A body of mass `200 g` oscillates about a horizontal axis at a distance of `20 cm` from its centre of gravity . If the length of the equivalent simple pendulum is `35cm`, find its moment of inertia about the point of suspension.

Explanation

To find the moment of inertia of a body oscillating about a horizontal axis, we can use the relationship between the time period of a simple pendulum and the moment of inertia. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Mass of the body, \( m = 200 \, \text{g} = 0.2 \, \text{kg} \) (since \( 1 \, \text{g} = 0.001 \, \text{kg} \))
- Distance from the center of gravity to the point of suspension, \( d = 20 \, \text{cm} = 0.2 \, \text{m} \) (since \( 1 \, \text{cm} = 0.01 \, \text{m} \))
- Length of the equivalent simple pendulum, \( L = 35 \, \text{cm} = 0.35 \, \text{m} \)

### Step 2: Write the formula for the time period of a simple pendulum
The time period \( T \) of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

### Step 3: Write the formula for the time period about the point of suspension
The time period \( T \) for a body oscillating about a point of suspension is given by:
\[ T = 2\pi \sqrt{\frac{I}{mgd}} \]
where \( I \) is the moment of inertia about the point of suspension, \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( d \) is the distance from the center of mass to the point of suspension.

### Step 4: Set the two expressions for \( T \) equal to each other
Since both expressions represent the same time period, we can set them equal:
\[ 2\pi \sqrt{\frac{L}{g}} = 2\pi \sqrt{\frac{I}{mgd}} \]
We can cancel \( 2\pi \) from both sides:
\[ \sqrt{\frac{L}{g}} = \sqrt{\frac{I}{mgd}} \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:
\[ \frac{L}{g} = \frac{I}{mgd} \]

### Step 6: Rearrange to find \( I \)
Rearranging the equation to solve for \( I \):
\[ I = mgd \cdot L \]

### Step 7: Substitute the known values
Now, substituting the known values into the equation:
- \( m = 0.2 \, \text{kg} \)
- \( g = 9.81 \, \text{m/s}^2 \)
- \( d = 0.2 \, \text{m} \)
- \( L = 0.35 \, \text{m} \)

\[ I = (0.2 \, \text{kg})(9.81 \, \text{m/s}^2)(0.2 \, \text{m})(0.35 \, \text{m}) \]

### Step 8: Calculate \( I \)
Calculating the values:
\[ I = 0.2 \times 9.81 \times 0.2 \times 0.35 \]
\[ I = 0.2 \times 9.81 \times 0.07 \]
\[ I = 0.2 \times 0.6867 \approx 0.13734 \, \text{kg m}^2 \]

### Step 9: Final result
The moment of inertia about the point of suspension is:
\[ I \approx 0.014 \, \text{kg m}^2 \]