Vectors Practice Questions - DPP for NEET

 Page 1


1. (d)
90
2 sin
2
vv
° æö
D=
ç÷
èø
1
2sin4522
2
v vv = °= ´=
22
2 21
60 30
pp
=´w=´´ = r cm/s
2. (b) 2 sin
2
vv
q æö
D=
ç÷
èø
10
2 5 sin 45
2
= ´ ´ °=
\ 
v
a
t
D
=
D
 
10/21
10 2
==
 m/s
2
3. (c) For motion of the particle from (0, 0) to (a, 0)
$
ˆˆ
(0) FKi aj FKaj =- + Þ =-
ur ur
Displacement 
ˆˆ ˆˆ
( 0) (0 0) r ai j i j ai = + - +=
r
$
So work done from (0, 0) to (a, 0) is given by
$
. .0 W F r Ka j ai = =-=
urr
$
For motion (a, 0) to (a, a)
$
() F K ai aj =-+
ur
$
 and displacement
 
$
ˆ ˆ ˆˆ
( ) ( 0) r ai aj ai j a j = + - +=
r
So work done from (a, 0) to (a, a)
$ $
2
. ( ).
urr
$
W F r K ai a j a j Ka = =- + =-
So total work done = 
2
Ka -
4. (d)
q
W
P
T
sin T q
cos T q
As the metal sphere is in equilibrium under the effect
of the three forces therefore
0 T PW + +=
ururuur
From the figure
cos TW q=
................... (i)
sin TP q=
................... (ii)
From equation (i) and (ii) we get
tan PW =q and 
2 22
T PW =+
5. (b) Time taken to cross the river along shortest possible
path is given by
22
=
-
d
t
vu
v = velocity of boat in still water
u = velocity of river water
d = width of river
22
151
60
5
\=
-u
Þ u = 3 km/h
6. (d) Here d = 320 m = 
320
km
1000
t = 4 min
v = 
5
3
u
Putting values in 
22
=
-
d
t
vu
, u = 60 m/min
7. (c)
12
sin sin sin150
P QR
==
q q°
Þ
 
1
1.93
sin sin150
R
=
q°
Þ
 
1
1.93 sin150 1.93 0.5
1
sin 0.9659
R
´ °´
= ==
q
P Q
R
150°
2
q
1
q
8. (b)
T
T cos 30°
T sin 30° 30 N
W
30°
Page 2


1. (d)
90
2 sin
2
vv
° æö
D=
ç÷
èø
1
2sin4522
2
v vv = °= ´=
22
2 21
60 30
pp
=´w=´´ = r cm/s
2. (b) 2 sin
2
vv
q æö
D=
ç÷
èø
10
2 5 sin 45
2
= ´ ´ °=
\ 
v
a
t
D
=
D
 
10/21
10 2
==
 m/s
2
3. (c) For motion of the particle from (0, 0) to (a, 0)
$
ˆˆ
(0) FKi aj FKaj =- + Þ =-
ur ur
Displacement 
ˆˆ ˆˆ
( 0) (0 0) r ai j i j ai = + - +=
r
$
So work done from (0, 0) to (a, 0) is given by
$
. .0 W F r Ka j ai = =-=
urr
$
For motion (a, 0) to (a, a)
$
() F K ai aj =-+
ur
$
 and displacement
 
$
ˆ ˆ ˆˆ
( ) ( 0) r ai aj ai j a j = + - +=
r
So work done from (a, 0) to (a, a)
$ $
2
. ( ).
urr
$
W F r K ai a j a j Ka = =- + =-
So total work done = 
2
Ka -
4. (d)
q
W
P
T
sin T q
cos T q
As the metal sphere is in equilibrium under the effect
of the three forces therefore
0 T PW + +=
ururuur
From the figure
cos TW q=
................... (i)
sin TP q=
................... (ii)
From equation (i) and (ii) we get
tan PW =q and 
2 22
T PW =+
5. (b) Time taken to cross the river along shortest possible
path is given by
22
=
-
d
t
vu
v = velocity of boat in still water
u = velocity of river water
d = width of river
22
151
60
5
\=
-u
Þ u = 3 km/h
6. (d) Here d = 320 m = 
320
km
1000
t = 4 min
v = 
5
3
u
Putting values in 
22
=
-
d
t
vu
, u = 60 m/min
7. (c)
12
sin sin sin150
P QR
==
q q°
Þ
 
