Dimensional formula of `(e^(4))/(epsilon_(0)^(2)m_(p)m_(e)^(2)c^(3)G)` is (Where `e=` charge , `m_(p)` and `m_(c )` are masses, `epsilon_(0)=` permittivity of free space, `c=` velocity of light and `G=` gravitational constant)

Options

A. `M`
B. `T`
C. `p`
D. `I^(4)T^(2)M`

Answer

To find the dimensional formula of the expression \(\frac{e^4}{\epsilon_0^2 m_p m_e^2 c^3 G}\), we will break down each component into its dimensional formula and then combine them according to the expression.

### Step 1: Identify the dimensions of each variable

1. **Charge (e)**: The dimensional formula for charge is given by:
\[ [e] = I \cdot T \]
where \(I\) is the dimension of electric current and \(T\) is time.

2. **Permittivity of free space (\(\epsilon_0\))**: The dimensional formula for \(\epsilon_0\) is:
\[ [\epsilon_0] = M^{-1} L^{-3} T^4 I^2 \]

3. **Mass of proton (\(m_p\)) and mass of electron (\(m_e\))**: Both have the same dimensional formula:
\[ [m_p] = [m_e] = M \]

4. **Velocity of light (c)**: The dimensional formula for velocity is:
\[ [c] = L T^{-1} \]

5. **Gravitational constant (G)**: The dimensional formula for \(G\) is:
\[ [G] = M^{-1} L^3 T^{-2} \]

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

Now, we can substitute these dimensional formulas into the expression:

\[ \frac{e^4}{\epsilon_0^2 m_p m_e^2 c^3 G} \]

Substituting the dimensions:

\[ = \frac{(I T)^4}{(M^{-1} L^{-3} T^4 I^2)^2 \cdot M \cdot M^2 \cdot (L T^{-1})^3 \cdot (M^{-1} L^3 T^{-2})} \]

### Step 3: Simplify the expression

Calculating the numerator:

\[ = I^4 T^4 \]

Calculating the denominator:

1. \(\epsilon_0^2\):
\[ (M^{-1} L^{-3} T^4 I^2)^2 = M^{-2} L^{-6} T^8 I^4 \]

2. \(m_p m_e^2\):
\[ M \cdot M^2 = M^3 \]

3. \(c^3\):
\[ (L T^{-1})^3 = L^3 T^{-3} \]

4. \(G\):
\[ M^{-1} L^3 T^{-2} \]

Now combine the denominator:

\[ = (M^{-2} L^{-6} T^8 I^4) \cdot M^3 \cdot (L^3 T^{-3}) \cdot (M^{-1} L^3 T^{-2}) \]

Combining the terms:

\[ = M^{-2 + 3 - 1} L^{-6 + 3 + 3} T^{8 - 3 - 2} I^4 = M^0 L^0 T^3 I^4 \]

### Step 4: Final expression

Now substituting everything back into the expression:

\[ = \frac{I^4 T^4}{M^0 L^0 T^3 I^4} = T^{4 - 3} = T^1 \]

Thus, the dimensional formula of the given expression is:

\[ \boxed{T} \]