July-2009

XtraEdge for IIT-JEE 1 JULY 2009

Dear Students, • It’s the question you dreamed about when you were ten years old. • It’s the question your parents nagged you about during high school. • It’s the question that stresses most of us out more and more the older

we get. “What do you want to be when you grow up?” There are people who are studying political science but hate politics, nursing majors who hate biology, and accounting majors who hate math. Obviously, a lot of people are confused about what exactly it is that they want to spend their life doing. Think about it. If you work for 10 hours each day, you’re going to end up spending over 50% of your awake life at work. Personally, I think it’s important that we spend that 50% wisely. But how can you make sure that you do? Here are some cool tips for how to decide what you really want to be when you grow up.

Relax and Keep an Open Mind: Contrary to popular belief, you don’t have to “choose a career” and stick with it for the rest of your life. You never have to sign a contract that says, “I agree to force myself to do this for the rest of my life”. It’s your life. You’re free to do whatever you want and the possibilities are endless. So relax, dream big, and keep an open mind.

Notice Your Passions: Every one of us is born with an innate desire to do something purposeful with our lives. We long to do something that we’re passionate about; something that will make a meaningful impact on the world.

Figure Out How to Use Your Passions for a Larger Purpose: You notice that this is one of your passions, so you decide to become a personal trainer. Making a positive impact on the world will not only ensure that you are successful financially, it will also make you feel wonderful. It’s a proven principle: The more you give to the world, the more the world will give you in return.

Figure Out How You Can Benefit Once you’ve figured out what your passions are and how you can use those passions to add value to the world &to yourself ,

It’s time to take the last step: figure out how you can make great success doing it. My most important piece of advice about this last step is to remember just that: it’s the last part of the decision process. I feel sorry for people who choose an occupation based on the average income for that field. No amount of money can compensate for a life wasted at a job that makes you miserable. However, that’s not to say that the money isn’t important. Money is important, and I’m a firm believer in the concept that no matter what it is that you love doing, there’s at least one way to make extraordinary money doing it. So be creative!

Simply discover your passions, figure out how to use your passions to make an impact on the world & to yourself.

Yours truly

Pramod Maheshwari, B.Tech., IIT Delhi

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WORRY IS A MISUSE OF IMAGINATION. Volume - 5 Issue - 1

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Volume-5 Issue-1 July, 2009 (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.

Xtra Edge Test Series for JEE-2010 & 2011

S

Success Tips for the Months

• " Always bear in mind that your own resolution to succeed is more important than any other thing."

• "God gave us two ends. One to sit on and one to think with. Success depends on which one you use; head you win -- tails, you lose."

• "The ladder of success is best climbed by stepping on the rungs of opportunity."

• "Success is getting what you want. Happiness is wanting what you get."

• "The secret of success in life is for a man to be ready for his opportunity when it comes."

• "I don't know the key to success, but the key to failure is trying to please everybody."

• "The secret of success is to be in harmony with existence, to be always calm… to let each wave of life wash us a little farther up the shore."

CONTENTS

INDEX PAGE

NEWS ARTICLE 3 IIT aspirants’ score cards get better IITs, IIMs need to revise curriculum, says Sibal

IITian ON THE PATH OF SUCCESS 7 Dr. Ashok Jhunjhunwala

KNOW IIT-JEE 8 Previous IIT-JEE Question

XTRAEDGE TEST SERIES 47

Class XII – IIT-JEE 2010 Paper

Class XI – IIT-JEE 2011 Paper

Regulars ..........

DYNAMIC PHYSICS 14

8-Challenging Problems [Set# 3] Students’ Forum Physics Fundamentals Capacitance - 1 Friction

CATALYST CHEMISTRY 28

Key Concept Reaction Mechanism Chemical Energetics Understanding : Organic Chemistry

DICEY MATHS 37

Mathematical Challenges Students’ Forum Key Concept 3-Dimensional Geometry Progression & Mathematical Induction

Study Time........

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


IIT aspirants’ score cards get better MUMBAI: It is a truth universally accepted that getting into an IIT is certainly no piece of cake — a feat achieved by few who can crack what is perceived to be the one of the toughest tests this side of the world. But records of students’ performances show that the number of candidates who scored negatively has dropped from 1.32 lakh in 2006 to 13,866 in 2008. The figures show that 45% of those who took the Joint Entrance Exam (JEE) in 2006 did not even score a ‘respectable’ zero. Three years ago, when 2.88 lakh took the JEE, almost half of them were in the minus zone. But in 2008, merely 4% of the candidates dropped below the thin red line, despite the fact that the number of students who gave the entrance test that year had increased to 3.11 lakh from 2.88 lakh in 2006.

A JEE 2009 chairman attributed the drop in negative scorers to the new objective-type pattern of the question paper as well as the restrictions imposed on the number of attempts. In September 2005, the C N R Rao panel recommended that the screening be done away with and a single objective test be introduced. The maximum number of attempts permitted was restricted to two. The move, introduced in JEE-2006 aimed at reducing exam stress, weaken the influence of coaching classes and “restore the sanctity of the school system”.

Till 2007, data from IIT shows that about 10% of the total aspirants were taking the JEE for the third time (or more). In 2005, for instance, of the 1.98 lakh aspirants,

19,806 had taken the entrance test at least two times earlier. Next year, of the total 2.99 lakh candidates, 23,931 students were sitting for the exam for the third time (or even higher). In other words, curtailing aspirants from taking as many shots as they wanted, changed more than just some rules on paper. Separate Right to Information applications filed at various IITs reveal some fresh trends: rock bottom exam scores (not qualifying marks) have been improving, and the number of those scoring negative aggregates is falling. While these trends may not say anything about students making it to the IITs, it shows that the majority of those in the ranks below are no longer taking the exam as another jolly ride.

Ropar to run IIT from polytechnic campus this year MOHALI: Classes of IIT, Punjab, currently being run in Delhi, will be held in the dusty town of Ropar from June this year. Initially, the campus will be set up at women’s polytechnic with the government deciding to shift its students to other colleges. For the next four-five years, women’s polytechnic at Ropar will house IIT. As engineering classes will start next month, hectic renovation work on the polytechnic campus is on these days. To provide campus residential facility, hostels and staff quarters too are being renovated, officials added. Apart from this, the state government is also going to hand over 500 acre land to the IIT authorities in the coming few days.

Minister for human resources development Arjun Singh had laid the foundation stone of IIT Ropar on February 24 this year. Out of total 513 acres land where the campus would come up some land owned by the village panchayat had come under dispute, but recently panchayat agreed to hand over the land to IIT.

Punjab technical education and industrial training secretary Tejinder Kaur said in the next few days, land for the IIT campus would be handed over so that construction could start. She said to ensure that IIT classes start from the polytechnic without any obstacle, no admission was held last year which this year the final year classes are going to be shifted to other colleges. “Our responsibility is to provide the land and transit campus for the IIT. Rest is the responsibility of IIT authorities, who will upgrade the labs as per their specifications,” she said.

Meanwhile, according to sources, last year 120 students were admitted to IIT Ropar in the streams of mechanical, electrical and computer sciences in Delhi.

CBSE toppers aim for IIT & NIT NAGPUR: IIT and NIT are the buzzwords for students of CBSE Standard XII, results of which were announced on Wednesday.

City science topper Antony Aron George of Bhavan’s BP Vidya Mandir (Civil Lines), who scored 96.4%, aims to work with missile guidance programme and military communication technology. “My path is very clear,” he says matter-of-factly. “I will first complete my


graduation from a renowned engineering college like any of the IITs or NITS and then pursue my studies in military communication. All the military equipment, hardware and related technology fascinate me a lot.”

Disha Murarka, who scored a whopping 97.2% to top the commerce stream in the city, aspires to become a chartered accountant. Though her father is a businessman, she feels that the CA profession is a dynamic, challenging and demanding one.

According to her, engineering and medicine fields have become very common. “The scope for this lucrative career is bright in an economically developing nation like ours. Such an option can be termed as challenging as well as rewarding for competent professionals in the field,” she feels. Disha also aspires to pursue a degree in business administration after finishing the CA course.

Another topper, Kushal Gupta from Modern School (Koradi branch), who scored 95.6%, toes Antony’s line. “My dream is to always pursue studies from these premier institutions. But my main aim is to become an electrical or electronics engineering and serve the nation,” he says.

Pooja Gavel of Bhavan’s BP Vidya Mandir too feels the same. “I wish to complete my studies from Birla Institute of Technology and Science, Pilani. I just want a good basic degree probably from NITS or BITS. I don’t want to go for IITs. But my ultimate aim is to don the Indian Administrative Services (IAS) and serve the society,” Pooja, who scored 92.6%, told TOI.

IIT-K gets ‘crack technology’ to help build safer cars Kanpur: The Indian Institute of Technology-Kanpur (IIT-K) has procured an Ultra High Speed

Camera from England which can capture 20 crore frames per second.

The camera, worth Rs 2 crore, is expected to help the institute make a breakthrough in their ongoing research on the development of cracks in objects of daily use, including vehicles.

The state-of-the-art instrument is the fastest in the world and the first of its kind in the country. It can capture 20 crore images per second, each of which have utmost clarity.

“Crack propagation is a complex issue which requires an extremely detailed study,” said Professor and Head of Department of Mechanical Engineering at IIT-K, Nalinaksh S Vyas. This camera is expected to help understand how cracks develop and how they spread. “And we can definitely get in-depth knowledge about the nature of each crack,” he said.

IITs, IIMs need to revise curriculum, says Sibal The Indian Institutes of Technology (IITs) and Indian Institutes of Management (IIMs) need to revise their curriculum to compete at the international level, newly-appointed Union Minister for Ministry of Human Resource Development (MHRD), Kapil Sibal, told mediapersons while taking charge today.

He also clarified that pending Bills, drafted by the previous government, would be pushed forward. The important Bills include the Foreign Education Bill and Right to Education Bill. If passed in Parliament, the Foreign Education Bill will allow foreign higher education providers to set up Indian campuses. At present, foreign universities are forbidden from offering degree courses in India by the MHRD that governs education. Sibal, however, did not provide any details in this regard.

Explaining his stance on revising the IIT/IIM curriculum, he said:

“Because of technology and automation, the economy has changed and it has become difficult for graduates to find jobs. So, there has to be a change in the school and university curriculum.”

Some IIMs agree there is a need to change the curriculum so that they are abreast with changing times. “Our education system has been slow in responding to the changes around us. For instance, management institutes like us are now questioned about the relevance of the management education that is being offered after the economic slowdown. So, for us to live upto what is expected from us, we will have to incorporate the changes,” said Devi Singh, Director, IIM-Lucknow.

However, another IIM professor countered: “IIMs have always upgraded the curriculum as and when required. We are conscious of the fact that we have to adjust and respond to the changing needs. However, if the ministry expects us to incorporate some changes, we will do it.”

Incidentally, the R C Bhargava (Chairman of Maruti Suzuki) IIM Review Committee report said that most of the IIM management development programmes (MDPs) consist of courses for comparatively junior PSU executives reflecting on the perception of business and industry about the value addition which IIMs can provide for senior managers. When contacted, Bhargava said: “I believe education is a very important ministry — among the top three ministries in India — but has not been given its due. Mr Sibal’s appointment gives a great degree of hope that the education sector will see a lot of reforms soon.”

Meanwhile, emphasising on the fact that this ministry is linked to youth, children and families and that it needs to be seen how it can take India forward, Sibal said: “For


us, the larger vision is of access and quality meaning that all should get good quality education with access to education. The challenges in this are private education institutions- both aided and unaided, government institutions and institutions of excellence.”

Commenting on the reservations in education because of the OBC quota, he said that all new policies have teething problems.

As for the recommendations made by the Prime Minister-appointed National Knowledge Commission (NKC) whose term ended in March, Sibal said, “NKC is pivotal for creating a knowledge economy but this does not mean that whatever it says should be embraced or adopted. If the need arises, their recommendations will be adopted.”

On the school education front, the minister said school fees are an area of great concern and that anyone in need, would receive financial aid so that no child is denied education because of lack of resources.

Teacher training is an ‘exceptionally serious’ problem and so is ragging in schools and colleges that will be looked at. Skill development is another area of concern for the ministry.

Explaining the similarities and proximity between MHRD and his previous ministry of Science and Technology, Sibal said, “There are synergies between the MHRD and Ministry of Science and Technology and these will be included in MHRD.”

New IITs get regular directors NEW DELHI: Functioning under temporary arrangement for over a year, the newly-created six IITs have now got their regular directors. Prof U B Desai, Prof M K Surappa, Prof Sudhir Kumar Jain, Prof Prem Kumar Kalra, Prof Madhusudan

Chakroborty and Prof Anil Bhowmick will head the IITs at Hyderabad, Ropar, Gandhinagar, Rajasthan, Bhubaneswar and Patna respectively. "The Union Cabinet has given its go ahead for the appointment of the directors who were selected by the search committee. We have issued appointment letters to the new directors last week," a senior HRD ministry official said. The search committee of ministry of HRD had earlier selected these academics to head the new IITs. The IIT Council, the highest decision making body of the elite institutes, had approved the appointment of directors in January this year. However, they could not be appointed pending Cabinet's approval. The government started six new IITs last year. Since they did not have campuses of their own, they were started on mentoring basis, an arrangement under which the old IITs had to accommodate the students of the new institutes. The directors of the old IITs were made acting directors of the new ones.

In a first, an IIT-IIMer as MP NEW DELHI: It is Indian education system's ultimate ticket to the corporate boardroom. According to one study, around 50% of all CEOs in India have those magic letters on their CVs — IIT-IIM. But for all their brilliance and achievement in the corporate world, not a single member of the IIT-IIM club had so far entered Parliament as a member.

Now, Prem Das Rai, elected to the 15th Lok Sabha as the lone member from Sikkim, has the unique distinction of being the first ever IIT-IIMer in the Lok Sabha. Rai, 54, hopes to contribute his bit in changing the way Parliament and MPs are viewed in the country. For the moment though, this IIT Kanpur (chemical engineering) and IIM Ahmedabad alumnus just wants to familiarize himself with the new job. As he puts it, ‘‘First, I need to look at the benchmarks of what constitutes a good MP. We are at a crossroads. People this time have voted for stability but they have also voted for better parliamentarians and parliamentary processes,’’ says the Sikkim Democratic Front MP, who early in his career chucked a cushy job as a multinational banker and later gave up opportunities in the US in favour of returning to his home state, Sikkim.

And what does an IIT-IIMer bring to Parliament? ‘‘I look at the entire IIT-IIM community as part of my support system. We have a very strong network. Through it I would be able to funnel a lot of intellectual capital. Then there is the skill set. In IITs and IIMs, you develop a certain way of learning, analysis and presentation. These would be handy in Parliament,’’ he says.

Rai, however, admits he is entering uncharted territory. ‘‘These are, as yet, just hypotheses. I don’t know how it will actually play out. Down the road you would be able to find out how I add value to governance.’’ Rai had been marked for brilliance early in life. As a boy growing up in the then sovereign kingdom of Sikkim, he was sent to an elite school in Mussoorie by the king as part of the royal policy of promoting bright kids. After school, he cracked the JEE to go directly to IIT and then to IIM.

‘‘Life in Sikkim was very feudal but in Mussoorie I learned that you


are no more and no less than anybody else. That’s a value I imbibed early,’’ he says.

In the midst of his BTech course, Sikkim merged with India and Rai become an Indian. After completing his MBA in 1978, Rai landed himself a ‘‘great job’’ at the American Bank in Calcutta. A meeting with B B Lal, the then Governor of Sikkim, changed the course of his life. ‘‘Lal told me, ‘Young man what are you doing in Calcutta? Come back, Sikkim needs you.’ That finally helped me make up my mind and I returned to join a state government enterprise with a three-fourth cut in salary.’’

41st convocation of the IIT-K KANPUR: The chief guest of the 41st convocation of the IIT-K, Jeet Bindra was accompanied by Prof RK Thereja and Prof Dhande in a press conference which was followed after the convocation ceremony.

Prof Dhande while talking to media persons said that for the first time in the history of IIT-K, an alumnus of the institute, JS Bindra has come as the chief guest on this occasion.

JS Bindra said, "Chevron Global Manufacturing has hired two students of the chemical engineering department from the institute in the campus recruitment programme. We have been sponsoring students and will continue to do the same for promoting ongoing research work in the institution."

"We will also decide upon how much to fund the IIT-K to promote research and education. The focus is on to attract high quality faculty from western shores to this institute to take ongoing research work at newer heights and this is only possible when the foreign faculty will have

access to quality labs in the institute", Bindra further added

SK Mitra new IIT-K Director KANPUR: An ex-IIT-ian and the senior most professor of the National Sugar Institute SK Mitra took over as the director of the institute, here on Monday.

A sugar technologist of proven ability, Mitra has rendered technical advice to large number of sugar factories across the country for capacity expansion, modernisation, co-generation and on various aspects to improve the overall technical efficiency of the plant.

Prof Mitra in an interview expressed that his first and foremost task would be to improve and up-grade the system and level of education and consultancy services being rendered by the institute besides making efforts to fill large number of posts lying vacant. He stressed upon carrying out research on areas which had a direct bearing on the sugar and alcohol industry and which might improve techno-economic situation of these industries.

Talking on the present sugar crisis, he informed that the situation had worsened due to shortfall in sugar production in other major sugar producing countries like Thailand, Pakistan etc and the sugar prices had increased rapidly in the global market. He stressed need for qualitative development of sugarcane, particularly, in the northern region of the country so as to bring it at par with that available in the other high sugar recovery areas like Maharashtra and Gujarat.

Regarding financial crisis which was reported earlier, Mitra informed that there was absolutely no such crisis during this financial year and the ministry had

sanctioned the desired budget. He said, "I am confident that with the type of support and guidance being provided by the ministry, the institute will again find its old glory."

Road from IIT to AIIMS blocked due to settlement New Delhi : The road connecting Indian Institute of Technology (IIT) to All India Institute of Medical Sciences (AIIMS) here has blocked for two days from today after it caved in, Delhi Metro Rail Corporation said.

The settlement over an area of 250 metre occurred in the morning hours after a storm water drain below the road was damaged due to heavy overnight rainfall, DMRC said in a release.

The damaged drain started leaking out rainwater and the loose soil above the under-construction tunnel has led to the settlement.

E Sreedharan, Managing Director, DMRC, personally inspected the affected portion and gave directions for its restoration to reduce public inconvenience soon, it said.

Normal traffic was expected to be restored on the stretch by Tuesday morning as the DMRC has already begun the restoration work round the clock.

Meanwhile, the traffic has been diverted through the Green Park market road. Vehicles coming from the IIT side were being diverted from the Green Park mosque point, while they can come back on the Aurobindo Marg through a road next to Hotel Parkland, it added.

Traffic marshals have been deputed at all major intersections to guide the vehicles. The traffic is, however, normal on the other side of the Aurobindo Marg from AIIMS to IIT, the release said.


Prof. Ashok Jhunjhunwala is Professor of the Department of Electrical Engineering, Indian Institute of Technology, Chennai, India and was department Chair till recently. He received his B.Tech degree from IIT, Kanpur, in 1975, and his MS and Ph.D degrees from the University of Maine, USA in 1977 and 1979, respectively. From 1979 to 1981, he was with Washington State University as Assistant Professor. Since 1981, . He has been a member of the faculty at IIT, Madras since 1981 and Department Chair until recently. Professor Jhunjhunwala has the unique distinction among academics for combining innovations in technology and business incubation with the social goal of sustainable development in India Dr.Jhunjhunwala leads the Telecommunications and Computer Networks group (TeNeT) at IIT Madras. This group is closely working with industry in the development of number of Telecommunications and Computer Network Systems. TeNeT group has incubated a number of R&D companies which work in partnership with TeNeT group to develop world class technologies. The products include corDECT Wireless in Local Loop system, Fibre Access Network, DSL Systems, Rural ATM [Approximately costing around Rs 40,000)and remote medical diagnostic Kit. The group has recently incubated a

company, which aims to install and operate telephone and Internet in every village in India. Dr. Jhunjhunwala is a Director in the Board of SBI. He is also a board member of several Telecom and IT companies in India, including Polaris, Sasken, Tejas Networks and HTL, NRDC, and IDRBT. He is a former board member of VSNL & BSNL. Dr.Jhunjhunwala has been awarded Padma Shri in the year 2002. He has been awarded Shanti Swarup Bhatnagar Award in 1998 and has received Dr.Vikram Sarabhai Research Award for the year 1997. He has also received Millennium Medal at Indian Science Congress in the year 2000 and Dr. P.Sheel Memorial Lecture Award of the Academy for the year 2001 by National Academy of Sciences, India. He is a Fellow of Indian National Academy of Engineering, Indian National Science Academy and National Academy of Science and a Governor of International Council for Computer Communications (ICCC). Dr.jhunjhunwala is a Board member of several Telecom and IT companies in India, including BSNL, VSNL, Polaris, Sasken, Tejas and HTL. He is on the board of several R&D and educational institutions. He is on several government bodies formulating and driving policies in the area of Telecom and Human Resource Development.

Dr. Ashok Jhunjhunwala Designation : Professor Education From : B.Tech degree from IIT,

Kanpur University of Maine, USA

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

"Success is a journey, not a destination."


PHYSICS

1. Masses M1, M2 and M3 are connected by strings of negligible mass which pass over massless and friction less pulleys P1 and P2 as shown in fig. The masses move such the portion of the string between P1 and P2 in parallel to the inclined plane and the portion of the string between P2 and M3 is horizontal. The masses M2 and M3 are 4.0 kg each and the coefficient of kinetic friction between the masses and the surfaces is 0.25. The inclined plane makes an angle of 37º with the horizontal. [IIT-1981]

P1

M1

M2 P2 M3

37º

If the mass M1 moves downwards with a uniform velocity, find

(a) the mass of M1 (b) The tension in the horizontal portion of the string (g = 9.8 m/sec2, sin 37º ≈ 3/5) Sol. (a) Applying Fnet = ma on M1 we get T – m1 . g = M1 × 0 = 0 ⇒ T = M1g ...(i) Applying Fnet = Ma on M2 we get T – (T´ + M2g sin θ – f) = M2 × a T = T´ + M2g sin θ + f = T´ + M2g sin θ + µN [Q f = µN = µM2 g cos θ] ∴ T = T´ + M2g sin θ + µM2g cos θ ...(ii)

P1

M1

M2

T´ P2

M2g M1g

T

T

V

θ

M2gcosθ

V

f

M2gsinθ

θ

T´

N

M3g Applying Fnet = Ma for M3 we get T´ – f ´ = M3 × 0 ⇒ T´ = f ´ = µN´ = µM3g ...(iii) Putting the value of T and T´ from (i) and (iii) in (ii)

we get M1g = µM3g + M2g sin θ – µ M2g cos θ M1 = 0.25 × 4 + 4 × sin 37º + 0.25 × 4 × cos 37º = 4.2 kg (b) The tension in the horizontal string will be T ´ = µM3g = 0.25 × 4 × 9.8 = 9.8 N

2. A 0.5 kg block slides from the point A (see fig.) on a horizontal track with an initial speed of 3 m/s towards a weightless horizontal spring of length 1 m and force constant 2 Newton/m. The part AB of the track is frictionless and the part BC has the coefficients of static and kinetic friction as 0.22 and 0.2 respectively. If the distances AB and BD are 2m and 2.14 m respectively, find the total distance through which the block moves before it comes to rest completely. (Take g = 10 m/s2) [IIT-1983]

A B D C Sol. K.E. of block = work against friction + P.E. of spring

21 mv2 = µk mg (2.14 + x) +

21 kx2

21 × 0.5 × 32 = 0.2 × 0.5 × 9.8(2.14 + x) +

21 2 × x2

2.14+ x + x2 = 2.25 ∴ x2 + x – 0.11 = 0

On solving we get x = – 1011

or x = 101 = 0.1 (valid answer)

Here the body stops momentarily. Restoring force at y = kx = 2 × 0.1 = 0.2 N Frictional force at y = µs mg × x = 0.22 × 0.5 × 9.8 = 1.078 N Since friction force > Restoring force the body will

stop here. ∴ The total distance travelled = AB + BD + DY = 2 + 2.14 + 0.1 = 4.24 m.

