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BITSAT 2018
Exam
BITSAT
Year
2018
Subject
Physics
Chapter
Gravitation

BITSAT 2018 · Physics · Gravitation

A spherically symmetric gravitational systom of particles has mass...

A spherically symmetric gravitational systom of particles has mass density ρ = \left\{ρ0forr≤R0forr>R\begin{array}{ll}{ρ}_{0}\text{for}r \le R \\ 0\text{for}r > R\end{array}\right) where,{ρ}_{0}isaconstant.Atestmasscanundergocircularmotionundertheinfluenceofthegravitationalfieldofparticles.Itsspeedis a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speedvasafunctionofdistanceas a function of distancer ( 0 < r < \infty )$ from the centre of the system is represented by

A spherically symmetric gravitational systom of particles has mass density ρ = \ array ll ρ _ 0 forr ≤ R 0forr > RA spherically symmetric gravitational systom of particles has mass density ρ = \ array ll ρ _ 0 forr ≤ R 0forr > RA spherically symmetric gravitational systom of particles has mass density ρ = \ array ll ρ _ 0 forr ≤ R 0forr > RA spherically symmetric gravitational systom of particles has mass density ρ = \ array ll ρ _ 0 forr ≤ R 0forr > RA spherically symmetric gravitational systom of particles has mass density ρ = \ array ll ρ _ 0 forr ≤ R 0forr > R

Options

  1. AIf y=(x^ x )^ x then which one of the following is correct?
  2. BIf y=(x^ x )^ x then which one of the following is correct?
  3. CIf y=(x^ x )^ x then which one of the following is correct?
  4. DIf y=(x^ x )^ x then which one of the following is correct?

Correct answer

C. If y=(x^ x )^ x then which one of the following is correct?

Step-by-step solution

When r≤Rr \le R, then force on the test mass nn at the surface of the sphere =mg= mg

Force on the test mass at distance rr from the centre of sphere is

A spherically symmetric gravitational systom of particles has mass density ρ = \ array ll ρ _ 0 forr ≤ R 0forr > R

If r<Rr < R, then F=GMmR2⋅rR=GMmR3⋅rF = \dfrac{GMm}{R2} \cdot \dfrac{r}{R} = \dfrac{GMm}{R3} \cdot r

∴mv2r=GMmR3⋅r∴ \dfrac{mv^{2}}{r} = \dfrac{GMm}{R3} \cdot r

∴v\alphar∴ v\alphar

If r>Rr > R then F=GMmr2F = \dfrac{GMm}{r2}

∴Mv2r=GMmr2∴ \dfrac{Mv^{2}}{r} = \dfrac{GMm}{r2}

∴vα1r∴ v\alpha\dfrac{1}{r}

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