Chapter 2 : Physics and Mathematics - HC Verma Solution, Physics Class 11 Notes | EduRev

Physics Class 12

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JEE : Chapter 2 : Physics and Mathematics - HC Verma Solution, Physics Class 11 Notes | EduRev

 Page 1


2.1
SOLUTIONS TO CONCEPTS 
CHAPTER – 2
1. As shown in the figure, 
The angle between A
?
and B
?
= 110° – 20° = 90°
| A |
?
= 3 and | B |
?
= 4m
Resultant R = ? ? ? cos AB 2 B A
2 2
= 5 m
Let ? be the angle between R
?
and A
?
? = ?
?
?
?
?
?
? ?
?
?
90 cos 4 3
90 sin 4
tan
1
= tan
–1
(4/3) =  53°
? Resultant vector makes angle (53° + 20°) = 73° with x-axis.
2. Angle between A
?
and B
?
is ? = 60° – 30° =30°
| A
?
| and | B
?
| = 10 unit
R = 
2 2
10 10 2.10.10.cos30 ? ? ? = 19.3
? be the angle between R
?
and A
?
? = tan
–1 1
10sin30 1
tan
10 10cos30
2 3
?
? ? ? ? ?
?
? ?
? ?
? ? ? ? ? ? ?
= tan
–1
(0.26795) = 15°
? Resultant makes 15° + 30° = 45° angle with x-axis.
?
3. x component of A
?
= 100 cos 45° = 2 / 100 unit
x component of B
?
= 100 cos 135° = 2 / 100
x component of C
?
= 100 cos 315° = 2 / 100
Resultant x component = 2 / 100 – 2 / 100 + 2 / 100 = 2 / 100
y component of A
?
= 100 sin 45° = 2 / 100 unit
y component of B
?
= 100 sin 135° = 2 / 100
y component of C
?
= 100 sin 315° = – 2 / 100
Resultant y component = 2 / 100 + 2 / 100 – 2 / 100 = 2 / 100
Resultant = 100
Tan ? = 
component x 
component y 
= 1
? ? = tan
–1
(1) = 45°
The resultant is 100 unit at 45° with x-axis.
4. j 3 i 4 a
? ?
?
? ? , j 4 i 3 b
? ? ?
? ?
a) 
2 2
3 4 | a | ? ?
?
= 5
b) 16 9 | b | ? ?
?
= 5
c) 2 7 | j 7 i 7 | | b a | ? ? ? ?
? ? ?
?
d) 
ˆ ˆ ˆ ˆ
a b ( 3 4)i ( 4 3)j i j ? ? ? ? ? ? ? ? ?
?
?
2 2
| a b | 1 ( 1) 2 ? ? ? ? ?
?
?
.
x
y
? ?
R
?
B
?
A
?
20
x
y
? ?
B
?
A
?
30°
60°
315°
45°
135°
Page 2


