The earth may be regarded as a spherically shaped uniform core of density `rho_(1)` and radius `(R)/(2)` surrounded by a uniform shell of thickness `(R)/(2)` and density `rho_(2)`. Find the ratio of `(rho_(1))/(rho_(2))` if the value of acceleration due to gravity is the same at surface as at depth `(R)/(2)` from the surface.

Explanation

To solve the problem, we need to find the ratio of the densities \( \frac{\rho_1}{\rho_2} \) given that the acceleration due to gravity is the same at the surface of the Earth and at a depth of \( \frac{R}{2} \) from the surface.

### Step-by-Step Solution:

1. **Understanding the Structure of the Earth**:
- The Earth consists of a uniform core of radius \( \frac{R}{2} \) and density \( \rho_1 \).
- It is surrounded by a uniform shell of thickness \( \frac{R}{2} \) and density \( \rho_2 \).
- Therefore, the outer radius of the Earth is \( R \).

2. **Acceleration due to Gravity at the Surface**:
- The formula for gravitational acceleration at the surface of a sphere is given by:
\[ g_{\text{surface}} = \frac{GM}{R^2} \]
- The mass \( M \) of the Earth can be expressed as the sum of the mass of the core and the mass of the shell:
\[ M = M_{\text{core}} + M_{\text{shell}} = \left(\frac{4}{3} \pi \left(\frac{R}{2}\right)^3 \rho_1\right) + \left(\frac{4}{3} \pi \left(R^3 - \left(\frac{R}{2}\right)^3\right) \rho_2\right) \]

3. **Calculating Masses**:
- The mass of the core:
\[ M_{\text{core}} = \frac{4}{3} \pi \left(\frac{R}{2}\right)^3 \rho_1 = \frac{4}{3} \pi \frac{R^3}{8} \rho_1 = \frac{\pi R^3 \rho_1}{6} \]
- The mass of the shell:
\[ M_{\text{shell}} = \frac{4}{3} \pi \left(R^3 - \left(\frac{R}{2}\right)^3\right) \rho_2 = \frac{4}{3} \pi \left(R^3 - \frac{R^3}{8}\right) \rho_2 = \frac{4}{3} \pi \left(\frac{7R^3}{8}\right) \rho_2 = \frac{7\pi R^3 \rho_2}{6} \]

4. **Total Mass**:
- Therefore, the total mass \( M \) becomes:
\[ M = \frac{\pi R^3 \rho_1}{6} + \frac{7\pi R^3 \rho_2}{6} = \frac{\pi R^3}{6} (\rho_1 + 7\rho_2) \]

5. **Gravitational Acceleration at Depth \( \frac{R}{2} \)**:
- The gravitational acceleration at a depth \( d = \frac{R}{2} \) can be calculated using:
\[ g_{\text{depth}} = \frac{GM_{\text{core}}}{(R - d)^2} = \frac{G \cdot M_{\text{core}}}{\left(\frac{R}{2}\right)^2} \]
- Since \( M_{\text{core}} = \frac{\pi R^3 \rho_1}{6} \):
\[ g_{\text{depth}} = \frac{G \cdot \frac{\pi R^3 \rho_1}{6}}{\left(\frac{R}{2}\right)^2} = \frac{G \cdot \frac{\pi R^3 \rho_1}{6}}{\frac{R^2}{4}} = \frac{4G \pi R \rho_1}{6} \]

6. **Setting Accelerations Equal**:
- Since \( g_{\text{surface}} = g_{\text{depth}} \):
\[ \frac{GM}{R^2} = \frac{4G \pi R \rho_1}{6} \]
- Substituting \( M \):
\[ \frac{G \cdot \frac{\pi R^3}{6} (\rho_1 + 7\rho_2)}{R^2} = \frac{4G \pi R \rho_1}{6} \]
- Canceling \( G \) and \( \pi \):
\[ \frac{R (\rho_1 + 7\rho_2)}{6} = \frac{4R \rho_1}{6} \]
- Simplifying gives:
\[ \rho_1 + 7\rho_2 = 4\rho_1 \]
- Rearranging leads to:
\[ 3\rho_1 = 7\rho_2 \quad \Rightarrow \quad \frac{\rho_1}{\rho_2} = \frac{7}{3} \]

### Final Answer:
The ratio of the densities is:
\[ \frac{\rho_1}{\rho_2} = \frac{7}{3} \]