JEE Exam > JEE Notes > DC Pandey Solutions for JEE Physics > DC Pandey Solutions: Unit & Dimensions Vectors
DC Pandey Solutions: Unit & Dimensions Vectors | DC Pandey Solutions for JEE Physics PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
Section I
Single Correct Option
1. Pressure ( ) p =
Force
Area
\ [p] =
-
[MLT ]
[L ]
2
2
=
- -
[ ] ML T
1 2
Option (d) is correct.
2. W I Rt =
2
\ [R] = =
- - -
[ML T ]
[A T]
[ML T A ]
T
2
2
2 2 2 2
[ ]
…(i)
V L
dI
dt
= and W Vq =
\ L
W
q
dt
dI
=
[L] =
-
[ML T ][T]
[A T]
2
2
2
=
- -
[ML T A ]
2 2 2
Using Eq. (i)
[R] =
[ ]
[ ]
L
T
i.e., [ ] T
L
R
=
é
ë
ê
ù
û
ú
Option (c) is correct.
3. F av =6ph
\ [ ]
[ ]
[ ]
h =
F
av
=
-
-
[MLT ]
LLT
2
[ ]
1
=
- -
[ML T ]
1 1
Option (d) is correct.
4. f=Li
\ [f ] = [ ][ L i]
=
- -
[ML T A ] A
2 2 2
[ ]
=
- -
[ML T A ]
2 2 1
Option (a) is correct.
5. Linear impulse ( ) I F t = × D
[I] =
-
[MLT ][T]
2
=
-
[MLT ]
1
Option (c) is correct.
6. F G
m m
d
=
1 2
2
[G] = =
-
[ ][ ]
[ ]
[ML T ] [L ]
[M ]
2
1 2
2
2
F d
m m
2
=
- -
[M L T ]
3 1 2
Option (c) is correct.
7. F
i i
d
=
× m
p
0 1 2
4
\ [ m
0
] =
-
[ML T ][L]
[A ]
2
2
=
- -
[ML T A ]
2 2 2
Option (c) is correct.
8. [ ]
[L]
[L T ]
[T] k = =
-1
Option (c) is correct.
9. [ ] a = =
-
-
[ML T ]
[ T ]
[ML T ]
2
3
[ ] b = =
-
-
[ML T ]
[ T ]
[ML T ]
2
2
4
Option (c) is correct.
Units & Dimensions
Vectors
2
Page 2
Section I
Single Correct Option
1. Pressure ( ) p =
Force
Area
\ [p] =
-
[MLT ]
[L ]
2
2
=
- -
[ ] ML T
1 2
Option (d) is correct.
2. W I Rt =
2
\ [R] = =
- - -
[ML T ]
[A T]
[ML T A ]
T
2
2
2 2 2 2
[ ]
…(i)
V L
dI
dt
= and W Vq =
\ L
W
q
dt
dI
=
[L] =
-
[ML T ][T]
[A T]
2
2
2
=
- -
[ML T A ]
2 2 2
Using Eq. (i)
[R] =
[ ]
[ ]
L
T
i.e., [ ] T
L
R
=
é
ë
ê
ù
û
ú
Option (c) is correct.
3. F av =6ph
\ [ ]
[ ]
[ ]
h =
F
av
=
-
-
[MLT ]
LLT
2
[ ]
1
=
- -
[ML T ]
1 1
Option (d) is correct.
4. f=Li
\ [f ] = [ ][ L i]
=
- -
[ML T A ] A
2 2 2
[ ]
=
- -
[ML T A ]
2 2 1
Option (a) is correct.
5. Linear impulse ( ) I F t = × D
[I] =
-
[MLT ][T]
2
=
-
[MLT ]
1
Option (c) is correct.
6. F G
m m
d
=
1 2
2
[G] = =
-
[ ][ ]
[ ]
[ML T ] [L ]
[M ]
2
1 2
2
2
F d
m m
2
=
- -
[M L T ]
3 1 2
Option (c) is correct.
7. F
i i
d
=
× m
p
0 1 2
4
\ [ m
0
] =
-
[ML T ][L]
[A ]
2
2
=
- -
[ML T A ]
2 2 2
Option (c) is correct.
8. [ ]
[L]
[L T ]
[T] k = =
-1
Option (c) is correct.
9. [ ] a = =
-
-
[ML T ]
[ T ]
[ML T ]
2
3
[ ] b = =
-
-
[ML T ]
[ T ]
[ML T ]
2
2
4
Option (c) is correct.
Units & Dimensions
Vectors
2
10. E h = n
\ [ ]
[ML T ]
[T ]
[ML T ]
2
2
h = =
-
-
-
2
1
1
Angular momentum ( ) J
nh
=
2p
[ ] J h = =
-
[ ] [ML T ]
2 1
Option (b) is correct.
11. [Energy] =
-
[ML T ]
2 2
=
-
[M][LT ]
2 1
\ [Mass] =
-
[ ] Ev
2
Option (c) is correct.
12.
1
2
0
2
e = E Energy density =
Energy
Volume
\
1
2
0
2
2
3
e
é
ë
ê
ù
û
ú
=
-
E
[ML T ]
L
2
[ ]
=
- -
[ML T ]
1 2
Option (b) is correct.
13. [ ] a =[T ]
2
[ ]
[ ]
[ ][ ]
b =
- -
T
L ML T
2
1 2
\
a
b
é
ë
ê
ù
û
ú
=
-
[MT ]
2
Option (b) is correct.
14. Velocity gradient =
dv
dx
[Velocity gradient] =
-
[L T ]
[L]
1
=
-
[T ]
1
=
-
[ ] M L T
0 0 1
Option (a) is correct.
15. [Force] =
-
[MLT ]
2
\ [Mass] =
-
[F]
[L T ]
2
=
-
[FL T ]
2 1
Option (a) is correct.
16. Coefficient of friction (m)
=
Limitting frictional force
Normal force
\ [m] =[M L T ]
0 0 0
Option (b) is correct.
