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

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Fluid Mechanics

Consider a solid sphere of radius

R

and mass density

p (r) = p_{0} (1 - \frac{r ^{2}}{R ^{2}}), 0 < r \leq R

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

Answer: (B)

For the sphere to float, its weight equals the buoyant force. Its mass comes from integrating the given density,

m = \int_{0}^{R} ρ_{0} (1 - \frac{r ^{2}}{R ^{2}}) 4 π r^{2} d r = 4 π ρ_{0} [\frac{r ^{3}}{3} - \frac{r ^{5}}{5 R ^{2}}]_{0}^{R} = 4 π ρ_{0} R^{3} (\frac{1}{3} - \frac{1}{5}) = \frac{8 π ρ _{0} R ^{3}}{15}

. The buoyant force on the fully submerged sphere is

ρ_{L} \cdot \frac{4 π R ^{3}}{3} g

. Equating the two,

\frac{8 π ρ _{0} R ^{3}}{15} = ρ_{L} \frac{4 π R ^{3}}{3}

, which gives the liquid density

ρ_{L} = \frac{2 ρ _{0}}{5}

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Single correct option