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In This Issuen This ssueIn This Issue

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MAHENDRA JAIN

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C.S.V. / July/ 2009 / 533

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C.S.V. / July / 2009 / 535 / 1A

For welfare and progress, a life of discipline is very essential whether it be an individual, a group, a society or a nation. For that matter it may be a case of the world.

It may be mentioned that for

want of a discipline, most of our country men appear dejected, pulled

down or without any enthusiasm. They

seem to be having no goal of life.

It is a well known fact the three factors—Family, atmosphere, edu-

cational system and political thinking

go to form the personality of persons.

In the family the individual learns to live in discipline, to obey elders and to lead a cooperative and tolerant

life. These things go a long way in

making the individual useful to the society and the nation. Aristotle has

rightly said that “State is an enlarged

form of the family.” The present deve-

lopment of civilization owes much to the good traditions and development

of the families from generation to

generation.

To draw out the best, specially the latent powers of the child ought to be the main function of an educational system. To make a perfect man of an individual has also always,been considered the chief aim of education. The aim of the education is the all round development of the individual. This, like the family, goes a long way in making the individual an useful instrument in giving proper and designed shape to the nation. The present system of education, which

is the legacy of the British rule has been only partially useful in this respect. Although many people find faults in the present system of education yet we will say that the present system has its own merits. In case otherwise it should not have produced big persons like Raja Ram Mohan Roy, Swami Vivekanand,Lokmanya Tilak, Mahatma Gandhi and others. Looking at different aspects of the role in the development of the person and the state it may be very easily said that discipline in

family and education life has a great role to play in the formation of the person, society and the nation.

In the present set up of the society, politics and political thinking have acquired top priority. From the personal life the national interests all depend on politics. We can easily say that the life of the nation has become politically biased Herbert Spencer had forseen the present conditions when he wrote, “You may leave politics, the politics will not leave you.”

The multi party system in our

country has made confusion worse

confounded. The main reason is that the party leaders have very little

sense of discipline, they have little respect for the declared objects of the party and are only running after

seeking pelf and power. The result is

that they are hardly making any social

or national progress and are fast loosing values of life.

People are often heard to say

that the present politics has no place for the intelligent or the intellectual.

They often are unpalatable meta-

phors for the political system. Be as it

may, the political life of the country has failed to make any contribution

for discipline of the people at large.

We should pause a little to think

how far our family life, educational

system and political system have been successful in disciplining people and develop in them the human

quality. The answer may be dis-

appointing but we need not be dejected. History tells us that there

were times when things had gone

worse—But people of firm determination led a life of strict discipline

and thus impacted their neighbour-

hood. In the modern times Mahatma

Gandhi has been a glaring example of such people. He by his personal

example made many persons to lead

a disciplined life which ultimately formed a big discipline party.

Education means drawing out of

all latent powers and all round deve-

lopment of faculties of the human being. It means that certain rules and

regulations are to be followed where

proper education is concerned. This

is nothing but observing a certain discipline. On the path of meditation

or spirituality, some discipline, namely

Raj Yoga, Dhyana Yoga, Bhakti Yoga or Hath Yoga is to be followed. Where

development is concerned, discipline

plays a very important role.

They say that humanity has

become civilized through discipline.We forget that the persons whom we

call natives or man of jungle also

follow certain rules of conduct and are in a way leading a disciplined life. In

the community of abonimals, certain

rules of discipline and conduct are

followed. In short, where there is human society, there is discipline,

because without it, no life is possible.

When there is no discipline, there

is confusion, disorder and ultimately war, but in war also certain rules are

to be followed, the soldiers lead a

life of ‘do or die’ discipline and then fight the war. In short, in both, dis-

order and peace, discipline is indis-

pensable. Therefore, let us remember that discipline is the life blood of the

human society, without it, no life is possible.

●●●


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❥ Everything appears coloured to the jaundiced eye.

❥ However clever you are, there is someone cleverer than you.

❥ If you don’t crack the shell, you can’t eat the nut.❥ Using threats of suspension as a major strategy for maintaining

discipline does not go a long way.

❥ Sooner or later the man who wins is the man who thinks he can.

❥ Prefer a loss to a dishonest gain; the one brings pain at the mo-

ment, the other for all time.

❥ The humblest citizen of all land, when clad in the armour of righ-teous cause, is stronger than all the hosts of error.

❥ To be vanquished and yet not surrender, that is victory.

❥ What you cannot say before your enemy, do not say before yourfriend.

❥ Criticism breeds criticism.

❥ Is not country more important than community ?

❥ There are some students who are playing the fool at the back of

the class but such are surely the boys who roam the streets in the

end.

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❥ Be the best you can be.


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C.S.V. / July / 2009 / 537

Rebirth of Hubble SpaceTelescope

The Hubble space telescope, theobject of NASA’s fifth and the last

servicing mission, is a veritable time

machine that has revolutionalisedhumankind’s vision and comprehen-

sion of the universe. Put into orbit at

an altitude of 600 km by the shuttleDiscovery on April 25, 1990, Hubble

has transmitted more than 7,50,000

spectacular images and streams of

data from the ends of the Universe,opening a new era. But the Hubble

telescope, the fruit of a collaboration

between NASA and the EuropeanSpace Agency, had a troubled start

and did not become operational until

three years after its deployment.

Its lense in effect had to be fixed

because of a flaw in its shape, a sen-

sitive operation that was not carried

out until 1993 in the first shuttle-borneservice mission, which installed

corrective lenses. From that time o n

Hubble space telescope transmittedstupefying images of supernovas,

gigantic explosions that marked the

death of a star and revealedmysterious black holes in the centre

of virtually all galaxies.

Helping Hand : In this image taken fromNASA video, Hubble is captured by thespace shuttle Atlantis’ robotic arm as itbegins its mission to service the spacetelescope.

Thanks to these observations,delivered with 10 times clarity of the

most powerful telescopes on the

Earth, the astronomers have been

able to confirm that the Universe isexpanding at an accelerating rate and

to calculate its age with greater

precision as an estimated 13·7 billionyears. The universe’s accelerationis the result of an unknown force

dubbed dark energy that consti-tutes three-quarters of the Universeand counterbalances the force of

gravity. The rest of the cosmos is

composed of five per cent visiblematter and about 20% shadow

matter or anti-matter.

Among the other discoveries,credited to Hubble, figures the detec-

tion of the first organic molecule in

the atmosphere of a planet orbiting

another star and the fact that the

process of formation of planets andsolar systems is relatively common in

our galaxy, the Milky Way. Hubblealso has observed small proto-

galaxies that were emitting rays of

light when the Universe was less thana billion years old, the farthest back in

time that a telescope has been able

to peer so far.

Eye in the sky : A few of the stunningimages taken by the Hubble space tele-scope over the years.

In space, Hubble was capturedby the space shuttle Atlantis’ robotic

arm as it began its mission to servicethe space telescope. Two astronautsMike Massimino and Mike Good

emerged from the airlock of the

shuttle Atlantis and began work on

Space Telescope.

In addition to the partial revival of

the Advanced Camera for surveys,

they installed two new scientific instru-ments and a crucial science computer

as well as replaced gyroscopes and

batteries to sustain the Observatory’s

pointing and power systems. Theoverhauling prepared Hubble to

search for the oldest and the most

distant galaxies, map the large scalestructure of the Universe and study

the planet forming processes around

other stars.

The two new installed instruments

will enable Hubble to look out in timeas far as 500 to 600 million years

after the Universe’s birth with the big

bang.

Closer to home, Hubble has

observed radical changes in thedirection of Saturn’s winds and

revealed that Neptune has seasons.

Hubble has also examined mysterious

lightning flashes on Jupiter and takenastonishing pictures of Mars.

