Physicscatalyst Magazine 1

PC –MAGAZINE -01

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Contents

• A Physics Scientist Profile and Information • Latest News and Invention in Physics • A detailed Explanation on Center of Mass • Motivational Stories • Crossword Puzzle • CBSE Sample Paper • Institute Information • Motivational Quotes • Challenging questions in Mechanics

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A Physics Scientist Profile and Information

James Prescott Joule

James Prescott Joule was born on 24 December 1818 in Salford, Lancashire.He was the son of Benjamin Joule a wealthy brewer, and Alice Prescott Joule. He received his early education at home and later on he was sent with his elder brother Benjamin, to study with John Dalton at the Manchester Literary and Philosophical Society. After Dalton was struck with an illness Joule inherited a lot of his work. A room his house was set back just for his laboratory investigations, and experiments

Joule was not just a man of work even though in the perspective of most he could be seen as one. A lot of what shows that he was not just working all the time was that in 1847 he was married to Amelia Grimes. Although he did spend most of his honeymoon in the Alps studding the nature of a waterfall. Joule observed that the water at the temperature at the peak of the fall was cooler than the water at the bottom. This led him to believe that the water was heated from the movement of the water down the ledge. This was significant to his hypothesis of heat conservation. Later he also had children, a son Benjamin Author and a Daughter Alice Amelia. In 1854 Joule was struck with a tragedy when his son and his wife passed away, so Joule stayed a widower for the rest of his time.

Joule studied the nature of heat, and discovered its relationship to mechanical work . His first major research was concerned with determining the quantity of heat produced by an electric current and, in 1840, Joule discovered a simple law connecting the current and resistance with the heat generated.

In 1848 Joule published a paper on the kinetic theory of gases, in which he estimated the speed of gas molecules. From 1852 he worked with William Thomson (later Lord Kelvin) on experiments on thermodynamics. Their best known result is the Joule–Kelvin effect – the effect in which an expanding gas, under certain conditions, is cooled by the expansion. The SI derived unit of energy, the joule, is named after him.

Joule discovered the first law of thermodynamics, which is a form of the law of conservation of energy. He showed that work can be converted as heat with a ratio of one to the other, and work can be made into heat. His works on the conservation of heat is put into the first part of the first law of thermodynamics. This law says that energy can neither be created nor destroyed, but only transformed.

He Died on October 11, 1889 (aged 70) at Sale, Cheshire, England http://physicscatalyst.com

Latest News and Invention in Physics

A faster way of boarding planes

A Fermilab astrophysicist has used his analytical skills to come up with a more efficient way to board planes. Most people are familiar with the usual holdups when boarding: waiting for the person ahead to stow his or her luggage, and having to dislodge seated passengers to get to a center or window seat. Jason Steffen’s method minimizes the former and eliminates the latter. He proposes boarding passengers in all the window seats first, then center, then aisle, starting at the back of the plane and moving forward, and boarding by alternate rows, so passengers are spaced far enough apart to stow their luggage at the same time. In tests conducted by Steffen and Hollywood producer Jon Hotchkiss, Steffen’s method took half the time of the current block boarding method used by most airlines, in which passengers are assigned to groups or zones within the cabin. Now all he has to do is persuade the airlines to adopt it. Steffen has published his findings in the Journal of Air Transport Management

LEDs could multitask as data transmitters

Light bulbs could soon be used to broadcast wireless internet. Harald Haas of the UK’s Edinburgh University has been working on a revolutionary method of data transmission that makes use of light waves rather than wires or radio waves. Using LEDs, which are more efficient than standard light bulbs and can be switched on and off very quickly, he has found that he can vary the intensity of their output and pick up the signals with a simple receiver. With data rates of 100 megabits per second, Haas’s system relies on the fact that the human eye cannot detect the rapid flickering on and off of the LEDs— instead they appear to maintain a normal steady glow. Besides faster transmission capabilities, such a device would also have applications in the oil and gas industries, where radio waves can cause sparks, and for underwater robotic vehicles and submarines, where the electrically conductive salt water stifles radio waves

Curiosity landed on Mars on 6 August

For the first time since the Viking missions of the 1970s, the US has send a rover— Curiosity—to Mars for the primary purpose of finding the building blocks of life. The Viking landers returned a picture of a cold, dead Mars that could at no time have supported organic material. However, recent data from Mars-orbiting observatories and from the two geology-focused rovers, Spirit and Opportunity, have provided significant evidence for the existence of liquid water at various points of Martian history. Curiosity, which has landed in a crater that may contain extensive mineral deposits, will be able to http://physicscatalyst.com

5 drill deeper than previous rovers and perform a wide range of analyses on gathered samples. Because it was three times heavier than earlier rovers, a new, automatic landing system had been developed to deploy the robot. The landing, which used the world’s largest supersonic parachute and a sky crane, was referred to as “seven minutes of terror” because of the lack of direct control by NASA scientists over the process. However, soon after touching down, the $2.6 billion rover was able to send two pictures, one showing one of its wheels on the ground and one showing its shadow. The successful landing elicited cheers from the NASA employees in the control room at the Jet Propulsion Laboratory. For the next two years, Curiosity will explore the Martian surface, collect information about the planet’s geologic history, and look for possible signs of life, past or present.

