11th Standard - Physics Tutorial - NCERT - CBSE Pattern

We will cover below chapters.
  • 8 - Mechanical Properties of Solids
  • 9 - Mechanical Properties of Fluids
  • 10 - Thermal Properties of Matter
  • 11 - Thermodynamics
  • 12 - Kinetic Theory
  • 13 - Oscillations
  • 14 - Waves


1.1 Introduction

1.2 The international system of units

Standard units are length, time, mass, electric current, thermodynamic temperature, amount of substance, and luminous intensity. The units (e.g. metre, second, kilogram) for the fundamental or base quantities are called fundamental or base units. There are lot of other units which can be derived from base units. e.g. Frequency's unit is hertz which has been derived by taking reciprocal of second (also called as Inverse Second) ! Other examples of derived units are joule, newton, watt. Base units and derived units together are called as system of units. When taking measurements, errors may occur in readings. It's important to reduce the error rate as much as possible.

1.3 Significant figures

In measured and computed quantities proper significant figures only should be retained. Rules for determining the number of significant figures, carrying out arithmetic operations with them, and rounding off the uncertain digits must be followed.

1.4 Dimensions of physical quantities

time (T), length (L), mass (M), electric current (I), absolute temperature (Θ), amount of substance (N) and luminous intensity (J) are standard dimensions.

1.5 Dimensional formulae and dimensional equations

1.6 Dimensional analysis and its applications

Dimensional analysis can be used to check the dimensional consistency of equations, deducing relations among the physical quantities.


2.1 Introduction

Rectilineaer motion is the study of motion of objects along a straight line. In Kinematics, we study motion without taking into consideration the causes of motion.

2.2 Instantaneous velocity and speed

2.3 Acceleration

2.4 Kinematic equations for uniformly accelerated motion

v=v0+atv = v_0 + at
x=(v+v02)tx = (\frac{v + v_0}{2})t
x=v0t+12at2x = v_0t + \frac{1}{2} at^2
v2=v02+2axv^2 = v_0^2 + 2ax


3.1 Introduction

3.2 Scalars and vectors

If Vector a is at an angle of θ\theta, then we can split it into 2 components.
x component acosθa cos\theta
y component asinθa sin\theta

3.3 Multiplication of vectors by real numbers

3.4 Addition and subtraction of vectors - graphical method

Vectors can be added in a simple way if it is given in the form of i,j,k format.

3.5 Resolution of vectors

3.6 Vector addition - analytical method

When magnitude and angle between 2 vectors a and b is given, we can add vectors by using below 2 approaches. When 2 vectors are added, we get Resultant Vector!
  • a2+b2+2abcosθ\sqrt { a^2 + b^2 + 2ab \cdot cos\theta} and we can also get  tanα=bsinθa+bcosθtan \alpha = \frac { b sin\theta} {a + bcos\theta }
  • This is called as triangle law of addition or parallelogram law of addition

3.7 Motion in a plane

When motion happens in x as well as y direction, it is called as a motion in plane!

3.8 Motion in a plane with constant acceleration

3.9 Projectile motion

We can easily calculate any parameters by applying rectilinear equations of motion for horizontal and vertical components of vector. If Vector a is at an angle of θ\theta, then we can split it into 2 components.
x component acosθa cos\theta
y component asinθa sin\theta
So maximum height of object thrown at an angle of theta and with speed of v will be
hmax=vy22ah_{max} = \frac {v_y^2} {2a}
hmax=v2sin2θ2ah_{max} = \frac {v^2 \cdot sin^2\theta} {2a}
Time of flight will be Tflight=2vsinθgT_{flight} = \frac {2vsin\theta} {g}
Range of flight will be Rflight=v2sin2θgR_{flight} = \frac {v^2sin2\theta} {g}

3.10 Uniform circular motion

Vavg=ΔxΔt V_{avg} = \frac {\Delta x}{\Delta t}
Vinstataneous=limΔt0ΔxΔt=dxdt V_{instataneous} = \lim_{\Delta t \to 0} \frac {\Delta x}{\Delta t} = \frac {dx}{dt}
aavg=ΔvΔt a_{avg} = \frac {\Delta v}{\Delta t}
ainstataneous=limΔt0ΔvΔt=dvdt a_{instataneous} = \lim_{\Delta t \to 0} \frac {\Delta v}{\Delta t} = \frac {dv}{dt}
ωavg=ΔθΔt\omega_{avg} = \frac {\Delta \theta}{\Delta t}
ωinstataneous=limΔt0ΔθΔt=dθdt\omega_{instataneous} = \lim_{\Delta t \to 0} \frac {\Delta \theta}{\Delta t} = \frac {d\theta}{dt}
αavg=ΔωΔt\alpha_{avg} = \frac {\Delta \omega}{\Delta t}
αinstantaneous=limΔt0ΔωΔt=dωdt\alpha_{instantaneous} = \lim_{\Delta t \to 0} \frac {\Delta \omega}{\Delta t} = \frac {d\omega}{dt}

Linear and Angular Relation

We know below relation which states that angular displacement is equal to product of arc length and radius of circle.
dθ=dlrd\theta = \frac {dl}{r}
Above equation can be written as below
dl=dθrdl = {d\theta}{r}
vinstantaneous=dldt=dθrdt=dθdtr=ωrv_{instantaneous} = \frac {dl}{dt} = \frac {d\theta r}{dt} = \frac {d\theta}{dt} r = {\omega} r
aradial=acentripetal=v2r=(ωr)2r=ω2ra_{radial} = a_{centripetal} = \frac {v^2}{r} = \frac {(\omega r)^2}{r} = {\omega}^2 r
atangential=dvdt=d(ωr)dt=dωdtr=αra_{tangential} = \frac {dv}{dt} = \frac {d({\omega} r)}{dt} = \frac {d{\omega} }{dt} r = {\alpha} r
x, v and a have corresponding notations in circular motion theta (θ\theta), omega (angular speed) (ω\omega) and (α\alpha) respectively. The time taken by an object to make one revolution around a circle with radius R is known as its time period T and the number of revolution made in one second is called its frequency f (=1/T).
ω=2πT\omega = \frac {2\pi } {T}

v=ωr=2πTr=2πRfv = \omega r = \frac {2\pi } {T} r = 2\pi R f


4.1 Introduction

4.2 Aristotle's fallacy

Aristotle's view that a force is necessary to keep a body in uniform motion is wrong. A force is necessary in practice to counter the opposing force of friction.

4.3 The law of inertia

4.4 Newton's first law of motion

Galileo extrapolated simple observations on motion of bodies on inclined planes, and arrived at the law of inertia. Object will maintain it's state (motion or rest) until external force is applied! In simple terms, the First Law is "If external force on a body is zero, its acceleration is zero".