New York-Sized Ice ShelfCollapses in Antarctic

An area of ice shelf, almost the

size of New York City, broke into ice-

bergs in April 2009 after the collapseof an ice-bridge widely blamed on

global warming.

Warming Disaster (Top) A satelliteimage of the Wilkins Ice Shelf, taken onApril 27, shows icebergs covering anarea of 700 sq km that have broken off.

Professor Angelika Humbert, a

glaciologist of the University of

Muenster (Germany) analysed theEuropean Space Agency Satellite

images of the Shelf of Antarctic and


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C.S.V. / July / 2009 / 538

came to this conclusion. She (Prof.

Humbert) said about 700 sq km of

ice–bigger than Singapore or Bahrainand almost of the size of New York

City–broke off the Wilkins and

shattered into a mass of icebergs.

She said 370 sq km of the ice

had cracked in recent days from the

Shelf, the latest of about 10 shelves

on the Antarctic Peninsula to retreatin a trend linked by the U.N. Climate

Panel to global warming.

The new icebergs added to 330

sq km of ice that broke up earlier in

April 2009 with the shattering of an ice

bridge apparently pinning the Wilkins

in place between Charkot island and

the Antarctic Peninsula.

Nine other ice shelves have

receded or collapsed around the

Antarctic Peninsula during the past 50years, often abruptly like the Larsen A

in 1995 or the Larsen B in 2002. The

trend is widely blamed on climate

change caused by heat-trapping

gases from burning fossil fuels.

Most Distant Object inthe Universe Spotted

Astronomers have spotted the

most distant object in the universe,

which is self-destructing star that

exploded 13·1 billlion light years from

Earth. It detonated just 640 million

years after the big bang, around the

end of the cosmic ‘dark ages’, when

the first stars and galaxies were

lighting up space. The object is a

gamma-ray burst—the brightest type

of stellar explosion. Gamma-rays

bursts occur when massive spinning

stars collapse to form black holes and

spew out jets of gas at nearly the

speed of light.

Stars ‘Eat Up’ Planets

Cannibalism is rampant in our

universe. Stars ‘eat’ the exoplanetsthat venture near them. The new

study has revealed that the

exoplanets are doomed to prematuredeaths even before they could get

close to be ripped apart by the host

star’s gravity, a finding that may helpexplain why few exoplanets are found

next to host stars. The research team

is lead by Professor Brian Jackson of

the University of Arizona (U.S.A.).

In accordance with this research,

a star’s gravity can put a nearby

planet on a ‘fast track’ to spirallinginto the star and may also cause the

planet to lose much of its atmosphere.

More than 300 exoplanets have

been catalogued to date. Many are

situated close to the host stars. Butthe closestin ones are commonly

found some 0·05 astronomical units

(AU) from their host stars. But, no

one is sure why the planets seem to

pile up there. Very close to a star, at

a boundary called the Roche limit,

planets are dismembered by the

star’s gravity. But the migration of

planets seems to stop there ? Some

models suggest gas and dust in the

disc around a star could drag theplanets inward.

Forthcoming SpaceTelescopes to Peek into

Future

A couple of space telescopes,

that are going to be launched very

shortly, will answer some of the

biggest questions of the universe.

Scientists hope that the probes will

answer questions such as how did we

get to, where we are now, and where

are we likely to end up.

Each telescope is designed to

probe the deepest reaches of space

to unravel the origins of matter, from

the earliest beginnings of the

universe, some 13·7 billion years ago

to the creation of the stars, galaxies

and planets. One of the telescopes

called, ‘Planck’, will study in

unprecedented detail of the ancient

‘fossilized’ radiation left over as a relic

of the big bang. The analysis could

help to explain how the universe

formed through a process of rapid

expansion in the first fractions of a

second after the big bang itself.

The other space telescope to be

launched is ‘Herschel’. It will concen-

trate on the invisible, infrared radiation

emitted by the star-forming regions of

the galaxies on the hope of explaininghow stellar objects from stars like the

Sun to planets such as Earth, can

form from clouds of cosmic gas, dust

and debris. Scientists involved in thetwin missions hope that the data

gleaned from instruments on board

each space telescope will enablethem to fill in the remaining mysteries

of how the universe came into exis-

tence, how it evolved and how it is

likely to end—if indeed it ever will.

Flowers May Bloom onJupiter’s Icy Moon

Scientists have suggested that

spacecraft should hunt for signs of life

on Jupiter’s ice-covered moon,

Europa, since it would be detectable

there in the form of blooming flowers.

Life could be visible from orbiting

spacecraft, however, if it made a hole

in cracks in Europa’s shell that con-

nect the surface to the interior,physicist and futurist, Professor

Freeman Dyson reported. Such life

might take the form of flowers that

focus sunlight on the interior of the

plant.

Europa flowers could be detect-

able through retro-reflection, an

optical effect that is seen in light

reflected from animals’ eyes.

●●●

UPKAR PRAKASHAN, AGRA-2E-mail : [email protected] Website : www.upkar.in

(Useful for Various Competitive Exams.)Useful for Various Competitive Exams.)

(Useful for Various Competitive Exams.)

Byy

Dr. Alok Kumarr. Alok Kum ar

By

Dr. Alok Kumar

Code No.ode No 16301630

Rs.s 40/-40/

Code No. 1630

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9/110C.S.V. / July / 2009 / 540

Wajed Mia—Nuclear scientist

and Bangladesh Prime Minister

Sheikh Hasina’s husband, Wajed Mia

(66), died of prolonged illness and

multiple complications in Dhaka on

May 9, 2009.

General Elections 2009—India

voted decisively for continuity and

stability in the general election to the

15th Lok Sabha, giving the Congress-

led United Progressive Alliance

(UPA) another five-year term in office.

People of India spoke and spoke with

great clarity. In contrast to 2004, the

UPA won close to 260 of the total 543

seats did not need the support of Left

Parties. The allies of Congress are—

Nationalist Congress Party, the All

India Trinamool Congress, theDravida Munnetra Kazhagam, the

National Conference, and the All

India Majlis-e-Ittehadul Muslimeen.

Now, Manmohan Singh is the first

Prime Minister since Indira Gandhi to

have two full terms.

Parties and their Vote Share

Party

Total

seats

Change

from

2004

Vote

share

(%)

Congress 206 61 29·67

BJP 116 – 22 19·29

JD (U) 20 12 1·58

CPM 16 – 27 5·52

CPI 4 – 6 1·46

BSP 21 2 6·27

AIADMK 9 9 1·79

DMK 18 2 1·91

TC 19 17 3·43

NCP 9 0 2·24

SP 23 – 13 3·44

TDP 6 1 1·53

SS 11 – 1 1·67

RLD 5 2 0·49

SAD 4 – 4 0·92

NC 3 1 0·13

RJD 4 – 20 1·31

LJP 0 – 4 0·48

TRS 2 – 3 0·63

BJD 14 3 1·35

AGP 1 – 1 0·45

INLD 0 0 0·33

JD(S) 3 0 0·89

JMM 2 – 3 0·43IUML 2 1 0·23

IND 9 2 4·16

Others 16 – 9 8·4

The triumph of the Congress was

actually an aggregation of specific

successes across different states.

The party retained its base in Andhra

Pradesh, cut its losses in Madhya

Pradesh, recovered lost ground in

West Bengal, Keral and Rajasthanand combined well with its allies in

Maharashtra and Tamil Nadu.

The BJP and Left parties are the

big losers in the current general

election.

Ashok Chawla (New Finance

Secretary )—Economic Affairs Secre-

tary, Ashok Chawla, took over as the

Union Finance Secretary, succeedingArun Ramanathan who retired. Mr.

Chawla is an IAS officer of 1973

batch of Gujarat cadre.

