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Page 1: 01 Physical world and measurement 01 · Target’s “Absolute Physics Vol - I” is compiled according to the notified syllabus for NEET-UG & JEE (Main), which in turn has been framed

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Page 2: 01 Physical world and measurement 01 · Target’s “Absolute Physics Vol - I” is compiled according to the notified syllabus for NEET-UG & JEE (Main), which in turn has been framed

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© Target Publications Pvt. Ltd. No part of this book may be reproduced or transmitted in any form or by any means, C.D. ROM/Audio Video Cassettes or electronic, mechanical

including photocopying; recording or by any information storage and retrieval system without permission in writing from the Publisher.

Printed at: Repro India Ltd., Navi Mumbai

P.O. No. 133970

TEID: 12870_JUP

Solutions/hints to Topic Test available in downloadable PDF format at

www.targetpublications.org/tp12870

For all Agricultural, Medical, Pharmacy and Engineering Entrance Examinations held across India.

Salient Features • Exhaustive coverage of MCQs subtopic wise. • ‘4039’ MCQs including questions from various competitive exams. • Includes solved MCQs from NEET-UG, JEE (Main), MHT-CET and

various entrance examinations from year 2015 to 2018. • Various competitive exam questions are exclusively covered. • Concise theory for every topic. • Hints provided wherever relevant. • Topic test at the end of each chapter. • Important inclusions: Knowledge bank and Googly questions.

Absolute

PHYSICS Vol. I

NEET – UG & JEE (Main)

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Target’s “Absolute Physics Vol - I” is compiled according to the notified syllabus for NEET-UG & JEE (Main), which in turn has been framed after reviewing various state syllabi as well as the ones prepared by CBSE, NCERT and COBSE. The book comprises of a comprehensive coverage of Theoretical Concepts & Multiple Choice Questions. The flow of content & MCQ’s is planned keeping in mind the weightage given to a topic as per the NEET-UG & JEE (Main) exam. MCQs in each chapter are a mix of questions based on high order thinking, theory, numerical, graphical, multiple concepts. The level of difficulty of the questions is at par with that of various competitive examinations like CPMT, JEE, AIEEE, TS EAMCET (Med. and Engg.), BCECE, Assam CEE, AP EAMCET (Med. and Engg.) and the likes. Also to keep students updated, questions from most recent examinations such as AIPMT/NEET, JEE (Main), MHT-CET, K CET, GUJ CET, WB JEEM, of years 2015, 2016, 2017 and 2018 are covered exclusively. Unique points are represented in the form of Notes at the end of theory section, Formulae are

collectively placed after notes for quick revision and Shortcuts are included to save time of students

while dealing with rigorous questions.

An additional feature of Knowledge Bank is introduced to give students glimpse of various interesting concepts related to the subtopic.

Googly Questions are specifically prepared to develop thinking skills required to answer any tricky or higher order question in students. These will give students an edge required to score in highly competitive exams.

Topic Test has been provided at the end of each chapter to assess the level of preparation of the student on a competitive level.

We are confident that this book will cater to needs of students of all categories and effectively assist them to achieve their goal. We welcome readers’ comments and suggestions which will enable us to refine and enrich this book further.

All the best to all Aspirants! Yours faithfully, Authors Edition: Second

Disclaimer This reference book is based on the NEET-UG syllabus prescribed by Central Board of Secondary Education (CBSE). We the publishers are making this reference book which constitutes as fair use of textual contents which are transformed by adding and elaborating, with a view to simplify the same to enable the students to understand, memorize and reproduce the same in examinations. This work is purely inspired upon the course work as prescribed by the National Council of Educational Research and Training (NCERT). Every care has been taken in the publication of this reference book by the Authors while creating the contents. The Authors and the Publishers shall not be responsible for any loss or damages caused to any person on account of errors or omissions which might have crept in or disagreement of any third party on the point of view expressed in the reference book. © reserved with the Publisher for all the contents created by our Authors. No copyright is claimed in the textual contents which are presented as part of fair dealing with a view to provide best supplementary study material for the benefit of students.

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No. Topic Name Page No.

1 Physical World and Measurement 1

2 Scalars and Vectors 47

3 Motion in One Dimension 74

4 Laws of Motion 117

5 Motion in two Dimensions 173

6 Work, Energy and Power 236

7 System of particles and Rotational Motion 288

8 Gravitation 356

9 Mechanical properties of solids: Elasticity 419

10 Mechanical properties of fluids: Viscosity 458

11 Mechanical properties of fluids: Surface Tension 495

12 Thermal properties of Matter: Heat 525

13 Thermodynamics 580

14 Kinetic Theory of Gases 619

15 Oscillations 650

16 Wave Mechanics 707

Note: ** marked section is not for JEE (Main)

Index

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47

Chapter 02 : Scalars and Vectors

B(x, y) A(x,y)

O

Y

X

Scalars: Physical quantities which have only magnitude

are called scalars. Example: length, mass, time, volume, distance,

speed, work, energy etc. Vectors: i. Physical quantities which have magnitude

as well as direction are called vectors. Example: Displacement, velocity,

acceleration, force, momentum, impulse etc. ii. A vector quantity is represented by putting

an arrow () over it. iii. The length of the arrow (on a convenient

scale) represents the magnitude of the vector and the direction in which the arrow points is the direction of the vector.

Position vector: i. A vector which gives the position of a

particle at a point with respect to the origin of chosen co-ordinate system is called position vector.

ii. Position vector gives the distance between origin and the position of an object at any instant in a plane.

iii. It gives the direction of object with reference to origin.

Displacement vector: i. A vector which shows how much and in

which direction an object has changed its position in a given interval of time is called displacement vector.

ii. It is the straight distance between initial and final position of an object.

iii. In the figure, if (x, y) and (x, y) are the co-ordinates of initial position A and final position B of a particle in XY plane, then displacement vector AB is given by,

AB = OB OA = (xx) i

+ (yy) j

Resultant vector: i. The resultant of two or more vectors is a

single vector which produces the same effect as produced by individual vectors together.

ii. Nature of resultant vector is same as that of the given vectors.

Equality of vectors: i. Two vectors are said to be equal, if they

have equal magnitude and same direction irrespective of their positions in space.

ii. If two vectors A

and B

are represented by two equal parallel lines drawn with same scale, having arrow heads in the same direction, then A

and B

are equal vectors, i.e., A

= B

Some important types of vectors: i. Negative vector: a. Negative vector of a given vector is a

vector of same magnitude but acting in a direction opposite to that of the given vector.

B A

b. The negative vectors are antiparallel vectors. ii. Co-initial vectors: Two or more vectors are

said to be co-initial, if their initial point is common.

Scalars and Vectors02 2.1 Scalars and vectors 2.2 Types of vectors 2.3 Resolution of vectors 2.4 Addition and subtraction of vectors

2.5 Multiplication of vectors by a real number and scalar

2.6 Scalar product (dot product) of vectors 2.7 Vector product (cross product) of vectors

Scalars and vectors2.1

Types of vectors 2.2P

O X

Y

A

B

A

B

O

R

P

Q

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48

48

Absolute Physics Vol - I (Med. and Engg.)

iii. Collinear vectors: Two or more vectors acting along the same

straight line are called collinear vectors. iv. Coplanar vectors: Those vectors which are

acting in the same plane are called coplanar vectors.

v. Zero vector: A vector having zero magnitude and

arbitrary direction (not known to us) is a zero vector.

Zero vector can be obtained by: a. multiplying a vector by zero. b. adding a vector to its own negative

vector. vi. Localised vector: Localised vector is that vector whose

initial point is fixed. It is also called fixed vector.

vii. Non-localised vector: Non-localised vector is that vector whose

initial point is not fixed. It is also called as free vector.

viii. Polar vectors: Those vectors which have their direction

along the direction of motion of a body are called Polar vectors. Polar vectors have a point of application.

ix. Axial vectors: Those vectors which are always along the

axis of rotation are called axial vectors. x. Unit vectors: a. A unit vector of the given vector is a

vector of unit magnitude and has the same direction as that of the given vector.

b. A unit vector of A

is written as A and is read as ‘A cap’ or ‘A hat’.

c. Since, magnitude of A

is | A

|,

A

= | A |

A

A = A

| A |

= vectormodulusof the vector

d. In Cartesian co-ordinate system, i

, j

and k

are the unit vectors along X-axis, Y-axis and Z-axis respectively.

e. A unit vector is unit less and dimensionless.

Resolution of a vector: i. The process of splitting a single vector

into two or more vectors in different directions which together produce same effect as produced by the single vector alone is called resolution of vector.

ii. The splitting vectors are called component vectors. When a vector A

is resolved into two rectangular components in XY plane,

it is given by A

= x yA A

= Ax i

+ Ay j

,

where xA

and yA

are rectangular components along X-axis and Y-axis respectively. Magnitude of this vector is given by

| A

| = 2 2x yA A

iii. When a vector A

is resolved into three rectangular components with the given vector along X, Y and Z axes,

A

= Ax i

+ Ay j

+ zA k

Magnitude of this vector is given by | A

| = 2 2 2x y zA +A +A

iv. If , , and are angles subtended by rectangular components with the given vector, then

cos = xAA

, cos = yAA

and cos = zAA

These values are called direction cosines of a vector.

cos2 + cos2 + cos2 = 1 Addition of vectors: i. When two or more vectors of same type

are taken in same order, then resultant of these vectors is called addition of vectors.

ii. If A

and B

are two vectors in same order, then their resultant vector R

is given by R

= A

+ B

. Magnitude of resultant vector is given by

| R

| = | A

| + | B

|

Resolution of vectors 2.3

Addition and subtraction of vectors 2.4

A B

C

a

and b

are collinear vectors.

b

a

B O A

xA

A

yA

Y

X

Resolution of a vector in two rectangular components

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49

Chapter 02 : Scalars and Vectors

Q

O

S

R

P

B

B

A

iii. Vectors of same nature alone can be added.

iv. Vector addition is commutative. i.e. A

+ B

= B

+ A

v. Vector addition is associative.

A

+ ( B

+ C

) = ( A

+ B

) + C

Vector addition by rectangular components: Consider in two dimension plane

A

= x yA i A j

B

= x yB i B j

R

= A

+ B

= x y x yA i A j B i B j

= x x y yA B i A B j

Also, R

= x yR i R j

Then, Rx = Ax + Bx and Ry = Ay + By In three dimensions,

A

= x y zA i A j A k

B

= x y zB i B j B k

R

= A

+ B

= x y y y z zA B i A B j A B k

= x y zR i R j R k

Rx = Ax + Bx Ry = Ay + By Rz = Az + Bz If vectors p, q and r are

p

= x y zp i p j p k

q

= x y zq i q j q k

r

= x y zr i r j r k

If R

= p

+ q

+ r

Then, Rx = px + qx + rx Ry = py + qy + ry Rz = pz + qz + rz Graphical and analytical method: There are various laws for addition of vectors

such as triangle law, parallelogram law, polygon law etc.

i. Triangle law of vector addition: If two vectors of the same type are

represented in magnitude and direction, by the two sides of a triangle taken in order, then their resultant is represented in magnitude and direction by the third side of the triangle drawn from the

starting point of the first vector to the end point of the second vector.

