xtra edge february 2010 - career point for iit-jee 1 february 2010 dear students, success is the...

XtraEdge for IIT-JEE 1 FEBRUARY 2010

Dear Students,

Success is the result of inspiration, aspiration, desperation and perspiration. Your luck generally changes as a result of the above inputs. There is always a fierce struggle on the road leading to your desired goal. No triumph comes without effort and changing obstacles into tools for success. For achieving your desired goal you must have a fierce and a strong desire to succeed.

Success begins in the mind. Whatever the mind of man can believe and conceive it can achieve. You should make a commitment to yourself to succeed. Do not believe that success is the result of either an accident or a fluke. Nothing is ever achieved without working for it with integrity, wisdom, commitment and a success oriented attitude. You have to accept responsibility for your decisions which will ultimately determine your destiny. The best way to succeed is to accept and learn from your mistakes. It is a fact that luck only follows hard work.

Master the fundamentals of whatever you are required to do keep on developing your character confidence, values, beliefs and personality by always keeping in view the examples and lives of successful people. You will notice that all successful people combine in themselves integrity, unselfishness, patience, understanding, conviction, courage, loyalty and self-esteem. They have both ability and character. Hence, they are successful in whatever they undertake.

No success is possible unless you believe that you can succeed. Positive faith in yourself is both vital and crucial for success. This attitude will help your competence reach visibly successful levels of performance and prepare you for hard work.

Look at the task and ask yourself whether your output can be further improved. There is little room at the top. The top is always rarefied and limited space. Only one person can be there at a time. Do not be the one who gets left out for want of persistence, determination and commitment.

Forever presenting positive ideas to your success. Yours truly

Pramod Maheshwari, B.Tech., IIT Delhi

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Owned & Published by Pramod Maheshwari, 112, Shakti Nagar, Dadabari, Kota & Printed by Naval Maheshwari, Published & Printed at 112, Shakti Nagar, Dadabari, Kota.

Editor : Pramod Maheshwari

Teachers open the door. You enter by yourself.

Volume - 5 Issue - 8

February, 2010 (Monthly Magazine)

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Volume-5 Issue-8 February, 2010 (Monthly Magazine)

NEXT MONTHS ATTRACTIONS

Key Concepts & Problem Solving strategy for IIT-JEE.

Know IIT-JEE With 15 Best Questions of IIT-JEE

Challenging Problems in Physics, Chemistry & Maths

Much more IIT-JEE News.

IIT-JEE Mock Test Paper with Solution

AIEEE & BIT-SAT Mock Test Paper with Solution

Success Tips for the Month

• If you haven nothing else to do, look about you and see if there is not something close at hand that you can improve !

• He has achieved success who has worked well, laughed often, and loved much.

• You always pass failure on the way to success.

• A journey of a thousand miles begins with a single step.

• Your success will be largely determined by your ability to concentrates single-mindedly on one thing at a time.

• Success is a journey, not a destination.

• Success comes in "Cans". Failure comes in "Can't".

• Success seems to be largely a matter of hanging on after others have let go.

CONTENTS

INDEX PAGE

NEWS ARTICLE 4 IIT Madras team wins New York competition Bio-energy centre launched at IIT-Kharagpur IIT-Kharagpur best technology school in India-survey

IITian ON THE PATH OF SUCCESS 8 Dr. Pradip K. Dutta

KNOW IIT-JEE 10 Previous IIT-JEE Question

XTRAEDGE TEST SERIES 57

Class XII – IIT-JEE 2010 Paper Class XII – IIT-JEE 2011 Paper Mock Test CBSE Pattern Paper-3 [Class # XII] Mock Test CBSE Pattern Paper-2 & 3 (Solution) [Class # XII]

Regulars ..........

DYNAMIC PHYSICS 15

8-Challenging Problems [Set# 10] Students’ Forum Physics Fundamentals Prism & Wave Nature of Light Waves & Doppler Effect CATALYST CHEMISTRY 34

Key Concept Carbonyl Compounds Co-ordination Compound & Metallurgy Understanding: Physical Chemistry

DICEY MATHS 42

Mathematical Challenges Students’ Forum Key Concept Integration Trigonometrical Equation

Study Time........

Test Time ..........


IIT Madras team wins New York competition New York: An entrepreneur team from the Indian Institute of Technology, Madras (IIT-M) has won the NYC Next Idea 2009-2010, an inaugural global business plan competition launched by New York City last...

Organs may soon be grown like nails at IIT Delhi New Delhi: "Science is a cemetery of dead ideas," said an eminent scientist but engineers at the Indian Institute of Technology, Delhi (IIT-D) believe in re-creating those dead ideas and harvest new. ...

India developing e-dog to sniff out explosives Thiruvananthapuram:

Indian scientists are developing an electronic device that will sniff out explosives like RDX, which remain undetected by existing security equipments."A prototype of the e-device.

India develops affordable nano sensors to detect heart attack Thiruvananthapuram: A team of Indian scientists and engineers has developed affordable sensors using nano materials to detect a heart attack quickly.

Bio-energy centre launched at IIT-Kharagpur

West Bengal: The country's first bio-energy centre was launched at the Indian Institute of Technology-Kharagpur (IIT-K) on Tuesday for undertaking research, teaching and technological implementati.

When there are no teachers to teach at IITs Mumbai: Luring requires no effort if the perks are high and in the field of employment what does one need if he is placed in one of the prestigious centrally funded technical institutions (collectivel).

'Share and Share Alike'- IIT Bombay's motto Mumbai: 'Share and share alike' is what the Indian Institute of Technology, Bombay (IIT-B) is spreading across the message to reach out to the smaller engineering colleges to share their exper.

DRDO, IIT-D joins hands for weather forecast system Chandigarh: In order to develop an indigenous capability and methodology for long term forecast of weather, the Defence Research and Development Organization (DRDO) in its first kind of venture.

IIT–Kharagpur best technology school in India – survey New Delhi: The premier Indian Institute of Technology (IIT)

Kharagpur is the best technology school in the country followed by IIT-Delhi, revealed a new survey released recently.

Crack the IIT code, it's too easy Kolkata: From next year, the IIT entrance test is likely to get simpler.

Concerned over the immense stress that IIT-JEE puts on thousands of students, the Union HRD ministry has set up a high-level panel to modify the test pattern.

Of the 1.5 lakh aspirants every year, only 3,500 make it to the seven IITs. To tone down the gruelling test, the ministry has formed a committee with teachers from IITs and representatives of the two +2 level national boards - CBSE and ICSE.

According to sources, the first change may be to limit questions to the +2 syllabus. "The HRD ministry feels many of the IIT-JEE questions are based on topics that are not taught at the +2 stage, and are, in fact, of a far advanced standard. This forces candidates to start preparing at least three years in advance - from Class IX itself. They overload themselves and this leads to depression, which sometimes leads to suicides," IIT-Kharagpur director SK Dubey told TOI.


Dubey and IIT-JEE chairman VK Tewari are on the committee, which is expected to submit its recommendations by July. The changes could be introduced from IIT-JEE, 2006.

The government believes the tougher syllabus forces students to neglect their board examinations for IIT-JEE. They also end up spending a lot on coaching classes that claim to be tailor-made for the entrance test, said Dubey.

The committee, therefore, is singling out topics that are... ...not taught in the +2 stage anywhere in the country.

These topics will probably be deleted. And every topic that is now included in IIT-JEE will be vetted by the representatives of CBSE and ICSE.

The two-tier exam system includes objective-type questions in the prelims and subjective ones in the finals. The panel will examine whether it should be replaced with a uniform system.

In the finals, candidates have to answer three papers - physics, chemistry and mathematics - through a gruelling six hours on a single day.

The ministry feels it is too taxing and has asked the committee to work out a better "fatigue and rest cycle."

The committee will also compare IIT-JEE question papers with those of the All India Engineering Entrance Examination (AIEEE) and other engineering entrance examinations conducted by various states to check if there is too wide a gap in their standards.

IITs move to hike fee, adopt IIM fee strategy New Delhi : Taking a cue from the Indian Institutes of Management, the IIT bosses are drawing a cautious plan to gradually equate their fee structure with that of the IIMs.

According to sources the exercise is to make the Indian Institutes Technology self-reliant and to cut dependence on state subsidy, which the IIT dons say, would gradually taper off in the coming years.

A panel set up by the IIT Council — the apex decision making body — headed by atomic energy chief Anil Kakodkar has been asked to draft the roadmap for gradual fee hikes, the sources said.

Drafting the fee hike roadmap for the IITs is one of the components of the mandate of the Kakodkar panel set up at the Council meeting on October 19. The Kakodkar panel has been asked to submit its report in six months.

The IIT Council, which met here on October 19, discussed the fee-hike possibility in view of the government starting a loan scheme with subsidised interest rate to help poor students in higher studies, sources said. The Kakodkar panel will also suggest how the IITs should increase the number of scholarships, fellowships and other financial aid to ensure that deserving but economically weak students do

not suffer from the hike, sources said.

The new fee-hike strategy aims at following the IIM practice of a gradual but regular fee hike supported by an increase in financial assistance for those students who cannot afford the new fee structure.

“The strategy of gradual fee hikes will allow us, for the first time, an opportunity to hike fees commensurate with rising costs,” an IIT director said.

The IITs had a fixed tuition fee of Rs 25,000 per annum for undergraduate and post-graduate science students for 10 years before the fees were doubled last year — to Rs 50,000 a year. But even with the new fee structure, the IITs earn only Rs 2 lakh for four years of undergraduate teaching, or Rs 1 lakh for two years of the masters in science programme from each student.

The top IIMs — which typically raise their fees each year — in contrast earn around 10 times as much through tuition fees from each student over comparable course lengths.

IIM Ahmedabad, for instance raised the fees for its two-year postgraduate diploma in management to Rs 12.5 lakh this year, from Rs 11.5 lakh last year.

The IIMs in Bangalore and Calcutta charge Rs 9.5 lakh and Rs 9 lakh for their two year postgraduate diploma courses respectively.

Six IITs figure among top ten technology institutes PTI, 14 January 2010, 12:19pm IST

KOLKATA: As many as six among the top 10 technology institutes in the country are from the


prestigious Indian Institute of Technology, a survey has revealed.

The top slot in the pecking order is occupied by the IIT- Kharagpur followed by IIT-Delhi, IIT-Madras, IIT-Kanpur, IIT- Roorkee and IIT-Guwahati, according to the 5th IDC-Dataquest T-School 2009 Survey of 111 engineering colleges across the country.

IIT-Bombay did not participate in the survey, global market intelligence provider IDC said.

The seventh best technology institute of the country is IIIT, Hyderabad followed by BITS Pilani, the survey report said.

IIIT, Hyderabad is the youngest T-School, set up in 1998, in the top 10 list.

Close behind BITS Pilani is the National Institute of Technology (NIT), Surathkal and the Institute of Technology (IT) of the Banaras Hindu University (BHU), Varanasi, it said.

IIT Bombay exhibits masterpieces From collapsible bamboo bar stools to workstations for people with cerebral palsy, the Degree Design Show by IIT-Bombays Industrial Design Centre has it all. A number of designs have been sanctioned and designed by corporates.

The youngsters behind the machines have designed a number of socially relevant products for the less privileged.

Sarabjit Singh Kalsis workstation for cerebral palsy patients has a circular grip at the base of the rotating chair that keeps it in place, a C-shaped table that helps the patient maintain the right posture and an abductor in the chair that separates the persons

legs, as crossing the legs is harmful to those with cerebral palsy.

IIT Kharagpur ties up with Taiwan school IIT Kharagpur ties up with Taiwan school The Indian Institute of Technology at Kharagpur (IIT-KGP) has is in the process of signing a memorandum of understanding with National Chiao Tung University (NCTU) of Taiwan which will facilitate research collaboration, joint research, joint mentoring for students and student and faculty exchange especially in the field of chip designing and fabrication. While the Indian market has facilities and trained manpower for chip designing, those for chip fabrication are not available. Although IIT-KGP has a VLSI (very large scale integration) designing laboratory, we are still trying to set up a chip laboratory.

Ajai Chowdhry to chair IIT Hyderabad's BoG New Delhi, India: Hardware, services and ICT system Integration company HCL Infosystems Ltd today announced that Ajai Chowdhry, the founder chairman and CEO of the company has been nominated by the Honorable President of India to be the Chairman of the Board of Governors, IIT Hyderabad.

Accepting his nomination, Ajai Chowdhry said it is a matter of immense pride and privilege to be associated with this esteemed knowledge body.

"I look forward to furthering IIT Hyderabad's objectives of setting new standards in engineering practice in India contribute actively to growth of India in the decades to come," he added.

Commenting on the appointment Prof. U. B. Desai, Director, IIT Hyderabad expressed confidence that under his Chairmanship IIT Hyderabad would become a world-class institute.

"We are very privileged and enthused that Ajai Chowdhry has been appointed as chairman BoG for IIT Hyderabad in its formative years. His vision will add new and challenging dimensions to IIT."

An engineer by training, Chowdhry is one of the six founder members of HCL, according to a press release. He is currently the Chairman of Confederation of Indian Industry's (CII) National Committee on Technology and innovation, besides being a part of the IT Hardware Task Force set up by the Prime Minister of India. Chowdhry also has chaired CII's National Committee for IT, ITES & E-Commerce, where he actively encouraged the deployment of IT in Indian SMEs to increase their productivity and to make them globally competitive.

Number of women at IITs triples in 5 years In the last five years, women’s presence in certain IITs has gone as high as 10 percent. In 2005,five per cent of the total number of those who cleared the JEE were women(381 out of 6,433). In 2009, this number has increased to 10 per cent (1,048 of 10,035).

“The number of applications from women has also increased and courses at IIT are no longer viewed as only-for-men. Even women are interested in technical fields,” said Anil Kumar, IIT-Bombay’s JEE chairperson.


IBM collaborates with engineering colleges to set up COE IBM India has collaborated with six engineering colleges of Tamil Nadu for setting up Centre of Excellence (COE) to promote high quality education by providing state-of the-art technologies in colleges with the objective of nurturing highly skilled computer professionals.The colleges will provide infrastructure and high end systems while IBM will extend its entire range of software suite free of charge. This will enable students to learn new skill sets on IBM software products including DB2, WebSphere, Lotus, Rational, and Tivoli.

HRD suggests Engineers Bill to standardize engineers In order to standardize the Indian engineers globally, the HRD ministry is planning to introduce the Engineers Bill, 2009, that would require obligatory certification of professional engineers. The bill aims to streamline the quality of engineers in the country and plans to set up the ICE, which will maintain a national and international record of professional engineers and associate professional engineers and will standardize the engineering profession.

To Set up a Campus in Quatar India’s premier technological institutes, Indian Institutes of Technology (IITs) has reportedly been given a nod by the central government to set up their first offshore campus in Qatar.

The new institute to be set up in Quatar will be called the International Institute of Technology and this will be set up all the IITs in a concerted effort. The entire process of setting up a new institute in Quatar will be co-ordinated by IIT Council, the top decision-making body for all IITs. Though the HRD ministry has initially opposed to IITs or Indian Institute of Managements venturing abroad, arguing that “elite” educational institutions must stay focused on India alone, the new dispensation in the ministry is keen that Brand India makes a mark abroad and for this reason IITs has got the go-ahead signal from the Indian Human Resource Development Ministry.

IIT-Qatar will be set up in Qatar Foundation’s Education City, which already has branch campuses of six noted universities.

Top IT companies continue to flock IITs With the economic recession taking backstage, top IT companies are quite happy to offer huge salaries to the IIT graduates . Training and Placement head at IIT Kharagpur, Suneel Srivastava, said that the highest number of offers (13) and the highest salary package of Rs 22 lakh was offered by Barclays Singapore. The institute also had about 27 pre-placement offers made to the students, about 10 of which were made by Reliance Industries (RIL).

IITs to reduce foreign student fee at PG level Under the chairmanship of HRD minister Kapil Sibal, the first IIT Council is eyeing for more foreign

admissions at the PG level by framing a policy that includes introducing scholarships for studying at IIT and fee reduction for PG students in the IITs.

Permitting IITs to create extra seats for foreign PG students to ensure that youth from other countries take part in R&D in a big way is also being considered by the IIT council.

IIT Kanpur to hold B-Plan Competition in Techkriti IIT Kanpur is going to organize its annual festival Techkriti on 11-14 February 2010. One of the flagship events of Techkriti will be ‘Ideas 10’, the international business plan competition. The event will witness participation from leading MBA schools, such as IIMs and XLRI as well as IITs and many international institutes like Stanford University, Cornell University, University of Purdue and the National University of Singapore. Cut throat competition from the best in the world will give participants tremendous exposure and will be an excellent test of their business acumen. There are high chances for the participants to start a company based on the B-plan.

IDEAS is the platform where the seeds of future business tycoons will be laid down. Ideas will have best prizes for Bio Business plan Competition, Web Business idea, Clean Energy Solutions and Social Business Plan. This year IDEAS has American Embassy as its co-sponsor and NEN, VC Hunt, Rajeev Kumar Foundation as its associates.


Dr. Pradip K. Dutta is the Corporate VP & MD of Synopsys (India) Private Limited, a wholly owned subsidiary of Synopsys Inc., a world leader in Electronic Design Automation (EDA) software. His primary focus is to position Synopsys as a leader in the Indian semiconductor eco-system and help foster its growth through partnership with government, academia and industry. Dr. Dutta has been heading the India operations since 2000, overseeing the growth from a little over 50 employees operating from one small office in Bangalore to more than 600 highly skilled employee base spread across Bangalore, Hyderabad and Noida.

Prior to joining Synopsys, Dr. Dutta started his career in the field of automotive electronics with General Motors in USA and held a variety of positions in engineering and management both in the US and in the Asian-Pacific region.

Dr. Dutta has earned his B.Tech in Electronics Engineering from IIT Kharagpur followed by MS and Ph.D. in Electrical Engineering from the University of Maryland, College Park under a US government fellowship grant from National Institute of Standards and Technology. He sits on the Advisory Board of the Govt. of West Bengal (IT Ministry) and is a member of Executive Committees of several industry associations

Dr. Pradip K. Dutta B.Tech -Electronics Engineering IIT Kharagpur, MS & Ph.D. Corporate Vice President & Managing Director, Synopsys (I) Pvt. Ltd.

Success StoryThis article contains story of a person who get succeed after graduation from different IIT's

Adventure : • Adventure is not outside man; it is within.

• There are two kinds of adventures: those who go truly hoping to find adventure and those who go secretly. hoping they won't.

• Life is either a daring adventure or nothing.

• Some people dream of worthy accomplishments while others stay awake and do them.

• Life is an adventure. The greatest pleasure is doing what people say you cannot do.


PHYSICS

1. A small ball of mass 2 × 10–3 kg having a charge of 1 µC is suspended by a string of length 0.8 m. Another identical ball having the same charge is kept at the point of suspension. Determine the minimum horizontal velocity which should be imparted to the lower ball so that it can make complete revolution. [IIT-2001]

Sol. This is a case of vertical of circular motion. The body undergoing vertical circular motion is moving under the action of three forces as shown

(i) mg (Gravitation pull) (ii) Electrostatic force of repulsion (iii) Tension of the string

PFe

mg

mgcosθ

mgsinθ

V

θTl θ

Vb Reference level for P.E. B

l

lcosθ

For the body to move in circular motion, a centripetal

force is required. Therefore at P

(T + mg cos θ) – Fe =r

mv2 … (i)

Applying conservation of mechanical energy. Total mechanical energy at B

= 21 2

bmV + mg (0) =21 2

bmV

Total mechanical energy at P

=21 mV2 + mg(l + l cos θ)

∴ 21 2

bmV =21 mV2 + mg(l + l cos θ) …(ii)

On using the eq. (i) and (ii) for the condition of just completing a circle we get for eq. (i)

T = 0, θ = 0º

∴ mg – Fe =l

2mv … (iii)

and 21 2

bmv =21 mv2 + mg(2l)

∴ 2bv = v2 + 4gl …(iv)

Putting the value of v from (iii) in (iv) we get

∴ 2bv = gl –

mFe l + 4gl

= 5gl –mFe l

vb = 5 × 10 × 0.8 –8.08.0

1010109 669

×××× −−

×002.08.0

= 2

21

rqKqFeQ

∴ vb = 5.86 m/s.

2. Shown in the figure is a container whose top and bottom diameters are D and d respectively. At the bottom of the container, there is a capillary tube of outer radius b and inner radius a.

DP

h

d

ρ

The volume flow rate in the capillary is Q. If the

capillary is removed the liquid comes out with a velocity of v0. The density of the liquid is given as in Fig. Calculate the coefficient of viscosity η. [IIT-2003]

Sol. When the tube is not there, using Bernoulli's theorem

P + P0 + 212

1ρν + ρgH = 2

021

ρν + P0

⇒ P + ρgH = )(21 2

120 ν−νρ

But according to equation of continuity

KNOW IIT-JEE By Previous Exam Questions


A1v1 = A2v2 or v1 =1

22

AvA

∴ P + ρgH = 21

ρ

−

2

01

220 v

AAv

P + ρgH = 21 2

0vρ

−

2

1

2

AA

1

Here P + ρgH = ∆P According to Poisseuille's equation

Q =lη

∆π8

a)P( 4 ∴ η =

lQ8a)P( 4∆π

∴ η =lQ8

a)gHP( 4ρ+π =lQ8

π ×

−ρ

2

1

220 A

A1v

21 × a4

Where 1

2

AA =

2

2

Db

3. Two moles of an ideal monatomic gas is taken

through a cycle ABCA as shown in the P-T diagram. During the process AB, pressure and temperature of the gas vary such that PT = Constant. If T1 = 300 K, calculate [IIT-2000]

T

P

2P1

P1

T1 2T1

A

CB

(a) the work done on the gas in the process AB and (b) the heat absorbed or released by the gas in each

of the processes. Give answer in terms of the gas constant R.

Sol. (a) Number of moles, n = 2, T1 = 300 K During the process A → B PT = constant or P2V = constant = K (say)

Therefore P =VK .

Therefore, WA → B = ∫∫ =B

A

B

A

V

V

V

V VKdV.P dV

= ]VV[K2 AB −

= ]KVKV[2 AB −

= 2 ]V)VP(V)VP([ AA2ABB

2B −

(K = P2V)

= 2[PBVB – PAVA] = 2[nRTB – nRTA] = 2nR[T1 – 2T1] = (2) (2) R [300 – 600] = – 1200R

Therefore work done on the gas in the process AB is 1200 R.

(b) Heat absorbed/released in different processes. Since the gas is monatomic, therefore

CV =23 R and CP =

25 R and γ =

35 .

Process A – B :

∆U = nCV ∆T = (2)

R

23 (TB – TA)

= (2)

R

23 (300 – 600) = – 900 R

QA → B = WA → B + ∆U = (– 1200R) – (900R) QA → B = – 2100 R (Heat released)

4. A metal bar AB can slide on two parallel thick metallic rails separated by a distance l. A resistance R and an inductance L are connected to the rails as shown in the figure. A long straight wire carrying a constant current I0 is placed in the plane of the rails and perpendicular to them as shown. The bar AB is held at rest at a distance x0 from the long wire. At t = 0, it is made to slide on the rails away from the wire. Answer the following questions. [IIT-2002]

x0 B

l I0

A

R

L

(a) Find a relation among i, dtdi and

dtdφ , where i is

the current in the circuit and φ is the flux of the magnetic field due to the long wire through the circuit.

(b) It is observed that at time t = T, the metal bar AB is at a distance of 2x0 from the long wire and the resistance R caries a current i1. Obtain an expression for the net charge that has flown through resistance R from t = 0 to t = T.


(c) The bar is suddenly stopped at time T. The

current through resistance R is found to be 4i1 at

time 2T. Find the value of RL in terms of the

other given quantities. Sol. As the metal bar AB moves towards the right, the

magnetic flux in the loop ABCD increases in the downward direction. By Lenz's law to oppose this, current will flow in anticlockwise direction as shown in figure.

x

B

I0

A

R

L

i

D

C

V

Applying Kirchoff's loop law is ABCD we get

Einduced – iR – L

dtdi = 0 …(i)

⇒ –dtdφ = iR + L

dtdi

φ

−=dtdEinducedQ

Let AB be at a distance x from the long straight wire at any instant of time t during its motion. The magnetic field at that instant at AB due to long straight current carrying wire is

B =x2I00

πµ

The change in flux through ABCD in time dt is dφ = B (dA) = Bldx Therefore the total flux change when metal bar moves

from a distance x0 to 2x0 is

∆φ = ∫∫ πµ

=π

µ=

0

0

00

0

0

x2

x

x2xe

0000x2

x]x[log

2Idx

x2IdxB ll

=π

µ2I00 l loge 2 … (ii)

The charge flowing through resistance R in time T is

q = ∫∫

−=

T

0induced

T

0dt

dtdiLE

R1idt [from eq. (i)]

= ∫ ∫−T

0

i

0induced

1di

RLdtE

R1

= 1iRL)(

R1

−φ∆

q =

π

µ2log

2I

R1

e00 l

– 1iRL from eq. (ii)

(c) When the metal bar AB is stopped, the rate of change of magnetic flux through ABCD becomes zero.

From (i) iR = – Ldtdi

∫∫ =4/i

0

T2

T

1

idi

RLdt

T = –RL loge

1

1

i4/i

⇒ RL =

2log2T

e

5. In hydrogen-like atom (z = 11), nth line of Lyman

series has wavelength λ. The de-Broglie's wavelength of electron in the level from which it originated is also λ. Find the value of n? [IIT-2006]

Sol. nth line of lyman series means electron jumping from (n + 1)th orbit to Ist orbit. for an electron to move in (n + 1)λ

⇒ λ =)1n(

2+π × r =

)1n(2+π

= [0.529 × 10–10] z

)1n( 2+

⇒ λ1 =

)1n](10529.0[2Z

10 +×π − …(i)

Also we know that when electron jumps from (n + 1)th orbit to Ist orbit.

λ1 = RZ2

+− 22 )1n(

111

= 1.09 × 107 Z2

+− 2)1n(

11

From (i) and (ii)

)1n)(10529.0(2

Z10 +×π −

= 1.09 × 107 Z2

+− 2)1n(

11

on solving, we get n = 24.


CHEMISTRY

6. A graph is plotted between PVm along Y-axis and P along X-axis, where Vm is the molar volume of a real gas. Find the intercept along Y-axis. [IIT-2004]

Sol. The van der Waal equation (for one mole) of a real gas is

+ 2

mVaP (Vm – b) = RT

PVm – Pb + mV

a – 2mV

ab = RT

PVm = RT + Pb – mVa + 2

mVab ....(i)

To calculate the intercept P → 0, hence Vm → ∞ due to which the last two terms on the right side of the equation (i) can be neglected.

∴ PVm = RT + Pb When P = 0, intercept = RT 7. The solubility product of Ag2C2O4 at 25ºC is

1.29 × 10–11 mol3 l–3. A solution of K2C2O4 containing 0.1520 mole in 500 ml water is shaken at 25ºC with excess of Ag2CO3 till the following equilibrium is reached :

Ag2CO3 + K2C2O4 Ag2C2O4 + K2CO3 At equilibrium the solution contains 0.0358 mole of

K2CO3. Assuming the degree of dissociation of K2C2O4 and K2CO3 to be equal, calculate the solubility product of Ag2CO3. [IIT-1991]

Sol. Ag2CO3 + K2C2O4 → Ag2C2O4 + K2CO3 Moles at start Excess 0.1520 0 0 Moles after reaction 0.1520 – 0.0358 0.0358 0.0358 = 0.1162 Molar concentration of K2C2O4 or C2O4

2– left

unreacted = 5.0

1162.0 = 0.2324 moles l–1

[K2CO3] = [CO32–] at equilibrium =

5.00358.0

= 0.07156 moles l–1 Given that Ksp for Ag2C2O4 = 1.29 × 10–11 mol3 l–3 at

25ºC So, [Ag+]2[C2O4

2–] = 1.29 × 10–11

or [Ag+]2 × 0.2324 = 1.29 × 10–11

Hence [Ag+]2 = 2324.029.1 × 10–11

Then Ksp for

Ag2CO3 = [Ag+]2 [CO32–] =

2324.01029.1 11−× × 0.0716

= 3.794 × 10–12 mol3 l–3

8. Hydrogen peroxide acts both as an oxidising and as a reducing agent in alkaline solution towards certain first row transition metal ions. IIIustrate both these properties of H2O2 using chemical equations. [IIT-1998]

Sol. When H2O2 acts as oxidising agent, therefore, following reaction takes place :

H2O2 + 2e → 2OH– While regarding its action as reducing agent, the

following reaction takes place : H2O2 + 2OH– → O2 + 2H2O + 2e Examples of oxidising Character of H2O2 in alkaline

medium 2Cr(OH)3 + 4NaOH + 3H2O2 → 2Na2CrO4 + 8H2O Here Fe3+ (Fe is a first row transition metal) is

reduced to Fe2+. Example of reducing character of H2O2 in alkaline

medium 2K3Fe(CN)6 + 2KOH + H2O2 → 2K4[Fe(CN)6] +

2H2O + O2 Here Cr3+ (Cr is a first row transition metal) is

oxidised to Cr6+ 9. Write the structures of (CH3)3N and (Me3Si)3N. Are

they isostructural ? Justify your answer. [IIT-2005] Sol. (CH3)3N and (Me3Si)3N are not isostructural, the

former is pyramidal while the latter is trigonal planar. Silicon has vacant d orbitals which can accommodate lone pair of electrons from N(back bonding) leading to planar shape.

N

..

CH3CH3

H3C

N

SiMe3SiMe3

SiMe3

10. How is boron obtained from borax ? Give chemical

equations with reaction conditions. Write the structure of B2H6 and its reaction with HCl.

[IIT-2002] Sol. When hot concentrated HCl is added to borax

(Na2B4O7.10H2O) the sparingly soluble H3BO3 is formed which on subsequent heating gives B2O3 which is reduced to boron on heating with Mg, Na or K

Na2B4O7(anhydrous) + 2HCl(hot, conc.) → 2NaCl + H2B4O7 H2B4O7 + 5H2O → 4H3BO3 ↓

2H3BO3 → heatingstrong B2O3 + 3H2O B2O3 + 6K → 2B + 3K2O or


B2O3 + 6Na → 2B + 3Na2O or B2O3 + 3Mg → 2B + 3MgO

Structure of B2H6 B

H

H

B

H

H

H

H

1.19Å

97º

1.37Å

122º

1.77Å

Hydrogen bridge bonding (3C-2e bond)

B2H6 + HCl → B2H5Cl + H2 Normally this reaction takes place in the presence of

Lewis acid (AlCl3).

MATHEMATICS

11. The curve y = ax3 + bx2 + cx + 5, touches the x-axis at P(–2, 0) and cuts the y axis at a point Q, where its gradient is 3. Find a, b, c. [IIT-1994]

Sol. It is given that y = ax3 + bx2 + cx + 5 touches x-axis at P(–2, 0) which implies that x-axis is tangent at (–2, 0) and the curve is also passes through (–2, 0).

The curve cuts y-axis at (0, 5) and gradient at this point is given 3 therefore at (0, 5) slope of the tangent is 3.