1
1.93
sin sin150
R
=
q°
Þ
 
1
1.93 sin150 1.93 0.5
1
sin 0.9659
R
´ °´
= ==
q
P Q
R
150°
2
q
1
q
8. (b)
T
T cos 30°
T sin 30° 30 N
W
30°
12
DPP/ P 05
From the figure
T sin 30° = 30 ...(i)
T cos 30° = W ...(ii)
By solving equation (i) and (ii) we get
303 WN = and T = 60 N
9. (c) Relative velocity = 
ˆˆ ˆˆ ˆ ˆ
(3 4) (34) 68 i j i j ij + -- - =+
10. (c)
30°
90°
m
v
r
v
sin 30° = 
1
2
r
m
v
v
=
Þ
 
0.5
0.25
22
m
r
v
v= == m/s
11. (a) To cross the river in minimum time, the shift is given
by 
du
v
.
12. (d) Relative velocity = 10 + 5 = 15 m/s.
Time taken by the bird to cross the train =
120
8
15
= sec
13. (d)
vr = w´
r urr
$ $
$ $
3 4 1 18 13 2
5 66
i jk
i jk = - =- -+
-
$
$
14. (d) | | 3( .)
ur u r ur ur
A B AB ´=
sin 3 cos AB AB q=q
Þtan3 q=
Þ 60 q=°
Now |||| R AB =+
ur ur ur
22
2 cos A B AB = ++q
22
1
2
2
A B AB
æö
= ++
ç÷
èø
= 
2 2 1/2
() A B AB ++
15. (a)
$ $ $ $
(7 3 )( 3 5) r F i j k i jk t= ´ = + + - ++
r r ur
$$
$ $
$ $
7 3 1 14 38 16
3 15
i jk
i jk t= = -+
-
$
r
$
16. (b)| |.
ur ur ur ur
B AB A´=
Þ sin cos AB AB q=q
Þ tan1 q=
\
45 q=°
17. (a) .0 PQ =
u r ur
Þ 
2
2 30 aa - -=
Þ a = 3
18. (a)
21
u u r ur r
S rr =-
. W FS =
ur ur
= 
$ $ $ $
(4 3).(11 11 15) i j k i jk + + ++
$$
= (4 × 11 + 1 × 11 + 3 × 15) = 100 J
19. (c)
ˆˆ ˆ ˆ ˆˆ
3 2 , 35 A i j kB i jk = - + = -+
u r ur
, 
ˆ ˆˆ
24 C i jk = +-
ur
2 22
3 ( 2) 1 9 4 1 14 A = + - + = + +=
r
2 22
1 ( 3) 5 1 9 25 35 B= +- += + +=
r
222
2 1 ( 4) 4 1 16 21
r
C= + +- = ++=
As 
22
B AC =+ therefore ABC will be right angled
triangle.
20. (b)
12
12
.
cos
FF
FF
q=
rr
= 
ˆˆ ˆ ˆ ˆˆ
(5 10 20 ).(10 5 15 )
25 100 400. 100 25 225
i j k i jk + - --
+ + ++
50 50 300
525. 350
-+
=
Þ
 
1
cos
2
q=
45 \q=°
21. (a)
ˆˆ ˆˆ ˆ ˆ ˆˆ
4 32 r abc i j i jk i jk = + + = -- + - = +-
r
rrr
222
ˆ ˆˆ
ˆ
||
1 1 ( 1)
r i jk
r
r
r
+-
==
+ +-
 
