IIT JEE Advanced 2008 Paper 1

We have
ρ = ρ0 for r ≤ R
= 0 for r > R
For r ≤ R ; We have
Eg =

−GMr

R3

Therefore, the gravitational force experienced by a mass m is
Fg = mEg =

−GMmr

R3

The negative sign signifies an attractive force. Therefore, for circular motion, we have

GMmr

R3

=

mv2

r

where M =

4

3

πR3ρ0
⇒

4

3

πGmrρ0 =

mv2

r

⇒ V = √(

4

3

)πGρ0r2 (1)
⇒ V α r
For r > R : We have
Eg =

−GM

r2

Therefore,
Fg = mEg =

GMm

r2

=

mv2

r

GM

r2

(

4

3

πR3ρ0) =

mv2

r

From Eq. (1), we get
V = √

( 4 3 )πGρ0R3

r

⇒ V α

1

√r

Therefore, the correct plot is as shown in option (C):