17. q CV =
and V iR =
\ q iCR =
i t iCR =
Þ [CR] = = [ ] t [M L T A ]
0 0 0
Option (a) is correct.
18. F
q q
r
=
1
4
0
1 2
2
pe
\ Unit of e
0
= Newton-metre
2
/coulomb
2
.
Option (b) is correct.
19. Angular momentum ( ) J
nh
=
2p
I mr = S
2
\
h
I
J n
mr
=
2
2
p /
S
=
mvr
mr S
2
h
I
é
ë
ê
ù
û
ú
=
é
ë
ê
ù
û
ú
=
-
-
[LT ]
L
[T ]
1
1
= Frequency
Option (a) is correct.
20. v at
b
t c
= +
+
[ ] [ ] c T =
b
t c
v
+
é
ë
ê
ù
û
ú
=[ ]
or [ ] b = =
-
[LT ][T] [L]
1
[ ] [ ] at v = =
-
[LT ]
1
Þ [a] =
-
[LT ]
2
Option (a) is correct.
21. y A ct x = -
é
ë
ê
ù
û
ú
sin ( )
2p
l
= -
é
ë
ê
ù
û
ú
A ct
x
sin
2 2 p
l
p
l
2p
l
q
x
= (angle)
\ [ ] [ ] [ ] x = = l L
Further, y A = sinq
\ [ ] [ ] [ ] A y = = L
Option (a) is correct.
22. [ ] X =
- -
[M L T A ]
3 1 3 2
=
-
[TA ]
[ML T ]
2
2 2
6 | Mechanics-1
Page 3
Section I
Single Correct Option
1. Pressure ( ) p =
Force
Area
\ [p] =
-
[MLT ]
[L ]
2
2
=
- -
[ ] ML T
1 2
Option (d) is correct.
2. W I Rt =
2
\ [R] = =
- - -
[ML T ]
[A T]
[ML T A ]
T
2
2
2 2 2 2
[ ]
…(i)
V L
dI
dt
= and W Vq =
\ L
W
q
dt
dI
=
[L] =
-
[ML T ][T]
[A T]
2
2
2
=
- -
[ML T A ]
2 2 2
Using Eq. (i)
[R] =
[ ]
[ ]
L
T
i.e., [ ] T
L
R
=
é
ë
ê
ù
û
ú
Option (c) is correct.
3. F av =6ph
\ [ ]
[ ]
[ ]
h =
F
av
=
-
-
[MLT ]
LLT
2
[ ]
1
=
- -
[ML T ]
1 1
Option (d) is correct.
4. f=Li
\ [f ] = [ ][ L i]
=
- -
[ML T A ] A
2 2 2
[ ]
=
- -
[ML T A ]
2 2 1
Option (a) is correct.
5. Linear impulse ( ) I F t = × D
[I] =
-
[MLT ][T]
2
=
-
[MLT ]
1
Option (c) is correct.
6. F G
m m
d
=
1 2
2
[G] = =
-
[ ][ ]
[ ]
[ML T ] [L ]
[M ]
2
1 2
2
2
F d
m m
2
=
- -
[M L T ]
3 1 2
Option (c) is correct.
7. F
i i
d
=
× m
p
0 1 2
4
\ [ m
0
] =
-
[ML T ][L]
[A ]
2
2
=
- -
[ML T A ]
2 2 2
Option (c) is correct.
8. [ ]
[L]
[L T ]
[T] k = =
-1
Option (c) is correct.
9. [ ] a = =
-
-
[ML T ]
[ T ]
[ML T ]
2
3
[ ] b = =
-
-
[ML T ]
[ T ]
[ML T ]
2
2
4
Option (c) is correct.
Units & Dimensions
Vectors
2
10. E h = n
\ [ ]
[ML T ]
[T ]
[ML T ]
2
2
h = =
-
-
-
2
1
1
Angular momentum ( ) J
nh
=
2p
[ ] J h = =
-
[ ] [ML T ]
2 1
Option (b) is correct.
11. [Energy] =
-
[ML T ]
2 2
=
-
[M][LT ]
2 1
\ [Mass] =
-
[ ] Ev
2
Option (c) is correct.
12.
1
2
0
2
e = E Energy density =
Energy
Volume
\
1
2
0
2
2
3
e
é
ë
ê
ù
û
ú
=
-
E
[ML T ]
L
2
[ ]
=
- -
[ML T ]
1 2
Option (b) is correct.
13. [ ] a =[T ]
2
[ ]
[ ]
[ ][ ]
b =
- -
T
L ML T
2
1 2
\
a
b
é
ë
ê
ù
û
ú
=
-
[MT ]
2
Option (b) is correct.
14. Velocity gradient =
dv
dx
[Velocity gradient] =
-
[L T ]
[L]
1
=
-
[T ]
1
=
-
[ ] M L T
0 0 1
Option (a) is correct.
15. [Force] =
-
[MLT ]
2
\ [Mass] =
-
[F]
[L T ]
2
=
-
[FL T ]
2 1
Option (a) is correct.
16. Coefficient of friction (m)
=
Limitting frictional force
Normal force
\ [m] =[M L T ]
0 0 0
Option (b) is correct.
17. q CV =
and V iR =
\ q iCR =
i t iCR =
Þ [CR] = = [ ] t [M L T A ]
0 0 0
Option (a) is correct.
18. F
q q
r
=
1
4
0
1 2
2
pe
\ Unit of e
0
= Newton-metre
2
/coulomb
2
.
Option (b) is correct.
19. Angular momentum ( ) J
nh
=
2p
I mr = S
2
\
h
I
J n
mr
=
2
2
p /
S
=
mvr
mr S
2
h
I
é
ë
ê
ù
û
ú
=
é
ë
ê
ù
û
ú
=
-
-
[LT ]
L
[T ]
1
1
= Frequency
Option (a) is correct.
20. v at
b
t c
= +
+
[ ] [ ] c T =
b
t c
v
+
é
ë
ê
ù
û
ú
=[ ]
or [ ] b = =
-
[LT ][T] [L]
1
[ ] [ ] at v = =
-
[LT ]
1
Þ [a] =
-
[LT ]
2
Option (a) is correct.