4.5 Newton's second law of motion

Momentum (P) = mass x velocity
Momentum=mvMomentum = m \cdot v
The rate of change of momentum of a body is proportional to the applied force and takes place in the direction in which the force acts. Force = mass x acceleration
F=kdpdt=kmaF = k \frac {dp} {dt} = k m a
  • The second law is consistent with the First Law (F = 0 implies a = 0)
  • It is a vector equation
  • It is applicable to a particle, and also to a body or a system of particles, provided F is the total external force on the system and a is the acceleration of the system as a whole.
  • F at a point at a certain instant determines a at the same point at that instant. That is the Second Law is a local law; a at an instant does not depend on the history of motion.

4.6 Newton's third law of motion

Every action has equal and opposite reaction! Forces in nature always occur between pairs of bodies. Force on a body A by body B is equal and opposite to the force on the body B by A. Action and reaction forces are simultaneous forces. There is no cause-effect relation between action and reaction. Any of the two mutual forces can be called action and the other reaction. Action and reaction act on different bodies and so they cannot be cancelled out. The internal action and reaction forces between different parts of a body do, however, sum to zero.

Tips to solve the problems

Ideal string is massless and inextensible. But in real life, strings can have a mass and they can be extensible. If string is massless, Tension is constant between string of 2 blocks. If string has mass, tension is not constant. Below types of problems can be asked.
  • Block in horizontal(e.g. on table - Normal Force and gravitational force cancel out) and vertical position (hanging - gravitational force and Tension force involved here. No normal force!!)
  • Blocks on inclined surface - Split the force in vertical and horizontal components
  • Multiple blocks connected by string (Tension force) in horizontal and vertical position - watch
  • Blocks on surface with specific coefficient of static and kinetic friction (Frictional Force)
  • String or rope connected to hooks
  • Pully
  • Object inside Lift/Elevator
  • Lift Rope Tension
  • Pendulum in lift
  • Object in circular motion
  • Torque problems
  • Object moving in rotational motion - Watch pendulum video
  • Object inside Lift from non inertial frame (psuedo force)
  • Stack of blocks
  • Rotating pullye problems
  • Combination of all of above
Here are steps you need to follow to solve problems.
  • Draw FBD
  • If system is stationary, Net force is 0
  • If system is moving, Net force is the product of mass and acceleration. Net force can be calculated by considering total mass of entire system (e.g. all blocks).
  • Direction of acceleration is considered as positive
Questions to be answered
  • Force , mass, acceleration, time relationship
  • Time and force to move object of mass m by 1 meter
  • Force and displacement relationship

4.7 Conservation of momentum

The total momentum of an isolated system of particles is conserved. The law follows from the second and third law of motion. Momentum of 2 bodies is conserved.
m1v1+m2v2=0 m_1 \cdot v_1 + m_2 \cdot v_2 = 0
Impulse is the product of force and time which equals change in momentum. The notion of impulse is useful when a large force acts for a short time to produce a measurable change in momentum. Since the time of action of the force is very short, one can assume that there is no appreciable change in the position of the body during the action of the impulsive force. Impulse can be calculated using below formula when force is constant.
impulse=forcetime impulse = force \cdot time
Impulse can be calculated using below formula when force is changing over a period of time.
impulse=Fdt impulse = \int \vec{F} \, dt

4.8 Equilibrium of a particle

When sum of all forces from all directions is 0, body is said to be in an equilibrium! When a body is on inclined surface in equilibrium, then below rules apply.
mgsinθ=Tmg \cdot sin\theta = T
mgcosθ=Nmg \cdot cos\theta = N
Thrust (force) necessary to lift the rocket of mass m with acceleration a is given by below formula.
thrust=upwardforcemgthrust = upward force - mg
Tension in cable of lift moving upward is given by
Tension(T)=Weight(W)+Netforce(F)Tension (T) = Weight (W) + Net force (F)
Tension in cable of lift moving downward is given by
Tension(T)=Weight(W)Netforce(F)Tension (T) = Weight (W) - Net force (F)

4.9 Common forces in mechanics

here is the list of common forces.
  • Tension - If an object of mass m is hanging from the rope, tension will be equal to weight of object (mg) provided there is no acceleration in object.
  • Spring - Spring Force is given by below formula.
    force=kxforce = kx
    where k is spring constant and x is the length by which spring is stretched.
  • Contact
    • Normal
    • Friction - Static , limiting and Kinetic (sliding)
    • Rolling
  • Frictional force opposes (impending or actual) relative motion between two surfaces in contact. It is the component of the contact force along the common tangent to the surface in contact. Static friction fs opposes impending relative motion; kinetic friction fk opposes actual relative motion. They are independent of the area of contact and satisfy the following approximate laws.
    fs<=(fs)max=μsRf_s <= (f_s)_{max} = μ_s R
    fk<=μkRf_k <= μ_k R
    μsμ_s(co-efficient of static friction) and μkμ_k (co-efficient of kinetic friction) are constants characteristic of the pair of surfaces in contact. It is found experimentally that μkμ_k is less than μsμ_s .
  • Centripetal

4.10 Circular motion

centripetalforce=macentripetal force = m \cdot a
centripetalforce=mv2Rcentripetal force = m \cdot \frac {v^2}{R}
centripetalforce=mω2Rcentripetal force = m \cdot \omega^2 \cdot R
If road is banked at an angle, max velocity (assuming 0N frictional force) that car can travel is given by
v=grtanθv = \sqrt {gr \cdot tan\theta}
where g is gravitational acceleration and r is the radius of curve.

4.11 Solving problems in mechanics


5.1 Introduction

5.2 Notions of work and kinetic energy : The work-energy theorem

Change in the kinetic energy is the work done by the net force! This is called as work energy theorem!
Change in Kinetic Energy=work done by the net force\text {Change in Kinetic Energy} = \text {work done by the net force}
KfKi=Wnet K_f - K_i = W_{net}
A force is conservative if (i) work done by it on an object is path independent and depends only on the end points xi,xj {xi, xj} , or (ii) the work done by the force is zero for an arbitrary closed path taken by the object such that it returns to its initial position Work done by force is calculated as
work done=Forcedisplacement\text {work done} = Force \cdot displacement
Work done by multiple forces is calculated as
Total work done=Force1displacement1+Force2displacement2+...+Forcendisplacementn\text {Total work done} = Force_1 \cdot displacement_1 + Force_2 \cdot displacement_2 + ... + Force_n \cdot displacement_n

5.3 Work

Work is 0 if
  • Displacement is 0
  • or force is 0
  • or the force and displacement are mutually perpendicular
W=(Fcosθ)d=F.dW = (F cos\theta)d = F.d
So meaning of work in physics is different than what most people assume! Work and Energy have same units - Joule or newton meter. Other units of energy are erg, electron volt (eV), calorie (cal), kilowatt hour (kWh) Work done on object A by object B is not necessarily equal and opposite to the work done on object B by object A.

