Page 1
(1) (c) If r is the distance between m and (M – m), the gravita
tional force will be 
F = G
2
m(M m)
r

=
2
G
r
(mM – m
2
)
The force will be maximum if,
dF
dm
= 0
i.e,
d
dm
2
2
G
(mM m)
r
éù

êú
ëû
= 0
or
m1
M2
= (as M and r are constants)
(2) (c) m
g
=
3
kg , v =
c
2
0
22
m
m
1 (v /c)
=

=
2
2
3
c
1
4xc

=
3
3
2
kg
(3) (a) g' = g – R
e
w
2
(at equator l = 0)
If a body is weightless,
g' = 0 , g – R
e
w
2
= 0
Þ w =
g
R
=
3
10
6400 10 ´
= 1.25 × 10
3
rad/sec.
(4) (b) The apparent weight of person on the equator
(latitude l = 0) is given by
W' = W – m R
e
w
2
,
W' =
3
5
W =
3
5
mg
3
5
mg = mg – mRw
2
or mRw
2
= mg –
3
5
mg
w =
2g
5R
=
3
2 9.8
5
6400 10
´
´
rad/ sec
= 7.826 × 10
–4
rad/sec
(5) (c) According to question,
g' =
p
2
p
G 4M
R
´
on the planet and g =
e
2
e
GM
R
on the earth
Q R
p
= R
e
and M
p
= M
e
Now,
g'
g
= 4 Þ g' = 4g = 40 m/sec
2
Energy needed to lift 2 kg mass through 2m distance
= mg'h = 2 × 40 × 2 = 160 J
(6) (d) V
g
= V
g1
+ V
g2
=
12
12
Gm Gm
rr

= – 6.67 × 10
–11
23
10 10
0.5 0.5
éù
+ êú
êú
ëû
= – 1.47 × 10
–7
Joule/kg
(7) (c) The P .E. of the object on the surface of earth is
U
1
= –
R
GMm
The P .E. of object at a height R, U
2
= –
GMm
(R R) +
The gain in P E is U
2
– U
1
=
GMm
2R
=
1
2
mgR
2
GM
g on surface of earth
R
éù
=
êú
ëû
Q
(8) (c) Resultant force on particle '1'
F
r
=
2
F + F'
or F
r
=
2
2
2
Gm
2r
+
2
2
Gm
4r
=
2
mv
r
or v =
Gm 2 21
r4
æö
+
ç÷
èø
(9) (b) The resultant gravitational force on each particle pro
vides it the necessary centripetal force
\
2
mv
r
=
2 2 2o
F F 2F cos60 ++
=
3F
,
But r =
32
23 3
´=
l
l
\ v =
GM
l
(10) (b) The acceleration due to gravity on the surface of the
earth, in terms of mass M
e
and radius R
e
of earth, is
given by g =
e
2
e
GM
R
if M
m
be the mass of the moon, R
m
its radius, then the
acceleration due to gravity on the surface of the moon
will be given by g' =
m
2
m
GM
R
Dividing eq. (ii) by eq. (i), we get
g
g'
=
m
e
M
M
2
e
m
R
R
æö
ç÷
èø
=
2
1 41
80 15
æö
´=
ç÷
èø
\ g' = g/5.
(11) (b) The value of g at the height h from the surface of earth
2h
g' g1
R
æö
=
ç÷
èø
The value of g at depth x below the surface of earth
x
g' g1
R
æö
=
ç÷
èø
Page 2
(1) (c) If r is the distance between m and (M – m), the gravita
tional force will be 
F = G
2
m(M m)
r

=
2
G
r
(mM – m
2
)
The force will be maximum if,
dF
dm
= 0
i.e,
d
dm
2
2
G
(mM m)
r
éù

êú
ëû
= 0
or
m1
M2
= (as M and r are constants)
(2) (c) m
g
=
3
kg , v =
c
2
0
22
m
m
1 (v /c)
=

=
2
2
3
c
1
4xc

=
3
3
2
kg
(3) (a) g' = g – R
e
w
2
(at equator l = 0)
If a body is weightless,
g' = 0 , g – R
e
w
2
= 0
Þ w =
g
R
=
3
10
6400 10 ´
= 1.25 × 10
3
rad/sec.
(4) (b) The apparent weight of person on the equator
(latitude l = 0) is given by
W' = W – m R
e
w
2
,
W' =
3
5
W =
3
5
mg
3
5
mg = mg – mRw
2
or mRw
2
= mg –
3
5
mg
w =
2g
5R
=
3
2 9.8
5
6400 10
´
´
rad/ sec
= 7.826 × 10
–4
rad/sec
(5) (c) According to question,
g' =
p
2
p
G 4M
R
´
on the planet and g =
e
2
e
GM
R
on the earth
Q R
p
= R
e
and M
p
= M
e
Now,
g'
g
= 4 Þ g' = 4g = 40 m/sec
2
Energy needed to lift 2 kg mass through 2m distance
= mg'h = 2 × 40 × 2 = 160 J
(6) (d) V
g
= V
g1
+ V
g2
=
12
12
Gm Gm
rr

