Question:
The coefficient of static friction between the road and tyres of a car is \(0.4\). The maximum permissible speed of the car is \(10\text{ ms}^{-1}\) on a curved unbanked road. Then the maximum radius of curvature of the road is
\[
\text{(acceleration due to gravity }=10\text{ ms}^{-2}\text{)}
\]
The coefficient of static friction between the road and tyres of a car is \(0.4\). The maximum permissible speed of the car is \(10\text{ ms}^{-1}\) on a curved unbanked road. Then the maximum radius of curvature of the road is
\[ \text{(acceleration due to gravity }=10\text{ ms}^{-2}\text{)} \]
\[ \text{(acceleration due to gravity }=10\text{ ms}^{-2}\text{)} \]
Show Hint
For motion on a curved unbanked road,
\[
\frac{mv^2}{r}=\mu mg
\]
because static friction provides the necessary centripetal force.
Updated On: Jun 25, 2026
- \(10\sqrt{5}\text{ m}\)
- \(25\text{ m}\)
- \(20\sqrt{2}\text{ m}\)
- \(30\text{ m}\)
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The Correct Option is B
Solution and Explanation
Step 1: Identify the force providing centripetal force.
On a curved unbanked road, static friction provides the centripetal force required for circular motion.
Thus, \[ f=\frac{mv^2}{r} \] Maximum static friction is \[ f_{\max}=\mu mg \] Therefore, \[ \frac{mv^2}{r}=\mu mg \]
Step 2: Simplify the equation.
Cancelling \(m\), \[ \frac{v^2}{r}=\mu g \] Hence, \[ r=\frac{v^2}{\mu g} \]
Step 3: Substitute the given values.
Given: \[ v=10\text{ ms}^{-1},\quad \mu=0.4,\quad g=10\text{ ms}^{-2} \] So, \[ r=\frac{10^2}{0.4\times 10} \] \[ =\frac{100}{4} \] \[ =25\text{ m} \]
Step 4: Final conclusion.
Therefore, the maximum radius of curvature is \[ \boxed{25\text{ m}} \]
On a curved unbanked road, static friction provides the centripetal force required for circular motion.
Thus, \[ f=\frac{mv^2}{r} \] Maximum static friction is \[ f_{\max}=\mu mg \] Therefore, \[ \frac{mv^2}{r}=\mu mg \]
Step 2: Simplify the equation.
Cancelling \(m\), \[ \frac{v^2}{r}=\mu g \] Hence, \[ r=\frac{v^2}{\mu g} \]
Step 3: Substitute the given values.
Given: \[ v=10\text{ ms}^{-1},\quad \mu=0.4,\quad g=10\text{ ms}^{-2} \] So, \[ r=\frac{10^2}{0.4\times 10} \] \[ =\frac{100}{4} \] \[ =25\text{ m} \]
Step 4: Final conclusion.
Therefore, the maximum radius of curvature is \[ \boxed{25\text{ m}} \]
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