XtraEdge for IIT JEE by mahaveer saini

WORRY IS A MISUSE OF IMAGINATION.

Volume - 7 Issue - 6 December, 2011 (Monthly Magazine) Editorial / Mailing Office : 112-B, Shakti Nagar, Kota (Raj.) Tel. : 0744-2500492, 2500692, 3040000 e-mail : xtraedge@gmail.com

Editorial

Editor : Pramod Maheshwari [B.Tech. IIT-Delhi]

Cover Design Satyanarayan Saini

Layout Rajaram Gocher

Circulation & Advertisement Praveen Chandna Ph 0744-3040000, 9672977502

Subscription Himanshu Shukla Ph. 0744-3040000 © Strictly reserved with the publishers • No Portion of the magazine can be

published/ reproduced without the written permission of the publisher • All disputes are subject to the exclusive jurisdiction of the Kota Courts only. Every effort has been made to avoid errors or omission in this publication. Inr spite of this, errors are possible. Any mistake, error or discrepancy noted may be brought to our notice which shall be taken care of in the forthcoming edition, hence any suggestion is welcome. It is notified that neither the publisher nor the author or seller will be responsible for any damage or loss of action to any one, of any kind, in any manner, there from.

Unit Price ` 20/Special Subscription Rates 6 issues : ` 100 /- [One issue free ] 12 issues : ` 200 /- [Two issues free] 24 issues : ` 400 /- [Four issues free] Owned & Published by Pramod Maheshwari, 112, Shakti Nagar, Dadabari, Kota & Printed by Naval Maheshwari, Published & Printed at 112, Shakti Nagar, Dadabari, Kota.

Editor : Pramod Maheshwari XtraEdge for IIT-JEE

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

DECEMBER 2011

Volume-7 Issue-6 December, 2011 (Monthly Magazine) NEXT MONTHS ATTRACTIONS

CONTENTS INDEX

PAGE

Regulars ..........

Much more IIT-JEE News. Know IIT-JEE With 15 Best Questions of IIT-JEE

NEWS ARTICLE

Challenging Problems in Physics,, Chemistry & Maths

IIT-B'S Plan : Job support to Entrepreneurial Spirit IIT Students' Research Work to Be Available Online

Key Concepts & Problem Solving strategy for IIT-JEE. Xtra Edge Test Series for JEE- 2011 & 2012 CBSE Mock Test Paper

IITian ON THE PATH OF SUCCESS

Mr. Amitabha Ghosh

KNOW IIT-JEE

Previous IIT-JEE Question

Study Time........ DYNAMIC PHYSICS S

Success Tips for the Months • "The way to succeed is to double your error rate."

8-Challenging Problems [Set# 8] Students’ Forum Physics Fundamentals Ray Optics Fluid Mechanics & Properties of Matter

CATALYSE CHEMISTRY

• "Success is the ability to go from failure to failure without losing your enthusiasm."

• Along with success comes a reputation for wisdom. • They can, because they think they can. • Nothing can stop the man with the right mental attitude from achieving his goal; nothing on earth can help the man with the wrong mental attitude. • Keep steadily before you the fact that all true success depends at last upon yourself.

XtraEdge for IIT-JEE

Key Concept Carboxylic Acid Chemical Kinetics Understanding : Inorganic Chemistry

• "Success is the maximum utilization of the ability that you have." • We are all motivated by a keen desire for praise, and the better a man is, the more he is inspired to glory.

DICEY MATHS

Mathematical Challenges Students’ Forum Key Concept Monotonicity, Maxima & Minima Function

Test Time .......... XTRAEDGE TEST SERIES

Class XII – IIT-JEE 2012 Paper Class XI – IIT-JEE 2013 Paper Mock Test CBSE Pattern Paper -1 [Class # XII]

DECEMBER 2011

IIT-B'S Plan : Job support to Entrepreneurial Spirit

IIT Students' Research Work to Be Available Online

IIT-Bombay has hit upon an idea that could boost the spirit of entrepreneurship among its students. Its placement cell is weighing the option of helping students whose start-ups have not fired to be placed in jobs after two years of experimenting with their ideas.

Students of Indian Institutes of Technology (IITs) will soon share their research details for all institutes via a common website to be launched in November.

As part of the scheme, students keen on their own start-ups will be assigned mentors after graduating. These experts-either people who have successfully floated their own companies or those with enough exposure to new businesses-will evaluate their ideas to see if there's any potential. Once the ideas are approved, students can float their own companies. After two years, if a start-up fails to take off, the student-entrepreneur can participate in the regular placement process and get a job. Ravi Sinha, professor in charge of placements, said the idea, which is at a nascent stage, can give students the assurance to float their ideas without hesitation. "Very few start-up ideas on the campus turn out to be successful ventures. Often, many good ideas are not commercially-viable. So, students are apprehensive about floating their ideas," added Sinha. The office of Society for Innovation and Entrepreneurship ( SINE), which promotes entrepreneurship on the campus by promoting business incubation, and the Entrepreneurship cell (E-cell) is working with the placement cell to check the feasibility of the project. "We have been working on the initiative for a couple of months. If it works out, studententrepreneurs will be reasonably assured of getting a job through the institute's placement office," said Sinha. XtraEdge for IIT-JEE

The two day long 'Inter-IIT Gymkhana Summit 2011' being held at IITGandhinagar saw students of various consortium discuss issues common to all. The students included representatives from 10 IITs including Madras, Mandi, Rajasthan, Mandi, Hyderabad, with IIT Kanpur and IIT Bombay participating in online sessions. Among the things discussed, students particularly discussed methods to share research works of B.Tech, M.Tech and Phd students. Students are seen to repeat researches, being unaware of past researchers. Sharing of past researches will help avoid such repetitions.

Sport's Alumni Student Meet The 2nd official annual Alumni Sports Day was full of tremendous competition between the energy of the present generation and that of the alumni. The event was full of activities comprising of talks, alumni -students matches and Inter-Sports Arm Wrestling and Tug of war competition which was followed by dinner and group discussions in SAC Lawns. “Good, better, best. Never let it rest. Until your good is better and your better is best. ” this quote perfectly fit the enthralling display of experience, wit and never say die attitude of the alumni at this year’s Alumni Sports Day. Brainchild of the basketball fraternity of IIT Bombay, this was the 2nd official annual Alumni Sports Day which 3

witnessed tremendous competition between the energy of the present generation and that of the alumni. The event was full of activities comprising of talks, alumni -students matches and Inter-Sports Arm Wrestling and Tug of war competition which was followed by dinner and group discussions in SAC Lawns. The post match fundae sessions (be it related to the life post IITs or the prospects and benefits of sports in everyday life) were undoubtedly the highlight of the complete event. The commitment and dedication shown by the Old Machine Guns from IITB left the students sports fraternity speechless. Alumni showed the perfect blend of attitude, experience and sporting spirit with some exceptional performances in all the sports. Results of this year’s meet like last year were pretty much dominated by the alumni, winning in almost all the sports.

IITians' Nano Satellite Launched IIT-Kanpur has made the 3kg satellite 'Jugnu'. It will generate real time data on droughts, floods, vegetation and forestation. Jugnu, the Nano satellite designed by IIT Kanpur students has been launched via PSLV C-18 rocket from Shrihari Kota, Andhra Pradesh 11:00 hrs today, reports TNN. This is the country's first Nano satellite and weighs just 3 Kgs incorporated with PSLV C-18. The mission of developing Jugnu was done under the guidance of Prof. N.S vyas. this is extremely a complex process and this was a hard exercise. The Nano satellite has been designed ethnically. "We have done our best. Jugnu has undergone laborious tests like thermo DECEMBER 2011

vacuum tests, vibration tests and electronic magnetic interference test. Students have worked very hard in developing Jugnu. They have worked round-the-clock. We are hopeful of its successful launch," said Vyas. The payload of the satellite has uniquely designed camera for infra red remote sensing, a GPS receiver and an inertial measurement unit. Shatanu Agarwal, the team head of Jugnu said "About 25 minutes after the launch, Jugnu would will click the first image of its launch vehicle and its antennas would be deployed. It will then stabilize and start transmitting images to the ground station." After 115 minutes of the take off Jugnu was visible at the ground station. "Jugnu's beacon (blinking signal at all times all over the earth) will get switched on after 30 minutes of separation from the launch vehicle. Five seconds after its separation, Jugnu will click its image," said Shantanu. It took two years to develop this Nano satellite and as people working on it used non space grade, commercial off the shelf (COTS) to make the research on Jugnu affordable. Jugnu has minimum number of redundant systems when compared to other conventional satellites.

SC refuses to adjudicate on IITJEE selections NEW DELHI: The Supreme Court has refused to interfere with the ranking and selection procedure adopted for the IIT-JEE saying there was no arbitrariness or ulterior motives in fixing the methodology, says a PTI report. A bench of Justices R.V. Raveendran and A.K. Patnaik said courts would interfere with the procedure only if there was proven malafide, caprice or arbitrariness, which it said was lacking in the present system adopted by the Joint Admission Board, which conducts the exams. “The fact that the procedure was complicated would not make it XtraEdge for IIT-JEE

arbitrary or discriminatory,” said.

unreasonable or Justice Raveendran

The apex court passed the judgment while dismissing an appeal filed by aspirant Sanchit Bansal, son of an IIT Kharagpur professor, who appeared in IIT-JEE 2006 as a general category candidate. Sanchit had secured 75 marks in mathematics, 104 in physics and 52 in chemistry, aggregating to 231. The board had fixed the cut-off marks for admission at 37 for math, 48 for physics and 55 for chemistry and the aggregate cut-off at 154. As Sanchit did not secure the minimum cut-off in chemistry, he failed to qualify even though his aggregate was higher than required. He then challenged the procedure on the ground that candidates with aggregates lower than his were selected. Rejecting his plea, the court said: “For a layman, the above procedure may appear to be highly cumbersome and complicated. But the object of the aforesaid procedure for arriving at the cut-off marks is to select candidates well equipped in all the three subjects, with reference to their merit, weighed against the average merit of all the candidates who appeared in the examination.”

IIT Council free from HRD ministry clutches NEW DELH: The apex decision making body of the Indian Institutes of Technology (IITs) has broken free of the Human Resource Development (HRD) Ministry in a move that could be the first step towards allowing the IITs to govern themselves. Empowered by independent staff and with a identity of its own, the IIT Council will now no longer need the HRD ministry to take its administrative decisions under the move, government sources told HT. The final decision on the plan -- aimed at creating an IIT Council Secretariat -4

may be made at a meeting between senior HRD ministry officials and the IIT directors on May 5, the sources said. The Council alone is empowered to appoint IIT directors and take any policy decisions binding across the top engineering schools -- including their fees, reforms, administrative structure and any amendments to the aw governing the Institutes. “The IIT Council -the highest decision making body of the IITs -at present depends on the HRD ministry to even invite members for meetings or to prepare the Council's agenda. All this will change,“ a source said. The IITs have decided to approach D Udaya Kumar, assistant professor at IIT Guwahati and the designer of the new rupee symbol, to design a logo for the IIT Council. The move is a component of HRD minister Kapil Sibal's larger plan to increase the functional autonomy of the IITs. The IIT Council -consisting of all IIT directors and chairman of boards, other eminent academic administrators and scientists -- at present does not even have an office of its own. It will now have an office - a location has been identified in South Delhi's Chittaranjan Park. But this may only be the first step towards greater autonomy for the IITs, sources indicated. “Once the independent IIT Council is capable of handling matters, there is a possibility that we will empower it with far greater powers and withdraw from many administrative aspects of the IIT governance system,“ a source said. (Courtesy : Hindustan Times)

No permanent foreign faculty for IITs NEW DELHI: In a major setback to the Indian Institutes of Technology (IITs) plan, the Ministry for External Affairs (MEA) has rejected a proposal to liberalise visa norms to allow foreign teachers to take up permanent posts at the IITs. DECEMBER 2011

The MEA has refused to change the rules under which each foreign faculty member at the IITs needs to re-obtain a work visa every five years, top government and IIT sources have confirmed to HT. Human resource development minister Kapil Sibal had on September 11, 2010 announced the plan to allow the IITs to fill up to 10% of their permanent teaching posts with foreign faculty. The proposal -first reported by HT on September 2, 2010 -was approved by the IIT Council -the highest decision making body of the IITs -and is aimed at reducing a massive faculty crunch plaguing the IITs. But the MEA's refusal to allow foreign faculty to join with visas of longer duration than five years means that the IITs will not be able to offer permanent posts to foreign faculty. “We will need to continue to offer contractual appointments something we wanted to, and quite frankly, need to change,“ an IIT Director said. Each IIT is facing a faculty crunch between 15 and 40% with a total of over 1,000 faculty posts vacant across the premier engineering schools. The Institutes have over the past year however received a number of applications from foreign faculty, including Persons of Indian Origin (PIOs) keen to teach at the IITs. The IITs are arguing that permanent posts would help them lure the best of foreign teachers. All foreign teachers are at present required to teach as visiting or ad-hoc faculty. (Courtesy : Hindustan Times)

Sibal rejects steep fee hike for IIT students NEW DELHI: There will be no steep fee hike for the students of the Indian Institutes of Technology, according to a decision taken by the IIT Council on January 21. Chairing the IIT Council meet here, Human Resource Development Minister Kapil Sibal rejected the Anil Kakodkar committee proposal for fiveXtraEdge for IIT-JEE

fold increase in fee for undergraduate programme of the IITs. The Kakodkar committee, set up by the government in October 2009 to study the roadmap for the autonomy and future of the IITs, had recommended that the fee be raised from Rs 50,000 to Rs 2 to Rs 2.5 lakh per annum. As the committee report came for discussion at the 42nd meeting of the IIT Council Sibal rejected the fee hike proposal saying “such a hike would prove a deterrent to a large number of IIT aspirants,” a ministry official said. The Council asked the committee to rework the fee structure taking into account the aspirations of all sections. During the meeting, Sibal announced setting up 50 research parks at a cost of Rs 200 crore during the 12th Five Year Plan period. Under the programme, industry will undertake research on various subjects with the support of experts from the IITs. The research parks have been proposed to be set up on public-privatepartnership (PPP) model. One such research park has already come up in Chennai. The meeting took note of the fact that credit-based practices were being followed by the IITs to promote students from one semester to the next, and agreed that academic bodies of the IITs should consider acquisition of credits as a criteria for students and granting of degrees to bring uniformity. The issue came up following submission of Dhande committee report on uniform and homogeneous criteria for promoting students in the IITs. Kanpur IIT director Dhande, who headed two committees, presented reports on a “uniform criteria for promoting students from one semester to the next in the IITs and on the “requirement of infrastructure for research”. Both reports have been accepted. Each IIT at present has its own criteria for promotion. The Council decided that a panel for visitor’s nominee for a particular department would be created which all IITs could use for faculty selection.

“This will ensure timely selection of professors,” the Council noted. It also decided that the appointment of directors should be through advertisements so that a wider base was created. “It was decided that in principle approval may be granted for setting up an institute in Mauritius with the help of the IITs,” an official said. At the meeting, a presentation was made on adopting cyber security as part of the curriculum for the IITs. So it was decided that a committee be set up to develop a roadmap for the future and give a report in next three months. “The committee would involve all educational institutions as well as government departments,” the HRD Ministry official said. The meeting also could not discuss reform in the Joint Entrance Examination and curriculum as T Ramaswamy, secretary, Department of Science & Technology, was not present. Ramaswamy had prepared a report on the two issues. "The Indian Statistics Institute has arrived at a formula for equalising marks in all boards. If one board gives 90 percent as highest marks, and the other gives 75 percent, the marks will be equalized on the basis of a formula," Sibal said. Producing more research scholars was one of the key issues taken up during the meeting, with the council deciding to enhance the capacity of IITs to produce 10,000 Ph.D. graduates annually from around 1,000 presently and increase faculty strength from around 4,000 presently to 16,000 by 2020. IANS

Science Fact:•

The brain uses over a quarter of the oxygen used by the human body.

•

Your heart beats around 100000 times a day, 36500000 times a year and over a billion times if you live beyond 30. DECEMBER 2011

Success Story This article contains storie/interviews of persons who succeed after graduation from different IITs

Dr. Amitabha Ghosh â&#x20AC;˘

Post graduation in applied geology from IIT Kharagpur,

â&#x20AC;˘

Working at NASA

Dr Amitabha Ghosh was the only Asian on NASA's Mars Pathfinder mission. At present, he is a member of the Mars Odyssey Mission and the Mars Exploration Rover Mission.

How India can we develop science and technology sector : It should be treated as a business. There should be more private participation. We must have an external review to evaluate the system and make changes as science and technology can take the country forward.

During the Mars Pathfinder Mission, he conducted chemical analysis of rocks and soil on the landing site. The simple and unassuming 34-year-old planetary geologist has won several accolades, which include the NASA Mars Pathfinder Achievement Award in 1997 and the NASA Mars Exploration Rover Achievement Award in 2004.

We must check brain drain. About 80,000 students migrate to the US for further studies, and settle there. They find the facilities much better abroad. We need to reverse brain drain by enhancing and upgrading institutes in India.

The journey from India to NASA.

The state of space research in India :

It has been an intriguing experience. I was keen on geologic research data interpretation and solar system formation. During my geological research days in India, I had slept in railway stations while traveling to various places.

I don't want to make controversial statements. All I can say is India is not at the frontier of space research. We have made commendable progress but there is a long way to go. We can do much better. I would be glad to be of help in any way. Investment in research is investment in imagination. It is a matter of national pride and internal recognition. We need to allocate more funds to enhance research and development work.

After my post graduation in applied geology from IIT Kharagpur, I wrote a letter to a professor at NASA expressing a desire to work at the space agency.

My suggestions were approved, while I got an opportunity to work at NASA.

We need good educational institutes like IITs and IIMs, but IITians don't rule the world. You must remember that Microsoft co-founder (Bill Gates does not have a college degree.

I think one requires luck and to put in sincere effort to achieve one's goals. Being in the right place at the right time is also important.

Youngsters must look around for role models and see what it is that they are doing right. Individuals must make use of their inherent strengths to succeed.

I made certain suggestions; in fact, it was a critical letter. In India, you can never imagine criticising your professor.

How can India become a leading global player :

In Mumbai for the Pravasi Bharatiya Divas, he spoke about his work at NASA and his vision for India.

Globalisation will reap huge and long-term benefits and India must make the best use of the opportunities. At the PBD seminar, I found people presenting grandiose plans. Instead, we should look at the realities and immediate solutions.

The Vision for India : I feel there India has a great future. We have world-class companies. Today, companies like Infosys can be compared with world leaders like Oracle. Like the Information Technology revolution, we can have a science or space revolution. We have the potential to bring about revolutions in other sectors as well. XtraEdge for IIT-JEE

The private sector has to be actively involved in the development of the country and the government has to respond to the needs of the people. Fifteen years ago, we didn't have an Infosys, today we have many global companies. 6

DECEMBER 2011

KNOW IIT-JEE By Previous Exam Questions

PHYSICS 1.

A cubical box of side 1 meter contains Helium gas (atomic weight 4) at a pressure of 100 N/m2. During an observation time of 1 second, an atom travelling with the root-mean-square speed parallel to one of the edges of the cube, was found to make 500 hits with a particular wall, without any 25 collision with other atoms. Take R = J/mol-K 3 [IIT-2002] and k = 1.38 × 10–23 J/K (a) Evaluate the temperature of the gas. (b) Evaluate the average kinetic energy per atom. (c) Evaluate the total mass of helium gas in the box. Sol. The distance travelled by an atom of helium in 1 sec (times between two successive collision) 500 is 2m. Therefore root mean square speed

A non-viscous liquid of constant density 1000 kg/m3 flows in a streamline motion along a tube of variable cross section. The tube is kept inclined in the vertical plane as shown in figure. The area of cross section of the tube two points P and Q at heights of 2 metres and 5 metres are respectively 4 × 10–3 m2 and 8 × 10–3 m2. The velocity of the liquid at point P is 1 m/s. Find the work doen per unit volume by the pressure and the gravity forces at the fluid flows from point P to Q. [IIT-1997] Q

5m 2m

Sol. Given that ρ = 1000 Kg/m3, h1 = 2 m, h2 = 5 m A1 = 4 × 10–3 m2 , A2 = 8 × 10–3 m2 . v1 = 1 m/s Equation of continuity A1 = 4 × 10–3 m2, A2 = 8 × 10–3 m2 , v1 = 1 m/s Equation of continuity A1v1 = A1v2,

∴ v2 =

Crms =

A1v1 = 0.5 m/s A2

dis tan ce 2 = = 1000 m/s time 1 / 500

(a) But Crms =

According to Bernouilli's theorem, 1 (p1 – p2) = ρg (h2 – h1) – ρ (v22 – v12) 2 where (p1 – p2) = work done/vol. [by the pressure] ρg (h2 – h1) = work done/vol. [by gravity forces.] Now, work done/vol by gravity forces

⇒ 1000 =

3RT M

3 × 25 / 3 × T 4 × 10 – 3

⇒ T = 160 K

(b) Average kinetic energy of an atom of a 3 monoatomic gas = kT 2 3 kT = 3.312 × 10–12 Joules 2

= ρg (h2 – h1) = 103 × 9.8 × 3 = 29.4 × 103 J/m3 and

∴ Eav =

3 1 1 1  ρ(v22 – v12) = × 103  – 1 = – × 103 8 2 2 4  J/m3 = – 0.375 × 103 J/m3

m RT M

⇒ m = 0.3012 gm

∴ Work done/vol. by pressure = 29.4 × 103 – 0.375 × 103 J/m3 = 29.025 × 103 m3. XtraEdge for IIT-JEE

DECEMBER 2011

A thin equiconvex lens of glass of refractive index µ = 3/2 and of focal length 0.3 m in air is sealed into an opening at one end of a tank filled with water (µ = 4/3). On the opposite side of the lens, a mirror is placed inside the tank on the tank wall perpendicular to the lens axis, as shown in figure. The separation between the lens and the mirror is 0.8 m. A small object is placed outside the tank in front of the lens at a distance of 0.9 m from the lens along its axis. Find the position (relative to the lens) of the image of the [IIT-1997] object formed by the system. 0.8 m 0.9 m

2   ⇒ R = 0.3 m R Substituting the value of R in equation (iii) we get 4 3 – –1+ 2× 4/3 1 3 2 + = – (0.9) v 0.3

1 = (1.5 – 1) 0.3

⇒

1 4 + = 0.9 3v

–4 – 3 + 9 2 3 = 0.3 0.9

⇒ v = 0.2 m As shown in the figure this image will form as I" behind the mirror. But the ray will get reflected from the mirror in between and the final image formed will be I. Since CI" = 0.2 m and CM = 0.8 m ∴ MI" = 0.2 m

Sol.

⇒ MI = 0.4 m

∴ CI = 0.4 m A 5 m long cylindrical steel wire with radius 2 × 10–3 m is suspended vertically from a rigid support and carries a bob of mass 100 kg at the other end. If the bob gets snapped, calculate the change in temperature of the wire ignoring radiation losses. (For the steel wire : Young's modulus = 2.1 × 1011 Pa ; Density = 7860 kg/m3 ; Specific heat = 420 [IIT-2001] J/kg-K). Sol. When the mass of 100 Kg is attached, the string is under tension and hence in the deformed state. Therefore it has potential energy (U) which is given by the formula. 4.

v O

I M

C u

I''

The first refraction of the ray coming form the object is suffered on the left side of convex lens. For this we can apply the equation µg – µa µg µ = ...(i) – a + u V' R The image formed by this can be treated as a virtual object and the refracting surface now is the right of the convex lens. µw – µg µg µ – + w = ...(ii) v' v –R Adding (i) and (ii) µg µg µg – µa µa µw – + – + = + v' v' u v R µw – µg

U= =

Q = mc∆T ...(ii) From (i) and (iii) Since U = Q Therefore

–µ w – µ a + 2µ g µa µ + w = u v R But according to lens formula

⇒

1 = f

a g

( µ – 1)  R1 a g

XtraEdge for IIT-JEE



–

1 R2

–

1 (Stress) 2 × × πr2l 2 Y

1 (Mg / πr 2 ) 2 1 M 2g 2l × πr2l = ...(i) 2 Y 2 πr 2 Y This energy is released in the form of heat, thereby raising the temperature of the wire

⇒ –

( µ – 1)  R1

1 × stress × strain × volume 2

–R

1 = f

[Q MI = MI"]

∴ mc∆T = 1 R2

  

∴ ∆T =

  

1 M 2g 2l 2 πr 2 Y

1 M 2g 2l 2 πr 2 Ycm

Here m = mass of string = density × volume of string = ρ × πr2l 8

DECEMBER 2011

∴ ∆T = =

M 2g 2 1 2 (πr 2 ) 2 Ycp

⇒ ⇒

(100 × 10) 2 1 × –3 2 2 (3.14 × 2 × 10 ) × 2.1× 1011 × 420 × 7860

CHEMISTRY

= 0.00457º C 5.