5.4 Kinetic energy

5.5 Work done by a variable force

work=Fdx work = \int F dx

5.6 The work-energy theorem for a variable force

If we have a force - displacement graph, the work done is given by the area under graph!

5.7 The concept of potential energy

A force is conservative if (i) work done by it on an object is path independent and depends only on the end points xi,xj{xi, xj}, or (ii) the work done by the force is zero for an arbitrary closed path taken by the object such that it returns to its initial position!
F(x)=dV(x)dxF(x) = - \frac {dV(x)}{dx}
ViVf=ifF(x)dxV_i - V_f = \int_{i}^{f} F(x) dx

5.8 The conservation of mechanical energy

The principle of conservation of mechanical energy states that the total mechanical energy of a body remains constant if the only forces that act on the body are conservative. The gravitational potential energy of a particle of mass m at a height x about the earth’s surface is V(x) = m g x
V(x)=mgxV(x) = mgx
where the variation of g with height is ignored.

5.9 The potential energy of a spring

The elastic potential energy of a spring of force constant k and extension x is
V(x)=12kx2V(x) = \frac {1}{2} k x ^2
The scalar or dot product of two vectors A and B is written as A.B and is a scalar quantity given by :A.B = AB cos θ\theta, where θ\theta is the angle between A and B. It can be positive, negative or zero depending upon the value of θ\theta. The scalar product of two vectors can be interpreted as the product of magnitude of one vector and component of the other vector along the first vector. For unit vectors
i×i^=j^×j^=k^×k^=1 i \times \hat {i} = \hat {j} \times \hat {j} = \hat {k} \times \hat {k} = 1
i^×i^=j^×j^=k^×k^=0\hat {i} \times \hat {i} = \hat {j} \times \hat {j} = \hat {k} \times \hat {k} = 0
Scalar products obey the commutative and the distributive laws.

5.10 Power

5.11 Collisions


6.1 Introduction

A rigid body is one for which the distances between different particles of the body do not change, even though there are forces on them. A rigid body fixed at one point or along a line can have only rotational motion. A rigid body not fixed in some way can have either pure translational motion or a combination of translational and rotational motions. In rotation about a fixed axis, every particle of the rigid body moves in a circle which lies in a plane perpendicular to the axis and has its centre on the axis. Every Point in the rotating rigid body has the same angular velocity at any instant of time. In pure translation, every particle of the body moves with the same velocity at any instant of time. Angular velocity is a vector. Its magnitude is ω=dθdtω = \frac {d\theta} {dt} and it is directed along the axis of rotation. For rotation about a fixed axis, this vector ω has a fixed direction. The vector or cross product of two vector a and b is a vector written as a×ba \times b. The magnitude of this vector is absinθabsinθ and its direction is given by the right handed screw or the right hand rule. The linear velocity of a particle of a rigid body rotating about a fixed axis is given byv=ω×rv = ω \times r , where r is the position vector of the particle with respect to an origin along the fixed axis. The relation applies even to more general rotation of a rigid body with one point fixed. In that case r is the position vector of the particle with respect to the fixed point taken as the origin.

6.2 Centre of mass

The centre of mass of a system of n particles is defined as the point whose position vector is
Rcm=miriMR_{cm} = \frac {\sum m_ir_i} {M}
Rcm=m1r1+m2r2+....+mnrnm1+m2+....+mnR_{cm} = \frac {m_1r_1 + m_2r_2 + .... + m_nr_n} {m_1+m_2+....+m_n}
Xcm=m1x1+m2x2+....+mnxnm1+m2+....+mnX_{cm} = \frac {m_1x_1 + m_2x_2 + .... + m_nx_n} {m_1+m_2+....+m_n}
Ycm=m1y1+m2y2+....+mnynm1+m2+....+mnY_{cm} = \frac {m_1y_1 + m_2y_2 + .... + m_ny_n} {m_1+m_2+....+m_n}
The center of mass of a uniformly distributed body is at geometrical center. When mass is not given, you can still find the center of mass using formula - mass = area x density.
Xcm=A1x1+A2x2+....+AnxnA1+A2X_{cm} = \frac {A_1x_1 + A_2x_2 + .... + A_nx_n} {A_1 + A_2}
Ycm=A1y1+A2y2+....+AnynA1+A2Y_{cm} = \frac {A_1y_1 + A_2y_2 + .... + A_ny_n} {A_1 + A_2}
List of Centroids

6.3 Motion of centre of mass

Velocity of the center of mass of a system of particles is given by V=PMV = \frac {P} {M} , where P is the linear momentum of the system. The centre of mass moves as if all the mass of the system is concentrated at this point and all the external forces act at it. If the total external force on the system is zero, then the total linear momentum of the system is constant.

6.4 Linear momentum of a system of particles

6.5 Vector product of two vectors

6.6 Angular velocity and its relation with linear velocity

6.7 Torque and angular momentum

The turning effect of force is called as Torque (τ\tau ).
Torque=Magnitude of force×Perpendicular distance from axis of rotationTorque = \text {Magnitude of force} \times \text {Perpendicular distance from axis of rotation}
τ=Fr\tau = F r
If force is at an angle, you can use below formula
τ=Fsin(θ)r\tau = F sin(\theta) r
τ=F×r\vec{\tau} = \vec{F} \times \vec{r}
Direction of torque can be calculated using right hand thumb rule.
Momentofcouple=Magnitude of either force×Perpendicular distance between forcesMoment of couple = \text {Magnitude of either force} \times \text {Perpendicular distance between forces}
The angular momentum of the particle rotating about an axis gives measure of the turning motion of body.
L=rpL = r p
L=rpsin(θ)L = r p sin (\theta)
L=r×p \vec{L} = \vec{r} \times \vec{p}

Relation between torque and angular momentum is
τ=dLdt\vec{\tau} = \frac {\vec{dL}} {dt}
So rate of change of momentum is torque!

