A train of mass `10^(5)` kg rounds a curve of radius 100m at a speed of 20 m/s . If the track is not , then the thrust on the outer rail of the track is

Options

A. `2xx10^(5)N`
B. `3xx10^(5)N`
C. `4xx10^(5)N`
D. `5xx10^(5)N`

Explanation

To solve the problem of finding the thrust on the outer rail of the track when a train rounds a curve, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values**:
- Mass of the train (m) = \(10^5\) kg
- Radius of the curve (r) = 100 m
- Speed of the train (v) = 20 m/s

2. **Understand the concept of centripetal force**:
- When an object moves in a circular path, it requires a centripetal force to keep it in that path. This force is directed towards the center of the circle.

3. **Use the formula for centripetal force**:
- The formula for centripetal force (F_c) is given by:
\[ F_c = \frac{mv^2}{r} \]
- Here, \(F_c\) is the centripetal force, \(m\) is the mass, \(v\) is the speed, and \(r\) is the radius of the circular path.

4. **Substitute the values into the formula**:
- Plugging in the values we have:
\[ F_c = \frac{(10^5 \, \text{kg}) \cdot (20 \, \text{m/s})^2}{100 \, \text{m}} \]

5. **Calculate \(v^2\)**:
- First, calculate \(v^2\):
\[ v^2 = 20^2 = 400 \, \text{m}^2/\text{s}^2 \]

6. **Calculate the centripetal force**:
- Now substitute \(v^2\) back into the formula:
\[ F_c = \frac{(10^5) \cdot (400)}{100} \]

7. **Simplify the expression**:
- Calculate the numerator:
\[ 10^5 \cdot 400 = 4 \times 10^7 \]
- Now divide by 100:
\[ F_c = \frac{4 \times 10^7}{100} = 4 \times 10^5 \, \text{N} \]

8. **Conclusion**:
- The thrust on the outer rail of the track due to the centripetal force is:
\[ \text{Thrust} = 4 \times 10^5 \, \text{N} \]

### Final Answer:
The thrust on the outer rail of the track is \(4 \times 10^5\) N.

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