= – 6.67 × 10
–11
23
10 10
0.5 0.5
éù
+ êú
êú
ëû
= – 1.47 × 10
–7
Joule/kg
(7) (c) The P .E. of the object on the surface of earth is
U
1
= –
R
GMm
The P .E. of object at a height R, U
2
= –
GMm
(R R) +
The gain in P E is U
2
– U
1
=
GMm
2R
=
1
2
mgR
2
GM
g on surface of earth
R
éù
=
êú
ëû
Q
(8) (c) Resultant force on particle '1'
F
r
=
2
F + F'
or F
r
=
2
2
2
Gm
2r
+
2
2
Gm
4r
=
2
mv
r
or v =
Gm 2 21
r4
æö
+
ç÷
èø
(9) (b) The resultant gravitational force on each particle pro
vides it the necessary centripetal force
\
2
mv
r
=
2 2 2o
F F 2F cos60 ++
=
3F
,
But r =
32
23 3
´=
l
l
\ v =
GM
l
(10) (b) The acceleration due to gravity on the surface of the
earth, in terms of mass M
e
and radius R
e
of earth, is
given by g =
e
2
e
GM
R
if M
m
be the mass of the moon, R
m
its radius, then the
acceleration due to gravity on the surface of the moon
will be given by g' =
m
2
m
GM
R
Dividing eq. (ii) by eq. (i), we get
g
g'
=
m
e
M
M
2
e
m
R
R
æö
ç÷
èø
=
2
1 41
80 15
æö
´=
ç÷
èø
\ g' = g/5.
(11) (b) The value of g at the height h from the surface of earth
2h
g' g1
R
æö
=
ç÷
èø
The value of g at depth x below the surface of earth
x
g' g1
R
æö
=
ç÷
èø
54
DPP/ P 18
These two are given equal, hence
2hx
11
RR
æ öæö
 =
ç ÷ç÷
è øèø
On solving, we get x = 2h
(12) (a) If g be the acceleration due to gravity at the surface of
the earth, then its value at a height h above the earth's
surface will be 
2
e
g
g
h
1
R
= ¢
æö
+
ç÷
èø
Here
g'1
g9
=
\
2
e
11
9
h
1
R
=
æö
+
ç÷
èø
or 1 +
e
h
R
= 3
or h = 2 R
e
= 2 × 6400 = 12800 km.
(13) (c) Consider the case of a body of mass m placed on the
earth's surface (mass of the earth M and radius R). If g
is acceleration due to gravity , then
mg = G
e
2
Mm
R
or g =
e
2
GM
R
where G is universal constant of gravitation.
Now when the radius is reduced by 1%, i.e., radius
becomes 0.99 R, let acceleration due to gravity be g',
then g' =
e
2
GM
(0.99R)
From equation (A) and (B), we get
g'
g
=
2
2
R
(0.99R)
=
2
1
(0.99)
\ g' = g ×
2
1
0.99
æö
ç÷
èø
or g' > g
Thus , the value of g is increased.
(14) (a) Force of gravity at surface of earth,
F
1
= Gm M/R
2
.......... (1)
Force of gravity at height H is
F
2
= Gm M (R + H)
2
.......... (2)
Dividing (A) by (B) and Rearranging
H = R
1
2
F
1
F
æö

ç÷
èø
= 350 km where (F
2
= .9F
1
)
(15) (a) The extension in the length of spring is
x
=
mg
k
=
2
GMm
rk
,
\ x µ
2
1
r
, \
2
2
2
1
x R
x
(R h)
=
+
or x
2
= 1 ×
2
6400
7200
æö
ç÷
èø
= 0.79 cm .
(16) (d) In the position of solar eclipse, net force on earth
F
E
= F
M
+ F
S
In the position of lunar eclipse, net force on earth
F'
E
= F
S
– F
M
\ Change in acceleration of earth,
Df =
2
2GM
R
=
11 22
2 16
2 6.67 10 7.36 10
3.82 10

´´ ´´
´
= 6.73 × 10
5
m/s
2
(17) (c) Let M
e
be the mass of the earth. The work required
W = GM
e
m
ee
11
R Rh
éù

