In the formula `X =5YZ^(2)`, X and Z have dimensions of capacitance and magnetic field, respectively. What are the dimensions of Y in SI units?

Options

A. `[M^(-2)L^(-2)T^(6)A^(3)]`
B. `[M^(-2)L^(0) T^(-4)A^(-2)]`
C. `[M^(-3) L^(-2) T^(8) A^(4)]`
D. `[M^(-1) L^(-2) T^(4) A^(2)]`

Explanation

To find the dimensions of \( Y \) in the formula \( X = 5YZ^2 \), where \( X \) has the dimensions of capacitance and \( Z \) has the dimensions of magnetic field, we will follow these steps:

### Step 1: Identify the dimensions of \( X \) (Capacitance)
Capacitance \( C \) is defined as the charge \( Q \) per unit potential difference \( V \). The dimensions of charge \( Q \) are given by:
\[ [Q] = I \cdot T \]
where \( I \) is current (in amperes) and \( T \) is time (in seconds).

The potential difference \( V \) can be expressed in terms of work done \( W \) per unit charge:
\[ [V] = \frac{[W]}{[Q]} = \frac{[F \cdot L]}{[Q]} = \frac{[M \cdot L \cdot T^{-2} \cdot L]}{[Q]} = \frac{[M \cdot L^2 \cdot T^{-2}]}{[I \cdot T]} \]
Thus, the dimensions of \( V \) become:
\[ [V] = \frac{[M \cdot L^2 \cdot T^{-2}]}{[I \cdot T]} = [M \cdot L^2 \cdot T^{-3} \cdot I^{-1}] \]

Now, substituting into the capacitance formula:
\[ [C] = \frac{[Q]}{[V]} = \frac{[I \cdot T]}{[M \cdot L^2 \cdot T^{-3} \cdot I^{-1}]} = [M^{-1} \cdot L^{-2} \cdot T^4 \cdot I^2] \]

### Step 2: Identify the dimensions of \( Z \) (Magnetic Field)
The magnetic field \( B \) can be expressed as:
\[ [B] = \frac{[F]}{[I \cdot L]} = \frac{[M \cdot L \cdot T^{-2}]}{[I \cdot L]} = [M \cdot T^{-2} \cdot I^{-1}] \]

### Step 3: Substitute into the equation \( X = 5YZ^2 \)
From the equation \( X = 5YZ^2 \), we can express \( Y \) as:
\[ Y = \frac{X}{Z^2} \]

### Step 4: Substitute the dimensions into the equation
Now substituting the dimensions we found for \( X \) and \( Z \):
\[ [Y] = \frac{[C]}{[B]^2} = \frac{[M^{-1} \cdot L^{-2} \cdot T^4 \cdot I^2]}{([M \cdot T^{-2} \cdot I^{-1}])^2} \]

Calculating \( [B]^2 \):
\[ [B]^2 = [M^2 \cdot T^{-4} \cdot I^{-2}] \]

Now substituting this back into the equation for \( Y \):
\[ [Y] = \frac{[M^{-1} \cdot L^{-2} \cdot T^4 \cdot I^2]}{[M^2 \cdot T^{-4} \cdot I^{-2}]} = [M^{-1} \cdot L^{-2} \cdot T^4 \cdot I^2] \cdot [M^{-2} \cdot T^4 \cdot I^2] \]

### Step 5: Simplify the dimensions
Combining the dimensions:
\[ [Y] = [M^{-1-2} \cdot L^{-2} \cdot T^{4+4} \cdot I^{2+2}] = [M^{-3} \cdot L^{-2} \cdot T^8 \cdot I^4] \]

### Final Answer
Thus, the dimensions of \( Y \) in SI units are:
\[ [Y] = [M^{-3} \cdot L^{-2} \cdot T^8 \cdot I^4] \]