A Solid sphere of density `rho=rho_0(1-r^2/R^2), 0 ltrleR` just floats in a liquid then density of liquid is– (r is distance from centre of sphere)

Options

A. `2/5 rho_0`
B. `5/2 rho_0`
C. `3/5 rho_0`
D. `rho_0s`

Explanation

To solve the problem, we need to find the density of the liquid in which a solid sphere with a varying density floats. The density of the sphere is given by:

\[ \rho = \rho_0 \left(1 - \frac{r^2}{R^2}\right) \]

where \( r \) is the distance from the center of the sphere, \( R \) is the radius of the sphere, and \( \rho_0 \) is a constant density.

### Step-by-Step Solution:

1. **Volume of a Shell**:
We consider a thin spherical shell of radius \( r \) and thickness \( dr \). The volume \( dV \) of this shell is given by:
\[ dV = 4\pi r^2 dr \]

2. **Mass of the Shell**:
The mass \( dm \) of this shell can be calculated by multiplying its volume by the density of the sphere at that radius:
\[ dm = \rho \cdot dV = \left(\rho_0 \left(1 - \frac{r^2}{R^2}\right)\right) \cdot (4\pi r^2 dr) \]
Simplifying this gives:
\[ dm = 4\pi \rho_0 \left(r^2 - \frac{r^4}{R^2}\right) dr \]

3. **Total Mass of the Sphere**:
To find the total mass \( M \) of the sphere, we integrate \( dm \) from \( r = 0 \) to \( r = R \):
\[ M = \int_0^R dm = \int_0^R 4\pi \rho_0 \left(r^2 - \frac{r^4}{R^2}\right) dr \]
This can be split into two integrals:
\[ M = 4\pi \rho_0 \left(\int_0^R r^2 dr - \frac{1}{R^2} \int_0^R r^4 dr\right) \]

4. **Calculating the Integrals**:
- The integral \( \int_0^R r^2 dr = \frac{R^3}{3} \)
- The integral \( \int_0^R r^4 dr = \frac{R^5}{5} \)

Substituting these results back into the equation for \( M \):
\[ M = 4\pi \rho_0 \left(\frac{R^3}{3} - \frac{1}{R^2} \cdot \frac{R^5}{5}\right) \]
Simplifying this gives:
\[ M = 4\pi \rho_0 \left(\frac{R^3}{3} - \frac{R^3}{5}\right) = 4\pi \rho_0 R^3 \left(\frac{5 - 3}{15}\right) = \frac{8\pi \rho_0 R^3}{15} \]

5. **Density of the Liquid**:
For the sphere to float, the weight of the sphere must equal the buoyant force. The buoyant force is equal to the weight of the liquid displaced, which can be expressed as:
\[ \text{Weight of sphere} = \text{Density of liquid} \cdot \text{Volume of sphere} \cdot g \]
Setting the mass of the sphere equal to the buoyant force:
\[ M g = \sigma \cdot \left(\frac{4\pi R^3}{3}\right) g \]
Here, \( \sigma \) is the density of the liquid. Canceling \( g \) from both sides gives:
\[ M = \sigma \cdot \left(\frac{4\pi R^3}{3}\right) \]
Substituting \( M \):
\[ \frac{8\pi \rho_0 R^3}{15} = \sigma \cdot \left(\frac{4\pi R^3}{3}\right) \]

6. **Solving for Liquid Density**:
Dividing both sides by \( \frac{4\pi R^3}{3} \):
\[ \sigma = \frac{8\pi \rho_0 R^3}{15} \cdot \frac{3}{4\pi R^3} = \frac{2\rho_0}{5} \]

### Final Answer:
The density of the liquid is:
\[ \sigma = \frac{2\rho_0}{5} \]