Test – 10

PHYSICS
Vector

Q.1 Which one of the following pair cannot be the Q.7 The forces, which meet at one point but their
rectangular components of force vector of 10 N? lines of action do not lie on one plane, are
called
(a) 6 N & 8 N (b) 7 N & 51 N (a) non-coplanar non-concurrent forces
(c) 6 2 N & 2 7 N (d) 9 N & 1 N (b) non-coplanar concurrent forces
Q.2 The resultant of two vectors at an angle 150° (c) coplanar concurrent forces
is 10 units and is perpendicular to one vector. (d) coplanar non-concurrent forces
The magnitude of the smaller vector is 
Q.8 The vector OA where O is origin is given by
(a) 10 units (b) 10 3 units 
OA = 2iˆ +2j.
ˆ Now it is rotated by 45° anticlockwise
(c) 10 2 units (d) 5 3 units
about O. What will be the new vector?
Q.3 Which of the following is the unit vector
perpendicular to A and B? (a) 2 2 ˆj (b) 2 ˆj

ˆ B
A ˆ ˆ B
A ˆ (c) 2 ˆi (d) 2 2 ˆi
(a) (b)
ABsin  ABcos  Q.9 What is the angle between (P + Q) and (P × Q)?
   
AB AB 
(c) (d) (a) Zero (b)
AB sin  AB cos  2

Q.4 Let  be the angle between vectors A and 
 (c) (d) 
B . Which of the following figures correctly 4
represents the angle ?  
Q.10 If vectors A = 2iˆ + 3j+
ˆ pkˆ and B = 3i–8j
ˆ ˆ + 2kˆ are
perpendicular to each other, then value of p is
(a) (b) (a) 2 (b) –8
(c) –9 (d) 9

Q.11 When a force F acts on a particle of mass m,

(c) (d) the acceleration of particle becomes a. Now
Q.5 Condition under which vectors (a + b) and if two forces of magnitude 3F and 4F acts on
(a – b) are parallel is the particle simultaneously as shown in figure,
then the acceleration of the particle is
(a) a  b (b) | a || b |
(c) a  b (d) a is parallel to b
Q.6 A particle is moving in a circle of radius r
having centre at O, with a constant speed v.
The magnitude of change in velocity in
moving from A to B is
(a) a (b) 2a
(c) 5a (d) 8a
 
Q.12 Two vectors A and B are such that
   
| A + B |=| A – B | . The angle between the
two vectors will be
(a) 2v (b) 0 (a) 180° (b) 0°
(c) 3 v (d) v (c) 60° (d) 90°

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     Q.18 Assertion : Three vectors having magnitudes
Q.13 If | A + B |=| A |=| B | then angle between A
10, 10 and 25 cannot produce zero resultant.
and B will be
(a) 90° (b) 120° Reason : If three vectors are producing zero
resultant, then sum of magnitude of any two is
(c) 0° (d) 60°
  more than or equal to magnitude of third and
Q.14 Six vectors, a through f have the magnitudes difference is less than or equal to the magnitude
and directions indicated in the figure. Which of third.
of the following statements is true?
(a) If both Assertion and Reason are true and
the Reason is the correct explanation of the
Assertion.
(b) If both Assertion and Reason are true but
the Reason is not the correct explanation of
the Assertion.
   
(a) b  e  f (b) b  c  f (c) If Assertion is true but Reason is false.
   
(c) d  c  f (d) d  e  f (d) If both Assertion and Reason are false.
   Q.19 The angle between A and B is . The value
Q.15 Assertion : A  B is perpendicular to both A of the triple product A  (B × A) is

and B.
(a) A 2 B (b) Zero
 
Reason : The direction of A  B can be found (c) A 2 Bsin  (d) A 2 Bcos 
out by using right hand thumb rule.
Q.20 The change in a vector may occur due to
(a) If both Assertion and Reason are true and the
(a) Rotation of frame of reference
Reason is the correct explanation of the
Assertion. (b) Translation of frame of reference
(b) If both Assertion and Reason are true but (c) Rotation of vector
the Reason is not the correct explanation of (d) Both (a) & (c)
the Assertion.    
Q.21 If A = ˆi + 2kˆ and B = ˆi + ˆj – kˆ then A × B is
(c) If Assertion is true but Reason is false. equal to
(d) If both Assertion and Reason are false.
(a) 2iˆ  ˆj  kˆ (b) 2iˆ  ˆj  kˆ
Q.16 The angle between the two vectos