P. K. Barbora (New Vice Chief

of Air Staff )—Air Marshal P. K.

Barbora has been appointed as the

new Vice Chief of Air Staff. He will

assume the charge on June 1, 2009.

Currently, he has been serving as the

Air Officer Commanding-in-Chief of

Western Air Command.

Deepak Verma (New Judge,Supreme Court )—Justice Deepak

Verma (61), Chief Justice of Rajas-

than, has been appointed as the

Judge of Supreme Court. He hails

from Madhya Pradesh and will have a

tenure of about four years.

Balbir Singh Chauhan (New

Judge, Appex Court )—President

Pratibha Patil cleared the appointment

of Justice Balbir Singh, Chief Justiceof Orissa High Court, as the judge of

the Supreme Court.

Naveen Patnaik (CM, Orissa )—Biju Janta Dal President NaveenPatnaik was sworn-in as ChiefMinister of Orissa for the third conse-cutive term. Twenty other legislatorswere also sworn-in as the Ministers.Patnaik’s Party won 103 seats in the147 member Assembly.

Pawan Chamling (CM, Sikkim )—Pawan Chamling was sworn-in as

Chief Minister of Sikkim for a fourthsuccessive term, making him thelongest serving Chief Minister in thestate. Eleven other Ministers werealso sworn-in. Chamling’s SikkimDemocratic Front created history bywinning in all 32 Assembly seats.

D. D. Lapang (CM, Meghalaya )—A seven member Congress andUnited Democratic Party coalitionMinistry, headed by D. D. Lapang,was sworn-in at the Raj BhavanShillong.

Zuma (New President, S.A.)—Jacob Zuma, the indefatigable fighteragainst apartheid was sworn-in asliberated South Africa’s fourth Presi-dent.


10/110C.S.V. / July / 2009 / 541

Dr. Manmohan Singh—Presi-

dent Pratibha Patil administered the

oath of office and secrecy to Dr.

Manmohan Singh as the Prime

Minister of India alongwith his cabinet

colleagues in Rashtrapati Bhavan on

May 22, 2009. This is Dr. Singh’ssecond successive term.

President Pratibha Patil is administer-ing the oath of Office and Secrecy toDr. Singh

Prime Minister Manmohan Singh

was born on September 26, 1932, in

a village in the Punjab province of

undivided India. Dr. Singh completed

his Matriculation examinations from

the Punjab University in 1948. His

acadmic career took him from Punjab

to the University of Cambridge, UK,

where he earned a First Class

Honours degree in Economics in

1957. Dr. Singh followed this with a

D. Phil in Economics from Nuffield

College at Oxford University in 1962.

His book, ‘‘India’s Export Trends and

Prospects for Self-Sustained Growth’’

[Clarendon Press, Oxford, 1964] was

an early critique of India’s inward-

oriented trade policy.

In 1971, Dr. Singh joined the

Government of India as Economic

Advisor in the Commerce Ministry.

This was soon followed by his

appointment as Chief Economic

Advisor in the Ministry of Finance in

1972. Among the many Governmental

positions that Dr. Singh has occupied

are Secretary in the Ministry of

Finance; Deputy Chairman of the

Planning Commission; Governor of

the Reserve Bank of India; Advisor of

the Prime Minister; and Chairman of

the University Grants Commission.

In what was to become the

turning point in the economic history

of independent India, Dr. Singh spentfive years between 1991 and 1996 as

India’s Finance Minister. His role in

ushering in a comprehensive policy of

economic reforms is now recognized

worldwide. In the popular view of

those years in India, that period is

inextricably associated with Dr.

Singh.

Among the many awards and

honours conferred upon Dr. Singh in

his public career, the most prominent

are India’s second highest civilian

honour, the Padma Vibhushan (1987);

the Jawaharlal Nehru Birth Centenary

Award of the Indian Science Congress

(1995); the Asia Money Award for

Finance Minister of the Year (1993

and 1994); the Euro Money Award for

Finance Minister of the Year (1993),

the Adam Smith Prize of the Univer-

sity of Cambridge (1956), and the

Wright’s Prize for Distinguished

Performance at St. John’s College in

Cambridge (1955). Dr. Singh has also

been honoured by a number of otherassociations including the Japanese

Nihon Keizai Shimbun.

Dr. Singh and his wife Mrs.

Gursharan Kaur have three daugh-

ters.

During the last 26 years, LTTE

rewrote many of the standards of

terrorism. The Sri Lankan armed

forces won a comprehensive victory

over the LTTE in a military campaign

that began in the eastern province in

August 2006. With its entire top

leadership and thousands of fighting

cadres are killed in action, its military

structure, assets and capabilities are

destroyed, its political organization

decimated, the LTTE no longer exists.

Belying conventional wisdom, Sri

Lanka has found a military solution to

what used to be regarded as an

intractable armed secessionist andterrorist challenge. Over a quarter of

century, this war waged and claimed

tens of thousands of lives.

The images of terrified children,

women and men fleeing the tiny sliver

of coastal land in which they were

confined by the Tigers for use as

human shieled. Senior LTTE leaders

made a final hopeless stand for a lost

cause will continue to haunt the

memories of journalists and others

who witnessed these scenes.As the years went by and

numerous proposals for a negotiated

political solution fell by wayside, the

one thing that remained constant was

the LTTE’s uncompromising seces-

sionism and militarism, and the rising

graph of its terrorist crimes, which

included the assassination of the

former Indian Prime Minister, Rajiv

Gandhi, Sri Lankan President

Premdasa, a Sri Lankan Defence

Minister, a Foreign Minister and

countless others.

Now, in the post Prabhakaran

era, the Sri Lankan Government

needs to address two big tasks—

rehabilitation of hundreds of thou-

sands of Tamils who have been

through a prolonged nightmare and

crafting an enduring political solution

based on far-going devolution of

power to the Tamils in their areas of

historical habitation. India, which has

excellent relations with its southern

neighbour, can make a constructive

difference by coming up with a

massive rehabilitation package for the

North and encouraging Colombo to

fast-track the political solution.

SPORTS

Cricket

IPL-2—The final of Indian

Premier League cricket match wasplayed in Johannesburg between

Deccan Chargers and Royal Challen-

gers, Bangalore on May 24, 2009 at

the Wanderers. Anil Kumble was the

‘Captain of Royal Challengers,

Bangalore’ while Adam Gilchrist was

the Captain of Deccan Chargers. A

fighting unbeaten half century by

Herschelle Gibbs (53 runs not out)

took Deccan Chargers 143 for six

wicket.

But then it appeared initially to bean innings dominated by a bowler, leg

spinner Anil Kumble, who finished four

wicket for sixteen runs. Three of his

scalps were—Adam Gilchrist, A.

Symonds and Rohit Sharma. In a

brilliant play, Kumble picked himself

to bowl the first over after inviting

Chargers to bat. It was Captain versus

Captain when Kumble operated to the

in-form.

Finally, the Royal Challengers,

Bangalore could score only 137 runsfor 9 wickets and lost to Deccan

Chargers by six runs.

●●●


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Physics

1. What is the relation between v , ω and r (in vectorform) ?

➠ →→→→v =

→→→→ωωωω ××××

→→→→r

2. What does the plug marked infinity in a resistance

box have below it ?