An unknown resistance X is to be determined using resistance R1, R2 or R3. Their corresponding null points are A, B and C. Find which of the above will give the most accurate reading and [IIT-2005] why?

R = R1 or R2 or R3

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

r1 r2

where R is a constant r1 and r2 are variable. The maximum fraction error is

X r1 M

 5.0 × 10 −4 4.0 × 10 −3  =  1.0 × 10 −3 1.6 × 10 − 2 

R G B

r2 C

   

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

A R=R1 R=R2 R=R3

∆r ∆r ∆X = 1 + 2 X r1 r2

5.0 × 10 −4

Here ∆r1 = ∆r2 = y (say) then ∆X For to be minimum r1 × r2 should be max X [Q r1 + r2 = c (Constt.)] Let E = r1 × r2 ⇒ E = r1 × (r1 – c) dE = (r1 – c) + r1 = 0 ∴ dr1 XtraEdge for IIT-JEE

Initial rate (mol L–1s–1) 300 K 320 K –4 5.0 ×10 2.0 × 10–3 4.0 × 10–3 – –2 1.6 × 10 –

[B] mol L–1 3.0 ×10–5 6.0 × 10–5 6.0 × 10–5

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. All Null point, the wheat stone bridge will be balanced X R ∴ = r1 r2

⇒X=R

From the following data, form the reaction between A and B. [IIT-1994] [A] mol L–1 2.5 ×10–4 5.0 × 10–4 1.0 × 10–3

G A

c c ⇒ r2 = ⇒ r 1 = r2 2 2 R2 gives the most accurate value

r1 =

4.0 × 10 −3

or or or

 2.5 × 10 −4 =  −4  5.0 × 10

1 1 1 ×  = 8 4 2 1 1 =  2 2 m=1

   

 3.0 × 10 −5   6.0 × 10 −5 

   

DECEMBER 2011

(b) At T1 = 300 K, Rate of reaction k1 = [A]2 [B]1

Hence, difference in energy between first and second Bohr orbit for a H-atom is given by, ∆E = E n i – E n f = E2 – E1

5.0 × 10 −4 = [2.5 × 10 − 4 ]2 [3.0 × 10 −5 ]

=–

hc 6.626 × 10 −34 × 3 × 108 = λ 3.0 × 10 −8 = 6.626 × 10–18 J ....(i) We know that, for H-like atoms, En for H-like atom = En for H-atom × Z2 ∴ ∆E for H-like atom = Z2 × ∆E for H-atom 1 1 = –Z2 × 21.76 × 10–19  2 − 2  1  2 ...(ii) = 16.32 × 10–19 Z2 From eq. (i) and (ii), 16.32 × 10–19 Z2 = 6.626 × 10–18 or Z=2 Thus, hydrogen atom like species for Z = 2 is He+.

∆E =

2.0 × 10 −3 [2.5 × 10 − 4 ]2 [3.0 × 10 −5 ]

= 1.067 × 109 L2 mol–2 s–1 We know, 2.303 log or 2.303 log

E T −T  k2 = a  2 1  R  T1T2  k1

E a  320 − 300  1.067 × 109 =   8.314  320 × 300  2.67 × 108

or 2.303 × 0.6017 =

Ea  20    8.314  320 × 300 

2.303 × 0.6017 × 8.314 × 320 × 300 20 –1 = 55.3 kJ mol (d) According to Arrhenius equation,

An organic compound A, C8H4O3, in dry benzene in the presence of anhydrous AlCl3 gives compound B. The compound B on treatment with PCl5 followed by reaction with H2/Pd(BaSO4) gives compound C, which on reaction with hydrazine gives a cyclised compound D(C14H10N2). Identify A, B, C and D. Explain the formation of D from C. [IIT-2000] Sol. The given reactions are as follows. O O 8.

or Ea =

k = Ae − E a / RT or 2.303 log k = 2.303 log A –

Ea RT

At 300 K, 55.3 × 10 3 8.314 × 300 or 2.303 × 8.4265 = 2.303 log A – 22.17 19.4062 + 22.17 41.5762 = or logA = = 18.0531 2.303 2.303 A = Antilog 18.0531 = 1.13 × 1018 s–1

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

O + O

21.76 × 10 −19 n

AlCl3

OH O

C6H5

PCl5

Estimate the difference in energy between 1st and 2nd Bohr orbit for a H atom. At what minimum atomic no., a transition from n=2 to n = 1 energy level would result in the emission of X-rays with λ = 3.0×10–8 m. Which hydrogen atom like species does this atomic no. corresponds to ? [IIT-1993] Sol. (a) For H atom, Z=1 ni = 2 nf = 1 7.

En = –

21.76 × 10 −19

12 1 1 = – 21.76 × 10–19  2 − 2  = 16.32 ×10–19J 2 1  –8 (b) For λ = 3.0 × 10 m

= 2.67 × 108 L2 mol–2 s–1

21.76 × 10 −19

H2/Pd (BaSO4)

C C

C6H5

O H2NNH2 O

N N

H The formation of D from C may be explained as follows. C6H5 O O

O–

C6H5

NH2

NH2 +

NH2

NH2 O–

C6H5

O– N–H N–H OH C6H5

N N

XtraEdge for IIT-JEE

DECEMBER 2011

9. (a) Write the chemical reaction associated with the "brown ring test". (b) Draw the structures of [Co(NH3)6]3+, [Ni(CN)4]2– and [Ni(CO)4]. Write the hybridization of atomic orbital of the transition metal in each case. (c) An aqueous blue coloured solution of a transition metal sulphate reacts with H2S in acidic medium to give a black precipitate A, which is insoluble in warm aqueous solution of KOH. The blue solution on treatment with KI in weakly acidic medium, turns yellow and produces a white precipitate B. Identify the transition metal ion. Write the chemical reaction involved in the formation of A and B. [IIT-2000] Sol. (a) NaNO3 + H2SO4 → NaHSO4 + HNO3 2HNO3 + 6FeSO4 + 3H2SO4 → 3Fe2(SO4)3 + 2NO + 4H2O [Fe(H2O)6]SO4.H2O + NO Ferrous Sulphate → [Fe(H2O)5NO] SO4 + 2H2O (Brown ring) (b) In [Co(NH3)6]3+ cobalt is present as Co3+ and its coordination number is six. Co27 = 1s1, 2s22p6, 3s23p63d7, 4s2 Co3+ion = 1s2, 2s22p6, 3s23p63d6 3d 4s 4p Hence 3d

Ni N≡C

NH3

Ni OC

H3N 2+

2CuSO4 + 4KI → Cu2I2 + I2 + 2K2SO4 (B) white –

In [Ni(CN)4 nickel is present as Ni coordination numbers is four Ni28 =1s2, 2s22p6, 3s23p63d8, 4s2 Ni2+ ion = 1s2, 2s22p6, 3s23p63d8 3d 4s 2+ Ni ion = 3d

–

I2 + I → I3 (yellow solution) 10. An organic compound A, C6H10O, on reaction with CH3MgBr followed by acid treatment gives compound B. The compound B on ozonolysis gives compound C, which in presence of a base gives 1-acetyl cyclopentene D. The compound B on reaction with HBr gives compound E. Write the structures of A, B, C and E. Show how D is formed [IIT-2000] from C. Sol. The given reactions are as follows. CH3 OMgBr O CH3 Br CH3

NH3

CO CO

Co3+

2–

Its structure is as follows :

H3N

sp3 hybridization

Co H3N

Ni in Complex

d sp hybridization H3N

C≡N

In [Ni(CO)4, nickel is present as Ni atom i.e. its oxidation number is zero and coordination number is four.

Co3+ion in Complex ion 2

C≡N

N≡C

NH3

ion and its

CH3MgBr

H+ –H2O

(A)

(B)

COCH3 Base

Ni2+ion in Complex ion

(D)

HBr

COCH3 O

(E)

CH3 O O (C)

The conversion of C into D may involve the following mechanism.

dsp2 hybridization

Hence structure of [Ni(CN)4]2– is XtraEdge for IIT-JEE

DECEMBER 2011

CH2

B+ –BH+

Solve for x the following equation [IIT-1987] 2 log(2x +3)(6x + 23x + 21) = 4 – log(3x + 7) (4x2 + 12x + 9) Sol. log(2x + 3) (6x2 + 23x + 21) = 4 – log(3x + 7) (4x2 + 12x + 9) ⇒ log(2x + 3) (2x + 3).(3x + 7) = 4 – log(3x + 7) (2x + 3)2 ⇒ 1 + log(2x + 3) (3x + 7) = 4 – 2 log(3x + 7) (2x + 3) Put log(2x + 3) (3x + 7) = y 2 ⇒ y+ –3=0 y

COCH3

O–

12. BH+ –B

(C)

COCH3 –

COCH3 OH +B –BH+

COCH3 OH –OH– (D)

MATHEMATICS 11.

y2 – 3y + 2 = 0 ⇒ (y – 1) (y – 2) = 0 y = 1 or y=2 log(2x +3) (3x + 7) = 1 log(2x + 3) (3x + 7) = 2 3x + 7 = 2x + 3 or (3x + 7) = (2x + 3)2 x = – 4 or 3x + 7 = 4x2 + 12x + 9 4x2 + 9x + 2 = 0 4x2 + 8x + x + 2 = 0 (4x + 1)(x + 2) = 0 x = – 2, – 1/4 ∴ x = – 2, – 4, – 1/4 ...(i) 2 But, log exists only when, 6x + 23x + 21 > 0, 4x2 + 12x + 9 > 0, 2x + 3 > 0 and 3x + 7 > 0 ⇒ x > – 3/2 ...(ii) ∴ x = – 1/4 is the only solution.

⇒ ⇒ ⇒ or ⇒ ⇒

Let the three digit numbers A28, 3B9, and 62C, where A, B and C are integers between 0 and 9, be divisible by a fixed integer K. Show that the determinant A

C is divisible by K.

[IIT-1990]

Sol. We know, A28 = A × 100 + 2 × 10 + 8 3B9 = 3 × 100 + B × 10 + 9 and 62C = 6 × 100 + 2 × 10 + C Since; A28, 3B9 and 62C are divisible by K, therefore there exist positive integers m1, m2 and m3 such that, 100 × A + 10 × 2 + 8 = m1K 100 × 3 + 10 × B + 9 = m2K ...(i) and, 100 × 6 + 10 × 2 + C = m3K

∴

∆= 8 2

Find the equation of the normal to the curve y = (1 + x)y + sin–1 (sin2x) at x = 0. [IIT-1993] Sol. y = (1 + x)y + sin–1 (sin2x) (given) Let y = u + v, where u = (1 + x)y, v = sin–1 (sin2 x). Differentiationg dy du dv = + ...(i) dx dx dx Now, u = (1 + x)y take logarithm of both sides loge u = loge(1 + x)y ⇒ loge u = y loge(1 + x) 1 du y dy = ⇒ . {loge(1 + x)} + u dx 1+ x dx 13.

Applying R2 → 100R1 + 10R3 + R2 3

⇒

∆ = 100 A + 2 × 10 100 × 3 + 10 × B + 9 2 B 6 100 × 6 + 10 × 2 + C 2

A 3 6 = A28 3B9 62C , Using (i) 2 B 2 A

= m1 K 2

m3 K = K m1

3 m2 K

du dy  y  + log e (1 + x) =u 1 + x dx dx  

⇒

du dy  y  + log e (1 + x) = (1 + x)y  dx 1 + x dx 

...(ii)

v = sin–1sin2 x Again, ⇒ sin v = sin2x

∴ ∆ = mK, Hence determinant is divisible by K. XtraEdge for IIT-JEE

⇒

DECEMBER 2011

cos v

⇒

1 dv [2sin x cos x] = dx cos v

⇒

π/3

dv = 2 sin x cos x dx

⇒

2 sin x cos x 2

2π / 3

π/3

...(iii)

t dt 2 = 2π 2 t π / 3 1 + 3 tan 2

1 – sin x

∫

dy dy  y  2 sin x cos x + log e (1 + x) + = (1 + x)y  1 + dx x dx   1 – sin 4 x

dy y (1 + x) y –1 + 2 sin x cos x / 1 – sin 4 x = dx 1 – (1 + x) y ln (1 + x)

⇒

1/ 3

= ∴

dy =1 dx

π + 4x3 dx. π  – π / 3 2 – cos | x | +  3 

∫

[IIT-2004]

Sol. π/3

π/3

∫

 0,  a (Using f ( x)dx =  2 f ( x)dx,  –a  0 a

∫

π/3

∴

I= 2

∫ 0

{ 3 tan

(tan–1 3 – tan–1 1) =

–1

}

3 1 3

4π

1 tan–1   3 2

4π π + 4 x3 1 tan –1   . dx = π  3 2 – π / 3 2 – cos | x | +  3  

∫

πdx x 3dx + 4 π π   – π / 3 2 – cos | x | +  – π / 3 2 – cos | x | +  3 3  

∫

4π . 3

The position vectors of the vertices A, B and C of a tetrahedron ABCD are i + j + k, i and 3i, respectively. The altitude from vertex D to the opposite face ABC meets the median line through A of the triangle ABC at a point E. If the length of the side AD is 4 and the 2 2 volume of the tetrahedron is , find the position 2 vector of the point E for all its possible positions. [IIT-1996] i + 3i = 2i and Sol. F is mid-point of BC i.e., F = 2 AE ⊥ DE (given) A(i+j+k) D

π/3

Let, I=

15.

Thus, the required equation of the normal is y – 1 = (–1) (x – 0) i.e., y + x – 1 = 0.

Evaluate

4π

1 + 3u

π/3

dy 1(1 + 0)1–1 + 2 sin 0. cos 0 / (1 – sin 4 0) = dx 1 – (1 + 0)1 ln (1 + 0)

Again the slope of the normal is 1 =–1 m=– dy / dx

14.

2du

∫

= 2π

At x = 0, y = (1 + 0)y + sin–1 sin (0) = 1 ⇒

sec 2

2π / 3

Put these values in equation (i)

⇒

dt π , where x + =t 2 – cos t 3

∫

= 2π

2 sin x cos x

1 – sin v

∫ 2 – cos( x + π / 3)

I = 2π

E 1

f (– x) = – f ( x)  f (– x) = f ( x)   

B(i)

C(3i)

Let E divides AF in λ : 1. Then position vector of E is given by

πdx +0 π  2 – cos | x | +  3 

2 λi + 1(i + j + k ) 1 1  2λ + 1  = j+ k i+ λ +1 λ +1 λ +1  λ +1  Now, volume of the tetrahedron 1 = (area of the base) (height) 3

   π/3  3 x dx   is odd  as π    – π / 3 2 – cos | x | +     3 

∫

XtraEdge for IIT-JEE

F(2i)

⇒

2 2 1 = (area of the ∆ ABC) (DE) 3 3

DECEMBER 2011

But area of the ∆ ABC = =

1 → → | BC × BA | 2

WHAT ARE EARTHQUAKES?

1 |2i × (j + k)| = |i × j + i × k| 2

= |k – j| =

1 2 2 = ( 2) (DE) 3 3 ⇒ DE = 2 Since ∆ ADE is a right angle triangle, AD2 = AE2 + DE2 ⇒ (4)2 = AE2 + (2)2 ⇒ AE2 = 12 → 2λ + 1 1 1 But AE = i+ j+ k – (i + j + k) λ +1 λ +1 λ +1

Therefore,

= ⇒

Earthquakes like hurricanes are not only super destructive forces but continue to remain a mystery in terms of how to predict and anticipate them. To understand the level of destruction associated with earthquakes you really need to look at some examples of the past.

λ λ λ i– j– k λ +1 λ +1 λ +1

→

| AE | 2 =

Therefore, 12 =

1 (λ + 1) 2

[λ2 + λ2 + λ2] =

3λ2

If we go back to the 27th July 1976 in Tangshan, China, a huge earthquake racked up an official death toll of 255,000 people. In addition to this an estimated 690,000 were also injured, whole families, industries and areas were wiped out in the blink of a second. The scale of destruction is hard to imagine but earthquakes of all scales continue to happen all the time.

(λ + 1) 2

3λ2 (λ + 1) 2

⇒ 4(λ + 1)2 = λ2 ⇒ 4λ2 + 4 + 8λ = λ2 ⇒ 3λ2 + 8λ + 4 = 0 ⇒ 3λ2 + 6λ + 2λ + 4 = 0 ⇒ 3λ(λ + 2) + 2 (λ + 2) = 0 ⇒ (3λ + 2) (λ + 2) = 0 ⇒ λ = – 2/3, λ = – 2 Therefore, when λ = – 2/3, position vector of E is given by

So what exactly are they ? Well the earths outer layer is made up of a thin crust divided into a number of plates. The edges of these plates are referred to as boundaries and it’s at these boundaries that the plates collide, slide and rub against each other. Over time when the pressure at the plate edges gets too much, something has to give which results in the sudden and often violent tremblings we know as earthquakes.

1 1  2λ + 1  j+ k  i+ λ + 1 λ +1 λ + 1  

2.(–2 / 3) + 1 1 1 i+ j+ k – 2 / 3 +1 – 2 / 3 +1 – 2 / 3 +1

The strength of an earthquake is measured using a machine called a seismograph. It records the trembling of the ground and scientists are able to measure the exact power of the quake via a scale known as the richter scale. The numbers range from 1-10 with 1 being a minor earthquake (happen multiple times per day and in most case we don’t even feel them) and 7-10 being the stronger quakes (happen around once every 10-20 years). There’s a lot to learn about earthquakes so hopefully we’ll release some more cool facts in the coming months.

–4 / 3 + 1 1 1 i+ j+ k – 2+3 – 2+3 – 2+3 3 3 3 –4 + 3 3 i+ 1 j+ 1 k = 1/ 3 1/ 3 1/ 3 = – i + 3j + 3k and when λ = – 2, Position vector of E is given by, 2 × (–2) + 1 1 1 –4 + 1 i+ j+ k= i–j–k – 2 +1 – 2 +1 – 2 +1 –1 = 3i – j – k Therefore, – i + 3j + 3k and +3i – j – k are the answer. =

XtraEdge for IIT-JEE

DECEMBER 2011

Physics Challenging Problems

Set # 8

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

Solutions will be published in next issue Passage # 1 (Que. 1 to 3) 1.

When barium is irradiated by a light of λ = 4000Å all the photoelectrons emitted are bent in a circle of radius 50cm by a magnetic field of 5.26 × 10-6 T. Then (A) the kinetic energy of fastest photoelectrons is 0.6eV (B) work function of the metal is 2.5eV (C) the maximum velocity of photoelectrons is 0.46 × 106 m/s (D) the stopping potential for photoelectric effect is 0.6V

Suppose the potential energy between on electron ke 2 . and a proton at a distance r is given by 3r 3 Application of Bohr’s theory to hydrogen atom in this case shows that (A) Energy in nth orbit is proportional to n6 (B) Energy is proportional to m-3 (m = mass of electron) (C) Energy in nth orbit is proportional to n-2 (D) Energy is proportional to m3 (m = mass of electron)

In the circuit shown C1 = 6µF, C 2 = 3µF and battery E = 20V. The switch S1 is first closed. It is then opened and afterwards S2 is closed them the final charge on C2 is C2 C1

S2 S1

E = 20 V

(A) 120µC

(B) 80µC

(D) 20µC

A current in a coil of self-inductance 2H is increasing as i = 2 sin( t 2 ) . The amount of energy spent during the period when current changes from 0 to 2A is (A) 1J (B) 2J (C) 3J (D) 4J A particle of charge q and mass m is projected with a velocity v forwards a circular region having uniform magnetic field B perpendicular and into the plane of paper from point P as shown. R is the radius and O is the centre of the circular region. If the line OP makes an angle θ with the direction of v then the value of v so that particle passes through O is

q m

qBR m sin θ 2qBR (C) m sin θ

(A)

Passage # (Q. No. 6 to Q. No. 8) An ideal gas is taken round a cyclic process ABCA as shown. If the internal energy of the gas at point A is assumed zero while at B it is 50J. The heat absorbed by the gas in the process BC is 90J. 30 P(N/m2)

× × × × × × O× × × R × × v × θ× × P × × qBR (B) 2m sin θ 3qBR (D) 2m sin θ

XtraEdge for IIT-JEE

20 A

B E D O 3 1 2 V(m)3 Heat energy absorbed by the gas in process AB is (A) 50J (B) 70J (C)30J (D) None 10

Heat energy rejected by gas in process CA is (A) 180J (B) 140J (C) 220J (D) None

The net work done by gas in the complete cycle ABCA is (A) 20J (B) zero (C) 40J (D) -20J DECEMBER 2011

Solution

Physics Challenging Problems Qu e s tio ns we r e Pub lis he d in Nov emb er I ss ue B 0 cos r = A 0 (1 + cos φ + cos 2φ) B 0 sin r = (A 0 (sin φ + sin 2φ))

3R R I and I DC = I 4R 4R Option [C] is correct I AB =

∴ B 02 = A 02 (1 + 4 cos φ + 4 cos 2 φ) ∴ I θ = I 0 (1 + 4 cos φ + 4 cos 2 φ)

L.∆I = area under v – t curve 1 2[i − 0] = × 10 x 2 2 Option [A,C] is correct

∴ (I θ ) max = 9I 0

Iθ 1 1 = = (1 + 4 cos φ + 4 cos 2 φ) (I θ ) max 2 9 ⇒ cos φ = 0.56

also

3. By conservation of energy   2    gR   − GMm  1 GMm 1   + = mV 2 +  × m   2   R  2 R   2   R+  4  by conservation of angular momentum gR R  .R sin θ = mV R +  m. 4 2  Option [A] is correct

Set # 7

A → P; B → R ; C → Q; D → S v net =

5 v0 = 4

5  GM    4  r0 

5 GM −3GMm −GMm 1 + m× . = r0 2 4 r0 8r0 when particle is at maximum or minimum distance r, then energy conservation GMm 3GMm 1 …….(1) =− mv 2 − r 8r0 2 Angular momentum conservation mvr 5 …….(2) = m. v 0 .r0 4 ∴ total energy = −

2q 2  1 1 1 1  − + − + + ..... 4π ∈0 d  d 2d 3d 4d  Option [C] is correct U=

Passage Based Question: 5. Option [A] is correct 6. Option [D] is correct

Science Jokes

7. Option [A] is correct Let A and w be the amplitude and angular frequency of the wave incident on the three slit grating. Let φ be

A chemistry professor couldn't resist interjecting a little philosophy into a class lecture. He interrupted his discussion on balancing chemical equations, saying, "Remember, if you're not part of the solution, you're part of the precipitate!"

the phase difference between the diffracted waves from S1 and S2 and φ between those from S2 and S2. If θ is angle of diffraction then using. 2πd sin θ φ= λ A 1 = A 0 sin ωt

A 2 = A 0 sin(ωt + φ) A 3 = A 0 sin(ωt + 2φ) A = A 1 + A 2 + A 3 = A 0 [sin ωt (1 + cos φ + cos 2φ) + cos ωt (sin φ + sin 2 φ)]

1 couldn't resist interjecting a little philosophy into a class lecture. He interrupted his discussion on balancing chemical equations, saying, "Remember, if you're not part of the solution, you're part of the precipitate!".

