A particle of mass 200 g is moving in a circle of radius 2 m. The particle is just looping the loop. The speed of the particle and the tension in the string at highest point of the circule path are `(g=10ms^(-2))`

Options

A. `4 ms^(-1),5N`
B. `4.47ms^(-1),` zero
C. `2.47ms^(-1),` zero
D. `1 ms^(-1),` zero

Answer

To solve the problem, we need to determine the speed of the particle and the tension in the string at the highest point of its circular path. Let's break down the solution step by step.

### Step 1: Identify the Given Data
- Mass of the particle, \( m = 200 \, \text{g} = 0.2 \, \text{kg} \) (since 1 g = 0.001 kg)
- Radius of the circular path, \( r = 2 \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Analyze Forces at the Highest Point
At the highest point of the circular path, two forces act on the particle:
1. The weight of the particle, \( mg \), acting downwards.
2. The tension in the string, \( T \), also acting downwards.

The centripetal force required to keep the particle moving in a circle is provided by the sum of the tension and the weight of the particle. Therefore, we can write the equation for centripetal force as:
\[ T + mg = \frac{mv^2}{r} \]

### Step 3: Calculate the Speed at the Highest Point
For a particle just looping the loop, the minimum speed at the highest point can be derived from the condition that the tension \( T \) can be zero (the particle is just maintaining the circular motion). Thus, we can set \( T = 0 \):
\[ mg = \frac{mv^2}{r} \]
Cancelling \( m \) from both sides (since \( m \neq 0 \)):
\[ g = \frac{v^2}{r} \]
Rearranging gives:
\[ v^2 = gr \]
Substituting the values:
\[ v^2 = 10 \times 2 = 20 \]
Taking the square root:
\[ v = \sqrt{20} = 4.47 \, \text{m/s} \]

### Step 4: Calculate the Tension in the String
Now, substituting \( v \) back into the centripetal force equation to find the tension:
\[ T + mg = \frac{mv^2}{r} \]
Substituting \( v^2 = 20 \):
\[ T + mg = \frac{m \cdot 20}{r} \]
Substituting \( m = 0.2 \, \text{kg} \), \( g = 10 \, \text{m/s}^2 \), and \( r = 2 \, \text{m} \):
\[ T + (0.2 \times 10) = \frac{0.2 \cdot 20}{2} \]
Calculating \( mg \):
\[ T + 2 = \frac{4}{2} \]
\[ T + 2 = 2 \]
Thus:
\[ T = 2 - 2 = 0 \]

### Final Results
- Speed of the particle at the highest point: \( v = 4.47 \, \text{m/s} \)
- Tension in the string at the highest point: \( T = 0 \, \text{N} \)