The angular momentum of a system of n particles about the origin is
L=i=1nri×piL = \sum_{i=1}^{n} r_i \times p_i
The torque or moment of force on a system of n particles about the origin is
τ=1ri×Fi\tau = \sum_{1} r_i \times F_i
The force FiF_iacting on the ithi^{th} particle includes the external as well as internal forces. Assuming Newton's third law of motion and that forces between any two particles act along the line joining the particles, we can show τint=0\tau_{int} = 0 anddLdt=τext \frac {dL} {dt} = \tau_{ext}

6.8 Equilibrium of a rigid body

A rigid body is in mechanical equilibrium if
  • it is in translational equilibrium, i.e., the total external force on it is zero  Fi=0 \sum F_i = 0 and
  • it is in rotational equilibrium, i.e. the total external torque on it is zero :τi=ri×Fi=0 \sum \tau_i = \sum r_i \times F_i = 0
The centre of gravity of an extended body is that point where the total gravitational torque on the body is zero.

6.9 Moment of inertia

Like inertia in translational motion, there is a moment of inertia in rotational motion. Property of body to resist angular change is called as moment of inertia!
I=mr2I = m r^2
Moment of inertia is a scalar quantity and it depends on
  • Mass of body
  • Mass distribution
  • Size and shape
  • Axis of rotation
To find the moment of inertia from the center of mass, we can follow below steps.
  • Find center of mass. So you will know distance between particle and center of mass
  • Use moment of inertia formula

The moment of intertia of a rigid body about an axis is defined by the formula
I=miri2I = \sum m_i {r_i}^2
where rir_i is the perpendicular distance of the ith point of the body from the axis.

Perpendicular axis theorem
Parallel axis theorem

Moment of inertia of different shapes

You need to know below formulae
density=massTotalVolumedensity = \frac{mass}{Total Volume}
density=massTotalAreadensity = \frac{mass}{Total Area}
density=massTotalLengthdensity = \frac{mass}{Total Length}
So you can calculate moment of inertia using below formula.
dI=dm×distance2dI = dm \times {distance}^2
I=dI=dm×distance2I = \int dI = \int {dm \times {distance}^2}
You can find the formulae for all types of shapes at List of moments of Inertia
QuantityLinear MotionRotation Motion
Displacement Δx \Delta x Δθ \Delta \theta
Speed v v ω \omega
Acceleration a a α \alpha
Inertia m m I I
Kinetic Energy K=12mv2 K = \frac {1}{2} m v ^2 K=12Iω2 K = \frac {1}{2} I \omega^2
Momentum P=mv P = mv L=Iω L = I \omega
Force - Change in Momentum F=dPdt F = \frac{dP}{dt}
F=ma F = ma
τ=dLdt \tau = \frac{dL}{dt}
τ=Iα \tau = I \alpha
Work W=FΔx W = F \Delta x W=τΔθ W = \tau \Delta \theta
Impulse FΔt F \Delta t τΔt \tau \Delta t
τ=Iα\tau = I \alpha
L=IωL = I \omega
Conservation of angular momentum - If no external torque, angular momentum remains constant!

6.10 Kinematics of rotational motion about a fixed axis

The kinetic energy of rotation is
K=12Iω2K = \frac {1} {2} I ω^2

6.11 Dynamics of rotational motion about a fixed axis

6.12 Angular momentum in case of rotations about a fixed axis


7.1 Introduction

Newton's law of universal gravitation states that the gravitational force of attraction between any two particles of masses m1 and m2 separated by a distance r has the magnitude.
F=Gm1m2r2F = G \frac {m_1 m_2} {r^2}
where G is the universal gravitational constant. If multiple forces are acting, we can find resultant force by doing vector addition of all forces.

Below formula gives the gravitational force on object with mass m at a distance of r from the center of the Earth. M is the mass of earth and R is the radius of the Earth.
F=GmMrR3F = G \frac {m M r} {R^3}

7.2 Kepler's laws

Kepler's laws of planetary motion state that
  • All planets move in elliptical orbits with the Sun at one of the focal points
  • The radius vector drawn from the Sun to a planet sweeps out equal areas in equal time intervals. This follows from the fact that the force of gravitation on the planet is central and hence angular momentum is conserved.
  • The square of the orbital period of a planet is proportional to the cube of the semi-major axis of the elliptical orbit of the planet The period T and radius R of the circular orbit of a planet about the Sun are related by
T2=(4π2GMs)R3T^2 = ( \frac {4 {\pi}^2} {GM_s}) R^3
where Ms is the mass of the Sun. Most planets have nearly circular orbits about the Sun. For elliptical orbits, the above equation is valid if R is replaced by the semi-major axis, a. Kepler's law Kepler's third law can be proved by comparing below equations.
gravitationalForce=GmMr2=CentripetalForce=mrω2gravitationalForce = G \frac {mM}{r^2} = CentripetalForce = m r \omega^2
GmMr2=mr(2πT)2 G \frac {mM}{r^2} = m r ({ \frac {2 \pi} {T}})^2
T2r3 T^2 \propto r^3
When 2 objects with mass m1 and m2 are seperated by distance x, we can use gravitational formula to find the force between them. When a object with mass m is rotating around some point and if we know the radium, we can calculate centripetal force using centripetal force formula for circular motion. When 2 stars of mass m and 2m at a distance d rotate about common center of mass in free space, then Period T of revolution will be given by below formula which can be derived from Kepler's third law.
T=2πd33Gm T = 2 \pi \sqrt {\frac {d^3}{3Gm}}

7.3 Universal law of gravitation

Gravitational field at point located at a distance of x from the uniform ring is given by below formula.
fieldg=GMx(x2+R2)(32) field_g = \frac{GMx}{(x^2 + R^2)^(\frac{3}{2})}
where G is gravitational constant, M is the mass of ring, x is the distance of the point from the center of the uniform ring, R is the radius of ring,

7.4 The gravitational constant

7.5 Acceleration due to gravity of the earth

Acceleration due to gravity of earth is given by below formula.
g=GMR2g = \frac { GM } {R^2}

7.6 Acceleration due to gravity below and above the surface of earth

The acceleration due to gravity.
(a) at a height h above the earth's surface
g(h)=GME(RE+h)2 g(h) = \frac {GM_E} {(R_E + h)^2}
when h<<RE h << R_E, it is approximately equal to
g(h)=GME(RE)2(12hRE) g(h) = \frac {GM_E} {(R_E)^2} (1-\frac {2h} {R_E})
g(h)=g(o)(12hRE) g(h) = g(o) (1-\frac {2h} {R_E})
g(o)=GMERE2 g(o) = \frac {GM_E} {R_E^2}
(a) at depth d below the earth’s surface is
g(d)=GME(RE)2(1dRE)=g(o)(1dRE) g(d) = \frac {GM_E} {(R_E)^2} (1-\frac {d} {R_E}) = g(o) (1-\frac {d} {R_E})