Q. What is volume of a person who lost all his memory ? A. 1/3 πr2h Because he keeps on saying, “main CONE hu!"

A = B 0 sin(ωt + r ) where

XtraEdge for IIT-JEE

DECEMBER 2011

XtraEdge for IIT-JEE

DECEMBER 2011

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

PHYSICS 1.

Oxygen is used as working substance in an engine working on the cycle shown in Figure 2

Heat

gas during the process, 7 nRT0 Q23 = nCP(T3 – T2) = 2 Work done by gas during the process, W23 = nR(T3 – T2) = nRT0 Now considering adiabatic process 3 → 4, V3 = 2V0 , T3 = 2T0 V4 = 5V0 , T4 = ? γ–1 Using T.V = constant (2T0) (2V0)γ – 1 = T4(5V0)γ – 1 or T4 = 2 (0.4)0.4 T0 Work done by the gas during the process, P V – P4 V4 nR (T3 – T4 ) W24 = 3 3 = γ –1 γ –1

4 1

V Processes 1-2, 2-3, 3-4 and 4-1 are isothermal, isobaric, adiabatic and isochoric, respectively. If ratio of maximum to minimum volume of oxygen during the cycle is 5 and that of maximum to minimum absolute temperature is 2, assuming oxygen to be an ideal gas, calculate efficiency of the engine. Given, (0.4)0.4 = 0.693 and loge 5 1.6094 Sol. Volume of gas is minimum at state 2 during the cycle. Let it be V0. Then maximum volume of gas during the cycle will be equal to 5V0 which is at states 4 and 1. Therefore, V4 = V1 = 5V0. Temperature during the cycle is maximum at the end of isobaric process 2 → 3 i.e. state 3 and minimum at the end of isochoric cooling process 4 → 1 i.e. state 1. Let minimum absolute temperature be T0. Then T1 = T0 and T3 = 2T0. Since gas is Oxygen which is di-atomic, therefore, 7 5 7 Cv = R, Cp = R and γ = 5 2 2

= 5 nRT0 (1 – (0.4)0.4) T0 During isochoric process 4 → 1, no work is done by the gas and heat is rejected from the gas. Hence, W41 = 0 and Q41 is negative ∴ Net work done by the gas during the cycle, W = W12 + W23 + W34 + W41 = nRT0 {6 – loge 5 – 5 × (0.4)0.4} Heat supplied to the gas during heating process, 7 QS = Q23 = ( 6 – loge 5 – 5 × 0.40.4) = 0.2642 2 2 W or η = = ( 6 – loge 5 – 5 × 0.40.4) = 0.2642 7 Qs

Since, process 1 → 2 is isothermal, therefore, temperature during the process remains constant. Hence temperature T2 is also equal to T0. Considering n mole of the gas, Work done by the gas during isothermal process 1 → 2, V W12 = nRT1.log 2 = – nRT0 loge 5 V1

or η = 26.42 %

Ans.

Calculate speed of sound in a mixture containing n1 mole of He, n2 mole of H2 and n3 mole of CO2 at temperature T (i) when n1 = 1, n2 = 2, n3 = 3 and T = 434 K, (ii) when n1 = 3, n2 = 2, n3 = 1 and T = 415 K Gas constant, R = 8.3 J (mole)–1 K–1 Sol. Sound waves in gases are longitudinal waves and speed of longitudinal waves in a gas is given by

But for isothermal process Q = W, therefore, Q12 = – nRT0 loge 5 Now considering isobaric process 2 → 3 T V3 = 3 =2 or V3 = 2V0 T2 V2

XtraEdge for IIT-JEE

supplied

γRT . M Hence, to calculate speed of sound waves in the mixture, its mean molar mass M and adiabatic exponent γ must be known.

DECEMBER 2011

Molar mass of He is M1 = 4 gm/mole and its molar 3 R. specific heat at constant volume is C v1 = 2 5 These for H2 are M2 = 2 gm/mole and C v 2 = R 2 respectively and for CO2, M3 = 44 gm/mole and C v3 = 3R respectively. Mean molar mass of the mixture, M m + m 2 n 2 + M 3n 3 M= 1 1 n1 + n 2 + n 3 and

Sol. First, incident waves are received by the wall and then reflected back. These reflected waves are received by the source. First, consider incident waves received by the wall. Since, source and wall (observer) both are approaching towards each other, therefore, frequency received by the wall is given by,  v + v0  v+u  = n0  n1 = n0   = 272 Hz  v – v v–u s   Now waves are reflected by the wall. For reflected waves wall works as a sonic source of frequency n1 = 272 Hz, which is approaching towards the receiver R with velocity u. Receiver is also approaching the wall (source) with the same speed. Therefore, frequency n2 of reflected waves received by the receiver is given by

...(i)

Mean molar specific heat at constant volume, n1C v1 + n 2 C v 2 + n 3C v3 Cv = ...(ii) n1 + n 2 + n 3

(i) substituting n1 = 1, n2 = 2, n3 = 3 in equation (i) and (ii) 140 140 M= gm/mole or × 10–3 kg/mole 6 6 31 Cv = R 12 43 But Cp = Cv + R, therefore, Cp = R 12 C 43 ∴ Adiabatic exponent, γ = P = 31 Cv

∴

Speed of longitudinal waves =

 v + v0  v+u  = n1  n2 = n1  Ans.  = 289 Hz  v–u  v – vs  Since, receiver is approaching the wall with velocity u to receive reflected waves, therefore, velocity of wave propagation relative to the receiver is v' = (v + u) = 340 ms–1. Let wavelength of reflected waves by λ'. 20 Using, v' = n2λ' or λ' = m Ans. 17

γRT M

Both ends of a solid cylindrical glass rod of diameter d = 10 cm are made hemispherical. When a luminous object is placed on axis of the rod at a distance a = 20 cm from one end, its real image is obtained at a distance b = 40 cm from the other end. If the refractive index of glass µ = 1.5. Calculate length l of the rod. Sol. Since image is real, therefore, it is formed by convergence of transmitted rays. If object is placed on left of left end of the rod, image will be on right of the right end of rod. Since diameter of the rod is d and ends are made hemispherical, radius of curvature of each end is equal to d/2. First considering refraction at left end, d u = – a, µ1 = 1, µ2 = µ = 1.5, R = + , v = v1 (let) 2 l 4.

= 214140 ms–1 or 462.75 ms–1 Ans. (ii) Substituting n1 = 3, n2 = 2, n3 = 1 in equation (i) and (ii) M = 10 gm/mole or 10 × 10–3 kg/mole 25 37 R and Cp = Cv + R = R Cv = 12 12 Cp 37 γ= = Cv 25 Speed of longitudinal waves

γRT M

= 83 74 ms–1 or 714 ms–1 3.

Ans.

A sonic source of natural frequency n0 = 256 Hz and a receiver are moving along the same line with speed u = 10 ms–1 towards a reflecting surface. Their line of motion is normal to the surface and the surface is also approaching towards them with the same speed as shown in figure. If speed of sound in air is v = 330 ms–1, calculate frequency and wavelength of reflected waves received by the receiver. u u S

(µ) O

µ2 µ µ – µ1 – 1 = 2 v u R v1 = + 30 cm It means that image I1 formed after refraction at left end is on right of left end at a distance of 30 cm. Using formula,

R u

XtraEdge for IIT-JEE

DECEMBER 2011

This image works as an object for refraction at the other end. Then, for refraction at right end, µ1 = 1.5, µ2 = 1, v = + b = + 40 cm, d R=– = – 5 cm, u = ? 2 µ µ µ – µ1 Substituting in 2 – 1 = 2 v u R u = – 20 cm Negative sign indicates that object for right end is on its left or image I1 is on left of right end of the rod and at a distance of 20 cm. ∴ Length of the rod = v1 + 20 cm = 50 cm Ans. 5.

dφ 1 = π ωa2 B sin (ωt) dt 2 since, resistance of the circuit is negligible, therefore, Potential difference across capacitor is equal to induced emf in the circuit. ∴ Charge on the capacitor at time t is q = C.e. 1 = πωa2 CB sin (ωt) 2 dq 1 = π ω2 a2 CB cos (ωt) Ans. But current I = dt 2 Due to flow of current, semicircle experience a moment. Therefore, power is required to keep the semi circle rotating with constant angular velocity. In fact, power required to rotate the semicircle is equal to electrical power generated in the circuit. 1 ∴ Power required, P = e.I = π2ω3a4 CB2 sin (2ωt) 8 Ans.

e=–

A copper rod is bent into a semi-circle of radius a and at ends straight parts are bent along diameter of the semi-circle and are passed through fixed, smooth and conducting rings O and O' as shown in figure. A capacitor having capacitance C is connected to the rings. The system is located in a uniform magnetic field of induction B such that axis of rotation OO' is perpendicular to the field direction. At initial moment of time (t = 0), plane of semi-circle was normal to the field direction and the semi-circle is set in rotation with constant angular velocity ω. Neglecting resistance and inductance of the circuit, calculate current I flowing through the circuit at time t and instantaneous power required to rotate the semi-circle. O

Dimensional Formulae of Some Physical Quantities

×B C

Sol. When the copper rod is rotated, flux linked with the circuit varies with time. Therefore, an emf is induced in the circuit. At time t, plane of semi-circle makes angle ωt with the plane of rectangular part of the circuit. Hence, component of the magnetic induction normal to plane of semi circle is equal to B.cos ωt. ∴ Flux linked with semicircular part is 1 φ1 = πa2.B cos ωt 2 Let area of rectangular part of the circuit be A. Then flux linked with this part is φ2 = BA ∴ Total flux linked with the circuit is φ = φ1 + φ2 1 or φ = πa2 B.cos(ωt) + B.A 2 ∴ Induced emf in the circuit,

XtraEdge for IIT-JEE

Work (W) Stress

Dimensional Formulae [ML2T–2] [ML–1T–2]

Torque (τ)

[ML2T–2]

Moment of Inertia (I)

[ML2]

Coefficient of viscosity (η)

[ML–1T–1]

Gravitational constant (G)

[M–1L3T–2]

Specific heat (S)

[L2T–2θ–1]

Coeficient of thermal conductivity (K)

[MLT–3θ–1]

Universal gas constant (R)

[ML2T–2θ–1]

Potential (V)

[ML2T–3A–1]

Intensity of electric field (E)

[MLT–3A–1]

Permittivity of free space (ε0)

[M–1L–3T4A2]

Specific resistance (ρ)

[ML3T–3A2]

Magnetic Induction (B)

[MT–2A–1]

Planck's constant (h)

[ML2T–1]

Boltzmann's constant (k)

[ML2T–2θ–1]

Entropy (S)

[ML2T–2θ–1]

Decay constant (λ)

[T–1]

Bohr magnetic (µB)

[L2A]

Thermmionic current density (J)

[AL–2]

Physical Quantity

DECEMBER 2011

P HYSICS F UNDAMENTAL F OR IIT-J EE

Ray Optics KEY CONCEPTS & PROBLEM SOLVING STRATEGY

Reflection : Key Concepts : (a) Due to reflection, none of frequency, wavelength and speed of light change. (b) Law of reflection : Incident ray, reflected ray and normal on incident point are coplanar. The angle of incidence is equal to angle of reflection Incident n Reflected Ray Ray θ θ

Real n Object

Tangent at point P

P Convex surface

Plane surface

α α

P Virtual Object

Real Object

P Real Object

n n

If the plane mirror is rotated through an angle θ, the reflected ray and image is rotated through an angle 2θ in the same sense. If mirror is cut into a number of pieces, then the focal length does not change. The minimum height of mirror required to see the full image of a man of height h is h/2. Rest v

Object v

vsinθ

Virual Object

Object

v θ

vcosθ

If reflected beam or refracted beam from an optical instrument is diverging in nature, image is virtual. Object v

XtraEdge for IIT-JEE

θ θ

α α

A Tangent at point P

Some important points : In case of plane mirror For real object, image is virtual. For virtual object, image is real. The converging point of incident beam behaves as a object. If incident beam on optical instrument (mirror, lens etc) is converging in nature, object is virtual. If incident beam on optical instrument is diverging in nature, the object is real. The converging point of reflected or refracted beam from an optical instrument behaves as image. If reflected beam or refracted beam from an optical instrument is converging in nature, image is real.

Virual Object

αα Convex surface

For solving the problem, the reference frame is chosen in which optical instrument (mirror, lens, etc.) is in rest. The formation of image and size of image is independent of size of mirror. Visual region and intensity of image depend on size of mirror. P P'

n θ θ

n αα

Image

Rest

vsinθ

vcosθ Image

vm Image

2vm–v

DECEMBER 2011

vm Image

Object In rest

2vm

y x

x' y'

The mirror formula is v

Object

vm Image

360º θ

β=

where θ = angle between mirrors.

−v image size = u object size

1. If object and image both are real, β is negative.

360º If is even number, the number of images is θ n – 1.

2. If object and image both are virtual, β is negative. 3. If object is real but image is virtual, β is positive. 4. If object is virtual but image is real, β is positive. 5. Image of star; moon or distant object is formed at focus of mirror. If y = the distance of sun or moon from earth. D = diameter of moon or sun's disc f = focal length of the mirror d = diameter of the image

360º is odd number and object is placed on θ bisector of angle between mirror, then number of images is n – 1.

360º is odd and object is not situated on θ bisector of angle between mirrors, then the number of images is equal to n. (d) Law of reflection in vector form : Let eˆ1 = unit vector along incident ray. If

θ = the angle subtended by sun or moon's disc Then tan θ = θ =

D d = y f

eˆ 2 = unit vector along reflected ray

Here, θ is in radian.

nˆ = unit vector along normal on point of Incidence eˆ 2 = eˆ1 − 2(eˆ1.nˆ ) nˆ Then

Sun D

eˆ 2

x y'

XtraEdge for IIT-JEE

x' y'

Problem solving strategy : Image formation by mirrors Step 1: Identify the relevant concepts : There are two different and complementary ways to solve problems involving image formation by mirrors. One approach uses equations, while the other involves drawing a principle-ray diagram. A successful problem solution uses both approaches. Step 2: Set up the problem : Determine the target variables. The three key quantities are the focal length, object distance, and image distance; typically

(e) Spherical mirrors : It easy to solve the problems in geometrical optics by the help of co-ordinate sign convention. y y y x'

F d

n nˆ

eˆ1

1 1 1 + = v u f

Also, R = 2f These formulae are only applicable for paraxial rays. All distances are measured from optical centre. It means optical centre is taken as origin. The sign conventions are only applicable in given values. The transverse magnification is

2vm+v

(c) Number of images formed by combination of two plane mirrors : The images formed by combination of two plane mirror are lying on a circle whose centre is at the meeting points of mirrors. Also, object is lying on that circle. Here, n =

DECEMBER 2011

you'll be given two of these and will have to determine the third. Step 3: Execute the solution as follows : The principal-ray diagram is to geometric optics what the free-body diagram is to mechanics. In any problem involving image formation by a mirror, always draw a principal-ray diagram first if you have enough information. (The same advice should be followed when dealing with lenses in the following sections.) It is usually best to orient your diagrams consistently with the incoming rays traveling from left to right. Don't draw a lot of other rays at random ; stick with the principal rays, the ones you know something about. Use a ruler and measure distance carefully ! A freehand sketch will not give good results. If your principal rays don't converge at a real image point, you may have to extend them straight backward to locate a virtual image point, as figure (b). We recommend drawing the extensions with broken lines. Another useful aid is to color-code the different principal rays, as is done in figure(a) & (b). Q I 4 2 C P' F

2 4

Note that the same sign rules (given in section) work for all four cases in this chapter : reflection and refraction from plane and spherical surfaces. Step 4: Evaluate your answer : You've already checked your results by using both diagrams and equations. But it always helps to take a look back and ask yourself. "Do these results make sense ?".

Refraction : Laws of Refraction :

The incident ray, the refracted ray and normal on incidence point are coplanar. µ1 sin θ1 = µ2 sin θ2 = ... = constant. θ1

µ1 µ2 θ2

Snell's law in vector form : nˆ

v eˆ1

eˆ 2

(a) Q 2 4

eˆ 2 = unit vector along refracted.

Q' v

P' F

nˆ = unit vector along normal on incidence point.

Then µ1( eˆ1 × nˆ ) = µ2( eˆ 2 × nˆ )

4 (b)

Some important points :

1 1 1 + = and the s s' f s' y' magnification equation m = = − . The s y results you find using this equation must be consistent with your principal-ray diagram; if not, double-check both your calculation and your diagram. Pay careful attention to signs on object and image distances, radii or curvature, and object and image heights. A negative sign on any of these quantities always has significance. Use the equations and the sign rules carefully and consistently, and they will tell you the truth !

(a) The value of absolute refractive index µ is always greater or equal to one.

Check your results using Eq.

XtraEdge for IIT-JEE

µ2

Let, eˆ1 = unit vector along incident ray

1 1 3 2

µ1

(b) The value of refractive index depends upon material of medium, colour of light and temperature of medium. (c) When temperature increases, refractive index decreases. (d) Optical path is defined as product of geometrical path and refractive index. i.e., optical path = µx

DECEMBER 2011

(e) For a given time, optical path remains constant. i.e., µ1x1 = µ2x2 = ... constant ∴

dx dx µ1 1 = µ2 2 dt dt

∴

µ1c1 = µ2c2

∴

µ2 c = 1 µ1 c2

i.e.,

µ∝

The apparent depth due to a number of media is Σ

Critical angle : When a ray passes from denser medium (µ2) to rarer medium (µ1), then for 90º angle of refraction, the corresponding angle of incidence is critical angle.

1 c

Mathematically,

∴

c1 = fλ1,

∴

λ µ1 c = 2 = 2 µ2 c1 λ1

∴

1 µ∝ λ

c2 = fλ2

µ1 µ2

µ  δ = sin–1  2 sin i  – i  µ1 

for i < c

(ii) When angle of incidence is greater than critical angle, total internal reflection takes place. The corresponding deviation is

real depth apparent depth

δ = π – 2i

when i < c

The δ – i graph is :

When object is in rarer and observer is in denser medium : µ=

sin c =

(i) When angle of incidence is lesser than critical angle, refraction takes place. The corresponding deviation is

When observer is rarer medium and object is in denser medium : µ=

ti µi

The lateral shifting due to a slab is d = t sec r sin(i – r).

(where c1 and c2 are speed of light in respective mediums)

(f) The frequency of light does not depend upon medium.

Then

ti t t = 1 + 2 + ... µi µ1 µ2

(i) Critical angle depends upon colour of light, material of medium, and temperature of medium.

apparent position real position

(ii) Critical angle does not depend upon angle of incidence

 1 The shift of object due to slab is x = t 1 –   µ

(a) This formula is only applicable when observer is in rarer medium.

(b) The object shiftiness does not depend upon the position of object.

µ2 µ µ − µ1 – 1 = 2 v u r

The equivalent refractive index of a combination of a

Here,

π/2

Refractive surface formula,

number of slabs for normal incidence is µ =

Σt i t Σ i µi

Here, v = image distance, u = object distance, r = radius of curvature of spherical surface.

Σti = t1 + t2 + ...

(a) For plane surface , r = ∞ (b) Transverse magnification,

XtraEdge for IIT-JEE

DECEMBER 2011

(b) This formula is only applicable when medium on both sides of lens are same.

µv Im age size = 1 object size µ2u

proportional

square

(d) When lens is placed in a medium whose refractive index is greater than that of lens. i.e., µ1 > µ2. Then converging lens behaves as diverging lens and vice versa.

Lens : Lens formula : 1 1 1 – = v u f

(e) When medium on both sides of lens are not same. Then both focal lengths are not same to each other.

(a) Lens formula is only applicable for thin lens. (b) r = 2f formula is not applicable for lens. (c) m =

v image size = u object size

(f) If a lens is cut along the diameter, focal length does not change.

(d) Magnification formula is only applicable when object is perpendicular to optical axis.

(g) If lens is cut by a vertical, it converts into two lenses of different focal lengths.

(e) lens formula and the magnification formula is only applicable when medium on both sides of lenses are same.

i.e.,

(h) If a lens is made of a number of layers of different refractive index number of images of an object by the lens is equal to number of different media.

(f) f(+ve)

f(–ve)

(i)

(ii)

1 1 1 = + f f1 f2

(i) The minimum distance between real object and real image in is 4f. (j) The equivalent focal length of combination of two lenses is given by

f(+ve)

f(–ve)

1 1 1 d = + – F f1 f 2 f1f 2

(iv)

(iii)

co-axial

(k) If a number of lenses are in contact, then

f(–ve)

f(+ve)

(v)

(vi)

1 1 1 = + + ...... F f1 f 2

(l) (i) Power of thin lens, P =

(g) Thin lens formula is applicable for converging as well diverging lens. Thin lens maker's formula :  µ − µ1  1  =  2 f  µ1 

(ii) Power of mirror is P = –

1 1  −   r1 r2 

µ1

1 F

(m) If a lens silvered at one surface, then the system behaves as an equivalent mirror, whose power

µ1

P = 2PL + Pm

µ2

Here, PL = Power of lens  µ − µ1   =  2  µ1 

(a) Thin lens formula is only applicable for paraxial ray. XtraEdge for IIT-JEE

1 F

1 1  −   r1 r2 

DECEMBER 2011

Pm = Power of silvered surface = –

1 Fm

Now

1 1 1 + = 20 25 v1 v1 = 100 cm. As v1 is positive, hence the image is real. In the absence of convex mirror, the rays after reflection from concave mirror would have formed a real image I1 at distance 100 cm from the mirror. Due to the presence of convex mirror, the rays are reflected and appear to come from I2. (ii) For convex mirror, In this case, I1 acts as virtual object and I2 is the virtual image. The distance of the virtual object from the convex mirror is 100 – 50 = 50 cm. Hence u2 = –50 cm. As focal length of convex mirror is negative and hence f2 = –30/2 = –15 cm. Here we shall calculate the value of v2. Using the mirror formula, we have 1 1 1 − + = − 15 50 v 2 or v2 = –21.42 cm As v2 is negative, image is virtual. So image is formed behind the convex mirror at a distance of 21.43 cm.

Here, Fm = r2/2, where r2 = radius of silvered surface.

P = – 1/F Here, F = focal length of equivalent mirror.

Solved Examples Rays of light strike a horizontal plane mirror at an angle of 45º. At what angle should a second plane mirror be placed in order that the reflected ray finally be reflected horizontally from the second mirror. Sol. The situation is shown in figure G C D A S θ θ 1.

45º

1 1 1 + = f1 u1 v1

45º Q B The ray AB strikes the first plane mirror PQ at an angle of 45º. Now, we suppose that the second mirror SG is arranged such that the ray BC after reflection from this mirror is horizontal. From the figure we see that emergent ray CD is parallel to PQ and BC is a line intersecting these parallel lines. So, ∠DCE = ∠CBQ = 180º ∠DCN + ∠NCB + ∠CBQ = 180º θ + θ + 45º = 180º ∴ θ = 67.5º As ∠NCS = 90º, therefore the second mirror should be inclined to the horizontal at an angle 22.5º. P

There is a small air bubble in side a glass sphere (n = 1.5) of radius 10 cm. The bubble is 4 cm below the surface and is viewed normally from the outside (Fig.). Find the apparent depth of the air bubble. A

n1 = 1.5

Sol. The observer sees the image formed due to refraction at the spherical surface when the light from the bubble goes from the glass to air. Here u = – 4.0 m, R = – 10 cm, n1 = 1.5 and n2 = 1

50cm

P2 C

We have P1

25cm

r = 30 cm

r = 40 cm

[(n2/v) – (n1/u) = (n2 – n1)/R

(1/v) – (1.5/ –4.0 cm) = (1 – 1.5)/ (– 10 cm)

(1/v) = (0.5/10 cm) – (1.5/4.0 cm)

v = – 3.0 cm

Thus, the bubble will appear 3.0 cm below the surface.

(i) For concave mirror, u1 = 25 cm, f1 = 20 cm and v1 = ? XtraEdge for IIT-JEE

n2 = 1

O C

An object is placed exactly midway between a concave mirror of radius of curvature 40 cm and a convex mirror of radius of curvature 30 cm. The mirrors face each other and are 50 cm apart. Determine the nature and position of the image formed by the successive reflections, first at the concave mirror and then at the convex mirror. Sol. The image formation is shown in figure. 2.

