The distance between the rails of the track is `1.67` m. How much the outer rail be elecated for curve of `0.5`km radius, so that a train moving with speed `54kmh^(-1)` can take safe turn on track.

Options

A. 80 mm
B. 75 mm
C. 60 mm
D. 75 mm

Explanation

To solve the problem of how much the outer rail should be elevated for a train to safely navigate a curve, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Data:**
- Distance between the rails (d) = 1.67 m
- Radius of the curve (r) = 0.5 km = 500 m
- Speed of the train (v) = 54 km/h

2. **Convert Speed to m/s:**
- To convert km/h to m/s, we use the conversion factor: \(1 \text{ km/h} = \frac{1}{3.6} \text{ m/s}\).
- Therefore, \(v = 54 \times \frac{1}{3.6} = 15 \text{ m/s}\).

3. **Use the Formula for Elevation (h):**
- The formula for the elevation of the outer rail is given by:
\[ h = \frac{v^2}{g \cdot \tan(\theta)} \]
- For safe turning, we can relate \(\tan(\theta)\) to the centripetal acceleration:
\[ \tan(\theta) = \frac{v^2}{g \cdot r} \]
- Rearranging gives us:
\[ h = \frac{d \cdot v^2}{g \cdot r} \]

4. **Substitute the Values:**
- Here, \(g \approx 10 \text{ m/s}^2\) (acceleration due to gravity).
- Now substituting the values:
\[ h = \frac{1.67 \cdot (15)^2}{10 \cdot 500} \]

5. **Calculate \(h\):**
- First, calculate \(15^2 = 225\).
- Then, substitute:
\[ h = \frac{1.67 \cdot 225}{10 \cdot 500} = \frac{375.75}{5000} = 0.07515 \text{ m} \]
- Converting to millimeters:
\[ h = 0.07515 \times 1000 = 75.15 \text{ mm} \]

6. **Final Answer:**
- The outer rail should be elevated by approximately **75 mm** for the train to safely navigate the curve.