A = 3iˆ + 4jˆ + 5kˆ and B = 3iˆ + 4jˆ – 5kˆ will be (c) 2iˆ  3jˆ  kˆ (d) ˆi  ˆj  kˆ

(a) 0º (b) 45° Q.22 If three vectors along coordinate axes

(c) 90° (d) 180° represent the adjacent sides of a cube of
length b, then the unit vector along its
     
Q.17 Assertion : If P  Q  Q  R, then | P || R | . diagonal passing through the origin will be

Reason : The dot product of two vectors depends ˆi  ˆj  kˆ ˆi  ˆj  kˆ

only on magnitude of vectors because it is a scalar (a) (b)
2 36
quantity.
(a) If both Assertion and Reason are true and ˆi  ˆj  kˆ
the Reason is the correct explanation of the (c) ˆi  ˆj  kˆ (d)
3
Assertion.
Q.23 If resultant of two vectors having magnitude
(b) If both Assertion and Reason are true but
3 and 4 is 5. The magnitude of their cross
the Reason is not the correct explanation of
product is
the Assertion.
(a) Zero (b) 12
(c) If Assertion is true but Reason is false.
(c) 15 (d) 20
(d) If both Assertion and Reason are false.

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       Q.31 The ratio of maximum and minimum
Q.24 If | A × B |= (A  B) 3 then value of | A + B |
magnitudes of the resultant of two vectors
is  
A and B is 3:2. The relation between A and
(a) [A 2  B2  AB]1/ 2 B is

(b) [A2  B2  3AB]1/ 2 (a) A = 5B (b) 5A = B

(c) A = 3B (d) A = 4B
(c) [A 2  B2  AB]1/ 2  
Q.32 A vector, A points vertically upwards and B
 
(d) [A 2  B2 ]1/ 2 towards north. The vector product A × B is
Q.25 The area of the parallelogram whose sides are (a) Zero
represented by the vectors ˆi + 2kˆ and ˆi + ˆj – kˆ is (b) Along east
(c) Along west
(a) 3 5 units (b) 2 5 units (d) Vertically downward
(c) 17 units (d) 14 units Q.33 If a unit vector is represented by 0.5iˆ + 0.8jˆ + ck,
ˆ
Q.26 Two forces of magnitude 100 N each is then the value of c is
applied on a block. One force is acting
(a) 1 (b) 0.11
towards East and the other acting along 30°
East of North. The resultant of the two forces (c) (d) 0.39
0.22
(in N) is of magnitude:

(a) 50 3 (b) 100 3 Q.34 If θ is angle between vectors a = ˆi – ˆj and

b = 2iˆ – kˆ + ˆj, then tanθ =
(c) 100 2 (d) 100
Q.27 At what angle should the two force vectors (a) 3 (b) 11
2F and 2F act so that the resultant force (c) 5 (d) 2 3
is 10F : Q.35 The position vector r of a particle is given

(a) 37° (b) 30° by r = (Asinωt)iˆ + (Acosωt)j.
ˆ If v is velocity
(c) 45° (d) 60°  
of particle then v  r =
Q.28 The resultant of two forces 3p and 2p is R. (a) Zero
If the first force is doubled keeping same
(b) A2
direction, then the resultant is also doubled.
Find the angle between two forces. (c) A 2 sin 2 t
(a) 60° (b) 120° (d) A 2 cos 2 t
(c) 150° (d) 90° Q.36 A vector is not changed if
 (a) it is displaced parallel to itself
Q.29 A is a vector of magnitude 2.7 units due
east. What is the magnitude and direction of (b) it is rotated through an arbitrary angle

vector 4A? (c) it is cross-multiplied by a unit vector
(a) 4 units due east (b) 4 units due west (d) it is multiplied by an arbitrary scalar.
(c) 2.7 units due east (d) 10.8 units due east Q.37 Let A = ˆiAcosθ + ˆjAsinθ be any vector..
Q.30 Which of the following forces cannot be a Another vector B which is perpendicular to
resultant of 15 N force and 11 N force? A can be expressed as
(a) 2 N (b) 10 N
(a) ˆiBcos   ˆjBsin  (b) ˆiBsin   ˆjBcos 
(c) 14 N (d) 5 N
(c) ˆiBcos   ˆjBsin  (d) ˆiBsin   ˆjBcos 

