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8/15/2019 52313794 Competition Science Vision July09
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In This Issuen This ssueIn This Issue
Editor
MAHENDRA JAIN
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Edited, printed and published by MahendraJain for M/s. Pratiyogita Darpan, 2/11A,Swadeshi Bima Nagar, AGRA–2 andprinted by him at Pratiyogita DarpanPrinting Unit, 5 & 6, Bye pass Road, Agra.Phone : 4053333, 2531101, 2530966
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July 2009
Year—12 Issue—137
C.S.V. / July/ 2009 / 533
Regulars
Editorial 535
Science and Technology 537
Latest General Knowledge 539
Science Tips 543
Physics
Thermal Physics-II 545
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Correct Solution and Prize Winners of CSV Quiz No. 131 655
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C.S.V. / July / 2009 / 534
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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 edu- cational 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 deter- mination 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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T T H O U G H TS FOR THE M O N T H
❥ 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.
❥ Be neither saint, nor sophist-led, but a man.
❥ Praise is a debt, flattery is a present.
❥ Success is never ending, failure is never final.
❥ 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
Rs. 40/-
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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.
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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
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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 )
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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)
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(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
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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.
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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
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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
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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
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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
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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.
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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