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DPP - Daily Practice Problems
Chapter-wise Sheets
Date :
Start Time :
End Time :
PHYSICS
CP07
SYLLABUS : Gravitation
Max. Marks : 180
Marking Scheme : (+4) for correct & (–1) for incorrect answer
Time : 60 min.
INSTRUCTIONS : This Daily Practice Problem Sheet contains 45 MCQs. For each question only one option is correct.
Darken the correct circle/ bubble in the Response Grid provided on each page.
1.
2.
3.
The radius of a planet is 1/4th of Re and its acc. due to
gravity is 2g. What would be the value of escape velocity
on the planet, if escape velocity on earth is ve.
ve
ve
(a)
(b) v e 2
(c)
2 ve (d)
2
2
A projectile is fired vertically from the Earth with a velocity
kve where ve is the escape velocity and k is a constant less
than unity. The maximum height to which projectile rises, as
measured from the centre of Earth, is
R
R
R
R
(d)
(a)
(b)
(c)
2
k
k -1
1- k
1+ k2
A solid sphere of uniform
density and radius R applies a
A
gravitational force of attraction
equal to F1 on a particle placed
at A, distance 2R from the centre
of the sphere.
R
R
RESPONSE GRID
1.
2.
4.
A spherical cavity of radius R/2 is now made in the sphere
as shown in the figure. The sphere with cavity now applies
a gravitational force F2 on the same particle placed at A.
The ratio F2/F1 will be
(a) 1/2
(b) 3
(c) 7
(d) 1/9
A geostationary satellite is orbiting the earth at a height of
5R above that surface of the earth, R being the radius of the
earth. The time period of another satellite in hours at a height
of 2R from the surface of the earth is :
(a) 5
5.
(b) 10
(c)
6 2 (d)
6
2
A satellite of mass m is orbiting around the earth in a circular
orbit with a velocity v. What will be its total energy?
(a) (3/4) mv2
(b) (1/2) mv2
2
(c) mv
(d) – (1/2)m v2
3.
Space for Rough Work
4.
5.
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6.
The gravitational force of attraction between a uniform
sphere of mass M and a uniform rod of length l and mass m
oriented as shown is
m
M
r
7.
8.
GMm
r(r + l )
l
13.
GM
(c) Mmr2 + l (d) (r2 + l) mM
r2
If the gravitational force between two objects were
proportional to 1/R (and not as 1/R2) where R is separation
between them, then a particle in circular orbit under such a
force would have its orbital speed v proportional to
(a) 1/R2 (b) R0
(c) R1
(d) 1/R
A satellite of mass m revolves around the earth of radius R
at a height ‘x’ from its surface. If g is the acceleration due to
gravity on the surface of the earth, the orbital speed of the
satellite is
(a)
(b)
12.
æ gR 2 ö 1/ 2
gR
gR 2
÷
(b)
(c) gx (d) çç
(a)
÷
R
x
+
R-x
R+x
ø
è
9.
A body is projected up with a velocity equal to 3/4th of the
escape velocity from the surface of the earth. The height it
reaches from the centre of the earth is (Radius of the earth = R)
10R
16R
9R
10R
(a)
(b)
(c)
(d)
9
7
8
3
10. A Planet is revolving around the sun.
B
A
C
S
14.
15.
16.
1
changes and becomes an inverse cube law i.e. F µ
, but
r3
still remaining a central force. Then
(a) Kepler’s law of area still holds
(b) Kepler’s law of period still holds
(c) Kepler’s law of area and period still holds
(d) neither the law of area nor the law of period still holds
17. Four equal masses (each of mass M) are placed at the corners
of a square of side a. The escape velocity of a body from the
centre O of the square is
(a)
D
11.
Which of the following is correct option?