➠only air gap

3. What is the moment of inertia for elliptical lamina ?

➠ I = (1/4) Ma 2 (about minor axis)

and I = (1/4) Mb 2 (about major axis)

4. How many protons and neutrons does anα

-particle

possess ?➠Two protons and two neutrons

5. What is tractive force ?➠ F = (P/ v )

6. What is conserved in the case of a freely falling

body ?➠sum of kinetic and potential energies

7. What is the pressure-temperature law ?➠ (P1 /T1) = (P2 /T2)

8. An astronomical telescope is made of two lenses of

powers 5 D and 20 D. Its magnifying power for nor-

mal vision is

➠4

9. What is Poynting vector ?

➠ →→→→ S =

→→→→ E ××××

→→→→ H =

1

µµµµ0 (→→→→ E ××××

→→→→ B) = c 2 εεεε0 (

→→→→ E ××××

→→→→ B)

10. To which region does the electromagnetic radiation of

wavelength of the order of 1°A belong ?

➠ X-ray radiation

11. What is ratio of the reflected intensity and incident

intensity ?

➠

Ir

Ii = ( )n 1 – n 2

n 1 + n 2

2

12. A 4 µF condenser is charged to 400 volt and then itsplates are joined through a resistance. Heat produced

in the resistance is

➠ 0·32 joule

13. What is polarization vector ?

➠ (i) →→→→P =

QPA

, (ii) →→→→P = εεεε0 χχχχ Ed

14. How will you connect three capacitors of 3 µF each

so that the capacitance of the combination is 4·5 µF?

➠ Two in series and then one in their parallel15. What is the trajectory of a charged particle when it is

projected perpendicular to a magnetic field ?

➠ Circle in a plane perpendicular to the field

16. When a radioactive nucleus emits a β-particle, theneutron to proton ratio

➠

decreases17. What is the dimensional formula of Hubble’s constant

is

➠ [M0 L0 T –1]

18. In nuclear reactor what is the function of moderators ?

➠To slow down fast fission neutrons

19. What will be the force when dipoles are along the line

joining their centres ?

➠ µµµµ04ππππ ·

6M1 M2r 4

(along r )

20. What provides the centripetal force to enable an earth

satellite to move in a circular orbital ?➠The gravitational force of attraction between the

earth and the satellite

Chemistry

21. The esters of long chain fatty acids with long chain

alcohols are commercially known as

➠ Waxes

22. The lines in the spectrum of hydrogen atom in the

visible region are termed as➠Balmer series

23. Both mass and volume are extensive properties but

the ratio of mass of a sample to its volume is an

intensive property, known as

➠ Density

24. The idea of elliptical orbits was propounded by

➠Sommerfeld

25. The scientist who first pointed out that an element is

any substance that cannot be decomposed into a

simpler substance

➠ Robert Boyle’s

26. Certain materials like potassium emit electrons when

irradiated with visible light. This is known as

➠Photoelectric effect

27. The term isotope was introduced by

➠ Frederick Soddy

28. High lattice energy of an ionic compound is favoured

by

➠Small inter-ionic distance and high

charge on ions

29. 1 mol of O is equal to 16·0 gm and 1 mol of O2 will be

equal to

➠ 32·0 gm30. The chemical compounds which exist over a range of

chemical composition are known as

➠Berthollide compounds


13/110C.S.V. / July / 2009 / 544

31. When neutron is outside the nucleus, it is unstable

and it changes into a proton, and electron and ano-

ther elementry particle, known as

➠ Neutrino

32. The product of the net positive or negative charge

and distance between the two charged ends is known

as

➠Dipole-moment

33. When a neutron collides with a proton, a nucleus of➠ Deuterium is formed

34. Who discovered chlorine ?

➠C. W. Scheele (1774)

35. A cold glow given out by some substances is called

➠ Phosphorescence

36. Entire mountain ranges in Italy consist of mineral

dolomite. Chemically dolomite is

➠MgCO3·CaCO3

37. The crystals that can detect ultrasound and produce

ultrasound are known as

➠ Piezoelectric crystals

38. Fluoroapatite is commercially important as a source

of phosphate. The composition of fluoroapatite is

➠ [3{Ca3(PO4)2}·CaF2]

39. The SI unit of pressure is

➠ Pascal (Pa) [1 atm = 101,325 Pa = 101·325 kPa]

40. Aluminium articles are often given decorative finishby electrolysing dil. H2SO4 with the aluminium anode.

This process is known as

➠Anodising

Zoology

41. What is the name of the hormone that causes deposi-

tion of fat in breast and hips in female humans during

puberty ?

➠ Estrogen

42. A group of coelomate metazoans in which the first

embryonic opening is associated with the mouth is

➠Protostome

43. Which ion must be present for binding of the cross

bridges in muscles ?➠ Calcium

44. Most of the carbon dioxide is transported in the blood

stream of humans is

➠Bicarbonate ion

45. What is called the form of enzymes that are encoded

by different allelic genes ?

➠ Allozymes

46. A small calcareous granules found in the inner ear of

many mammals, is

➠Otolith

47. Over production of which neurotransmitter has beenassociated with the mental disorder, called

schizophrenia ?

➠ Dopamine

48. Ridges or folds found in the lining of vertebrate

stomach is called

➠Rugae

49. Where each restriction enzyme cleaves a molecule ?

➠ At a particular nucleotide sequence

50. A rod of bone or cartilage that forms the only ear

ossicle in amphibians, birds and reptiles is called

➠Collumelar auris

51. Where the spermatogenesis occurs ?

➠ Seminiferous tubules

52. In cerebrum, the roof of each paracoel is called

➠Pallium

53. Which hormone prevents dehydration of human

body ?

➠ ADH

54. Part of coelom in mammals containing lungs and

lined by pleura is

➠Viscera

55. Which area of human brain is responsible for arousaland wakefulness ?

➠ Reticular formation

56. Large marine mammals well adapted for aquatic life

are collectively known as

➠Cetacea

57. Which ion is most concentrated outside a resting

potential ?

➠ Sodium

58. The endocrine part of pancreas consists of

➠ Islets of Langerhans

59. Which kind of cells transmit the sensory impulses inhuman eye to optic nerve ?

➠ Ganglion cells

60. The cells from Graafian follicle that surround the ovu-

lated mammalian egg are known as

➠Cumulus cells

Botany

61. Who prepared an infectious extract from tobacco

plants that were suffering from mosaic disease ?➠ D. I. Ivanovsky

62. When does chromosome number becomes halved ?

➠ In meiosis during anaphase-I

63. What type of lysine, an amino acid, is ?

➠ Basic amino acid

64. What food is used by fungal partner made by algal

partner in a lichen ?

➠Mannitol

65. What refers to the number of death per unit time ?

➠ Mortality

66. What is the major role of phosphorus in plantmetabolism ?

➠To generate metabolic energy

(Continued on Page 612 )


14/110C.S.V. / July / 2009 / 545

ThermodynamicsThermodynamics is that branch of physics in which

heat is converted into other forms of energy and other

forms of energy are converted into heat. This branch

deals the transformation of heat into mechanical work and

the inter-relationship between them.

Thermodynamical Variables

Thermodynamical variables are those parameters

which define the thermodynamical system completely.

These are pressure (P), volume (V), temperature (T),

internal energy (U) and entropy (S). These are also called

thermodynamical coordinates.

Thermodynamical Equilibrium

A system is said to be in thermodynamical equili-

brium if the temperature of its various parts is the same

and equal to that of surroundings.

External Work done (W)

When a body is heated, it expands. This is opposed

by external atmospheric pressure. The work done against

external atmospheric pressure during expansion of a body

is called external work.

∴ Work done = Force × displacement

= Pressure × area × displacement

= Pressure × change in volume

or∆W = P ∆V

∆W = P (V2 – V1)

where,

V1 = Initial volume of gas, V2 = Final volume of gas

(i) If V2 > V1, then ∆W = +ve, then work is done by

the system(ii) If V2 < V1, then ∆W = –ve, then work is done on

the system

(iii) If V1 = V2 or V = constant, then ∆W = 0

● If pressure is constant, then work done

W = P(V2 – V1)

● If pressure and volume both are variable then the

work done

W = ∫ V1

V2P d V

= Area between P -V curve and volume axis

● If the system expands into vacuum (free expansion),

then

∆W = 0

P-V diagram or indicator diagram—A graph bet-ween pressure (P) and volume (V) is known as P-V

diagram or indicator diagram.

