A mass 2 kg is whirled in a horizontal circle by means of a string at an initial speed of 5 revolutions per minute . Keeping the radius constant the tension in the string is doubled. The new speed is nearly

Options

A. 14rpm
B. 10rpm
C. 2.25 rpm
D. 7 rpm

Answer

To solve the problem step by step, we need to analyze the relationship between tension, speed, and centripetal force in circular motion.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Mass (m) = 2 kg
- Initial speed (ω₁) = 5 revolutions per minute (rpm)
- Tension in the string is doubled.

2. **Convert Initial Speed to Radians per Second:**
- We need to convert the speed from revolutions per minute to radians per second for calculations.
- \( \text{1 revolution} = 2\pi \text{ radians} \)
- \( \text{1 minute} = 60 \text{ seconds} \)
- Therefore,
\[ ω₁ = 5 \, \text{rpm} = 5 \times \frac{2\pi \, \text{radians}}{60 \, \text{seconds}} = \frac{5 \times 2\pi}{60} = \frac{\pi}{6} \, \text{radians/second} \]

3. **Calculate Initial Tension:**
- The tension in the string provides the centripetal force required for circular motion.
- The formula for centripetal force is:
\[ T = m \cdot ω₁^2 \cdot r \]
- Here, \( T \) is the tension, \( m \) is the mass, \( ω₁ \) is the angular speed, and \( r \) is the radius (which remains constant).

4. **New Tension:**
- According to the problem, the new tension \( T' \) is double the initial tension:
\[ T' = 2T \]

5. **Set Up the Equation for New Tension:**
- For the new speed \( ω₂ \), we can write:
\[ T' = m \cdot ω₂^2 \cdot r \]
- Since \( T' = 2T \), we can substitute:
\[ 2(m \cdot ω₁^2 \cdot r) = m \cdot ω₂^2 \cdot r \]

6. **Cancel Common Terms:**
- Since mass \( m \) and radius \( r \) are constant and non-zero, we can cancel them out:
\[ 2ω₁^2 = ω₂^2 \]

7. **Solve for New Speed:**
- Now, we can express \( ω₂ \):
\[ ω₂ = \sqrt{2} \cdot ω₁ \]
- Substituting \( ω₁ = \frac{\pi}{6} \):
\[ ω₂ = \sqrt{2} \cdot \frac{\pi}{6} \]

8. **Convert Back to Revolutions per Minute:**
- To convert \( ω₂ \) back to rpm:
\[ ω₂ \text{ (in rpm)} = ω₂ \cdot \frac{60}{2\pi} = \sqrt{2} \cdot \frac{\pi}{6} \cdot \frac{60}{2\pi} = \sqrt{2} \cdot 5 = 5\sqrt{2} \]
- Approximating \( \sqrt{2} \approx 1.414 \):
\[ ω₂ \approx 5 \cdot 1.414 \approx 7.07 \text{ rpm} \]

### Final Answer:
The new speed is approximately **7 rpm**.