A uniform disc rotating freely about a vertical axis makes `90` rpm. A small piece of wax of mass `m` gram falls vertically on the disc and sticks to it at a distance `r cm` from the axis. If number of rotations per minute reduces to `60`, find the moment of inertia of the disc.

Options

A. `m^(2)`
B. `(3)/(2)mr^(2)`
C. `2mr^(2)`
D. `3mr^(2)`

Answer

To solve the problem, we will use the principle of conservation of angular momentum. The steps are as follows:

### Step 1: Understand the initial and final conditions
- The initial angular velocity of the disc, \( \omega_i \), is given as 90 rpm.
- The final angular velocity of the disc after the wax sticks to it, \( \omega_f \), is given as 60 rpm.

### Step 2: Convert angular velocities from rpm to rad/s
To apply the conservation of angular momentum, we need to convert the angular velocities from revolutions per minute (rpm) to radians per second (rad/s).

\[ \omega_i = 90 \, \text{rpm} = 90 \times \frac{2\pi}{60} \, \text{rad/s} = 9.42 \, \text{rad/s} \]

\[ \omega_f = 60 \, \text{rpm} = 60 \times \frac{2\pi}{60} \, \text{rad/s} = 6.28 \, \text{rad/s} \]

### Step 3: Set up the conservation of angular momentum equation
According to the conservation of angular momentum:

\[ L_i = L_f \]

Where:
- \( L_i \) is the initial angular momentum.
- \( L_f \) is the final angular momentum.

The initial angular momentum \( L_i \) can be expressed as:

\[ L_i = I \cdot \omega_i \]

Where \( I \) is the moment of inertia of the disc.

The final angular momentum \( L_f \) can be expressed as:

\[ L_f = (I + m r^2) \cdot \omega_f \]

Where \( m \) is the mass of the wax and \( r \) is the distance from the axis.

### Step 4: Write the equation using the conservation of angular momentum
Setting the initial and final angular momentum equal gives us:

\[ I \cdot \omega_i = (I + m r^2) \cdot \omega_f \]

### Step 5: Substitute the known values
Substituting the values of \( \omega_i \) and \( \omega_f \):

\[ I \cdot 9.42 = (I + m r^2) \cdot 6.28 \]

### Step 6: Rearranging the equation
Expanding the right side:

\[ I \cdot 9.42 = I \cdot 6.28 + m r^2 \cdot 6.28 \]

Rearranging gives:

\[ I \cdot 9.42 - I \cdot 6.28 = m r^2 \cdot 6.28 \]

\[ I (9.42 - 6.28) = m r^2 \cdot 6.28 \]

### Step 7: Solve for the moment of inertia \( I \)
Calculating \( 9.42 - 6.28 \):

\[ I \cdot 3.14 = m r^2 \cdot 6.28 \]

Thus,

\[ I = \frac{m r^2 \cdot 6.28}{3.14} \]

Since \( 6.28 / 3.14 = 2 \):

\[ I = 2 m r^2 \]

### Final Answer
The moment of inertia of the disc is:

\[ I = 2 m r^2 \]