A B D C Rough L 2m

2.14mx

Y 3. A small sphere rolls down without slipping from the

top of a track in a vertical plane. The track in a vertical plane. The track has an elevated section and a horizontal part, The horizontal part is 1.0 meter above the ground level and the top of the track is 2.4 metres above the ground. Find the distance on the

KNOW IIT-JEE By Previous Exam Questions


ground with respect to the point B(which is vertically below the end of the track as shown in fig.) where the sphere lands. During its flight as a projectile, does the sphere continue to rotate about its centres of mass ? Explain. [IIT-1987]

A

B

1.0m

2.4

m

Sol. Applying law of conservation of energy at point D

and point A P.E. at D = P.E. at A + (K.E.)T + (K.E.)R (K.E.)T = Translational K.E.

mg (2.4) = mg (1) + 21 mv2 +

21 Iω2

(K.E.)R = Rotational K.E. Since the case is of rolling without slipping

D

2.4m

1m A

B C ∴ v = rω

∴ ω = rv where r is the radius of the sphere Also

I = 52 mr2

∴ mg(2.4) = mg(1) + 21 mv2 +

21 ×

52 mr2 × 2

2

rv

⇒ v = 4.43 m/s After point A, the body takes a parabolic path. The

vertical motion parameters of parabolic motion will be

uy = 0 S = ut + 21 at2

Sy = 1m 1 = 4.9 ty2

ay = 9.8 m/s2

∴ ty = ? ty = 9.4

1 = 0.45 sec

Applying this time in horizontal motion of parabolic path, BC = 4.43 × 0.45 = 2m

During his flight as projectile, the sphere continues to rotate because of conservation of angular momentum.

4. Two square metal plates of side 1 m are kept 0.01 m

apart like a parallel plate capacitor in air in such a way that one of their edges is perpendicular to an oil surface in a tank filled with an insulating oil. The plates are connected to a battery of emf 500 V. The

plates are then lowered vertically into the oil at a speed of 0.001 ms–1. Calculate the current drawn from the battery during the process. (Dielectric constant of oil = 11, ε0 = 8.85 × 10–12C2N–1m–1)

[IIT-1994] Sol. The adjacent figure is a case of parallel plate

capacitor. The combined capacitance will be

v

+

1–x

x1m

d C = C1 + C2

= d

)1x(k 0 ×ε +

d]1)x1[(0 ×−ε

C = d0ε

[kx + 1 – x]

After time dt, the dielectric rises by dx. The new equivalent capacitance will be

C + dC = C1´ + C2´

= d

k 0ε[(x + dx) × 1] +

d]1)dxx1[0 ×−−ε

dC = Change of capacitance in time dt

= d0ε [kx + kdx + 1 – x – dx – kx – 1 + x]

= d0ε

(k – 1)dx

dtdC =

d0ε

(k – 1)dtdx =

d0ε

(k – 1)v ...(i)

where v = dtdx

We know that q = CV

dtdq = V

dtdC ...(ii)

⇒ I = Vd0ε (k – 1)v

From (i) and (ii)

I = 01.0

1085.8500 12−×× (11 – 1) × 0.001

= 4.425 × 10–9 Amp. 5. Two resistors, 400 ohms, and 800 ohms are

connected in series with a 6-volt battery. It is desired to measure the current in the circuit. An ammeter of a 10 ohms resistance is used for this purpose. What will be the reading in the ammeter? Similarly, If a voltmeter of 10,000 ohms resistance is used to


measure the potential difference across the 400-ohms resistor, What will be the reading in the voltmeter.

[IIT-1982] Sol. Applying Kirchoff's law moving in clockwise

direction starting from battery we get

A 10Ω

400Ω 800Ω

6 volt + 6 – 10I – 400 I – 800 I = 0 ∴ 6 = 1210 I

∴ I = 1210

6 = 4.96 × 10–3 A

The voltmeter and 400 Ω resistor are in parallel and hence p.d. will be same

∴ 10,000 I1 = 400 I2 ...(i) Applying Kircoff's law in loop ABCDEA starting

from A in clockwise direction. – 400 I2 – 800 I + 6 = 0 ∴ 6 = 400 I2 + 800 (I1 + I2) ∴ 6 = 400 I2 + 800(0.04 I2 + I2) From (i) putting the value of I1 ∴ 6 = 1232 I2

V

I400Ω 800Ω

6 volt

10,000Ω

B

F G

C

ID

A E

∴ I2 = 4.87 × 10–3 Amp. Potential drop across 400 Ω resistor = I2 × 400 = 4.87 × 10–3 × 400 = 1.948 volt ≈ 1.95 volt ∴ The reading measured by voltmeter = 1.95 volt

CHEMISTRY 6. How many grams of silver could be plated out on a

serving tray by electrolysis of solution containing silver in +1 oxidation state for a period of 8.0 hours at a current of 8.46 amperes ? What is the area of the tray if the thickness of the silver plating is 0.00254 cm? Density of silver is 10.5 g cm–3. [IIT-1997]

Sol. Given that, t = 8.0 hrs = 8 × 3600 s

I = 8.46 A Thickness = 0.00254 cm Density = 10.5 g cm–3 M = 108 g mol–1 p = 1 (for Ag) F = 96500 C mol–1 We know, according to Faraday's first law of

electrolysis,

m = FpMIt =

1965003600846.8108

××××

= 272.684 g Also, m = Density × Volume = Density × Area × Thickness

∴ Area = ThicknessDensity

m×

= 00254.05.10684.272

×cm2 = 10224.37 cm2

7. From the following data, form the reaction between

A and B. [IIT-1994]

Initial rate (mol L–1s–1) [A]

mol L–1 [B]

mol L–1 300 K 320 K 2.5 ×10–4 3.0 ×10–5 5.0 ×10–4 2.0 × 10–3

5.0 × 10–4 6.0 × 10–5 4.0 × 10–3 – 1.0 × 10–3 6.0 × 10–5 1.6 × 10–2 –

Calculate (a) the order of reaction with respect to A and with respect to B, (b) the rate constant at 300 K, (c) the energy of activation, (d) the pre exponential factor.

Sol. Rate of reaction = k[A]l [B]m where l and m are the order of reaction with respect

to A and B respectively. From the given data, we obtain following expressions :

5.0 × 10–4 = k[2.5 × 10–4]l [3.0 × 10–5]m ..(i) 4.0 × 10–3 = k[5.0 × 10–4] l [6.0 × 10–5]m ...(ii) 1.6 × 10–2 = k[1.0 × 10–3]l [6.0 × 10–5]m ..(iii) From eq. (ii) and eq. (iii), we get

2

3

106.1100.4

−

−

×× =

l

××

−

−

3

4

100.1100.5

or 0.25 = (0.5)l or (0.5)2 = (0.5) l or l = 2 From eq. (i) and eq. (ii), we get

3

4

100.4100.5

−

−

×

× = m

5

52

4

4

100.6100.3

100.5105.2

××

××

−

−

−

−

or 81 =

41 ×

m

21

or 21 =

m

21


or m = 1 (b) At T1 = 300 K,

k1 = 12 ]B[]A[reactionofRate

= ]100.3[]105.2[

100.5524

4

−−

−

×××

= 2.67 × 108 L2 mol–2 s–1 (c) At T2 = 320 K,

k2 = 12 ]B[]A[reactionofRate

= ]100.3[]105.2[

100.2524

3

−−

−

×××

= 1.067 × 109 L2 mol–2 s–1

We know, 2.303 log 1

2

kk =

−

21

12a

TTTT

RE

or 2.303 log 8

9

1067.210067.1

×× =

×−

300320300320

314.8Ea

or 2.303 × 0.6017 =

×30032020

314.8Ea

or Ea = 20

300320314.86017.0303.2 ××××

= 55.3 kJ mol–1 (d) According to Arrhenius equation, k = RT/EaAe−

or 2.303 log k = 2.303 log A – RTEa

At 300 K,

2.303 log (2.67 × 108) = 2.303 log A – 300314.8

103.55 3

××

or 2.303 × 8.4265 = 2.303 log A – 22.17

or logA = 303.2

17.224062.19 + = 303.25762.41 = 18.0531

A = Antilog 18.0531 = 1.13 × 1018 s–1 8. An organic compound CxH2yOy was burnt with twice

the amount of oxygen needed for complete combustion to CO2 and H2O. The hot gases, when cooled to 0 ºC and 1 atm pressure, measured 2.24 L. The water collected during cooling weighed 0.9 g. The vapour pressure of pure water at 20 ºC is 17.5 mm Hg and is lowered by 0.104 mm Hg when 50 g of the organic compound is dissolved in 1000 g of water. Give the molecular formula of the organic compound. [IIT-1983]

Sol. The combustion reaction is CxH2yOy + x O2 → x CO2 + y H2O To start with, the amount of O2 taken is 2x. Hence,

after the combustion reaction, we will be left with the following amounts.

Amount of oxygen left unreacted = x Amount of carbon dioxide = x

Amount of water = y When this mixture is cooled to 0 ºC and 1 atm, we

will be left with oxygen and carbon dioxide. Hence, the amount 2x occupies the given volume of 2.24 L at STP. Hence,

Amount x = 1Lmol4.22L)2/24.2(−

= 0.05 mol

Now, Mass of water collected = 0.9 g Amount of water collected,

y = 1molg18g9.0

−= 0.05 mol

Thus, the empirical formula of the compound is C0.05H2 × 0.05 O0.05, i.e. CH2O.

Now, according to Raoult's law

–*pp∆ = x2

i.e. mmHg5.17mmHg104.0 =

)molg18/g1000()M/g50()M/g50(

1−+

Solving for M, we get M = 150.5 g mol–1 Number of repeating units of CH2O in the molecular

formula = 16212

5.150++

≈ 5

Hence, Molecular formula of the compound is C5H10O5.

9. An organic compound containing C, H and O exists

in two isomeric forms A and B. A mass of 0.108 g of one of the isomers gives on combustion 0.308 g of CO2 and 0.072 g of H2O. A is insoluble in NaOH and NaHCO3 while B is soluble in NaOH. A reacts with concentrated HI to give compounds C and D. C can be separated from D by the ethanolic AgNO3 solution and D is soluble in NaOH. B reacts readily with bromine to give compound E of molecular formula, C7H5OBr3. Identify A, B, C, D and E with justification and give their structures. [IIT-1991]

Sol. We have Percent of carbon in the compound

= 2CO

C

MM

compound

CO

mm

2 × 100

=

4412

108.0308.0 (100) = 77.78

Percent of hydrogen in the compound

= 2HO

H

MM2

compound

OH

mm

2 × 100

=

182

108.0072.0 (100) = 7.41

Percent of oxygen in the compound = 100 – (77.78 + 7.41) = 14.81 The ratios of atoms in the compound are


C : H : O : : 12

78.77 : 141.7 :

1681.14 : : 6.48 : 7.41 :

0.926 : : 7 : 8 : 1 Hence, Empirical formula of the compound is

C7H8O. Since the isomer B on reacting with bromine water

gives compound E(C7H5OBr3), the molecular formula of A and B will be the same as the empirical formula derived above, since both contain the same number of carbon atoms. As E is obtained from B by the substitution of hydrogen with bromine and since there is high carbon content in B, the compounds A and B must be aromatic. Now, since compound A is insoluble in NaOH and NaHCO3 and compound B is soluble in NaOH, it may be concluded that B is a phenolic compound and A is an ether. Hence, the structures of A and B are

OCH3

(A) anisol

OH

(B) m-cresol

CH3

The bromination of B gives

OH OH

CH3CH3

bromination BrBr

Br The reaction of compound A with HI is

OCH3

(A)

OH

(D)

+ CH3I HI

(C) The compound C can be separated from D by use of

ethanolic AgNO3 solution as it is soluble in it whereas D will remain insoluble. The compound D will be soluble in NaOH as it is phenol. Hence, the structures of A, B, C, D and E are

OCH3

;

anisol

OH

(B) m-cresol

CH3 ; CH3I ;

(C)

OH

(D)

;

OH

CH3

BrBr

Br(E)

(A)

10. A scarlet compound A is treated with concentrated

HNO3 to give chocolate brown precipitate B. The precipitate is filtered and the filtrate is neutralised with NaOH. Addition of KI to the resulting solution gives a yellow precipitate C. The precipitate B on warming with concentrated HNO3 in the presence of Mn(NO3)2 produces a pink-coloured solution due to the formation of D. Identify A, B, C and D. Write the reaction sequence. [IIT-1995]

Sol. The given reactions may be summarized as follows.

A

scarlet compound

conc. HNO3 B filtered

(i) neutralized with NaOH(ii) KI

CYellow

ppt. chocolate brownMn(NO3)2

HNO3

pink coloured solution

Filtrate

The compound B must be a powerful oxidizing agent

which converts Mn2+ to the pink coloured MnO4– ion.

Normally, PbO2 is used for this purpose. The compound C may be PbI2 which is yellow in

colour. The given reactions may be explained as follows.

scarlet)A(

43OPb + 4HNO3 →

.pptbrownchocolate

)B(2PbO + 2Pb(NO3)2 + 2H2O

Pb(NO3)2 + 2KI →

.pptyellow)C(2PbI + 2KNO3

)B(

2PbO5 + 2Mn(NO3)2 + 4HNO3 →

colouredpink)D(

24 )MnO(Pb + 4Pb(NO3)2 + 2H2O

MATHEMATICS

11. Let λ and α be real. Find the set of all values of λ for which the system of linear equations

λx + (sin α)y + (cos α)z = 0 x + (cos α)y + (sin α)z = 0 – x + (sin α)y – (cos α)z = 0 has a non-trivial solution. for λ = 1, find all values

of α. [IIT-1993] Sol. Given, λx + (sin α)y + (cos α)z = 0 x + (cos α)y + (sin α)z = 0 –x + (sin α)y – (cos α)z = 0 has non-trivial solution. ⇒ ∆ = 0

⇒ α−α−

ααααλ

cossin1sincos1cossin

= 0

⇒ λ(–cos2α – sin2α) – sin α(– cos α + sin α) + cos α(sin α + cos α) = 0 ⇒ –λ + sin α cos α + sin α cos α – sin2α + cos2α = 0 ⇒ λ = cos 2α + sin 2α

we know, – 22 ba + ≤ a sin θ + b cos θ ≤ 22 ba +

∴ – 2 ≤ λ ≤ 2 ...(1) Again when λ = 1, cos 2α + sin 2α = 1

or 2

1 cos 2α + 2

1 sin 2α = 2

1


⇒ cos

π

−α4

2 = cos4π

∴ 2α – 4π = 2bπ ±

4π

⇒ 2α = 2nπ – 4π +

4π

2α = 2nπ + 4π +

4π

∴ α = nπ or nπ + π/4 12. A bag contains 12 red balls and 6 white balls. Six

balls are drawns one by one without replacement of which at least 4 balls are white. Find the probability that in the next two drawns exactly one white ball is drawn (leave the answer in nCr). [IIT-2004]

Sol. Using Baye's theorem; P(B/A) =

∑

∑

=

=3

1ii

3

1iii

)A(P

)A/B(P).A(P

where A be the event at least 4 white balls have been drawn.

A1 be the event exactly 4 white balls have been drawn. A2 be the event exactly 5 while balls have been drawn.

A3 be the event exactly 6 white balls have been drawn B be the event exactly 1 white ball is drawn from two draws.

∴ P(B/A) =

618

66

012

618

56

112

618

46

212

212

11

111

618

56

112

212

12

110

618

46

212

CC.C

CC.C

CC.C

CC.C.

CC.C

CC.C.

CC.C

++

+

= )C.CC.CC.C(C

)C.C.C.C()C.C.C.C(

66

012

56

112

46

212

212

11

111

56

112

112

110

46

212

++

+

13. Let C1 and C2 be two circles with C2 lying inside C1. A circle C lying inside C1 touches C1 internally and C2 externally. Identify the locus of the centre to C.

[IIT-2001] Sol. Let the given circles C1 and C2 have centres O1 and

O2 and radii r1 and r2 respectively. Let the variable circle C touching C1 internally, C2

externally have a radius r and centre at O.

O2

C2 C1

O C

O1

r2

r

r1

Now, OO2 = r + r2 and OO1 = r1 – r ⇒ OO1 + OO2 = r1 + r2 which is greater than O1O2 as O1O2 < r1 + r2 (Q C2 lies inside C1) ⇒ locus of O is an ellipse with foci O1 and O2. 14. Tangents are drawn from any point on the hyperbola

9x2

– 4y2

= 1 to the circle x2 + y2 = 9. Find the

locus of mid-point of the chord of contact. [IIT-2005] Sol. Let any point on the hyperbola is (3 sec θ, 2 tan θ) ∴ Chord of contact of the circle x2 + y2 = 9 with

respect to the point (3 sec θ, 2 tan θ) is, (3 sec θ)x + (2 tan θ) y = 9 ...(1) let (x1, y1) be the mid-point of the chord of contact ⇒ Equation of chord in mid-point form is xx1 + yy1 = x1

2 + y12 ...(2)

Since (1) and (2) are identically equal

∴ 1x

sec3 θ = 1y

tan2 θ = 21

21 yx

9+

⇒ sec θ = )yx(3

x921

21

1

+ and tan θ =

)yx(2y9

21

21

1

+

Thus eliminating 'θ' from above equation, we get

221

21

21

)yx(9x81+

– 221

21

21

)yx(4y81+

= 1

∴ Required locus 9

x2 –

4y2

= 81

)yx( 222 +

15. A plane is parallel to two lines whose direction ratios

are (1, 0, –1) and (–1, 1, 0) and it contains the point (1, 1, 1). If it cuts co-ordinate axis at A, B, C. Then find the volume of the tetrahedron OABC.[IIT-2004]

Sol. Let the equation of plane through (1, 1, 1) having a, b, c as d.r's of normal to plane,

a(x – 1) + b(y – 1) + c(z – 1) = 0 and plane is parallel to straight line having d.r's.

(1, 0, – 1) and (–1, 1, 0) ⇒ a – c = 0 and –a + b = 0 ⇒ a = b = c ∴ Equation of plane, x – 1 + y – 1 + z – 1 = 0

or 3x +

3y +

3z = 1. Its intercept on coordinate axes

are A(3, 0, 0), B(0, 3, 0), C(0, 0, 3). Hence, the volume of tetrahedron OABC

= ]cba[61 rrr

= 300030003

61 =

627 =

29 cubic units


Passage # 1 (Q. 1 to Q. 3) A region in space contains a total positive charge Q

that is distributed spherically such that the volume charge density

ρ(r) = α for r ≤ 2R ; ρ(r) = 2α

−

Rr1 for

2R ≤ r ≤ R

Where α is a positive constant ρ(r) = 0 for r ≥ R 1. What fraction of total charge is contained in the

region r ≥ R/2 (A) 4/15 (B) 8/15 (C) 7/15 (D) None 2. If an electron is placed at the centre and slightly

displaced it will execute SHM. Find the time period of oscillation assuming x < R/2

(A) Qe8mR

22

0πεπ (B)

Qe8mR15

22

0πεπ

(C) Qe8mR72

20πε

π (D) None of these

3. The electric field in a region R/2 < r < R is -

(A)

−

εα

+ε

αRr1

3r2

r24R

002

3 (B)

−

εα

+ε

αRr1

6r2

r8R

002

3

(C)

−

εα

+ε

αRr1

3r2

r16R

002

3 (D) None of these

Passage # 2 (Q. 4 to Q. 6) A non-conducting vessel containing n moles of a

diatomic gas is fitted with a conducting piston. The cross-sectional area, thickness and thermal conductivity of piston are A, l and K respectively. The right side of piston is open to atmosphere at temperature T0. Heat is supplied to the gas by means of an electric heater at a constant rate q.

l

gas

To atmosphere

4. Temperature of gas as a function of time is: (If initial temperature is T0)

(A) T = T0 + KAql [1 – e–(2KAt/7nRl)]

(B) T = T0 + KAql [1 – e–(7nRl/2KAt)]

(C) T = T0 + KAql [1 – e–(2KAt/7nRl)]

(D) None of these 5. Maximum temperature of gas -

(A) Tmax = T0 + KAql (B) Tmax = T0 + KA

ql [1 – e]

(C) Tmax = T0 + lq

KA (D) Tmax = KAql

6. The ratio of the maximum volume to the minimum

volume is -

(A) 1 + 0KAT

ql (B) 1 + lq

KAT0

(C) 0KAT

ql (D) lq

KAT0

7. A constant voltage is applied between two ends of a

uniform conducting wire. If both the length and radius of the wire are doubled -

(A) the heat produced in the wire will be doubled (B) the electric field across the wire will be doubled (C) the heat produced will remain unchanged (D) the electric field across the wire will become half 8. A solenoid is connected to a source of constant emf

for a long time. A soft iron piece is inserted into it. Then -

(A) self inductance of the solenoid gets increased (B) flux linked with the solenoid increases hence

steady state current gets decrease (C) energy stored in the solenoid gets increased (D) magnetic moment of the solenoid increased

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 Sharma Director Academics, Jodhpur Branch

Physics Challenging Problems

Solut ions wi l l be publ ished in next issue

Set # 3


Passage # 1 (Q. 1 to Q. 5) 1. Yes 2. Magnetic moment = i A = iπr2 = feπr2

= r2

vπ

e.πr2 = 2

evr

As mvr = hπ2

h so, magnetic moment

M = 21 e.

m2.h.n

π

=

π me

4h n

Quanta of M.m =

π me.

4h for n = 1

=

π me

4h

M.m = Quanta of magnetic moment.

3. n

n

ML =

2/evrmvr = 2

em

⇒ n

n

LM =

21 .

me

Qty is e/m, specific charge of electron.

4. Thomson's experiment, SI unit is kg

coulomb .

5. For electron me , for deutron

m2e

me = C and

m2e = P

So P = C/2 Passage # 2 (Q. 6 to Q. 8)

6. As. ρP =

0MRT

⇒ R

M0 = ρP .T

Y = mX

Slope m = Pρ = tan 45º = 1

So, ρP = 1 for the gas

7. Velocity v = 0M

RTγ = ργP as

ρP = 1 constant so,

velocity is temperature independent. 8. No, speed is not temperature dependent.

Solution Physics Challenging Problems

Set # 2

8 Quest ions were Publ ished in June Issue

Regents Physics You Should Know Mechanics

1. Weight (force of gravity) decreases as you move away from the earth by distance squared.

2. Mass and inertia are the same thing.

3. Constant velocity and zero velocity means the net force is zero and acceleration is zero.

4. Weight (in newtons) is mass x acceleration (w = mg). Mass is not weight!

5. Velocity, displacement [s], momentum, force and acceleration are vectors.

6. Speed, distance [d], time, and energy (joules) are scalar quantities.

7. The slope of the velocity-time graph is acceleration.

8. At zero (0) degrees two vectors have a resultant equal to their sum. At 180 degrees two vectors have a resultant equal to their difference. From the difference to the sum is the total range of possible resultants.


1. A particle is projected up from the bottom of an

inclined plane of inclination α with velocity v0. If it returns to the point of projection after an elastic, impact with the plane, find the total time of motion of the particle.

