Chapter 11 : Gravitation - HC Verma Solution, Physics Class 11 Notes | EduRev

HC Verma and Irodov Solutions

JEE : Chapter 11 : Gravitation - HC Verma Solution, Physics Class 11 Notes | EduRev

 Page 1


11.1
SOLUTIONS TO CONCEPTS 
CHAPTER 11
1. Gravitational force of attraction,
F = 
2
r
GMm
= 
2
11
) 1 . 0 (
10 10 10 67 . 6 ? ? ?
?
= 6.67 × 10
–7
N
2. To calculate the gravitational force on ‘m’ at unline due to other mouse.
OD
F = 
2 2
) r / a (
m 4 m G ? ?
=  
2
2
a
Gm 8
OI
F = 
2 2
) r / a (
m 2 m G ? ?
=  
2
2
a
Gm 6
OB
F = 
2 2
) r / a (
m 2 m G ? ?
=  
2
2
a
Gm 4
OA
F = 
2 2
) r / a (
m m G ? ?
=  
2
2
a
Gm 2
Resultant 
OF
F = 
2
2
2
2
2
2
a
Gm
36
a
Gm
64
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= 10
2
2
a
Gm
Resultant 
OE
F = 
2
2
2
2
2
2
a
Gm
4
a
Gm
64
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= 
2
2
a
Gm
5 2
The net resultant force will be,
F = 5 20
a
Gm
2
a
Gm
20
a
Gm
100
2
2
2
2
2
2
2
2
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= ? ? 5 40 120
a
Gm
2
2
2
?
?
?
?
?
?
?
?
?
= ) 6 . 89 120 (
a
Gm
2
2
2
?
?
?
?
?
?
?
?
?
= 4 . 40
a
Gm
2
2
= 
2
2
a
Gm
2 4
3. a) if ‘m’ is placed at mid point of a side 
then 
OA
F = 
2
2
a
Gm 4
in OA direction
OB
F  = 
2
2
a
Gm 4
in OB direction
Since equal & opposite cancel each other
oc
F = 
? ? ? ?
2
3
2
a 2 / r
Gm
= 
2
2
a 3
Gm 4
in OC direction
Net gravitational force on m = 
2
2
a
Gm 4
b) If placed at O (centroid)
the 
OA
F = 
) r / a (
Gm
3
2
= 
2
2
a
Gm 3
F
D
B A
E
C
2m m
4m
3m
m
B
A
C
m
m
m m
O
B
A
C
m
m
m m
O
Page 2


11.1
SOLUTIONS TO CONCEPTS 
CHAPTER 11
1. Gravitational force of attraction,
F = 
2
r
GMm
= 
2
11
) 1 . 0 (
10 10 10 67 . 6 ? ? ?
?
= 6.67 × 10
–7
N
2. To calculate the gravitational force on ‘m’ at unline due to other mouse.
OD
F = 
2 2
) r / a (
m 4 m G ? ?
=  
2
2
a
Gm 8
OI
F = 
2 2
) r / a (
m 2 m G ? ?
=  
2
2
a
Gm 6
OB
F = 
2 2
) r / a (
m 2 m G ? ?
=  
2
2
a
Gm 4
OA
F = 
2 2
) r / a (
m m G ? ?
=  
2
2
a
Gm 2
Resultant 
OF
F = 
2
2
2
2
2
2
a
Gm
36
a
Gm
64
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= 10
2
2
a
Gm
Resultant 
OE
F = 
2
2
2
2
2
2
a
Gm
4
a
Gm
64
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= 
2
2
a
Gm
5 2
The net resultant force will be,
F = 5 20
a
Gm
2
a
Gm
20
a
Gm
100
2
2
2
2
2
2
2
2
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= ? ? 5 40 120
a
Gm
2
2
2
?
?
?
?
?
?
?
?
?
= ) 6 . 89 120 (
a
Gm
2
2
2
?
?
?
?
?
?
?
?
?
= 4 . 40
a
Gm
2
2
= 
2
2
a
Gm
2 4
3. a) if ‘m’ is placed at mid point of a side 
then 
OA
F = 
2
2
a
Gm 4
in OA direction
OB
F  = 
2
2
a
Gm 4
in OB direction
Since equal & opposite cancel each other
oc
F = 
? ? ? ?
2
3
2
a 2 / r
Gm
= 
2
2
a 3
Gm 4
in OC direction
Net gravitational force on m = 
2
2
a
Gm 4
b) If placed at O (centroid)
the 
OA
F = 
) r / a (
Gm
3
2
= 
2
2
a
Gm 3
F
D
B A
E
C
2m m
4m
3m
m
B
A
C
m
m
m m
O
B
A
C
m
m
m m
O
Chapter 11
11.2
OB
F = 
2
2
a
Gm 3
Resultant F
?
