NEET-UG / JEE (Main) Challenger Physics Vol. - 1 · NEET – UG & JEE (Main) Salient Features • Exhaustive coverage of MCQs under each sub-topic. • ‘1819’ MCQs including questions

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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: India Printing Works, Mumbai

P.O. No. 131112

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

Challenger

PHYSICS Vol. I

NEET – UG & JEE (Main)

Salient Features • Exhaustive coverage of MCQs under each sub-topic. • ‘1819’ MCQs including questions from various competitive exams. • Includes solved MCQs from NEET, MHT-CET, JEE (Main) and various

entrance examinations from year 2015 to 2018. • Concise theory for every topic. • Hints provided wherever deemed necessary. • Test papers for thorough revision and practice. • Important inclusions: Mind Over Matter and Problems to Ponder.

TEID: 12820_JUP

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Target’s ‘Challenger Physics: Vol-I’ is a compact guidebook, extremely handy for preparation of various competitive exams like NEET, JEE (Main).

Features of each chapter: Theoretical Concepts presented in the form of pointers, tables and diagrams that form a vital part of any

competitive examination.

Multiple Choice Questions segregated into two sections. Concept Building Problems – Contains questions of various difficulty range and pattern. Practice Problems – Contains ample questions for thorough revision.

Formulae section for each chapter according to its relevance for quick revision.

Shortcuts section to help students save time while dealing with lengthy questions. Mind Over Matter: Tips/clues to help students understand and focus on the key-concept that is involved in

solving certain brain-racking questions.

Problems to Ponder: Multiple questions, passages, MCQs / non- MCQs of different pattern created with theprimary objective of helping students to understand the application of various concepts of Physics.

Two Model Test Papers are included to assess the level of preparation of the student on a competitive level.

MCQs have been created and complied with the following objective in mind – to help students solve complex problems which require strenuous effort and understanding of multiple-concepts. The MCQs are a mix of questionsbased 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 CBSE, AIIMS,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, MHT-CET, K CET, GUJ CET, WB JEEM, JEE (Main) of years 2015, 2016, 2017 and 2018 are covered exclusively.

The journey to create a complete book is strewn with triumphs, failures and near misses. If you think we’ve nearlymissed something or want to applaud us for our triumphs, we’d love to hear from you.

Please write to us on : [email protected]

A book affects eternity; one can never tell where its influence stops.

From, PublisherEdition: 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 28

3 Motion in One Dimension 54

4 Laws of motion 94

5 Motion in two Dimensions 135

6 Work, Energy and Power 173

7 System of particles and Rotational motion 204

8 Gravitation 249

9 Mechanical properties of solids: Elasticity 296

10 Mechanical properties of fluids: Viscosity 327

11 Mechanical properties of fluids: Surface Tension 359

12 Thermal properties of Matter: Heat 381

13 Thermodynamics 419

14 Kinetic Theory of Gases 447

15 Oscillations 472

16 Wave Mechanics 510

Model Test Paper - I 547

Model Test Paper – II 551

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

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Why Challenger Series? Gradually, every year the nature of competitive entrance exams is inching towards conceptual

understanding of topics. Moreover, it is time to bid adieu to the stereotypical approach of solving a problem using a single conventional method. To be able to successfully crack the NEET and JEE (Main) examination, it is imperative to develop skills such as data interpretation, appropriate time management, knowing various methods to solve a problem, etc. With Challenger Series, we are sure, you’d develop all the aforementioned skills and take a more holistic approach towards problem solving. The way you’d tackle advanced level MCQs with the help of hints, tips, shortcuts and necessary practice would be a game changer in your preparation for the competitive entrance examinations.

What is the intention behind the launch of Challenger Series? The sole objective behind the introduction of Challenger Series is to severely test the student’s

preparedness to take competitive entrance examinations. With an eclectic range of critical and advanced level MCQs, we intend to test a student’s MCQ solving skills within a stipulated time period.

What do I gain out of Challenger Series? After using Challenger Series, students would be able to: a. assimilate the given data and apply relevant concepts with utmost ease. b. tackle MCQs of different pattern such as match the columns, diagram based questions, multiple

concepts and assertion-reason efficiently. c. garner the much needed confidence to appear for various competitive exams. Can the Questions presented in Problems to Ponder section be a part of the NEET/JEE (Main)

Examination? No, the questions would not appear as it is in the NEET/JEE (Main) Examination. However, there are fair chances that these questions could be covered in parts or with a novel question construction.

Why is then Problems to Ponder a part of this book? The whole idea behind introducing Problems to Ponder was to cover an entire concept in one question.

With this approach, students would get more variety and less repetition in the book.

Best of luck to all the aspirants!

Frequently Asked Questions

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28

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

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

O

Y

X

Scalars: Physical quantities which have only magnitude

are called scalars. Vectors: Physical quantities which have magnitude as

well as direction and obey the laws of vector addition are called vectors.

[Note: Certain physical quantities which have both magnitude and direction but do not obey laws of vector addition are considered to be scalars.]

Position vector: 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.

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, 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.

i. Equal vectors:

Two vectors are said to be equal, if they have equal magnitude and same direction irrespective of their positions in space.

ii. Negative vector: Negative vector of a given vector is a

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

iii. 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. iv. Unit vector: 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= A

| A |

= vectormodulusof thevector

c. In cartesian co-ordinate system, i , j and k are the unit vectors along

X-axis, Y-axis and Z-axis respectively.

d. A unit vector is unitless and dimensionless.

Scalars and Vectors02

P

O X

Y

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

Scalars and vectors2.1

Types of vectors2.2

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29

Chapter 02: Scalars and Vectors

v. Co-initial vectors: Two or more vectors are said to be

co-initial, if their initial point is common. vi. Collinear vectors: Two or more vectors acting along the

same straight line are called collinear vectors.

vii. Coplanar vectors: Those vectors which are acting in the

same plane are called Coplanar vectors. viii. Localised vector: Localised vector is that vector whose

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

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

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

x. 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.

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

axis of rotation are called axial vectors. 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

Magnitude of this vector is given by

| A

| = 2 2x yA A

iii. In three dimensions,

A

= Ax i

+ Ay j

+ zA k

,

| 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. 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

| ii. Vectors of same nature alone can be added. iii. Vector addition is commutative and

associative

i.e., A

+ B

= B

+ A

and

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 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

A B

C

a

and b

are collinear vectors.

b a

B O A

O

R

P

Q

xA

A

yA

Y

X

Resolution of a vector in two rectangular components

Resolution of vectors 2.3

Addition and subtraction of vectors 2.4

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30

30


Q

O

S

R

P

B

B

A

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 and Rz = Az + Bz 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.

From figure,

| 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. Also, the direction of R

with respect to A

is given by angle as,

tan = BsinA B cos

iii. Polygon law of vector addition: If 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

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. For subtraction of two vectors A

and B

inclined at an angle ,

| R

| = 2 2A B 2ABcos(180 )

| R

| = 2 2A B 2ABcos

tan = BsinθA Bcosθ

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

i.e., A

B

B

A

A

B C

A B

C

i. The multiplication of a vector A

by a scalar or a real number n, gives 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 positive or negative.

n A

= n A

and n A

= n A

.

ii. The unit of n A

is the same as that of A

if n is a real number, but it is different if n is a scalar.