Sol. To return to the point of projection after one elastic collision the particle must meet the plane along x-axis at right angle. Hence, for the motion of particle along x-axis

vx – 0xv = (– g sin α)t0

or, 0 – v0 cos θ = (– g sin α)t0

or, t0 = αθ

singcosv0 ...(1)

For the motion of the particle along y-axis vy – (–

0yv ) = (g cos α)t0

or, vy + 0yv = (g cos α)t0

or, v0 sin θ + v0 sin θ = (g cos α)t0

∴ t0 = α

θcosg

sinv2 0 ...(2)

v0

y 0yv = v0 sin θ

t = 0

θα

g sinα

g cosα

vx=0 Plane

x

t = t0

0xv = v0 cos θ

Equation (1) and (2),

αθ

singcosv0 =

αθ

cosgsinv2 0

cot θ = 2 tan α

cos θ = α+

α2tan41

tan2

and sin θ = α+ 2tan41

1

∴ Time of light for to and fro motion of the particle

T = 2t0 = α

θsing

cosv2 0

= αsing

v2 0 × α+

α2tan41

tan2 = α+ 2

0

sin31g

v4

2. (i) A uniform ladder of length L and weight w rests against a vertical wall and makes an angle θ with the horizontal ground. If the coefficient of friction at the point of contact of the ladder with the wall and the ground is µ, show that the greatest height x, measured along the ladder from the foot to which a man of weight W may climb without the ladder slipping is given by

Lx =

)µ1(W)wW(µ

2++ (µ + tan θ) –

W2w

(ii) If the wall be smooth and coefficient of friction between ladder and ground be 0.25, show that

x = 4L

+

Ww1 tan θ –

W2w

Sol. The different forces acting on the ladder are shown in figure.

A µR1w

W

R1L/2

L/2

R2

µR2

B

OD C

G

(i) Resolving the forces horizontally and vertically,

we get R2 = µR1 ...(1) R1 + µR2 = W + w ...(2) From equations (1) and (2), R2 = µ[W + w – µR2] or R2[1 + µ2] = µ [W + w]

∴ R2 = 2µ1]wW[µ

++ ...(3)

and R1 = 2µ1wW

++ ...(4)

Taking moment about point A, we get R2 × BO + µR2 × AO = w × AC + W × AD R2L sin θ + µR2L cos θ = w(L/2) cos θ + W.x cos θ

∴ x =

−+θ

2wLLµRtanLR

W1

22 ...(5)

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

Students' ForumPHYSICS


Substituting the values of R2, from eq. (3) in eq. (5), we get

x =

−

++

+θ+

+2

wLµ1

)wW(µtanLµ1

)wW(µW1

2

2

2

or Lx =

−+θ

++

2w)µ(tan

µ1)wW(µ

W1

2

or Lx =

)µ1(W)wW(µ

2++ (µ + tan θ) –

W2w

(ii) As wall is smooth, µR2 = 0, also µ = 0.25 Substituting µR2 = 0, in eq. (5),

we get : x = W1

−θ

2wLtanLR 2

Now R2 = µR1 = µ(W + w) = 0.25(W + w)

∴ x = W1

−θ+

2wLtanL)wW(25.0

=

+

Ww1

4L tan θ –

W2wL

3. A point charge 40 pC is placed at the centre of a

sphere of radius 0.5 m. For this data, is Gauss's law consistent with coulomb's law ?

Sol. Let q1 = 40 pC = 40 × 10–12C Place a test charge q2 at the surface of sphere. Then it experience a force,

F = 04

1πε 2

21

rqq , given by Coulomb's law.

Field at the surface,

E = 2q

F = 04

1πε 2

1

rq

= 2

129

)5.0()1040(109 −××× = 1.44 N/C

This field is same at all points on the surface of the sphere. Since the field lines are radial, field cuts the surface normally.

∴ Flux φE = EA = E(4πr2) = 1.44 × [4π × (0.5)2] = 4.5 N-m2/C (ε0φE)on surface = (8.85 × 10–12)4.5 = 40 × 10–12C = 40 pC In this situation ε0φE = ΣQ Hence, Gauss law is consistent with coulomb's law. 4. Find the electric potential, at any point on the axis of

a uniformly charged circular disc, whose surface charge density is σ, radius, a.

Sol. Let us consider a small elemental thin ring of width dy.

Area of the ring = 2πy dy

Charge on this elemental ring = (2πy dy)σ Again, we can consider that this ring is divided into a

large number of small elements. Each such element e is at the same distance from P. Hence, potential produced by this ring of width dy at the point P is given by dV, where

dV = 220 yr

dyy24

1

+

σππε

e

y

O

a r P

22 yr +

Again, each ring as we go from centre to rim,

produces different contributions. Since the distance of each ring from P changes as y increases from 0 to a, hence, total potential produced by the whole ring,

V = ∫ =

a

0ydV

= ∫ πεπσa

0 042 .

22 yr

ydy

+

= ∫+ε

σ a

0 220 yr

ydy2

Put )yr( 22 + = P ∴ r2 + y2 = p2 or 2ydy = 2pdp

∴ ∫+ 22 yr

ydy = ∫ ppdp

= ∫dp = p = 22 yr +

∴ ∫+ε

σ a

0 220 yr

ydy2

= a

0

22

0yr

+

εσ

=

−+

εσ rar

222

0

∴ V =

−+

εσ rar

222

0

As a special case, if r >> a

22 ar + = r2/12

ra1

+ ≈ r

+ 2

2

ra

211

or, V = 02ε

σ .r2

a 2 ×

ππ =

r4q

0πε

[Q πa2 = A and Aσ = q] i.e., the result is the same as if all the charge is

concentrated at the centre of the ring.


5. Two concentric shells of radii R and 2R are shown in the figure. Initially a charge q is imparted to the inner shell. After the keys K1 and K2 are alternately closed n times each, find the potential difference between the shells.

Sol. When K1 is closed first time, outer sphere is earthed and its potential becomes zero. Let the charge on it be q1´.

V1´ = Potential due to charge on inner sphere and that due to charge on outer sphere.

∴ 0 =

+πε R2

´qR2q

41 1

0

or, q1´ = – q When K2 is closed first time, the potential V2´ on

inner sphere becomes zero as it is earthed. Let the new charge on inner sphere be q2´.

∴ 0 = 04

1πε R

´q2 + 04

1πε

−

R2q

∴ q2´ = q/2

R

2R

K1K2 Now when K1 will be closed second time charge on

outer sphere will be –q2´ i.e., –q/2. Similarly, when K1 will be closed nth time, charge on

outer sphere will be –q/2n–1 as each charge will be reduced to half the previous value.

After closing K2 nth time charge on inner shell will be negative of half the charge on outer shell i.e.,

−

n2q and potential on it will be zero.

For potential of outer shell

V0 = R2

)2/q(4

1 n

0

−πε

+ R2

)2/q(4

1 1n

0

−

πε

= R24)21(q

1n0

−πε+−

= + R24

q1n

0+πε

Potential difference = V0 – Vi

= R24

q1n

0+πε

– 0

= R24

q1n

0+πε

FRACTIONAL DISTILLATION

OF AIR

Did you know that the air we breathe isn’t just oxygen, infact it’s made up of a number of different gases such as nitrogen, oxygen, carbon dioxide, argon, neon and many others. Each of these gases carry useful properties so separating them from the air around us is extremely beneficial.

The process is called fractional distillation and consists of two steps, the first relies on cooling the air to a very low temperature (i.e. converting it into a liquid), the second involves heating it up thus allowing each gas within the mixture to evaporate at its own boiling point. The key to success here is that every element within air has its own unique boiling temperature. As long as we know these boiling temperatures we know when to collect each gas.

So what are the real world benefits of separating and extracting these gases? Well liquid oxygen is used to power rockets, oxygen gas is used in breathing apparatus, nitrogen is used to make fertilizers, the nitric acid component of nitrogen is used in explosives.

The other gases all have their own uses too, for example argon is used to fill up the empty space in most light bulbs (thanks to its unreactive nature). Carbon dioxide is used in fire extinguishers and is great for putting out fires in burning liquids and electrical fires. There really are too many uses to list but suffice it to say that fractional distillation is an extremely useful process for humans the world over.


Capacitance : Whenever charge is given to a conductor of any

shape its potential increases. The more the charge (Q) given to the conductor the more is its potential (V) i.e. Q ∝ V

⇒ Q = CV where C is constant of proportionality called

capacitance of the conductor C = Q/V, C = Q SI unit of capacitance is farad (F) and 1 F = 1

coulomb/volt (1CV–1) Energy stored in a charged capacitor :

W = 20CV

21 =

C2Q2

= 21 QV0

Capacitance of an isolated sphere : Let a conducting sphere of radius a acquire a

potential V when a charge Q is given to it. The potential acquired by the sphere is

V = a4

Q

0πε ⇒ C =

VQ = 4πε0a

Charge sharing Between two charged conductors :

C1

V1

C2

V2

q1 = C1V1 q2 = C2V2

(Initially)

C1

V

C2

V

q´1 = C1V q´2 = C2V

(Finally)

V = 21

2211

CCVCVC

++

There is always a loss in energy during the sharing process as some energy gets converted to heat.

Loss = – ∆U =

+ 21

21

CCCC

21 (V1 – V2)2

Capacitor or Condenser : An arrangement which has capability of collecting

(and storing) charge and whose capacitance can be varied is called a capacitor (or condenser)

The capacitance of a capacitor depends. (a) directly on the size of the conductors of the

capacitor.

(b) directly on the dielectric constant K of the medium between the conductors.

(c) inversely on the distance of separation between the conductor.

Principle of a condenser : Consider a conducting plate A which is given a

charge Q such that its potential rises to V. Then C = Q/V Let us place another identical conducting plate B

parallel to it such that charge is induced on plate B (as shown in figure).

++++++++

++++++++

A

Q

If V– is the potential at A due to induced negative charge on B and V+ is the potential at A due to induced positive charge on B, then

++++++++

++++++++

A++++++++

– – – – – – – –

B

C´ = ´V

Q = −+ −+ VVV

Q

Since V´ < V (as the induced negative charge lies closer to the plate A in comparison to induced positive charge).

⇒ C´ > C Further, if B is earthed from the outer side (see

figure) then Vn = V – V– as the entire positive charge flows to the earth. So

C" = nVQ =

−− VVQ ⇒ Cn >> C

So, if an identical earthed conductor is placed in the viscinty of a charged conductor then the capacitance of the charged conductor increases appreciably. This is the principle of a parallel plate capacitor.

Capacitor-1

PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


Parallel Plate Capacitor :

–σ

A B

+ σ

d

+ + + + + + + +

– – – – – – – –

+ + + + + + + +

A B

A = Area of plated = Separation between the plates

E =εσ =

0kεσ

It consists of two metallic plates A and B each of area

A at separation d. Plate A is positively charged and plate B is earthed. If K is the dielectric constant of the material medium and E is the field that exists between the two plates, then

E = εσ =

0Kεσ

=σ=

Aqand

dVEQ

⇒ dV =

AKq

0ε

⇒ C = Vq =

dAK 0ε

If medium between the plates is air or vacuum, then K = 1

⇒ C0 = dA0ε

Special Case I : When the space between the parallel plate capacitor

is partly filled with a dielectric of thickness t(<d) If no slab is introduced between the plates of the

capacitor, then a field E0 given by E0 = 0ε

σ , exists in

a space d. E=

KE0

+σ

dt

On inserting the slab of thickness t, a field E = KE0

exists inside the slab of thickness t and a field E0 exists in remaining space (d – t). If V is total potential then

V = E0(d – t) + Et

⇒ C = Vq =

−−

ε

K11td

A0

Special Case II : When the space between the parallel plate capacitor

is partly filled by a conducting slab of thickness t(<d).

It no conducting slab is introduced between the

plates, then a field E0 = 0ε

σ exists in a space d. If C0

be the capacitance (without the introduction of

conducting slab), then C0 = dA0ε

+σ

dt

E= 0

E0

On inserting the slab, field inside it is zero and so a

field E0 = 0ε

σ now exists in a space (d – t)

⇒ V = E0(d – t)

⇒ V = 0ε

σ (d – t)

⇒ V = 0A

qε

(d – t)

⇒ C = Vq =

tdA0

−ε

⇒ C =

−

ε

dtt1d

A0

⇒ C =

−

dt1

C0

Since d – t < d ⇒ C > C0 i.e. Capacitance increases on insertion of conducting

slab between the plates of capacitor.

Charge induced on a dielectric :

++++++++

––––––––

–qp

+ + + + + + + +

– – – – – – – –

+qp

–q+q E0

Ep

E = E0 – Ep

Resultant dielectric field within the plates is E = E0 – Ep


⇒ E = 0

1ε

(σ – σp) ...(1)

Also E = 0Kε

σ ...(2)

Compare (1) and (2), we get

0

1ε

(σ – σp) = 0Kε

σ

⇒ σp = σ

−

K11

⇒ Aqp =

−

K11

Aq

⇒ qp = q

−

K11

Spherical capacitor : B

C2

A

aC1

b

let C1 be the capacitance in between the two

conductors and C2 be capacitance out side both. To find C1 : Imagine the outer surface of B to be earthed. Then –q

is the charge induced on the inner surface of B. If V is the potential difference between the two

surfaces, then

V = Ka4

q

0πε +

Kb4q

0πε−

⇒ V = K4

q

0πε

−

b1

a1

⇒ C = Vq = 4πε0K

− abab ...(1)

To find C, Imagine A to be made open circuited (i.e. made non

conducting), then C2 = 4πε0Kb ...(2) Case I : When battery is connected to B and A is

earthed. Then C1 and C2 are in parallel ⇒ C = C1 + C2

⇒ C = 4πε0K

− abab + 4πε0Kb

⇒ C = 4πε0K

− abb2

Case II : When battery is connected to A, then C1 and C2 are in series.

⇒ C1 =

1C1 +

2C1

⇒ C1 =

abab −

K41

0πε +

Kb41

0πε

⇒ C1 =

+

−πε

1a

abKb4

1

0

⇒ C1 =

πε ab

Kb41

0

⇒ C = 4πε0Ka Case III : When battery connected to A and B is

earthed. Then C2 can be omitted as it will not receive any charge.

So, C = C1

⇒ C = 4πε0K

− abab

Case IV : When battery connected to B and A is open circuited (or made non conducted) then C1 can be omitted (as it is open circuited). So,

C = C2 ⇒ C = 4πε0Kb

Cylindrical capacitor : Let inner cylinder be given a charge per unit length

of λ

=

l

q . A charge – q is induced on length l at

inner surface of outer cylinder

ba q –q

l

E = r2 0πε

λ for a < r < b

⇒ – drdV =

Kr2 0πελ

⇒ ∫surfaceouter

surfaceinner

dV = – ∫=

=πε

λbr

ar0 rdr

K2

⇒ Vinner surface – Vouter surface = K2 0πε

λ loge

ab

Since, inner surface is at higher potential and outer at lower potential, so


–q

ba+q

Gaussian surface

⇒ Vouter surface – Vinner surface = K2 0πε

λ loge

ab

⇒ Vinner surface – Vouter surface = K2

q

0lπεloge

ab

⇒ C = surfaceoutersurfaceinner VV

q−

=

πε

ablog

K2

e

0l

⇒ C =

πε

ablog

K2

e

0l

Solved Examples

1. A capacitor of 20 µF and charged to 500 volt is connected in parallel with another capacitor of 10 µF charged to 200 volt. Find the common potential.

Sol. Charge on one capacitor q1 = C1V1 ∴ q1 = 20 × 10–6 × 500 = 0.01 coulomb Charge on second capacitor q2 = 10 × 10–6 × 200 = 0.002 coulomb The charge on the two capacitors q = q1 + q2 = 0.01 + 0.002 = 0.003 coulomb Total capacity C = C1 + C2 = 20 × 10–6 + 10 × 10–6 = 30 × 10–6 Farad. Common potential = q/C

= 61030012.0

−× = 400 Volt.

2. A battery of 10V is connected to a capacitor of

capacity of 0.1 F. The battery is now removed and this capacitor is connected to a second uncharged capacitor. If the charge distributes equally on these two capacitors, find the total energy stored in the two capacitors. Further, compare this energy with the initial energy stored in the first capacitor.

Sol. The initial energy stored in the first capacitor.

U0 = 21 CV2

= 21 × 0.1 × (10)2 = 5.0 J

When this capacitor is connected to the second uncharged capacitor, the charge distributes equally. This shows that the capacitance of the second capacitor is also C. The voltage across each capacitor will be V/2. If U be the energy stored in the two capacitors, then

U = 21 C

2

2V

+

21 C

2

2V

= 41 CV2 = 2.5 J

0U

U = 0.55.2 =

21

3. Two isolated metallic solid spheres of radii R and 2R are charged such that both of these have same charge density σ. The spheres are located far away from each other, and connected by a thin conducting wire. Find the new charge density on the bigger sphere.

Sol. Charge on smaller sphere Q1 = 4πR2 . σ Charge on bigger sphere Q2 = 4π(2R)2σ = 16πR2σ ∴ Total charge Q = Q1 + Q2 = 20πR2σ ...(1) Capacitances of two spherical conductors are C1 = 4πε0R and C2 = 4πε0(2R) ∴ Total capacitance C = C1 + C2 = 12πε0R ...(2) After connection, the common potential V is given by

V = CQ =

R12R20

0

2

πεσπ =

03R5ε

σ

New charge on bigger sphere Q2

´ = C2V

= 4πε0R(2R) × (5Rσ/3ε0) = 3R40 2σπ

Surface density

σ2´ = areasurface

´Q2 = 2

2

)R2(4

3R40

π

σπ

= 65

σ.

4. A 8 µF capacitor C1 is charged to V0 = 120 volt. The

charging battery is then removed and the capacitor is connected in parallel to an uncharged 4 µF capacitor C2 (a) what is the potential difference V across the combination ? (b) What is the stored energy before and after the switch S is thrown ?

S

V0 C1 C2


Sol. (a) Let q0 be the charge on C1 initially. Then q0 = C1 V0 when C1 is connected to C2 in parallel, the charge q0

is distributed between C1 and C2. Let q1 and q2 be the charges on C1 and C2 respectively. Now let V be the potential difference across each condenser.

Now q0 = q1 + q2 or C1V0 = C1V + C2V

∴ V = 21

1

CCC+

V0 = µF4µF8

µF8+

(120 V)

= 80 volt. (b) Initial energy stored

U0 = 21 C1V0

2

= 21 (8 × 10–6) (120)2

= 5.76 × 10–2 Joule Final energy stored

U = 21 C1V2 +

21 C2V2

= 21 (8 × 10–6)(80)2 +

21 (4 × 10–6)(80)2

= 3.84 × 10–2 joule Final energy is less than the initial energy. The loss

of energy appears as heat in connecting wires. 5. Calculate the capacitance of a parallel plate

condenser, with plate area A and distance between plates d, when filled with a dielectric whose dielectric constant varies as

ε(x) = ε0 + βx 0 < x < 2d

ε(x) = ε0 + β(d – x) 2d < x < d

For what value of β would the capacity of the condenser be twice that when it is without any dielectric.

Sol. The capacitance in series is given by

Ć

1 = 1C

1 + 2C

1

∴ Ć

1 = A1 ×

−β+ε

+β+ε ∫∫

d

2/d 0

2/d

0 0 )xd(dx

xdx

= βA

1 [log(ε0 + βx) 2/d0 – log(ε0 + β(d – x) d

2/d ]

=

β+ε−ε−

ε−

β+ε

β 2dlogloglog

2dlog

A1

0000

=

ε−

β+ε

β 00 log2dlog

A2

= βA

2 log

εβ+ε

0

0 2/d

The capacitance C of a condenser without dielectric is given by

C = d

A 0ε

According to the question, C´ = 2C

∴ 0A

2ε

log

εβ+ε

0

0 2/d = A2

d

0ε

β = d

4 0ε log

εβ+ε

0

0 2/d

Physics Facts

1. Due to gravity, the maximum speed a raindrop during a rain with falling speed can hit you is about 18 miles per hour (29 kilometers per hour).

2. The speed of light in meters is 299,792,458 meters per second. And how on Earth are you going to remember that? The number can be remembered from the number of letters in each word of the following phrase: "We guarantee certainty, clearly referring to this light mnemonic."

(The speed of light in miles per second is 186,282.397051221, or in miles per hour, 670,616,629.384395).

3. In air, at a temperature of 32 degrees Fahrenheit/0 degrees Celsius (freezing point of water) the speed of sound travels 1,087 feet (331 meters) per second. (It travels faster at higher temperatures).

(In 64 degrees Fahrenheit [18 degrees Celsius] the speed of sound travels 1,123 feet [342 meters] per second).

4. If an object floats on water, it displaces the water equal to its mass, but if the object sinks, it displaces water equal to its volume.

5. A calorie is defined as the amount of energy needed to raise one gram of water one degree Celsius (or from 14.5 degrees Celsius to 15.5 degrees Celsius).


Friction : Whenever there is a relative motion between two

surfaces in contact with each other, an opposing force comes into play which forbids the relative motion of two bodies. This opposing force is called the force of friction.

Ex. : If a book on a table slides from left to right along the surface of a table, a frictional force directed from right to left acts on the book.

Frictional force may also exist between the surfaces when there is no relative motion. Frictional forces arise due to molecular interactions.

Static and Kinetic Friction : The frictional force between two surface before the

relative motion actually starts is called static frictional force or static friction, While the frictional force between two surfaces in contact and in relative motion is called kinetic frictional force or kinetic friction.

Static friction is a self adjusting force and it adjusts both in magnitude and direction automatically. Its magnitude is always equal to external effective applied force, tending to cause the relative motion and its direction is always opposite to that of external applied force.

So, when a body is not in motion or equilibrium, then Force of static Friction = Applied External Force Limiting friction, coefficients of friction and angle of friction : Consider a block resting on a rough horizontal

surface. The forces acting on the block are its weight mg downwards and normal reaction N acting upward. Such that N = mg.

NR

f

θ

mg

P(<f))

M

Now suppose a force Fapp is applied to the block to

the right, then there will arise a frictional force f directed to the left (opposite to direction of applied

force), which prevents the motion of the block. Let the resultant of N

rand F

r be R

r which makes an

angle θ with normal reaction Nr

. Resolving Rr

along Nr

and Fr

, we get R cos θ = N and R sin θ = f For equilibrium N = mg and f = Fapp If we increase the pull Fapp continuously, the force of

friction increases and a stage comes when the body is just on the state of moving. This state is called limiting equilibrium. Under this condition the frictional force is maximum and is equal to applied force.

Limiting Friction : The maximum value of static frictional force exerted

between two surfaces in contact parallel to surfaces for a given normal force between when the body is on the verge of motion them is called limiting friction.

Angles of Friction : Angle of friction (θ) is the angle which the resultant

of force of static friction (f) and normal (N) makes with the normal reaction

The Coefficient of Friction (µ) : It is defined as the ratio of limiting friction F to the

normal reaction N between two surface in contact, i.e., µ = F/N ...(3) from figure, tan θ = F/N ...(4) Equation (3) and (4) µ = tan θ Static and Kinetic Regions : If a graph is plotted between applied force and

frictional force, the graph is obtained. In figure AC is limiting or (maximum) static friction and BD is kinetic friction. Obviously, kinetic friction is less than static friction.

If relative motion is absent and is at the verge of start µ = µs, the coefficient of static friction but if relative motion is present µ = µk, the coefficient of kinetic friction.

The coefficient of friction depends on the (a) strength of molecular interaction between the

surfaces in contact, (b) roughness of the two surface in contact. Whenever we are dealing with problem involving

friction we can follow the following analysis flow chart.

Friction

PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


Read the problem Carefully

Find the value of applied force Fapp and limiting force

of Static Friction (fs)

If Fapp < fs then body does not move and the force of friction

f = Fapp

if Fapp > fs then body moves

If Fapp = fs the body is on the verge of motion (still in

equilibrium)

EITHER with Constant Velocity

OR with an Acceleration a

On a Level Track

On an inclined Plane

Fapp – fk = 0 or Fapp = fk

On a Level Track

On an inclined Plane

Fapp – fk = ma

Applied Pull = fk fk = mg sinθ Fapp – µkmg = ma

or Fapp = m(a + µkg) Fapp – µkmg cosθ = ma

or Fapp = m(a + µkg cosθ)

Laws of static and kinetic friction : (a) The force of limiting friction is directly

proportional to normal reaction for the same two surfaces in contact and acts opposite to direction of pull.