21. y A ct x = -
é
ë
ê
ù
û
ú
sin ( )
2p
l
= -
é
ë
ê
ù
û
ú
A ct
x
sin
2 2 p
l
p
l
2p
l
q
x
= (angle)
\ [ ] [ ] [ ] x = = l L
Further, y A = sinq
\ [ ] [ ] [ ] A y = = L
Option (a) is correct.
22. [ ] X =
- -
[M L T A ]
3 1 3 2
=
-
[TA ]
[ML T ]
2
2 2
6 | Mechanics-1
=
[ ][ ]
[ ]
t i
2
Work
\ X is resistance. [ Q W i Rt =
2
]
23. F i j k
®
= - + 2 3 4
^ ^ ^
r i j k
®
= + + 3 2 3
^ ^ ^
\ t
® ® ®
= ´ r F =
-
½
½
½
½
½ ½
½
½
½
½
½ ½
i j k
^ ^ ^
3 2 3
2 3 4
or t
®
= - - 17 6 13 i j k
^ ^ ^
24. ( ) ( ) ( ) 0.5 0.8
2 2 2
1 + + = c
or 0.25 0.64 + + = c
2
1
or c
2
1 = -0.89
c = 0.11
Option (b) is correct.
25. | | | | A B A B
® ® ® ®
+ = -
( ) ( ) ( ) ( ) A B A B A B A B
® ® ® ® ® ® ® ®
+ × + = - × -
A B A B
2 2 2 2
2 2 + + × = + - ×
® ® ® ®
A B A B
i.e., A B
® ®
× =0
\ Angle between A
®
and B
®
= ° 90
26. ( ) ( ) A B A B
® ® ® ®
+ × - = 0
A A B A B B A B
® ® ® ® ® ® ® ®
× + × - × - × = 0
A B
2 2
0 - =
A B = ±
| | | | A B
® ®
=
Option (d) is correct.
27. Work ( ) = ×
® ®
F s is a scalar quantity.
Option (d) is correct.
28. Speed =
®
| | v
Option (d) is correct.
29. | | A
®
= 3, | | B
®
= 5 and angle between A
®
and B
®
is 60°.
\ A B A B
® ® ® ®
× = ° | || |cos 60
=
æ
è
ç
ö
ø
÷ ( ) ( ) 3 5
1
2
= 7.5
Option (b) is correct.
30. A B C
® ® ®
+ =
\ ( ) ( ) A B A B CC
® ® ® ® ®®
+ × + = ×
or AA AB BB CC
® ® ® ® ®® ®®
× + × + × = × 2
or A B C
2 2 2
2 + × + =
® ®
A B
or A B
® ®
× =0
or | || |cos A B
® ®
= q 0
or cosq =0
or q
p
=
2
Option (d) is correct.
31. Magnetic field intensity.
Option (d) is correct.
32. P Q R
® ® ®
+ =
( ) ( ) P Q P Q R R
® ® ® ® ® ®
+ × + = ×
P Q R
2 2 2
2 + + × =
® ®
P Q
12 5 2 13
2 2 2
+ + × =
® ®
P Q
P Q
® ®
× = 0
\ Angle between P
®
and Q
®
=
p
2
Option (b) is correct.
33. Option (b) is correct.
34. P Q R
® ® ®
+ + = 0
\ P Q R
® ® ®
+ = -
or ( ) ( ) ( ) ( ) P Q P Q R R
® ® ® ® ® ®
+ × + = - × -
or P P Q Q P Q R R
® ® ® ® ® ® ® ®
× + × + × = × 2
or P Q R
2 2 2
2 + + × =
® ®
P Q …(i)
Let Q P
2 2
= and R P = 2
Thus, Eq. (i) takes the form
P P PQ P
2 2 2
2 2 + + = cos q
or 2 0 PQcos q =
Units & Dimensions Vectors 7
Page 4
Section I
Single Correct Option
1. Pressure ( ) p =
Force
Area
\ [p] =
-
[MLT ]
[L ]
2
2
=
- -
[ ] ML T
1 2
Option (d) is correct.
2. W I Rt =
2
\ [R] = =
- - -
[ML T ]
[A T]
[ML T A ]
T
2
2
2 2 2 2
[ ]
…(i)
V L
dI
dt
= and W Vq =
\ L
W
q
dt
dI
=
[L] =
-
[ML T ][T]
[A T]
2
2
2
=
- -
[ML T A ]
2 2 2
Using Eq. (i)
[R] =
[ ]
[ ]
L
T
i.e., [ ] T
L
R
=
é
ë
ê
ù
û
ú
Option (c) is correct.
3. F av =6ph
\ [ ]
[ ]
[ ]
h =
F
av
=
-
-
[MLT ]
LLT
2
[ ]
1
=
- -
[ML T ]
1 1
Option (d) is correct.
4. f=Li
\ [f ] = [ ][ L i]
=
- -
[ML T A ] A
2 2 2
[ ]
=
- -
[ML T A ]
2 2 1
Option (a) is correct.
5. Linear impulse ( ) I F t = × D
[I] =
-
[MLT ][T]
2
=
-
[MLT ]
1
Option (c) is correct.
6. F G
m m
d
=
1 2
2
[G] = =
-
[ ][ ]
[ ]
[ML T ] [L ]
[M ]
2
1 2
2
2
F d
m m
2
=
- -
[M L T ]
3 1 2
Option (c) is correct.
7. F
i i
d
=
× m
p
0 1 2
4
\ [ m
0
] =
-
[ML T ][L]
[A ]
2
2
=
- -
[ML T A ]
2 2 2
Option (c) is correct.
8. [ ]
[L]
[L T ]
[T] k = =
-1
Option (c) is correct.
9. [ ] a = =
-
-
[ML T ]
[ T ]
[ML T ]
2
3
[ ] b = =
-
-
[ML T ]
[ T ]
[ML T ]
2
2
4
Option (c) is correct.