Q

O P

B

A

R

B

C

S D

T

R

O A

Q

P

Multiplication of vectors by a scalar 2.5

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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 of two vectors is a

scalar quantity. 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 = A A ( = 0) iii. For orthogonal vectors, = 90 and

A

B

= AB cos 90 = 0.

iv. i

. i

= j

. j

= k

. k

= 1

v. i j

= j k

= k i

= 0 vi. The distributive law holds good for scalar

product

A B C

= A B

+ A C

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. Â B ABsin n

. Where, n

is a unit vector perpendicular to the plane containing A

and B

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

i.e., A B B A

[ sin() = sin]

ii. A A

= 0 ( sin = 0)

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

vi. A B

=

x y z

x y x

ˆ ˆ î j kA A AB B B

= i (AyBz – AzBy) – j (AxBz –AzBx) + k (AxBy – AyBx) 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

Formulae

Scalar product (dot product) of vectors 2.6

Vector product (cross product) of vectors

2.7

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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 n ii. ˆ ˆ ˆ ˆ ˆ î j = j j = k k = 0 iii. ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ î 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. The magnitude of the resultant of A

and B

varies between A B to A + B.

2. If | A

B

| = | A

B

|, then the angle between A

and B

is 90º. 3. 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

4. If the resultant of A

and B

is perpendicular to A

, then the angle between A

and B

is,

= cos–1 AB

5. 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.

6. If two vectors represents 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.

7. The minimum number of non-coplanar vectors

whose sum can be zero is four. 8. 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 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.

2. Which of the following is a scalar? (A) Electric field (B) Electric dipole moment (C) Work (D) Torque


Shortcuts

Concept Building Problems

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3. If two physical quantities have same dimensions then,

(A) both should be scalar quantities. (B) both should be vector quantities. (C) one can be scalar and other can be

vector quantity. (D) both can either be scalar or vectors but

will be numerical multiple of each other. 4. Electric current has both magnitude and

direction, still it is considered as scalar quantity because

(i) electric current does not obey laws of vector addition.

(ii) electric current cannot be resolved into components unlike other vector quantities.

(iii) electric current is not variable along the bends.

(A) Only (ii) (B) (i) and (ii) (C) (ii) and (iii) (D) All of the above. 5. Assertion: Rotations of a body are not

categorized as vectors. Reason: Rotation is a type of angular

displacement which, in order to be specified, needs the direction of the axis of rotation as well as the angle of rotation about the axis.

(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. 6. 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. 7. 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.

8. 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

9. 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 10. Walking of a person on the road is an example of (A) scalars. (B) unit vector. (C) resolution of vectors. (D) null vector. 11. 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.

12. Given vector A

= 3 i

+ 3 3 j

, then the angle

between A

and Y-axis is (A) 30 (B) 35 (C) 60 (D) 45 13. 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

14. If A

= i

– 4 j

– 8 k

, then the direction

cosines of the vector A

are

(A) 19

, 89 and 4

9

(B) 19

, 29

and 49

(C) 19

, 49 and 8

9

(D) 19 , 4

9 and 8

9

Types of vectors2.2


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15. 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

16. If two vectors inclined at each other and their

resultant form a triangle such that each vector has same magnitude as that of resultant vector, then angle between the given vectors is

(A) 60 (B) 90 (C) 120 (D) 45

17. If two vectors Q

and PQ2

are shown in the

figure, then the figure representing vector QR 2 P2

is

(A) (B) (C) (D) 18. The components of the sum of two vectors

2 i

– 3 k

and –2 i

+3 j

along x and y directions respectively are

(A) 0 and 3 (B) 4 and 6 (C) 0 and 6 (D) 4 and 3 19. What vector must be added to the sum of two

vectors i

+ 2 j

+ 2 k

and 2 i

– j

– 2 k

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

(A) 2 i

+ j

(B) –2 i

+ j

– k

(C) –2 i

– j

– k

(D) –2 i

– j

20. If a

= –2 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

(A) 13

( i

+ j

k

)

(B) 12

( i

+ j

k

)

(C) 13

( i

j

+ k

)

(D) 12

( i

+ j

+ k

) 21. If a

, b

, c

are three consecutive vectors

forming a triangle, then a

+ b

+ c

is (A) (B) 0 (C) 1 (D) –1 22. If ˆ ˆ Â 3i 2 j k,

ˆ ˆ ˆB 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 23. 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

c

b

a

R

P Q

E D

C F

A B

O

R

R

The key to crack this question lies incomprehending that, negative vector hasdirection opposite to that of given vector.

Mind over Matter

QPQ

2

RR


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24. If three forces 1F

= –3 i

4 j

+ 5 k

,

2F

= 2 i

+ 4 j

and 3F

= ˆ î 5k 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. 25. A carrom board of 50 cm 50 cm is shown

below. Assuming side B as +ve Y-axis and C as +ve X-axis, If striker shot from middle of the border from side C, obeying laws of reflection with respect to X-axis, stops exactly opposite to its original position at side A, then the position vector of a point at which the striker hits along side B is

(Given: Border of carrom board is 8 cm away from its edge.)

(A) 25i 25 j cm

(B) 50 i 8 j cm

(C) 25 j cm

(D) 25i 42 j cm

26. A vector a

makes an angle 30° and b

makes an angle 120 with the X-axis. The magnitude of these vectors are 9 unit and 12 unit respectively. The magnitude of resultant vector is

(A) 23 unit (B) 4 unit (C) 15 unit (D) 3 unit 27. If magnitude of the resultant of two vectors is

equal to the sum of their individual magnitudes, then the vectors are

(A) perpendicular (B) coinitial (C) antiparallel (D) parallel 28. Two equal vectors have a resultant equal to

either of them. The angle between them is (A) 30 (B) 60 (C) 90 (D) 120 29. Two forces of equal magnitude, F = 10 N are

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

(A) 10 cos (B) 5 cos2

(C) 20 cos2 (D) 10 cos

2

30. Two forces 3 N and 2 N are at an angle such that the resultant is R. The second force is now increased to 4 N and the resultant remains same. The value of is

(A) c

6 (B)

c

4

(C) c (D) c

2

31. The magnitudes of two forces are in the ratio

7 : 8 and the angle between their directions is 60. If their resultant force is 26 N, then their magnitudes respectively will be

(A) 12 N, 20 N (B) 15 N, 25 N (C) 14 N, 16 N (D) 21 N, 24 N 32. The angle between two vectors A and B is

60. Vector R is the resultant of the two vectors. If R makes an angle 30 with A, then

(A) A = 2B (B) 2A = B (C) A = B (D) AB = 1 33. 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] (A) 45 (B) 180 (C) 0 (D) 90 34. 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 35. 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 36. 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 of A

B

C

A

D B

SAMPLE C

ONTENT

36

36


37. Which of the plot best represents resultant R

of 30 units at 45 and 20 units at 280? (Angles given are measured counter – clockwise

from the positive X-axis) (A) (B) (C) (D) 38. Vectors A 5i 4 j a k

and B 10i 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 39. Assertion: Angle between i k

and i

is 4 .

