If 'C' and 'V' represent capacity and voltage respectively then what are the dimensions of `lambda`, where `(C )/(V )=lambda` ?

Options

A. `[M^(-2)L^(-3)I^(2)T^(6)]`
B. `[M^(-3)L^(-4)L^(-4)I^(3)T^(7)]`
C. `[M^(-1)L^(-3)I^(-2)T^(-7)]`
D. `[M^(-2)L^(-4)I^(3)T^(7)]`

Answer

To find the dimensions of \( \lambda \) where \( \lambda = \frac{C}{V} \), we will start by determining the dimensions of capacitance \( C \) and voltage \( V \).

### Step 1: Determine the dimensions of capacitance \( C \)

Capacitance \( C \) is defined as the charge \( Q \) per unit voltage \( V \):
\[ C = \frac{Q}{V} \]
Thus, we can express \( C \) in terms of its dimensions:
\[ C = \frac{[Q]}{[V]} \]

### Step 2: Determine the dimensions of voltage \( V \)

Voltage \( V \) can be defined as the work done per unit charge. The work done \( W \) has dimensions of energy, which is given by:
\[ W = F \cdot d \]
where \( F \) is force and \( d \) is distance. The dimensions of force \( F \) are:
\[ [F] = [M][L][T^{-2}] \]
and the dimensions of distance \( d \) are:
\[ [d] = [L] \]
Thus, the dimensions of work \( W \) are:
\[ [W] = [F][d] = [M][L][T^{-2}][L] = [M][L^2][T^{-2}] \]
Now, substituting this into the voltage definition:
\[ V = \frac{W}{Q} = \frac{[M][L^2][T^{-2}]}{[Q]} \]

### Step 3: Substitute the dimensions of \( V \) into the capacitance equation

Now we can substitute the dimensions of \( V \) back into the capacitance equation:
\[ C = \frac{[Q]}{[V]} = \frac{[Q]}{\frac{[M][L^2][T^{-2}]}{[Q]}} = \frac{[Q]^2}{[M][L^2][T^{-2}]} \]

### Step 4: Substitute the dimensions of \( C \) and \( V \) into \( \lambda \)

Now we can express \( \lambda \) in terms of its dimensions:
\[ \lambda = \frac{C}{V} = \frac{\frac{[Q]^2}{[M][L^2][T^{-2}]}}{\frac{[M][L^2][T^{-2}]}{[Q]}} = \frac{[Q]^2}{[M][L^2][T^{-2}]} \cdot \frac{[Q]}{[M][L^2][T^{-2}]} = \frac{[Q]^3}{[M^2][L^4][T^{-4}]} \]

### Step 5: Express \( Q \) in terms of current \( I \)

We know that charge \( Q \) can be expressed in terms of current \( I \) and time \( t \):
\[ Q = I \cdot t \]
Thus, substituting this into our expression for \( \lambda \):
\[ \lambda = \frac{(I \cdot t)^3}{[M^2][L^4][T^{-4}]} = \frac{I^3 \cdot t^3}{[M^2][L^4][T^{-4}]} \]

### Step 6: Combine dimensions

Now, we can simplify the dimensions of \( \lambda \):
\[ \lambda = \frac{I^3 \cdot T^3}{M^2 \cdot L^4 \cdot T^{-4}} = \frac{I^3 \cdot T^7}{M^2 \cdot L^4} \]

### Final Result

Thus, the dimensions of \( \lambda \) are:
\[ \lambda = [M^{-2}][L^{-4}][T^7][I^3] \]