7.7 Gravitational potential energy

The gravitational force is a conservative force, and therefore a potential energy function can be defined. The gravitational potential energy associated with two particles separated by a distance r is given by
V=Gm1m2r V = - \frac {Gm_1m_2} {r}
where V is taken to be zero at r → ∞. The total potential energy for a system of particles is the sum of energies for all pairs of particles, with each pair represented by a term of the form given by above equation. This prescription follows from the principle of superposition. If an isolated system consists of a particle of mass m moving with a speed v in the vicinity of a massive body of mass M, the total mechanical energy of the particle is given by
E=12mv2GMmr E = \frac {1} {2} m v^2 - \frac {GMm} {r}
That is, the total mechanical energy is the sum of the kinetic and potential energies. The total energy is a constant of motion. If m moves in a circular orbit of radius a about M, where M>>m M >> m, the total energy of the system is
E=GMm2a E = \frac {-GMm} {2a}
with the choice of the arbitrary constant in the potential energy given in the point 5., above. The total energy is negative for any bound system, that is, one in which the orbit is closed, such as an elliptical orbit. The kinetic and potential energies are
K=GMm2a K = \frac {GMm} {2a}
V=GMma V = \frac {-GMm} {a}
If the sphere is rolling without slipping and encounters the inclined surface, distance by which it climbs on the surface is given by below formula.
710v2gsin(θ)\frac {7} {10} \frac {v^2} {g sin (\theta)}
This formula is derived by equating the potential energy at the highest point on inclined surface with the sum of kinetic energy due to translation motion and rotational motion.
mglsin(θ)=Et+Ermgl \cdot sin(\theta) = E_t + E_r
mglsin(θ)=12mv2+12Iω2mgl \cdot sin(\theta) = \frac {1}{2} m v^2 + \frac {1}{2} I \omega^2
Moment of inertia of solid sphere is given by below formula.
I=25mr2I = \frac {2}{5} m r^2
and angular velocity is given by below formula
ω=vtr\omega = \frac {v_t}{r}

7.8 Escape speed

The escape speed from the surface of the earth is
ve=2GMERE=2gRE v_e = \sqrt \frac {2GM_E} {R_E} = \sqrt {2gR_E}
and has a value of
11.2kms111.2 km s^-1
. If a particle is outside a uniform spherical shell or solid sphere with a spherically symmetric internal mass distribution, the sphere attracts the particle as though the mass of the sphere or shell were concentrated at the centre of the sphere. If a particle is inside a uniform spherical shell, the gravitational force on the particle is zero. If a particle is inside a homogeneous solid sphere, the force on the particle acts toward the centre of the sphere. This force is exerted by the spherical mass interior to the particle.

7.9 Earth satellites

7.10 Energy of an orbiting satellite

Mechanical Properties of Solids

8.1 Introduction

8.2 Stress and strain

Stress is the restoring force per unit area and strain is the fractional change in dimension! 3 types of stress
  • Tensile Stress or Compressive Stress
  • Shearing Stress
  • Hydraulic Stress

8.3 Hooke's law

Hook's law says that "For small deformations, stress is directly proportional to strain!"
stress=modulus of Elasticity×strainstress = modulus\ of\ Elasticity \times strain
3 types of Elastic modulus
  • Young's Modulus
  • Shear Modulus
  • Bulk Modulus
Elastomers (type of solids ) does not obey Hooke's law.

8.4 Stress-strain curve

8.5 Elastic moduli

Young's Modulus

When an object is under tension or compression, the Hooke's law is given by formula F/A = YΔL/L where ΔL/L is the tensile or compressive strain of the object, F is the magnitude of the applied force causing the strain, A is the cross-sectional area over which F is applied (perpendicular to A) and Y is the Young's modulus for the object. The stress is F/A.

Shear Modulus

A pair of forces when applied parallel to the upper and lower faces, the solid deforms so that the upper face moves sideways with respect to the lower. The horizontal displacement ΔL of the upper face is perpendicular to the vertical height L. This type of deformation is called shear and the corresponding stress is the shearing stress. In this kind of deformation the Hooke's law is given by formula F/A = G x ΔL/L where ΔL is the displacement of one end of object in the direction of the applied force F, and G is the shear modulus.

Bulk Modulus

When an object undergoes hydraulic compression due to a stress exerted by a surrounding fluid, the Hooke's law is given by formula p = B (ΔV/V), where p is the pressure (hydraulic stress) on the object due to the fluid, ΔV/V (the volume strain) is the absolute fractional change in the object's volume due to that pressure and B is the bulk modulus of the object.

8.6 Applications of elastic behaviour of materials

Mechanical Properties of Fluids

9.1 Introduction

A liquid is incompressible and has a free surface of its own. Liquids are often considered to be incompressible because they have very little volume change in response to pressure. This is a result of the intermolecular forces and the arrangement of particles in a liquid. A gas is compressible and it expands to occupy all the space available to it

9.2 Pressure

If F is the normal force exerted by a fluid on an area A then the average pressure Pav is defined as the ratio of the force to area
Pav=FAP_{av} = \frac {F}{A}
Unit of pressure is pascal. Other units of pressure are Nm2Nm^{-2},atm, bar, torr. 1 mm of hg is equal to 1 torr. Pascal's Law states that "Pressure in a fluid at rest is same at all points which are at the same height. A change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and the walls of the containing vessel." The pressure in a fluid varies with depth h according to the expression
P=Pa+ρghP = P_a + \rho g h
where ρ is the uniform density of the fluid.

9.3 Streamline flow

The volume of an incompressible fluid passing any point every second in a pipe of non uniform crossection is the same in the steady flow. v A = constant ( v is the velocity and A is the area of crossection) The equation is due to mass conservation in incompressible fluid flow.
Nature of flow is deteremined by Reynolds number!
R=ρvDηR = \frac {\rho v D} {\eta}
where ρ {\rho} is a fluid density, v is a velocity of fluid, D is the diameter of tap, η {\eta} is the viscocity.
  • If R<1000 R < 1000 , flow is steady
  • If 1000<R<20001000 < R < 2000 , flow is unsteady
  • If R>2000R > 2000 , flow is turbulent

9.4 Bernoulli's principle

Bernoulli's principle states that as we move along a streamline, the sum of the pressure (P), the kinetic energy per unit volume ( ρv22\rho \frac {v^2}{2}) and the potential energy per unit volume (ρgy) remains a constant.
P+ρv22+ρgy=constantP + \rho \frac {v^2}{2} + \rho gy = constant
There is no fluid which have zero viscosity, so the above statement is true only approximately. The viscosity is like friction and converts the kinetic energy to heat energy.

9.5 Viscosity

The ratio of the shear stress to the time rate of shearing strain is known as coefficient of viscosity - η. Stokes' law states that the viscous drag force F on a sphere of radius a moving with velocity v through a fluid of viscosity is
F=6πηavF = 6 \pi ηav

9.6 Surface tension

Surface tension is a force per unit length (or surface energy per unit area) acting in the plane of interface between the liquid and the bounding surface. Molecules have more energy on the surface!