(a) The time taken in travelling DAB is less than that for
BCD
(b) The time taken in travelling DAB is greater than that
for BCD
(c) The time taken in travelling CDA is less than that for
ABC
(d) The time taken in travelling CDA is greater than that
for ABC
The acceleration due to gravity on the planet A is 9 times
the acceleration due to gravity on planet B. A man jumps to
a height of 2m on the surface of A. What is the height of
jump by the same person on the planet B?
2
2
m
m
(a)
(b)
(c) 18 m
(d) 6 m
3
9
RESPONSE
GRID
6.
11.
16.
7.
12.
17.
If suddenly the gravitational force of attraction between the
earth and a satellite revolving around it becomes zero, then
the satellite will
(a) continue to move in its orbit with same speed
(b) move tangentially to the original orbit with same speed
(c) become stationary in its orbit
(d) move towards the earth
Mass M is divided into two parts xM and (1 – x )M. For a
given separation, the value of x for which the gravitational
attraction between the two pieces becomes maximum is
3
1
(a)
(b)
(c) 1
(d) 2
5
2
The potential energy of a satellite, having mass m and
rotating at a height of 6.4 × 106 m from the earth surface, is
(a) – mgRe
(b) – 0.67 mgRe
(c) – 0.5 mgRe
(d) – 0.33 mgRe
If the radius of the earth were to shrink by 1%, with its mass
remaining the same, the acceleration due to gravity on the
earth’s surface would
(a) decrease by 1%
(b) decrease by 2%
(c) increase by 1%
(d) increase by 2%
Suppose the law of gravitational attraction suddenly
18.
19.
4
2GM
a
(b)
8 2 GM
a
(c)
4GM
a
(d)
4 2 GM
a
If the gravitational force had varied as r –5/2 instead of r–2;
the potential energy of a particle at a distance ‘r’ from the
centre of the earth would be directly proportional to
(a) r -1
(b) r - 2
(c) r -3 / 2
(d) r -5 / 2
A particle of mass ‘m’ is kept at rest at a height 3R from the
surface of earth, where ‘R’ is radius of earth and ‘M’ is mass
of earth. The minimum speed with which it should be
projected, so that it does not return back, is (g is acceleration
due to gravity on the surface of earth)
1
(a)
8.
13.
18.
Space for Rough Work
æ GM ö 2
ç
÷
è R ø
1
æ GM ö 2
(b) ç
÷
è 2R ø
9.
14.
19.
1
æ gR ö 2
(c) ç
÷
è 4 ø
10.
15.
1
æ 2g ö 2
(d) ç
÷
è 4 ø
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20. The ratio between the values of acceleration due to gravity
at a height 1 km above and at a depth of 1 km below the
Earth’s surface is (radius of Earth is R)
R–2
R
R–2
(b)
(c)
(d) 1
R –1
R –1
R
21. The weight of an object in the coal mine, sea level and at the
top of the mountain, are respectively W1, W2 and W3 then
(a) W1< W2 > W3
(b) W1= W2 = W3
(c) W1< W2 < W3
(d) W1> W2 > W3
22. The period of moon's rotation around the earth is nearly 29
days. If moon's mass were 2 fold its present value and all
other things remain unchanged, the period of moon's rotation
would be nearly
O
(c)
1
rad sec –1
800
1
1
rad sec–1
(d) 8 rad sec –1
(c)
80
A body weighs 72 N on the surface of the earth. What is the
gravitational force on it due to earth at a height equal to half
the radius of the earth from the surface?
(a) 32 N (b) 28 N
(c) 16 N
(d) 72 N
A body weighs W newton at the surface of the earth. Its
weight at a height equal to half the radius of the earth, will be
W
2W
4W
8W
(a)
(b)
(c)
(d)
2
3
9
27
A shell of mass M and radius R has a point mass m placed at
a distance r from its centre. The graph of gravitational
potential energy U(r) vs distance r will be
25.
26.
27.
(a)
O
29.
31.
r
(b) O
20.
25.
30.
21.
26.