India set for Mars mission in 2013

The Indian government has approved a mission to Mars in what would be the country's first visit to the red planet. The news comes just four years after India launched its maiden mission to the Moon – Chandrayaan-1 – and days after NASA landed the Curiosity car-sized rover on Mars.

The £70m mission will be launched in November 2013 from India's spaceport at the Satish Dhawan Space Centre on the island of Sriharikota using the Polar Satellite Launch Vehicle. The mission, which will orbit Mars and study the planet's geology and climate, has already been allocated £26m in the country's science budget.

Details of the new mission remain scarce as the Indian Space Research Organisation (ISRO) remains tight-lipped about the probe. However, it is expected that more information will be released on 15 August, when India's prime minister delivers the country's Independence Day speech.

The Mars orbiter is expected to weigh 500 kg, with a scientific payload of around 25 kg, and be placed in a highly elliptical orbit around the planet. "Not knowing the payload specifics, it is hard to judge [the mission]," says Jeffrey Plescia, a Mars researcher at Johns Hopkins University in the US. "But the Chandrayaan mission had a diverse set of instruments that provided a great deal of unique data. If they do something similar [to the Mars mission], then it could be a substantial contribution to science."

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CENTER OF MASS

• Consider a body consisting of large number of particles whose mass is equal to the total mass of all the particles. When such a body undergoes a translational motion the displacement is produced in each and every particle of the body with respect to their original position. • If this body is executing motion under the effect of some external forces acting on it then it has been found that there is a point in the system , where if whole mass of the system is supposed to be concentrated and the nature the motion executed by the system remains unaltered when force acting on the system is directly applied to this point. Such a point of the system is called centre of mass of the system. • Hence for any system Centre of mass is the point where whole mass of the system can be supposed to be concentrated and motion of the system can be defined in terms of the centre of mass. • Consider a stationary frame of reference where a body of mass M is situated. This body is made up of n number of particles. Let m i be the mass and ri be the position vector of i'th particle of the body. • Let C be any point in the body whose position vector with respect to origin O of the frame of reference is Rc and position vector of point C w.r.t. i'th particle is rci as shown below in the figure.

• From triangle OCP ri=Rc+rci (1) multiplying both sides of equation 1 bt m i we get miri=m iRc+m irci taking summation of above equation for n particles we get http://physicscatalyst.com

If for a body

then point C is known as the centre of mass of the body.

• Hence a point in a body w.r.t. which the sum of the product of mass of the particle and their position vector is equal to zero is equal to zero is known as centre of mass of the body.

Position of centre of mass (i) Two particle system

• Consider a system made up of two particles whose mass are m 1 and m 2 and their respective position vectors w.r.t. origin O be r1 and r2 and Rcm be the position vector of centre of mass of the system as shown below in the figure. So from equation

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• If M=m 1+m 2=total mass of the system , then

(ii) Many particle system

• Consider a many particle system made up of number of particles as shown below in the figure. Let m 1 , m 2 , m 3 , ...... , m n be the masses of the particles of system and their respective position vectors w.r.t. origin are r1 , r2 , r3 , ...... , rn.

• Also position vector of centre of mass of the system w.r.t. origin of the refance frame be Rcm then from equation 2

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• Because of the definition of centre of mass

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Position vector of centre of mass in terms of co-ordinate components

• Let in a system of many particles the co-ordinates of centre of mass of the system be (X cm ,Y cm ,Z cm ) then position vector of centre of mass would be Rcm =X cm i+Y cm j+Z cm k (9) and position vectors of various particles would be • Putting the values from equation 9 and 10 in equation 6, 7 and 8 we get

• If in any system there are infinite particles of point mass and are distributed continuously also if the distance between them is infinitesimally small then summation in equations 6, 11, 12 and 13 can be replaced by integration. If r is the position vector of very small particle of mass dm of the system then position vector of centre of the system would be

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and the value of its co-ordinates would be

• If ρ is the density of the system then dm=ρdV where dV is the very small volume element of the system then,

• The centre of mass of a homogeneous body (body having uniform distribution of mass) must coincide with the geometrical centre of the body. In other words we can say that if the homogeneous body has a point , a line or plane of symmetry , then its centre of mass must lie at this point , line or plane of symmetry. • The centre of mass of irregular bodies and shape can be found using equations 14, 15, 16.

Center of Mass of a System of Objects

We now know how to find the center of mass of a collection of point particles as well as that of a continuous solid object. What happens when we want to find the Center of mass of a collection of solid objects By definition, the center of mass of the system is found by integrating the position vector over all of the mass in the system. Since the system is made of two objects, the total integral is just the sum of two separate integrals, one for each object. Let us http://physicscatalyst.com

12 assume a system which contains two solid objects X and Y of any shape,. The center of mass would be given by

1   R =  rdm + rdm CM ∫ ∫  M total  a b 

If we multiply and divide each of these integrals by the mass of the object, we certainly haven't changed anything, but we can now see that the numerator just becomes the mass of each object times the position of its center of mass.