If OP = A

and PQ = B

are two vectors acting at angle , then magnitude of resultant vector is given by

| R

| = 2 2A B 2ABcos ii. Parallelogram law of vector addition: a. If two vectors are represented in

magnitude and direction by the adjacent sides of the parallelogram, then their resultant vector is given by the diagonal of the parallelogram passing through the point of intersection of given vectors.

b. From the figure, magnitude of resultant

vector is given by

| R

| = 2 2A +B +2ABcos

where = angle between A

and B

c. The angle which resultant vector R

subtends with vector A

is given by

tan = BsinA B cos

iii. Polygon law of vector addition: If a number of vectors are represented in

magnitude and direction by the sides of a polygon, taken in order, then their resultant is represented in magnitude and direction by the closing side of the polygon taken in the opposite order.

From the figure, R

= A

+ B

+ C

+ D

Q

O P

B

A

R

B

C

S D

T

R

O A

Q

P

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50

Absolute Physics Vol - I (Med. and Engg.)

Subtraction of vectors: i. Subtraction of a vector B

from vector A

is the addition of vector B

to the vector

A

.

A

B

= A

+ B

ii. All the laws of vector addition such as triangle law, parallelogram law etc., are equally applicable to subtraction of vectors also.

For subtraction of two vectors A

and B

inclined at an angle ,

| R

| = 2 2A B 2ABcos(180 )

| R

| = 2 2A B 2ABcos

tan =

Bsin 180 θA + Bcos 180 θ

= BsinθA Bcosθ

iii. Subtraction of vectors follow neither commutative law nor associative law.

A

B

B

A

A

B C

A B

C

i. The multiplication of a vector A

by a real number n becomes another vector n A

. Its magnitude becomes n times the magnitude of the given vector. Its direction is same or opposite as that of A

, depending on whether n is a positive or negative real number.

n A

= n A

and n A

= n A

.

ii. For example, if a vector A is multiplied by a real number n = 2, we get 2 A

, which is a vector, acting in the direction of A

and having magnitude twice as A

. iii. If vector A

is multiplied by real number n = 2, then we get 2 A

, which is also a vector acting in the opposite direction of A

and having magnitude twice as that of A

. iv. The unit of n A

is the same as that of A

. Multiplication of a vector by a scalar: i. When a vector A

is multiplied by a scalar ‘s’, it becomes ‘sA’

whose magnitude is s

times the magnitude of A

.

ii. The unit of A

is different from the unit of ‘sA’

. For example, If A

= 10 newton and s = 5 second, then

s A

= 5 second 10 newton = 50 Ns. Scalar product (dot product) of two vectors: i. The scalar product of two non-zero

vectors is defined as the product of the magnitudes of the two vectors and cosine of the angle between them.

ii. The scalar product of two vectors A

and B

is given by A

B

= AB cos . iii. The scalar product is represented by

putting a ‘dot’ between the two vectors. iv. The scalar product of two vectors is a

scalar quantity. Example:

a. Work done, W = F

S

b. Power, P = F

V

c. Electric current, I = J

A

where J

is current density vector

and A

is area vector.

d. Magnetic flux, = B

A

where

B

= magnetic induction,

A

= area vector. Properties of scalar product: i. The scalar product of two vectors is

commutative over multiplication.

A

B

= B

A

….( AB cos = BA cos)

ii. A

A

= A2 or A = AA This is true because in this case = 0.

iii. If two vectors A

and B

are perpendicular to each other, then = 90 and A

B

= AB cos 90 = 0. iv. Scalar product of two same unit vectors is

unity. i

i

= j

j

= k

k

= 1 v. Scalar product of two different unit

vectors is zero. i j

= j k

= k i

= 0 vi. The distributive law holds good for scalar

product

A B C

= A B

+ A C

Scalar product (dot product) of vectors2.6

Multiplication of vectors by areal number and scalar

2.5

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Chapter 02 : Scalars and Vectors

vii. A

= x y zA i A j A k

B

= x y zB i B j B k

A.B

= x y z x y zA i A j A k . B i B j B k

= AxBx + AyBy + AzBz Vector product (cross product) of two vectors: i. The vector product of two vectors is a

third vector whose magnitude is equal to the product of magnitudes of the two vectors and sine of the angle between them.

ii. Cross product of two vectors A

and B

is given by ˆA B (AB sin ) n

, where A and B are respectively the

magnitudes of vectors A

and B

and n is a unit vector perpendicular to the plane containing A

and B

. iii. Cross product of two vectors is a vector

quantity, so it is also called vector product of vectors.

Example:

a. Torque,

= r F

, where r

is the

position vector and F

is the force vector.

b. Linear velocity,

v

= r

where

is the angular frequency

vector and r

is the position vector. c. Angular momentum,

L

= r p

where r

is the position vector and p

is the linear momentum vector. Properties of vector product: i. Vector product is anti commutative,

i.e., (A B)

= (B A)

….( sin() = sin)

ii. A A

= 0, i.e., the vector product of a vector by itself is zero. This is because, in this case, = 0 and hence sin = 0.

Hence, the condition for two vectors to be parallel ( = 0) or anti parallel ( = 180) is that their vector product should be zero.

If A

B

= 0, it means A

is zero or B

= 0 or the angle between them is 0 or 180.

iii. The distributive law holds good for vector products.

A

( B

+ C

) = A

B

+ A

C

iv. ( A) B A ( B) (A B)

; where is a real number.

v. For three orthogonal unit vectors, i , j and k

ˆ ˆi j = ˆ ˆj i = k , ˆ ˆj k = ˆ ˆk j = i , ˆ ˆk i = ˆ ˆi k = j ˆ ˆi i = ˆ ˆj j = ˆ ˆk k = 0

vi. x y z

x y z

ˆ ˆ ˆi j k(A B) A A A

B B B

= i (AyBz – AzBy) – j

(AxBz –AzBx) + k (AxBy – AyBx) 1. Finite angular displacement is a scalar quantity

because it does not obey the laws of vector addition.

2. The resultant of two vectors of unequal magnitudes can never be a null vector.

3. The magnitude of rectangular components of a vector is always less than the magnitude of the vector.

4. Division of vectors is not allowed as directions cannot be divided.

5. Minimum number of collinear vectors whose resultant can be zero is two.

6. A vector can have any number, even infinite number of components (minimum 2 components).

7. Displacement, velocity, linear momentum and force are polar vectors.

8. Angular velocity, angular acceleration, angular momentum and torque are axial vectors.

9. A quantity having magnitude and direction is not necessarily a vector. For example, time and electric current. These quantities have magnitude and direction but they are scalars. This is because they do not obey the laws of vector addition.

10. Scalars are added, subtracted or divided algebraically.

11. Vectors are added and subtracted geometrically.

Vector product (cross product) of vectors

2.7

Notes

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12. Moment of inertia is not a vector quantity because clockwise or anticlockwise direction is not associated with it.

13. Moment of inertia is not a scalar quantity because for the same body, its values are different for different orientations of the axis of rotation.

14. Moment of inertia is considered as tensor. 1. Magnitude of resolution of a vector: i. In two rectangular components, R = 2 2

x yR + R ii. In three rectangular components, R = 2 2 2

x y zR + R + R

2. Resultant of addition of two vectors: | R

| = 2 2A + B + 2AB cos θ 3. Direction of resultant vector:

= tan1 B sin θA + B cos θ

4. Commutative law of vector addition: A

+ B

= B

+ A

5. Associative law of vector addition: A

+ ( B

+ C

) = ( A

+ B

) + C

6. Distributive law of multiplication over addition: i. A

( B

+ C

) = A

B

+ A

C

ii. A

( B

+ C

) = A

B

+ A

C

7. Distributive law of multiplication over

subtraction: i. A

( B

C

) = A

B

A

C

ii. A

( B

C

) = A

B

A

C

8. Angle of inclination of resultant with positive

direction of X-axis:

= tan1 y

x

RR

9. Scalar (dot) product of two vectors: i. A

B

= AB cos

ii. i

i

= j

j

= k

k

= 1

iii. i

j

= j

k

= k

i

= 0 10. Vector (cross) product of two vectors: i. A

B

= AB sin ii. ˆ ˆ ˆ ˆ ˆ ˆi j = j j = k k = 0 iii. ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆi j = k, j k = i, k i = j

11. Unit vector perpendicular to the cross product:

n = A BABsin

12. Direction cosine of a vector :

i. cos = xRR

ii cos = yRR

iii. cos = zRR

1. Commutative and associative laws are true for

vector addition but are not true for subtraction of vectors.

2. Two vectors can be added by using either

triangle law or parallelogram law of vector addition.

3. The magnitude of the resultant of A

and B

varies between A B to A + B. 4. If | A

B

| = | A

B

|, then the angle between A

and B

is 90º. 5. Angle between two vectors can be determined by using dot or cross product of two vectors.

Angle between vectors ()

Dot product

A B

Cross product

A B

0 AB 0 90 0 AB 180 –AB 0

6. If the resultant of A

and B

is perpendicular to

A

, then the angle between A

and B

is,

= cos–1 AB

7. If two vectors represent the two adjacent sides of a parallelogram, then the area of the parallelogram is equal to magnitude of the cross product of the two vectors.

8. If two vectors represent the two sides of a triangle, then the area of triangle is equal to half the magnitude of the cross product of the two vectors.

9. The minimum number of non-coplanar vectors whose sum can be zero is four.

n

Formulae

Shortcuts

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10. To find the volume of a parallelopiped formed with three vectors as its sides, scalar triple product is used.

i. A B C =

x y z

x y z

x y z

A A AB B BC C C

= Ax(ByCz – BzCy) – Ay (BxCz – BzCx) + Az (BxCy – ByCx)

= [ABC] = Volume of parallelopiped ii. Scalar triple product remains same if

vectors are taken in same order.

i.e; A B C = A B C

= C A B

iii. A B C = B A C

= A C B

iv. Scalar triple product is zero for coplanar vectors as well as if any two vectors are equal, parallel or collinear.

1. Which of the following is a vector? (A) Pressure (B) Surface tension (C) Moment of inertia (D) Torque 2. Which one of the following statements is true?

