In the formula X=3YZ^(2), X and Z have dimension of capacitance and magnetic induction respectively. The dimensions of Y in MKSQ system are

Options

A. 1. `[M^(-3)L^(-2)T^(4)Q^4]`
B. 2. `[M^(-2)L^(-1)T^(5)Q^3]`
C. 3. `[M^(-1)L^(-2)T^(4)Q^4]`
D. 4. `[M^(-3)L^(-1)T^(4)Q^4]`

Explanation

To solve the problem step by step, we need to find the dimensions of Y in the equation \( X = 3YZ^2 \), where \( X \) has the dimensions of capacitance and \( Z \) has the dimensions of magnetic induction.

### Step 1: Identify the dimensions of \( X \) (Capacitance)
Capacitance \( C \) is defined as:
\[ C = \frac{Q}{V} \]
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 has the dimensions of energy, which is:
\[ W = F \cdot d = m \cdot a \cdot d = m \cdot \frac{L}{T^2} \cdot L = mL^2T^{-2} \]
Thus, substituting this into the equation for voltage, we get:
\[ V = \frac{mL^2T^{-2}}{Q} \]
Now substituting \( V \) back into the capacitance formula:
\[ C = \frac{Q}{\frac{mL^2T^{-2}}{Q}} = \frac{Q^2}{mL^2T^{-2}} = \frac{Q^2T^2}{mL^2} \]
Thus, the dimensions of capacitance \( C \) are:
\[ [X] = \frac{Q^2}{mL^2T^{-2}} = M^{-1}L^{-2}T^{2}Q^{2} \]

### Step 2: Identify the dimensions of \( Z \) (Magnetic Induction)
Magnetic induction \( B \) can be defined as:
\[ B = \frac{F}{Qv} \]
Where \( F \) is force, \( Q \) is charge, and \( v \) is velocity. The dimensions of force \( F \) are:
\[ F = m \cdot a = m \cdot \frac{L}{T^2} = mL T^{-2} \]
And the dimensions of velocity \( v \) are:
\[ v = \frac{L}{T} \]
Thus, substituting these into the equation for magnetic induction:
\[ B = \frac{mLT^{-2}}{Q \cdot \frac{L}{T}} = \frac{mLT^{-2}}{QLT^{-1}} = \frac{m}{QT} \]
So, the dimensions of magnetic induction \( B \) are:
\[ [Z] = M T^{-2} Q^{-1} \]

### Step 3: Rearranging the equation to find \( Y \)
From the equation \( X = 3YZ^2 \), we can rearrange it to find \( Y \):
\[ Y = \frac{X}{Z^2} \]

### Step 4: Substitute the dimensions of \( X \) and \( Z \)
Substituting the dimensions we found:
\[ [Y] = \frac{[X]}{[Z]^2} = \frac{M^{-1}L^{-2}T^{2}Q^{2}}{(M T^{-2} Q^{-1})^2} \]
Calculating \( [Z]^2 \):
\[ [Z]^2 = (M T^{-2} Q^{-1})^2 = M^2 T^{-4} Q^{-2} \]
Now substituting back into the equation for \( Y \):
\[ [Y] = \frac{M^{-1}L^{-2}T^{2}Q^{2}}{M^2 T^{-4} Q^{-2}} = M^{-1}L^{-2}T^{2}Q^{2} \cdot M^{-2} T^{4} Q^{2} \]
This simplifies to:
\[ [Y] = M^{-3}L^{-2}T^{6}Q^{4} \]

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

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