Check by dimensions whether the equation `tan theta=(rg)/(v^(2))` is correct.
Here, r = radius of curvature of path, g = acceleration due to gravity
v = velocity of body moving on the curved path, `theta`= angle of banking of the road

Answer

To check the dimensional correctness of the equation \( \tan \theta = \frac{rg}{v^2} \), we will analyze both sides of the equation in terms of their dimensions.

### Step 1: Analyze the Left-Hand Side (LHS)
The left-hand side of the equation is \( \tan \theta \).

- The tangent function is a trigonometric function that is dimensionless. Therefore, the dimensions of \( \tan \theta \) are:
\[ [\tan \theta] = M^0 L^0 T^0 \]

### Step 2: Analyze the Right-Hand Side (RHS)
The right-hand side of the equation is \( \frac{rg}{v^2} \).

- **Radius of curvature \( r \)** has dimensions of length:
\[ [r] = L \]

- **Acceleration due to gravity \( g \)** has dimensions of length per time squared:
\[ [g] = LT^{-2} \]

- **Velocity \( v \)** has dimensions of length per time:
\[ [v] = LT^{-1} \]

Now, we need to find the dimensions of \( v^2 \):
\[ [v^2] = (LT^{-1})^2 = L^2 T^{-2} \]

### Step 3: Combine the Dimensions on the RHS
Now we can substitute the dimensions into the RHS:
\[ \frac{rg}{v^2} = \frac{L \cdot (LT^{-2})}{L^2 T^{-2}} \]

This simplifies to:
\[ \frac{L^2 T^{-2}}{L^2 T^{-2}} = M^0 L^0 T^0 \]

### Step 4: Compare Dimensions
Now we compare the dimensions of both sides:
- LHS: \( [\tan \theta] = M^0 L^0 T^0 \)
- RHS: \( \frac{rg}{v^2} = M^0 L^0 T^0 \)

Since both sides have the same dimensions, we conclude that the equation is dimensionally correct.

### Final Conclusion
Thus, we have verified that the equation \( \tan \theta = \frac{rg}{v^2} \) is dimensionally correct.

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