2.1
SOLUTIONS TO CONCEPTS 
CHAPTER – 2
1. As shown in the figure, 
The angle between A
?
and B
?
= 110° – 20° = 90°
| A |
?
= 3 and | B |
?
= 4m
Resultant R = ? ? ? cos AB 2 B A
2 2
= 5 m
Let ? be the angle between R
?
and A
?
? = ?
?
?
?
?
?
? ?
?
?
90 cos 4 3
90 sin 4
tan
1
= tan
–1
(4/3) =  53°
? Resultant vector makes angle (53° + 20°) = 73° with x-axis.
2. Angle between A
?
and B
?
is ? = 60° – 30° =30°
| A
?
| and | B
?
| = 10 unit
R = 
2 2
10 10 2.10.10.cos30 ? ? ? = 19.3
? be the angle between R
?
and A
?
? = tan
–1 1
10sin30 1
tan
10 10cos30
2 3
?
? ? ? ? ?
?
? ?
? ?
? ? ? ? ? ? ?
= tan
–1
(0.26795) = 15°
? Resultant makes 15° + 30° = 45° angle with x-axis.
?
3. x component of A
?
= 100 cos 45° = 2 / 100 unit
x component of B
?
= 100 cos 135° = 2 / 100
x component of C
?
= 100 cos 315° = 2 / 100
Resultant x component = 2 / 100 – 2 / 100 + 2 / 100 = 2 / 100
y component of A
?
= 100 sin 45° = 2 / 100 unit
y component of B
?
= 100 sin 135° = 2 / 100
y component of C
?
= 100 sin 315° = – 2 / 100
Resultant y component = 2 / 100 + 2 / 100 – 2 / 100 = 2 / 100
Resultant = 100
Tan ? = 
component x 
component y 
= 1
? ? = tan
–1
(1) = 45°
The resultant is 100 unit at 45° with x-axis.
4. j 3 i 4 a
? ?
?
? ? , j 4 i 3 b
? ? ?
? ?
a) 
2 2
3 4 | a | ? ?
?
= 5
b) 16 9 | b | ? ?
?
= 5
c) 2 7 | j 7 i 7 | | b a | ? ? ? ?
? ? ?
?
d) 
ˆ ˆ ˆ ˆ
a b ( 3 4)i ( 4 3)j i j ? ? ? ? ? ? ? ? ?
?
?
2 2
| a b | 1 ( 1) 2 ? ? ? ? ?
?
?
.
x
y
? ?
R
?
B
?
A
?
20
x
y
? ?
B
?
A
?
30°
60°
315°
45°
135°
Chapter-2
2.2
5. x component of OA = 2cos30° = 3
x component of BC = 1.5 cos 120° = –0.75
x component of DE = 1 cos 270° = 0
y component of OA = 2 sin 30° = 1
y component of BC = 1.5 sin 120° = 1.3
y component of DE = 1 sin 270° = –1
R
x
= x component of resultant = 0 75 . 0 3 ? ? = 0.98 m
R
y
= resultant y component = 1 + 1.3 – 1 = 1.3 m
So, R = Resultant = 1.6 m
If it makes and angle ? with positive x-axis
Tan ? = 
component x 
component y 
= 1.32
? ? = tan
–1
1.32
?
6. | a |
?
= 3m | b |
?
= 4
a) If R = 1 unit ? ? ? ? cos . 4 . 3 . 2 4 3
2 2
= 1
? = 180°
b) ? ? ? cos . 4 . 3 . 2 4 3
2 2
= 5
? = 90°
c) ? ? ? cos . 4 . 3 . 2 4 3
2 2
= 7
? = 0°
Angle between them is 0°.
?
7. K
ˆ
4 J
ˆ
5 . 0 i
ˆ
2 AD ? ? ? = 
ˆ ˆ
6i 0.5j ?
AD = 
2 2
DE AE ? = 6.02 KM
Tan ? = DE / AE = 1/12
??= tan
–1
(1/12)
The displacement of the car is 6.02 km along the distance tan
–1
(1/12) with positive x-axis.
8. In ?ABC, tan ? = x/2 and in ?DCE, tan ? = (2 – x)/4 tan ? = (x/2) = (2 – x)/4 = 4x
? 4 – 2x = 4x
? 6x = 4 ? x = 2/3 ft
a) In ?ABC, AC = 
2 2
BC AB ? = 
2
10
3
ft
b) In ?CDE, DE = 1 – (2/3) = 4/3 ft
CD = 4 ft. So, CE = 
2 2
DE CD ? = 
4
10
3
ft
c) In ?AGE, AE = 
2 2
GE AG ? = 2 2 ft.
9. Here the displacement vector k
ˆ
3 j
ˆ
4 i
ˆ
7 r ? ? ?
?
a) magnitude of displacement = 74 ft
b) the components of the displacement vector are 7 ft, 4 ft and 3 ft.
2m
D
A
E
B
x
O 
y
1m
1.5m
90° 
30° 60°
6m
E  A
D 
B
0.5 km
? ?
C 
2m
4m
0.5 km
2–x 
G   
A
D 
B
BC = 2 ft
AF = 2 ft
DE = 2x
x 
C 
E    
F    
r   
z
Y 
Page 3


