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 __________, __________.

Step 1: Identify the given quantities and their relations.

The given formula is:

$$X = 3YZ^{2}$$

Since $3$ is a dimensionless constant, the dimensional formula for $Y$ is:

$$[Y] = [X][Z]^{-2}$$

Here, $X$ represents capacitance and $Z$ represents magnetic induction.

Step 2: Determine the dimensions of capacitance $X$.

Capacitance $C$ is defined as charge $Q$ divided by potential difference $V$.

$$[X] = [C] = \frac{[Q]}{[V]}$$

Since potential difference is work done per unit charge ($V = \frac{W}{Q}$):

$$[X] = \frac{[Q]^{2}}{[W]}$$

Using the dimensions of work $[W] = M L^{2} T^{-2}$:

$$[X] = \frac{Q^{2}}{M L^{2} T^{-2}}$$ $$[X] = M^{-1} L^{-2} T^{2} Q^{2}$$

Step 3: Determine the dimensions of magnetic induction $Z$.

Magnetic induction $B$ is derived from the magnetic force formula:

$$F = q v B$$ $$[Z] = [B] = \frac{[F]}{[q][v]}$$

Using the dimensions of force $[F] = M L T^{-2}$ and velocity $[v] = L T^{-1}$:

$$[Z] = \frac{M L T^{-2}}{Q \cdot L T^{-1}}$$ $$[Z] = M T^{-1} Q^{-1}$$

Step 4: Calculate the dimensions of $Y$.

Substitute the dimensions of $[X]$ and $[Z]$ into the relation $[Y] = [X][Z]^{-2}$.

$$[Y] = \left(M^{-1} L^{-2} T^{2} Q^{2}\right) \left(M T^{-1} Q^{-1}\right)^{-2}$$ $$[Y] = M^{-1-2} L^{-2} T^{2+2} Q^{2+2}$$ $$[Y] = M^{-3} L^{-2} T^{4} Q^{4}$$

Answer: $M^{-3} L^{-2} T^{4} Q^{4}$