A stonetied to the end of a string 100 cm long is whirled in a horizontal circle with a constant speed . If the stone makes 14 revolutions in 25 s, what is the acceleration of the stone ?

Options

A. `(88/25)^2 ms^(-2)`
B. `(25/88)^2 ms^(-2)`
C. `(88/25) ms^(-2)`
D. `(25/88) ms^(-2)`

Explanation

To find the acceleration of the stone being whirled in a horizontal circle, we can follow these steps:

### Step 1: Identify the given data
- Length of the string (radius of the circle), \( r = 100 \) cm = \( 1 \) m
- Number of revolutions, \( n = 14 \)
- Time taken for those revolutions, \( t = 25 \) s

### Step 2: Calculate the frequency of revolutions
The frequency \( f \) (number of revolutions per second) can be calculated as:
\[ f = \frac{n}{t} = \frac{14}{25} \text{ revolutions per second} \]

### Step 3: Convert frequency to angular velocity
The angular velocity \( \omega \) in radians per second can be calculated using the formula:
\[ \omega = 2\pi f \]
Substituting the value of \( f \):
\[ \omega = 2\pi \left(\frac{14}{25}\right) = \frac{28\pi}{25} \text{ radians per second} \]

### Step 4: Calculate the centripetal acceleration
Centripetal acceleration \( a_c \) can be calculated using the formula:
\[ a_c = \omega^2 r \]
Substituting the values of \( \omega \) and \( r \):
\[ a_c = \left(\frac{28\pi}{25}\right)^2 \times 1 \]
Calculating \( \omega^2 \):
\[ \omega^2 = \left(\frac{28\pi}{25}\right)^2 = \frac{784\pi^2}{625} \]
Thus, the centripetal acceleration is:
\[ a_c = \frac{784\pi^2}{625} \text{ m/s}^2 \]

### Step 5: Calculate the numerical value
Using \( \pi \approx 3.14 \):
\[ a_c \approx \frac{784 \times (3.14)^2}{625} \approx \frac{784 \times 9.8596}{625} \approx \frac{7724.7936}{625} \approx 12.36 \text{ m/s}^2 \]

### Final Answer
The acceleration of the stone is approximately \( 12.36 \text{ m/s}^2 \).

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