A body of mass 1 kg tied to one end of string is revolved in a horizontal circle of radius 0.1 m with a speed of `3 "revolution"//sec` , assuming the effect of gravity is negligible, then linear velocity, acceleration and tension in the string will be

Options

A. `1.88 m//s, 35.5 m//s^(2), 35.5 N`
B. `2.88 m//s, 45.5 m//s^(2), 45.5 N`
C. `3.88 m//s, 55.5 m//s^(2), 55.5 N`
D. None of these

Answer

To solve the problem step by step, we will calculate the linear velocity, centripetal acceleration, and tension in the string for a body of mass 1 kg revolving in a horizontal circle of radius 0.1 m at a speed of 3 revolutions per second.

### Step 1: Calculate the Linear Velocity

1. **Understanding the motion**: The body completes 3 revolutions in 1 second. The distance covered in one revolution is the circumference of the circle, which is given by the formula:
\[ \text{Circumference} = 2 \pi r \]
where \( r \) is the radius of the circle.

2. **Calculate the circumference**:
\[ \text{Circumference} = 2 \pi (0.1) = 0.2 \pi \text{ meters} \]

3. **Calculate the total distance covered in 3 revolutions**:
\[ \text{Total distance} = 3 \times \text{Circumference} = 3 \times 0.2 \pi = 0.6 \pi \text{ meters} \]

4. **Calculate the linear velocity (v)**:
Since the time taken for 3 revolutions is 1 second, the linear velocity is:
\[ v = \frac{\text{Total distance}}{\text{Time}} = \frac{0.6 \pi}{1} = 0.6 \pi \text{ m/s} \]

Approximating \( \pi \) as \( 3.14 \):
\[ v \approx 0.6 \times 3.14 \approx 1.884 \text{ m/s} \approx 1.88 \text{ m/s} \]

### Step 2: Calculate the Centripetal Acceleration

1. **Using the formula for centripetal acceleration (a)**:
\[ a = \frac{v^2}{r} \]

2. **Substituting the values**:
\[ a = \frac{(1.88)^2}{0.1} \]

3. **Calculating \( v^2 \)**:
\[ (1.88)^2 \approx 3.5344 \]

4. **Calculating acceleration**:
\[ a = \frac{3.5344}{0.1} = 35.344 \text{ m/s}^2 \approx 35.34 \text{ m/s}^2 \]

### Step 3: Calculate the Tension in the String

1. **Using the formula for tension (T)**:
The tension in the string provides the centripetal force required to keep the body moving in a circle:
\[ T = m \cdot a \]

2. **Substituting the values**:
\[ T = 1 \cdot 35.344 \approx 35.344 \text{ N} \approx 35.34 \text{ N} \]

### Summary of Results

- Linear Velocity \( v \approx 1.88 \text{ m/s} \)
- Centripetal Acceleration \( a \approx 35.34 \text{ m/s}^2 \)
- Tension in the String \( T \approx 35.34 \text{ N} \)