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. `(5)/(sqrt2)`rpm
B. 10 rpm
C. `10sqrt2`rpm
D. `5sqrt2`rpm

Explanation

To solve the problem step by step, we will use the relationship between tension, mass, angular velocity, and the number of revolutions per minute (RPM).

### Step 1: Understand the relationship between tension and RPM
The tension \( T \) in the string is related to the mass \( m \), angular velocity \( \omega \), and radius \( r \) of the circular motion. The formula for tension is given by:

\[ T = m \omega^2 r \]

Where \( \omega \) (angular velocity) can be expressed in terms of RPM as:

\[ \omega = 2\pi n \]

Here, \( n \) is the number of revolutions per minute.

### Step 2: Initial conditions
Given:
- Mass \( m = 2 \, \text{kg} \)
- Initial speed \( n_1 = 5 \, \text{RPM} \)

We can calculate the initial tension \( T_1 \):

\[ T_1 = m (2\pi n_1)^2 r \]

### Step 3: New conditions after doubling the tension
When the tension is doubled, we have:

\[ T_2 = 2T_1 \]

Using the relationship between tension and RPM, we can express the new tension in terms of the new RPM \( n_2 \):

\[ T_2 = m (2\pi n_2)^2 r \]

### Step 4: Set up the equation
Since \( T_2 = 2T_1 \), we can write:

\[ m (2\pi n_2)^2 r = 2 \cdot m (2\pi n_1)^2 r \]

### Step 5: Cancel out common terms
We can cancel \( m \) and \( r \) from both sides:

\[ (2\pi n_2)^2 = 2 \cdot (2\pi n_1)^2 \]

### Step 6: Simplify the equation
This simplifies to:

\[ n_2^2 = 2 \cdot n_1^2 \]

### Step 7: Substitute the initial RPM
Substituting \( n_1 = 5 \):

\[ n_2^2 = 2 \cdot (5)^2 = 2 \cdot 25 = 50 \]

### Step 8: Solve for \( n_2 \)
Taking the square root gives us:

\[ n_2 = \sqrt{50} = 5\sqrt{2} \approx 7.07 \, \text{RPM} \]

### Conclusion
The new speed \( n_2 \) is approximately \( 7.07 \, \text{RPM} \).

### Final Answer
The new speed is nearly \( 7.07 \, \text{RPM} \).

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