31.
r
The largest and the shortest distance of the earth from the
sun are r1 and r2. Its distance from the sun when it is at
perpendicular to the major-axis of the orbit drawn from the sun
(a) (r1 + r2)/4
(b) (r1 + r2)/(r1 – r 2)
(c) 2r1 r2 /(r1 + r2)
(d) (r1 + r2)/3
A planet is moving in an elliptical orbit around the sun. If T,
V, E and L stand respectively for its kinetic energy,
gravitational potential energy, total energy and magnitude
of angular momentum about the centre of force, then which
of the following is correct ?
(a) T is conserved
(b) V is always positive
(c) E is always negative
(d) L is conserved but direction of vector L changes
continuously
The earth is assumed to be sphere of radius R. A platform is
arranged at a height R from the surface of Earth. The escape
velocity of a body from this platform is kv, where v is its
escape velocity from the surface of the earth. The value of k is
1
1
1
(a)
(b)
(c)
(d)
2
3
2
2
m
A solid sphere of mass M and radius R is
h
A
surrounded by a spherical shell of same
mass M and radius 2R as shown. A small
B
particle of mass m is released from rest
R
from a height h [ << R] above the shell.
There is a hole in the shell.
2R
What time will it enter the hole at A ?
(a)
2
hR 2
GM
(b)
2hR 2
GM
3hR 2
hR 2
(d)
GM
GM
32. A body starts from rest from a point distance R0 from the
centre of the earth. The velocity acquired by the body when
it reaches the surface of the earth will be (R represents radius
of the earth).
(c)
U (r )
U (r )
RESPONSE
GRID
30.
(b)
r
O
U (r )
U (r )
28.
(a)
(a) Zero
(d)
GMm
–
R
(a)
(b) 29 / 2 days
29 2 days
(c) 29 × 2 days
(d) 29 days
23. The mean radius of earth is R, its angular speed on its own
axis is w and the acceleration due to gravity at earth's surface
is g. What will be the radius of the orbit of a geostationary
satellite ?
(a) (R2g / w2)1/3
(b) (Rg / w2)1/3
2
2
1/3
(c) (R w / g)
(d) (R2g / w)1/3
24. In order to make the effective acceleration due to gravity
equal to zero at the equator, the angular velocity of rotation
of the earth about its axis should be (g = 10 ms–2 and radius
of earth is 64000 km)
r
(a)
æ1
1 ö
÷
2 G M çç ÷
R
R
0ø
è
(b)
æ 1
1ö
2 G M çç
- ÷÷
R
R
ø
è 0
(c)
æ1
1 ö
÷
G M çç ÷
R
R
0ø
è
(d)
æ1
1 ö
÷
2 G M çç ÷
è R R0 ø
22.
27.
32.
Space for Rough Work
23.
28.
24.
29.
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33.
A satellite of mass M is moving in a circle of radius R under a
centripetal force given by (–k/R2), where k is a constant. Then
k
(a) The kinetic energy of the particle is
R
12
(b) The total energy of the particle is æç - k ö÷
2R
è
ø
39.
40.
æ k ö
(c)
(a)
(b)
gR
2.5gR (d)
7.1gR
R
37. A satellite revolves around the earth of radius R in a circular
orbit of radius 3R. The percentage increase in energy
required to lift it to an orbit of radius 5R is
(a) 10 %
(b) 20 %
(c) 30 %
(d) 40 %
38. A (nonrotating) star collapses onto itself from an initial radius
Ri with its mass remaining unchanged. Which curve
a
in figur e best gives th e
gravitational acceleration ag on
d
the surface of the star as a
b
function of the radius of the
a
star during the collapse
g
(a) a
(b) b
33.
38.
43.
RESPONSE
GRID
(c) c
(d) d
34.
39.
44.
