A circular railway track of radius r is banked at angle `theta` so that a train moving with speed v can safely go round the track. A student writes: `tan theta =rg//v^(2)`. Why this relation is not correct?
(i) Equality of dimensions does not guarantee correctness of the relation .
(ii) Dimensionally correct relation may not be numerically correct.
iii) The relation is dimensionally incorrect.

Options

A. (i )& (ii)
B. (ii) & (iii)
C. (iii) & (i)
D. (i) , (ii) & (iii)

Answer

To analyze why the relation \( \tan \theta = \frac{rg}{v^2} \) is not correct, we can break down the reasoning step by step.

### Step 1: Understanding the Terms
- **Radius (r)**: This is a measure of length, so its dimension is \( [L] \).
- **Acceleration due to gravity (g)**: This is also a measure of length per time squared, so its dimension is \( [L][T]^{-2} \).
- **Speed (v)**: This is a measure of length per time, so its dimension is \( [L][T]^{-1} \).

### Step 2: Analyzing the Right Side of the Equation
The right side of the equation is \( \frac{rg}{v^2} \).
- The dimension of \( rg \) is:
\[ [r][g] = [L] \cdot [L][T]^{-2} = [L^2][T]^{-2} \]
- The dimension of \( v^2 \) is:
\[ [v^2] = ([L][T]^{-1})^2 = [L^2][T]^{-2} \]
- Therefore, the dimension of \( \frac{rg}{v^2} \) becomes:
\[ \frac{[L^2][T]^{-2}}{[L^2][T]^{-2}} = [1] \quad \text{(dimensionless)} \]

### Step 3: Analyzing the Left Side of the Equation
The left side of the equation is \( \tan \theta \).
- The tangent of an angle is also dimensionless, \( [1] \).

### Step 4: Dimensional Consistency
Since both sides of the equation are dimensionless, the relation is dimensionally correct. However, dimensional correctness does not imply that the relation is physically valid or numerically correct.

### Step 5: Evaluating the Statements
1. **Equality of dimensions does not guarantee correctness of the relation**: This is true. Just because the dimensions match does not mean the physical relationship is valid.
2. **Dimensionally correct relation may not be numerically correct**: This is also true. The numerical values may not correspond to the physical scenario.
3. **The relation is dimensionally incorrect**: This is false. The relation is dimensionally correct.

### Conclusion
The correct reason why the relation \( \tan \theta = \frac{rg}{v^2} \) is not correct is that while it is dimensionally consistent, it does not represent a valid physical relationship.

### Final Answer
The correct option is (i) and (ii).

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