= 
2
1
a
Gm 3
2
a
Gm 3
2
2
2
2
2
2
2
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= 
2
2
a
Gm 3
Since 
OC
F = 
2
2
a
Gm 3
, equal & opposite to F, cancel 
Net gravitational force = 0
4.
CB
F = j
ˆ
60 sin
a 4
Gm
i
ˆ
60 cos
a 4
Gm
2
2
2
2
?
CA
F = j
ˆ
60 sin
a 4
Gm
i
ˆ
60 cos
a 4
Gm
2
2
2
2
?
?
F
?
= 
CB
F + 
CA
F
= j
ˆ
60 sin
a 4
Gm 2
2
2
?
= 
2
r
a 4
Gm 2
3
2
2
?
= 
2
2
3
a 4
Gm r
5. Force on M at C due to gravitational attraction.
CB
F = j
ˆ
R 2
Gm
2
2
CD
F = i
ˆ
R 4
GM
2
2
?
CA
F = j
ˆ
45 sin
R 4
GM
j
ˆ
45 cos
R 4
GM
2
2
2
2
?
?
So, resultant force on C,
?
C
F =  
CA
F + 
CB
F + 
CD
F
= j
ˆ
2
1
2
R 4
GM
i
ˆ
2
1
2
R 4
GM
2
2
2
2
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
? ?
?F
C
= ? ? 1 2 2
R 4
GM
2
2
?
For moving along the circle, F
?
= 
R
mv
2
or ? ? 1 2 2
R 4
GM
2
2
? = 
R
MV
2
  or V = 
?
?
?
?
?
?
?
?
?
4
1 2 2
R
GM
6.
? ?
2
h R
GM
?
= 
6 2
22 11
10 ) 1000 1740 (
10 4 . 7 10 67 . 6
? ?
? ? ?
?
= 
6
11
10 2740 2740
10 358 . 49
? ?
?
= 
13
11
10 75 . 0
10 358 . 49
?
?
= 65.8 × 10
–2
= 0.65 m/s
2
7. The linear momentum of 2 bodies is 0 initially. Since gravitational force is internal, final momentum is 
also zero.
So (10 kg)v
1
= (20 kg) v
2
Or v
1
= v
2
…(1)
Since P.E. is conserved
Initial P.E. = 
1
20 10 10 67 . 6
11
? ? ? ?
?
= –13.34×10
–9
J
When separation is 0.5 m,
M
B
C
A
D
B
C
A
R
Page 3


11.1
SOLUTIONS TO CONCEPTS 
CHAPTER 11
1. Gravitational force of attraction,
F = 
2
r
GMm
= 
2
11
) 1 . 0 (
10 10 10 67 . 6 ? ? ?
?
= 6.67 × 10
–7
N
2. To calculate the gravitational force on ‘m’ at unline due to other mouse.
OD
F = 
2 2
) r / a (
m 4 m G ? ?
=  
2
2
a
Gm 8
OI
F = 
2 2
) r / a (
m 2 m G ? ?
=  
2
2
a
Gm 6
OB
F = 
2 2
) r / a (
m 2 m G ? ?
=  
2
2
a
Gm 4
OA
F = 
2 2
) r / a (
m m G ? ?
=  
2
2
a
Gm 2
Resultant 
OF
F = 
2
2
2
2
2
2
a
Gm
36
a
Gm
64
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= 10
2
2
a
Gm
Resultant 
OE
F = 
2
2
2
2
2
2
a
Gm
4
a
Gm
64
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= 
2
2
a
Gm
5 2
The net resultant force will be,
F = 5 20
a
Gm
2
a
Gm
20
a
Gm
100
2
2
2
2
2
2
2
2
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= ? ? 5 40 120
a
Gm
2
2
2
?
?
?
?
?
?
?
?
?
= ) 6 . 89 120 (
a
Gm
2
2
2
?
?
?
?
?
?
?
?
?
= 4 . 40
a
Gm
2
2
= 
2
2
a
Gm
2 4
3. a) if ‘m’ is placed at mid point of a side 
then 
OA
F = 
2
2
a
Gm 4
in OA direction
OB
F  = 
2
2
a
Gm 4
in OB direction
Since equal & opposite cancel each other
oc
F = 
? ? ? ?
2
3
2
a 2 / r
Gm
= 
2
2
a 3
Gm 4
in OC direction
Net gravitational force on m = 
2
2
a
Gm 4
b) If placed at O (centroid)
the 
OA
F = 
) r / a (
Gm
3
2
= 
2
2
a
Gm 3
F
D
B A
E
C
2m m
4m
3m
m
B
A
C
m
m
m m
O
B
A
C
m
m
m m
O
Chapter 11
11.2
OB
F = 
2
2
a
Gm 3
Resultant F
?
