The angular frequency of a fan increases uniformly from 30 rpm to 60 rpm in `pi` second. A dust particle is present at a distance of 20 cm from axis of rotation. The tangential acceleration of the particle is :

Options

A. `0.2 ms^(-2)`
B. `0.4 ms^(-2)`
C. `0.6 ms^(-2)`
D. `2m s^(-2)`

Explanation

To find the tangential acceleration of a dust particle on a fan that is accelerating uniformly, we can follow these steps:

### Step 1: Convert angular velocities from rpm to radian per second
The initial angular velocity (\( \omega_i \)) is 30 rpm and the final angular velocity (\( \omega_f \)) is 60 rpm. We need to convert these to radians per second.

\[ \omega_i = 30 \, \text{rpm} \times \frac{2\pi \, \text{radians}}{60 \, \text{seconds}} = \pi \, \text{radians/second} \]

\[ \omega_f = 60 \, \text{rpm} \times \frac{2\pi \, \text{radians}}{60 \, \text{seconds}} = 2\pi \, \text{radians/second} \]

### Step 2: Calculate angular acceleration (\( \alpha \))
Angular acceleration is defined as the change in angular velocity over time. The time interval (\( \Delta t \)) is given as \( \pi \) seconds.

\[ \alpha = \frac{\omega_f - \omega_i}{\Delta t} = \frac{2\pi - \pi}{\pi} = \frac{\pi}{\pi} = 1 \, \text{radians/second}^2 \]

### Step 3: Calculate the tangential acceleration (\( a_t \))
The tangential acceleration is related to the angular acceleration and the radius (distance from the axis of rotation). The distance is given as 20 cm, which we convert to meters:

\[ r = 20 \, \text{cm} = 0.2 \, \text{m} \]

The formula for tangential acceleration is:

\[ a_t = r \cdot \alpha \]

Substituting the values we have:

\[ a_t = 0.2 \, \text{m} \cdot 1 \, \text{radians/second}^2 = 0.2 \, \text{m/s}^2 \]

### Final Answer
The tangential acceleration of the particle is \( 0.2 \, \text{m/s}^2 \).

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