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

Explanation

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 will follow these steps:

### Step 1: Identify the dimensions of \( X \) (Capacitance)
Capacitance \( C \) is defined as:
\[ C = \frac{Q}{V} \]
Where \( Q \) is charge and \( V \) is voltage. The dimensions of voltage \( V \) can be expressed as:
\[ V = E \cdot d \]
Where \( E \) is electric field and \( d \) is distance. The electric field \( E \) has dimensions of force per unit charge:
\[ E = \frac{F}{Q} = \frac{m \cdot a}{Q} = \frac{m \cdot L \cdot T^{-2}}{Q} \]
Thus, substituting \( E \) into the voltage equation gives:
\[ V = \frac{m \cdot L \cdot T^{-2}}{Q} \cdot L = \frac{m \cdot L^2 \cdot T^{-2}}{Q} \]
Now substituting \( V \) back into the capacitance formula:
\[ C = \frac{Q}{V} = \frac{Q}{\frac{m \cdot L^2 \cdot T^{-2}}{Q}} = \frac{Q^2}{m \cdot L^2 \cdot T^{-2}} \]
Thus, the dimensions of capacitance \( C \) are:
\[ [X] = M^{-1} L^{-2} T^{4} Q^{2} \]

### Step 2: Identify the dimensions of \( Z \) (Magnetic Induction)
Magnetic induction \( B \) can be related to the force on a charge moving in a magnetic field:
\[ F = Q \cdot v \cdot B \]
Where \( F \) is force, \( Q \) is charge, and \( v \) is velocity. Rearranging gives:
\[ B = \frac{F}{Q \cdot v} \]
Substituting the dimensions of force \( F = M \cdot L \cdot T^{-2} \) and velocity \( v = L \cdot T^{-1} \):
\[ [B] = \frac{M \cdot L \cdot T^{-2}}{Q \cdot (L \cdot T^{-1})} = \frac{M \cdot L \cdot T^{-2}}{Q \cdot L \cdot T^{-1}} = \frac{M}{Q \cdot T} \]
Thus, the dimensions of magnetic induction \( B \) are:
\[ [Z] = M^{1} T^{-1} Q^{-1} \]

### Step 3: Substitute the dimensions into the formula
From the formula \( X = 3YZ^2 \), we can express the dimensions of \( Y \):
\[ [Y] = \frac{[X]}{[Z]^2} \]
Substituting the dimensions we found:
\[ [Y] = \frac{M^{-1} L^{-2} T^{4} Q^{2}}{(M^{1} T^{-1} Q^{-1})^2} \]
Calculating \( [Z]^2 \):
\[ [Z]^2 = M^{2} T^{-2} Q^{-2} \]
Now substituting this into the equation for \( Y \):
\[ [Y] = \frac{M^{-1} L^{-2} T^{4} Q^{2}}{M^{2} T^{-2} Q^{-2}} = M^{-1-2} L^{-2} T^{4+2} Q^{2+2} = M^{-3} L^{-2} T^{6} Q^{4} \]

### Final Answer
Thus, the dimensions of \( Y \) in the MKSQ system are:
\[ [Y] = M^{-3} L^{-2} T^{6} Q^{4} \]