= 
2
1
a
Gm 3
2
a
Gm 3
2
2
2
2
2
2
2
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= 
2
2
a
Gm 3
Since 
OC
F = 
2
2
a
Gm 3
, equal & opposite to F, cancel 
Net gravitational force = 0
4.
CB
F = j
ˆ
60 sin
a 4
Gm
i
ˆ
60 cos
a 4
Gm
2
2
2
2
?
CA
F = j
ˆ
60 sin
a 4
Gm
i
ˆ
60 cos
a 4
Gm
2
2
2
2
?
?
F
?
= 
CB
F + 
CA
F
= j
ˆ
60 sin
a 4
Gm 2
2
2
?
= 
2
r
a 4
Gm 2
3
2
2
?
= 
2
2
3
a 4
Gm r
5. Force on M at C due to gravitational attraction.
CB
F = j
ˆ
R 2
Gm
2
2
CD
F = i
ˆ
R 4
GM
2
2
?
CA
F = j
ˆ
45 sin
R 4
GM
j
ˆ
45 cos
R 4
GM
2
2
2
2
?
?
So, resultant force on C,
?
C
F =  
CA
F + 
CB
F + 
CD
F
= j
ˆ
2
1
2
R 4
GM
i
ˆ
2
1
2
R 4
GM
2
2
2
2
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
? ?
?F
C
= ? ? 1 2 2
R 4
GM
2
2
?
For moving along the circle, F
?
= 
R
mv
2
or ? ? 1 2 2
R 4
GM
2
2
? = 
R
MV
2
  or V = 
?
?
?
?
?
?
?
?
?
4
1 2 2
R
GM
6.
? ?
2
h R
GM
?
= 
6 2
22 11
10 ) 1000 1740 (
10 4 . 7 10 67 . 6
? ?
? ? ?
?
= 
6
11
10 2740 2740
10 358 . 49
? ?
?
= 
13
11
10 75 . 0
10 358 . 49
?
?
= 65.8 × 10
–2
= 0.65 m/s
2
7. The linear momentum of 2 bodies is 0 initially. Since gravitational force is internal, final momentum is 
also zero.
So (10 kg)v
1
= (20 kg) v
2
Or v
1
= v
2
…(1)
Since P.E. is conserved
Initial P.E. = 
1
20 10 10 67 . 6
11
? ? ? ?
?
= –13.34×10
–9
J
When separation is 0.5 m,
M
B
C
A
D
B
C
A
R
Chapter 11
11.3
–13.34 × 10
–9
+ 0 = 
) 2 / 1 (
10 34 . 13
9 ?
? ?
+ (1/2) × 10 v
1
2
+ (1/2) × 20 v
2
2
…(2)
? – 13.34 × 10
–9
= -26.68 ×10
–9
+ 5 v
1
2
+ 10 v
2
2
? – 13.34 × 10
–9
= -26.68 ×10
–9
+ 30 v
2
2
? v
2
2
= 
30
10 34 . 13
9 ?
?
= 4.44 × 10
–10
? v
2
= 2.1 × 10
–5
m/s.
So, v
1
= 4.2 × 10
–5
m/s.
8. In the semicircle, we can consider, a small element of d ? then R d ? = (M/L) R d ? = dM.
F = 
2
LR
m GMRd ?
dF
3
= 2 dF since = 
LR
GMm 2
sin ? d ???
?F = 
?
?
? ?
2 /
0
d sin
LR
GMm 2
??? ? ?
2 /
0
cos
LR
GMm 2
?
? ? ?
? = –2 ) 1 (
LR
GMm
? = 
LR
GMm 2
= 
A / L L
GMm 2
?
= 
2
L
GMm 2 ?
9. A small section of rod is considered at ‘x’ distance mass of the element = (M/L). dx = dm
dE
1
= 
? ?
2 2
x d
1 ) dm ( G
?
?
= dE
2
Resultant dE = 2 dE
1
sin ?
= 2 × 
? ?
? ?
2 2
2 2
x d
d
x d
) dm ( G
?
?
?
= 
? ? ? ?
?
?
?
?
?
?
? ?
? ?
2 2 2 2
x d x d L
dx d GM 2
Total gravitational field 
E = 
? ?
?
?
2 / L
0
2 / 3
2 2
x d L
dx Gmd 2
Integrating the above equation it can be found that, 
E = 
2 2
d 4 L d
GM 2
?
?
10. The gravitational force on ‘m’ due to the shell of M
2
is 0.
M is at a distance 
2
R R
2 1
?
Then the gravitational force due to M is given by 
= 
2 / 2 1
1
R R (
m GM
?
= 
2
2 1
1
) R R (
m GM 4
?
11. Man of earth M = (4/3) ?R
3
?
