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2013-2015

Notes for School Exams

Physics XI

Quick Revision Units & dimensions Vectors Kinematics Laws of Motion Work, Energy & Power Centre of Mass & Collision

P. K. Bharti, B. Tech., IIT Kharagpur

© 2007 P. K. Bharti

All rights reserved.

www.vidyadrishti.org

Quick Revision Author: P. K. Bharti (B. Tech., IIT Kharagpur) www.vidyadrishti.org

P a g e | 2 CONCEPT: JB-20, 3rd Floor, Near Jitendra Cinema, Sec 4, Bokaro Mb: 7488044834

Important questions for 1st terminal Examinations

1. Different type of errors and derivations on product and exponents

Units and dimensions

2. Least count 3. Significant digits 4. Finding dimensions of physical quantity 5. Numerical on checking correctness of an equation

using dimensional analysis 6. Numerical on derivation of formula using

dimensional analysis

7. Triangle law of vector addition and parallelogram law of vector addition with proof.

Vectors

8. Numerical on component of vectors, vector addition, subtraction, dot product and cross product.

9. To prove triangular inequalities |a+b| < |a| + |b| etc.

10. To prove sin sin sin

a b cA B C

= =

11. To prove one of these using calculus method

Kinematics

2

2 2

1 2

2

v u at

s ut at

v u as

= +

= +

= +

Distance covered in nth second

12. Projectile motion derivations Equation of trajectory (parabolic path) Maximum height Range Time of flight Speed at a particular instant Time of ascent = time of descent Condition for maximum range etc. Particle thrown horizontally from a height

13. Numerical on relative motion and projectile motion 14. Simple problem on graphs

15. Newton’s 2nd law of motion in terms of linear momentum

Newton’s Laws of Motion

16. To prove Newton’s 2nd law is the real law of motion 17. Numerical on Conservation of Linear Momentum 18. Numerical on Free Body Diagram, Newton’s 2nd Law

and pseudo force

19. Graph of friction

Friction

20. Laws of limiting friction 21. Friction is necessary evil 22. Friction increases even after polishing a surface

23. Cause of friction 24. Rolling friction 25. Angle of repose and angle of friction 26. Numerical

27. Derivation of centripetal acceleration

Circular Motion 2

cvar

=

28. Banking of roads without and with friction 29. Numerical

30. Derivation of Work-Kinetic Energy theorem

Work, Energy and Power

31. Derivation of spring potential energy 32. Numerical on Conservation of Mechanical Energy 33. Numerical on ballistic pendulum 34. Conservative force 35. Potential, kinetic and mechanical energy graph

36. Derivation of centre of mass of semicircular ring and semicircular disc

Centre of Mass & Collision

37. Perfectly elastic collisions and different cases 38. Oblique collision and different cases 39. Coefficient of restitution

40. Derivation of moment of inertia of simple bodies (please note that it is not in NCERT and should not be asked in schools exam)

Rotational Motion

41. Derivation of parallel and perpendicular axis theorem 42. To prove

o tω ω α= + 21

2ot tθ ω α= +

2 20 2ω ω αθ= +

43. Geometrical meaning of angular momentum 44. Conservation of angular momentum 45. Deduction of Kepler’s 2nd law from conservation of

angular momentum 46. Numerical on pure rolling, radius of gyration, torque

equation

47. Newton’s law of gravitation

Gravitation

48. Variation in acceleration due to gravity with height, depth and rotation

49. Gravitational potential energy 50. Escape velocity 51. Orbital velocity 52. Kepler’s Laws 53. Gravitational field and potential 54. Geostationary and polar satellites 55. Gravitational field and potential (please note that it is

not in NCERT and should not be asked in schools exam)



56. Definitions of all types of stress and strains

Mechanical properties of solids

57. Hooke’s law 58. Stress vs strain graph (imp) 59. Elastic potential energy

60. Variation of pressure with depth in hydrostatic

Mechanical properties of fluids

61. Pascal law 62. Archimedes’ Principle 63. Bernoulli’s Theorem (imp) 64. Speed of efflux: Torricelli’s theorem 65. Venturimeter 66. Stoke’s law and terminal velocity 67. Reynold’s number 68. Capillary rise 69. Drops and bubbles 70. Reasoning on surface tension (eg why droplets are

spherical etc)

71. Zeroth law of thermodynamics

Thermodynamics

72. 2 3β γα = =

73. Assumptions of Kinetic theory of gases (imp)

74. 21 3

P ρ υ= (imp)

75. Mean free path 76. Degrees of freedom 77. Equipartition of energy 78. Work done by a gas for different processes and

graphically (imp) 79. Cp

– Cv

= R

80. Why Cp > Cv

81. constant PV γ = (imp) 82. Why is adiabatic curve steeper than isothermal

curve? 83. Carnot Engine & its efficiency (imp) 84. Second law of thermodynamics 85. Newton’s Law of cooling 86. Conduction

87. SHM

Oscillations (SHM)

88. Energy in SHM 89. Spring mass system 90. Simple pendulum 91. Damped SHM 92. Resonance 93. Physical Pendulum 94. Oscillations of a liquid column in a U-tube 95. Oscillations of a body dropped in a tunnel along the

diameter of the earth 96. Oscillation of a floating cylinder 97. Oscillation of a ball in the neck of an air chamber

98. Types of waves

Waves

99. Travelling wave & standing wave 100. Standing waves on string with both end fixed and one

end fixed 101. Standing waves on open and closed organ pipe 102. Speed sound (Newton) and Laplace’s correction

(imp) 103. Beats (imp) 104. Doppler effect (imp)

Email:

All the best!

From P. K. Bharti, B. Tech. (IIT Kharagpur) HOD Physics @ Concept, Bokaro Centre JB-20, Near Jitendra Cinema, Sec4, Bokaro Ph: 7488044834

[email protected]

mailto:[email protected]�



Units

Measurement of any physical quantity involves comparison with a certain basic, arbitrarily chosen, internationally accepted reference standard called unit.

Fundamental or base units

The units for the fundamental or base quantities are called fundamental or base units.

Derived units

The units of all other physical quantities can be expressed as combinations of the base units. Such units obtained for the derived quantities are called derived units. System of units

A complete set of these units, both the base units and derived units, is known as the system of units.

The International system of units

CGS (centimetre, gram and second)

The base units for length, mass and time in CGS system were centimetre, gram and second respectively. FPS (foot, pound and second)

The base units for length, mass and time in FPS system were foot, pound and second respectively. MKS (metre, kilogram and second)

The base units for length, mass and time in MKS system were metre, kilogram and second S. I. Units (Système Internationale d’ Unites)

The system of units which is at present internationally accepted for measurement is the Système Internationale d’ Unites (French for International System of Units), abbreviated as SI. We shall follow the SI units in our syllabus. Base SI Units Sl. No. Quantity SI unit Symbol

1. Length metre m 2. Mass kilogram kg 3. Time second s 4. Electric Current ampere A 5. Temperature kelvin K 6. Amount of substance mole mol 7. Luminous Intensity candela cd

Two more units Sl. No.

Quantity SI unit Symbol

1. Plane angle radian rad 2. Solid angle steradian sr

Dimensions

The dimensions of a physical quantity are the powers (or exponents) to which the base quantities are raised to represent that quantity.

Sl. No.

Bases Quantity SI unit Dimension

1. Length m [L] 2. Mass kg [M[ 3. Time s [T] 4. Electric Current A [A] 5. Temperature K [K] 6. Amount of substance mol [mol] 7. Luminous Intensity cd [cd]

Application of dimensional analysis

Following are the three main uses of dimensional analysis: (a) To check the correctness of a given physical relation.

It is also known as principle of homogeneity of dimensions.

(b) To convert a physical quantity from one system of units to another.

(c) To derive a relationship between different physical quantities.

a) Principle of homogeneity of dimensions: According to this principle, a physical equation will be dimensionally correct if the dimensions of all the terms occurring on both sides of the equation are the same. To check the dimensional correctness of a physical equation we make use of the principle of homogeneity of dimensions. If the dimensions of all the terms on the two sides of the equation are same, then the equation is dimensionally correct. Example. Let us check the dimensional accuracy of the equation of motion,

212

s ut at= +

Dimensional of different terms are [ ] [ ][ ] [ ] [ ]

[ ]

1

2 2 2

12

s L

ut LT T L

at LT T L

−

−

=

⇒ = = ⇒ = =

As all the terms on both sides of the equations have the same dimensions, so the given equation is dimensionally correct.



b) To convert a physical quantity from one system of units to another: It is based on the fact that the magnitude of a physical quantity remains the same, whatever may be the system of units. If u1 and u2 are the units of measurement of a physical quantity Q and n1 and n2 are the corresponding numerical values, then Q = n1u1 = n2u2 Let M1, L1 and T1 be the sizes of fundamental units of mass, length and time in one system; and M2, L2, T2 be corresponding units in another system. If the dimensional formula of quantity Q be Ma Lb Tc, then

1 1 1 1

2 2 2 2

1 1 1 1 2 2 2 2

1 1 12 1

2 2 2

and

or .

a b c

a b c

a b c a b c

a b c

u M L Lu M L L

n M L L n M L L

M L Tn nM L T

=

=

∴ =

=

This equation can be used to find the numerical value in the second or new system of units.

Example. Let us convert one joule into erg. Joule is SI unit of energy and erg is the CGS unit of energy. Dimensional formula of energy is ML2T–2. ∴ a = 2, b = 2, c = –2.