2.1
SOLUTIONS TO CONCEPTS 
CHAPTER – 2
1. As shown in the figure, 
The angle between A
?
and B
?
= 110° – 20° = 90°
| A |
?
= 3 and | B |
?
= 4m
Resultant R = ? ? ? cos AB 2 B A
2 2
= 5 m
Let ? be the angle between R
?
and A
?
? = ?
?
?
?
?
?
? ?
?
?
90 cos 4 3
90 sin 4
tan
1
= tan
–1
(4/3) =  53°
? Resultant vector makes angle (53° + 20°) = 73° with x-axis.
2. Angle between A
?
and B
?
is ? = 60° – 30° =30°
| A
?
| and | B
?
| = 10 unit
R = 
2 2
10 10 2.10.10.cos30 ? ? ? = 19.3
? be the angle between R
?
and A
?
? = tan
–1 1
10sin30 1
tan
10 10cos30
2 3
?
? ? ? ? ?
?
? ?
? ?
? ? ? ? ? ? ?
= tan
–1
(0.26795) = 15°
? Resultant makes 15° + 30° = 45° angle with x-axis.
?
3. x component of A
?
= 100 cos 45° = 2 / 100 unit
x component of B
?
= 100 cos 135° = 2 / 100
x component of C
?
= 100 cos 315° = 2 / 100
Resultant x component = 2 / 100 – 2 / 100 + 2 / 100 = 2 / 100
y component of A
?
= 100 sin 45° = 2 / 100 unit
y component of B
?
= 100 sin 135° = 2 / 100
y component of C
?
= 100 sin 315° = – 2 / 100
Resultant y component = 2 / 100 + 2 / 100 – 2 / 100 = 2 / 100
Resultant = 100
Tan ? = 
component x 
component y 
= 1
? ? = tan
–1
(1) = 45°
The resultant is 100 unit at 45° with x-axis.
4. j 3 i 4 a
? ?
?
? ? , j 4 i 3 b
? ? ?
? ?
a) 
2 2
3 4 | a | ? ?
?
= 5
b) 16 9 | b | ? ?
?
= 5
c) 2 7 | j 7 i 7 | | b a | ? ? ? ?
? ? ?
?
d) 
ˆ ˆ ˆ ˆ
a b ( 3 4)i ( 4 3)j i j ? ? ? ? ? ? ? ? ?
?
?
2 2
| a b | 1 ( 1) 2 ? ? ? ? ?
?
?
.
x
y
? ?
R
?
B
?
A
?
20
x
y
? ?
B
?
A
?
30°
60°
315°
45°
135°
Chapter-2
2.2
5. x component of OA = 2cos30° = 3
x component of BC = 1.5 cos 120° = –0.75
x component of DE = 1 cos 270° = 0
y component of OA = 2 sin 30° = 1
y component of BC = 1.5 sin 120° = 1.3
y component of DE = 1 sin 270° = –1
R
x
= x component of resultant = 0 75 . 0 3 ? ? = 0.98 m
R
y
= resultant y component = 1 + 1.3 – 1 = 1.3 m
So, R = Resultant = 1.6 m
If it makes and angle ? with positive x-axis
Tan ? = 
component x 
component y 
= 1.32
? ? = tan
–1
1.32
?
6. | a |
?
= 3m | b |
?
= 4
a) If R = 1 unit ? ? ? ? cos . 4 . 3 . 2 4 3
2 2
= 1
? = 180°
b) ? ? ? cos . 4 . 3 . 2 4 3
2 2
= 5
? = 90°
c) ? ? ? cos . 4 . 3 . 2 4 3
2 2
= 7
? = 0°
Angle between them is 0°.
?
7. K
ˆ
4 J
ˆ
5 . 0 i
ˆ
2 AD ? ? ? = 
ˆ ˆ
6i 0.5j ?
AD = 
2 2
DE AE ? = 6.02 KM
Tan ? = DE / AE = 1/12
??= tan
–1
(1/12)
The displacement of the car is 6.02 km along the distance tan
–1
(1/12) with positive x-axis.
8. In ?ABC, tan ? = x/2 and in ?DCE, tan ? = (2 – x)/4 tan ? = (x/2) = (2 – x)/4 = 4x
? 4 – 2x = 4x
? 6x = 4 ? x = 2/3 ft
a) In ?ABC, AC = 
2 2
BC AB ? = 
2
10
3
ft
b) In ?CDE, DE = 1 – (2/3) = 4/3 ft
CD = 4 ft. So, CE = 
2 2
DE CD ? = 
4
10
3
ft
c) In ?AGE, AE = 
2 2
GE AG ? = 2 2 ft.
9. Here the displacement vector k
ˆ
3 j
ˆ
4 i
ˆ
7 r ? ? ?
?
a) magnitude of displacement = 74 ft
b) the components of the displacement vector are 7 ft, 4 ft and 3 ft.
2m
D
A
E
B
x
O 
y
1m
1.5m
90° 
30° 60°
6m
E  A
D 
B
0.5 km
? ?
C 
2m
4m
0.5 km
2–x 
G   
A
D 
B
BC = 2 ft
AF = 2 ft
DE = 2x
x 
C 
E    
F    
r   
z
Y 
Chapter-2
2.3
10. a
?
is a vector of magnitude 4.5 unit due north.
a) 3| a
?
| = 3 ? 4.5 = 13.5
3 a
?
is along north having magnitude 13.5 units.
b) –4| a
?
| = –4 ? 1.5 = –6 unit
–4 a
?
is a vector of magnitude 6 unit due south.
11. | a
?
| = 2 m, | b
?
| = 3 m
angle between them ? = 60°
a) ? ? ? ? 60 cos | b | | a | b a
?
?
?
?
= 2 ? 3 ? 1/2 = 3 m
2
b) ? ? ? ? 60 sin | b | | a | | b a |
?
?
?
?
= 2 ? 3 ? 3 / 2 = 3 3 m
2
.
12. We know that according to polygon law of vector addition, the resultant 
of these six vectors is zero.
Here A = B = C = D = E = F (magnitude)
So, Rx = A cos ? + A cos ?/3 + A cos 2 ?/3 + A cos 3 ?/3 + A cos 4 ?/4 + 
A cos 5 ?/5 = 0
[As resultant is zero. X component of resultant R
x
= 0]
= cos ? + cos ?/3 + cos 2 ?/3 + cos 3 ?/3 + cos 4 ?/3 + cos 5 ?/3 = 0
Note : Similarly it can be proved that, 
sin ? + sin ?/3 + sin 2 ?/3 + sin 3 ?/3 + sin 4 ?/3 + sin 5 ?/3 = 0 ?
13. a 2i 3 j 4k; b 3i 4 j 5k ? ? ? ? ? ?
? ? ? ? ? ? ?
?
? ? ? cos ab b a
?
?
?
ab
b a
cos
1
?
?
?
? ?
?
?
1 1
2 2 2 2 2 2
2 3 3 4 4 5 38
cos cos
1450
2 3 4 3 4 5
? ?
? ? ? ? ? ? ?
?
? ?
? ? ? ? ? ?
14. 0 ) B A ( A ? ? ?
? ? ?
(claim)
As, n
ˆ
sin AB B A ? ? ?
? ?
AB sin ? n
ˆ
is a vector which is perpendicular to the plane containing A
?
and B
?
, this implies that it is 
also perpendicular to A
?
. As dot product of two perpendicular vector is zero.
Thus 0 ) B A ( A ? ? ?
? ? ?
. ?
15.
ˆ ˆ ˆ
A 2i 3j 4k ? ? ?
?
, 
ˆ ˆ ˆ
B 4i 3j 2k ? ? ?
?
ˆ ˆ ˆ
i j k
A B 2 3 4
4 3 2
? ?
? ?
?
ˆ ˆ ˆ ˆ ˆ ˆ
i(6 12) j(4 16) k(6 12) 6i 12j 6k ? ? ? ? ? ? ? ? ? .
16. Given that A
?
, B
?
and C
?
are mutually perpendicular
A
?
× B
?
is a vector which direction is perpendicular to the plane containing A
?
and B
?
.
Also C
?
is perpendicular to A
?
and B
?
? Angle between C
?
and A
?
× B
?
is 0° or 180° (fig.1)
So, C
?
× ( A
?
× B
?
) = 0
The converse is not true.
For example, if two of the vector are parallel, (fig.2), then also 
C
?
× ( A
?
× B
?
) = 0
So, they need not be mutually perpendicular.
A 1
60° = ?/3 ?
A 2
A 3
A 4
A 5
A 6
(A B) ?
? ?
A
?
B
?
C
?
    