Thermal Properties of Matter

10.1 Introduction

10.2 Temperature and heat

10.3 Measurement of temperature

Thermometer can be used to measure the temperature. Thermometer uses thermometric property! Celcius temperature can be converted into Farenheit temperature using below formula.
tf=(95)tc+32t_f = (\frac {9} {5}) t_c + 32

10.4 Ideal-gas equation and absolute temperature

The ideal gas equation is given by below formula.
where μ is the number of moles and R is the universal gas constant. The Kelvin absolute temperature scale (T ) has the same unit size as the Celsius scale (Tc ), but differs in the origin.
Tc=T273.15T_c = T - 273.15

10.5 Thermal expansion

The coefficient of linear expansion (αlα_l ) and volume expansion (αvα_v ) are defined by below formula.
ΔLL=αlΔT\frac{\Delta L}{L } = \alpha_l \cdot \Delta T
ΔVV=αvΔT\frac{\Delta V}{V } = \alpha_v \cdot \Delta T
αv=3αl\alpha_v = 3 \alpha_l

10.6 Specific heat capacity

Specific heat capacity of a substance is given by below formula.
s=1mΔQΔTs = \frac{1}{m } \cdot \frac{\Delta Q}{\Delta T }
The molar specific heat capacity of a substance is given by below formula.
C=1μΔQΔTC = \frac{1}{ μ } \cdot \frac{\Delta Q}{\Delta T }
where μ is the number of moles of the substance!

10.7 Calorimetry

The latent heat of fusion (LfL_f ) is the heat per unit mass required to change a substance from solid into liquid at the same temperature and pressure. The latent heat of vaporisation (LvL_v )is the heat per unit mass required to change a substance from liquid to the vapour state without change in the temperature and pressure.

10.8 Change of state

10.9 Heat transfer

Three modes of heat transfer are
  • conduction
  • convection
  • radiation
In conduction, heat is transferred between neighbouring parts of a body through molecular collisions, without any flow of matter. For a bar of length L and uniform cross section A with its ends maintained at temperatures TCT_C and TDT_D, the rate of flow of heat H is given by formula.
H=KATCTDLH = K A \frac{T_C - T_D}{ L }
where K is the thermal conductivity of the material of the bar.

10.10 Newton's law of cooling

Newton's Law of Cooling says that the rate of cooling of a body is proportional to the excess temperature of the body over the surroundings.
dQdt=k(T2T1)\frac{dQ}{ dt } = -k (T2-T1)
Where T1 is the temperature of the surrounding medium and T2 is the temperature of the body.


11.1 Introduction

11.2 Thermal equilibrium

11.3 Zeroth law of thermodynamics

The zeroth law of thermodynamics states that "two systems in thermal equilibrium with a third system separately are in thermal equilibrium with each other". The Zeroth Law leads to the concept of temperature.

11.4 Heat, internal energy and work

Internal energy of a system is the sum of kinetic energies and potential energies of the molecular constituents of the system. It does not include the over-all kinetic energy of the system. Heat and work are two modes of energy transfer to the system. Heat is the energy transfer arising due to temperature difference between the system and the surroundings. Work is energy transfer brought about by other means, such as moving the piston of a cylinder containing the gas, by raising or lowering some weight connected to it. Internal energy (U) is given by below formula!
U=nCvTU = n C_v T
Cv=f2RC_v = \frac{f}{2} R
So U can be defined as
U=n(f2)RTU = n (\frac{f}{2}) RT
But We know below formula.
So U can be defined as
U=(f2)nRTU = (\frac{f}{2}) nRT
U=(f2)PVU = (\frac{f}{2}) PV
So if internal energy is given by specific equation, we can calculate degree of freedom and atomicity of gas.
Degree of freedom is given below
  • Monoatomic gas - 3
  • Diatomic gas - 5
  • Triatomic Linear - 5
  • Triatomic Non Linear - 6 and more
  • Polyatomic - 6 and more
So assuming total work done is 0, total internal energy will be zero.
Ut=(f2)P1V1+(f2)P2V2=(f2)P1(V1+V2)U_t = (\frac{f}{2}) P_1V_1 + (\frac{f}{2}) P_2V_2 = (\frac{f}{2}) P_1(V_1 + V_2)
If 2 identical gases are mixed together, below equations hold.
Pf=p1v1+p2v2v1+v2P_f = \frac{p_1v_1 + p_2v_2}{v_1+v_2}
Tf=n1T1+n2T2n1+n2T_f = \frac{n_1T_1 + n_2T_2}{n_1+n_2}
If 2 different gases are mixed together, equations can be derived by below formula.

11.5 First law of thermodynamics

The first law of thermodynamics is the general law of conservation of energy applied to any system in which energy transfer from or to the surroundings (through heat and work) is taken into account. Formula is
ΔQ=ΔU+ΔW\Delta Q = \Delta U + \Delta W
where ΔQ is the heat supplied to the system, ΔW is the work done by the system and ΔU is the change in internal energy of the system

11.6 Specific heat capacity

Specific heat capacity of a substance is given by below formula.
s=1mΔQΔTs = \frac{1}{m } \cdot \frac{\Delta Q}{\Delta T }
The molar specific heat capacity of a substance is given by below formula.
C=1μΔQΔTC = \frac{1}{ μ } \cdot \frac{\Delta Q}{\Delta T }
where μ is the number of moles of the substance! For a solid, the law of equipartition of energy gives
C=3RC = 3 R
Calorie is the old unit of heat. 1 calorie is the amount of heat required to raise the temperature of 1 g of water from 14.5 °C to 15.5 °C and 1 cal = 4.186 J. For an ideal gas, the molar specific heat capacities at constant pressure and volume satisfy the relation
CpCv=RC_p - C_v = R
where R is the universal gas constant

11.7 Thermodynamic state variables and equation of state

Equilibrium states of a thermodynamic system are described by state variables. The value of a state variable depends only on the particular state, not on the path used to arrive at that state. Examples of state variables are pressure (P ), volume (V ), temperature (T ), and mass (m ). Heat and work are not state variables. An Equation of State (like the ideal gas equation PV = μ RT ) is a relation connecting different state variables.