SI CGS M1 = 1 kg = 1000 g M2 = 1 g L1 = 1 m = 100 cm L2 = 1 cm T1 = 1 s T2 = 1 s n1 = 1 (joule) n2 = ? (erg)

1 1 12 1

2 2 2

1 2 2

3 4 7

100 100 1 11 1 1

1 10 10 10

a b cM L Tn nM L T

−

=

= = × × =

∴ 1 joule = 107 erg. c) To derive the relationship among physical quantities.

By making use of the homogeneity of dimensions, we can derive an expression for a physical quantity if we know the various factors on which it depends.

Example: Consider a simple pendulum, having a bob attached to a string, that oscillates under the action of the force of gravity. Suppose that the period of oscillation of the simple pendulum depends on mass m of the bob, length l of the pendulum and acceleration due to gravity g at the place. Drive the expression for its time period using method of dimensions. Solution. Let us assume that

( ) or ... 1a b c a b cT m l g T Km l g∝ =

where K is a dimensions of various quantities are [T] = T, [m] = M, [l] = L, [g] = LT–2

Substituting these dimensions in eqn. (i), we get T = [M]a [L]b [LT–2]c or M0L0T1 = MaLb + cT–2c Equating the exponents of M, L and T on both sides, we get a = 0, b + c = 0, –2c = 1

On solving, 1 10, , 2 2

a b c= = = −

0 1/2 1/2 lT Km l g Kg

−∴ = =

From experiments, K = 2 π.

Therefore 2 .lTg

π=

Limitations of dimensional analysis:

(a) The method does not give any information about the dimensionless constant K.

(b) It fails when a physical quantity depends on more than three physical quantities.

(c) It fails when a physical quantity 21. ., 2

e g s ut at = +

is the sum or difference of two or more quantities. (d) It fails to derive relationship which involve trigonometric,

logarithmic or exponential functions. (e) Sometimes, it is difficult to identify the factors on which

the physical quantity depends. The method becomes more complicated when dimensional constants like G, h, etc. are involved.

ERRORS

Some important terms

Error in a measurement: The error in a measurement is equal to the difference between the true value and the measured value of the quantity.

Different types of errors: True value: If a1, a2, a3, … an be the n measured values a physical quantity, then its true value, a is given by the arithmetic mean,

1 2 3

1

... 1 or .n

nmean i

i

a a a aa a a

n n =

+ + + += = ∑

(a) Absolute error: The magnitude of the difference between the true value of the quantity measured and the individual measured value is called absolute error. If we take arithmetic mean a as the true value, then the absolute errors in the individual measured values will be

1 1 2 2

3 3

n n

a a a a a a

a a a a a a

∆ = − ∆ = −

∆ = − ∆ = −



The ∆a calculated above may be positive in certain cases and negative in some other cases. But absolute error |∆a| will always be positive.

(b) Mean or final absolute error: The arithmetic mean of the positive magnitudes of all the absolute errors is called mean absolute error. It is given by

1 2

1

... 1 nn

ii

a a aa a

n n =

∆ + ∆ + + ∆∆ = = ∆∑

Thus the final result of the measure of a physical quantity can be expressed as .a a a= ± ∆

(c) Relative error: The ratio of the mean absolute error to the true value of the measured quantity is called relative error.

Relative error, aaa

δ ∆=

(d) Percentage error: The relative error expressed in percent is called percentage error.

Percentage error 100%aa

∆= ×

COMBINATION OF ERRORS (a) Error in the sum or difference of two quantities: Let Δ

A and ΔB be the absolute errors in the two quantities A and B respectively. Then Measured value of A = A ± Δ A Measured value of B = B ± Δ B Consider the sum Z = A + B or difference Z = A − B, The error ΔZ in Z is then given by or Δ Z = Δ A + Δ B Hence the rule: The maximum possible error in the sum or difference of two quantities is equal to the sum of the absolute errors in the individual quantities.

(b) Error in the product or quotient of two quantities: Consider the product, Z = AB or Z = A/B The maximum fractional error in Z is

Z A BZ A B

∆ ∆ ∆= +

Hence the rule: The maximum fractional error in the product or quotient of two quantities is equal to the sum of the fractional errors in the individual quantities.

(c) Errors in the power of a quantity:

If ,p q

r

A BZC

= then maximum fractional error in Z is given by

Z A B Cp q rZ A B C

∆ ∆ ∆ ∆= + +

The percentage error in Z is given by

100 100 100 100Z A B Cp q rZ A B C

∆ ∆ ∆ ∆× = × + × + ×

Derivation of general rule using differentiation. We have

p q

r

A BZC

=

Taking logarithms, we get log Z = p log A + q log B – r log C On differentiating both sides, we get dZ dA dB dCp q rZ A B C

= + −

Writing the above equation in terms of fractional errors. Z A B Cq

Z A B C∆ ∆ ∆ ∆

± = ± ± ±

The maximum permissible error in Z is given by

.Z A B Cp q rZ A B C

∆ ∆ ∆ ∆= + +



VECTORS

AB

Representation of a vector (Arrow over letters or bold letters)

= AB = P

= P

AB

Magnitude of a vector (with modulus or without bold) = |AB| = P

=|P | = AB = P

Note: Magnitude is the length of the vector.

, , ,i j k n

Unit Vector A vector whose magnitude is unity (1 unit) is called a unit vector. Generally it is represented by cap over letter. e.g.,

etc.

Unit Vector in the direction of a given vector A

i

Unit Vectors along coordinate axes:

Unit vector along X-axis = Unit vector along Y- axis = j

Unit vector along Z- axis = k

Clearly, 1i j k= = =

Suppose the coordinates of tip of position vector

Representation of a position vector in unit vector form (3D)

r

is (x, y, z), then that vector is represented in unit vector form as

r xi y j zk= + +

Here, x = x-component of vector r

y = y-component of vector r

z = z-component of vector r

The magnitude (or length) of this vector is

2 2 2r r x y z= = + +

Component of a vectorComponent of a vector

A

along a direction making an angle θ with it is A cos θ.

A

Rectangular component of a vector Let a vector makes an angle of θ with positive direction of x- axis. Then, X-component of A

= A cosθ

Y-component of A

= A sinθ

We can write A

in the unit vector notation as cos sin A A i A jθ θ= +

Let a vector in unit vector form as

Magnitude and direction of vector (when given in unit vector form in 2D)

A xi y j= +

Here, x = x-component of vector A

y = y-component of vector A

The magnitude (or length) of this vector is

2 2A A x y= = +

Angle with positive direction of x-axis (Use trigonometry)

1tan yx

α −=

Vector Addition

1 1 1 1

2 2 2 2

1 2

1 2 1 2 1 2

...........

Then,

...

( ... ) ( ... ) ( ... ) .

n n n n

n

n n n

A a i b j c k

A a i b j c k

A a i b j c k

R A A A

R a a a i b b b j c c c k

= + +

= + +

= + +

= + + +

= + + + + + + + + + + +

Method 1 : Analytically

Here, R

= resultant (i.e., vector sum of all vectors)

B P

Note: A is tail and B is tip (or head) of AB =P A

AAA

=

A

A

x

y

z

i

j

k

x

y

z

r

(x, y, z) x

y

α α α α



Vector Addition

AB BC AC+ =

Method 2: Triangle law of vector addition (when tail of one vector is on the head of another vector)

Statement: It states that if two vectors are represented in magnitude and direction by the two sides of a triangle taken in the same order, then the resultant is represented in magnitude and direction by the third side of the triangle taken in opposite order.

Clearly, (See graphical method for explanation)

R P Q= +

Magnitude of R

:

2 2 2 cosR R P Q PQ θ= = + +

Note θ is the angle between vectors &P Q

(Explained in class)

Angle α which R

makes with P

:

sintancos

QP Q

θαθ

=+

Proof of triangle law of vector addition

Draw a perpendicular CM on AB. Clearly

BC = Q cos θ and CM = Q sin θ.

Now using Pythagoras theorem in right angled triangle

ACM, we have

( )( ) ( )( ) ( )

( )

( )

22 2 2 2

2 22

2 2 2 2 2 2

2 2 2 2 2

2 2 2

2 2

cos sin

cos 2 cos sin

cos sin 2 cos

2 cos

cos sin 1

AC AM CM AB BM CM

R P Q Q

R P Q PQ Q

R P Q PQ

R P Q PQ

θ θ

θ θ θ

θ θ θ

θ

θ θ

= + = + +

⇒ = + +

⇒ = + + +

⇒ = + + +

⇒ = + +

+ =

Again, in right angled triangle ACM

tan

sintancos

perpendicular CM CMbase AM AB BM

QP Q

α

θαθ

= = =+

=+

Vector Addition

R P Q= +

Method 3: Parallelogram law of vector addition (when two vectors are placed tail to tail i.e., co-initial vector)

Statement: It states that if two vectors acting simultaneously are represented in magnitude and direction by the two adjacent sides of a parallelogram taken in the same order, then the resultant is represented in magnitude and direction by the diagonal of the parallelogram passing through that point.

Magnitude of R

2 2 2 cosR R P Q PQ θ= = + +

Note θ is the angle between vectors &P Q

Angle α which R

makes with P

: sintancos

QP Q

θαθ

=+

Angle β which R

makes with Q

: sintancos

PQ P

θβθ

=+

Clearly, α β θ+ =

Proof is same as that of triangle law of vector addition.

NOTE: In general, we should use parallelogram /triangle law for adding two vectors only. For addition of more than two vectors we should use analytical method. Special cases of vector addition

2 2 2 cosR R P Q PQ θ= = + +

Case 1: Maximum magnitude of resultant

Clearly R is maximum when cos θ is maximum. Maximum value of cos θ is 1 when θ = 0o.

