In the formula X=3YZ^2, X and Z have dimeniosn of capacitance and magnetic induction respectively. The dimensions of Y in MKSQ system are

Options

A. `[M^(-3)L^(-2)T^(4)Q^4]`
B. `[M^(-2)L^(-1)T^(5)Q^3]`
C. `[M^(-1)L^(-2)T^(4)Q^4]`
D. `[M^(-3)L^(-1)T^(4)Q^4]`

Explanation

To find the dimensions of \( Y \) in the formula \( X = 3YZ^2 \), where \( X \) has dimensions of capacitance and \( Z \) has dimensions of magnetic induction, we will follow these steps:

### Step 1: Identify the dimensions of \( X \) (Capacitance)
The dimensions of capacitance \( C \) are given by:
\[ [C] = M^{-1}L^{-2}I^{2}T^{4} \]

### Step 2: Identify the dimensions of \( Z \) (Magnetic Induction)
The dimensions of magnetic induction \( B \) are given by:
\[ [B] = M^{1}L^{0}I^{-1}T^{-2} \]

### Step 3: Write the equation for \( Y \)
From the equation \( X = 3YZ^2 \), we can express \( Y \) as:
\[ Y = \frac{X}{Z^2} \]

### Step 4: Substitute the dimensions into the equation
Substituting the dimensions of \( X \) and \( Z \) into the equation:
\[ [Y] = \frac{M^{-1}L^{-2}I^{2}T^{4}}{(M^{1}L^{0}I^{-1}T^{-2})^2} \]

### Step 5: Calculate \( Z^2 \)
Calculating \( Z^2 \):
\[ Z^2 = (M^{1}L^{0}I^{-1}T^{-2})^2 = M^{2}L^{0}I^{-2}T^{-4} \]

### Step 6: Substitute \( Z^2 \) back into the equation for \( Y \)
Now substituting \( Z^2 \) back into the equation for \( Y \):
\[ [Y] = \frac{M^{-1}L^{-2}I^{2}T^{4}}{M^{2}L^{0}I^{-2}T^{-4}} \]

### Step 7: Simplify the dimensions
Now we simplify the dimensions:
\[ [Y] = M^{-1-2}L^{-2-0}I^{2-(-2)}T^{4-(-4)} = M^{-3}L^{-2}I^{4}T^{8} \]

### Step 8: Final dimensions of \( Y \)
Thus, the dimensions of \( Y \) in the MKSQ system are:
\[ [Y] = M^{-3}L^{-2}I^{4}T^{8} \]

### Conclusion
The dimensions of \( Y \) in the MKSQ system are \( M^{-3}L^{-2}I^{4}T^{8} \).

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