DPP for NEET: Daily Practice Problems: Gravitation- 1 (Solutions)

 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.