( )22 2max 2R P Q PQ P Q P Q= + + = + = +

maxR P Q∴ = + when two vectors lies in same direction.

2 2 2 cosR R P Q PQ θ= = + +

Case 2: Minimum magnitude of resultant

Clearly R is minimum when cos θ is minimum. Minimum value of cos θ is – 1 when θ = 180o.

( )22 2min 2R P Q PQ P Q P Q= + − = − = −

minR P Q∴ = − when two vectors are in opposite direction.

P

R

Q

θ

β

α

B A

C

P

R

Q

θ α

M

P

Q

P

Q



Therefore,

Case 3: When two vectors have equal magnitude

If P = Q, then parallelogram will be a rhombus. We know that diagonal of a rhombus is equally inclined to each sides.

2θα β= =

( )P Q P Q− = + −

Vector subtraction

i.e. vector subtraction is obtained by taking negative of another vector and adding it to first vector. For example: To subtract Q from P, we take negative of vector Q by reversing its direction. Then add P and –Q by parallelogram law or triangle law. Shortcut: Angle between P and Q = θ Therefore, angle between P and – Q = π – θ

( )( )2 2

2 2

2 cos

2 cos

P Q P Q

P Q PQ

P Q PQ

π θ

θ

− = + −

= + + −

= + −

2 2 2 cosP Q P Q PQ θ∴ − = + −

Angle α which R

makes with P

: sintan

cosQ

P Qθα

θ=

−

| || | co sA B A B θ⋅ =

Dot Product or Scalar Product

(Dot product)

where θ is the angle between A and B. Dot product of two vectors gives a scalar quantity.

Using definition of dot product we can find the angle between two vectors.

cos| || |

A BA B

θ ⋅=

(Angle between two vectors)

1 1 1

1 2 1 2 1 2

2 2 2

A a i b j c k

A B a a b b c cB a i b j c k

= + + ⇒ ⋅ = + += + +

Dot product of two vectors (when given in unit vector form):

| || | sin A B A B nθ× =

Cross Product or Vector Product

Cross product gives a vector quantity.

Here, n is a unit vector perpendicular to both & .A B

Clearly, direction of A B×

is

along n , i.e., perpendicular to both & .A B

Direction of A B×

is obtained by using “Right hand thumb

rule”.

( ) ( ) ( )

1 1 1

2 2 2

1 1 1

2 2 2

1 2 2 1 1 2 2 1 1 2 2 1

&

A a i b j c k

B a i b j c k

i j kA B a b c

a b c

A B i b c b c j a c a c k a b a b

= + +

= + +

× =

⇒ × = − − − + −

Cross product of two vectors (in unit vector form):

Important points

Let us consider two vectors

1 1 1 2 2 2 & A a i b j c k B a i b j c k= + + = + +

1. They are parallel if

1 1 1

2 2 2

a b ca b c

= =

2. They are perpendicular, if 0A B⋅ =

1 2 1 2 1 2 0a a b b c c+ + =

3. A unit vector perpendicular to both &A B

is .| |A BA B

××

1. Component of A

along A BBB⋅

=

Vector form: Component of A

along B =

A B BB B

⋅

2. Physical meaning of A B×

A B×

= area of parallelogram

P

Q

Q−

θ

π θ−

R

α

P

R

Q

θ

β

α



KINEMATICS Distance Actual length of the path Denoted by or s r Scalar S.I. unit: m Displacement Shortest distance between two points Denoted by or s r

.

Vector S.I. unit: m Average Speed (between two instants of time) Distance travelled per unit time Denoted by v∆

distanceavg speed = time

svt

∆⇒ ∆ =

∆

We can use this formula for constant speed also. Scalar S.I. unit: m/s Instantaneous Speed (at a particular instant of time) Denoted by v

0lim

t

s dsv vt dt∆ →

∆= ⇒ =

∆

NOTE: Speed means instantaneous speed. This is main formula for speed. We can use this formula for any case of speed. Average Velocity (between two instants of time) Displacement per unit time Denoted by v∆

displacementavg velocity =

timervt

∆⇒ ∆ =

∆

We can use this formula for constant velocity also. Vector S.I. unit: m/s Instantaneous velocity (at a particular instant of time) Denoted by v

0lim

t

r d rv vt dt∆ →

∆= ⇒ =

∆

NOTE: • Velocity means instantaneous velocity. This is main

formula for velocity. We can use this formula for any case of velocity.

• Velocity is tangential to path • Magnitude of instantaneous velocity = instantaneous

speed v v=

Average Acceleration (between two instants of time) Change in velocity per unit time Denoted by a∆

change in velocityavg acceleration= time

vat

∆⇒ ∆ =

∆

We can use this formula for constant acceleration also. Vector S.I. unit: m/s2 Instantaneous Acceleration (at a particular instant of time) Denoted by a

0lim

t

v dva at dt∆ →

∆= ⇒ =

∆

We can also use dv dva vdt dr

= =

NOTE: Acceleration means instantaneous acceleration. These are main formulae for acceleration. We can use these formulae for any case of acceleration. Graphs Case 1: Average value (From slope of chord)

Form:xyt

∆∆ =

∆

Suppose xyt

∆∆ =

∆ and you want y∆ between two instants

1 2 and t t graphically. Draw a chord AB between 1 2 and t t . Suppose this chord (sometimes after extension of chord) makes an θ angle with positive t axis. Then slope of this chord gives average of y i.e. y∆ between 1 2 and t t .

tany θ∆ = Using this method we can find average speed from distance time graph average velocity from displacement time graph average acceleration from velocity time graph

t

x

A

B

t1 t2 θ



Case 2: Instantaneous value (From slope of tangent)

Form: dxydt

=

Suppose dxydt

= and you want y at an instant t graphically.

Draw a tangent at t . Suppose this tangent (sometimes after extension of chord) makes an θ angle with positive t axis. Then slope of this tangent gives instantaneous value of y at t .

tany θ= Using this method we can find instantaneous speed from distance time graph instantaneous velocity from displacement time graph instantaneous acceleration from velocity time graph Case 3: Change in value (From area)

Form: 2 2 2

1 1 1

2 1

y t t

y t t

dy xdt y y xdt= ⇒ − =∫ ∫ ∫

Suppose 2 2 2

1 1 1

2 1

y t t

y t t

dy xdt y y xdt= ⇒ − =∫ ∫ ∫ and you want 2 1y y−

between two instants 1 2 and t t graphically. Draw two lines AB and CD parallel to x − axis at 1 2 and t t . Then area of the region between curve, AB, CD and t axis gives instantaneous

2 1y y− between 1 2 and t t . 2

1

2 1

area of the region between curve, , and axis

t

t

y y xdt

AB CD t

− =

= −

∫

Using this method we can find change in velocity from acceleration time graph change in position (i.e. displacement) from velocity time

graph distance from speed time graph

Uniformly accelerated motion: constanta =

1. v u at= +

[ ] [ ]00 0

( 0) ...(i)

v t tv t

uu

dvadt

dv adt a dt v a t

v u a tv u at

=

⇒ = = ⇒ =

⇒ − = −⇒ = +

∫ ∫ ∫

2. 21 2

s ut at= +

From (i), we have

( ) [ ] [ ]2

0 00 0 0

2

2

1 ...(ii)2

ts ts t

v u atds dsu at vdt dt

tds u at dt s u t a

s ut at

= +

⇒ = + =

⇒ = + ⇒ = +

⇒ = +

∫ ∫

3. 2 2 2v u as= + We know that

[ ]2 2 2

00

2 2

2 2 2

2 ...(iii)

vs vs

u u

dv dva vdt ds

dva vds

v v uads vdv a s as

v u as

= =

∴ =

⇒ = ⇒ = ⇒ = −

⇒ = +

∫ ∫

4. Distance travelled in nth second

( )

[ ] [ ]

( ) ( )

( )

0 1

2

0 11

22

From (i)

211 1 2

1 2 1 ...(ii)2

s n

nn

s n

nn

v u at

ds u at ds u at dtdt

ts u t a

s u n n a n n

s u a n

−

−−

= +

⇒ = + ⇒ = +

⇒ = +

⇒ = − − + − −

⇒ = + −

∫ ∫

t

x

t

θ

t

x

t1 t2

A C

B D



Sign convention

You can take your axis along any side at your will. All the physical quantities which are along that axis are positive and quantities opposite to axis direction are negative.

Motion in a plane (when acceleration is constant)

Motion in a plane (two-dimensions) can be treated as two separate simultaneous one-dimensional motions with constant acceleration along two perpendicular directions. Along x axis

2

2 2

1 2

2

x x x

x x

x x x

v u a t

x u t a t

v u a x

= +

= +

= +

Along y-axis

2

2 2

1 2

2

y y y

y y

y y y

v u a t

y u t a t

v u a y

= +

= +

= +

Note we have assumed that acceleration is constant.