A
?
B
?
C
?
    
Page 4


2.1
SOLUTIONS TO CONCEPTS 
CHAPTER – 2
1. As shown in the figure, 
The angle between A
?
and B
?
= 110° – 20° = 90°
| A |
?
= 3 and | B |
?
= 4m
Resultant R = ? ? ? cos AB 2 B A
2 2
= 5 m
Let ? be the angle between R
?
and A
?
? = ?
?
?
?
?
?
? ?
?
?
90 cos 4 3
90 sin 4
tan
1
= tan
–1
(4/3) =  53°
? Resultant vector makes angle (53° + 20°) = 73° with x-axis.
2. Angle between A
?
and B
?
is ? = 60° – 30° =30°
| A
?
| and | B
?
| = 10 unit
R = 
2 2
10 10 2.10.10.cos30 ? ? ? = 19.3
? be the angle between R
?
and A
?
? = tan
–1 1
10sin30 1
tan
10 10cos30
2 3
?
? ? ? ? ?
?
? ?
? ?
? ? ? ? ? ? ?
= tan
–1
(0.26795) = 15°
? Resultant makes 15° + 30° = 45° angle with x-axis.
?
3. x component of A
?
= 100 cos 45° = 2 / 100 unit
x component of B
?
= 100 cos 135° = 2 / 100
x component of C
?
= 100 cos 315° = 2 / 100
Resultant x component = 2 / 100 – 2 / 100 + 2 / 100 = 2 / 100
y component of A
?
= 100 sin 45° = 2 / 100 unit
y component of B
?
= 100 sin 135° = 2 / 100
y component of C
?
= 100 sin 315° = – 2 / 100
Resultant y component = 2 / 100 + 2 / 100 – 2 / 100 = 2 / 100
Resultant = 100
Tan ? = 
component x 
component y 
= 1
? ? = tan
–1
(1) = 45°
The resultant is 100 unit at 45° with x-axis.
4. j 3 i 4 a
? ?
?
? ? , j 4 i 3 b
? ? ?
? ?
a) 
2 2
3 4 | a | ? ?
?
= 5
b) 16 9 | b | ? ?
?
= 5
c) 2 7 | j 7 i 7 | | b a | ? ? ? ?
? ? ?
?
d) 
ˆ ˆ ˆ ˆ
a b ( 3 4)i ( 4 3)j i j ? ? ? ? ? ? ? ? ?
?
?
2 2
| a b | 1 ( 1) 2 ? ? ? ? ?
?
?
.
x
y
? ?
R
?
B
?
A
?
20
x
y
? ?
B
?
A
?
30°
60°
315°
45°
135°
Chapter-2
2.2
5. x component of OA = 2cos30° = 3
x component of BC = 1.5 cos 120° = –0.75
x component of DE = 1 cos 270° = 0
y component of OA = 2 sin 30° = 1
y component of BC = 1.5 sin 120° = 1.3
y component of DE = 1 sin 270° = –1
R
x
= x component of resultant = 0 75 . 0 3 ? ? = 0.98 m
R
y
= resultant y component = 1 + 1.3 – 1 = 1.3 m
So, R = Resultant = 1.6 m
If it makes and angle ? with positive x-axis
Tan ? = 
component x 
component y 
= 1.32
? ? = tan
–1
1.32
?
6. | a |
?
= 3m | b |
?
= 4
a) If R = 1 unit ? ? ? ? cos . 4 . 3 . 2 4 3
2 2
= 1
? = 180°
b) ? ? ? cos . 4 . 3 . 2 4 3
2 2
= 5
? = 90°
c) ? ? ? cos . 4 . 3 . 2 4 3
2 2
= 7
? = 0°
Angle between them is 0°.
?
7. K
ˆ
4 J
ˆ
5 . 0 i
ˆ
2 AD ? ? ? = 
ˆ ˆ
6i 0.5j ?
AD = 
2 2
DE AE ? = 6.02 KM
Tan ? = DE / AE = 1/12
??= tan
–1
(1/12)
The displacement of the car is 6.02 km along the distance tan
–1
(1/12) with positive x-axis.
8. In ?ABC, tan ? = x/2 and in ?DCE, tan ? = (2 – x)/4 tan ? = (x/2) = (2 – x)/4 = 4x
? 4 – 2x = 4x
? 6x = 4 ? x = 2/3 ft
a) In ?ABC, AC = 
2 2
BC AB ? = 
2
10
3
ft
b) In ?CDE, DE = 1 – (2/3) = 4/3 ft
CD = 4 ft. So, CE = 
2 2
DE CD ? = 
4
10
3
ft
c) In ?AGE, AE = 
2 2
GE AG ? = 2 2 ft.
9. Here the displacement vector k
ˆ
3 j
ˆ
4 i
ˆ
7 r ? ? ?
?
a) magnitude of displacement = 74 ft
b) the components of the displacement vector are 7 ft, 4 ft and 3 ft.
2m
D
A
E
B
x
O 
y
1m
1.5m
90° 
30° 60°
6m
E  A
D 
B
0.5 km
? ?
C 
2m
4m
0.5 km
2–x 
G   
A
D 
B
BC = 2 ft
AF = 2 ft
DE = 2x
x 
C 
E    
F    
r   
z
Y 
Chapter-2
2.3
10. a
?
is a vector of magnitude 4.5 unit due north.
a) 3| a
?
| = 3 ? 4.5 = 13.5
3 a
?
is along north having magnitude 13.5 units.
b) –4| a
?
| = –4 ? 1.5 = –6 unit
–4 a
?
is a vector of magnitude 6 unit due south.
11. | a
?
| = 2 m, | b
?
| = 3 m
angle between them ? = 60°
a) ? ? ? ? 60 cos | b | | a | b a
?
?
?
?
= 2 ? 3 ? 1/2 = 3 m
2
b) ? ? ? ? 60 sin | b | | a | | b a |
?
?
?
?
= 2 ? 3 ? 3 / 2 = 3 3 m
2
.
12. We know that according to polygon law of vector addition, the resultant 
of these six vectors is zero.
Here A = B = C = D = E = F (magnitude)
So, Rx = A cos ? + A cos ?/3 + A cos 2 ?/3 + A cos 3 ?/3 + A cos 4 ?/4 + 
A cos 5 ?/5 = 0
[As resultant is zero. X component of resultant R
x
= 0]
= cos ? + cos ?/3 + cos 2 ?/3 + cos 3 ?/3 + cos 4 ?/3 + cos 5 ?/3 = 0
Note : Similarly it can be proved that, 
sin ? + sin ?/3 + sin 2 ?/3 + sin 3 ?/3 + sin 4 ?/3 + sin 5 ?/3 = 0 ?
13. a 2i 3 j 4k; b 3i 4 j 5k ? ? ? ? ? ?
? ? ? ? ? ? ?
?
? ? ? cos ab b a
?
?
?
ab
b a
cos
1
?
?
?
? ?
?
?
1 1
2 2 2 2 2 2
2 3 3 4 4 5 38
cos cos
1450
2 3 4 3 4 5
? ?
? ? ? ? ? ? ?
?
? ?
? ? ? ? ? ?
14. 0 ) B A ( A ? ? ?
? ? ?
(claim)
As, n
ˆ
sin AB B A ? ? ?
? ?
AB sin ? n
ˆ
is a vector which is perpendicular to the plane containing A
?
and B
?
, this implies that it is 
also perpendicular to A
?
. As dot product of two perpendicular vector is zero.
Thus 0 ) B A ( A ? ? ?
? ? ?
. ?
15.
ˆ ˆ ˆ
A 2i 3j 4k ? ? ?
?
, 
ˆ ˆ ˆ
B 4i 3j 2k ? ? ?
?
ˆ ˆ ˆ
i j k
A B 2 3 4
4 3 2
? ?
? ?
?
ˆ ˆ ˆ ˆ ˆ ˆ
i(6 12) j(4 16) k(6 12) 6i 12j 6k ? ? ? ? ? ? ? ? ? .
16. Given that A
?
, B
?
and C
?
are mutually perpendicular
A
?
× B
?
is a vector which direction is perpendicular to the plane containing A
?
and B
?
.
Also C
?
is perpendicular to A
?
and B
?
? Angle between C
?
and A
?
× B
?
is 0° or 180° (fig.1)
So, C
?
× ( A
?
× B
?
) = 0
The converse is not true.
For example, if two of the vector are parallel, (fig.2), then also 
C
?
× ( A
?
× B
?
) = 0
So, they need not be mutually perpendicular.
A 1
60° = ?/3 ?
A 2
A 3
A 4
A 5
A 6
(A B) ?
? ?
A
?
B
?
C
?
    
A
?
B
?
C
?
    