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   (a) x-axis
Q.38 When two vector a and b are added, the
magnitude of the resultant vector is always (b) y-axis
(a) greater than (a + b) (c) z-axis
(b) less than or equal to (a + b) (d) Object does not move
(c) less than (a + b) Q.45 Find the resultant of three vectors OA, OB
and OC shown in the following figure.
(d) equal to (a + b)
(Radius of the circle is R)
  
Q.39 Given : a + b + c = 0. Out of the three vectors
  
a, b and c two are equal in magnitude. The
magnitude of the third vector is 2 times that
of either of the two having equal magnitude.
The angles between the vectors are:
(a) 90º, 135º, 135º (b) 30º, 60º, 90º
(c) 45º, 45º, 90º (d) 45º, 60º, 90º
 (a) 2R (b) R(1  2)
Q.40 Vector A is of length 2 cm and is 60º above
 (c) R 2 (d) R( 2  1)
the x-axis in the first quadrant. Vector B is
of length 2 cm and 60º below the x-axis in Q.46 Six forces, 9.81 N each, acting at a point are
 
the fourth quadrant. The sum A + B is a coplanar. If the angles between neighboring
vector of magnitude - forces are equal, then the resultant is
(a) 2 along + y-axis (b) 2 along + x-axis (a) 0 N (b) 9.81 N
(c) 1 along – x axis (d) 2 along – x axis (c) 2 × 9.81 N (d) 3 × 9.81 N.
Q.41 A particle projected from origin moves in x-y Q.47 Two vectors A and B are such that A + B = C
 and A2 + B2 = C2. If  is the angle between A
plane with a velocity v = 3iˆ + 6xjˆ , where î and and B, then the value of  is
ĵ are the unit vectors along x and y axis. Find 2
the equation of path followed by the particle (a)  (b)
3
1
(a) y = x2 (b) y = 2 
x (c) 0 (d)
1 2
(c) y = 2x2 (d) y =  
x Q.48 A + B can also be written as
Q.42 If a vector 2iˆ + 3jˆ + 8kˆ is perpendicular to    
(a) A  B (b) B  A
the vector 4jˆ – 4iˆ + αk,
ˆ then the value of  is    
(c) B  A (d) B.A
1 Q.49 Which of the following represents a unit vector?
(a) –1 (b)  
2 A
|A|
(a)  (b) 
1 A |A|
(c)  (d) 1  
2 |A|
A
Q.43 The resultant of A and B makes an angle  (c)  (d) 
A |A|
with A and  with B, then
Q.50 A vector is added to an equal and opposite
(a)  <  (b)  > , if A < B vector of similar nature, forms a
(c)  < , if A = B (d)  < , if A < B (a) Unit vector (b) Position vector
Q.44 Three forces given by vectors 2iˆ + 2j,
ˆ 2iˆ – 2jˆ (c) Null vector (d) Displacement vector
and –4iˆ are acting together on a point object
at rest. The object moves along the direction.

4
 Solution
1. (d) 7. (b)
Coplanar forces are those forces which lies in a
The vector magnitude = A x 2  A y2 same plane & concument forces are those forces
which acts on a single point.
Vector magnitude = 10
8. (a)
But (d) option gives the magnitude

 92  12  82 10 [by trial method check options]

2. (b)
 R 2  A 2  B2 .....(a)
R = 10 OA  2iˆ  2ˆj

Perpendicular | OA | 4  4  2 2
Also tan 30 
Base On rotating by an angle of 45º anticlockwise it
will lie along y-axis.
1 R
 
3 A So, A  2 2 ˆj
From equation (a) 9. (b)
P × Q is perpendicular to the plane formed by P
A  10 3 and Q.
(10) 2  (10 3) 2  B2 P + Q lies in this plane. Hence, P + Q is
perpendicular to P × Q.
B = 20
10. (d)
 Smaller vector is 10 3 units  
If two vectors A and B ,  to each other then,
 