The kinetic friction is also proportional to normal reaction and acts opposite to direction of instantaneous relative motion. The kinetic friction is less than the static friction.

(b) The force of limiting (or static) friction is independent of area of contact of bodies as long as normal reaction remains the same.

The kinetic friction (to a good approximation) is independent of velocity, provided the velocity is neither too large nor too small.

Angle of repose (α) This is concerned with an inclined plane on which a

block rests, exerting its weight on the plane. The angle of repose α is the angle which an inclined

plane makes with the horizontal such that a body placed on it is on the verge of motion (is just about to loose the state of rest).

Under this condition the forces acting on the block are: (a) its weight mg, downward, (b) normal reaction N, normal to plane, (c) a force of friction fs, parallel and tangential to

plane upward. Taking α as angle of inclination of the plane with the

horizontal and resolving mg, parallel and

perpendicular to inclined plane, then for equilibrium, we get

N = mg cos α and fs = mg sin α

⇒ tan α = Nfs

N

mg mg cos α

fs

mg sinα

Tendency to slide

α Frictional force on a bicycle in motion : (a) When a wheel is rotated about its axle without

sliding, the frictional force acting on it is the rolling friction and it acts opposite to the direction of tendency of motion of a points of its contacts with the ground. In case the wheel rotates clockwise and frictional force (f) on wheel is forward. In case the wheel rotates anticlockwise, the frictional force (f) on wheel is backward.

(b) When the bicycle is pedalled, the force exerted on the rear wheel through the pedal-chain-axle system is in backward direction, therefore force of friction on rear wheel is forward. The front wheel of cycle moves by itself in forward direction, hence the force of friction of front-wheel is in backward direction.


(c) When the bicycle is not pedalled, no external force is being exerted, both wheels move forward by itself due to inertia and so the net frictional force on both wheels is in backward direction.

Solved Examples

1. A block of mass 5 kg is placed on a slope which makes an angle of 20º with the horizontal and is given a velocity of 10 m/sec up the slope. Assuming that the coefficient of sliding friction between the block and the slope is 0.20, find how far the block travels up the slope ? Take g = 10 m/sec2.

Sol. This situation is shown in fig. R

mg mg cos 20º

u = 10 m/s

mg sin 20º

x

20º The component of the weight perpendicular to plane = mg cos 20º = 5 × 10 × 0.9397 = 46.98 N The component of the weight parallel to the plane = mg sin 20º = 5 × 10 × 0.3420 = 17.10 N From figure R = mg cos 20º = 46.98 N Here the coefficient of kinetic friction = 0.2 Thus the frictional force X = 0.2 × 46.98 = 9.39 N The frictional force will be downward because the

motion is in the upward direction. The resultant force parallel to the plane is given by = X + mg sin 20º = 9.39 + 17.10 = 26.49 N From Newton's law F = ma, i.e., 26.49 = 5 × a

∴ a = 549.26 = 5.29 m/s2 downward

When the block is given a velocity 10 m/s in the upward direction we have

u = 10 m/s, v = 0, a = – 5.9 m/s2. (Taking the direction up the plane as positive) Let s be the distance traveled by the block. Using the formula v2 = u2 + 2a s, we have 0 = (10)2 – 2 × 5.29 × s

or s = 29.52

100×

= 9.45 m.

2. A block is projected up with 10 m/s along a fixed inclined plane of inclination 37º with the horizontal. If the time of ascend from the point of projection is half the time of descend to the same point, find the distance travelled by the block during the up and down journey.

Sol. Let µ, t1 and t2 be the coefficient of friction between the plane and the block, time of ascend and time of descend respectively.

The retardation while going up

a1 = g (sin θ + µ cos θ) = 10

+

54µ

53

The acceleration while descending

a2 = g(sin θ – µ cos θ) = 10

−

54µ

53

Now, s = distance of ascend = distance of descend. As final velocity is zero, we have 0 = u – a1t1 or u = a1t1

Now s = a1t12 –

21 a1t1

2 = 21 a1t1

2

s = a1t12 =

21 a2t2

2 and t2 = 2t1

∴

1

2

aa =

2

2

1

tt

or

+

−

54µ

53

54µ

53

= 2

21

Solving we get µ = (9/20)

Again a1 = 10

×+

54

209

53 = 9.6 m/sec2

∴ s = 1

2

a2u =

6.92)10( 2

× = 5.21 meter

So total distance = 2s = 10.42 metre 3. A block weighing 20 nt is at rest on a horizontal

table. The coefficient of static friction between block and table is 0.50. (a) What is the magnitude of the horizontal force that will just start the block moving ? (b) What is the magnitude of a force acting upward 60º from the horizontal that will just start the block moving ? (c) If the force acts down at 60º from the horizontal how large can it be without causing the block to move ?

Sol. (a) As shown in fig. the horizontal force F that will just start the block moving is equal to the maximum force of static friction. Thus,

R

W

F µR

F = µR = µW = 0.50 × 20 nt. = 10.0 nt. (b) The forces acting on the block are shown in fig.

R

W

Fcos θ µR

F sin θ F

θ

The applied force is inclined at an angle θ in the

upward direction. Its horizontal and vertical


components are F cos θ and F sin θ respectively. In equilibrium.

F cos θ = µR and F sin θ + R = W or R = (W – F sin θ) ∴ F cos θ = µ(W – F sin θ) = µ W – µF sin θ F (cos θ + µ sin θ) = µW

or F = θ+θ sinµcos

µW

Here µ = 0.50, W = 20 nt. and θ = 60º

F = º60sin5.0º60cos

2050.0+

× = 866.05.050.0

10×+

= 10.72 nt. (c) In this case,

R

W

Fcos θ

µR F sin θ

F θ

F cos θ = µR and R = W + F sin θ Solving we get,

F = θ−θ sinµcos

µW = 866.05.050.0

2050.0×−

×

= 149.2 nt. 4. Two blocks, m1 = 2kg and m2 = 4kg, are connected

with a light string that runs over a frictionless peg to a hanging block with a mass M as shown in fig. (a). The coefficient of sliding friction between block m2 and the horizontal surface at the speeds involved is µk = 0.2. The coefficient of static friction between the two blocks is µS = 0.4. What is the maximum mass M for the hanging block if the block m1 is not to slip on block m2 while m2 is sliding over the surface ?

Sol. The relevant free body diagrams are shown in fig.(b) Using two body system, we have

m1 m2

ms = 0.4 N1

m1g M

µk = 0.2 f1 f

T

N

(m1+m2)g Mg

T+

a

(a) (b) N – (m1 + m2)g = 0 ...(1) T – F = (m1 + m2)a ...(2) For hanging block Mg – T = Ma ...(3) From eqs. (2) and (3), Mg – f = (M + m1 + m2)a But f = µkN = µk(m1 + m2)g [Q using eq. (1)] ∴ Mg – µk(m1 + m2)g = (m + m1 + m2)a

or a = )mmM(

g)mm(µM

21

21k

+++− ...(4)

From free body diagram of mass m1, we have N1 – m1g = 0 and f1 = m1a It should be noticed that the force f1 accelerates m1 to

the right. Just before slipping occurs, we find

1

1

Nf = µS or µS =

gmam

1

1 = ga

∴ µS = )mmM(

)mm(µM

21

21k

+++− ....(5)

Solving eq. (5) for M, we have

M = S

21kS

µ1)mm(µµ(

−++

or M = )4.01(

)kg4kg2)(2.04.0(−

++ = 6 kg.

5. In fig.(a) the blocks A, B and C weight are 3kg, 4kg

and 8kg respectively. The coefficient of sliding friction between any two surfaces is 0.25. A is held at rest by a massless rigid rod fixed to the wall, while B and C are connected by a light flexible cord passing around a fixed frictionless pulley. Find the force P necessary to drag C along the horizontal surface to the left at a constant speed. Assume that the arrangement shown in the diagram, B on C and A on B is maintained all the throughout.

A

B

CP

Sol. When block C moves towards left, B moves towards right, while A is fixed. There would be a tension T in the string. Under this condition, let us consider the frictional forces between different surfaces.

Frictional force between A and B = µR = 0.25 × 3 Frictional force between C and B = µR = 0.25 (3 + 4) = 0.25 × 7 Frictional force between C and surface = 0.25(3 + 4 + 8) = 0.25 × 15 Considering fig. (b)

0.25(3 + 4)

C P 0.25 × 15TB

0.25 × 3T

0.25(3 + 4)

Fig (b) Tension in the string = Frictional forces at upper and

lower surfaces of block B or T = 0.25 × 3 + 0.25 × 7 = 2.5 kg wt. For block C, P = T + Frictional force between C and B + Frictional

force between C and surface = 2.5 + 0.25 × (3 + 4) + 0.25 × (15) = 8 kg wt. = 8 × 9.8 = 78.4 newton


Elimination reactions : The elimination reactions are reverse of addition

reactions. In these reactions two atoms or group attached to the adjacent carbon atoms of the substrate molecule are eliminated to form a multiple bond. In these reactions a atom or group from α-carbon atom and a proton from the β-carbon are eliminated.

– C – C –

X α –HX – C = C –

β

H In eliminations reactions, the presence of one

hydrogen on the β-carbon atom is necessary. In general the elimination reactions are divided into two types, i.e., bimolecular elimination reactions (E2) and unimolecular elimination reactions (E1).

Bimolecular elimination reactions (E2) : In these elimination reactions, the rate of elimination

depends on the concentration of the substrate and the nucleophile and the reaction is of second order. It is represented as E2. Like SN2 reaction, the E2 reaction is also one step process. In these reactions abstraction of proton from the β-carbon atom and the expulsion of an atom or group from the α-carbon atom occur simultaneously. The mechanism of this reaction is represented as follows:

R – CH – CH2

B HB:

β α

Transition state

R — CH — CH2

X

δ+

δ –

RCH = CH2 + BH + X⊕ Θ

H

X

The above reaction is a one step process and passes

through a transition state. This reaction is also known as 1, 2-elimination or simply β-elimination. In these reactions, the two groups to be eliminated (i.e., H and X) are trans to each other and hence E2 reactions are generally trans elimination.

The second-order elimination reaction may also proceed in two steps (as in E1 elimination which will be discussed subsequently). In this mechanism, the base removes the hydrogen in the first step to form an intermediate carbanion. In the second step, the intermediate carbanion looses the leaving group. The second step is slow and is rate determining step.

– C – C – + OEt

HFast – C – C –

Br

Θ

Br

(First step)–

– C – C –Slow

– C = C – + Br–

Br

(Second step)Θ

The rate of this reaction is dependent on the

carbanion (conjugate base of the substrate). So this mechanism is called ElcB mechanism (Elimination, Unimolecular from conjugate base).

E1cB mechanism is not common for the E2 reactions. The carbanion mechanism occurs only where the carbanion from the substrate is stabilized and where the leaving group is a poor leaving group. A typical example, which follows E1cB mechanism is the formation of 1,1-dichloro-2,2-difluoroethene from 1,1-dichloro-2,2,2-trifluoroethane in presence of sodium ethoxide.

CHCl2 – CF3

C2H5ONaCl2C – CF3 –F–

Cl2C = CF2

1,1-Dichloro-2,2,2-trifluoroethane

Carbanion 1,1-Dichloro-2,2-difluoroethane

–

In the above case the carbanion is strongly stabilized

due to –I effect of halogens. Also F– is a poor leaving group.

A distinction between the E2 and E1cB mechanism can be made by tracer experiments. Thus, the reaction of 1-bromo-2-phenylethane (this substrate was selected as Ph group is expected to increase the acidity of β-hydrogen and also to stabilize the carbanion) with C2H5OD gives back the starting 1-bromo-2-phenylethane. If the carbanion mechanism had operated, the deuterium would have been found in the recovered 1-bromo-2-phenylethane, which is not the case.

C6H5CH2CH2Br + C2H5O–

C6H5CHCH2Br + C2H5OH1-Bromo-2-phenylethane

C6H5CHCH2Br + C2H5OD

C6H5CHCH2Br + OC2H5

D

In case the above reaction is allowed to go to

completion, the product obtained will be

Organic Chemistry

Fundamentals

REACTION MECHANISM

KEY CONCEPT


PhCH – CH2Br Ph – CH – CH2Br

C2H5OD

Fast

Ph – CH – CH2Br

D

OEt

PhCD – CH2BrPhCD = CH2 + BrStyrene

– H EtO–

The styrene obtained does not contain any deuterium

(contrary to what has been shown in the above E1cB mechanism). So in the above reaction E2 mechanism operates.

The E2 mechanism is supported by the following evidences.

(i) During elimination, there is no rearranged product obtained. This is due to the fact that E2 is a single step process and does not involve the formation of intermediate carbocation (the carbocations are known to undergo rearrangement).

(ii) The E2 mechanism finds support from isotope labeling experiments. Dehydrohalogenation of unlabelled 1-bromopropane is seven times faster than the dehydrohalogenation of CH3CD2CH2Br.

CH3CH2CH2Br → 2E CH3CH = CH2

CH3CD – CH2 E2 CH3CD = CH2

Br

D In E2 mechanism a hydrogen (from CH3CH2CH2Br)

or a deuterium (from CH3CD2CH2Br) has to be abstracted. It is known that the C – D bond is stronger than the C – H bond and requires more energy to be broken. Therefore, rate of elimination in CH3CD2CH2Br should be slower. In fact, it has been found that in the unlabelled alkyl halides the elimination rate is seven times more than in labelled alkyl halides.

Unsymmetrical substrate which has hydrogen attached to two different β-carbons can affored two alkenes. For example, 2-bromobutane on dehydrohalogenation may give 1-butene or 2-butene.

CH3 – CHCH2CH3

Br–HBr

CH2 = CHCH2CH3 + CH3CH=CHCH3

2-Bromobutane

1-Butene 2-Butene In a similar way, decomposition of sec-butyl-

trimethylammonium hydroxide may give a mixture of two alkenes. The question arises as to which alkene will be obtained in major amount in the above

dehydrohalogenation. The orientation of the reaction is determined by Hafmann and Saytzeff Rule.

Hofmann Rule : This rule is applicable for those substrates in which α-carbon atom is attached to a positively charged atom. According to this rule, in the elimination reaction of positively charged species, the major product will be the alkene which is least substituted.

CH3CH2NCH2CH2CH3

CH3

CH3

OH–

+ Heating

CH2 = CH2 + CH3CH2CH2N(CH3)2

CH3CH2S(CH3)2

+ C2H5O CH2 = CH2 + S(CH3)2

–

Saytzeff Rule : In case of unsymmetrical alkyl

halides, for example in 2-bromobutane, the course of elimination is determined by Saytzeff Rule. According to this rule, hydrogen is eliminated preferentially from the carbon atom which has less number of hydrogen atoms and so the highly substituted alkene is the major product.

CH3CH2 – CH – CH3

Br

2-Bromobutane

alk.KOH

CH3CH = CHCH3 + CH3CH2CH = CH22-Butene(80%) 1-Butene (20%)

CH3CH2 – C – CH3

CH3

2-Bromo-2-methylbutane

C2H5O–

Br

CH3CH = C – CH3 + CH3CH2C = CH2

CH3 CH3

2-methyl-2-butene (71%)

2-methyl-1-butene(29%)

The formation of highly substituted alkene can be explained as follows.

The transition states of less substituted and more substituted alkenes from an alkyl halide are represented as shown below:

CH3CH2CH — CH2

H OR

Br

δ–

δ– T.S. of less substituted alkene

CH3CH — CHCH3

RO H

Br

δ–

δ– T.S. of more substituted alkene

Both the transition states have partial double bond character. However, the transition state leading to more stable alkene is more stabilized and is of lower energy. Thus, the more stable alkene is formed as the major product.


less substitued alkene more substitued alkenepredominant product

Energy diagram for a typical E2 reaction, showing why the more substituted alkene predominates

Reaction progress

E

RX + base

Hofmann rule can be understood by considering the

mechanism of elimination reaction of quaternary ammonium hydroxide.

H – C – CH2 – N – CH2CH2CH3

Route a

H

H

CH3

CH3 B–

β´ β´´

+

CH2 = CH2 + (CH3)2N(CH2)2CH3 Another possibility is :

CH3CH2 – N – CH2 – C – CH3

Route b

H

H

CH3

CH3 B–

β´ +

(CH3)2NCH2CH3 + CH3CH = CH2

β´´

In the above reaction the strong electron-withdrawing

group makes the hydrogens of the β-carbons more acidic for facile abstraction by the base. In this compound, with alternate β-hydrogens (marked β´ and β´´), the β" hydrogen are less acidic due to +I effect of the adjacent methyl group. Hence β´- hydrogen is relatively more acidic and is removed to give the alkene (ethene) by route a.

In elimination reactions steric effect also plays an important role. Thus, dehydrohalogenation of alkyl halide using the bulky base leads to the formation of terminal alkene as the major product.

CH3CH2CHCH3

Br t-BuO

–

CH3CH2CH=CH2 + CH3CH=CHCH32-Bromobutane (73%) (27%)

Unimolecular elimination reactions (E1) : In these reactions the rate of elimination is dependent

only on the concentration of the substrate and is independent of the concentration of the nucleophile and the reaction is of first order, (E1). Like SN1

reaction the E1 reaction is also a two step process. The first step is the slow ionization of alkyl halide to give the carbocation. The second step involves the fast abstraction of a proton from the adjacent β-carbon atom giving rise to the formation of an alkene.

CH3 — C — X

CH3

CH3

Slow CH3 — C + X–CH3

CH3

+

Carbocation

CCH3

CH3

Fast CH3C = CH2 + BH

CH3

2-Methylpropene

CH2 – H:B

–

Carbocation In case the substrate is such that more than one

alkenes can be formed, that alkene will predominate which has larger number of alkyl groups on the double bonded carbon (this is as per Saytzeffs rule. This can be visualised since the substituted alkyl groups will stabilise the alkene by hyperconjugation.

CH3 — C — C — CH3

CH3

H

Br

H

–HBrCH3 — C == C — CH3

CH3

H

+ CH3 — CH — CH == CH2

CH3

2-Bromo-3-methylbutane 2-Methyl-2-butene (major)

3-Methyl-1-butene (minor) The acid catalysed dehydration of alcohols also

follows E1 mechanism.

(CH3)3COH H2SO4 (CH3)3C—OH2

⊕ –H2O (CH3)3C⊕

t-Butyl alcohol

H3C — C

CH3

CH2 – H

CH3 — C == CH2

CH3

2-Methylpropene

⊕

In the E1 mechanism the rate of reaction is

determined by the rate of formation of carbocation, which in turn depends on the stability of carbocation. Due to the formation of carbocation, these may undergo rearrangements. This has been experimentally confirmed.

(CH3)3CCHCH3 C2H5OH

(CH3)3CCH = CH2 + CH.3C = CCH3

Br

CH3

CH3


Enthalpy of reaction : Since the enthalpy of reaction is defined as the

enthalpy change for unit extent of reaction, the amounts of reactants consumed and products formed will be equal to the corresponding stoichiometric numbers expressed in mol. For example, for the reaction

2N2O5(g) → 4NO2(g) + O2(g) the enthalpy of reaction is the enthalpy change when

2 mol of N2O5 dissociates to give 4 mol of NO2 and 1 mol of O2. It may be noted that

Enthalpy of reaction refers to the entire chemical equation and not to any particular reactant or product.

The enthalpy of a reaction may be computed using the expression.

∆rH = ∑products

mBHV (B) – ∑tstanreac

mB H|V| (B)

where the symbol Σ represents summation over the indicated substances (product or reactant) and VB is the stoichiometric number of the substance B in the balanced chemical equation. For example, for the reaction

Fe2O3(s) + 3H2(g) → 2Fe(s) + 3H2O(1) we have

∆rH = ∑products

mBHV (B) – ∑tstanreac

mB H|V| (B)

= [2Hm(Fe,s) + 3Hm(H2O, 1)] – [Hm(Fe2O3, s + 3Hm(H2,g)] Enthalpy of formation : It is not possible to determine the absolute value of

the enthalpy of a substance. However, based on the following convention, the relative values of standard molar enthalpies of formation of various substances can be built.

The standard enthalpy of formation of every element in its stable state of aggregation at one bar pressure and at specified temperature is assigned a zero value.

The specified temperature is usually taken as 25 ºC. A few examples are ∆fHº(O2, g) = 0 ∆fHº(C, graphite) = 0 ∆fHº(C, diamond) ≠ 0 ∆fHº(Br2, 1) = 0

∆fHº(S, rhombic) = 0 ∆fHº(S, monoclinic) ≠ 0 ∆fHº(P, white) = 0 ∆fHº(P, black) ≠ 0 Definition of Enthalpy of Formation : The standard

enthalpy of formation of a substance is defined as follows:

The standard enthalpy of formation of a compound is the change in the standard enthalpy when one mole of the compound is formed starting from the requisite amounts of elements in their stable states of aggregation.

The formation of one mole of the compound implies that the compound appears as product with stoichiometric number equal to one.

The chemical equations corresponding to enthaly of formation of a few substances are given below.

Enthalpy of formation of HBr(g) : The chemical equation to be referred is

21 H2(g) +

21 Br2(1) → HBr(g)

∆fHº(HBr, g) = ΣvBHmº(B)

= Hmº(HBr, g) – 21 Hmº (H2, g) –

21 Hmº(Br2, 1)

Enthalpy of formation of SO2(g) : The chemical equation to be referred is

S(rhombic) + O2(g) → SO2(g) ∆fHº(SO2, g) = Hmº(SO2, g) – Hmº(S, rhombic) – Hmº(O2, g) Enthalpy of formation of SO3(g) : The chemical

equation to be referred is

S (rhombic) + 23 O2(g) → SO3(g)

∆fHº(SO3, g) = Hmº(SO3'g) – Hmº(S, rhombic)

– 23 Hmº(O2, g)

Lattice Energy of a crystal (Born-Haber Cycle) : The lattice energy is defined as the energy required to

completely separate one mole of a solid ionic compound into gaseous ions.

The larger the lattice energy the more stable the ionic compound and the more tightly the ions held.

The larger energy cannot be measured directly. However, this can be determined from the Born-

Physical Chemistry

Fundamentals

CHEMICAL ENERGETICS

KEY CONCEPT


Haber cycle. Consider the following sequence of steps for the formation of NaCl crystals from Na(s) and Cl2(g)

(i) Vaporization of Na(s) Na(s) → Na(g) ∆rH1 (ii) Ionization of Na(g) Na(g) → Na+(g) + e– ∆rH2 (iii) Dissociation of chlorine

21 Cl2(g) → Cl(g) ∆rH3

(iv) Formation of Cl–(g) Cl(g) + e– → Cl–(g) ∆rH4 (v) Condensation of Na+(g) and Cl–(g) Na+(g) + Cl–(g) → NaCl(s) ∆rH5

Net change: Na(s) + 21 Cl2)(g) → NaCl(s) ∆rH6

According to Hess's law, we can write ∆rH6 = ∆rH1 + ∆rH2 + ∆rH3 + ∆rH4 + ∆rH5 Except ∆rH5 all of these changes of enthalpy can be

determined experimentally. Hence, ∆rH5 can be determined from the above relation. The lattice energy is the negative of ∆rH5 value.

Enthalpy of combustion : Enthalpy of combustion of a given compound is

defined as follows : It is the enthalpy change when one mole of this

compound combines with the requisite amount of oxygen to give products in their stable forms.

