A bus turns a slippery road having coefficient of friction of `0*5` with a speed of 10 m/s .The minimum radius of the arc in which bus turns is [ Take g = 10 `m/s^(2)] `

Options

A. 4m
B. 10m
C. 15m
D. 20m

Answer

To find the minimum radius of the arc in which the bus turns on a slippery road, we can use the concepts of centripetal force and friction. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Coefficient of friction (μ) = 0.5
- Speed of the bus (v) = 10 m/s
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Write the expression for the force of friction
The force of friction (F_friction) that provides the necessary centripetal force for the bus to turn is given by:
\[ F_{\text{friction}} = \mu \cdot N \]
where N is the normal force. For a bus on a flat road, the normal force (N) is equal to the weight of the bus (mg):
\[ F_{\text{friction}} = \mu \cdot mg \]

### Step 3: Write the expression for centripetal force
The centripetal force (F_c) required to keep the bus moving in a circular path is given by:
\[ F_c = \frac{mv^2}{r} \]
where m is the mass of the bus, v is the speed, and r is the radius of the arc.

### Step 4: Set the forces equal to each other
For the bus to turn without slipping, the frictional force must equal the centripetal force:
\[ \mu \cdot mg = \frac{mv^2}{r} \]

### Step 5: Cancel out the mass (m)
Since the mass (m) appears on both sides of the equation, we can cancel it out:
\[ \mu g = \frac{v^2}{r} \]

### Step 6: Rearrange the equation to solve for r
Rearranging the equation to find the radius (r):
\[ r = \frac{v^2}{\mu g} \]

### Step 7: Substitute the known values
Now, substitute the known values into the equation:
- v = 10 m/s
- μ = 0.5
- g = 10 m/s²

Thus,
\[ r = \frac{(10)^2}{0.5 \cdot 10} \]
\[ r = \frac{100}{5} \]
\[ r = 20 \, \text{meters} \]

### Final Answer
The minimum radius of the arc in which the bus turns is **20 meters**.

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