Motion in a plane (when acceleration is not constant)

Along x axis

x

x xx x

dxvdtdv dv

a vdt dx

=

= =

Along y axis

y

y yy y

dyvdtdv dv

a vdt dy

=

= =

Projectile motion

When a particle is thrown obliquely near the earth’s surface, it moves along a curved path. Such a particle is called a projectile & its motion is called projectile motion. Its motion is an example of 2D motion. We shall break equation of motion along x and y axes. Some points to note about projectile motion • Acceleration in y direction, ay = – g (Negative g,

because g is opposite to our +ve y axis.) • Acceleration in x direction ax = 0 (because there is no

acceleration in x direction). • Velocity in x direction, vx = u cos𝛉𝛉 is always same

(because ax = 0); i.e., vx does not change with time. • At maximum height H, velocity is parallel to x-axis

(because velocity is tangential to path). Therefore, y component of velocity at maximum height, vy = 0

To prove:

Equation of trajectory: 2

2 2tan 2 cos

gxy xu

θθ

= −

Maximum height:2 2sin

2uH

gθ

=

Horizontal Range:2 sin 2uR

gθ

=

Time of Flight:2 sinuT

gθ

=

Data to be used: Initial velocity: u at an angle θ with horizontal (x-axis) x – component of initial velocity: cosxu u θ=

y – component of initial velocity: sinyu u θ=

Acceleration:

( ) acceleration due to gravity along negative -axisa g y= −

x – component of acceleration:

( )0 no acceleration component along -axisxa x=

y – component of acceleration: ya g= −

Position at time t : ( ),x y y – component of velocity at maximum height: vy = 0


2 2

1tan 2 cos

gxy xu

θθ

= −

Along x – axis

( )

21 2cos cos and 0

...(1)cos

x x

x x

x u t a t

x u t u u axt

u

θ θ

θ

= +

⇒ = = =

⇒ =

Along y – axis

( )

2

2

2

2

2 2

1 2

1sin sin and 2

1sin cos 2 cos

from (i) cos

1tan 2 cos

y y

y y

y u t a t

y u t gt u u a g

x xy u gu u

xtu

gxy xu

θ θ

θθ θ

θ

θθ

= +

⇒ = − = = −

⇒ = −

=

⇒ = −

...(A)

x

y

u

𝜽𝜽 R

H



Maximum height:2 2sin

2uH

gθ

=

At maximum height H, velocity is parallel to x-axis (because velocity is tangential to path). Therefore, y component of velocity at maximum height, vy = 0. Along y – axis

( )( )

2 2

22

2 2

2

0 sin 2

sin , for max height and

sin2

y y y

x y

v u a y

u gH

u u y H a g

uHg

θ

θ

θ

= +

⇒ = −

= = = −

⇒ =

Time of Flight:2 sinuT

gθ

=

For time of flight we should use y = 0 and t =T, because displacement along y – axis for complete motion is zero.

( )

2

2

1 2

10 sin 0, , sin and 2

1sin 02

2 sin0 or,

y y

y y

y u t a t

u T gT y t T u u a g

u gT T

uT Tg

θ θ

θ

θ

= +

⇒ = − = = = = −

⇒ − =

⇒ = =

2 sinNeglecting 0, we get, uT T

gθ

= =

Horizontal Range:2 sin 2uR

gθ

=

Putting x = R, 0xa = , and t = 2 sinuT

gθ

= along x-axis we

get,

( )

( )

2

2

2

1 2cos

2sin cos2 sincos =

sin 2= sin 2 2sin cos

x xx u t a t

R u TuuR u

g g

uRg

θ

θ θθθ

θ θ θ θ

= +

⇒ =

⇒ =

⇒ =

Condition for maximum range

We know that 2 sin 2uR

gθ

=

Clearly range will be maximum when 0 0sin 2 1 2 90 45θ θ θ= ⇒ = ⇒ =

Therefore, 2

maxuRg

=

To prove there are two angles of projection for same range

We know that 2 sin 2uR

gθ

=

Also from Trigonometry,

( )( )

0

0

sin 2 sin 180 2

sin 2 sin 2 90

θ θ

θ θ

= −

⇒ = −

Therefore there are two angles 𝛉𝛉 and 900 – 𝛉𝛉 for same range.

To prove time of ascent = time of descent For upward motion upto maximum height:

Let the required time be t1. We have, sin , , 0y y yu u a g vθ= = − =

Now we have,

1

1

0 sinsin ...(i)

y y yv u a tu gtut

g

θθ

= +

⇒ = −

⇒ =

For downward motion from maximum height to ground:

Let the required time be t2. We have, 2 2sin0, ,

2y yuu a g y H

gθ

= = − = − = −

Now we have, 2

2 22

2

2

1 2

sin 1 2 2sin= ...(ii)

y yy u t a t

u gtg

utg

θ

θ

= +

⇒ − = −

⇒

Clearly from (i) and (ii), we have t1 = t2. Hence, time of ascent = time of descent Relative Motion: Terminology

12

12

12

1=object, 2 = observer

= relative position of particle 1 wrt particle 2

= relative velocity of particle 1 wrt particle 2

= relative acceleration of particle 1 wrt particle 2

x

v

a

General equation

12 10 02

12 10 20

m m m

m m m

= +

⇒ = −

i.e., Relative motion of 1 wrt 2

= (Relative motion of 1 wrt 0)

– (Relative motion of 2 wrt 0)

where O is a third object wrt which we know the motion of 1

and 2.



Projectile given horizontal projection

Data to be used: Initial velocity: u in horizontal direction x – component of initial velocity: xu u=

y – component of initial velocity: 0yu =

Projection height above the ground: H Acceleration:

( ) -axis is downwarda g y=

x – component of acceleration:

( )0 no acceleration component along -axisxa x=

y – component of acceleration: ya g= −

Position at time t : ( ),x y


2 2gxyu

=

Along x – axis

( )

21 2 and 0

...(1)

x x

x x

x u t a t

x ut u u axtu

= +

⇒ = = =

⇒ =

Along y – axis

( )

2

2

2

2

2

1 2

1 0 and 21 from (i) 2

...(A)2

y y

y y

y u t a t

y gt u a g

x xy g tu u

gxyu

= +

⇒ = = =

⇒ = =

⇒ =

Time of Flight: 2HTg

=

For time of flight we should use y = H and t =T, because displacement along y – axis for complete motion is H.

( )

2

2

1 2

1 , , 0 and 22

y y

y y

y u t a t

H gT y H t T u a g

HTg

= +

⇒ = = = = =

⇒ =

Horizontal Range: 2HR ug

=

Putting x = R, xu u= , 0xa = , and t = 2HTg

= along x-axis

we get, 21

2

2

x xx u t a t

R u T

HR ug

= +

⇒ =

⇒ =

Speed at time t: 2HR ug

=

Along x – axis

...(i)

x x x

x

v u a tv u= +

⇒ =

Along y – axis

...(ii)y y y

y

v u a tv gt= +

⇒ =

Speed at time t

y

x

u

H

R



Laws of Motion

Newton’s 1st law of motion

• If the vector sum of all the forces acting on a particle is zero then and only then the particle remains unaccelerated. a = 0 if and only if F = 0.

Newton’s 1st law gives definition of force.

• Inertial frame of reference

• A frame of reference in which Newton’s 1st Law is valid

is called an inertial frame of reference. It is non-accelerating frame. i.e., a=0. In simple language, an inertial frame of reference is such a frame which has no acceleration.

• Linear Momentum

Momentum, of a particle is defined to be the product of its mass m and velocity v, and is denoted by p: p = m v Denoted by p

or p

Vector S.I. unit: =kg-m/s

• Newton’s 2nd law of motion

The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts. According to the Second Law

d pFdt

d pF kdt

∝

⇒ =

where k is a constant of proportionality. In S. I. system, we choose k = 1. Hence,

d pFdt

=

For a particle of fixed mass m, d p dmv dvm madt dt dt

= = =

Thus, Newton’s second law can be written as d pF madt

= =

Newton’s 2nd law measures force.

Conditions for Newton’s 2nd Law to hold 1. Inertial frame for reference 2. Particle or particle like object 3. Speed of particles well below speed of light c 4. The second law of motion is a local relation which

means that force F at a point in space (location of the particle) at a certain instant of time is related to a at that point at that instant. Acceleration here and now is determined by the force here and now.

• Impulse

From Newton’s 2nd law, we know that

2 1

d pF Fdt d pdt

Fdt p p

= ⇒ =

⇒ = −

∫ ∫

∫

Here, Fdt∫

is known as impulse which is equal to change

in linear momentum. We can write for a constant force

2 1

2 1

Fdt F dt F t p p

F t p p

= = ∆ = −

⇒ ∆ = −

∫ ∫

A large force acting for a short time to produce a finite change in momentum is called an impulsive force.

Conservation of linear Momentum

• The total momentum of an isolated system of interacting particles is conserved.

• Isolated system means, net external force acting on the system is zero.

Newton’s 3rd law of motion

There is an equal and opposite reaction to every action.

Conditions for Newton’s 3rd Law to hold 1. Forces always occur in pairs. Force on a body A by B is

equal and opposite to the force on the body B by A. 2. There is no cause effect relation implied in the third law.

The force on A by B and the force on B by A act at the same instant.

3. Action and reaction forces act on different bodies, not on the same body. Consider a pair of bodies A and B. According to the third law,

AB BAF F= −

(force on A by B) = – (force on B by A) In scalar form AB BAF F=

NOTE:

1. If you are considering the system of two bodies as a whole, and AB BAF F

are internal forces of the system (A

+ B). They add up to give a null force. Internal forces in a body or a system of particles thus cancel away in pairs. This is an important fact that enables the second law to be applicable to a body or a system of particles

To prove Newton’s 2nd law is the real law of motion

• To prove that Newton’s 1st law is contained in Newton’s 2nd law

• Newton’s 2nd Law states that net external force F

exerted on a particle is equal to mass m times acceleration a

of

the particle, i.e., F ma=

• If 0F =

then 0 0ma a= ⇒ =

, which is Newton’s 1st law i.e., if the net external force acting on a particle is zero then and only then the particle remains unaccelerated.