Reason: i k

is equally inclined both to X

and Z axes.

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


(C) Assertion is True, Reason is False. (D) Assertion is False, Reason is True. 40. The vector sum of two velocities is

perpendicular to their vector differences. In this case, the velocities

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

magnitude. (C) are at an angle of 90 with each other. (D) are equal to each other in direction. 41. If a vector 2i

+ 3 j

+ 6k

is perpendicular to

the vector 6i

+ 8 j

+ k

, then the value of α is

(A) 12

(B) 12 (C) 1 (D) –2

42. Given A

= 5i

+ 3 j

and B

= i

– j

. The

magnitude of component of vector A

along

B

is

(A) 12

unit (B) 2 unit

(C) 2 unit (D) 12

unit 43. If vectors A

= cost i

+ sint j

and

B

= cos ωt2

i

+ sin ωt2

j

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 = π4ω

(C) t = π2ω

(D) t = πω

The key to crack this question lies in comprehending that, perpendicular vectorssatisfy the orthogonality condition.

Mind over Matter

+X

+Y

–XR

–Y

+X

+Y

–XR

–Y

+X

+Y

–X

R

–Y

+X

+Y

–X

R

–Y

The key to crack this question lies in comprehending that, the

component of A

along B

will be A cos if is the

angle between A

and B

.

Mind over Matter

B

A cos

A sin

A

Multiplication of vectors by a scalar2.5

Scalar product (dot product) of vectors2.6

SAMPLE C

ONTENT

37


44. In a triangle ABC, the sides AB and AC are represented by the vectors ˆ ˆ ˆ3i j k and ˆ ˆ î 2j k respectively. Calculate the angle

∠ABC. [WB JEE 2018]

(A) 1 5cos11

(B) 1 6cos11

(C) 1 590 cos11

(D) 1 5180 cos11

45. A vector A

points vertically downward and

B

points towards east. The vector product

A

B

is (A) zero. (B) along north. (C) along south. (D) vertically upward. 46. According to right handed screw rule, when a

right handed screw is placed with its axis perpendicular to plane containing two vectors

1A

and 2A

, is rotated from 1A

to 2A

through angle , then the sense of advancement of tip of the screw is in direction of 3A

. The advancement of the tip takes place for (A) 0 < 90 (B) 0 90 (C) 0 < 360 (D) 0 360

47. The value of ( P

+ Q

) ( Q

P

) is (A) 0 (B) P2 – Q2

(C) P Q

(D) 2 P Q

48. Two adjacent sides of a parallelogram are represented by the two vectors ˆ î +2j and

ˆ ˆ5i +2j . What is the area of parallelogram?

(A) 8 sq. unit (B) 8 3 sq. unit (C) 3 8 sq. unit (D) 192 sq. unit 49. 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 2k

and the radius vector

r 4 j 3k

, then v

is

(A) 29 units (B) 31 units (C) 37 units (D) 41 units 50. The moment of the force, ˆ ˆ ˆF 4i 5j 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 51. In the figure, the vectors from origin to the

points A and B are A 3i 6 j 2k

and

B

= 2i j 2k

respectively. The area of the triangle OAB is

(A) 5 17

2sq.unit (B) 2 17

5 sq.unit

(C) 3 175

sq.unit (D) 5 173

sq.unit 52. If a vector A

makes angles , and with the X, Y and Z-axes respectively, then the vector cos i

+ cos j

+ cos k

is a (A) null vector. (B) unit vector. (C) resultant vector. (D) displacement vector.

Miscellaneous

The key to crack this question lies incomprehending that, the direction of any

vector A B C

is always perpendicular to the

plane containing vectors B

and C

.

Mind over Matter

1A

2A

3A

= 1A

2A


2.7

A

O B

A 3i 6 j 2k

B 2i j 2k

SAMPLE C

ONTENT

38

38


53. Obtain the direction cosines of vector ( A

B

),

if A

= 2 i

+3 j

+ k

, B

= 2 i

– j

+3 k

.

(A) 0, 15

, 25

(B) 0, 25

, 15

(C) 0, 0, 15

(D) 15

, 0, 15

54. 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.



(C) Assertion is True, Reason is False. (D) Assertion is False, Reason is True. 55. 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 56. If A B C

; where, A

, B

and C

are unit

vectors, then A B

=

(A) 2 (B) 3

(C) 12

(D) 5 57. The resultant of two vectors having magnitudes

4 and 5 is 1. What is their cross product? (A) 6 (B) 3 (C) 1 (D) 0 58. For what value of ‘x’ are A

= i

– 7 j

+ 4 k

,

B

= x j

+ 3 k

and C

= 6 i

+3 j

–11 k

coplanar?

(A) 35117

(B) 277

(C) 5132

(D) 727

59. The angle between the vectors A

and B

is .

The value of the triple product A B A

is

(A) A2B (B) zero (C) A2B sin (D) A2B cos

60. Three vectors satisfy the relations B A

= 0

and B C

= 0, then B

is parallel to

(A) A C

(B) A C

(C) C

(D) A

61. Scalar product of two vectors is 2 and the

magnitude of their vector product is equal to 2 3 , then the angle between them will be

(A) 30 (B) 45 (C) 60 (D) 90 62. A sail boat sails 3 km due east, 4 km 41 north

of east, and finally an unknown displacement. If the final displacement of the boat from the starting point is 12 km due west, determine the third displacement. (Take cos 41 = 0.75, sin 41 = 0.66)

(A) 18i 2.64 j

(B) 0.66i 12 j

(C) 12i

(D) 2.64i 12 j

63. 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 64. What operations should be performed on

vector A

= i

+ j

to obtain B

= i

j

?

(A) Taking cross product of A

with ( j

i

) and multiplying with – 1.

(B) Taking cross product of A

with k

and dividing with i

.

(C) Taking cross product of A

with k

and dividing with .

(D) Taking cross product of A

with its orthonormal vector.

B

A

O

a

a

d

a

2.0 m3.0 m

37X

Y

SAMPLE C

ONTENT

39


65. A cable, attached to a rigid support is acted upon by tension in the cable and other two forces as shown in the diagram below. If the resultant force is (260 i

+ 560 j

) units what would be the tension in the cable?

(A) 310 units (B) 365.12 units (C) 620.1 units (D) 490 units 1. Which of the following is a vector? (A) Pressure (B) Surface tension (C) Moment of inertia (D) Torque 2. Two physical quantities, one of which is

vector and other is scalar, having same dimensions are

(A) work and energy. (B) angular momentum and Planck’s

constant. (C) pressure and power. (D) impulse and momentum. 3. Assertion: A scalar quantity does not vary

from point to point in space. Reason: A scalar quantity has the same value

for observers with different orientations of axes at same instant.



