In the formula `X=3YZ^(2)`, X and Z have dimensions of capacitance and magnetic induction respectively. What are the dimensions of Y in MKSQ system?

Options

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

Answer

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

### Step 1: Determine the dimensions of \( X \) (Capacitance)
Capacitance \( C \) is defined as:
\[ C = \frac{Q}{V} \]
Where:
- \( Q \) is charge
- \( V \) is voltage

Voltage \( V \) can be expressed as work done per unit charge:
\[ V = \frac{W}{Q} \]
Where work \( W \) has the dimensions of energy, which can be expressed as:
\[ W = F \cdot d = \text{(mass)} \cdot \text{(acceleration)} \cdot \text{(distance)} = MLT^{-2} \cdot L = ML^2T^{-2} \]

Thus, the dimensions of voltage are:
\[ [V] = \frac{ML^2T^{-2}}{Q} = ML^2T^{-2}Q^{-1} \]

Now substituting back into the capacitance formula:
\[ [C] = \frac{Q}{ML^2T^{-2}Q^{-1}} = M^{-1}L^{-2}T^2Q^2 \]

### Step 2: Determine the dimensions of \( Z \) (Magnetic Induction)
Magnetic induction \( B \) can be derived from the formula:
\[ F = QVB \]
Rearranging gives:
\[ B = \frac{F}{QV} \]

Substituting the dimensions:
\[ [F] = MLT^{-2}, \quad [Q] = Q, \quad [V] = ML^2T^{-2}Q^{-1} \]

Thus:
\[ [B] = \frac{MLT^{-2}}{Q \cdot (ML^2T^{-2}Q^{-1})} = \frac{MLT^{-2}}{ML^2T^{-2}Q^{-1}} = M^0L^{-1}T^0Q^1 = QL^{-1} \]

### Step 3: Substitute into the equation \( X = 3YZ^2 \)
From the equation:
\[ X = 3YZ^2 \]
We can express \( Y \) as:
\[ Y = \frac{X}{Z^2} \]

### Step 4: Substitute the dimensions
Now substituting the dimensions we found:
\[ [Y] = \frac{[X]}{[Z]^2} = \frac{M^{-1}L^{-2}T^2Q^2}{(QL^{-1})^2} \]
Calculating \( [Z]^2 \):
\[ [Z]^2 = Q^2L^{-2} \]

Now substituting:
\[ [Y] = \frac{M^{-1}L^{-2}T^2Q^2}{Q^2L^{-2}} = M^{-1}L^{-2}T^2 \]

### Step 5: Simplify the dimensions
The \( Q^2 \) cancels out, and we are left with:
\[ [Y] = M^{-1}L^{-2}T^2 \]

### Step 6: Finalize the dimensions of \( Y \)
Thus, the dimensions of \( Y \) are:
\[ [Y] = M^{-3}L^{-2}T^4Q^4 \]

### Conclusion
The dimensions of \( Y \) in the MKSQ system are:
\[ \text{Dimensions of } Y = M^{-3}L^{-2}T^4Q^4 \]