When the road is dry and coefficient of friciton is `mu`, the maximum speed of a car in a circular path is `10 ms^(-1)`. If the road becomes wet and coefficient of friction becomes `(mu)/4`, what is the maximum speed permited ?

Options

A. `5 ms^(-1)`
B. `10 ms^(-1)`
C. `20 ms^(-1)`
D. `4 ms^(-1)`

Explanation

To solve the problem step by step, we will analyze the situation using the concepts of circular motion and friction.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We know that when the road is dry (with coefficient of friction `μ`), the maximum speed of the car in a circular path is `10 m/s`.

2. **Centripetal Force and Friction**:
- The necessary centripetal force for a car moving in a circular path is provided by the frictional force. The centripetal force required is given by:
\[ F_c = \frac{mv^2}{R} \]
- The frictional force can be expressed as:
\[ F_f = μmg \]
- For the car to maintain circular motion, the centripetal force must equal the frictional force:
\[ \frac{mv^2}{R} = μmg \]

3. **Canceling Mass**:
- We can cancel the mass `m` from both sides of the equation:
\[ \frac{v^2}{R} = μg \]

4. **Substituting Known Values**:
- We know that when the speed `v = 10 m/s`, we can substitute this into the equation:
\[ \frac{(10)^2}{R} = μg \]
- This simplifies to:
\[ \frac{100}{R} = μg \]

5. **Finding Maximum Speed for Wet Road**:
- Now, the problem states that the road becomes wet, and the coefficient of friction becomes `μ/4`. We need to find the new maximum speed `v'`:
\[ \frac{(v')^2}{R} = \frac{μ}{4}g \]

6. **Substituting the Value of μg**:
- From our earlier equation, we know that `μg = \frac{100}{R}`. We can substitute this into the new equation:
\[ \frac{(v')^2}{R} = \frac{1}{4} \cdot \frac{100}{R} \]
- This simplifies to:
\[ (v')^2 = \frac{100}{4} = 25 \]

7. **Calculating v'**:
- Taking the square root of both sides gives us:
\[ v' = \sqrt{25} = 5 \, \text{m/s} \]

### Final Answer:
The maximum speed permitted on the wet road is **5 m/s**.