Fluid Mechanics

JEE Main level, single correct, medium

15 of 192

Question 15 of 192

Consider a solid sphere of radius RR and mass density ρ(r)=ρ0(1−r2R2),0<r≤R.\rho \left(r\right)={\rho }_{0}\left(1-\dfrac{{r}^{2}}{{R}^{2}}\right),0\lt r\leq R. The minimum density of a liquid in which it will float is:

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Answer: (C) 2ρ05\dfrac{2{\rho }_{0}}{5}

The density of any point inside the sphere is
ρ=ρ0(1−r2R2)0<r≤R\rho ={\rho }_{0}\left(1-\dfrac{{r}^{2}}{{R}^{2}}\right)0\lt r\leq R
weight = buoyant force
mg = B
mg=ρLVgmg={\rho }_{L}Vg
∫ρ(4πr2dr)g=ρL43πR3g\int \rho \left(4\pi {r}^{2}dr\right)g={\rho }_{L}\dfrac{4}{3}\pi {R}^{3}g
∫0Rρ0(1−r2R2)4πr2dr=ρL43πR3\int _{0}^{R}{\rho }_{0}\left(1-\dfrac{{r}^{2}}{{R}^{2}}\right)4\pi {r}^{2}dr={\rho }_{L}\dfrac{4}{3}\pi {R}^{3}
∫0Rρ04π(r2−r4R2)dr=ρ04π(r33−r55R2)0R=ρL43πR3\int _{0}^{R}{\rho }_{0}4\pi \left({r}^{2}-\dfrac{{r}^{4}}{{R}^{2}}\right)dr={\rho }_{0}4\pi {\left(\dfrac{{r}^{3}}{3}-\dfrac{{r}^{5}}{5{R}^{2}}\right)}_{0}^{R}={\rho }_{L}\dfrac{4}{3}\pi {R}^{3}
25ρ0=ρL\dfrac{2}{5}{\rho }_{0}={\rho }_{L}

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