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 Z represent frequency then choose the correct alternative

Options

A. The dimension of X is `[ML^(1) T^(-3)]`
B. The dimension of Y is `[M^(0)LT^(-1)]`
C. The dimension of `beta` is `[M^(0)L^(-1)T^(0)]`
D. The dimension of `alpha` is `[M^(0)L^(1) T^(0)]`

Explanation

To solve the problem, we need to analyze the given expression for the physical quantity \(X\) and determine the dimensions of the various components involved.

The equation given is:

\[ X = YFe^{-\beta r^2} + ZW \sin(\alpha r) \]

where:
- \(r\) is distance,
- \(F\) is force,
- \(W\) is work,
- \(Y\) and \(Z\) are unknown physical quantities,
- \(\alpha\) and \(\beta\) are positive constants.

### Step 1: Determine the dimensions of \(X\)

Since \(X\) is expressed as the sum of two terms, both terms must have the same dimensions. Therefore, we can analyze each term separately.

### Step 2: Analyze the first term \(YFe^{-\beta r^2}\)

1. **Dimensions of Force \(F\)**:
\[ [F] = M L T^{-2} \]

2. **Dimensions of \(e^{-\beta r^2}\)**:
The exponential function is dimensionless. Therefore, the dimensions of \(\beta r^2\) must also be dimensionless. This implies:
\[ [\beta] [r^2] = M^0 L^0 T^0 \]
Since \(r\) has dimensions of length \([r] = L\), we have:
\[ [r^2] = L^2 \implies [\beta] = L^{-2} \]

3. **Combining dimensions**:
Thus, the dimensions of the first term \(YFe^{-\beta r^2}\) can be expressed as:
\[ [Y][F] = [Y](M L T^{-2}) \implies [Y] = \frac{[X]}{M L T^{-2}} \]

### Step 3: Analyze the second term \(ZW \sin(\alpha r)\)

1. **Dimensions of Work \(W\)**:
\[ [W] = M L^2 T^{-2} \]

2. **Dimensions of \(\sin(\alpha r)\)**:
The sine function is also dimensionless, which means \(\alpha r\) must be dimensionless:
\[ [\alpha][r] = M^0 L^0 T^0 \implies [\alpha] = L^{-1} \]

3. **Combining dimensions**:
Thus, the dimensions of the second term \(ZW \sin(\alpha r)\) can be expressed as:
\[ [Z][W] = [Z](M L^2 T^{-2}) \implies [Z] = \frac{[X]}{M L^2 T^{-2}} \]

### Step 4: Equate dimensions of both terms

Since both terms must have the same dimensions, we can equate the dimensions of \(X\) from both terms:

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

2. From the second term:
\[ [X] = [Z](M L^2 T^{-2}) \]

### Step 5: Determine the dimension of \(Z\) given it represents frequency

Given that \(Z\) represents frequency, we know:
\[ [Z] = T^{-1} \]

### Step 6: Substitute and solve for dimensions of \(X\)

Substituting \(Z\) into the equation for dimensions:
\[ [X] = T^{-1}(M L^2 T^{-2}) = M L^2 T^{-3} \]

### Step 7: Conclusion

The dimensions of \(X\) can be expressed as:
\[ [X] = M L^2 T^{-3} \]