Consider a solid sphere of radius R and mass density ρ(r) = ρ₀ (1 - (r²)/(R²)), 0 < r ≤ R The minimum density of a liquid in which it will float is: (a) (2ρ₀)/(3) (b) (2ρ₀)/(5) (c) (ρ₀)/(5) (d) (ρ₀)/(3)

Answer: (b). It floats if the liquid density is at least the average density of the sphere. M = ∫₀R ρ₀(1 - (r²)/(R²)) 4π r² dr = 4πρ₀(R³/3 - R³/5) = (8πρ₀ R³)/(15) ρavg = (M)/(4/3π R³) = (8ρ₀)/(15) × 3/4 = (2ρ₀)/(5) So the answer is (b) (2ρ₀)/(5).

Consider a solid sphere of radius R and mass density ρ(r) = ρ₀ (1 -… | JEE Main 2020 (8 Jan, Shift 1)

Q14JEE Main 2020 · 8 JAN · SHIFT 1Easy

Archimedes' Principle & Floatation

Consider a solid sphere of radius R and mass density

$\rho(r) = \rho_0 \left(1 - \frac{r^2}{R^2}\right),\ 0 < r \leq R$

The minimum density of a liquid in which it will float is:

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Answer: (b)

It floats if the liquid density is at least the average density of the sphere.
$M = \int_0^R \rho_0\left(1 - \dfrac{r^2}{R^2}\right) 4\pi r^2\,dr = 4\pi\rho_0\left(\dfrac{R^3}{3} - \dfrac{R^3}{5}\right) = \dfrac{8\pi\rho_0 R^3}{15}$
$\rho_{avg} = \dfrac{M}{\frac{4}{3}\pi R^3} = \dfrac{8\rho_0}{15} \times \dfrac{3}{4} = \dfrac{2\rho_0}{5}$
So the answer is (b) $\dfrac{2\rho_0}{5}$ .

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Archimedes' Principle & Floatation question from JEE Main 2020 (8 Jan, Shift 1)

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