[NCERT Exemplar] (A) A scalar quantity is the one that is

conserved in a process. (B) A scalar quantity is the one that can

never take negative values. (C) A scalar quantity is the one that does not

vary from one point to another in space. (D) A scalar quantity has the same value for

observers with different orientation of the axes.

3. Which of the following is a scalar? (A) Electric field (B) Angular momentum (C) Frequency (D) Torque 4. Which of the following is a vector quantity? (A) Charge on proton. (B) Velocity of air. (C) Mass of radium nucleus. (D) Energy of thermal neutron. 5. Two physical quantities, one of which is vector

and other is scalar, having same dimensions are (A) work and energy (B) work and torque (C) Pressure and power (D) Impulse and momentum

6. A vector which shows how much and in which direction an object has changed its position in a given interval of time is called

(A) unit vector (B) position vector (C) resultant vector (D) displacement vector 7. Which of the following represents a vector of

magnitude 3 units? (A) A

= 3 unit (B) A

= 2 unit (C) | A |

= 3 unit

(D) | A |

= 3 unit 8. From the given pair of vectors, select the equal

vectors.

(A) 2 unit

3 unit

(B) 1unit

1unit

(C) 2 unit

1.5unit

(D) 2 unit

2 unit

9. Assertion: Position of a particle in a given

plane has unique vector representation. Reason: Position vector is non-localised type

of vector. (A) Assertion is True, Reason is True;

Reason is a correct explanation for Assertion.

(B) Assertion is True, Reason is True; Reason is not a correct explanation for Assertion.

(C) Assertion is True, Reason is False. (D) Assertion is False, Reason is True. 10. Axial vectors direct along the axis of rotation

in accordance to (A) associative properties of vectors. (B) Fleming’s Right hand rule. (C) right hand screw rule. (D) Fleming’s left hand rule. 11. Angular displacement is (A) scalar (B) vector (C) axial vector (D) polar vector 12. Angular momentum is _______. (A) scalar (B) free vector (C) axial vector (D) polar vector 13. The magnitude and direction of a given vector

is indicated by (A) axial vectors. (B) unit vectors. (C) polar vectors. (D) none of the above.

Scalars and vectors2.1Types of vectors 2.2

Multiple Choice Questions

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Absolute Physics Vol - I (Med. and Engg.)

14. If A

= 2 i

+ 4 j

– 4 k

, then unit vector in the

direction of A

is

(A) 3 j

(B) ( i

+ 2 j

– 2 k

)

(C) (i j k)12

(D) (i j k)3

15. A unit vector is represented as

ˆ ˆ ˆ0.8i bj 0.4k . Hence the value of ‘b’ must

be [MHT CET 2018] (A) 0.4 (B) 0.6 (C) 0.2 (D) 0.2 16. The expression 1 1ˆ ˆi j

2 2

is a

(A) Unit vector (B) Null vector (C) Vector of magnitude 2

(D) Vector of magnitude 12

17. Walking of a person on the road is an example

of (A) scalars (B) unit vector (C) resolution of vectors (D) null vector

18. The component of a vector r

along x-axis will have a maximum value if

[K CET 2016]

(A) r

is along + ve x-axis.

(B) r

is along + ve y-axis.

(C) r

is along ve y-axis.

(D) r

makes an angle of 45 with the x-axis.

19. The vector projection of a vector 3 i

+ 4 k

on Y-axis is [R PMT 2007]

(A) Five (B) Four (C) Three (D) Zero 20. Vector A

makes equal angles with X, Y and Z axes. Value of its components in terms of magnitude of A

will be

(A) A3

(B) A2

(C) 3 A (D) 3A

21. Given vector A

= 2 i

+ 3 j

, then the angle

between A

and Y-axis is [C PMT 1993]

(A) tan1 32

(B) tan1 23

(C) sin1 23

(D) cos1 23

22. If A

= 2 i

+ 4 j

– 5 k

, then the direction

cosines of the vector A

are

(A) 245

, 445

and 545

(B) 145

, 245

and 345

(C) 445

, 0 and 445

(D) 345

, 245

and 345

23. If a vector P

makes angles , and with the X, Y and Z axes respectively, then

sin2 + sin2 + sin2 = (A) 0 (B) 1 (C) 2 (D) 3 24. If P

= Q

, then which of the following is NOT correct?

(A) P

= Q

(B) | P

| = | Q

|

(C) P Q

= Q P

(D) P

+ Q

= P Q

25. Can the resultant of two vectors be zero?

[IIIT 2000] (A) Yes, when the two vectors are same in

magnitude and direction. (B) No. (C) Yes, when the two vectors are same in

magnitude but opposite in sense. (D) Yes, when the two vectors are same in

magnitude making an angle of 2π3

with

each other. 26. If A

= 2 i

+ j

, B

= 3 j

k

and C

= 6 i

2k

,

then value of A

2 B

+ 3 C

would be

(A) 20 i

+ 5 j

+ 4 k

(B) 20 i

5 j

4 k

(C) 4 i

+ 5 j

+ 20 k

(D) 5 i

+ 4 j

+ 10 k

Resolution of vectors 2.3

Addition and subtraction of vectors 2.4

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SAMPLE C

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55

Chapter 02 : Scalars and Vectors

27. ( P

+ Q

) is a unit vector along X-axis.

If P

= i

j

+ k

, then Q

is

(A) i

+ j

– k

(B) j

– k

(C) i

+ j

+ k

(D) j

+ k

28. What vector must be added to the sum of two

vectors i

– 2 j

+ 2 k

and 2 i

+ j

– k

, so that the resultant may be a unit vector along X-axis?

[BHU 1990]

(A) 2 i

+ j

– k

(B) –2 i

+ j

– k

(C) 2 i

– j

– k

(D) –2 i

– j

– k

29. The vector that must be added to the sum of

vectors i

3 j

+ 2 k

and 3 i

+ 6 j

7 k

so that the resultant vector is a unit vector along the Y-axis is

(A) 4 i

+ 2 j

+ 5 k

(B) –4 i

2 j

+ 5 k

(C) 3 i

+ 4 j

+ 5 k

(D) Null vector 30. The unit vector parallel to the resultant of the

vectors A

= 4 i

+ 3 j

+ 6 k

and

B

= i

+ 3 j

8 k

is [EAMCET 2000]

(A) 17

(3 i

+ 6 j

2 k

)

(B) 17

(3 i

+ 6 j

+ 2 k

)

(C) 149

(3 i

+ 6 j

2 k

)

(D) 149

(3 i

6 j

+ 2 k

) 31. If a

= 4 i

– j

, b

= 3 i

+ 2 j

and c

= k

, then the unit vector r along the direction of sum of these vectors will be

[Kerala CET (Engg.) 2010]

(A) r

= 13

( i

+ j

k

)

(B) r

= 12

( i

+ j

k

)

(C) r

= 13

( i

j

+ k

)

(D) r

= 12

( i

+ j

+ k

) 32. The magnitude of displacement vector with

end points (4, – 4, 0) and (– 2, – 2, 0) must be

(A) 6 (B) 5 2 (C) 4 (D) 2 10

33. If a

, b

, c

are three consecutive vectors

forming a triangle, then a

+ b

+ c

is (A) (B) 0 (C) 1 (D) –1 34. If ˆ ˆ ˆ ˆ ˆ ˆA 3i 2 j k, B i 3j 5k

and ˆ ˆ ˆC 2i j 4k

form a right angled triangle then out of the following which one is satisfied? [MHT CET 2018]

(A) A B C

and A2 = B2 + C2

(B) A B C

and B2 = A2 + C2

(C) B A C

and B2 = A2 + C2

(D) B A C

and A2 = B2 + C2 35. For a regular hexagon ABCDEF, what will be

the value of AB + AC + AD + AE + AF , if O is the centre of the regular hexagon?

(A) zero (B) 2 AO (C) 4 AO (D) 6 AO 36. A particle is simultaneously acted by two

forces equal to 4 N and 3 N. The net force on the particle is [C PMT 1979]

(A) 7 N (B) 5 N (C) 1 N (D) Between 1 N and 7 N 37. Maximum and minimum magnitudes of the

resultant of two vectors of magnitudes P and Q are in the ratio 3:1. Which of the following relations is true?

(A) P = 2Q (B) P = Q (C) PQ = 1

(D) P = Q2

c

b

a

R

P Q

E D

C F

A B

O

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ONTENT

56

56

Absolute Physics Vol - I (Med. and Engg.)

38. If three forces 1F

= 3 i

4 j

+ 5 k

,

2F

= –3 i

+ 4 j

and 3F

= ˆ5k act on a body, then the direction of resultant force on the body is

(A) along X-axis (B) along Y-axis (C) along Z-axis (D) in indeterminate form 39. Two forces 3 N and 2 N are at an angle such

that the resultant is R. The first force is now increased to 6 N and the resultant became 2R. The value of is [HP PMT 2000]

(A) 30º (B) 60º (C) 90º (D) 120º 40. If A

= 4 i

3 j

and B

= 6 i

+ 8 j

, then

magnitude and direction of A

+ B

will be (A) 5, tan1(3/4) (B) 5 5 , tan1 (1/2) (C) 10, tan1(5) (D) 25, tan1 (3/4) 41. If two forces each of magnitude 4 units act at a

point and the angle between them is 120, then the magnitude and direction of the sum of the two forces are

(A) 4, = tan1 (1.73) (B) 4, = tan1 (0.73) (C) 2, = tan1 (1.73) (D) 6, = tan1 (0.73) 42. The resultant of two forces, one double the

other in magnitude, is perpendicular to the smaller of the two forces. The angle between the two forces is [K CET 2002]

(A) 60 (B) 120 (C) 150 (D) 90 43. Two forces, F1 and F2 are acting on a body.

One force is double that of the other force and the resultant is equal to the greater force. Then the angle between the two forces is

(A) cos1 12

(B) cos1 12

(C) cos1 14

(D) cos1 14

44. If | A

+ B

| = | A

| + | B

|, then angle between A

and B

will be [CBSE PMT 2010] (A) 90° (B) 120° (C) 0° (D) 60°

45. Two equal vectors have a resultant equal to either of them. The angle between them is

[AIIMS 2001] (A) 60 (B) 90 (C) 100 (D) 120 46. Three vectors A

, B

and C

are related as

A

+ B

= C

. If vector C

is perpendicular to

vector A

and the magnitude of C

is equal to

the magnitude of A

, what will be the angle between vectors A

and B

? (A) 45º (B) 90º (C) 135º (D) 180º 47. The resultant of two vectors A and B is

perpendicular to the vector A and its magnitude is equal to half the magnitude of vector B. The angle between A and B is

(A) 120° (B) 150° (C) 135° (D) 180 48. The resultant of P

and Q

is perpendicular to

P

. What is the angle between P

and Q ?