1  rdm rdm  1 R =  M + M  = M R + M R cm  a ∫ b ∫  ()a cm,a b cm,b M total  a M a b M b  M total

Therefore, we have just arrived at a simple procedure for finding the center of mass of a System of solid objects. Namely, we just treat each object as a point particle with all of It’s mass located at its center of mass! That’s all there is to it!

Example-1 Discrete Distribution

Consider the following masses and their coordinates which make up a "discrete mass" rigid body. m1=5 kg r1=3 i-2k m2=10 kg r1=-3i+2 j-2k m3=5 kg r1=i+2 k Find the center of mass of this mass system

Solution

So,

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m1 x1 + m2 x2 + m3 x3 5X 3 +10X (− )3 + 5X1 xcm = = = − 5. m1 + m2 + m3 5 +10 + 5

m1 y1 + m2 y2 + m3 y3 5X 0 +10X 2 + 5X 0 ycm = = = 1 m1 + m2 + m3 5 +10 + 5

m1 z1 + m2 z2 + m3 z3 5X (− )2 +10X )2( + 5X 2 zcm = = = 1 m1 + m2 + m3 5 +10 + 5

Rcm =-.5 i+j+k

Example -2 Continuous distribution

A straight rod of length L has one of its ends at origin and other at (L, 0). If the mass per unit length of rod is Ax + B .Find the centre of mass (a) L(2AL + 3B)/(3AL + 6B) (b) L(AL + 3B)/(AL + 6B) (c) L(AL - B)/(AL + 2B) (d) none of the above Ans. (a)

∫ xdm ∫ x(Ax + B)dx xcm = = ∫ dm ∫ (Ax + B)dx

L 2( AL + 3B) x = cm 3( AL + 6B)

Dynamics of Center of Mass

Velocity of Center of Mass

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p is the vector sum of linear momentum of various particles of the system or it is the total linear momentum of the system.

• If no external force is acting on the system then its linear momentum remains constant. Hence in absence of external force

• In the absence of external force velocity of centre of mass of the system remains constant or we can say that centre of mass moves with the constant velocity in absence of external force. • Hence from equation 18 we came to know that the total linear momentum of the system is equal to the product of the total mass of the system and the velocity of the centre of mass of the system which remains constant. • Thus in the absence of the external force it is not necessary that momentum of individual particles of system like p1 , p2 , p3 ...... pn etc. remains constant but their vector sum always remains constant.

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Acceleration of centre of mass

• Differentiating equation 7 we get

• If m iai = Fi which is the force acting on the i'th particle of the system then

• Net force acting on the i'th is

Fi=Fi(ext) +Fi(int)

• Here internal force is produced due to the mutual interaction between the particles of the system. Therefore, from Newton's third law of motion

is the total external force acting on the system since internal force on the system

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because of mutual interaction between the particles of the system become equal to zero because of the action reaction law. • Hence from equation 23 it is clear that the centre of mass of the system of particles moves an if the whole mass of the system were concentrated at it. This result holds whether the system is a rigid body with particles in fixed position or system of particles with internal motions.

Center Of Mass equation

Integrating the equation of the motion of center of mass, we get Center of Mass equation which is defined as

 1 2   M totalVcm  = Fexternal .dlcm  2  ∫

Left hand size denotes the change in kinetic energy of the Center of the mass . It is calculated as if all the mass is concentrated at the center of mass.

Right hand side calculates the work done by the external forces on the center of mass. It is calculated as if all the forces are acting on the center of Mass

Example -3

Two objects, one having twice the mass of the other, are initially at rest. Two forces, one twice as big as the other, act on the objects in opposite directions as shown below

Find the acceleration of the center of mass of the system Solution

Now acceleration of the center of mass of the system is given by

Fext,net a cm = M total

Taking right direction as positive,

Fext =2F-F

Mtotal =2M+M

So http://physicscatalyst.com

F a = cm 3M

Direction is towards right

Example 4

An astronaut has just finished fixing a space telescope using a big instrument whose mass is one tenth as big as his mass. You realize you have no way to get back to your spaceship which is 10 meters away from you, so you throw the instrument as hard as you can in a direction away from the spaceship which causes you to move in the opposite direction, toward the spaceship. When you finally reach the space ship, how far away are you from the instrument?

Solution

Key concept : The key concept needed to answer this question is that the acceleration of the center of mass of a system will be zero if the external force on the system is zero.

In this question, we define the system to be astronaut and the instrument, and the center of mass of the system is initially at rest a distance of 10 meters from your spaceship. Since there are no external forces acting on the system and the center of mass is initially at rest, the location of the center of mass of the system can never change! If we choose the initial location of the center of mass to be at x = 0, the center of mass will always be at x = 0.