11.8 Thermodynamic processes

A quasi-static process is an infinitely slow process such that the system remains in thermal and mechanical equilibrium with the surroundings throughout. In a quasi-static process, the pressure and temperature of the environment can differ from those of the system only infinitesimally. In an isothermal expansion of an ideal gas from volume V1 to V2 at temperature T the heat absorbed (Q) equals the work done (W ) by the gas, each given by
Q=W=μRT×ln(V2V1)Q = W = μ R T \times ln (\frac{V_2}{V_1})
In an adiabatic process of an ideal gas
PVγ=constantPV^γ = constant
γ=CpCvγ = \frac{C_p}{C_v}
Work done by an ideal gas in an adiabatic change of state from (P1, V1, T1) to (P2, V2, T2) is
W=μR(T1T2)γ1)W = \frac{μR(T1-T2)}{γ-1})

11.9 Second law of thermodynamics

The second law of thermodynamics disallows some processes consistent with the First Law of Thermodynamics. It states

Kelvin-Planck statement - "No process is possible whose sole result is the absorption of heat from a reservoir and complete conversion of the heat into work."

Clausius statement - "No process is possible whose sole result is the transfer of heat from a colder object to a hotter object."

The Second Law implies that no heat engine can have efficiency η equal to 1 or no refrigerator can have co-efficient of performance α equal to infinity.

11.10 Reversible and irreversible processes

A process is reversible if it can be reversed such that both the system and the surroundings return to their original states, with no other change anywhere else in the universe. Spontaneous processes of nature are irreversible. The idealised reversible process is a quasi-static process with no dissipative factors such as friction, viscosity, etc

11.11 Carnot engine

Carnot engine is a reversible engine operating between two temperatures T1 (source) and T2 (sink). The Carnot cycle consists of two isothermal processes connected by two adiabatic processes. The efficiency of a Carnot engine is given by
η=1T2T1η = 1 - \frac{T_2}{T_1}
No engine operating between two temperatures can have efficiency greater than that of the Carnot engine.
  • If Q>0 Q > 0 , heat is added to the system
  • If Q<0 Q < 0, heat is removed to the system
  • If W>0 W > 0, Work is done by the system
  • If W<0 W < 0, Work is done on the system

Kinetic Theory

12.1 Introduction

12.2 Molecular nature of matter

12.3 Behaviour of gases

12.4 Kinetic theory of an ideal gas

The ideal gas equation connecting pressure (P), volume (V) and absolute temperature (T ) is
PV=μRT=kBNTPV = μ RT = k_B NT
where μ is the number of moles and N is the number of molecules. R and kBk_B are universal constants. Real gases satisfy the ideal gas equation only approximately, more so at low pressures and high temperatures. Kinetic theory of an ideal gas gives the relation
P=13nmv2P = \frac {1}{3} nm \vec {v^2}
where n is number density of molecules, m the mass of the molecule and v2\vec {v^2} is the mean of squared speed. Combined with the ideal gas equation it yields a kinetic interpretation of temperature.
12mv2=32kBT\frac {1}{2} m \vec {v^2} = \frac {3}{2} k_B T
vrms=(v2)12=3kBTmv_{rms} = ({\vec {v^2}})^{\frac {1}{2}} = \sqrt{\frac {3 k_B T}{m}}
This tells us that the temperature of a gas is a measure of the average kinetic energy of a molecule, independent of the nature of the gas or molecule. In a mixture of gases at a fixed temperature the heavier molecule has the lower average speed.

12.5 Law of equipartition of energy

The translational kinetic energy
E=32kBNTE = \frac {3}{2} k_B N T
PV=23EPV = \frac {2}{3} E
The law of equipartition of energy states that if a system is in equilibrium at absolute temperature T, the total energy is distributed equally in different energy modes of absorption, the energy in each mode being equal to ½ kBTk_B T. Each translational and rotational degree of freedom corresponds to one energy mode of absorption and has energy ½ kBTk_B T. Each vibrational frequency has two modes of energy (kinetic and potential) with corresponding energy equal to 2 × ½ kBTk_B T = kBTk_B T. Using the law of equipartition of energy, the molar specific heats of gases can be determined and the values are in agreement with the experimental values of specific heats of several gases. The agreement can be improved by including vibrational modes of motion. The mean free path l is the average distance covered by a molecule between two successive collisions -
l=12nπd2l = \frac {1}{\sqrt 2 n \pi d^2}
where n is the number density and d the diameter of the molecule.

12.6 Specific heat capacity

12.7 Mean free path


13.1 Introduction

13.2 Periodic and oscilatory motions

All Oscillatory motions are periodic but not vice versa. e.g. To and Fro motion and pendulum are oscillatory motions as well as periodic motion.

13.3 Simple harmonic motion

Important terms
  • Amplitude A - Max displacement from mean point
  • Time Period T - Time to complete one cycle
  • Frequency f - Number of cycles per second
  • Angular Frequency - Number of radians per second ω=2πT\omega = \frac {2\pi} {T}   This can also be written as   ω=2π×Frequency\omega = {2\pi} \times {Frequency}
  • Phase - Position and direction of particle at specific point
Simple Harmonic Motion is a special case of Oscillatory Motion. Acceleration is always directed to mean position and is directly proportional to displacement from mean position.

13.4 Simple harmonic motion and uniform circular motion

Restoring force is directly proportional to the displacement.
F=kxF = - kx
where k is spring factor or force constant. Restoring force is always towards the mean position! But force is a product of mass and acceleration!
ma=kxma = - kx
α=kmx\alpha = - \frac {k} {m} x
So acceleration is also directly proportional to the displacement!
speedmax=Aωspeed_{max} = A \omega
ω=km\omega = \sqrt{ \frac {k} {m}}
potentialEnergy=kx22potentialEnergy = \frac {kx^2} {2}
kineticEnergy=kx22kineticEnergy = \frac {kx^2} {2}
v=ωA2x2v = \omega \sqrt {A^2 - x^2}
ω=2πT\omega = \frac {2 \pi} {T}
x=Asin(ωt)x = A sin(\omega t)
x=Asin(ωt+ϕ)x = A sin(\omega t + \phi)
x=x0+Asin(ωt+ϕ)x = x_0 + A sin(\omega t + \phi)
Here is a differential equation of SHM
d2xdt2+ω2x=0\frac{{d^2x}}{{dt^2}} + \omega^2 x = 0

13.5 Velocity and acceleration in simple harmonic motion

13.6 Force law for simple harmonic motion

13.7 Energy in simple harmonic motion

13.8 The Simple Pendulum

Any pendulum undergoes simple harmonic motion when the amplitude of oscillation is small.