0a =

if and only if 0F =

.



• To prove that Newton’s 3rd law is contained in Newton’s 2nd law

• Let us consider two isolated bodies A and B interacting with each other only. Now from Newton’s 2nd law we have,

• Force on B by A

= rate of change in linear momentum of B, i.e.,

BAB

d pFdt

⇒ =

• Force on A by B

= rate of change in linear momentum of A, i.e.,

ABA

d pFdt

⇒ =

• Sum of these forces,

( )B AAB BA A B

d p d p dF F p pdt dt dt

⇒ + = + = +

• As no external forces acting on our system, rate of change of total momentum must be zero, i.e,

( ) 0A Bd p pdt

+ =

.

• Hence,

0AB BAF F+ =

AB BAF F⇒ = −

, which is Newton’s 3rd law.

Important forces

1. Weight

Direction of weight is always vertically downwards, however the orientation of body may be. It originates from centre of mass of a body.

2. Normal force

• When a body presses against a surface, the surface deforms and pushes on the body with a force N that is perpendicular to the surface . This force N is called Normal force.

• Normal force comes into existence when there is some contact between two bodies.

• Number of normal forces acting on a body = Number of contacts.

• In general Normal reaction is perpendicular to the surface of contact.

3. Tension

• When a cord (or a rope, cable, or other such object) is attached to a body and pulled taut, the cord pulls on the body with a force T directed away from the body and along the cord.

• This force T is called as tension force. • Tension can never push a body. • Tension is always pulling in nature. • It means direction of tension is always along the rope

such that it pulls the body (does not push). Tension in all part of the string will be same, only when, the string is

• Massless, and • Inextensible.

Massless means mass of string is negligible compared to mass of block. • For a massless or frictionless pulley, tension in two parts

of the string will have same magnitude T = T1=T

2, even if

the bodies are accelerating. Cord should be massless and unstretchable.



Free Body Diagram (F.B.D.)

F.B.D. is nothing, but a diagram showing all the forces acting on a body.

Lami’s Theorem

• When three coplanar forces A, B and C act on a particle such that particle remains in equilibrium, then, Lami’s Theorem states that:

sin sin sinA B Cα β γ

= =

where A, B and C are the magnitude of forces A, B and C respectively. α = angle opposite to force A β = angle opposite to force B γ = angle opposite to force C

Pseudo force

• To write Newton’s law correctly in non-inertial frame we apply a force ma opposite to the acceleration of frame. This ma is called as pseudo force. Here m is the mass of the particle (not the mass of frame) whose FBD we are drawing in non-inertial frame.

• Pseudo force = (mass of body of interest) x (acceleration of non-inertial frame)

SSttaattiicc FFrriiccttiioonn

• The magnitude of the static Friction fs is given by

fs ≤ µsN

• µs = a constant known as coefficient of static friction between the body and the surface of contact. It depends upon roughness of the surface. More rough the surface is more will be µs .

• N = Normal force on the object from the surface.

• Magnitude of static friction = magnitude of applied force.

• Limiting Friction = Maximum value of Static Friction Limiting Friction = fs, max = µsN.

KKiinneettiicc FFrriiccttiioonn

The magnitude of the kinetic friction fk is given by

fk = µkN

µk = a constant known as coefficient of kinetic friction between the body and the surface of contact. It depends upon roughness of the surface. More rough the surface is more will be µk .

N = Normal force on the object from the surface.

GGrraapphh ooff ffrriiccttiioonn vvss.. aapppplliieedd ffoorrccee

• From O to A particle remains stationary. Thus static friction acts in part OA.

• Here, static friction = applied force, therefore angle =45o. • Length AB represents maximum friction, i.e., Limiting

friction, fs, max = µsN . • After A, point start moving thus in this part kinetic

friction acts. • Length CD represents kinetic friction which is almost

constant; fk = µkN in this part.



Important points The coefficients of friction depend on the nature of the

surface. The frictional force is nearly independent of the contact

area between the objects. The kinetic friction force is usually less than the

maximum static friction force (Limiting friction). • Static friction opposes impending motion. The term

impending motion means motion that would take place (but does not actually take place) under the applied force, if friction were absent.

• Note that it is not motion, but relative motion that the frictional force opposes.

Contact Force (R) and Angle of Friction (λ):

• Resultant of friction and normal force is called contact force R.

• The angle between the resultant contact force and the normal is called angle of friction λ.

ANGLE OF REPOSE The angle of repose is defined as the angle of the inclined plane at which a body placed on it just begins to slide. Consider an inclined plane, whose inclination with horizontal is gradually increased till the body placed on its surface just begins to slide down. If θ is the inclination at which the body just begins to slide down, then θ is called the angle of repose The body is under the action of the following forces: 1. The weight M g of the body acting vertically downwards. 2. The limiting friction f in upward direction along the

inclined plane, which in magnitude is equal to the component of the weight M g acting along the inclined plane i.e. f = M g sin θ (i)

3. The normal reaction N acting at right angle to the inclined plane in upward direction, which is equal to the component of weight acting perpendicular to the inclined plane i.e. N = M g cos θ (ii)

Dividing equation (i) by (ii), we have

sin tancos

f MgN Mg

θ θθ

= =

Since

f NfN

µ

µ

=

∴ =

Thus, tan θ = µ Therefore, coefficient of limiting friction is equal to the tangent of the angle of repose.

CIRCULAR MOTION

When a particle moves along a circular path, its motion is said to be circular motion. At any instant of time velocity v is always tangent to the circular path. (Why tangent? Because, velocity is tangent to its path at each instant of time.) Therefore, velocity is always perpendicular to radius during circular motion.

(Because, from Mathematics, we know that a tangent of a circle is perpendicular (⊥) to radius.)

v ⊥ R

Acceleration in circular motion

In a circular motion, when the particle’s velocity is v, there is an acceleration a. This acceleration a has two components: • Tangential acceleration

(at) • Centripetal acceleration

or Radial acceleration or Normal acceleration (ac or an or ar )

Tangential acceleration at

: Its direction is along tangent.

tdvadt

= (magnitude of tangential accln )

• Centripetal acceleration or Radial acceleration or

Normal acceleration (ac or an or ar ) Its direction along tangent towards the center.

2

cvaR

= (magnitude of centripetal accln )

M g cos θ

f N

M g

M g sin θ

θ

θ



Therefore, net acceleration a,

2 2t ca a a= + (magnitude of net accln )

tan t

n

aa

θ = (direction of a with an )

NOTE: Speed v may or may not be uniform.

Uniform Circular Motion

• When a particle moves in a circle with constant speed v, its motion is said to be uniform circular motion.

• It means magnitude of velocity vector v is constant, but the direction of v changes continuously. (NOTE: Here speed remains same but velocity changes.)

• It means velocity is changing with time (because velocity is a vector, which changes when either direction or magnitude changes). It means there is acceleration.

• Thus a particle moving in a circle, undergoes an acceleration.

• For uniform circular motion speed, v = constant. • Therefore, tangential acceleration becomes

0tdvadt

= =

• Therefore, there is only centripetal acceleration, which is 2

cvaR

=

• Thus, net acceleration a is radial acceleration here. • Since, net acceleration direction is towards the centre

here, it has got a special name centripetal acceleration. • Thus, during uniform circular motion, • net acceleration =centripetal (radial) acceleration

2

cva aR

= =

Derivation of centripetal acceleration

Change in velocity ∆v when velocity changes from v1 to v2 is given by ∆v = v2 – v1.

This change in velocity is along the centre of the circle.

For small angle we can use sr

θ ∆∆ =

Also v

vθ

∆∆ =

From (i) and (ii), we have,

vs sv vr v r

∆∆ ∆= ⇒ ∆ =

Dividing both sides by time

v v st r t

∆ ∆ = ∆ ∆

Now c

va

t

∆=

∆

and s vt

∆=

∆. Hence,

2

.c

c

va vr

var

=

⇒ =

Angular variables

Angular velocity ω and angular acceleration α

For the time being, it is sufficient to know that angular velocity represents the change in angle with time and angular acceleration is the change in angular velocity per unit time. Physically, if angular velocity is higher means the body rotates faster. Relation between linear and angular variables • Linear displacement (l) and angular displacement (θ)

lR

θ =

• Linear velocity (v) and angular velocity (ω) v Rω=

• Linear acceleration (l) and angular acceleration (θ)

ta Rα=

SI units of angular variables SI unit of angular displacement θ: rad (radian) SI unit of angular velocity ω: rad/s

Another unit of ω is rpm (revolution per minute) 21 rpm rad/s60π

=

SI unit of angular acceleration α: rad/s2

Centripetal acceleration in terms of angular velocity ω Since, v Rω= Therefore, centripetal acceleration,

22

cva RR

ω= =

v1

v2

r

r

∆r

∆𝜽𝜽

–v1

v2 ∆v

∆𝜽𝜽

v2 ∆v

v1



Centripetal Force

We have just seen that a particle in a uniform circular motion has centripetal acceleration directed towards its centre. Thus, from Newton’s 2nd law, a force of magnitude F = mac will act on the particle which will be directed towards the centre. This force F is known as centripetal force. Centripetal Force

22

cmvF ma m R

Rω= = =

NOTE: Centripetal force direction is always towards the centre of the circular path.

Vehicle moving on a circular track with uniform speed v

• Suppose a vehicle is taking a circular level turn of radius R with uniform speed v. We are assuming vehicle to be a particle like object.