The location of the center of mass of the system is determined from its definition

(M ast + M inst )xcm = M ast xast + M inst xinst

Now x cm =0

M ast xast = −M inst xinst

Since the mass of the instrument is 1/10 of mass of the astronaut, the instrument will always be ten times as far away from the center of mass as you are, and it will always be on the opposite side of the center of mass from you,

So for 10 m move by astronaut, the instrument will move by 100m. Final distance of the instrument from the spaceship will be 110 m http://physicscatalyst.com

Example 5

A mass of man m standing on a block of mass M. The system is at rest. The man moves relative in x direction to the block with velocity v' and then stops. Find the velocity of centre of mass Solution: As no external force, there will be no change in center of mass velocity. Initially center of mass is at rest and it will remain in rest

Example 6

Find the displacement of block relative to ground if the displacement of the man with respect to block is x i Solution

Rcm = constant

Rcm = 0

∑ mi ri = 0 ---(1)

− mxi = M + m Some Other Important things about Center Of mass

A) A concept similar to center of mass is Center Of gravity (CG) . A CG is a point in the body where the total force of gravity can be considered to act. The force of gravity acts on all the parts of the body but for purpose of determining the translational motion of the body as a whole, we consider the sum of the gravity of the all the parts acting on the CG..

When we are talking about a uniform gravitational field then both the center of mass as well as the center of gravity will be the same point. But if we consider a non-uniform field it is not always true. An object can rotate due to the torque produced by a non- uniform gravitational field.

B) For symmetrical shaped objects like sphere, uniform cylinder, rectangular solid, the CM lies at the geometrical center of the objects. CM of the object can be outside the objects also in some cases, example donut.

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C) The technique of a high jumper is another important use of center of mass. A high jumper bends his body in a certain way so that the center of mass does not clear the bar, but the body clears it.

Motivational Stories

IT IS THE LITTLE THINGS THAT MAKE A BIG DIFFERENCE

There was a man taking a morning walk at or the beach. He saw that along with the morning tide came hundreds of starfish and when the tide receded, they were left behind and with the morning sun rays, and they would die. The tide was fresh and the starfish were alive. The man took a few steps, picked one and threw it into the water. He did that repeatedly. Right behind him there was another person who couldn't understand what this man was doing. He caught up with him and asked, "What are you doing? There are hundreds of starfish. How many can you help? What difference does it make?" This man did not reply, took two more steps, picked up another one, threw it into the water, and said, "It makes a difference to this one."

Focus and Concentrate

An ancient Indian sage was teaching his disciples the art of archery. He put a wooden bird as the target and asked them to aim at the eye of the bird. The first disciple was asked to describe what he saw. He said, "I see the trees, the branches, the leaves, the sky, the bird and its eye.." The sage asked this disciple to wait. Then he asked the second disciple the same question and he replied, "I only see the eye of the bird." The sage said, "Very good, then shoot." The arrow went straight and hit the eye of the bird.

What is the moral of the story?

Unless we focus, we cannot achieve our goal. It is hard to focus and concentrate, but it is a skill that can be learned.

There may be days when you get up in the morning and things aren’t the way you had hoped they would be.

That’s when you have to tell yourself that things will get better. There are times when people disappoint you and let you down.

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But those are the times when you must remind yourself to trust your own judgments and opinions, to keep your life focused on believing in yourself.

There will be challenges to face and changes to make in your life, and it is up to you to accept them.

Constantly keep yourself headed in the right direction for you. It may not be easy at times, but in those times of struggle you will find a stronger sense of who you are.

So when the days come that are filled with frustration and unexpected responsibilities, remember to believe in yourself and all you want your life to be.

Because the challenges and changes will only help you to find the goals that you know are meant to come true for you.

Keep Believing in Yourself! Crossword Puzzle

1 2 3

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ACROSS DOWN

4. Second law of Newton's 1. Quantity to measure the hotness of 5. The unit of charge the object 6. This measure how fast are we 2. Rocket propulsion is based on this moving 3. Pressure is inversely proportional to 7. Quantity use to measure energy Volume at constant temperature 9. It revolves around nucleus 5. It is used in fan 8. This person invented electric bulb

CBSE XII Sample Paper

General Instruction:

1. Answer all questions 2. Internal choices are provided for some questions 3. Question numbers 1 to 8 are very short answer questions and carry 1 mark each. 4. Question numbers 8 to 18 are short answer questions and carry 2 marks each.

5. Question numbers 19 to 27 are also short answer questions and carry 3 marks each. 6. Question numbers 28 to 30 are long answer questions and carry 5 marks each.

7. Use log tables if necessary.

Very Short Answer type questions

Question 1

What are the Energy losses in Transformer

Question 2

Which has a greater resistance for same voltage 220 V

a) 1 KW electric Heater b) 100W filament bulb

Question 3

Find the equivalent resistance in the circuit http://physicscatalyst.com

Question 4

Find the voltage across 470ohm resistor in previous question

Question 5

What is the average value of AC during a half cycle and full cycle?