êú
+
ëû
=
e
ee
GM mh
R (R h) +
=
2
e
ee
gR mh
R (R h) +
[\ GM
e
= gR
e
2
]
=
e
mgh
h
1
R
æö
+
ç÷
èø
(18) (a) The P .E of the mass at d/2 due to the earth and moon is
U = – 2
1
GMm
d
– 2
2
GMm
d
or U = –
2Gm
d
(M
1
+ M
2
) (Numerically)
1
2
m V V
e
2
= U ÞV
e
=
12
G
2 (M M)
d
+
(19) (d) Let m be the mass of the body . The gravitational poten
tial energy of the body at the surface of the earth is
U = –
e
e
R
m GM
The potential energy at a height 10 R
e
above the sur
face of the earth will be
U' = –
) 10
e e
e
R (R
m GM
+
Earth
O
1
R
1
P
m
M
1
R
2
O
2
M
2
d
Moon
Page 3
(1) (c) If r is the distance between m and (M – m), the gravita
tional force will be 
F = G
2
m(M m)
r

=
2
G
r
(mM – m
2
)
The force will be maximum if,
dF
dm
= 0
i.e,
d
dm
2
2
G
(mM m)
r
éù

êú
ëû
= 0
or
m1
M2
= (as M and r are constants)
(2) (c) m
g
=
3
kg , v =
c
2
0
22
m
m
1 (v /c)
=

=
2
2
3
c
1
4xc

=
3
3
2
kg
(3) (a) g' = g – R
e
w
2
(at equator l = 0)
If a body is weightless,
g' = 0 , g – R
e
w
2
= 0
Þ w =
g
R
=
3
10
6400 10 ´
= 1.25 × 10
3
rad/sec.
(4) (b) The apparent weight of person on the equator
(latitude l = 0) is given by
W' = W – m R
e
w
2
,
W' =
3
5
W =
3
5
mg
3
5
mg = mg – mRw
2
or mRw
2
= mg –
3
5
mg
w =
2g
5R
=
3
2 9.8
5
6400 10
´
´
rad/ sec
= 7.826 × 10
–4
rad/sec
(5) (c) According to question,
g' =
p
2
p
G 4M
R
´
on the planet and g =
e
2
e
GM
R
on the earth
Q R
p
= R
e
and M
p
= M
e
Now,
g'
g
= 4 Þ g' = 4g = 40 m/sec
2
Energy needed to lift 2 kg mass through 2m distance
= mg'h = 2 × 40 × 2 = 160 J
(6) (d) V
g
= V
g1
+ V
g2
=
12
12
Gm Gm
rr

= – 6.67 × 10
–11
23
10 10
0.5 0.5
éù
+ êú
êú
ëû
= – 1.47 × 10
–7
Joule/kg
(7) (c) The P .E. of the object on the surface of earth is
U
1
= –
R
GMm
The P .E. of object at a height R, U
2
= –
GMm
(R R) +
The gain in P E is U
2
– U
1
=
GMm
2R
=
1
2
mgR
2
GM
g on surface of earth
R
éù
=
êú
ëû
Q
(8) (c) Resultant force on particle '1'
F
r
=
2
F + F'
or F
r
=
2
2
2
Gm
2r
+
2
2
Gm
4r
=
2
mv
r
or v =
Gm 2 21
r4
æö
+
ç÷
èø
(9) (b) The resultant gravitational force on each particle pro
vides it the necessary centripetal force
\
2
mv
r
=
2 2 2o
F F 2F cos60 ++
=
3F
,
But r =
32
23 3
´=
l
l
\ v =
GM
l
(10) (b) The acceleration due to gravity on the surface of the
earth, in terms of mass M
e
and radius R
e
of earth, is
given by g =
e
2
e
GM
R
if M
m
be the mass of the moon, R
m
its radius, then the
acceleration due to gravity on the surface of the moon
will be given by g' =
m
2
m
GM
R
Dividing eq. (ii) by eq. (i), we get
g
g'
=
m
e
M
M
2
e
m
R
R
æö
ç÷
èø
=
2
1 41
80 15
æö
´=
ç÷
èø
\ g' = g/5.
(11) (b) The value of g at the height h from the surface of earth
2h
g' g1
R
æö
=
ç÷
èø
The value of g at depth x below the surface of earth
x
g' g1
R
æö
=
ç÷
èø
54
DPP/ P 18
These two are given equal, hence
2hx
11
RR
æ öæö
 =
ç ÷ç÷
è øèø
On solving, we get x = 2h
(12) (a) If g be the acceleration due to gravity at the surface of
the earth, then its value at a height h above the earth's
surface will be 
2
e
g
g
h
1
R
= ¢
æö
+
ç÷
èø
Here
g'1
g9
=
\
2
e
11
9
h
1
R
=
æö
+
ç÷
èø
or 1 +
e
h
R
= 3
or h = 2 R
e
= 2 × 6400 = 12800 km.
(13) (c) Consider the case of a body of mass m placed on the
earth's surface (mass of the earth M and radius R). If g
is acceleration due to gravity , then
mg = G
e
2
Mm
R
or g =
e
2
GM
R
where G is universal constant of gravitation.
Now when the radius is reduced by 1%, i.e., radius
becomes 0.99 R, let acceleration due to gravity be g',
then g' =
e
2
GM
(0.99R)
From equation (A) and (B), we get
g'
g
=
2
2
R
(0.99R)
=
2
1
(0.99)
\ g' = g ×
2
1
0.99
æö
ç÷
èø
or g' > g
Thus , the value of g is increased.
(14) (a) Force of gravity at surface of earth,
F
1
= Gm M/R
2
.......... (1)
Force of gravity at height H is
F
2
= Gm M (R + H)
2
.......... (2)
Dividing (A) by (B) and Rearranging
H = R
1
2
F
1
F
æö