Now, dxdy = 3ax2 + 2bx + c

since x-axis is tangent at (–2, 0) therefore

2xdx

dy

−=

= 0

⇒ 0 = 3a(–2)2 + 2b(–2) + c ⇒ 0 = 12a – 4b + c ...(1) again slope of tangent at (0, 5) is 3

⇒ )5,0(dx

dy = 3

⇒ 3 = 3a(0)2 + 2b(0) + c ⇒ 3 = c ...(2) Since, the curve passes through (–2, 0), we get 0 = a(–2)3 + b(–2)2 + c(–2) + 5 0 = –8a + 4b – 2c + 5 ...(3) from (1) and (2), we get 12a – 4b = –3 ...(4) from (3) and (2), we get –8a + 4b = 1 ...(5) adding (4) and (5), we get 4a = –2

⇒ a = –1/2 Putting a = –1/2 in (4), we get 12(–1/2) – 4b = –3 ⇒ –6 – 4b = –3 ⇒ –3 = 4b ⇒ b = –3/4 Hence, a = –1/2, b = –3/4 and c = 3 12. Prove that :

sin x + 2x ≥ π

+ )1x(x3 , ∀ x ∈

π

2,0

(justify the inequality, if any used). [IIT-2004]

Sol. Let f(x) = sin x + 2x – π

+ )1x(x3

f´(x) = cos x + 2 –π+ )3x6(

f´ (x) = –sin x – π6 < 0 for all x ∈

π

2,0

∴ f´(x) is decreasing for all x ∈

π

2,0

⇒ f´(x) > 0 as, x < π/2 ⇒ f´(x) > f´(π/2) ∴ f(x) is increasing Thus, when x ≥ 0 f(x) ≥ f(0)

sin x + 2x – π

+ )1x(x3 ≥ 0

or sin x + 2x ≥ π

+ )1x(x3

13. A window of perimeter (including the base of the

arch) is in the form of a rectangle surrounded by a semi-circle. The semi-circular portion is fitted with coloured glass while the rectangular part is fitted with clear glass. The clear glass transmits three times as much light per square meter as the coloured glass does.

What is the ratio for the sides of the rectangle so that the window transmits the maximum light?[IIT-1991]

Sol. Let '2b' be the diameter of the circular portion and 'a' be the lengths of the other sides of the rectangle.

Total perimeter = 2a + 4b + πb = K (say) ...(1) Now, let the light transmission rate (per square

metre) of the coloured glass be L and Q be the total amount of transmitted light.


Coloured

glass

Clear glassa a

Then, Q = 2ab(3L) + 21

πb2(L)

Q = 2L πb2 + 12ab

Q = 2L πb2 + 6b (K – 4b – πb)

Q = 2L 6Kb – 24b2 – 5πb2

dbdQ =

2L 6K – 48b – 10πb = 0

⇒ b = π+1048

K6 ...(2)

and 2

2

dbQd =

2L –48 + 10πLa

Thus, Q is maximum and from (1) and (2), (48 + 10π) b = 6K and K = 2a + 4b + πb ⇒ (48 + 10π) b = 62a + 4b + πb

Thus, the ratio = ab2 =

π+66

14. With usual notation, if in a triangle ABC

11

cb + = 12

ac + = 13

ba + , then prove that

7

Acos = 19

Bcos = 25

Ccos [IIT-1984]

Sol. Let 11

cb + = 12

ac + = 13

ba + = λ

⇒ (b + c) = 11λ, c + a = 12λ, a + b = 13λ ⇒ 2(a + b + c) = 36λ or a + b + c = 18λ Now, b + c = 11λ and a + b + c = 18λ ⇒ a = 7λ c + a = 12, and a + b + c = 18λ ⇒ b = 6λ a + b = 13λ and a + b + c = 18λ ⇒ c = 5λ

∴ cos A = bc2

acb 222 −+

= 2

222

)30(2492536

λ

λ−λ+λ = 51

cos B = ac2

bca 222 −+

= 2

222

70364925

λλ−λ+λ =

3519

cos C = ab2

cba 222 −+

= 2

222

84253649

λλ−λ+λ =

75

∴ cos A : cos B : cos C = 51 :

3519 :

75 = 7 : 19 : 25

15. Let ABC be a triangle with incentre I and inradius r.

Let D, E, F be the feet of the perpendiculars from I to the sides BC, CA and AB respectively. If r1, r2 and r3 are the radii of circles inscribed in the quadrilaterals AFIE, BDIF and CEID respectively, prove that :

1

1

rrr−

+ 2

2

rrr−

+ 3

3

rrr−

= )rr)(rr)(rr(

rrr

321

321

−−−

[IIT-2000] Sol. The quadrilateral HEKJ is a square because all four

angles are right angle and JK = JH.

A/2 A/2

Circle Circle

H

E F

B D C

K r1

r1

90º 90º

A

90º

J

Therefore, HE = JK = r1 and IE = r (given) ⇒ IH = r – r1 Now, in right angled triangle IHJ, ∠JIH = π/2 – A/2 [Q ∠IEA = 90º, ∠IAE = A/2 and ∠JIH = ∠AIE] in

triangle JIH

tan(π/2 – A/2) = 1

1

rrr−

⇒ cot A/2 = 1

1

rrr−

Similarly, cot B/2 = 2

2

rrr−

and cot C/2 = 3

3

rrr−

adding above results, we obtain

cot2A + cot

2B + cot

2C = cot

2A cot

2B cot

2C

⇒ 1

1

rrr−

+ 2

2

rrr−

+ 3

3

rrr−

= )rr)(rr)(rr(

rrr

321

321

−−−


Passage # 1 (Q. No. 1 to 4) A circular coil is placed in a time varying

magnetic field as shown in figure. The numeral relation between the radius and the resistance of

the coil is - r =πR where r = radius of coil

R = resistance of the coil

the magnetic field is perpendicular to the plane of paper and inwards and given by B = B0 + B1t2 where B0 is a positive and B1 is a negative constant. The numeral values of B0 and B1 are 1 and 2 respectively.

r

→B

× × × × × × × × × × × × × × × × × × × ×

Q.1 The value of induced electric field in the circular coil

(A) Depends only on radius of circular coil in linear manner

(B) Independent of the radius of circular coil (C) Varies nonlinearly with respect to time (D) Numeral value depends on the radius of

circular coil and for given radius it varies linearly with respect to time

Q.2 Induced current in the coil at t = 0 (A) 4 amp. (B) zero (C) 1 amp. (D) π amp.

Q.3 RMS value of the induced current for the time interval 0 ≤ t ≤ 2s

(A) 8 amp. (B) 4 amp. (C) 8/ 3 amp. (D) 8/3 amp.

Q.4 Induced charge for time period as stated above (A) 8x Faraday (B) 2x Faraday (C) x Faraday (D) 4x Faraday

Here x =96500

1

Passage # 2 (Q. No. 5 to 8) A rod ab of mass M, resistance R and length L is

supported by two supports SP-1 and SP-2 against gravity At t = 0 both the supports get removed and the rod starts falling under the gravity in a very long spread uniform magnetic field perpendicular to the plane of paper and directed inwards then

SP-1

t = 0

SP-2

x = 0

B→

L

×××××××××

× × × × × × × × ×

× × × × × × × × ×

××××

××××

a b

SP-1 and SP-2 Rigid supports

Q.5 Expression for the speed at time t is

(A) kg (1 – e–kt) (B)

kg .e–kt

(C) gk (1 – e–kt) (D)

gk .e–kt

where k =MR

LB 22

Q.6 Time after which the acceleration of rod is 37% of the maximum acceleration

(A) 22LBMR (B) 22LB

MR21

(C) 0.37 22LBMR (D) 0.63 22LB

MR

This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety of possible twists and turns of problems in physics that would be very helpful in facing IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and we hope that this section would prove a rich resource for practicing challenging problems and enhancing the preparation level of IIT JEE aspirants.

By : Dev SharmaDirector Academics, Jodhpur Branch

Physics Challenging Problems

Solutio ns wi ll b e pub lish ed in n ext issue

Set #10


Q.7 Time after which the velocity of the rod is 63% of terminal velocity

(A) 22LBMR (B) 22LB

MR21

(C) 0.37 22LBMR (D) 0.63 22LB

MR

Q.8 Expression for a real velocity at any instant t

(A) KgL (1 – e–kt) (B) gL(1 – e–kt)

(C) gK(1 – e–kt) (D) KL (1 – e–kt)

Where K =MR

LB 22

ELECTRONIC NOSE

NASA researchers are developing an exquisitely sensitive artificial nose for space exploration.

Onboard the space station, astronauts are surrounded by ammonia. It flows through pipes, carrying heat generated inside the station (by people and electronics) outside to space. Ammonia helps keep the station habitable.

But it's also a poison. And if it leaks, the astronauts will need to know quickly. Ammonia becomes dangerous at a concentration of a few parts per million (ppm). Humans, though, can't sense it until it reaches about 50 ppm.

Ammonia is just one of about forty or fifty compounds necessary on the shuttle and space station, which cannot be allowed to accumulate in a closed environment.

And then there's fire. Before an electrical fire breaks out, increasing heat releases a variety of signature molecules. Humans can't sense them either until concentrations become high.

Astronauts need better noses!

That's why NASA is developing the Electronic Nose, or ENose for short. It's a device that can learn to recognize almost any compound or combination of compounds. It can even be trained to distinguish between Pepsi and Coke. Like a human nose, the ENose is amazingly versatile, yet it's much more sensitive.

"ENose can detect an electronic change of 1 part per million," says Dr. Amy Ryan who heads the project at JPL. She and her colleagues are teaching the ENose to recognize those compounds - like ammonia - that cannot be allowed to accumulate in a space habitat.

Here's how it works: ENose uses a collection of 16 different polymer films. These films are specially designed to conduct electricity. When a substance - such as the stray molecules from a glass of soda - is absorbed into these films, the films expand slightly, and that changes how much electricity they conduct.

Because each film is made of a different polymer, each one reacts to each substance, or analyte, in a slightly different way. And, while the changes in conductivity in a single polymer film wouldn't be enough to identify an analyte, the varied changes in 16 films produce a distinctive, identifiable pattern.

Electronic Noses are already being used on Earth. In the food industry, for example, they can be used to detect spoilage. There's even an Electronic Tongue, which identifies compounds in liquids. The ENose needs to be able to detect lower concentrations than these devices.

E-Nose

Right now, Ryan is working on a stand-alone version of ENose. "Everything is in one package," she explains: polymer films, a pump to pull air (and everything in the air) through the device, computers to analyze data, the energy source. The noses could simply be posted, like smoke detectors, at various points around the habitat.


1. Question is based on Ampere's circuital law

As line integral of magnetic field over the closed

loop ∫ µ=→→

net01 i.d.B

here inet = 4 – 3 = 1 amp.

εµ=

ε=µ=µ=∫

→→

00

2

02001 .

1cAs.c1)1(d.B

where c is the speed of light

so line integral of magnetic field over the closed

loop abcda 0

20s c1d.Bε

=µ=∫→→

As we know that surface integral of magnetic

field over the closed loop, ∫→→

sd.B is always zero

according to Gauss law for magnetostatics.

Option D is correct.

2. Positively charged particle will turn towards left

and moves anticlockwise. In mirror it is seen

clockwise. Just opposite for negatively charged

particle.

Radius of circular path r =qBmv

To avoid hitting the upper plate d ≥ r, d ≥ qBmv

A positively charged particle will never hit the

upper plate if d ≥qBmv it rotates anticlockwise but

if viewed in mirror it is clockwise.


3. Electric force on particle eF→

= →

± Eq

= iqE)i(qE 00 ±=−±

Magnetic force on particle )Bv(qF m→→→

×±=

= )]k(Bjv[q 00 ×±

= )kj(Bqv 00 ×±

mF→

= )i(B.v.q 00±

As mF→

and eF→

are in opposite direction so either

the particle is positively charged or negatively it

can go undeviated if

|F||F| em→→

=

Option C is correct

4. As |F||F| em→→

=

qv0.B0 = qE0

v0 =0

0

BE

= s/m)6.19(

Height achieved by the particle

h =g2

v2=

)8.9(26.19 = 1m

As from mirror point of view focal length

f = 20cm (Negative for concave mirror)

Using mirror formula. 1/v + 1/u = 1/f

601

6032

201

301

v1

201

301

v1

−=−

=−=⇒−

=−

+

So image is at, v = -60cm

Height achieved by the particle behaves like

object so mirror forms it's image and the

magnification

m = v/u = –3060

−− = 2

As height achieved by the charged particle

upward = Height of object = 1m

Solution Physics Challenging Problems

Set # 9

8 Questions were Pu blished in January Issue


Height achieved by the charged particle

downward = Height of image

as m =HOOHOI So HOI = m (HOO) = 2(1) = 2m

Option D is correct

5. (1). Initial current is 18 amp.

Total energy stored in the inductor

=21 Li2 =

21 (2/3)(18)2

=31 (18)2 = 108 Joule

L = L1 + L2 = 1/3 + 1/3 = 2/3(As two inductor

are in series)

(1). Total energy dissipated in resistors = 108 J

(2). Time constant

The equivalent circuit is

a

1/3H 1/3H

1Ω 3Ω

1Ω

6Ω b

τ =eq

eq

RL

between terminals a and b.

=3/53/2 =

52 = 0.4

[2.Time constant of the circuit = 0.4 sec.]

(3). Potential drop across R1 initially

v = i.R1 = 18(1) = 18 volt

(4). Using current division formula.

Current passing through the R2 initially is

= 12 amp.

So potential drop v = i.R = R(1) = 12 volt.

6. As the particles are having same de-Broglie wave

lengths so

λA = λB

APh =

BPh

PA = PB Means momentum of both the particles is

same.

As the radius of circular path in magnetic field is

r = p/qB so rA = B.q

P

A

A , rB =B.q

P

B

B

Particle moving left Particle moving right Positively charged Negatively charged radius of path less - Particle B electron - Particle A radius of path is more as shown in figure

Particle A should have more charge as compared

to electron so it is doubly ionized Helium atom.


7. 1. Force on each and every coil is zero

2. Torque on coil of option A and option B is

maximum because angle between →M and

→B is

90º.

τmax. = MB τm ≠ 0

3. Torque on coil of option C and option D is zero

because angle between →M and

→B is zero

So τmin. = MB sin 0 = 0

4. Coil in option C and option D are having the

tendency of compression.

8. Conceptual problem

1. Gauss law for electrostatics

net0

se q.1d.Eε

=φ→→

∫ = µc2qnet

µε=

00

2 1c

2. Gauss law for magnetistatics

0d.E sm =φ→→

∫ , Magnetic monopole is impossible

3. Ampere's circuital law

→→

∫ 1d.B = µ0inet =0

2C1ε

.inet

4. Faraday's law of electromagnetic induction

Induced emf e =→→

∫ ed.E


1. AB is a horizontal diameter of a ball of mass m = 0.4 kg and radius R = 0.10 m. At time t = 0, a sharp impulse is applied at B at angle of 45º with the horizontal, as shown in Fig. So that the ball immediately starts to move with velocity v0 = 10 ms–1.

B A

45º

(i) Calculate the impulse

If coefficient of kinetic friction between the floor and the ball is µ = 0.1, calculate

(ii) Velocity of ball when it stops sliding,

(iii) time t at that instant,

(iv) horizontal distance traveled by the ball upto that instant,

(v) Angular displacement of the ball about horizontal diameter perpendicular to AB, upto that instant, and

(vi) energy lost due to friction. (g = 10 ms–2)

Sol. Since, the impulse applied is sharp and its line of action does not pass through centre of mass of the sphere, therefore, (just after application of impulse), sphere starts to move, both translationally and rotationally,. Translational motion is produced by horizontal component of the impulse, while rotational motion is produced by moment of the impulse. Let the impulse applied be J.

lαO

ma

µN N

mg

Then its horizontal component, J. cos 45º = Initial horizontal momentum (m.v0) of

the ball.

∴ J = 4 2 kg ms–1 Ans. (i)

Moment of inertia of ball about centroidal axis is

I =52 mR2 = 1.6 × 10–3 kg m2

Initial angular momentum of ball (about centre) = J (R. sin 45) or Iω0 = J.R. sin 45º ∴ ω0 = 250 rad sec–1 (clockwise) Now sphere slides on floor (to the left). Therefore,

friction on it acts towards right. Considering free body diagram of sphere Fig. (while it is sliding).

(Note: Since, the sphere is sliding on the floor, therefore, A is not an instantaneous axis of rotation. Hence, we can not take moments about A)

For vertical forces, N = mg …(1) For horizontal forces, µN = ma or a = µg = 1 ms–2 Now taking moments (about O) of forces acting on

sphere, µN . R = Iα …(2) From equations (1) and (2) α = 25 rad/sec2

(anticlockwise) Let sliding continue for a time 't'. At that instant, translational velocity, v = v0 – at or v = (10 – t)ms–1 (towards left) and angular velocity, ω = (–ω0)+ αt (anticlockwise) or ω = (25t – 250) rad s–2 But when sliding stops, v = rω ∴ (10 – t) = 0.1 (25t – 205) or t = 10 sec Ans. (iii) ∴ At that instant v = 10 – t = 0 Ans. (ii) Considering leftward translational motion of ball (for

first 10 second),

Distance moved by the ball is s = v0t – 21 at2

or s = 50 m Ans. (iv) Now considering clockwise rotational motion of the

ball (about its centroidal axis),

Angular displacement, θ = ω0t – 21

αt2

or θ = 1250 radian (clockwise) Ans. (v)

Expert’s Solution for Question asked by IIT-JEE Aspirants

Students' ForumPHYSICS


Since, the ball has stopped, it means whole of its initial kinetic energy is lost against friction and that is

ω+ 2

020 I

21mv

21

∴ Energy lost against friction = 70 joule Ans. (vi)

2. A rectangular tank having base 15 cm × 20 cm is filled with water (density ρ = 1000 kg m–3) upto 20 cm height. One end of an ideal spring of natural length h0 = 20 cm and force constant K = 280 Nm–1 is fixed to the bottom of a tank so that spring remains vertical. This system is in an elevator moving downwards with acceleration a = 2 ms–2. A cubical block of side l = 10 cm and mass m = 2 kg is gently placed over the spring and released gradually, as shown in Fig.

a

20cm

K

(i) Calculate compression of the spring in

equilibrium position.

(ii) If block is slightly pushed down from equilibrium position and released, calculate frequency of its vertical oscillations. (g = 10 ms–2)

Sol. Let, in equilibrium position, compression of spring be x. Liquid of volume l2x is displaced from its original position and level of liquid in tank rises as shown in

Fig. This rise in level, ∆ x =2

2

Axl

l−

where A = 15 cm × 20 cm (Base area of tank)

x

K

∆xIncreased level of water surface

Original level of water surface

Fig. : 1 or ∆x = 0.5x ∴ Mass of water displaced by the block = l2 (x + ∆x)ρ = 15x kg

Upthrust exerted by water = apparent weight of water displaced,

∴ Upthrust F1 = 1.5x (g – a) = 120.x newton Upward force exerted by spring F2 = Kx = 280.x. Considering free body diagram of the block, (Fig.)

mg

m.a

(F1 + F2)

Fig. : 2 Mg – (F1 + F2) = ma Substituting values of F1 and F2, x = 0.04 m = 4 cm Ans. (i) If the block is slightly pushed downward by dx, both

F1 and F2 increase. Increase in F1 is dF1 = 120 dx Increase in F2 is dF2 = 280.dx restoring force on block = increase in F1 + increase in

F2 = dF1 + dF2 = (120 dx +280.dx) = 400.dx

or Restoring acceleration =m

dx.400 = 200.dx

Since, restoring acceleration ∝ displacement (dx) Therefore, block performs SHM. Hence, frequency,

f =π2

1ntdisplaceme

onaccelerati =π21

200

=π

25 per second Ans. (ii)

3. A two way switch S is used in the circuit shown in Fig. First, the capacitor is charged by putting the switch in position 1.

Calculate heat generated across each resistor when switch is in position 2.

2

0.1F1

S

60V+ – 10Ω

4Ω

6Ω

3Ω


Sol. Initially the switch was in position 1. Therefore, initially potential difference across capacitor was equal to emf of the battery i.e. 60 volt.

∴ Initially energy stored in the capacitor was

U =21 CV2 =

21 × 0.1 × 602 J

= 180 J q

4Ω

+ –

i6Ω

3Ω

i1

i

i2

When switch is shifted to position 2, capacitor begins

to discharge and energy stored in it is dissipated in the form of heat across resistances. Let at some instant discharging current through the capacitor be i as shown in Fig.

According to Kirchhoff's laws, i1 + i2 = i … (1) 6i1 – 3i2 = 0 or i2 = 2i1 … (2) From above two equations,

i1 =3i and i2 =

32 i

But thermal power generated in a resistance R is P = i2R where i is current flowing through it. Therefore, heat generated P1, P2 and P3 across 4Ω, 6Ω and 3Ω resistances is in ration 2

221

2 i3:i6:i4

or P1 : P2 : P3 = 4 :32 :

34 = 6 : 1 : 2

But total heat generated is P1 + P2 + P3 = U ∴ Heat generated across 4Ω is P1 = 120 J Ans. Heat generated across 6Ω is P2 = 20 J Ans.

Heat generated across 3Ω is P3 = 40 J Ans. Since, during discharging, no current flows through

10Ω, therefore heat generated across it is equal to zero. Ans.

4. Calculate magnetic induction at point O if the wire carrying a current I has the shape shown in (i) Fig. (a) and (ii) Fig.(b).

z

yR

OR

x Fig. (a)

z

y

45º

O R

x

Fig. (b)

The radius of the curved part of the wire is equal to R and linear parts of the wire are very long.

Sol. (i) Current carrying wire shown in figure (a) can be considered in four parts.

(1) A straight part along y-axis. Since, point O lies on its axis, therefore magnetic

induction due to it is →

1B = 0. (2) A semi-circular part in y – z plane. Magnetic induction due to it is

→

2B =R2

I)2/1(µ0 (– i ) = –R4Iµ0 i

(3) One fourth circle in x – y plane. Magnetic induction due to it is

→

3B =R2

I)4/1(µ0 (– k ) = –R8Iµ0 k

(4) A straight part in x – y plane carrying current along negative y-directions.

Magnetic induction due to it is

→

4B =R4Iµ0

π(– k ) = –

R4Iµ0

πk

∴ →B =

→

1B + →

2B +→

3B +→

4B

= –R4Iµ0 i –

R8Iµ0

π(π+2) k Ans. (i)

(ii) Circuit segment shown in figure (b) can be considered in three parts.

I

OR45º

r

z

y

45º


(1) A circular loop in y – z plane. Since, this loop is made of uniform wire, therefore, magnetic induction at O due to it is B1 = 0

(2) A straight part, parallel to x-axis. Magnetic

induction due to it is B2 = R4Iµ0

π(– k ) = –

R4Iµ0

πk .

(3) A straight part in y – z plane. Perpendicular distance of O from axis of this straight part is r = R cos 45º as shown in fig (a). Angles subtended by lines joining O and ends of this straight part with perpendicular drawn from O are α = – 45º and β = 90º.

Magnetic induction at O due to this part is

B3 = r4Iµ0

π(sin α + sin β)

or →

3B =R4Iµ0

π( 2 – 1) i

∴ →B =

→

1B + →

2B +→

3B

=R4Iµ0

π( 2 – 1) i –

R4Iµ0

πk Ans. (ii)

5. In a Young's double slit experiment a parallel light beam containing wavelength λ1 = 4000 Å and λ2 = 5600 Å is incident on a diaphragm having two narrow slits. Separation between the slits is d = 2 mm. If distance between diaphragm and screen is D = 40 cm, calculate

(i) distance of first black line from central bright fringe and

(ii) distance between two consecutive black lines. Sol. When a monochromatic light of wavelength λ is used

to obtain interference pattern in Young's double slit

experiment, fringe width is given by ω =dDλ where

D is distance of screen from slits and d is distance between the slits.

Hence, fringe width for light of wavelength λ1,

ω1 = dD1λ

∴ ω1 = 80 µm and fringe width for light of wavelength, λ2,

ω2 = dD2λ = 112 µm

Since, the incident light beam has both the wavelength λ1 and λ2, therefore, interference patterns are formed on the screen for both the wavelengths. A black line is formed at the position where dark fringes are formed for both of the wavelengths.

Let first black line be formed at distance y from central bright fringe. Let at this position there be mth dark fringe of wavelength λ1 and nth dark fringe of wavelength λ2.

∴ Distance of first black line, from central bright line,

y =

−

21m ω1 =

−

21n ω2 … (1)

or 1n21m2

−− =

1

2

ωω …(2)

For first black line, y should be minimum possible which corresponds to least possible integer values of m and n.

Hence, 1n21m2

−− =

57 or m = 4, n = 3

∴ Position of first black line

y =

−

21m ω1 = 280 µm Ans. (i)

Since, interference pattern is always symmetric about central bright fringe, therefore, there are two first black lines one is at height y from central bright fringe and the other at a depth y from it.

Hence, distance between two consecutive black lines = 2y = 560 µm Ans. (ii)

SCIENCE TIPS • By seeing a glowing electric bulb can you say if it is

being fed by A.C. or D.C. No

• What does the sudden burst of a cycle tyre represent? Adiabatic process

• What happens to the velocity and wavelength of light when it enters a denser medium ?

Both decrease

• The skylab space station did not have a safe landing. Why ?

Because its remote control system failed

• What happens when even a small bird hits a flying aeroplane ? It causes heavy damage

• When does the lunar eclipse occur ? It occurs when the earth comes

in between the moon and the sun


Prism :

(i) Deviation 'δ' produced by the prism,

B C

A

Normal Normal

i i' QP r r'

δ

δ = i + i' – A and A = r + r' (ii) For minimum deviation 'δm'

i = i' and r = r' and also PQ||BC and the refractive index for the material of prism is given by

µ =

δ+

2Asin

2Asin m

(iii) δ – i graph for prism

δm

i

δ

(iv) For not transmitting the ray from prism,

µ > cosec

2A

(v) For grazing incidence i = 90º and for grazing emergence i' = 90º. For maximum deviation i = 90º or i' = 90º

(vi) The limiting angle of prism = 2C when i = i' = 90º If the angle of prism A > 2C, then the rays are totally

reflected.

(vii) Right-angled prism : These prisms are used to turn a light beam to 90º or 180º. These are usually made of crown glass for which

µg = 1.5 and C = tan–1

gµ1 = 42º.

Such prisms are used in binoculars and submarine periscopes.

(viii) Deviation produced by a thin prism δ = (µ – 1)A (ix) Angular dispersion D = δv – δR = (µV – µR)A Where V and R stand for violet and red colours

respectively. Mean deviation δY = (µY – 1)A where µY is the refractive index of mean yellow

colour.

(x) Dispersive Power, ω =deviationMeandispersionAngular

=Y

RV

δδ−δ

ω =1Y

RV

−µµ−µ where µY =

2RV µ+µ

(xi) Pair of prisms (or crossed prism) : Two thin prisms of different material when placed crossed, i.e., with their refracting edges parallel and pointing in opposite directions as shown in figure, produce a total deviation δ given by δ = δ1~ δ2

A

Crownglass

Flint glass

A'

where δ1 and δ2 are the mean deviations produced by the first and second prism respectively.

Total angular dispersion D = D1 ~ D2 where D1 and D2 are the angular dispersions

produced by respective prisms. (xii) Dispersion without deviation : If the angle of two

prisms A and A' are so adjusted that the deviation produced by the mean ray by the first prism is equal and opposite to that produced by the second prism, then the total final beam will be parallel to the

Prism & Wave Nature of Light

PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


incident beam and there will be dispersion without deviation.

Here, δ = δ1 – δ2 = 0 or δ1 = δ2 i.e., (µ1 – 1)A = (µ2 – 1)A' This combination produces total angular dispersion. D = D1 – D2 = (µ1V – µ1R) A – (µ2V – µ2R)A' (xiii) Deviation without dispersion : If the combination is such that D = D1 ~ D2 = 0 or D1 = D2 or (µ1V – µ1R) A = (µ2V – µ2R)A' The combination is said to be achromatic and the

total mean deviation will be δ = δ1 ~ δ2 = (µ1 – 1)A ~ (µ2 – 1)A' Wave nature of light :: Wave front : A point source produces a spherical wave front

∝

r1A or

∝ 2r

11

Where A = Amplitude, I = intensity and r = distance of point of observation from source. A line source produces a cylindrical wave front

∝

r1A or

∝

r1I .

Wave front is locus of points in the same phase. A distance source produce a plane wave front. Wave front for a parallel beam of light is plane. The angle between ray and wave front is 90º Huygen's principle: Huygen's principle is a geometrical method to find

secondary wave front produced by a primary wave front.

Thin lines shows the rays of light. Dotted line shows the wavefronts.

Interference of light : (a) Redistribution of light energy i.e. alternate

maximum and minima). Conditions for two light waves producing

interference is that (i) Wave should be of same wavelength/frequency. (ii) Waves should be travelling in the same direction. (iii)Wave should have a constant phase difference For the above conditions the two source must be

coherent and that is possible when we make two sources out of a single source of light.

For monochromatic light we get alternate maxima and minima of same colour. For white light we get white central fringe flanked by coloured fringes because fringe width of different colour is different due to different wavelengths.

(b) Resultant intensity at a point is

I = I1 + I2 + 2 21 II cos φ

When I1 = I2 = I0 then I = 4I0 cos2 φ/2 For constructive interference φ = ± 2nπ and ∆x = ±nλ

Imax = ( 1I + 2I )2 ∝ (A1 + A2)2 [Q I ∝ A2]

For destructive interference

φ = (2n + 1)π and ∆x =

−

21n λ

Imin = ( 1I – 2I )2 ∝ (A1 – A2)2

⇒ min

max

II

= )II()II(

21

21

−

+ = 2

21

221

)AA()AA(

−+

The energy remains conserved during the process of interference.

P

αS1

S2

Intensity of light at any point P as shown the

figure I = I0cos2

λαπ tand

(c) The fringe width β = dDλ

Angular width θ = Dβ =

dλ

⇒ θ does not depend on D


(d) When the source of light is placed asymmetrical with respect to the slits then the central maxima also shifts.

α

S2

x y

Dx Dy

θ

S1

S

yD

y = xD

x and θ = – α

(e) If young's double slit experiment is done in a liquid of refractive index medium µ then the fringe width β = β/µ

(f)

PS1

S2

t

O´ O

If a transparent sheet of thickness t is placed in

front of upper slit then the central maxima shift upside. The new optical path becomes µt instead of t and the increase in optical path is (µ – 1)t.

The shift = dD (µ – 1)t =

λβ (µ – 1)t

(g) Interference in thin films :

r r r r

21i

t

Transmitted rays For reflected rays interference

Maxima 2 µt cos r = (2n – 1)2λ

Minima 2 µt cos r = nλ

Diffraction : Bending of light through an aperture / corner when

the dimension of aperture is comparable to the wavelength of light is called diffraction.

Fraunhoffer diffraction at a single slit Condition for minima : a sin θn = n λ Condition of secondary maxima :

a sin θn = 22

1n λ

+ Where n = n = 1, 2 ...

Width of central maxima = 2λD/a P

O θa

Angular width of central maxima = 2λ/a Angular width of secondary maxima = λ/a

Intensity at any point P = I0

2sin

αα

where α = λπ (a sin θ)

The ratio of intensities of secondary maxima are

221 ,

611 ,

1211 , ...

For a path difference of λ, the phase difference is 2π radian.

Polarisation :

I0

Unpolarisedlight

Polarisedlight

I cos2θI = I0/2

rip Medium 1

Medium 2 ½ µ

ip = Angle of polarization, ip + r = 90º, µ = tan ip


Solved Examples

1. (i) A ray of light incident normally on one of the

faces of a right-angled isosceles prism is found to be totally reflected. What is the minimum value of the refractive index of the material of prism ?

(ii) When the prism is immersed in water, trace the path of the emergent ray for the same incident ray indicating the values of all the angles (µω = 4/3)

Sol. (i) According to the problem, ∠A = 90º, ∠B = ∠C = 45º.

At face BC, incident ray PQ is totally reflected therefore i ≥ C fig.

PBA

R r=iN

i45º

Q

N´

C Here i = 45º, ∴ Cmax = 45º; ∴ µ = 1/sin C

or µmin = (1/sin Cmax) = (1/sin 45º) = 2 = 1.414

(ii) When the prism is immersed in water, then for normal incident ray, the ray passes undeviated up to PQ and becomes incident at face BC at angle of incidence 45º(fig.) The ray travels from glass to water, therefore from Snell's law,

rsinisin =

1

2

µµ , we have

rsinº45sin =

g

w

µµ

= gµw

∴ sin r = wgµ

º45sin = wµ

º45sin × µg

= 2

1 ×3/4

414.1 = 43 = 0.75

∴ r = sin–1(0.75) = 48º36

P BA

R

90º45º

45º

Q r

C

45ºR

The path of light ray is shown in fig.

