In the formula : X= 3YZ², X and Z have dimensions of capacitance and magnetic induction respectively. What are the dimensions of Y in MKSQ system?

The correct option is [M⁻³L⁻²T⁴Q⁴].

To find the dimensions of $Y$ in the MKSQ system from the given relation:
$X=3Y{Z}^2$
we first need to determine the dimensional formulas for $X$ (capacitance) and $Z$ (magnetic induction).

1. Finding the dimensions of capacitance ($X$):
Capacitance ($C$) is defined as the ratio of electric charge ($Q$) to electric potential ($V$):
$C=\frac{Q}{V}$
Since electric potential is work done ($W$) per unit charge, we have:
$V=\frac{W}{Q}$
Substituting this into the capacitance equation gives:
$C=\frac{Q^2}{W}$
The dimensional formula for work (energy) is:
$\left[W\right]=\left[M{L}^2{T}^{-2}\right]$
Therefore, the dimensional formula for capacitance $X$ is:
$\left[X\right]=\frac{\left[Q^2\right]}{\left[M{L}^2{T}^{-2}\right]}=\left[M^{-1}{L}^{-2}{T}^2{Q}^2\right]$

2. Finding the dimensions of magnetic induction ($Z$):
The magnetic force ($F$) on a charge ($Q$) moving with velocity ($v$) in a magnetic field ($B$) is given by:
$F=QvB$
Thus, the magnetic induction $B$ can be written as:
$B=\frac{F}{Qv}$
The dimensional formula for force is $\left[ML{T}^{-2}\right]$ and for velocity is $\left[L{T}^{-1}\right]$.
Therefore, the dimensional formula for magnetic induction $Z$ is:
$\left[Z\right]=\frac{\left[ML{T}^{-2}\right]}{\left[Q\right]\left[L{T}^{-1}\right]}=\left[M{T}^{-1}{Q}^{-1}\right]$

3. Determining the dimensions of $Y$:
From the given relation:
$X=3Y{Z}^2$
Since 3 is a dimensionless constant, we can write:
$\left[X\right]=\left[Y\right]{\left[Z\right]}^2$
Rearranging to solve for the dimensions of $Y$:
$\left[Y\right]=\frac{\left[X\right]}{{\left[Z\right]}^2}$
Substitute the derived dimensions of $X$ and $Z$:
$\left[Y\right]=\frac{\left[M^{-1}{L}^{-2}{T}^2{Q}^2\right]}{{\left[M{T}^{-1}{Q}^{-1}\right]}^2}$
Simplifying the denominator:
$\left[Y\right]=\frac{\left[M^{-1}{L}^{-2}{T}^2{Q}^2\right]}{\left[M^2{T}^{-2}{Q}^{-2}\right]}$
Combining the terms:
$\left[Y\right]=\left[M^{-1-2}{L}^{-2}{T}^{2-(-2)}{Q}^{2-(-2)}\right]$
$\left[Y\right]=\left[M^{-3}{L}^{-2}{T}^4{Q}^4\right]$