Area under P-V diagram = Work done.

Area = Work done by thegas in path AB

A

P

V

B

Cyclic Process (or Closed Path)

Cyclic process is that process in which the system

returns to its original state (P, V, T) after doing work or

after work being done on it.

The work done on the system or work done by the

system depends upon the area of cycle. If the cycle traced

in clockwise direction then the network is done by the gas

and if the cycle is traced in anticlockwise direction the network is done on the gas.

Conversion of units—Work is measured in joule or

erg and heat is measured in kilo calorie or calorie. In the

relation W = JH, J is conversion factor.

In C.G.S. system J = 4·2 × 107 erg/cal

In M.K.S. system J = 4·2 × 103 joule/k cal

= 4·2 joule/cal

In F.P.S. system

J = 778 foot-pound/B.Th.u.

Example 1. When a body falls from a great height

(e.g., water in a waterfall), potential energy is finally

converted into heat energy. Here the temperature

increases slightly.

mgh = J × ms ∆t

∴ ∆t =gh

Js

Example 2. When a bullet is fired at a target, kinetic

energy is converted into heat energy and the temperature

increases too much.

HereIf bullet does not melt

and

If bullet melts

1

2 mv 2 = J (ms ∆t )

1

2 mv 2 = J (ms ∆t + m L)


15/110


16/110C.S.V. / July / 2009 / 547 / 3

(d) For isobaric change—In this process

P = Constant.

Q = m Cp ∆t

∆U = m Cv ∆t

Here, Cp is specific heat of gas at constant pressure.

Therefore, in isobaric change the heat supplied to the

system is used partly in changing the volume and partly in

changing temperature.

(e) For isolated system—An isolated system is one

which is completely cut off from the surroundings,

therefore, Q = 0, and there is no change in internal energy

i.e., ∆U = 0 and hence, ∆W = 0. So system does not

perform any work.

(f) For cyclic process—For this process change in

internal energy ∆U = 0.

Hence, Q = W

i.e., whole of the heat supplied to the system is used

in doing work against external pressure.

Specific Heat of GasesSpecific heat of a gas depends on the condition of

pressure and volume of the gas during its heating.

Accordingly specific heat of a gas may be anything from

zero to infinity.

In general two modes of heating a gas has been

selected. They are (a) At constant pressure (b) At cons-

tant volume.

Accordingly there are two specific heats in case of

gases.

(i) Specific heat at constant pressure (Cp )

(ii) Specific heat at constant volume (Cv )

Cp > Cv —In case of Cv , volume V = Constant

⇒ ∆V = 0

Hence, work done by gas W = P ∆V = 0 but in case of

Cp , pressure P = constant. Therefore, when gas is heated

its volume increases and some work (W = P ∆V) is done

by the gas. Hence extra amount of heat should be givento the gas to do this work. So Cp > Cv .

Cp – Cv = Extra work done

= P ∆V

= PV2 – PV1= RT2 – RT1

= R (T + 1) – RT

= R

∴ Cp – Cv = R

It is known as Mayer’s formula

For one gm mole gas

Cp – Cv ~ – 2 cal/mole-K

Molar Specific Heat

It is equal to specific heat multiplied by the molecular

weight M.

Thus, Cp = M × Cp

and Cv = M × Cv

Values of Specific Heats Cp and Cv for Gases

(a) For monoatomic gases (e.g., He, Ne, Ar etc.)

Cv =3

2 R ~ – 3 cal/mole-K

Cp =5

2 R ~ – 5 cal/mole-K

γ =Cp C

v

= 1·67

(b) Diatomic gases (e. g., H2, O2, N2 etc.)

Cv =5

2 R

≈ 5 cal/mol-K

Cp =7

2 R

= 7 cal/mole-K

γ =Cp Cv

= 1·4

(c) Polyatomic gases

Cv = 3 R = 6 cal/mole-KCp = 4R ~ – 8 cal/mole-K

γ =Cp Cv

= 1·33

● For n moles of gas

∆U = n Cv ∆T

and ∆Q = n Cp ∆T

∴ ∆W = ∆Q – ∆U = n R ∆T

Nature of Internal Energy

Every thermodynamic system has some internal

energy which is characteristic of its state. It consists of

kinetic energy due to molecular motion and potential

energy due to molecular attraction.

We know that monoatomic molecules undergo only

translational motion, i.e., the centre of mass of the mole-

cule moves [fig. (a)]. Hence, these molecules have kinetic

energy due to translational motion.

Translational Motion(a)

Rotational Motion(b)

Vibrational Motion

(c)

Diatomic and polyatomic molecules undergo not only

the translational motion inside the substance; but also

rotate about the axis passing through the centre of massof the molecule [fig. (b)] and also vibrate relative to each

other [fig. (c)]. Thus, in diatomic and polyatomic molecules,

in addition to translational motion, there is also internal


17/110C.S.V. / July / 2009 / 548

rotational motion and vibrational motion. Hence, these

molecules in addition to translational kinetic energy, have

rotational kinetic energy and vibrational kinetic energy

also.

Thus, the internal energy of a substance consists of :

(i) The translational kinetic energy of molecules.

(ii) The internal rotational and vibrational kinetic

energies of molecules (if they are polyatomic).

(iii) The potential energy of the molecules due tointer-atomic forces.

Important Points to Note

1. Ideal gases—In case of ideal gases there is no

molecular attraction between the molecules. Hence, they

have no potential energy. Thus, the internal energy of an

ideal gas is only the kinetic energy of its molecules.

2. Real gases—In real compressed gases the

molecules come closer and so exert appreciable force on

one another. Hence, potential energy also adds to their

internal energy. Since, potential energy is negative, it

follows that internal energy of a compressed gas is less

than its internal energy in rarefied state at the same

temperature.

3. Liquids—Molecules in liquids are very close to

one another exert stronger forces and possess sufficient

potential energy. But their translation motion is very

limited in comparison to gas molecules. Since potential

energy is negative, the internal energy of the liquids is

very small compared to the internal energy of the gas at

the same temperature.

4. Solids—In solids molecules are fixed in definite

positions in a lattice. These molecules vibrate to and fro

about these positions but can not leave these positionspermanently. These vibrations are called lattice vibrations.

In solids the potential energy of molecules is very large.

Since this is negative, the internal energy of solids is less

than that of liquids.

5. Translational K. E. of molecules—According to

the kinetic theory, the translational kinetic energy of the

molecules (and not the whole internal energy of the

substance) is directly proportional to the absolute

temperature of the substance. Hence the temperature of

the substance rises on increasing the translational kinetic

energy of its molecules.

Second Law of ThermodynamicsThe first law of thermodynamics states the equi-

valence heat and mechanical work when one is

completely converted into the other. It simply tells that

when ever work is obtained an equivalent amount of heat

is used up, or vice-versa . It does not say anything either

about the limitation in the conversion of heat into work or

about the condition necessary for such a conversion.

The quest for deciding these points led to the

formulation of Second Law of Thermodynamics. This

law is generalisation of certain experiences and

observations and is concerned with the direction in which

energy transfers take place. This law has been stated invarious forms but all the statements are equivalent. Below

are given two simple forms of this law. According to one

statement :

‘It is impossible to convert ‘all’ the heat extracted

from a hot body into work’

According to a second statement :

‘It is impossible to transfer heat from a cold body

to a hot body without expenditure of work by an

external agency’.

