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Step-by-Step Solution
Step 1: Write down the given density function
The density of the solid sphere varies with the radial distance r as:
$\rho(r) = \rho_0 \Bigl(1 - \frac{r^2}{R^2}\Bigr), \quad 0 \le r \le R$
Step 2: Express the total mass of the sphere
The mass of the sphere, $M$, is found by integrating the density over its volume. In spherical coordinates, a thin spherical shell at radius r has volume element $dV = 4\pi r^2\,dr$. Hence,
$M = \int_{\text{Volume}} \rho(r)\,dV
= \int_{0}^{R} \rho_0 \Bigl(1 - \frac{r^2}{R^2}\Bigr) \, 4\pi r^2 \,dr.$
Step 3: Perform the integration
Take $\rho_0$ and $4\pi$ outside the integral:
$M = 4\pi \rho_0 \int_{0}^{R} \Bigl(r^2 - \frac{r^4}{R^2}\Bigr) dr
= 4\pi \rho_0 \Bigl[\int_{0}^{R} r^2\,dr - \frac{1}{R^2}\int_{0}^{R} r^4\,dr \Bigr].$
Now compute each integral separately:
$\displaystyle \int_{0}^{R} r^2\,dr = \frac{R^3}{3},$
$\displaystyle \int_{0}^{R} r^4\,dr = \frac{R^5}{5}.$
Substitute these back into the expression for M:
$M = 4\pi \rho_0 \Bigl[\frac{R^3}{3} - \frac{1}{R^2} \cdot \frac{R^5}{5}\Bigr]
= 4\pi \rho_0 \Bigl[\frac{R^3}{3} - \frac{R^3}{5}\Bigr]
= 4\pi \rho_0 R^3 \Bigl(\frac{1}{3} - \frac{1}{5}\Bigr)
= 4\pi \rho_0 R^3 \cdot \frac{2}{15}
= \frac{8\pi \rho_0 R^3}{15}.$
Step 4: Calculate the average density of the sphere
The volume of the sphere is $V = \frac{4}{3}\pi R^3.$ Hence, its average density is:
$\rho_{\text{avg}} = \frac{M}{V}
= \frac{ \frac{8\pi \rho_0 R^3}{15} }{ \frac{4}{3}\pi R^3 }
= \frac{8}{15} \cdot \frac{3}{4} \rho_0
= \frac{2}{5}\rho_0.$
Step 5: Floating condition and final answer
A solid will float in a fluid if its average density is less than or equal to the fluid’s density. Therefore, the minimum density of the liquid in which the sphere will just float is:
$\boxed{\frac{2\rho_0}{5}}.$
Reference Image (Provided)