(A) cos1 PQ

(B) cos1 P

Q

(C) sin1 PQ

(D) sin1

PQ

49. For getting a resultant displacement of 10 m,

two displacement vectors having magnitudes 6 m and 8 m should be joined

(A) parallel to each other (B) anti-parallel to each other (C) at an angle of 60 (D) perpendicular to each other 50. Two forces of magnitudes 8 N and 15 N are

acting at a point. If the resultant force is 17 newton, then the angle between two forces is

(A) 30 (B) 45 (C) 60 (D) 90 51. Two forces of equal magnitude F are at a

point. If is the angle between two forces, then magnitude of the resultant force will be

[EAMCET 95]

(A) 2F cos 2

(B) F cos 2

(C) 2F cos

(D) Fcos2 2

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SAMPLE C

ONTENT

57

Chapter 02 : Scalars and Vectors

52. Two forces are such that the sum of their magnitudes is 18 N and their resultant is perpendicular to the smaller force and magnitude of resultant is 12 N. Then the magnitudes of the forces are [AIEEE 2002]

(A) 12 N, 6 N (B) 13 N, 5 N (C) 10 N, 8 N (D) 16 N, 2 N 53. The maximum resultant of two vectors is

26 unit and minimum resultant is 16 unit, then the magnitude of each vector is

(A) 21, 5 (B) 13, 13 (C) 20, 6 (D) 13, 6 54. Two forces of magnitudes 3 N and 4 N act on

a body. The ratio of magnitudes of minimum and maximum resultant force on the body is

(A) 17

(B) 34

(C) 43

(D) 7 55. A vector a

makes an angle 30° and b

makes an angle 120 with the X-axis. The magnitudes of these vectors are 3 unit and 4 unit respectively. The magnitude of resultant vector is

(A) 3 unit (B) 4 unit (C) 5 unit (D) 1 unit 56. Assertion: The magnitude of resultant vector

of two given vectors can never be less than the magnitude of any of the given vectors.

Reason: The resultant vector is the vector sum of two vectors.

(A) Assertion is True, Reason is True; Reason is a correct explanation for Assertion.

(B) Assertion is True, Reason is True; Reason is not a correct explanation for Assertion.

(C) Assertion is True, Reason is False. (D) Assertion is False, Reason is True. 57. Which pair of the following forces will never

give resultant force of 2 N? [HP PMT 1999] (A) 2 N and 2 N (B) 1 N and 1 N (C) 1 N and 3 N (D) 1 N and 4 N 58. Which of the following vectors can’t be equal

to the resultant of 5 and 10? [AFMC 1996] (A) 2 (B) 5 (C) 7 (D) 8 59. If more than three forces are acting on a heavy

rigid body such that the body is in balanced state, then all the forces are

[ICS (Prelims) 2000] (A) collinear (B) coplanar (C) acting in random direction (D) represented by the sides of a polygon of

vectors

60. Assertion: If | P Q |

= | P Q |

, then angle

between P

and Q

is 2 .

Reason: P Q

= Q P

(A) Assertion is True, Reason is True;

Reason is a correct explanation for Assertion.

(B) Assertion is True, Reason is True; Reason is not a correct explanation for Assertion.

(C) Assertion is True, Reason is False. (D) Assertion is False, Reason is True. 61. If A

+ B

= A

B

, then vector B

must be (A) zero vector (B) unit vector (C) non-zero vector (D) equal to A

62. The vectors A,

B

and C

are such that

| A | | B |,| C | 2 | A |

and A B C

= 0. The

angles between A

and B

, B

and C

respectively are [TS EAMCET (Med.) 2015]

(A) 45, 90 (B) 90, 135 (C) 90, 45 (D) 45, 135 63. Three vectors P

, Q

, R

are such that

| P |

= | Q |

, | R |

= 2 | P |

and P

+ Q

+ R

= 0.

The angle between P

and Q

, Q

and R

and P

and R

will be respectively. (A) 90º, 135º, 135º (B) 90º, 45º, 45º (C) 45º, 90º, 90º (D) 45º, 135º, 135º 64. Let the angle between two non-zero vectors A

and B

be 120º and its resultant be C

, then the correct statement is

(A) C must be equal to | A B |

(B) C must be greater than | A B |

(C) C must be less than | A B |

(D) C may be equal to | A B |

65. Two vectors A

and B

lie in a plane. Another vector C

lies outside the plane. The resultant

of these three vectors i.e., A

+ B

+ C

(A) can be zero. (B) cannot be zero. (C) lies in the plane of A

+ B

(D) lies in the plane containing A

B

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SAMPLE C

ONTENT

58

58

Absolute Physics Vol - I (Med. and Engg.)

66. Sum of magnitude of two forces is 25 N. The resultant of these forces is normal to the smaller force and has a magnitude of 10 N. Then the two forces are

[TS EAMCET (Engg.) 2015] (A) 14.5 N, 10.5 N (B) 16 N, 9 N (C) 13 N, 12 N (D) 20 N, 5 N 67. The magnitudes of two forces are in the ratio

3 : 5 and the angle between their directions is 60. If their resultant force is 28 N, then their magnitude will be

(A) 12 N, 20 N (B) 15 N, 25 N (C) 18 N, 30 N (D) 21 N, 28 N 68. The resultant of the three vectors shown in

figure and the angle made by the resultant with X-axis is [Assam CEE 2015]

(A) 10 m and 37 (B) 8.6 m and 35.5 (C) 5 3 m and 37

(D) None of these 69. The resultant of two forces acting in opposite

directions is 10 N. If these forces act perpendicular to each other, their resultant will be 50 N. The magnitude of these forces will be

(A) 40 N, 30 N (B) 50 N, 25 N (C) 18 N, 30 N (D) 21 N, 28 N 70. The maximum and minimum magnitudes of

the resultant of two given vectors are 17 units and 7 unit respectively. If these two vectors are at right angles to each other, the magnitude of their resultant is

[Kerala CET (Engg.) 2000] (A) 18 (B) 16 (C) 14 (D) 13 71. The angle between two vectors A and B is .

Vector R is the resultant of the two vectors. If R

makes an angle 2 with A, then [AFMC 1994]

(A) A = 2B (B) A = B2

(C) A = B (D) AB = 1 72. A body is at rest under the action of three

forces, two of which are 1F

= 4 i

, 2F

= 6 j

, the third force is [AMU 1996]

(A) 4 i

+ 6 j

(B) 4 i

6 j

(C) 4 i

+ 6 j

(D) 4 i

6 j

73. Two vectors A

and B

have components Ax, Ay, Az and Bx, By, Bz respectively. If A

+ B

= 0 , then [Orissa JEE 2010] (A) Ax = Bx , Ay = By , Az = Bz

(B) Ax = Bx , Ay = By , Az = Bz

(C) Ax = Bx , Ay = By , Az = Bz

(D) Ax = Bx , Ay = By , Az = Bz 74. The components of the sum of two vectors

2 i

+ 3 j

and 2 j

+ 3 k

along x and y directions respectively are [J & K CET 2010]

(A) 2 and 5 (B) 4 and 6 (C) 2 and 6 (D) 4 and 3 75. A body moves 6 m north, 8 m east and 10 m

vertically upwards. What is its resultant displacement from initial position?

[DCE 2000] (A) 10 2 m (B) 10 m

(C) 102

m (D) 10 2 m

76. If P

is multiplied by a real number 3, then new vector after multiplication will be represented by

(A) P

(B) 3 P

(C) 3 P

(D) P

77. If an arbitrary number ‘2’ is multiplied with

vector A

, then (A) the magnitude of vector will be doubled

and direction will be same. (B) the magnitude of vector will be doubled

and direction will be opposite. (C) the magnitude of vector and its direction

remains constant. (D) magnitude will be half and direction

remains constant. 78. Vectors A 5i a j 3k

and B 10i 8 j bk

are parallel to each other, then values of ‘a’ and ‘b’ are

(A) 4, 6 (B) 4, – 6 (C) – 4, 6 (D) – 4, – 6 79. The velocity of a particle is v

= 6 i

+ 2 j

2 k

. The component of the velocity of a particle

parallel to the vector a

= i

+ j

+ k

in vector form is

(A) 6 i

+ 2 j

+ 2 k

(B) 2 i

+ 2 j

+ 2 k

(C) 2 i

2 j

+ 2 k

(D) 6 i

+ 2 j

– 2 k

2.0 m3.0 m

37X

Y

Multiplication of vectors bya real number and scalar

2.5

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SAMPLE C

ONTENT

59

Chapter 02 : Scalars and Vectors

80. If A

= 4 i

3 j

and B

= 7 i

+ 24 j

, then the

vector parallel to A

and with same magnitude as B

will be

(A) 15 i

+ 20 j

(B) 20 i

15 j

(C) 5 i

25 j

(D) 8 i

+ 6 j

81. Force F

and displacement x

are vector quantities. Work = F x

will be (A) a scalar (B) a vector (C) both scalar and vector (D) Zero vector 82. Work done by the force is represented by

W = F

s

. In special circumstances, F and s are not equal to zero but the work done is zero. This concludes that

(A) F and s are in same direction. (B) F and s are in opposite direction. (C) F and s are perpendicular to each other. (D) F and s are inclined at 45. 83. Assertion: Angle between j k

and j

is 4 .

Reason: j k

is equally inclined both to j

and k

, which are mutually perpendicular. (A) Assertion is True, Reason is True;

Reason is a correct explanation for Assertion.

(B) Assertion is True, Reason is True; Reason is not a correct explanation for Assertion.

