RHS:
- LHS: \( [M^0 L^0 T^1] \)
- RHS: \( [M^0 L^0 T^1] \)

Since both sides are equal, the formula is dimensionally correct.

### Conclusion
The formula for the time period of a compound pendulum is dimensionally correct.

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