Units & Dimensions
Vectors
2
10. E h = n
\ [ ]
[ML T ]
[T ]
[ML T ]
2
2
h = =
-
-
-
2
1
1
Angular momentum ( ) J
nh
=
2p
[ ] J h = =
-
[ ] [ML T ]
2 1
Option (b) is correct.
11. [Energy] =
-
[ML T ]
2 2
=
-
[M][LT ]
2 1
\ [Mass] =
-
[ ] Ev
2
Option (c) is correct.
12.
1
2
0
2
e = E Energy density =
Energy
Volume
\
1
2
0
2
2
3
e
é
ë
ê
ù
û
ú
=
-
E
[ML T ]
L
2
[ ]
=
- -
[ML T ]
1 2
Option (b) is correct.
13. [ ] a =[T ]
2
[ ]
[ ]
[ ][ ]
b =
- -
T
L ML T
2
1 2
\
a
b
é
ë
ê
ù
û
ú
=
-
[MT ]
2
Option (b) is correct.
14. Velocity gradient =
dv
dx
[Velocity gradient] =
-
[L T ]
[L]
1
=
-
[T ]
1
=
-
[ ] M L T
0 0 1
Option (a) is correct.
15. [Force] =
-
[MLT ]
2
\ [Mass] =
-
[F]
[L T ]
2
=
-
[FL T ]
2 1
Option (a) is correct.
16. Coefficient of friction (m)
=
Limitting frictional force
Normal force
\ [m] =[M L T ]
0 0 0
Option (b) is correct.
17. q CV =
and V iR =
\ q iCR =
i t iCR =
Þ [CR] = = [ ] t [M L T A ]
0 0 0
Option (a) is correct.
18. F
q q
r
=
1
4
0
1 2
2
pe
\ Unit of e
0
= Newton-metre
2
/coulomb
2
.
Option (b) is correct.
19. Angular momentum ( ) J
nh
=
2p
I mr = S
2
\
h
I
J n
mr
=
2
2
p /
S
=
mvr
mr S
2
h
I
é
ë
ê
ù
û
ú
=
é
ë
ê
ù
û
ú
=
-
-
[LT ]
L
[T ]
1
1
= Frequency
Option (a) is correct.
20. v at
b
t c
= +
+
[ ] [ ] c T =
b
t c
v
+
é
ë
ê
ù
û
ú
=[ ]
or [ ] b = =
-
[LT ][T] [L]
1
[ ] [ ] at v = =
-
[LT ]
1
Þ [a] =
-
[LT ]
2
Option (a) is correct.
21. y A ct x = -
é
ë
ê
ù
û
ú
sin ( )
2p
l
= -
é
ë
ê
ù
û
ú
A ct
x
sin
2 2 p
l
p
l
2p
l
q
x
= (angle)
\ [ ] [ ] [ ] x = = l L
Further, y A = sinq
\ [ ] [ ] [ ] A y = = L
Option (a) is correct.
22. [ ] X =
- -
[M L T A ]
3 1 3 2
=
-
[TA ]
[ML T ]
2
2 2
6 | Mechanics-1
=
[ ][ ]
[ ]
t i
2
Work
\ X is resistance. [ Q W i Rt =
2
]
23. F i j k
®
= - + 2 3 4
^ ^ ^
r i j k
®
= + + 3 2 3
^ ^ ^
\ t
® ® ®
= ´ r F =
-
½
½
½
½
½ ½
½
½
½
½
½ ½
i j k
^ ^ ^
3 2 3
2 3 4
or t
®
= - - 17 6 13 i j k
^ ^ ^
24. ( ) ( ) ( ) 0.5 0.8
2 2 2
1 + + = c
or 0.25 0.64 + + = c
2
1
or c
2
1 = -0.89
c = 0.11
Option (b) is correct.
25. | | | | A B A B
® ® ® ®
+ = -
( ) ( ) ( ) ( ) A B A B A B A B
® ® ® ® ® ® ® ®
+ × + = - × -
A B A B
2 2 2 2
2 2 + + × = + - ×
® ® ® ®
A B A B
i.e., A B
® ®
× =0
\ Angle between A
®
and B
®
= ° 90
26. ( ) ( ) A B A B
® ® ® ®
+ × - = 0
A A B A B B A B
® ® ® ® ® ® ® ®
× + × - × - × = 0
A B
2 2
0 - =
A B = ±
| | | | A B
® ®
=
Option (d) is correct.
27. Work ( ) = ×
® ®
F s is a scalar quantity.
Option (d) is correct.
28. Speed =
®
| | v
Option (d) is correct.
29. | | A
®
= 3, | | B
®
= 5 and angle between A
®
and B
®
is 60°.
\ A B A B
® ® ® ®
× = ° | || |cos 60
=
æ
è
ç
ö
ø
÷ ( ) ( ) 3 5
1
2
= 7.5
Option (b) is correct.
30. A B C
® ® ®
+ =
\ ( ) ( ) A B A B CC
® ® ® ® ®®
+ × + = ×
or AA AB BB CC
® ® ® ® ®® ®®
× + × + × = × 2
or A B C
2 2 2
2 + × + =
® ®
A B
or A B
® ®
× =0
or | || |cos A B
® ®
= q 0
or cosq =0
or q
p
=
2
Option (d) is correct.
31. Magnetic field intensity.
Option (d) is correct.
32. P Q R
® ® ®
+ =
( ) ( ) P Q P Q R R
® ® ® ® ® ®
+ × + = ×
P Q R
2 2 2
2 + + × =
® ®
P Q
12 5 2 13
2 2 2
+ + × =
® ®
P Q
P Q
® ®
× = 0
\ Angle between P
®
and Q
®
=
p
2
Option (b) is correct.
