A stone of mass `0.3 kg` attched to a `1.5 m` long stirng is whirled around in a horizontal cirlcle at a speed of 6 m/s The tension in the string is

Options

A. 10 N
B. 20 N
C. 7.2 N
D. none of these

Answer

To find the tension in the string when a stone is whirled in a horizontal circle, we can follow these steps:

### Step 1: Identify the given values
- Mass of the stone (m) = 0.3 kg
- Length of the string (which is the radius of the circular path, r) = 1.5 m
- Speed of the stone (v) = 6 m/s

### Step 2: Understand the relationship between tension and centripetal force
The tension in the string provides the necessary centripetal force to keep the stone moving in a circular path. The formula for centripetal force (F_c) is given by:

\[ F_c = \frac{mv^2}{r} \]

Where:
- \( m \) = mass of the stone
- \( v \) = speed of the stone
- \( r \) = radius of the circular path

### Step 3: Substitute the values into the formula
Now, we can substitute the known values into the centripetal force formula to find the tension (T):

\[ T = \frac{mv^2}{r} \]

Substituting the values:

\[ T = \frac{0.3 \, \text{kg} \times (6 \, \text{m/s})^2}{1.5 \, \text{m}} \]

### Step 4: Calculate \( v^2 \)
First, calculate \( v^2 \):

\[ v^2 = 6^2 = 36 \, \text{m}^2/\text{s}^2 \]

### Step 5: Substitute \( v^2 \) back into the equation
Now substitute \( v^2 \) back into the equation for tension:

\[ T = \frac{0.3 \, \text{kg} \times 36 \, \text{m}^2/\text{s}^2}{1.5 \, \text{m}} \]

### Step 6: Perform the multiplication
Calculate the numerator:

\[ 0.3 \times 36 = 10.8 \, \text{kg} \cdot \text{m}^2/\text{s}^2 \]

### Step 7: Divide by the radius
Now divide by the radius:

\[ T = \frac{10.8}{1.5} \]

### Step 8: Calculate the final tension
Perform the division:

\[ T = 7.2 \, \text{N} \]

### Conclusion
The tension in the string is \( 7.2 \, \text{N} \).