• Therefore, it will have a centripetal acceleration 2v

R

directed towards the centre of the turning. • This centripetal acceleration on the vehicle must be

provided by some external force. • Let us see the forces acting here. Forces are :

N (upward) mg (downward)

• But both of N and mg are in vertical direction. No force in radial direction.

• But from Newton’s 2nd law there must be some external force in the radial inward direction to provide centripetal acceleration. Which force is this??

• This external force is static friction f from the road on the vehicle. Thus, friction force provides the necessary centripetal acceleration.

• Therefore, 2

...(i)cmvf ma

R= =

Condition for safe turn • Static friction force provides the necessary centripetal

acceleration here. Therefore, f ≤ µsN .

• Also, along verticle direction, N = mg

• Therefore, f ≤ µsN f ≤ µsmg …(ii) • Therefore, from (i) & (ii)

2

smv mg

Rµ≤

2

(Condition for safe turn)svRg

µ⇒ ≥

• Thus for safe turn, coefficient of static friction µs must be

greater than or equal to2v

Rg.

• In day to day life, when speed of a vehicle taking a turn increase sufficiently, this condition may be violated.

• Therefore, friction force is not very reliable for turning on a circular road.

• This also shows that for a given value of sµ and R, there is a maximum speed of circular motion of the car possible, namely

max sv Rgµ=

• Banking of roads

• The angle through which the outer edge of the road is raised above the inner edge is called angle of banking.

Banking of road analysis without friction

• Let the banking angle of the road be θ . • Forces are:

mg (downward) N (perpendicular to road surface)

• Now , we have to resolve N into radial and tangential components.

• Along radial direction: • Necessary centripetal accln is provided by N sinθ.

Therefore, 2

sin ...(i)mvNR

θ =

• Thus, sin θ component of normal force provides the necessary centripetal accln to vehicles in this case. This force is quite greater than friction force. So, banked roads are safer for vehicles than plane roads during turning.

• Along vertical: N cos θ = mg …(ii)

• Dividing (1) by (2), we get: 2

tan vRg

θ =

N

f

mg

N

mg

𝜽𝜽

N sin 𝜽𝜽

N cos 𝜽𝜽

ca ca

ca 𝜽𝜽



21tan (angle of banking)v

Rgθ −⇒ =

• Thus, angle of banking depends upon velocity of the vehicle and radius of the road.

• Banking of road is designed in such a way, that most of vehicles may go with their average speed safely.

Banking of road analysis with friction

• Let us first call the speed of the vehicle as videal when there was no fiction from road.

• Here, we can have two cases: CASE 1: If the speed is well below a videal, vehicle will tend to skid downward the road. Therefore, friction f will act in upward the inclination of the road. CASE 2: If the speed is well above videal, vehicle will tend to skid upward. Therefore, friction will f act in downward the inclination of road.

• Case 1: To find minimum speed • In this case friction will act upward the incline as the

vehicle will have a tendency to move downward the incline. Let us draw FBD.

• Now we have to resolve f and N into radial and vertical components.

• Along vertical: N cos θ + f sin θ – mg =0

N cos θ + 𝛍𝛍 N sin θ – mg = 0 …(1) (because, limiting friction f =𝛍𝛍 N)

• Along radial direction: 2

sin cos mvN fR

θ θ− =

2

sin cos (2)mvN NR

θ µ θ⇔ − =

• Solving (1) and (2) simultaneously, we get, 1/2

tan1 tan

v Rg µ θµ θ

−= + −

• This is the minimum speed with which a vehicle can take circular turn on a banked road safely.

• Case 2: To find maximum speed

• In this case friction will act upward the incline as the vehicle will have a tendency to move downward the incline.

• Let us draw FBD.

• Now we have to resolve f and N into radial and vertical components.

• Along vertical: N cos θ + f sin θ – mg =0

N cos θ + 𝛍𝛍 N sin θ – mg = 0 …(1) (because, limiting friction f =𝛍𝛍 N)

• Along radial direction: 2

sin cos mvN fR

θ θ+ =

2

sin cos (2)mvN NR

θ µ θ⇔ + =

• Solving (1) and (2) simultaneously, we get, 1/2

tan1 tan

v Rg µ θµ θ

+= −

• Hence, maximum speed for safe turn on a banked road is given by the expression

1/2

maxtan

1 tanv Rg µ θ

µ θ +

= −

Bending of a cyclist

A cyclist provides himself the necessary centripetal force by leaning inward on a horizontal track, while going round a curve. Consider a cyclist of weight M g taking a turn of radius r with velocity v. In order to provide the necessary centripetal force, the cyclist leans through angle θ inwards as shown in Fig. The cyclist is under the action of the following forces: (a) The weight M g acting vertically downward at the centre

of gravity of cycle and the cyclist. (b) The reaction R of the ground on cyclist. It will act along a

line making angle θ with the vertical. The vertical component R cos θ of the normal reaction R will balance the weight of the cyclist, while the horizontal component R sin θ will provide the necessary centripetal force to the cyclist.

( )

( )2

cos ... 1

and sin ... 2

R Mg

MvRr

θ

θ

=

=

Dividing equation (2) by (1), we have

N

mg

𝜽𝜽

N sin 𝜽𝜽

N cos 𝜽𝜽

ca 𝜽𝜽

𝜽𝜽

f cos 𝜽𝜽

fsin 𝜽𝜽

f

N

mg

𝜽𝜽

N sin 𝜽𝜽

N cos 𝜽𝜽

ca

𝜽𝜽

𝜽𝜽

f cos 𝜽𝜽

fsin 𝜽𝜽

f



( )

2

2

sin /cos

tan ... 3

R Mv rR Mg

vrg

θθ

θ

=

⇒ =

Therefore, the cyclist should bend through an angle

21tan v

rgθ −=

So as to have the necessary centripetal force while going round a curved path. It follows that the angle through which cyclist should bend will be greater, if (a) the radius of the curve is small i.e. the curve is

sharper and (b) the velocity of the cyclist is large.

It may be pointed out that for the same reasons, an ice skater or an aeroplane has to bend inwards, while taking a turn.

WORK ENERGY & POWER

Work: When a force F is applied to a particle, and because of the application of this force if there is a displacement or component of net displacement in the direction of applied force, then we say that force F do work on that particle. Work done W by a constant force F in a displacement r at an angle 𝜽𝜽:

. cos W F r Fr θ= =

Two important conditions about this formula: i. Force should be constant.

ii. Valid for particles. We can apply this for those bodies, which are particle like. That is when a force is applied to a particle like object, each part of it moves with a constant velocity. It means that the body is in pure translation motion. No rotation is there in any part of the body

⋅ Units of work

⋅ Work done is a scalar quantity. ⋅ S. I. Unit : joule = J.

1 J = 1 Nm = 1 kgm2/s2. ⋅ CGS Unit: erg

1 erg = 1 dyne cm = 1 gcm2/s2

⋅ 1 J = 107erg.

Work done by a variable force

Work done by a variable force F

in displacing a particle from

position vector 1r

to 2r

is given by

2

1

.r

r

W F dr= ∫

Work done by spring force

Let us find the work done by spring of spring constant k If the spring is displaced from 1 2. to x x x x= =

2 2

2 1

2 21 2

1 = ( )2

x x

x x

W Fdx kx dx k x x= = − −∫ ∫

Special case: When spring is initially at normal position. Putting 1 2=0 to x x x= we get

( )2 2 21 2

1 12 2

W k x x kx= − = −

Spring potential energy: U = - W = 212

kx

Energy: The ability to do work is called energy. S. I. Unit of energy: J Kinetic Energy(KE or K or T) =The energy due to motion of a particle or body.

212

K mv=

where v = velocity magnitude (speed). Kinetic energy depends on frame of reference.

Work Kinetic Energy Theorem

r

M g

R sin θ

θ

θ

R R cos θ

VERTICAL



It states that work done by all the forces acting on a body is equal to the change produced in the kinetic energy of the body. Suppose that a body in initially at rest and net force F

is

applied on the body to displace along its own direction. Then, work done

( ). . 0W F ds F ds θ= = =∫ ∫

According to Newton’s second law of motion, F = m a, where a is acceleration produced (in the direction of force) on applying the force. Therefore,

( )

W Fds mads F ma

dv dv dvW mv ds mvdv a vds dt ds

= = =

⇒ = = = =

∫ ∫

∫ ∫

Therefore, work done by the force in order to increase its velocity from u (initial velocity) to υ (final velocity) is given by

2

2 2

2 1

2

1 12 2

vv v

u u u

vW mvdv m vdv m

W mv mu

W K K

= = =

⇒ = −

⇒ = −

∫ ∫

Hence, work done on a body by all forces is equal to the change in its kinetic energy.

Conservative Field

A force is said to be conservative force if work done by it is path independent and depends only on net change of position and not on the particular path followed in reaching the new position.

or

A force is said to be conservative force if work done by it in a closed path is zero. Examples of conservative forces are

1. Gravitational force 2. Spring force 3. Electrostatic force

Potential Energy

Energy associated with position or configuration is known as potential energy. It is denoted by U. Note that the potential energy is a property of system of two or more particles rather than of either particle alone.

Gravitational potential energy

gU mgh=

We have to take reference level in problems where gravitational potential energy is zero. Gravitational potential energy above reference level is taken as positive and it is negative below reference level.

Spring potential energy

212sU kx=

Mechanical Energy: Sum of kinetic energy and potential energy is known as mechanical energy.

ME = K + U

Conservative Field

A field is said to be conservative if:

1. A system where all the forces acting are conservative in nature

OR 2. A system where some forces are conservative and some

are non-conservative such that work done by all non-conservative forces = 0.