Question 6

The maximum KE of the electrons emitted in a photocell is 10eV. What is the stopping potential?

Question 7

What do you understand by the transformation ratio of the transformer?

Question 8

If M is the mass of the nucleus and A its atomic mass, the what is the packing fraction

Short Answer type questions

Question 9

An electron of charge e moves in a circular orbit of radius R around the nucleus. The magnetic field due to the orbital motion of the electron at the site of the nucleus is B 0.

Find out the angular velocity of the electron in terms of B 0 ,e and R

Question 10

The half life of radioactive sample is 30 sec.

Find out following a)the decay constant b) Time taken for the sample to decay to ¾ of its initial value http://physicscatalyst.com

Question 11

A cylindrical wire is stretched to increase its length by 10%. Calculate the % increase in the resistance of this wire

Question 12

State Faraday’s law of electromagnetic induction. Express it mathematically.

Question 13

The binding energies of the nuclei A n and B 2n are respectively X and Y joules per nucleon, if 2X -Y < 0 ,then find the energy released in the nuclear reaction

An + An − > B 2n

Question 14

The output voltage of an ideal transformer connected to 240 ac mains is 24 V. When the transformer is used to light a bulb with rating 40V, 40W, calculate a) The current in the bulb b) Current in the primary coil of the circuit.

Question 15

What is Lorentz Force? If the charged particles moves parallel to magnetic field, what will be the force on it?

Question 16

Two straight and parallel wires X and Y are present. .The current in X is I 0 a)when the wire Y is brought close to wire X,what will be the direction of the induced current in Y b) when the wire Y is being taken away from X,what will be the direction of induced current in Y

Question 17

What is photoelectric cells? Explain the working of Photovoltaic cell http://physicscatalyst.com

Question 18

In a common –emitter amplifier the load resistance of the output circuit is 1000 times the load resistance of the input circuit . If α =.98 ,the calculate the voltage gain

Question 19

Heavy water is a suitable moderator in Nuclear reactor why?

Question 20

What is an antenna and how it works? Name the types of antenna

Question 21

A circular coil of N turns and radius R 0 is kept normal to a magnetic field given by

B = B0 cosωt

Deduce the expression for the EMF induced in this coil. State the rule which help us determine the direction of induced current

Question 22

Explain the meaning of photo electric work function by giving necessary equation?

Question 23

Write down the properties of Magnetic lines of forces ? What is the unit of Magnetic Field

Question 24

Two concentric coplanar circular loops made of wire with resistance per unit length 10 -4 Ω/m have diameters .2 m and 2 m respectively.

A time varying potential difference V=(4+2.5 t) V is applied to larger loop.

Find the induced current in the small loop

Question 25

An alternating voltage given by

V (t) = 10 2 sin(2500t + 450 ) http://physicscatalyst.com

It is applied across a series combination of Resistance R=3K ohm and Capacitor of capacitance C=.1μF

Find out following things from the above given values i)the peak value and rms value of the current in the circuit ii)the phase difference between current and voltage iii)the power factor of the circuit

Question 26

There are three charges on a straight line

One Positive Charge q

Two Negative Charges -Q

Find the value of q/Q so that the entire system is in equilibrium?

Will this equilibrium be stable

Question 27

What are coherent sources ? How does the width of interference fringes in Young’s double slit experiment change when a)the distance between the slits and screen is increased b)Frequency of the source is decreased

Question 28

Device the formula for the electric field due to a circular loop of positive charge i.e., charged ring at an axial point

Question 29

What is a transistor ? Explain the construction and working of p-n-p transistor with the help of neat diagram?

Question 30 i)States Bohr’s postulates for hydrogen atom. Explain the origin of Balmer series of spectral lines http://physicscatalyst.com

26 ii) The first line of Balmer series of a hydrogen like atom has a wavelength of 1640 Å. Find the ionization energy of the atom

Given

h=6.6X10 -34 Js and c=3.0X10 8 m/s

Solutions

Solution 1

1) copper losses 2) Eddy current losses 3) Hysteresis losses 4) Flux losses

Solution 2

Resistance of Heater

V 2 220X 220 R = = = 48 4. ohm 1 P 1000

Resistance of Bulb

V 2 220X 220 R = = = 484ohm 2 P 100

So it is clear bulb has higher resistance

Solution 3

The equivalent resistance is found by combining the 820 and 680 resistors in parallel, and

Then adding the 470 resistor in series with that parallel combination.

