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JEE Main 20179 AprilOnline
Exam
JEE Main
Year
2017
Date
9 April
Mode
Online
Subject
Physics
Chapter
Gravitation

JEE Main 2017 · 9 April · Online · Physics · Gravitation

The mass density of a spherical body is given by ρ(r) = (k)/(r) for...

The mass density of a spherical body is given by ρ(r)=krρ\left(r\right) = \dfrac{k}{r} for r≤Rr \le R and ρ(r)=0ρ\left(r\right) = 0 for r>R,r > R, where rr is the distance from the center. The correct graph that describes qualitatively the acceleration, aa of a test particle as a function of rr is:The mass density of a spherical body is given by ρ(r) = (k)/(r) for r ≤ R and ρ(r) = 0 for r > R, where r is theThe mass density of a spherical body is given by ρ(r) = (k)/(r) for r ≤ R and ρ(r) = 0 for r > R, where r is theThe mass density of a spherical body is given by ρ(r) = (k)/(r) for r ≤ R and ρ(r) = 0 for r > R, where r is theThe mass density of a spherical body is given by ρ(r) = (k)/(r) for r ≤ R and ρ(r) = 0 for r > R, where r is theThe mass density of a spherical body is given by ρ(r) = (k)/(r) for r ≤ R and ρ(r) = 0 for r > R, where r is the

Options

  1. AWhat is the mass of Jupiter compared to the Sun, given that Io, one of Jupiter's satellites, has an orbital period of
  2. BWhat is the mass of Jupiter compared to the Sun, given that Io, one of Jupiter's satellites, has an orbital period of
  3. CWhat is the mass of Jupiter compared to the Sun, given that Io, one of Jupiter's satellites, has an orbital period of
  4. DWhat is the mass of Jupiter compared to the Sun, given that Io, one of Jupiter's satellites, has an orbital period of

Correct answer

A. What is the mass of Jupiter compared to the Sun, given that Io, one of Jupiter's satellites, has an orbital period of

Step-by-step solution

The mass density of a spherical body is given by ρ(r) = (k)/(r) for r ≤ R and ρ(r) = 0 for r > R, where r is the

For Inside point (r≤R)\left(r \le R\right)

g=E=Gr2∫0r(4πr2dr)(Kr)=GKr2×4πr22=2\piGK=constantg = E = \dfrac{G}{{r}^{2}}\mathop{∫}_{0}^{r}\left(4\pi{r}^{2}dr\right)\left(\dfrac{K}{r}\right) = \dfrac{GK}{{r}^{2}} \times \dfrac{4\pi{r}^{2}}{2} = 2\piGK = constant

For outside point \left(r > R\right)<br/><br/><br /><br />E = gE = \dfrac{GM}{{r}^{2}}E = \dfrac{G}{{r}^{2}}\mathop{∫}_{0}^{R}( 4\pi{r}^{2}d{r}^{'} )( \dfrac{K}{{r}^{'}} )E = \dfrac{4\piKG}{{r}^{2}}{[ \dfrac{{{r}^{'}}^{2}}{2} ]}_{0}^{R}<br/><br/><br /><br />E = \dfrac{2\piKG}{{r}^{2}}{R}^{2}<br/><br/><br /><br />E ∝ \dfrac{1}{{r}^{2}}.
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