pc â€“magazine -01

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http://physicscatalyst.com

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

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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

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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 mi 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 mi we get miri=miRc+mirci taking summation of above equation for n particles we get

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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 m1 and m2 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=m1+m2=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 m1 , m2 , m3 , . . . . . . . . . . . . . , mn 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 (Xcm,Ycm,Zcm) then position vector of centre of mass would be Rcm=Xcmi+Ycmj+Zcmk (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

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assume a system which contains two solid objects X and Y of any shape,. The center of

mass would be given by

+= ∫∫

batotal

CM rdmrdmM

R1

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.

( )bcmbacma

totalb b

b

a a

a

total

cm RMRMMM

rdmM

M

rdmM

MR ,,

11+=

+= ∫∫

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=3i-2k

m2=10 kg r1=-3i+2j-2k

m3=5 kg r1=i+2k

Find the center of mass of this mass system

Solution

So,

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5.5105

15)3(1035

321

332211 −=++

+−+=

++

++=

XXX

mmm

xmxmxmxcm

15105

0521005

321

332211 =++++

=++

++=

XXX

mmm

ymymymycm

15105

25)2(10)2(5

321

332211 =++

++−=

++

++=

XXX

mmm

zmzmzmzcm

So

Rcm=-.5i+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)

∫∫

∫∫

+

+==

dxBAx

dxBAxx

dm

xdmxcm

)(

)(

)63(

)32(

BAL

BALLxcm +

+=

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 miai = 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

cmexternalcmtotal dlFVM .2

1 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

total

netext

M

F ,

cma =

Taking right direction as positive,

Fext=2F-F

Mtotal=2M+M

So

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M

Facm

3=

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

instinstastastcminstast xMxMxMM +=+ )(

Now xcm=0

So

instinstastast xMxM −=

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

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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 xi Solution

Rcm = constant

0=∆ cmR

∑ =∆ 0ii rm ---(1)

=mM

mx

+− i

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

4

5

6

7

8

9

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ACROSS

4. Second law of Newton's

5. The unit of charge

6. This measure how fast are we

moving

7. Quantity use to measure energy

9. It revolves around nucleus

DOWN

1. Quantity to measure the hotness of

the object

2. Rocket propulsion is based on this

3. Pressure is inversely proportional to

Volume at constant temperature

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

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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 B0.

Find out the angular velocity of the electron in terms of B0 ,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

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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 An and B2n are respectively X and Y joules per

nucleon, if 2X -Y < 0 ,then find the energy released in the nuclear reaction

nnn BAA 2>−+

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 I0

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

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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 R0 is kept normal to a magnetic field given by

tBB ωcos0=

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

)452500sin(210)( 0+= ttV

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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

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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=3.0X108 m/s

Solutions

Solution 1

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

Solution 2

Resistance of Heater

ohmX

P

VR 4.48

1000

2202202

1 ===

Resistance of Bulb

ohmX

P

VR 484

100

2202202

2 ===

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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ohmReq 842470680

1

820

11

=+

+=−

Solution 4

The current delivered by the battery will be given by

AXR

VI

eq

210425.1

842

12 −===

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

V470=IR=6.7 V

Solution 5

Half cycle = π

02i

Full cycle =zero

Solution 6

0

2

2

1eVmv =

So V0=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

v

rT

π2=

Therefore current would be

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πω

π 22

e

r

evI ==

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

0

0

0

0

042 R

e

R

IB

πωµµ

==

Or

e

BR

0

004

µπ

ω =

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

2/1

693.