Chapter-2
2.4
17. The particle moves on the straight line PP’ at speed v. 
From the figure, 
n
ˆ
sin v ) OP ( v OP ? ? ? = v(OP) sin ? ˆ n = v(OQ) ˆ n ?
It can be seen from the figure, OQ = OP sin ? = OP’ sin ?’
So, whatever may be the position of the particle, the magnitude and 
direction of v OP
?
? remain constant.
? v OP
?
? is independent of the position P. ?
18. Give 0 ) B v ( q E q F ? ? ? ?
?
?
? ?
? ) B v ( E
?
?
?
? ? ?
So, the direction of B v
?
?
? should be opposite to the direction of E
?
. Hence, 
v
?
should be in the positive yz-plane.
Again, E = vB sin ? ? v = 
? sin B
E
For v to be minimum, ? = 90° and so v
min
= F/B
So, the particle must be projected at a minimum speed of E/B along +ve z-axis ( ? = 90°) as shown in the 
figure, so that the force is zero. ?
19. For example, as shown in the figure,
A B ?
? ?
B
?
along west
B C ?
? ?
A
?
along south
C
?
along north
A B ?
? ?
= 0 ? A B B C ? ? ?
? ? ? ?
B C ?
? ?
= 0 But B C ?
? ?
20. The graph y = 2x
2
should be drawn by the student on a graph paper for exact 
results.
To find slope at any point, draw a tangent at the point and extend the line to meet 
x-axis. Then find tan ? as shown in the figure.
It can be checked that, 
Slope = tan ? = ) x 2 (
dx
d
dx
dy
2
? = 4x
Where x = the x-coordinate of the point where the slope is to be measured. ?
21. y = sinx
So, y + ?y = sin (x + ?x)
?y = sin (x + ?x) – sin x
= sin
3 100 3
? ? ? ? ?
? ?
? ?
? ?
= 0.0157. ?
22. Given that, i = 
RC / t
0
e i
?
? Rate of change of current = 
RC / t
0
RC / i
0
e
dt
d
i e i
dt
d
dt
di
? ?
? ? = 
t / RC 0
i
e
RC
?
?
?
When a) t = 0, 
RC
i
dt
di ?
?
b) when t = RC, 
RCe
i
dt
di ?
?
c) when t = 10 RC, 
10
0
RCe
i
dt
di ?
?
V
?
Q
? ?
O    
? ?
P P ?
? ?
x 
E
?
B
?
V
?
    
y
C
?
B
?
A
?
    