P I

DECEMBER 2011

A convex lens focuses a distance object on a screen placed 10 cm away from it. A glass plate (n = 1.5) of thickness 1.5 is inserted between the lens and the screen. Where should the object be placed so that its image is again focused on the screen ? Sol. The situation when the glass plate is inserted between the lens and the screen, is shown in fig. The lens forms the image of object O at point I1 but the glass plate intercepts the rays and forms the final image at I on the screen. The shift in the position of image after insertion of glass plate Screen

DEEPEST LAKE IN THE WORLD ?

Lake Baikal (Baikal) in Siberia, Russia is the deepest lake in the world measuring 1620m deep at its deepest point. This makes it not only deep but also the oldest lake in the world estimated to be around 25 million years old. At over 636 kilometers long and 80 kilometers wide this fresh water lake holds over 20 percent of all the fresh water in the world and is second in size only to the Caspian Sea (the caspian is called a sea but is technically a lake).

10 cm

1   1  I1I = t 1 −  = (1.5 cm) 1 −  = 0.5 cm.  n  1.5 

Thus, the lens forms the image at a distance of 9.5 cm from itself. Using 1 1 1 – = , we get v u f

To put things into perspective the lake is so big that if all the rivers in the world flowed into its basin it would take almost 1 year to fill. We all know Siberia isn’t the warmest of places so you can imagine what a phenomenal site it is when in the winter months the lake freezes over holding ice up to 115 meters thick. Now that’s a lot of ice!

1 1 1 1 1 = – = – u v f 9.5 10

or u = – 190 cm. i.e. the object should be placed at a distance of 190 cm. from the lens.

WHICH IS THE HIGHEST WATERFALL IN THE WORLD ?

A candle is placed at a distance of 3 ft from the wall. Where must a convex lens of focal length 8 inches be placed so that a real image is formed on the wall ? Sol. According to formula for refraction though a lens 36 – v

f = 8" d = 3 ft = 36"

1 1 1 – = v u f or

1 1 1 = + v 36 − v 8

1 1 1 = – v − (36 − v) 8

36 − v + v 1 = v(36 − v) 8

or, v2 – 36 v + 8 × 36 = 0 or v = 12" or

The highest waterfall in the world is the Angel Falls in Venezuela. At a towering height of 979m did you know that each drop of water takes 14 seconds to fall from the top to the bottom. The water flows from the top of a “Tepui” which is a flat topped mountain with vertical sides. The waterfall which despite being known to the local indians for thousands of years was originally called the “Churun Meru” but for some reason they were renamed by an American bush pilot called Jimmy Angel, who noticed them in 1935 whilst flying over the area looking for gold.

24" = 1 ft or 2 ft.

∴ u = 24" or 12" = 2 ft or 1 ft Hence, lens should be placed at either 1 ft or 2 ft away from the wall.

XtraEdge for IIT-JEE

DECEMBER 2011

P HYSICS F UNDAMENTAL F OR IIT-J EE

Fluid Mechanics & Properties of Matter KEY CONCEPTS & PROBLEM SOLVING STRATEGY Fluid dynamics :

Fluid Mechanics :

1 2 p v + gh + = a constant ρ 2 for a streamline flow of a fluid (liquid or gas). Here, v is the velocity of the fluid, h is its height above some horizontal level, p is the pressure and ρ is the density. p1

Fluid statics :

Bernoulli's Theorem :

Pressure at a point inside a Liquid : p = p0 + ρgh

where p0 is the atmospheric pressure, ρ is the density of the liquid and h is the depth of the point below the free surface. p0 h

v2 h2 v2 > v1

Pressure is a Scalar : The unit of pressure may be atmosphere or cm of mercury. These are derived units. The absolute unit of pressure is Nm–2. Normal atmospheric pressure, i.e, 76 cm of mercury, is approximately equal to 105 Nm–2. Thrust : Thrust = pressure × area. Thrust has the unit of force. Laws of liquid pressure (a) A liquid at rest exerts pressure equally in all directions. (b) Pressure at two points on the same horizontal line in a liquid at rest is the same. (c) Pressure exerted at a point in a confined liquid at rest is transmitted equally in all directions and acts normally on the wall of the containing vessel. This is called Pascal's law. A hydraulic press works on this principle of transmission of pressure. The principle of floating bodies (law of flotation) is that W = W´, that is, weight of body = weight of displaced liquid or buoyant force. The weight of the displaced liquid is also called buoyancy or upthrust. Hydrometers work on the principle of floating bodies. This principle may also be applied to gases (e.g., a balloon). Liquids and gases are together called fluids. The important difference between them is that liquids cannot be compressed, while gases can be compressed. Hence, the density of a liquid is the same everywhere and does not depend on its pressure. In the case of a gas, however, the density is proportional to the pressure. XtraEdge for IIT-JEE

p2 < p1

According to this principle, the greater the velocity, the lower is the pressure in a fluid flow. It would be useful to remember that in liquid flow, the volume of liquid flowing past any point per second is the same for every point. Therefore, when the cross-section of the tube decreases, the velocity increases. Note : Density = relative density or specific gravity × 1000 kg m–3. Surface tension and surface energy : Surface Tension : The property due to which a liquid surface tends to contract and occupy the minimum area is called the surface tension of the liquid. It is caused by forces of attraction between the molecules of the liquid. A molecule on the free surface of a liquid experiences a net resultant force which tends to draw it into the liquid. Surface tension is actually a manifestation of the forces experienced by the surface molecules. If an imaginary line is drawn on a liquid surface then the force acting per unit length of this line is defined as the surface tension. Its unit is, therefore, newton / metre. This force acts along the liquid surface. For curved surfaces, the force is tangent to the liquid surface at every point. Surface Energy : A liquid surface possesses potential energy due to surface tension. This energy per unit area of the surface is called the surface energy of the liquid. Its units is joule per square metre. The surface energy of a liquid has the same numerical values as the surface tension. The surface 28

DECEMBER 2011

tension of a liquid depends on temperature. It decreases with rise in temperature. Excess of Pressure : Inside a soap bubble or a gas bubble inside a liquid, there must be pressure in excess of the outside pressure to balance the tendency of the liquid surface to contract due to surface tension.

The upward force by which a liquid surface is pulled up in a capillary tube is 2πrTcos θ, and the downward force due to the gravitational pull on the mass of liquid in the tube is (πr2h + v)ρg, where v is the volume above the liquid meniscus. If θ = 0º, the meniscus is hemispherical in shape. Then v = difference between the volume of the cylinder of radius r and height r and the volume of the hemisphere of radius r

1 1 p(excess of pressure) = T  +  in general  r1 r2 

= πr3 –

where T is surface tension of the liquid, and r1 and r2 are the principal radii of curvature of the bubble in two mutually perpendicular directions. For a spherical soap bubble, r1 = r2 = r and there are two free surfaces of the liquid.

When θ ≠ 0, we cannot calculate v which is generally very small and so it may be neglected. For equilibrium (πr2h + v) ρg = 2πrT cos θ When a glass capillary tube is dipper in mercury, the meniscus is convex, since the angle of contact is obtuse. The surface tension forces now acquire a downward component, and the level of mercury inside the tube the falls below the level outside it. the relation 2T cos θ = hρgr may be used to obtain the fall in the mercury level. Problem Solving Strategy Bernoulli's Equations : Bernoulli's equation is derived from the work-energy theorem, so it is not surprising that much of the problem-solving strategy suggested in W.E.P. also applicable here. Step 1: Identify the relevant concepts : First ensure that the fluid flow is steady and that fluid is incompressible and has no internal friction. This case is an idealization, but it hold up surprisingly well for fluids flowing through sufficiently large pipes and for flows within bulk fluids (e.g., air flowing around an airplane or water flowing around a fish). Step 2: Set up the problem using the following steps Always begin by identifying clearly the points 1 and 2 referred to in Bernoulli's equation. Define your coordinate system, particular the level at which y = 0. Make lists of the unknown and known quantities 1 1 in Eq. p1 + ρgy1 + ρv12 = p2 + ρgy2 + ρv22 2 2 (Bernoulli's equation) The variables are p1, p2, v1, v2, y1 and y2, and the constants are ρ and g. Decide which unknowns are your target variables. Step 3: Execute the solutions as follows : Write Bernoulli's equation and solve for the unknowns. In some problems you will need to use the continuity equation, Eq. A1v1 = A2v2 (continuity equation, incompressible fluid), to get a relation between the two speeds in terms of cross-sectional areas of pipes

4T ∴ p= r For a gas bubble inside a liquid, r1 = r2 = r and there is only one surface.

∴

2T r

For a cylindrical surface r1 = r and r2 = ∞ and there are two surfaces. 2T r Angle of Contact : The angle made by the surface of a liquid with the solid surface inside of a liquid at the point of contact is called the angle of contact. It is at this angle that the surface tension acts on the wall of the container.

∴

The angle of contact θ depends on the natures of the liquid and solid in contact. If the liquid wets the solid (e.g., water and glass), the angle of contact is zero. In most cases, θ is acute (figure i). In the special case of mercury on glass, θ is obtuse (figure ii). θ

θ fig. (i)

fig. (ii)

Rise of Liquid in a Capillary Tube : In a thin (capacity) tube, the free surface of the liquid becomes curved. The forces of surface tension at the edges of the liquid surface then acquire a vertical component. T θ meniscus θ T θ

XtraEdge for IIT-JEE

2 3 1 3 πr = πr 3 3

DECEMBER 2011

or containers. Or perhaps you will know both speeds and need to determine one of the areas. You may also dV = Av (volume flow rate) to find need to use Eq. dt the volume flow rate. Step 4: Evaluate your answer : As always, verify that the results make physical sense. Double-check that you have used consistent units. In SI units, pressure is in pascals, density in kilograms per cubic meter, and speed in meters per second. Also note that the pressures must be either all absolute pressure or all gauge pressures.

F (c) Shear strain = φ

Properties of matter :

Shear strain

Key Concepts : Stress : The restoring force setup inside the body per unit area is known as stress. Restoring forces : If the magnitude of applied deforming force at equilibrium = F

Stress-strain graph : From graph, it is obvious that in elastic limit, stress is proportional to strain. This is known as Hooke's law.

∴ Stress ∝ Strain ∴ Stress = E .strain

F Then, Stress = A In SI system, unit of stress is N/m2. Difference between pressure and stress : (a) Pressure is scalar but stress is tensor quantity. (b) Pressure always acts normal to the surface, but stress may be normal or tangential. (c) Pressure is compressive in nature but stress may be compressive or tensile. Strain :

Strain =

∴ E=

Stress

where E is proportionality dimensional constant known as coefficient of elasticity. Plastic region Breaking B C strength

change in dimension original dimension

∆L (a) Longitudinal strain = L L F

Elastic limit

Strain

Types of coefficient of elasticity :

(a) Young's modulus = Y = F

∴

Longitudinal strain is in the direction of deforming force but lateral strain is in perpendicular direction of deforming force. Poisson ratio : σ=

stress strain

logitudinal stress longitudinal strain

F FL = ∆L A∆L A L

∆d/D lateral strain = ∆L/L longitudinal strain

Here ∆d = change in diameter. (b) Volumetric strain =

∆L

∆V V

(b) Bulk modulus = B =

volumetric stress volumetric strain

Compressibility = 1/B XtraEdge for IIT-JEE

DECEMBER 2011

Surface tension :

shear stress F = Aφ shear strain

F L Here L = length of imaginary line drawn at the surface of liquid. and F = force acting on one side of line (shown in figure) (a) Surface tension does not depend upon surface area. (b) When temperature increases, surface tension decreases. (c) At critical temperature surface tension is zero.

(d) For isothermal process, B = P. F φ

φ F

(e) For adiabatic process, B = γP (f)

Adiabatic bulk modulus =γ Isothermal bulk modulus

(g) Esolid > Eliquid > Egas

(h) Young's modulus Y and modulus of rigidity η exist only for solids. (i) Bulk modulus B exist for solid, liquid and gas. (j) When temperature increases, coefficient of elasticity (Y, B, η) decreases. (k)

L F

1 9 3 + = η B Y

Rise or fall of a liquid in a capillary tube :

(l) Y = 2(1 + σ)η (m) Poisson's ratio σ is unitless and dimensionless. Theoretically,

1 –1 < σ < 2

Practically,

1 0<σ< 2

Here

1 1 1 × load × extension = Fx = kx2 2 2 2 = stress × strain × volume For twisting motion,

(a) For a drop of radius R, W = 4πR2T (b) For a soap bubble, W = 8πR2T Excess pressure :

1 × torque × angular twist 2

1 1 τ × θ = cθ2 2 2 Elastic energy density,

(a) For drop, P =

1 1 × stress × strain J/m3 = Y × strain2J/m3 2 2

4T R

Viscosity : (a) Newton's law of viscous force :

F = – ηA

1 F∆L 2

where

1 × stress × strain 2 (c) Breaking weight = breaking stress × area

(b) Work done per unit volume =

XtraEdge for IIT-JEE

2T R

(b) For soap bubble, P =

Thermal stress = Yα∆θ and Thermal strain = α∆θ Work done in stretching a wire : (a) W =

θ = angle of contact. r = radius of capillary tube

ρ = density of liquid For a given liquid and solid at a given place, hr = constant Surface energy : Surface energy density is defined as work done against surface tension per unit area. It is numerically equal to surface tension. W = work = surface tension × area

(n) Thermal stress = Yα∆θ (o) Elastic energy stored,

2T cos θ rρg

dv dy

dv = velocity gradient dy

A = area of liquid layer η = coefficient of viscosity The unit of coefficient of viscosity in CGS is poise. 31

DECEMBER 2011

(b) SI unit of coefficient of viscosity = poiseuille = 10 poise. (c) In the case of liquid, viscosity increases with density. (d) In the case of gas, viscosity decreases with density. (e) In the case of liquid, when temperature increases, viscosity decreases. (f) In the case of gas, when temperature increases, viscosity increases. Poiseuille's equation : V=

Sol. Draw a horizontal line through the mercury-glycerine surface. This is a horizontal line in the same liquid at rest namely, mercury. Therefore, pressure at the points A and B must be the same.

(1 – h)

Pπr 4 8ηL

Pressure at B = p0 + 0.1 × (1.3 × 1000) × g Pressure at A = p0 + h × 800 × g + (0.1 – h) × 13.6 × 1000g ∴ p0 + 0.1 × 1300 × g = p0 + 800gh + 1360g – 13600 × g × h ⇒ 130 = 800h + 1360 – 13600h 1230 ⇒ h= = 0.096 m = 9.6 cm 12800

where V = the volume of liquid flowing per second through a capillary tube of length L and radius r η = coefficient of viscosity and P = pressure difference between ends of the tube Stoke's law : The viscous force acting on a spherical body moving with constant velocity v in a viscous liquid is

A liquid flows out of a broad vessel through a narrow vertical pipe. How are the pressure and the velocity of the liquid in the pipe distributed when the height of the liquid level in the vessel is H from the lower end of the length of the pipe is h ? Sol. Let us consider three points 1, 2, 3 in the flow of water. The positions of the points are as shown in the figure. Applying Bernoulli's theorem to points 1, 2 and 3 2.

F = 6πηrv where r = radius of spherical body Determination of η :

η=

10 cm

2r 2 (ρ − σ)g 9v

where r = radius of spherical body moving with constant velocity v in a viscous liquid of coefficient of viscosity η and density ρ

•1

and σ = density of spherical body Critical velocity (v0) : v0 =

kη ρr

where k = Reynold's number for narrow tube, k ≈ 1000. (a) For stream line motion, flow velocity v < v0. (b) For turbulant motion, flow velocity v > v0.

•3

p0 1 2 1 p + v 1 + gH = + v 22 + g (h – x) ρ 2 ρ 2 p 1 = 0 + v 32 + 0 ρ 2 By continuity equation v 1A1 = A2v 2 = A2v 3 Since A1 >> A2,v1 is negligible and v2 = v3 = n (say). p0 p 1 ∴ + gH = 2 + v2 + g (h – x) ρ ρ 2 p0 1 2 + v = ρ 2

A vertical U-tube of uniform cross-section contains mercury in both arms. A glycerine (relative density 1.3) column of length 10 cm is introduced into one of the arms. Oil of density 800 kg m–3 is poured into the other arm until the upper surface of the oil and glycerine are at the same horizontal level. Find the length of the oil column. Density of mercury is 13.6 × 103 kg m–3.

XtraEdge for IIT-JEE

•2

Solved Examples 1.

∴

v = 2gH

(i)

DECEMBER 2011

= 4.99 × 10–3 cm Also, elastic limit for copper = 1.5 × 109 dynes/cm2 If d' is the minimum diameter, then maximum stress F 4F = on the wire = πd '2 / 4 πd' 2

p0 p + gH = 2 + gH + g (h – x) ρ ρ ⇒ p0 + p2 + ρg (h – x) (ii) ⇒ p2 = p0 – ρg (h – x) Thus pressure varies with distance from the upper end of the pipe according to equation (ii) and velocity is a constant and is given by (i).

and

Calculate the difference in water levels in two communicating tubes of diameter d = 1 mm and d = 1.5 mm. Surface tension of water = 0.07 Nm–1 and angle of contact between glass and water = 0º. 2T cos θ Sol. Pressure at A = p0 – r2 3.

πd' 2

or d'2 =

= 1.5 × 109 4F 9

A uniform horizontal rigid bar of 100 kg in supported horizontally by three equal vertical wires A, B and C each of initial length one meter and cross-section 1 mm2. B is a copper wire passing through the centre of the bar; A and C are steel wires and are arranged symmetrically one on each side of B YCu = 1.5 × 1012 dynes / cm2, Ys = 2 × 1012 dynes/cm2. Calculate the tension in each wire and extension. Sol. The situation is shown in figure. Because the rod is horizontally supported, hence extensions in all the wires must be equal i.e., strains in all the wires are equal as initial lengths are also equal.

As Y = Let this pressure difference correspond to h units of the liquid. 1 1 Then 2T cos θ  −  = ρgh  r1 r2 

A mass of 5 kg is suspended from a copper wire of 5 mm diameter and 2 m in length. What is the extension produced in the wire ? What should be the minimum diameter of the wire so that its elastic limit is not exceed ? Elastic limit for copper = 1.5 × 109 dynes/cm2. Y for copper = 1.1 × 1012 dynes/cm2. Sol. Given that Y = 1.1 × 1012 dynes/cm2, L = 2m = 200 cm, d = 5 mm = 0.5 cm or r = d/2 = 0.25 cm, F = 5.0 × 1000 × 980 dynes.

or l =

πr 2 Y

FCu / A Strain

Hence,

YCu =

and

Ys =

∴

YCu FCu 1.5 3 = = = YS FS 2 4

… (1)

Fs / A Strain

… (2) or 4FCu = 3FS

...(3)

FCu = (3/11) × 100g = (3/11) × 100 Kgwt = 27.28 Kgwt and FS = (4/3) FCu = (4/3) × (3/11) × 100g = (400/11)g = 36.36 Kgwt Extension in each wire,

5.0 × 1000 × 980 × 200

3.142 × (0.25) 2 × 1.1× 10 2

XtraEdge for IIT-JEE

∴

πr 2l

According to figure, we can write 2FS + FCu = 100 g or 2 × (4/3) FCu + FCu = 100 g or [(8/3) + 1] FCu = 100 g

FL FL

B Cu

100 Kg

  

Stress Strain

A S

2 × 0.07  1 1  −   = 4.76 mm −3 −3 1000 × 9.8  1× 10 1.5 × 10 

∴ h=

4 × 5.0 × 1000 × 980

1 1 ∴ pressure difference = 2T cos θ  −   r1 r2 

2T cos θ  1 1  − ⇒ h= ρg  r1 r2

π × 1.5 × 10 3.142 × 1.5 × 109 –4 = 41.58 × 10 d' = 0.0645 cm.

(Q pressure inside a curved surface is greater than that outside) 2T cos θ Pressure at B = p0 – r1

Hence,

FCu L 27280 × 980 × 100 = = 0.178 cm AYCu 10 − 2 × 1.5 × 1012

DECEMBER 2011

KEY CONCEPT

Organic Chemistry Fundamentals

CARBOXYLIC ACID (1.27Å) which is nearly intermediate between C O and C—O bond length values. This proves resonance in carboxylate anion.

Acidity of carboxylic acids. Fatty acids are weak acids as compared to inorganic acids. The acidic character of fatty acids decreases with increase in molecular weight. Formic acid is the strongest of all fatty acids. The acidic character of carboxylic acids is due to resonance in the acidic group which imparts electron deficiency (positive charge) on the oxygen atom of the hydroxyl group (structure II). O– O R

I Non-equivalent structures

O H

O–

O+H

–

C O

Resonance hybrid of carboxylate ion

Due to equivalent resonating structures, resonance in carboxylate anion is more important than in the parent carboxylic acid. Hence carboxylate anion is more stabilised than the acid itself and hence the equilibrium of the ionisation of carboxylic acids to the right hand side. RCOOH RCOO– + H+ The existence of resonance in carboxylate ion is supported by bond lengths. For example, in formic acid, there is one C=O double bond (1.23 Å) and one C—O single bond (1.36Å), while in sodium formate both of the carbon-oxygen bond lengths are identical XtraEdge for IIT-JEE

Alkoxide ion (No resonance)

Relative acidic character of carboxylic acids with common species not having —COOH group. RCOOH > Ar—OH > HOH > ROH > HC CH > NH3 > RH Effect of Substituents on acidity. The carboxylic acids are acidic in nature because of stabilisation (i.e., dispersal of negative charge) of carboxylate ion. So any factor which can enhance the dispersal of negative charge of the carboxylate ion will increase the acidity of the carboxylic acid and vice versa. Thus electron-withdrawing substitutents (like halogens, —NO2, —C6H5, etc.) would disperse the negative charge and hence stabilise the carboxylate ion and thus increase acidity of the parent acid. On the other hand, electron-releasing substituents would increase the negative charge, destabilise the carboxylate ion and thus decrease acidity of the parent acid.

Resonating forms of carboxylate ion (Equivalent structures) (Resonance more important)

Na+

Sodium formate

Alcohol (No resonance)

The positive charge (electron deficiency) on oxygen atom causes a displacement of electron pair of the O—H bond towards the oxygen atom with the result the hydrogen atom of the O—H group is eliminated as proton and a carboxylate ion is formed. Once the carboxylate ion is formed, it is stabilised by means of resonance. O O– R

–

It is important to note that although carboxylic acids and alcohols both contain –OH group, the latter are not acidic in nature. It is due to the absence of resonance (factor responsible for acidic character of –COOH) in both the alcohols as well as in their corresponding ions (alkoxide ions). R—O—H R—O– + H+

Formic acid

II (Resonance less important) O– R

O X

–

The substituent X withdraws electrons, disperses negative charge, stabilises the ion and hence increases acidity

O Y

–

The substituent Y releases electrons, intensifies negative charge, destabilises the ion and hence decreases acidity

DECEMBER 2011

Now, since alkyl groups are electron-releasing, their presence in the molecule will decrease the acidity. In general, greater the length of the alkyl chain, lower shall be the acidity of the acid. Thus, formic acid (HCOOH), having no alkyl group, is about 10 times stronger than acetic acid (CH3COOH) which in turn is stronger than propanoic acid (CH3CH2COOH) and so on. Similarly, following order is observed in chloro acids. Cl Cl Cl

pKa

CO2H > Cl

Comparison of nucleophilic substitution (e.g., hydrolysis) in acid derivatives. Let us first study the mechanism of such reaction. O R

R C Nu + Z (where Z= —Cl, —OCOR, —OR, —NH2 and Nu= A nucleophile) Nucleophilic substitution in acid derivatives O O OH

H CO2H > H

R CO2H

δ–

R C

δ–

δ+ Acid chloride

XtraEdge for IIT-JEE

δ+

H+

(where R' = H or alkyl group) Nucleophilic addition on aldehydes and ketones The (i) step is similar to that of nucleophilic addition in aldehydes and ketones and favoured by the presence of electron withdrawing group (which would stabilise the intermediate by developing negative charge) and hindered by electron-releasing group. The (ii) step (elimination of the leaving group Z) depends upon the ability of Z to accommodate electron pair, i.e., on the basicity of the leaving group. Weaker bases are good leaving groups, hence weaker a base, the more easily it is removed. Among the four leaving groups (Cl–, –OCOR, –OR, and –NH2) of the four acid derivatives, Cl– being the weakest base is eliminated most readily. The relative order of the basic nature of the four groups is – NH2 > –OR > –O.COR > Cl– Hence acid chlorides are most reactive and acid amides are the least reactive towards nucleophilic acyl substitution. Thus, the relative reactivity of acid derivatives (acyl compounds) towards nucleophilic substitution reactions is ROCl > RCO.O.COR > RCOOR > RCONH2 Acid Acid Esters Acid chlorides anhydrides amides OH– being stronger base than Cl–, carboxylic acids (RCOOH) undergo nucleophilic substitution (esterfication) less readily than acid chlorides.