For example, the standard enthalpy of combustion of methane at 298.15 K is –890.36 kJ mol–1. This implies the following reaction :

CH4(g) + 2O2(g) → CO2(g) + 2H2O(1) ∆Hº = –890.36 kJ mol–1 The standard enthalpy of combustion of methane at

298.15 K may be written as ∆CHº(CH4, g, 298.15 K) = –890.36 kJ mol–1

Enthalpy of Neutralization : Enthalpy of neutralization is defined as the enthalpy

change when one mole of H+ in dilute solution combines with one mole of OH– to give rise to undissociated water, i.e.,

H+(aq) + OH–(aq) → H2O(1); ∆neutH ≈ – 55.84 kJ mol–1 In this reaction, there is always a release of heat

because of the bond formation H – OH. Whenever one mole of a strong monoprotic acid (HCl, HNO3) is mixed with the one mole of a strong base (NaOH, KOH), the above neutralization reaction takes place, since these acids and bases are present in the completely dissociated from in dilute solutions. The corresponding enthalpy change is of the order of –55.8 kJ mol–1

In general, the enthalpy change of the reaction H+.nH2O + OH–. nH2O → H2O(1)

depends on the value of n and may be visualized mixing HCl.nH2O and NaOH. nH2O. The reaction is

HCl.nH2O + NaOH . nH2O → NaCl. nH2O + H2O(1) The enthalpy change of the above reaction is ∆rH = ∆fH(NaCl . nH2O) + ∆fH(H2O,1) – ∆fH(HCl.nH2O) – ∆fH(NaOH . nH2O) For different values of n, the values are n = 100; ∆rH = [– 407.07 – 285.83 –(– 165.93) – (–469.65)] kJ mol–1 = – 57.32 kJ mol–1 n = 200; ∆rH = [–406.92 – 285.83 – (– 166.27) – (–469.61)] kJ mol–1 = –56.87 kJ mol–1 n = ∞; ∆rH = [–407.27 – 285.83 –(–167.16) –(–470.10)] kJ mol–1 = 55.84 kJ mol–1 When n = ∞, the neutralization reaction may be

written as H+(aq) + OH–(aq) → H2O (1) ∆rH = – 55.84 kJ mol–1 Enthalpy of Formation of ions : We have seen that H+(aq) + OH–(aq) → H2O(1) ∆rHº = –55.84 kJ mol–1 For this reaction, we write ∆rHº = ∆fHº(H2O, 1) – ∆fHº(H+, aq) + ∆fHº(OH–, aq) Hence at 25 ºC, we get ∆fHº(H+, aq) + ∆fHº(OH–, aq) = ∆fHº(H2O, 1) – ∆rHº = [–285.83 – (–55.84)] kJ mol–1 = –229.99 kJ mol–1 By convention, the standard enthalpy of formation of

H+(aq) is taken to be zero. Thus ∆fHº(OH–, aq) = –229.99 kJ mol–1 Bond Enthalpies : Bond enthalpy of a given bond is defined as follows : The bond enthalpy is the average of enthalpies

required to dissociate the said bond present in different gaseous compounds into free atoms or radicals in the gaseous state.

The term bond enthalpy may be distinguished from the term bond dissociation enthalpy which is defined as follows :

The bond dissociation enthalpy is the enthalpy required to dissociate a given bond of some specific compound.

The distinction these two terms may be more evident if described in terms of a simple example, say of the O–H bond. The enthalpy of dissociation of the O–H bond depends on the nature of molecular species from which the H atom is being separated. For example, in the water molecule.


H2O(g) → H(g) + OH(g) ∆rHº = 501.87 kJ mol–1 However, to break the O–H bond in the hydroxyl

required a different quantity of heat OH(g) → O(g) + H(g) ∆rHº = 423.38 kJ mol–1 The bond enthalpy, ∈OH, is defined as the average of

these two values; that is

∈OH = 2

molkJ38.423molkJ87.501 11 −− +

= 462.625 kJ mol–1 In the case of diatomic molecules, such as H2, the

bond enthalpy and bond dissociation enthalpy are identical because each refers to the reaction

H2(g) → 2H(g) ∈H–H = ∆rHº = 435.93 kJ mol–1 Thus, the bond enthalpy given for any particular pair

of atoms is the average value of the dissociation enthalpies of the bond for a number of molecules in which the pair of atoms appears.

Estimation of Enthalpy of a reaction from Bond Enthalpies : Let the enthalpy change for the reaction C2H4(g) + HCl(g) → C2H5Cl(g) be required from the bond enthalpy data. This may be

calculated as follows:

∆H =

atoms gaseous into reactantsbreak

torequired Enthalpy+

atoms gaseous thefrom products form

torequired Enthalpy

= [4∈C–H + ∈C=C + ∈H–Cl] + [–5∈C–H – ∈C–C – ∈C–Cl] = (∈C=C + ∈H–Cl) – (∈C–H + ∈C–C + ∈C–Cl) Relation between energy and enthalpy of a reaction: For a chemical equation, the expression of ∆rH is

∆rH = ∑B

Bv Hm(B)

where vB is the stoichiometric number of B in the chemical equation (it is positive for products and negative for reactants).

The molar enthalpy of B is given as Hm = Um + pVm Substituting this in the previous expression, we get

∆rH = [ ]BmmB

B )pV()B(Uv +∑

= ∑B

Bv Um(B) + ∑B

Bv (pVm)B

Using the fact that pVm ≈ 0 for one mole of solid or liquid pVm = RT for one mole of gaseous species

We get ∆rH = ∑B

Bv Um(B) + ( ∑B

Bv )RT

= ∆rU + (∆vg)RT where ∆vg is the change in the stoichiometric number

of gaseous species in going from reactants to products.

Galena

Galena is the natural mineral form of lead sulfide. It is the most important lead ore mineral.

Galena is one of the most abundant and widely distributed sulfide minerals. It crystallizes in the cubic crystal system often showing octahedral forms. It is often associated with the minerals sphalerite, calcite and fluorite. Galena deposits often contain significant amounts of silver as included silver sulfide mineral phases or as limited solid solution within the galena structure. These argentiferous galenas have long been the most important ore of silver in mining. In addition zinc, cadmium, antimony, arsenic and bismuth also occur in variable amounts in lead ores. Selenium substitutes for sulfur in the structure constituting a solid solution series. The lead telluride mineral altaite has the same crystal structure as galena. Within the weathering or oxidation zone galena alters to anglesite (lead sulfate) or cerussite (lead carbonate). Galena exposed to acid mine drainage can be oxidized to anglesite by naturally occurring bacteria and archaea, in a process similar to bioleaching [3] Galena uses : One of the earliest uses of galena was as kohl, which in Ancient Egypt, was applied around the eyes to reduce the glare of the desert sun and to repel flies, which were a potential source of disease.[4] Galena is a semiconductor with a small bandgap of about 0.4 eV which found use in early wireless communication systems. For example, it was used as the crystal in crystal radio sets, in which it was used as a point-contact diode to detect the radio signals. The galena crystal was used with a safety pin or similar sharp wire, which was known as a "cat's whisker". Making such wireless sets was a popular home hobby in the North of England during the 1930s. Derbyshire was one of the main areas where Galena was mined. Scientists that were linked to this application are Karl Ferdinand Braun and Sir Jagdish Bose. In modern wireless communication systems, galena detectors have been replaced by more reliable semiconductor devices, though silicon point-contact microwave detectors still exist in the market.


1. Compound (A), C3H6Cl2, on reduction with LiAlH4 gives propane. Treatment of (A) with aqueous alkali followed by oxidation gives (B) C3H4O4 which gives effervescence with NaHCO3. Esterification of (B) with ethanol gives (C), C7H12O4, which is well known synthetic reagent. When (B) is heated alone, the product is ethanoic acid, but while heating with soda-lime it gives methane. Compound (B) on reduction with LiAlH4 gives a diol which on reaction with SOCl2 gives back compound (A). Identify all the compounds and give balanced equation of the reactions.

Sol. Compound (B) gives effervescence with NaHCO3 solution. Hence it is a dicarboxylic acid, since it on heating alone gives acetic acid and with soda-lime CH4, it means two –COOH in it are at different carbon atoms.

CH2

COOH

COOH

2NaHCO3 CH2

COONa

COONa + 2CO2 + 2H2O

CH3COOH + CO2 CH4 + 2CO2

∆

∆ Soda-lime

Acid (B) can be prepared from (A), C3H6Cl2, which

should be 1, 3-dichloro propane.

CH2

CH2Cl

CH2Cl 2NaOH(aq.)

CH2

CH2OH

CH2OH (–2NaCl) Propane 1,3-diol

3[O] CH2

COOH

COOH (B)

+ H2O

(A)

Esterification of (B) with ethanol gives malonic ester

which is a synthetic reagent of high importance.

CH2

COOH

COOH ∆ CH2

COOC2H5

COOC2H5–2H2OMalonic ester(B)

+ C2H5OH

CH2

COOH

COOH CH2

CH2OH

CH2OH (B)

LiAlH4

–2H2O

CH2

CH2Cl

CH2Cl2SOCl2

–2SO2; –2HCl

(A) Hence,

(A) CH2

CH2Cl

CH2Cl (B)

CH2

COOH

COOH

2. An organic compound (A) contains 69.42% C, 5.78% H and 11.57% N. Its vapour density is 60.50. It evolves NH3 when boiled with KOH. On heating with P2O5, it gives a compound (B) containing 81.55%C, 4.85% H and 13.59% N. On reduction with Na and C2H5OH (B) gives a base, which reacts with HNO2 giving off N2 and yielding an alcohol (C), which can be oxidised to benzoic acid. What are (A) to (C)? Explain the reactions involved.

Sol. Empirical formula of (A) :

Element % At. wt. Relative no. of atoms

Simplest ratio

C 69.42 12 12

42.69 = 5.78 82.078.5 = 7

H 5.78 1 178.5 = 5.78

82.078.5 = 7

N 11.57 14 14

57.11 = 0.82 82.082.0 = 1

O 13.23 16 16

23.13 = 0.82 82.082.0 = 1

Empirical formula of (A) = C7H7NO Empirical formula wt. = 121 Molecular wt. of compound = 60.5 × 2 = 121

n = .wtEmpirical.wtMolecular =

121121 = 1

Hence, Mol. formula = Empirical formula = C7H7NO Since compound (A) on heating with KOH evolves

NH3, hence it is an amide. C7H7NO

∆ →KOH NH3 + C6H5COOK

Empirical formula and molecular formula of (B) :

Element % At. wt.

Relative no. of atoms

Simplest ratio

C 81.55 12 12

55.81 = 6.8 97.08.6 = 7

H 4.85 1 185.4 = 4.85

97.085.4 = 5

N 13.59 14 14

59.13 = 0.97 97.097.0 = 1

Empirical formula of (B) = C7H5N Since compound (B) is obtained by heating (A) with

P2O5

)A(77 NOHC

∆ → 52OP C7H5N + H2O

UNDERSTANDINGOrganic Chemistry


This is a dehydration reaction showing that (A) is amide, while (B) is a cyanide.

CONH2

(A)

C7H7NO =

The various reactions are as follows.

CONH2

(A)

KOH

Boil

COONa

+ NH3

P2O5 ∆; –H2O

C≡N

(B)

4[H]C2H5OH + Na

CH2NH2

Benzyl amine

HNO2 –N2; – H2O

CH2OH

Benzyl alcohol

(C)

2[O]COOH

+ H2O

3. Two isomeric compounds (A) and (B) have the molecular formula C7H9N. (A) being soluble in water, the solution being alkaline to litmus It does not undergoes diazotization, but show carbylamine reaction and mustard oil reaction, it reacts with acetyl chloride and acetic anhydride. Its product with benzene sulphonyl chloride dissolves in KOH. (B) on the other hand, does not dissolve in water, but undergoes diazotization. Its product with C6H5SO2Cl dissolves in KOH. Its salt undergo hydrolysis in aqueous solution showing an acidic test. What are (A) and (B) ?

Sol. As both (A) and (B) give products with C6H5SO2Cl, which are soluble in KOH, they contain –NH2 group. (B) can be diazotized so contains – NH2 in the nucleus. (A) cannot be diazotized, hence contains –NH2 in the side chain. The number of carbon and hydrogen atoms also indicates aromatic character.

On the basis of above considerations we may show that (A) is benzylamine and (B) o–, m– or p-toludine.

CH2NH2

Benzylamine (A)

CHCl3 + 3KOH

CH2OH

CH2NC

CS2 + HgCl2 CH2NCS

CH2NHSO2C6H5C6H5SO2Cl

– HCl

NaNO2 + HCl

KOH

–H2OCH2 N–SO2C6H5

K Soluble

C6H5CH2NH2 + HOH → C6H5CH2N+H3OH–

C6H4

CH3

NH2

NaNO2 + HCl C6H4 CH3 N2Cl (B)

C6H5SO2Cl–HCl

C6H4CH3

NHSO2C6H5

KOH C6H4 CH3

NKSO2C6H5

Soluble 4. An organic compound (A) of molecular weight

140.5, has 68.32% C, 6.4% H and 25.26% Cl. Hydrolysis of (A) with dilute acid gives compound (B), C8H10O. (B) can be oxidised under milder conditions to (C), C8H8O. (C) forms a phenyl hydrazone (D) with C6H5NHNH2 and gives positive iodoform test. What are (A) to (D) ?

Sol. (i) Calculation of empirical formula of (A)

Element % Relative no. of atoms

Simplest ratio

C 68.32 12

32.68 = 5.59 71.059.5 = 8

H 6.4 14.6 = 6.40

71.040.6 = 9

Cl 25.26 5.3526.25 = 0.71

71.071.0 = 1

Empirical formula = C8H9Cl Empirical formula wt. = 140.5 Molecular weight = Emp. formula weight Hence, Molecular formula = Empirical formula = C8H9Cl (ii) Given that

)A(98 ClHC →HOH

)B(98 OHHC

OH

]O[

2−→

)C(88 OHC

(iii) (C) reacts with C6H5NHNH2 to give C8H8=N.NHC6H5, hence (C) contains a C = O group.

(iv) (C) on heating with I2 + NaOH gives CHI3, hence (C) contains –COCH3 group. Thus (C) is

– C – CH3

O

(v) Oxidation of (B) gives (C), hence (B) is a

secondary alcohol, i.e.,

– CH – CH3

OH (vi) (B) is obtained by the hydrolysis of (A), hence it

is :

– CH.CH3

Cl1-Chloro-1-phenyl ethane

Now, different reactions are as follows :


– CH.CH3

Cl (A)

– CH–CH3

OH (B)

HOH/H+

–HCl

– COCH3

(C)

[O]–H2O

– C = O + H2 N . NH . C6H5

CH3 (C) – C=N.NH.C6H5

–H2O CH3

– C – CH3 + 3I2 + 4NaOH → CHI3 + 3NaI

O (C) – COONa

+ 3H2O +

∆

Yellow ppt

5. An organic compound (A), C10H15N, undergoes

carbylamine reaction but no diazotization. It reacts with HNO2 giving off N2 and a compound (B), C10H14O. (B) reacts with Lucas reagent immediately, but no colour in Victor meyer's test. (B) on heating with conc. H2SO4 eliminates water to give (C), C10H12, which decolourises Br2/CCl4 and cold dilute neutral KMnO4 solution. (C) on ozonolysis gives (D), C7H6O and (E), C3H6O. Compound (E) on heating with I2 and NaOH produced yellow precipitate and sodium acetate. Compound (D) reacts with conc. NaOH to give (F) and (G). Compound (G) on heating with sodalime gives benzene. Compound (F) gives a red colour with ceric ammonium nitrate, and on oxidation and heating the product with sodalime produced benzene. What are (A) to (G) ?

Sol. The given data are :

C10H15N HNO2

–N2;–H2O (A) C10H14O Conc. H2SO4

∆;–H2O(B)C10H12

(C)(I) O3

(II) H2/Pd C7H6O + C3H6O

(D) (E)

C7H6O Conc. NaOH

(D) (F) + (G) Sodalime C6H6

[O]Product ∆ C6H6

Sodalime

)E(63 OHC → +NaOHI2 2 CHI3 ↓ + CH3COONa

+ 3NaI + 3H2O Since (C) decolourise Br2/CCl4 and KMnO4 colour,

hence it has C=C bond. Its ozonolysis gives (D) and (E). Among these (D) undergoes Cannizaro's reaction, while (E) gives iodoform test, hence (D) is benzaldehyde and (E) acetone. Now joining (D) and (E), the structure of (C) can be determined.

–2[O]

C=O + O =C

HCH3

CH3

C=C

HCH3

CH3

(D) (E)

(C) Since (C) is produced from (B), which is a t-alcohol,

as it gives Lucas test immediately, hence (B) is.

C–CH

H CH3

CH3

(B) OH

As (B) is obtained by the action of HNO2 on (A),

hence (A) would be

C–CH

HCH3

CH3

(A)NH2

Chemistry Facts

1. An element's name must be approved by the International Union of Pure and Applied Chemistry, or I.U.P.A.C., in Geneva, Switzerland.

2. The heaviest element gas is radon at room temperature. (There may be heavier ones, but they are compunds not atoms). It was discovered by Friedrich Ernst Dorn in Germany in 1900, but he first called it niton, until 1923.

3. The lightest gas is hydrogen, it is also the lightest of all elements.

4. The element with the highest melting/freezing point is carbon at 6,381 degrees Fahrenheit (3,527 degrees Celsius).

5. The element with the highest boiling point is rhenium at 10,105 degrees Fahrenheit (5,596 degrees Celsius).

6. The element with the lowest melting/freezing point is helium at -458 degrees Fahrenheit (-272.2 degrees Celsius).


1. Let y = f(x) be a curve satisfying

dxdy – y ln 2 = 2sin x (cos x – 1). ln2, then

(A) y is bounded when x → ∞ (B) f(x) = 2sin x + c . 2x, where c is an arbitrary

constant (C) y = 2sinx, y is bounded when x → ∞ (D) f(x) = 2sinx does not have any solution if y is not

bounded. 2. In a right angled triangle the length of its hypotenuse

is four times the length of the perpendicular drawn from its orthocentre on the hypotenuse. The acute angles of the triangle can be

(A) 6π ,

3π (B)

8π ,

83π

(C) 6π ,

3π (D)

12π ,

125π

3. Let a, b ∈ R such that 0 < a < 1 and 0 < b < 1. The

values of a and b such that the complex number z1 = –a + i, z2 = –1 + bi and z3 = 0 form an equilateral triangle are

(A) 3,32 − (B) 32,32 −−

(C) 32,3 − (D) None of these 4. If c1 is a fixed circle and c2 is a variable circle with

fixed radius. The common transverse tangents to c1 and c2 are perpendicular to each other. The locus of the centre of variable circle is :

(A) circle (B) ellipse (C) hyperbola (D) parabola 5. The length of the latus rectum of the parabola 169 (x – 1)2 + (y – 3)2 = (5x – 12y + 17)2 is –

(A) 1314 (B)

1356 (C)

1328 (D) None

6. Evaluate : ∫ −+

x3cos21x4cosx5cos dx

7. Find all the real values of a, for which the roots of the equation x2 – 2x – a2 + 1 = 0 lie between the roots of equation x2 – 2(a + 1) x + a(a – 1) = 0

8. Given the base of a triangle and the sum of its sides

prove that the locus of the centre of its incircle is an ellipse.

9. A bag contains 7 tickets marked with the number 0,

1, 2, 3, 4, 5, 6 respectively. A ticket is drawn and replaced. Then the chance that after 4 drawings the sum of the numbers drawn is 8, is –

10. A polynomial is x of degree greater than 3 leaves

remainders 2, 1 and – 1 when divided by (x – 1), (x + 2) and (x + 1) respectively. What would be the remainder if the polynomial is divided by (x2 – 1) (x + 2) ?

`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 Maheshwari Joint Director Academics, Career Point, KotaSolut ions wi l l be publ ished in next i ssue

3Set

Puzzle : Bags of Marbles

• You have three bags, each containing two marbles. Bag A contains two white marbles, Bag B contains two black marbles, and Bag C contains one white marble and one black marble.

• You pick a random bag and take out one marble.

• It is a white marble.


1. 1st box can be filled in 4 ways. Next each box can be filled in 3 ways (except the ball

of colour in previous box). Hence the required no. of ways = 4 × 35 = 972 2. Given |A| ≠ 0; AA–1 = I ⇒ (AA–1)T = IT (A–1)TAT = I (as A is symmetric) (A–1)T A = I so by the definition of inverse A–1 = (A–1)T Hence A–1 is also symmetric. 3. The normal to hyperbola at the point P(a sec θ, b tan θ) is ax cos θ + by cot θ = a2 + b2 If it passes through (h, k) then a h cos θ + b k cot θ = a2 + b2 ...(1)

Let z = eiθ = cos θ + i sin θ then put cos θ = z2

1z2 +

and cot θ = i

−+

1z1z

2

2 in the equation (1).

ahz4 + 2(i bk – (a2 + b2)) z3 + 2(i bk + (a2 + b2))z – ah = 0 z1, z2, z3, z4 are its four solutions so

Σ z1z2 = 0 = Σ )(i 21e θ+θ = 0 Σ (cos (θ1 + θ2) + i sin (θ1 + θ2)) = 0 Hence Σ cos(θ1 + θ2) = Σ sin(θ1 + θ2) = 0 4. Planes are – x – 2y – 2z + 9 = 0 ....(1) and 4x – 3y + 12z + 13 = 0 ...(2) The plane bisecting the angle b/w these planes

containing origin is

3

9z2y2x +−−− = + 13

13z12y3x4 ++−

i.e. 25x + 17y + 62z – 78 = 0 ...(3) If θ be the angle between (1) & (3) then

cos θ = 475861

⇒ tan θ = 61

1037 < 1

Hence plane given by (3) is bisecting the acute angle between given two planes also. Hence the conclusion holds true.

5. Let I2 = ∫ −−)b(f

)a(f

221 )a))y(f(( dy

Let f–1(y) = x ⇒ f(x) = y

I2 = ∫ −b

a

22 )ax( f´(x) dx

Using by parts

I2 = ( )ba22 )x(f)ax( − – ∫b

ax2 f(x) dx

= (b2 – a2) f(b) – ∫b

ax2 f(x) dx

= ∫b

ax2 f(b) dx – ∫

b

ax2 f(x) dx

= ∫b

ax2 (f(b) – f(x)) dx

Hence 2

1

II =

21

6. y + y1 = 2

⇒ y = 1

x + x1 = 25 +

⇒ x2 + 2x1

= ( 5 + 2) – 2 = 5

x4 + 4x1 = 5 – 2

⇒ x8 + 8x1 = 9 – 2

⇒ x16 + 16x1 = 49 – 2

⇒ 47 + 1 + 1 = 49

MATHEMATICAL CHALLENGES SOLUTION FOR JUNE ISSUE (SET # 2)


7. Let the radius of S2 is r

6

6

6

6rr

2 r + r + 6 = 2 6

r = 6

+

−

1212

= 6(3 – 22 )

= 18 – 212 8. S1 = 2 + 4 + 6 + .... + 120

= 260 (2 + 120)

= 30 × 122 = 3660 S2 = 7 + 14 + 21 + ..... + 119

= 2

17 (7 + 119)

= 17 × 63 = 1071 S3 = 14 + 28 + ..... + 102

= 28 (14 + 112)

= 4 × 126 = 504

Ans. = 2

121120× – 3660 – 1071 + 504

= 7260 – 4731 + 504 = 2529 + 504 = 3033 9. Here F(x) is even function so f(x) = F(–x) = F(x) ⇒ f(–x) = f(x) g(x) = –F(x) = – f(x) = –f(–x) h(x) = –F(–x) = – F(x) = –f(x) Ans. (C) 10. f(x) + h(x) = f(x) – f(x) = 0 g(x) – h(x) = – f(x) + f(x) = 0 F(x) + f(x) ≠ 0 f(x) – g(x) = f(x) + f(x) ≠ 0 Ans. (B)

Know about "Root two"

1.41

is also called Pythagoras' constant. is the ratio of diagonal to side length in a square. ≈ 1.4142135623 7309504880 1688724209 6980785696 7187537694 8073176679 7379907324 7846210703 8850387534 3276415727 3501384623 0912297024 9248360558 5073721264 4121497099...