(C) Assertion is True, Reason is False. (D) Assertion is False, Reason is True. 84. If ˆ ˆ ˆA 2i 3j 8k

is perpendicular to ˆ ˆ ˆB 4j 4i k

, then the value of ‘’ is [CBSE PMT 2005, K CET 2017]

(A) 12

(B) 12

(C) 1 (D) –1 85. Three vectors ˆ ˆ ˆA ai j k

; ˆ ˆ ˆB i bj k

and ˆ ˆ ˆC i j ck

are mutually perpendicular

( i , j and k are unit vectors along X, Y and Z axis respectively). The respective values of a, b and c are [WB JEE 2017]

(A) 0, 0, 0 (B) 1 1 1, ,2 2 2

(C) 1, –1, 1 (D) 1 1 1, ,2 2 2

86. Given A

= 2 i

+ 3 j

and B

= i

+ j

. The

component of vector A

along B

is

(A) 12

i j

(B) 32

i j

(C) 52

i j

(D) 72

i j

87. The position vectors of four points A, B, C

and D are a

= 2 i

+ 3 j

+4 k

, b

= 3 i

+5 j+7 k

,

c

= i

+ 2 j

+ 3 k

and d

= 3 i

+ 6 j

+ 9 k

respectively. Then vectors AB

and CD

are (A) coplanar (B) collinear (C) perpendicular (D) antiparallel 88. A force F

= 3 i

+ c j

+ 2 k

acting on a particle

causes a displacement s

= – 4 i

+ 2 j

– 3 k

in its own direction. If the work done is 6 J, then the value of c will be [D PMT 1997]

(A) 12 (B) 6 (C) 1 (D) 0 89. The vectors P

= a i

+ a j

+ 3 k

and

Q

= a i

– 2 j

– k

are perpendicular to each other. The positive value of a is

[AFMC 2000; AIIMS 2002] (A) 3 (B) 4 (C) 9 (D) 13 90. If vectors A

= cost i

+ sint j

and

B

= cos t i2

+ sin t j2

are functions of time,

then the value of t at which they are orthogonal to each other is

[AIPMT Re-Test 2015]

(A) t = 0 (B) t = 4

(C) t = 2

(D) t =

91. Consider three vectors A i j 2k,

B i j k

and C 2i 3 j 4k

. A vector X

of the form A B

( and are numbers) is

perpendicular to C

. The ratio of and is [WB JEEM 2014]

(A) 1 : 1 (B) 2 : 1 (C) 1 : 1 (D) 3 : 1

Scalar product (dot product) of vectors 2.6

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SAMPLE C

ONTENT

60

60

Absolute Physics Vol - I (Med. and Engg.)

92. When A

B

= | A

| | B

|, then [Orissa JEE 2003]

(A) A

and B

are perpendicular to each other

(B) A

and B

act in the same direction

(C) A

and B

act in the opposite directions

(D) A

and B

can act in any direction 93. If | 1V

+ 2V

| = | 1V

2V

| and 2V is finite, then [C PMT 1989]

(A) V1 is parallel to V2

(B) 1V

= 2V

(C) V1 and V2 are mutually perpendicular

(D) | 1V

| = | 2V

| 94. If the magnitude of sum of two vectors is

equal to the magnitude of difference of the two vectors, the angle between these vectors is

[NEET P-I 2016; Similar in CBSE PMT 1991, 92, 2000,

WB JEE 2016] (A) 45 (B) 180 (C) 0 (D) 90 95. A force i

+ 2 j

– 3 k

N displaces a body from position vector of point (2, 4, –1) m to the position vector of point (5, 2, –4) m. The work done is

(A) 8 J (B) 9 J (C) 10 J (D) 7 J 96. If i

, j

and k

are unit vectors along x, y and z-axes respectively, then the angle between

the vectors i j k

and i

is given by [Assam CEE 2015,

Similar in MH-CET 2014] (A) = cos–1 1

3

(B) = sin–1 13

(C) = cos–1 32

(D) = sin–1 32

97. The angle between vectors

A

= 3 i

+ 4 j

+ 5 k

and B

= 6 i

+ 8 j

+ 10 k

is

(A) zero (B) 45 (C) 90 (D) 180 98. In a triangle ABC, the sides AB and AC are

represented by the vectors ˆ ˆ ˆ3i j k and ˆ ˆ ˆi 2 j k respectively. Calculate the angle

∠ABC. [WB JEE 2018]

(A) 1 5cos11

(B) 1 6cos11

(C) 1 590 cos11

(D) 1 5180 cos11

99. A vector A

points vertically upward and B

points towards north. The vector product A

B

is (A) Zero (B) Along west (C) Along east (D) Vertically downward 100. If angle between the vectors A

and B

is θ, which one of the following relations is correct?

(A) A B B A

(B) A B AB

(C) A B ABcos

(D) A B B A

101. If is the force and r

is the position vector then value of torque is

(A) r

F

(B) r

F

(C) | r

| | F

| (D) | r

| / | F

| 102. In a clockwise system, [CPMT 1990] (A) j k i

(B) i i = 0

(C) j j 1

(D) k j =1

103. The torque of the force F

= (2 i

– 3 j

+ 4 k

)

acting at the point r

= (3 i

+ 2 j

+ 3 k

) about the origin is [CBSE PMT 1995]

(A) 6 i

– 6 j

+ 12 k

(B) 17 i

– 6 j

– 13 k

(C) – 6 i

+ 6 j

– 12 k

(D) –17 i

+ 6 j

+ 13 k

104. The moment of the force, ˆ ˆ ˆF 4i 5 j 6k

at (2, 0, –3), about the point (2, –2, –2), is given by [NEET (UG) 2018]

(A) ˆ ˆ ˆ8i 4j 7k (B) ˆ ˆ ˆ4i j 8k (C) ˆ ˆ ˆ7i 8j 4k (D) ˆ ˆ ˆ7i 4j 8k

F

Vector product (cross product)of vectors

2.7

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105. A force F = 5i + 2 j 5k

acts on a particle

whose position vector is r = i 2 j k

. What is the torque about the origin? [K CET 2014]

(A) 8i+10 j+12k

(B) 8i+10 j 12k

(C) 8i 10 j 8k

(D) 10i 10 j k

106. The value of ( A

+ B

) ( A

B

) is [R PET 1991, 2002; BHU 2002]

(A) 0 (B) A2 – B2 (C) B

A

(D) 2( B

A

) 107. The position of a particle is given by

r

= ( i

+ 2 j

– k

), momentum p

= (3 i

+ 4 j

2 k

). The angular momentum is perpendicular to

[EAMCET (Engg.) 1998] (A) X-axis (B) Y-axis (C) Z-axis (D) Line at equal angles to all the three axes 108. What is the unit vector perpendicular to the

vectors, 2 i

+ 2 j

k

and 6 i

– 3 j

+ 2 k

?

(A) i 10 j 18k5 17

(B) i 10 j 18k5 17

(C) i 10 j 18k5 17

(D) i 10 j 18k5 17

109. Scalar product of two vectors is 2 3 and the

magnitude of their vector product is equal to 2, then the angle between them will be

(A) 30 (B) 45 (C) 60 (D) 90 110. The modulus of the vector product of two

vectors is 13

times their scalar product. The

angle between vectors is

(A) 6 (B)

2 (C)

4 (D)

3

111. Two adjacent sides of a parallelogram are

represented by the two vectors i

+ 2 j

+ 3 k

and 3 i

2 j

+ k

. What is the area of parallelogram? [AMU 1997]

(A) 8 (B) 8 3 (C) 3 8 (D) 192

112. The position vectors of radius are 2 i

+ j

+ k

and 2 i

–3 j

+ k

while those of linear

momentum is 2 i

+ 3 j

– k

. Then the angular momentum is [BHU 1997]

(A) 2 i

– 4 k

(B) 4 i

+ 8 k

(C) 2 i

– 4 j

+ 2 k

(D) 4 i

– 8 k

113. The linear velocity of a rotating body is given

by v

=

r

, where

is the angular velocity

and r

is the radius vector. The angular velocity

of a body is

= i

– 2 j

+ 2 k

and the radius

vector r

= 4 j

– 3 k

, then | v |

is (A) 29 units (B) 31 units (C) 37 units (D) 41 units 114. In the figure, the vectors from origin to the

points A and B are a

= 3 i

– 6 j

+ 2 k

and

b

= 2 i

+ j

– 2 k

respectively. The area of the triangle OAB is

(A) 5 17

2 sq.unit (B) 2 17

5sq.unit

(C) 3 175

sq.unit (D) 5 173

sq.unit 115. The three coterminous edges of a

parallelopiped are a

= 2 i

–6 j

+ 3 k

, b

= 5 j

,

c

= –2 i

+ k

. The volume of parallelopiped is (A) 36 cubic unit (B) 40 cubic unit (C) 45 cubic unit (D) 54 cubic unit 116. The X-component of the resultant of several

vectors a. is equal to the sum of the x-components

of the vectors b. may be smaller than the sum of the

magnitudes of the vectors c. may be greater than the sum of the

magnitudes of the vectors d. may be equal to the sum of the

magnitudes of the vectors (A) a, c, d (B) a, b, c (C) a, b, d (D) b, c, d 117. An object is subjected to a force in the north-

east direction. To balance this force, a second force should be applied in the direction

[K CET 1994] (A) North-East (B) South (C) South-West (D) West

Miscellaneous

O B

A (3 i

– 6 j

+ 2 k

)

(2 i

+ j

– 2 k

)

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118. Choose the WRONG statement (A) The division of vector by scalar is valid. (B) The multiplication of vector by scalar is

valid. (C) The multiplication of vector by another

vector is valid by using vector algebra. (D) The division of a vector by another

vector is valid by using vector algebra. 119. If electric current is assumed as vector

quantity, then (A) charge conservation principle fails. (B) charge conservation principle does not

fail. (C) Coulomb’s law fails. (D) both Coulomb’s law and charge

conservation principle fail. 120. A vector a

is turned without a change in its length through a small angle d. The values of | Δ a |

and a are respectively (A) 0, ad (B) ad, 0 (C) 0, 0 (D) ad, ad 121. The position vector of a particle is determined

by the expression r

= 3t2 i

+ 4t2 j

+ 7 k

. The distance traversed in first 10 s is

[D PMT 2002] (A) 500 m (B) 300 m (C) 150 m (D) 100 m 122. Position of a particle in a rectangular

co-ordinate system is (3, 2, 5). Then its position vector will be

(A) 3 i

+ 5 j

+ 2 k

(B) 3 i

+ 2 j

+ 5 k

(C) 5 i

+ 3 j

+ 2 k

(D) 2 i

+ 3 j

+ 5 k

123. Obtain the direction cosines of vector

( A

B

), if A

= 2 i

+3 j

+ k

, B

= 2 i

+2 j

+3 k

.

(A) 0, 15

, 25

(B) 0, 25

, 15

(C) 0, 0, 15

(D) 15

, 0, 15

124. Five equal forces of 10 N each are applied at one point and all are lying in one plane. If the angles between them are equal, the resultant force will be [CBSE PMT 1995]

(A) Zero (B) 10 N (C) 20 N (D) 10 2 N 125. Assertion: Minimum number of non-equal

vectors in a plane required to give zero resultant is three.

Reason: If P Q R 0

, then they must be coplanar.

(A) Assertion is True, Reason is True; Reason is a correct explanation for Assertion.

(B) Assertion is True, Reason is True; Reason is not a correct explanation for Assertion.

(C) Assertion is True, Reason is False. (D) Assertion is False, Reason is True. 126. If two vectors 2 i

+ 3 j

k

and 4 i

6 j

+ k

are parallel to each other, then the value of will be

(A) 0 (B) 2 (C) 3 (D) 4 127. If the sum of two unit vectors is a unit vector,

then magnitude of difference is [CBSE PMT 1989; C PMT 1995]

(A) 2 (B) 3

(C) 12

(D) 5 128. Assertion: P

P

= 0

. Also, P

P

= 0

.