B
?
? ?
?x 
y=2x
2
?y 
x 
y = sinx
y 
Page 5


2.1
SOLUTIONS TO CONCEPTS 
CHAPTER – 2
1. As shown in the figure, 
The angle between A
?
and B
?
= 110° – 20° = 90°
| A |
?
= 3 and | B |
?
= 4m
Resultant R = ? ? ? cos AB 2 B A
2 2
= 5 m
Let ? be the angle between R
?
and A
?
? = ?
?
?
?
?
?
? ?
?
?
90 cos 4 3
90 sin 4
tan
1
= tan
–1
(4/3) =  53°
? Resultant vector makes angle (53° + 20°) = 73° with x-axis.
2. Angle between A
?
and B
?
is ? = 60° – 30° =30°
| A
?
| and | B
?
| = 10 unit
R = 
2 2
10 10 2.10.10.cos30 ? ? ? = 19.3
? be the angle between R
?
and A
?
? = tan
–1 1
10sin30 1
tan
10 10cos30
2 3
?
? ? ? ? ?
?
? ?
? ?
? ? ? ? ? ? ?
= tan
–1
(0.26795) = 15°
? Resultant makes 15° + 30° = 45° angle with x-axis.
?
3. x component of A
?
= 100 cos 45° = 2 / 100 unit
x component of B
?
= 100 cos 135° = 2 / 100
x component of C
?
= 100 cos 315° = 2 / 100
Resultant x component = 2 / 100 – 2 / 100 + 2 / 100 = 2 / 100
y component of A
?
= 100 sin 45° = 2 / 100 unit
y component of B
?
= 100 sin 135° = 2 / 100
y component of C
?
= 100 sin 315° = – 2 / 100
Resultant y component = 2 / 100 + 2 / 100 – 2 / 100 = 2 / 100
Resultant = 100
Tan ? = 
component x 
component y 
= 1
? ? = tan
–1
(1) = 45°
The resultant is 100 unit at 45° with x-axis.
4. j 3 i 4 a
? ?
?
? ? , j 4 i 3 b
? ? ?
? ?
a) 
2 2
3 4 | a | ? ?
?
= 5
b) 16 9 | b | ? ?
?
= 5
c) 2 7 | j 7 i 7 | | b a | ? ? ? ?
? ? ?
?
d) 
ˆ ˆ ˆ ˆ
a b ( 3 4)i ( 4 3)j i j ? ? ? ? ? ? ? ? ?
?
?
2 2
| a b | 1 ( 1) 2 ? ? ? ? ?
?
?
.
x
y
? ?
R
?
B
?
A
?
20
x
y
? ?
B
?
A
?
30°
60°
315°
45°
135°
Chapter-2
2.2
5. x component of OA = 2cos30° = 3
x component of BC = 1.5 cos 120° = –0.75
x component of DE = 1 cos 270° = 0
y component of OA = 2 sin 30° = 1
y component of BC = 1.5 sin 120° = 1.3
y component of DE = 1 sin 270° = –1
R
x
= x component of resultant = 0 75 . 0 3 ? ? = 0.98 m
R
y
= resultant y component = 1 + 1.3 – 1 = 1.3 m
So, R = Resultant = 1.6 m
If it makes and angle ? with positive x-axis
Tan ? = 
component x 
component y 
= 1.32
? ? = tan
–1
1.32
?
6. | a |
?
= 3m | b |
?
= 4
a) If R = 1 unit ? ? ? ? cos . 4 . 3 . 2 4 3
2 2
= 1
? = 180°
b) ? ? ? cos . 4 . 3 . 2 4 3
2 2
= 5
? = 90°
c) ? ? ? cos . 4 . 3 . 2 4 3
2 2
= 7
? = 0°
Angle between them is 0°.
?
7. K
ˆ
4 J
ˆ
5 . 0 i
ˆ
2 AD ? ? ? = 
ˆ ˆ
6i 0.5j ?
AD = 
2 2
DE AE ? = 6.02 KM
Tan ? = DE / AE = 1/12
??= tan
–1
(1/12)
The displacement of the car is 6.02 km along the distance tan
–1
(1/12) with positive x-axis.
8. In ?ABC, tan ? = x/2 and in ?DCE, tan ? = (2 – x)/4 tan ? = (x/2) = (2 – x)/4 = 4x
? 4 – 2x = 4x
? 6x = 4 ? x = 2/3 ft
