The breaking tension of a string is 10 N. A particle of mass 0.1 kg tied to it is rotated along a horizontal circle of radius 0.5 meter. The maximum speed with which the particle can be rotated without breaking the string is

Options

A. `sqrt(5) m//sec`
B. `sqrt((50)) m//sec`
C. `sqrt((500)) m//sec`
D. `sqrt((1000)) m//sec`

Explanation

To find the maximum speed with which a particle can be rotated without breaking the string, we can follow these steps:

### Step 1: Understand the Forces Acting on the Particle
The particle is undergoing circular motion, which means there is a centripetal force acting towards the center of the circle. This centripetal force is provided by the tension in the string.

### Step 2: Write the Equation for Centripetal Force
The centripetal force (\(F_c\)) required to keep the particle moving in a circle is given by the formula:
\[ F_c = \frac{mv^2}{r} \]
where:
- \(m\) = mass of the particle (0.1 kg)
- \(v\) = speed of the particle
- \(r\) = radius of the circle (0.5 m)

### Step 3: Set the Centripetal Force Equal to the Tension
Since the maximum tension in the string is equal to the centripetal force required to keep the particle moving in a circle, we can write:
\[ T = \frac{mv^2}{r} \]
where \(T\) is the breaking tension of the string (10 N).

### Step 4: Rearrange the Equation to Solve for \(v^2\)
Rearranging the equation gives us:
\[ v^2 = \frac{Tr}{m} \]

### Step 5: Substitute the Known Values
Now, substitute the known values into the equation:
- \(T = 10 \, \text{N}\)
- \(r = 0.5 \, \text{m}\)
- \(m = 0.1 \, \text{kg}\)

So,
\[ v^2 = \frac{10 \times 0.5}{0.1} \]

### Step 6: Calculate \(v^2\)
Calculating the right-hand side:
\[ v^2 = \frac{5}{0.1} = 50 \]

### Step 7: Take the Square Root to Find \(v\)
Now, take the square root to find the maximum speed \(v\):
\[ v = \sqrt{50} \approx 7.07 \, \text{m/s} \]

### Final Answer
The maximum speed with which the particle can be rotated without breaking the string is approximately \(7.07 \, \text{m/s}\).

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