14.1 Introduction

3 types of waves
  • Mechanical waves - need medium e.g. sound waves, waves on a string, water waves, sound waves, seismic waves
  • Electromagnetic waves - may not need medium (Medium is optional). But can travel via medium as well!
  • Matter waves - need medium - e.g. De-broglie wave in Quantum Mechanics

14.2 Transverse and longitudinal waves

In transverse waves, oscillations happen perpendicular to the direction of propogation of wave. In longitudinal waves, oscillations happen in the same direction of propogation of wave.
distance=velocity×Time distance = velocity \times Time
In any wave, distance travelled in one cycle is called as lambda. Time to complete one cycle is called as Period T.
λ=VT \lambda = V T
Number of cycles completed in 1 period is called as frequency.
frequency=Vλ frequency = \frac {V } {\lambda}
Waves appears to be in motion because particles perform simple harmonic motion by maintaining phase difference between them.
phaseDifference=2πλ×pathDifference phaseDifference = \frac { 2\pi } {\lambda} \times pathDifference
AngularWaveNumber(k)=2πλ AngularWaveNumber (k) = \frac { 2\pi } {\lambda}
AngularFrequency(ω)=2πT AngularFrequency (\omega) = \frac { 2\pi } {T}
Equation of wave going in right direction
y=A×sin[kxωt] y = A \times sin [kx - \omega t]
Here A is amplitude, k (AngularWaveNumber(k)=2πλ AngularWaveNumber (k) = \frac { 2\pi } {\lambda}) is the angular wave number, x is the distance of specific particle and t is time. So wave is a function of x and time. So given the value of distance of particle in direction (x), time(t), Amplitude (A), Wavelength (λ{\lambda})! In a mechanical wave, many particles are making SHM at the same time. That's why it depends on x and t as Every particle has different x at given time t. In SHM, where only one particle is involved, then we can define it as a function which depends only on time (Please note that x is constant in such case).

Equation of wave going in left direction
y=A×sin[kx+ωt] y = A \times sin [kx + \omega t]
Velocity of wave is given by below formula.
v=ωk v = \frac {\omega } {k}
The tension in a wave-carrying wire can be determined based on the wave velocity and the properties of the wire, specifically the linear mass density (μ) of the wire. The linear mass density represents the mass per unit length of the wire. To find the tension (T) in the wire, you can use the following formula:
T=μv2 T = μ * v^2
Particle velocity is changing but wave velocity is unchanged! Velocity of particle = - WaveVelocity * dy/dx

Maximum particle velocity is at
2πA2 \pi A
Intensity of wave is given by below formula.
I=12ρvω2A2I = \frac {1} {2} \rho v \omega^2 A^2
Hard boundry vs Soft Boundry In hard boundry, phase difference is pi and path difference is lamda/2

14.3 Displacement relation in a progressive wave

14.4 The speed of a travelling wave

14.5 The principle of superposition of waves

14.6 Reflection of waves

14.7 Beats

Electric and Magnetic field

QuantityElectric FieldMagnetic Field
Symbol EE BB
Unit volt / meter Tesla (T) or Gauss (G)
Field creator Charge (Coloumb) Pole strength (Ampere meter)
Field E=(14πϵ0)qr2E = (\frac {1} { 4 \pi \epsilon_0 }) \frac { q} { r^2}
where ϵ0\epsilon_0 is Vacuum permittivity.
B=(μ04π)mr2B = (\frac {\mu_0} { 4 \pi }) \frac { m} { r^2}
where μ0\mu_0 is The vacuum magnetic permeability, also known as the magnetic constant, is the magnetic permeability in a classical vacuum.
Force FE=(14πϵ0)q1q2r2F_E = (\frac {1} { 4 \pi \epsilon_0 }) \frac { q_1 q_2} { r^2} FB=(μ04π)m1m2r2F_B = (\frac {\mu_0} { 4 \pi }) \frac { m_1 m_2} { r^2}
Dipole Moment P=qrP = q r M=mrM = m r
Field for Dipole
( l<<rl << r )
Eaxis=(14πϵ0)2Pr3E_{axis} = (\frac {1} { 4 \pi \epsilon_0 }) \frac { 2P} { r^3}
Eequator=(14πϵ0)Pr3E_{equator} = (\frac {1} { 4 \pi \epsilon_0 }) \frac { P} { r^3}
Baxis=(μ02π)Mr3B_{axis} = (\frac {\mu_0} { 2 \pi }) \frac { M} { r^3}
Bequator=(μ04π)Mr3B_{equator} = (\frac {\mu_0} { 4 \pi }) \frac { M} { r^3}
Flux Unit of Electric flux is volt meter!
ϕE=ES=EScosθ\phi_E = E \cdot S = E S cos \theta
Unit of Magnetic flux is Weber or Tesla meter2Tesla \text{ } {meter}^2 . Magnetic flux density of one wb per meter square is 1 tesla.
ϕB=BS=BScosθ\phi_B = B \cdot S = B S cos \theta

Modern Physics


mp=1860×me m_p = 1860 \times m_e
Proton +1.6×1019+1.6 \times 10^ {-19} 1.67×10271.67 \times 10^ {-27}
Electron 1.6×1019-1.6 \times 10^ {-19} 9.11×10319.11 \times 10^ {-31}
Newtron No charge 1.67×10271.67 \times 10^ {-27}

De Broglie Wavelength

Louis de Broglie assumed that for particles (like electron) the same relations are valid as for the photon.
E=hfE = h f
c=λfc = \lambda f
E=hcλ=pcE = \frac { h c} {\lambda } = pc
where E and p are energy and momentum of photon! ff and λ\lambda are frequency and wavelength of photon! h is planck's constant. c is speed of light!

The relationship between electric potential (also known as electric potential energy) and the electron wavelength is described by the de Broglie wavelength of electrons, which is a fundamental concept in quantum mechanics.

The de Broglie wavelength (λ\lambda) of a particle, such as an electron, is given by:

λ=hp\lambda = \frac{h}{p}


  • λ\lambda is the de Broglie wavelength.
  • hh is Planck's constant.
  • pp is the momentum of the particle.

Now, let's consider the relationship between electric potential and the de Broglie wavelength of an electron. When an electron is in an electric field and is subjected to an electric potential difference (voltage), it gains kinetic energy due to the electric field. The kinetic energy KK of the electron can be related to its charge qq and the electric potential difference (VV ) through the equation:

K=qVK = q \cdot V

In quantum mechanics, the momentum of a particle is related to its kinetic energy ( K ) by the equation:

p=2mKp = \sqrt{2 \cdot m \cdot K}


  • pp is the momentum of the electron.
  • mm is the mass of the electron.

Combining these equations, we can express the de Broglie wavelength of the electron in terms of its charge, the electric potential difference, and its mass:

λ=h2mqV\lambda = \frac{h}{\sqrt{2 \cdot m \cdot q \cdot V}}

This equation relates the electron's de Broglie wavelength to the electric potential ( V ) it experiences. It shows that as the electron gains more kinetic energy (due to a higher electric potential difference), its de Broglie wavelength will change accordingly.