A ceiling fan rotates about its own axis with some angular velocity. When the fan is switched off, the angular velocity becomes `(1/4)`th of the original in time 't' and 'n' revolutions are made in that time. The number of revolutions made by the fan during the time interval between switch of and rest are (Angular retardation is uniform)

Options

A. `(4n)/(15)`
B. `(8n)/(15)`
C. `(16n)/(15)`
D. `(32n)/(15)`

Answer

To solve the problem step-by-step, we will use the equations of rotational motion.

### Step 1: Identify the given data
- Initial angular velocity: \( \omega_0 \)
- Final angular velocity after time \( t \): \( \omega = \frac{1}{4} \omega_0 \)
- Time taken: \( t \)
- Number of revolutions made during time \( t \): \( n \)

### Step 2: Use the equation of motion for angular velocity
The equation of motion for angular velocity in rotational dynamics is given by:
\[ \omega^2 = \omega_0^2 + 2\alpha\theta \]
Where:
- \( \alpha \) is the angular retardation (negative because the fan is slowing down)
- \( \theta \) is the angular displacement in radians

### Step 3: Substitute the known values
Substituting \( \omega = \frac{1}{4} \omega_0 \) and \( \theta = 2\pi n \) into the equation:
\[ \left(\frac{1}{4} \omega_0\right)^2 = \omega_0^2 + 2(-\alpha)(2\pi n) \]
This simplifies to:
\[ \frac{1}{16} \omega_0^2 = \omega_0^2 - 4\pi n \alpha \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ 4\pi n \alpha = \omega_0^2 - \frac{1}{16} \omega_0^2 \]
\[ 4\pi n \alpha = \frac{16}{16} \omega_0^2 - \frac{1}{16} \omega_0^2 = \frac{15}{16} \omega_0^2 \]
Thus, we can express \( \alpha \) in terms of \( n \):
\[ \alpha = \frac{15 \omega_0^2}{64 \pi n} \]

### Step 5: Find the total number of revolutions until rest
Using the equation again for the total angular displacement until the fan comes to rest:
\[ 0 = \omega_0^2 + 2(-\alpha)\theta' \]
Where \( \theta' \) is the total angular displacement until rest. Rearranging gives:
\[ \theta' = \frac{\omega_0^2}{2\alpha} \]

### Step 6: Substitute \( \alpha \)
Substituting the expression for \( \alpha \):
\[ \theta' = \frac{\omega_0^2}{2 \left(\frac{15 \omega_0^2}{64 \pi n}\right)} = \frac{64 \pi n}{30} = \frac{32 \pi n}{15} \]

### Step 7: Convert angular displacement to revolutions
To convert \( \theta' \) from radians to revolutions:
\[ \text{Number of revolutions} = \frac{\theta'}{2\pi} = \frac{\frac{32 \pi n}{15}}{2\pi} = \frac{32n}{30} = \frac{16n}{15} \]

### Final Answer
Thus, the total number of revolutions made by the fan during the time interval between switch off and rest is:
\[ n' = \frac{16n}{15} \]