(C) Assertion is False, Reason is True. (D) Assertion is True, Reason is False.

4. Angular momentum is _______. (A) scalar (B) free vector (C) axial vector (D) polar vector 5. 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.

6. The vector projection of a vector 5 j

– 3 k

on X-axis is

(A) five. (B) six. (C) three. (D) zero. 7. In a tug of war Ramesh and Suresh manage to

hold rope straight. When a weight is suspended in the middle of the rope, the rope deviates from its horizontal position. To make it horizontal,

(A) Ramesh and Suresh should both apply equal amount of extra force.

(B) Ramesh and Suresh should apply unequal amount of extra force.

(C) either of the person should pull with original force while another person should pull with extra force.

(D) none of the Above. 8. 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 9. Which vector at an angle 30 to the X-axis has

a x-component of 6 units? ( i

and j

are unit vectors along X and Y axes respectively.)

(A) 6 i

(B) 6i 6 3 j

(C) 66i j3

(D) All of the above.

90 cm

650 units6

1.5 m

400 units

8 10 12 5 13

The key to crack this question lies incomprehending that, the rectangularcomponents of given forces can be determinedby the Pythagorean triplets.

Mind over Matter

Practice Problems


Types of vectors2.2


SAMPLE C

ONTENT

40

40


10. Vector representation of the resultant vector of vectors A

and B

shown in the figure is (A) (Asinα +B sinβ) i

+ (Acosα +B cosβ) j

(B) (Asinα +B sinβ) i

– (Acosα +B cosβ)k

(C) (Acosα +B cosβ)k

+ (Asinα +B sin ) i

(D) (Acosβ +B cosα) i

(Asinβ +B sinα) j

11. Assertion: The magnitude of resultant vector of two given vectors is always greater than the magnitude of any of the given vectors.

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



(C) Assertion is True, Reason is False. (D) Assertion is False, Reason is True. 12. If A

+ B

= A

B

, then vector B

must be (A) zero vector (B) unit vector

(C) non-zero vector (D) equal to A

13. Choose the incorrect statement. (A) When magnitudes of two vectors are

changed without changing the angle between them, only magnitude of resultant vector changes.

(B) When angle between two vectors is changed without changing their magnitudes only direction of resultant vector changes.

(C) When angle between two vectors is changed without changing their magnitudes, magnitude and direction of resultant vector change.

(D) The magnitude of resultant vector of two vectors can be less than the magnitude of any one of the vectors when the vectors are inclined at 105 with each other.

14. Maximum and minimum magnitudes of the resultant of two vectors of magnitudes P and Q are in the ratio 4:1. Which of the following relations is true?

(A) 3P = 2Q (B) 3P = 5Q

(C) PQ = 1 (D) 5P = Q3

15. The resultant of two forces acting in opposite directions is 20 N. If these forces act perpendicular to each other, their resultant will be 100 N. The magnitude of these forces will be

(A) 40 N, 30 N (B) 100 N, 50 N (C) 80 N, 60 N (D) 21 N, 28 N 16. The resultant of two vectors P

and Q

is R

. If

P

is doubled, then R

becomes doubled. When P

is reversed, R

again doubles, then the ratio of P : Q : R is,

(A) 1 : 3 : 1 (B) 5 : 1 : 10 (C) 1 : 2 5 : 4 5 (D) 1 : 10 : 5 17. 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 |

18. The magnitude of displacement vector with

end points (3, – 4) and (– 1, – 2) must be (A) 6 (B) 2 5 (C) 4 (D) 2 10 19. The position vector of a particle is determined

by the expression r

= 2t2 i

+ 5t2 j

+ 7 k

. The distance traversed in first 5 s is

(A) 500 m (B) 300 m (C) 150.25 m (D) 134.63 m 20. If gravity also has component equal to its

vertical component along Z-direction, then body of mass 10 kg would weigh as

(A) 98 2 N

(B) 10( i

+ j

) N

(C) 10( j

+ k

) N

(D) 98( k

+ j

) N

X

Z

B

A


SAMPLE C

ONTENT

41


21. Sum of three vectors P

, Q

and 2 i j k

is

unit vector along positive X-axis. If P

is doubled, Q

is kept same and the third vector

is changed to 4 i j 2k

then, the sum of

three vectors is unit vector along positive Y-axis. The value of P

and Q

will be,

(A) P 2 i 2 j 2k, Q 4 i 8 j

(B) P 4 i 2k, Q 9i 6 j 9k

(C) P 3i j k, Q 2i 2 j

(D) P 3i j k,Q 2 i 2 j

22. If the sum of two velocities 1 2v v

of a body

is given by 6 i 9 j 2k

and their difference

1 2v v

is given by 9 i 3 j 2k

, then ratio of

magnitudes of v1 to that of v2 would be (A) 5.24 (B) 2.46 (C) 4.80 (D) 3.98 23. 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) 15i

+ 20 j

(B) 20 i

15 j

(C) 5 i

25 j

(D) 8i

+ 6 j

24. Two vectors A

and B

acting at a point are such that if A

is reversed, the direction of the resultant is turned through 90. Then,

(A) A B

(B) A B

(C) A B

(D) A 3 B

25. The angle between vectors

A 3i 4 j 5k

and B 6i 8 j 10k

is (A) zero (B) 45 (C) 90 (D) 180

26. What is the unit vector perpendicular to the

vectors, 2i 2 j k

and 6i 3 j 2k

?

(A) i 10 j 18k

5 17

(B) i 10 j 18k

5 17

(C) i 10 j 18k

5 17

(D) i 10 j 18k

5 17

27. The three coterminous edges of a

parallelopiped are a 2i 6 j 3k

, b 5 j

,

c 2i k

. The volume of parallelopiped is (A) 36 cubic unit. (B) 40 cubic unit. (C) 45 cubic unit. (D) 54 cubic unit. 28. What is the angle between P

and Q

, if P

and

Q

are the adjacent sides of a parallelogram drawn from a common point and the area of

the parallelogram is PQ

2 ?

(A) 15 (B) 30 (C) 45 (D) 60

29. A force F 5 3 i 5 j N

acts on a body kept

in a certain co-ordinate system (XY). The same force, when applied in a co-ordinate system (XY) inclined at angle 18 with the XY system, has component in Y direction as

(A) 10 sin (30) N. (B) 10 sin (12) N. (C) 10 cos (48) N. (D) 10 cos (30) N.

Miscellaneous

The key to crack this question lies incomprehending that, the change in orientationof co-ordinate system does not affect theactual magnitude and direction of a vector.