As an illustration we take the case of a heat engine.Here the working substance takes heat Q1 from the hot

body (source), converts a part of it

into work W and gives the rest Q2

to a cold body (sink)

No engine has ever been

designed which may convert ‘all’

the heat Q taken from the source

into work W without giving any

heat to the sink. For obtaining

continuous work a sink is

necessary. In other words, all the

heat taken from a body cannot be

converted into work.

→W

Hot Body(Source)

Cold Body(Sink)

Q 2

Q 1

● A refrigerator is a heat engine running in the reverse

direction. In it, the working substance (a gas) takes

in heat from a cold body and gives out to the hotter

body (external atmosphere). For doing this it uses

electrical energy. No refrigerator has yet been

designed which may transfer heat from a cold body to

a hot body without using an external source of

energy. It implies that it is impossible for a self-acting

machine, unaided by any external agency, to transfer

heat from a cold body to a hot body.

Efficiency of Heat Engine

η =Amount of heat converted into mechanical work

Amount of heat taken from the source

=Q1 – Q2

Q1

= 1 –Q2Q1

Carnot engine (Reversible cycle)

For a reversible cycle

Q1Q2

=T1T2

∴ η = 1 –T2T1

Refrigerator

← Mechanical

work

Q1

Q 2

W = Q1 – Q2

Hot BodyT K1

Cold BodyT K2

The efficiency of rever-sible thermodynamic cycle(Carnot cycle) depends not onthe nature of the gas, but onlyon the temperature rangebetween which it operates.

Carnot’s theorem—Noengine can be more efficientthan a reversible engineworking between the sametemperatures.

W = Amount of mechani-cal work given from outsidewhich changes into amount(Q1 – Q2) of heat.


18/110C.S.V. / July / 2009 / 549

At a Glance

Efficiency of Engines

Steam engine—ηs = 17% (Max.)

Petrol engine—ηp = 44% (Max.)

Diesel engine—ηd = 55% (Max.)

Thus, ηd > ηp > ηs

Electric engine

η = 90% (Max.)

The coefficient of performance of the refrigerator.

β =Q2W

=Q2

Q1 – Q2

For a reversible cycle

Q1Q2

=T1T2

∴ β =T2

T1 – T2

Relation between ββββ and ηηηη

β =T2

T1 – T2 =

1

T1T2

– 1

and η = 1 –T2T1

or

T2T1 = 1 – η

orT1T2

=1

1 – η

∴ β =1

1

1 – η – 1

β =1 – η

η

OBJECTIVE QUESTIONS01. The volume of a gas expands by

0·25 m3 at constant pressure of

103 Nm –2. The work done is

equal to—

(A) 2·5 erg (B) 250 joule

(C) 250 watt (D) 250 newton

02. An ideal monoatomic gas is

taken round the cycle ABCDA as

shown in the P-V diagram. The

work done during the cycle is—

P

(2P, V) (2P, 2V)

A

D

(P, V) (P, 2V)

B

C

↑

V →

(A) PV (B) 2 PV

(C)

1

2 PV (D) Zero

03. 1 gm water at 100°C becomes

1671 c.c. steam at 100°C and at

1 atmosphere pressure when

540 cal heat is supplied. The

external work done is nearly—

(A) 2268 J (B) Zero

(C) 169 J (D) 2100 J

04. If 10 moles of oxygen gas is

heated at constant volume from

20°C to 40°C. The change ininternal energy of the gas is—

(A) 1400 cal (B) 1000 cal

(C) 400 cal (D) 1000 kilo cal

05. A bullet moving with a uniform

velocity v stops suddenly after

hitting the target and the whole

mass melts. If the mass of the

bullet be m , specific heat S, initial

temperature 25°C, melting point

475°C and latent heat L. Then—

(A) m L = m S (475 – 25) + 1/2

mv 2/J

(B) m S (475 – 25) + m L = 1/2

mv 2/J

(C) m S (475 – 25) + m L = 2J/ mv 2

(D) m S (475 – 25) = m L + 2J/ mv 2

06. A waterfall is 84 m high. Assum-

ing that half of the kinetic energy

of the falling water gets converted

into heat, the rise in temperature

of water is—

(A) 0·098°C (B) 0·98°C

(C) 9·8°C (D) 0·0098°C

07. 1 gm coal gives 2 kilo cal of heateffectively on burning. The coal

costs 14 paise per kg. The cost to

produce 1 kWh electrical energy

is—

(A) 60 paise (B) 6 paise

(C) 1 paisa (D) 100 paise

08. During the adiabatic expansion

of 2 mole of a gas, the internal

energy of the gas is found to

decrease by 2 joule. The work

done during the process on thegas will be equal to—

(A) 1 joule (B) – 1 joule

(C) 2 joule (D) – 2 joule

09. If the amount of heat given to a

system be 35 joule and the

amount of work done by the

system be – 15 joule, then the

change in the internal energy of

the system is—

(A) – 50 joule (B) 20 joule

(C) 30 joule (D) 50 joule

10. A gas expands from 50 litre

volume to 250 litre at 105 N/m2

atm pressure. Calculate the workdone by the gas—

(A) 2 × 107 J (B) 2 × 104 J

(C) 2 cal (D) Zero

11. The kinetic energy of gas mole-

cules will be half the value at

room temperature (27°C), when

temperature becomes—

(A) 327°C (B) 123°C

(C) – 123°C (D) – 27°C

12. The average energy associated

per molecule for a gas whose

molecules have n degrees of

freedom is—

(A)1

2 nk T (B)

nk T

N

(C)1

2 nk T

N(D)

3

2 k T

13. The efficiency of Carnot engine

working between the source atabsolute temperature T1 and sink

at absolute temperature T2 is—

(A) T2T1

(B) 1 – T1T2

(C) 1 –T2T1

(D)T1T2

– 1


19/110C.S.V. / July / 2009 / 550

14. A Carnot engine is working bet-

ween temperature 527°C and

27°C. Its efficiency will be—

(A) 62·5% (B) 37·5%

(C) 50% (D) 25%

15. Theoretically the efficiency of

Carnot engine is 100%, when the

temperature of the sink is—

(A) 0°C (B) 0K

(C) 0°F (D) 0°R

16. By opening the door of a refrige-

rator which is inside the room—

(A) The room can be cooled to a

certain degree

(B) Room can be cooled to the

temperature of the refrige-

rator

(C) Ultimately room is slightlywarmed

(D) The room is neither cooled

or warmed

17. One gram of ice, when melts,

requires 336 joule of heat. The

increase in internal energy will

be—

(A) Equal to 336 J

(B) More than 336 J

(C) Less than 336 J

(D) Equal to zero

18. A perfect gas is contained in a

cylinder kept in vacuum. The

cylinder suddenly bursts. The

temperature of the gas—

(A) Becomes 0 K

(B) Remains unchanged

(C) Becomes more than before

(D) Becomes less than before

19. A perfect gas is heated in anisothermal way. The heat will be

used to—

(A) Do external work

(B) Increase temperature

(C) Increase internal energy

(D) Decrease internal energy

20. In which process will the change

in internal energy be equal to the

work done ?