2. The cross-section of a glass prism has the form of an isosceles triangle. One of the equal faces is coated with silver. A ray is normally incident on another unsilvered face and being reflected twice emerges through the base of the prism perpendicular to it. Find the angles of the prism.

Sol. Suppose refracting angle of prism be α and other two base angles of the isosceles prism be β. The light ray PQ, incident normally on the face AB, is refracted undeviated along QR. The refracted ray QR strikes the silvered face AC and gets reflected from it. The reflected ray RS now strikes the face AB from where it is again reflected along ST and emerges perpendicular to base BC.

A

B C

P

R S

Q

90º

N´2

N2

N´1i = α

ββ

θ=β

N1

T

i

α

It follows from fig. that angle of incidence on face

AC = i = α and also angle of incidence of face AB = θ = β As N2 N2 is parallel to PR, hence θ = 2i

i.e., β = 2α

Also α + 2β = 180º or α + 2(2α) = 180º

or α = 36º so β = 2α = 72º

3. Two coherent light sources A and B with separation 2λ are placed on the x-axis symmetrically about the origin. They emit light of wavelength λ. Obtain the positions of maximum on a circle of large radius, lying in the xy-plane and with centre at the origin.

Sol. Distance between two coherent light sources = 2λ.

Consider the interference of waves at some point C of the circumference of circle.

BC = )cosr2r( 22 θλ−λ+

AC = )cosr2r( 22 θλ+λ+

∴ Path difference = AC – BC = λ (For Maxima)

It is clear from figure, that

Path difference = AC – BC = AP + OM = 2λ cos θ


Y

X

C

MrP

θ θOA B

λ λ

∴ 2λ cos θ = λ or cos θ = 1/2

∴ Possible values of angle θ = 60º, 120º, 180º, 240º, 300º, 360º.

4. Two point coherent sources are on a straight line

d = nλ apart. The distance of a screen perpendicular to the line of the sources is D >> d from the nearest source. Calculate the distance of the point on the screen where the first bright fringe is formed.

Sol. Consider any point P on the screen at a distance x from O. Then

dO

D S2 S1

21D = D2 + x2 or D1 = D

2/1

2

2

Dx1

+ = D

+ 2

2

D2x1 ;

∴ D1 = D + D2

x 2

Similarly, D2 = (D + d) +)dD(2

x 2

+

∴ D2 – D1 = (D + d) +)dD(2

x 2

+– D –

D2x2

= d +2

x 2

−

+ D1

dD1 = d – d

)dD(D2x2

+

O D

D2

S1 S2

D1 x

P

d For the point O, D2 – D1 = d = nλ (given).

Thus there is brightness at O of nth order. Since the path difference decreases, the other fringes will be of lower order. The next bright fringe will be of (n – 1)th order. Hence for the next bright fringe

D2 – D2 = (n – 1)λ

d – d)dD(D2

x 2

+= (n – 1)λ

nλ – nλ)nD(D2

x2

λ+= (n – 1)λ

∴ x =n

)nD(D2 λ+

5. One slit of a Young's experiment is covered by a glass plate (n = 1.4) and the other by another glass plate (n' = 1.7) of the same thickness t. The point of central maximum on the screen, before the plates were introduced, is now occupied by the previous fifth bright fringe. Find the thickness of the plates (λ = 4800 Å)

Sol. Path of the wave from slit S1 = D1 + n't – t Path of the wave from slit S2 = D2 + nt – t ∴ Path difference = D2 + nt – t – D1 – n't + t = (D2 – D1) + (n – n')t

O′

D

D2

S1

S2

D1

Od

x

But D2 – D1 = Dxd

∴ Path difference =Dxd + (n – n')t

Let O' be the point where paths difference is zero.

∴ Dxd = (n' – n)t

or, x =dD (n' – n) t =

λβ− t)n'n(

λ

=βdD

Q

Given that x = 5β ∴ 5β = λ

β− t)n'n(

or, t =n'n

5−λ or, t =

4.17.11048005 10

××× −

= 8 µm.


Key Concepts : 1. Equation of a harmonic wave is y = a sin(kx ± ωt ± φ). Here y is measure of disturbance from zero level.

y may represent as electric field, magnetic field, pressure etc. Also K = 2π/λ = wave number.

Note : The positive sign between kx and ωt shows that the wave propagates is the +x direction. If the wave travels in the –x direction then negative sign is used between kx and ωt.

2. Particle Velocity :

v = dtdy = aω cos (kx ± ωt ± φ)

∴ Maximum particle velocity = aω = velocity amplitude Particle velocity is different from wave velocity.

The wave velocity v = vλ. 3. Particle acceleration :

A = dtdv =

2

2

dtyd = – aω2 cos(kx + ωt ± φ) = – ω2y

Max acceleration = acceleration amplitude = –ω2a

4. Velocity of transverse wave on a string = mT

Where m = mass per unit length = ρ × 4D2π

Where ρ = density of the wire material and D = diameter of wire More the tension, more is the velocity 5. A wave, after reflection from a free end, suffers

change of π. A wave, after reflection from a free end, suffers no

change in the phase.

6. Velocity of sound in a fluid = ρB

For air B = γP ∴ v = ργP =

MRTγ

Velocity of sound in general follows the order Vsolid > Vliquid > Vgas

⇒ Velocity of sound ∝ T

Also velocity of sound ∝ M/γ

and velocity of sound ∝ ρ/1 .

But velocity of sound does not depend on pressure because P/ρ becomes constant.

Velocity of sound depend on the frame of reference. 7. (a) According to principle of superposition

→y =

→1y +

→2y + ….

(b) Interference of waves y1 = A1 sin(kx – ωt) y2 = A2 sin(kx – ωt + φ) For constructive interference φ = 2nπ n = 0, 1, 2, ……

(i) Imax = ( 1I + 2I )2

(waves should be in same phase) (ii) Amax = A1 + A2 For destructive interference φ = (2n + 1)π n = 0, 1, 2, ……

(i) Imin = ( 1I – 2I )2

(waves should be in opposite phase) (ii) Amin = A1 – A2

(c) I = I1 + I2 + 2 21II cos φ

Where φ is the phase difference between the two waves.

8. Beats : When two waves of same amplitude with slight difference in frequency (<10), traveling in the same direction superpose, beats are produced.

The equation for beats is

y =

ω−ω2

t)cos(A2 21 sin

ω−ω2

21 t

Where amplitude at a given location

= 2Acos

ω−ω

221 t

The above expression shows that amplitude change with time.

Beat frequency = no of maxima / minima per second = v1 – v2

Waves & Doppler Effect

PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


9. Standing waves (stationary) When two waves of same amplitude and frequency

moving in opposite direction superimpose, standing waves are produced.

Nodes are the point where the displacement is always zero.

The amplitudes of different particles different and is maximum at antinodes.

The equation of standing waves is y = [2A sin kx]cos ωt where amplitude = 2A sin kx The above expression shows that the amplitude is

different for different values of x and varies sinusoidally.

For a node to occur at position x, y = 0 ⇒ kx = 0 For an antinode two occur at position x, y should

be max ⇒ kx = π/2 , …. In terms of pressure ∆P = ∆P0 cos kx cos ωt. 10. For standing waves on strings (and both end open

organ pipe)

Fundamental frequency v0 = 0

vλ

= l2

v

First mode of vibration v1 = 1

vλ

= 2

l2v

= 2v0 = 2nd harmonic

nth mode of vibration vn = n

l2v = nv0

= nth harmonic where v = mT for string.

Also more the tension in the same string, higher is the value of v0

11. For closed organ pipe :

Fundamental frequency v0 = 0

vλ

= l4

v

First mode of vibration v1 = 1

vλ

= 3

l4v = 3v0

= Third harmonic

nth mode of vibration vn = (2n + 1)l4

v

where n = 1, 2, ….. In case where end correction is taken replace l by

(l + e) 12. (a) Intensity of sound at a distance r from a point

source is I = 2r4Pπ

where P = power of source.

(b) For a line source I = lr

Pπ

where l is the length of source

(c) I = 21

ρv(4π2v2)A2 = ρν2

)amplitudeessure(Pr 2

13. Doppler's effect :

v = v0

±±

s

L

vvvv vL = velocity of listener

The above formula is valid when vs < v Replace v by (v ± vm) if it is given that the

medium also moving. When listener and source are not moving along

the line joining the two, then the component of velocity along the line joining the two are taken as velocity of listener or source.

14. If the source and listener are on the same vehicle and the sound is reflected from a stationary object towards which the vehicle is approaching then the frequency of sound as heard by the observer is

v´ = v0

++

s

L

vvvv

15. For a path difference of λ, the phase difference is 2π for harmonic waves.

16. For a transverse wave the energy per unit length possessed by a string is given as

ld

dE = m(4π2f2)A2cos2 (kx – ωt)

17. Equation for a wave pulse is y = f(x + vt) 18. When a wave on reaching on interface is partly

reflected and partly transmitted then for no power loss.

Pi = Pt + Pr where Pi = Power of incident wave Pt = Power of transmitted wave Pr = Power of reflected wave.

Also in this case Ar =

+−

12

12

vvvv Ai; At =

+ 21

2

vvv2 Ai

Where Ai, Ar and At are amplitudes of incident reflected and transmitted waves v1 is the velocity in the medium of incidence and v2 is the velocity in the medium where transmitted wave is present.

Problem Solving Strategy : Mechanical Waves Identify the relevant concepts : Wave problems fall

into two broad categories. Kinematics problems are concerned with describing wave motion; they involve wave speed v, wave length λ(or wave number k), frequency f (or angular frequency ω), and amplitude A. They may also involve the position, velocity, and acceleration of individual particles in the medium. Dynamics problems also use concepts from Newton's laws such as force and mass. In this chapter we'll encounter problems that involve the relation of wave speed to the mechanical properties of the wave medium. We'll get into these relations.

As always, make sure that you identify the target variable(s) for the problem. In some cases it will be


the wavelength, frequency, or wave speed; in other cases you'll be asked to find an expression for the wave function.

Set up the problem using the following steps : Make a list of the quantities whose value are

given. To help you visualize the situation, you'll find it useful to sketch graphs of y versus x (fig. a) and of y versus (fig. b). Label your graphs with the values of the known quantities.

y

Wave displacementversus coordinate x

at time t = 0

Wavelength λ

A

A

(a)

y Wave displacement

versus time t at coordinate x = 0

Period T

A

A

(b)

t t

Decide which equations you'll need to use. If any

two of v, f, and λ are given, you'll need to use eq. v = λf (periodic wave) to find the third quantity. If the problem involves the angular frequency ω and / or the wave number k, you'll need to use the definitions of those quantities and eq. (ω = vk). You may also need the various forms of the wave function given in Eqs.

y(x, t) = A cos

−ω t

vx = A cos 2πf

− t

vx ,

y(x, t) = A cos 2π

−

λ Ttx

and y(x, t) = A cos (kx – ωt). If the wave speed is not given, and you don't have

enough information to determine it using v = λf, you may be able to find v using the relationship between v and the mechanical properties of the system.

Execute the solution as follows : Solve for the unknown quantities using the equations you've selected. In some problems all you need to do is find the value of one of the wave variables.

If you're asked to determine the wave function, you need to know A and any two of v, λ and f(or v, k and ω). Once you have this information, you can use it in eq. (ω = vk). You may also need the various forms of the wave function given in Eqs.

y(x, t) = A cos

−ω t

vx = A cos 2πf

− t

vx ,

y(x, t) = A cos 2π

−

λ Ttx and y(x, t) = A cos (kx – ωt)

to get the specific wave function for the problem at hand. Once you have that, you can find the value of y

at any point (value of x) and at any time by substituting into the wave function.

Evaluate your answer : Look at your results with a critical eye. Check to see whether the values of v, f, and λ (or v, ω, and k) agree with the relationships given in eq. . v = λf or w = vk. If you've calculated the wave function, check one or more special cases for which you can guess what the results ought to be.

Problem Solving Strategy : Standing waves Identify the relevant concepts : As with traveling

waves, it's useful to distinguish between the purely kinematic quantities, such as wave speed v, wavelength λ, and frequency f, and the dynamic quantities involving the properties of the medium, such as F and µ for transverse waves on a string. Once you decide what the target variable is, try to determine whether the problem is only kinematic in nature or whether the properties of the medium are also involved.

Set up the problem using the following steps : In visualizing nodes and antinodes in standing

waves, it is always helpful to draw diagrams. For a string you can draw the shape at one instant and label the nodes N and antinodes A. The distance between two adjacent nodes or two adjacent antinodes is always λ/2, and the distance between a node and the adjacent antinode is always λ/4.

Decide which equation you'll need to use. The wave function for the standing wave is almost always useful ex. y(x, t) = (ASW sin kx) sin ωt.

You can compute the wave speed if you know either λ and f (or, equivalently, k = 2π/λ and ω = 2πf) or the properties of the medium (for a string. F and µ.)

Execute the solution as follows: Solve for the unknown quantities using the equations you've selected. Once you have the wave function, you can find the value of the displacement y at any point in the wave medium (value of x) and at any time. You can find the velocity of a particle in the wave medium by taking the partial derivative of y with respect to time. To find the acceleration of such a particle, take the second partial derivative of y with respect to time.

Evaluate your answer : Compare your numerical answers with your diagram. Check that the wave function is compatible with the boundary conditions (for example, the displacement should be zero at a fixed end).

Problem Solving Strategy : Sound Intensity Identify the relevant concepts : The relationships

between intensity and amplitude of a sound wave are rather straightforward. Quite a few other quantities are involved in these relationships, however, so it's particularly important to decide which is your target variable.


Set up the problem using the following steps : Sort the various physical quantities into

categories. The amplitude is described by A or pmax, and the frequency f can be determined from ω, k, or λ. These quantities are related through the wave speed v, which in turn is determined by the properties of the medium: B and ρ for a liquid; γ, T, and M for a gas.

Determine which quantities are given and which are the unknown target variables. Then start looking for relationships that take you where you want to go.

Execute the solution as follows: Use the equations you've selected to solve for the target variables. Be certain that all of the quantities are expressed in the correct units. In particular, if temperature is used to calculate the speed of sound in a gas, make sure that it is expressed in Kelvins (Celsius temperature plus 273.15).

Evaluate your answer: There are multiple relationships among the quantities that describe a wave. Try using an alternative one to check your results.

Problem Solving Strategy : Doppler Effect Identify the relevant concepts : The Doppler effect is

relevant whenever the source of waves, the wave detector (listener), or both are in motion.

Set up the problem using the following steps : Establish a coordinate system. Define the positive

direction to be the direction from the listener to the source, and make sure you know the signs of all relevant velocities. A velocity in the direction from the listener toward the source is positive; a velocity in the opposite direction is negative. Also, the velocities must all be measured relative to the air in which the sound is traveling.

Use consistent notation to identify the various quantities: subscript S for source, L for listener.

Determine which unknown quantities are your target variables.

Execute the solutions :

Use eq. fL = S

L

vvvv

++ fS to relate the frequencies at

the source and the listener, the sound speed, and the velocities of the source and the listener. If the source is moving, you can find the wavelength measured by the listener using Eq.

λ = Sfv –

S

S

fv =

S

S

fvv − or λ =

S

S

fvv + .

When a wave is reflected from a surface, either stationary or moving, the analysis can be carried out in two steps. In the first, the surface plays the role of listener; the frequency with which the

wave crests arrive at the surface is fL. Then think of the surface as a new source, emitting waves with this same frequency fL. Finally, determine what frequency is heard by a listener detecting this new wave.

Evaluate your answer: Ask whether your final result makes sense. If the source and the listener are moving towards each other, fL > FS; if they are moving apart, fL < fS. If the source and the listener have no relative motion, fL = fS.

1. A stationary wave is given by

y = 5 sin 3xπ cos 40 πt

where x and y are in cm and t is in seconds. (a) What are the amplitude and velocity of the

component waves whose superposition can give rise to this vibration ?

(b) What is the distance between the nodes ? (c) What is the velocity of a particle of the string at

the position x = 1.5 cm when t = 9/8 s ? Sol. Using the relation 2 sin C cos D = sin (C + D) +

sin(C – D)

y = 5 sin 3xπ cos 40 πt =

25 × 2 sin

3xπ cos 40πt

⇒ y =

π−

π+

π+

π t403xsint40

3xsin

25

= 25 .sin

π

+π3xt40 –

25 sin

π

−π3xt40

= 25 .sin

π

+π3xt40 +

25 sin

π+

π−π

3xt40

Thus, the given stationary wave is formed by the superposition of the progressive waves

y1 = 25 sin

π

+π3xt40 and y2 =

25 sin

π+

π−π

3xt40

Comparing each wave with the standard form of the progressive wave

y = a sin

α+

λπ

−ω2t ; a =

25 = 2.5 cm

ω = 40π or n = 20

and λπ2 =

32π or λ = 6 cm = 0.06 m

∴ c = nλ = 20 × 0.06 = 1.2 ms–1

Distance between the nodes = 2λ =

206.0 = 0.03 m

Q y = 5 sin3xπ cos 40 πt

Solved Examples


v = dtdy = – 5 × 40π sin

3xπ sin 40 πt

⇒ v = – 200 π sin3xπ sin 40πt

∴ At x = 1.5 cm and t = 89 s

v = – 200π sin 45π = 0 2. An engine blowing a whistle of frequency 133 Hz

moves with a velocity of 60 m s–1 towards a hill from which an echo is heard. Calculate the frequency of the echo heard by the driver. (Velocity of sound in air = 340 ms–1.)

Sol. The 'image' of the source approaches the driver at the same speed. Here, the image or echo is the source.

∴ vs = + 60 ms–1, v0 = – 60 ms–1

n´ = s

0

vcvc

−−

× n

∴ n´ = 60340

)60(340−−− × 133 = 190 Hz

3. A source of sound of frequency 1000 Hz moves to

the right with a speed of 32 ms–1 relative to the ground. To its right is reflecting surface moving to the left with a speed of 64 ms–1 relative to the ground. Take the speed of sound in air to be 332 m s–1 and find

(a) the wavelength of the sound emitted in air by the source

(b) the number of waves per second arriving at the reflecting surface

(c) the speed of the reflected waves, and (d) the wavelength of the reflected waves Sol. (a) Due to the motion of the source, the wavelength

(and hence, the frequency) is actually changed from λ to λ´ such that if n = actual frequency

λ´ = nvc S−

= 1000

32332 − = 0.3 m

(b) The number of waves arriving at the reflecting surface is the same as the number of waves received by an observer moving towards the source. This is given by the apparent frequency.

n = S

0

vcvc

−− × n =

32332)64(332

−−− × 1000 = 1320 Hz

(c) Same as that of the incident wave because the speed of a wave depends only on the characteristics of the medium.

∴ speed of the reflected wave = 332 ms–1 (d) To calculate the wavelength of the reflected wave,

we may consider the source to be stationary and emitting waves of wavelength 0.3 m. If the reflector were stationary, waves in a tube of length c would reach the reflector and the same number of reflected waves would be contained in a tube of the same length, so the wavelength of the reflected wave

would also be the same as that of the incident wave. But when the reflector moves towards the source with speed vref´ it would reflect additional waves contained in vref and the total number of waves reflected would be contained in a tube of length c – vref. If λ´is the changed wavelength of the wave due to the motion of the source

λ ´´ =

λ

+λ

−´

v´

c)vc( refref

or λ´´ = ref

ref

vcvc

+− × λ =

6433264332

+− × 0.3 = 0.2 m

4. Find the ratio of the fundamental frequencies of two

identical strings after one of them is stretched by 2% and the other by 4%.

Sol. n = mT

21l

. If l0 be the initial length and f be

fractional increase in length, l = l0 + fl0. Since tension is proportional to the increase in length,

T = k × fl0 where k is a constant.

m = 00 f

Mll +

where M is the mass of the string

∴ n = )f1(2

1

0 +l )f1(M/kf

0

0

+l

l =

)f1(Mfk

21 0

0 +l

l

Since l0, k and M are constants n ∝ f1

f+

∴ 2

1

nn =

)f1(f)f1(f

12

21

++ =

)02.01(04.0)04.01(02.0

++ = 0.71

5. An open organ pipe has a fundamental frequency of

300 Hz. The first overtone of a closed organ pipe has the same frequency as the first overtone of the open pipe. How long is each pipe ? The velocity of sound in air = 350 ms–1.

Sol. For a closed pipe n = l4

c and 3n, 5n, 7n, ... are the

overtones. For an open pipe n = l2

c and 2n, 3n, 4n,

... are the overtones.

⇒ l = n2c =

3002350×

= 0.58 m

The frequency of the first overtone = 2n = 2 × 300 = 600 Hz ∴ the frequency of the first overtone of the closed

pipe = 600 = 3n ∴ n = 200 Hz

∴ 200 = l4

350 or l = 2004

350×

= 0.44 m


The Diels-Alder reaction : α, β-Unsaturated carbonyl compounds undergo an

exceedingly useful reaction with conjugated dienes, known as the Diels-Alder reaction. This is an addition reaction in which C-1 and C-4 of the conjugated diene system become attached to the doubly bonded carbons of the unsaturated carbonyl compound to form a six membered ring.

C

C

C

C

C

C C

O

+C

C C

C

C

C

C

O

Diene Dienophile (Greek: diene-loving)

Adduct Six-membered ring

A concerted, single-step mechanism is almost

certainly involved; both new carbon-carbon bonds are partly formed in the same transition state, although not necessarily to the same exent. The Diels-Alder reaction is the most important example of cycloaddition, Since reaction involves a system of four π electrons (the diene) and a system of two π electrons (the dienophile), it is known as a [4 + 2] cycloaddition.

The Diels-Alder reaction is useful not only because a ring is generated, but also because it takes place so readily for a wide variety of reactants. Reaction is favored by electron-withdrawing substituents in the dienophile, but even simple alkenes can react. Reaction often takes place with the evolution of heat when the reactants are simply mixed together. A few examples of the Diels-Alder reaction are:

HC

HC

CH2

CH2

C

C C

O

+

1,3-Butadiene Maleic anhydride

H

H

O C

O

C

O

OC

O

benzene, 20 ºCquantitative

cis-1,2,3,6-Tetrahydrophthalicanhydride

HC

HC

CH2

CH2

HC

HCC O

+

1,3-Butadiene Acrolein

H

CHO100 ºCquantitative

H 1,2,3,6- Tetrahydrobenzaldehyde

HC

HC

CH2

CH2

+

1,3- Butadiene

p-Benzoqpuinone

benzene,35ºCquantitative

O

O

O

O

1,3-butadiene,100 ºC

O

O

5,8,9,10-Tetrahydro-1,4-naphthoquinone

1,4,5,8,11,12,13,14 -Octahydro-9,10-anthraquinone

Cannizzaro reaction : In the presence of concentrated alkali, aldehydes

containing no α-hydrogens undergo self-oxidation-and -reduction to yield a mixture of an alcohol and a salt of a carboxylic acid. This reaction, known as the Cannizzaro reaction, is generally brought about by allowing the aldehyde to stand at room temperature with concentrated aqueous or alcoholic hydroxide. (Under these conditions an aldehyde containing α-hydrogens would undergo aldol condensation faster)

2HCHO → NaOH%50 CH3OH + HCOO– Na+ Formaldehyde Methanol Sodium formate

O2N CHO 35%NaOH

p-Nitrobenzaldehyde

O2N CH2OH

p-Nitrobenzl alcohol

O2N COO–Na+

Sodium p-nitrobenzoate

+

Organic Chemistry

Fundamentals

CARBONYL COMPOUNDS

KEY CONCEPT


In general, a mixture of two aldehydes undergoes a Cannizzaro reaction to yield all possible products. If one of the aldehydes is formaldehyde, however, reaction yields almost exclusively sodium formate and the alcohol corresponding to the other aldehyde:

ArCHO + HCHO → NaOH.conc

ArCH2OH + HCOO–Na+ Such a reaction is called a crossed Cannizzaro

reaction. For example : CHO

+ HCHO

Anisaldehyde (p-Methoxy

benzaldehyde)

p-Methoxybenzyl alcohol

OCH3

conc. NaOHCH2OH

+ HCOO–Na+

OCH3

Evidence, chiefly from kinetics and experiments with

isotopically labeled compounds, indicates that even this seemingly different reaction follows the familiar pattern for carbonyl compounds: nucleophilic addition. Two successive additions are involved: addition of hydroxide ion (step 1) to give intermediate I;

Ar–C=O + OH– Ar–C–O–

H H

OH I

(1)

Ar–C = O + Ar–C–O–

H H

OHI

(2)

Ar–C–O– + Ar–C=O

H

OHH+H+ –H+

ArCH2OH ArCOO– and addition of a hydride ion from I (step 2) to a

second molecule of aldehyde. The presence of the negative charge on I aids in the loss of hydride ion.

Reduction : Aldehydes can be reduced to primary alcohols, and

ketones to secondary alcohols, either by catalytic hydrogenation or by use of chemical reducing agents like lithium aluminum hydride, LiAlH4. Such reduction is useful for the preparation of certain alcohols that are less available than the corresponding carbonyl compounds, in particular carbonyl compounds that can be obtained by the aldol condensation. For example :

O

LiAIH4 H+

H OH

Cyclopentanone Cyclopentanol

alcoholButyln2223

Ni,H

deacetaldehy ofoncondensati aldol From

hydeCrotonaldeButenal2

3 OHCHCHCHCHCHCHOCHCH 2

−− →=

CH=CHCHO

3-PhenylpropenalCinnamaldehyde

From aldol condensation of benzaldehyde and acetaldehyde

9-BBN

CH=CHCH2OH

3-Phenyl-2-propen-1-olCinnamyl alcohol

HOCH2CH2NH2

To reduce a carbonyl group that is conjugated with a

carbon-carbon double bond without reducing the carbon-carbon double bond, too, requires a regioselective reducing agent.

Aldehydes and ketones can be reduced to hydrocarbons by the action (a) of amalgamated zinc and concentrated hydrochloric acid, the Clemmensen reduction; or (b) of hydrazine, NH2NH2, and a strong base like KOH or potassium tertbutoxide, the Wolff-Kishner reduction. These are particularly important when applied to the alkyl aryl ketones obtained from Friedel – Crafts acylation, since this reaction sequence permits, indirectly, the attachment of straight alkyl chains to the benzene ring. For example

OH

OH

CH3(CH2)4COOH, ZnCl2

OH

CO(CH2)4CH3

OH

Zn(Hg),HCl

OH

CH2(CH2)4CH3

OH

4-n-Hexy lresorcinolUsed as an antiseptic

Alcohols are formed from carbonyl compounds, smoothly and in high yield, by the action of such compounds as lithium aluminum hydride, LiAlH4. 4R2C=O + LiAlH4 → (R2CHO)4AlLi

→ OH2 4R2CHOH + LiOH + Al(OH)3.


Tetragonal distortion of octahedral complexes (Jahn-Teller distortion) : The shape of transition metal complexes are affected

by whether the d orbitals are symmetrically or asymmetrically filled.

Repulsion by six ligands in an octahedral complex splits the d orbitals on the central metal into t2g and eg levels. It follows that there is a corresponding repulsion between the d electrons and the ligands. If the d electrons are symmetrically arranged, they will repel all six ligands equally. Thus the structure will be a completely regular octahedron. The symmetrical arrangements of d electrons are shown in Table.

Symmetrical electronic arrangements :

Electronic configuration

t2g eg

d5

d6

d8

d10 All other arrangements have an asymmetrical

arrangement of d electrons. If the d electrons are asymmetrically arranged, they will repel some ligands in the complex more than others. Thus the structure is distorted because some ligands are prevented from approaching the metal.

as closely as others. The eg orbitals point directly at the ligands. Thus asymmetric filling of the eg orbitals in some ligands being repelled more than others. This causes a significant distortion of the octahedral shape. In contrast the t2g orbitals do not point directly at the ligands, but point in between the ligand directions. Thus asymmetric filling of the t2g orbitals has only a very small effect on the stereochemistry. Distortion caused by asymmetric filling of the t2g orbitals is usually too small to measure. The electronic arrangements which will produce a large distortion are shown in Table.

The two eg orbitals 22 yxd − and 2zd are normally

degenerate. However, if they are asymmetrically filled then this degeneracy is destroyed, and the two

orbitals are no longer equal in energy. If the 2zd orbital contains one.

Asymmetrical electronic arrangements :

Electronic configuration

t2g eg

d4

d7

d9 more electron than the 22 yxd − orbital then the ligands

approaching along +z and –z will encounter greater repulsion than the other four ligands. The repulsion and distortion result in elongation of the octahedron along the z axis. This is called tetragonal distortion. Strictly it should be called tetragonal elongation. This form of distortion is commonly obsered.

If the 22 yxd − orbital contains the extra electron, then

elongation will occur along the x and y axes. This means that the ligands approach more closely along the z-axis. Thus there will be four long bonds and two short bonds. This is equivalent to compressing the octahedron along the z axis, and is called tetragonal compression, and it is not possible to predict which will occur.

For example, the crystal structure of CrF2 is a distorted rutile (TiO2) structure. Cr2+ is octahedrally surrounded by six F–, and there are four Cr–F bonds of length 1.98 – 2.01 Å, and two longer bonds of length 2.43 Å. The octahedron is said to be tetragonally distorted. The electronic arrangement in Cr2+ is d4. F– is a weak field ligand, and so the t2g level contains three electrons and the eg level contains one electron. The 22 yxd − orbital has four lobes whilst

the 2zd orbital has only two lobes pointing at the ligands. To minimize repulsion with the ligands, the single eg electron will occupy the 2zd orbital. This is equivalent to splitting the degeneracy of the eg level so that 2zd is of lower energy, i.e. more stable, and

22 yxd − is of higher energy, i.e. less stable. Thus the

Inorganic Chemistry

Fundamentals

CO-ORDINATION COMPOUND & METALLURGY

KEY CONCEPT


two ligands approaching along the +z and –z directions are subjected to greater repulsion than the four ligands along +x, –x, +y and –y. This causes tetragonal distortion with four short bonds and two long bonds. In the same way MnF3 contains Mn3+ with a d4 configuration, and forms a tetragonally distorted octahedral structure.

Many Cu(+II) salts and complexes also show tetragonally distorted octahedral structures. Cu2+ has a d9 configuration :

t2g

eg

To minimize repulsion with the ligands, two

electrons occupy the 2zd orbital and one electron

occupies the 22 yxd − orbital. Thus the two ligands

along –z and –z are repelled more strongly than are the other four ligands.

The examples above show that whenever the 2zd and

22 yxd − orbitals are unequally occupied, distortion

occurs. This is know as Jahn–Teller distortion. Leaching : It involves the treatment of the ore with a suitable

reagents as to make it soluble while impurities remain insoluble. The ore is recovered from the solution by suitable chemical method. For example, bauxite ore contains ferric oxide, titanium oxide and silica as impurities. When the powdered ore is digested with an aqueous solution of sodium hydroxide at about 150ºC under pressure, the alumina (Al2O3) dissolves forming soluble sodium meta-aluminate while ferric oxide (Fe2O3), TiO2 and silica remain as insoluble part.

Al2O3 + 2NaOH → 2NaAlO2 + H2O Pure alumina is recovered from the filtrate NaAlO2 + 2H2O → Al(OH)3 + NaOH

2Al(OH)3 )autoclave(Ignited → Al2O3 + 3H2O

Gold and silver are also extracted from their native ores by Leaching (Mac-Arthur Forrest cyanide process). Both silver and gold particles dissolve in dilute solution of sodium cyanide in presence of oxygen of the air forming complex cyanides.

4Ag + 8NaCN + 2H2O + O2 → 4NaAg(CN)2 + 4NaOH Sod. argentocyanide 4Au + 8NaCN + 2H2O + O2 → 4NaAu(CN)2 + 4NaOH Sod. aurocyanide Ag or Au is recovered from the solution by the

addition of electropositive metal like zinc.

