A stone tied to the end of string 100cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 22s, then the acceleration of the stone is

Options

A. `16ms^(-2)`
B. `4ms^(-2)`
C. `12ms^(-2)`
D. `8ms^(-2)`

Explanation

To solve the problem step by step, we will find the centripetal acceleration of the stone being whirled in a horizontal circle.

### Step 1: Identify the Given Values
- Length of the string (radius, r) = 100 cm = 1 m
- Number of revolutions (n) = 14
- Time taken for these revolutions (t) = 22 s

### Step 2: Calculate the Time Period (T)
The time period (T) is the time taken for one complete revolution. It can be calculated using the formula:
\[ T = \frac{t}{n} \]
Substituting the values:
\[ T = \frac{22 \, \text{s}}{14} \approx 1.57 \, \text{s} \]

### Step 3: Calculate the Angular Velocity (ω)
The angular velocity (ω) can be calculated using the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of T:
\[ \omega = \frac{2\pi}{1.57} \approx 4 \, \text{rad/s} \]

### Step 4: Calculate the Centripetal Acceleration (a_c)
The centripetal acceleration (a_c) can be calculated using the formula:
\[ a_c = r \omega^2 \]
Substituting the values of r and ω:
\[ a_c = 1 \, \text{m} \times (4 \, \text{rad/s})^2 \]
\[ a_c = 1 \, \text{m} \times 16 \, \text{(rad/s)}^2 \]
\[ a_c = 16 \, \text{m/s}^2 \]

### Final Answer
The acceleration of the stone is \( 16 \, \text{m/s}^2 \).

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