Ri
2 pMR 2
4 pMR 2
pMR 3
MR 2 p
(b)
(c)
(d)
5T
5T
T
T
A satellite is launched into a circular orbit of radius R around
the earth. A second satellite is launched into an orbit of
radius 1.01 R. The period of second satellite is larger than
the first one by approximately
(a) 0.5%
(b) 1.0%
(c) 1.5%
(d) 3.0%
A uniform spherical shell gradually shrinks maintaining
its shape. The gravitational potential at the centre
(a) increases
(b) decreases
(c) remains constant
(d) cannot say
The depth d at which the value of acceleration due to gravity
1
becomes times the value at the surface of the earth, is
n
[R = radius of the earth]
(a)
æ kö
(c) The kinetic energy of the particle is ç - ÷
è Rø
(d) The potential energy of the particle is ç ÷
è 2R ø
34. The change in the value of ‘g’ at a height ‘h’ above the surface
of the earth is the same as at a depth ‘d’ below the surface of
earth. When both ‘d’ and ‘h’ are much smaller than the radius
of earth, then which one of the following is correct?
h
3h
(a) d =
(b) d =
(c) d = h
(d) d =2 h
2
2
35. Two identical geostationary satellites are moving with equal
speeds in the same orbit but their sense of rotation brings
them on a collision course. The debris will
(a) fall down
(b) move up
(c) begin to move from east to west in the same orbit
(d) begin to move from west to east in the same orbit
36. A diametrical tunnel is dug across the Earth. A ball is dropped
into the tunnel from one side. The velocity of the ball when
it reaches the centre of the Earth is .... (Given : gravitational
3 GM
potential at the centre of Earth = –
)
2 R
If the earth is treated as a sphere of radius R and mass M;
its angular momentum about the axis of its rotation with
period T, is
41.
42.
æ n -1 ö
R
R
æ n ö
(b) R ç
(c)
(d) R ç
÷
÷
2
n
è n ø
è n +1ø
n
Radius of moon is 1/4 times that of earth and mass is 1/81
times that of earth. The point at which gravitational field
due to earth becomes equal and opposite to that of moon, is
(Distance between centres of earth and moon is 60R, where
R is radius of earth)
(a) 5.75 R from centre of moon
(b) 16 R from surface of moon
(c) 53 R from centre of earth
(d) 54 R from centre of earth
If earth is supposed to be a sphere of radius R, if g30 is
value of acceleration due to gravity at lattitude of 30° and g
at the equator, the value of g – g30 is
3 2
1 2
(a) 1 w2 R (b)
(d)
w R (c) w 2 R
w R
4
2
4
What is the minimum energy required to launch a satellite of
mass m from the surface of a planet of mass M and radius R
in a circular orbit at an altitude of 2R?
(a)
43.
44.
45.
(a)
R
35.
40.
45.
5GmM
(b)
6R
36.
41.
2GmM
(c)
3R
GmM
(d)
2R
37.
42.
DAILY PRACTICE PROBLEM DPP CHAPTERWISE CP07 - PHYSICS
Total Questions
Attempted
Incorrect
Cut-off Score
45
Total Marks
Correct
Net Score
45
Qualifying Score
Success Gap = Net Score –Space
Qualifying
for RoughScore
Work
Net Score = (Correct × 4) – (Incorrect × 1)
Space for Rough Work
180
60
GmM
2R
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PHYSICS
SOLUTIONS
DAILY PRACTICE
PROBLEMS
1.
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(a) The escape velocity on the earth is defined as
ve =
r
So v P =
2g P R P =
2 ´ 2g e ´ R e / 4 =
x
ve
r+l
2
r +l
(c) Applying conservation of energy principle, we get
F=
1
2GM GMm
GMm
=mk 2
2
R
R
r
2
2
k
1
1 1 1 k
Þ
- =- Þ = R R
r
r R R
1 1
R
Þ = (1 - k 2 ) Þ r =
r R
1- k 2
(d) The gravitational force due to the whole sphere at A
point is
=
T22
R13
R23
24 ´ 24
T22
T22 =
=
(6 R )3
(3 R )3
=8
7.
8.
K.E =
9.
