petal force required for circular motion. This prevents the train from slipping or tipping over.

The formula for the elevation (h) of the outer rail required for safe turning at a velocity (v) on a curve of radius (r) and width (d) can be derived from balancing the forces acting on the train:

$$h = \frac{v^2}{g \cdot r} \cdot d$$

where

$h$ is the height by which the outer rail should be raised,

$v$ is the speed of the train,

$r$ is the radius of the curve,

$d$ is the distance between the rails, and

$g$ is the acceleration due to gravity ($9.8 \, \text{m/s}^2$).

Firstly, we need to convert the speed from km/h to m/s.

$$ v = 48 \, \text{km/h} = \frac{48 \times 1000}{3600} = 13.33 \, \text{m/s} $$

Now, we can plug the values into our equation:

$$ h = \frac{(13.33)^2}{9.8 \times 400} \times 1 $$

Let's calculate this step by step:

Squaring the speed: $(13.33)^2 = 177.7$

Multiplying by the distance between rails and dividing by the product of gravity and the radius:

$$ h = \frac{177.7}{9.8 \times 400} = \frac{177.7}{3920} = 0.0453 \, \text{meters} $$

Converting meters to centimeters (since $1 \, \text{meter} = 100 \, \text{centimeters}$),

$$ h = 0.0453 \, \text{m} \times 100 = 4.53 \, \text{centimeters} $$

Therefore, the correct option is D) 4.5 cm.

Circular Motion

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