T=λ

Or

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 3/4th 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 L1 ,A1 and R1 be the final length , cross-sectional area and resistance

Now since length is increased by 10%

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LLXLL 1.1100

101 =+=

Since Volume remains same

1.11

11

AA

LAAL

=

=

Let a be the specific resistance of the material of the wire, then resistance of the wire

before and after will be given by

A

La

A

LaR

A

LaR

21.11

11 ==

=

So percentage increase in resistance is

%211001100 11 =

−=

−X

R

RX

R

RR

Solution 13

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

Solution 14

The Current in the bulb or secondary is

AV

Pi

S

s

s 140

40===

Now for the transformer ,we know that

PPss IVIV =

Or Ip=.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

30


Solution 18

4998.1

98.

1=

−=

−=

αα

β

Therefore Voltage gain is given by

3

1

2 1049100049 XXR

RVgain === β

Solution 20

Suppose the initial quantity of radium is N0 .Then the quantity left after n half lives will

be

n

""

=2

10

Now here N=25% of N0=N0/4

So

n

""

=2

1

40

0

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

= 444 10210*1*210*2 −−− == πππR

Current in the larger loop will be

4102

)5.24(−

+==

πt

R

VI

The magnetic Field due to this current in the larger loop at the common center will be

4

0

4

00

104

)5.24(

102*1*2

)5.24(

2 −−

+=

+==

πµ

πµµ tt

R

IB

31


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

20)2

4

0 104

)5.24(10(

104

)5.24( ttBA

+=

+== −

−

µπ

πµ

φ

Induced EMF in the smaller loop

05.62 µφ==

dt

dE

Now resistance of the smaller loop

= 544 10210*1.*210*2 −−− == πππR

So induced current

AR

EI 25.1

102

5.625

0 ===−πµ

Solution 25

The voltage is given by

)452500sin(210)( 0+= ttV

Comparing it with the standard equation

)sin(0 φω += tVV

We get

serad

VoltV

/2500

2100

=

=

ω

Now the impedance of the circuit is given by

2

2 1

+=C

RZω

Substituting the above values

Z=5000Ω

32


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

AXZ

V

Z

Vi rms

rms

30 1022

−===

AXii rms

3

0 10222 −==

The phase difference between the current and voltage is

)3/4(tan/1

tan 11 −− =

=R

Cωφ

The power factor of the circuit is

6.cos ==Z

Rφ

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

So

2

2

4r

Q -

2r

Qq=0

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

F1=2

2

)(4 xa

Q

+

The force of repulsion from negative charge

F2=2

2

)2( xa

Q

+

It is clear that F2 > F1

33


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

.....5,4,3,1

2

1122

=

−= nn

Rλ

For first line,n=3

36

5

3

1

2

1122

Rr =

−=λ

λ536

=R

Ionization energy is given by

=Rhc

Substituting the values ,we get

E=54.4eV

34


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.”

35


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.

36


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.

37


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.

38


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

39


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 MA collides elastically with a stationary body of mass MB. After the

collision ,the bodies move with angle θ1 and θ2 with the original direction of Mass MA.

Prove the following things by considering events in center of mass frame

a) If MA =MB ,then θ1+ θ2=π/2

b) If MA > MB ,then maximum value of θ1 is given by

A

B

M

M=1sinθ

Question-3

A particle of mass M is acted upon by a force that has an initial magnitude F0 When t=0

and decrease at a uniform rate until when t=t1,its magnitude is zero. It is moving across

the X-axis. Find the velocity and coordinate of the particle when t=t2 assuming that t2 >

t1. Assume xt=0=0 and (dx/dt)t=0=0

40


Answer

Velocity=F0t1/2M

Displacement=(F0t1/2M)(t2-t1/3)

Question -4

A wheel of radius r rolls without slip along the x-axis. With constant speed. v0.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=v0[1-cos(v0t/r)]

dy/dt=v0sin(v0t/r)

d2x/dt2=(v02/r)sin(v0t/r)

d2y/dt2=(v02/r)cos(v0t/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 Ax2.Show that the

limiting velocity is VL=(F/A)1/2.

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+m2g2)]/k v=√(2gL+mg2/k) t=√(2L/g) + √(m/k)tan-1{-√(2KL/mg)}

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