A circular disc of radius carries surface charge density `=sigma_(0)(1-(r )/(R ))` , where `sigma_(0)` is a constant and is the distance from the center of the disc Electric flux through a lange spherical surface that endcloses the charged disc completely is `phi_(0)`. electric flux through another spherical surface of radius `(R ) /(4)` and consentri with the disc is `phi` . Then the ratio `(phi_(0))/(phi)` is __________.

Explanation

To solve the problem, we need to find the ratio of the electric flux through a large spherical surface that encloses a charged disc completely (denoted as \(\phi_0\)) to the electric flux through another spherical surface of radius \(R/4\) that is concentric with the disc (denoted as \(\phi\)).

### Step-by-Step Solution:

1. **Understanding the Surface Charge Density**:
The surface charge density on the disc is given by:
\[ \sigma(r) = \sigma_0 \left(1 - \frac{r}{R}\right) \]
where \(r\) is the distance from the center of the disc and \(R\) is the radius of the disc.

2. **Finding the Total Charge on the Disc**:
To find the total charge \(Q\) on the disc, we need to integrate the charge density over the area of the disc. We can consider the disc as composed of infinitesimally thin rings:
- The area of an elemental ring at radius \(r\) and thickness \(dr\) is:
\[ dA = 2\pi r \, dr \]
- The charge \(dQ\) on this ring is:
\[ dQ = \sigma(r) \cdot dA = \sigma_0 \left(1 - \frac{r}{R}\right) \cdot 2\pi r \, dr \]
- Thus, the total charge \(Q\) on the disc is:
\[ Q = \int_0^R dQ = \int_0^R \sigma_0 \left(1 - \frac{r}{R}\right) \cdot 2\pi r \, dr \]

3. **Calculating the Integral**:
- Expanding the integral:
\[ Q = 2\pi \sigma_0 \int_0^R \left(r - \frac{r^2}{R}\right) dr \]
- Evaluating the integral:
\[ \int_0^R r \, dr = \frac{R^2}{2}, \quad \int_0^R r^2 \, dr = \frac{R^3}{3} \]
- Therefore:
\[ Q = 2\pi \sigma_0 \left( \frac{R^2}{2} - \frac{1}{R} \cdot \frac{R^3}{3} \right) = 2\pi \sigma_0 \left( \frac{R^2}{2} - \frac{R^2}{3} \right) = 2\pi \sigma_0 \cdot \frac{3R^2 - 2R^2}{6} = \frac{\pi \sigma_0 R^2}{3} \]

4. **Calculating Electric Flux \(\phi_0\)**:
The electric flux through the large spherical surface that encloses the charged disc is given by Gauss's law:
\[ \phi_0 = \frac{Q}{\epsilon_0} = \frac{\frac{\pi \sigma_0 R^2}{3}}{\epsilon_0} \]

5. **Calculating Electric Flux \(\phi\) through the smaller sphere**:
For the smaller sphere of radius \(R/4\), we find the charge enclosed within this radius:
- The charge \(Q'\) enclosed by the smaller sphere can be calculated similarly, but we integrate only up to \(R/4\):
\[ Q' = 2\pi \sigma_0 \int_0^{R/4} \left(1 - \frac{r}{R}\right) r \, dr \]
- Evaluating this integral gives:
\[ Q' = 2\pi \sigma_0 \left( \int_0^{R/4} r \, dr - \frac{1}{R} \int_0^{R/4} r^2 \, dr \right) \]
- Carrying out the integrals, we find:
\[ Q' = 2\pi \sigma_0 \left( \frac{(R/4)^2}{2} - \frac{1}{R} \cdot \frac{(R/4)^3}{3} \right) = 2\pi \sigma_0 \left( \frac{R^2}{32} - \frac{R^2}{192} \right) = 2\pi \sigma_0 \cdot \frac{6R^2 - R^2}{192} = \frac{10\pi \sigma_0 R^2}{192} \]
- Thus, the electric flux through the smaller sphere is:
\[ \phi = \frac{Q'}{\epsilon_0} = \frac{\frac{10\pi \sigma_0 R^2}{192}}{\epsilon_0} \]

6. **Finding the Ratio \(\frac{\phi_0}{\phi}\)**:
Now, we can find the ratio:
\[ \frac{\phi_0}{\phi} = \frac{\frac{\frac{\pi \sigma_0 R^2}{3}}{\epsilon_0}}{\frac{\frac{10\pi \sigma_0 R^2}{192}}{\epsilon_0}} = \frac{\frac{1}{3}}{\frac{10}{192}} = \frac{192}{30} = \frac{32}{5} = 6.4 \]

### Final Answer:
The ratio \(\frac{\phi_0}{\phi}\) is \(6.4\).