Question:
Consider a solid sphere of radius R and mass density
\(\rho(r) = \rho_0\left(1 - \frac{r^2}{R^2}\right)\), \(0 \lt r \le R\)
The minimum density of a liquid in which it will float is :
Consider a solid sphere of radius R and mass density
\(\rho(r) = \rho_0\left(1 - \frac{r^2}{R^2}\right)\), \(0 \lt r \le R\)
The minimum density of a liquid in which it will float is :
\(\rho(r) = \rho_0\left(1 - \frac{r^2}{R^2}\right)\), \(0 \lt r \le R\)
The minimum density of a liquid in which it will float is :
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The sphere floats if the liquid density is at least the sphere's average density; integrate the density over shells to get the mass.
Updated On: Nov 5, 2026
- \(\frac{2\rho_0}{3}\)
- \(\frac{2\rho_0}{5}\)
- \(\frac{\rho_0}{5}\)
- \(\frac{\rho_0}{3}\)
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The Correct Option is B
Solution and Explanation
Mass \(M = \int_0^R \rho_0\left(1 - \frac{r^2}{R^2}\right)4\pi r^2\,dr = 4\pi\rho_0\left(\frac{R^3}{3} - \frac{R^3}{5}\right) = \frac{8\pi\rho_0R^3}{15}\).
Volume \(V = \frac{4}{3}\pi R^3\).
Average density \(\bar\rho = \frac{M}{V} = \frac{8\pi\rho_0R^3/15}{4\pi R^3/3} = \frac{2\rho_0}{5}\).
The sphere floats if the liquid density \(\ge \bar\rho\), so the minimum density is \(\frac{2\rho_0}{5}\).
Answer: (2) \(\frac{2\rho_0}{5}\)
Volume \(V = \frac{4}{3}\pi R^3\).
Average density \(\bar\rho = \frac{M}{V} = \frac{8\pi\rho_0R^3/15}{4\pi R^3/3} = \frac{2\rho_0}{5}\).
The sphere floats if the liquid density \(\ge \bar\rho\), so the minimum density is \(\frac{2\rho_0}{5}\).
Answer: (2) \(\frac{2\rho_0}{5}\)
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