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−1  1 1  Req =  +  + 470 = 842ohm  820 680 

Solution 4

The current delivered by the battery will be given by

V 12 I = = = .1 425X10−2 A Req 842

The voltage across the 470 ohm resistor can be found using ohm’s law

V470 =IR=6.7 V

Solution 5

2i Half cycle = 0 π

Full cycle =zero

Solution 6

1 mv 2 = eV 2 0

So V 0=10 V

Solution 9

An electron moving in a circular orbit is equivalent to a current carrying loop. Therefore current in that loop will be given by

I=e/T

Where T is the time period of the motion of electron around the nucleus

Now

2πr T = v

Therefore current would be http://physicscatalyst.com

ev eω I = = 2πr 2π

Now we know that Magnetic field at the center of a current carrying loop is given by

0 I 0eω B0 = = 2R0 4πR0

4πR B ω = 0 0 0e

Solution 8

In n-type of semi-conductor, the conduction is due to electrons while p-type conductor has holes for conduction. The mobility of electron are higher for electrons then holes so that whys n-type are more conductive then p –type semiconductor

Solution 10 a) Disintegration constant

.693 λ = T 2/1

λ=.0231 s -1 b) By definition of half life

½ of the initial mass remains undistintegrated in 30 sec

¼ of the initial mass remains undistintegrated in next 30 sec

Therefore 3/4 th of the initial mass disintegrate in 60sec

Solution 11

Let us assume L , A and R is the initial length , cross-sectional area and resistance

And Let L 1 ,A 1 and R 1 be the final length , cross-sectional area and resistance

Now since length is increased by 10% http://physicscatalyst.com

10 L = L + LX = 1.1 L 1 100

Since Volume remains same

AL = A1L1 A A = 1 1.1

Let a be the specific resistance of the material of the wire, then resistance of the wire before and after will be given by

L R = a A

L1 L R1 = a = .1 21a A1 A

So percentage increase in resistance is

R − R  R  1 X100 =  1 −1X100 = 21% R  R 

Solution 13

Energy released =2nY-nX-nX=2n(Y-X)

Solution 14

The Current in the bulb or secondary is

Ps 40 is = = = 1A VS 40

Now for the transformer ,we know that

Vs I s = VP I P

Or I p=.1 A

Solution 16 a)the induced current will be opposite to current in X b) The Induced current will be in same direction as X http://physicscatalyst.com

Solution 18

α .98 β = = = 49 1− α 1− .98

Therefore Voltage gain is given by

R2 3 Vgain = β = 49X1000 = 49X10 R1

Solution 20

Suppose the initial quantity of radium is N 0 .Then the quantity left after n half lives will be

n  1  = 0    2 

Now here N=25% of N 0=N 0/4

n 0  1  = 0   4  2 

Or n=2

Therefore time of disintegration =half-life X number of half lives

=1600X2=3200 years

Solution 24

The resistance of the Larger loop

= 2πR *10−4 = 2π *1*10−4 = 2π10−4

Current in the larger loop will be

V 4( + 5.2 t) I = = R 2π10 −4

The magnetic Field due to this current in the larger loop at the common center will be I 4( + 5.2 t) 4( + 5.2 t) B = 0 = 0 = 0 2R 2*1* 2π10−4 4π10−4 http://physicscatalyst.com

Since the smaller loop is very small compared to bigger loop, the magnetic field at the center can be assumed to be the magnetic field in the smaller loop.

So flux through smaller loop

4( + 5.2 t) 4( + 5.2 t) φ = BA = 0 (π10− )2 = 0 102 4π10−4 4

Induced EMF in the smaller loop

dφ E = = 62 5. dt 0

Now resistance of the smaller loop

= 2πR *10−4 = 2π 1.* *10−4 = 2π10−5

So induced current

E 62 5. I = = 0 = .1 25A R 2π10−5

Solution 25

The voltage is given by

V (t) = 10 2 sin(2500t + 450 )

Comparing it with the standard equation

V = V0 sin(ωt +φ)

We get

V = 10 2Volt 0 ω = 2500rad / se

Now the impedance of the circuit is given by

2  1  Z = R 2 +    ωC 

Substituting the above values

Z=5000Ω http://physicscatalyst.com

The rms Value of the current in the circuit is given by

Vrms V0 −3 irms = = = 2X10 A Z Z 2

−3 i0 = irms 2 = 2 2X10 A

The phase difference between the current and voltage is

 /1 ωC  φ = tan −1   = tan −1 )3/4(  R 

The power factor of the circuit is

R cosφ = = 6. Z

Solution 26

For the charge q to be in equilibrium, the charges –Q should be at equal distance from it in opposite direction. For equilibrium of Charge Q, the sum of forces acting on it should also be zero. Let assume r be the distance between the charges

Q 2 Qq - =0 4r 2 r 2

Hence q=Q/4

It does not depend on the distance r.

The equilibrium position is not stable. Since when charge –Q is shifted along left by a distance x,The force of attraction from positive charge

Q 2 F1= (4 a + x) 2

The force of repulsion from negative charge

Q 2 F2= 2( a + x) 2

It is clear that F 2 > F 1 http://physicscatalyst.com

So the charge Q will move still farther from the position of equilibrium

Similarly if the charge –Q is move toward right.The force of attraction will be more then Force of repulsion and it will move toward the center.

Now if the charge q is moved right or left ,the force of attraction on one side will more than the force of attraction of other side, so it will return to equilibrium position.

