A mass 2 kg describes a circle of radius 1 m on a smooth horizontal table at a uniform speed .If is joined to the centre of the circle by a string, which can just withstand 32 N, then the greatest number of revolution per minute , perfomed by the mass would be

Options

A. 38
B. 4
C. 76
D. 16

Explanation

To find the greatest number of revolutions per minute that a mass can perform while being attached to a string that can withstand a maximum tension of 32 N, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass (m) = 2 kg
- Radius (r) = 1 m
- Maximum Tension (T) = 32 N

2. **Understand the Relationship between Tension and Centripetal Force:**
- The tension in the string provides the centripetal force required for circular motion.
- The formula for centripetal force (F_c) is given by:
\[ F_c = m \omega^2 r \]
- Here, \( \omega \) is the angular velocity in radians per second.

3. **Set Up the Equation:**
- Since the maximum tension the string can withstand is 32 N, we can set up the equation:
\[ T = m \omega^2 r \]
- Substituting the known values:
\[ 32 = 2 \cdot \omega^2 \cdot 1 \]

4. **Solve for Angular Velocity (\( \omega \)):**
- Rearranging the equation gives:
\[ \omega^2 = \frac{32}{2} = 16 \]
- Taking the square root:
\[ \omega = \sqrt{16} = 4 \text{ rad/s} \]

5. **Convert Angular Velocity to Revolutions per Minute (RPM):**
- To convert from radians per second to revolutions per minute, use the conversion factor:
\[ \text{RPM} = \omega \cdot \frac{60}{2\pi} \]
- Substituting the value of \( \omega \):
\[ \text{RPM} = 4 \cdot \frac{60}{2\pi} = \frac{240}{2\pi} = \frac{120}{\pi} \]

6. **Calculate the Final Value:**
- Using \( \pi \approx 3.14 \):
\[ \text{RPM} \approx \frac{120}{3.14} \approx 38.19 \]

7. **Final Answer:**
- The greatest number of revolutions per minute performed by the mass is approximately **38.19 RPM**. The closest answer is **38 RPM**.