Consider three quantities: `x = (E)/( b) , y = (1)/(sqrt( mu_(0) epsilon _(0))) , and z = (l) /( C R)`.
Here , `l` is the length of a wire , `C` is the capacitance , and `R` is a resistance. All other symbols have usual meanings. Then

Options

A. ` x and y ` have the same dimensions.
B. `x and z` have the same dimensions.
C. `y and z` have the same dimensions.
D. None of the above three pairs have the same dimensions.

Explanation

To solve the problem, we need to determine the dimensions of the three quantities \( x \), \( y \), and \( z \) given in the question. Let's analyze each quantity step by step.

### Step 1: Determine the dimensions of \( x = \frac{E}{b} \)

Here, \( E \) represents electric field strength, and \( b \) represents magnetic field strength.

1. **Electric Field \( E \)**:
- The electric field \( E \) is defined as force per unit charge. The dimension of force is \( [F] = [M][L][T^{-2}] \) and the dimension of charge is \( [Q] \).
- Therefore, the dimension of electric field \( E \) is:
\[ [E] = \frac{[F]}{[Q]} = \frac{[M][L][T^{-2}]}{[Q]} = [M][L][T^{-2}][Q^{-1}] \]

2. **Magnetic Field \( b \)**:
- The magnetic field \( b \) (or magnetic flux density) is defined as force per unit charge per unit velocity. The dimension of velocity is \( [L][T^{-1}] \).
- Therefore, the dimension of magnetic field \( b \) is:
\[ [b] = \frac{[F]}{[Q][v]} = \frac{[M][L][T^{-2}]}{[Q][L][T^{-1}]} = [M][T^{-2}][Q^{-1}][L^{-1}] \]

3. **Combining the dimensions**:
- Now substituting the dimensions into \( x \):
\[ [x] = \frac{[E]}{[b]} = \frac{[M][L][T^{-2}][Q^{-1}]}{[M][T^{-2}][Q^{-1}][L^{-1}]} = [L][T^{-1}] \]
- Thus, the dimension of \( x \) is \( [L][T^{-1}] \) (which represents velocity).

### Step 2: Determine the dimensions of \( y = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \)

1. **Permeability \( \mu_0 \)**:
- The dimension of permeability \( \mu_0 \) is given by:
\[ [\mu_0] = \frac{[F]}{[Q^2][L]} = \frac{[M][L][T^{-2}]}{[Q^2][L]} = [M][T^{-2}][Q^{-2}] \]

2. **Permittivity \( \epsilon_0 \)**:
- The dimension of permittivity \( \epsilon_0 \) is given by:
\[ [\epsilon_0] = \frac{[Q^2]}{[F][L]} = \frac{[Q^2]}{[M][L][T^{-2}]} = [M^{-1}][L^{-3}][T^{4}][Q^{2}] \]

3. **Combining the dimensions**:
- Now substituting the dimensions into \( y \):
\[ [y] = \frac{1}{\sqrt{[\mu_0][\epsilon_0]}} = \frac{1}{\sqrt{[M][T^{-2}][Q^{-2}][M^{-1}][L^{-3}][T^{4}][Q^{2}]}} = \frac{1}{\sqrt{[L^{-3}][T^{2}]}} = [L][T^{-1}] \]
- Thus, the dimension of \( y \) is also \( [L][T^{-1}] \).

### Step 3: Determine the dimensions of \( z = \frac{l}{CR} \)

1. **Length \( l \)**:
- The dimension of length \( l \) is simply:
\[ [l] = [L] \]

2. **Capacitance \( C \)**:
- The dimension of capacitance \( C \) is given by:
\[ [C] = \frac{[Q]}{[V]} = \frac{[Q]}{[E]} = \frac{[Q]}{[M][L][T^{-2}][Q^{-1}]} = [M^{-1}][L^{-2}][T^{4}][Q^{2}] \]

3. **Resistance \( R \)**:
- The dimension of resistance \( R \) is given by:
\[ [R] = \frac{[V]}{[I]} = \frac{[E]}{[Q][T^{-1}]} = \frac{[M][L][T^{-2}][Q^{-1}]}{[Q][T^{-1}]} = [M][L][T^{-3}][Q^{-2}] \]

4. **Combining the dimensions**:
- Now substituting the dimensions into \( z \):
\[ [z] = \frac{[l]}{[C][R]} = \frac{[L]}{[M^{-1}][L^{-2}][T^{4}][Q^{2}][M][L][T^{-3}][Q^{-2}]} = [L][L^{2}][M^{-1}][T^{-4}][Q^{-2}][M][L][T^{3}][Q^{2}] \]
- Simplifying gives:
\[ [z] = [L][T^{-1}] \]
- Thus, the dimension of \( z \) is also \( [L][T^{-1}] \).

### Conclusion

All three quantities \( x \), \( y \), and \( z \) have the same dimension of \( [L][T^{-1}] \), which represents velocity.