(A) Isothermal process

(B) Adiabatic process

(C) Isochoric process

(D) Isobaric process

21. Work done in an adiabatic

change for a perfect gas depends

only on—

(A) Change in volume

(B) Change in pressure

(C) Change in temperature

(D) Change in heat content

22. An ideal Carnot engine whoseefficiency is 40% receives heat at

500 K, If the efficiency is to be

50%, the intake temperature for

the same exhaust temperature

is—

(A) 900 K (B) 800 K

(C) 700 K (D) 600 K

23. A given mass of a gas expands

from the state A to the state B by

three paths 1, 2 and 3 as shown

in figure. If W1, W2 and W3respectively be the work done by

the gas along the three paths

then—

AP

3

2

1

V

B

→

→

(A) W1 > W2 > W3(B) W1 < W2 < W3

(C) W1 = W2 = W3

(D) W1 < W2 and W1 > W3

24. The specific heat of hydrogen

gas at constant pressure is

Cp = 3·4 × 103 calorie/kg°C and

at constant volume is Cv = 2·4 ×

103 calorie/kg°C. If one kilogram

hydrogen gas is heated from

10°C to 20°C at constant pres-

sure the external work done on

the gas to maintain it at constant

pressure is—

(A) 103 calorie

(B) 5 × 103 calorie

(C) 104 calorie

(D) 105 calorie

25. The differential form of first law of

thermodynamics is—

(A) d Q = d U – d W

(B) d Q + d U = d W

(C) d Q = d U + d W

(D) d Q + d U + d W = 0

26. Which of the following is not

thermodynamic function ?

(A) Enthalpy

(B) Work done

(C) Gibbs energy

(D) Internal energy

27. An ideal heat engine exhausting

heat at 77°C is to have a 30%efficiency. It must take heat at—

(A) 127°C (B) 327°C

(C) 227°C (D) 673°C

28. A Carnot’s engine first works

between 200°C and 0°C and

then between 0°C and – 200°C.

The ratio of its efficiencies in

these two cases is—

(A) 1·000 (B) 0·722

(C) 0·577 (D) 0·340

29. In a Carnot’s engine, the tempe-

rature of the source is found to

be 727°C and that of sink to be

27°C. The approximate effi-

ciency of the engine is—

(A) 0·7 (B) 0·9

(C) 0·4 (D) 1

30. A Carnot’s engine takes 300

calorie of heat at 500 K and

rejects 150 calorie of heat to the

sink. The temperature of the sink

is—

(A) 1000 K (B) 750 K

(C) 250 K (D) 125 K

31. A lead bullet of mass 21 gm hits

a hard target with a velocity 200

m/s. The total amount of heat

produced would be—

(A) 100 cal (B) 1000 cal

(C) 500 cal (D) 2000 cal

32. A gas is compressed at a cons-tant pressure of 50 N/m2 from

volume of 10 m3 to a volume of

4 m3. Energy of 100 J is then

added to the gas by heating. Its

internal energy is—

(A) Increased by 400 J

(B) Increased by 200 J

(C) Increased by 100 J

(D) Decreased by 200 J

33. In a thermodynamic process the

pressure of a fixed mass of thegas is changed in such a manner

that the gas releases 20 J of heat

and 8 J of work is done on the


20/110C.S.V. / July / 2009 / 551

gas. If the initial energy of the

gas was 30 J, then the final

internal energy will be—

(A) 2 J (B) 42 J

(C) 18 J (D) 58 J

34. Find the change in internal

energy of the system when a

system absorbs 2 kilo-calorie ofheat and at the same time does

500 joule of work—

(A) 7900 J (B) 8200 J

(C) 5600 J (D) 6400 J

35. The efficiency of a Carnot’s

engine working between steam

point and ice point is—

(A) 16·8% (B) 26·81%

(C) 36·8% (D) 46·8%

36. The coefficient of performance ofa refrigerator working between

– 10°C and 20°C is—

(A) 8·77 (B) 6·77

(C) 7·77 (D) 10·77

37. From what minimum height a

block of ice has to be dropped in

order that it may melt completely

on hitting the ground ? (L is the

latent heat of ice and J is joules

constant)

(A) mgh (B) mgh /J

(C) JL/ g (D) J/Lg

38. A Carnot’s engine takes in 3000

k cal of heat from a reservoir at

627°C and it gives it to a sink at

27°C. The work done by the

engine is—

(A) 4·2 × 106 J

(B) 8·4 × 106 J

(C) 16·8 × 10

6

J(D) Zero

39. An ideal heat engine workingbetween temperature T1 and T2

has efficiency η. If both the tem-

peratures are raised by 100 K

each, the new efficiency of the

heat engine will be—

(A) Equal to η

(B) Greater than η

(C) Less than η(D) Greater or less than η de-

pending upon the nature of

working substances

40. A Carnot’s engine operates with

a source at 500 k and sink at

375 k. The engine takes in 600 k

cal of heat in one cycle. The heat

rejected to the sink per cycle is—

(A) 250 k cal (B) 350 k cal

(C) 480 k cal (D) 550 k cal

41. The P-V diagram shows the

thermodynamic behaviour of an

ideal gas. The work done in the

complete cycle ABCDA is—

1

A B

D C

2

2

4

6

8

10

12

3 4 5 6

P ( 1 0 5 N

/ m 2 )

V (litre)

(A) 6000 J

(B) 5000 J, done by the gas

(C) 5000 J, done on the gas

(D) 6 × 106 J done by the gas

42. The figure shows the changes in

a thermodynamical system as it

goes from A → B → C → A. It is

given thatUA = 0, UB = 30 J and heat given

to the system in the process

B → C is 50 J.

1

BA

ED

C

0

30

60

90

2 3

P r e s s u r e P ( N / m 2 )

Volume V (m3)

Which of the following inference

from it is not correct ?

(A) Internal energy of the system

in state C is 80 J

(B) Heat given to the system in

process A → B is 90 J

(C) Heat taken out from the

system in process C → A is – 200 J

(D) Work done in complete cycle

ABCA is 120 J

ANSWERS WITH HINTS


21/110C.S.V. / July / 2009 / 554

Thermal RadiationHeat travelling by the process of radiation is called

radiant heat or thermal radiation. When heat is propagated

by radiation, no material medium is necessary for the

transmission and if there is any medium it is not necessary

that it should first get itself heated (as in case of conduc-

tion and convection) before it could assist the propagation

of thermal radiation. Thermal radiation has following pro-

perties :

(1) Thermal radiation travels through empty space with

the velocity of light.

(2) Thermal radiation exhibits properties of light. The only

difference is that its average wavelength is greaterthan that of visible light. Therefore, the thermal radia-

tion is called infrared radiation.

At a Glance

Some DefinitionsTotal energy density—The total energy density of radia-

tions at any point is the total radiant energy per unit volumearound that point for all the wavelengths taken together. It isgenerally expressed by u , its unit is joule m – 3.

Spectral energy density—The spectral energy densityfor a particular wavelength is the energy per unit volume perunit range of wavelength. This is denoted by u λ.

Total emissive power—The total emissive power of abody is the radiant energy emitted per unit time per unit surfacearea of the body for all wavelengths taken together. It isdenoted by E.

Spectral emissive power—The spectral emissive powerof a body at a particular wavelength is the radiant energyemitted per unit time per unit surface area of the body within aunit wavelength range. It is denoted by Eλ.

Absorptive power—The absorptive power of a body at aparticular temperature and for a particular wavelength isdefined as the ratio of the radiant energy absorbed per unitsurface area per unit time to the total energy incident on thesame area of the body in unit time within a unit wavelengthrange. It is denoted by a λ.

From these definitions

u = ∫ ∞

0

u λd λ and E = ∫ ∞

0

Eλd λ

Black Body and Black Body Radiation

A perfectly black-body is one which absorbs all the

heat radiations, of whatever wavelength, incident on it. It

neither reflects nor transmits any of the incident radiation

and, therefore, appears black whatever be the colour of

incident radiation.

Let a black-body be placed in an isothermal enclosure.

The body will emit the full radiation of the enclosure after it

is in thermal equilibrium with the enclosure. These radia-tions are independent of the nature of the substance.

Clearly the radiation from an isothermal enclosure is

identical with that from a black-body at the same tem-

perature. Hence, the heat-radiations in an isothermalenclosure are termed as black-body radiation.