33. Option (b) is correct.
34. P Q R
® ® ®
+ + = 0
\ P Q R
® ® ®
+ = -
or ( ) ( ) ( ) ( ) P Q P Q R R
® ® ® ® ® ®
+ × + = - × -
or P P Q Q P Q R R
® ® ® ® ® ® ® ®
× + × + × = × 2
or P Q R
2 2 2
2 + + × =
® ®
P Q …(i)
Let Q P
2 2
= and R P = 2
Thus, Eq. (i) takes the form
P P PQ P
2 2 2
2 2 + + = cos q
or 2 0 PQcos q =
Units & Dimensions Vectors 7
or cosq =0
or q = ° 90
\ Angle between P
®
and Q
®
is 90°
P Q R
® ® ®
+ + = 0
\ P R Q
® ® ®
+ = -
or ( ) ( ) ( ) ( ) P R P R Q Q
® ® ® ® ® ®
+ × + = - × -
or P R PR Q
2 2 2
2 + + f = cos
or 2
2 2 2
PR Q P R cosf = - -
or 2
2
PR R cosf = -
or 2P R cosf = -
or 2 2 P P cosf = -
or cosf = -
1
2
\ f = ° 135
\ Angle between P
®
and R
®
is 135°.
Option (a) is correct.
35. Angle ( ) f between P Q
® ®
+ and P Q
® ®
-
tan
sin
cos
f =
+
Q
P Q
q
q
Angle f¢ between P Q
® ®
- and P
®
tan
sin ( )
cos ( )
f¢ =
+
+ +
Q
P Q
p q
p q
=
-
-
Q
P Q
sin
cos
q
q
tan[ ( )]
tan tan
tan tan
f + -f¢ =
f - f¢
+ f f¢ 1
=
+
-
-
-
-
+
×
-
Q
P Q
Q
P Q
Q
P Q
Q
sin
cos
( sin )
cos
sin
( cos )
( s
q
q
q
q
q
q
1
in )
( cos )
q
q P Q -
=
+
2
2
2 2
PQ
P Q
sin
cos
q
q
This implies that angle between P Q
® ®
+ and
P Q
® ®
- will vary from 0 to p.
Option (b) is correct.
36. R P Q PQ
2 2 2
2 = + + cos q
for R P Q = =
P P P PP
2 2 2
2 = + + cos q
or cosq = -
1
2
or q = ° 120
Option (b) is correct.
37. W = ×
® ®
F s
= + × + ( ) ( )
^ ^ ^ ^
3 4 3 4 i j i j
=25 J
Option (b) is correct.
38. P Q i j k i j k
® ®
× = + + × - - ( ) ( )
^ ^ ^ ^ ^ ^
a a a 3 2
= - - a a
2
2 3
For P Q
® ®
^ , P Q
® ®
× =0
i.e., a a
2
2 3 0 - - =
or ( )( ) a a - + = 3 1 0
Þ a =3
Other value is - ive.
Option (d) is correct.
39. If a vector makes angles a, b and g with
the co-ordinate axes, then
cos cos cos
2 2 2
1 a b g + + =
Now,
3
7
9
49
2
æ
è
ç
ö
ø
÷ = ,
6
7
36
49
2
æ
è
ç
ö
ø
÷ = ,
2
7
4
49
2
æ
è
ç
ö
ø
÷ =
and
9
49
36
49
4
49
1 + + =
\ Option (a) is correct.
40. A i j
®
= - 4 3
^
and B i j
®
= + 8 8
^
$
\ A B C i j
® ® ®
+ = = + 12 5
^ ^
8 | Mechanics-1
P
90°
Q
R
135°
135°
®
®
®
– Q
Q
Q P –
Q P +
P
q
f
f'
®
®
®
®
®
®
®
Page 5
Section I
Single Correct Option
1. Pressure ( ) p =
Force
Area
\ [p] =
-
[MLT ]
[L ]
2
2
=
- -
[ ] ML T
1 2
Option (d) is correct.
2. W I Rt =
2
\ [R] = =
- - -
[ML T ]
[A T]
[ML T A ]
T
2
2
2 2 2 2
[ ]
…(i)
V L
dI
dt
= and W Vq =
\ L
W
q
dt
dI
=
[L] =
-
[ML T ][T]
[A T]
2
2
2
=
- -
[ML T A ]
2 2 2
Using Eq. (i)
[R] =
[ ]
[ ]
L
T
i.e., [ ] T
L
R
=
é
ë
ê
ù
û
ú
Option (c) is correct.
3. F av =6ph
\ [ ]
[ ]
[ ]
h =
F
av
=
-
-
[MLT ]
LLT
2
[ ]
1
=
- -
[ML T ]
1 1
Option (d) is correct.
4. f=Li
\ [f ] = [ ][ L i]
=
- -
[ML T A ] A
2 2 2
[ ]
=
- -
[ML T A ]
2 2 1
Option (a) is correct.
5. Linear impulse ( ) I F t = × D
[I] =
-
[MLT ][T]
2
=
-
[MLT ]
1
Option (c) is correct.
6. F G
m m
d
=
1 2
2
[G] = =
-
[ ][ ]
[ ]
[ML T ] [L ]
[M ]
2
1 2
2
2
F d
m m
2
=
- -
[M L T ]
3 1 2
Option (c) is correct.
7. F
i i
d
=
× m
p
0 1 2
4
\ [ m
0
] =
-
[ML T ][L]
[A ]
2
2
=
- -
[ML T A ]
2 2 2
Option (c) is correct.
8. [ ]
[L]
[L T ]
[T] k = =
-1
Option (c) is correct.
9. [ ] a = =
-
-
[ML T ]
[ T ]
[ML T ]
2
3
[ ] b = =
-
-
[ML T ]
[ T ]
[ML T ]
2
2
4
Option (c) is correct.
Units & Dimensions
Vectors
2
10. E h = n
\ [ ]
[ML T ]
[T ]
[ML T ]
2
2
h = =
-
-
-
2
1
1
Angular momentum ( ) J
nh
=
2p
[ ] J h = =
-
[ ] [ML T ]
2 1
Option (b) is correct.
11. [Energy] =
-
[ML T ]
2 2
=
-
[M][LT ]
2 1
\ [Mass] =
-
[ ] Ev
2
Option (c) is correct.
