`X=3YZ^(2)` find dimension of `Y` in (MKSA) system, if `X` and `Z` are the dimension of capacity and magnetic field respectively

Options

A. `M^(-1)L^(-2)T^(-4)A^(-1)`
B. `ML^(-2)`
C. `M^(-3)L^(-2)T^(-4)A^(4)`
D. `LML^(2)T^(-2)`

Answer

To find the dimension of \( Y \) in the equation \( X = 3YZ^2 \), where \( X \) represents capacity and \( Z \) represents magnetic field, we can follow these steps:

### Step 1: Identify the dimensions of \( X \) and \( Z \)

- The dimension of capacity \( X \) is given as:
\[ [X] = M^{-1} L^{-2} T^{4} A^{2} \]

- The dimension of magnetic field \( Z \) is given as:
\[ [Z] = M^{1} T^{-2} A^{-1} \]

### Step 2: Write the equation in terms of dimensions

From the equation \( X = 3YZ^2 \), we can ignore the constant \( 3 \) since it is dimensionless. Thus, we can express the dimensions as:
\[ [X] = [Y][Z]^2 \]

### Step 3: Substitute the dimensions into the equation

Substituting the dimensions we identified:
\[ M^{-1} L^{-2} T^{4} A^{2} = [Y] \cdot (M^{1} T^{-2} A^{-1})^2 \]

### Step 4: Simplify the right side

Calculating \( [Z]^2 \):
\[ [Z]^2 = (M^{1} T^{-2} A^{-1})^2 = M^{2} T^{-4} A^{-2} \]

Now substituting back into the equation gives:
\[ M^{-1} L^{-2} T^{4} A^{2} = [Y] \cdot (M^{2} T^{-4} A^{-2}) \]

### Step 5: Rearranging to solve for \( [Y] \)

To isolate \( [Y] \), we can rearrange the equation:
\[ [Y] = \frac{M^{-1} L^{-2} T^{4} A^{2}}{M^{2} T^{-4} A^{-2}} \]

### Step 6: Simplify the dimensions

Now we simplify the right side:
\[ [Y] = M^{-1 - 2} L^{-2} T^{4 - (-4)} A^{2 - (-2)} = M^{-3} L^{-2} T^{8} A^{4} \]

### Conclusion

Thus, the dimension of \( Y \) in the MKSA system is:
\[ [Y] = M^{-3} L^{-2} T^{8} A^{4} \]