The relation `tan theta = v^(2)// r g` gives the angle of banking of the cyclist going round the curve . Here `v` is the speed of the cyclist , `r` is the radius of the curve , and `g` is the acceleration due to gravity . Which of the following statements about the relation is true ?

Options

A. It is both dimensionally as well as numerically correct.
B. It is neither dimensionally correct correct.
C. It is dimensionally correct but not numerically.
D. It is numerically correct but not dimensionally.

Explanation

To determine the correctness of the relation \( \tan \theta = \frac{v^2}{rg} \), we will analyze both its numerical and dimensional correctness step by step.

### Step 1: Understand the Relation
The relation given is \( \tan \theta = \frac{v^2}{rg} \). Here:
- \( \theta \) is the angle of banking.
- \( v \) is the speed of the cyclist.
- \( r \) is the radius of the curve.
- \( g \) is the acceleration due to gravity.

### Step 2: Analyze the Left Side
The left side of the equation is \( \tan \theta \). The tangent of an angle is a ratio of two lengths (opposite/adjacent) and is therefore dimensionless. Thus, we can denote its dimensions as:
- \( [\tan \theta] = M^0 L^0 T^0 \) (unitless)

### Step 3: Analyze the Right Side
Now, we will analyze the right side of the equation \( \frac{v^2}{rg} \):
- The unit of speed \( v \) is \( \text{LT}^{-1} \).
- Therefore, \( v^2 \) has units of \( \text{L}^2 \text{T}^{-2} \).

Next, we analyze \( rg \):
- The unit of radius \( r \) is \( \text{L} \).
- The unit of acceleration due to gravity \( g \) is \( \text{LT}^{-2} \).
- Thus, the units of \( rg \) are \( \text{L} \times \text{LT}^{-2} = \text{L}^2 \text{T}^{-2} \).

### Step 4: Combine the Right Side
Now we can combine the right side:
\[ \frac{v^2}{rg} = \frac{\text{L}^2 \text{T}^{-2}}{\text{L}^2 \text{T}^{-2}} = 1 \]
This shows that the right side is also dimensionless.

### Step 5: Conclusion
Since both sides of the equation are dimensionless, we conclude that:
1. The relation is dimensionally correct.
2. The relation is also numerically correct as it accurately describes the relationship between the angle of banking, speed, radius, and gravitational acceleration.

Thus, the correct statement about the relation \( \tan \theta = \frac{v^2}{rg} \) is:
**Both dimensionally and numerically correct.**

### Final Answer
**Option A: Both dimensionally and numerically correct.**