If `e, E_0, h and C` respectively represents electronic change, permittively of free space, planks constant and speed of light then `(e^2)/(E_0 hC)` has the dimensions of
A) angle `" "` B) relative density
C) strain `" "` D) current

Options

A. A ,B
B. D
C. A, B, C
D. A, B, C , D

Answer

To solve the problem, we need to find the dimensions of the expression \(\frac{e^2}{E_0 h C}\), where \(e\) is the electronic charge, \(E_0\) is the permittivity of free space, \(h\) is Planck's constant, and \(C\) is the speed of light. We will determine the dimensions of each quantity involved and then simplify the expression step by step.

### Step 1: Determine the dimensions of each quantity

1. **Electronic Charge (\(e\))**:
- The dimension of electric charge is given by the relation \(Q = I \cdot T\), where \(I\) is current and \(T\) is time.
- Therefore, the dimension of \(e\) is:
\[ [e] = [I][T] = A \cdot T \]

2. **Permittivity of Free Space (\(E_0\))**:
- The permittivity of free space has the unit \(\text{F/m}\) (farads per meter), which can be expressed as:
\[ [E_0] = \frac{C^2}{N \cdot m^2} = \frac{A^2 \cdot T^2}{kg \cdot m^2} = M^{-1} L^{-3} T^4 A^2 \]

3. **Planck's Constant (\(h\))**:
- The unit of Planck's constant is joule-seconds, which can be expressed as:
\[ [h] = J \cdot s = kg \cdot m^2 \cdot s^{-1} = M L^2 T^{-1} \]

4. **Speed of Light (\(C\))**:
- The speed of light has the dimension of length over time:
\[ [C] = L T^{-1} \]

### Step 2: Substitute the dimensions into the expression

Now we substitute the dimensions into the expression \(\frac{e^2}{E_0 h C}\):

\[ \frac{e^2}{E_0 h C} = \frac{(A \cdot T)^2}{(M^{-1} L^{-3} T^4 A^2)(M L^2 T^{-1})(L T^{-1})} \]

### Step 3: Simplify the expression

1. **Calculate \(e^2\)**:
\[ e^2 = (A \cdot T)^2 = A^2 \cdot T^2 \]

2. **Calculate the denominator**:
\[ E_0 h C = (M^{-1} L^{-3} T^4 A^2)(M L^2 T^{-1})(L T^{-1}) \]
- Combine the dimensions:
\[ = M^{-1} L^{-3} T^4 A^2 \cdot M L^2 T^{-1} \cdot L T^{-1} = M^{0} L^{-3 + 2 + 1} T^{4 - 1 - 1} A^2 = M^{0} L^{0} T^{2} A^{2} \]

3. **Final expression**:
\[ \frac{e^2}{E_0 h C} = \frac{A^2 T^2}{M^{0} L^{0} T^{2} A^{2}} = \frac{A^2 T^2}{A^{2} T^{2}} = 1 \]

### Conclusion

The dimensions of the expression \(\frac{e^2}{E_0 h C}\) simplify to \(M^0 L^0 T^0\), which is dimensionless.

### Answer

The correct option is **C) strain**, as strain is a dimensionless quantity.