6.
ò
x -2 dx =
r
GMm é x -2+1 ù
l êë -2 + 1úû r
GMm -1 r + l
GMm é 1 ù
GMm
éx ù = =
l ë ûr
l êë x úû r
r (r + l )
k Mv 2
=
. Hence v µ R0
R
R
(b)
F=
(d)
mv 2
GmM
GM
=
g= 2
(R + x ) (R + x ) 2 also
R
\
mv 2
æ GM ö R 2
= mç
è R 2 ÷ø (R + x) 2
(R + x)
\
R2
mv 2
= mg
(R + x )
(R + x ) 2
(b)
v=
æ gR 2 ö
gR 2
÷
Þ v = çç
÷
R+x
èR+xø
3
ve
4
2
K.E. =
24 ´ 24
= 72 = 36 × 2
8
1
mv 2
2
r
1
dx
x2
1/ 2
=8
(d) Total energy = – K E =
ò
r +l
r +l
\ v2 =
9 æ 2GM ö
m
32 çè R ÷ø
GMm
9 GMm
K.E. =
; P.E. = 16 R
R
7 GMm
Total energy = K.E. + P.E. = 16 R
Let the height above the surface of earth be h, then
PE
2
1
2
\ Total energy = - mv
2
(a) The force of attraction between sphere and shaded
æm ö
ç dx ÷
l
ø
position dF = GM è
x2
1
1 æ3 ö
9
mv 2 = m ç v e ÷ =
mve2
2
2 è4 ø
32
=
T2 = 6 2
5.
r +l
r+l
GM e m o , where m is the assumed rest mass at
0
(2R ) 2
point A.
In the second case, when we made a cavity of radius
(R/2), then gravitational force at point A is
GM e m o
F2 =
\ F2/F1= 1/9
( R + R / 2) 2
(c) According to Kepler’s law of period T2 µ R3
=
GMm
GMm
dx =
lx 2
l
GMm
l
=-
F1 =
T12
ò
r
Þ
4.
m
M
1
GMm
GMm
=mk 2 v e2 2
R
r
3.
l
2g e R e
Where Re & ge are the radius & acceleration due to
gravity of earth.
Now for planet gP=2ge, RP=Re/4
2.
DPP/CP07
GMm
h
7 GMm
GMm
16R
=\ h=
16 R
h
7
(a) When closer to the sun, velocity of planet will be greater.
So time taken in covering a given area will be less.
P.E. = -
10.
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11. (c) Applying conservation of total mechanical energy
principle
1 2
mv = mg A h A = mg B hB
2
Þ g A hA = g B hB
æ
ö
GMm ÷ 1
8 2 GM
ç
+ mv2 = 0 Þ v 2 =
\ -4
a
ç
÷ 2
a
è
2ø
18.
(c) The potential energy for a conservative force is defined
as
F=
æg ö
Þ hB = ç A ÷ hA = 9 × 2 = 18 m
èg ø
r r r
- dU
or U = - ò F.dr
dr
B
12.
13.
14.
15.
GM e m
gR 2 m = - gR e m = -0.5 mgR
=- e
e
2
Re + h
2R e
r
Ur =
(where, GMe = gRe2 and h
= Re)
(d) Acceleration due to gravity on earth's surface
r
2
dr =
-GM1M 2
r
…… (ii)
ò
GM1M 2
¥
Þ Ur µ
M
19.
r
5/2
dr =
-2 GM1M 2
3 r 3/2
1
+3/ 2
r
(b) As we know, the minimum speed with which a body is
projected so that it does not return back is called escape
speed.
Ve =
2GM
2GM
2GM
=
=
4R
r
R+h
1
20.
æ GM ö 2
=ç
÷ (Q h = 3R)
è 2R ø
(a) Acceleration due to gravity at a height h above the
earth’s surface is
dA
L
=
dt
2m
æ 2h ö
g h = g ç1 - ÷
è
Rø
For central forces, torque = 0
\ L = constant
Acceleration due to gravity at a depth d below the
earth’s surface is
dö
æ
g d = g ç1 - ÷
è Rø
\
17.