Man of the imaginary sphere, having 
Radius = x, M ? = (4/3) ?x
3
?
or 
M
M ?
= 
3
3
R
x
? Gravitational force on F = 
2
m
m M G ?
or F = 
2 3
3
x R
m GMx
= 
3
R
GMmx
?
R
? ?
? ?
M
d ? ? d ? ?
m
L
a
dx
x
dE 1
O
M
? ?
d
dE 2
x
? ?
m
R 1
M 1
R 2
m 2
x
m
Page 4


11.1
SOLUTIONS TO CONCEPTS 
CHAPTER 11
1. Gravitational force of attraction,
F = 
2
r
GMm
= 
2
11
) 1 . 0 (
10 10 10 67 . 6 ? ? ?
?
= 6.67 × 10
–7
N
2. To calculate the gravitational force on ‘m’ at unline due to other mouse.
OD
F = 
2 2
) r / a (
m 4 m G ? ?
=  
2
2
a
Gm 8
OI
F = 
2 2
) r / a (
m 2 m G ? ?
=  
2
2
a
Gm 6
OB
F = 
2 2
) r / a (
m 2 m G ? ?
=  
2
2
a
Gm 4
OA
F = 
2 2
) r / a (
m m G ? ?
=  
2
2
a
Gm 2
Resultant 
OF
F = 
2
2
2
2
2
2
a
Gm
36
a
Gm
64
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= 10
2
2
a
Gm
Resultant 
OE
F = 
2
2
2
2
2
2
a
Gm
4
a
Gm
64
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= 
2
2
a
Gm
5 2
The net resultant force will be,
F = 5 20
a
Gm
2
a
Gm
20
a
Gm
100
2
2
2
2
2
2
2
2
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= ? ? 5 40 120
a
Gm
2
2
2
?
?
?
?
?
?
?
?
?
= ) 6 . 89 120 (
a
Gm
2
2
2
?
?
?
?
?
?
?
?
?
= 4 . 40
a
Gm
2
2
= 
2
2
a
Gm
2 4
3. a) if ‘m’ is placed at mid point of a side 
then 
OA
F = 
2
2
a
Gm 4
in OA direction
OB
F  = 
2
2
a
Gm 4
in OB direction
Since equal & opposite cancel each other
oc
F = 
? ? ? ?
2
3
2
a 2 / r
Gm
= 
2
2
a 3
Gm 4
in OC direction
Net gravitational force on m = 
2
2
a
Gm 4
b) If placed at O (centroid)
the 
OA
F = 
) r / a (
Gm
3
2
= 
2
2
a
Gm 3
F
D
B A
E
C
2m m
4m
3m
m
B
A
C
m
m
m m
O
B
A
C
m
m
m m
O
Chapter 11
11.2
OB
F = 
2
2
a
Gm 3
Resultant F
?
= 
2
1
a
Gm 3
2
a
Gm 3
2
2
2
2
2
2
2
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= 
2
2
a
Gm 3
Since 
OC
F = 
2
2
a
Gm 3
, equal & opposite to F, cancel 
Net gravitational force = 0
4.
CB
F = j
ˆ
60 sin
a 4
Gm
i
ˆ
60 cos
a 4
Gm
2
2
2
2
?
CA
F = j
ˆ
60 sin
a 4
Gm
i
ˆ
60 cos
a 4
Gm
2
2
2
2
?
?
F
?
= 
CB
F + 
CA
F
= j
ˆ
60 sin
a 4
Gm 2
2
2
?
= 
2
r
a 4
Gm 2
3
2
2
?
= 
2
2
3
a 4
Gm r
5. Force on M at C due to gravitational attraction.
CB
F = j
ˆ
R 2
Gm
2
2
CD
F = i
ˆ
R 4
GM
2
2
?
CA
F = j
ˆ
45 sin
R 4
GM
j
ˆ
45 cos
R 4
GM
2
2
2
2
?
?
So, resultant force on C,
?
C
F =  
CA
F + 
CB
F + 
CD
F
= j
ˆ
2
1
2
R 4
GM
i
ˆ
2
1
2
R 4
GM
2
2
2
2
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
? ?
?F
C
= ? ? 1 2 2
R 4
GM
2
2
?
For moving along the circle, F
?
= 
R
mv
2
or ? ? 1 2 2
R 4
GM
2
2
? = 
R
MV
2
  or V = 
?
?
?
?
?
?
?
?
?
4
1 2 2
R
GM
6.
? ?
2
h R
GM
?
= 
6 2
22 11
10 ) 1000 1740 (
10 4 . 7 10 67 . 6
? ?
? ? ?
?
= 
6
11
10 2740 2740
10 358 . 49
? ?
?
= 
13
11
10 75 . 0
10 358 . 49
?
?