a) In ?ABC, AC = 
2 2
BC AB ? = 
2
10
3
ft
b) In ?CDE, DE = 1 – (2/3) = 4/3 ft
CD = 4 ft. So, CE = 
2 2
DE CD ? = 
4
10
3
ft
c) In ?AGE, AE = 
2 2
GE AG ? = 2 2 ft.
9. Here the displacement vector k
ˆ
3 j
ˆ
4 i
ˆ
7 r ? ? ?
?
a) magnitude of displacement = 74 ft
b) the components of the displacement vector are 7 ft, 4 ft and 3 ft.
2m
D
A
E
B
x
O 
y
1m
1.5m
90° 
30° 60°
6m
E  A
D 
B
0.5 km
? ?
C 
2m
4m
0.5 km
2–x 
G   
A
D 
B
BC = 2 ft
AF = 2 ft
DE = 2x
x 
C 
E    
F    
r   
z
Y 
Chapter-2
2.3
10. a
?
is a vector of magnitude 4.5 unit due north.
a) 3| a
?
| = 3 ? 4.5 = 13.5
3 a
?
is along north having magnitude 13.5 units.
b) –4| a
?
| = –4 ? 1.5 = –6 unit
–4 a
?
is a vector of magnitude 6 unit due south.
11. | a
?
| = 2 m, | b
?
| = 3 m
angle between them ? = 60°
a) ? ? ? ? 60 cos | b | | a | b a
?
?
?
?
= 2 ? 3 ? 1/2 = 3 m
2
b) ? ? ? ? 60 sin | b | | a | | b a |
?
?
?
?
= 2 ? 3 ? 3 / 2 = 3 3 m
2
.
12. We know that according to polygon law of vector addition, the resultant 
of these six vectors is zero.
Here A = B = C = D = E = F (magnitude)
So, Rx = A cos ? + A cos ?/3 + A cos 2 ?/3 + A cos 3 ?/3 + A cos 4 ?/4 + 
A cos 5 ?/5 = 0
[As resultant is zero. X component of resultant R
x
= 0]
= cos ? + cos ?/3 + cos 2 ?/3 + cos 3 ?/3 + cos 4 ?/3 + cos 5 ?/3 = 0
Note : Similarly it can be proved that, 
sin ? + sin ?/3 + sin 2 ?/3 + sin 3 ?/3 + sin 4 ?/3 + sin 5 ?/3 = 0 ?
13. a 2i 3 j 4k; b 3i 4 j 5k ? ? ? ? ? ?
? ? ? ? ? ? ?
?
? ? ? cos ab b a
?
?
?
ab
b a
cos
1
?
?
?
? ?
?
?
1 1
2 2 2 2 2 2
2 3 3 4 4 5 38
cos cos
1450
2 3 4 3 4 5
? ?
? ? ? ? ? ? ?
?
? ?
? ? ? ? ? ?
14. 0 ) B A ( A ? ? ?
? ? ?
(claim)
As, n
ˆ
sin AB B A ? ? ?
? ?
AB sin ? n
ˆ
is a vector which is perpendicular to the plane containing A
?
and B
?
, this implies that it is 
also perpendicular to A
?
. As dot product of two perpendicular vector is zero.
Thus 0 ) B A ( A ? ? ?
? ? ?
. ?
15.
ˆ ˆ ˆ
A 2i 3j 4k ? ? ?
?
, 
ˆ ˆ ˆ
B 4i 3j 2k ? ? ?
?
ˆ ˆ ˆ
i j k
A B 2 3 4
4 3 2
? ?
? ?
?
ˆ ˆ ˆ ˆ ˆ ˆ
i(6 12) j(4 16) k(6 12) 6i 12j 6k ? ? ? ? ? ? ? ? ? .
16. Given that A
?
, B
?
and C
?
are mutually perpendicular
A
?
× B
?
is a vector which direction is perpendicular to the plane containing A
?
and B
?
.
Also C
?
is perpendicular to A
?
and B
?
? Angle between C
?
and A
?
× B
?
is 0° or 180° (fig.1)
So, C
?
× ( A
?
× B
?
) = 0
The converse is not true.
For example, if two of the vector are parallel, (fig.2), then also 
C
?
× ( A
?
× B
?
) = 0
So, they need not be mutually perpendicular.
A 1
60° = ?/3 ?
A 2
A 3
A 4
A 5
A 6
(A B) ?
? ?
A
?
B
?
C
?
    
