The peak electric field produced by the radiation coming from the 80 W bulb at a distance of 10 m is `(x)/(10) sqrt((mu_(0)c )/(pi))(V)/(m)`. The efficiency of the bulb is 10% and it is a point source. The value of x is____________.

Answer

To solve the problem, we need to find the peak electric field produced by the radiation from an 80 W bulb at a distance of 10 m, given that the bulb has an efficiency of 10%.

### Step-by-Step Solution:

1. **Calculate the Effective Power Output:**
The bulb has a power rating of 80 W, but it operates at 10% efficiency. Therefore, the effective power output (P) that contributes to the light is:
\[ P = 80 \, \text{W} \times 0.10 = 8 \, \text{W} \]

2. **Determine the Intensity of the Radiation:**
The intensity (I) of the radiation at a distance (r) from a point source is given by the formula:
\[ I = \frac{P}{A} \]
where \( A \) is the surface area of a sphere with radius \( r \):
\[ A = 4\pi r^2 \]
Substituting \( r = 10 \, \text{m} \):
\[ A = 4\pi (10)^2 = 400\pi \, \text{m}^2 \]
Now, substituting the values into the intensity formula:
\[ I = \frac{8 \, \text{W}}{400\pi} = \frac{1}{50\pi} \, \text{W/m}^2 \]

3. **Relate Intensity to Electric Field:**
The intensity of an electromagnetic wave can also be expressed in terms of the peak electric field \( E_0 \):
\[ I = \frac{1}{2} \epsilon_0 c E_0^2 \]
where \( \epsilon_0 \) is the permittivity of free space and \( c \) is the speed of light.

4. **Express \( \epsilon_0 \) in terms of \( \mu_0 \):**
We know that:
\[ c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \implies \epsilon_0 = \frac{1}{\mu_0 c^2} \]
Substituting this into the intensity equation gives:
\[ I = \frac{1}{2} \left(\frac{1}{\mu_0 c^2}\right) c E_0^2 = \frac{1}{2\mu_0 c} E_0^2 \]

5. **Equate the Two Expressions for Intensity:**
Now we equate the two expressions for intensity:
\[ \frac{1}{50\pi} = \frac{1}{2\mu_0 c} E_0^2 \]
Rearranging gives:
\[ E_0^2 = \frac{2}{\mu_0 c} \cdot \frac{1}{50\pi} \]

6. **Solve for \( E_0 \):**
Taking the square root:
\[ E_0 = \sqrt{\frac{2}{50\pi}} \sqrt{\frac{1}{\mu_0 c}} = \frac{1}{5\sqrt{\pi}} \sqrt{\frac{2}{\mu_0 c}} \]

7. **Express in the Required Form:**
The problem states that the peak electric field can be expressed as:
\[ E_0 = \frac{x}{10} \sqrt{\frac{\mu_0 c}{\pi}} \, \text{V/m} \]
To match this with our expression, we have:
\[ \frac{1}{5\sqrt{\pi}} \sqrt{\frac{2}{\mu_0 c}} = \frac{x}{10} \sqrt{\frac{\mu_0 c}{\pi}} \]

8. **Finding \( x \):**
By comparing both sides, we can find \( x \):
\[ x = 2 \]

### Final Answer:
The value of \( x \) is **2**.