= 65.8 × 10
–2
= 0.65 m/s
2
7. The linear momentum of 2 bodies is 0 initially. Since gravitational force is internal, final momentum is 
also zero.
So (10 kg)v
1
= (20 kg) v
2
Or v
1
= v
2
…(1)
Since P.E. is conserved
Initial P.E. = 
1
20 10 10 67 . 6
11
? ? ? ?
?
= –13.34×10
–9
J
When separation is 0.5 m,
M
B
C
A
D
B
C
A
R
Chapter 11
11.3
–13.34 × 10
–9
+ 0 = 
) 2 / 1 (
10 34 . 13
9 ?
? ?
+ (1/2) × 10 v
1
2
+ (1/2) × 20 v
2
2
…(2)
? – 13.34 × 10
–9
= -26.68 ×10
–9
+ 5 v
1
2
+ 10 v
2
2
? – 13.34 × 10
–9
= -26.68 ×10
–9
+ 30 v
2
2
? v
2
2
= 
30
10 34 . 13
9 ?
?
= 4.44 × 10
–10
? v
2
= 2.1 × 10
–5
m/s.
So, v
1
= 4.2 × 10
–5
m/s.
8. In the semicircle, we can consider, a small element of d ? then R d ? = (M/L) R d ? = dM.
F = 
2
LR
m GMRd ?
dF
3
= 2 dF since = 
LR
GMm 2
sin ? d ???
?F = 
?
?
? ?
2 /
0
d sin
LR
GMm 2
??? ? ?
2 /
0
cos
LR
GMm 2
?
? ? ?
? = –2 ) 1 (
LR
GMm
? = 
LR
GMm 2
= 
A / L L
GMm 2
?
= 
2
L
GMm 2 ?
9. A small section of rod is considered at ‘x’ distance mass of the element = (M/L). dx = dm
dE
1
= 
? ?
2 2
x d
1 ) dm ( G
?
?
= dE
2
Resultant dE = 2 dE
1
sin ?
= 2 × 
? ?
? ?
2 2
2 2
x d
d
x d
) dm ( G
?
?
?
= 
? ? ? ?
?
?
?
?
?
?
? ?
? ?
2 2 2 2
x d x d L
dx d GM 2
Total gravitational field 
E = 
? ?
?
?
2 / L
0
2 / 3
2 2
x d L
dx Gmd 2
Integrating the above equation it can be found that, 
E = 
2 2
d 4 L d
GM 2
?
?
10. The gravitational force on ‘m’ due to the shell of M
2
is 0.
M is at a distance 
2
R R
2 1
?
Then the gravitational force due to M is given by 
= 
2 / 2 1
1
R R (
m GM
?
= 
2
2 1
1
) R R (
m GM 4
?
11. Man of earth M = (4/3) ?R
3
?
Man of the imaginary sphere, having 
Radius = x, M ? = (4/3) ?x
3
?
or 
M
M ?
= 
3
3
R
x
? Gravitational force on F = 
2
m
m M G ?
or F = 
2 3
3
x R
m GMx
= 
3
R
GMmx
?
R
? ?
? ?
M
d ? ? d ? ?
m
L
a
dx
x
dE 1
O
M
? ?
d
dE 2
x
? ?
m
R 1
M 1
R 2
m 2
x
m
Chapter 11
11.4
12. Let d be the distance from centre of earth to man ‘m’ then 
D = 
?
?
?
?
?
?
?
?
?
4
R
x
2
2
= (1/2)  
2 2
R x 4 ?
M be the mass of the earth, M ? the mass of the sphere of radius d/2.
Then M = (4/3) ?R
3
?
M ? = (4/3) ?d
3
?
or 
M
M ?
= 
3
3
R
d
? Gravitational force is m,
F = 
2
d
m m G ?
= 
2 3
3
d R
Mm Gd
= 
3
R
GMmd
So, Normal force exerted by the wall = F cos ?.
= 
d 2
R
R
GMmd
3
? = 
2
R 2
GMm
      (therefore I think normal force does not depend on x) ?
13. a) m ? is placed at a distance x from ‘O’.
If r < x , 2r, Let’s consider a thin shell of man
dm = 
3
2
x
3
4
r ) 3 / 4 (
m
? ?
?
= 
3
3
r
mx
Thus 
?
dm = 
3
3
r
mx
Then gravitational force F = 
2
x
m md G
= 
2
3 3
x
r / Gmx
= ?
3
r
Gmx
b) 2r < x < 2R, then F is due to only the sphere.
? F = 
? ?
2
r x
m Gm
?
?
c) if x > 2R, then Gravitational force is due to both sphere & shell, then due to shell,
F = 
? ?
2
R x
m GM
?
?
due to the sphere = 
? ?
2
r x
m Gm
?
?
So, Resultant force = 
? ?