A
?
B
?
C
?
    
Chapter-2
2.4
17. The particle moves on the straight line PP’ at speed v. 
From the figure, 
n
ˆ
sin v ) OP ( v OP ? ? ? = v(OP) sin ? ˆ n = v(OQ) ˆ n ?
It can be seen from the figure, OQ = OP sin ? = OP’ sin ?’
So, whatever may be the position of the particle, the magnitude and 
direction of v OP
?
? remain constant.
? v OP
?
? is independent of the position P. ?
18. Give 0 ) B v ( q E q F ? ? ? ?
?
?
? ?
? ) B v ( E
?
?
?
? ? ?
So, the direction of B v
?
?
? should be opposite to the direction of E
?
. Hence, 
v
?
should be in the positive yz-plane.
Again, E = vB sin ? ? v = 
? sin B
E
For v to be minimum, ? = 90° and so v
min
= F/B
So, the particle must be projected at a minimum speed of E/B along +ve z-axis ( ? = 90°) as shown in the 
figure, so that the force is zero. ?
19. For example, as shown in the figure,
A B ?
? ?
B
?
along west
B C ?
? ?
A
?
along south
C
?
along north
A B ?
? ?
= 0 ? A B B C ? ? ?
? ? ? ?
B C ?
? ?
= 0 But B C ?
? ?
20. The graph y = 2x
2
should be drawn by the student on a graph paper for exact 
results.
To find slope at any point, draw a tangent at the point and extend the line to meet 
x-axis. Then find tan ? as shown in the figure.
It can be checked that, 
Slope = tan ? = ) x 2 (
dx
d
dx
dy
2
? = 4x
Where x = the x-coordinate of the point where the slope is to be measured. ?
21. y = sinx
So, y + ?y = sin (x + ?x)
?y = sin (x + ?x) – sin x
= sin
3 100 3
? ? ? ? ?
? ?
? ?
? ?
= 0.0157. ?
22. Given that, i = 
RC / t
0
e i
?
? Rate of change of current = 
RC / t
0
RC / i
0
e
dt
d
i e i
dt
d
dt
di
? ?
? ? = 
t / RC 0
i
e
RC
?
?
?
When a) t = 0, 
RC
i
dt
di ?
?
b) when t = RC, 
RCe
i
dt
di ?
?
c) when t = 10 RC, 
10
0
RCe
i
dt
di ?
?
V
?
Q
? ?
O    
? ?
P P ?
? ?
x 
E
?
B
?
V
?
    
y
C
?
B
?
A
?
    
B
?
? ?
?x 
y=2x
2
?y 
x 
y = sinx
y 
Chapter-2
2.5
23. Equation i = 
RC / t
0
e i
?
i
0
= 2A, R = 6 ? 10
–5
?, C = 0.0500 ? 10
–6
F = 5 ? 10
–7
F
a) i = 
3 7
0.3
0.3
6 0 5 10 0.3
2
2 e 2 e amp
e
?
? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ? ?
? ? ? ? .
b)
t / RC 0
i di
e
dt RC
?
?
? when t = 0.3 sec ?
( 0.3 / 0.3)
di 2 20
e Amp / sec
dt 0.30 3e
?
?
? ? ?
c) At t = 0.31 sec, i = 
( 0.3 / 0.3)
5.8
2e Amp
3e
?
? .
24. y = 3x
2
+ 6x + 7
? Area bounded by the curve, x axis with coordinates with x = 5 and x = 10 is 
given by,
Area = 
?
y
0
dy = 
10
2
5
(3x 6x 7)dx ? ?
?
= 
?
10 10
3 2
10
5
5 5
x x
3 5 7x
3 3
? ?
? ?
? ?
? ?
= 1135 sq.units.
25. Area = 
?
y
0
dy = 
0
0
sinxdx [cos x]
?
?
? ?
?
= 2 
26. The given function is y = e
–x
When x = 0, y = e
–0
= 1
x increases, y value deceases and only at x = ?, y = 0.
So, the required area can be found out by integrating the function from 0 to ?.
So, Area = 
x x
0
0
e dx [e ] 1
?
? ? ?
? ? ?
?
.
27. bx a
length
mass
? ? ? ?
a) S.I. unit of ‘a’ = kg/m and SI unit of ‘b’ = kg/m
2
(from principle of 
homogeneity of dimensions)
b) Let us consider a small element of length ‘dx’ at a distance x from the 
origin as shown in the figure.
? dm = mass of the element = ? dx = (a + bx) dx
So, mass of the rod = m = dx ) bx a ( dm
L
0
? ?
? ? = 
L
2 2
0
bx bL
ax aL
2 2
? ?
? ? ?
? ?
? ?
?
28.
dp
dt
= (10 N) + (2 N/S)t
momentum is zero at t = 0
? momentum at t = 10 sec will be
dp = [(10 N) + 2Ns t]dt
p 10 10
0 0 0
dp 10dt (2tdt) ? ?
? ? ?
= 
?
10
2
10
0
0
t
10t 2
2
?
?
?
?
= 200 kg m/s.
5
y = 3x
2
+ 6x + 7
10
x
y
y = sinx
y
x
y
x
x =1
O
y
x
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