Question

A car moves at a speed of \(600\,\text{km/h}\) on a frictionless banked road with \(\theta=30^{\circ}\). Take \(g=10\,\text{m/s}^{2}\). Find the radius of the road.

• A. \(2.4\,\text{km}\)
• B. \(3.2\,\text{km}\)
• C. \(4.8\,\text{km}\)
• D. \(6.4\,\text{km}\)

Hint:

For a frictionless banked road, always use \[ \boxed{\tan\theta=\frac{v^{2}}{rg}} \] or equivalently, \[ \boxed{r=\frac{v^{2}}{g\tan\theta}}. \] Before substituting values, always convert the speed from km/h to m/s using \[ \boxed{1~\text{km/h}=\frac{5}{18}~\text{m/s}}. \] Also remember: \[ \boxed{\text{Higher speed } \Rightarrow \text{ Larger radius for the same banking angle}.} \]

Answer 1

The Correct Option is C

Concept: A banked road is designed so that a vehicle can negotiate a curve safely without relying on friction. On a frictionless banked road, the horizontal component of the normal reaction provides the required centripetal force, while the vertical component balances the weight of the vehicle. The relation between the speed, radius of the curve and the banking angle is \[ \boxed{\tan\theta=\frac{v^{2}}{rg}} \] where \[ v=\text{Speed of the vehicle}, \] \[ r=\text{Radius of the circular path}, \] \[ g=\text{Acceleration due to gravity}, \] \[ \theta=\text{Angle of banking}. \] Rearranging, \[ \boxed{r=\frac{v^{2}}{g\tan\theta}} \] This formula is applicable only when the road is frictionless.

Step 1: Write the given data.
Given, \[ v=600\,\text{km/h}, \] \[ \theta=30^{\circ}, \] \[ g=10\,\text{m/s}^{2}. \] Convert the speed into SI units. \[ v = 600\times\frac{5}{18} = 166.67\,\text{m/s}. \] Also, \[ \tan30^{\circ} = \frac{1}{\sqrt{3}} \approx0.577. \]

Step 2: Apply the banking road formula.
Using \[ r=\frac{v^{2}}{g\tan\theta}, \] we get \[ r = \frac{(166.67)^{2}} {10\times0.577}. \]

Step 3: Calculate the numerator.
\[ (166.67)^{2} = 27777.8. \] Hence, \[ r = \frac{27777.8} {5.77}. \]

Step 4: Evaluate the radius.
\[ r \approx 4814\,\text{m}. \] Converting into kilometres, \[ r = \frac{4814}{1000} = 4.814\,\text{km}. \] Therefore, \[ \boxed{r\approx4.8\,\text{km}.} \] Hence, the correct answer is \[ \boxed{\textbf{Option (C)}}. \]

Verification: Alternatively, \[ r = \frac{(166.67)^2} {10/\sqrt{3}} = 27777.8\times\frac{\sqrt{3}}{10} \approx4811\,\text{m}, \] which again gives \[ \boxed{r\approx4.8\,\text{km}.} \] Thus, the result is verified.