Solution 30

For balmer series of a hydrogen-like atom ,we have

1  1 1  = R − , n = 5,4,3 ..... λ  2 2 n 2 

For first line,n=3

1  1 1  5R = r −  = λ  2 2 32  36

36 R = 5λ

Ionization energy is given by

=Rhc

Substituting the values ,we get

E=54.4eV

http://physicscatalyst.com

Motivational Quotes

Strength does not come from physical capacity. It comes from an indomitable will."

Mahatma Gandhi

"What this power is I cannot say; all I know is that it exists and it becomes available only when a man is in that state of mind in which he knows exactly what he wants and is fully determined not to quit until he finds it. " Alexander Graham Bell

"First we form habits, then they form us. Conquer your bad habits or they will conquer you."

Rob Gilbert

"It is better to conquer yourself than to win a thousand battles. Then the victory is yours. It cannot be taken from you, not by angels or by demons, heaven or hell." Buddha

Would you like me to give you a formula for success? It's quite simple, really. Double your rate of failure. You are thinking of failure as the enemy of success. But it isn't at all. You can be discouraged by failure or you can learn from it, so go ahead and make mistakes. Make all you can. Because remember that’s where you will find success."

- Thomas J. Watson

Thomas Edison tried two thousand different materials in search of a filament for the light bulb. When none worked satisfactorily, his assistant complained, “All our work is in vain. We have learned nothing.”

Edison replied very confidently, “Oh, we have come a long way and we have learned a lot. We know that there are two thousand elements which we cannot use to make a good light bulb.” http://physicscatalyst.com

Institute Information

Indian Institute of Technology, Delhi

The Indian Institute of Technology Delhi (abbreviated IIT Delhi or IITD) is a public engineering institution located in Delhi, India. It is one of the IITs along with other Indian Institutes of Technology institutions in India. The IITS are institutions of national importance established through Act of Parliament for fostering excellence in education. There are fifteen IITs at present, located in Bhubaneswar, Chennai, Delhi, Gandhinagar, Guwahati, Hyderabad, Indore, Jodhpur, Kanpur, Kharagpur, Mandi, Mumbai, Patna, Ropar and Roorkee. Over the years, IITs have created world class educational platforms dynamically sustained through internationally recognized research based on excellent infrastructural facilities. The faculty and alumni of IITs continue to make significant impact in all sectors of society, both in India .and abroad

LOCATION

IIT Delhi is located in Hauz Khas, South Delhi. The campus of 320 acres is surrounded by the beautiful Hauz Khas area and monuments such as the Qutub Minar and Lotus Temple. The campus is also close to other educational institutions such as the Jawaharlal Nehru University, All India Institute of Medical Sciences, National Institute of Fashion Technology, National Council of Educational Research and Training (NCERT) and Indian Statistical Institute.

Multi-Storey Building facing the front lawns the inside of the campus resembles a city, with gardens, lawns, residential complexes and wide roads. The campus has its own water supply and backup electricity supply along with shopping complexes to cater to the daily needs of residents.

FACLITIES

INFRASTRUCTURE

IIT, Delhi has well equipped with all the modern facilities. The classrooms are well furnished. The IIT Library System consists of a Central Library and 18 departmental libraries which collectively support the teaching, research and extension programmes of the Institute.

http://physicscatalyst.com

HOSTELS

There are 13 hostels (eleven for boys and two for girls). There are also apartments for married students. All the hostels are named after mountain ranges in India - Karakoram, Aravali, Himadri, Jwalamukhi, Kailash, Kumaon, Nilgiri, Satpura, Girnar, Shivalik, Udaygiri, Vindhyachal and Zanskar, of which Himadri and Kailash are for girls. The residential apartments are named after ancient Indian universities - Takshila and Nalanda. . Each Hostel has its distinct culture of sports and cultural activities.

ACTIVITES

The Student Activity Center or SAC is a part of the Student Recreation Zone in IIT Delhi. The SAC is for the extracurricular activities of the students. The SAC consists of a gymnasium, swimming pool, pool and billiards rooms, squash courts, table tennis rooms, a badminton court, a music room, a fine arts room, a robotics room and a committee room used to organize quizzing and debating events. The SAC also has an Open Air Theater where concerts are hosted

PROGRAMS

Undergraduate programs

IIT Delhi offers Bachelor of Technology in various areas:

Chemical Engineering

Civil Engineering

Computer Science and Engineering

Electrical Engineering

Electrical Engineering (Power)

Engineering Physics

Mechanical Engineering

Production and Industrial Engineering

Textile Technology.

http://physicscatalyst.com

Admission Procedure

Each year IIT-JEE is conducted by any one of the old IITs in a round robin fashion

The age limit for appearing in IIT-JEE is 25 years

For candidates belonging to SC, ST and PD categories, the relaxed age limit is 30 years

A candidate can take the JEE two times at the most

Students who are selected for admission to an IIT cannot attempt the examination again in the future

The exam is set by the JEE Committee (consisting of a group of faculty members drawn from the admitting colleges) under the tightest security

As of 2009,around 8295 seats are available in various IIT

It is one of the toughest engineering entrance exams in the world with a success rate of around 1 in 45

Candidates who qualify in the IIT-JEE can apply for admission to the BArch (Bachelor of Architecture), BDes (Bachelor of Design), BTech (Bachelor of Technology), Dual Degree (Integrated Bachelor of Technology and Master of Technology) and Integrated MSc (Master of Sciences) courses in the various institutes.