12.
1
2
0
2
e = E Energy density =
Energy
Volume
\
1
2
0
2
2
3
e
é
ë
ê
ù
û
ú
=
-
E
[ML T ]
L
2
[ ]
=
- -
[ML T ]
1 2
Option (b) is correct.
13. [ ] a =[T ]
2
[ ]
[ ]
[ ][ ]
b =
- -
T
L ML T
2
1 2
\
a
b
é
ë
ê
ù
û
ú
=
-
[MT ]
2
Option (b) is correct.
14. Velocity gradient =
dv
dx
[Velocity gradient] =
-
[L T ]
[L]
1
=
-
[T ]
1
=
-
[ ] M L T
0 0 1
Option (a) is correct.
15. [Force] =
-
[MLT ]
2
\ [Mass] =
-
[F]
[L T ]
2
=
-
[FL T ]
2 1
Option (a) is correct.
16. Coefficient of friction (m)
=
Limitting frictional force
Normal force
\ [m] =[M L T ]
0 0 0
Option (b) is correct.
17. q CV =
and V iR =
\ q iCR =
i t iCR =
Þ [CR] = = [ ] t [M L T A ]
0 0 0
Option (a) is correct.
18. F
q q
r
=
1
4
0
1 2
2
pe
\ Unit of e
0
= Newton-metre
2
/coulomb
2
.
Option (b) is correct.
19. Angular momentum ( ) J
nh
=
2p
I mr = S
2
\
h
I
J n
mr
=
2
2
p /
S
=
mvr
mr S
2
h
I
é
ë
ê
ù
û
ú
=
é
ë
ê
ù
û
ú
=
-
-
[LT ]
L
[T ]
1
1
= Frequency
Option (a) is correct.
20. v at
b
t c
= +
+
[ ] [ ] c T =
b
t c
v
+
é
ë
ê
ù
û
ú
=[ ]
or [ ] b = =
-
[LT ][T] [L]
1
[ ] [ ] at v = =
-
[LT ]
1
Þ [a] =
-
[LT ]
2
Option (a) is correct.
21. y A ct x = -
é
ë
ê
ù
û
ú
sin ( )
2p
l
= -
é
ë
ê
ù
û
ú
A ct
x
sin
2 2 p
l
p
l
2p
l
q
x
= (angle)
\ [ ] [ ] [ ] x = = l L
Further, y A = sinq
\ [ ] [ ] [ ] A y = = L
Option (a) is correct.
22. [ ] X =
- -
[M L T A ]
3 1 3 2
=
-
[TA ]
[ML T ]
2
2 2
6 | Mechanics-1
=
[ ][ ]
[ ]
t i
2
Work
\ X is resistance. [ Q W i Rt =
2
]
23. F i j k
®
= - + 2 3 4
^ ^ ^
r i j k
®
= + + 3 2 3
^ ^ ^
\ t
® ® ®
= ´ r F =
-
½
½
½
½
½ ½
½
½
½
½
½ ½
i j k
^ ^ ^
3 2 3
2 3 4
or t
®
= - - 17 6 13 i j k
^ ^ ^
24. ( ) ( ) ( ) 0.5 0.8
2 2 2
1 + + = c
or 0.25 0.64 + + = c
2
1
or c
2
1 = -0.89
c = 0.11
Option (b) is correct.
25. | | | | A B A B
® ® ® ®
+ = -
( ) ( ) ( ) ( ) A B A B A B A B
® ® ® ® ® ® ® ®
+ × + = - × -
A B A B
2 2 2 2
2 2 + + × = + - ×
® ® ® ®
A B A B
i.e., A B
® ®
× =0
\ Angle between A
®
and B
®
= ° 90
26. ( ) ( ) A B A B
® ® ® ®
+ × - = 0
A A B A B B A B
® ® ® ® ® ® ® ®
× + × - × - × = 0
A B
2 2
0 - =
A B = ±
| | | | A B
® ®
=
Option (d) is correct.
27. Work ( ) = ×
® ®
F s is a scalar quantity.
Option (d) is correct.
28. Speed =
®
| | v
Option (d) is correct.
29. | | A
®
= 3, | | B
®
= 5 and angle between A
®
and B
®
is 60°.
\ A B A B
® ® ® ®
× = ° | || |cos 60
=
æ
è
ç
ö
ø
÷ ( ) ( ) 3 5
1
2
= 7.5
Option (b) is correct.
30. A B C
® ® ®
+ =
\ ( ) ( ) A B A B CC
® ® ® ® ®®
+ × + = ×
or AA AB BB CC
® ® ® ® ®® ®®
× + × + × = × 2
or A B C
2 2 2
2 + × + =
® ®
A B
or A B
® ®
× =0
or | || |cos A B
® ®
= q 0
or cosq =0
or q
p
=
2
Option (d) is correct.
31. Magnetic field intensity.
Option (d) is correct.
32. P Q R
® ® ®
+ =
( ) ( ) P Q P Q R R
® ® ® ® ® ®
+ × + = ×
P Q R
2 2 2
2 + + × =
® ®
P Q
12 5 2 13
2 2 2
+ + × =
® ®
P Q
P Q
® ®
× = 0
\ Angle between P
®
and Q
®
=
p
2
Option (b) is correct.
33. Option (b) is correct.
34. P Q R
® ® ®
+ + = 0
\ P Q R
® ® ®
+ = -
or ( ) ( ) ( ) ( ) P Q P Q R R
® ® ® ® ® ®
+ × + = - × -
or P P Q Q P Q R R
® ® ® ® ® ® ® ®
× + × + × = × 2
or P Q R
2 2 2
2 + + × =
® ®
P Q …(i)
Let Q P
2 2
= and R P = 2
Thus, Eq. (i) takes the form
P P PQ P
2 2 2
2 2 + + = cos q
or 2 0 PQcos q =
Units & Dimensions Vectors 7
or cosq =0
or q = ° 90
\ Angle between P
®
and Q
®
is 90°
P Q R
® ® ®
+ + = 0
\ P R Q
® ® ®
+ = -
or ( ) ( ) ( ) ( ) P R P R Q Q
® ® ® ® ® ®
+ × + = - × -
or P R PR Q
2 2 2
2 + + f = cos
or 2
2 2 2
PR Q P R cosf = - -
or 2
2
PR R cosf = -
or 2P R cosf = -
or 2 2 P P cosf = -
or cosf = -
1
2
\ f = ° 135
\ Angle between P
®
and R
®
is 135°.
