In the formula `X = 3Y Z^(2)` , `X` and `Z` have dimensions of capacitance and magnetic induction respectively . What are the dimensions of Y in MESQ system ?

Options

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

Explanation

To find the dimensions of \( Y \) in the formula \( X = 3Y Z^2 \), where \( X \) has the dimensions of capacitance and \( Z \) has the dimensions of magnetic induction, we can follow these steps:

### Step 1: Determine the dimensions of \( X \) (Capacitance)
The dimension of capacitance \( C \) is given by the formula:
\[ C = \frac{Q}{V} \]
Where \( Q \) is charge and \( V \) is voltage. Voltage \( V \) can be expressed as:
\[ V = \frac{W}{Q} \]
Where \( W \) is work done. Work done can be expressed as:
\[ W = F \cdot d \]
Where \( F \) is force and \( d \) is displacement. The dimension of force \( F \) is:
\[ F = m \cdot a = m \cdot \frac{L}{T^2} = M L T^{-2} \]
Thus, the dimension of work \( W \) becomes:
\[ W = F \cdot d = (M L T^{-2}) \cdot L = M L^2 T^{-2} \]
Substituting back into the expression for voltage:
\[ V = \frac{W}{Q} = \frac{M L^2 T^{-2}}{Q} \]
So, the dimension of capacitance \( C \) is:
\[ C = \frac{Q}{V} = \frac{Q}{\frac{M L^2 T^{-2}}{Q}} = \frac{Q^2}{M L^2 T^{-2}} = M^{-1} L^{-2} T^2 Q^2 \]

### Step 2: Determine the dimensions of \( Z \) (Magnetic Induction)
The dimension of magnetic induction \( B \) can be expressed as:
\[ B = \frac{F}{I \cdot L} \]
Where \( I \) is current and \( L \) is length. Current \( I \) can be expressed as:
\[ I = \frac{Q}{T} \]
Thus, substituting this into the expression for \( B \):
\[ B = \frac{F}{\left(\frac{Q}{T}\right) \cdot L} = \frac{F \cdot T}{Q \cdot L} \]
Substituting the dimension of force:
\[ B = \frac{M L T^{-2} \cdot T}{Q \cdot L} = \frac{M}{Q \cdot T^2} \]

### Step 3: Substitute into the equation \( X = 3Y Z^2 \)
From the equation \( X = 3Y Z^2 \), we can express \( Y \) as:
\[ Y = \frac{X}{Z^2} \]
Ignoring the dimensionless constant \( 3 \), we can substitute the dimensions we found:
\[ \text{Dimension of } Y = \frac{\text{Dimension of } X}{(\text{Dimension of } Z)^2} \]
Substituting the dimensions:
\[ \text{Dimension of } Y = \frac{M^{-1} L^{-2} T^2 Q^2}{\left(\frac{M}{Q T^2}\right)^2} \]
Calculating \( Z^2 \):
\[ Z^2 = \left(\frac{M}{Q T^2}\right)^2 = \frac{M^2}{Q^2 T^4} \]
Now substituting this back into the equation for \( Y \):
\[ \text{Dimension of } Y = \frac{M^{-1} L^{-2} T^2 Q^2}{\frac{M^2}{Q^2 T^4}} = M^{-1} L^{-2} T^2 Q^2 \cdot \frac{Q^2 T^4}{M^2} = M^{-1 - 2} L^{-2} T^{2 + 4} Q^{2 + 2} \]
This simplifies to:
\[ \text{Dimension of } Y = M^{-3} L^{-2} T^6 Q^4 \]

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