2
r x
m Gm
?
?
+ 
? ?
2
R x
m GM
?
?
14. At P
1
, Gravitational field due to sphere M = 
? ?
2
a a 3
GM
?
= 
2
a 16
GM
At P
2
, Gravitational field is due to sphere & shell,
= 
2
) a a 4 a (
GM
? ?
+ 
2
) a a 4 (
GM
?
= ?
?
?
?
?
?
?
25
1
36
1
a
GM
2
= 
2
a
GM
900
61
?
?
?
?
?
?
15. We know in the thin spherical shell of uniform density has gravitational field at its internal point is zero.
At A and B point, field is equal and opposite and cancel each other so Net field is 
zero.
Hence, E
A
= E
B
16. Let 0.1 kg man is x m from 2kg mass and (2 – x) m from 4 kg mass.
?
2
x
1 . 0 2 ?
= –
2
) x 2 (
1 . 0 4
?
?
R/2
m
O
d
x
? ?
x
n
F
R/2
? ?
d
m
R
O
M
r
49
P 1
a
P 2
a
a
B
A
B
A
Page 5


11.1
SOLUTIONS TO CONCEPTS 
CHAPTER 11
1. Gravitational force of attraction,
F = 
2
r
GMm
= 
2
11
) 1 . 0 (
10 10 10 67 . 6 ? ? ?
?
= 6.67 × 10
–7
N
2. To calculate the gravitational force on ‘m’ at unline due to other mouse.
OD
F = 
2 2
) r / a (
m 4 m G ? ?
=  
2
2
a
Gm 8
OI
F = 
2 2
) r / a (
m 2 m G ? ?
=  
2
2
a
Gm 6
OB
F = 
2 2
) r / a (
m 2 m G ? ?
=  
2
2
a
Gm 4
OA
F = 
2 2
) r / a (
m m G ? ?
=  
2
2
a
Gm 2
Resultant 
OF
F = 
2
2
2
2
2
2
a
Gm
36
a
Gm
64
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= 10
2
2
a
Gm
Resultant 
OE
F = 
2
2
2
2
2
2
a
Gm
4
a
Gm
64
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= 
2
2
a
Gm
5 2
The net resultant force will be,
F = 5 20
a
Gm
2
a
Gm
20
a
Gm
100
2
2
2
2
2
2
2
2
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= ? ? 5 40 120
a
Gm
2
2
2
?
?
?
?
?
?
?
?
?
= ) 6 . 89 120 (
a
Gm
2
2
2
?
?
?
?
?
?
?
?
?
= 4 . 40
a
Gm
2
2
= 
2
2
a
Gm
2 4
3. a) if ‘m’ is placed at mid point of a side 
then 
OA
F = 
2
2
a
Gm 4
in OA direction
OB
F  = 
2
2
a
Gm 4
in OB direction
Since equal & opposite cancel each other
oc
F = 
? ? ? ?
2
3
2
a 2 / r
Gm
= 
2
2
a 3
Gm 4
in OC direction
Net gravitational force on m = 
2
2
a
Gm 4
b) If placed at O (centroid)
the 
OA
F = 
) r / a (
Gm
3
2
= 
2
2
a
Gm 3
F
D
B A
E
C
2m m
4m
3m
m
B
A
C
m
m
m m
O
B
A
C
m
m
m m
O
Chapter 11
11.2
OB
F = 
2
2
a
Gm 3
Resultant F
?
= 
2
1
a
Gm 3
2
a
Gm 3
2
2
2
2
2
2
2
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= 
2
2
a
Gm 3
Since 
OC
F = 
2
2
a
Gm 3
, equal & opposite to F, cancel 
Net gravitational force = 0
4.
CB
F = j
ˆ
60 sin
a 4
Gm
i
ˆ
60 cos
a 4
Gm
2
2
2
2
?
CA
F = j
ˆ
60 sin
a 4
Gm
i
ˆ
60 cos
a 4
Gm
2
2
2
2
?
?
F
?
= 
CB
F + 
CA
F
= j
ˆ
60 sin
a 4
Gm 2
2
2
?
= 
2
r
a 4
Gm 2
3
2
2
?
= 
2
2
3
a 4
Gm r
5. Force on M at C due to gravitational attraction.
CB
F = j
ˆ
R 2
Gm
2
2
CD
F = i
ˆ
R 4
GM
2
2
?
CA
F = j
ˆ
45 sin
R 4
GM
j
ˆ
45 cos
R 4
GM
2
2
2
2
?
?
So, resultant force on C,
?
C
F =  
CA
F + 
CB
F + 
CD
F
= j
ˆ
2
1
2
R 4
GM
i
ˆ
2
1
2
R 4
GM
2
2
2
2
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
? ?