Decreasing order of aliphatic acids (i) O2NCH2COOH > FCH2COOH > ClCH2COOH > BrCH2COOH (ii) HCOOH > CH3COOH > (CH3)2 CHCOOH > (CH3)3CCOOH (iii) CH3CH2CCl2COOH > CH3CHCl.CHCl.COOH > ClCH2CHClCH2COOH (iv) F3CCOOH > Cl3CCOOH > Br3CCOOH Benzoic acid is somewhat stronger than simple aliphatic acids. Here the carboxylate group is attached to a more electronegative carbon (sp2 hybridised) than in aliphatic acids (sp3 hybridised). HCOOH > C6H5COOH > CH3COOH. Nucleophilic substitution at acyl carbon : It is important to note that nucleophilic substitution (e.g., hydrolysis, reaction with NH3, C2H5OH, etc.) in acid derivatives (acid chlorides, anhydrides, esters and amides) takes place at acyl carbon atom (difference from nucleophilic substitution in alkyl halides where substitution takes place at alkyl carbon atom). Nucleophilic substitution in acyl halides is faster than in alkyl halides. This is due to the presence of > CO group in acid chlorides which facilitate the release of halogen as halide ion. O

H 4.76

H 2.86

pKa

C O

1.48

> Cl

(ii) Elimination step

CO2H

Z + Nu

Cl 0.70

O (i) Addition step

δ–

Alkyl chloride

DECEMBER 2011

KEY CONCEPT

Physical Chemistry Fundamentals

CHEMICAL KINETICS behaviour is a signal that the reaction has a complex mechanism. The temperature dependence of some reactions is non-Arrhenius, in the sense that a straight line is not obtained when ln k is plotted against 1/T. However, it is still possible to define an activation energy at any temperature as  d ln k  .......(ii) Ea = RT2    dT 

The temperature dependence of reaction rates : The rate constants of most reactions increase as the temperature is raised. Many reactions in solution fall somewhere in the range spanned by the hydrolysis of methyl ethanoate (where the rate constant at 35ºC is 1.82 times that at 25ºC) and the hydrolysis of sucrose (where the factor is 4.13). (a) The Arrhenius parameters : It is found experimentally for many reactions that a plot of ln k against 1/T gives a straight line. This behaviour is normally expressed mathematically by introducing two parameters, one representing the intercept and the other the slope of the straight line, and writing the Arrhenius equaion.

This definition reduces to the earlier one (as the slope of a straight line) for a temperature-independent activation energy. However, the definition in eqn.(ii) is more general than eqn.(i), because it allows Ea to be obtained from the slope (at the temperature of interest) of a plot of ln k against 1/T even if the Arrhenius plot is not a straight line. Non-Arrhenius behaviour is sometimes a sign that quantum mechanical tunnelling is playing a significant role in the reaction. (b) The interpretation of the parameters : We shall regard the Arrhenius parameters as purely empirical quantities that enable us to discuss the variation of rate constants with temperature; however, it is useful to have an interpretation in mind and write eqn.(i) as

Ea ......(i) RT The parameter A, which corresponds to the intercept of the line at 1/T = 0(at infinite temperature, shown in figure), is called the pre-exponential factor or the 'frequency factor'. The parameter Ea, which is obtained from the slope of the line (–Ea/R), is called the activation energy. Collectively the two quantities are called the Arrhenius parameters.

ln k = ln A –

ln A

k = Ae − E a / RT .......(iii) To interpret Ea we consider how the molecular potential energy changes in the course of a chemical reaction that begins with a collision between molecules of A and molecules of B(shown in figure).

ln k

Slope = –Ea/R

Potential energy

1/T A plot of ln k against 1/T is a straight line when the reaction follows the behaviour described by the Arrhenius equation. The slope gives –Ea/R and the intercept at 1/T = 0 gives ln A.

The fact that Ea is given by the slope of the plot of ln k against 1/T means that, the higher the activation energy, the stronger the temperature dependence of the rate constant (that is, the steeper the slope). A high activation energy signifies that the rate constant depends strongly on temperature. If a reaction has zero activation energy, its rate is independent of temperature. In some cases the activation energy is negative, which indicates that the rate decreases as the temperature is raised. We shall see that such

XtraEdge for IIT-JEE

Ea Reactants

Products Progress of reaction

A potential energy profile for an exothermic reaction. The height of the barrier between the reactants and products is the activation energy of the reaction

DECEMBER 2011

XtraEdge for IIT-JEE

DECEMBER 2011

CAREER POINT

Correspondence & Test Series Courses Information Courses of IIT-JEE For Class 12th/ 12th Pass Students COURSE →

For Class 11th Students All India Postal All Study Foundation India Test Material Test Series IITSeries Package JEE 2013 IIT-JEE 2013 IIT-JEE 2013 (At Center) Class 11th Class 11th Class 11th Students Students Students

Study Material Package IIT-JEE 2012

All India Test Series IIT-JEE 2012 (At Center)

Postal All India Test Series IIT-JEE 2012

Major Test Series IIT-JEE 2012 (At Center)

Postal Major Test Series IIT-JEE 2012 (By Post)

CP Ranker's Package IIT-JEE 2012

Eligibility for Admission

Class 12th or 12th Pass Students

Medium

English or Hindi

English

English or Hindi

English

Test Center

Postal

visit our website or contact us

Postal

visit our website or contact us

Postal

visit our website or contact us

Postal

Immediate Dispatch Immediate Dispatch 9800

Immediate Dispatch 15 Aug 11 onwards 4500

Immediate Dispatch 15 Aug 11 onwards 2500

INFORMATION ↓

Issue of Application Immediate Dispatch Kit Date of Dispatch/ Immediate Dispatch First Test Course Fee 9000

Immediate Dispatch

Immediate Dispatch Immediate Dispatch Immediate Dispatch Immediate Dispatch

15 Aug 11 onwards

3600

1500

15 Jan 2012 onwards 1000

15 Jan 2012 onwards 700

Immediate Dispatch 900

Courses of AIEEE For Class 12th/ 12th Pass Students

COURSE →

For Class 11th Students Study Material Package AIEEE 2013

Study Material Package AIEEE 2012

All India Test Series AIEEE 2012 (At Center)

Postal All India Test Series AIEEE 2012

Major Test Series AIEEE 2012 (At Center)

Postal Major Test Series AIEEE 2012 (By Post)

CP Ranker's Package AIEEE 2012

Eligibility for Admission

Class 12th or 12th Pass Students

Class 11th Students

Medium

English or Hindi

Test Center

Postal

visit our website or contact us

Postal

visit our website or contact us

Postal

Immediate Dispatch

Immediate Dispatch Immediate Dispatch Immediate Dispatch Immediate Dispatch

Immediate Dispatch

15 Aug 11 onwards

9000

3600

1500

INFORMATION ↓

Issue of Application Kit Date of Dispatch/ First Test Course Fee

15 Jan 2012 onwards 1000

15 Jan 2012 onwards 700

Immediate Dispatch 900

Immediate Dispatch Immediate Dispatch 9800

Courses of Pre-Medical For Class 12th/ 12th Pass Students COURSE →

For Class 11th Students

Study Material Package AIPMT 2012

All India Test Series AIPMT 2012 (At Center)

Postal All India Test Series AIPMT 2012

Major Test Series AIPMT 2012 (At Center)

Postal Major Test Series AIPMT 2012 (By Post)

CP Ranker's Package AIPMT 2012

Study Material Package AIPMT 2013

All India Foundation Test Series AIPMT 2013 (At Center)

Postal All India Test Series AIPMT 2013

Eligibility for Admission

Class 12th or 12th Pass Students

Class 11th Students

Medium

English or Hindi

English

English or Hindi

English

Test Center

Postal

visit our website or contact us

Postal

visit our website or contact us

Postal

visit our website or contact us

Postal

Immediate Dispatch

Immediate Dispatch Immediate Dispatch Immediate Dispatch Immediate Dispatch

Immediate Dispatch

15 Aug 11 onwards

9000

3600

1500

Immediate Dispatch Immediate Dispatch 9800

Immediate Dispatch 15 Aug 11 onwards 4500

Immediate Dispatch 15 Aug 11 onwards 2500

INFORMATION ↓

Issue of Application Kit Date of Dispatch/ First Test Course Fee

15 Jan 2012 onwards 1000

15 Jan 2012 onwards 700

Immediate Dispatch 900

• For details about all correspondence & test Series Course Information, Please visit our website.: www.careerpointgroup.com

XtraEdge for IIT-JEE

DECEMBER 2011

XtraEdge for IIT-JEE

DECEMBER 2011

XtraEdge for IIT-JEE

DECEMBER 2011

ratio of the two rates, and therefore of the two rate constants :

As the reaction event proceeds, A and B come into contact, distort, and begin to exchange or discard atoms. The reaction coordinate is the collection of motions, such as changes in interatomic distances and bond angles, that are directly involved in the formation of products from reactants. (The reaction coordinate is essentially a geometrical concept and quite distinct from the extent of reaction.) The potential energy rises to a maximum and the cluster of atoms that corresponds to the region close to the maximum is called the activated complex. After the maximum, the potential energy falls as the atoms rearrange in the cluster and reaches a value characteristic of the products. The climax of the reaction is at the peak of the potential energy, which corresponds to the activation energy Ea. Here two reactant molecules have come to such a degree of closeness and distortion that a small further distortion will send them in the direction of products. This crucial configuration is called the transition state of the reaction. Although some molecules entering the transition state might revert to reactants, if they pass through this configuration then it is inevitable that products will emerge from the encounter. We also conclude from the preceding discussion that, for a reaction involving the collision of two molecules, the activation energy is the minimum kinetic energy that reactants must have in order to form products. For example, in a gas-phase reaction there are numerous collisions each second, but only a tiny proportion are sufficiently energetic to lead to reaction. The fraction of collisions with a kinetic energy in excess of an energy Ea is given by

[P2 ] k = 2 [P1 ] k1

This ratio represents the kinetic control over the proportions of products, and is a common feature of the reactions encountered in organic chemistry where reactants are chosen that facilitate pathways favouring the formation of a desired product. If a reaction is allowed to reach equilibrium, then the proportion of products is determined by thermodynamic rather than kinetic considerations, and the ratio of concentration is controlled by considerations of the standard Gibbs energies of all the reactants and products. The kinetic isotope effect The postulation of a plausible mechanism requires careful analysis of many experiments designed to determine the fate of atoms during the formation of products. Observation of the kinetic isotope effect, a decrease in the rate of a chemical reaction upon replacement of one atom in a reactant by a heavier isotope, facilitates the identification of bond-breaking events in the rate-determining step. A primary kinetic isotope effect is observed when the ratedetermining step requires the scission of a bond involving the isotope. A secondary isotope effect is the reduction in reaction rate even though the bond involving the isotope is not broken to form product. In both cases, the effect arises from the change in activation energy that accompanies the replacement of an atom by a heavier isotope on account of changes in the zero-point vibrational energies. First, we consider the origin of the primary kinetic isotope effect in a reaction in which the ratedetermining step is the scission of a C–H bond. The reaction coordinate corresponds to the stretching of the C–H bond and the potential energy profile is shown in figure. On deuteration, the dominant change is the reduction of the zero-point energy of the bond (because the deuterium atom is heavier). The whole reaction profile is not lowered, however, because the relevant vibration in the activated complex has a very low force constant, so there is little zero-point energy associated with the reaction coordinate in either isotopomeric form of the activated complex.

the Boltzmann distribution as e − E a / RT . Hence, we can interpret the exponential factor in eqn(iii) as the fraction of collision that have enough kinetic energy to lead to reaction. The pre-exponential factor is a measure of the rate at which collisions occur irrespective of their energy. Hence, the product of A and the exponential factor,

e − E a / RT , gives the rate of successful collisions.

Potential energy

Kinetic and thermodynamic control of reactions : In some cases reactants can give rise to a variety of products, as in nitrations of mono-substituted benzene, when various proportions of the ortho-, meta-, and para- substituted products are obtained, depending on the directing power of the original substituent. Suppose two products, P1 and P2, are produced by the following competing reactions :

A + B → P1 Rate of formation of P1 = k1[A][B] A + B → P2 Rate of formation of P2 = k2[A][B] The relative proportion in which the two products have been produced at a given state of the reaction (before it has reached equilibrium) is given by the XtraEdge for IIT-JEE

C–H C–D

Ea(C–H) Ea(C–D)

Reaction coordinate

DECEMBER 2011

UNDERSTANDING

Inorganic Chemistry

An inorganic halide (A) gives the following reactions. (i) The cation of (A) on raction with H2S in HCl medium, gives a black ppt. of (B). (A) neither gives ppt. with HCl nor blue colour with K4Fe(CN)6. (ii) (B) on heating with dil.HCl gives back compound(A) and a gas (C) which gives a black ppt. with lead acetate solution. (iii) The anion of (A) gives chromyl chloride test. (iv) (B) dissolves in hot dil. HNO3 to give a solution, (D). (D) gives ring test. (v) When NH4OH solution is added to (D), a white precipitate (E) is formed. (E) dissolves in minimum amount of dil. HCl to give a solution of (A). Aqueous solution of (A) on addition of water gives a whitish turbidity (F). (vi) Aqueous solution of (A) on warming with alkaline sodium stannite gives a black precipitate of a metal (G) and sodium stannate. The metal (G) dissolves in hydrochloride acid to give solution of (A). Identify (A) to (G) and give balanced chemical equations of reactions. Sol. Observation of (i) indicates that cation (A) is Bi3+ because it does not give ppt. with HCl nor blue colour with K4Fe(CN)6, hence it is neither Pb2+ nor Cu2+. Since anion of (A) gives chromyl chloride test, hence it contains Cl– ions. Thus, (A) is BiCl3. Its different reactions are given below : (i) 2BiCl3 + 3H2S → Bi2S3 + 6HCl (A) (B) (ii) Bi2S3 + 6HCl → 3H2S + 2 BiCl3 (B) (C) (A)

Bi + NaSnO3 + H2O + 3NaCl (G) Black ppt.

∆ 2Bi + 6HCl → 2BiCl3 + 3H2 (G) (A) Hence, (A) is BiCl3, (B) is Bi2S3, (C) is H2S, (D) is Bi(NO3)2, (E) is Bi(OH)3, (F) is BiOCl and (G) is Bi

A colourless solid (A) on heating gives a white solid (B) and a colourless gas (C). (B) gives off reddishbrown fumes on treating with H2SO4. On treating with NH4Cl, (B) gives a colourless gas (D) and a residue (E). The compound (A) on heating with (NH4)2SO4 gives a colourless gas (F) and white residue (G). Both (E) and (G) impart bright yellow colour to Bunsen flame. The gas (C) forms white powder with strongly heated Mg metal which on hydrolysis produces Mg(OH)2. The gas (D) on heating with Ca gives a compound which on hydrolysis produces NH3. Identify compounds (A) to (G) giving chemical equations involved. Sol. The given information is as follows : 2.

(i) A Heat  → B + C Colourless Solid Colourless Solid gas ∆ (ii) B + H2SO4 → Reddish brown gas ∆ D + E (iii) B + NH4Cl → Colourless gas

∆ (iv) A + (NH4)2SO4 → F + G olourless gas White Residue (v) E and G imparts yellow colour to the flame.

∆

(iii) Bi2S3 + 8HNO3 → 2Bi(NO3)3 + 2NO (B) (D) + 3S + 4H2O (iv) Bi(N O3)3 + 3NH4OH → (D) Bi(OH)3 ↓ + 3NH4NO3 (E) White ppt.

 → White powder (vi) C + Mg Heat 2O H  → Mg(OH)2

∆

Bi(OH)3 + 3HCl → BiCl3 + 3H2O Dil. (A) BiCl3 + H2O → BiOCl + 2HCl (A) (F) Bismuth oxychloride (White turbidity) (v) BiCl3 + 2NaOH +Na2[SnO2] → (A) XtraEdge for IIT-JEE

H O

2 (vii) D + Ca Heat  → Compound   → NH3 Information of (v) indicates that (E) and (G) and also (A) are the salts of sodium because Na+ ions give yellow coloured flame. Observations of (ii) indicate that the anion associated with Na+ in (A) may be NO3–. Thus, the compound (A) is NaNO3. The reactions involved are as follows :

DECEMBER 2011

H2S + 2HNO3 → 2NO2 + 2H2O + S (D) (C) White turbidity CuSO4 + H2S → CuS ↓ + H2SO4 (E) (B) (F) Black ppt. 3CuS + 8HNO3 → Dil. 3Cu(NO3)2 + 2NO + 3S + 4H2O Cu++ + 4NH3 → [Cu(NH3)4]2+ (G) Blue colour [Cu(NH3)4]2+ + 4CH3COOH → Cu2+ + 4CH3COONH4 Cu2+ + [Fe(CN)6]4– → Cu2[Fe(CN)6] (H) Chocolate colour CuSO4(aq) + BaCl2(aq) → BaSO4 ↓ + CuCl2 (E) White ppt. Insuluble in HNO3 Hence, (A) is FeS, (B) is H2S, (C) is HNO3, (D) is S, (E) is CuSO4, (F) is CuS, (G) is [Cu(NH3)4]SO4 and (H) is Cu2[Fe(CN)6]

∆ (i) 2NaNO3 → 2NaNO2 + O2 ↑ (A) (B) (C) (ii) 2NaNO2 + H2SO4 → Na2SO4 + 2HNO2 (B) Dil. 3HNO2 → HNO3 + H2O + 2NO↑ 2NO + O2 → 2NO2 ↑ Reddish brown Fumes (iii) NaNO2 + NH4Cl → NaCl + N2 ↑ + 2H2O (B) (E) (D)

∆ Na2SO4 + 2NH3 (iv) 2NaNO3 + (NH4)2SO4 → (A) (G) (F) 2HNO3 ∆ 2 (v) O2 + 2Mg → 2MgO   → Mg(OH)2 (C) H O

∆ (vi) N2 + 3Ca → Ca3N2 (D) Ca3N2 + 6H2O → 3Ca(OH)2 + 2NH3 ↑ Hence, (A) is NaNO3, (B) is NaNO2, (C) is O2, (D) is N2, (E) is NaCl, (F) is NH3 and (G) is Na2SO4.

(i) An inorganic compound (A) is formed on passing a gas (B) through a concentrated liquor containing Na2S and Na2SO3. (ii) On adding (A) into a dilute solution of AgNO3, a white ppt. appears which quickly changes into black coloured compound (C). (iii) On adding two or three drops of FeCl3 into excess of solution of (A), a violet coloured compound (D) is formed. This colour disappears quickly. (iv) On adding a solution of (A) into the solution of CuCl2, a white ppt. is first formed which dissolves on adding excess of (A) forming a compound (E). Identify (A) to (E) and give chemical equations for the reactions at steps (i) to (iv) Sol. (i) The compound (A) appears to be Na2S2O3 from its method of preparation given in the problem. Na2S + Na2SO3 + I2 → 2NaI + Na2S2O3 (B) (A) or Na2SO3 + 3Na2S + 3SO2 → 3Na2S2O3 (B) (A) (ii) White ppt. of Ag2S2O3 is formed which is hydrolysed to black Ag2S Na2S2O3 + 2AgNO3 → 2NaNO3 + Ag2S2O3 ↓ White ppt Ag2S2O3 + H2O → Ag2S + H2SO4 (C) (iii) A violet ferric salt is formed. 3Na2S2O3 + 2FeCl3 → Fe2(S2O3)3 + 6NaCl (D)(violet) (iv) 2CuCl2 + 2Na2S2O3 → 2CuCl + Na2S4O6 + 2NaCl White ppt. 4.

A black coloured compound (A) on reaction with dil. H2SO4 gives a gas (B) which on passing in a solution of an acid (C) gives a white turbidity (D). Gas (B) when passed through an acidified solution of a compound (E), gives ppt.(F) which is soluble in dilute nitric acid. After boiling this solution an excess of NH4OH is added, a blue coloured compound (G) is produced. To this solution, on addition of CH3COOH and aqueous K4[Fe(CN)6], a chocolate ppt. (H) is produced. On addition of an aqueous solution of BaCl2 to aqueous solution of (E), a white ppt. insoluble in HNO3 is obtained. Identify compounds (A) to (H). Sol. From the data on compounds (G) and (H), it may be inferred that the compound (E) contains cupric ions (Cu2+), i.e., (E) is a salt of copper. Since the addition of BaCl2 to (E) gives a white ppt. insoluble in HNO3, it may be said that the anion in the salt is sulphate ion (SO42–). Hence, (E) is CuSO4. The gas (B) which is obtained by adding dil. H2SO4 to a black coloured compound (A), may be H2S since it can cause precipitation of Cu2+ ions in acidic medium. The black coloured compound (A) may be ferrous sulphide (iron pyrite). Hence, the given observation may be explained from the following equations. Fe S + H2 SO4 → FeSO4 + H2S (A) Dil. (B) 3.

XtraEdge for IIT-JEE

DECEMBER 2011

2CuCl + Na2S2O3 → Cu2S2O3 + 2NaCl 3Cu2S2O3 + 2Na2S2O3 → Na4[Cu6(S2O3)5] (E) or 6CuCl + 5Na2S2O3 → Na4[Cu6(S2O3)5] + 6NaCl (E) Hence, (A) is Na2S2O3, (B) is I2 or SO2, (C) is Ag2S, (D) is Fe2(S2O3)3 and (E) is Na4[Cu6(S2O3)5]. 5.

SCIENCE TIPS • What is the expression for growing current, in LR R  − t ® I = I0  1 − e L  circuit ?     • What is the range of infrared spectrum ? ® This covers wavelengths from 10–3 m down to 7.8 × 10–7 m

A white amorphous powder (A) on heating yields a colourless, non-combustible gas (B) and solid (C). The latter compound assumes a yellow colour on heating and changes to white on cooling. 'C' dissolves in dilute acid and the resulting solution gives a white precipitate on adding K4Fe(CN)6 solution. 'A' dissolves in dilute HCl with the evolution of gas, which is identical in all respects with 'B'. The gas 'B' turns lime water milky, but the milkiness disappears with the continuous passage of gas. The solution of 'A', as obtained above, gives a white precipitate (D) on the addition of excess of NH4OH and and passing H2S. Another portion of the solution gives initially a white precipitate (E) on the addition of sodium hydroxide solution, which dissolves on futher addition of the base. Identify the compounds A, B, D [IIT-1979] and E.

• What is the nature of graph between electric field and potential energy (U) ? ® The nature of the graph will be parabola having symmetry about U-axis • Why no beats can be heard if the frequencies of the two interfering waves differ by more than ten ? ® this is due to persistence of hearing • Why heating systems based on steam are more efficient than those based on circulation of hot ® This is because steam water ? has more heat than water a the same temperature • Can the specific heat of a gas be infinity ? ® Yes

∆

ZnCO3 → ZnO + CO2 (A) (B) (ii) ZnO + 2HCl → H2O + ZnCl2 (C) (soluble) (iii) 2ZnCl2 + K4[Fe(CN)6] → 4KCl + Zn2[Fe(CN)6]↓ (white ppt) (iv) ZnCO3 + HCl → CO2 + ZnCl2 (A) (soluble) (v) CO2 + Ca(OH)2 → CaCO3 + H2O (B) (Milky) (vi) CaCO3 + CO2 + H2O → Ca(HCO3)2 (soluble)

Sol. (i)

• What is the liquid ascent formula for a capillary ? r 2T cos θ ®h= – 3 γpg where h is the height through which a liquid of density ρ and surface tension T rises in a capillary tube of radius r • What is the expression for total time of flight (T) 2u sin θ ®T= for oblique projection ? g • The space charge limited current iP in the diode ® iP = k Vp3/2 value is given by • What is an ideal gas ?

