A physical quantity X depends on another physical quantities as `X=YFe^(-beta r^(2))+ZW sin (alpha r)` where r, F and W represents distance, force and work respectively & Y and Z are unknown physical quantities and `alpha, beta` are positive contsnats.
If T represent velocity then `dim (X)` is equal to

Options

A. `ML^(2) T^(-3)`
B. `M^(-1)L^(2) T^(-3)`
C. `ML^(2)T^(-2)`
D. None of these

Explanation

To find the dimension of the physical quantity \( X \) given by the equation:

\[ X = Y F e^{-\beta r^2} + Z W \sin(\alpha r) \]

we will analyze each term in the equation step by step.

### Step 1: Identify the dimensions of the components

1. **Distance \( r \)**: The dimension of distance is given by:
\[ [r] = L^1 \]

2. **Force \( F \)**: The dimension of force can be derived from Newton's second law \( F = ma \), where \( m \) is mass and \( a \) is acceleration. Thus:
\[ [F] = [m][a] = M^1 L^1 T^{-2} \]

3. **Work \( W \)**: The dimension of work is given by \( W = F \cdot d \) (force times distance):
\[ [W] = [F][d] = (M^1 L^1 T^{-2})(L^1) = M^1 L^2 T^{-2} \]

### Step 2: Analyze the exponential term

The term \( e^{-\beta r^2} \) must be dimensionless. Therefore, the exponent \( -\beta r^2 \) must also be dimensionless. Since \( r^2 \) has dimensions \( L^2 \), \( \beta \) must have dimensions of \( L^{-2} \):
\[ [\beta] = L^{-2} \]

### Step 3: Analyze the sine term

The term \( \sin(\alpha r) \) must also be dimensionless. This means that \( \alpha r \) must be dimensionless. Since \( r \) has dimensions \( L^1 \), \( \alpha \) must have dimensions of \( L^{-1} \):
\[ [\alpha] = L^{-1} \]

### Step 4: Determine the dimensions of \( Y \) and \( Z \)

1. **For the term \( Y F e^{-\beta r^2} \)**:
- Since \( e^{-\beta r^2} \) is dimensionless, the dimensions of \( Y F \) must equal the dimensions of \( X \).
- Therefore:
\[ [Y][F] = [X] \]
- We already have \( [F] = M^1 L^1 T^{-2} \). Let \( [Y] = Y_d \):
\[ [Y_d][M^1 L^1 T^{-2}] = [X] \]

2. **For the term \( Z W \sin(\alpha r) \)**:
- Again, since \( \sin(\alpha r) \) is dimensionless, the dimensions of \( Z W \) must equal the dimensions of \( X \).
- Therefore:
\[ [Z][W] = [X] \]
- We have \( [W] = M^1 L^2 T^{-2} \). Let \( [Z] = Z_d \):
\[ [Z_d][M^1 L^2 T^{-2}] = [X] \]

### Step 5: Equate the dimensions

From both terms, we can set the dimensions of \( X \) equal:

1. From \( Y F \):
\[ [Y][F] = [Y_d][M^1 L^1 T^{-2}] = [X] \]

2. From \( Z W \):
\[ [Z][W] = [Z_d][M^1 L^2 T^{-2}] = [X] \]

### Step 6: Solve for dimensions of \( X \)

To find the dimension of \( X \), we can express it in terms of \( Y \) and \( F \) or \( Z \) and \( W \).

Assuming \( Y \) has dimensions \( L^1 T^{-1} \) (velocity):
\[ [Y] = L^1 T^{-1} \]
Then:
\[ [Y][F] = (L^1 T^{-1})(M^1 L^1 T^{-2}) = M^1 L^2 T^{-3} \]

Thus, the dimension of \( X \) is:
\[ [X] = M^1 L^2 T^{-3} \]

### Final Result

The dimension of \( X \) is:
\[ \text{dim}(X) = M^1 L^2 T^{-3} \]

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