Hence, P

P

= P

P

. Reason: Two vectors are equal if their

magnitudes are equal and have the same direction.

(A) Assertion is True, Reason is True; Reason is a correct explanation for Assertion.

(B) Assertion is True, Reason is True; Reason is not a correct explanation for Assertion.

(C) Assertion is True, Reason is False. (D) Assertion is False, Reason is True. 129. The resultant of two vectors having

magnitudes 2 and 3 is 1. What is their cross product?

(A) 6 (B) 3 (C) 1 (D) 0 130. The vector sum of two forces is perpendicular

to their vector differences. In this case, the forces [AIEEE 2002; CBSE PMT 2003]

(A) are equal to each other in magnitude. (B) are not equal to each other in

magnitude. (C) cannot be predicted. (D) are equal to each other in direction.

B

A

O

a

a

d

a

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131. For what value of ‘x’ are A

= i

– 2 j

+ 3 k

,

B

= x j

+ 3 k

and = 7 i

+ 3 j

– 11 k

coplanar?

(A) 3621

(B) 5132

(C) 5132

(D) 3621

132. For any two vectors A

and B

, if

A B | A B |

, the magnitude of C

= A

+ B

is equal to

(A) 2 2A B (B) A + B

(C) 2 2 ABA B2

(D) 2 2A B 2 AB 133. If | A

B

| = 3 A

. B

, then the value of

| A

+ B

| is [CBSE PMT 2004]

(A) 122 2 ABA B

3

(B) A + B

(C) 1

2 2 2A B 3AB

(D) 1

2 2 2A B AB 134. The angle between the vectors A

and B

is . The value of the triple product A

( B

A

) is [CBSE PMT 1991, 2005]

(A) A2B (B) Zero (C) A2B sin (D) A2B cos 135. If a

and b

are two parallel vectors, then the

value of ( a

+ b

) ( a

– ) is [BHU 2002]

(A) 2 ( b

a

) (B) – 2 ( b

a

)

(C) ( b

a

) (D) a

b

136. Three vectors satisfy the relations A

B

= 0

and A

C

= 0, then A

is parallel to [K CET 2003]

(A) B

C

(B) B

C

(C) C

(D) B

1. (D) 2. (D) 3. (C) 4. (B) 5. (B) 6. (D) 7. (C) 8. (D) 9. (C) 10. (C) 11. (D) 12. (C) 13. (D) 14. (D) 15. (D) 16. (A) 17. (C) 18. (A) 19. (D) 20. (A) 21. (B) 22. (A) 23. (C) 24. (D) 25. (C) 26. (B) 27. (B) 28. (B) 29. (B) 30. (A) 31. (A) 32. (D) 33. (B) 34. (B) 35. (D) 36. (D) 37. (A) 38. (D) 39. (D) 40. (B) 41. (A) 42. (B) 43. (C) 44. (C) 45. (D) 46. (A) 47. (B) 48. (B) 49. (D) 50. (D) 51. (A) 52. (B) 53. (A) 54. (A) 55. (C) 56. (D) 57. (D) 58. (A) 59. (D) 60. (B) 61. (A) 62. (B) 63. (A) 64. (B) 65. (B) 66. (A) 67. (A) 68. (B) 69. (A) 70. (D) 71. (C) 72. (D) 73. (D) 74. (A) 75. (A) 76. (C) 77. (B) 78. (C) 79. (B) 80. (B) 81. (A) 82. (C) 83. (A) 84. (A) 85. (B) 86. (C) 87. (B) 88. (A) 89. (A) 90. (D) 91. (A) 92. (C) 93. (C) 94. (D) 95. (A) 96. (A) 97. (A) 98. (A) 99. (B) 100. (D) 101. (A) 102. (A) 103. (B) 104. (D) 105. (A) 106. (D) 107. (A) 108. (C) 109. (A) 110. (A) 111. (B) 112. (B) 113. (A) 114. (A) 115. (B) 116. (C) 117. (C) 118. (D) 119. (A) 120. (B) 121. (A) 122 (B) 123. (A) 124. (A) 125. (B) 126. (B) 127. (B) 128. (D) 129. (D) 130. (A) 131. (B) 132. (D) 133. (D) 134. (B) 135. (A) 136. (A) 2. Scalar quantity is defined only by its

magnitude. Hence it has the same value for observers with different orientations of the axes.

9. Position vector is a localised vector. i.e., it

has a fixed initial point. Hence, position of particle is represented by a unique vector.

10. 13. Both the magnitude and direction of vector are

indicated using rectangular components and unit vectors of the given vector. In which, unit

C

b

Answers to MCQ's

Hints to MCQ's

Axial vector

Axis of rotation

Right hand screw rule

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vectors indicate only the direction. Axial vector and polar vectors are just two different types of vectors which can possess any given magnitude and direction.

14. Unit vector in the direction of A

, A = A

| A |

= 22 2

(2i j 4k)

2 4 4

= (i j k)

3

15. Unit vector = ˆ ˆ ˆ0.8i bj 0.4k

2 220.8 b 0.4 = 1 0.64 + b2 + 0.16 = 1 0.80 + b2 = 1 b2 = 1 – 0.8 = 0.2 b = 0.2 16. P

= 12

i

+ 12

j

| P |

= 2 21 1

2 2

= 1

It is a unit vector. 17. A person while walking pushes the road with

his feet backward by a force F at an angle with ground. The road in reaction exerts an equal and opposite force R (= F) on the feet as shown in figure. R is resolved into two rectangular components.

Vertical component R sin balances the weight of the man while the horizontal component R cos helps the man to walk forward provided it is greater than the force of friction.

18. Component of vector r

along x-axis is r cos. xr

= r cos Now rx will have maximum value if cos = 1

= cos1(1) = 0

component of r

along x-axis will have maximum value if r

is along +ve x-axis. 19. As the multiple of j

in the given vector is zero, this vector lies in xz-plane and projection of this vector on Y-axis is zero.

20. Let the components of A

make angles , and with X, Y and Z axes respectively, then = =

cos2 + cos2 + cos2 = 1 3cos2 = 1 or cos = 1

3

Ax = Ay = Az = A cos = A3

21. tan = 2

3

= tan1 23

22. A

= 2 i

+ 4 j

– 5 k

| A

| = 2 2 2(2) + (4) + ( 5) = 45

cos = 245

, cos = 445

, cos = 545

23. sin2 + sin2 + sin2 = 1 cos2 + 1 cos2 + 1 cos2 = 3 (cos2 + cos2 + cos2) = 3 1 = 2 24. P Q

P Q

Sum of two vectors cannot be equal to sum of

their unit vectors. 26. A

2 B

+ 3 C

= (2i j)

2 (3j k)

+ 3 (6i 2k)

= 2i

+ j

6 j

+ 2k

+18i

6k

= 20 i

5j

4k

28. ( i

– 2 j

+ 2 k

) + (2 i

+ j

– k

) + R

= i

Required vector, R

= –2 i

+ j

– k

29. Unit vector along Y-axis is j. So, the required

vector,

R

= j

– [( i

3 j

+ 2 k

) + (3 i

+ 6 j

–7 k

)]

= – 4 i

2 j

+ 5 k

R

R cos

F

Rsin

r sin

r cos x

y

r

O

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30. Resultant of vectors A

and B

,

R

= A

+ B

= 4 i

+ 3 j

+ 6 k

i

+ 3 j

8 k

R

= 3 i

+ 6 j

2 k

R

= R

| R |

= 2 2 2

3i+ 6 j 2k3 + 6 + ( 2)

= 3i+ 6 j 2k

7

31. r

= a

+ b

+ c

= 4 i

j

3 i

+ 2 j

k

= i

+ j

k

r

= r| r |

= 2 2 2

i j k1 1 ( 1)

= i j k

3

32. r

= 2r

1r

= (–2 i

– 2 j

+ 0 k

) (4 i

– 4 j

+ 0 k

)

r

= –6 i

+ 2 j

+ 0 k

| r

| = 2 2 2( 6) + (2) + 0 = 36 + 4 = 40 2 10 33. Let PQ and QR represent a

and b

in the same order, then according to triangle law of

vector addition, the resultant of a

and b

is represented by PR

i.e. P R = a

+ b

Since PR = RP = c

,

c

= a

+ b

a b c

= 0 34. Here, B C

= ˆ ˆ ˆ ˆ ˆ ˆi 3j 5k 2i j 4k

= ˆ ˆ ˆ3i 2 j k A

As, ˆ ˆ ˆA 3i 2 j k

,

A

= 9 4 1 14 ….(i)

Similarly,

B

= 1 9 25 35 ….(ii)

C

= 4 1 16 21 ….(iii)

From equations (i), (ii) and (iii), we get, B2 = A2 + C2

35. AB + AC + AD + AE + AF = AB + AD DC + AD + AD DE

+ AF

= 3 AD + AB DE + DC AF

= 3 AD …. AB = DE , and DC = AF

3 2AO 6AO

36. If two vectors A

and B

are given, then the resultant Rmax = A + B = 7 N and Rmin = 4 3 = 1 N i.e., net force on the particle is between 1 N and 7 N.

37. According to the problem, P + Q = 3 and P Q = 1 By solving, we get P = 2 and Q = 1

PQ

= 2 or P = 2Q 38. R

= 1 2 3F F F

= (3 – 3) i

+ (–4 + 4) j

+ (5 – 5) k

= 0 Hence, the direction is indeterminate. 39. A = 3 N, B = 2 N R = 2 2A B 2ABcos R = 9 4 12cos .…(i) Now, A = 6 N, B = 2 N then 2R = 36 + 4 + 24 cos θ .…(ii)

From (i) and (ii) we get, cos = 12

= 120º 40. A

+ B

= 4 i

3 j

+ 6 i

+ 8 j

= 10 i

+ 5 j

| A

+ B

| = 5 5

tan = 510

= 12

or = tan1 12

41. R = 2 24 + 4 + 2× 4× 4cos120° = 32 16 = 4

= tan–1 4×sin120°4 + 4×cos120°

= tan–1 (1.73) 42. tan = 2F sin θ

F + 2F cos θ= (as = 90°)

F + 2F cos = 0

or cos = – 12

or = 120 44. Resultant of two vectors A

and B

can be given by, R

= A

+ B

| R

| = | A

+ B

| = 2 2A B 2ABcos

If = 0, then | R

| = A + B = | A

| + | B

|

2F R

= 90°F

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45. Since, R = 2 2A B 2ABcos A = B = R A2 = 2A2 + 2A2 cos

cos = – 12

= cos 120

= 120 46. Since A

+ B

= C

, vector C

is the resultant

of vector A

and B

. Using the triangle law of

vector addition, we have = 45º ( A

= C

) 47. B

2= 2 2A B 2AB cos .…(i)

tan 90 = BsinθA Bcosθ+

A + B cos = 0

cos = AB

Hence, from (i), 2B

4 = A2 + B2 2A2

A = B32

cos = AB

= – 32

= 150 48. tan 90 = Qsin

P Q cos

P + Q cos = 0

cos = PQ or = cos1 P

Q

52. | A

| + | B

| = 18 .…(i) 12 = 2 2A B 2ABcos .…(ii)

tan = Bsin

A Bcos

= tan 90

or cos = – AB

.…(iii)

By solving (i), (ii) and (iii), A = 13 N and B = 5 N 54. min

max

RR

= 4 34 3

= 17

55. = 120 30 = 90 R = 2 23 4 2 3 4cos90 = 5 unit 56. The magnitude of the resultant vector of two

given vectors can be less than the magnitude of individual vectors if the angle between two vectors is in between

2 to 3

2 .