NH OH

4 → 2HCl + ZnS↓ (vii) ZnCl2 + H2S   (white) (viii) ZnCl2 + 2NaOH → 2NaCl + Zn(OH)2↓ (white) (ix) Zn(OH)2 + 2NaOH → Na2ZnO2 + H2 sod. ziniate (soluble)

XtraEdge for IIT-JEE

® An ideal gas is one in which intermolecular forces are absent

• Can a rough sea be calmed by pouring oil on its surface ? ® Yes • What is the expression for fringe width (β) in Young's double slit experiment? ® β=Dλ/d where D is the distance between the source and screen and d is distance between two slits

DECEMBER 2011

Set

`tà{xÅtà|vtÄ V{tÄÄxÇzxá

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 Solutions will be p ublished in next issue Joint Director Academics, Career Point, Kota 1.

Show that the six planes through the middle point of each edge of a tetrahedron perpendicular to the opposite edge meet in a point.

Show that an equilateral triangle is a triangle of maximum area for a given perimeter and a triangle of minimum perimeter for a given area.

Let az2 + bz + c be a polynomial with complex coefficients such that a and b are non zero. Prove that the zeros of this polynomial lie in the region b c + |z|≤ a b

m2In = 2n(2n – 1) In–1 – 4n(n – 1) In–2. 9.

Find the sum to infinite terms of the series 5 9 11 3 7 + + + + + ........ ∞ 36 400 900 4 144

10. ABC is a triangle inscribed in a circle. Two of its sides are parallel to two given straight lines. Show that the locus of foot of the perpendicular from the centre of the circle on to the third side is also a circle, concentric to the given circle.

MEMORABLE POINTS • The vector relation between linear velocity and

An isosceles triangle with its base parallel to the x2 y2 major axis of the ellipse + = 1 is a2 b2 circumscribed with all the three sides touching the ellipse. Find the least possible area of the triangle.

→

® v =ω× r

angular velocity is

• In the case of uniform circular motion the angle between →

→

ω and r is always

→

® 90º(hence | v | = ωr

• The relation between Faraday constant F, Avogadro number N and the electronic charge e is ® F = Ne

If one of the straight lines given by the equation ax2 + 2hxy + by2 = 0 coincides with one of those given by a′x2 + 2h′xy + b′y2 = 0 and the other lines represented by them be perpendicular, show that ha´b´ h´ab´ = b´− a´ b−a

• Depolariser used in Lechlanche cell is ® Manganese dioxide

• The absorption or evolution of heat at a junction of two dissimilar metals when a current is passed is known as ® Peltier effect

Prove that  n  m  n   m + 1  n  m + 2     +     +     + ......... 0 n  1   n   2  n  .... to (n + 1) terms  n  m  n  m  n  m =     +     2 +     22 + ........  0  0  1  1   2  2  ..... to (n + 1) terms

XtraEdge for IIT-JEE

∫

If n ≥ 2 and In = (1 − x 2 ) n cos mx dx, then show that −1

Prove that if the graph of the function y = f (x), defined throughout the number scale, is symmetrical about two lines x = a and x = b, (a < b), then this function is a periodic one.

• The part of the human ear where sound is transduced ® Cochlea is the • Similar trait resulting from similar selection pressure acting on similar gene pool is termed ® Parallel evolution

• Group of related species with the potential, directly or indirectly, of forming fertile hybrids with one ® Coenospecies another is called

DECEMBER 2011

MATHEMATICAL CHALLENGES SOLUTION FOR NOVEMBER ISSUE (SET # 7)

Let the line be y = 2x + c  9 − c 9 + 2c  , Point A   3   6  2c − 3 + c − 6  , Point B   −3   −3  c + 6 5c + 12  Point C  ,  3   3 1  2c − 3 c + 6  mid point of B & C is .  + , 3  2  −3 1  − c + 6 5c + 12   9 − c 2c + 9  , + =    2 +3 3  3   6 which is point A, so AB and AC are equal.

  h − mk   2k  2 2hn  2  y – 2   xy + 1 +  x = 0 1 −  β − mα   β − mα   β − mα  It must represent y2 – x2 = 0 h − mk so = 0 ⇒ m = h/k ...(1) β − mα 2k 2hm and 1– = –1 – β − mα β − mα ⇒ β – mα – 2k = –β + mα – 2hm ⇒ –β + mα + k – hm = 0 ⇒ –β + k + h/k(α – h) = 0 (using (1) in it) ⇒ k2 – βy + αh – h2 = 0 so required locus is x2 – y2 – αx + βy = 0

1 AC 1 AB . . + AB AB AC AC 1 1 = AB + AC AB2 AC 2 1 1 = (AD + DB) + (AD + DC) AB 2 AC 2 DB DC 1   1 + =  +  AD + 2 2 BD.DC CD.CB AC   AB

a +b =

 DB DC  1 1   1  + + =   AD +  2 2  BC BD CD AC   AB  

1   1 = AD .  +   BD.DC CD.CB 

AD  1 1  . +  CD  BD CD 

AD DC + BD . BC BD.CD

|a +b | =

x →0

f ( x ) − f ( 0) tan x ≤ lim x →0 x x

⇒ |f (0)| ≤ 1 1 1 1 ⇒ a1 + a 2 + a 3 + ..... + a n ≤ 1 2 3 n n

⇒

∑i

≤1

i =1

1 . AD

1 AD

XtraEdge for IIT-JEE

 π π As |f (x)| ≤ |tan x| for ∀ x ∈  − ,   2 2 so f (0) = 0 so |f (x) – f (0)| ≤ |tan x| divides both sides by |x| f ( x ) − f ( 0) tan x ≤ ⇒ x x

⇒ lim

AD AD AD 1 = = = . 2 BD.CD AD AD AD

so it is vector along AB with magnitude

(h,k) Q

The line PQ always passes through (α, β) so it is y –β = m(x – α) Let the circle be x2 + y2 – 2hx – 2ky = 0 Joint equation of OP and OQ. ( y − mx) x2 + y2 – 2 (hx + ky) =0 β − mα P

Let the number is xyz, here x < y and z < y. Let y = n, then x can be filled in (n – 1) ways. (i.e. from 1 to (n – 1)) and z can be filled in n ways (i.e. from 0 to (n – 1)) DECEMBER 2011

3 (tan A . tan B . tan C)1/3 and for a triangle tan A + tan B + tan C = tan A . tan B . tan C so (tan A . tan B . tan C)2/3 ≥ 3

here 2 ≤ n ≤ 9 so total no. of 3 digit numbers with largest middle digit 9

∑

n(n − 1) =

n=2

∑

n2 –

n=2

∑n

⇒ tan A . tan B . tan C ≥ 3 3 ⇒ tan2A + tan2B + tan2C ≥ 3(tan A. tan B tan C)2/3 ≥ 3.3 a2 b2 c2 so from (1), 2 + 2 + 2 ≥ 144. r1 r2 r3

n=2

9.10.19 9.10 – 6 2 = 285 – 45 = 240

240 9 × 10 × 10 8 = 30 4 = 15

required probability =

Z1, Z2, Z3 are centroids of equilateral triangles ACX, ABY and BCZ respectively. Z − Z A iπ/6 Z1 – ZA = (ZC – ZA) 1 e ZC − ZA

(–1,log23)

∫ log

2 (2

− x) dx +

Z1 – ZA = (ZC – ZA)

−1

∫ (2 − 2

) dy +

−1

1 π 4

1  3 i  − ...(2) 2  3  2 i 1 (ZC + ZB – 2ZA) So, Z1 – Z2 = (ZC – ZB) + 2 2 3 ...(3) 1 similarly Z2 – Z3 = (ZA – ZC) 2 i + (ZA + ZC – 2ZB) ..(4) 2 3 To prove ∆xyz as equilateral triangle, we prove that (Z3 – Z2)eiπ/3 = Z1 – Z2 1 So, (Z3 – Z2)eiπ/3 = ( (ZC – ZA) 2 1 3  i (ZA + ZC – 2ZB))  + – i 2 2  2 3   1 i (ZC + ZB – 2ZA) = (ZC – ZB) + 2 2 3 = Z1 – Z2

E M

∠BMC = 2∠BAC = 2∠BMD BD BC BC a = = = so tan A = 4r1 MD 2MD 4r1 so so

a2 r12 a2

= tan2A +

r12 r22 r32 2 = 16 (tan A + tan2B + tan2C) Now as tan A + tan B + tan C ≥

XtraEdge for IIT-JEE

...(1)

Z2 – ZA = (ZB – ZA)

π e2 e +2+ sq units 27 4 A F

1  3 i  + 2  3  2

similarly,

π 1 2 + 2 log 2 3 + 2 – + = log 2 3 − 2ln2 ln 2 4

= – log2

Z3 z

(3/2,–1) 0

(1,0) (0,–1)

ZB (–1,0)

The region bounded by the curve y = log2(2 – x) and the inequality (x – |x|)2 + (y – |y|)2 ≤ 4 is required area is

...(1)

DECEMBER 2011

0 a

1 − r cos u

∫ 1 − 2r cos u + r

Tr = 2

∫

du.

1 − 2r cos u + r 2 − r 2 + 1   du  

 1− r2 1 +  1 − 2r cos u + r 2 0 = a + (1 – r2) ×

1 − 2r cos u + r 2

0 a

∫

∫ (1 + r

sec 2 u / 2 2

)(1 + tan 2 u / 2) − 2r (1 − tan 2 u / 2) a

∫ (1 + r )

= a + (1 – r2)

sec 2 u / 2 2

=a+

FRACTIONAL DISTILLATION OF AIR

...(1)

1− r2 (1 + r ) 2

tan 2 u / 2 + (1 − r ) 2

sec 2 u / 2 du

∫

tan 2 u / 2 +

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.

(1 − r ) 2 (1 + r ) 2

Let tan u/2 = t tan a / 2

1− r2

so, Tr = a +

(1 + r )

∫

2 dt t2 +

Now and

2(1 − r ) 1 + r  −1  1 + r  tan  t (1 + r ) 2 1 − r   1− r 2

=a+

1− r 1+ r

   0

lim Tr = a –

2(1 + r )(r − 1) π = a–π (1 + r )(r − 1) 2

lim Tr = a +

2(1 − r )(r + 1) π = a+π (1 + r )(r − 1) 2

r →1

r →1+

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.

tan a / 2

and (from (1)) T1 =

∫ du = a

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.

Hence

lim Tr, T1,

r →1+

lim Tr form an A.P. with

r →1−

common difference π. 10. Let α, β, γ be the three real roots of the equation without loss of generality, it can be assumed that α ≤ β ≤ γ. so x2 + ax2 + bx + c = (x – γ) (x2 + (a + γ) x + (γ2 + aγ + b)) where – γ (γ2 + aγ + b) = c, as γ is the root of given equation, so x2 + (a + γ) x + (γ2 + aγ + b) = 0 must have two roots i.e. α and β. So its discriminant is non negative, thus (γ + a)2 – 4(γ2 + aγ + b) ≥ 0 ⇒ 3γ2 + 2aγ – a2 + 4b ≤ 0

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.

− a + 2 a 2 − 3b 3 so greatest root is also less than or equal to so γ ≤

− a + 2 a 2 − 3b . 3

XtraEdge for IIT-JEE

DECEMBER 2011

Students' Forum MATHS

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

The line of intersection OC of the planes OAC and OBC, i.e., y + z = 0 & x + y = 0, is given by x = – y = z, y–0 x–0 z–0 i.e., = = . ...(ii) 1 –1 1 The line of intersection OB of the planes OBC and OAB, i.e., x + y = 0 and z + x = 0, is given by – x = y = z, x–0 y–0 z–0 i.e., = = . ...(iii) –1 1 1 The line BC is the intersection of the planes O (0,0,0) 1,1,–1 P

A man parks his car among n cars standing in a row, his car not being parked at an end. On his return he finds that exactly m of the n cars are still there. What is the probability that both the cars parked on two sides of his car, have left? Sol. Clearly, his car is at one of the crosses (×). |× × × . . . × × ×| The number of the ways in which the remaining m–1 cars can take their places (excluding the car of the man) = n–1Cm–1 {Q there are n – 1 places for the m – 1 cars}. The number of ways in which the remaining m – 1 cars can take places keeping the two places on two sides of his car vacant = n–3Cm–1 ∴ the required probability 1.

n( E ) = = n( S )

n –3

C m –1

n –1

C m –1

A C

(n – 3) ! (m – 1) ! (n – m) ! × (m – 1)!(n – m – 2) ! (n – 1) !

(n – m)(n – m – 1) . (n – 1)(n – 2)

(0,0,a)

x + y = 0 and x + y + z = a ⇒ y = – x, z = a z–a x y . ...(iv) = = ⇒ 0 1 –1 Similarly, CA has the equations y + z = 0, x + y + z = a x–a z y ⇒ = = ...(v) 0 1 –1 AB has the equation z + x = 0, x + y + z = a y–a x z = ⇒ = 0 –1 1 Let PQ be the shortest distance between OA and BC, and its line has the direction cosines l, m, n.

Show that the shortest distance between any two opposite edges of the tetrahedron formed by the planes y + z = 0, z + x = 0, x + y = 0 and x + y + z = a 2a is . 6

Sol. Clearly, the planes y + z = 0, z + x = 0 and x + y = 0 pas through the origin O(0, 0, 0,). O C A

Then using perpendicularity, l .1 + m. 1 + n . (–1) = 0 and l . 1 + m (–1) + n . 0 = 0 ⇒ l + m – n = 0 and l – m = 0

B The line of intersection OA of the planes OAC and OAB, i.e., y + z = 0 and z + x = 0, is given by x = y = – z, i.e.,

z–0 x–0 y–0 = = . 1 1 –1

XtraEdge for IIT-JEE

Q 1,–1,0

⇒

l m n = = = 1 1 2

l2 + m2 + n2 2

1 +1 + 2

1 6

...(i) 49

DECEMBER 2011

∴ l=

,m=

,n=

∴ the shortest distance PQ = projection of OC on the line PQ

= |(x2 – x1) l + (y2 – y1)m + (z2 – z1)n| = ( 0 – 0)

+ (0 – 0)

+ ( a – 0)

Similarly, other shortest distances between opposite edges are 1

(0 – 0)

+ ( a – 0)

6 2

and (a – 0)

2 6

+ (0 – 0)

' i.e.,

6 1

' i.e.,

2a 6

x – y +1

From (ii), x + y =

y '–x' 2

2 x'

...(iv)

2 x + 1 = 0.

2x + 1 =

2 ( x'+ y ' )

1 2

2y – 1 =

 x'+ y ' 1   x'+ y ' 1   –  + 6 –  2 2  2  2

β/2 β α d

From the question, we have to find PC'. Clearly, PQ = PA = d and OC' = OC = AC – r, r being the radius of the wheel. AC ; ∴ AC = d tan α In the ∆CAP, tan α = d ∴ OC + r = d tan α; ∴ OC' = d tan α – r

2 (y' – x')

1 2

 y '–x' 1   +  2  2

∴ C'Q = 2

 y '– x' 1  +  = 1 +  2  2

OC' 2 –OQ 2 =

∴ PC' = PQ + C'Q = d +

(d tan α – r ) 2 – r 2 (d tan α – r ) 2 – r 2 ...(i)

Also, from the ∆OAP, r β β = ; ∴ r = d tan . tan d 2 2

( x'+ y ' ) 2 1 x'+ y ' + – 2 4 2

  y ' 2 – x' 2 1 1 + 6 – + ( x'+ y '– y '+ x' ) 4 2 2   2

XtraEdge for IIT-JEE

∴ Putting these in the equation of the locus, we get

x'2 – 2y'2 – 2 x' + 1 = 0 ∴ the changed equation of the locus is (obtained by dropping dashes)

...(iii)

2 y'

⇒

(iv) – (iii) ∴ y=

–

...(ii)

12 + 12

x'+ y '

4y'2 – 2x'2 + 2 2 x' = 2

x– y

From (i), x – y + 1 =

∴ x=

A wheel with diameter AB touches the horizontal ground at A. The rod BC is fixed at B, ABC being vertical. A man from a point P on the ground at a distance d from A, finds that the angle of elevation of C is α. The wheel turns about the fixed centre O of the wheel such that C turns away from the man and its angle of elevation is β when it is about to disappear. Find PC when C is about to disappear. Sol. When C is about to disappear, let it be at C' and then PC' touches the wheel at Q.

...(i)

12 + (–1) 2

(iii) + (iv) ⇒

2 x' + 3y'2 – 3x'2 + 3 2 x' = 2

Find the changed equation of the locus x2 + 6xy + y2 = 1 when the lines x + y = 0 and x – y + 1 = 0 are taken as the new x and y axes respectively. Sol. Here the lines x + y = 0 and x – y + 1 = 0 are perpendicular to each other.

and y' =

x'2 + y'2 –

x2 – 2y2 –

So, take x' =

1 1 1 {(x' + y')2 + (y' – x')2} + + (y' – x' – x' – y') 2 2 2 3 3 + . 2x' = 1 + 3(y'2 – x'2) – 2 2

Hence, the problem.

( y '– x' ) 2 1 y '–x' + + =1 2 4 2

∴ PC' = d +

d 2 tan 2 α – 2d tan α . d tan

β 2

DECEMBER 2011

{from (i)}

–4

 β  = d 1 + tan α. 1 – 2 cot α . tan  . 2  

5. Sol.

ck = ck –1

∑

k =1

 ck   =  k –1 

∑ k (n – k + 1) n

∑ k{(n + 1)

 n – k +1  k 

∑ k . 3

k =1

– 2(n + 1)k + k 2 }

∴ I=

∑

k – 2(n + 1)

k =1

k2 +

∑

k =1

n(n + 1)(2n + 1) n(n + 1) = (n + 1) . – 2(n + 1) . 6 2

n 2 (n + 1) 2 4

= 6

(a) (b)

Sol. (a) I =

∫ =

1 z4

1 4  + c = – 1 + 4  + c  x 

∫x ∫

x +3 x

+ z2

. 6z5dz

∫

z6 dz z +1

∫

( z 6 – 1) + 1 dz z +1

∫

( z 3 + 1)( z 3 – 1) dz + 6 z +1

=x–

(1 + x 4 ) 3 / 4 x

∫ z +1

– z 4 + z 3 – z 2 + z – 1) dz + 6log (1 + z)

 z6 z5 z 4 z3 z 2  + + – – z  + 6 log (1+z)+ c =6 – 5 4 3 2   6

dx 2

∫z

= 6 ( z 2 – z + 1)( z 3 – 1) dz +6log(1 + z)

1 n(n + 1) 2 . (n + 2) 12

Evaluate

1 . +c 4 1 4

∫ = 6 ∫ (z

2 n(n + 1) 2  n .(n + 1) – (2n + 1) +  2 3 2 

n(n + 1) 2 6(n + 1) – 4(2n + 1) + 3n . = 2 6

∫

The LCM of 2 and 3 = 6 So, Put x = z6; then dx = 6z5 dz

k =1

1 –3 / 4 –1 dz = – z dz 4 4

(b) Here we have second and third roots of x.

k =1

= (n + 1)2

1 dz 4

=–

(1 + x 4 ) 4 + c. =– x

k =1

∑ k . c 3

1 3/ 4

1 z4

= –

n – k +1 = k

∴

∫z

=–

(k – 1) !(n – k + 1) ! n! . k !( n – k ) ! n!

C k –1

= z,

dx = dz, i.e.,

 c  k . k  . If n ∈ N and ck = Ck find the value of  c k –1  k =1 n

∴ I= n

Putting 1 +

 β  = d 1 + tan 2 α – 2 tan α. tan  2  

6 5/ 6 3 x + x 2 / 3 – 2x1/2 + 3x1/3 – 6x1/6 5 2

+ 6 log (1 + x1/6) + c

dx  1  x 2 .x 3  4 + 1 x 

∫

1  1 + 4   x 

XtraEdge for IIT-JEE

3/ 4

dx x5

DECEMBER 2011

MATHS

MONOTONICITY, MAXIMA & MINIMA Mathematics Fundamentals or negative according as f (x) is monotonic increasing or decreasing at x = a. So at x = a, function f (x) is

Monotonic Functions : A function f (x) defined in a domain D is said to be (i) Monotonic increasing :

monotonic increasing ⇔ f ´(a) > 0

 x < x 2 ⇒ f ( x1 ) ≤ f ( x 2 ) ⇔  1 ∀ x1, x2 ∈ D  x1 > x 2 ⇒ f ( x1 ) ≥ f ( x 2 ) y

monotonic decreasing ⇔ f ´(a) < 0 (ii) In an interval : In [a, b], f (x) is

monotonic increasing ⇔ f '(x) ≥ 0 monotonic decreasing ⇔ f '(x) ≤ 0

∀ x ∈ (a, b)

constant ⇔ f '(x) = 0 O

Note : (i) In above results f ´(x) should not be zero for all values of x, otherwise f (x) will be a constant function. (ii) If in [a, b], f ´(x) < 0 at least for one value of x and f ´(x) > 0 for at least one value of x, then f (x) will not be monotonic in [a, b]. Examples of monotonic function : If a functions is monotonic increasing (decreasing ) at every point of its domain, then it is said to be monotonic increasing (decreasing) function. In the following table we have example of some monotonic/not monotonic functions Monotonic Monotonic Not increasing decreasing monotonic x3 1/x, x > 0 x2 x|x| 1 – 2x |x| x –x e e ex + e–x log x log2x sin x sin h x cosec h x, x > 0 cos h x [x] cot h x, x > 0 sec h x

 x < x 2 ⇒ f ( x1 ) >/ f ( x 2 ) i.e., ⇔  1 ∀ x1, x2 ∈ D  x1 > x 2 ⇒ f ( x1 ) </ f ( x 2 ) (ii) Monotonic decreasing :  x < x 2 ⇒ f ( x1 ) ≥ f ( x 2 ) ⇔  1 ∀ x1, x2 ∈ D  x1 > x 2 ⇒ f ( x1 ) ≤ f ( x 2 ) y

 x < x 2 ⇒ f ( x1 ) </ f ( x 2 ) i.e., ⇔  1 ∀ x1, x2 ∈ D / f ( x2 )  x1 > x 2 ⇒ f ( x1 ) > A function is said to be monotonic function in a domain if it is either monotonic increasing or monotonic decreasing in that domain. Note : If x1 < x2 ⇒ f (x1) < f (x2) ∀ x1, x2 ∈ D, then f (x) is called strictly increasing in domain D and similarly decreasing in D. Method of testing monotonicity : (i) At a point : A function f (x) is said to be monotonic increasing (decreasing) at a point x = a of its domain if it is monotonic increasing (decreasing) in the interval (a – h, a + h) where h is a small positive number. Hence we may observer that if f (x) is monotonic increasing at x = a then at this point tangent to its graph will make an acute angle with x-axis where as if the function is monotonic decreasing there then tangent will make an obtuse angle with x-axis. Consequently f ´(a) will be positive

XtraEdge for IIT-JEE

Properties of monotonic functions : If f (x) is strictly increasing in some interval, then in that interval, f – 1 exists and that is also strictly increasing function. If f (x) is continuous in [a, b] and differentiable in (a, b), then

f ´(c) ≥ 0 ∀ c ∈ (a, b) ⇒ f (x) is monotonic increasing in [a, b] f ´(c) ≥ 0 ∀ c ∈ (a, b) ⇒ f (x) is monotonic decreasing in [a, b] 52

DECEMBER 2011

If both f (x) and g (x) are increasing (or decreasing) in [a, b] and gof is defined in [a, b], then gof is increasing. If f (x) and g (x) are two monotonic functions in [a, b] such that one is increasing and other is decreasing then gof, it is defined, is decreasing function. Maximum and Minimum Points : The value of a function f (x) is said to be maximum at x = a if there exists a small positive number δ such f (a) > f (x) that y

(a)

(b)

(c)

Least value = min. {f (a), f (b), f (c)} where x = c is a point such that f´(c) = 0. If a continuous function has only one maximum (minimum) point, then at this point function has its greatest (least) value. Monotonic functions do not have extreme points. Conditions for maxima and minima of a function Necessary condition : A point I = a is an extreme point of a function f (x) if f ´(a) = 0, provided f ´(a) exists. Thus if f ´(a) exists, then x = a is an extreme point ⇒ f ´(a) = 0 or f ´(a) ≠ 0 ⇒ x = a is not an extreme point But its converse is not true i.e. f ´(a) = 0 ⇒ / x = a is an extreme point. For example if f (x) = x3, then f ´(0) = 0 but x = 0 is not an extreme point. Sufficient condition : For a given function f (x), a point x = a is a maximum point if f ´(a) = 0 and f´´(a) < 0 a minimum point if f´(I) = 0 and f ´´(a) > 0 not an extreme point if f ´(a) = 0 = f ´´(a) and f ´´´(a) ≠ 0. Note : If f ´(a) = 0, f ´´(a) = 0, f ´´´(a) = 0 then the sign of f(4)(a) will determine the maximum or minimum point as above. Working Method : Find f ´(x) and f ´´(x). Solve f ´(x) = 0. Let its roots be a, b, c, ... Determine the sign of f ´´(x) at x = a, b, c, .... and decide the nature of the point as mentioned above. Properties of maxima and minima : If f (x) is continuous function, then Between two equal values of f (x), there lie atleast one maxima or minima. Maxima and minima occur alternately. For example if x = –1, 2, 5 are extreme points of a continuous function and if x = –1 is a maximum point then x = 2 will be a minimum point and x = 5 will be a maximum point. When x passes a maximum point, the sign of dy/dx changes from + ve to – ve, where as when x passes through a minimum point, the sign of f ´(x) changes from –ve to + ve. If there is no change in the sign of dy/dx on two sides of a point, then such a point is not an extreme point. If f (x) is maximum (minimum) at a point x = a, then 1/f (x), [f (x) ≠ 0] will be minimum (maximum) at that point. If f (x) is maximum (minimum) at a point x = a, then for any λ ∈ R, λ + f (x), log f (x) and for any k > 0, k f (x), [f (x)]k are also maximum (minimum) at that point.

