A physical quantity of the dimension of length that can be formed out of `c,G` and `(e^(2))/(4 pi epsilon_(0))` is [`c` is velocity of light `G` is universal constant of gravitation, e is charge

Options

A. `1/c^(2)[G(e^(2))/(4piepsilon_(0))]^(1/2)`
B. `c^(2)[G(e^(2))/(4piepsilon_(0))]^(1/2)`
C. `1/c^(2)[G(e^(2))/(4piepsilon_(0))]^(1/2)`
D. `(1)/(c)G[e^(2))/(4piepsilon_(0))`

Answer

To solve the problem of finding a physical quantity of the dimension of length that can be formed out of \( c \), \( G \), and \( \frac{e^2}{4 \pi \epsilon_0} \), we will follow these steps:

### Step 1: Identify the dimensions of each quantity
1. **Speed of light \( c \)**:
- Dimension: \( [c] = [L][T]^{-1} \)
- This means \( c \) has dimensions of length per time.

2. **Gravitational constant \( G \)**:
- The formula for gravitational force is \( F = \frac{G m_1 m_2}{r^2} \).
- Rearranging gives us \( G = \frac{F r^2}{m_1 m_2} \).
- Dimensions of force \( F \) are \( [F] = [M][L][T]^{-2} \).
- Therefore, \( [G] = \frac{[M][L][T]^{-2} \cdot [L]^2}{[M]^2} = [M]^{-1}[L]^3[T]^{-2} \).

3. **Electrostatic force from \( \frac{e^2}{4 \pi \epsilon_0} \)**:
- The formula for electrostatic force is \( F = \frac{e^2}{4 \pi \epsilon_0 r^2} \).
- Rearranging gives us \( \frac{e^2}{4 \pi \epsilon_0} = F r^2 \).
- Using the dimensions of force as above, we have:
- Dimensions of \( \frac{e^2}{4 \pi \epsilon_0} \) are \( [F][L]^2 = [M][L][T]^{-2} \cdot [L]^2 = [M][L]^3[T]^{-2} \).

### Step 2: Formulate the expression for length
We can express length \( L \) in terms of \( c \), \( G \), and \( \frac{e^2}{4 \pi \epsilon_0} \):
\[ L = c^x G^y \left(\frac{e^2}{4 \pi \epsilon_0}\right)^z \]

### Step 3: Substitute the dimensions into the equation
Substituting the dimensions we found:
\[ [L] = [L]^{x} [M]^{y} [L]^{3z} [M]^{1} [L]^{3} [T]^{-2} \]
This simplifies to:
\[ [L] = [M]^{y + 1} [L]^{x + 3z + 3} [T]^{-2} \]

### Step 4: Set up the equations based on dimensions
To find \( x \), \( y \), and \( z \), we equate the powers of \( M \), \( L \), and \( T \) to those of length:
1. For \( M \): \( 0 = y + 1 \) → \( y = -1 \)
2. For \( L \): \( 1 = x + 3z + 3 \)
3. For \( T \): \( 0 = -2 \) → This equation is always satisfied.

### Step 5: Solve for \( x \) and \( z \)
Substituting \( y = -1 \) into the equation for \( L \):
\[ 1 = x + 3z + 3 \]
This simplifies to:
\[ x + 3z = -2 \quad \text{(Equation 1)} \]

### Step 6: Express \( z \) in terms of \( x \)
From \( y = -1 \) and substituting into the equation for \( L \):
\[ z = -1 \quad \text{(since \( y = z \))} \]
Substituting \( z = -1 \) into Equation 1:
\[ x + 3(-1) = -2 \implies x - 3 = -2 \implies x = 1 \]

### Step 7: Final expression for length
Now substituting \( x = 1 \), \( y = -1 \), and \( z = -1 \) back into the length equation:
\[ L = c^1 G^{-1} \left(\frac{e^2}{4 \pi \epsilon_0}\right)^{-1} \]
This can be rewritten as:
\[ L = \frac{c}{G \cdot \frac{e^2}{4 \pi \epsilon_0}} = \frac{4 \pi \epsilon_0 c}{G e^2} \]

### Conclusion
Thus, a physical quantity of the dimension of length that can be formed out of \( c \), \( G \), and \( \frac{e^2}{4 \pi \epsilon_0} \) is:
\[ L = \sqrt{\frac{c^2 G}{\frac{e^2}{4 \pi \epsilon_0}}} \]