The IITs were created to train scientists and engineers, with the aim of developing a skilled workforce to support the economic and social development of India after independence in 1947.

Postgraduate programs

IIT Delhi offers many postgraduate programs (M. Tech., M.S.(R), M. Des., MBA under various departments and centers. Each department or centre may provide one or many courses. e.g. the department of civil engineering in IIT Delhi provides M. Tech. courses in environmental engineering & management, structural engineering, transportation engineering, construction engineering and management, rock mechanics etc. where as the department of biotechnology provides only one postgraduate course i.e. M.S. (R) in biotechnology.

http://physicscatalyst.com

Admission Procedure

The admission to these programs is carried out mainly based on Graduate Aptitude Test in Engineering (GATE). Students securing very good scores in this exam are called for personal interview. Out of tens of lakhs (more than 1200000 students appeared in GATE 2012) of students appearing GATE every year only the topmost end people are shortlisted for the interview by IIT Delhi.

PLACEMENTS

IIT, Delhi offers full placements to all the students in every stream. Some of the recruitment companies are

CISCO

Amazon

Havells

Intel Technogy

IBM Corporation

IOCL

GE India

Cummins

Hindustan Unilever Ltd.

KPMC

Monitor Group

Accenture

Essex

NOTABLE ALUMINI

Manvinder Singh Banga, Former Chairman, Unilever

Yogesh Chander Deveshwar, Chairman, ITC

Chetan Bhagat, Indian novelist http://physicscatalyst.com

Vinod Khosla, one of the co-founders of Sun Microsystems

Raghuram Rajan, Professor of Finance, Chicago Booth

Padmasree Warrior, Chief Technology Officer of Cisco Systems

Rohit Pande CEO & Co-founder of Classteacher Learning Systems

Rajat Gupta, first Indian-born CEO of a global corporation, served as managing director of McKinsey & Company

CHALLENGING PROBLEMS IN MECHANICS

Question -1

A plane whose Mass is 2400 Kg makes an emergency landing on a short runway. With its engine off ,it lands on the runway at 120 m/s. A hook on the plane snages a cable attached to a 300 Kg sandbag and drags the sandbag along. If the coefficient of friction between the sandbag and the runway is .5 and if the plane brakes gives an additional retarding force of 300N,how far does the plane go before comes to a stop

Question -2

A body of Mass M A collides elastically with a stationary body of mass M B. After the collision ,the bodies move with angle θ 1 and θ 2 with the original direction of Mass M A.

Prove the following things by considering events in center of mass frame a) If M A =M B ,then θ 1+ θ2=π/2 b) If M A > M B ,then maximum value of θ 1 is given by

M B sinθ1 = M A

Question-3

A particle of mass M is acted upon by a force that has an initial magnitude F 0 When t=0 and decrease at a uniform rate until when t=t 1,its magnitude is zero. It is moving across the X-axis. Find the velocity and coordinate of the particle when t=t 2 assuming that t 2 > t1. Assume x t=0 =0 and (dx/dt) t=0 =0

http://physicscatalyst.com

Answer

Velocity=F 0t1/2M

Displacement=(F 0t1/2M)(t 2-t1/3)

Question -4

A wheel of radius r rolls without slip along the x-axis. With constant speed. v 0.Find out the motion(velocity and acceleration ) of the point A on the rim of the wheel which starts from the origin O.Take Y axis as perpendicular at X axis at origin

Answer dx/dt=v 0[1-cos(v 0t/r)] dy/dt=v 0sin(v 0t/r)

2 2 2 d x/dt =(v 0 /r)sin(v 0t/r) 2 2 2 d y/dt =(v 0 /r)cos(v 0t/r)

Question-5

A body moves under the action of a constant force F through fluid that opposes the motion with a force proportional to the square of the velocity that is Ax 2.Show that the 1/2 limiting velocity is V L=(F/A) .

Question -6

A Bungee Jumper is attached to one end of a long elastic rope. The other end of the elastic rope is fixed to a high bridge. The Jumper steps off the bridge and falls from rest towards the river below. He does not hit the river below. The mass of the jumper is M and length of unstretched rope is L.Force constant of the rope is K. and gravitational field strength is g.Mass of rope is negligible ,air resistance is negligible. 1.Find out the distance y dropped by the jumper before coming instantaneously to rest for the first time 2.Maximum speed attained by the jumper during this drop 3.The time taken during the drop before coming to rest for the first time

Answer y=[KL+mg+√(2mgKL+m 2g2)]/k v=√(2gL+mg 2/k) t=√(2L/g) + √(m/k)tan 1{√(2KL/mg)} http://physicscatalyst.com