GM1M 2
(Q U¥ = 0)
If we bring the mass from the infinity to the centre of
earth, then we obtain work, ‘so it has negative
(gravitational force do work on the object) sign &
potential energy decreases. But if we bring the mass
from the surface of earth to infinite, then we must do
work against gravitational force & potential energy of
the mass increases.
GM M
Now in equation (i) if F = 51/ 2 2 instead of
r
GM1M 2
then
F=
r2
F µ xM ´ (1 – x )M = xM 2 (1 - x )
dF
=0
For maximum force,
dx
dF
= M 2 - 2 xM 2 = 0 Þ x = 1/ 2
Þ
dx
(c) Mass of the satellite = m and height of satellite from
earth (h) = 6.4 × 106 m.
We know that gravitational potential energy of the
satellite at height
R2
This implies that as radius decreases, the acceleration
due to gravity increases.
Dg
DR
DR
= -2
= -1%
But
g
R
R
('–' sign is due to shrinking of earth)
Dg
\
= -2 ´ (-1%) = 2%
g
(a) According to kepler's law of area
ò
¥
(a)
g=G
16.
r
or U r =
(b) Due to inertia of motion it will move tangentially to the
original orbit with same velocity.
h =-
…… (i)
¥
dA
= constant
dt
æ 2h ö
1- ÷
g h çè
R ø (R - 2h)
=
=
Now,
dö
gd
(R - d)
æ
1
çè
÷
Rø
(b) Potential energy of particle at the centre of square
æ
ö
ç GMm ÷
= -4 ç
a ÷
ç
÷
è 2 ø
As h = 1 km, d = 1 km
\
gh R - 2
=
gd R - 1
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9.8m/sec2.
21. (a) At the surface of earth, the value of g =
If
we go towards the centre of earth or we go above the
surface of earth, then in both the cases the value of g
decreases.
Hence W1=mgmine, W2=mgsea level, W3=mgmoun
So W1< W2 > W3 (g at the sea level = g at the suface
of earth)
22. (d) Time period does not depend upon the mass of satellite
23. (a) T =
24. (b)
2 1 1 r1 + r2
= + =
R r1 r2
r1 r2
3/ 2
Sun
R
major axis
R=
gR2
gR 2
or r 3 = 2
w
w
=
30.
(a) Here, v =
equator l = 0 and g ' = 0
g
1 rad
\ 0 = g - w2 R Þ w =
=
R 800 s
25. (a) mg = 72 N (body weight on the surface)
GM
Solving k =
31.
R2
Rö
æ
çè R + ÷ø
2
2
(a) g =
R
,
2
GM
=
t=
4 GM
9 R2
32.
R
,
2
(b)
4
4
´ g = mg
9
9
4
= ´ 72 = 32 N
9
26. (c) At a height h,
æ R ö
Þ mg ' = mg ç
è R + h ÷ø
æ R ö
Þ W' = Wç
è R + h ÷ø
Here, h = R/2
=
2GM
.
R+R
2
GM
2R 2
1 2
gt
2
[ Q U= 0]
2h
hR 2
=2
g
GM
R
GMm
R0
2
P.E. = ò
\
= m´
(R + h)2
(2R)2
2GM
and kv =
R
1
r
é1
1 ù
dr = -GMm ê ú
ë R R0 û
The K.E. acquired by the body at the surface =
4 GM
mg ¢ = m ´
9 R2
R2
G (2M)
From h =
Body weight at height H =
g' =g
2 r1 r2
r1 + r2
(c) In a circular or elliptical orbital motion, torque is always
acting parallel to displacement or velocity. So, angular
momentum is conserved. In attractive field, potential
energy is negative. Kinetic energy changes as velocity
increase when distance is less. So, option (c) is correct.
