Which of the following product of `e , h , mu , G` ( where ` mu` is permeability ) be taken so that the dimensions of the product are same as that of the speed of light ?

Options

A. `he^(-2) mu^(-1) G^(0)`
B. `h^(2) e G^(0) mu`
C. `h^(0)e^(2) G^(-1) mu`
D. `he^(-2) mu^(0)`

Explanation

To solve the problem of determining which product of \( e, h, \mu, G \) gives dimensions equivalent to the speed of light \( c \), we will follow these steps:

### Step 1: Understand the dimensions of speed of light
The speed of light \( c \) has dimensions given by:
\[ [c] = L^1 T^{-1} \]

### Step 2: Write the relationship
We assume that the speed of light can be expressed in terms of the variables \( e, h, \mu, G \) as:
\[ c \propto e^a h^b \mu^c G^d \]
where \( a, b, c, d \) are the powers we need to determine.

### Step 3: Write the dimensional formulas
Next, we need the dimensional formulas for each of the quantities:
- Electric charge \( e \): \( [e] = A T \)
- Planck's constant \( h \): \( [h] = M L^2 T^{-1} \)
- Permeability \( \mu \): \( [\mu] = M L T^{-2} A^{-2} \)
- Gravitational constant \( G \): \( [G] = M^{-1} L^3 T^{-2} \)

### Step 4: Write the dimensions in terms of \( a, b, c, d \)
Now, we can express the dimensions of the right-hand side:
\[ [e^a] = (A T)^a = A^a T^a \]
\[ [h^b] = (M L^2 T^{-1})^b = M^b L^{2b} T^{-b} \]
\[ [\mu^c] = (M L T^{-2} A^{-2})^c = M^c L^c T^{-2c} A^{-2c} \]
\[ [G^d] = (M^{-1} L^3 T^{-2})^d = M^{-d} L^{3d} T^{-2d} \]

### Step 5: Combine the dimensions
Combining these, we get:
\[ [e^a h^b \mu^c G^d] = M^{b+c-d} L^{2b+c+3d} T^{a-b-2c-2d} A^{a-2c} \]

### Step 6: Set up equations
We equate the dimensions from both sides:
- For \( M \): \( 0 = b + c - d \) (Equation 1)
- For \( L \): \( 1 = 2b + c + 3d \) (Equation 2)
- For \( T \): \( -1 = a - b - 2c - 2d \) (Equation 3)
- For \( A \): \( 0 = a - 2c \) (Equation 4)

### Step 7: Solve the equations
From Equation 4, we have:
\[ a = 2c \]

Substituting \( a = 2c \) into Equation 3 gives:
\[ -1 = 2c - b - 2c - 2d \implies -1 = -b - 2d \implies b + 2d = 1 \quad (Equation 5) \]

Now substituting \( a = 2c \) into Equation 1 gives:
\[ 0 = b + c - d \quad (Equation 1) \]

Now we have two equations (Equation 1 and Equation 5):
1. \( b + c - d = 0 \)
2. \( b + 2d = 1 \)

From Equation 1, we can express \( c \):
\[ c = d - b \]
Substituting into Equation 5:
\[ b + 2d = 1 \implies b + 2(d - b) = 1 \implies b + 2d - 2b = 1 \implies -b + 2d = 1 \implies b = 2d - 1 \]

Substituting \( b = 2d - 1 \) back into Equation 1:
\[ 2d - 1 + c - d = 0 \implies c = -d + 1 \]

### Step 8: Find values of \( a, b, c, d \)
Now we can express everything in terms of \( d \):
- \( b = 2d - 1 \)
- \( c = -d + 1 \)
- \( a = 2(-d + 1) = -2d + 2 \)

### Step 9: Solve for \( d \)
Substituting \( d = 0 \) gives:
- \( b = 1 \)
- \( c = 1 \)
- \( a = -2 \)

### Final expression
Thus, we find:
\[ c \propto e^{-2} h^1 \mu^{-1} G^0 \]
This means the correct product is:
\[ c \propto \frac{h}{e^2 \mu} \]

### Final Answer
The correct option is \( e^{-2} h^1 \mu^{-1} \).