If `X` is capacitance and `Y` is the magnetic field which are related by `X=2aY^(2)`. Dimension of a will be:-

Options

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

Answer

To find the dimensions of \( a \) in the equation \( X = 2aY^2 \), where \( X \) is capacitance and \( Y \) is magnetic field, we will follow these steps:

### Step 1: Identify the dimensions of capacitance \( X \)
The capacitance \( X \) is defined as:
\[ X = \frac{Q}{V} \]
where \( Q \) is charge and \( V \) is voltage.

The dimensions of charge \( Q \) are:
\[ [Q] = A \cdot T \]
The dimensions of voltage \( V \) can be expressed as:
\[ [V] = \frac{[Energy]}{[Charge]} = \frac{[ML^2T^{-2}]}{[Q]} = \frac{[ML^2T^{-2}]}{[AT]} = [ML^2T^{-3}A^{-1}] \]
Thus, the dimensions of capacitance \( X \) are:
\[ [X] = \frac{[Q]}{[V]} = \frac{[AT]}{[ML^2T^{-3}A^{-1}]} = [A^2T^4M^{-1}L^{-2}] \]

### Step 2: Identify the dimensions of magnetic field \( Y \)
The magnetic field \( Y \) can be defined in terms of force per unit charge and velocity:
\[ [Y] = \frac{[F]}{[Q][v]} = \frac{[MLT^{-2}]}{[Q][L T^{-1}]} = \frac{[MLT^{-2}]}{[AT][L]} = [MT^{-2}A^{-1}] \]

### Step 3: Substitute the dimensions into the equation
The equation given is:
\[ X = 2aY^2 \]
Rearranging gives us:
\[ a = \frac{X}{2Y^2} \]
Since the constant \( 2 \) has no dimensions, we focus on the dimensions of \( a \):
\[ [a] = \frac{[X]}{[Y]^2} \]

### Step 4: Substitute the dimensions of \( X \) and \( Y \)
Substituting the dimensions we found:
\[ [a] = \frac{[A^2T^4M^{-1}L^{-2}]}{([MT^{-2}A^{-1}])^2} \]
Calculating \( [Y]^2 \):
\[ [Y]^2 = [M^2T^{-4}A^{-2}] \]
Now substituting back into the equation for \( a \):
\[ [a] = \frac{[A^2T^4M^{-1}L^{-2}]}{[M^2T^{-4}A^{-2}]} = [A^{2+2}T^{4+4}M^{-1-2}L^{-2}] = [A^4T^8M^{-3}L^{-2}] \]

### Final Result
Thus, the dimensions of \( a \) are:
\[ [a] = [M^{-3}L^{-2}T^{4}A^{4}] \]