A boy ties a stone of mass 100 g to the end of a 2 m long string and whirls it around in a horizontal plane. The string can withstand the maximum tension of 80 N. If the maximum speed with which the stone can revolve is `K/pi` rev./min. The value of K is :
(Assume the string is massless and unstretchable)

Options

A. 400
B. 300
C. 600
D. 800

Answer

To solve the problem, we need to determine the maximum speed at which the stone can revolve around the boy while being tied to a string. The maximum tension that the string can withstand is given as 80 N.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the stone, \( m = 100 \, \text{g} = 0.1 \, \text{kg} \) (since 1 g = 0.001 kg)
- Length of the string (radius of the circular motion), \( r = 2 \, \text{m} \)
- Maximum tension in the string, \( T_{\text{max}} = 80 \, \text{N} \)

2. **Understand the Relationship Between Tension and Circular Motion:**
- The tension in the string provides the centripetal force required to keep the stone moving in a circle. The centripetal force \( F_c \) is given by the formula:
\[ F_c = \frac{m v^2}{r} \]
- Here, \( v \) is the linear speed of the stone.

3. **Set Up the Equation Using Maximum Tension:**
- Since the maximum tension is equal to the maximum centripetal force, we can write:
\[ T_{\text{max}} = \frac{m v^2}{r} \]
- Substituting the known values:
\[ 80 = \frac{0.1 v^2}{2} \]

4. **Solve for \( v^2 \):**
- Rearranging the equation:
\[ 80 \times 2 = 0.1 v^2 \]
\[ 160 = 0.1 v^2 \]
\[ v^2 = \frac{160}{0.1} = 1600 \]

5. **Calculate \( v \):**
- Taking the square root:
\[ v = \sqrt{1600} = 40 \, \text{m/s} \]

6. **Convert Linear Speed to Angular Speed:**
- The relationship between linear speed \( v \) and angular speed \( \omega \) is given by:
\[ v = r \omega \]
- Rearranging for \( \omega \):
\[ \omega = \frac{v}{r} = \frac{40}{2} = 20 \, \text{rad/s} \]

7. **Convert Angular Speed to Revolutions per Minute:**
- To convert from radians per second to revolutions per minute, we use the conversion factor \( \frac{1 \, \text{rev}}{2\pi \, \text{rad}} \) and \( 60 \, \text{s/min} \):
\[ \text{Revolutions per minute} = \omega \times \frac{60}{2\pi} = 20 \times \frac{60}{2\pi} = \frac{1200}{2\pi} = \frac{600}{\pi} \]

8. **Identify \( K \):**
- From the problem, we have \( K/\pi \) revolutions per minute, where \( K = 600 \).

### Final Answer:
The value of \( K \) is \( 600 \).