Relation between potential energy and work done in a conservative field:

When a particle is displaced from position 1 to 2 in a conservative field by the application of a conservative force, work done by conservative force is given by

W = – ΔU = – (U2 – U

1)

As work done by non-conservative forces in a conservative field is zero, therefore, work done by total forces in a conservative field is given by

WT = – ΔU = – (U

2 – U

1)

Conservation of Mechanical Energy in a conservative field: Suppose a particle is displaced from point 1 to 2 in a conservative field. Then,

Change in Potential Energy + Change in K.E. = 0 ΔU+ ΔK = 0

U1 + K

1 = U

2 + K

2

Power

Average power: Average power is defined as work done per unit time. Denoted by P∆

work doneAverage Power = time

WPt

∆⇒ ∆ =

∆

We can use this formula for constant power also. Instantaneous Power (Power) Denoted by P

dWPdt

=

Power is a scalar quantity S.I. unit: Watt Another unit: horsepower (hp) 1 hp = 746 W

Collision



1. Elastic collision. Those collisions, in which both momentum and kinetic energy of the system are conserved, are called elastic collisions. The collision between atomic and subatomic particles are elastic in nature. In daily life, the collisions between two glass or preferably ivory balls may be taken as elastic collisions. Characteristics of elastic collisions: (a) The momentum is conserved. (b) Total energy is conserved. (c) The kinetic energy is conserved. (d) The mechanical energy is not converted into any

other form (sound, heat, light) of energy. (e) Forces involved during the interaction are of

conservative nature. 2. Inelastic collision. Those collisions, in which the

momentum of the system is conserved but the kinetic energy is not conserved, are called inelastic collisions. Most of collisions in every-day life are inelastic collisions.

3. Perfectly inelastic collision. Those collisions, in which the colliding particles stick together after the collision and then move with a common velocity, are called perfectly inelastic collision. Mud thrown on the wall and sticking to it, a bullet fired into wooden block and remaining embedded in it, are the examples of perfectly inelastic collision. Characteristics of inelastic collisions: (a) The momentum is conserved. (b) The total energy is conserved. (c) Loss in the kinetic energy is maximum. (d) A part of whole of the mechanical energy may be

converted into other forms (heat, light, sound) of energy.

(e) Some or all of the forces involved are non-conservative in nature.

It may be pointed out that in all types of collisions, (a) momentum is conserved; (b) total energy is conserved and (c) it is the kinetic energy which may or not be

conserved.

ELASTIC COLLISION IN ONE DIMENSION

The collision between two bodies is said to be head-on or in one dimension, if the colliding bodies continue to move along the same straight line after the collision. Consider two perfectly elastic bodies A and B of masses M1

and M2 moving along the same straight line with velocities u1 and u2 respectively. The two bodies will collide, only if u1 > u2. The two bodies undergo a head-on collision and continue moving along the same straight line with velocities υ1 and υ2 along the same direction. The two bodies will separate after the collision, only if υ2 > υ1.

As in elastic collision momentum is conserved, we have M1 u1 + M2 u2 = M1 υ1 + M2 υ2

⇒ M1(u1 – υ1) = M2(υ2 – u2) (i)

Since, kinetic energy is also conserved in an elastic collision, we have

( ) ( ) ( )

2 2 2 21 1 2 2 1 1 2 2

2 2 2 21 1 1 2 2 2

1 1 1 1 2 2 2 2

– – ...

M u M u M M

M u M u ii

υ υ

υ υ

+ = +

⇒ =

Dividing equation (ii) by equation (i), we have

( )

2 2 2 21 1 2 2

1 1 2 2

1 1 2 2

1 2 2 2

...

u uu u

u uu u iii

υ υυ υ

υ υυ υ

− −=

− −⇒ + = +

⇒ − = −

From equation (iii), it follows that in one dimensional elastic collision, the relative velocity of approach (u1 – u2) before collision is equal to the relative velocity of separation (υ2 – υ1) after collision. Let us first and velocity of body A after collision. From equation (iii), we have υ2 = u1 – u2 + υ1 Substituting for υ2 in equation (i), we get

( )

( ) ( )( ) ( )

1 1 2 2 1 1 2 1 2 1

1 1 2 2 1 1 2 1 2 2 2 1

1 2 1 2 2 1 2 1

1 2 1 2 21

1 2

2

2 ...

M u M u M M u uM u M u M M u M u MM M u M u M M

M M u M uiv

M M

υ υυ υ

υ

υ

+ = + − +

⇒ + = + − +

⇒ − + = −

− +⇒ =

+

Again from equation (iii), have υ1 = υ2 – u1 + u2 Substituting for υ1 in equation (1.22), we have

( ) ( )2 1 2 1 12

1 2

2 ...

M M u M uv

M Mυ

− +=

+

Let us calculate the final velocities of the two bodies after collision in the following special cases:

A u1

M1 M2 u2

B

BEFORE COLLISION

A υ 1

M1 M2 υ 2

B

AFTER COLLISION



1. When the two bodies are of equal masses: Let us consider that M1 = M2 = M (say) From equation (iv), we have

( ) 1 2 21

1 1

2 0 2

2M M u Mu Mu

M M Mu

υ

υ

− + += =

+⇒ =

Also, from equation (v), we have ( ) 2 1 1

2

2 1

2 0 2

2M M u Mu Mu

M M Mυ

υ υ

− + += =

+⇒ =

Thus, if two bodies of equal masses suffer elastic collision in one dimension, then after the collision, the bodies will exchange their velocities.

2. When the target body is at rest: In this case, the body B is at rest i.e. u2 = 0. Then, substituting u2 = 0 in equations (iv) and (v), we have

( ) ( )

( )

1 21 1

1 2

12 1

1 2

...

2 ...

M Mu vi

M MM u vii

M M

υ

υ

−=

+

=+

When the target body B is at rest, let us find the final velocities of the two bodies in the following subcases:

(a) When the two bodies are of equal masses: Setting M1 = M2 = M in equations (vi) and (vii), we get υ1 = 0 and υ2 = u1 Therefore, when body A collides against body B of equal mass at rest, the body A comes to rest and the body B moves on with the velocity of the body A. This is sometimes observed, when one of the two boys playing with glass balls, shoots a stationary glass ball with a ball with the help of his fingers; his own glass ball comes to rest, while the stationary ball of the other boy starts moving with the same velocity.

(b) When the mass of body B is negligible as compared to that of A: When M2 < < M1, then in equations (vi) and (vii), M2 can be neglected as compared to M1 i.e. M1 – M2 ≈ M1 and M1 + M2 ≈ M1. Therefore, we have

1 11 1 1 2 1 1

1 1

2 and 2M Mu u u u

M Mυ υ= = = =

Therefore, when a heavy body A collides against a light body B at rest, the body A should keep on moving with same velocity and the body B will come in motion with velocity double that of A. Thus, in principle, if a moving truck (heavy body) collides against a stationary drum, then the truck would keep on moving with the same velocity, while the drum would come in motion with a velocity double the velocity of the truck.

(c) When the mass of body B is very large as compared to that of A: When M2 > > M1, then in equations (vi) and (vii), M1 can be neglected in comparison to M2 i.e. M1 – M2 ≈ M2 and M1 + M2 ≈ –M2. Therefore, we have

( )2 11 1 1 2 1 2 1

2 2

2 and 0M Mu u u M M

M Mυ υ

−= = − = ≈ >>

Therefore, when a light body A collides against a heavy body B at rest, the body A should start moving with equal velocity in opposite direction, while the body B should practically remain at rest. This result is in accordance with the observation that when a rubber ball hits a stationary wall, the wall remains at rest, while the ball bounces back with the same speed.

COEFFICIENT OF RESTITUTION

The coefficient of restitution is defined as the ratio of the velocity of separation to the velocity of approach of the colliding particles. It is denoted by e. According to definition, the coefficient of restitution is given by

velocity of separation velocity of approach

e =

The coefficient of restitution can be used to distinguish between the three types of collision as below: 1. For elastic collision, e = 1

i.e. in an elastic collision, velocity of separation is equal to velocity of approach.

2. For inelastic collision, 0 < e < 1 i.e. in an inelastic collision, the two particles possess non-zero velocity of separation, which is always less than the velocity of approach.

3. For perfectly inelastic collision, e = 0 i.e. in a perfectly inelastic collision, the two bodies do not get separated and move with a common velocity.



ELASTIC COLLISION IN TWO DIMENSIONS

Consider two perfectly elastic bodies A and B of masses M1 and M2 moving along the same straight line (say X-axis) with velocities u1 and u2. If the body A is moving with a velocity greater than that of B i.e. if u1 > u2, then two bodies will collide. After the collision, the two bodies A and B travel with velocities making angles θ1 and θ2 with the incident direction (X-axis) as shown in Fig.

