The breaking tension of a string if 50 N. A body of mass 1 kg is tied to one end of a 1m long string and whirled in a horizontal circle, The maximum speed of the body should be

Options

A. `5sqrt2m//s`
B. `10m//s`
C. `7.5m//s`
D. `5m//s`

Answer

To solve the problem step by step, we will determine the maximum speed of a body of mass 1 kg tied to a string with a breaking tension of 50 N, while it is whirled in a horizontal circle of radius 1 m.

### Step 1: Understand the Forces Acting on the Body
In a horizontal circular motion, the only force providing the centripetal acceleration is the tension in the string. The weight of the body does not affect the horizontal motion.

### Step 2: Write the Formula for Centripetal Force
The centripetal force \( F_c \) required to keep the body moving in a circle is given by the formula:
\[ F_c = \frac{m v^2}{r} \]
Where:
- \( m \) = mass of the body (1 kg)
- \( v \) = speed of the body (unknown)
- \( r \) = radius of the circle (1 m)

### Step 3: Set the Maximum Tension Equal to Centripetal Force
The maximum tension in the string is 50 N, which will be equal to the centripetal force at maximum speed:
\[ F_c = T_{\text{max}} = 50 \, \text{N} \]
Thus, we have:
\[ \frac{m v^2}{r} = 50 \]

### Step 4: Substitute Known Values into the Equation
Substituting the known values into the equation:
\[ \frac{1 \cdot v^2}{1} = 50 \]
This simplifies to:
\[ v^2 = 50 \]

### Step 5: Solve for Maximum Speed \( v \)
Taking the square root of both sides to find \( v \):
\[ v = \sqrt{50} \]
We can simplify \( \sqrt{50} \):
\[ v = \sqrt{25 \cdot 2} = 5\sqrt{2} \, \text{m/s} \]

### Final Answer
The maximum speed of the body should be:
\[ v = 5\sqrt{2} \, \text{m/s} \approx 7.07 \, \text{m/s} \]