Option (a) is correct.
35. Angle ( ) f between P Q
® ®
+ and P Q
® ®
-
tan
sin
cos
f =
+
Q
P Q
q
q
Angle f¢ between P Q
® ®
- and P
®
tan
sin ( )
cos ( )
f¢ =
+
+ +
Q
P Q
p q
p q
=
-
-
Q
P Q
sin
cos
q
q
tan[ ( )]
tan tan
tan tan
f + -f¢ =
f - f¢
+ f f¢ 1
=
+
-
-
-
-
+
×
-
Q
P Q
Q
P Q
Q
P Q
Q
sin
cos
( sin )
cos
sin
( cos )
( s
q
q
q
q
q
q
1
in )
( cos )
q
q P Q -
=
+
2
2
2 2
PQ
P Q
sin
cos
q
q
This implies that angle between P Q
® ®
+ and
P Q
® ®
- will vary from 0 to p.
Option (b) is correct.
36. R P Q PQ
2 2 2
2 = + + cos q
for R P Q = =
P P P PP
2 2 2
2 = + + cos q
or cosq = -
1
2
or q = ° 120
Option (b) is correct.
37. W = ×
® ®
F s
= + × + ( ) ( )
^ ^ ^ ^
3 4 3 4 i j i j
=25 J
Option (b) is correct.
38. P Q i j k i j k
® ®
× = + + × - - ( ) ( )
^ ^ ^ ^ ^ ^
a a a 3 2
= - - a a
2
2 3
For P Q
® ®
^ , P Q
® ®
× =0
i.e., a a
2
2 3 0 - - =
or ( )( ) a a - + = 3 1 0
Þ a =3
Other value is - ive.
Option (d) is correct.
39. If a vector makes angles a, b and g with
the co-ordinate axes, then
cos cos cos
2 2 2
1 a b g + + =
Now,
3
7
9
49
2
æ
è
ç
ö
ø
÷ = ,
6
7
36
49
2
æ
è
ç
ö
ø
÷ = ,
2
7
4
49
2
æ
è
ç
ö
ø
÷ =
and
9
49
36
49
4
49
1 + + =
\ Option (a) is correct.
40. A i j
®
= - 4 3
^
and B i j
®
= + 8 8
^
$
\ A B C i j
® ® ®
+ = = + 12 5
^ ^
8 | Mechanics-1
P
90°
Q
R
135°
135°
®
®
®
– Q
Q
Q P –
Q P +
P
q
f
f'
®
®
®
®
®
®
®
C
C i j ^
^ ^
| |
= =
+
+
®
C
12 5
12 5
2 2
= +
12
13
5
13
i j
^ ^
Option (b) is correct.
41. A i j k
®
= + - 2 3 2
^ ^
$
, B i j k
®
= + + 5
^ ^
$
n ,
C i j k
®
= - + +
^ ^
$
2 3
\ Vectors A B C
® ® ®
, ,and will be coplanar if
their scalar triple product is zero i.e.,
( ) A C B
® ® ®
´ × = 0
$ $ $
( )
^ ^ ^
i j k
i j k 2 3 2
1 2 3
5 0 -
-
½
½
½
½
½
½
½
½
½
½
× + + = n
or ( ) ( )
^ ^ ^ ^ ^ ^
13 4 7 5 0 i j k i j k - + × + + = n
or 65 4 7 0 - + = n
or n = 18
Option (a) is correct.
42. Option (a) is correct.
43. ( ) ( ) a b a b
® ® ® ®
+ ´ -
= ´ + ´ - ´ - ´
® ® ® ® ® ® ® ®
a a b a a b b b
= - ´ - ´ -
® ® ® ®
0 0 a b a b
=- ´
® ®
2( ) a b = ´
® ®
2( ) b a
Option (a) is correct.
44. A i j k
®
= + + 3 4 5
^ ^ ^
B i j k
®
= + - 3 4 5
^ ^ ^
cos
( )
| || |
q=
×
® ®
® ®
A B
A B
=
+ -
+ +
9 16 25
3 4 5
2 2 2
=0
Þ q = ° 90
Option (c) is correct.
45. A B + =7
A B - =3
\ B=2 N
Option (c) is correct.
46. Angle between A i j
®
= + 2 3
^ ^
and B i j
®
= +
^ ^
q=
×
® ®
® ®
A B
|A||B|
=
+
+ ×
2 3
2 3 2
2 2
=
×
5
13 2
=
5
26
Component of A
®
along i j
^ ^
+
C i j
®
= +
5
26
2 3 ( )
^ ^
| | C
®
=
5
2
Option (a) is correct.
47. R P P P P
2 2 2
3 2 2 3 2 = + + ´ ´ ( ) ( ) cos q
or R P P
2 2 2
13 12 = + cos q …(i)
Further
( ) ( ) ( ) cos 2 6 2 2 6 2
2 2 2
R P P P P = + + ´ ´ q
or 4 40 24
2 2 2
R P P = + cos q …(ii)
Dividing Eq. (ii) by Eq. (i),
10 6 13 12
2 2 2 2
P P P P + = + cos cos q q
or 6cosq = - P
or q = ° 120
Option (b) is correct.
