NEET Exam > NEET Notes > Physics Class 11 > HC Verma Solutions: Chapter 11 - Gravitation
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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
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FAQs on HC Verma Solutions: Chapter 11 - Gravitation - Physics Class 11 - NEET
| 1. What is the formula for gravitational force between two objects? |  |
Ans. The formula for gravitational force between two objects is given by Newton's law of universal gravitation: F = G * (m1 * m2) / r^2, where F is the gravitational force, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between their centers.
| 2. How does the gravitational force between two objects change with distance? | |
Ans. The gravitational force between two objects decreases as the distance between them increases. According to the inverse square law, the force is inversely proportional to the square of the distance. Therefore, if the distance is doubled, the force becomes one-fourth of its original value.
| 3. What is the difference between mass and weight? | |
Ans. Mass is the amount of matter in an object and is constant regardless of the location, while weight is the force of gravity acting on an object and varies with the location. Mass is measured in kilograms (kg), while weight is measured in newtons (N).
| 4. How does the value of the gravitational constant affect the strength of gravitational force? | |
Ans. The gravitational constant (G) determines the strength of the gravitational force between two objects. A higher value of G leads to a stronger force, while a lower value of G leads to a weaker force. The value of G is approximately 6.67430 × 10^-11 N m^2/kg^2.
| 5. Does the gravitational force between two objects depend on their sizes? | |
Ans. No, the gravitational force between two objects does not depend on their sizes. It only depends on the masses of the objects and the distance between their centers. The size or shape of the objects does not affect the gravitational force.
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