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 Y respresent displacement then `dim((alphaYZ)/(betaF))` is equal to

Options

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

Explanation

To solve the problem, we need to determine the dimensional formula for the expression \(\frac{\alpha YZ}{\beta F}\) given the dependencies and definitions provided in the question.

### Step-by-Step Solution:

1. **Identify the dimensions of each variable**:
- \(Y\) represents displacement, so its dimension is:
\[ [Y] = L \]
- \(F\) represents force, so its dimension is:
\[ [F] = MLT^{-2} \]
- \(W\) represents work, so its dimension is:
\[ [W] = ML^2T^{-2} \]

2. **Determine the dimensions of constants**:
- The term \(\sin(\alpha r)\) must be dimensionless, which implies that \(\alpha r\) is dimensionless. Since \(r\) has the dimension of length \(L\), we have:
\[ [\alpha] = L^{-1} \]
- The term \(e^{\beta r^2}\) is also dimensionless, leading to:
\[ [\beta r^2] = 1 \implies [\beta] = L^{-2} \]

3. **Determine the dimension of \(Z\)**:
- From the expression \(X = YFe^{-\beta r^2} + ZW \sin(\alpha r)\), we find that both terms must have the same dimension. The dimension of the first term \(YF\) is:
\[ [YF] = [Y][F] = L(MLT^{-2}) = ML^2T^{-2} \]
- The dimension of the second term \(ZW\) must also equal \(ML^2T^{-2}\):
\[ [ZW] = [Z][W] = [Z](ML^2T^{-2}) = ML^2T^{-2} \]
- To find the dimension of \(Z\):
\[ [Z] = \frac{ML^2T^{-2}}{ML^2T^{-2}} = 1 \implies [Z] = M^0L^0T^0 \]

4. **Substitute dimensions into the expression**:
- Now we can substitute the dimensions into the expression \(\frac{\alpha YZ}{\beta F}\):
\[ \frac{\alpha YZ}{\beta F} = \frac{[L^{-1}][L][M^0L^0T^0]}{[L^{-2}][MLT^{-2}]} \]

5. **Calculate the dimensions**:
- The numerator becomes:
\[ [\alpha YZ] = [L^{-1}][L][1] = L^{0} = 1 \]
- The denominator becomes:
\[ [\beta F] = [L^{-2}][MLT^{-2}] = ML^{-1}T^{-2} \]
- Therefore, we have:
\[ \frac{1}{ML^{-1}T^{-2}} = M^{-1}L^{1}T^{2} \]

6. **Final Result**:
- The dimension of \(\frac{\alpha YZ}{\beta F}\) is:
\[ [\frac{\alpha YZ}{\beta F}] = M^{-1}L^{1}T^{2} \]