One of the earliest numerical approximation of 2 was found on a Babylonian clay tablet (from the Yale Babylonian Collection), dated approximately to between 1800 B.C. and 1600 B.C. The annotations on this tablet give an impressive numerical approximation in four sexagesimal figures: 1 + 24/60 + 51/602 + 10/603 = 1.41421296... ≈ (Pn+1 - Pn)/Pn (P = Pell numbers) ≈ 17/12 ≈ 99/70 ≈ 1.0110101000001001111...2 = 2sinus(45°) = 2cosinus(45°) = 1 + (1 / (2 + (1 / (2 + (1 / (2 + ... )))))) = ( i + i i) / i

If you want to have some fun with 2: start with the very rough approximation 7/5. Then (7+5+5)/(7+5) = 17/12 (17+12+12)/(17+12) = 41/29 (41+29+29)/(41+29) = 99/70 (99+70+70)/(99+70) = 239/169 ... continuing closer approximations of 2 - posted by Larry Bickford - Writing numbers using only square roots of 2: 3 = -log2log2 ( ( 2)) 4 = -log2log2 ( ( ( 2))) 5 = -log2log2 ( ( ( ( 2)))) 6 = -log2log2 ( ( ( ( ( 2))))) ... etc. ISO paper sizes are all based on a single aspect ratio of the square root of two, or approximately 1:1.4142. Basing paper upon this ratio was conceived by Georg Lichtenberg in 1786, and at the beginning of the 20th century, Dr Walter Porstmann turned Lichtenberg's idea into a proper system of different paper sizes.


1. Prove that

(99)50 – 99.9850 + 2.198.99 (97)50 – ........ + 99 = 0

Sol. 9950 – 99.9850 + 2.198.99 9750 – ............ + 99

= 9950 – 99C1(98)50 + 99C2(97)50 – ...... + 99C98 . 1 = 99C09950 – 99C1(99 – 1)50 + 99C2(99 – 2)50 – ....... + 99C98 (99 – 98)50 – 99C99(99 – 99)50

= 9950 [99C0 – 99C1 + 99C2 – ....... + 99C98 – 99C99] + 50C1 9949[99C1 – 2. 99C2 + ...... + 3.99C3 ...] + ....

= 0 + 0 + 0 .............. = 0 2. Find the least value of n for which

(n – 2)x2 + 8x + n + 4 > sin–1(sin 12)

+ cos–1(cos 12) ∀ x ∈R when n∈ N. Sol. we have sin–1(sin 12) + cos–1(cos 12)

= sin–1(sin (4π – (4π – 12)))

+ cos–1(cos (4π – (4π – 12)))

= –(4π – 12) + 4π – 12 = 0

So that, (n – 2)x2 + 8x + n + 4 > 0 ∀ x ∈ R

⇒ n – 2 > 0

⇒ n ≥ 3 and 82 – 4(n – 2) (n + 4) < 0

or n2 + 2n – 24 > 0 ⇒ n > 4

⇒ n ≥ 5

⇒ n = 5 3. If parameters p, r, q are in H.P. and d be the length of

perpendicular from origin to any member of family of lines x r(p + q – 2pq) – 2pq (y – 5r) – 3 pqr = 0

then show that |d| ≤ 2

7 .

Sol. Given family of line is x r (p + q – 2pq) – 2pq (y – 5r) – 3 pqr = 0 dividing by pqr, we get

x

−+ 2

p1

q1 – 2

− 5

ry – 3 = 0

x

− 2

r2 – 2

− 5

ry – 3 = 0

(7 – 2x) + r2 (x – y) = 0

⇒ Given family of lines passes through fixed point

27,

27 . Equation of line in normal form

x cos α + y sin α = p passes through

27,

27

⇒ cos α + sin α

= 7p2 but – 2 ≤ cos α + sin α ≤ 2

⇒ 27p2

≤

⇒ |p| ≤ 2

7

4. Prove that if sin2x + sin2y < 1 for all x ∈R, y ∈R,

then (sin–1(tanx . tany)) ∈

ππ

−2

,2

.

Sol. Consider (tan x . tan y)2 = ycos.xcosysin.xsin

22

22∀ x, y ∈ R

= ycos.xcos

ycosxcosysin.xsin22

2222 − + 1

= ycos.xcos

ysin.xsin–ysinxsin1–ysin.xsin22

222222 ++ + 1

= ycos.xcos

1–ysin.xsin22

22 + 1

using sin2x + sin2y – 1 < 0, ∀ x, y ∈ R

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

Students' Forum

MATHS


= tan2x . tan2y < 1

= tan x tany ∈ (–1, 1)

= sin–1(tan x . tan y) ∈

ππ

−2

,2

5. If x5 – x3 + x = a than P.T. x6 ≥ 2a – 1

Sol. Given that x(x4 – x2 + 1) = 1x

)1x(x2

6

+

+ = a

Suppose that x > 0, ⇒ a > 0

⇒ x6 + 1 = a

+x

1x2 = a

+

x1x ≥ 2a

⇒ x6 ≥ 2a – 1 .....(i)

if x ≤ 0 ⇒ a ≤ 0

2a – 1 < 0 ≤ x6 ...(ii)

from equation (i) and (ii)

x6 ≥ 2a – 1 6. Let f(x) satisfies the differential equation

x . dx

)x(df + f(x) = g(x) where f(x) and g(x) are

continuous functions and f(x) is a decreasing function

∀ x > 0. Prove that x . g(x) < ∫x

0)x(g dx ∀ x > 0.

Sol. Given that x . dx

)x(df + f(x) = g(x)

⇒ dx

)x(df = x

)x(f)x(g − < 0

⇒ g(x) < f(x) ∀ x > 0 ...(1) Again x . df(x) + f(x) . dx = g(x) dx

⇒ ∫ +x

0)x(fx(d ) = ∫

x

0)x(g dx

⇒ x . f(x) = ∫x

0)x(g dx

f(x) = ∫x

0)x(g

x1 dx

⇒ g(x) < ∫x

0)x(g

x1 dx (using eqn(1))

⇒ x . g(x) < ∫x

0)x(g dx ∀ x > 0

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• The term 'black hole' was coined in 1968 when John Wheeler described how an in-falling object 'becomes dimmer millisecond by millisecond...light and particles incident from outside...go down the black hole only to add to its mass and increase its gravitational attraction.'

• The 'Red Planet' isn't really red at all, Nasa photographs indicate that it is more of a tan or butterscotch colour.

• The International Space Station orbits at 248 miles above the Earth.

• The axis of orbit of the planet Uranus is tilted at a 90 degree angle.

• Astronomers have discovered over 10,000 asteroids - but put them together and they would be smaller than the Moon.

• Have you ever seen a ring around the moon? Folklore has it that this means bad weather is coming.


Coordinates of a point :

x y

P (x,y,z)

z

N

O

Y

Z

L

M

X

x-coordinate = perpendicular distance of P from yz-plane y-coordinate = perpendicular distance of P from zx-plane z-coordinate = perpendicular distance of P from xy-plane Coordinates of a point on the coordinate planes and axes: yz-plane : x = 0 zx-plane : y = 0 xy-plane : z = 0 x-axis : y = 0, z = 0 y-axis : y = 0, x = 0 z-axis : x = 0, y = 0 Distance between two points : If P(x1, y1, z1) and Q(x2, y2, z2) are two points, then

distance between them

PQ = 221

221

221 )zz()yy()xx( −+−+−

Coordinates of division point : Coordinates of the point dividing the line joining two

points P(x1, y1, z1) and Q(x2, y2, z2) in the ratio m1 : m2 are

(i) in case of internal division

++

++

++

21

1221

21

1221

21

1221mm

zmzm,mm

ymym,mm

xmxm

(ii) in case of external division

−−

−−

−−

21

1221

21

1221

21

1221

mmzmzm

,mm

ymym,

mmxmxm

Note: When m1, m2 are in opposite sign, then division will be external.

Coordinates of the midpoint: When division point is the mid-point of PQ, then

ration will be 1 : 1; hence coordinates of the mid-point of PQ are

+++

2zz,

2yy,

2xx 212121

Coordinates of the general point : The coordinates of any point lying on the line joining

points P(x1, y1, z1) and Q(x2, y2, z2) may be taken as

++

++

++

1kzkz,

1kyky,

1kxkx 121212

which divides PQ in the ratio k : 1. This is called general point on the line PQ.

Division by coordinate planes : The ratios in which the line segment PQ joining

P(x1, y1, z1) and Q(x2, y2, z2) is divided by coordinate planes are as follows : (i) by yz-plane : –x1/x2 ratio (ii) by zx-plane : – y1/y2 ratio (iii) by xy-plane : –z1/z2 ratio Coordinates of the centroid : (i) If (x1, y1, z1); (x2, y2, z2) and (x3, y3, z3) are

vertices of a triangle then coordinates of its centroid are

++++++3

zzz,

3yyy

,3

xxx 321321321

(ii) If (xr, yr, zr); r = 1, 2, 3, 4 are vertices of a tetrahedron, then coordinates of its centroid are

+++++++++4

zzzz,

4yyyy

,4

xxxx 432143214321

Direction cosines of a line [Dc's] : The cosines of the angles made by a line with

positive direction of coordinate axes are called the direction cosines of that line.

Let α, β, γ be the angles made by a line AB with positive direction of coordinate axes then cos α, cos β, cos γ are the direction cosines of AB which are generally denoted by l, m, n. Hence

l = cos α, m = cos β, n = cos γ

3-DIMENSIONAL GEOMETRY

Mathematics Fundamentals

MATH


x-axis makes 0º, 90º and 90º angles with three coordinate axes, so its direction cosines are cos 0º, cos 90º, cos 90º i.e. 1, 0, 0. Similarly direction cosines of y-axis and z-axis are 0, 1, 0 and 0, 0, 1 respectively. Hence

dc's of x-axis = 1, 0, 0 dc's of y-axis = 0, 1, 0 dc's of z-axis = 0, 0, 1 Relation between dc's ∴ l2

+ m2 + n2 = 1 Direction ratios of a line [DR's] : Three numbers which are proportional to the

direction cosines of a line are called the direction ratios of that line. If a, b, c are such numbers which are proportional to the direction cosines l, m, n of a line then a, b, c are direction ratios of the line. Hence

⇒ l = ± 222 cba

a

++,

m = ± 222 cba

b

++, n = ±

222 cba

c

++

Direction cosines of a line joining two points : Let ≡ (x1, y1, z1) and Q ≡ (x2, y2, z2); then (i) dr's of PQ : (x2 – x1), (y2 – y1), (z2 – z1)

(ii)dc's of PQ : PQ

zz,PQ

yy,PQ

xx 121212 −−−

i.e., 2

12

122

12

122

12

12

)xx(

zz,)xx(

yy,)xx(

xx

−Σ

−

−Σ

−

−Σ

−

Angle between two lines : Case I. When dc's of the lines are given If l1, m1, and l2,m2 n2 are dc's of given two lines, then

the angle θ between them is given by cos θ = l1l2 + m1m2 + n1n2

sin θ = 21221

21221

21221 )nn()nmnm()mm( llll −+−+−

The value of sin θ can easily be obtained by the following form :

sin θ = 2

22

112

22

112

22

11nn

nmnm

mm

l

l

l

l++

Case II. When dr's of the lines are given If a1, b1, c1 and a2, b2, c2 are dr's of given two lines,

then the angle θ between them is given by

cos θ = 22

22

22

21

21

21

212121

cbacba

ccbbaa

++++

++

sin θ = 22

22

22

21

21

21

21221

cbacba

)baba(

++++

−Σ

Conditions of parallelism and perpendicularity of two lines : Case I. When dc's of two lines AB and CD, say l1,

m1,n1 and l 2, m2, n2 are known AB || CD ⇔ l 1 = l 2, m1 = m2, n1 = n2 AB ⊥ CD ⇔ l 1 l 2 + m1m2 + n1n2 = 0. Case II. When dr's of two lines AB and CD, say : a1,

b1, c1 and a2, b2, c2 are known

AB || CD ⇔ 2

1

2

1

2

1cc

bb

aa

==

AB ⊥ CD ⇔ a1a2 + b1b2 + c1c2 = 0. Area of a triangle : Let A(x1, y1, z1); B(x2, y2, z2) and C(x3, y3, z3) are

vertices of a triangle. Then dr's of AB = x2 – x1, y2 – y1, z2 – z1 = a1, b1, c1 (say)

and AB = 21

21

21 cba ++

dr's of BC = x3 – x2, y3 – y2, z3 – z2 = a2, b2, c2(say)

and BC = 22

22

22 cba ++

Now sin B = 22

21

21221

aa

)cbcb(

ΣΣ

−Σ

= BC.AB

)cbcb( 21221 −Σ

∴ Area of ∆ABC = 21 AB. BC sin B

= 21221 )cbcb(

21

−Σ

Projection of a line segment joining two points on a line : Let PQ be a line segment where P ≡ (x1, y1, z1) and

Q ≡ (x2, y2, z2); and AB be a given line with dc's as l, m, n. If P'Q' be the projection of PQ on AB, then

P'Q' = PQ cos θ where θ is the angle between PQ and AB. On

replacing the value of cos θ in this, we shall get the following value of P'Q'.

P'Q' = l (x2 – x1) + m(y2 – y1) + n (z2 – z1) Projection of PQ on x-axis : a = |x2 – x1| Projection of PQ on y-axis : b = |y2 – y1| Projection of PQ on z-axis : c = |z2 – z1|

Length of line segment PQ = 222 cba ++

* If the given lines are l

α−x = m

y β− = n

z γ− and

´´x

l

α− = ´m

ý β− = ń

´z γ− , then condition for

intersection is


If the given lines are l

α−x = m

y β− = n

z γ− and

´´x

l

α− = ´m

ý β− = ń

´z γ− , then condition for

intersections is

ń´m

nm´–´´–

l

l

γγβ−βαα = 0

Plane containing the above two lines is

ń´m´

nm–zy–x

l

l

γβ−α = 0

Condition of coplanarity if both the lines are in general form: Let the lines be ax + by + cz + d = 0 = a´x + bý + c´z + d´ and αx + βy + γz + δ = 0 = α´x + βý + γ´z + δ´

These are coplanar if

´´´´

´dc´bádcba

δγβαδγβα

= 0

Reduction of non-symmetrical form to symmetrical form: Let equation of the line in non-symmetrical form be' a1x + b1y + c1z + d1 = 0; a2x + b2y + c2z + d2 = 0. To find the equation of the line in symmetrical form,

we must know (i) its direction ratios (ii) coordinates of any point on it.

Direction ratios : Let l, m, n be the direction ratios of the line. Since the line lies in both the planes, it must be perpendicular to normals of both planes. So

a1l + b1m + c1n = 0; a2l + b2m + c2n = 0 From these equations, proportional values of l, m, n

can be found by cross-multiplication as

1221 cbcb −

l = 1221 acac

m−

= 1221 baba

n−

Point on the line : Note that as l, m, n cannot be zero simultaneously, so at least one must be non-zero. Let a1b2 – a2b1 ≠ 0, then the line cannot be parallel to xy-plane, so it intersect it. Let it intersect xy-plane in (x1,y1, 0). Then

a1x1 + b1y1 + d1 = 0 and a2x1 + b2y1 + d2 = 0 Solving these, we get a point on the line. Then its

equation becomes

1221

1

cbcbxx

−− =

1221

1

acacyy

−− =

1221 baba0z

−−

or 1221

1221

1221

cbcbbabadbdbx

−−−

−=

1221

1221

1221

acacbabaadady

−−−

−=

1221 baba0z

−−

Note : If l ≠ 0, take a point on yz –plane as (0, t1, z1) and if m ≠ 0, take a point on xz-plane as (x1, 0, z1)

Skew lines : The straight lines which are not parallel and non-coplanar i.e. non-intersecting are called skew lines.

If ∆ = ń´m´

nm–zy–x

l

l

γβ−α ≠ 0, the lines are skew.

Shortest distance : Suppose the equation of the lines

are l

α−x = m

y β− = n

z γ−

and ´

´xl

α− = ´m

ý β− = ń

´z γ− . Then

S.D. = 2)n´m´mn(

)m´´m)(ńń´)(()n´m´mn´)((

−Σ

−−β−β+−α−α llll

= ń´m´

nm´–´´–

l

l

γγβ−βαα

Some results for plane and straight line: (i) General equation of a plane : ax + by + cz + d = 0 where a, b, c are dr's of a normal to this plane. (ii) Equation of a straight line :

General form :

=+++=+++

0dzcybxa0dzcybxa

2222

1111

(In fact it is the straight line which is the intersection of two given planes)

Symmetric form : czz

byy

axx 111 −

=−

=−

where (x1, y1, z1) is a point on this line and a, b, c are its dr's

(iii) Angle between two planes : If θ be the angle between planes a1x + b1y c1z + d1 = 0

and a2x + b2y + c2z + d2 = 0, then

cos θ = 22

22

22

21

21

21

212121

cbacba

ccbbaa

++++

++

(In fact angle between two planes is the angle between their normals.)

Further above two planes are

parallel ⇔ 2

1

2

1

2

1

cc

bb

aa

==

perpendicular ⇔ a1a2 + b1b2 + c1c2 = 0


Arithmetic Progression (AP) AP is a progression in which the difference

between any two consecutive terms is constant. This constant difference is called common difference (c.d.) and generally it is denoted by d.

Standard form: Its standard form is a + (a + d) + (a + 2d) +.......... General term : Tn = a + (n – 1) d If Tn = l then it should be noted that

(i) d

a1n)ii(1nad −

+=−−

=ll

Note: cab2APinarec,b,a +=⇔

Sum of n terms of an AP :

)a(2nSn l+=

where l is last term (nth term). Replacing the value of l, it takes the form

]d)1n(a2[2nSn −+=

Arithmetic Mean : (i) If A be the AM between two numbers a and b,

then )ba(21A +=

(ii) The AM of n numbers a1, a2,..............,an

= n1 (a1 + a2 +........+ an)

(iii) n AM's between two numbers If A1, A2,....., An be n AM's between a and b then a A1, A2,....., An, b is an AP of (n + 2) terms. Its common

difference d is given by

Tn+2 = b = a + (n + 1)d ⇒ d = 1nab

+−

so A1 = a + d, A2 = a + 2d,....., An = a + nd. Sum of n AM's between a and b ∴ ΣAn = n(A) Assuming numbers in AP : (i) When number of terms be odd Three terms : a –d, a, a + d

Five terms : a – 2d, a–d, a, a + d, a + 2d ................ ....... ....... ....... ....... (ii) When number of terms be even Four terms: a – 3d, a – d, a + d, a + 3d Six terms : a –5d, a – 3d, a –d, a + d, a+3d, a + 5d ............... ........ ...... ...... ...... ...... Geometrical Progression (GP) : A progression is called a GP if the ratio of its each

term to its previous term is always constant. This constant ratio is called its common ratio and it is generally denoted by r.

Standard form : Its standard form is a + ar + ar2 +......... General term : Tn = arn–1

a, b, c are in GP ⇔ bc

ab

= ⇔ b2 = ac

Sum of n terms of a GP : The sum of n terms of a GP a + ar + ar2

+....... is given by

Sn =

>−−

=−

−

<−−

=−−

1rwhen,1rar

1r)1r(a

1rwhen,r1ra

r1)r1(a

n

n

l

l

when l = Tn. Sum of an infinite GP : (i) When r > 1, then rn → ∞, so Sn → ∞ Thus when

r > 1, the sum S of infinite GP = ∞ (ii) When | r | < 1, then rn → 0, so

S = r1

a−

(iii) When r = 1, then each term is a so S = ∞. Geometric Mean : (i) If G be the GM between a and b then

G = ab (ii) G.M. of n numbers a1, a2 ......, an = (a1a2a3

.....an)1/n (iii) n GM’s between two numbers ⇒ r = (b/a)1/n+1

PROGRESSION & MATHEMATICAL INDUCTION

Mathematics Fundamentals

MATH

XtraEdge for IIT-JEE JULY 2009 46

Product of n GM's between a and b Product of GM's = (ab)n/2 = Gn Assuming numbers in GP : (i) When number of terms be odd Three terms : a/r, a, ar Five terms : a/r2, a/r, a, ar, ar2 ............... .. ..... ..... ..... ..... ..... (ii) When number of terms be even Four terms : a/r3, a/r, ar, ar3 Six terms : a/r5, a/r3, a/r, ar, ar3, ar5

Arithmetic-Geometric Progression : If each term of a progression is the product of the

corresponding terms of an AP and a GP, then it is called arithmetic-geometric progression (AGP). For example:

a, (a + d)r, (a + 2d)r2 ....... Tn = [a + (n – 1)d] rn–1

Sn = 2

1n

)r1()r1(dr

r1a

−

−+

−

− –

r1r]d)1n(a[ n

−−+

S∞ = 2)r1(dr

r1a

−+

− | r | < 1

Harmonic Progression : A progression is called a harmonic progression

(HP) if the reciprocals of its terms are in AP.

Standard form : d2a

1da

1a1

++

++ +.............

General term : d)1n(a

1Tn −+=

∴ a, b, c are in HP ⇔ c1

a1

b2

+= ⇔ b = ca

ac2+

Harmonic Mean : (i) If H be a HM between two numbers a and b, then

ba

ab2H+

= or b1

a1

H2

+=

(ii) To find n HM's between a and b we first find n AM's between 1/a and 1/b, then their reciprocals will be the required HM's.

Relations between AM, GM and HM : G2

= AH A > G > H, when a, b > 0. If A and AM and GM respectively between two

positive numbers, then those numbers are

2222 GAA,GAA −−−+ Some Important Results : If number of terms in an AP/GP/HP is odd then

its mid term is the AM/GM/HM between the first and last term.

If number of terms in an AP/GP/HP is even the AM/GM/HM of its two middle terms is equal to the AM/GM/HM between the first and last term.

a, b, c are in AP, GP and HP ⇔ a = b = c a, b, c are in AP and HP ⇒ a, b,c are in GP. a, b, c are in AP

⇔ ab1,

ca1,

bc1 are in AP. ⇔ bc, ca, ab are in

HP. a, b, c are in GP ⇔ a2, b2, c2 are in GP. a, b, c are in GP ⇔ loga, logb, logc are in AP. a, b, c are in GP ⇔ logam logbm, logcm are in HP. a, b, c d are in GP ⇔ a + b, b + c, c + d are in GP. a, b, c are in AP ⇔ αa, αb, αc

are in GP (α ∈ R0) Principle of Mathematical Induction : It states that any statement P(n) is true for all

positive integral values of n if (i) P(1) is true i.e., it is true for n = 1. (ii) P(m) is true ⇒ P(m + 1) is also true i.e., if the statement is true for n = m then it must

also be true for n = m + 1. Some Formula based on the Principle of Induction :

Σn = 1 + 2 + 3 +....... + n = 2

)1n(n +

(Sum of first n natural numbers) Σ(2n – 1) = 1 + 3 + 5 + ... + (2n – 1) = n2 (Sum of first n odd numbers) Σ2n = 2 + 4 + 6 + ...... + 2n = n(n + 1) (Sum of first n even numbers)

Σn2 = 12 + 22 + 32 +.......+ n2 =

6)1n2()1n(n ++

(Sum of the squares of first n natural numbers)

Σn3 = 13 + 23 + 33 +.......+ n3 =

4)1n(n 22 +

(Sum of the cubes of first n natural numbers) Application in Solving Objective Question : For solving objective question related to natural

numbers we find out the correct alternative by negative examination of this principle. If the given statement is P(n), then by putting n = 1, 2, 3, ..... in P(n), we decide the correct answer.

We also use the above formulae established by this principle to find the sum of n terms of a given series. For this we first express Tn as a polynomial in n and then for finding Sn, we put Σ before each term of this polynomial and then use above results of Σn, Σn2, Σn3 etc.


a

PHYSICS

Questions 1 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. The V – T diagram of an ideal gas for the process

A → B → C is a shown in the figure. Select the correct alternative.