?F
C
= ? ? 1 2 2
R 4
GM
2
2
?
For moving along the circle, F
?
= 
R
mv
2
or ? ? 1 2 2
R 4
GM
2
2
? = 
R
MV
2
  or V = 
?
?
?
?
?
?
?
?
?
4
1 2 2
R
GM
6.
? ?
2
h R
GM
?
= 
6 2
22 11
10 ) 1000 1740 (
10 4 . 7 10 67 . 6
? ?
? ? ?
?
= 
6
11
10 2740 2740
10 358 . 49
? ?
?
= 
13
11
10 75 . 0
10 358 . 49
?
?
= 65.8 × 10
–2
= 0.65 m/s
2
7. The linear momentum of 2 bodies is 0 initially. Since gravitational force is internal, final momentum is 
also zero.
So (10 kg)v
1
= (20 kg) v
2
Or v
1
= v
2
…(1)
Since P.E. is conserved
Initial P.E. = 
1
20 10 10 67 . 6
11
? ? ? ?
?
= –13.34×10
–9
J
When separation is 0.5 m,
M
B
C
A
D
B
C
A
R
Chapter 11
11.3
–13.34 × 10
–9
+ 0 = 
) 2 / 1 (
10 34 . 13
9 ?
? ?
+ (1/2) × 10 v
1
2
+ (1/2) × 20 v
2
2
…(2)
? – 13.34 × 10
–9
= -26.68 ×10
–9
+ 5 v
1
2
+ 10 v
2
2
? – 13.34 × 10
–9
= -26.68 ×10
–9
+ 30 v
2
2
? v
2
2
= 
30
10 34 . 13
9 ?
?
= 4.44 × 10
–10
? v
2
= 2.1 × 10
–5
m/s.
So, v
1
= 4.2 × 10
–5
m/s.
8. In the semicircle, we can consider, a small element of d ? then R d ? = (M/L) R d ? = dM.
F = 
2
LR
m GMRd ?
dF
3
= 2 dF since = 
LR
GMm 2
sin ? d ???
?F = 
?
?
? ?
2 /
0
d sin
LR
GMm 2
??? ? ?
2 /
0
cos
LR
GMm 2
?
? ? ?
? = –2 ) 1 (
LR
GMm
? = 
LR
GMm 2
= 
A / L L
GMm 2
?
= 
2
L
GMm 2 ?
9. A small section of rod is considered at ‘x’ distance mass of the element = (M/L). dx = dm
dE
1
= 
? ?
2 2
x d
1 ) dm ( G
?
?
= dE
2
Resultant dE = 2 dE
1
sin ?
= 2 × 
? ?
? ?
2 2
2 2
x d
d
x d
) dm ( G
?
?
?
= 
? ? ? ?
?
?
?
?
?
?
? ?
? ?
2 2 2 2
x d x d L
dx d GM 2
Total gravitational field 
E = 
? ?
?
?
2 / L
0
2 / 3
2 2
x d L
dx Gmd 2
Integrating the above equation it can be found that, 
E = 
2 2
d 4 L d
GM 2
?
?
10. The gravitational force on ‘m’ due to the shell of M
2
is 0.
M is at a distance 
2
R R
2 1
?
Then the gravitational force due to M is given by 
= 
2 / 2 1
1
R R (
m GM
?
= 
2
2 1
1
) R R (
m GM 4
?
11. Man of earth M = (4/3) ?R
3
?
Man of the imaginary sphere, having 
Radius = x, M ? = (4/3) ?x
3
?
or 
M
M ?
= 
3
3
R
x
? Gravitational force on F = 
2
m
m M G ?
or F = 
2 3
3
x R
m GMx
= 
3
R
GMmx
?
R
? ?
? ?
M
d ? ? d ? ?
m
L
a
dx
x
dE 1
O
M
? ?
d
dE 2
x
? ?
m
R 1
M 1
R 2
m 2
x
m
Chapter 11
11.4
12. Let d be the distance from centre of earth to man ‘m’ then 
D = 
?
?
?
?
?
?
?
?
?
4
R
x
2
2
= (1/2)  
2 2
R x 4 ?
M be the mass of the earth, M ? the mass of the sphere of radius d/2.
Then M = (4/3) ?R
3
?
M ? = (4/3) ?d
3
?
or 
M
M ?
= 
3
3
R
d
? Gravitational force is m,
F = 
2
d
m m G ?
= 
2 3
3
d R
Mm Gd
= 
3
R
GMmd
So, Normal force exerted by the wall = F cos ?.
= 
d 2
R
R
GMmd
3
? = 
2
R 2
GMm
      (therefore I think normal force does not depend on x) ?
13. a) m ? is placed at a distance x from ‘O’.
If r < x , 2r, Let’s consider a thin shell of man
dm = 
3
2
x
3
4
r ) 3 / 4 (
m
? ?