ˆ ˆˆ
3
i jk +-
=
22. (a)
D C
A B
300 m
400 m
Displacement 
AC AB BC =+
uuu r uuu r u uu r
Page 3


1. (d)
90
2 sin
2
vv
° æö
D=
ç÷
èø
1
2sin4522
2
v vv = °= ´=
22
2 21
60 30
pp
=´w=´´ = r cm/s
2. (b) 2 sin
2
vv
q æö
D=
ç÷
èø
10
2 5 sin 45
2
= ´ ´ °=
\ 
v
a
t
D
=
D
 
10/21
10 2
==
 m/s
2
3. (c) For motion of the particle from (0, 0) to (a, 0)
$
ˆˆ
(0) FKi aj FKaj =- + Þ =-
ur ur
Displacement 
ˆˆ ˆˆ
( 0) (0 0) r ai j i j ai = + - +=
r
$
So work done from (0, 0) to (a, 0) is given by
$
. .0 W F r Ka j ai = =-=
urr
$
For motion (a, 0) to (a, a)
$
() F K ai aj =-+
ur
$
 and displacement
 
$
ˆ ˆ ˆˆ
( ) ( 0) r ai aj ai j a j = + - +=
r
So work done from (a, 0) to (a, a)
$ $
2
. ( ).
urr
$
W F r K ai a j a j Ka = =- + =-
So total work done = 
2
Ka -
4. (d)
q
W
P
T
sin T q
cos T q
As the metal sphere is in equilibrium under the effect
of the three forces therefore
0 T PW + +=
ururuur
From the figure
cos TW q=
................... (i)
sin TP q=
................... (ii)
From equation (i) and (ii) we get
tan PW =q and 
2 22
T PW =+
5. (b) Time taken to cross the river along shortest possible
path is given by
22
=
-
d
t
vu
v = velocity of boat in still water
u = velocity of river water
d = width of river
22
151
60
5
\=
-u
Þ u = 3 km/h
6. (d) Here d = 320 m = 
320
km
1000
t = 4 min
v = 
5
3
u
Putting values in 
22
=
-
d
t
vu
, u = 60 m/min
7. (c)
12
sin sin sin150
P QR
==
q q°
Þ
 
1
1.93
sin sin150
R
=
q°
Þ
 
1
1.93 sin150 1.93 0.5
1
sin 0.9659
R
´ °´
= ==
q
P Q
R
150°
2
q
1
q
8. (b)
T
T cos 30°
T sin 30° 30 N
W
30°
12
DPP/ P 05
From the figure
T sin 30° = 30 ...(i)
T cos 30° = W ...(ii)
By solving equation (i) and (ii) we get
303 WN = and T = 60 N
9. (c) Relative velocity = 
ˆˆ ˆˆ ˆ ˆ
(3 4) (34) 68 i j i j ij + -- - =+
10. (c)
30°
90°
m
v
r
v
sin 30° = 
1
2
r
m
v
v
=
Þ
 
