Consider a solid sphere of radius R and mass density ρ r = ρ 0 1 - r 2 R 2 , 0 < r ≤ R . The minimum density of a liquid in which it will float is:

The density of any point inside the sphere is ρ = ρ 0 1 - r 2 R 2 0 < r ≤ R weight = buoyant force mg = B m g = ρ L V g ∫ ρ 4 π r 2 d r g = ρ L 4 3 π R 3 g ∫ 0 R ρ 0 1 - r 2 R 2 4 π r 2 d r = ρ L 4 3 π R 3 ∫ 0 R ρ 0 4 π r 2 - r 4 R 2 d r = ρ 0 4 π r 3 3 - r 5 5 R 2 0 R = ρ L 4 3 π R 3 2 5 ρ 0 = ρ L

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PhysicsFluidBernoulli's theorem and applicationsHard2 minPYQ_2020

PhysicsHardsingle choice

Consider a solid sphere of radius $R$ and mass density $ρ (r) = ρ_{0} (1 - \frac{r^{2}}{R^{2}}), 0 < r \leq R .$ The minimum density of a liquid in which it will float is:

Options:

Answer:

Solution:

The density of any point inside the sphere is

$ρ = ρ_{0} (1 - \frac{r^{2}}{R^{2}}) 0 < r \leq R$

weight = buoyant force
mg = B

$m g = ρ_{L} V g$
$\int ρ (4 π r^{2} d r) g = ρ_{L} \frac{4}{3} π R^{3} g$
$\int_{0}^{R} ρ_{0} (1 - \frac{r^{2}}{R^{2}}) 4 π r^{2} d r = ρ_{L} \frac{4}{3} π R^{3}$
$\int_{0}^{R} ρ_{0} 4 π (r^{2} - \frac{r^{4}}{R^{2}}) d r = ρ_{0} 4 π {(\frac{r^{3}}{3} - \frac{r^{5}}{5 R^{2}})}_{0}^{R} = ρ_{L} \frac{4}{3} π R^{3}$
$\frac{2}{5} ρ_{0} = ρ_{L}$

Stream:JEESubject:PhysicsTopic:FluidSubtopic:Bernoulli's theorem and applications

⏱ 2mℹ️ Source: PYQ_2020

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