(A) Pressure of the gas first increases then remain constant

(B) Pressure of the gas first decreases then remains constant

(C) Pressure of the gas remains constant throughout (D) Nothing can be said about the pressure of the gas

from this graph V

A

B

C

TO

2. A steam turbine cycle is shown diagrammatically below.

Streamwork out

heat out Q2

Condenser water

heat out Q1

If heat only enters and leaves through the boiler and

condenser, then, when the system is working steadily, heat Q1 is absorbed in the boiler in the same time that heat Q2 is rejected in the condenser. The fraction of heat converted into work is -

(A) 1

21

QQQ − (B)

1

12

QQQ −

(C) 2

21

QQQ − (D)

21

21

QQQQ

+−

3. A solid ball is immersed in a liquid. The coefficients

of volume expansion of temperature of the ball and liquid are 8 × 10–6 and 3 × 10–6 per ºC respectively. The percentage change in upthrust when temperature is increased by 100 ºC is -

(A) 0.5 % (B) 0.11 % (C) 1.1 % (D) 0.05 %

IIT-JEE 2010

XtraEdge Test Series # 3

Based on New Pattern

Time : 3 Hours Syllabus : Physics : Calorimetry, K.T.G., Thermodynamics, Heat Transfer, Thermal expansion, Transverse wave, Sound wave, Doppler's effect, Atomic Structure, Radioactivity, X-ray, Nuclear Physics, Matter Waves, Photoelectric Effect, Practical Physics. Chemistry : Chemical Equilibrium, Acid Base, Ionic Equilibrium, Classification & Nomenclature, Isomerism , Hydrogen Family, Boron Family & Carbon Family, S-block elements, Nitrogen Family, Oxygen Family, Halogen Family & Noble Gas, Salt Analysis, Metallurgy, Co-ordination Compounds, Transitional Elements. Mathematics: Point, Straight line, Circle, Parabola, Ellipse, Hyperbola, Vector, 3-D, Probability, Determinants, Matrices.

Instructions : Section - I • Question 1 to 9 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct

answer and -1 mark for wrong answer. • Question 10 to 14 are multiple choice questions with multiple (one or more) is correct answer. +4 marks will be

awarded for correct answer and -1 mark for wrong answer. • Question 15 to 20 are passage based single correct type questions. +4 marks will be awarded for correct answer and

-1 mark for wrong answer. Section - II • Question 21 to 22 are Column Matching type questions. +6 marks will be awarded for the complete correctly

matched answer and No Negative marks for wrong answer. However, 1 mark will be given for a correctly marked answer in any row.


4. Water – wave generators S1 and S2 generate waves of equal wavelength and velocity, but there is a phase difference of π between the two generators. At a point P (equidistant from S1 and S2) S1 by itself produce an oscillation of amplitude 2a and S2 by itself produces an oscillations of amplitude a. When both generators are switched on, which one of the graphs below correctly describes the resultant oscillations at P ?

(A)

y

t O a

a (B)

y

t O a

(C)

y

t O aa

3a

3a

(D)

y

tO

3a

–3a

5. A plane wave of sound traveling in air is incident upon a plane surface of a liquid. The angle of incidence is 60º. The speed of sound in air is 300 m/s and in liquid it is 600 m/s. Assume Snell's law to be valid for sound waves.

(A) The wave will refract into liquid away from normal (B) The wave refract into liquid towards the normal (C) The wave will reflect back into air (D) None of these

6. A gun emitting radio waves is aimed at an approaching car, the speed and frequency of radio waves are v0 and f0. The reflected wave from the car is detected by the gun and a frequency change ∆f is recorded. The approaching velocity of the car is

(A)

∆+

0ff1 v0 (B) 2

∆

0ff v0

(C) 2

vf

f 0

0

∆ (D)

∆−

0ff1 v0

7. Suppose the potential energy between electron and

proton at a distance r is given by – 3

2

r3ke . Application of

Bohr's theory to hydrogen atom in this case shows that (i) Energy in the nth orbit is proportional to n6 (ii) Energy is proportional to m3 (m : mass of electron) (A) Only (i) is correct (B) Only (ii) is correct (C) Both (i) and (ii) are correct (D) None of these

8. In a sample of hydrogen like atoms all of which are in ground state, a photon beam containing photons of various energies is passed. In absorption spectrum, five dark lines are observed. The number of bright lines in the emission spectrum will be (assume that all transitions take place)

(A) 5 (B) 10 (C) 15 (D) None of these

9. In the given nuclear reaction 90X232 → 82Y208 + n1α + n2β + v n1 α-particles and n2-particles are emitted. The values

of n1 and n2 are (A) 4 and 6 (B) 3 and 3 (C) 6 and 4 (D) 6 and 0 Questions 10 to 14 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. 10. The maximum spectral radiancy of a black body

corresponds to a wavelength of λ. If the temperature is changed so that the maximum spectral radiancy

now corresponds to a wavelength of 23

λ. It follows

that (A) the new temperature is 3/2 times the old

temperature (B) the new temperature is 2/3 times the old

temperature (C) the power radiated by the body change by a

factor of 16/81 (D) the power radiated by the body changed by a

factor of 81/16 11. Choose the correct statement(s) about the hydrogen

like atom. (A) Velocity of electron is directly proportional to its

mass (B) The time period of revolution of electron in nth

orbit is directly proportional to n3 (C) The binding energy of deuterium is more than

that of hydrogen (D) The angular momentum of an electron revolving

in the first orbit of helium

12. According to Einstein's theory of photoelectric effect (A) Light propagates through space in the form of

photons (B) Energy of photon is proportional to frequency (C) The number of photons incident per unity area

depends on the intensity (D) Part of a photon can not be absorbed by the

electron

13. Choose the correct statement(s) about X-rays. (A) The intensity of Kα radiation is more than that of Kβ (B) The intensity of Kα radiation is less than that of Kβ (C) The frequency of Kα radiation is more than that

of Kβ (D) The frequency of Kα radiation is less than that of

Kβ


14. The decay constant of a radioactive substance is 0.173 per year. Which of the following statements are correct

(A) About 63% of the radioactive substance will decay in 1/0.173 years

(B) 1/4 th of the radioactive substance will be left after 8 years

(C) Half life of the radioactive substance is 1/0.173 years

(D) All the above statements are correct

This section contains 2 paragraphs, each has 3 multiple choice questions. (Questions 15 to 20) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Passage : I (No. 15to 17) A narrow tube is bent in the form of a circle of radius

R, as shown in the fig. Two small holes S and D are made in the tube at the positions right angle to each other. A source placed at S generates a wave of intensity I0 which is equally divided into two parts: one part travels along the longer path, while the other travels along the shorter path. Both the part waves meet at the point D where a detector is placed.

R

D

S

15. The maximum intensity produced at D is given by- (A) 4I0 (B) 2I0 (C) I0 (D) 3I0

16. The maximum value of λ to produce a maximum at

D is given by-

(A) πR (B) 2πR (C) 2Rπ (D)

23

πR

17. The maximum value of λ to produce a minimum at D is given by -

(A) πR (B) 2πR (C) 2Rπ (D)

23

πR

Passage : II (No. 18to 20) de Broglie considered matter waves as travelling

waves. Suppose that a particle of mass 'm' is confined to a narrow region. The given particle moves in one dimension back and forth with a speed v. This is called the problem of a particle in a box. If we consider the matter wave associated with the particle, such back and forth motion of waves will result in a 'standing wave'.

m v

a Our understanding of standing waves tells us that a

standing wave will be set up such that

λa2 = n

where a is the length of the box and 'n' an integer. Using de Broglie's expression for wavelength of matter wave, it can be easily proved that energy (kinetic energy) of the given particle will be quantised.

18. Energy of the particle can be expressed as

(A) 22

22

ah4mn (B) 2

222

m8ahn (C) 2

22

ma8hn (D) 2

222

a4hhn

19. Minimum energy that will be required to excite the

particle from the ground state to a higher state will be

(A) 22

2

ah4m3 (B) 2

22

m8ah3 (C) 2

2

ma8h3 (D) 2

22

a4hm3

20. Referring to fig. given below quantum number of the

particle confined to the box is

(A) 6 (B) 4 (C) 2 (D) 1 The section contains 2 questions (Questions 21 to 22). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S, then the correctly 4 × 4 matrix should be as follows :

DCBA

PPPP

Q Q Q Q

R R R R

S S S S Q P R S

21. Match Column-I with Column-II : Column –I Column II (A) Lyman series (P) Visible region (B) Balmer series (Q) UV region (C) Pfund series (R) IR region (D) Light emitted by (S) Line emission spectrum lamp


22. Match Column-I with Column-II : Column –I Column II (A) Wave character of (P) Photoelectric effect radiation (B) Photon character of (Q) Compton effect radiation (C) Interaction of a photon (R) Diffraction with an electron, such that photon energy is equal to or slightly greater than the binding energy of electron, is more likely to result in (D) Interaction of a photon (S) Interference with an electron, such that photon energy is much greater than the binding energy of electron, is more likely to result in

CHEMISTRY

Questions 1 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. Ammonium carbamate decomposes as NH2COONH4(s) 2NH3(g) + CO2(g) For the reaction, Kp = 2.9 × 10–5 atm3. If we start

with 1 mole of the compound, the total pressure at equilibrium would be

(A) 0.0766 atm (B) 0.0582 atm (C) 0.0388 atm (D) 0.0194 atm 2. A solution which is 10–3 M each in Mn2+, Fe2+, Zn2+

and Hg2+ is treated with 10–16 M sulphide ion. If Ksp of MnS, FeS, ZnS and HgS are 10–15, 10–23, 10–20 and 10–54 respectively, which one will precipitate first ?

(A) FeS (B) MgS (C) HgS (D) ZnS

3. The pKa of acetyl sacylic acid (aspirin) is 3.5. The pH of gastric juice in human stomach is about 2-3 and pH in the small intestine is about 8. Aspirin will be

(A) unionised in the small intestine and in the stomach

(B) completely ionised the small intestine and in the stomach

(C) ionised in the stomach and almost unionised in the small intestine

(D) ionised in the small intestine and almost unionised in the stomach

4. The Diels-Alder reaction between 1, 3-

cyclohexadiene and acrylonitrile gives the adduct,

CN Its IUPAC name is (A) Bicyclo [2.2.2] oct-2-en-5 nitrile (B) Bicyclo [2.2.2] oct-5-en-2-carbonitrile (C) 3-Cyanobicyclo [2.2.2] oct-5-ene (D) 2-Cyanobicyclo [2.2.2] oct -5-ene

5. A metal salt solution forms a yellow precipitate with potassium chromate in acetic acid, a white precipitate with dilute sulphuric acid but gave no precipitate with sodium chloride or iodide. The white precipitate obtained when sodium carbonate is added to the metal salt solution will consist of

(A) Lead carbonate (B) Basic lead carbonate (C) Barium carbonate (D) Strontium carbonate 6. When a substance A reacts with water it produces a

combustible gas B and a solution of substance C in water. When another substance D reacts with this solution of C, it also produces the same gas B on warming but D can produce gas B on reaction with dilute sulphuric acid at room temperature. A imparts a deep golden yellow colour to smokeless flame of Bunsen burner. A, B, C and D respectively are

(A) Na, H2, NaOH, Zn (B) K, H2, KOH, Al (C) Ca,H2, Ca(OH)2, Sn (D) CaC2, C2H2, Ca(OH)2, Fe 7. XeF6 dissolves in anhydrous HF to give a good

conducting solution which contains (A) H+ and XeF7

– ions (B) HF2– and XeF5

+ ions (C) HXeF6

+ and F– ions (D) None of these

8. A yellow metallic powder is burnt in a stream of fluorine to obtain a colourless gas X which is thermally stable and chemically inert. Its molecule has octahedral geometry. Another colourless gas Y with same constituent atoms as that of X is obtained when sulphur dichloride is heated with sodium fluoride. Its molecule has trigonal bipyramidal geometry. Gases X and Y are respectively

(A) SF4 and S2F2 (B) SF6 and SF4 (C) NaF and NaCl (D) SF4 and SF6

9. In the silver plating of copper, K[Ag(CN)2] is used instead of AgNO3. The reason is

(A) A thin layer of Ag is formed on Cu (B) More voltage is required (C) Ag+ ions are completely removed from solution (D) Less availability of Ag+ ions, as Cu cannot

displace Ag from [Ag(CN)2]– ion


Questions 10 to 14 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct.

10. Which of the following metals cannot be obtained by electrolysis of their aqueous solutions of their salts ?

(A) Ag (B) Mg (C) Al (D) Cu

11. In solid CuSO4.5H2O (A) Cu2+ ion is coordinate bonded to four water

molecules (B) One H2O molecule is H-bonded (C) It has four types of bonds : Ionic, covalent,

coordinate and H-bond (D) Five molecules of H2O are coordinate bonded to Cu2+ 12. Which of the following statement(s) is/are correct for

group-2 metals ? (A) On descending down the group, the lattice

energy as well as hydration energy decreases (B) Only BeF2 is soluble whereas MgF2, CaF2, SrF2

and BaF2 are insoluble (C) BeCl2 is insoluble whereas MgCl2, CaCl2, SrCl2

and BaCl2 are soluble (D) BeSO4 is soluble whereas BaSO4 is insoluble

13. Which of the following is/are correct statements about ozone?

(A) O3 is an unstable, dark blue diamagnetic gas (B) The central oxygen in O3 is sp2 hybridised (C) It causes the tailing of mercury (D) It does not react with KOH 14. Salts 'A' and 'B' on reaction with dil H2SO4 liberate

gases 'X' and 'Y' respectively. Both turn lime water milky and milkyness disappears when excess of gases are passed. 'X' has pungent suffocating smell and turns K2Cr2O7 (acidified) paper green whereas 'Y' is colourless, odourless gas. Salts 'A' and 'B' are

(A) Na2CO3 (B) Na2SO3 (C) Na2S (D) (COO)2(NH4)2


Passage : I (No. 15 to 17) One of the most common antibiotics is penicillin G

(benzylpenicillinic acid), which has the following structure :

C H3C H3C

C H C – OH

N — C S — C — C — N — C — CH2

HO

H H O

O

It is a weak monoprotic acid HP H+ + P– Ka = 1.64 × 10–3 where HP denotes the parent acid and P– the

conjugate base. Penicillin G is produced by growing molds in fermentation tanks at 25ºC and a pH range of 4.5 to 5.0. The crude form of this antibiotic is obtained by extracting the fermentation broth with an organic solvent in which the acid is soluble. In one stage of purification, the organic extract of the crude penicillin G is treated with buffer solution at pH = 6.5

15. What is the ratio of the conjugate base of penicillin G to the acid at this pH ?

(A) 5.2 × 103 (B) 3.2 × 106 (C) 1.8 × 103 (D) 5.6 × 104 16. Penicillin G is not suitable for oral administration,

but the sodium salt (NaP) is because it is soluble. Calculate the pH of a 0.12 M NaP solution formed when a tablet containing the salt is dissolved in a glass of water

(A) 6.07 (B) 7.93 (C) 0.44 (D) 13.56 17. Which hydrogen is acidic ? (A) H attached to N (B) H attached to C (C) H attached to O (D) all H are acidic Passage : II (No. 18to 20) A bluish green coloured compound'A' on heating

gives two products 'B' and 'C'. A metal 'D' is deposited on passing H2 through heated 'B'. The compound 'A' and 'B' are insoluble in water. 'B' is black in colour, dissolves in HCl and on treatment with K4[Fe(CN)6] gives a chocolate brown ppt of compound 'E' 'C' is colourless, odourless gas and turns lime water milky.

18. Compound 'A' is (A) CuSO4 (B) CuCO3 (C) FeSO4 (D) CrCl3

19. The compound 'B' and 'C' are respectively (A) CuS, SO2 (B) CuO, CO2 (C) FeO, H2S (D) Cr2O3, CO

20. The compound 'D' and 'E' are respectively (A) Cu, Cu2[Fe(CN)6] (B) Fe, Cu2[Fe(CN)6] (C) Cr, CuCO3 (D) Zn, CuO

The section contains 2 questions (Questions 21 to 22). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S, then the correctly 4 × 4 matrix should be as follows :


D C B A

P P P P

Q Q Q Q

R R R R

S S S S Q P R S

21. Column –I Column II (A) Al2O3 (P) Electrolytic reduction (B) ZnO (Q) Reduction with coke (C) MnO2 (R) Reduction with carbon monoxide (D) NaCl (S) Reduction with Al 22. Column –I Column II (A) Haemoglobin (P) Anti cancer agent (B) Chlorophyll (Q) Complex of cobalt (C) Vit B12 (R) Complex of iron (D) cis-Platin (S) Complex of Mg

MATHEMATICS

Questions 1 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

1. If 3k2kk

3k2kk

3k2kk

zzzyyyxxx

++

++

++

= (x – y)(y – z)(z – x)(1/x + 1/y + 1/z) then

(A) k = – 3 (B) k = –1 (C) k = 1 (D) k = 3 2. There are two balls in an urn whose colours are not

known (each ball can be either white or black). A white ball is put into the urn. A ball is drawn from the urn. The probability that it is white is

(A) 1/4 (B) 1/3 (C) 2/3 (D) 1/6

3. If two vertices of a triangle are (–2, 3) and (5, – 1), orthocentre lies at the origin and centroid on the line x + y = 7, then the third vertex lies at

(A) (7, 4) (B) (8, 14) (C) (12, 21) (D) None

4. If the pairs of lines x2 + 2xy + ay2 = 0 and ax2 + 2xy + y2 = 0 have exactly one line in common

then the joint equation of the other two lines is given by

(A) 3x2 + 8xy – 3y2 = 0 (B) 3x2 + 10xy + 3y2 = 0 (C) y2 + 2xy – 3x2 = 0 (D) x2 + 2xy – 3y2 = 0

5. If F1 = (3, 0), F2 = (–3, 0) and P is any point on the curve 16x2 + 25y2 = 400, then PF1 + PF2 equals

(A) 8 (B) 6 (C) 10 (D) 12

6. The angle between the lines whose direction cosines are given by the equations

l2 + m2 – n2 = 0, l + m + n = 0 is (A) π/6 (B) π/4 (C) π/3 (D) π/2 7. Equation of the line of shortest distance between the

lines 2x =

3y

− =

1z and

32x − =

51y

−− =

22z + is

(A) 3(x – 21) = 3y + 92 = 3z – 32

(B) 3/1

)3/62(x − = 3/131y − =

3/1)3/31(z +

(C) 3/121x − =

3/1)3/92(y − =

3/1)3/32(z +

(D) 3/12x − =

3/13y + =

3/11z −

8. The value of a for which the volume of

parallelopiped formed by the vectors i + aj + k, j + ak and ai + k is minimum is

(A) –3 (B) 3 (C) 1/ 3 (D) – 3 9. An equation of a plane containing the lines r = a1 + tb1 and r = a2 + tb2 where [a2 – a1, b1, b2] = 0

is (A) [r – a1, b1, b2] = 0 (B) [r – a2, b1, b2] = 0 (C) [r – a2, a1, b2] = 0 (D) [r – a, a2, b2] = 0 Questions 10 to 14 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. 10. Let λ and α be real. Let S denote the set of all values

of λ for which the system of linear equations λx + (sin α)y + (cos α)z = 0 x + (cos α) y + (sin α)z = 0 –x + (sin α)y – (cos α)z = 0 has a non-trivial solution then S contains

(A) (–1, 1) (B) [– 2 , –1]

(C) [1, 2 ] (D) (– 2, 2) 11. An equation of a circle which touches the y-axis at

(0, 2) and cuts off an intercept 3 from the x-axis is (A) x2 + y2 + 4x – 5y + 4 = 0 (B) x2 + y2 + 5x – 4y + 4 = 0 (C) x2 + y2 – 5x – 4y + 4 = 0 (D) x2 + y2 – 5x + 4y + 4 = 0

12. A straight line touches the rectangular hyperbola 9x2 – 9y2 = 8 and the parabola y2 = 32x. An equation of the line is

(A) 9x + 3y – 8 = 0 (B) 9x – 3y + 8 = 0 (C) 9x + 3y + 8 = 0 (D) 9x – 3y – 8 = 0


13. An equation of the line passing through 3i – 5j + 7k and perpendicular to the plane 3x – 4y + 5z = 8 is

(A) 3

3x − = 45y

−+ =

57z −

(B) 3

3x − = 54y

−+ =

75z −

(C) r = 3i – 5j + 7k + λ(3i – 4j + 5k) (D) r = 3i – 4j – 5k + µ(3i + 5j + 7k) 14. If a × b = 2i + 2j – k, a + b = i – 2j + 4k then

(A) a = 41 i +

41 j + k; b =

43 i –

413 j – 5k

(B) a = j + 2k; b = i – 4j – 6k

(C) a = –41 i +

47 j + 2k; b =

45 i –

419 j – 7k

(D) a = 21 i –

21 j; b =

43 i –

413 j – 4k

This section contains 2 paragraphs, each has 3 multiple choice questions. (Questions 15 to 20) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Passage : I (No. 15to 17) A square matrix A of size n × n is said to be

unimodular if det(A) = 1. 15. If A unimodular, then which of the following is not

necessarily unimodular. (A) –A (B) A–1 (C) adj A (D) ωA, where ω is cube root of unity 16. If A and B unimodular matrices, then adjoint of APB

is (A) A (adj P)B (B) B(adj P)A (C) B–1(adj P)A–1 (D) A–1 (adj P)B–1 17. If A is a matrix such that AÁ = In, then (A)A is unimodular (B) –A is unimodular (C) I + A is unimodular (D) None of these

Passage : II (No. 18 to 20) A chess match between two grandmasters X and Y is

won by whoever first wins a total of two games. X´s chances of wining, drawing or losing any particular game are a, b, c respectively. The games are independent and a + b + c = 1

18. The probability that Y wins the match after the 4th game is

(A) 3bc2 (b + 2a) (B) bc2(3b + a) (C) 2ac2(b + c) (D) abc(2a + 3b)

19. The probability that X wins the match is

(A) 3

23

)ca(caa

++ (B) 3

23

)ca(ca3a

++

(C) 3

3

)ca(a+

(D) None of these

20. The probability that Y wins the match is

(A) 3

23

)cb(cbb

++ (B) 3

23

)ca(ac3c

++

(C) 3

3

)cb(c+

(D) None of these

The section contains 2 questions (Questions 21 to 22). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S, then the correctly 4 × 4 matrix should be as follows :

DCBA

PPPP

Q Q Q Q

R R R R

S S S S Q P R S

21. Let f(x) = xcotxcos1xeccosxcosxcos

11xsec

22

222

2

then

Column –I Column II

(A) period of f(x) (P) 323π

(B) maximum value of f(x) (Q) π

(C) ∫π 4/

0)x(f dx –

41 (R) 1

(D) minimum value of f(x) (S) 0 22. A coin has probability p of showing head when

tossed n times. Let pn denote the probability that no two (or more) consecutive heads occur, then

Column –I Column II (A) p1 (P) 1 – 2p2 + p3 (B) p2 (Q) 1 – p2 (C) p3 (R) 1 (D) pn (S) (1 – p)pn–1 + p(1 – p)pn–2


PHYSICS

Questions 1 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

1. In the phenomenon of interference - (A) two waves with a phase difference are

superimposed (B) two waves with a constant phase difference are

superimposed (C) two waves with same frequency are

superimposed with a constant phase difference (D) None of these

2. Mark the correct wave function -

(A) y = f )vtx( ± (B) y = f vtx ± (C) y = f(x2 ± v2t2) (D) None of these 3. The equation of a harmonic wave is given by- (A) y = A cos (kx ± ωt) (B) y = A sin [(kx)2 ± (ωt)2] (C) y = Ae–(x±vt)

(D) All of the above

4. Beats frequency is defined as - (A) the difference in frequencies (B) the number of times intensities becomes

maximum or minimum in one second

(C) the average of frequencies (D) both (A) and (B) 5. The speed of sound in air is v. Both the source and

observer are moving towards each other with equal speed u. The speed of wind is w from source to

observer. Then, the ratio

0ff of the apparent

frequency to the actual frequency is given by -

(A) uvuv

−+ (B)

uwvuwv

−+++

(C) uwvuwv

−−++ (D)

uwvuwv

−−+−

6. The specific latent heat of vaporisation of water at

20ºC is appreciably greater than the value at 100ºC. This is because

(A) the specific latent heat at 20ºC includes the energy necessary to raise the temperature of one kilogram of water from 20ºC to 100ºC

(B) more work must be done in expanding the water vapour against atmospheric pressure at 20ºC than at 100ºC

(C) the molecules in the liquid are more tightly bound to one another at 20ºC than at 100ºC

(D) the root mean square speed of the vapour molecules is less at 20ºC than at 100ºC

IIT-JEE 2011

XtraEdge Test Series # 3

Based on New Pattern

Time : 3 Hours Syllabus : Physics : Calorimetry, K.T.G.,Thermodynamics, Heat Transfer, Thermal expansion, Transverse wave, Sound wave, Doppler's effect. Chemistry : Chemical Equilibrium, Acid Base, Ionic Equilibrium, Classification & Nomenclature, Isomerism , Hydrogen Family, Boron Family & Carbon Family, S-block elements. Mathematics: Point, Straight line, Circle, Parabola, Ellipse, Hyperbola, Vector, 3-D Instructions : Section - I • Question 1 to 9 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct

answer and -1 mark for wrong answer. • Question 10 to 14 are multiple choice questions with multiple (one or more) is correct answer. +4 marks will be

awarded for correct answer and -1 mark for wrong answer. • Question 15 to 20 are passage based single correct type questions. +4 marks will be awarded for correct answer and

-1 mark for wrong answer. Section - II • Question 21 to 22 are Column Matching type questions. +6 marks will be awarded for the complete correctly

matched answer and No Negative marks for wrong answer. However, 1 mark will be given for a correctly marked answer in any row.