Since the collision is perfectly elastic, the kinetic energy must be conserved. Therefore,

( )2 2 2 21 1 2 2 1 1 2 2

1 1 1 1 ...2 2 2 2

M u M u M M iυ υ+ = +

Momentum is a vector quantity. As the two bodies move along different directions after the collision, the momentum of the two bodies is separately conserved along X-axis and Y-axis. The component of momentum of body A after collision along X-axis = M1 υ1 cos θ1 The component of momentum of body B after collision along X-axis = M2 υ2 cos θ2 Applying the law of conservation of momentum along X-axis, we have M1 u1 + M2 u2 = M1 υ1 cos θ1 + M2 υ2 cos θ2 (ii) The component of momentum of body A after collision along Y-axis = M1 υ1 sin θ1 (along OY) The component of momentum of body B after collision along Y-axis = M2 υ2 sin θ2 (along OY’) = – M2 υ2 sin θ2 (along OY) Before collision, the component of momentum of body A or of body B along Y-axis is zero. Therefore, applying the law of conservation of momentum along Y-axis, we have 0 + 0 = M1 υ1 sin θ1 + (–M2 υ2 sin θ2) or M1 υ1 sin θ1 = M2 υ2 sin θ2 …(iii)

Special cases:

(a) Glancing collision. For such collisions, 1 0θ = and

2 90 .θ =

From equations (1) and (2), we get u1 = v1 and v2 = 0

K.E. of the target particle 22 2

1 02

m v= =

Hence in a glancing collision, the incident particle does not lose any kinetic energy and is scattered almost undeflected.

(b) Head-on collision. In such a collision, the target particle moves in the direction of the incident particle, i.e., θ2 = 0o. Then equations (1) and (2) take forms: m1u1 = m1v1 cos θ1 + m2v2 and 0 = m1v1 sin θ1 Equation (3) for the kinetic energy remains unchanged.

(c) Elastic collision of two identical particles. As the two particles are identical, so m1 = m2 = m (say). By conservation of K.E. for elastic collision,

2 2 2 2 2 21 1 2 1 1 2

1 1 1 or 2 2 2

mu mv mv u v v= + = +

By conversation of linear momentum,

( ) ( )2 2 2

1 1 2 1 1 2

1 1 1 2 1 2

1 1 1 2 2 1 2 2

2 2 21 1 2 1 2

2 2 2 2 21 1 1 2 1 2 1

or

. .

. . . .

or 2 .

or 2 .

o

mu mv mv u v v

u u v v v v

v v v v v v v v

u v v v v

u u v v v v u

= + = +

∴ = + +

= + + +

= + +

= + + =

1 2r . 0. v v =

This shows that the angle between 1v

and 2v

is 90o.

Hence two identical particles move at right angles to each other after elastic collision in two dimensions.

u1

M1 M2

u1

B

A

θ1

X’ A B

υ2

M1

M2

υ1

θ2

Y

Y’

X



Physics Classes by Pranjal Sir (Admission Notice for XI & XII - 2014-15)

Batches for Std XIIth

Batch 1 (Board + JEE Main + Advanced): (Rs. 16000) Batch 2 (Board + JEE Main): (Rs. 13000) Batch 3 (Board): (Rs. 10000) Batch 4 (Doubt Clearing batch): Rs. 8000

About P. K. Bharti Sir (Pranjal Sir)

• B. Tech., IIT Kharagpur (2009 Batch) • H.O.D. Physics, Concept Bokaro Centre • Visiting faculty at D. P. S. Bokaro • Produced AIR 113, AIR 475, AIR 1013 in JEE -

Advanced • Produced AIR 07 in AIEEE (JEE Main)

Address: Concept, JB 20, Near Jitendra Cinema, Sec 4, Bokaro Steel City Ph: 9798007577, 7488044834 Email: [email protected] Website: www.vidyadrishti.org

Physics Class Schedule for Std XIIth (Session 2014-15) by Pranjal Sir

Sl. No. Main Chapter Topics Board level JEE Main Level JEE Adv Level Basics from XIth Vectors, FBD, Work, Energy, Rotation,

SHM 3rd Mar to 4th Apr 14

1. Electric Charges and Fields

Coulomb’s Law 5th & 6th Apr 5th & 6th Apr 5th & 6th Apr Electric Field 10th & 12th Apr 10th & 12th Apr 10th & 12th Apr Gauss’s Law 13th & 15th Apr 13th & 15th Apr 13th & 15th Apr Competition Level NA 17th & 19th Apr 17th & 19th Apr

2. Electrostatic Potential and Capacitance

Electric Potential 20th & 22nd Apr 20th & 22nd Apr 20th & 22nd Apr Capacitors 24th & 26th Apr 24th & 26th Apr 24th & 26th Apr Competition Level NA 27th & 29th Apr 27th & 29th Apr, 1st, 3rd

& 4th May PART TEST 1 Unit 1 & 2 4th May NA NA

NA 11th May 11th May 3.

Current Electricity Basic Concepts, Drift speed, Ohm’s

Law, Cells, Kirchhoff’s Laws, Wheatstone bridge, Ammeter, Voltmeter, Meter Bridge, Potentiometer etc.

6th, 8th, 10th, 13th May



Competition Level NA 15th & 16th May 15th, 16th, 17th, 18th & 19th May

PART TEST 2 Unit 3 18th May NA NA NA 20th May 20th May

SUMMER BREAK 21st May 2013 to 30th May 2013 4. Moving charges and

Magnetism Force on a charged particle (Lorentz force), Force on a current carrying wire, Cyclotron, Torque on a current carrying loop in magnetic field, magnetic moment

31st May, 1st & 3rd Jun



Biot Savart Law, Magnetic field due to a circular wire, Ampere circuital law, Solenoid, Toroid

5th, 7th & 8th Jun 5th, 7th & 8th Jun 5th, 7th & 8th Jun

Competition Level NA 10th & 12th Jun 10th, 12th, 14th & 15th Jun

PART TEST 3 Unit 4 15th Jun NA NA NA 22nd Jun 22nd Jun

mailto:[email protected]�

http://www.vidyadrishti.org/�



5. Magnetism and Matter 17th, 19th & 21st Jun

17th, 19th & 21st Jun

Not in JEE Advanced Syllabus

6.

Electromagnetic Induction

Faraday’s Laws, Lenz’s Laws, A.C. Generator, Motional Emf, Induced Emf, Eddy Currents, Self Induction, Mutual Induction

24th, 26th & 28th Jun



Competition Level NA 29th Jun & 1st Jul 29th Jun, 1st, 3rd & 5th Jul

PART TEST 4 Unit 5 & 6 6th Jul NA NA NA 13th Jul 13th Jul

7. Alternating current AC, AC circuit, Phasor, transformer, resonance,

8th, 10th & 12th Jul



Competition Level NA 15th July 15th & 17th July 8. Electromagnetic Waves 19th & 20th July 19th & 20th July Not in JEE Advanced

Syllabus PART TEST 5 Unit 7 & 8 27th Jul 27th Jul 27th Jul Revision Week Upto unit 8 31st Jul & 2nd

Aug 31st Jul & 2nd Aug

31st Jul & 2nd Aug

Grand Test 1 Upto Unit 8 3rd Aug 3rd Aug 3rd Aug 9.

Ray Optics

Reflection 5th & 7th Aug 5th & 7th Aug 5th & 7th Aug Refraction 9th & 12th Aug 9th & 12th Aug 9th & 12th Aug Prism 14th Aug 14th Aug 14th Aug Optical Instruments 16th Aug 16th Aug Not in JEE Adv

Syllabus Competition Level NA 19th & 21st Aug 19th, 21st, 23rd, 24th Aug

10.

Wave Optics

Huygens Principle 26th Aug 26th Aug 26th Aug Interference 28th & 30th Aug 28th & 30th Aug 28th & 30th Aug Diffraction 31st Aug 31st Aug 31st Aug Polarization 2nd Sep 2nd Sep 2nd Sep Competition Level NA 4th & 6th Sep 4th, 6th, 7th, 9th, 11th Sep

PART TEST 6 Unit 9 & 10 14th Sep 14th Sep 14th Sep REVISION ROUND 1 (For JEE Main & JEE Advanced Level): 13th Sep to 27th Sep

Grand Test 2 Upto Unit 10 28th Sep 28th Sep 28th Sep

DUSSEHRA & d-ul-Zuha Holidays: 29th Sep to 8th Oct 11.

Dual Nature of

Radiation and Matter Photoelectric effect etc 9th & 11th Oct 9th & 11th Oct 9th & 11th Oct

Grand Test 3 Upto Unit 10 12th Oct 12th Oct 12th Oct 12.

Atoms 14th & 16th Oct 14th & 16th Oct 14th & 16th Oct

13. Nuclei 18th & 19th Oct 18th & 19th Oct 18th & 19th Oct X-Rays NA 21st Oct 21st & 25th Oct

PART TEST 7 Unit 11, 12 & 13 26th Oct NA NA 14. Semiconductors Basic Concepts and Diodes, transistors,

logic gates 26th, 28th, 30th Oct & 1st Nov

26th, 28th, 30th Oct & 1st Nov

Not in JEE Adv Syllabus

15.

Communication System 2nd & 4th Nov 2nd & 4th Nov Not in JEE Adv Syllabus

PART TEST 8 Unit 14 & 15 9th Nov 9th Nov NA Unit 11, 12 & 13 Competition Level NA 8th, 9th & 11th

Nov 8th, 9th, 11th, 13th & 15th Nov

PART TEST 9 Unit 11, 12, 13, X-Rays NA 16th Nov 16th Nov Revision Round 2

(Board Level) Mind Maps & Back up classes for late registered students

18th Nov to Board Exams



Revision Round 3 (XIth portion for JEE)

18th Nov to JEE 18th Nov to JEE 18th Nov to JEE

30 Full Test Series Complete Syllabus Date will be published after Oct 2014



arget IIT-JEE 2012 - Conceptvidyadrishti.com/Quick_revision_mechanics_XI_2014.pdf · Quick Revision Author: P. K. Bharti (B. Tech., IIT Kharagpur) Page rd| 2 CONCEPT: JB-20, 3 Floor,

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