?
= 
3
3
r
mx
Thus 
?
dm = 
3
3
r
mx
Then gravitational force F = 
2
x
m md G
= 
2
3 3
x
r / Gmx
= ?
3
r
Gmx
b) 2r < x < 2R, then F is due to only the sphere.
? F = 
? ?
2
r x
m Gm
?
?
c) if x > 2R, then Gravitational force is due to both sphere & shell, then due to shell,
F = 
? ?
2
R x
m GM
?
?
due to the sphere = 
? ?
2
r x
m Gm
?
?
So, Resultant force = 
? ?
2
r x
m Gm
?
?
+ 
? ?
2
R x
m GM
?
?
14. At P
1
, Gravitational field due to sphere M = 
? ?
2
a a 3
GM
?
= 
2
a 16
GM
At P
2
, Gravitational field is due to sphere & shell,
= 
2
) a a 4 a (
GM
? ?
+ 
2
) a a 4 (
GM
?
= ?
?
?
?
?
?
?
25
1
36
1
a
GM
2
= 
2
a
GM
900
61
?
?
?
?
?
?
15. We know in the thin spherical shell of uniform density has gravitational field at its internal point is zero.
At A and B point, field is equal and opposite and cancel each other so Net field is 
zero.
Hence, E
A
= E
B
16. Let 0.1 kg man is x m from 2kg mass and (2 – x) m from 4 kg mass.
?
2
x
1 . 0 2 ?
= –
2
) x 2 (
1 . 0 4
?
?
R/2
m
O
d
x
? ?
x
n
F
R/2
? ?
d
m
R
O
M
r
49
P 1
a
P 2
a
a
B
A
B
A
Chapter 11
11.5
or 
2
x
2 . 0
= –
2
) x 2 (
4 . 0
?
or 
2
x
1
= 
2
) x 2 (
2
?
or (2 – x)
2
= 2 x
2
or 2 – x = 2 x or  x(r
2
+ 1) = 2
or x = 
414 . 2
2
= 0.83 m from 2kg mass.
17. Initially, the ride of ? is a 
To increase it to 2a,
work done = 
a
Gm
a 2
Gm
2 2
? = 
a 2
Gm 3
2
18. Work done against gravitational force to take away the particle from sphere,
= 
1 . 0 1 . 0
1 . 0 10 G
?
? ?
= 
1
11
10 1
1 10 67 . 6
?
?
?
? ?
= 6.67 × 10
–10
J
19. E
?
= (5 N/kg) i
ˆ
+ (12 N/kg) j
ˆ
a) F
?
= E
?
m
= 2kg [(5 N/kg) i
ˆ
+ (12 N/kg) j
ˆ
] = (10 N) i
ˆ
+ (12 N) j
ˆ
F
?
= 576 100 ? = 26 N
b) V
?
= E
?
r
At (12 m, 0), V
?
= – (60 J/kg) i
ˆ
V
?
= 60 J
At (0, 5 m), V
?
= – (60 J/kg) j
ˆ
V
?
= – 60 J
c) ? V
?
= 
?
) 5 , 2 , 1 (
) 0 , 0 (
mdr E
?
= ? ? ? ?
) 5 , 12 (
) 0 , 0 (
r j
ˆ
) N 24 ( i
ˆ
) N 10 ( ?
= – (120 J i
ˆ
+ 120 J i
ˆ
) = 240 J
d) ? v = – ? ?
? ?
? ? m 5 , 0
0 , m 12
) j
ˆ
N 24 i
ˆ
N 10 ( r ?
= –120 j
ˆ
+ 120 i
ˆ
= 0  
20. a) V = (20 N/kg) (x + y)
R
GM
= L
M
MLT
2 ?
or 
L
M T L M
1 2 3 1 ? ?
= 
M
T ML
2 2 ?
Or M
0
L
2
T
–2
= M
0
L
2
T
–2
? L.H.S = R.H.S 
b) 
) y , x (
E
?
= – 20(N/kg) i
ˆ
– 20(N/kg) j
ˆ
c) F
?
= E
?
m
= 0.5kg [– (20 N/kg) i
ˆ
– (20 N/kg) j
ˆ
= – 10N i
ˆ
- 10 N j
ˆ
? | F |
?
= 100 100 ? = 10 2 N
21. E
?
= 2 i
ˆ
+ 3 j
ˆ
The field is represented as 
tan ?
1
= 3/2
Again the line 3y + 2x = 5 can be represented as 
tan ?
2
= – 2/3
m
1
m
2
= –1
Since, the direction of field and the displacement are perpendicular, is done by the particle on the line.
m
m
m
a
a
a
100g
10cm
10kg
? ? ? ?
2j
3j
5/2
5/3
? 2
Read More
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