0.5
0.25
22
m
r
v
v= == m/s
11. (a) To cross the river in minimum time, the shift is given
by 
du
v
.
12. (d) Relative velocity = 10 + 5 = 15 m/s.
Time taken by the bird to cross the train =
120
8
15
= sec
13. (d)
vr = w´
r urr
$ $
$ $
3 4 1 18 13 2
5 66
i jk
i jk = - =- -+
-
$
$
14. (d) | | 3( .)
ur u r ur ur
A B AB ´=
sin 3 cos AB AB q=q
Þtan3 q=
Þ 60 q=°
Now |||| R AB =+
ur ur ur
22
2 cos A B AB = ++q
22
1
2
2
A B AB
æö
= ++
ç÷
èø
= 
2 2 1/2
() A B AB ++
15. (a)
$ $ $ $
(7 3 )( 3 5) r F i j k i jk t= ´ = + + - ++
r r ur
$$
$ $
$ $
7 3 1 14 38 16
3 15
i jk
i jk t= = -+
-
$
r
$
16. (b)| |.
ur ur ur ur
B AB A´=
Þ sin cos AB AB q=q
Þ tan1 q=
\
45 q=°
17. (a) .0 PQ =
u r ur
Þ 
2
2 30 aa - -=
Þ a = 3
18. (a)
21
u u r ur r
S rr =-
. W FS =
ur ur
= 
$ $ $ $
(4 3).(11 11 15) i j k i jk + + ++
$$
= (4 × 11 + 1 × 11 + 3 × 15) = 100 J
19. (c)
ˆˆ ˆ ˆ ˆˆ
3 2 , 35 A i j kB i jk = - + = -+
u r ur
, 
ˆ ˆˆ
24 C i jk = +-
ur
2 22
3 ( 2) 1 9 4 1 14 A = + - + = + +=
r
2 22
1 ( 3) 5 1 9 25 35 B= +- += + +=
r
222
2 1 ( 4) 4 1 16 21
r
C= + +- = ++=
As 
22
B AC =+ therefore ABC will be right angled
triangle.
20. (b)
12
12
.
cos
FF
FF
q=
rr
= 
ˆˆ ˆ ˆ ˆˆ
(5 10 20 ).(10 5 15 )
25 100 400. 100 25 225
i j k i jk + - --
+ + ++
50 50 300
525. 350
-+
=
Þ
 
1
cos
2
q=
45 \q=°
21. (a)
ˆˆ ˆˆ ˆ ˆ ˆˆ
4 32 r abc i j i jk i jk = + + = -- + - = +-
r
rrr
222
ˆ ˆˆ
ˆ
||
1 1 ( 1)
r i jk
r
r
r
+-
==
+ +-
 
ˆ ˆˆ
3
i jk +-
=
22. (a)
D C
A B
300 m
400 m
Displacement 
AC AB BC =+
uuu r uuu r u uu r
DPP/ P 05
13
AC = 
2 2 22
(AB) (BC) (400) (300) 500m + = +=
Distance = AB+BC =400+300=700 m
23. (a)
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ
A 3i 2j k,B i 3j 5k,C 2i j 4k = - + = - + = -+
rr r
2 22
A 3 ( 2) 1 9 4 1 14 = + - + = + +=
r
2 22
B 1 ( 3) 5 1 9 25 35 = +-+= + + =
r
222
C 2 1 ( 4) 4 1 16 21 = + +- = ++=
uv
As 
22
B AC =+ therefore ABC will be right angled
triangle.
24. (a)AB0sin 00 ´ = \ q= \q= °
r r
Two vectors will be parallel to each other.
25. (a), 26 (b), 27. (b)
ˆ ˆˆ
2 11
1 11
r r
i jk
AB ´=
 = 
ˆ ˆˆ
( 1 1) (2 1) (2 1) i jk -- -+ -
= 
ˆ ˆ
jk -+
Unit vector perpendicular to A
r
 and B
r
 is 
ˆ ˆ
2
jk
æö
-+
ç÷
èø
.
Any vector whose magnitude is k (constant) times
ˆ ˆˆ
(2) i jk ++ is parallel to A
r
 so, unit vector 
ˆ ˆˆ
2
6
i jk ++
is parallel to A
r
.
28. (b) A B AB + =-
rr rr
Þ
A
2
 + B
2
 + 2AB cosq = A
2
 + B
2
 + 2AB cosq
Hence cos q = 0 which gives q = 90°
Also vector addition is commutative.
Hence A B BA +=+
rr rr
29. (a) Let P
r
 and 
Q
r
 are two vectors in opposite direction,
then their sum () P Q PQ +- =-
rr rr
If PQ =
r r
 then sum equal to zero.
30. (d) The resultant of two vectors of unequal magnitude
given by 
22
2 cos R A B AB = ++q cannot be zero
for any value of q.