57. If two vectors A and B are given, then range

of their resultant can be written as (A B) R (A + B).

i.e. Rmax = A + B and Rmin = A B

If B = 1 and A = 4 then their resultant will lie in between 3 N and 5 N.

Hence it can never be 2 N. 62. As A B C 0,

it means A,

B

and C

form a closed triangle and hence from triangle law, Resultant is zero.

Also, As | A | | B|,

And | C | 2 | A |

A B

and angle between B and C is 180 45 = 135 66. (25 – x)2 = 102 + x2

625 + x2 50x = 100 + x2 x = 10.5 N 25 – x = 14.5 N 67. Let A and B be the two forces. Then A = 3x; B = 5x; R = 28 N and = 60 Thus, A

B= 3

5

Now, R = 2 2A B 2ABcos 28 = 2 2 03x 5x 2 3x 5x cos60

28 = 2 2 29x 25x 15x = 7x

x = 287

= 4

Forces are; A = 3 4 = 12 N and B = 5 4 = 20 N 68. R A B C

From figure we have, A

= 4 i 3 j

….(i)

B

= 3 i

….(ii)

C

= 2 j

….(iii)

Resultant is given by R

= A

+ B

+ C

R

= (4 i

+ 3 j

) + 3 i

+ 2 j

R

= 7 i

+ 5 j

Magnitude of resultant vector is

| R |

= 49 25

C

A

B

C

A

B

135

10 25 – x

x

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| R |

= 74

| R |

= 8.6 m and angle with X-axis is

= 35.5 69. As per given condition; A B = 10 ….(i) and 2 2A B = 50 ….(ii) From (i), A = B + 10 Putting this value in (ii), we get (B + 10)2 + B2 = 502. On solving we get, B = 30 N or 40 N. Therefore, A = 40 N or 30 N. 70. Rmax = A + B = 17 when = 0 Rmax = A B = 7 when = 180 by solving we get, A = 12 and B = 5 Now when = 90 then R = 2 2A + B R = 2 2(12) + (5) = 169 = 13 71. The angle which the resultant R makes with

A is given by tan = Bsin θ

A + Bcosθ

tan2

= BsinA Bcos

….2

sin

2

cos2

=

θ θ2Bsin cos2 2

A + Bcosθ

which gives, A + B cos = 2B cos2 2

A + B 22cos 12

= 2B cos2

2

which gives A = B. 72. As the body is at rest,

The net force acting on the body ( netF

) is zero

netF

= 1F

+ 2F

+ 3F

0 = 1F

+ 2F

+ 3F

3F

= 1F

2F

3F

= 4 i

6 j

73. Given that A

+ B

= 0 (Ax i

+ Ay j

+ Az k

) + (Bx i

+ By j

+ Bz k

) = 0

(Ax + Bx) i

+ (Ay+ By) j

+ (Az + Bz) k

= 0 Ax + Bx = 0; ….(i) Ay + By = 0; ….(ii) Az + Bz = 0; ….(iii)

From the above equations, we get Ax = Bx , Ay = By , Az = Bz 74. Let

A

= 2 i

+ 3 j

and B

= 2 j

+ 3 k

A

+ B

= (2 i

+ 3 j

) + (2 j

+ 3 k

)

= 2 i

+ (3 + 2) j

+ 3 k

= 2 i

+ 5 j

+ 3 k

x – component of ( A

+ B

) = 2 i

y component of ( A

+ B

) = 5 j

75. r

= x i

+ y j

+ z k

| r

| = 2 2 2x + y + z

r = 2 2 26 +8 +10 = 10 2 m 78. A

and B

are parallel to each other. This implies A

= m B

. Comparing X-component, m = 1

2. Comparing Y-component, a = – 4 and

comparing Z-component b = 6. 80. As the required vector, say C

, is parallel to A

,

C

= n A

where, n is a real number,

C

= 4n i 3n j

Now, | C |

= | B |

2 24n 3n = 2 27 24 = 25 5n = 25 n = 5

C

= 4 5 i

3 5 j

= 20 i

15 j

83. If angle between j

and j k

is , then

cos = j ( j k)

| j | | j k |

= 1

1 2= 1

2

= 4 .

84. Given: A

= 2 i

+ 3 j

+ 8 k

, B

= – 4 i

+4 j

+ k

A B

= 0 ….[ A B

]

A B

= (2 i

+ 3 j

+ 8 k

)(– 4 i

+ 4 j

+ k

) = 2 ( 4) + 3 4 + 8 = 0 = 1

2

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85. As the vectors are mutually perpendicular,

A B B C A C 0

ˆ ˆ ˆ ˆ ˆ ˆa i j k i bj k 0

a + b + 1 = 0 ….(i) Similarly, 1 + b + c = 0 ....(ii) a + 1 + c = 0 ....(iii) Adding equations (i), (ii) and (iii), we get, 2(a + b + c) + 3 = 0

a + b + c = 32

1 + c = 32 ....[from (i)]

c = 12

Substituting in equation (ii) and (iii), we get,

a = b = 12

86. The component of vector A

along vector

B

= ( A

B

) B

where, B

= B

| B |

and | B |

is

the magnitude of vector B

.

Now ( A

B

) = (2 i

+ 3 j

) ( i

+ j

)

= 2 i

i

+ 2 i

j

+ 3 j

i

+ 3 j

= 2 + 0 + 0 + 3 = 5

Also, B

= B

| B |

= i j

| i j |

=

2 2

i j1 1

= i+ j

2

5B A B B i j2

87. AB

= (3 2) i (5 3) j (7 4)k

= i 2 j 3k

Similarly,

CD

= 2 i 4 j 6k

AB CD

= |AB||CD| cos

cos =2 2 2 2 2 2

(2) (8) (18)(1) (2) (3) (2) (4) (6)

= 1

= 0 The vectors are collinear if they are scalar

multiples of each other. 88. W = F

. s

= (3 i

+ c j

+ 2 k

).(–4 i

+ 2 j

– 3 k

) = 12 + 2c 6 Work done = 6 J ….(given) 12 + 2c 6 = 6 or c = 12

89. P Q

= 0 a2 2a 3 = 0 or a = 3 90. Vectors are orthogonal i.e. A B

= 0

cost cos t2

+ sin t sin t2

= 0

cos tt2

= 0

or cos t2

= 0

t2 =

2

t =

91. A B C 0,

[ (i j 2k) (i j k)].[(2 i 3 j 4k)] 0

2( + ) 3 ( ) + 4( 2) = 0 9 + 9 = 0 : = 1 : 1 92. A

B

= AB cos

Given, A

B

= A B

i.e. cos = 1 or = 180 93. Given, 1 2| V V |

= 1 2| V V |

,

then 21 2| V V |

= 21 2| V V |

or

1 2V V

1 2V + V

= 1 2V V

1 2V V

On solving we get, 4V1V2 cos = 0 or = 90 94. As A

+ B

= A

B

,

A2 + B2 + 2AB cos = A2 + B2 2AB cos 4AB cos = 0, i.e. cos = 0 = cos 90 = 90 97. A

B

= AB cos ….(i)

= (3 i

+ 4 j

+ 5 k

)(6 i

+ 8 j

+ 10 k

) = 3 6 + 4 8 + 5 10 = 100 ….(ii) The magnitudes of A and B are A = 2 2 2(3) + (4) + (5) = 50 ….(iii)

B = 2 2 2(6) (8) (10) = 200 ….(iv)

Substituting the values from (ii), (iii) and (iv)

in (i), we have 100 = 50 200 cos = 100 cos cos = 1 or = zero.

Page 27: 01 Physical world and measurement 01 · Target’s “Absolute Physics Vol - I” is compiled according to the notified syllabus for NEET-UG & JEE (Main), which in turn has been framed

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Chapter 02 : Scalars and Vectors

98. ˆ ˆ ˆAB 3i j k

, ˆ ˆ ˆAC i 2 j k

ˆ ˆ ˆ ˆ ˆ ˆCB AB AC 3i j k i 2j k

= ˆ ˆ2i j

∠ABC is angle between AB

and CB

, Consider,

AB CB

= sB oA CB c

....(i)

ˆ ˆ ˆ ˆ ˆAB CB 3i j k 2i j 6 1 5

2 2 2AB 3 1 1 11

and 2 2CB 2 (1) 5

5 = 11 5 cos ....[from (i)]

cosθ = 511

1 5cos11

99. Direction of vector A is along Z-axis (z-axis is r to plane of paper) A

= a k

Direction of vector B is

towards north

B

= b j

Now A

B

= a k

b j

= ab( i

)

The direction of A

B

is along west. 100. Since, cross product of two vectors is

anti-commutative.

103.

= r

F

= i j k3 2 32 3 4

= [(2 4) (3 (3))] i

[(4 3) (2 3)] j

+ [(3 (3)) (2 2)] k

= 17 i

6 j

13 k

104. ˆ ˆ ˆ ˆ ˆr 2i 3k 2i 2 j 2k

= ˆ ˆ2j k

r F

= ˆ ˆ ˆ ˆ ˆ2 j k 4i 5 j 6k

= ˆ ˆ ˆi j k0 2 14 5 6

= ˆ ˆ ˆi 12 5 j 0 4 k 0 8

= ˆ ˆ ˆ7i 4j 8k

105.

