Question

A car is driven on the banked road of radius of curvature 20 m with maximum safe speed. In order to increase its safety speed by 20%, without changing the angle of banking, the increase in the radius of curvature will be [Assume friction is same on the road]}

• A. 28.8 m
• B. 14.4 m
• C. 8.8 m
• D. 4.8 m

Hint:

For constant banking, the radius must be increased by the square of the speed factor ($1.2^2 = 1.44$).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The maximum safe speed of a car on a banked road depends on the radius of curvature, acceleration due to gravity, the angle of banking, and the coefficient of friction. Since only the radius is varied to change the speed, we look for a proportionality between speed and radius.
Step 2: Key Formula or Approach:
The maximum safe speed \(v\) is given by:
\[ v = \sqrt{rg \left( \frac{\mu + \tan\theta}{1 - \mu\tan\theta} \right)} \]
For a constant \(\mu\), \(g\), and \(\theta\), we have:
\[ v \propto \sqrt{r} \quad \text{or} \quad v^2 \propto r \]
Step 3: Detailed Explanation:
Let the initial speed be \(v_1\) and initial radius be \(r_1 = 20 \text{ m}\).
The new speed is \(v_2 = v_1 + 20% \text{ of } v_1 = 1.2v_1\).
Using the proportionality:
\[ \frac{r_2}{r_1} = \left( \frac{v_2}{v_1} \right)^2 \]
\[ \frac{r_2}{20} = \left( \frac{1.2v_1}{v_1} \right)^2 \]
\[ \frac{r_2}{20} = (1.2)^2 = 1.44 \]
\[ r_2 = 20 \times 1.44 = 28.8 \text{ m} \]
The increase in the radius of curvature \(\Delta r\) is:
\[ \Delta r = r_2 - r_1 = 28.8 - 20 = 8.8 \text{ m} \]
Step 4: Final Answer:
The increase in the radius of curvature will be 8.8 m.