Also then the point x = a is called a maximum point for the function f (x). Similarly the value of f (x) is said to be minimum at x = b if there exists a small positive number δ such that f (b) < f (x) ∀ x ∈ (b – δ, b + δ) Also then the point x = b is called a minimum point for f (x) Hence we find that : (i) x = a is a maximum point of f (x)  f ( a ) – f (a + h) > 0 ⇔   f ( a ) – f (a – h) > 0 (ii) x = b is a minimum point of f(x)  f (b) – f (b + h) > 0 ⇔   f (b) – f (b – h) > 0 (iii) x = c is neither a maximum point nor a minimum point  f (c ) – f (c + h )    ⇔  and  have opposite signs.  f (c ) – f (c – h )    Where h is a very small positive number. Note : The maximum and minimum points are also known as extreme points. A function may have more than one maximum and minimum points. A maximum value of a function f (x) in an interval [a, b] is not necessarily its greatest value in that interval. Similarly a minimum value may not be the least value of the function. A minimum value may be greater than some maximum value for a function. The greatest and least values of a function f (x) in an interval [a, b] may be determined as follows : Greatest value = max. {f (a), f (b), f (c)}

XtraEdge for IIT-JEE

DECEMBER 2011

MATHS

FUNCTION Mathematics Fundamentals f + g, f – g, fg, f /g, fog

Definition of a Function :

Let A and B be two sets and f be a rule under which every element of A is associated to a unique element of B. Then such a rule f is called a function from A to B and symbolically it is expressed as

and they are defined as follows : (f + g) (x) = f (x) + g(x) (f – g) (x) = f (x) – g(x)

f :A→B

(f g) (x) = f (x) f (g)

f A → B

(f /g) (x) = f (x)/g(x) (fog) (x) = f [g(x)]

Function as a Set of Ordered Pairs

Formulae for domain of functions :

Every function f : A → B can be considered as a set of ordered pairs in which first element is an element of A and second is the image of the first element. Thus

Df ± g = Df ∩ Dg Dfg = Df ∩ Dg Df/g = Df ∩ Dg ∩ {x |g(x) ≠ 0}

f = {a, f (a) /a ∈ A, f (a) ∈ B}.

Dgof = {x ∈ Df | f(x) ∈ Dg}

Domain, Codomain and Range of a Function :

If f : A → B is a function, then A is called domain of f and B is called codomain of f. Also the set of all images of elements of A is called the range of f and it is expressed by f (A). Thus

= Df ∩ {x |f (x) ≥ 0}

Algebraic and Transcendental Functions : Algebraic functions : If the rule of the function consists of sum, difference, product, power or roots of a variable, then it is called an algebraic function. Transcendental Functions : Those functions which are not algebraic are named as transcendental or non algebraic functions.

f (A) ⊂ B.

Note : Generally we denote domain of a function f by Df and its range by Rf. Equal Functions :

Two functions f and g are said to be equal functions if

domain of f = domain of g codomain of f = codomain of g

f (x) = g(x) ∀ x. Algebra of Functions :

Even and Odd Functions : Even functions : If by replacing x by – x in f (x) there in no change in the rule then f (x) is called an even function. Thus f (x) is even ⇔ f (– x) = f (x) Odd function : If by replacing x by – x in f (x) there is only change of sign of f (x) then f (x) is called an odd function. Thus

If f and g are two functions then their sum, difference, product, quotient and composite are denoted by

XtraEdge for IIT-JEE

Classification of Functions

f (A) = {f (a) |a ∈ A} obviously

(g(x) ≠ 0)

f (x) is odd ⇔ f (– x) = – f (x) 54

DECEMBER 2011

Explicit and Implicit Functions :

Function

Explicit function : A function is said to be explicit if its rule is directly expressed (or can be expressed( in terms of the independent variable. Such a function is generally written as

sin x, cos x, sec x, cosec x,

2π

tan x, cot x

sinnx, cosn x, secn x, cosecn x

2π if n is odd

y = f (x), x = g(y) etc.

π if n is even

Implicit function : A function is said to be implicit if its rule cannot be expressed directly in terms of the independent variable. Symbolically we write such a function as

f (x, y) = 0, φ(x, y) = 0 etc. 4.

Continuous and Discontinuous Functions :

Discontinuous Functions : A function is said to be discontinuous if it has a gap or break in its graph atleast at one point. Thus a function which is not continuous is named as discontinuous.

π∀n∈N

|sin x|, |cos x|, |sec x|, |cosec x|

|tan x|, |cot x|,

|sin x| + |cos x|, sin4x + cos4x

π 2

|tan x| + |cot x|

π 2

x – [x]

Period of f (x) = T ⇒ period of f (ax + b)= T/|a| Period of f1(x) = T1, period of f2(x) = T2 ⇒ period of a f1(x) + bf2(x) ≤ LCM {T1, T2}

Increasing and Decreasing Functions : Increasing Functions : A function f (x) is said to be increasing function if for any x1, x2 of its domain

Kinds of Functions : One-one/ Many one Functions :

A function f : A → B is said to be one-one if different elements of A have their different images in B.

x1 < x2 ⇒ f (x1) ≤ f (x2) or x1 > x2 ⇒ f (x1) ≥ f (x2) Decreasing Functions : A function f (x) is said to be decreasing function if for any x1, x2 of its domain

Thus ⇒  a≠b  f is one-one ⇔  or  f (a) = f (b) ⇒ 

x1 < x2 ⇒ f (x1) ≥ f(x2) or x1 > x2 ⇒ f (x1) ≤ f (x2)

f (a) ≠ f (b) a=b

A function which is not one-one is called many one. Thus if f is many one then atleast two different elements have same f -image.

Periodic Functions :

A functions f (x) is called a periodic function if there exists a positive real number T such that

Onto/Into Functions : A function f : A → B is said to be onto if range of f = codomain of f

f (x + T) = f (x) ∀ x Also then the least value of T is called the period of the function f (x).

Thus f is onto ⇔ f (A) = B Hence f : A → B is onto if every element of B (co-domain) has its f –preimage in A (domain).

Period of f (x) = T ⇒ Period of f (nx + a) = T/n

A function which is not onto is named as into function. Thus f : A → B is into if f (A) ≠ B. i.e.,

Periods of some functions : XtraEdge for IIT-JEE

tann x, cotnx

|sec x| + |cosec x|

Continuous functions : A functions is said to be continuous if its graph is continuous i.e. there is no gap or break or jump in the graph.

Period

DECEMBER 2011

if there exists atleast one element in codomain of f which has no preimage in domain.

Domain and Range of some standard functions : Function

Note :

Domain

Range

Polynomial function

Identity function x

total number of functions which can be defined from A to B = nm.

Constant function c

{c}

total number of one-one functions from A to B

Reciprocal function 1/x

Total number of functions : If A and B are finite sets containing m and n elements respectively, then

 n P =  m  0

m≤n

x2, |x|

R+ ∪ {0}

m>n

x3, x |x|

Signum function

{–1, 0, 1}

x + |x|

R+ ∪ {0}

x – |x|

R– ∪ {0}

[x]

x – [x]

[0, 1)

[0, ∞)

total number of onto functions from A to B (if m ≥ n) = total number of different n groups of m elements. Composite of Functions :

Let f : A → B and g : B → C be two functions, then the composite of the functions f and g denoted by gof, is a function from A to C given by gof : A → C, (gof) (x) = g[f (x)].

R+

log x

R+

The following properties of composite functions can easily be established.

sin x

[–1, 1]

cos x

[–1, 7]

Composite of functions is not commutative i.e.,

tan x

R – {± π/2, ± 3π/2, ...} R

cot x

R – {0, ± π. ± 2π, .....

sec x

R – (± π/2, ± 3π/2, .....

R – (–1, 1)

cosec x

R – {0, ±π, ± 2π, ......} R –(–1, 1)

Properties of Composite Function :

fog ≠ gof Composite of functions is associative i.e. (fog)oh = fo(goh) Composite of two bijections is also a bijection. Inverse Function :

If f : A → B is one-one onto, then the inverse of f i.e., f –1 is a function from B to A under which every b ∈ B is associated to that a ∈ A for which f (a) = b. Thus

–1

sinh x

cosh x

[1, ∞)

tanh x

(–1, 1)

coth x

R –[1, –1]

sech x

(0, 1]

cosech x

–1

: B → A,

f –1(b) = a ⇔ f (a) = b.

sin x

[–1, 1]

[–π/2, π/2]

cos–1x

[–1, 1]

[0, π]

–1

(–π/2, π/2}

–1

(0, π)

tan x cot x –1

sec x –1

cosec x

XtraEdge for IIT-JEE

R –(–1, 1)

[0, π] – {π/2}

R – (–1, 1)

(– π/2, π/2] – {0}

DECEMBER 2011

XtraEdge for IIT-JEE

DECEMBER 2011

Based on New Pattern

IIT-JEE 2012 XtraEdge Test Series # 8

Time : 3 Hours Syllabus :

Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus

Instructions : Section - I • Question 1 to 6 are passage based questions. +4 marks will be awarded for correct answer and –1 mark for wrong answer. • Question 7 to 9 are Reason and Assertion type question with one is correct answer. +4 marks and –1 mark for wrong answer.

• •

Question 10 to 15 are Numerical Response Question (single digit Ans. type) +4 marks will be awarded for correct answer and –1 mark for wrong answer. Question 16 to 18 are Numerical Response Question (four digit Ans. type) +6 marks will be awarded for correct answer and –1 mark for wrong answer. (C) If the speed of water is 4 m/sec then direction is which 3 he should row his boat is cos–1   w.r.t. bank 4 (D) If the speed of water is 4 m/sec then direction in 3 which he should row his boat is sin–1   w.r.t bank 4

PHYSICS This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 1 to 6) Each question has 4 choices (A), (B), (C) and (D) out of which ONE OR MORE THAN ONE may be correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and – 1 mark for each wrong answer.

The velocity of boatman is still water is 3m/sec and river is flowing at 2 m/sec. To cross the river in minimum time. (A) The boatman should move at an angle 2 θ = sin–1   w.r.t bank 3 (B) The boatman should move at an angle 2 θ = cos–1   w.r.t bank 3 (C) the batman should move at an angle 90º w.r.t bank (D) None of these

Two boats A & B move away from a buoy anchored at the middle of river along the mutually perpendicular 1 straight lines : the boat A along the river & boat B across the river. Having moved off an equal distance from the buoy the boats returned. Then the ratio of times of boats tA/tB. If the velocity of each boat w.r.t flow of water ‘x’ times greater than the stream velocity tA/tB is x x (A) (B) 2 x −1 x2 +1 1 1 (D) (C) 2 x −1 x2 +1

Passage # 1 (Ques. 1 to 3) →

There is a river which is flowing at the rate v r . If a →

boatman starts to row his boat at the speed v br with respect to river then the velocity of boatman with →

→

respect to ground can be given by, v b = v br + v r . Now a boat can travel at a speed of 3m/s in still water. A boatman has to across the river to reach the other side. Now give the answer of following question 1.

The boatman wants to cross the river in such a way that he should cover the shortest possible distance. (A) If the speed of water is 2m/sec then direction in which 2 he should row his boat is cos–1   w.r.t. bank 3 (B) If the speed of water is 2m/sec then direction in which 2 he should row his boat is sin–1   w.r.t bank 3

XtraEdge for IIT-JEE

DECEMBER 2011

Passage # 2 (Ques. 4 to 6) If two masses A & B are drawn in their attached  0.2t  2 cables with a =  m / sec where t is in second.  3 

→ →

displacement. Hence w = F ⋅ S = zero This section contains 6 questions (Q.10 to 15). +4 marks will be given for each correct answer and –1 mark for each wrong answer. The answer to each of the questions is a SINGLE-DIGIT INTEGER, ranging from 0 to 9. The appropriate bubbles below the respective question numbers in the OMR have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following : X Y Z W 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9

D C E A

The speed of block ‘E’ when it reaches a height of h = 4 m starting from rest is (A) 4 m/sec (B) 2 m/sec (C) 1 m/sec (D) None of these

The approx speed of pulley D is (A) – 2.5 m/sec (B) – 3.5 m/sec (C) – 4.5 m/sec (D) – 5.5 m/sec

If vA = KvD then K, is (A) 2 (B) 3

(D) 5

This section contains 3 questions numbered 7 to 9, (Reason and Assertion type question). Each question contains Assertion and Reason. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and – 1 mark for each wrong answer. The following questions given below consist of an "Assertion" (A) and "Reason" (R) Type questions. Use the following Key to choose the appropriate answer. (A) If both (A) and (R) are true, and (R) is the correct explanation of (A). (B) If both (A) and (R) are true but (R) is not the correct explanation of (A). (C) If (A) is true but (R) is false. (D) If (A) is false but (R) is true. 7.

Assertion (A) : A lighter and a heavier bodies moving with same momentum and experiencing same retarding force have equal stopping times. Reason (R) : For a given force and momentum, stopping time is independent of mass.

10.

A particle starts oscillating simple harmonically from its equilibrium position then the KE of the particle is ‘n’ times the P.E. of particle at the time T/12, find the value of n (T : time period)

11.

A bird flies for 4 sec with a velocity of |t – 2| m/sec in a straight line where t is in second the distance covered by the bird is ….. (in m)

12.

An object of mass 0.2 kg executes SHM along the xaxis with a frequency 25/π Hz. At the position x = 0.04 m, the object has K.E., 0.5 J and P.E., 0.4 J. The amplitude of oscillation in cm will be …... (PE is zero at mean position)

13.

Difference between nth & (n + 1)th Bohr's radius of H atom is equal to its (n – 1)th Bohr's radius. The value of n is ……..

14.

In the arrangement shown in the figure m1 = 1kg m2 = 2 kg. Pulley are mass less & strings are light for what value of M, the mass m1 moves with constant velocity (in kg) M

Assertion (A) : The shortest wavelength of x-rays emitted from x-ray tube is independent of voltage applied to tube. Reason (R) : wavelength of characteristic spectrum depends upon the atomic number of target.

XtraEdge for IIT-JEE

Assertion (A) : Work done by the static friction is always zero. Reason (R) : when the body is stationary, there is no

DECEMBER 2011

15.

A block of mass 1 kg is attached to one end of a spring of force constant k = 20 N/m. The other end of the spring is attached to a fixed rigid support. This spring block system is made to oscillate on a rough horizontal surface with µ = 0.04. The initial displacement of block from the (Eq.) mean position is a = 30 cm. How many times the block will pass from the mean position before coming to rest ? (g = 10 m/sec2)

This section contains Numerical response type questions (Q. 16 to 18). +6 marks will be given for each correct answer and –1 mark for each wrong answer. Answers to this Section are to be given in the form of nearest integerin four digits. Please follow as per example : (i.e. for answer : 1492.2 write 1492; 491.8 write 0492; 92.5 write 0092; 2.1 write 0002)

16.

235 U reactor takes 30 days to consume 4 kg of fuel If 92 and each fission gives 185 MeV of unstable energy then find the output power [× 107 W]

17.

A bullet in fired at a plank of wood with a speed of 200 m/sec. After passing through the plank, its speed is reduced to 150 m/sec. Another bullet of same mass & size but moving with a speed of 100 m/sec is fired at the same plank. What would be the speed of bullet after passing through the plank ? Assume that the resistance offered by plank is same for both the bullets.

18.

N N

N–H, In this compound HN3 (hydrazoic acid),

oxidation state of N1, N2 and N3 are (A) 0, 0, 3 (B) 0, 0, –1 (C) 1, 1, –3 (D) –3, –3, –3 3.

Equivalent weight of chlorine molecule in the equation 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O (A) 42.6 (B) 35.5 (C) 59.1 (D) 71

Passage # 2 (Ques. 4 to 6) Secondary and tertiary alcohols always give E1 reaction in dehydration. Primary alcohols whose βcarbon is 3º or 4º also give E1 reaction. However, the primary alcohols whose β-carbon is 1º or 2º give E2 reaction. Dehydrating agents like conc. H2SO4, Al2O3 anhydrous ZnCl2 are used. The reactivity of alcohols for elimination reaction lies in following sequence : Tertiary alcohol > secondary alcohol > primary alcohol Electron attracting groups present in alcohols increase the reactivity for dehydration. Greater is the –I effect of the group present in alcohol, more will be its reactivity. Both E1 and E2 mechanism give the product according to Saytzeff's rule, i.e., major product is the most substituted alkene. OH

If the end of the cord A is pulled down with 2m/sec then the velocity of block will be : ( × 10–1 m/sec)

2 m/sec B

CH3 – CH – CH – CH3

CHEMISTRY

conc. H2SO4 Above 413K

CH3 CH3 – C = CH – CH3 + CH3 – CH – CH = CH2 CH CH3 CH3 Major product Minor product

Passage # 1 (Ques. 1 to 3) Redox reactions are those in which oxidation and reduction take place simultaneously. Oxidising agent can gain electron whereas reducing agent can lose XtraEdge for IIT-JEE

electron easily. The oxidation state of any element can never be in fraction. If oxidation number of any element comes out be in fraction, it is average oxidation number of that element which is present in different oxidation states. The oxidation state of Fe in Fe3O4 is (A) 2 and 3 (B) 8/3 (C) 2 (D) 3

Arrange the reactivity of given four alcohols in decreasing order for dehydration. OH OH NO2 NO2 (a)

(b)

DECEMBER 2011

NO2

Assertion (A) : The value of van der Waals constant 'a' is larger for ammonia than for nitrogen. Reason (R) : Hydrogen bonding is present in ammonia.

Assertion (A) : 3-hydroxy - butan-2-one on treatment with [Ag(NH3)2]⊕ cause precipitation of silver. Reason (R) : [Ag(NH3)2] ⊕ oxidises 3-hydroxy butan-2-one to butan-2-3-dione

Assertion (A) : HBr adds to 1,4-pentadiene at a faster rate than to 1,3-pentadiene Reason (R) : 1,4-pentadiene is less stable than 1,3-pentadiene.

NO2

(d) (B) d > c > b > a (D) b > c > a > d

CH3

In the reaction, CH3 – C – CH2OH

conc. H2SO4

CH3 the product obtained will be : (A) CH2 = C – CH2 – CH3 CH3 (B) CH3 – C = CH – CH3

This section contains 6 questions (Q.10 to 15). +4 marks will be given for each correct answer and –1 mark for each wrong answer. The answer to each of the questions is a SINGLE-DIGIT INTEGER, ranging from 0 to 9. The appropriate bubbles below the respective question numbers in the OMR have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following : X Y Z W 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9

CH3 CH3

Which of the following dehydration product is/are correct ? CH2OH conc.H2SO4 ∆

(A) CH3 CH3

(B) CH3 – C — CH – CH3 CH3

CH3 conc. H2SO4 ∆

CH3 – C – CH = CH2 CH3

conc. H2SO4 (C) CH3CH2CH2CH2OH ∆ CH3 – CH = CH – CH3 CH3 CH3 conc.H2SO4 CH 3 (D) ∆ CH3 OH

10. Equal volumes of 0.02 M AgNO3 & 0.02 M HCN were mixed. If the [Ag+] at equilibrium was 10–n. Find n. Given Ka(HCN) = 4 × 10–10, Ksp(AgCN) = 4 × 10–16.

11.

Haemoglobin contains 0.25% iron by weight. The molecular weight of haemoglobin is 89600. Calculate the number of iron atoms per molecule of haemoglobin.

12. Two liquids A and B form an ideal solution at temperature T. When the total vapour pressure above the solution is 600 torr, the amount of A in the vapour phase is 0.35 and in the liquid phase is 0.70. What is the vapour pressure of pure A ? Express your answer after divide actual answer by 100. 13. The value of x in the complex Hx[Co(CO)4] is

DECEMBER 2011

14. Calculate the emf of the cell Cd|Cd2+ (0.10M)1|| H+(0.20M)| Pt, H2(0.5 atm) [Given: EºCd2+/Cd = – 0.403 V, 2.303RT = 0.0591] F Round off your answer after multiplying actual answer by 10.

15. Calculate enthalpy change (in calories) adiabatic compression of one mole of an ideal monoatomic gas against constant external pressure of 2 atm starting from initial pressure of 1 atm and initial temperature of 300 K. (R = 2 cal/mol degree) Give your answer after divide actual answer by 100. This section contains Numerical response type questions (Q. 16 to 18). +6 marks will be given for each correct answer and –1 mark for each wrong answer. Answers to this Section are to be given in the form of nearest integerin four digits. Please follow as per example : (i.e. for answer : 1492.2 write 1492; 491.8 write 0492; 92.5 write 0092; 2.1 write 0002)

16.

If a student attempts 3 particular questions-one from section A and two from section B, the probability that he will get marks in only two questions is (assuming all ways to answer a question to be equally likely) 31 23 23 (A) (B) (C) (D) None 900 900 484

The probability that a student gets 10 marks if he attempts only 4 questions is 1 1   5  15 

 1  (C) 16   60 

(A)

A current of 4 A flows in a coil when connected to a 12V dc source. If the same coil is connected to a 12V, 50 rad/s ac source a current of 2.4 A flows in the circuit. Also find the power developed in the circuit if a 2500 µF capacitor is connected in series with the coil.

(B)

4 1   3  11 

(D)

1 1    4  15 

Passage # 2 (Ques. 4 to 6) a

∫

17. A capacitor of capacity 2 µF is charged to a potential difference of 12V. It is then connected across an inductor of inductance 0.6 mH. What is the current in the circuit at a time when the potential difference across the capacitor is 6.0 V ?