3. (c) A.B = 0
Unit vector perpendicular to A and B,
   
 2iˆ +3j+   ˆ 2kˆ   0
ˆ pkˆ . 3iˆ – 8j+

AB AB 6 – 24 + 2p = 0
n̂  
| A  B | AB sin 
–18 + 2p = 0
4. (c) p=9
To find angle between vectors, they will be joined 11. (c)
either head to head or tail to tail.
5. (d) Fnet  32  42  5F
(d) Their cross product should be zero. So, Fnet  ma '
(a + b) × (a – b) = 0
5F
 2(a × b) = 0  a' 
m
So, a is parallel to b.
6. (d) a '  5a

 θ  60o  F
 ΔV  = 2V sin = 2×V×sin   a
  2  2  m
1 
= 2×V×  V =| V |
2

5
12. (d) 17. (d)
     
| A  B || A  B | P  Q | P || Q | cos 
 
A 2  B2  2ABcos   A 2  B 2  2ABcos  Q  R | Q || R | cos 
Squaring both the sides,  
2 2 2 2
It doesnot mean | P || R | dot product of 2
 A  B  2ABcos   A  B  2ABcos 
vectors depends on magnitude of vectors and
4ABcos   0 angle between them.
 cos   0 18. (a)
   90 Since sum of magnitude of two vector is smaller
13. (b) than third one so it is not possible to get resultant
    zero because the two would not be enough to
| A  B || A || B | complensate 3rd one.
 
| A  B | A 2  B2  2ABcos  19. (b)
In scalar triple product of vectors, the positions
A 2  A 2  A 2  2A 2 cos 
of dot and cross can be interchanged, i.e.
1
  cos     120 A  (B  A)  (A  B)  A  (A  A)  B
2
14. (d) but A  A  0

 A  (B  A)  0
20. (c)
Change in a vector may occur due to rotation of vector
and not due to rotation of frame of reference.
   21. (c)
def
15. (a) ˆi ˆj kˆ
 
By right hand thumb rule. AB  1 0 2
    1 1 1
A  B is perpendicular to both A & B.
16. (c)  ˆi(0  2)  ˆj( 1  2)  k(1
ˆ  0)
Angle between two vectors is given as from dot
 2iˆ  3jˆ  kˆ
product A  B | A || B | cos 
22. (d)
AB
cos  
AB Diagonal vector, A  biˆ  bjˆ  bkˆ
Here, A  3iˆ  4ˆj  5kˆ or A  b 2  b 2  b 2  3b
B  3iˆ  4ˆj – 5kˆ ˆi  ˆj  kˆ
A
 A 
 A  (3)2  (4) 2  (5)2  50 |A| 3
23. (b)
B  (3)2  (4)2  (5)2  50  
We know for two vector A & B, makes an angle  then
and A  B  (3iˆ  4ˆj  5k)
ˆ  (3iˆ  4ˆj  5k)
ˆ
 
 9  16  25  0 A × B = ABsin  and,
 
0 A + B = A 2  B2  2ABcos 
 cos   0
50  50
5  32  42  2  3  4cos 
   90

6
 25 = 9 + 16 + 24 cos  28. (b)
24 cos 0  90º R2 = 9p2 + 4p2 + 12p2 cos  ... (i)
  2 2 2 2
4R = 36p + 4p + 24 p cos  ... (ii)
So, A × B = AB sin90º
From equation (i) & (ii) equating value of R2
=3×4×1 13p2 + 12p2 cos  = 10p2 + 6p2 cos 
= 12 units 3p2 = – 6p2 cos 
24. (a)
    1
cos   
| A  B| (A.B) 3 2
ABsin   (ABcos ) 3   120
29. (d)
tan   3 
A  2.7 ˆi
   60 
  Vector 4A
| A  B | A 2  B2  2ABcos 
ˆ  10.8iˆ or 10.8 units due east.
 4(2.7i)
 A 2  B2  2AB  cos 60 30. (a)
1 Rmax A + B = 26N
 1 2
  A 2  B2  2AB   Rmin A – B = 4N
 2
So, the value of resultant force in between 4 N to 26 N
25. (d)
31. (a)
     
a  (iˆ  2k),
ˆ (iˆ  ˆj  k)
ˆ b
| A  B |max  A  B, | A  B |min  A  B
Then Therefore
 