To make effective acceleration due to gravity zero at
g¢ =
r2
29.
g ' = g - w2 R cos 2 l
At a height H =
Instant position
of satellite
r1
r = (gR2 / w2)1/3
g=
(c) Applying the properties of ellipse, we have
2 pr
2pr
2 pr 3 / 2 2p
=
=
=
v0
w
( gR 2 / r )1/ 2
gR 2
Hence, r
or,
28.
é1 1 ù
1 2
mv = -GMm ê ú
2
ë R R0 û
æ 1
1ö
v = 2G M ç
- ÷
è R0 R ø
2
33.
mv 2
k
k
2
=
or mv =
2
R
R
R
1
k
mv 2 =
2
2R
In case of satellites P.E = – 2 K.E
Kinetic energy =
2
4
W
9
27. (c) Gravitational P.E. = m × gravitational potential
U = mV, so the graph of U will be same as that of V for
a spherical shell.
(b)
and T.E = P. E + K. E
\ W' =
k
k
k
- =2R R
2R
(d) Variation of g with altitude is,
Total energy =
34.
é 2h ù
g h = g ê1 - ú ;
ë Rû
1
m v2
2
t.me/Magazines4all
DPP/ CP07
S-34
variation of g with depth is,
é dù
g d = g ê1 - ú
ë Rû
Equating gh and gd, we get d = 2h
35.
36.
(a) The total momentum will be zero and hence velocity
will be zero just after collisiion. The pull of earth will
make it fall down.
(b) Loss in potential energy = Gain in kinetic energy
GMm æ 3 GMm ö 1
2
-ç÷ = mv
R
2
R
2
ø
è
Þ
37.
(d)
38.
(b)
39.
gµ
T1 = 2 p
R3
(1.01R )3
, T2 = 2p
GM
GM
T2 - T1
´ 100 = 1.5%
T1
41.
43.
(a) The gravitational potential at the centre of uniform
spherical shell is equal to the gravitational potential at
the surface of shell i.e.,
V=
-GM
, where a is radius of spherical shell
a
g
æ dö
æ dö
g ' = g ç1 - ÷ Þ = g ç1 - ÷
n
è Rø
è Rø
(d)
R/4 Moon
E earth = E moon
Þ
GM
x
2
=
GM / 81
(60R - x) 2
60 R
x
1
1
Þ x = 9 (60R - x)
44.
R
Earth
Þ x = 54 R from centre of earth.
(b) Acceleration due to gravity at lattitude’ l ’ is given by
g l = g e - R e w 2 cos 2 l
At equator, l = 90° Þ cos l = cos 90° = 0
or g l = g e = g (as given in question)
2
2
At 30°, g 30 = g - Rw cos 30 = g -
4p MR 2
5T
(R + h) 3
T = 2p
GM
(b)
æ n -1 ö
Þd =ç
÷R
è n ø
1
R2
R decreasing g increase hence, curve b represents
correct variation.
(d) Angular momentum, L = Iw; moment of inertia of sphere
along the axis passing through centre of mass,
2
2p
2
I = MR and w =
.
5
T
(c)
42.
GMm 1
GM
= mv 2 Þ v =
= gR
2R
2
R
Putting these values, L =
40.
Now, if the shell shrinks then its radius decrease then
density increases, but mass is constant. so from above
expression if a decreases, then V increases.
3
Rw 2
4
3
Rw 2
4
(a) As we know,
or, g - g 30 =
45.
-GMm
r
and orbital velocity, v0 = GM / R + h
1
GMm 1 GM GMm
Ef = mv02 = m
2
3R
2 3R
3R
GMm æ 1 ö - GMm
=
ç - 1÷ =
3R è 2 ø
6R
-GMm
Ei =
+K
R
Ei = E f
Gravitational potential energy =
Therefore minimum required energy, K =
5GMm
6R
EBD_7156
t.me/Ebooks_Encyclopedia27.