Mind over Matter

Multiplication of vectors by a scalar2.5

Scalar product (dot product) of vectors 2.6


2.7

SAMPLE C

ONTENT

42

42


30. Match the vector operations given in column I with their magnitudes in column II.

Column I Column II

(i) i i j j

(a) 0

(ii) k i j j i k j k i

(b) 1

(iii) i j i j k k k

(c) 2

(iv) j k j i j i

(d) 3 (A) i b, ii c, iii a, iv d (B) i b, ii d, iii c, iv a (C) i d, ii c, iii a, iv b (D) i a, ii c, iii d, iv b 31. If A

= 2 i

+ j

, B

= j

+ 3 k

and C

= 6 i

2 k

,

then value of 2 A

B

+ 3 C

would be

(A) 20 i j+ 9k

(B) 22 i j 9k

(C) 4 i 5 j 20k

(D) 22 i 4 j 10k

32. The expression 1 3ˆ î j2 2

is a

(A) unit vector. (B) null vector. (C) vector of magnitude 2 .

(D) vector of magnitude 12

.

33. The position vectors of four points A, B, C and D are a

= i

+ 3j

+ 7k

, b

= 2i

+ 5 j

+ 9k

,

c

= i

+ 2 j

– 4k

and d

= 3i

+ 6 j

– 9k

respectively.

Then vectors AB

and CD

are (A) coplanar (B) collinear (C) perpendicular (D) parallel 34. Consider three vectors A i j k

,

B i j 2k

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

35. 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.



(C) Assertion is True, Reason is False. (D) Assertion is False, Reason is True. 36. If A.B 3 A B

, then the value of

| A

+ B

| is

(A) 122 2 ABA B

3

(B) A + B

(C) 1

2 2 2A B 3AB

(D) 1

2 2 2A B AB 37. If P

and Q

are the nonzero positive vectors making angle of 60 with each other, then the angle between additive inverse of P

with Q

will be

(A) zero (B) 90 (C) 120 (D) 180 38. If ay i

b

, where a, b are functions of x and t

respectively, then which of the following is a vector?

(i) Differential of y

with respect to x.

(ii) Differential of y

with respect to t.

(iii) Integral of y

with respect to x.

(iv) Integral of y

with respect to t. (A) (i) and (iii). (B) (ii) and (iii). (C) (ii) and (iv). (D) (i), (ii), (iii) and (iv). 39. Three vectors AB , BC and CD add together

to give rise to resultant vector 8 i 7 j

. if

BC 4 i

and CD 4 j

, by what angle AB must be aligned with vertical?

[Take tan(3652) = 0.75] (A) 3652 (B) 436 (C) 538 (D) 61

SAMPLE C

ONTENT

43


1. (D) 2. (C) 3. (C) 4. (D) 5. (B) 6. (C) 7. (C) 8. (D) 9. (D) 10. (C) 11. (A) 12. (A) 13. (A) 14. (C) 15. (D) 16. (C) 17. (D) 18. (A) 19. (D) 20. (A) 21. (B) 22. (B) 23. (D) 24. (D) 25. (C) 26. (C) 27. (D) 28. (D) 29. (C) 30. (C) 31. (C) 32. (C) 33. (D) 34. (A) 35. (B) 36. (B) 37. (C) 38. (C) 39. (A) 40. (A) 41. (D) 42. (C) 43. (D) 44. (A) 45. (C) 46. (A) 47. (D) 48. (A) 49. (A) 50. (D) 51. (A) 52. (B) 53. (B) 54. (B) 55. (B) 56. (B) 57. (D) 58. (B) 59. (B) 60. (A) 61. (C) 62. (A) 63. (B) 64. (C) 65. (C) 1. (D) 2. (B) 3. (C) 4. (C) 5. (D) 6. (D) 7. (D) 8. (C) 9. (C) 10. (B) 11. (D) 12. (A) 13. (B) 14. (B) 15. (C) 16. (D) 17. (B) 18. (B) 19. (D) 20. (D) 21. (D) 22. (B) 23. (B) 24. (A) 25. (A) 26. (C) 27. (B) 28. (C) 29. (B) 30. (B) 31. (B) 32. (A) 33. (C) 34. (A) 35. (D) 36. (C) 37. (C) 38. (D) 39. (C) 1. Scalar quantity is defined only by its magnitude.

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

3. Dimensions of measurement are independent of the scalar or vector nature of quantity. Also, it does not depend on numerical value. Hence, two quantities one being scalar and other being vector can have same dimensions. Eg.:

Work and torque are quantities having same dimensions but work is scalar quantity and torque is a vector quantity.

4. When electric currents are added it follows Kirchhoff’s law of addition and not law of vector addition. Current doesn’t change along length or bend and it cannot be resolved into components. All these properties belong to scalar quantity. Therefore, electric current is considered as scalar quantity.

5. Rotations though a type of angular displacement are not vectors because finite rotations do not obey commutative law of vector additions. This means, resultant orientation of body after successive rotations 1 and 2 is different than orientation after successive rotations 2 and 1.

6. Position vector is a localised vector. i.e., it has a fixed initial point. Hence, position of particle is represented by a unique vector.

7.

8. Unit vector in the direction of A

, A = A

A

= 22 2

(2i j 4k)

2 4 4

= (i j k)

3

9. 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 10. 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.

Answer Key


Practice Problems


Hints

Axial vector

Axis of rotation

Right hand screw rule

R

R cos

F

Rsin

SAMPLE C

ONTENT

44

44


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.

11. 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. 12. tan = x

y

RR

= 13

= tan1 13

= 30 13. 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

14. A

= i

– 4 j

– 8 k

| A

| = 2 2 21 4 8 = 9

cos = 19

, cos = 49 , cos = 8

9

15. P Q

P Q

Sum of two vectors cannot be equal to sum of

their unit vectors. 16. Let two vectors be A

and B

and R

be resultant vector.

A

= B

= R

implies three form equilateral

triangle.

Therefore, angle between A

and B

is 120

18. Let

A

= 2 i

– 3 k

and B

= –2 i

+ 3 j

A B

= (2 i – 3 k

) + (–2 i

+ 3 j

)

= (2 2) i

+ 3 j

– 3 k

= 3 j

– 3 k

x – component of ( A

+ B

) = 0 i

y component of ( A

+ B

) = 3 j

19. ( i

+ 2 j

+ 2 k

) + (2 i

– j

– 2 k

) + R

= i

Required vector, R

= –2 i

– j

20. r

= a

+ b

+ c

= –2 i

j

+ 3 i

+ 2 j

k

= i

+ j

k

r = r

r

= 22 2

ˆ ˆ î + j k

1 1 1

=

ˆ ˆ î + j k3

21. According to triangle law of vector addition,

the resultant of a

and b

is represented by PR

i.e., PR = a

+ b

Since, PR = RP = c

,

c

= a

+ b

a b c 0

22. Here, B C

= ˆ ˆ ˆ ˆ ˆ î 3j 5k 2i j 4k

= ˆ ˆ ˆ3i 2 j k A

As, ˆ ˆ Â 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

23. 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

r sin

r cos x

y

r

O

A

R

60 120 B

SAMPLE C

ONTENT

45


24. netF

= 1 2 3F F F

= –3 2 1 i –4 4 j 5 – 5 k

= 0 i 0 j 0k

25. Co-ordinates of initial point I are (25, 8). Similarly, for final point F co-ordinates are,

(25, 42). As laws of reflection are obeyed with respect

to X-axis, striker must hit at point N or N to reach point F.

Hence, along side B, co-ordinates of point N are (0, 25).

i.e., position vector of striker at N = 25 j cm

26. = 120 30 = 90 R = 2 29 12 2 9 12 cos90 = 15 unit 27. Resultant of two vectors A

and B

can be given by, R

= A

+ B

R

= A B

= 2 2A B 2AB cos

If = 0, then R

= A + B = A

+ B

28. Since, R = 2 2A B 2AB cos A = B = R A2 = 2A2 + 2A2 cos

cos = – 12

= cos 120

= 120 29. R = 2 2A B 2AB cos Here, A = B = F R = 2 22F 2F cos

= 22F 1 cos

= 2F cos2 .... 2(1 cos ) 2cos

2

= 2 10 cos2

= 20 cos2

30. A = 3 N, B = 2 N R = 2 2A B 2ABcosθ R = 9 4 12cosθ .…(i)

Now, A = 3 N, B = 4 N then, R = 9 16 24cosθ .…(ii) From equations (i) and (ii) we get, cos = 1 = c 31. Let A and B be the two forces. Then A = 7x, B = 8x, R = 26 N and = 60

Now, R = 2 2A B 2AB cos

26 = 2 27x 8x 2 7x 8x cos60

26 = 2 2 249x 64x 56x = 13x

x = 2613

= 2

Forces are; A = 7 2 = 14 N and B = 8 2 = 16 N 32. The angle which the resultant R makes with

A is given by

tan = BsinθA Bcosθ

Now, = 60 and = 30 = 2

tan θ2

= BsinθA Bcosθ

θsin2θcos2

=

θ θ2Bsin cos2 2

A+Bcosθ

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

A + B2

= 2B 34

which gives A = B. 33. As A B

= A B

,

A2 + B2 + 2AB cos = A2 + B2 2AB cos

4AB cos = 0, i.e. cos = 0 = cos 90, = 90 34. (25 – x)2 = 102 + x2

625 + x2 50x = 100 + x2 x = 10.5 N 25 – x = 14.5 N

N

I

N

F

O

10 25 – x

x

SAMPLE C

ONTENT

46

46


Alternate Method: Using shortcut no. 4

cos = AB

R = 2 2 2A B 2A

Also, A + B = 25 N Hence, on solving we get, A = 10.5 N, B = 14.5 N 35. 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 36. Use shortcut no. 7 37. From figures, = 45 + (360 280) = 125, XOA = 80 Using parallelogram law of vectors addition,

As B

= 30 units and A

= 20 units,

tan = BsinA Bcos

=

30 sin12520 30 cos125

= 8.799 8.8 = tan1 (8.8) > tan1 3 = 60

60 < < 90

As is measured with respect to A

, option (C) satisfies the condition.

38. A

and B

are parallel to each other. This implies A

= m B

. Comparing X-component,

m = 12

. Comparing Y-component, b = –8 and

comparing Z-component, a = 3.

39. If angle between i

and i k

is , then

cos = i i k

i i k

= 1

1 2= 1

2

= π4

. 40. Let the two velocities be 1v

and 2v

according to given condition we have,

1 2 1 2v v v v

= 0

….(orthogonality condition)

1 1 1 2 2 1 2 2v v v v v v v v

= 0

1 1 2 2v v v v

= 0

1 1 2 2v v v v

21v cos = 2

2v cos

21v = 2

2v v1 = v2 41. Let A

= 2i

+ 3 j

+ 6k

and B

= 6i

+ 8 j

+ k

,

A B

= 0 ….( A B

)

A B

= ( 2i

+ 3 j

+ 6k

)( 6i

+ 8 j

+ k

) = 2 (6) + 3 8 + 6 = 0 = 2 42. Since,

A B

= AB cos

A cos = A B

B

Component of A

along B

will be,

A cos = A B

B

Now,

A B

= 5i 3 j i j

= 5 + (–3) = 2

B

= 2

A cos = A B

B

= 2 22 unit

C

A

B

C

A

B

135

O

A

B

X

Y

–X

–Y

O 45280

B

A

SAMPLE C

ONTENT

47


43. Vectors are orthogonal

i.e., A B

= 0

cost. cos ωt2

+ sin t.sin ωt2

= 0

cos tt2

= 0 or cos t2

= 0

t2 =

2

t =

44. ˆ ˆ ÂB 3i j k

, ˆ ˆ ÂC 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 ÂB 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

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

A

= ak

Direction of vector B is

towards east

B

= bi

Now A

B

= ak bi

= ab j

The direction of A

B

is along south. 46. Advancement of the tip is in direction of 3A

.

As, 3A

is perpendicular to the plane

containing 1A

and 2A

, 3 1 2A A A

3A

= A1 A2 sin

sin has maximum value at = 90.

For > 90, the direction of 3A

gets reversed.

Also for = 0, 3A

= 0

which means advancement of screw is not possible.

Hence, for advancement of tip of the screw, 0 < 90.

47. P Q Q P

= P Q P P Q Q Q P

= P Q 0 0 Q P

= P Q P Q

= 2 P Q

48. Area of parallelogram = A B

= i 2 j

5i 2 j

= i j k1 2 05 2 0

= 0i 0 j 8 k

Magnitude = 0 0 64 = 8

49. v

=

r

= i j k1 2 20 4 3

= i 6 8 j 3 4k

= 2i 3 j 4k

v

= 2 2 22 3 4 = 29 units 50. ˆ ˆ ˆ ˆ ˆr 2i 3k 2i 2 j 2k

= ˆ ˆ2j k

r F

= ˆ ˆ ˆ ˆ ˆ2 j k 4i 5j 6k

= ˆ ˆ î j k0 2 14 5 6

= ˆ ˆ î 12 5 j 0 4 k 0 8

= ˆ ˆ ˆ7i 4j 8k 51. 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

= 10i 10 j 15k

N

E

S

W

j

i

SAMPLE C

ONTENT

48

48


a b

= 2 2210 10 15

= 425 = 5 17

Area of OAB = 12

a b

= 5 172

sq.unit. 52. is the angle subtended by A

with +ve X-axis cos = xA

A

Similarly, cos = yA

A

,

cos = zA

A

given vector can be written as,

yx zAA A Ai j k AA A A A

53. R

= A B

= 4 j

– 2 k

, R

= 16 4 = 2 5

Direction cosines are,

cos = 0 02 5

cos = 4 22 5 5

cos = 2 12 5 5

54. 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.

55. From the figure, OA

= a and OB

= a

Also, from triangle rule, OB

– OA

= AB

= a

| a

| = AB

Since d = arcradius

or AB = ad

So | a

| = ad a means change in magnitude of vector i.e.

OB OA

a a = 0 Hence, a = 0 56. C A B

or C2 = A2 + B2 + 2AB cos = 1 + 1 + 2 cos

Since it is given that C

is also a unit vector, therefore, 1 = 1 + 1 + 2 cos

cos = 12

or = 120

Now the difference vector is, D

= A

B

or D2 = A2 + B2 – 2AB cos = 1 + 1 2cos(120) D2 = 2 2(1/2) = 2 + 1 = 3 | D

| = 3 57. R = 2 24 5 2 4 5 cos = 1 By solving, we get = 180 Cross product = 4 5 sin 180 = 0 58. Using shortcut no. 8(iv) If three vectors are coplanar, then ( A

B

)C

= 0

i.e., 1 7 40 x 36 3 11

= 0

1 ( 11x 9) + 7(0 18) + 4(0 6x) = 0 11x 9 126 24x = 0

35x = 135 or x = 277

59. Let A B A

= A

C

Here, C

= B

A

which is perpendicular to

both vectors A

and B

A C

= 0 60. B A

= 0; B C

= 0 B

is perpendicular to A

as well as C

Now, let D A C

The direction of D

is perpendicular to the plane containing A

and C

Hence, B

is parallel to D

that is, B

is parallel to A C

61. A B

= A B

cos = 2 ….(i)

A B

= A B

sin = 2 3 ….(ii)

Dividing equation (ii) by equation (i) we get, tan = 2 3

2 = 3

= tan1( 3 ) = 60

Ay Az Ax

X

ZD A C

C

Y

A

SAMPLE C

ONTENT

49


62. Given displacements are,

3 km due east 3i

and 4 km 41 north of east components of second displacement

along X-axis = 4 cos(41) = 4 34

= 3i

along Y-axis = 4 sin(41) = 4 0.66 = 2.64 j

second displacement = 3i 2.64 j

Let third displacement be x i y j

3i 3i 2.64 j x i y j

= 12i

comparing RHS and LHS, 3 + 3 + x = –12 and 2.64 + y = 0

the third displacement is 18i 2.64 j

. 63. R A B C

From figure we have,

A

= 4i 3 j

….(i)

B

= 3i

….(ii)

C

= 2 j

….(iii)

Resultant is given by R

= A

+ B

+ C

R

= ( 4i

+ 3 j

) + 3i

+ 2 j

R

= 7i

+ 5 j

Magnitude of resultant vector is R

= 49 25

R

= 74

R

= 8.6 m and angle with X-axis can be

given by dot product, R i

= R i cos

7 i 5 j i

= 8.6 1 cos

cos = 7

8.6

= 35.5

64. A

A

= 0

. Hence, option (A) is incorrect. Division of vectors is not defined. Hence, option (B) is incorrect. Cross product of two orthonormal vectors is zero. Hence, option (D) is incorrect.

A

k

= ( i

+ j

) k

= 2 ( i

k

) + ( j

k

)

= 2 j

+ i

A k

= i

j

.

Hence, option (C) is correct. 65. From the given Pythagorean triplets, For 10 units of force, 8 units act along –ve

Y-axis while 6 units act along –ve X-axis. Hence, 400 units of force, can be resolved into two vectors of –320 units along Y-axis and –240 units along X-axis.

Similarly, resolving 3 force vectors into their

respective rectangular components, Magnitude X component Y component

T Tx Ty 400 240 320 650 600 250 R 260 –560

Where, R

= Rx i

+Ry j

Now comparing x components, 260 = 240 + 600 + Tx Tx = 620 units Comparing y component, –560 = 250 320 + Ty Ty = 10 units T = 2 2

x yT T

T = 2 2620 10 T = 620.1 units.

41o

4sin(41o)

4 cos(41o) E

N

650

400

T

8x

10x = 400

6x

650 units

1.5 m = 150 cm

400 units

90 cm

T 120

cm

SAMPLE C

ONTENT

50

50


3. A scalar quantity may vary from point to point in

space such as density of air. Density of air varies according to altitude. Hence, assertion is false.

But its value at a fixed point remains same for observers with different orientations of axes at same instant.

For example, mass of a body measured by observers with different orientation of axes is unique. Hence, reason is true.

5. Both the magnitude and direction of vector are indicated using rectangular components and unit vectors of the given vector. In which, unit 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.

6. As the multiple of i

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

7. Weight of the body is vertical force and cannot be balanced by horizontal forces.

8. sin2 + sin2 + sin2 = 1 cos2 + 1 cos2 + 1 cos2 = 3 (cos2 + cos2 + cos2) = 3 1 = 2 9. In option (A), angle of vector with X-axis is,

tan = 06

= 0

= 0 In option (B), angle of vector with X-axis is,

tan = 6 36

= 3

= 60 In option (C), angle of vector with X-axis is,

tan = 6 13 6 3

= 30 10. Resolving the given vectors in their

rectangular components, we get, A

= (Asinα) i

(Acosα)k

B

= (Bsinβ) i

(Bcosβ)k

R

= A

+ B

= Asinα+Bsinβ i

Acosα+Bcosβ k

11. 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 .

14. According to the problem, P + Q = 4 and P Q = 1 By solving, we get P = 2.5 and Q = 1.5

PQ = 5

3 or 3P = 5Q

15. As per given condition; A B = 20 ….(i) and 2 2A B = 100 ….(ii) From (i), A = B + 20 Substituting this in equation (ii), we get (B + 20)2 + B2 = 1002. On solving we get, B = 60 N or 80 N. Therefore, A = 80 N or 60 N. 16. R

= P

+ Q

R = 2 2P Q 2PQcosθ

….( is angle between P

and Q

) i.e., R2 = P2 + Q2 + 2PQ cos ….(i) The resultant of 2P and Q is 2R 4R2 = 4P2 + Q2 + 4PQ cos ….(ii) Also, resultant of (P) and Q is 2R 4R2 = P2 + Q2 + 2PQ cos(180 ) = P2 + Q2 2PQ cos ....(iii) Solving equations (i), (ii) and (iii), we get, Q = 10 P and R = 5 P P : Q : R = 1 : 10 : 5 17. As angle between two vectors taken in same

order is greater than 90, = 180 – 120 = 60

| C

| = 2 2A B AB and

| A B

| = 2 2A B AB

| C

| > | A B

| 18. r

= 2r

1r

= (– i

– 2 j

) (3 i

– 4 j

)

r

= –4 i

+ 2 j

| r

| = 2 24 2

= 16 4 = 20 = 2 5

Practice Problems

B sin

A sin

B

B cos

A

A cos

Z

X

120 =60BC

A

SAMPLE C

ONTENT

51


19. r

= 2t 2 i

+ 5t 2 j

+ 7 k

At t = 0, 1r

= 7 k

At t = 5 s, 2r

= 50 i

+ 125 j

+ 7 k

,

r

= 2r

1r

= 50 i

+ 125 j

| r

| = | 2r

1r

|= 2 250 125 = 134.63 m 20. Let vertical component be y

i.e.,

yg

= 9.8 j

m/s2.

Now if it has component along z

i.e.,

zg

= 9.8 k

m/s2.

Resultant g

= (9.8 j

+ 9.8 k

) m/s2

Weight of body m kg = 9.8m ( j

+ k

) N

Weight of body 10 kg = 98 ( j

+ k

) N 21. P

+ Q

+ 2i

+ j

– k

= i

P

+ Q

= i

– 2i

– j

+ k

P

+ Q

= – i

– j

+ k

….(i)

2 P

+ Q

+ 4 i j 2k j

2 P

+ Q

= j 4 i j 2k

2 P

+ Q

= 4 i 2k

….(ii) Solving equations (i) and (ii),

Q 2 i 2 j

Substituting this value in equation (i), we get,

P 2 i 2 j i j k

P 3i j k

22. 1 2v v

= 6i 9 j 2k

....(i)

1 2v v

= 9 i 3 j 2k

....(ii) Solving equations (i) and (ii) simultaneously,

2 1v

= 15i 12 j

and 2 2v

= 3i 6 j 4k

1v

= 12

15i 12 j

and 2v

= 12

3i 6 j 4k

1

2

v

v

=

2 2

2 2 2

15 12

3 6 4

= 369

61= 2.46

23. 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

24. Let, R A B

and S A B

According to given condition, S R

R S 0

A B A B 0

–A2 + B2 = 0

A B

25. A B

= AB cos ….(i)

= 3i 4 j 5k 6i 8 j 10k

= 3 6 + 4 8 + 5 10 = 100 ….(ii) The magnitudes of A and B are A = 2 2 23 4 5 = 50 ….(iii)

B = 2 2 26 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. 26. A 2i 2 j k

and B 6i 3 j 2k

C A B

= 2i 2 j k 6i 3 j 2k

= i j k2 2 16 3 2

= i 10 j 18k

Unit vector perpendicular to both A

and B

u = 2 22

i 10 j 18k

1 10 18

= i 10 j 18k

5 17

27. Using shortcut no. 8

Volume of parallelopiped = A B C

A B

=i j k2 6 30 5 0

= i 0 15 j 0 k 10 0

= 15i 10k

SAMPLE C

ONTENT

52

52


A B C 15i 10k 2i k

= 30 + 10 = 40 cubic unit.

28. Area of parallelogram = P Q

P Q

= PQ sin = PQ

2

sin = 12

= 45 29. Given: F 5 3 i 5 j

tan = Y

X

F 1F 3

= 30 Now, when the co-ordinate system is inclined

at angle = 18 with the X-direction, force F

will make angle ( – ) with X-direction.

FY = F

sin ( – )

but, 2 2F 5 3 5

= 10 N

FY = 10 sin (30 – 18) = 10 sin (12) N 30. Using properties of dot product and cross

product, we get magnitude of

i i j j

= 1 0 = 1

k i j j i k j k i

= 1 – (–1) + 1 = 3

i j i j k k k

= 0 + 1 + 1 = 2

j k j i j i

= 0 – 0 = 0 31. 2A

B

+ 3C

= 2 2 i j

j 3k

+ 3 6i 2k

= 4 i 2 j j 3k 18i 6k

= 22 i j 9k

32. P

= 12

i

+ 32

j

| P |

= 221 3

2 2

= 1

It is a unit vector. 33. AB

= 2 1 i 5 3 j 9 7 k

= i 2 j 2k

Similarly, CD

= 2i 4 j 5k

AB CD

= |AB||CD|cos cos =

2 2 2 2 2 2

2 8 100

1 2 2 2 4 5

= 90 The vectors are perpendicular to each other. 34. A B C 0,

i j k i j 2k 2 i 3 j 4k

= 0

2( + ) 3 ( ) + 4( 2) = 0 9 – 9 = 0 : = 1 : 1 35. 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

). 36. A

. B

= 3 | A

B

| AB cos = 3 AB sin tan = 1

3 = 30

cos = cos 30o = 32

Now, | R

| = | A

+ B

| = 2 2A B 2AB cos

= 2 2 3A B 2AB

2

=

12 2 2A B 3AB

37. If P

is multiplied by negative real number, then we get a vector acting in the

opposite direction of P

and changing angle by 180.

Y Y

F

X

X

P

Q60

120

P

Q

SAMPLE C

ONTENT

53


38. Integration and differentiations are the mathematical operations which work on magnitude of the given function or physical quantity. Direction associated with it remains unchanged.

Hence, integration and differentiation upon a vector gives answer in vector form only.

39. Assuming point A as origin, three vectors can be represented as given below.

Let R AB BC CD

AB R BC CD

= 8 i 7 j 4 i 4 j

= 4 i 3 j

Now, tan = Y component of ABX component of AB

= tan–134

= 3652

Angle that AB would make with vertical = 90 – = 538 1. How scalar or vector nature of a physical

quantity is related with units and dimensions of measurement of the quantity?

Ans: A physical quantity is completely expressed with the help of the numerical value and unit.

The numerical value is only magnitude if the quantity is scalar and magnitude and direction if quantity is vector.

In short, quantity = (scalar / vector) units. While dimensions of measurements are

independent of scalar / vector nature of quantity as well as system of units in which it is expressed. The dimensions of the quantity indicate powers to which the fundamental quantities are raised to.

2. Physical quantities mass and moment of inertia, both are completely expressed using only magnitude. Yet one has to take care while adding moment of inertia unlike mass. Masses 28 kg, 52 kg and 132 kg are added to give total mass of 212 kg. But adding moments of inertia 0.4 kg/m2 and 0.25 kg/m2 needs careful consideration. Why?

Is there any other such scalar quantity where more than pure arithmetic addition needs to be considered?

Ans: Mass is simply added arithmetically as its value is independent of reference frame or choice of axes of orientations.

Though moment of inertia is expressed completely using magnitude, its value depends on the axis along which the value is calculated. For a given orientation of a body its moment of inertia is constant, but as the orientation of the body changes, value of moment of inertia changes as well.

Hence, while adding values of moments of inertia, a care has to be taken to calculate values with respect to specific orientation and then add them using simple arithmetic.

Another physical quantity which is scalar but needs to be added considering physical laws is angular speed.

3. Uniform circular motion is a type of motion in which, a particle moves along a circular path with constant speed. However, the direction of its velocity is different at different points (see figure). Determine the change in velocities of the particle at points A and B.

Ans: As given, magnitude of velocity of the particle

is same at all points on the circle.

A Bv v v

....(i)

Now change in velocity will be,

A Bd v v v

Squaring both the sides,

(dv)2 = 2

A Bv v

= A B A Bv v v v

= 2 2A A B Bv 2v v cos v

= 2v2 – 2v2 cos ....[from (i)] = 2v2(1 – cos)

= 2v2 22sin2

.... 21 cos 2sin2

dv = 2vsin2

B C

D

X A

Y

Problems To PonderA

Av

Bv

B

SAMPLE C

ONTENT

www.targetpublications.org/neet-ug-jee-main-challenger-physics-vol-1.html

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NEET-UG / JEE (Main) Challenger Physics Vol. - 1 · NEET – UG & JEE (Main) Salient Features • Exhaustive coverage of MCQs under each sub-topic. • ‘1819’ MCQs including questions