2NaAg(CN)2 + Zn → Na2Zn(CN)4 + 2Ag ↓ 2NaAu(CN)2 + Zn → Na2Zn(CN)4 + 2Au ↓ Soluble complex Special Methods : Mond's process : Nickel is purified by this method.

Impure nickel is treated with carbon monoxide at 60–80º C when volatile compound, nickel carbonyl, is formed. Nickel carbonyl decomposes at 180ºC to form pure nickel and carbon monoxide which can again be used.

Impure nickel + CO 60–80ºCNI(CO)4

Ni + 4CO

180ºC

Gaseous compound

Zone refining or Fractional crystallisation : Elements such as Si, Ge, Ga, etc., which are used as

semiconductors are refined by this method. Highly pure metals are obtained. The method is based on the difference in solubility of impurities in molten and solid state of the metal. A movable heater is fitted around a rod of the impure metal. The heater is slowly moved across the rod. The metal melts at the point of heating and as the heater moves on from one end of the rod to the other end, the pure metal crystallises while the impurities pass on the adjacent melted zone.

Molten zonecontainingimpurity

Pure metalMoving circular

heater

Impure zone

Different metallurgical processes can be broadly

divided into three main types. Pyrometallurgy : Extraction is done using heat

energy. The metals like Cu, Fe, Zn, Pb, Sn, Ni, Cr, Hg, etc., which are found in nature in the form of oxides, carbonates, sulphides are extracted by this process.

Hydrometallurgy : Extraction of metals involving aqueous solution is known as hydrometallurgy. Silver, gold, etc., are extracted by this process.

Electrometallurgy : Extraction of highly reactive metals such as Na, K, Ca, Mg, Al, etc., by carrying electrolysis of one of the suitable compound in fused or molten state.


1. Calculate ∆rU, ∆rH and ∆rS for the process 1 mole H2O (1,293 K, 101.325 kPa) → 1 mol H2O (g, 523 K, 101.325 kPa) Given the following data : Cp,m (1) = 75.312 J K–1 mol–1 ; Cp,m(g) = 35.982 J K–1 mol–1

∆vapH at 373 K, 101.325 kPa = 40.668 kJ mol–1 Sol. The changes in ∆rU, ∆rH and ∆rS can be calculated

following the reversible paths given below. Step I: 1 mole H2O(1,293 K, 101.325 kPa) → 1 mole H2O(1,373 K, 101.325 kPa) qp = ∆rH = Cp,m(1) ∆T = (75.312 J K–1 mol–1 ) (80 K) = 6024.96 J mol–1

∆rS = Cp,m ln 1

2

TT

= (75.312 J K–1 mol–1) × 2.303 × log

K293K373

= 18.184 J K–1 mol–1

∆rU = ∆rH – p∆rV ~– ∆rH Step II: 1 mol H2O(1,373 K, 101.325 kPa) → 1 mol H2O (g, 373K, 101.325 kPa) qp = ∆vapH = 40.668 kJ mol–1

∆rS = K373molJ40668 1–

= 109.03 J K–1 mol–1

∆rU = ∆rH – p∆rV = 40668 J mol–1 – (101.325 kPa)

×

K273K373)moldm414.22( 1–3

= 40668 J mol–1 – 3 103 J mol–1 = 37565 J mol–1 Step III: 1 mol H2O(g, 373 K, 101.325 kPa) → 1 mol H2O(g, 523 K, 101.325 kPa) ∆rH = Cp,m (g) ∆T = (35.982 J K–1 mol–1) (150 K) = 5397.3 J mol–1

∆rS = Cp,m (g) ln 1

2

TT

= (35.982 J K–1 mol–1) × 2.303 × log

K373K523

= (35982 J K–1mol–1) × 2.303 × 0.1468 = 12.164 J K–1 mol–1 ∆rU = ∆rH – R(∆T) = 5397.3 J mol–1 – (8.314 J K–1 mol–1) (150 K) = 5397.3 J mol–1 – 1247.1 J mol–1 = 4 150.2 J mol–1 Thus ∆Utotal

= (6024.96 + 37565 + 4150.2) J mol–1

= 47740.16 J mol–1 ∆Htotal = (6024.96 + 40668 + 5397.3) J mol–1

= 52090.26 J mol–1 ∆Stotal = (18.184 + 109.03 + 12.164) J K–1 mol–1

= 139.378 J K–1 mol–1 2. It is possible to supercool water without freezing. 18

g of water are supercooled to 263.15 K(–10ºC) in a thermostat held at this temperature, and then crystallization takes place.

Calculate ∆rG for this process. Given: Cp(H2O,1) = 75.312 J K–1 mol–1

Cp (H2O,s) = 36.400 J K–1 mol–1 ∆fusH (at 0ºC) = 6.008 kJ mol–1 Sol. The process of crystallization at 0ºC and at 101.325

kPa pressure is an equilibrium process, for which ∆G = 0. The crystallization of supercooled water is a spontaneous phase transformation, for which ∆G must be less than zero. Its value for this process can be calculated as shown below.

The given process H2O(1, – 10ºC) → H2O(s, –10ºC) is replaced by the following reversible steps. (a) H2O(1, – 10ºC) → H2O(1, 0ºC) ...(1)

∆rH1 = ∫K15.273

K15.263m,p )1(C dT

UNDERSTANDINGPhysical Chemistry


= (75.312 J K–1 mol–1 ) (10 K) = 753.12 J mol–1

∆rS1 = ∫K15.273

K15.263

m,p

R)1(C

dT

= (75.312 J K–1mol–1) × ln

K15.263K15.273

= 2.809 J K–1 mol–1 (b) H2O(1, 0ºC) → H2O(s, 0ºC) ...(2) ∆rH2 = – 6.008 kJ mol–1

∆rS2 = – )K15.273(

)molJ6008( 1– = – 21.995 J K–1 mol–1

(c) H2O(s, 0ºC) → H2O(s, –10ºC) ...(3)

∆rH3 = ∫K15.263

K15.273m,p )s(C dT

= (36.400 J K–1 mol–1)(–10 K) = – 364.0 J mol–1

∆rS3 = ∫K15.263

K15.273

m,p

T)s(C

dT

= (36.400 J K–1 mol–1) ×ln

K15.273K15.263

= – 1.358 J K–1 mol–1 The overall process is obtained by adding Eqs. (1),

(2) and (3), i.e. H2O(1, –10ºC) → H2O(s, –10ºC) The total changes in ∆rH and ∆rS are given by ∆rH = ∆rH1 + ∆rH2 + ∆rH3 =(753.12 – 6008 – 364.0) J mol–1

= – 5618.88 J mol–1 ∆rS = ∆rS1 + ∆rS2 + ∆rS3 = (2.809 – 21.995 – 1.358) J K–1 mol–1 = – 20.544 J K–1 mol–1 Now ∆rG of this process is given by ∆rG = ∆rH – T∆rS = – 5618.88 J mol–1 – (263.15 K)( –20.544 J K–1 mol–1 ) = – 212.726 J mol–1

3. Given a solution that is 0.5 M CH3COOH. To what volume at 25ºC must one dm3 of this solution be diluted to (a) double the pH; (b) double the hydroxide-ion concentration ?

Given that Ka = 1.8 × 10–5 M. Sol. If α is the degree of dissociation of acetic acid of

concentration c then the concentrations of various species in the solution are

CH3COOH + H2O CH3COO– + H2O+ c(1 – α) cα cα With these concentrations, the equilibrium constant

becomes

Ka = ]COOHCH[

]OH][COOCH[

3

3–

3+

= )–1(c)c)(c(

ααα = cα2

or = α = c

Ka

Substituting the values, we have

α = )M5.0(

)M108.1( 5–× = 6 × 10–3

The concentration of hydrogen ions is given as [H3O+] = cα = (0.5 M) (6 × 10–3) = 3 × 10–3 M Hence, pH = – log [H3O+]/M = – log (3 × 10–3) = 2.52 (a) To double the pH Thus pH = 5.04 Since pH = – log [H3O+]/M therefore [H3O+]/M= 10–pH . Substituting the value of

pH , we have [H3O+]/M = 10–5.04 = 0.912 × 10–5 = 9.12 × 10–6 Thus, c1α = 9.12 × 10–6 M In dilution, α will increase, and its value will not be

negligible in comparison to one. Thus, we shall have to use the expression

Ka= )–1(c)c(

1

21

αα =

αα–1

c 21 =

ααα

–1)c( 1 =

αα×

–1)M1012.9( 6–

or (1.8 × 10–5 M) (1 – α) = ( 9.12 × 10–6 M) α which gives (9.12 × 10–6 M + 1.8 × 10–5 M) α = 1.8 × 10–5 M

or a = M1012.27

M108.16–

5–

×× = 0.6637

Since c1α = 9.12 × 10–6 M and α = 0.6637 Therefore,

c1 = 6637.0

M1012.9 6–× = 1.374 × 10–5 M

Volume to which the solution should be diluted

= 1c

cV = )M10374.1()dm1()M5.0(

5–

3

× = 3.369 × 104 dm3

(b) To double the hydroxyl-ion concentration Since [H3O+] in 0.5 M acetic acid is 3 × 10–3 M,

therefore

[OH–] = )M103(

)M100.1(3–

214–

×

×


In the present case, the concentration of hydroxyl becomes

[OH–] = )M103(

)M100.1(23–

214–

×

×

which gives

[H3O+] = 2

)M103( 3–× = 1.5 × 10–3 M

For the concentration, we can use Ka = c2α

2 = (c2α) (α)

or α = )c(

K

2

a

α =

)M105.1()M108.1(

3–

5–

×

× = 1.2 × 10–2

Thus, c2 = 2–

3–

102.1)M105.1(

×

× = 1.25 × 10–1 M = 0.125 M

Volume to which the solution should be diluted

= 2c

cV = )M125.0(

)dm1)(M5.0( 3 = 4 dm3

4. The freezing point of an aqueous solution of KCN

containing 0.189 mol kg–1 was – 0.704 ºC. On adding 0.095 mol of Hg(CN)2, the freezing point of the solution became –0.530ºC. Assuming that the complex is formed according to the equation

Hg(CN)2 + x CN– → –x2x)CN(Hg +

Find the formula of the complex. Sol. Molality of the solution containing only KCN is

m = f

f

K)T(–∆

= )molkgK86.1(

)K704.0(1–

= 0.379 mol kg–1

This is just double of the given molality ( = 0.189 mol kg–1) of KCN, indicating complete dissociation of KCN. Molality of the solution after the formation of the complex

m = f

f

K)T(–∆ =

)molkgK86.1()K530.0(

1– = 0.285 mol kg–1

If it be assumed that the whole of Hg(CN)2 is converted into complex, the amounts of various species in 1 kg of solvent after the formation of the complex will be

n(K+) = 0.189 mol, n(CN–) = (0.189 – x) mol

))CN(Hg(n –x2x+ = 0.095 mol

Total amount of species in 1 kg solvent becomes ntotal = [0.189 + (0.189 – x) + 0.095] mol = (0.473 – x) mol Equating this to 0.285 mol, we get (0.473 – x) mol = 0.285 mol i.e. x = (0.473 – 0.285) = 0.188

Number of CN– units combined = mol095.0mol188.0 = 2

Thus, the formula of the complex is –24)CN(Hg .

5. From the standard potentials shown in the following

diagram, calculate the potentials º1E and º

2E .

BrO3– 0.54 V BrO– 0.45 V

21 Br2

1.07 V Br–

0.17 V

E2º

E1º

Sol. The reaction corresponding to the potential Eº1 is

BrO3– + 3H2O + 5e– =

21 Br2 + 6OH– ...(1)

This reaction can be obtained by adding the following two reduction reactions:

BrO3– + 2H2O + 4e– = BrO– + 4OH– ...(2)

BrO– + H2O + e– = 21 Br2 + 2OH– ...(3)

Hence the free energy change of reaction (1) will be

º)1(reactionG∆ = º

)2(reactionG∆ + º)3(reactionG∆

Replacing ∆Gºs in terms of potentials, we get – 5FE1º = – 4F(0.54 V) – 1F (0.45 V) = (–2.61 V) F

Hence E1º = 5

V61.2 = 0.52 V

Now the reaction corresponding to the potential E2º is BrO3

– + 2H2O + 6e– = Br– + 6OH– ...(4) This reaction can be obtained by adding the

following three reactions. BrO3

– + 2H2O + 4e– = BrO– + 4OH– (Eq.2)

BrO– + H2O + e– = 21 Br2 + 2OH– (Eq.3)

21 Br2 + e– = Br– ...(5)

Hence º

)4(reactionG∆ = º)2(reactionG∆ + º

)3(reactionG∆

+ º)5(reactionG∆

or – 6F(E2º) = – 4F(0.54 V) – 1F(0.45 V) – 1F (1.07 V) = (– 3.68 V) F

or E2º = 668.3 = 0.61 V.


1. Let f(x) = sinx and

g(x) =

π>π≤≤≤≤

x;2/xsinx0for;xt0);t(fmax

2 Discuss

the continuity and differentiability of g(x) in (0, ∞) 2. Is the inequality sin2x < x sin(sinx) true for

0 < x < π/2 ? Justify your answer. 3. A shop sells 6 different flavours of ice-cream. In how

many ways can a customer choose 4 ice-cream cones if

(i) they are all of different flavours; (ii) they are not necessarily of different flavours; (iii) they contain only 3 different flavoures; (iv) they contain only 2 or 3 different flavoures ? 4. Using vector method, show that the internal

(external) bisector of any angle of a triangle divides the opposite side internally (externally) in the ratio of the other two sides containing the triangle.

5. Prove that (a) cos x + nC1 cos 2x + nC2 cos 3x + ............. + nCn

cos(n + 1)x = 2n. cosnx/2. cos

+ x

22n

(b) sin x + nC1 sin 2x + nC2 sin 3x + .............. + nCn

sin(n + 1)x = 2n . cosn x/2 . sin

+ x

22n

6. In a town with a population of n, a person sands two

letters to two sperate people, each of whom is asked

to repeat the procedure. Thus, for each letter

received, two letters are sent to separate persons

chosen at random (irrespective of what happened in

the past). What is the probability that in the first k

stages, the person who started the chain will not

receive a letter ?

7. Prove the identity :

∫ −x

0

zzx 2e dz = ∫ −x

0

4z4x 22ee dz, deriving for the

function f(x) = ∫ −x

0

zzx 2e dz a differential equation

and solving it.

8. Prove that ∫ θθsecnsin dθ

= –1n

)1ncos(2−

θ− – ∫ θθθ dsec)2–nsin( dθ.

Hence or otherwise evaluate

∫π

θθθ2/

0 cos3sin5cos dθ.

9. Find the latus rectum of parabola 9x2 – 24 xy + 16y2 – 18x – 101y + 19 = 0. 10. A circle of radius 1 unit touches positive x-axis and

positive y-axis at A and B respectively. A variable line passing through origin intersects the circle in two

points in two points D and E. Find the equation of the

lines for which area of ∆ DEB is maximum.

`tàxÅtà|vtÄ VtÄÄxÇzxá This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety of possible twists and turns of problems in mathematics that would be very helpful in facing IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and we hope that this section would prove a rich resource for practicing challenging problems and enhancing the preparation level of IIT JEE aspirants.

By : Shailendra MaheshwariJoint Director Academics, Career Point, KotaSo lu tions will be p ublished in next issue

10Set


1. as φ (a) = φ (b) = φ (c) so by Rolle’s theorem there must exist at least a point

x = α & x = β each of intervals (a, c) & (c, b) such that φ′(α) = 0 & φ′(β) = 0. Again by Rolle’s theorem, there must exist at least a point x = µ such that α < µ < β where φ′(µ) = 0

so )ca()ba(

)a(f2−−

+ )ab()cb(

)b(f2−−

+ )bc()ac(

)c(f2−−

- f ′′ (µ) = 0

so )ca()ba(

)a(f−−

+ )ab()cb(

)b(f−−

+ )bc()ac(

)c(f−−

= 21 f ′′ (µ)

where a < µ < b. 2. Required probability

1 . 65 .

65 .

65 ........

65 .

61 =

2r

65 −

.

61 (r – 2) times

Note : any number in 1st loss same no. does not in 2nd (any other comes). Now 3rd is also diff. (and in same r − 2 times) Now (r − 1)th & r th must be same. 3. 2s = a + b + c ON = − BN + BO Let BN = x 2BN + 2CN + 2AR = 2s x + (a − x) + (b − a + x) = s x = s − b

R

O C

N B

M

A

r

I (h,k)

so h = ON = 2a − (s − b)

= 2

b2as2 ++− = 2

cb − & r = k.

so r = k = s∆ =

s)cs)(bs)(as(s −−−

r = k = s

)cs)(bs)(as(s −−−

2sk = )cba)(cba)(as(s −++−−

= )x2a)(x2a)(as(s +−−

2sk = )h4a)(as(s 22 −− required locus is 4s2y2 = A(a2 – 4x2)

⇒ s2y2 + Ax2 = 4

Aa2

where A is = s (s – a) here h2 < as so it is an ellipse 4. f (0) = c f (1) = a + b + c & f (−1) = a − b + c solving these,

a = 21 [f (1) + f (−1) − 2 f (0)] ,

b = 21 [f (1) − f (−1)] & c = f (0)

so f (x) = 2

)1x(x + f (1) + (1− x2) f (0) + 2

)1x(x −

f (−1) 2 | f (x) | < | x | | x + 1 | + 2 | 1 − x2 | + |x| | x − 1| ; as | f (1) | , | f (0) |, | f (−1) | ≤ 1.

2 | f (x) | ≤ | x | (x + 1) + 2 (1 − x2) + | x | (1 − x) as x ∈ [−1, 1]

so 2 | f (x) | ≤ 2 (|x| + 1 − x2) ≤ 2 . 45

so | f (x) | ≤ 45

Now as g (x) = x2 f (1/x) = 21 (1 + x) f (1)

+ (x2 − 1) f (0) + 21 (1 − x) f (−1)

so 2 | g (x) | ≤ | x + 1 | + 2 | 1 − x2 | + | 1 − x| ⇒ 2 | g (x) | ≤ x + 1 + 2 (1 − x2) | + 1 − x ; as x ∈ [−1, 1] ⇒ 2 | g (x) | < − 2x2 + 4 ≤ 4. ⇒ |g (x) | ≤ 2.

MATHEMATICAL CHALLENGES SOLUTION FOR JANUARY ISSUE (SET # 9)


5. Oil bed is being shown by the plane A′ PQ. θ be the angle between the planes A′ PQ & A′ B′ C′ Let A′ B′ C′ be the x − y plane with x axis along A′ C′ and origin at A′. The P.V.s of the various points are defined as follows

B

A C

B´

A´ P

Q

C´

point C′ : b i , point B′ : cos A i + c sin A j , point Q :

b i – z k , point P : i cos A i + c sin A j – y k normal vector to the plane A′ B′ C′ = 1n

r = bc sin A k

normal vector to the plane A’PQ = 2nr

= cz sin A i + (by - cz cos A) j + bc sin A k

so cos θ = |n||n|

n.n

11

21rr

rr

= 2/12222222 ]Asincb)Acosczby(Asinzc[Asinbc

+−+

cos θ = 2/12222222 )]Acosbycz2ybzc(Asincb[Asincb

−++

so tan θ = Asinbc

]Acosbycz2ybzc[ 2/12222 −+

so tan θ . sin A = Acosbcyz2

cy

bz

2

2

2

2−+

6. ∫ +−

x5cos21x7cosx8cos .

x5sin2x5sin2 dx

= ∫ ++−−)x10sinx5(sin2

x2sinx12sinx3sinx13sin dx

= ∫ +−+

)x10sinx5(sin2x12sin–x3sinx2sinx13sin dx

= ∫−

2x5cos

2x15sin.2.2

2x9cos

2x15sin2

2x11cos

2x15sin2

dx

= ∫−

2x5cos2

2x9cos

2x11cos

dx

= ∫−

2x5cos2

2xsinx5sin2

dx

= − 2 ∫

2xsin

2x5sin dx

= ∫

−

2x4cos

2x6cos dx

= ∫ − dx)x2cosx3(cos

= 3

x3sin − 2

x2sin + C

7. 2

2

dxyd = 2 ∫

x

0

dt)t(f

integrate using by parts method

dxdy = 2

− ∫∫

x

0

x

0

dx)x(f.xdt)t(fx

= 2

−∫

x

0

dt)t(f)tx(

again integrating,

y = 2

−−− ∫∫∫ dx0dt)t(fxdt)t(f)tx(x

x

0

x

0

x

0

= 2

+−− ∫∫∫ dx)x(f

2xdt)t(f

2xdt)t(f)tx(x

x

0

2x

0

2x

0

= ∫ −x

0

2 dt)t(f)xtx(2 − ∫x

0

2 dt)t(fx + ∫x

0

2 dt)t(ft

y = ∫ +−x

0

22 dt)t(f)txt2x( = ∫ −x

0

2 dt)t(f)tx(

8. To prove that αα

+

/1

1ba <

ββ

+

/1

1ba

Let ba = c > 0

so (cα + 1)1/α < (cβ + 1)1/β. Let f (x) = (cx + 1)1/x ; x > 0

f ′(x) = (cx + 1)1/x ln (cx + 1)

− 2x

1

+x1 (cx + 1) x

1 –1. cx ln c


= 2

1x1

x

x)1c(

−+ ]cnc)1c(n)1c([ xxxx ll +++− < 0

so f (x) is decreasing function so f (α) < f (β). Hence proved. 9. Point P (x, 1/2) under the given condition are length

PB = OB

B (t, 1)

A (t – 1)

CO

P θ

rθ = t ; so θ = t

from ∆PAB : 2

PB = PA sin 2θ

⇒ PB = 2 sin 2t ........(1)

Now ∠ PBC = 2θ =

2t ;

so from ∠ PCB ; 2θ =

2t

so from ∆ PCB ; PB

2/1 = sin 2t ........(2)

from (1) & (2) PB = 1 ; so θ = t = π/3

thus | PB |2 = (t − x)2 + 41 = 1.

| t − x | = 23 ; t − x =

23 ; as t > x

so x = 3π −

23

10. Let xn = 1n − + 1n + be rational, then

nx

1 = 1n1n

1++−

is also rational

nx

1 = 2

1n1n −−+ is also rational

1n + − 1n − is also rational

as 1n + + 1n − & 1n + − 1n − are rational

so 1n + + 1n − must be rational i.e. (n + 1) & (n – 1) are perfect squares. This is not possible as any two perfect squares differe

at least by 3. Hence there is not positive integer n for

which 1n − + 1n + is a rational.

TRUE OR FALSE

1. The thrust exerted by a liquid on the base of a vessel does not depend on the mass of the liquid but depends on the area of the base and height of the liquid.

2. The path of one projectile as seen from another projectile is a straight line.

3. The arithmetic logic shift unit (ALSU) is combinational circuit that performs a number of arithmetic, logic and shift micro-operations.

4. A thin circular disc of mass M and radius R is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with an angular velocity ω. Another disc of

the same dimensions but of mass 4M is gently

placed on the first disc coaxially. The angular velocity of the system now is 2ω/ 2 .

5. The rms speed of oxygen molecules (O2) at a certain temperature T (absolute) is v. If the temperature is doubled and oxygen gas dissociates into atomic oxygen, the rms speed remains unchanged.

Sol. 1. [True] 2. [True] Let u1 and θ1 be the initial speed and angle

of projection of the projectile and u2 and θ2 be the corresponding quantities, respectively, for the other projectile.

Then the coordinates of one as seen from the other projectile are

x = (u1 cos θ1 – u2 cos θ2) t, y = (u1 sin θ1 – u2 cos θ2) t

∴ yx =

2211

2211

sinusinucosucosu

θ−θθ−θ = m (say)

or x = my, which is the equation of a straight line.

3. [True]

4. [False]

+

4MM

21 R2 ω′ =

21 MR2ω

⇒ ω′ =54

ω

5. [False] vrms = const.MT

'rmsv = const.

2/MT2

= const. × 2MT = 2 vrms


1. An urn containing '14' green and '6' pink ball. K (< 14, 6) balls are drawn and laid a side, their colour being ignored. Then one more ball is drawn. Let P(E) be the probability that it is a green ball, then 20 P(E) = ..............

Sol. Let Ei denote the event that out of the first k balls drawn, i balls are green. Let A denote the event that (k + 1)th ball drawn is also green.

P(Ei) = k

20ik

6i

14

CCC −× 0 ≤ i ≤ k

and P(A/Ei) = k20i14

−−

Now P(A) = k20j14

C

CCk

0j k20

ik6

j14

−−

+×

∑=

−

Also (1 + x)14 – 1 (1 + x)6 = (14–1C0 + 14–1C1x +.......+ 14 – 1C14 – 1 x14–1) (6C0 + 6C1x + .......+ 6C6x6)

⇒ ∑=

−− +

k

0jjk

6j

114 )CC( = co-efficient of xk

∴ P(E) = 2014

14614

=+

∴ 20P(Ε) = 14 2. If f(x + y + z) = f(x) + f(y) + f(z) with f(1) = 1 and f(2) = 2 and x, y, z ∈ R, then evaluate

3

n

1rn n

)r3(f)r4(

lim∑

=∞→

is equal to__________

Sol. f(3) = 3f(1) = 3, f(4) = f(2 + 1 + 1) = 2 + 1 + 1 = 4 and so on. In general, we get f(r) = r for r ∈ N

⇒ 3n3

n

1rn n6

)1n2()1n(n12lim

n

)r3(f)r4(

lim++

=∞→

=∞→

∑

= 4

3. Six points (xi , yi), i = 1, 2,..., 6 are taken on the circle

x2 + y2 = 4 such that ∑

=

=6

1ii 8x and ∑

=

=6

1ii 4y . The

line segment joining orthocentre of a triangle made by any three points and the centroid of the triangle made by other three points passes through a fixed points (h, k), then h + k is_________

Sol. Let ∑=

α=6

1iix and ∑

=

β=6

1iiy .

Let O be the orthocentre of the triangle made by (x1, y1), (x2, y2) and (x3, y3)

⇒ O is (x1 + x2 + x3, y1 + y2 + y3) ≡ (α1, β1) Similarly let G be the centroid of the triangle made

by other three points

⇒ G is

++++3

yyy,

3xxx 654654

⇒ G is

β−βα−α

3,

311 .

The point dividing OG in the ratio 3 : 1 is

βα

4,

4 ≡ (2, 1) ⇒ h + k = 3

4. Let P(x) = x4 + ax3 + bx2 + cx + d where a, b, c, d are

constants. If P(1) = 10, P(2) = 20 and P(3) = 30,

compute 10

)8(P)2(P −+

Sol. Let Q(x) = P(x) – 10 x Q(1) = P(1) – 10 = 0 Q(2) = P(2) – 20 = 0 Q(3) = P(3) – 30 = 0 ∴ Q(x) is divisible by (x – 1) (x – 2) (x – 3) But Q (x) is a 4th degree polynomial ∴ Q(x) = (x – 1) (x – 2) (x – 3) (x – K) ∴ P(x) = (x – 1) (x – 2) (x – 3) (x – K) + 10x P(12) = (11) (10) (9) (12 – K) + 120 P (–8) = (–9) (–10) (–11) (–8 – K) – 80

Expert’s Solution for Question asked by IIT-JEE Aspirants

Students' Forum

MATHS


∴ 10

)8(P)12(P −+

= 10

80)K8(990120)K12(990 −+++−

= 10

80K9907920K99012011880 −++−+

= 10

1984010

784012000=

+ = 1984

5. If A be the area bounded by y = f(x), y = f–1(x) and

line 4x + 4y – 5 = 0 where f(x) is a polynomial of 2nd degree passing through the origin and maximum value of 1/4 at x = 1, then 96A is equal to______

Sol. Let f(x) = ax2 + bx

41 = a + b ...(1)

f '(x) = 2ax + b ⇒ 2a + b = 0 ...(2)

P Q

AC

B y = x

Y

O X

From (1) and (2),

a = – 41 , b =

21

f(x) = 4

xx2 2−

Since 4x + 4y – 5 = 0 passes through

A

41,1 and B

1,

41 so area bounded is

OAB = 2 × OAC = 2 [area (OCP) + area(CPQA) – OAQ]

= 2

∫−

−−××++××1

0dx

4

2xx2

8

51

2

1

4

1

8

5

8

5

8

5

2

1

= 2

−×+

61

163

87

12825 =

9637 = A, then 96A = 37

6. If sin–1 x ∈

π

2,0 , then the value of

tan

+ −−−−

2))x(cos(sinsin))x(sin(coscos 1111

is___

Sol. As sin–1x ∈

π

2,0 and cos–1x =

2π – sin–1x

⇒ cos–1 x ∈

π

2,0

⇒ sin(cos–1 x) = cos (sin–1x) = 2x1

1

−

Thus, cos–1(sin(cos–1x)) + sin–1(cos(sin–1x)) = 2π .

⇒ required value = tan 4π = 1

Do you know

• The smallest bone in the human body is the stapes or stirrup bone located in the middle ear. It is approximately .11 inches (.28 cm) long.

• The longest cells in the human body are the motor neurons. They can be up to 4.5 feet (1.37 meters) long and run from the lower spinal cord to the big toe.

• There are no poisonous snakes in Maine.

• The blue whale can produce sounds up to 188 decibels. This is the loudest sound produced by a living animal and has been detected as far away as 530 miles.

• The largest man-made lake in the U.S. is Lake Mead, created by Hoover Dam.

• The poison arrow frogs of South and Central America are the most poisonous animals in the world.

• A new born blue whale measures 20-26 feet (6.0 - 7.9 meters) long and weighs up to 6,614 pounds (3003 kg).

• The first coast-to-coast telephone line was established in 1914.

• The Virginia opossum has a gestation period of only 12-13 days.

• The Stegosaurus dinosaur measured up to 30 feet (9.1 meters) long but had a brain the size of a walnut.


Integration :

If dxd f(x) = F(x), then ∫ )x(F dx = f(x) + c, where

c is an arbitrary constant called constant of integration.

1. ∫ dxxn = 1n

x 1n

+

+(n ≠ –1)

2. ∫ dxx1 = log x

3. ∫ dxex = ex

4. ∫ dxa x = alog

a

e

x

5. ∫ dxxsin = – cos x

6. ∫ dxxcos = sin x

7. ∫ dxxsec2 = tan x

8. ∫ dxxeccos 2 = – cot x

9. ∫ sec x tan x dx = sec x

10. ∫ cosec x cot x dx = – cosec x

11. ∫ sec x dx = log(sec x + tan x) = log tan

π

+42

x

12. ∫ cosec x dx = – log (cosec x + cot x) = log tan

2x

13. ∫ tan x dx = – log cos x

14. ∫ cot x dx = log sin x

15. ∫− 22 xa

dx = sin–1

ax = – cos–1

ax

16. ∫ + 22 xadx =

a1 tan–1

ax = –

a1 cot–1

ax

17. ∫− 22 axx

dx = a1 sec–1

ax = –

a1 cosec–1

ax

18. ∫ − 22 ax1 =

a21 log

axax

+− , when x > a

19. ∫ − 22 xa1 dx =

a21 log

xaxa

−+ , when x < a

20. ∫− 22 ax

dx = log

−+ 22 axx = cos h–1

ax

21. ∫+ 22 ax

dx = log

++ 22 axx = sin h–1

ax

22. ∫ − 22 xa dx = 21 x 22 xa − +

21 a2 sin–1

ax

23. ∫ − 22 ax dx = 21 x 22 ax −

– 21 a2log

−+ 22 axx

24. ∫ + 22 ax dx = 21 x 22 ax +

+ 21 a2 log

++ 22 axx

25. ∫ )x(f)x´(f dx = log f(x)

26. ∫ )x(f)x´(f dx = 2 )x(f

Integration by Decomposition into Sum : 1. Trigonometrical transformations : For the

integrations of the trigonometrical products such as sin2x, cos2x, sin3x, cos3x, sin ax cos bx, etc., they are expressed as the sum or difference of the sines and cosines of multiples of angles.

2. Partial fractions : If the given function is in the form of fractions of two polynomials, then for its integration, decompose it into partial fractions (if possible).

Integration of some special integrals :

(i) ∫ ++ cbxaxdx

2

This may be reduced to one of the forms of the above formulae (16), (18) or (19).

INTEGRATION Mathematics Fundamentals

MA

TH


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(ii) ∫++ cbxax

dx2

This can be reduced to one of the forms of the above formulae (15), (20) or (21).

(iii) ∫ ++ cbxax2 dx

This can be reduced to one of the forms of the above formulae (22), (23) or (24).

(iv) ∫ +++

cbxaxdx)qpx(

2 , ∫++

+

cbxax

dx)qpx(2

For the evaluation of any of these integrals, put px + q = A differentiation of (ax2 + bx + c) + B

Find A and B by comparing the coefficients of like powers of x on the two sides.

1. If k is a constant, then

∫ dxk = kx and ∫ dx)x(fk = k ∫ dx)x(f

2. ∫ ± dx)x(f)x(f 21 = ∫ dx)x(f1 ± ∫ dx)x(f2

Some Proper Substitutions :

1. ∫ f(ax + b) dx, ax + b = t

2. ∫ f(axn + b)xn–1dx, axn + b = t

3. ∫ fφ(x) φ´(x) dx, φ(x) = t

4. ∫ dx)x(f)x´(f , f(x) = t

5. ∫ − 22 xa dx, x = a sin θ or a cos θ

6. ∫ + 22 xa dx, x = a tan θ

7. ∫ +−

22

22

xaxa dx, x2 = a2 cos 2θ

8. ∫ ± xa dx, a ± x = t2

9. ∫ +−

xaxa dx, x = a cos 2θ

10. ∫ − 2xax2 dx, x = a(1 – cos θ)

11. ∫ − 22 ax dx, x = a sec θ

Substitution for Some irrational Functions :

1. ∫ ++ bax)qpx(dx , ax + b = t2

2. ∫+++ cbxax)qpx(

dx2

, px + q = t1

3. ∫ +++ bax)rqxpx(dx

2, ax + b = t2

4. ∫++ cax)rpx(

dx22

, at first x = t1 and then a + ct2 = z2

Some Important Integrals :

1. To evaluate ∫ β−α− )x)(x(dx , ∫

−β

α−x

x dx,

∫ −βα− )x)(x( dx. Put x = α cos2θ + β sin2θ

2. To evaluate ∫ + xcosbadx , ∫ + xsinba

dx ,

∫ ++ xsincxcosbadx

Replace sin x =

+

2xtan1

2xtan2

2 and cos x =

+

−

2xtan1

2xtan1

2

2

Then put tan 2x = t.

3. To evaluate ∫ +++

xsincxcosbaxsinqxcosp dx

Put p cos x + q sin x = A(a + b cos x + c sin x) + B. diff. of (a + b cos x + c sin x) + C A, B and C can be calculated by equating the

coefficients of cos x, sin x and the constant terms.

4. To evaluate ∫ ++ xsincxcosxsinb2xcosadx

22 ,

∫ + bxcosadx2 , ∫ + xsinba

dx2

In the above type of questions divide Nr and Dr by cos2x. The numerator will become sec2x and in the denominator we will have a quadratic equation in tan x (change sec2x into 1 + tan2x).

Putting tan x = t the question will reduce to the form

∫ ++ cbtatdt

2

5. Integration of rational function of the given form

(i) ∫ +++

424

22

akxxax dx, (ii) ∫ ++

−424

22

akxxax dx, where

k is a constant, positive, negative or zero. These integrals can be obtained by dividing

numerator and denominator by x2, then putting

x – x

a 2 = t and x +

xa 2

= t respectively.

Integration of Product of Two Functions :

1. ∫ f1(x) f2(x) dx = f1(x) ∫ f2(x) dx – [ ]∫ ∫ dx)x(f)x(f( 2'1 dx


Proper choice of the first and second functions : Integration with the help of the above rule is called integration by parts, In the above rule, there are two terms on R.H.S. and in both the terms integral of the second function is involve. Therefore in the product of two functions if one of the two functions is not directly integrable (e.g. log x, sin–1x, cos–1x, tan–1x etc.) we take it as the first function and the remaining function is taken as the second function. If there is no other function, then unity is taken as the second function. If in the integral both the functions are easily integrable, then the first function is chosen in such a way that the derivative of the function is a simple functions and the function thus obtained under the integral sign is easily integrable than the original function.

2. ∫ + )cbxsin(eax dx

= 22

ax

bae+

[a sin (bx + c) – b cos (bx + c)]

= 22

ax

ba

e

+sin

−+ −

abtancbx 1

3. ∫ + )cbxcos(eax dx

= 22

ax

bae+

[a cos (bx + c) + b sin(bx + c)]

= 22

ax

ba

e

+cos

−+ −

abtancbx 1

4. ∫ ekxkf(x) + f´(x) dx = ekxf(x)

5. ∫ xloge = x(logex – 1) = x loge

ex

Integration of Trigonometric Functions : 1. To evaluate the integrals of the form

I = ∫ sinmx cosnx dx, where m and n are rational

numbers. (i) Substitute sin x = t, if n is odd; (ii) Substitute cos x = t, if m is odd; (iii) Substitute tan x = t, if m + n is a negative even

integer; and

(iv) Substitute cot x = t, if 21 (m + 1) +

21 (n – 1) is

an integer.

2. Integrals of the form ∫R (sin x, cos x) dx, where R is

a rational function of sin x and cos x, are transformed into integrals of a rational function by the substitution

tan 2x = t, where –π < x < π. This is the so called

universal substitution. Sometimes it is more

convenient to make the substitution cot2x = t for

0 < x < 2π. The above substitution enables us to integrate any

function of the form R (sin x, cos x). However, in practice, it sometimes leads to extremely complex rational functions. In some cases, the integral can be simplified by –

(i) Substituting sin x = t, if the integral is of the form

∫R (sin x) cos x dx.

(ii) Substituting cos x = t, if the integral is of the form

∫ R (cos x) sin x dx.

(iii) Substituting tan x = t, i.e. dx = 2t1dt+

, if the

integral is dependent only on tan x. Some Useful Integrals :

1. (When a > b) ∫ + xcosbadx

= 22 ba

2

−tan–1

+−

2xtan

baba + c

2. (When a < b) ∫ + xcosbadx

= –22 ab

1

−log

baaxtanab

baaxtanab

++−

+−−

3. (when a = b) ∫ + xcosbadx =

a1 tan

2x + c

4. (When a > b) ∫ + xsinbadx

= 22 ba

2

− tan–1

−

+

22 ba

b2xtana

+ c

5. (When a < b) ∫ + xsinbadx

= 22 ab

1

− log

22

22

abb2xtana

abb2xtana

−++

−−+

+ c

6. (When a = b) ∫ + xsinbadx =

a1 [tan x – sec x] + c


Functions with their Periods :

Function Period

sin (ax + b), cos (ax + b), sec (ax + b), cosec (ax + b)

2π/a

tan(ax + b), cot (ax + b) π/a

|sin (ax + b)|, |cos (ax + b)|, |sec (ax + b)|, |cosec (ax + b)|

π/a

|tan (ax + b)|, |cot (ax + b)| π/2a

Trigonometrical Equations with their General Solution:

Trgonometrical equation General Solution

sin θ = 0 θ = nπ

cos θ = 0 θ = nπ + π/2

tan θ = 0 θ = nπ

sin θ = 1 θ = 2nπ + π/2

cos θ = 1 θ = 2nπ

sin θ = sin α θ = nπ + (–1)n α

cos θ = cos α θ = 2nπ ± α

tan θ = tan α θ = nπ + α

sin2θ = sin2α θ = nπ ± α

tan2θ = tan2α θ = nπ ± α

cos2θ = cos2α θ = nπ ± α

*coscossinsin

α=θα=θ

θ = 2nπ + α

*tantansinsin

α=θα=θ

θ = 2nπ + α

*coscostantan

α=θα=θ

θ = 2nπ + α

* If α be the least positive value of θ which satisfy two given trigonometrical equations, then the general value of θ will be 2nπ + α.

Note : 1. If while solving an equation we have to square it,

then the roots found after squaring must be checked whether they satisfy the original equation or not. e.g. Let x = 3. Squaring, we get x2 = 9, ∴ x = 3 and – 3 but x = – 3 does not satisfy the original equation x = 3.

2. Any value of x which makes both R.H.S. and L.H.S. equal will be a root but the value of x for which ∞ = ∞ will not be a solution as it is an indeterminate form.

3. If xy = xz, then x(y – z) = 0 ⇒ either x = 0 or

y = z or both. But xy =

xz ⇒ y = z only and

not x = 0, as it will make ∞ = ∞. Similarly, if ay = az, then it will also imply y = z only as a ≠ 0 being a constant.

Similarly, x + y = x + z ⇒ y = z and x – y = x – z ⇒ y = z. Here we do not take x = 0 as in the above because x is an additive factor and not multiplicative factor.

4. When cos θ = 0, then sin θ = 1 or –1. We have to verify which value of sin θ is to be chosen which

satisfies the equation cos θ = 0 ⇒ θ =

+

21n π

If sin θ = 1, then obviously n = even. But if sin θ = –1, then n = odd.

Similarly, when sin θ = 0, then θ = nπ and cos θ = 1 or –1.

If cos θ = 1, then n is even and if cos θ = –1, then n is odd.

5. The equations a cos θ ± b sin θ = c are solved as follows :

Put a = r cos α, b = r sin α so that r = 22 ba + and α = tan–1 b/a.

The given equation becomes

r[cos θ cos α ± sin θ sin α] = c ;

cos (θ ± α) = rc provided

rc ≤ 1.

TRIGONOMETRICAL EQUATION

Mathematics Fundamentals

MA

TH

XtraEdge for IIT-JEE FEBRUARY 2010 56

Relation between the sides and the angle of a triangle: 1. Sine formula :

a

Asin = b

Bsin = c

Csin = R21

Where R is the radius of circumcircle of triangle ABC.

2. Cosine formulae :

cos A = bc2

acb 222 −+ , cos B = ac2

bca 222 −+ ,

cos C = ab2

cba 222 −+

It should be remembered that, in a triangle ABC If ∠A = 60º, then b2 + c2 – a2 = bc If ∠B = 60º, then a2 + c2 – b2 = ac If ∠C = 60º, then a2 + b2 – c2 = ab 3. Projection formulae : a = b cos C + c cos B, b = c cos A + a cos C c = a cos B + b cos A Trigonometrical Ratios of the Half Angles of a Triangle:

If s = 2

cba ++ in triangle ABC, where a, b and c

are the lengths of sides of ∆ABC, then

(a) cos2A =

bc)as(s − , cos

2B =

ac)bs(s − ,

cos2C =

ab)cs(s −

(b) sin2A =

bc)cs)(bs( −− ' sin

2B =

ac)cs)(as( −− ,

sin2C =

ab)bs)(as( −−

(c) tan2A =

)as(s)cs)(bs(

−−− ,

tan2B =

)bs(s)cs)(as(

−−− , tan

2C

)cs(s)bs)(as(

−−−

Napier's Analogy :

tan2

CB − = cbcb

+− cot

2A , tan

2AC − =

acac

+− cot

2B

tan2

BA − = baba

+− cot

2C

Area of Triangle :

∆ = 21 bc sin A=

21 ca sin B =

21 ab sin C

∆ =)CBsin(CsinBsina

21 2

+=

)ACsin(AsinCsinb

21 2

+=

)BAsin(BsinAsinc

21 2

+

sin A = bc2

)cs)(bs)(as(s −−− = bc2∆

Similarly sin B = ca2∆ & sin C =

ab2∆

Some Important Results :

1. tan2A tan

2B =

scs − ∴ cot

2A cot

2B =

css−

2. tan2A + tan

2B =

sc cot

2C =

∆c (s – c)

3. tan2A – tan

2B =

∆− ba (s – c)

4. cot2A + cot

2B =

2Btan

2Atan

2Btan

2Atan +

= cs

c−

cot2C

5. Also note the following identities : Σ(p – q) = (p – q) + (q – r) + (r – p) = 0 Σp(q – r) = p(q – r) + q(r – p) + r(p – q) = 0 Σ(p + a)(q – r) = Σp(q – r) + aΣ(q – r) = 0 Solution of Triangles : 1. Introduction : In a triangle, there are six

elements viz. three sides and three angles. In plane geometry we have done that if three of the elements are given, at least one of which must be a side, then the other three elements can be uniquely determined. The procedure of determining unknown elements from the known elements is called solving a triangle.

2. Solution of a right angled triangle : Case I. When two sides are given : Let the

triangle be right angled at C. Then we can determine the remaining elements as given in the following table.

Given Required

(i) a, b tanA =

ba , B = 90º – A, c =

Asina

(ii) a, c sinA =

ca , b = c cos A, B = 90º – A

Case II. When a side and an acute angle are given – In this case, we can determine

Given Required

(i) a, A B = 90º – A, b = a cot A, c =

Asina

(ii) c, A B = 90º – A, a = c sin A, b = c cos A


PHYSICS

Questions 1 to 6 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. A cylinder of radius R is floating in a liquid as

shown. The work done in submerging the cylinder completely in the liquid of density ρ is –

R

L/3

L

(A) 92

ρπR2L2g (B) 188

ρπR2L2g

(C) 31 ρπR2L2g (D)

92 ρR2L2g

2. An electron with a kinetic energy of 100 eV

enters the space between the plates of plane capacitor made of two dense metal grids at an angle of 30° with the plates of capacitor and leaves this space at an angle of 45° with the plates. What is the potential difference of the capacitor –

30°

45°

(A) 100 V (B) 50 V (C) 150 V (D) 200 V

3. A plane mirror is inclined at an angle θ with the horizontal surface. A particle is projected with velocity v at angle α. Image of the particle is observed from the frame of the particle projected path of the image as seen by the particle is –

θ α v

(A) parabolic path (B) straight line (C) circular path (D) helical path

4. The amplitude of wave disturbance propagating in

the positive x-axis is given by y = 1x2–x

12 +

at

t = 2 sec and y = 5x2x

12 ++

at t = 6 sec, where

x and y are in meters. Velocity of the pulse is - (A) 1 m/s in positive x-direction (B) + 2 m/s in negative x-direction (C) 0.5 m/s in negative x-direction (D) 1 m/s in negative x-direction

XtraEdge Test Series # 10

IIT-JEE 2010

Based on New Pattern

Time : 3 Hours Syllabus : Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus Instructions : Section - I • Question 1 to 6 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct

answer and -1 mark for wrong answer. • Question 7 to 10 are multiple choice questions with multiple (one or more than one) correct answer. +4 marks and

-1 mark for wrong answer. • Question 11 to 16 are passage based questions with multiple (one or more than one) correct answer. +5 marks will be

awarded for correct answer and -1 mark for wrong answer. Section - II • Question 17 to 19 are Numerical type questions. +6 marks will be awarded for correct answer and No Negative

marks for wrong answer.


5. A straight conductor of mass m and carrying a current i is hinged at one end and placed in a plane perpendicular to the magnetic field B as shown in figure. At any moment if the conductor is let free, then the angular acceleration of the conductor will be (neglect gravity) –

× × × × × × × × × × × × × × × ×

B

i L

Hinged end

(A) m2iB3 (B)

miB

32

(C) m2

iB (D) mB2

i3

6. The velocity of a body moving on a straight line

in v = v0 τ−

t

e , then the total distance moved by it before it stops -

(A) v0τ (B) 2 v0τ (C) 3v0τ (D) None of these Questions 7 to 10 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. 7. A solid is heated up and ∆H vs ∆θ (∆H : Heat

given, ∆θ : change in temperature) is plotted as shown in figure. Material exist in only one phase in –

B

C

D

F

A ∆θ

∆H

E

(A) AB (B) BC (C) CD (D) EF

8. A container having dimension 5 m × 4 m × 3 m is

accelerated along its breadth in horizontal. Container is filled with water upto the height of 1.5 m. Container is accelerated with 7.5 m/s2. If g = 10 m/s2

and density of water is 103 kg/m3 - A B

D C

3m

4 m

1.5m

(A) Gauge pressure at point C is 104 Pascal (B) Gauge pressure at point D is 3 × 104 Pascal (C) Gauge pressure at the middle of the base is

1.5 × 104 Pascal (D) Remaining value of liquid inside the

container is 20 m3 9. All capacitors were initially uncharged –

12Ω

15Ω

15Ω

10 µF

5 µF

10 Ω

50 V (A) Battery current just after closing of switch S is

3.42 A (B) Battery current just after closing of switch S is

0.962 A (C) Battery current after long time of closing of

switch S is 3.42 A (D) Battery current after long time of closing of

switch S is 0.962 A 10. R = 10 Ω and ε = 13 V and voltmeter and

ammeter are ideal then –

V

A

3Ω

ε

R

c 6V

a

b

8V

(A) Reading of ammeter is 2.4 A (B) Reading of ammeter is 8.4 A (C) Reading of voltmeter is 8.4 V (D) Reading of voltmeter is 27 V This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 11 to 16) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Passage : I (No. 11 to 13)

A thin super conducting (zero resistance) ring is held above a vertical, cylindrical metallic rod as shown in figure. The axis of symmetry of the ring is the same as that of the rod. The cylindrically symmetrical magnetic field around the ring can be described approximately in terms of the vertical


and radial components of the magnetic field vector as Bz = B0(1 – αz) and Br = B0βr where B0, α and β are constants and z and r are the vertical and radial position co-ordinates respectively. Initially the ring has no current flowing in it. When released, it starts to move downward with its axis still vertical. Consider the effect of self induction also.

Br

11. As ring will move downward magnetic flux through the ring -

(A) will increase (B) will decrease (C) will remain constant (D) will increase first and then decrease

12. As ring will move downward after it release current induced in ring -

(A) will increase (B) will decrease (C) remain constant (D) will oscillate 13. Lorentz force acting on the ring due to induced

current is - (A) vertical and constant (B) horizontal and constant (C) vertical and depend on vertical displacement

of ring (D) horizontal and depend on vertical displacement

of ring Passage : II (No. 14 to 16)

Medical researchers and technicians can track the characteristic radiation patterns emitted by certain inherently unstable isotopes as they spontaneously decay into other elements. The half-life of a radioactive isotope is the amount of time necessary for one-half of the initial amount of its nuclei to decay. The decay curves of

isotopes 39Y90 and 39Y91 are graphed below as functions of the ratio of N, the number nuclei remaining after a given period, to N , the initial number of nuclei.

1.00.90.80.70.60.50.40.30.20.1

1 2 3 4 5 6

39Y90

N/N0

Time (days)

1.00.90.80.70.60.50.40.30.20.1

30 60 90 120 150 180

39Y91

N/N0

Time (days) 14. The half-life of 39Y90 is approximately – (A) 2.7 days (B) 5.4 days (C) 27 days (D) 58 days 15. What will the approximate ratio of 39Y90 to 39Y91

be after 2.7 days if the initial samples of the two isotopes contain equal numbers of nuclei ?

(A) 1 : 1 (B) 1 : 2 (C) 2 : 1 (D) 10 : 1 16. Approximately how many 39Y91 nuclei will exist

after three half-lives have passed, if there are 1,000 nuclei to begin with ?

(A) 50 (B) 125 (C) 250 (D) 500

Numerical response questions (Q. 17 to 19). Answers to this Section are to be given in the form of nearest integer-in four digits. Please follow as per example : (i.e. for answer : 1492.2 write 1492; 491.8 write 0492; 92.5 write 0092; 2.1 write 0002)

17. A thermometer of mass 50 gm and specific heat 0.4 cal/gm/ºC reads 10ºC. It is then inserted into 1 kg of water and reads 40ºC in thermal equilibrium. The temperature of water before insertion of thermometer in 10 ºC is (Neglect other heat losses).

18. A uniform ball of radius R = 10 cm rolls without

slipping between two rails such that the horizontal distance is d = 16 cm between two contact points of the rail to the ball. If the angular velocity is 5 rad/s, then find the velocity of centre of mass of the ball in cm/s.

19. A wedge of mass M = 2 m0 rests on a smooth

horizontal plane. A small block of mass m0 rests over it at left end A as shown in figure. A sharp impulse is applied on the block, due to which it starts moving to the right with velocity v0 = 6 m/s. At highest point of its trajectory, the block


collides with a particle of same mass m0 moving vertically downwards with velocity v = 2 m/s and gets stuck with it. If the combined mass lands at the end point A of the body of mass M, calculate length l in cm. Neglect friction, take g = 10 m/s2.

l

20 cm

B

m0

A

CHEMISTRY

Questions 1 to 6 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

1. At 25º C, for the reaction Br2(l) + Cl2(g) 2BrCl (g) Kp = 2.032 . At the same temperature the vapour

pressure of Br2(l) is 0.281 atm. Pure BrCl(g) was introduced into a closed container of adjustable volume. The total pressure was kept 1 atm. and the temperature at 25ºC. What is the fraction of BrCl originally present that has been converted into Br2 and Cl2 at equilibrium, assuming that the gaseous species behave ideally ?

(A) 0.357 (B) 0.667 (C) 0.2 (D) None of these 2. The activation energy of a non-catalysed reaction

at 37ºC is 83.68 kJ mol–1 and the activation energy of the same reaction catalysed by an enzyme is 25.10 kJ mol–1. What is approximate ratio of the rate constants of the enzyme catalysed and the non-catalysed reactions ?

(A) 1022 (B) 1010 (C) 1020 (D) 106

3. Consider the following reaction sequence

3

23

AlCl

CHCCl)CH(||O

→ A →–

22 OH/NHNH B

3

3

AlCl

CClCH||O

→ C →HCN D →+H/OH2 E

end product (E) is -

(A) CH3–CH–CH2

| CH3

CH2–CH3

(B)

CH3–CH–CH2 |CH3

C–COOH|

|

OH

CH3

(C) CH3–CH–CH2

|CH3

CH–COOH| CH3

(D)

CH3–CH–CH2 |CH3

C–COOH| CH3

| Cl

4. (A) light blue coloured compound on heating will convert into black (B) which reacts with glucose gives red compound (C) and (A) reacts with ammonium hydroxide in excess in presence of ammonium sulphate give blue compound (D). What is (A) ?

(A) CuO (B) CuSO4 (C) Cu(OH)2 (D) [Cu(NH3)4] SO4 5. 0.80 g of impure (NH4)2 SO4 was boiled with 100

ml of a 0.2 N NaOH solution till all the NH3(g) evolved. The remaining solution was diluted to 250 ml. 25 ml of this solution was neutralized using 5 ml of a 0.2 N H2SO4 solution. The percentage purity of the (NH4)2SO4 sample is.

(A) 82.5 (B) 72.5 (C) 62.5 (D) 17.5

6. Which of the following is incorrect ? (A) The kinetic energy of the gas molecules is

higher above TC, is considered as super critical fluid

(B) At this temperature (TC) the gas and the liquid phases have different critical densities

(C) At the Boyle temperature the effects of the repulsive and attractive inter molecular forces just offset each other

(D) In the Maxwell's distribution curve of velocities the fraction of molecules have different velocities are different at a given temperature

Questions 7 to 10 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. 7. Consider the following compound CH3–CH(OH)–CH=CH–CH3 Which of the following is/are correct ? (A) Cis form is optically active (B) Trans form is optically active (C) Total number of stereo isomers are six (D) Trans form is optically inactive


8. Which of the following is/are correct ? (A) Trialkyl phosphine oxides are more stable

than the corresponding amine oxides due to pπ-dπ back bonding

(B) (SiH3)3N is less basic than (CH3)3N (C) PBr5 exist in ionic form as [PBr4]+ [Br]– in

solid state (D) CO, CNR, PR3 and NO all are the π acid

ligands 9. Which of the following statements is/are correct ? (A) The conductance of one cm3 of a solution is

called specific conductance (B) Specific conductance increases while molar

conductivity decreases on progressive dilution

(C) The limiting equivalent conductivity of weak electrolyte cannot be determine exactly by extraplotation of the plot of Λeq against c

(D) The conductance of metals is due to the movement of free electrons

10. Select the correct statement (s) (A) Radial function [R(r)] a part of wave function

is dependent on quantum number n only (B) Angular function depends only on the

direction and is independent to the distance from the nucleus

(C) ψ2 (r, θ, φ) is the probability density of finding the electron at a particular point in space

(D) Radial distribution function (4πr2R2) gives the probability of the electron being present at a distance r from the nucleus

This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 11 to 16) Each question has 4 choices (A), (B), (C) and (D) out of which ONE OR MORE THAN ONE is correct.

Passage : I (Que. No. 11 to 13)

Ascorbic acid, C6H8O6, also known as vitamin C is a dibasic acid undergoes dissociation as

C6H8O6 C6H7O6– + H+ ; K1 = 8 × 10–5

C6H7O6– C6H6O6

2– + H+ ; K2 = 1 × 10–12 The ascorbic acid is readily oxidised to

dehydroascorbic acid as

O

HOOH

HO

O

OH

→

O

O OH

HO

O

O

+ 2H+ + 2e

The estimation of ascorbic acid in a sample is made by titrating its solution with KIO3 solution which acts

as an intermediate and in presence of 1M HCl solution, the first excess of iodate gives blue colour with starch due to the redox change given below

3C6H8O6 + IO3– → 3C6H6O6 + I– + 3H2O

IO3– + 5I– + 6H+ → 3I2 + 3H2O

excess (generated in reaction)

However, if 5M HCl is used, the redox change occurs as follows : C6H8O6 + IO3

– + H++Cl– C6H6O6 +ICl + 2H2O 11. The 250 mL sample of fruit juice collected by

crushing a fruit is supposed to have only one ingradient which can react with KIO3 is taken in 500 mL measuring flask and 250 mL of 2M HCl is added. A 50 mL solution is now pipette out and titrated against intermediate KIO3 of concentration 4 × 10–3 M. It was found that 1 mL of KIO3 were used. The molarity and strength of ascorbic acid are.

(A) 4.8 × 10–4 M, 84.5 mg/litre (B) 9.6 × 10–4 M, 169 mg/litre (C) 4.8 × 10–4 M, 84.5 × 10–3 g/litre (D) 9.6 × 10–4 M, 169 g/litre

12. The deactivation of ascorbic acid follows first order kinetics. The 25 mL sample of juice is kept for 2 months and after 2 months one titration with same KIO3 in presence of 1 M HCl solution requires 0.5 mL of KIO3 solution. The average life of fruit juice is -

(A) 60 day (B) 50 day (C) 86.5 day (D) 120 day 13. The degree of dissociation of ascorbic acid

solution is - (A) 0.40 (B) 0.33 (C) 0.20 (D) 0.15 Passage : II (Que. No. 14 to 16)

BiCl3

NH3

H2SBlack ppt.(U)

Red substance(T)

water White turbidity(P)

Alkali +Na2SnO2

Boil withdil HNO3KI solution Black ppt.(R)

Conc.H2SO4

(Q)

Yellow solution(S)

14. Black ppt. (R) is - (A) Bi2O3 (B) Na2SnO3 (C) Bi(OH)3 (D) Bi 15. (Q) is - (A) Bi2(SO4)3 (B) Bi2O3 (C) Bi2O5 (D) Both (A) & (B)


16. Yellow solution(s) is because of the formation of - (A) ppt of BiI3 (B) I2 in aqueous solution (C) KI3 (D) All of these Numerical response questions (Q. 17 to 19). Answers to this Section are to be given in the form of nearest integer-in four digits. Please follow as per example : (i.e. for answer : 1492.2 write 1492; 491.8 write 0492; 92.5 write 0092; 2.1 write 0002)

17. A saturated solution of iodine in water contains 0.33g I2 per dm3. More than this can dissolve in a KI solution as a result of the reaction I2 + I– I3

–. A 0.10 M KI solution actually dissolves 12.5g I2 per dm3, most of which is converted into I3

–. Assuming the concentration of I2 in all saturated solution is same, calculate the equilibrium constant for the above reaction.

18. A 0.138-g sample of solid magnesium (molar

mass = 24.30g mol–1) is burned in a constant volume bomb calorimeter that has a heat capacity of 1.77 kJ ºC–1. The calorimeter contains 300 mL of water (density 1g mL–1) and its temperature is raised by 1.126ºC. The numerical value for the enthalpy of combustion of the solid magnesium at 298 K in kJ mol–1 is.

19. Two liquids A and B form an ideal solution at

temperature T. When the total vapour pressure above the solution is 450 torr, the amount fraction of A in the vapour phase is 0.35 and in the liquid phase is 0.70. The sum of the vapour pressures of pure A and pure B at temperature T is.

MATHEMATICS

Questions 1 to 6 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. If the last term in the binomial expansion of

n3/1

212

− is

8log

3/5

3

31

, then the 5th term

from the beginning is - (A) 210 (B) 420 (C) 105 (D) none of these

2. The matrix product

−121

[1 2 – 1]

(A) is not defined (B) equals [–1]

(C) equals

141

(D) is not invertible

3. The domain of definition of

f(x) =

+−

5x1xlog 4.0 ×

36x1

2 − is

(A) (– ∞, 0) ~ – 6 (B) (0, ∞) ~ 1, 6 (C) (1, ∞) ~ 6 (D) [1, ∞) ~ 6

4. The coordinates of the point on the parabola y2 = 8x which is at minimum distance from the circle x2 + (y + 6)2 = 1 are

(A) (2, – 4) (B) (18, –12) (C) (2, 4) (D) none of these

5. If

I=

+

∫

πxcos

21cos3xcos

21sin2e

0

|x|cos sinxdx,

then I equals - (A) e7 cos (1/2)

(B) e7 [cos(1/2) – sin(1/2)] (C) 0 (D) none of these

6. The solution of the differential equation

2

2

dxyd = sin 3x + ex + x2 when y1(0) = 1 and

y(0) = 0 is -

(A) 9

x3sin− + ex +12x 4

+31 x – 1

(B) 9

x3sin− + ex +12x 4

+31 x

(C) 3

x3cos− + ex +12x 4

+31 x + 1

(D) none of these Questions 7 to 10 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. 7. The determinant

∆ =0cbba

cbcbbaba

+α+α+α+α

is equal to zero if -

(A) a, b, c are in A.P. (B) a, b, c are in G.P. (C) a, b, c are in H.P. (D) α is a root of ax2 + 2bx + c = 0


8. For two events A and B, if P(A) = P(A \ B) = 1/4 and P(B \ A) = 1/2, then (A) A and B are independent (B) A and B are mutually exclusive (C) P (A′ \ B) = 3/4 (D) P (B′ \ A′) = 1/2

9. The

→ 38

0x x1xlim (where [x] is greatest integer

function) is - (A) a nonzero real number (B) a rational number (C) an integer (D) zero

10. The solution of xy2

1yxdxdy 22 ++

= satisfying

y(1) = 1 is given by - (A) a system of hyperbola (B) a system of circles (C) y2 = x(1 + x) – 1 (D) (x – 2)2 + (y – 3)2 = 5 This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 11 to 16) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.


C: x2 + y2 = 9, E:4

y9

x 22+ = 1, L: y = 2x

11. P is a point on the circle C, the perpendicular PL

to the major axis of the ellipse E meets the ellipse

at M, then PLML is equal to -

(A) 1/3 (B) 2/3 (C) 1/2 (D) none of these 12. If L represents the line joining the point P on C to

its centre O, then equation of the tangent at M to the ellipse E is -

(A) x + 3y = 3 5 (B) 4x + 3y = 3 5

(C) x + 3y + 3 5 = 0 (D) 4x + 3 + 5 = 0 13. Equation of the diameter of the ellipse E

conjugate to the diameter represented by L is - (A) 9x + 2y = 0 (B) 2x + 9y = 0 (C) 4x + 9y = 0 (D) 4x – 9y = 0 Passage : II (No. 14 to 16)

Integrals of class of functions following a definite pattern can be found by the method of reduction

and recursion. Reduction formulas make it possible to reduce an integral dependent on the index n > 0, called the order of the integral, to an integral of the same type with a smaller index. Integration by parts helps us to derive reduction formulas.

14. If In = ∫ + n22 )ax(dx then In + 1 +

2a1.

n2n21− In is

equal to -

(A) n22 )ax(x

+ (B) 1n222 )ax(

1an2

1−+

(C) n222 )ax(x.

an21

+ (D)

1n222 )ax(x

an21

++

15. If In, –m = ∫ xcosxsin

m

ndx then In,–m +

1m1n

−− In–2, 2–m, is

equal to -

(A) xcosxsin

1m

1n

−

− (B)

xcosxsin

)1m(1

1m

1n

−

−

−

(C) xcosxsin

)1n(1

1m

1n

−

−

− (D)

xcosxsin

1m1n

1m

1n

−

−

−−

16. If un = ∫++ cbx2ax

x2

ndx, then

(n + 1)aun+1 + (2n + 1)bun + nc un–1 is equal to -

(A) xn–1 cbxax2 ++ (B) cbxax

x2

2n

++

−

(C) cbxax

x2

n

++ (D) cbxaxx 2n ++


17. If sec A tan B + tan A sec B = 91, then the value of (sec A sec B + tan A tan B)2 is equal to …

18. If ∫π

+

2/

0 2

2

)xsin1(xcosx dx =

498A

π – π2 then A is …

19. Two circles are inscribed and circumscribed about

a square ABCD, length of each side of the square is 32. P and Q are two points respectively on these circles, then Σ(PA)2 + Σ(QA)2 is equal to …


PHYSICS

Questions 1 to 6 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

1. Ice point and steam point on a particular scale reads 10º and 80º respectively. The temperature on ºF scale when temperature on new scale is 45º is -

(A) 50º F (B) 112ºF (C) 122ºF (D) 138ºF 2. In a system of four unequal particles located in an

arrangement of non-linear, coplanar system - (A) The centre of mass must lie within the closed

figure formed by joining the extreme particles by straight line

(B) The centre of mass may lie within or outside the closed figure formed by joining the extreme particles by straight lines

(C) The centre of mass must lie within or at the edge of at least one of the triangles formed by any three particles.

(D) None of these

3. A body starts slipping on a smooth track from point A and leaves the track from point B as shown. The part OB of track is straight at angle 37º with horizontal. The maximum height of body from ground when it is in air is : (g = 10 m/s2)

37ºO ground

u = 0

H1 = 15mH2 = 10m

B

A

(A) 16.8 m (B) 13.6 m (C) 11.8 m (D) None of these

4. A particle is moving along x-axis and graph between velocity of the particle and position is given in figure :

6

2

4

v (m/s)

x (m) Acceleration of particle at x = 2 m is – (A) 2 m/s2 (B) 1 m/s2 (C) 4 m/s2 (D) 3 m/s2 5. Select the incorrect statement – (A) It is possible to transfer heat to a gas without

raising its temperature (B) It is possible to raise temperature of gas

without transfer heat to gas

IIT-JEE 2011

XtraEdge Test Series # 10

Based on New Pattern

Time : 3 Hours Syllabus : Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus Instructions : Section - I • Question 1 to 6 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct

answer and -1 mark for wrong answer. • Question 7 to 10 are multiple choice questions with multiple (one or more than one) correct answer. +4 marks and

-1 mark for wrong answer. • Question 11 to 16 are passage based questions with multiple (one or more than one) correct answer. +5 marks will be

awarded for correct answer and -1 mark for wrong answer. Section - II • Question 17 to 19 are Numerical type questions. +6 marks will be awarded for correct answer and No Negative

marks for wrong answer.


(C) It is possible to raise temperature by expanding the volume at keeping pressure constant

(D) It is not possible to calculate the work done by gas if we do not know the initial and final value of pressure and volume when process is isobaric

6. Select the incorrect statement – (A) The pressure on the bottom of a vessel filled

with liquid does not depend upon the area of liquid surface

(B) Buoyancy occurs because, as the depth in a fluid increase, the pressure increases

(C) The output piston of a hydraulic press cannot exceed the input piston's work

(D) The pressure of atmosphere at sea level corresponds to 101.3 millibar

Questions 7 to 10 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. 7. For two different gases x and y, having degrees

of freedom f1 and f2 and molar heat capacities at constant volume

1VC and 2VC respectively, the

lnP versus lnV graph is plotted for adiabatic process as shown, then –

x

y

lnV

lnP

(A) f1 > f2 (B) f2 > f1 (C)

2VC < 1VC (D)

21 VV CC >

8. Which of the following is valid wave equation

traveling on string - (A) Ae–b(x + vt) (B) A sec (kx – ωt)

(C) 2)x/t1(x11++

(D) A sin (x2 – vt2)

9. Two sphere of same radius and material, one solid

and one hollow are heated to same temperature and kept in a chamber maintained at lower temperature at t = 0 - (A) Rate of heat loss of the two sphere will be

same at t = 0 (B) Rate of temperature loss of the two sphere

will be same at t = 0 (C) Rate of heat loss of solid sphere will be more

than hollow sphere at t > 0 (D) Rate of temperature loss of the two sphere

may be same at t > 0

10. A stick is tied to the floor of the water tank with a string as shown. The length of stick is 2 m and its area of cross-section is 10–3 m2. If specific gravity of stick is 0.25 and g = 10 m/s2 , then –

(A) tension in the string is 5 N (B) buoyancy force acting on stick is 10 N (C) length of stick immersed in water is 1 m (D) tension in the string is zero This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 11 to 16) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.


A completely inelastic collision takes place between two body A and B of masses m and 2 m moving respectively with speed v each as shown.

The collision is oblique and before collision A is moving along positive x-axis while B is moving at angle θ with x-axis as shown. Then -

A vr

2m

θ

B

m x

y

vr

11. Speed of composite body after collision is -

(Α) θ+ cos453v (B)

3v

θ+ sin45

(C) θ+ cos543v (D) None of these

12. The angle α that the velocity vector of composite

body makes with x-axis after collision is -

(A)

θ+θ−

cos21sin2tan 1 (B)

θ+θ−

sin21cos2tan 1

(C)

θ+θ−

sin21sin2tan 1 (D) None of these

13. The loss in kinetic energy in collision is -

(A) 6

mv2 2(1 – cos θ) (B)

6mv2 2

(1 – sin θ)

(C) 6

mv4 2(1 – cos θ) (D)

6mv4 2

(1 – sin θ)


Passage : II (No. 14 to 16)

A disc of radius 20 cm is rolling with slipping on a flat horizontal surface. At a certain instant, the velocity of its centre is 4 m/s and its angular velocity is 10 rad/s. The lowest contact point is O.

4 m/s

10 rad/s

P C

O

14. Velocity of point O is -

(A) 0 (B) 2 m/s (C) 4 m/s (D) 8 m/s

15. Velocity of point P is -

(A) 2 5 m/s (B) 5 2 m/s

(C) 2 2 m/s (D) 8 m/s 16. The distance of instantaneous center of rotation

from the point O is -

(A) 0.2 m below (B) 0.2 m above (C) 0.4 m below (D) 0.4 m above


17. The upper edge of a gate in a dam runs along the water surface. The gate is 2.00 m high and 4.00 m wide and is hinged along a horizontal line through its center. The torque about the hinge arising from the force due to the water is (n × 104 Nm). Find value of n.

2 m

18. A longitudinal wave of frequency 220 Hz travels

down a copper rod of radius 8.00 mm. The average power in the wave is 6.50 µW. The amplitude of the wave is n × 10–8 m. Find n. (Density of copper is 8.9 × 103 kg/m3, young's modulus of copper Ycu = 1.1 × 1011 Pa).

19. A piston-cylinder device with air at an initial temperature of 30ºC undergoes an expansion process for which pressure and volume are related as given below

P (kPa) 100 25 6.25 V (m3) 0.1 0.2 0.4 The work done by the system is n × 103 J. Find n.

CHEMISTRY

Questions 1 to 6 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. The equilibrium constant for the reaction in

aqueous solution H3BO3 + glycerin (H3BO3 – glycerin) is

0.90. How many moles of glycerin should be added per litre of 0.10 M H3BO3 so that 80% of the H3BO3 is converted to the boric acid glycerin complex ?

(A) 4.44 (B) 4.52 (C) 3.62 (D) 0.08 2. Liquid NH3 ionises to a slight extent. At a certain

temperature it's self ionization constant(KSIC) is 10–30. The number of +

4NH ions are present per 100 cm3 of pure liquid are -

(A) 1 × 10–15 (B) 6.022 × 108 (C) 6.022 × 107 (D) 6.022 × 106

3. The preparation of SO3(g) by reaction

SO2(g) + 21 O2(g) SO3(g) is an exothermic

reaction. If the preparation follows the following temperature-pressure relationship for its % yield, then for temperature T1, T2 and T3 which of the following is correct -

T3

T2

T1

5040302010

1 2 3 4P(atm)

% y

ield

(A) T1 > T2 > T3 (B) T3 > T2 > T1 (C) T1 = T2 = T3 (D) None is correct


4. Which of the following would be optically inactive ?

Cl

C

CH3

C

Cl

CH3

H

H

CH3

OHH

CH3

OHH

(I) (II)

H

CH3

CH3

OH

OHH

(III) (A) Only I (B) Only II (C) Only II & III (D) I, II & III

5. A molecule may be represented by three structures having energies Q1, Q2 and Q3 respectively. The energies of these structures follow the order Q1 > Q2 > Q3 respectively. If the experimental bond energy of the molecule is QE, the resonance energy is -

(A) (Q1 + Q2 + Q3) – QE (B) QE – Q3 (C) QE – Q1 (D) QE – Q2

6. In a compound NC

C C

M(CO)3

NC C4H3 The number of sigma and pi bonds respectively

are - (A) 19, 11 (B) 19, 10 (C) 13, 11 (D) 19, 14 Questions 7 to 10 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. 7. Select the correct statement(s) about the

compound NO[BF4]. (A) It has 5σ and 2π bond (B) Nitrogen-oxygen bond length is higher than

nitric oxide (C) It is a diamagnetic species (D) B-F bond length in this compound is lower

than in BF3

8. Which of the following process is/are associated

with change of hybridization of the underlined compound ?

(A) Solidification PCl5 vapour (B) SiF4 vapour is passed through liquid HF (C) B2H6 is dissolved in THF (D) Al(OH)3 ppt. dissolved in NaOH

9. S, T and U are the aqueous chlorides of the elements X, Y and Z respectively. X, Y and Z are in the same period of the periodic table. U gives a white precipitate with NaOH but this white precipitate dissolves as more NaOH is added. When NaOH is added to T, a white precipitate forms which does not dissolve when more base is added. S does not give precipitate with NaOH.

Which of the following statements are correct ? (A) The three elements are metals (B) The electronegativity decreases from X to Y

to Z (C) X, Y and Z could be sodium, magnesium and

aluminium respectively (D) The first ionization increases from X to Y to Z 10. Which of the following enol form dominate over

keto form ?

(A)

O

H

N (B)

O

O

(C)

OO O

O

(D)

O

This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 11 to 16) Each question has 4 choices (A), (B), (C) and (D) out of which ONE OR MORE THAN ONE is correct.

Passage : I (Que. No. 11 to 13)

Alkenes undergo electrophilic addition reaction with Hg(OAc)2, BH3 and H2O. In all these cases reaction is regioselective reaction. BH3 gives addition reaction via formation of four centred cyclic transition state. Hg(OAc)2 gives addition reaction via formation of bridge carbocation as reaction intermediate whereas water gives addition reaction via formation of classical carbocation.

11. Alkene can be converted into alcohol by which of the following reagents -

(A) Hg(OAc)2/HOH followed by NaBH4 (B) BH3/THF followed by H2O2/NaOH (C) H2O/H2SO4 (D) None of these

12. In the gives reaction :

3

23

CH|

CHCCH =− → ]X[

3

23

CH|

OHCHCHCH −−

[X] is/are -


(A) Hg(OAc)2/HOH followed by NaBH4 (B) BH3 followed by H2O2/NaOH (C) H2O/H2SO4 (D) None of these

13. In the given reaction

3

23

3

CH|

CHCHCCH|CH

=−− → ]X[

33

33

CHCH||

CHCH—CCH|

HO

−−

[X] is/are - (A) H2O/H2SO4 (B) Hg(OAc)2/HOH followed by NaBH4 (C) BH3 followed by H2O2/NaOH (D) None of these

Passage : II (Que. No. 14 to 16)

Heat of neutralization is heat evolved when 1g equivalent of acid and 1g equivalent of base react together to form salt and water. Heat of neutralization is –57.1 kJ mol–1 for strong acid and strong base. In case of weak acid or weak base, it is less than 57.1 kJ mol–1.

14. 400 ml of 0.1 M NaOH is mixed with 300 ml of 0.1 M H2SO4. The heat evolved will be -

(A) 2.284 kJ (B) 1.713 kJ (C) 9.59 kcal (D) 7.1946 kcal 15. A solution of 200 ml of 1 M KOH is added to 200

ml of 1M HCl and the mixture is well shaken. This rise in temperature T1 is noted. The experiment is repeated by using 100 ml of each solution and increase in temperature T2 is again noted. Which of the following is/are incorrect -

(A) T1 = T2 (B) T2 is twice as large as T1 (C) T1 is twice as large as T2 (D) T1 is four times as large as T2 16. Which of the following will not produce

maximum energy except one? (A) Ba(OH)2 + H2SO4 (B) NH4OH + HCl (C) (COOH)2 + NaOH (D) H3PO4 + NaOH Numerical response questions (Q. 17 to 19). Answers to this Section are to be given in the form of nearest integer-in four digits. Please follow as per example : (i.e. for answer : 1492.2 write 1492; 491.8 write 0492; 92.5 write 0092; 2.1 write 0002)

17. Calculate the % of free SO3 in an oleum that is labelled '109% H2SO4'.

18. A mixture of NH3(g) and N2H4(g) is placed in a sealed container at 300 K. The total pressure is 0.5 atm. The container is heated to 1200 K at which time both substances decompose completely according to the equations

2NH3(g) → N2(g) + 3H2(g) and N2H4(g) → N2(g) + 2H2(g) After decomposition is complete, the total

pressure at 1200 K is found to be 4.5 atm. Find the mole % of N2H4 in the original mixture.

19. A 200 g sample of hard water is passed through the column of cation exchange resin, in which H+ is exchanged by Ca2+. The outlet water of column required 50 ml of 0.1 M NaOH for complete neutralization. What is the hardness of Ca2+ ion in ppm ?

MATHEMATICS

Questions 1 to 6 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. If w is an imaginary cube root of unity, then value

of the expression 2(1 + w) (1 + w2) + 3(2 + w) (2 + w2) + ….. + (n + 1) (n + w) (n + w2) is

(A) 41 n2(n + 1)2 + n (B)

41 n2(n + 1)2 – n

(C) 41 n(n + 1)2 – n (D) none of these

2. In a triangle PQR, R∠ = π/4. If tan (P/3)

and tan(Q/3) are the roots of the equation ax2 + bx + c = 0, then -

(A) a + b = c (B) b + c = 0 (C) a + c = b (D) b = c

3. Sum of all three digit numbers (no digit being zero) having the property that all digits are perfect squares, is -

(A) 3108 (B) 6210 (C) 13986 (D) none of these

4. In a triangle ABC,

bcr1 +

car2 +

abr3 is equal to

(A) R21 –

r1 (B) 2R – r

(C) r – 2R (D) r1 –

R21

5. If an ellipse slides between two perpendicular straight lines, then the locus of its centre is -

(A) a parabola (B) an ellipse (C) a hyperbola (D) a circle

6. If the lines whose vector equations are r = a + tb, r = c + t'd are coplanar then -

(A) (a – b). c × d = 0 (B) (a – c). b × d = 0 (C) (b – c). a × d = 0 (D) (b – d). a × c = 0


Questions 7 to 10 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct.

7. For a positive integer n, let

a(n) = 1 +21 +

31 +

41 + … +

1)2(1

n −. Then -

(A) a(n) < n (B) a(n) >2n

(C) a(2n) > n (D) a(2n) < 2n 8. If log2 (32x – 2 + 7) = 2 + log2 (3x – 1 + 1) then x is - (A) 0 (B) 1 (C) 2 (D) none of these 9. Let P (a sec θ, b tan θ) and Q (a sec φ, b tan θ)

where θ + φ = π/2, be two points on the hyperbola x2/a2 – y2/b2 = 1. If (h, k) is the point of intersection of normals at P and Q, then k is equal to -

(A) a

ba 22 + (B) –

+a

ba 22

(C) b

ba 22 + (D) –

+b

ba 22

10. If a, b, c are three unit vectors such that

a × (b × c) =21 b and c being non parallel then -

(A) angle between a and b is π/2 (B) angle between a and c is π/4 (C) angle between a and c is π/3 (D) angle between a and b is π/3 This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 11 to 16) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.


α, β, γ, δ are angles in I, II, III and IV quadrant respectively and no one of them is an integral multiple of π/2. They form an increasing arithmetic progression.

11. Which statement are true - (A) cos (α + δ) > 0 (B) cos (α + δ) = 0 (C) cos (α + δ) < 0 (D) none of these

12. Which statement are true - (A) sin (β + γ) = sin (α + δ) (B) sin (β – γ) = sin (α – δ) (C) tan 2(α – β) = tan (β – δ) (D) cos (α + γ) = cos 2β

13. If α + β + γ + δ = θ and α = 70º (A) 400º < θ < 580º (B) 470º < θ < 650º (C) 680º < θ < 860º (D) 540º < θ < 900º Passage : II (No. 14 to 16)

P is a point on the circle C1 : q2 (x2 + y2) = a2p2 Q is a point on the circle C2 : x2 + y2 = a2

14. If the coordinates of P are (h, k) then the locus of the point which divides the join of PQ in the ratio p : q is a circle C3, whose centre is at the point -

(A)

++ qpkq,

qphp (B)

++ qpk,

qph

(C)

++ qpkq,

qphq (D)

++ qpkp,

qphp

15. Locus of the centre of C3 as P moves on the circle

C1 is a circle C4

(A) concentric with C1 (B) concentric with C2

(C) having radius equal to the radius of C3 (D) having area equal to the area of C1

16. If the point (p, q) lies on the line y = 2x, then the

4

1

CofradiusCofradius is equal to -

(A) 2/3 (B) 3/2 (C) 3 (D) 1/3


17. If zn = n)3i1( + , find the value of 3 lm (z5 4z ).

18. If x = tan

− −−

174sin

251cos 11 then 2x

90 is

equal to.

19. If Q is the foot of the perpendicular from the

point P(4, –5, 3) on the line 3

5x − =42y

−+ =

56z −

then 100(PQ)2 is equal to.


PHYSICS 1. Write the formula for the force 'F' experienced by a

particle carrying a charge 'q' moving with velocity 'v' in a uniform magnetic field 'B'. Under what condition is this force zero ?

2. Two metals A and B have a work function 4eV and

10eV respectively. Which metal has a higher threshold wavelength ?

3. Why is the transmission of signals using ground

waves restricted to frequencies less than about 1500 kHz ?

4. Name the phenomenon responsible for the reddish

appearance of the sun at sunrise and sunset. 5. Why is the penetrating power of gamma rays very

large ? 6. What are the two main considerations that have to be

kept in mind while designing the 'objective' of an astronomical telescope ?

7. Is Young's experiment interference or diffraction experiment ?

8. Light passes from air into glass. Which of the

following quantities namely, velocity, frequency and wavelength change during the process ?

9. Draw the graphs showing variation of resistivity with

temperature for (i) nichrome and (ii) silicon. 10. The circuit shown in the diagram contains a battery

'B', a rheostat 'Rh' and identical lamps P and Q. What will happen to the brightness of the lamps, if the resistance through the rheostat is increased ? Give reasons.

B

QP

Rh

General Instructions : Physics & Chemistry • Time given for each subject paper is 3 hrs and Max. marks 70 for each. • All questions are compulsory. • Marks for each question are indicated against it. • Question numbers 1 to 8 are very short-answer questions and carrying 1 mark each. • Question numbers 9 to 18 are short-answer questions, and carry 2 marks each. • Question numbers 19 to 27 are also short-answer questions, and carry 3 marks each. • Question numbers 28 to 30 are long-answer questions and carry 5 marks each. • Use of calculators is not permitted.

General Instructions : Mathematics • Time given to solve this subject paper is 3 hrs and Max. marks 100. • All questions are compulsory. • The question paper consists of 29 questions divided into three sections A, B and C. Section A comprises of 10 questions of one mark each. Section B comprises of 12 questions of four marks each. Section C comprises of 7 questions of six marks each. • All question in Section A are to be answered in one word, one sentence or as per the exact requirement of the question. • There is no overall choice. However, internal choice has been provided in 4 questions of four marks each and

2 question of six marks each. You have to attempt only one of the alternatives in all such questions. • Use of calculators is not permitted.

MOCK TEST PAPER-3

CBSE BOARD PATTERN

CLASS # XII

SUBJECT : PHYSICS , CHEMISTRY & MATHEMATICS


11. A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60º with the normal to the coil. Calculate the magnitude of counter torque that must be applied to prevent the coil from turning.

12. A uniform magnetic field exists normal to the plane

of the paper over a small region of space. A rectangular loop of wire is slowly moved with a uniform velocity across the field as shown. Draw the graph showing the variation of

(i) magnetic flux linked with the loop and (ii) the induced e.m.f. in the loop with time.

× × × × × × × × × ×

× × × × × × × × × × × × × × × × × × × × × × × × × × ×

Stage-1 Stage-2 Stage-3 OR A bar magnet is dropped so that it falls vertically

through the coil C. The graph obtained for the voltage produced across the coil versus time is as shown in figure (b) (i) Explain the shape of the graph and (ii) why is the negative peak longer than the positive peak ?

v R Coil C

Magnet

(a) (b)

p.d/mV Time/ms

13. Violet light is incident on a thin convex lens. If this

light is replaced by red light, explain with reason, how the power of the lens would change.

14. (a) Draw a graph showing the variation of potential

energy of a pair of nucleons as a function of their separation. Indicate the regions in which nuclear force is

(i) attractive, and (ii) repulsive. (b) Write two characteristic features of nuclear force

which distinguish it from the Coulomb force. 15. Distinguish between 'point to point' and 'broadcast'

communication modes. Give one example of each. 16. What does the term LOS communication mean ?

Name the types of waves that are used for this communication. Which of the two-height of transmitting antenna and height of receiving antenna-can affect the range over which this mode of communication remains effective ?

17 For a CE-transistor amplifier, the audio signal voltage across the collector resistance of 2kΩ is 2V. Suppose the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1kΩ.

18. Give the logic symbol for an OR gate. Write its truth

table. Draw the output wave form for input wave forms shown for this gate.

A

B

(Inputs)

19. Define mutual inductance of a pair of coils. Deduce

an expression for the mutual inductance between a pair of coils having number of turns N1 and N2 wound over an air core.

20. In the given circuit, the potential difference across the

inductor L and resistor R are 120V and 90V respectively and the rms value of current is 3A. Calculate (i) the impedance of the circuit and (ii) the phase angle between the voltage and the current.

~

R L

21. With the help of a labelled circuit diagram, explain

how an n-p-n transistor is used to produce self-sustained oscillations in an oscillator.

OR Draw a labelled circuit diagram to show how an n-p-

n transistor can be used as an amplifier in common emitter configuration. For the given input waveform,

, draw the corresponding output waveform.

22. Two point charges 5 × 10-8C and –2 × 10-8C are

separated by a distance of 20cm in air as shown in the figure.

5×10–6C –2×10–8C

A B 20 cm

(i) Find at what distance from point A the electric potential would be zero.

(ii) Also calculate the electrostatic potential energy of the system.


23. Which constituent radiation of the electromagnetic spectrum is used

(i) in radar, (ii) to photograph internal parts of a human body, and (iii) for taking photographs of the sky during light

and foggy conditions ? Give one reason for your answer in each case. 24. In the potentiometer circuit shown, the balance (null)

point is at X. State with reason, where the balance point will be shifted when

(i) Resistance R is increased, keeping all parameters unchanged.

(ii) Resistance S is increased, keeping R constant. (iii) Cell P is replaced by another cell whose e.m.f. is

lower than that of cell Q.

G

P R

S

A B

Q

X

25. The work function of caesium is 2.14eV. Find (i) the

threshold frequency for caesium, and (ii) the wavelength of incident light if the photocurrent is brought to zero by stopping potential of 0.60V.

26. Complete the following decay process for β-decay of

Phosphorus 32: .......SP32

15 +→ The graph shows how the activity of a radioactive

nucleus changes with time. Using the graph, determine (i) half-life of the nucleus and (ii) its decay constant.

0 50 100 150 200 250Time/s

80

60

40

20Act

ivity

/Bq

27. In Young's double-slit experiment, explain with

reason what happens to the interference fringes, when (i)widths of the slits are increased,

(ii) mono-chromatic light source is replaced by a white light source, and (iii) one of the slits is closed.

28. Draw electric field lines between the plates of a

parallel plate capacitor with (i) air and (ii) dielectric as the medium. A parallel plate capacitor with air as

dielectric is connected to a power supply and charged to a potential difference V0. After disconnecting from power supply, a sheet of insulating material is inserted between the plates completely filling the space between them. How will its (i) capacity, (ii) electric field and (iii) energy change ? Given that the capacity of capacitor with air as medium is C0 and permittivity for air and medium are ε and ε0 respectively.

OR Derive an expression for the electric potential at a

point along the axial line of an electric dipole. At a point due to a point charge, the values of electric field and electric potential are 32 NC-1 and 16JC-1 respectively. Calculate (i) magnitude of the charge and (ii) distance of the charge from the point of observation.

29. (i) With the help of a schematic sketch of a cyclotron

explain its working principle. Mention its two applications. What is the important

limitation encountered in accelerating a light elementary particle such as electron to high energies.

(ii) A particle of mass m and charge q moves at right angles to a uniform magnetic field. Plot a graph showing the variation of the radius of the circular path described by it with the increase in its (a) charge, (b) kinetic energy, where, in each case other factors remain constant. Justify your answer.

OR (i) Using Biot-Savart's law derive an expression for

the magnetic field due to a current carrying loop at a point along the axis of the loop.

(ii) A long straight conductor carries a steady current I. The current is distributed uniformly across its cross-section of radius 'a'. Plot a graph showing the variation of magnetic field 'B' produced by the conductor with the distance 'r' from the axis of the conductor in the region (i) r < a and (ii) r > a.

30. How would you estimate rough focal length of a

converging lens ? Draw a ray diagram to show image formation by a diverging lens. Using this diagram, derive the relation between object distance 'u' image distance 'v' and focal length 'f' of the lens, Sketch the graph between 1/u and 1/v for this lens.

OR Define magnifying power of an optical telescope.

Draw a ray diagram for an astronomical refracting telescope in normal adjustment showing the paths through the instrument of three rays from a distant object. Derive an expression for its magnifying power. Write the significance of diameter of the objective lens on the optical performance of a telescope.


CHEMISTRY

1. What are F- centres ? 2. 2.46 g of NaOH (molar mass = 40) are dissolved in

water and solution is made to 100 cm3. Calculate molarity of solution.

3. How are gold and Pt sol prepared. 4. Why is PbO2 and PbCl2 are good oxidising agent? 5. Arrange the following in decreasing order (a) F2, Cl2, Br2, I2 [Bond energy.] (b) MF, MCl, MBr, MI [Ionic character] 6. Account for the following (a) NH3 has higher boiling point than PH3. (b) H3PO3 is diprotic acid. 7. Why Ce+3 can be easily oxidised to Ce+4? 8. What is oxoprocess? For what purpose is it used. 9. A decimolar solution of K4[Fe(CN)6] is 50%

dissociate at 300 K. Calculate the osmotic pressure of solution (R = 0.0821 L atm K–1 mol–1).

10. º

Zn/Zn 2E + = – 0.76 V. Write the reactions occuring at the electrodes when coupled with SHE (Standard Hydrogen Electrode).

11. Convert : (a) 2-propanol to chloroform (b) Acetone to iodoform 12. Complete the following reactions

(a) OHCH

|OHCH

2

2+

Conc3HNO →

(b)

OHCH|CHOH|

OHCH

2

2

Heat

KHSO4 →

13. Convert : (a) Aniline to benzonitrile (b) Aniline to phenyl isocyanide. 14. Name two principal ways by which cell obtain

energy for the synthesis of ATP.

15. Name the vitamins, the deficiency of which causes the following disease.

(a) Beri-beri (b) Night blindness (c) Poor coagulation of blood (d) Pernicous anaemia

16. KF has NaCl structure. What is the distance between K+ and F– in KF, if the density is 2.48 g cm–3?

17. Derive the relationship between activation energy

and rate constant.

18. Define heterogeneous catalysis. Give four example. 19. Complete the following : (a) XeF4 + SbF5 → (b) Cl2 + NaOH → (Cold & dil.) (c) F2 +

)Hot(2OH →

(d) F2 + )conc&Hot(

NaOH →

(e) XeF6 + KF → (f) BrO3

– + F2 + 2OH– →

20. Discuss : (a) Catenation (b) Thermal stability of hydride (c) Reducing power of hydrides with respect to group

15, 16 and 17. 21. Work out the following chemical equations : (i) In moist air copper corrodes to produce green

layer on its surface. (ii) Chlorination of Ca(OH)2 produces bleaching

powder. (iii) Copper sulphate from metallic copper. 22. Using VBT predict the shape and magnetic behaviour

of (i) [Ni(CO)4] (ii) [NiCl4]2–

23. Mention one method of preparation of following

organometallics (a) Zeise's salt (b) Dibenzene chromium (c) n-butyl lithium 24. What is Lucas reagent ? For what purpose is it used

and how ? 25. (a) What is Buna-S ? Name the monomers used in its

preparation. Mention its use. (b) Differentiate between elastomer and fibres on the

basis of intermolecular forces. (c) Give an example of step growth polymer.


26. Give an example of (a) Triphenyl methane dye (b) Azo dye (c) Anthraquinone dye 27. Using IUPAC names write the formula for the

following : (a) Tetrahydrozincate (II) (b) Hexammine cobalt (III) sulphate (c) Potassium tetracyanonickelate (II) (d) Potassium tetrachloro palladate (II) (e) Potassium tri (oxalato) chromate (III) (f) Diammine dichloro platinum (II) 28. Calculate the EMF of the following cell at 298 K

Fe/Fe+2(0.1 M) || Ag+(0.1 M) | Ag (s). [Given º

Fe/Fe 2E + = – 0.44 V, ºAg/Ag

E + = + 0.80V]

R = 8.31 J/K/mol, F = 96500 C. 29. In a reaction between A & B, the initial rate of

reaction was measured for different initial concentration of A & B as given below :

A/M 0.20 0.20 0.40 B/M 0.20 0.10 0.05

r0/Ms–1 5.07 × 10–5 5.07 × 10–5 7.6 × 10–5

What is the order of reaction with respect to A & B. 30. Draw the structure of all isomeric form of alcohol of

molecular formula C5H12O and give their IUPAC names. Classify them as primary, secondary and tertiary alcohols.

MATHEMATICS

Section A

1. Let f : R –

−

53 → R be a function defined as

f(x) = 3x5

x2+

, find f–1 : Range of f → R –

−

53

2. Write the range of one branch of sin–1x, other than

the Principal Branch.

3. If A =

− xcosxsin

xsinxcos, find x, 0 < x <

2π when

A + A´ = I

4. If B is a skew symmetric matrix, write whether the matrix (ABA´) is symmetric or skew symmetric.

5. On expanding by first row, the value of a third order

determinant is a11A11 + a12A12 + a13A13. Write the expression for its value on expanding by 2nd column. Where Aij is the cofactor of element aij.

6. Write a value of ∫ ++

xsinlogxxcot1 dx.

7. Write the value of ∫π

++

2/

0xsin53xcos53log dx

8. Let →a and

→b be two vectors such that |

→a | = 3 and

| →b | =

32 and →

a × →b is a unit vector. Then what is

the angle between →a and

→b ?

9. Write the value of i .( j × k ) + j .( k × i ) + k .( j × i )

10. For two non zero vectors →a and →

b write when

|→a +

→b | = |

→a | + |

→b | holds.

Section B

11. Show that the relation R in the set A = x|x ∈W, 0 ≤ x ≤ 12| given by R = (a, b) : (a – b) is a multiple of 4 is an

equivalence relation. Also find the set of all elements related to 2.

OR Let * be a binary operation defined on N × N, by

(a, b) * (c, d) = (a + c, b + d). Show that * is commutative and associative. Also find the identity element for * on N × N, if any.

12. Solve for x :

tan–1

−−

2x1x + tan–1

++

2x1x =

4π , |x| < 1

13. If a, b and c are real numbers and

accbbacbbaacbaaccb

+++++++++

= 0

Show that either a + b + c = 0 or a = b = c.


14. If f(x) =

<+−−

=+

<+−−

5xif,b|5x|

5x5xif,ba

5xif,a|5x|

5x

is a continuous function. Find a, b.

15. If xy + yx = log a, find dxdy .

16. Use lagrange's Mean Value theorem to determine a point P on the curve y = 2x − where the tangent is parallel to the chord joining (2, 0) and (3, 1).

17. Evaluate : ∫ −− )bxcos()axcos(1 dx

OR

Evaluate : ∫ ++ 2/xe.

xcos1xsin2 .dx

18. If →a and

→b are unit vectors and θ is the angle

between them, then prove that cos2θ =

21 |

→a +

→b |.

OR If are the diagonals of a parallelogram with sides,

→a and

→b in d the area of parallelogram in terms of

and hence find the area with →

1d = i + 2 j + 3 k and →

2d = 3i – 2 j + k.

19. Find the shortest distance between the lines, whose equations are

3

8x − =16

9y−

+ =7

z10−

− & 315x − =

16y258

−− =

55z

−−

20. A bag contains 50 tickets numbered 1, 2, 3, ..... , 50 of which five are drawn at random and arranged in ascending order of the number appearing on the tickets (x1 < x2 < x3 < x4 < x5). Find the probability that x3 = 30

21. Show that the differential equation 2yex/ydx + (y – 2xex/y)dy = 0 is homogeneous and find its particular solution given

that x = 0 when y = 1. OR Find the particular solution of the differential

equation dydx + y cot x = 2x + x2cot x, x ≠ 0 given

that y = 0, when x = 2π

22. From the differential equation representing the family of ellipses having foci on x-axis and centre at origin.

Section C

23. A letter is known to have come from either TATANAGAR or CALCUTTA. On the envelope just two consecutive letters TA are visible. What is the probability that the letter has come from

(i) Tata Nagar (ii) Calcutta OR Find the probability distribution of the number of

white balls drawn in a random draw of 3 balls without replacement from a bag containing 4 white and 6 red balls. Also find the mean and variance of the distribution.

24. Find the distance of the point (3, 4, 5) from the plane x + y + z = 2 measured parallel to the line 2x = y = z.

25. Using integration, compute the area bounded by the

lines x + 2y = 2, y – x = 1 and 2x + y = 7 OR Find the ratio of the areas into which curve y2 = 6x

divides the region bounded by x2 + y2 = 16

26. Evaluate : ( )∫

+

−

22

xtan

x1

e1

dx

27. A point the hypotenuse of a right triangle is at a

distance 'a' and 'b' from the sides of the triangle. Show that the minimum length of the hypotenuse is

[ ] 2/33/23/2 ba + . 28. Using elementary tranformations, find the inverse of

the matrix

−−−

052503231

29. A furniture firm manufactures chairs and tables, each requiring the use of three machines A, B and C. Production of one chair requires 2 hours on machine A, 1 hour on machine B and 1 hour on machine C. Each table requires 1 hour each on machine A and B and 3 hours on machine C. The profit obtained by selling one chair is Rs. 30 while by selling one table the profit is Rs. 60. The total time available per week on machine A is 70 hours, on machine B is 40 hours and on machine C is 90 hours. How many chairs and tables should be made per week so as to maximize profit ? Formulate the problem as L.P.P. and solve it graphically.


PHYSICS 2. For diamagnetic materials like Bi 3. Zero

4. For an electron, ÅV27.12

=λ Å9.31027.12

==

5. If a thin foil is introduced parallel to plates than

capacity remains reflected. 6. Core is laminated to present eddy current losses 7. Photodiode

8. Maxwell’s fourth equation is Ampere's law and

according to it dt

did.B E

000θ

µ∈+µ=→→

∫ l

10. Let the charge on β particle be -e then charge on

deutron is +e and on α -particle is +2e

2e

a a

a –e e

ake2

ake2

ake.E.P

222−+

−=

a

keE.P2

−=

a2

ke2a2

ke2a2

keE.P222

F −+−

=

= a2

ke2−

a2

keE.PE.PW2

iF =−=

11. AB Y

A B Y0 0 1 0 1 1 1 0 0 1 1 1

A

B Y

A B Y0 0 0 0 1 0 1 0 0 1 1 0

13. No. of field lines emitted by a charge = 0

q∈

for a proton = 12

19

10854.8106.1

−

−

×

×

14. The magnetic force on sides AB and BC due to magnetic field of current carrying wires is equal and opposite thus they balance each other while on AC it is towards the wire. Thus the loop will move towards the conductor

B

C

A

i2i1

15.

C B

C

C

A5V

= A B2C 2C

+Q +Q QQ

5V

Charge remains same in series combination

thus P.D. on volt5.2)V5(21

C2QC2 ===

P.D. across AB = 5 + 2.5 = 7.5 V

16.

R

RR

R

A B

R

R R

R

X

MOCK TEST PAPER SOLUTION FOR PAPER – 2 PUBLISHED IN JANUARY ISSUE


2R A B

2R

2R

2R

X =

A

R =

R B X

= VA – VX = IR ...(1) VX – VB = IR = 10 V ...(2) Equation (1) + (2) VA – VB = 2IR = 20 V 17. V = IR

Amp1010100

RVI ===

Ω===⇒= 21010

2100IVZ

ZVI

2

1210

10ZRcos ===θ

18. Einstein’s equation hv = hv0 + KEmax

or hv = hv0 + eV v = frequency of incident light v0 = Threshold frequency,

V = Stopping potential hv = 5 eV, hv0 = 2 eV 5 = 2 + eV St. p. = 3 eV Stopping potential = – 3 volt 19. (i) t = 1 sec. I ∝ w (width of slit) t = 1 sec y = x I1 = I2 a1 = a2 = a [I ∝ a2 ] amax = 2a,

Imax = 4a2 amin = 0, Imin = a2

(ii) t = 4 sec 2

21

21

min

maxII

IIII

−

+=

14

II

2

1 = = 9 : 1

20. According to shell’s law As light ray pass from rare to denser medium it bands

towards the normal and when passes from denser to rarer bends away from normal

Thus µ3 > µ1

µ2 > µ3

Medium-2 is most dense µ2 > µ3 > µ1

21. (i) tan ip = aµw = 4/3 ip = tan–1(4/3) (ii) θ = 90º (iii) Reflected ray is plane polarized

(iv) pw

ac itan11sin =

µ=θ

22.

7Ω

03.V 1Ω

2Ω

As the diode is in forward bias total resistance = 2 + 1 = 3Ω

i = Rν =

33.0 = 0.1 amp

V0 = i(7) = 0.7 volt

23. =

Input Output

.24 (i) Current Flowing in the circuit.

10V 3Ω

RL=2Ω

amp25

10i ==

For battery -1 V = E + Ir = 10 + 2 × 1 = 12 volt For battery -2 V = E – Ir = 20 – 2 × 2 = 16 volt (ii) Because battery - 1 is in charging state while

battery -2 is in discharging state

25. When current flows through metallic spring, current is in same direction thus due to magnetic force difference coils are attracted towards each other and spring gets strinked


26. Inconsistency in Ampere’s law According to ampere’s law

id.B 0µ=→→

∫ l

and the relation is valid only when the electric field at the surface does not changes with time and this law tells is that an electric current produces magnetic field. If there exists an electric current as well as changing electric field. The resultant magnetic field is given by

θ

∈µ+µ=∧→→

∫ dtdid.B E

000l

Suppose that for a parallel plate capacitor.

A

QE0∈

= (electric field between plates)

Flux of the field through given area

00

EQA

AQ

∈=×

∈=θ

dt

di E0d

θ=∈

= dtdQ

dtdQ

0 =∈

id = ic

id = displacement current, ic = conduction current

27. 0

netq∈

=φ

for a charge placed at corner of cube,

08q∈

=φ

∴ thus for given system

02

)85764321(∈

−++−−+−=φ

02

1∈−

=φ

28.

B

L'

E

C K

V∞

VBB

LC2

1ƒπ

=

CHEMISTRY

1. Co-ordination no. of Ca+2 = 8 F– = 4 2. [Cr(H2O)5SO4]Br Pentaaqua sulphato chromium (III) bromide. 3. (a) 4-nitro –1–methoxy benzene. (b) 4-bromo-3, 3, 4-trimethyl hex-1-ene-2, 5-diol. 4. Aromatic ketones are less reactive, so they do not

react with NaHSO3. 5. Sulphanilic exist as zwitter ion, so they are

amphoteric in nature. 6. Ethylene glycol & Phthalic acid. 7. Substance which remove the excess acid and raise the

pH to appropriate level in stomach are Antacids. Eg Lansoprazole.

8. Carbohydrate having chiral carbon, so they optically

active. 9. Given r+ = 95 pm r– = 181 pm.

−

+

rr

= 18195 = 0.524

Since, r+/r– lies between 0.414 to 0.732, ∴ A X has FCC (NaCl type structure) structure so, the co-ordination number of each ion = 6. 10. In MgO, co-ordination number of Mg+2 is 6 and that

of O2– is also 6 due to NaCl structure. In TlCl, co-ordination number Tl+ is 8 and that of Cl–

is also 8 due to CsCl type structure. 11. Mole fraction of solute is defined as ratio of number

of moles of solute to the total number of moles of solute and solvent

xB = BA

B

nnn+

=

B

B

A

A

BB

MW

MW

M/W

+

xA = BA

A

nnn+

=

B

B

A

A

AA

MW

MW

M/W

+

or xA + xB = 1 Where xA = mole fraction of solvent xB = mole fraction of solute


12. With increasing voltage, the sequence of deposition of metal is

)80.0(

Ag+

+ > )79.0(

22Hg

+

+ > )34.0(

2Cu+

+ > )37.2(

2Mg−

+

Mg+2 will not be reduced because its reduction potential value is much lower than water (–0.834).

13. In Cr2O7

–2, all the six normal Cr–O bonds are equivalent and two bridged Cr– O bonds are equivalent.

Cr

O O–O

O

Cr

OO–

O

14. CuSO4 + 2KCN → Cu(CN)2 + K2SO4 2Cu(CN)2 → Cu2(CN)2 + C2N2 (cyanogen) Cu2 (CN)2 + 6 KCN → 2K3 [Cu(CN)4] 15. Two pairs.

CH3 – CH – CH2 – CH3

CH3

Br2

CH2Br – CH – CH2.CH3

CH3

CH3 – CH – CH – CH3

CH3 Br

*

*

16. Fe(s) → Fe+2(aq) + 2e– (oxidation) O2(g) + 4H+ + 4e– → 2H2O (Reduction) Atmospheric oxidation occurs as following

2Fe+2 (aq) + 2H2O(l)+ 21 O2(g)

→ Fe2O3(s) + 4H+(aq)

17. (a)

OHBr2/CS2

0ºC

OHBr

+

OH

Br

Mono halogen derivative form. (b) C2H5OC2H5 + 2HI →Heat 2C2H5I + H2O 18. Fehling, Tollen's, Schiff's reagent react only with

aldehyde. Grignard reagent react with both aldehyde & ketones.

R – C – H + R´MgBr

O

R – C – H

OMgBr

R´

R – C – H

OH

R´

R – C – R + R´MgBr

O

R – C – R

OMgBr

R´

R – C – R

OH

R´ 19. Iodoform test is given by only those compounds

which having – C – CH3

O

or – CH – CH3

OH

group.

Therefore, pentanone-2 give iodoform test

CH3 – CH2 – CH2 – C – CH3

O 20. From the Question, I → Is most basic due to lone pair present in sp3

hybridised orbital to available for donation. II → lone pair present in sp2 orbital but disperse to

small extent. III → lone pair present is sp3 orbital but possess

electronegative atom. IV → lone pair e– present on 'N is in p-orbital to

form a part of aromatic sextet. (least basicity) so, order of basicity = I > III > II > IV

21. HX H+ + X– n = 2

from the formula i = 1

n1 α+α−

= 1

2.022.01 ×+− = 1.2

from the formula ∆Tf = kf . m. i = 1.2 × 1.8 × 0.2 = 0.432 so, the freezing point of solution = 0 – 0.432 = – 0.432 ºC 22. We know that Surface volume = area × thick ness

= 80 × 10005.0 cm3

= 0.04 cm3 Mass of silver deposited = volume × density = 0.04 × 10.5 = 0.042 g The cell reaction is Ag+ + e– → Ag

since Ew =

FQ =

Ft.i

∴ 108042.0 =

96500t3×

or t = 125.1 sec


23. From the below graph we can say that with increase of temperature, then occurs a decrease in rate of physisorption.

t

x/m

Where x/m = Mass of gas adsorbed per unit mass of

adsorbent. t = Temperature 24. In H3BO3, 'B´ atom having 6e– (e-deficient) it is a

Lewis acid with one vacant p-orbital and no d-orbital thus, it can accomodate only one e– pair in its outer most shell.

H2O: + B – OH H2O → B – OH

OH

OH

[B(OH)4]– + H+

OH

OH

25. (a) The colour of transition metal compound is

obtained due to unpaired e– which gives d-d-transition.

VOCl2 & CuCl2 → V+4 & Cu+2 both having one unpaired e–, so gives same colour in

aqueous medium. (b) [CuSO4 + 2KCN → Cu(CN)2 + K2SO4] × 2 2Cu(CN)2 → Cu2(CN)2 + (CN)2 Cu2(CN)2 + 6KCN → 2K3 Cu(CN)4 2CuSO4 + 10KCN → 2K3Cu(CN)4 + 2K2SO4 + (CN)2 26.

CHO CHO

CHO CHO

OH–/100ºC

Intra molecular Cannizaro reaction

COO– CH2OH

CH2OH COO–

H+/H2O

COOH CH2OH

CH2OH COOH

Product

27. (i) In aspartame following four functional groups are present.

(a) (–NH2) Amine (b) –COOH (Carboxylic acid)

(c)

(– C – NH –)

(Amide)

O

(d)

(– C – O –)

(Ester)

O

(ii) Its zwitter ion is

H3N – CH – CONH – CH – COOCH3

CH2 – C6H5

CH2 – COO–

+

(iii)

H2N – CH – C – NH – CH – COOCH3

CH2COOH

CH2C6H5

Hydrolysis

H2N – CH – COOH

CH2COOH

H2N – CH – COOH

CH2– C6H5

+

(a) (b)

O

28. (i) From the equation k = A.e–Ea/RT

or log k = log A – RT303.2

Ea

Comparing this equation with the given equation

R303.2

Ea = 1.25 × 104

or Ea = 1.25 × 104 × 2.303 × 8.314 = 2.39 × 105 J/mol = 239 kJ/mol (ii) In the question, the unit of rate constant is s–1,

therefore, the reaction is first order.

∴ t1/2 = k693.0

or k = 2/1t

693.0 = 60256

693.0×

= 4.51 × 10–5 s–1 substituting this value in the given expression, we get

log(4.51 × 10–5) = 14.34 – T

1025.1 4×

or –4.346 = 14.34 – T

1025.1 4×


⇒ T

1025.1 4× = 14.34 + 4.346 = 18.686

T = 686.18

1025.1 4× = 669 K

29. CaO + H2O →

)A(2)OH(Ca

NH3 + CO2 + H2O → )B(

34 )HCO(NH

NH4(HCO3) + NaCl → Na(HCO3) + )D(4ClNH

2NaHCO3 →∆

)C(32CONa + H2O + CO2

Ca(OH)2 + 2NH4Cl → )E(

2CaCl + 2NH3 + 2H2O

30. Ozonolyis of 'A' to acetone and aldehyde indicated the presence of the following structure in the molecule 'A' (alkene).

C = CHR

H3C

H3C O3 C = O + RCHO

H3C

H3C

(ketone) (aldehyde)

RCHO → ]O[

)B(RCOOH → P/Br2

)C(

compoundBromo → OH2 Hydroxy acid

Hydroxy acid can be determined by following reaction

C = O

H3C

H3C HCN C

H3C

H3C

(D)

OH

CN

H2O/H+ C H3C

H3C

OH

COOH

from the above, bromo compound 'C' is –

(C)

C H3C

H3C

Br

COOH

'C' is formed by bromination of (B) so 'B' is

C

H3C

H3C

H

COOH compound 'B' is formed by oxidation of an aldehyde,

so the structure of the aldehyde is

C

H3C

H3C

H

CHO The aldehyde and acetone are formed by the

ozonolysis of alkene 'A'. So, the structure of alkene

C = C H3C

H3C

ozonolysis

CH3H H

(A)

CH3

C H3C

CHO

H

H3C

(Aldehyde)

+ O = C CH3

CH3

(Ketone)

C H3C

H3C

[O]

CHO

HC

H3C

COOH

H

H3CBr2/P

C H3C

COOH

Br

H3C

Hydrolysis C H3C

COOH

OH

H3C

(B)

(C)(D)

MATHEMATICS

Section A

1. (a) ∴ for every value of x there is unique y

2. π

3. 135

4. 3

5. (1, 2)

6. π/6

7. (1, –7, 2) or their any multiple

8. 8

x8 + c

9. k5j3

11i3 ++

10. order of AB is 2 × 2 order of BA is 3 × 3

Section B

11. f(x) = 3

1x2 − , x ∈R

To show f is one-one


Let x1, x2 ∈ R s.t. x1 ≠ x2 ⇒ 2x1 ≠ 2x2

⇒ 2x1 – 1 ≠ 2x2 – 1

⇒ 3

1x2 1 − ≠ 3

1x2 2 −

⇒ f(x1) ≠ f(x2) ⇒ f is one-one To show f is onto

Let y = 3

1x2 − , y ∈R(codomain of f)

or 3y = 2x – 1

or x = 2

1y3 + ∈ R

∴ for all y ∈ R (codomain of f), there exist

x = 2

1y3 + ∈ R (codomain of f), such that

f(x) = f

+

21y3 =

3

12

1y32 −

+

= y

⇒ every element in codomain of f has its pre-image in the domain of f,

⇒ f is onto. To find f–1

Let f(x) = y, x = 2

1y3 +

⇒ f–1(y) = x ⇒ f–1(y) = 2

1y3 +

∴ f–1 : R → R given by f–1(y) = 2

1y3 +

OR

(i) a*b = 2

ba + , a, b ∈ N

∀ a, b ∈ N 2

ba + may or may not belong to N.

∴ a*b is not always natural no. ∴ '*' is not a binary operation on N

(ii) a*b =2

ba + , a, b ∈ Q

∀ a, b ∈ Q; 2

ba + ∈ Q

⇒ a*b ∈ Q ⇒ '*' is a binary operation on Q

(iii) For a*b = 2

ba + , a, b ∈ Q

a*b = 2

ba + = 2

ab + = b*a

⇒ * is commutative

(iv) (a*b)*c =

+

2ba *C

∀ a, b, c, ∈ Q

= 2

c2

ba+

+

= 4

c2ba ++

a*(b*c) = a *

+

2cb

= 2

2cba

+

+=

4cba2 ++

(a * b) * c ≠ a * (b * c) ∀ a, b, c, ∈ Q

∴ '*' is not associative,

12. Let sin–1

135 = x & cos–1

53 = y

⇒ sin x = 135 & cos y =

53

& cos x = 1312 & sin y =

54

⇒ tan x = 135 & tan y =

34

tan(x + y) = ytanxtan1ytanxtan

−+

tan(x + y) = 1663

⇒ x + y = tan–1

1663

⇒ sin–1

1312 + cos–1

53 = tan–1

1663

13. Let = A =

−−

2162

A = IA

−−

2162

=

1001

A

R1 ⇔ R2

⇒

−−

6221

=

0110

A

R2 → R2 – 2R1

⇒

−−

2021

=

− 2110

A

R1 → R1 – R2

⇒

− 2001

=

−

−21

31A

R2 → – 21 R2


⇒

1001

=

−

−

121

31A

∴ A–1 =

−

−

121

31

OR Operate R1 → aR1, R2 → bR2, R3 → cR3

ababbabaaccacacabccbcbbc

22

22

22

−+++−+++−

= abcabccbabcca

bacbcabcabcbaabcacabcababc

abc1

22

22

22

−+++−+++−

Take, a, b, c common from C1, C2, C3 respectively

= abacbcbcac

babcacbcababacacabbc

−+++−+++−

R1 → R1 + R2 + R3

= abacbcbcac

babcacbcabacbcabacbcabacbcab

−+++−+

++++++

= (ab + bc + ca)abacbcbcac

babcacbcab111

−+++−+

C1 → C1 – C3, C2 → C2 – C3

= (ab + bc + ca)ababacbcabbcac

babc)cabcab(0100

−+++++++−

On expanding by R1 we get = (ab + bc + ca)3 14. Being a polynomial function f(x) is continuous at all

point for x < 1, 1 < x < 2 and x ≥ 2. Thus the possible points of discontinuity are x = 1 and x = 2

To check continuity at x = 1

=∴

≠==

−=−=

=+=

+−

−

→→

→+→

→→

1at x continuousnot is )x(f

)x(flim)1(f)x(flim nce,si3)1(f12xlim)x(flim

32xlim)x(flim

1x1x

1x1x

1x1x

To check continuity at x = 2 )x(flim

2x −→= 2xlim

2x−

−→ = 0

)x(flim2x +→

= 2x

lim→

x – 2 = 0

f(2) = 0 since )x(flim

2x −→)x(flim

2x +→ = f(2) = 0

∴ f(x) is continuous at x = 2 ∴ The only point of discontinuity is x = 1 15. xpyp = (x + y)p + q Take log on both sides p log x + q log y = (p + q) log (x + y)

xp +

yq .

dxdy =

+

++

dxdy1

yxqp

or xp –

yxqp

++ =

dxdy

−

++

yq

yxqp

or )yx(x

qxpxpypx+

−−+ =

+

−−+)yx(y

qyqxqypydxdy

or x

qxpy − =

−y

qxpydxdy

or xy =

dxdy

OR

y = tan–1

−++

−++22

22

x1x1

x1x1

Put x2 = cos θ

y = tan –1

θ−−θ+

θ−+θ+

cos1cos1cos1cos1

= tan–1

θ−

θ

θ+

θ

2sin

2cos

2sin

2cos

= tan–1

θ−

θ+

2tan1

2tan1

= tan–1

θ

+π

24tan

y = 4π +

2θ or y =

4π +

21 cos–1(x2)

dxdy = –

− 4x1

x221 =

4x1

x

−

−

16. ∫ −+++

)5x()3x()4x)(1x(

22

22dx

Consider

)5x()3x()4x)(1x(

22

22

−+++ =

)5t)(3t()4t)(1t(

−+++ where t = x2


= 1 + )5t)(3t(

19t7−+

+

Consider

)5t)(3t(

19t7−+

+ = 3t

A+

+ 5t

B−

A = 41 , B =

427

∴ ∫ −+++

)5x()3x()4x)(1x(

22

22dx

= ∫ dx + ∫ ∫ −+

+ 5xdx

427

3xdx

41

22

= x +34

1 tan–1

3x +

5827 log

5x5x

+

− + c

17.

r

αh

Let r = radius of cone formed by water at any time h = height of cone formed by water at any time

Given α = tan–1

21

∴ tan α = 21

Also tan α = hr ⇒ h = 2r

Volume of this cone

v = 31

πr2h = 2

2h

3

π h

v = 12π h3

⇒ dtdv =

12π (3h2)

dtdh =

4π h2

dtdh

But dtdv = 5 m3/minute

∴ 5 = 4π h2

dtdh

or dtdh = 2)10(

20π

when h = 10 m

or dtdh =

π51 m/minute

18. For ∫ −2

1

2 dx)1x3(

a = 1, b = 2, h = n1

as n → ∞, h → 0 f(x) = 3x2 – 1

∫b

a

dx)x(f = 0h

lim→

h[f(a) + f(a + h) + ... + f(a + (n – 1)h)]

∫ −2

1

2 dx)1x3( = 0h

lim→

h[3n + 3h2 (12 + 22 + .. + (n – 1)2)

+ 6h(1 + 2 + .. + (n – 1) – n)]

=

−+

−−+

→)1n)(n(h3

6)1n2)(1n)(n(h3n2hlim 2

0h

=

−

+

−−+

→ hh1

h1h3

h6)h2)(h1(h3

h2hlim 3

2

0h

= 2 + 21 (2) + 3 = 6

19. Given I = ∫π 2/

0

dxxsinlog

I = ∫π 2/

0

dxxcoslog

=− ∫∫

a

0

a

0

dx)x(fdx)xa(fQ

∴ 2I = ∫π

−2/

0

dx)2logdxx2sin(log

I = ∫π

π−

2/

0

2log4

dxx2sinlog21 ....(1)

Consider

I1 = ∫π 2/

0

dxx2sinlog = ∫π

0

dttsinlog21

[Put 2x = t, dx = 2dt ; x = 0 ⇒ t = 0; x =

2π ⇒ t = π]

= ∫π 2/

0

dttsinlog2.21

=−=∫ ∫

a2

0

a

0

)x(f)xa2(fifdx)x(f2dx)x(fQ

= ∫π 2/

0

dttsinlog


I1 = ∫π 2/

0

dxxsinlog ...(2)

From (1) and (2)

π−=

π=−

π−= ∫

π

2log2

I

2log4

–I21I

2log4

dxxsinlog21I

2/

0

20. Given line

5

1x − = 2

y3 − = 4

1z +

or, 5

1x − = 23y

−− =

4)1(z −− ....(i)

is passing through (1, 3, –1) and has D.R. 5, –2, 4. Equations of line passing through (3, 0, –4) and

parallel to given line is

5

3x − = 20y

−− =

44z + ...(ii)

Vector equations of line (i) & (ii)

→r = i + 3 j – k + λ(5 i – 2 j + 4 k )

→r = 3 i – 4 j + µ (5 i – 2 j + 4 k )

∴ →

2a – →

1a = 2 i – 3 j – 3 k

→b = 222 )4()2()5( +−+

= 45 = 53

Also →b ×

−

→→

12 aa = 332

425kji

−−−

= 18 i + 23 j – 11 k

∴

−×

→→→

12 aab = 222 )11()23()18( ++ = 974

∴ Distance between two parallel lines.

= →

→→→

−×

b

aab 12

= 45

974 units

21.

E D

F

A B

C

a→

b→

From fig. →

DE = –→a ;

→EF = –

→b

→

AC = →

AB + →

BC = →a +

→b

→

AD = →

BC2 = →b2

→

AD = →

AC + →

CD

⇒ →

CD = →

AD – →

AC = →b –

→a

→

FA = – →

CD = →a –

→b

→

CE = →

CD +→

DE = →b – 2

→a

→

AE = →

AD + →

DE = 2→b –

→a

22. P(Correct forecast) = 31

P(Incorrect forecast) = 32

P (At least three correct forecast for four matches) = P(3 correct) + P(4 correct)

= 4C3 3

31

1

32

+ 4C4

4

31

=

818 +

811 =

91

OR Let E : Candidate Reaches late A1 = Candidate travels by bus A2 : Candidate travels by scooter A3 : Candidate travels by other modes of transport

P(A1) = 103 , P(A2) =

101 , P(A3) =

53

P(E/A1) = 41 , P(E/A2) =

31 , P(E/A3) = 0

∴ By Baye's Theorem P(A1/E) =

)A/E(P)A(P)A/E(P)A(P)A/E(P)A(P)A/E(P)A(P

332211

11

++

= 0

301

403

41

103

++

× =

139


Section C

23. Given

2312

P

−

−35

23 =

−1221

Let R =

2312

then |R| = 1

S =

−

−35

23 then |S| = –1

Q =

−1221

Since R and S are non-singular matrices ∴ R–1 and S–1 exist

R–1 = |R|

AdjR =

−2312

S–1 = |S|

AdjS =

3523

Now given

==

===

=

−−

−−−

−

−−

−−

11

111

1

11

11

QSRPQSRPSS

QRPSQRPS)RR(QR)RPS(R

QRPS

(Q R–1R = I I.P = P)

∴

−−

=

−

−

−=

22371525

3523

1221

2312

P

24. f(x) = sin 2x – x –2π < x <

2π

f´(x) = 2 cos 2x – 1

f´(x) = 0 ⇒ cos 2x = 21

or 2x = 3π ,

3π or x = –

6π ,

6π

f´´(x) = –4 sin 2x

f´´(x) = 32 > 0 at x = – 6π

⇒ x = –6π is point of local minima

f´´(x) = 32 < 0 at x = 6π

⇒ x = 6π is point of local maxima

∴ Local minimum value is

f

π

−6

= 2

3− + 6π

Local maximum value is

f

π

6 =

23 –

6π

OR Let h = length of cylinder r = radius of semi-circular ends of cylinder

v = 21

πr2h

S = Total surface area of half circular cylinder = 2(Area of semi circular ends) + Curved surface area of half circular cylinder + Area of rectangular base.

=

π 2r

212 +

21 (2πrh) + 2rh

= πr2 + (π + 2)rh

= πr2 + (π + 2)r. 2rv2

π

drds = 2πr –

π+π )2(v2

2r1

drds = 0 ⇒ r3 =

2v)2(

π+π

2

2

drSd = 2π +

π+π )2(v2 . 3r

2 > 0

∴ S is minimum when

r3 = 2v)2(

π+π =

π

π+π hr

21)2( 2

2

⇒ r = π+π

22 .h ∴

r2h =

2+ππ

Which is required result.

25. f(x)

>−≤+−−

2x,2x2x2)2x(

2

or f(x) =

>−≤−

2x,2x2xx4

2

To sketch the graph of above function following tables are required.

For f(x) = 4 – x, x ≤ 2 & for f(x) = x2 – 2, x ≥ 2

x –1 0 1 2 y 5 4 3 2

Also f(x) = x2 – 2 represent parabolic curve.

x 2 3 4 5 6 y 2 7 14 23 34


y

C

A B

Ox = 0 2 x = 4

D

y = 4 – x y = x2 – 2

Area = ∫4

0

dx)x(f = ∫ −2

0

dx)x4( + ∫ −4

2

2 dx)2x(

= 4x – 4

2

32

0

2x2

3x

2x

−+

= 6 + 344 =

362 sq. units

On the graph ∫4

0

dx)x(f represents the area bounded

by x-axis the lines x = 0; x = 4 and the curve y = f(x). i.e. area of shaded region shown in fig.