Area | a  b | AB 3
  2A  2B  3A  3B  A  5B.
AB 2
ˆi ˆj kˆ
32. (c)
1 0 2   
Area A  B  C along west.
1 1 1
(Using right hand thumb rule)
 ˆi(0  2)  ˆj( 1  2)  k(1
ˆ  0) 33. (b)
Unit vector can be found by dividing a vector
 2iˆ  3jˆ  kˆ with its magnitude i.e.
So, area  4  9  1  14 units. A
A
|B|
26. (b)
Let we represent the unit vector by n̂.
Fnet  (100)2  (100)2  2  100  100  cos 60
We also know that the modulus of unit vector is 1
 100 3 N i.e., | nˆ | 1

27. (c)  | nˆ || 0.5iˆ  0.8ˆj  ckˆ | 1

R 2  A 2  B2  2ABcos 
or (0.5)2  (0.8)2  c 2  1
2 2 2
( 10F)  (2F)  ( 2F)  2  2F  2Fcos 
or 0.25  0.64  c 2  1
1
cos   or 0.89  c 2  1
2
  45
or c2  1  0.89  0.11  c  0.11

7
34. (b)
 
a  ˆi  ˆj, b  2iˆ  kˆ  ˆj
 
a .b  ab cos 

2  1  2  6 cos 

1 2  3 2 cos 

1 B
 cos   
H
2 3  90°, 135°, 135°
P 11 40. (b)
So, tan   
B 1

 tan   11
35. (a)

r  (A sin t)iˆ  (A cos t)ˆj

dr
v  A cos tiˆ  A sin tjˆ Only horizontal along + x-axis
dt
   2cos 60° + 2 cos 60° = 2
v  r  A 2  sin t cos t  A 2  sin t cos t
41. (a)
 
v r  0 Method 1: Method 2:
36. (a)  
V = 3iˆ + 6xjˆ Vx ˆi + Vy ˆj= V
On displacing a vector parallel to itself neither
magnitude changes nor direction, therefore vector  dx dy
also V = ˆi  ˆj Vx  3 s
remains same. dt dt
37. (b) dx
Dot product of two perpendicular vectors should   3, Vy  6x
dt
be zero.

 B  ˆiBsin  – ˆjBcos   dx   3dt We know

ˆ cos   ˆjA sin )  (iBsin

ˆ dy Vy
[ (iA   ˆjBcos )  0] x  3t  tan  
dx Vx
38. (b)
  dy dy 6x
a  b | a  b | a  b  6x 
dt dx 3
39. (a)
y x

dy  6x  dt 
0

dy  2xdx
0

 dy   63tdt y  xy  x 2

By vector translation t2
 
dy 18 tdt 18
2
y  9t 2
x2
 9
9
y  x2

8
42. (c) 47. (d)
 Here, A + B = C and A2 + B2 = C2
a  2iˆ  3jˆ  8kˆ
  (A + B)2 = (C)2
b  4iˆ  4ˆj  kˆ
| A |2  | B |2 2 | A || B | cos  | C |2
 
a .b  0
| C |2 2 | A || B | cos  | C |2
8  12  8  0
4  8  0  2 | A || B | cos   0

1  cos   0

2 
43. (b)  
2
Consider the figure, 48. (c)
   
A B BA
Vector addition is commutative.
49. (b)
Any vector which is divided by its magnitude gives
Resultant is inclined towards a vector having unit vector
greater magnitude. 
A
  Â
 |A
44. (d) |
Net force in x direction = 2 + 2 – 4 = 0
50. (c)
Net force in y direction = 2 – 2 = 0
A vector which is added to its opposite vactor
 F   gives null vector.
Object does not move  
a  ( a)  0
45. (b)

R net  R  R 2  R 2  R  2R  R( 2  1)
46. (a)

Coplanar all in a single plane (xy plane)

6 = 360°
 = 60°
Check now every component concels out net = 0