7. An ideal gas of volume 5 × 10–3 m3 and at pressure 1.0 × 105 Pa is supplied with 70 J of energy. The volume increases to 1.7 × 10–3 m3, the pressure remaining constant. The internal energy of the gas is

(A) increased by 90 J (B) increased by 70 J (C) increased by 50 J (D) decreased by 50 J 8. The spectrum of a black body at two temperatures

27ºC and 327ºC is shown in the figure. Let A1 and A2 be the areas under the two curves respectively. The value of A2/A1 is

27ºC 327ºC1

2

Wavelength

Intensity

(A) 1 : 16 (B) 4 : 1 (C) 2 : 1 (D) 16 : 1 9. Identify the diagram which correctly represents the

heat inflow and outflow of the system.

(A)

O V

D Q3 Q1

Q4

Q2 B C

A

P

(B)

O V

D Q3 Q1

Q4

Q2 A B

C

P

(C)

O V

C Q3 Q1

Q4

Q2 A B

D

P

(D)

O V

C Q3 Q1

Q4

Q2A B

D

P

Questions 10 to 14 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct.

10. A body X at an original temperature 100 ºC and another body at an original temperature 0 ºC are placed in an evacuated enclosure, the walls of which are maintained at 10 ºC, Which one of the following statements is consistent with Prevost's Theory ?

(A) X emits but does not absorb heat (B) Y absorbs but does not emit heat (C) The final temperature of the bodies will be the

mean of their initial temperature (i.e. 50ºC) (D) The walls of the enclosure radiate heat to both X

and Y.

11. Let v , vrms and vp respectively denote the mean speed, root mean square speed, and most probable speed of the molecules in an ideal monatomic gas at absolute temperature T. The mass of a molecule is m. Then

(A) no molecule can have a speed greater than 2 vms

(B) no molecule can have speed less than vp/ 2 (C) vo < v < vrms (D) the average kineitc energy of a molecule is

3/4 m 2pv

12. According to kinetic theory of gases, the pressure exerted on the walls of a container

(A) is directly proportional to the square of the root mean square speed of the atom

(B) is inversely proportional to the volume of the container

(C) is directly proportional to the number of atoms in the gas.

(D) all the above

13. Choose the correct statement(s) related to the variation of sound intensity with distance.

(A) For a point source it is inversely proportional to distance

(B) For a point source it is inversely proportional to the square of distance

(C) For a line source it is inversely proportional to distance

(D) For a line source it is inversely proportional to the square of distance

14. The principle of superposition of waves is valid (A) when amplitude of wave is very much less than

the wavelength (B) for any magnitude of wavelength (C) when velocity of wave is much larger than the

particle's velocity (D) when velocity of wave is much smaller than the

particle's velocity.


This section contains 2 paragraphs, each has 3 multiple choice questions. (Questions 15 to 20) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Passage : I (No. 15 to 17) Five moles of helium are mixed with two moles of

hydrogen to form a mixture. Take molar mass of helium M1 = 4g and that of hydrogen M2 = 2g

15. The equivalent value of γ is (A) 1.59 (B) 1.53 (C) 1.56 (D) None of these 16. If the internal energy of He sample of 100J and that

of the hydrogen sample is 200 J, then the internal energy of the mixture is

(A) 900 J (B) 128.5 J (C) 171.4 J (D) 300 J 17. Identify, which pair of state parameters can

completely describe the system (A) P and V (B) P and ρ (C) P and U (D) all the above Passage : II (No. 18 to 20) The fig. represents the instantaneous picture of a

transverse harmonic wave traveling along the negative x-axis. Choose the correct alternative(s) related to the movement of the nine points shown in the figure.

B

A C

D E

F G

H x

y

O

18. The points moving upward are : (A) E (B) C (C) F (D) G 19. The points moving downward are : (A) O (B) B (C) D (D) H 20. The stationary points are : (A) O (B) B (C) D (D) H The section contains 2 questions (Questions 21 to 22). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements

(P, Q, R, S) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S, then the correctly 4 × 4 matrix should be as follows :

DCBA

PPPP

Q Q Q Q

R R R R

S S S S Q P R S

21. Match the followings : Column –I Column II

(A) Adiabatic bulk modulus (P) –VP

(B) Slope of P-V graph in (Q) 1

2−γ

isothermal process (C) Degree of freedom (R) γP

(D) Molar heat capacity at (S) 1−γ

γ

constant pressure divided by R 22. Following is given the equation of a stationary wave

(all in SI units) y = (0.06) sin(2πx) cos(5πt) Column –I Column II (A) Amplitude of constituent wave (P) 0.06 (B) Position of node at x = ... m (Q) 0.5 (C) Position of antinode at x = .... m (R) 0.25 (D) Amplitude at x = 3/4 m (S) 0.03

CHEMISTRY

Questions 1 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. If 0.2 mol of H2(g) and 2.0 mol of S(s) are mixed in a

1 dm3 vessel at 90ºC, the partial pressure of H2S(g) formed according to the reaction

H2(g) + S(s) H2S, Kp = 6.8 × 10–2 would be (A) 0.19 atm (B) 0.38 atm (C) 0.6 atm (D) 0.072 atm 2. The vapour density of N2O4 at a certain temperature

is 30. What is the percentage dissociation of N2O4 at this temperature ?

(A) 53.3% (B) 106.6 % (C) 26.7 % (D) None of these


3. A saturated solution of H2S in 0.1 M HCl at 25ºC contains a S2– ion concentration of 10–23 mol L–1. The solubility products of some sulphides are :

CuS = 10–44, FeS = 10–14, MnS = 10–15, CdS = 10–25. If 0.01 M solutions of these salts in 1 M HCl are saturated with H2S, which of these will be precipitated ?

(A) All (B) All excepts MnS (C) All except MnS and FeS (D) Only CuS 4. The following reactions are known to occur in the

body CO2 + H2O H2CO3 H+ + HCO3

– If CO2 escapes from the system (A) pH will decrease (B) hydrogen ion concentration will diminish (C) H2CO3 concentration will be altered (D) the forward reaction will be promoted 5. The IUPAC name of the spiro compound,

CH3 is (A) 2-Methylspiro [5. 4] deca – 1, 6-diene (B) 2-Methylspiro [4. 5] deca-1, 6-diene (C) 8-Methylspiro [4. 5] deca-1, 7-diene (D) 3-Methylspiro [5. 4] deca-3, 7-diene 6. The total number of acyclic isomers including the

stereoisomers (geometrical and optical) with the molecular formula C4H7Cl is

(A) 12 (B) 11 (C) 10 (D) 9 7. The correct increasing order of extent of hydrolysis

in the following is (A) CCl4 < MgCl2 < AlCl3 < SiCl4 < PCl5 (B) CCl4 < AlCl3 < MgCl2 < PCl5 < SiCl4 (C) AlCl3 < MgCl2 < CCl4 < PCl5 < SiCl4 (D) SiCl4 < MgCl2 < AlCl3 < PCl5 < CCl4 8. Hydrated aluminium chloride is ionic and soluble in

water giving (A) Al3+ and Cl– ions (B) [Al(H2O)6]3+ and Cl– ions (C) [AlCl2 (H2O)4]+ and [AlCl4(H2O)2]– ions (D) None of these 9. A salt on treatment with dil. HCl gives a pungent

smelling gas and a yellow precipitate. The salt gives green flame test and a yellow precipitate with potassium chromate. The salt is

(A) NiSO4 (B) BaS2O3 (C) PbS2O3 (D) CuSO4

Questions 10 to 14 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. 10. For the gas phase reaction C2H4 + H2 C2H6 ∆H = –136.8 kJ mol–1 carried out in a vessel, the equilibrium concentration

of C2H4 can be increased by (A) increasing the temperature (B) decreasing the pressure (C) removing some H2 (D) adding some C2H6 11. Which of the following statement(s) is (are) correct? (A) The pH of 1.0 × 10–8 M solution of HCl is 8 (B) The conjugate base of H2PO4

– is HPO42–

(C) Autoprotolysis constant of water increases with temperature

(D) When a solution of a weak monoprotic acid is titrated against a strong base, at half-neutralisation point pH = (1/2)pKa

12. Which of the following statement(s) is are correct

when a mixture of NaCl and K2Cr2O7 is gently warmed with concentrated H2SO4 ?

(A) A deep red vapour is evolved (B) A vapour when passed into NaOH solution gives

a yellow solution of Na2CrO4 (C) Chlorine gas is evolved (D) Chromyl chloride is formed 13. Which of the following statements are correct ? (A) B2H6 reacts with excess of ammonia at low

temperature to form an ionic substance (B) B2H6 reacts with excess of ammonia at high

temperature to form a white slippery solid called boron nitride

(C) Boron nitride has a layer structure (D) Borazine is an inorganic benzene but has – ve

and + ve charges in it on B and N respectively 14. Which of the following statements is/are correct ? (A) Boric acid is a hydrogen bonded molecule (B) Al2O3 is amphoteric while B2O3 is acidic (C) Boric acid can combine with CuO to give

metaborate and borax bead test (D) Boric acid is a lewis acid This section contains 2 paragraphs, each has 3 multiple choice questions. (Questions 15 to 20) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.


Passage : I (No. 15to 17) Alkali and alkaline earth metals along with hydrogen

and helium constitute s-block elements. They have low ionization enthalpies and hence exhibit charcteristic flame colourration. They have highly negative electrode potentials and hence are strong reducing agents. Their solutions in liquid ammonia are conducting and also act as strong reducing agents. Being stronger reducing agents than hydrogen, they are usually prepared by electrolysis of their fused chlorides. Their oxides are basic and the basic strength increases down the group. The solubility of carbonates and sulphates of alkali and alkaline earth metals show opposite trends. The carbonate of alkaline earth metals and lithium carbonate decompose on heating while the carbonates of other alkali metals do not decompose on heating. The bicarbonates of both alkali and alkaline earth metals on heating give carbonates.

15. The basic chararcter of the oxides, MgO, SrO, K2O,

NiO and Cs2O increases in the order : (A) MgO > SrO > K2O > NiO > Cs2O (B) Cs2O < K2O < MgO < SrO < NiO (C) NiO < MgO < SrO < K2O < Cs2O (D) K2O < NiO < MgO < SrO < Cs2O 16. Which of the following are arranged in increasing

order of solubilites ? (A) CaCO3 < KHCO3 < NaHCO3 (B) NaHCO3 < KHCO3 < CaCO3 (C) KHCO3 < NaHCO3 < CaCO3 (D) CaCO3 < NaHCO3 < KHCO3 17. The following compounds have been arranged in

order of their increasing thermal stabilities. Identify the correct order :

K2CO3(I), MgCO3(II), CaCO3(III), BeCO3(IV) (A) I < II < III < IV (B) IV < II < III < I (C) IV < II < I < III (D) II < IV < III < I Passage :II (No. 18to 20) The expression for the reaction quotient, Q, is similar

to that for equilibrium constant K. The value of Q for the given composition of a reaction mixture helps us to know whether the reaction will move forward or backward or remain in equilibrium. It also helps to predict the effect of pressure on the direction of the gaseous reaction. In certain reactions, addition of inert gas also favours either the formation of reactants or products. The value of equilibrium constant of a reaction changes with change of temperature and the change is given by van't Hoff equation, d ln Kp/dT = ∆Hº/RT2 where enthalpy change, ∆Hº, is taken as constant in the small temperature range.

18. The reaction N2(g) + 3H2(g) 2NH3(g) is in equilibrium. Now the reaction mixture is compressed to half the volume

(A) More of ammonia will be formed (B) Ammonia will dissociate back into N2 and H2 (C) There will be no effect on equilibrium (D) Equilibrium constant of the reaction will change

19. For the above reaction in equilibrium, helium gas was added but the mixture was allowed to expand to keep the pressure constant. Then

(A) More of ammonia will be formed (B) Ammonia will dissociate back into N2 and H2 (C) There will be no effect on equilibrium (D) Equilibrium constant of the reaction will change

20. Which of the following will be correct ? (A) Plot of ln Kp versus 1/T2 will be linear with +ve

slope (B) Plot of ln Kp versus 1/T will be linear with +ve

slope (C) Plot of ln Kp versus 1/T2 will be linear with – ve

slope (D) Plot of ln Kp versus 1/T will be linear with – ve

slope The section contains 2 questions (Questions 21 to 22). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S, then the correctly 4 × 4 matrix should be as follows :

DCBA

PPPP

Q Q Q Q

R R R R

S S S S Q P R S

21. Column –I Column II (A) KHCO3 (P) Exists in solid state (B) NaHCO3 (Q) Soluble in water (C) LiHCO3 (R) Hydrogen bending (D) NH4.HCO3 (S) Dimeric anion

22. Column –I Column II (A) pKb of X– (Ka of HX = 10–6) (P) 6.9 (B) pH of 10–8 M HCl (Q) 8 (C) pHof 10–2M acetic acid solution (R) 3.3 (Ka = 1.6 × 10–5) (D) pH of a solution obtained by (S) 3.4 mixing equal volumes of solution with pH 3 & 5.


MATHEMATICS

Questions 1 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. If the circumcentre of a triangle lies at the origin and

the centroid is the middle point of the line joining the points (a2 + 1, a2 + 1) and (2a, – 2a); then the orthocentre lies on the line

(A) y = (a2 + 1)x (B) y = 2ax (C) x + y = 0 (D) (a – 1)2x – (a + 1)2y = 0 2. The point (4, 1) undergoes the following

transformation successively. (i) reflection about the line y = x (ii) translation through a distance 2 units along the

positive direction of x-axis. (iii) rotation through an angle π/4 about the origin in

the anticlockwise direction. (iv) reflection about x = 0 The final position of the given point is

(A) (1/ 2 , 7/2) (B) (1/2, 27 )

(C) (1/ 2 , 7/ 2 ) (D) (1/2, 7/2) 3. C1 is a circle with centre at the origin and radius

equal to r and C2 is a circle with centre at (3r, 0) and radius equal to 2r. The number of common tangents and can be drawn to the two circles are

(A) 1 (B) 2 (C) 3 (D) 4 4. An equation of the chord of the circle x2 + y2 = a2

passing through the point (2, 3) farthest from the centre is

(A) 2x + 3y = 13 (B) 3x – y = 3 (C) x – 2y + 4 = 0 (D) x – y + 1 = 0 5. Equation of a common tangent to the curves y2 = 8x

and xy = – 1 is (A) 3y = 9x + 2 (B) y = 2x + 1 (C) 2y = x + 8 (D) y = x + 2 6. If P is a point on the rectangular hyperbola

x2 – y2 = a2, C is its centre and S, S´ are the two foci, then SP.S´P =

(A) 2 (B) (CP)2 (C) (CS)2 (D) (SS´)2 7. The reflection of the point P(1, 0, 0) in the line

21x − =

31y

−+ =

810z + is

(A) (3, –4, –2) (B) (5, –8, –4) (C) (1, –1, –10) (D) (2, –3, 8)

8. If the planes x = cy + bz, y = az + cx and z = bx + ay pass through a line, then a2 + b2 + c2 + 2abc is equal to

(A) –1 (B) 0 (C) 1 (D) None 9. The values of k for which the points A(1, 0, 3),

B(–1, 3, 4), C(1, 2, 1) and D(k, 2, 5) are coplanar, are (A) 1 (B) 2 (C) 0 (D) –1 Questions 10 to 14 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. 10. If a chord of the circle x2 + y2 – 4x – 2y – c = 0 is

trisected at the points (1/3, 1/3) and (8/3, 8/3), then

(A) Length of the chord = 27 (B) c = 20 (C) radius of the circle 25 (D) c = 25 11. The Cartesian equation of the curve whose

parametric equation is x = 2t – 3 and y = 4t2 – 1 is given by

(A) (x + 3)2 – y – 1 = 0 (B) x2 + 6x – y + 8 = 0 (C) (y + 1)2 + x + 3 = 0 (D) y2 + 6x – 2y + 4 = 0 12. Equation of a tangent passing through (2, 8) to the

hyperbola 5x2 – y2 = 5 is (A) 3x – y + 2 = 0 (B) 3x + y + 14 = 0 (C) 23x – 3y – 22 = 0 (D) 3x – 23y + 178 = 0

13. The coordinates of a point on the line 2

1x − = 31y

−+ = z

at a distance 144 from the point (1, –1, 0) are (A) (9, –13, 4)

(B) ( 148 + 1, –12 14 – 1, 144 ) (C) (–7, 11, –4)

(D) (– 148 + 1, 1412 – 1, – 144 ) 14. The point of intersection of the lines l1 : r(t) = (i – 6j + 2k) + t(i + 2j + k) l2 : R(u) = (4j + k) + u(2i + j + 2k) (A) at the tip of r(7) (B) at the tip of R(4) (C) (8, 8, 9) (D) at the tip of R(2)



Passage : I (No. 15 to 17) C1 : x2 + y2 – 4 = 0 C2 : x2 + y2 – 14x + 40 = 0 15. P is a point on C1 farthest from the circle C2. An

equation of a pair of tangents from P to C2 is (A) x2 + 8y2 – 4x + 4 = 0 (B) x2 – 8y2 + 4x + 4 = 0 (C) x2 – 8y2 + 4x – 4 = 0 (D) None of these 16. Angle of intersection θ between the tangents from P

to C2 is given by

(A) tan–1(4/9) (B) tan–1( 22 /3)

(C) tan–1( 24 /9) (D) tan–1(2) 17. Equation of the pair of lines through the centre of C2

perpendicular to the pair of tangents from P to C2 is (A) 72x2 – 9y2 – 1008x + 3538 = 0 (B) 72x2 + 9y2 + 1008x – 3538 = 0 (C) 72x2 – 9y2 – 26x – 3346 = 0 (D) None of these Passage : II (No. 18 to 20) P(2, 3, –4), b = 2i – j + 2k 18. Vector equation of a plane passing through the point

P perpendicular to the vector br

is (A) r.(2i – j + 2k) = –7 (B) r.(2i – j + 2k) = 7 (C) r.(2i + 3j – 4k) = –7 (D) r.(2i + 3j – 4k) = 7 19. Cartesian equation of a plane π passing through the

point with position vector b and perpendicular to the

vector OP , O being the origin is (A) 2x – y + 2z + 7 = 0 (B) 2x – y + 2z – 7 = 0 (C) 2x + 3y – 4z + 7 = 0 (D) 2x + 3y – 4z – 7 = 0 20. Sum of the lengths of the intercepts made by the

plane π on the coordinate axes is (A) 14 (B) 91/12 (C) 9/7 (D) 5/7 The section contains 2 questions (Questions 21 to 22). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct

matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S, then the correctly 4 × 4 matrix should be as follows :

DCBA

PPPP

Q Q Q Q

R R R R

S S S S Q P R S

21. y2 = 4ax be the equation of a parabola, then Column –I Column II (A) yy1 = 2a(x + x1) (P) Equation of the normal at (x1, y1) (B) xy1 = 2a(y1 – y) + x1y1 (Q) Equation of the focal chord through (x1, y1) (C) xy1 = y(x1 – a) + ay1 (R) Equation of the line through (x1, y1) and the point of intersection of axis with the directrix (D) (x + a)y1 = (x1 + a)y (S) Equation of the tangent at (x1, y1) 22. Column –I Column II

(A) 3

2x − = 4

7y − =2

5z + (P) Perpendicular to the

plane 3x + 4y + 2z = 1

(B) 3

1x + = 4

2y − = 2

7z − (Q) Passes through

(2, 7, –5)

(C) 1

5x − = 3

2y + =4

2z − (R) direction cosines are

2/ 30 , 5/ 30 , 1/ 30

(D) 2

1x − = 5

1y + =1

1z + (S) lies in the plane

7x – y – z = 35

Ability

• We can accomplish almost anything win tin our ability if we but think we can.

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• Our work is the presentation of our capabilities.

• The wind and the waves are always on the side of the ablest navigator.


XtraEdge Test Series ANSWER KEY

PHYSICS Ques 1 2 3 4 5 6 7 8 9 10 Ans A A D A C C C C C B,C Ques 11 12 13 14 15 16 17 18 19 20 Ans B,C,D A,B,C,D A,D A,B B A B C C B 21 A → Q, S B → P,S C → R,S D → P, S 22 A → P,S B → P,Q C → P D → Q

CHEMISTRY

Ques 1 2 3 4 5 6 7 8 9 10 Ans B C D B C A B B D B,C Ques 11 12 13 14 15 16 17 18 19 20 Ans A,B,C A,B,C,D A,B ,C A,B A B C B B A 21 A → P B → Q, R C → S D → P 22 A → R B → S C → Q D → P

MATHEMATICS Ques 1 2 3 4 5 6 7 8 9 10 Ans B C D B C C A C A A,B,C Ques 11 12 13 14 15 16 17 18 19 20 Ans B,C B,C A,C A,B ,C A C D A B B 21 A → Q B → R C → P D → S 22 A → R B → Q C → P D → S

pH

YIIT- JEE 2011 (July issue)

SI PHYSICS

Ques 1 2 3 4 5 6 7 8 9 10 Ans C B A B B C C D B D Ques 11 12 13 14 15 16 17 18 19 20 Ans C,D D B,C A,D C D D D C B 21 A → R B → P C → Q D → S 22 A → S B → Q C → R D → P

CHEMISTRY Ques 1 2 3 4 5 6 7 8 9 10 Ans B A C B B A A C B A,B,C ,D Ques 11 12 13 14 15 16 17 18 19 20 Ans B,C A,B ,D A,B ,C ,D A,B ,C,D C D B A B D 21 A → P,Q,R,S B → P, Q, R C → Q D → P 22 A → Q B → P C → S D → R

MATHEMATICS Ques 1 2 3 4 5 6 7 8 9 10 Ans D C C A D B B C D A,B ,C Ques 11 12 13 14 15 16 17 18 19 20 Ans A,B A,C A,C A,B,C B C A A C B 21 A → S B → P C → Q D → R 22 A → Q B → P C → S D → R

IIT- JEE 2010 (July issue)

July-2009

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