48. tan
sin
cos
q
a
a
=
+
Q
P Q
As q = ° 90 , tana = ¥
\ P Q + = cosa 0
i.e., cosa = -
P
Q
= -
P
P 2
Units & Dimensions Vectors 9
q = 90°
R
P
Q = 2P
Read More
FAQs on DC Pandey Solutions: Unit & Dimensions Vectors - DC Pandey Solutions for JEE Physics
| 1. What are units and dimensions in physics? |  |
Ans. Units and dimensions in physics refer to the measurement and description of physical quantities. Units are used to measure and express the magnitude of a physical quantity, such as length, time, or mass. Dimensions, on the other hand, represent the nature of the physical quantity being measured, such as whether it is a scalar or a vector quantity.
| 2. How do we convert units in physics? | |
Ans. To convert units in physics, you can use conversion factors. A conversion factor is a ratio of two equivalent quantities expressed in different units. By multiplying the given quantity by the appropriate conversion factor, you can convert it into the desired unit. For example, to convert meters to kilometers, you can use the conversion factor 1 kilometer = 1000 meters.
| 3. What are vectors in physics? | |
Ans. Vectors in physics are quantities that have both magnitude and direction. They are used to represent physical quantities that require both a numerical value and a specific direction to fully describe them. Examples of vector quantities include displacement, velocity, force, and acceleration.
| 4. How do we perform vector addition and subtraction? | |
Ans. Vector addition and subtraction involve combining or separating vectors to find their resultant or difference, respectively. To perform vector addition, you can use the head-to-tail method or the parallelogram method. In the head-to-tail method, you place the tail of one vector at the head of another vector and draw a vector from the tail of the first vector to the head of the second vector. The resultant vector is the vector that connects the tail of the first vector to the head of the second vector. To perform vector subtraction, you can add the negative of the vector you want to subtract to the original vector.
| 5. How do we resolve vectors into their components? | |
Ans. Resolving vectors into their components involves breaking down a vector into its perpendicular components along two or more axes. This is often done to simplify vector calculations or to analyze vector quantities in different directions. To resolve a vector, you can use trigonometry. For example, if you have a vector at an angle θ with the horizontal axis, you can find its horizontal and vertical components by multiplying the magnitude of the vector by the cosine and sine of θ, respectively.
About this Document
|
2.1K Views |
|
4.91/5 Rating |
|
Dec 23, 2024 Last updated |
Document Description: DC Pandey Solutions: Unit & Dimensions Vectors for JEE 2024 is part of DC Pandey Solutions for JEE Physics preparation.
The notes and questions for DC Pandey Solutions: Unit & Dimensions Vectors have been prepared according to the JEE exam syllabus. Information about DC Pandey Solutions: Unit & Dimensions Vectors covers topics
like and DC Pandey Solutions: Unit & Dimensions Vectors Example, for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for DC Pandey Solutions: Unit & Dimensions Vectors.
Introduction of DC Pandey Solutions: Unit & Dimensions Vectors in English is available as part of our DC Pandey Solutions for JEE Physics
for JEE & DC Pandey Solutions: Unit & Dimensions Vectors in Hindi for DC Pandey Solutions for JEE Physics course.
Download more important topics related with notes, lectures and mock test series for JEE
Exam by signing up for free. JEE: DC Pandey Solutions: Unit & Dimensions Vectors | DC Pandey Solutions for JEE Physics
Description
Full syllabus notes, lecture & questions for DC Pandey Solutions: Unit & Dimensions Vectors | DC Pandey Solutions for JEE Physics - JEE | Plus excerises question with solution to help you revise complete syllabus for DC Pandey Solutions for JEE Physics | Best notes, free PDF download
Information about DC Pandey Solutions: Unit & Dimensions Vectors
In this doc you can find the meaning of DC Pandey Solutions: Unit & Dimensions Vectors defined & explained in the simplest way possible. Besides explaining types of
DC Pandey Solutions: Unit & Dimensions Vectors theory, EduRev gives you an ample number of questions to practice DC Pandey Solutions: Unit & Dimensions Vectors tests, examples and also practice JEE
tests
|
Explore Courses for JEE exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Free
,Important questions
,ppt
,Extra Questions
,DC Pandey Solutions: Unit & Dimensions Vectors | DC Pandey Solutions for JEE Physics
,DC Pandey Solutions: Unit & Dimensions Vectors | DC Pandey Solutions for JEE Physics
,practice quizzes
,study material
,Semester Notes
,Exam
,Objective type Questions
,mock tests for examination
,video lectures
,Viva Questions
,past year papers
,Previous Year Questions with Solutions
,shortcuts and tricks
,Sample Paper
,DC Pandey Solutions: Unit & Dimensions Vectors | DC Pandey Solutions for JEE Physics
,MCQs
,Summary
;
Additional Information about DC Pandey Solutions: Unit & Dimensions Vectors for JEE Preparation
DC Pandey Solutions: Unit & Dimensions Vectors Free PDF Download
The DC Pandey Solutions: Unit & Dimensions Vectors is an invaluable resource that delves deep into the core of the JEE exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the DC Pandey Solutions: Unit & Dimensions Vectors now and kickstart your journey towards success in the JEE exam.
Importance of DC Pandey Solutions: Unit & Dimensions Vectors
The importance of DC Pandey Solutions: Unit & Dimensions Vectors cannot be overstated, especially for JEE aspirants.
This document holds the key to success in the JEE exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
DC Pandey Solutions: Unit & Dimensions Vectors Notes
DC Pandey Solutions: Unit & Dimensions Vectors Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to DC Pandey Solutions: Unit & Dimensions Vectors.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, DC Pandey Solutions: Unit & Dimensions Vectors Notes on EduRev are your ultimate resource for success.
DC Pandey Solutions: Unit & Dimensions Vectors JEE Questions
The "DC Pandey Solutions: Unit & Dimensions Vectors JEE Questions" guide is a valuable resource for all aspiring students preparing for the
JEE exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study DC Pandey Solutions: Unit & Dimensions Vectors on the App
Students of JEE can study DC Pandey Solutions: Unit & Dimensions Vectors alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the DC Pandey Solutions: Unit & Dimensions Vectors,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of DC Pandey Solutions: Unit & Dimensions Vectors is prepared as per the latest JEE syllabus.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup