1. Units, Measurements & Motion

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ISBN : 9789386320018

Product Name : Units, Measurements & Motion for JEE Main

& Advanced (Study Package for Physics)

Product Description : Disha's Physics series by North India's

popu-lar faculty for IIT-JEE, Er. D. C. Gupta, have achieved a lot of

ac-claim by the IIT-JEE teachers and students for its quality and

in-depth coverage. To make it more accessible for the students Disha

now re-launches its complete series in 12 books based on chapters/

units/ themes. These books would provide opportunity to students to

pick a particular book in a particular topic.

Unit, Measurement & Motion for JEE Main & Advanced (Study

Pack-age for Physics) is the 1st book of the 12 book set.

These topics are taken from our Book:

UNITS, MEASUREMENTS & MOTION

• The book provides basic concepts related to Mathematics useful in Physics.

• The chapters provide detailed theory which is followed by Important Formulae, Strategy to solve problems

and Solved Examples.

• Each chapter covers 5 categories of New Pattern practice exercises for JEE - MCQ 1 correct, MCQ more than

1 correct, Assertion & Reason, Passage and Matching based Questions.

• The book provides Previous years’ questions of JEE (Main and Advanced). Past years KVPY questions are

also incorporated at their appropriate places.

• The present format of the book would be useful for the students preparing for Boards and various competitive

exams.

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0. Mathematics Used in Physics 1-16

1. Units and Measurements 17-60

Definitions Explanations and Derivations 18 1.1 Fundamental quantities 18

1.2 Derived quantities 18

1.3 The SI system of units 18 1.4 Definitions of SI units 19 1.5 Advantages of SI system 20 1.6 Dimensions of a physical quantity 20

1.7 Order of magnitude 22

1.8 Rules of significant figures 22 1.9 Errors in measurement 26 1.10 Indirect methods of measuring

large distances 31

1.11 Indirect method of measuring

small distances 33

1.12 Vernier callipers and ccrew gauge 34 Exercise 1.1 Level 1 (Single correct option)

Exercise 1.1 Level 2 (Single correct option) Exercise 1.2 (more than one correct options)

Exercise 1.3 (Assertion and Reasoning type questions) Exercise 1.4 (Passage & Matrix)

Exercise 1.5 (Past years JEE-(Main and Advance) Hints and Solutions (Solution of all exercises)

2. Vectors 61-92

Definitions Explanations and Derivations 62 2.1 Scalar quantity or scalar 62 2.2 Vector quantity or vector 62

2.3 Vectors operations 65

2.4 Addition or subtraction of two vectors 65 2.5 Addition or subtraction of more than

two vectors 68

2.6 Product of two vectors 73 2.7 Geometrical interpretation of scalar

triple product 77

Exercise 2.1 Level 1 (Single correct option) Exercise 2.1 Level 2 (Single correct option) Exercise 2.2 (more than one correct options)

3. Motion in a Straight Line 93-146 3.1 Concept of a point object 94 3.2 Rest and motion are relative terms 94

3.3 Motion 94

3.4 Motion parameters 94

3.5 Equations of motion 97

3.6 Study of motion by graphs 105

3.7 Relative velocity 112

3.8 Motion with variable acceleration 118 3.9 Problems based on maxima and minima 118 Exercise 3.1 Level 1 (Single correct option)

4. Motion in a Plane 147-202

4.1 Introduction 148

4.2 Position vector and displacement 148

4.3 Average velocity 148

4.4 Average acceleration 149 4.5 Motion in a plane with constant

acceleration 150

4.6 Relative velocity in two dimensions 151

4.7 Projectile motion 157

4.8 Projection up on an inclined plane 169 4.9 Projection down the inclined plane 170 4.10 Motion along a curved path 171 4.11 Constraint relations 179 Exercise 4.1 Level 1 (Single correct option)

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1.9 E

RRORS IN MEASUREMENT

Every measurement is limited by the reliability of the measuring instrument and skill of the person making the measurement. If we repeat a particular measurement, we usually do not get the same result every time. This imperfection in measurement can be expressed in two ways :

Accuracy and precision

Accuracy refers to the closeness of observed values to its true value of the quantity while precision refers to closeness between the different observed values of the same quantity. High precision does not mean high accuracy. The difference between accuracy and precision can be understand by the following example : Suppose three students are asked to find the length of a rod whose length is known to be 2.250 cm. The observations are given in the table.

Student Measurement-1 Measurement-2 Measurement-3 Average length A. 2.25 cm 2.27 cm 2.26 cm 2.26 cm B. 2.252 cm 2.250 cm 2.251 cm 2.251 cm C. 2.250 cm 2.250 cm 2.251 cm 2.250 cm

It is clear from the above table, that the observations taken by student A are neither precise nor accurate. The observations of student B are more precise. The observations of student C are precise as well as accurate.

Error : Each instrument has its limitation of measurement. While taking the observation, some uncertainty gets introduced in the observation. As a result, the observed value is somewhat different from true value. Therefore,

Error = True value – Observed value

Systematic errors : The errors which tend to occur of one sign, either positive or negative, are called systematic errors. Systematic errors are due to some known cause which follow some specified rule. We can eliminate such errors if we know their causes. Systematic errors may occur due to zero error of an instrument, imperfection in experimental techniques, change in weather conditions like temperature, pressure etc.

Random errors : The errors which occur randomly and irregularly in magnitude and sign are called random errors. The cause of random errors are not known. If a person repeat the observations number of times, he may get different readings every time. Random errors have almost equal chances for positive and negative sign. Hence the arithmetic mean of large number of observations can be taken to minimize the random error.

Mean value of a quantity : Since the probability of occurrence of positive and negative errors are same, so the arithmetic mean of all observations can be taken as the true value of a observed quantity. If a₁, a₂, ...a_n are the observed values of a quantity, then its true value a can be given by

a = amean a1 a2 ..._n an + + + = = 1 1 n i i a n

å

₌ The absolute errors in individual observations are:

1 1 a a a D = -2 2 a a a D = ... n n a a a D = -The mean absolute error is defined as

a

D = | a1| | a2| ... | an|

n

D + D + + D

Chapter 1

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= ` 1 | | n i i a n ₌ 1

_å

_D

Thus the final result of the observed quantity can be expressed as a = a± Da. It is clear from above that any observed value can by (a- D £ £a) a (a+ D .a)

Relative or fractional error : The ratio of the mean absolute error to the true value of the quantity is called relative error.

Thus relative error = D_aa

Percentage error : If relative error is expressed in percentage is called percentage error. Thus percentage error a 100

a

D = ´

Note:

_{Absolute error has the unit of quantity. But relative error has no unit.} Combination of errors

Let we want to get the volume of sphere, V = 4 3

3pr. There involves multiplication of radius three times.

The measurement of radius has some error, then what will be error in calculating the volume of sphere? The error in final result depends on the individual measurement as well as the mathematical operations involved in calculating the result. Following rules are used to evaluate maximum possible error in any computed quantity.

1. Error in addition

Let Z = X + Y. Suppose ± Dx be absolute errors in X and ± Dy be the absolute error in Y, then we have

Z + Dz = (X ± Dx) + (Y ± Dy) = (X + Y) ± (Dx + Dy)

\ Dz = (Z + Dz) – Z

= ± (Dx + Dy)

Note:

The maximum value of Dx or Dy can be least count of the instrument used. Example : x = 2.20 cm, Dx will be 0.01 cm.

RULE : The maximum possible error in the addition of quantities is equal to the sum of their absolute error. % error in Z, z Z D × 100 = ± êé_ëD + D_{X Y}x₊ yùú_û × 100 2. Error in subtraction Let Z = X – Y Z + Dz = (X ± Dx) – (Y ± Dy) = (X – Y) ± (Dx _m Dy) \ Dz = (Z +Dz) – Z = ± Dx _mDy For maximum possible error Dx and yD must be of same sign.

\ Dz = ± (Dx + Dy)

RULE : The maximum possible error in subtraction of quantities is equal to the sum of their absolute errors.

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MECHANICS % error in Z, z Z D × 100 = ± êé_ëD + D_{X Y}x_- yùú_û. 3. Error in product Let Z = XY Z + Dz = (X ± Dx) (Y ± Dy) = XY ± DxY ± XDy ± DxDy

\ Dz = (Z + Dz) – Z

= ± (DxY + XDy) ± DxDy

If Dx and Dy are both small, their product be very small, therefore we can neglect it.

\ Dz = ± (DxY + XDy)

The maximum fractional error in Z,

z Z

D

= ±éê_ëD_Xx+D_Yyùú_û

and maximum percentage error in Z,

z Z

D

× 100 = ±éê_ëD_Xx+D_Yyùú_û × 100

RULE : The maximum fractional error in the product is equal to the sum of the fractional errors in the individual quantities.

Note:

The product Dx Dy can not be neglected if the errors in x and y are order of 10% or more. The product can be neglected, if the error in x and y are 1% or little more than this (say 2 to 3%). 4. Error in quotient or division

Let Z = X Y Then Z + Dz = X_Y_{± D}± D_yx = 1 1 x X X y Y Y D æ _± ö ç ÷ è ø D æ _± ö ç ÷ è ø = 1 1 1 X x y Y X Y -D D æ _± öæ _± ö ç ÷ç ÷ è øè ø = _YX_èçæ1±D_Xxöæ÷ç_øè1mD_Yyö÷_ø or Z + Dz = Z 1 x 1 y X Y D D æ _± öæ ö ç ÷ç ÷ è øè m ø 1 + z Z D = _èçæ1±D_Xx_{ø è}÷ çö æ1mD_Yyö÷_ø = 1 ± x X D m D_Yy ± x X D . y Y D

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As the term x X D and y Y D

are small, so their product can be neglected. The maximum fractional error is given by z Z D = ± æçD_Xx+D_Yyö÷ è ø

And maximum possible percentage error in Z,

\ z

Z D

× 100 = ± æçD_Xx+D_Yyö÷ è ø× 100

RULE : The maximum fractional error in the quotient is equal to the sum of their individual fractional errors.

5. Error in the power of a quantity

Let Z = Xn Z + Dz = (X ± Dx)n = Xn ₁ n x X D æ _± ö ç ÷ è ø ; Z 1 n x X D æ _± ö ç ÷ è ø or Z z Z + D = æç1±nD_Xxö÷ è ø or 1 + z Z D = 1 ± n x X D \ D_Zz = ± n x X D

Maximum percentage error in Z z Z D

× 100 = ± n æç_èD_Xx´100ö÷_ø

RULE : The fractional error in the quantity with n powers is n times the fractional error in that quantity.

Note:

Here n may have any value. It may be a whole number, fraction, positive or negative.

General case : If Z =

a b c

X Y

W , the maximum possible fractional error in Z,

z Z D

= ± é_ëêaD_Xx+bD_Yy+cD_Wwùú_û

The maximum possible percentage error z Z D × 100 = ±é_ê D + D + D ù_ú ë û x y w a b c X Y W × 100

The above used algebraic method in many operations become difficult to operate. In such situations we can used differential method to find the error.

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30

MECHANICS 1. Let Z = a b c X Y k W where k is a constant.

Taking logarithms of both sides of equation, we get

ln Z = ln k + a ln X + b ln Y – c ln W Now differentiating partially the above expression, we have

z Z d = a x b y c w X Y W d ₊ d _- d

We can write above equation by writing D in place of d; z Z D = a x b y c w X Y W D ₊ D _- D

Errors calculated by above equation, is known as mathematical error. But our interest is in finding the maximum possible error.

\ D_Zz× 100 = ± ê_ëéaD_Xx+bD_Yy+cD_Wwùú_û×100

2. Let Z = ₍_{X Y}W₊ ₎

Taking logarithms of both sides of above equation, we have ln Z = ln W – ln (X + Y) Differentiating partially, we get

( ) ( ) z w x y Z W X Y d ₌d _-d + + = d_Ww-(d + d_{X Y}x₊ y)

(a) The maximum possible error in Z z Z D = ±êéD_Ww+₍D + D_{X Y}x₊ y₎ùú ë û (b) For Z = W X Y -z Z D = ±_ëêéD_Ww+D + D_{X Y}x_- yùú_û 3. Let Z = XY U V+

Taking logarithms of both sides of above equation, we have ln Z = ln X + ln Y – ln (U + V) Differentiating partially, we get

z Z d = x y (u v) X Y U V d d d + + -+ = x y ( u v) X Y U V d d d + d + -+

The maximum possible error in Z

(a) z

Z

D

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(b) For Z = _{U V}XY_- , z

Z

D

=±_ëêéD_Xx+D_Yy+D + D_{U V}u_- vùú_û

4. Let Z = sinx Differentiating partially, we get

dz = cos x dx or Dz = cos x Dx and z Z D = cos 1 sin2 sin sin x x x x x x -D = D or z Z D = 1 z2 Z - _Dx

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4.7 P

ROJECTILE

MOTION

When a particle is projected obliquely near the earth surface, it moves simultaneously in horizontal and vertical directions. Motion of such a particle is called projectile motion. Since there is no force acting in horizontal direction, the velocity remains constant in this direction. In vertical direction gravitational pull of earth produces the acceleration.

Assumptions used in projectile motion

(i) Neglecting the effect of air resistance on the projectile.

(ii) Assuming the acceleration due to gravity is constant at each point of projectile. (iii) Neglecting the effect of curvature of earth.

We will discuss the following types of projections in details :

Type 1 : Projectile fired at some Type 2 : Horizontal projection Type 3 : Projectile fired from

angle with the horizontal. some height

Type 4 : Projection on an inclined plane Type 5 :Projection down the inclined plane

Figure. 4.21

ANALYSIS OF PROJECTILE OF TYPE 1

Let us consider a particle is projected with initial velocity u at an angle q with the horizontal (called angle of projection). The velocity u has two rectangular components:

(i) The horizontal component ucosq, which remains constant throughout the motion of particle.

(ii) The vertical component usinq, which changes with time due to effect of gravity. Thus we have initial velocity ur u = uxˆi +uyˆj or uur = ucos i + sin jqˆ u qˆ Figure. 4.22

Chapter 4 Motion in a Plane

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Figure. 4.24 Figure. 4.23

Velocity at any time t : Using first equation of motion in vertical direction, we have

v_y = u_y – gt = u sinq – gt

\Velocity at any time t,urv = v i + v j_xˆ _yˆ

or urv = ucos i + ( sin - )jqˆ u q gtˆ.

Velocity at any height : At any height h

v_x = u_x = u cosq …(i)

and v_y2 _{= u}

y2 – 2gh

= (u sinq)2 _{– 2gh} _…(ii)

Squaring (i) and adding with equation (ii), we get

v = _u2_-₂_{gh .} Position at any time t

Position of particle at any time t, is given by ur r = xˆi + jyˆ where x = u cosq t …(1) and y = u sinq t – 1 2gt2 \ rr = ucosq +tˆi ( sinu q -t ₂1gt2)ˆj or r = 2 2 1 2 ( cos ) sin 2 æ ö q +_ç q - _÷ è ø u t u t gt = ut 2 _sin 1 2 q æ ö +_ç _÷ -è ø gt gt u u and tanf = y_x or f = tan–1 2 1 sin 2 cos é _{q -} ù ê ú ê _q ú ê ú ë û u t gt ut = tan–1 2 sin 2 cos q -é ù ê _q ú ë û u gt u .

The angle of elevation f of the highest point of the projectile; t = usin_g q \ tanf = sin 2 sin 2 cos q q - ´ q u u g g u or tanf = tan 2 q Equation of trajectory We have, x = u cosq t or t = æç_è_u_cosx _qö÷_ø

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159

Motion in a Plane and y = u sinq t – 1 2gt2 = u sinq æç_è_u_cosx _qö÷_ø – 1 2g 2 cos æ ö ç ÷ è qø x u y = 2 2 2 tan . 2 cos q -q gx x u

On comparing this equation with general equation of parabola, y = ax ± bx2_{, we find that path of}

projectile is parabolic in nature.

Time of flight (T) : Total time of motion of particle in air is called time of flight.

The displacement in vertical direction (y-axis) becomes zero in whole time of motion. So we have y = u_y t – 1

2ayt2 or 0 = u_yT – 1

2gT2

which gives _T = 2uy = 2 sinu q.

g g

Maximum height attained (H): The maximum vertical distance achieved by particle is called maximum height.

At the highest point of projection v_y = 0, so we have, v_y2 _{= u} y2 – 2gh or 0 = u_y2_{– 2gH} Figure. 4.25 which gives _H = 2 ₂ ₂ sin 2 2 q = y u _u g g .

Horizontal range (R) : The horizontal distance moved by particle in total time of flight is called horizontal range.

Horizontal range, R = u_x × T = u_x × 2uy g = 2u ux y g = 2 cosu usin g q q or R = 2_{sin 2} . q u g

For maximum range, sin2q = 1 or 2q = 90° or q = 45°.

Thus R_max = 2 u g . Corresponding, H = 2_{sin 45}2 2 u g ° = 2 . 4 u g

There are two angles of projection for same range:

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R' = 2_{sin 2(90} ₎ u g ° -q = 2_sin(180 _{2 )} u g ° - q = 2_{sin 2} u g q = R.

Thus, for a given velocity of projection, a projectile has the same range for angle of projection q and (90° – q).

Time of flight for angle of projection q,

T₁ = 2 sinu_g q and time of flight for angle of projection (90° – q),

T₂ = 2 sin(90u _g ° -q) = 2u co_gsq. Multiplying T₁ and T₂, we get

\ T₁T₂ = 2 sinu_g q × 2 cosu_g q or T₁T₂ = 2 2 u sin 2 g g æ _qö ç ÷ è ø or T₁T₂ = 2R_g .

More about projectile motion

1. If t₁ is the time taken by projectile to reach a point P at height h and t₂ is the time taken from point P to ground level, then

t₁ + t₂ = T = 2 sinu_g q

or u sinq = (1 2)

2

g t +t

. The height of point P,

h = u sinq t₁ – 1 2 gt12 = (1 2) 2 g t +t t₁ – 1 2 gt12 or h = 1 2 g t1 t2.

2. Change in momentum : Change in momentum between two positions of projectile is given by ΔPuur = Puurf -uurP = (vi m urf -urv ) .i

(a) Between point of projection and highest point

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161

Motion in a Plane

i

ur

v = u cosq ˆi + u sinq ˆj

and f

ur

v _{= u cosq ˆi}

\ ΔPuur = m [(u cosq ˆi ) – (u cosq ˆi + u sinq ˆj)]

or DP = mu sin q

(b) For the complete projectile motion i r

v = u cosq ˆi + u sinq ˆj f

r

v _{= u cosq ˆi – u sinq ˆj}

\ ΔPuur = m [(u cosq ˆi – u sinq ˆj) – (u cosq ˆi + u sinq ˆj)]

or DP = 2 mu sin q

ANALYSIS OF PROJECTILE OF TYPE 2 : HORIZONTAL PROJECTION

Let a particle be projected horizontally with initial velocity u from height h. Velocity at any time t

We have, v_x = u and v_y = u_y + gt or = 0 + gt \ rv = uˆi-gtˆj and v = u2+( )g t 2 Also tana = gt u .

Position at any time t

Taking point of projection as the origin, the position vector at any time t r ur = x ˆi – y ˆj. where x = ut and y = 1 2gt2 \ _rr = ˆ 1 2ˆ 2 -uti gt j Displacement s = r = 2 2 1 2 ( ) 2 u t _{+ ç}æ_è g t ö_÷_ø Equation of trajectory We have, x = ut or t = x u and y = – 1 2gt2 = – 1 2 g 2 x u æ ö ç ÷ è ø Figure. 4.27 Figure. 4.28

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Figure. 4.29 or y = 2 2 1 2 - gx u Time of flight (T) We have, h = u_yt +1 2 ayt2 or h = 0 + 1 2 g T2 which gives T = 2h_{g .} Horizontal range (R) R = u_x × T = u 2h g . The average acceleration in total time of flight is g downward.

ANALYSIS OF PROJECTILE OF TYPE 3

Method - I :

Let us consider a particle is projected with initial velocity 100 m/s at an angle 30° with the horizontal. The height of projection is 100 m.

Time of flight (T) We have y = u_y t + 1 2 g t2 or 100 = – 100 sin 30° T + 1 2 × 10 × T2 or T2_{– 10 T – 20 = 0} which gives T = 2 10 ( 10) 4 1 ( 20) 2 ± - ´ ´ -= 11.71 s

(consider only positive value) The horizontal range (R)

R = u_x × T = 100 cos 30° × 11.71 = 100 × 3

2 × 11.71 = 1014 m.

Method - II :

Taking point of projection as the origin, the coordinates of point of strike are (R, – 100 m). We have, y = x tanq – ₂ 2₂ 2 cos gx u q Here y = – 100 m and x = R \ – 100 = R tan 30° – 2 2 2 10 2(100) cos 30 R ° or R2_{– 866 R – 150000 = 0} Figure. 4.30

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163

Motion in a Plane which gives, R = 2 866 (866) 4 150000 2 ± + ´

= 1014 m (consider only positive value)

Time of flight, T = cos R u q = 11.71 s.

F

ORMULAE

U

SED Projectile Type 1 1. u_x = u cos q, u_y= u sin q a_x = 0, a_y = –g .

2. Position after time t

x = u cos q t, y = u sin q t – 1₂gt2 3. Equation of trajectory y = x tan q – ₂ 2₂ 2 cos gx u q.

4. Time of flight, T = 2 sinu

g q . 5. Maximum height, H = 2_sin2 2 u g q . 6. Horizontal range, R = 2_{sin 2} u g q

7. Maximum range, R_max =

2

u

g , for q = 45°

Projectile Type 2 1. Position after time t

x = ut, y = 1 2gt2 2. Equation of trajectory, y = 2 2 2 gx u .

3. Velocity after time t, v = _u2₊_(gt)2

4. Time to hit the ground, T = 2h_g

5. Horizontal range, R = uT = u 2h_g . y x u O q _R H u h x R