33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

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72 NEET-AIPMT Chapterwise Topicwise Solutions Physics
Mm 4
FR
G
F
e = = ×
(R e+R
4
2
e / 2)
9 = ×
9 72 23. (b)
= 32 N
15. (a) : Force between the two masses, F = −G mm 24. (a) : Acceleration due to gravity
4R
2 g = G
M 3
43π
R × ρ 4
× = G
= G × πR
× ρ
R 2 2
R
3
2
This force will provide the necessary centripetal force for
GMm mv 2 GM
∴ = ; v =
…(i) the masses to go around a circle, then
2
r r r
2
Time period of the planet is given by
Gmm mv 2 Gm 1 Gm
= ⇒ v = ⇒ v =
2
2 2 2 2
4R
R 4R
2 R
r
r r
T 2π 2
T
v
4π v
4π
,
2 ⎛ GM
(Using (i)) 16. (a) : Given : mass (m) = 6 × 10 24 kg;
2 3
r ⎠
⎟
2 4 r
T = π angular velocity (w) = 2 × 10 –7 rad/s and
…(ii) radius (r) = 1.5 × 10 8 km = 1.5 × 10 11 m
GM
According to question, T 2 = Kr 3
Force exerted on the earth = mRw 2
…(iii)
= (6 × 10 24 ) × (1.5 × 10 11 ) × (2 × 10 –7 ) 2
Comparing equations (ii) and (iii), we get
= 36 × 10 21 N
2
4π
K = , ∴ GMK = 4π
2
17. (b) : Centripetal force (F) = mv 2
and the gravitational
GM R
force (F) = GMm GMm
= (where R 2 → R). Since
2
10. (a) : For a point inside the earth i.e. r < R
R R
2
mv GMm
GM
E 3
R
r
= therefore v= GM . Thus velocity v is
R R
independent of R.
where M and R be mass and radius of the earth
respectively.
At the centre, r = 0
18. (d) : If universal gravitational constant becomes ten
times, then G′ = 10 G.
So, acceleration due to gravity increases. i.e, (d) is the
\ E = 0
wrong option.
For a point outside the earth i.e. r > R,
19. (a) : Gravitational force acting on particle of mass m is
GM
GM
pm
E =− F =
r 2
( Dp
/ 2)
2
On the surface of the earth i.e. r = R,
Acceleration due to gravity experienced by the particle is
F GMp
4GMp
GM
g = = =
E =−
R 2
m
2 2
( Dp
/ 2)
Dp
The variation of E with distance r from the centre is as
GM G ⎛ 4 3 ⎞
shown in the figure.
20. (d) : g = = × r
Gr
r r ⎝
⎜
⎠
⎟ = 4
π ρ × πρ
2 2 3 3
11. (b) : Gravitational field due to the thin spherical shell g ′ 3R
=
Inside the shell, (for r < R) F = 0
g R ⇒ g ′ = 3 g.
On the surface of the shell, (for r = R)
21. (d) : From equation of acceleration due to gravity.
GM
3
F =
GMe
G( 4/ 3)πRe
2
ge
= =
ρ
R
2
2 e
Re
Re
Outside the shell, (for r > R)
ge ∝Reρe
GM
F =
r 2
Acceleration due to gravity of planet gp ∝Rpρp
The variation of F with distance r from the centre is as R e
r e
= R p
r p
⇒ R e
r e
= R p
2r e
shown in the figure.
⇒ R p
= 1
12. (b) : The gravitational force does not depend upon
2 R (Q R e = R)
the medium in which objects are placed.
22. (b) : Initial velocity of the mass on both the planets
is same.
13. (b)
ie .., 2g′ h′ = 2gh
14. (b) : F G Mm
surface = 2
2× g ′ × h′ = 2× 9g ′ × 2 ⇒ 2h′ = 36
R e ⇒ h′ = 18 m