ç÷
èø
= 350 km where (F
2
= .9F
1
)
(15) (a) The extension in the length of spring is
x
=
mg
k
=
2
GMm
rk
,
\ x µ
2
1
r
, \
2
2
2
1
x R
x
(R h)
=
+
or x
2
= 1 ×
2
6400
7200
æö
ç÷
èø
= 0.79 cm .
(16) (d) In the position of solar eclipse, net force on earth
F
E
= F
M
+ F
S
In the position of lunar eclipse, net force on earth
F'
E
= F
S
– F
M
\ Change in acceleration of earth,
Df =
2
2GM
R
=
11 22
2 16
2 6.67 10 7.36 10
3.82 10

´´ ´´
´
= 6.73 × 10
5
m/s
2
(17) (c) Let M
e
be the mass of the earth. The work required
W = GM
e
m
ee
11
R Rh
éù

êú
+
ëû
=
e
ee
GM mh
R (R h) +
=
2
e
ee
gR mh
R (R h) +
[\ GM
e
= gR
e
2
]
=
e
mgh
h
1
R
æö
+
ç÷
èø
(18) (a) The P .E of the mass at d/2 due to the earth and moon is
U = – 2
1
GMm
d
– 2
2
GMm
d
or U = –
2Gm
d
(M
1
+ M
2
) (Numerically)
1
2
m V V
e
2
= U ÞV
e
=
12
G
2 (M M)
d
+
(19) (d) Let m be the mass of the body . The gravitational poten
tial energy of the body at the surface of the earth is
U = –
e
e
R
m GM
The potential energy at a height 10 R
e
above the sur
face of the earth will be
U' = –
) 10
e e
e
R (R
m GM
+
Earth
O
1
R
1
P
m
M
1
R
2
O
2
M
2
d
Moon
DPP/ P 18
55
\ Increase in potential energy
U' – U = –
e
e
GMm
11R
+
e
e
GMm
R
æö
ç÷
èø
=
e
e
GMm 10
11R
This increase will be obtained from the initial kinetic
energy given to the body . Hence if the body be thrown
with a v velocity then
1
2
mv
2
=
e
e
GMm 10
11R
Þ
e
e
20Gm
v
11R
=
Substituting the given values, we get
11 24
6
20 (6.67 10 ) (6 10 )
v
11 (6.4 10 )

æö
´ ´ ´´
=
ç÷
´´ èø
= 1.07 × 10
4
m/s.
(20) (b)
? (
2
GmMm
F
r
,
<
For maximum force
dF
0
dm
<
Þ
2
22
d GmM Gm
0
dm
rr
æö
÷ ç
÷ ç ,<
÷ ç
÷ ç ÷ ç
èø
m1
M 2m0
M2
Þ , <Þ<
(21) (b)
22
g ' g R cos =wl
For weightlessness at equator 0andg'0 l==
\
2
g 1 rad
0gR
R 800s
=w Þw==
(22) (a) k represents gravitational constant which depends only
on the system of units.
(23) (a) All statements except (4) are wrong.
(24) (a) V alue of g decreases when we go from poles to equator.
(25) (b)
r
2
r
1
Gravitational PE at perihelion
11
GMm/r asr <, is
minimum Therefore, PE is minimum.
(26) (c) Total energy = constant.
(27) (c) As Pluto moves away, displacement has component
opposite to air force, hence work done is –negative.
(28) (b) For two electron
g
43
e
F
10
F

= i.e. gravitational force is
negligible in comparison to electrostatic force of
attraction.
(29) (c) The universal gravitational constant G is totally
different from g.
2
FR
G
Mm
=
The constant G is scalar and posses the dimensions
132
MLT.

éù
ëû
2
GM
g
R
=
g is a vector and has got the dimensions
02
MLT.

éù
ëû
It is not a universal constant.
(30) (a) As the rotation of earth takes place about polar axis
therefore body placed at poles will not feel any cen
trifugal force and its weight or acceleration due to grav
ity remains unaffected.
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