A car is moving on a circular path and takes a turn. If `R_(1)` and `R_(2)` be the reactions on the inner and outer wheels, respectively, then

Options

A. `R_(1)=R_(2)`
B. `R_(1)lt_(2)`
C. `R_(1)gtR_(2)`
D. `R_(1)geR_(2)`

Answer

To solve the problem, we need to analyze the forces acting on the car as it moves along a circular path. Let's break down the solution step by step:

### Step 1: Identify the Forces Acting on the Car
When a car moves in a circular path, two primary forces act on it:
- The gravitational force (weight) acting downwards, which is equal to \( mg \), where \( m \) is the mass of the car and \( g \) is the acceleration due to gravity.
- The normal reaction forces \( R_1 \) and \( R_2 \) acting on the inner and outer wheels, respectively.

### Step 2: Set Up the Equation for Vertical Forces
Since the car is in equilibrium in the vertical direction, the sum of the vertical forces must equal zero. Therefore, we can write:
\[ R_1 + R_2 = mg \quad \text{(Equation 1)} \]

### Step 3: Analyze the Centripetal Force
As the car turns, it requires a centripetal force to maintain its circular motion. The centripetal force \( F \) can be expressed as:
\[ F = \frac{mv^2}{R} \]
where \( v \) is the speed of the car and \( R \) is the radius of the circular path.

### Step 4: Consider the Torque About the Center of Gravity
To compare \( R_1 \) and \( R_2 \), we need to consider the torques about the center of gravity of the car. The torque due to the centripetal force \( F \) and the reactions \( R_1 \) and \( R_2 \) can be expressed as:
- Torque due to \( F \): \( F \cdot h \) (where \( h \) is the perpendicular distance from the center of gravity to the line of action of \( F \))
- Torque due to \( R_1 \): \( R_1 \cdot a \) (where \( a \) is the distance from the center of gravity to the inner wheel)
- Torque due to \( R_2 \): \( R_2 \cdot a \) (acting in the opposite direction)

Balancing the torques gives us:
\[ F \cdot h + R_1 \cdot a = R_2 \cdot a \]

### Step 5: Rearranging the Torque Equation
Rearranging the torque equation, we find:
\[ R_2 \cdot a = F \cdot h + R_1 \cdot a \]
\[ R_2 = \frac{F \cdot h}{a} + R_1 \]

### Step 6: Conclusion
Since \( F \) and \( h \) are both positive quantities, the term \( \frac{F \cdot h}{a} \) is also positive. Therefore, we can conclude that:
\[ R_2 = R_1 + \frac{F \cdot h}{a} \quad \Rightarrow \quad R_2 > R_1 \]

Thus, the reaction on the outer wheel \( R_2 \) is greater than the reaction on the inner wheel \( R_1 \).

### Final Answer
The correct conclusion is that \( R_2 > R_1 \).

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