= r

F

= i j k1 2 15 2 5

= i(10 2) j( 5 5) + k(2 10)

= 8i 10 j +12k

106. ( A

+ B

) ( A

B

) = A

A

A

B

+ B

A

B

B

= 0 A

B

+ B

A

0 = B

A

+ B

A

= 2( B

A

)

107. L

= r

p

= i j k1 2 13 4 2

= – j

– 2 k

i.e. the angular momentum is perpendicular to X-axis. 108. A

= 2 i

+ 2 j

k

and B

= 6 i

3 j

+ 2 k

C

= A

B

= (2 i

+ 2 j

k

) (6 i

3 j

+ 2 k

)

= i j k2 2 16 3 2

= i

10 j

18 k

Unit vector perpendicular to both A

and B

u = 2 2 2

i 10 j 18k1 ( 10) ( 18)

= i 10 j 18k

5 17

109. A B

= | A | | B|

cos = 2 3 ….(i)

| A B |

= | A | | B|

sin = 2 ….(ii) Dividing (ii) by (i) we get,

tan = 22 3

= 13

= tan1(1/ 3 ) = 30 111. Area of parallelogram = A

B

= ( i

+ 2 j

+ 3 k

) (3 i

2 j

+ k

)

= i j k1 2 33 2 1

= (8) i

+ (8) j

– (8) k

Magnitude = 64 64 64 = 8 3 112. Radius vector

r

= 2 1r r

= (2 i

–3 j

+ k

) – (2 i

+ j

+ k

)

r

= 4 j

N

E

S

W

j

i

Page 28: 01 Physical world and measurement 01 · Target’s “Absolute Physics Vol - I” is compiled according to the notified syllabus for NEET-UG & JEE (Main), which in turn has been framed

SAMPLE C

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70

Absolute Physics Vol - I (Med. and Engg.)

70

Linear momentum, p

= 2 i

+ 3 j

– k

Angular momentum is given by

L

= r

p

= i j k0 4 02 3 1

= 4 i

+ 8 k

113. v

=

r

= i j k1 2 20 4 3

= i

(6 – 8) – j

(–3) + 4 k

= –2 + 3 j

+ 4 k

| v

| = 2 2 2( 2) (3) 4 = 29 units 114. Given, OA a 3i 6 j 2k

and

OB b 2i j 2k

( a

b

) = i j k3 6 22 1 2

= (12 2) i

+ (4 + 6) j

+ (3 + 12) k

= 10 i

+ 10 j

+ 15 k

| a

b

| = 2 2 210 10 15 = 425 = 5 17

Area of OAB = 12

| a

b

| = 5 172

sq.unit 115. Volume of parallelopiped = ( A

B

)C

A

B

= i j k2 6 30 5 0

= i

(0 15) j

(0) + k

(10 0)

= 15 i

+ 10 k

( A

B

)C

= (15 i

+ 10 k

)( 2 i

+ k

) = 30 + 10 = 40 cubic unit 117. Direction of second force should be at 180°. 120. From the figure, | OA | = a and | OB | = a

Also, from triangle rule, OB – OA = AB = a

| a

| = AB

Since d = arcradius

or AB = a.d

So, | a

| = ad

a means change in magnitude of vector i.e. | OB | | OA |

a a = 0 Hence, a = 0 121. r

= 3t 2 i

+ 4t 2 j

+ 7 k

At t = 0, 1r

= 7 k

At t = 10 s, 2r

= 300 i

+ 400 j

+ 7k

,

r

= 2r

1r

= 300 i

+ 400 j

| r

| = | 2r

1r

|= 2 2(300) (400) = 500 m 122. If a point has coordinates (x, y, z), then its

position vector = x i

+ y j

+ z k

. 123. A B

= j

2 k

, | R |

= 1 4 = 5 Direction cosines are,

cos = 05

cos = 15

cos = 25

124. If the angle between all the forces is equal and

the forces lie in one plane, then the resultant force will be zero.

125. For giving a zero resultant, it should be possible to represent the given vectors along the sides of a closed polygon with the minimum number of sides of a polygon to be three.

126. Let A

= 2 i

+ 3 j

k

and B

= 4 i

6 j

+ k

A

and B

are parallel to each other

1

1

ab

= 2

2

ab

= 3

3

ab

i.e. 24

= 36

= 1

or = 2 127. Let 1n and 2n be the two unit vectors, then

the sum is Sn = 1n + 2n 2

sn = 21n + 2

2n + 2n1n2 cos = 1 + 1 + 2 cos Since it is given that nS is also a unit vector,

therefore 1 = 1 + 1 + 2 cos

cos = 12

or = 120

Now the difference vector is, dn = 1n 2n or 2

dn = 21n + 2

2n 2n1n2 cos = 1 + 1 2cos(120)

i

Page 29: 01 Physical world and measurement 01 · Target’s “Absolute Physics Vol - I” is compiled according to the notified syllabus for NEET-UG & JEE (Main), which in turn has been framed

SAMPLE C

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Chapter 02 : Scalars and Vectors

2dn = 2 2(1/2) = 2 + 1 = 3

nd = 3 128. If P

P

= 0

and P

P

= 0

, then

P

P

P

P

because P

P

is perpendicular

to P

and ( P

P

) is collinear with P

. 129. R = 2 22 3 2 2 3 cos = 1 By solving, we get = 180 Cross product = 2 3 sin 180 = 0 130. Let the two forces be 1F

and 2F

according to given condition we have,

1 2F F

1 2F F

= 0

….(orthogonality condition)

1 1 1 2 2 1 2 2F F F F F F F F

= 0

1 1 2 2F F F F

= 0

1 1F F

= 2 2F F

2

1F cos= 22F cos

21F = 2

2F F1 = F2 131. If three vectors are coplanar, then

( A

B

)C

= 0

i.e.,1 2 30 x 37 3 11

= 0

1 ( 11x 9) + 2(0 21) + 3(0 7x) = 0 11x 9 42 21x = 0

32x = 51 or x = 5132

132. If A B | A B |

| A

|| B

| cos = | A

|| B

| sin tan = 1 or = 45

Magnitude of C

= 2 2A B 2AB cos

= 2 2 1A B 2AB2

= 2 2A B 2 AB 133. | A

B

| = 3 A

. B

AB sin = 3 AB cos tan = 3 = 60

cos = cos 60 = 12

Now, | R

| = | A

+ B

| = 2 2A B 2ABcos

= 2 2 1A B 2AB2

= 1

2 2 2A B AB 134. Let A

( B

A

) = A

Here, C

= B

A

which is perpendicular to

both vectors A

and B

A

C

= 0 135. ( a

+ b

) ( a

– b

)

= a

a

– a

b

+ b

– b

b

….(i) Cross product of parallel vectors is zero,

Therefore, a

a

= b

b

= 0

a

b

= (ab sin ) = – [(ba sin ) ]

= – b

a

a

b

= – b

a

Substituting the values in relation (i), we get

( a

+ b

) ( a

– b

) = 2( b

a

) 136. A

B

= 0; A

C

= 0

A

is perpendicular to B

as well as C

Now, let D B C

The direction of D

is perpendicular to the

plane containing B

and C

Hence, A

is parallel to D

that is, A

is parallel to B C

C

a

n n

X

ZD B C

C

Y

B

Page 30: 01 Physical world and measurement 01 · Target’s “Absolute Physics Vol - I” is compiled according to the notified syllabus for NEET-UG & JEE (Main), which in turn has been framed

SAMPLE C

ONTENT

72

Absolute Physics Vol - I (Med. and Engg.)

72

1. If a unit vector is represented by

0.5 i

0.8 j

+ c k

, then the value of c is (A) 0.01 (B) 0.11 (C) 0.39 (D) 1 2. If n is the unit vector in the direction of A

, then

(A) n = A

| A |

(B) n = A

| A

|

(C) n = | A |

A

(D) n = n A

3. Out of the following sets of forces, the

resultant of which set of forces can never be zero?

(A) 15, 15, 15 (B) 15, 30, 60 (C) 25, 15, 30 (D) 15, 30, 30 4. How many minimum number of vectors of

equal magnitude are required to produce zero resultant?

(A) 2 (B) 3 (C) 4 (D) More than 4 5. Any vector in an arbitrary direction can (A) always be replaced by two or three

parallel vectors which have the original vector as their resultant.

(B) always be replaced by two or three mutually perpendicular vectors which have the original vector as their resultant.

(C) always be replaced by two or three arbitrary vectors which have the original vector as their resultant.

(D) not be resolved into component vectors. 6. Find the resultant of three vectors, OA , OB

and OC shown in the following figure. Radius of the circle is R.

(A) 2R

(B) R(1 + 2 )

(C) R 2

(D) R( 2 – 1) 7. Consider a vector F

= 4 i 3 j . Another

vector that is perpendicular to F

is (A) ˆ ˆ4i + 3j (B) 6 i

(C) 7 k (D) ˆ ˆ3i 4j

8. For what value of x will the two vectors A

= 2 i + 2 j – x k and B

= 2 i j – 3 k be perpendicular to each other?

(A) x = 23

(B) x = 3

2

(C) x = 43

(D) x = 2

3 9. If A

= 3 i + 4 j and B

= 6 i + 8 j , then which of the following is NOT true?

(A) | A B |

= 10

(B) | A |

| B |

= 12

(C) | A.B |

= 50

(D) | A |

= 5 10. If c

= a

b

, then

(A) the direction of c

changes when the

angle between a

b

increases up to (0-180)

(B) the direction of c

changes, when the

angle between a

and b

decreases up to ( > 0° )

(C) the direction of c

does not change,

when the angle between a

and b

increases

(D) the direction of c

changes when angle

between a

and b

increases. 11. A

and B

are two vectors and θ is the angle between them. If | A

B

| = 3 ( A

B

), the value of θ is

(A) 30 (B) 45 (C) 60 (D) 90 12. Vector A

is along + X-axis and the vector B

is such that A

B

= 0, then B

will be

(A) 4 j

(B) – 4 i

(C) – i j

(D) j k

Topic Test

B C

O A 45o

45o

Page 31: 01 Physical world and measurement 01 · Target’s “Absolute Physics Vol - I” is compiled according to the notified syllabus for NEET-UG & JEE (Main), which in turn has been framed

SAMPLE C

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Chapter 02 : Scalars and Vectors

13. If A

= i

j

+ k

, then unit vector in the

direction of A

is

(A) 3 j

(B) ( i

j

+ k

)

(C) (i j k)2

(D) (i j+ k)3

14. If a vector ( A

) of magnitude 7 units is multiplied by 2, then the new vector quantity has a magnitude of

(A) 14 units in direction of A

(B) 14 units in opposite direction of A

(C) 7 units in direction of A

(D) 7 units in opposite direction of A

15. Vectors A 5 i 4 j a k

and B 10 i b j 6k

are parallel to each other, then values of ‘b’ and ‘a’ are

(A) 8, 3 (B) –8, –3 (C) –8, 3 (D) 8, –3 1. (B) 2. (A) 3. (B) 4. (A) 5. (C) 6. (B) 7. (C) 8. (A) 9. (A) 10. (C) 11. (C) 12. (B) 13. (D) 14. (B) 15. (C)

Answer to Topic Test