A fan of moment of inertia 0.6 kg `xx` `"metre"^2` is to run upto a working speed of 0.5 revolution per second. Indicate the correct value of the angular momentum of the fan

Options

A. `0.6pi kg xx ("metre"^(2))/("sec") `
B. `6pi kg xx ("metre"^(2))/("sec") `
C. `3pi kg xx ("metre"^(2))/("sec") `
D. `(pi)/(6) kg xx ("metre"^(2))/("sec") `

Answer

To find the angular momentum of the fan, we can use the formula for angular momentum (L):

\[ L = I \cdot \omega \]

where:
- \( L \) is the angular momentum,
- \( I \) is the moment of inertia,
- \( \omega \) is the angular speed in radians per second.

### Step 1: Identify the moment of inertia (I)
The moment of inertia of the fan is given as:

\[ I = 0.6 \, \text{kg} \cdot \text{m}^2 \]

### Step 2: Convert the angular speed (ω) from revolutions per second to radians per second
The angular speed is given as:

\[ \text{Angular speed} = 0.5 \, \text{revolutions per second} \]

To convert revolutions per second to radians per second, we use the fact that one revolution is equal to \( 2\pi \) radians:

\[ \omega = 0.5 \, \text{revolutions/second} \times 2\pi \, \text{radians/revolution} = \pi \, \text{radians/second} \]

### Step 3: Calculate the angular momentum (L)
Now we can substitute the values of \( I \) and \( \omega \) into the angular momentum formula:

\[ L = I \cdot \omega = 0.6 \, \text{kg} \cdot \text{m}^2 \cdot \pi \, \text{radians/second} \]

Calculating this gives:

\[ L = 0.6\pi \, \text{kg} \cdot \text{m}^2/\text{s} \]

### Step 4: Final Result
The angular momentum of the fan is:

\[ L \approx 1.884 \, \text{kg} \cdot \text{m}^2/\text{s} \quad (\text{using } \pi \approx 3.14) \]

Thus, the correct value of the angular momentum of the fan is approximately \( 1.884 \, \text{kg} \cdot \text{m}^2/\text{s} \).