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JEE Main 20208 JanuaryMorning Shift
Exam
JEE Main
Year
2020
Date
8 January
Shift
Morning Shift
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Main 2020 · 8 January · Morning Shift · Physics · Mechanical Properties of Fluids

Consider a solid sphere of radius R and mass density ρ(r) = ρ _ 0...

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

Options

  1. Aρ03\dfrac{{ρ}_{0}}{3}
  2. Bρ05\dfrac{{ρ}_{0}}{5}
  3. C2ρ05\dfrac{2{ρ}_{0}}{5}
  4. D2ρ03\dfrac{2{ρ}_{0}}{3}

Correct answer

C. 2ρ05\dfrac{2{ρ}_{0}}{5}

Step-by-step solution

The density of any point inside the sphere is

ρ=ρ0(1−r2R2)0<r≤Rρ = {ρ}_{0}\left(1 - \dfrac{{r}^{2}}{{R}^{2}}\right)0 < r \le R

weight = buoyant force
mg = B

mg=ρLVgmg = {ρ}_{L}Vg
∫ρ(4πr2dr)g=ρL43πR3g∫ ρ\left(4\pi{r}^{2}dr\right)g = {ρ}_{L}\dfrac{4}{3}\pi{R}^{3}g
∫0Rρ0(1−r2R2)4πr2dr=ρL43πR3\mathop{∫}_{0}^{R}{ρ}_{0}\left(1 - \dfrac{{r}^{2}}{{R}^{2}}\right)4\pi{r}^{2}dr = {ρ}_{L}\dfrac{4}{3}\pi{R}^{3}
∫0Rρ04π(r2−r4R2)dr=ρ04π(r33−r55R2)0R=ρL43πR3\mathop{∫}_{0}^{R}{ρ}_{0}4\pi\left({r}^{2} - \dfrac{{r}^{4}}{{R}^{2}}\right)dr = {ρ}_{0}4\pi{\left(\dfrac{{r}^{3}}{3} - \dfrac{{r}^{5}}{5{R}^{2}}\right)}_{0}^{R} = {ρ}_{L}\dfrac{4}{3}\pi{R}^{3}
25ρ0=ρL\dfrac{2}{5}{ρ}_{0} = {ρ}_{L}

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