A non-homogeneous sphere of radius R has the following density variation :
`rho{{:(rho_(0),rleR//3),(rho_(0)//2,(R//3) lt r le (3R//4)),(rho_(0)//8,(3R//4) lt r le R):}`
The gravitational field at a distance 2R from the centre of the sphere is

Options

A. `0.1piGRrho_(0)`
B. `0.2piGRrho_(0)`
C. `0.3piGRrho_(0)`
D. `0.4piGRrho_(0)`

Explanation

To solve the problem of finding the gravitational field at a distance of \(2R\) from the center of a non-homogeneous sphere with a given density variation, we can follow these steps:

### Step 1: Understand the Density Variation
The sphere has three different density regions:
1. For \(0 \leq r \leq \frac{R}{3}\), the density \(\rho = \rho_0\).
2. For \(\frac{R}{3} < r \leq \frac{3R}{4}\), the density \(\rho = \frac{\rho_0}{2}\).
3. For \(\frac{3R}{4} < r \leq R\), the density \(\rho = \frac{\rho_0}{8}\).

### Step 2: Calculate the Mass of Each Region
The mass of each region can be calculated using the formula \(m = \rho \cdot V\), where \(V\) is the volume of the sphere segment.

#### Mass \(m_1\) for \(0 \leq r \leq \frac{R}{3}\):
\[ V_1 = \frac{4}{3} \pi \left(\frac{R}{3}\right)^3 = \frac{4}{3} \pi \frac{R^3}{27} = \frac{4\pi R^3}{81} \]
\[ m_1 = \rho_0 \cdot V_1 = \rho_0 \cdot \frac{4\pi R^3}{81} = \frac{4\pi R^3 \rho_0}{81} \]

#### Mass \(m_2\) for \(\frac{R}{3} < r \leq \frac{3R}{4}\):
\[ V_2 = \frac{4}{3} \pi \left(\left(\frac{3R}{4}\right)^3 - \left(\frac{R}{3}\right)^3\right) \]
Calculating \(V_2\):
\[ V_2 = \frac{4}{3} \pi \left(\frac{27R^3}{64} - \frac{R^3}{27}\right) \]
Finding a common denominator (which is 1728):
\[ = \frac{4}{3} \pi \left(\frac{27 \cdot 27R^3 - 64R^3}{1728}\right) = \frac{4}{3} \pi \left(\frac{729R^3 - 64R^3}{1728}\right) = \frac{4}{3} \pi \left(\frac{665R^3}{1728}\right) \]
\[ = \frac{4 \cdot 665 \pi R^3}{5184} \]
\[ m_2 = \frac{\rho_0}{2} \cdot V_2 = \frac{\rho_0}{2} \cdot \frac{4 \cdot 665 \pi R^3}{5184} = \frac{1330 \pi R^3 \rho_0}{5184} \]

#### Mass \(m_3\) for \(\frac{3R}{4} < r \leq R\):
\[ V_3 = \frac{4}{3} \pi R^3 - V_1 - V_2 \]
Calculating \(V_3\):
\[ = \frac{4}{3} \pi R^3 - \left(\frac{4\pi R^3}{81} + \frac{4 \cdot 665 \pi R^3}{5184}\right) \]
Finding a common denominator and simplifying will yield \(m_3\).

### Step 3: Total Mass of the Sphere
The total mass \(M\) of the sphere is:
\[ M = m_1 + m_2 + m_3 \]

### Step 4: Calculate the Gravitational Field at \(2R\)
Using the formula for gravitational field \(g = \frac{GM}{r^2}\):
\[ g = \frac{G \cdot M}{(2R)^2} = \frac{G \cdot M}{4R^2} \]

### Step 5: Substitute the Total Mass
Substituting \(M\) into the gravitational field equation gives:
\[ g = \frac{G \cdot \left(m_1 + m_2 + m_3\right)}{4R^2} \]

### Step 6: Final Calculation
After calculating \(M\) and substituting it into the gravitational field equation, we can simplify to find the final expression for \(g\).