In practice no substance possesses strictly the pro-

perties of a black-body. Lamp-black and the platinum black

are the nearest approach to a black-body. However, the

bodies showing close approximation to a perfectly black-

body have been constructed e.g., Ferry’s black body and

Wien’s black-body.

Kirchhoff’s Law

It states that the ratio of the emissive power to the

absorptive power for a given wavelength at a given tem-

perature is the same for all bodies and is equal to theemissive power of a perfectly black-body at that tempera-

ture. Expressed in symbols, it is

e λa λ

= Eλ

Pressure of Radiation

The radiation possesses the properties of light. Like

light it exerts a small but definite pressure on the surface

on which it is incident.

For normal incidence on the surface, the pressure of

radiation is equal to the energy density, i.e.

p = u =I

c

The density of radiation u is simply the amount of

radiation contained in unit volume and is, therefore, equal

to I/ c where I is the intensity of radiation and c is the

velocity of light.

For diffuse radiation

Pressure =1

3 × Energy density

p =1

3 u =

1

3

I

c

Stefan-Boltzmann LawStefan’s law states that the rate of emission of radiant

energy by unit area of a perfectly black-body is directly

proportional to the fourth power of its absolute tempera-

ture. In symbols

E = σT4

where σ is a constant and is called Stefan’s constant. The

unit of σ is Jm – 2 s – 1K – 4 or Wm – 2 K – 4.

The law in the above form refers to the emission only

and not to the net loss of heat by the body after exchange

with the surroundings. The law can be extended to repre-

sent the net loss of heat and may be enunciated asfollows :

A black-body at absolute temperature T surroundedby another black-body at absolute temperature T0 not only


22/110C.S.V. / July / 2009 / 555

loses an amount of energy σT4 but also gains σT04, thus,

the amount of heat lost by the former per unit time is given

by

E = σ (T4 – T04)

The law is known as Stefan-Boltzmann’s law as

Boltzmann deduced it thermodynamically in 1884 and

showed that the law strictly applies to emission from a

perfectly black-body.

Newton’s Law of Cooling

It states that the rate of loss of heat from a body is

proportional to the mean excess temperature of the body

over the temperature of its surroundings provided that the

temperature excess is small, i.e.

Rate of loss heat from the body ∝ Mean temp. difference

Consider a hot body of mass m , specific heat s and at

temperature θ1. Its temperature falls from θ1 to θ2 in a

time-interval t , when the temperature of surroundings is θ0.

Then

Rate of loss of heat from the body = ms (θ1 – θ2)

t

During cooling, average temperature of the body

θ =θ1 + θ2

2, so the average temperature-difference bet-

ween the body and its surroundings is (θ – θ0). According

to Newton’s law, we have

ms (θ1 – θ2)

t ∝ (θ – θ0)

or ms

(θ1 – θ2)

t = k (θ – θ0)

where k is a proportionality constant.

Derivation of Newton’s Law from Stefan’s Law

Consider that a hot body at temperature T is

surrounded by a medium at temperature T′. According to

Stefan’s law the net rate of loss of heat by the body is

e σ (T4 – T′4)

where e is emissive power of the body.

Further suppose that temperature T of the body is

only slightly higher than the temperature T′ of its

surroundings both at

T – T′ = ∆T

or T = (T ′ + ∆T)

so the rate of loss of heat is

e σ [(T′ + ∆T)4 – T′4] = e σ [ ]T′4 ( )1 + ∆TT′4 – T′4

Since ∆T is very small compared to T ′, hence

( )1 + ∆TT′4

= 1 + 4∆T

T′

by the binomial theorem neglecting higher powers of ∆T/T′

∴ Rate of loss of heat

= e σ [ ]T′4 ( )1 + 4∆TT′ – T′4

= e σ (4T′3) ∆T

∝ ∆T

The rate of cooling of a body depends upon the

energy radiated by it. Hence, rate of cooling of a body is

proportional to the mean temperature difference between

the body and its surroundings.

Thus, Newton’s law is only a special case of Stefan’s

law for small temperature differences.

Spectral Distribution of Black Body Radiation

A perfectly black-body is a full radiator, i.e., it emits

radiation of all possible wavelengths. Lummer and

Pringsheim studied the spectral

distribution of energy (i.e., energy

distribution among different wave-

lengths) in the radiation of a black

body at different temperatures.

Spectral distribution curves so

obtained are shown in the figure.

T2

T3

T1

Eλ

λ

T3 > T2 > T1

These curves have the same general shape for all

temperatures and give the following information regarding

the characteristics of black body radiation :

(i) At a given temperature T, with increase in wave-

length λ, the energy Eλ first increases, reaches a maxi-

mum and then decreases. It means that for a given tem-

perature, the radiant energy emitted by a black-body is

maximum for a particular wavelength.

(ii) As the temperature increases, the peak of the

curve shifts towards shorter wavelength side, i.e., the

maximum value of Eλ is obtained at smaller value of λ.Wien in 1896 established the following relation bet-

ween temperature T and wavelength λm corresponding to

maximum emission.

λm T = constant = b (suppose)

This is called Wien’s displacement law.

The constant b is called Wien’s constant and has the value

0·2896 cm K or 0·2896 × 10 – 2 mK

(iii) As the temperature rises, the area enclosed by the

curve goes on increasing. This area represents total energy

E (for all wavelengths) emitted by a black body at that

temperature. The areas enclosed by different curves, when

measured, are found to be proportional to the fourth power

of corresponding absolute temperatures. Thus

E ∝ T4

This is Stefan’s law.

(iv) Wien also proved that at a temperature T, themaximum emitted energy (Eλ)m corresponding to wave-

length λ is proportional to the fifth power of that tempera-

ture (T5). Thus

(Eλ)m ∝ T5

This shows that on raising temperature, the maximum

energy (Eλ)m emitted corresponding to wavelength λ

increases very rapidly.


23/110C.S.V. / July / 2009 / 556

SOME TYPICAL SOLVED EXAMPLES

Example 1. A black-body radiates heat energy at a

rate 1·45 ×××× 103 Js – 1 m – 2 at a temperature of 127°°°°C. At

what temperature will it radiate heat at the rate of

1·17 ×××× 105 Js – 1 m – 2 ?

Solution :

Example 2. At what temperature a perfectly black-

body of area 104 m2 would radiate energy at the rate of

90·72 Wm – 2 ? (Given : σσσσ = 5·67 ×××× 10 – 8 Wm – 2 K – 4)

Solution

Example 3. Estimate the temperatures at which a

body would appear red and blue. The corresponding

wavelengths of maximum emission are λλλλm = 7500°°°°A

and 5000 °°°°A respectively.

(Given : Wien’s constant b = 0·3 cm-K)

Solution

OBJECTIVE QUESTIONS

1. A hot body will radiate heat most

rapidly its surface is—

(A) White and polished

(B) White and rough

(C) Black and polished

(D) Black and rough2. The best black-body is—

(A) A metal coated with a black

dye

(B) A lamp of charcoal heated to

a high temperature

(C) A glass surface coated with

coal-tar

(D) A hollow enclosure black-

ened inside and having a

small hole

3. The colour of a star indicatesits—

(A) Weight (B) Size

(C) Distance (D) Temperature

4. Three stars A, B and C appear

green, red and blue respectively.

The star having minimum tem-

perature is—

(A) A

(B) B

(C) C

(D) All are at the same tempera-ture

5. Two sphere A and B of the same

material having radii 1m and 4m

are at temperatures 4000 K and

2000 K. Which will emit more

energy per second ?

(A) A

(B) B

(C) Equal for both(D) None of these

6. Choose the wrong statement—

(A) Black surface is a better

absorber of radiation than a

white one

(B) Rou

52313794 Competition Science Vision July09

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