26. (1 – x2)dxdy – xy = x2

or dxdy – 2x1

x−

.y = 2

2

x1x−

P = – 2x1x

−, Q = 2

2

x1x−

I.F. = ∫Pdxe = ∫ −

− dxx1x

2e = )x1log(

21 2

e−

= 2x1− ∴ Solution of diff. equation is

2x1y − = ∫ −−

dxx1.x1

x 22

2

= ∫

−−

−

22

x1x1

1 dx

= sin–1x –

+− − xsin

21x1x 12 + c

2x1y − = 21 sin–1x – x 2x1− + c

When x = 0, y = 2 ⇒ 2 = c ∴ Solution is

2x1y − = 21 sin–1x – x 2x1− + 2

27. The given line is

3

6x − = 2

7y − = 27z

−− = λ (say) ...(i)

Let N be the foot of the perpendicular from P(1, 2, 3) to the given line

P(1, 2, 3)

A N B Coordinates of N = (3λ + 6, 2λ + 7, –2λ + 7) D.R. of NP 3λ + 5, 2λ + 5, – 2λ + 4 D.R. of AB 3, 2, –2 Since NP ⊥ AB ∴ 3(3λ + 5) + 2(2λ + 5) – 2(–2λ + 4) = 0 or λ = –1 ∴ Coordinates of foot of perpendicular N are (3, 5, 9) Equation of plane containing line (i) and point (1, 2, 3)

is Equation of plane containing point (6, 7, 7) & (1, 2, 3)

and parallel to line with D.R. 3, 2, –2 is

2234557z7y6x

−−−−−−−

= 0

or, 18x – 22y + 5z + 11 = 0 28. Given x P(x) 0 0 1 k 2 4k 3 2k 4 k Σpi = 8k

But Σpi = 1 ⇒ k = 81

∴ Probability distribution is

221

814

49

43

413

21212

81

81

811

0000xpxppx 2

iiiiii

Probability of getting admission in two colleges = 21

Mean = µ = Σpixi = 8

19


Variance = σ2 = Σpixi2 – µ2 =

851 –

2

819

=

6447

OR

4 W 3 B

W

W

B BW

RA

2 W 2 B

1W 1B

Three cases arise, when 2 balls from bag A are

shifted to bag B. Case 1 : If 2 white balls are transferred from bag A.

P(WAWA) = 74 .

62 =

72

Case 2 : If 2 black balls are transferred from bag A

P(BABA) = 73 .

62 =

71

Case 3 : If 1 white and 1 black ball is transferred from bag A

P(WABA) = 2

63.

74 =

74

(a) Probability of drawing 2 white balls from bag B = P(WAWA).P(WBWB) + P(BABA).P(WBWB) + P(WABA).P(WB.WB)

=

53.

62

72 +

51.

62

71 +

52.

63

74 =

215

(b) Probability of drawing 2 black balls from B

=

51.

62

72 +

43.

64

71 +

52.

63

74 =

214

(c) Probability of drawing 1 white and 1 black ball from bag B

=

52.2.

64

72 +

54.

62.2

71 +

53.

63.2

74 =

74

29.

A

P RQ

B

x y

40 – y 40 – x

50–(60–x – y)

60 –x – y

Let x no. of packets from kitchen A are transported to P and y of packets from kitchen A to Q. Then only 60 – x – y packets can be transported to R from A.

Similarly from B, 40 – x packets can be transported to P and 40 – y to Q. Remaining requirement of R i.e. 50 –(60 – x – y) can be transported from B to Q.

∴ Constraints are

≥≥≥++−

≥−−≥−≥−

0y,0x0yx10

0yx600y400x40

Objective function is : Minimise. z = 5x + 4y + 3(60 – x – y) + 4(40 – x) + 2(40 – y) + 5(x + y – 10) ∴ L.P.P. is To Minimise. z = 3x + 4y + 370 subject to constraints

≥≥≥+≤+

≤≤

0y,0x10yx60yx

40y40x

y

O

B

F E(40, 0)xx´

y´

C

x + y = 60(0, 40)

(10, 0)

x+y = 10 D(40, 20)A(0, 10)

Feasible Region is ABCDEFA with corner points A(0, 10) z = 3(0) + 4(10) + 370 = 410 B(0, 40) z = 3(10) + 4(40) + 370 = 530 C(20, 40) z = 3(20) + 4(40) + 370 = 590 D(40, 20) z = 3(40) + 4(20) + 370 = 570 E(40, 0) z = 3(40) + 4(0) + 370 = 490 F (10, 0) z = 3(10) + 4(0) + 370 = 400 ∴ x = 10, y = 0 gives minimum cost of transportion. Thus No. of packets can be transported as follows

A B P 10 30 Q 0 40 R 50 0

Minimum cost of transportation is Rs. 400


PHYSICS

2. hv = hv0 + K.E max v = frequency of incident light v0 = threshold frequency. 5. γ-rays have maximum penetrating power and

minimum ionising power. 8. Wavelength of the light increases because light ray

travels from denser to rarer medium as it bends away from normal.

Velocity and wavelength of light changes as it passes from air to glass.

11. τ = MB sinθ M = NIA = 30 × 6 × 3.14 × (0.08)2 τ = 3. 62 × 1 × sin 60º τ = 3.135 N-M 14. (b) (i) Nuclear force is a short range force (ii) It does not varies as inverse of the square of the

distance 19. Then magnetic flux produced in a coil is associated

with another coil and change in magnetic flux of a coil induces emf in other coil, this phenomenan is called mutual inductance.

Derivation

r1S1

S2r2

suppose a current i is passed through s1 then B = µ0n1i(n1 = no. of turns/length in s1) Then flux through each turn of s2 Bπr1

2 = µ0n1iπr12

φtotal = n2l(µ0n1iπr12)

φtotal = µ0n1n2liπr12

M = µ0n1n2πr1lI

20. Z = rms

rms

iE =

rms

2R

2

ivv +l

= i = 90º

22.

q1(r–x) q2

x

r

)xr(

kqx

kq 21

−=

xr

qxq 21

−= ;

x105 8−× =

)x20(102 8

−×− −

100 – 5x = –2x ; x = 3

100 cm

(ii) U = rqkq 21 = 12

59

10201025109

−

−

×××××−

= –4.5 × 105 J

23. (i) Microwaves are used in radars because of their short wavelength.

(ii) x-rays are used because of their penetration power

24. (i) Null point is shifted towards B (ii) Null point is shifted towards A (iii) Null point is not obtained

25. (i) φ = hv0

v0 = hφ

= 34

19

106.6106.114.2−

−

×××

= 5.18 × 1014 Hz

(ii) λhc = hv0 + eV0 = 2.14 + 0.60

λhc = 2.74 eV

λ = 19

834

106.174.2103106.6−

−

××××× = 5 × 10–7

26. (i) P3215 → S32

16 + –1eº + ν

(antineutrino)

(ii) half life T1/2 = 50 sec. λ = 2/1T

693.0

27. (i) width of interference fringes will reduce. (ii) if white light is used then due to overlapping of

pattern central fringe will be white with red edges. (iii) No interference pattern is obtained.

MOCK TEST PAPER SOLUTION FOR PAPER – 3 PUBLISHED IN THIS ISSUE


28.

(r + a)

p –q2a

(r – a)

r

q

Vp = )ar(

kq−

– )ar(

kq+

= kq

++

− ar1

ar1 = kq

− 22 ara2

Vp = 22 arkp−

a << r, Vp = 2rxp

(b) E = 32 N/C V = 16J/C

E = dV

d = EV = 0.5 m

dkq = 16

q = kd16

q = 91095.016

××

CHEMISTRY

1. They are due to presence of an unpaired e–. It absorbs light from visible regions and radiates complementary colour.

2. M = B

B

Mw ×

solutionof.vol100

= 10040100046.2

××

= 0.615 mole/litre 3. In this method, electrodes are made up of metal

whose sol has to be prepared dipped in dispersion medium like water. When an electric current is passed, a lot of heat convert metal into vapours which on cooling form colloidal solution.

4. Due to inert pair effect Pb+2 is more stable than

Pb+4.

5. (a) energy bond oforder ecreasingd

IFBrCl 2222 → >>>

(b) character ionic oforder ecreasingd

MIMBrMClMF → >>>

6. (a) NH3 molecules are associated with intermolecular H-bonding where as PH3 is not.

(b) H3PO3 is diprotic acid because it has two replacable hydrogen i.e.

P

O

HHOOH

7. It is because Ce+4 has stable electronic

configuration.

8. CH2 = CH2 + CO + H2

[Co(CO)4]2 CH3CH2CHOH2/Ni

CH3CH2CH2OH

It is used to convert alkene to higher aldehydes and alcohols.

9. K4(Fe(CN)6] → 4K+ + [Fe(CN)6]4– n = 5 πV = i nRT

π = iVn RT

π = iCRT where 'C' is molarity.

α = 1n1i

−− ⇒

10050 =

151i

−− ⇒ i = 3

π = 3 × 101 × 0.0821 × 300

= 7.386 atm

10. At anode Zn → Zn+2 + 2e– At cathode 2H+ + 2e– → H2(g)

11. (a) CH3 – CH – CH3

OH

Cl2 CH3 – C – CH3

O

3Cl2Ca(OH)2 + CCl3 – C – CH3

OCHCl3 + (CH3COO)2Ca

Chloroform

(b)

CH3 – C – CH3

O

(Acetone) + 3I2 + 4NaOH →

CHI3 + 3NaI + CH3COONa + 2H2O

12. (a) CH2OH

CH2OH+ 2HNO3

CH2ONO2

CH2ONO2+ 2H2O

Glycol dinitrate

(b)

CH2OH

CHOH

CH2

CH + 2H2O

CH2OH

KHSO4

HeatCHO

Acrolein (prop-2-en-1-al)


13. (a) C6H5NH2 Cº50

HClNaNO2

−

+ → C6H5N2+Cl–

KCNCuCN → C6H5C≡N + N2

(b) C6H5NH2 + CHCl3 + 3KOH → C6H5N ——→ C + 3KCl + 3H2O

Phenyl isocyanide 14. The principal ways by which cells obtain energy for

synthesis of ATP are – (a) Photo synthesis (b) Catabolism of nutrients such as carbohydrate, proteins & lipid. 15. (a) Beri-beri is caused by deficiency of Vit.-B. (b) Night blindness is caused by deficiency of Vit.-

A. (c) Vit.-K (d) Vit. – B12 16. a = ?; d = 2.48 g cm–3 N0 = 6.023 × 1023, z = 4

d = 0

3 NaMz

××

or a3 = dN

Mz

0 ×× =

2310023.648.2584

×××

= 2310023.648.2232

×× =

937.14232 × 10–23

= 15.53 × 10–23 = 155.3 × 10–24 a = (155.3)1/3 × 10–8 = 5.375 × 10–8 = 5.375 × 10–8 × 1010 pm = 537.5 pm. a = 2(r+ + r–)

Distance between K+ & F– = 2a

= 2

5.537 = 268.75 pm

17. Arrhenius equation is k = Ae–Ea/RT Where A = Frequency factor Ea = Activation Energy R = 8.314 J/K/mol T = Temperature in Kelvin

lnk = lnA – RTEa

lnk1 = lnA – 1

a

RTE

lnk2 = lnA – Ea/RT2

ln1

2

kk =

REa

−

21 T1

T1

= log1

2

kk =

−

21

12a

TTTT

303.2E

18. When catalyst and reactions are in different physical states. It is called heterogenous catalysis.

(a) 2SO2(g) + O2(g) V2O5 2SO3(g) [Contact process]

(b) N2(g) + 3H2(g) Fe 2NH3(g)

[Haber's process] (c) 4NH3(g) + 5O2(g) →Pt 4NO + 6H2O [Ostwald process ] (d)

CH = CH2

CH3

TiCl4 + Al(C2H5)3

Ziglar Natta catalyst CH – CH2

CH3

Polymer

n

19. (a) XeF4 + SbF5 → [XeF3]+ [SbF6]– (b) Cl2 + NaOH → NaCl + NaClO + H2O (c) 2F2 + 2H2O → 4HF + O2 (d) 2F2 + 4NaOH → 4NaF + O2 + 2H2O (e) XeF6 + KF → K+[XeF7]– (f) BrO3

– + F2 + 2OH– → BrO4– + 2F– + H2O

20. (a) Tendency to show catenation process decreases down the group due to increase in atomic size. In group 15, (P), in 16 (S), show catenation to maximum extent because 'N' & 'O' form multiple bond.

21. (i) 2Cu + O2 → 2CuO CuO + H2O → Cu(OH)2 Cu(OH)2 + CO2 → CuCO3 + H2O Green layer is due to formation of basic copper

carbonate. (ii) Ca(OH)2 + Cl2 → CaOCl2 + H2O (Bleaching powder) (iii) Cu +

.)conc(42SOH2 →Heat CuSO4 + 2H2O + SO2

22. (i) Ni(28) ⇒ [Ar] 4s2 3d8 Ni (0) ⇒ [Ar] 4s0 3d10

[Ni(CO)4] =

sp3 hybridisation ⇒ Tetrahedral ⇒ Diamagnetic (ii) Ni+2 = [Ar] 4s03d8

sp3

'Cl' does not cause pairing of electron because it is a weak field ligand.

[NiCl4]2– is Tetrahedral & Paramagnetic.

23. (a) CH2 = CH2 + K2[PtCl4] → K[PtCl3(C2H4) + KCl Zeise's salt


(b) 2C6H6 + Cr → [(C6H6)2Cr] dibenzene chromium

(vapours) (c) FeCl2 + 2C5H5MgBr → [(C5H5)2FCl] + 2Mg(Br)Cl Ferrocene

24. Lucas reagent is a mixture of conc. HCl and Anhyd. ZnCl2. It is used to distinguish between 1º, 2º and 3º alcohols.

Primary alcohols do not react with Lucas reagent at room temperature. Secondary alcohols react with Lucas reagent and turbidity (milkyness) appears after 5-minutes. 3º Alcohols react immediately forming milkyness.

CH3 – CH – CH3 + HCl

OH (conc)

ZnCl2(Anhy.)

CH3 – CH – CH3 + H2O

Cl 2-chloropropane

CH3 – C – CH3 + HCl

OH (conc)

ZnCl2(Anhy.)

CH3 – C – CH3 + H2O

Cl 2-chloro-2-methylpropane

CH3

CH3

25. (a) Buna-S is synthetic rubber. Its monomers are

butadiene and styrene, Na is polymerising agent. It is used for making automobile tyres.

(b) Elastomers have less force of attraction as compared to fibres. Elastomers regain their shape after the stress removed.

Eg. Buna-S, Vulcanised rubber are elastomers whereas nylon, terylene are example of fibers.

(c) Nylon - 66 26. (a) Malachite green (b) Methyl Orange (c) Alizarine

27. (a) [Zn(OH)4]2– (b) [Co(NH3)6]2(SO4)3 (c) K2[Ni(CN)4] (d) K2[PdCl4] (e) K3[Cr(C2O4)3] (f) [Pt (NH3)2Cl2] 28. Fe → Fe+2 + 2e– 2Ag+ + 2e– → 2Ag(s) Fe(s) + +

)aq(Ag2 → 2)aq(Fe+ + 2 Ag(s)

Ecell = ºcellE –

n0591.0 log 2

2

]Ag[]Fe[

+

+

= [ ]ºFe/Fe

ºAg/Ag 2EE ++ − –

20591.0 log 2)1.0(

1.0

= + 0.80 – (– 0.44) – 2

0591.0 log 10

= + 1.24 – 0.0295 Ecell = + 1.205 V

29. dtdx = k[A]x[B]y

5.07 × 10–5 = k[0.2]x[0.2]y ...(1) 5.07 × 10–5 = k[0.2]x[0.1]y ...(2) on dividing (1) & (2), we get 1 = 2y ⇒ 20 = 2 y i.e. y = 0 5.07 × 10–5 = k[0.2]x[0.1]y 7.60 × 10–5 = k[0.4]x [0.05]y

5.1

1 = y21 ⇒ 2/12

1 = x21

x = 21 = 0.5

The order of reaction is 0.5 with respect to (A) and zero with respect to (B).

30. (a) CH3CH2CH2CH2CH2OH Pentan-1-ol (1º alcohol) (b) CH3 – CH2 – CH – CH2 – OH

CH3

2-Methyl butan-1-ol (1º alcohol)

(c) CH3 – CH – CH2 – CH2 – OH

CH3


(d)

CH3 – C – CH2 – OH

CH3

2, 2-Dimethyl propan-1-ol (1º alcohol)

CH3

(e) CH3 – CH2 – CH2 – CH – CH3

OHPentan-2-ol (2º alcohol)

(f) CH3 – CH2 – CH – CH2 – CH3

OHPentan-3-ol (2º alcohol)

(g) CH3 – CH – CH – CH3

OH3-Methyl butan-2-ol (2º alcohol)

CH3

(h)

CH3 – C – CH2 – CH3


OH

CH3


MATHEMATICS

Section A

1. f–1(x) = x52

x3−

2.

ππ

23,

2 (or any other equivalent)

3. x = 3π

4. Skew symmetric

5. a12A12 + a22A22 + A32A32

6. log |x + log sin x| + c

7. Zero

8. 4π

9. 1

10. →a and →b are like parallel vectors.

Section B

11. (i) since (a – a) = 0 is a multiple of 4, ∀ a ∈A ∴ R is reflexive (ii) (a, b) ∈ R ⇒ (a – b) is a multiple of 4 ⇒ (b – a) is also a multiple of 4 ⇒ (b, a) ∈ R ∀ a, b ∈ A ⇒ R is Symmetric (iii) (a, b) ∈ R and (b, c) ∈ R ⇒ (a – b) = 4k, k ∈ z (b – c) = 4m, m ∈ z ∀ a, b, c ∈ A ⇒ (a – c) = 4(k + m), (k + m) ∈ Z ∴ (a, c) ∈ R ⇒ R is transitive Set of all elements related to 2 are 2, 6, 10 OR (i) ∀ a, b, c, d ∈ N, (a, b)* (c, d) = (a + c, b + d) = (c + a, d + b) = (c, d) * (a, b) ⇒ * is commutative (ii) [(a, b)*(c, d)]*(e, f) = (a + c, b + d) * (e, f) = ((a + c) + e, (b + d) + f) = (a + c + e, b + d + f) = (a + (c + e), b + (d + f)) ∀ a, b, c, d, e, f, ∈ N ⇒ = (a, b) * [(c, d) * (e, f)]* is associative (iii) Let (e, f) be the identity element, then (a, b) * (e, f) = (a, b) ⇒ (a + e, b + f) = (a, b) ⇒ e = 0, f = 0 but (0, 0) ∉ N × N So, identity element does not exist

12. We have tan–1

++

−−

−

++

+−−

2x1x.

2x1x1

2x1x

2x1x

= 4π

⇒ tan–1

+−−−−+−+

1x4x2xx2xx

22

22 =

4π

⇒ 3

4x2 2

−− = 1 ⇒ 2x2 = 1

⇒ x2 = 21 , x = ±

21

13. accbbacbbaacbaaccb

+++++++++

= 0

C1 → C1 + C2 + C3

⇒ 2(a + b + c)accb1cbba1baac1

++++++

= 0

R2 → R2 – R1, R3 → R3 – R1

⇒ 2(a + b + c)bcab0accb0baac1

−−−−++

= 0

⇒ 2(a + b + c) (–a2 – b2 – c2 + ab + bc + ca) = 0 ⇒ –(a + b + c) [(a – b)2 + (b – c)2 + (c – a)2] = 0 ⇒ a + b + c = 0 or a = b = c

14. )x(flim5x −→

= –1 + a

f(5) = a + b )x(flim

5x +→= 7 + b

⇒ – 1 + a = a + b = 7 + b ⇒ b = –1, a = 7

15. Let u = xy and v = yx ⇒ u + v = log a

⇒ dxdu +

dxdv = 0 ...(i)

log u = y log x

⇒ dxdu

u1 =

xy + log x

dxdy

⇒ dxdu = xy

+

dxdyxlog

xy

log v = x log y

⇒ dxdv

v1 =

yx

dxdy + log y

⇒ dxdv = yx

+ ylog

dxdy

yx

(i) ⇒ yxy–1 + xylog xdxdy + xyx–1

dxdy + yx log y = 0

⇒ dxdy = –

++

−

−

xlog.xy.xylog.yx.y

y1x

x1y


16. (i) Since (x – 2) ≥ 0 in [2, 3] so f(x) = 2x − is continuous

(ii) f ´(x) = 2x2

1−

exists for all x ∈(2, 3)

∴ f(x) is differentiable in (2, 3) Thus lagrang's mean value theorem is applicable; ∴ There exists at least one real number in (2, 3)

such that

f´(c) = 23

)2(f)3(f−−

or 2c2

1−

= 1

0)1( − ⇒ 2c2 − = 1

c = 2 + 41 = 2.25 ∈(2, 3)

⇒ LMV is verified and the req. point is (2.25, 0.5)

17. I = ∫ −− )bxcos()axcos(1 dx

= ∫ −−−−−

− )bxcos()axcos())bx()axsin((

)absin(1 dx

= ∫ −−−−

)]bxtan()ax[tan()absin(

1 dx

= )absin(

1−

[log | sec(x – a)| – log |sec (x – b)|]

= )absin(

1−

−−

)bxsec()axsec(log + c

OR

I = ∫ ++

xcos1xsin2 ex/2 dx

= ∫

++

+ xcos1xsin

xcos12 ex/2 dx

= ∫

+

2xcos2

2xcos

2xsin2

2xcos

222

ex/2 dx

= ∫

+

2xtan

2xsec2 ex/2 dx

2 tan2x .ex/2 + c

18. 2

ba→→

+ = 2

ba

+

→→=

→2a +

→2b + 2

→a .

→b

= |→a |2 + |

→b |2 + 2|

→a | |

→b | cos θ

= 1 + 1 + 2.1.1. cos θ

= 2(1 + cos θ) = 2.2cos2

2θ

⇒ 2

ba41 →→

+ = cos2

2θ

⇒ cos2θ =

→→+ ba

21

OR

Let ABCD be the parallelogram with sides →a and

→b

A B

CD

bd2

d1

a

∴ →

1d = →

AC = →a +

→b and

→2d =

→a –

→b

Now →→

× 21 dd =

−×

+

→→→→baba = 2

→→× ba

⇒ area | | gm = →→

× ba = →→

× 21 dd21

When →

1d = i + 2 j + 3 k and →

2d = 3 i – 2 j + k

⇒ →

1d ×→

2d = 123321kji

− = 8 i + 8 j – 8 k

∴ area of || gm =

++ 222 888

21 = 34 sq.u

19. The given equations can be written as

3

8x − = 16

9y−

+ = 710z − and

315x − =

829y − =

55z

−−

The shortest distance between two lines

→r =

→1a + λ

→1b and

→r =

→2a + µ

→2b is given by

S.D. = →→

→→→→

×

×

−

21

2112

bb

bb.aa

Here →

1a = (8, –9, 10), →

2a = (15, 29, 5)

⇒ →→

− 12 aa = (7, 38, –5)

and →

1b = (3, –16, 7) and →

2b = (3, 8, –5)

⇒ →

1b × →

2b = 583

7163kji

−−

= 24 i + 36 j + 72 k

∴ S.D. = 518412965763601368168

++

−+ = 84

1176 = 14 units.


20. Since x3 = 30 ∴ x1, x2 < 30 and x4, x5 > 30 ∴ Required Probability is = 29C2 . 1C1 . 20C2/50C5

=

1.2.3.4.546.47.48.49.501.219.20.

11.

1.228.29

= 15134

551

21. Here dydx = y/x

y/x

ye2yxe2 −

Let F(x, y) = y/x

y/x

ye2yxe2 −

then F(λx, λy) = )ye2(

)yxe2(y/x

y/x

λ−λ

F(x, y) is a homogeneous function of degree zero, thus the given differential equation is a homogeneous differential equation

Put x = vy to get dydx = v + y

dydv

∴ v + ydydv =

v

v

e21ve2 −

or ydydv = v

vv

e2ve21ve2 −− = – ve2

1

∴ 2ev dv = – y

dy ⇒ 2ev + log|y| = c

x = 0, y = 1 or, 2ex/y + log|y| = c ⇒ c = 2 ∴ 2ex/y + log |y| = 2 OR

Here integrating factor = ∫ xdxcote = elog sinx = sin x

∴ the solution of differential equation is given by y.sin x = ∫ + )xcotxx2( 2 sin x dx

= ∫ dxxsinx2 + ∫ dxxcosx 2

= ∫ dxxsinx2 + x2 sin x – ∫ dxxsinx2 + c

= x2 sin x + c ...(1) Substituting y = 0 and x = π/2, we get

0 = 4

2π + c or c = –4

2π

∴ (i) ⇒ y sin x = x2 sin x – 4

2π

or y = x2 – 4

2π cosec x

22. Equation of the said family is

2

2

ax +

2

2

by = 1

Differentiating w.r.t. x, we get

2ax2 + 2b

y2dxdy = 0 or

dxdy

xy = – 2

2

ab

=−

+

=

−+

0dxdyy

dxdyx

dxydxyor

0dxdy

x

ydxdyx

dxyd

xy

2

2

2

22

2

Section C

23. Let E1 : Letter has come from tatanagar ∴ P(E1)=1/2 E2 : Letter has come from calcutta P(E2) = 1/2 A : Obtaining two consecutive letters "TA"

∴ P(A|E1) = 82 =

41

Total possible TA, AT, TA, AN, NA, AG, GA, AR = 8, favourable = 2

P(A|E2) = 71 Total possibilities CA, AL, LC, CU,

UT, TT, TA = 7 favourable = 1

∴ P(E1|A) = )E|A(P)E(P)E|A(P)E(P

)E|A(P)E(P

2211

11

+

=

71.

21

41.

21

41.

21

+ =

117

∴ P(E2|A) = 1 – 117 =

114

OR Let x = Number of while balls.

====

====

====

====

301

8.9.102.3.4

CC)3x(P

103

8.9.103.6.3.4

CC.C

)2x(P

21

8.9.103.6.3.4

CC.C)1x(P

61

8.9.104.5.6

CC

)0x(P

310

34

310

16

24

310

26

14

310

36

Thus we have

x P(x) xP(x) x2P(x) 0 1/6 0 0 1 1/2 1/2 1/2 2 3/10 6/10 12/10 3 1/30 3/30 9/30 1 36/30 60/30 = 2

Mean = ΣxP(x) = 3036 =

1518 =

56 = 1.2

Variance = Σx2P(x) – [ΣxP(x)2]2

= 2 – 2536 =

2514 or 0.56


24. AB is parallel to the line 2x = y = z

A(3, 4, 5)

B

x + y + z = 2

2x= y = z

or 2/1

x = 1y =

1z

or 1x =

2y =

2z

⇒ Equation of AB is 1

3x − = 2

4y − = 2

5z −

For some value of λ, B is (λ + 3, 2λ + 4, 2λ + 5) B lies on the plane ∴ λ + 3 + 2λ + 4 + 2λ + 5 – 2 = 0 ⇒ 5λ = –10 ⇒ λ = 2 ∴ B is (1, 0, 1)

ΑΒ = 222 )15()04()13( −+−+− = 6 units. 25. Solving the equations in pairs to get the vertices of

∆ as (0, 1), (2, 3) and (4, – 1) For correct figure

y

3 B (2, 3)

A (0, 1)

2x + y = 7

x + 2y = 20

–1 C(4, –1)

x

–x + y = 1

Required area

= ∫−

−3

1

dy)y7(21 – ∫

−

−1

1

dy)y22( – ∫ −3

0

dy)1y(

= 3

1

2

2yy7

21

−

− – [ ]11

2yy2 −− – 3

1

2y

2y

−

= 12 – 4 – 2 = 6 sq. U OR

A2 A1

(0, 0) (2, 0) (4, 0)

Area A1 = 2

−+∫ ∫

2

0

4

2

2 dxx16dxx6

= 2

+−+

−4

2

122

0

2/3

4xsin8x16

2x

3x2.6

= 3

1634 π+ sq. U.

A2 = Area of circle – shaded area

16π – 3

1634 π+ = 3

3432 −π

∴ 2

1

AA =

34323416

+π

+π = 3834

−π

+π

26. I = ∫− xtan 1

e 22 )x1(1

+dx

Put x = tan θ to get I = ∫ θθθ dcos.e 2

= ∫ θθ+θ d)2cos1(e21

= 21 eθ+ ∫ θθθ d2cos.e

21

= 21 eθ+ I1 ...(1)

I1 = ∫ θθθ d2cos.e21

= [ ]∫ θθ−−θ θθ de.2sin22cos.e21

= [ ]∫ θθ−θ+θ θθθ de.2cos2e.2sin22cos.e21

= 21 [eθcos 2θ + 2 sin 2θ eθ] –

4.21

∫ θθ θde.2cos

I1 = 21 eθcos 2θ + sin 2θ eθ – 4I1

⇒ I1 = 101 eθ cos 2θ +

51 sin 2θ eθ

Putting in (i) we get

I = 21 eθ +

101 eθ cos 2θ +

51 sin 2θ eθ + c

= 101 eθ [5 + cos 2θ + 2sin 2θ] + c

= xtan 1e

101 −

.

++

+−

+ 22

2

x1x4

x1x15 + c

27.

A

L

B

a P

b

Mθ C


Let ∠C = θ. ∴ AC = AP + PC = S (say) ∴ S = a sec θ + b cosec θ

∴ θd

ds = a sec θ tan θ – b cosec θ cot θ

θd

ds = 0 ⇒ θθ

2cossina =

θθ

2sincosb

or tan3θ = ab or tan θ =

3/1

ab

2

2

dsd

θ = a[sec3θ + secθ tan2θ]

+ b[cosec3θ + cosec θ cot2 θ] Which is +ve as a, b > 0 and θ is acute

∴ S is minimum when tan θ = 3/1

ab

∴ Minimum S = AC = θ+ 2tan1a +

θ+ 2cot1b

= 3/2

ab1a

+ +

3/2

ba1b

+

= a2/3 3/23/2 ba + + b2/3 3/23/2 ab + = (a2/3 + b2/3)3/2

28. Let A =

−−−

052503231

Writing A = IA

or

−−−

052503231

=

100010001

A.

R2 → R2 + 3R1, R3 → R3 – 2R1

−−−

4101190231

=

− 102013001

A

R1 → R1 + 3R3

−−

4101190

1001 =

−

−

102013305

A

R2 → R2 + 8R3

− 41021101001

=

−−−

1028113305

A

R3 → R3 + R2

250021101001

=

−−−

91158113305

A

R3 → 251 R3

100010

1001 =

−

−−

259

251

2515

8113305

A

R1 → R1 – 10R3

100010001

=

−

−

−−

259

251

2515

2511

254

2510

2515

25101

A

∴ A–1 =

−

−

−−

259

251

2515

2511

254

2510

2515

25101

or

−−

−−

911511410151025

251

29. Let number of chairs = x number of tables = y ∴ LPP is Maximise P = 30x + 60y

Subject to

≤+≤+≤+

90y3x40yx70yx2

x ≥ 0, y ≥ 0 For correct graph

20 35 40 60 80 100

203040

607080

n (15, 25)

(30, 10)C

y–40 P = 30(x + 2y) P(A) = 30(60) PB = 30(65) PC = 30(50) PD = 30(35) ∴ For Max Profit (30 × 65) No. of chairs = 15 No. of tables = 25


XtraEdge Test Series ANSWER KEY

PHYSICS

Ques. 1 2 3 4 5 6 7 8 9 10 Ans. D B C C A A A,C,D B,C A,D B ,D Ques. 11 12 13 14 15 16 17 18 19 Ans. C D C A B B 0041 0030 0040

CHEMISTRY

Ques. 1 2 3 4 5 6 7 8 9 10 Ans. A B B C A B A,B,C A,B,C ,D A,C,D B,C,DQues. 11 12 13 14 15 16 17 18 19 Ans. A,C C B D A C 0707 0602 1200

MATHEMATICS

Ques. 1 2 3 4 5 6 7 8 9 10 Ans. A D C A D A B,D A,C,D B ,C,D A,C Ques. 11 12 13 14 15 16 17 18 19 Ans. B A,C B C B D 8282 0996 7168

PHYSICS

Ques. 1 2 3 4 5 6 7 8 9 10 Ans. C D C C D D B,C A,C A,C A,B,C Ques. 11 12 13 14 15 16 17 18 19 Ans. A A C B A A 0003 0003 0005

CHEMISTRY

Ques. 1 2 3 4 5 6 7 8 9 10 Ans. B C A D B D A,C A,B,D A,C A,B,C ,DQues. 11 12 13 14 15 16 17 18 19 Ans. A,B,C B A A,C B,C,D B,C,D 0040 0025 0500

MATHEMATICS

Ques. 1 2 3 4 5 6 7 8 9 10 Ans. A A C D D B A,B,C ,D B,C D A,C Ques. 11 12 13 14 15 16 17 18 19 Ans. A A,C,D C C A,B,C B 1536 8410 1828

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