∫

f ( x) dx = ( f ( x) + f (− x)) dx

−a

If f(x) is odd, then

(i)

∫ f ( x) dx = 0

−a a

(ii) If f(x) is even, then

18. A ball of mass 100 g is projected vertically upwards from the ground with a velocity of 49 m/s. At the same lime another identical ball is dropped from a height of 98 metre to fall freely along the same path as that followed by the first ball. After some time the two balls collide and stick together and finally fall to ground. Find the time of flight of the masses. (g = 9.8 m/s2)

∫

f ( x) dx = 2 f ( x) dx

−a

This is one of the important property for the integrable function f(x) 1

∫ (x

+ x | cos x |)(sin −1 x 2 ) dx is equal to :

−1

(A) 1

(B) – 1

(D) 2

1/ 4

MATHEMATICS

  16 x 2 + 16 x + 41 + 40 x 2 + x + 1 + −1 / 4 

∫

 1 16 x 2 + 16 x + 41 − 40 x 2 + x + 1  cosx dx = k sin , 4  then the value of k is : (A) 10 (B) 0 (C) 20 (D) 30 π/ 4

Passage # 1 (Ques. 1 to 3) An objective test contains two sections : A and B each consisting of 10 questions. In section A, only one choice out of 4 choices is correct and student is awarded 1 mark for every correct answer. In section XtraEdge for IIT-JEE

B, one or more than one choice is (are) correct out of 4 choices and the student is awarded 3 marks if he (she) ticks only the correct choice and all the correct choices. There is no negative marking. In how many ways can a student answer to any question of section B (A) 11 (B) 5C2 + 5C4 (C) 24 – 1 (D) 15

∫ ( 1 + sin 2 x +

)

1 − sin 2 x dx is equal to

−π / 4

(A)

(B) 2 2

(D)

4 2

DECEMBER 2011

10.

PQ is perpendicular to AB and CD and PQ = 15 units, then possible position vectors of Q are x1iˆ + x2 ˆj + x3 kˆ and y1iˆ + y 2 ˆj + y3 kˆ . Find the value 3

∫ f (t ) dt is also periodic.

12.

Reason (R) : If α(x) is a differentiable and periodic function, then α′(x) is also periodic. Assertion (A) : The locus represented by xy + yz = 0 is a pair of perpendicular planes. Reason (R) : If a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 are perpendicular then a1a2 + b1b2 + c1c2 = 0

Assertion (A) : Locus of center of a variable circle touching two circles (x – 1)2 + (y – 2)2 = 25 and (x – 2)2 + (y – 1)2 =16 is an ellipse. Reason (R) : If a circle S2= 0 lies completely inside the circle S1= 0 then locus of center of a variable circle S = 0 which touches both the circles is an ellipse.

x −1 y −3 z−4 = = in the 3 5 2 plane 2x – y + z + 3 = 0 be L. A plane 7x + By + Cz + D = 0 is such that it contains the line L and perpendicular to the plane 2x – y + z + 3 = 0 then find the value of (B + C + D)/10. A circle touches the hypotenuse of a right angled triangle at its middle point and passes through the middle point of shorter side. If 3 unit and 4 unit be the length of the sides and 'r' be the radius of the circle, then find the value of '3r'.

13. The remainder when 2740 is divided by 12 14. Let f (x) = x3 – x2 – 3x – 1 and h(x) = f (x)/g(x) where h is a function such that (a) it is continuous every where except when x = – 1 (b) lim h(x) = ∞ and (c) lim h(x) = 1/2 x →∞

x → –1

FInd lim (4h(x) + f (x) + 2g(x)). x →0

15. In a triangle ABC, if sin A cos B = 1/4 and 3 tan A = tan B, then cot2A = This section contains Numerical response type questions (Q. 16 to 18). +6 marks will be given for each correct answer and –1 mark for each wrong answer. Answers to this Section are to be given in the form of nearest integerin four digits. Please follow as per example : (i.e. for answer : 1492.2 write 1492; 491.8 write 0492; 92.5 write 0092; 2.1 write 0002)

11. Let image of the line

∑ (x + y ) . i =1

Assertion (A) : Let f (x) be an even function which

is periodic, then g (x) =

The position vectors of two points A and C are 9iˆ − ˆj + 7kˆ and 7iˆ − 2 ˆj + 7 kˆ respectively. The point of intersection of the lines containing vectors AB = 4iˆ − ˆj + 3kˆ and CD = 2iˆ − ˆj + 2kˆ is P. If vector

16.

The vertices B and C of a triangle ABC lie on the lines 3y = 4x and y = 0 respectively and the side BC passes through the point (2/3, 2/3). If ABOC is a rhombus, O being the origin. If co-ordinates of vertex A is (α, β), then find the value of 5(α + β).

17.

Number of different words that can be formed using all the letters of the word "DEEPMALA", I two vowels are together and the other two are also together but separated from the first two.

18.

If a complex number z satisfies the conditions  4i 1+ i 2 – 3i  z – 4i   = 1 and z =  x + – 1 + i – 2i 3 + 4i  , z + 4i  – 2 – 3i – 3 + 4i 0   then x = DECEMBER 2011

XtraEdge for IIT-JEE

DECEMBER 2011

Based on New Pattern

IIT-JEE 2013 XtraEdge Test Series # 8

Time : 3 Hours Syllabus : Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus

Instructions : Section - I • Question 1 to 6 are passage based questions. +4 marks will be awarded for correct answer and -1 mark for wrong answer. • Question 7 to 9 are Reason and Assertion type question with one is correct answer. +4 marks and –1 mark for wrong answer.

• •

Passage # 2 (Ques. 4 to 6) A body of mass 1 kg moving along x-axis has velocity 4 m/s at x = 0. The acceleration and potential energy of body varies as shown in diagrams.

Passage # 1 (Ques. 1 to 3) In espresso coffee machines steam is passed into milk at room temperature for a brief time interval. Some of the steam condenses and the temperature or milk rises. Since the time for which the steam is passed is brief, one can ignore the heat lost to the environment and assume that the usual assumption of calorimetry : Heat lost = Heat gain is valid. 1.

a(m/s2)

U(J) 120

2 4

Steam at 100ºC is passed into milk to heat it. The amount of heat required to heat 150 g of milk from room temperature (20ºC) to 80ºC is (specific heat of capacity of milk = 4.0 kJ kg–1 K–1 specific latent heat of steam = 2.2 MJ kg–1 , specific heat capacity of water = 4.2 × 103 J kgK–1) (A) 3.6 × 104 J (B) 3.6 × 103 J 2 (C) 3.6 × 10 J (D) None of these How many grams of steam condensed into water in above question (A) 1.57 g (B) 15.7 g (C) 157 g (D) None of these

XtraEdge for IIT-JEE

If some of heat is allowed to escape to surrounding (temperature of surrounding is 20ºC) then this amount of steam (mentioned in question 22) is increase the temperature to (A) greater than 80ºC (B) less than 80ºC (C) equal to 80ºC (D) can't say anything

x(m)

–120

Work done by conservative forces when body moves from x = 0m to x = 8m is – (A) 0 J (B) 120 J (C) 240 J (D) – 240 J

Work done by external forces when body moves from x = 0m to x = 8m is – (A) 120 J (B) – 120 J (C) – 112 J (D) None of these

The change in kinetic energy when body moves from x = 0m to x = 8m is – (A) 256 J (B) 240 J (C) 128 J (D) 120 J

DECEMBER 2011

10.

If y = 4x2 – 4x + 7. Find the minimum value of 'y'.

11. A block of mass 1 kg starts slipping on a circular track of radius 2m and it is observed that when θ = 60º its speed is 4 m/s as shown in figure. Assuming O r = 2m size of block to be negligible and coefficient of θ friction between block and track is 0.5 frictional force (in N) on block when v θ = 60º is (g = 10 m/s2) – 12.

A projectile is fired with speed 'u' at angle 60º with horizontal. Velocity of projectile when it makes an angle 120º with initial direction of velocity is u' then ratio b/w u : u' is .

13.

Assertion (A) : In the flow-tube as the cross-section area decreases the flow velocity increases. Reason (R) : In ideal fluid flow the total energy per unit mass remains constant.

A block of mass 'm' is hanged vertically by a wire. Potential energy stored in wire u1. Potential energy stored in wire is u2. When mass hanged is doubled. Ratio u2 : u1 is.

14.

Assertion (A) : When a spring is elongated work done by spring is negative but when it compressed work done by spring is positive. Reason (R) : Work done by spring is path independent.

An electrically heating coil is placed in calorimeter containing 360 gm of H2O at 10ºC. The coil consumes energy at the rate of 90 W. The water equivalent of calorimeter and the coil is 40 g. The temperature of water after 10 minutes will be n/8.5 then find the value of 'n' -

15.

Variation of pressure at certain point is space is given by : P = P0 is 2πt cos 212 πt sin (220πt – π/2) The Beat Frequency is -

Assertion (A) : If in some case work done by a force is path independent then it must be conservative. Reason (R) : Work done by conservative forces in a round trip must be zero.

XtraEdge for IIT-JEE

Y 0 1 2 3 4 5 6 7 8 9

Z 0 1 2 3 4 5 6 7 8 9

16.

W 0 1 2 3 4 5 6 7 8 9

17.

A conical container of radius R = 1m and height H = 5m is filled completely with liquid. There is a hole at the bottom of container of area π × 10–3 m2 (see figure). Time taken to empty the conical container (in sec) is…… ………….. Take g = 10 m/s2 –

A pendulum of length l is given a horizontal velocity kgl at the lowest point of

vertical circular path as shown. In the subsequent motion the string gets slag at a certain point and the pendulum bob strikes the point of suspensión then the value of k is – 66

O l v = kgl

DECEMBER 2011

Instead of developing a catalyst, why didn’t Haber increase the rate of the reaction by raising the temperature ? (A) The Haber process is exothermic, so raising the temperature would have lowered the rate of the reaction (B) The Haber process is exothermic, so raising the temperature would have reduced the yield of the reaction (C) Higher temperatures would have caused an increased in pressure lowering the yield of the reaction (D) Higher temperatures might have decomposed the hydrogen

Which of the following graphs could also accurately reflect the establishment of equilibrium between nitrogen, hydrogen, and ammonia ?

y A B x

+1m

CHEMISTRY

Concentration

This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 1 to 6) Each question has 4 choices (A), (B), (C) and (D) out of which ONE OR MORE THAN ONE may be correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and – 1 mark for each wrong answer. Passage # 1 (Ques. 1 to 3) The Haber Process shown below : N2(g) + 3H2(g) 2 NH3(g) In 1912 Fritz Haber developed the Haber process for making ammonia from nitrogen and hydrogen. His development was crucial for the German war effort of world War Ι, providing the Germans with ample fixed nitrogen for the manufacture of explosives. The Haber process takes place at 500° C and 200 atm. It is an exothermic reaction .The graph below shows the change in concentrations of reactants and products as the reaction progresses.

(A)

Time

(B)

N2 NH3 H2 Time

(C)

H2 NH3 N2 Time

(D)

N2 NH3 H2

A B

Time

Passage # 2 (Ques. 4 to 6)

Entropy is measure of degree of randomness. Entropy is directly proportional to temperature. Every system tries to acquire maximum state of randomness or disorder. Entropy is measure of unavailable energy. Unavailable energy = Entropy × Temperature The ratio of entropy of vapourisation and boiling point of substance remains almost constant.

Time Even at the high temperatures, the conversion of nitrogen and hydrogen to ammonia was slow. In order to make the process industrially efficient Fritz Haber used a metal oxide catalyst. Which of the following did not accomplished by use of the metal oxide catalyst ? (A) The rate of production of ammonia increased (B) The energy of activation was raised (C) The equilibrium shifted to the right (D) activation energy equals to zero

XtraEdge for IIT-JEE

H2 NH3 N2

Conc

–2m

Conc

A small body is released from point A of the smooth parabolic path y = x2. Where y is vertical axis and x is horizontal axis at ground as shown. The body leaves the surface from point B. If g = 10 m/s2 then the total horizontal distance travelled by body before it hits ground is –

Conc

18.

Which of the following process have ∆S = – ve ? (A) Adsorption (B) Dissolution of NH4Cl in water (C) H2 → 2H (D) 2NaHCO3(s) → Na2CO3 + CO2 + H2O DECEMBER 2011

Observe the graph and identify the incorrect statement(s)

Entropy

Β ∆Svap

Α ∆Sfusion

T1 T2 Temperature (A) T1 is melting point, T2 is boiling point (B) T1 is boiling point, T2 is melting point (C) ∆Sfusion is more than ∆Svap (D) T2 is lower than T1

The law of Thermodynamics invented by Nernst, which helps to determine absolute entropy is (A) Zeroth law (B) 1st law nd (C) 2 law (D) 3rd law

10. For the Ca atom calculate total No. of e– which have m = –1

Assertion (A) : At zero degree Kelvin the volume occupied by a gas is negligible. Reason (R) : All molecular motion ceases at 0 K.

Assertion (A) : Compressibility factor for hydrogen varies with pressure with positive slope at all pressures. Reason (R) : Even at low pressures, repulsive forces dominate hydrogen gas.

11. The molecular formula of a non-stoichiometric tin oxide containing Sn (II) and Sn (IV) ions is Sn4.44O8. Therefore, the molar ratio of Sn (II) to Sn (IV) is approximately

How much volume (in mL) 0.001 M HCl should we add to 10 cm3 of 0.001 M NaOH to change its pH by one unit ?

13.

What is the sum of total electron pairs (b.p. + l.p.) present in XeF6 molecule ?

14. The number of geometrical isomers of CH3CH=CH–CH=CH–CH=CHCl is. 15. No. of π bond in the compound H2CSF4 is. This section contains Numerical response type questions (Q. 16 to 18). +6 marks will be given for each correct answer and –1 mark for each wrong answer. Answers to this Section are to be given in the form of nearest integerin four digits. Please follow as per example : (i.e. for answer : 1492.2 write 1492; 491.8 write 0492; 92.5 write 0092; 2.1 write 0002)

Assertion (A) : Enthalpy of graphite is lower than that of diamond. Reason (R) : Entropy of graphite is greater than that of diamond.

XtraEdge for IIT-JEE

12.

16.

The ion An+ is oxidised to AO3– by MnO4–, changing to Mn2+ in acidic solution. Given that 2.68 × 10–3 mol of An+ requires 1.61 × 10–3 mol of MnO4–. What is the value of n ?

17.

Standard heat of formation of HgO(s) at 298 K and at constant pressure is –90.8 kJ mol–1. Excess of HgO(s) absorbs 41.84 kJ of heat. Calculate the mass of Hg (in g) that can be obtained at constant volume and at 298 K. (Hg = 200.6g mol–1) DECEMBER 2011

18.

A near ultraviolet photon of 300 nm is absorbed by a gas and then re-emitted as two photons. One photon is red with wavelength 760 nm. What would be the wavelength (in nm) of the second photon ?

MATHEMATICS

(B)

3x–y+4+2 3 =0

(C)

3x–y–4+2 3 =0

(D)

3x–y–4–2 3 =0

Assertion (A) : If a, b > 0 and a3 + b3 = a – b, then a2 + b2 < 1 1 (a + b) Reason (R) : If a, b > 0, then ab < 2

Assertion (A) : The line bx – ay = 0 will not meet the x2 y2 hyperbola 2 – 2 = 1 (a > b > 0) a b Reason (R) : The line y = mx + c does not meet the x2 y2 hyperbola 2 – 2 = 1 if c2 = a2 m2 – b2. a b

Assertion (A) : The solution set of the inequality

This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 1 to 6) Each question has 4 choices (A), (B), (C) and (D) out of which ONE OR MORE THAN ONE may be correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and – 1 mark for each wrong answer. Passage # 1 (Ques. 1 to 3) A circle C1 of radius 2 units rolls on the outerside of the circle C2 : x2 + y2 + 4x = 0, touching it externally. 1. If the line joining the centres of C1 and C2 makes an angle of 60º with the x-axis, equation of a common tangent to them is -

(A) x +

3y–2=0

Area of the quadrilateral formed by a pair of tangents from a point on C3 to the circle C2 with a pair of radii at the points of contact of the tangents is (A) 2 3 sq. units

(B) 4 3 sq. units

(C)

(D) 3 3 sq. units

3 sq. units

If the line joining the centres of C1 and C2 is perpendicular to the x-axis; equation of the chord of contact of the tangents drawn from the centre of C2 to the circle C1 is (B) y + 2 = 0 (A) y – 2 = 0 (C) y – 3 = 0 (D) y + 3 = 0

 x 2 + x  < 0 is (– 4, – 3) ∪ (8, ∞). log 0.7  log 6  x + 4   Reason (R) : For x > 0, loga x is an increasing function if a > 1 and a decreasing function if 0 < a < 1.

Passage # 2 (Ques. 4 to 6) AL, BM and CN are diameter of the circumcircle of a triangle ABC. ∆1, ∆2, ∆3, and ∆ are the areas of the triangles BLC, CMA, ANB and ABC respectively, 4. ∆1 is equal to (A) 2R2 sin A cos B cos C (B) 2R2 sin A sin B cos C (C) 2R2 cos A cos B sin C (D) 2R2 sin A sin B sin C 5.

If BL2 + CM2 + AN2 = x and CL2 + AM2 + BN2 = y then (A) x + y = 0 (B) x – y = 0 (C) x + y = 24R2 – 2(a2 + b2 + c2) (D) none of these

∆1 + ∆2 + ∆3 is equal to (A) 2∆ (C) ∆

XtraEdge for IIT-JEE

(B) 3∆ (D) none of these 69

DECEMBER 2011

X 0 1 2 3 4 5 6 7 8 9

10.

11.

Y 0 1 2 3 4 5 6 7 8 9

Z 0 1 2 3 4 5 6 7 8 9

Elements Named for Places

0 1 2 3 4 5 6 7 8 9

This is an alphabetical list of element toponyms or elements named for places or regions. Ytterby in Sweden has given its name to four elements: Erbium, Terbium, Ytterbium and Yttrium. •

Americium : America, the Americas

•

Berkelium : University of California at Berkeley

•

Californium : State of California and University of California at Berkeley

•

Copper : probably named for Cyprus

•

Darmstadtium : Darmstadt, Germany

•

Dubnium : Dubna, Russia

•

Erbium : Ytterby, a town in Sweden

•

Europium : Europe

•

Francium : France

•

Gallium : Gallia, Latin for France. Also named for Lecoq de Boisbaudran, the element's discoverer (Lecoq in Latin is gallus)

•

Germanium : Germany

•

Hafnium : Hafnia, Latin for Copenhagen

•

Hassium : Hesse, Germany

•

Holmium : Holmia, Latin for Stockholm

•

Lutetium : Lutecia, ancient name for Paris

•

Magnesium : Magnesia prefecture in Thessaly, Greece

•

Polonium : Poland

•

Rhenium : Rhenus, Latin for Rhine, a German province

•

Ruthenium : Ruthenia, Latin for Russia

The number of natural numbers which are smaller than 2.108 and which can be written by means of the digits 1 and 2 is ............. .

•

Scandium : Scandia, Latin for Scandinavia

•

Strontium : Strontian, a town in Scotland

1 ( 3 – i), find the smallest value of positive 2 integer n for which (z89 + i97)94 = zn.

•

Terbium : Ytterby, Sweden

•

Thulium : Thule, a mythical island in the far north (Scandinavia?)

•

Ytterbium : Ytterby, Sweden

•

Yttrium : Ytterby, Sweden

If two of the lines represented by x4 + x3 y + cx2 y2 – xy3 + y4 = 0 bisect the angle between the other two, then the value of |c| is For n > 3, a, b ∈ R, let n

n(n – 1)...(n – r + 1) (a – r) (b – r) S(n, a, b)= (–1) r r! r =0

∑

 1 1 Find S  5, ,   2 7 12.

Find the degree of the remainder when x2007 – 1 is divided by (x2 + 1) (x2 + x + 1).

13.

Let l be the length of the interval satisfying the inequality 1 log 6 (7). log6(x + 2) (x + 4) + log1/6 (x + 2) < 2 Find the value of l.

14.

The number of pairs (x, y) satisfying the equation sin x + sin y = sin (x + y) and |x| + |y| = 1 is.

15.

If a = ( 0, 1, – 1) and c = (1, 1, 1) are given vectors, the |b|2 where b satisfies a × b + c = 0 and a . b = 3 is.

16.

17.

18.

If z =

If α + β = γ and tan γ = 22, a is the arithmetic and b is the geometric mean respectively between tan α and a3 tan β, then the value of is equal to (1 – b 2 ) 3

XtraEdge for IIT-JEE

DECEMBER 2011

MOCK TEST PAPER-1 CBSE BOARD PATTERN CLASS # XII

SUBJECT : PHYSICS , CHEMISTRY & MATHEMATICS Solutions will be p ublished in next issue General Instructions : Physics & Chemistry • Time given for each subject paper is 3 hrs and Max. marks 70 for each. • All questions are compulsory. Marks for each question are indicated against it. • Question numbers 1 to 8 are very short-answer questions and carrying 1 mark each. • Question numbers 9 to 18 are short-answer questions, and carry 2 marks each. • Question numbers 19 to 27 are also short-answer questions, and carry 3 marks each. • • Question numbers 28 to 30 are long-answer questions and carry 5 marks each. Use of calculators is not permitted. • General Instructions : Mathematics • Time given to solve this subject paper is 3 hrs and Max. marks 100. • All questions are compulsory. • The question paper consists of 29 questions divided into three sections A, B and C. Section A comprises of 10 questions of one mark each. Section B comprises of 12 questions of four marks each. Section C comprises of 7 questions of six marks each. All question in Section A are to be answered in one word, one sentence or as per the exact requirement of the question. • • There is no overall choice. However, internal choice has been provided in 4 questions of four marks each and 2 question of six marks each. You have to attempt only one of the alternatives in all such questions. Use of calculators is not permitted. •

PHYSICS 1.

What is the phase difference between two points on a wave front

Why stars seems to be twinkling ?

Name the characteristics of electromagnetic wave that (i) increases (ii) remains constant in the electromagnetic spectrum as one moves from infrared to ultraviolet region.

What is total energy of an electron revolving in first orbit of Hydrogen atom .

What are digital and analog signals ?

Name any two basic properties of electric charge.

Out of the two bulbs marked 25 W and 100 W which one has higher resistance ?

What is displacement current ?.

A thin glass prism has a minimum deviation δm in air. State with reason, how the angle of minimum deviation will change if the prism is immersed in a liquid of refractive index greater than 1

XtraEdge for IIT-JEE

10. Explain how the interference fringe width in young's double slits experiment will change if (i) separation between the two slits is decreased (ii) wavelength of light increased. 11. In an atom, two electrons moves around the nucleus in a circular orbit of radius R and 4R. Calculate the ratio of speeds of two electrons. 12. Which type of semiconductor is better out of p-type and n-type ? 13. Relate input frequency and output frequency of a half wave rectifier and a full wave rectifier 14. Define the term modulation index for an AM wave. What would be the modulation index for an AM wave for which the maximum amplitude is a while the minimum amplitude is 'b' ? 15. Distinguish between point to point and broadcast communication modes. Give one example of each 16. Why does the conductivity of a semi conductor change with the rise in temperature ?

DECEMBER 2011

17. An electric dipole is held in uniform electric field. (i) Show that no translatory force acts on it. (ii) Derive an expression for the torque acting on it.

Why objective lens of an astronomical telescope is of large size ? what defects can be remoned by using reflecting telescope in place of refracting telescope ? IF length of an astronomical telescope is 100 cm and its magnifying power is 19 then calculate focal lengths of objective and eye-piece lens.

18. Define angle of dip at a given place. What is the value of angle of dip on the equator ? 19. When a Uranium nucleus (U238) originally at rest decays by emitting alpha particle having speed u. Find the recoil speed of residual nucleus.

29. Explain with the help of a labelled diagram, the principle, construction and working of a transformer. OR An a.c. generator consists of a coil of 50 turns and area 2.5 m2 rotating at an angular speed of 60 rad s–1 in a uniform magnetic field B = 0.30 T between two fixed pole pieces. The resistance of the circuit including that of the coil is 500 Ω. (i) Find the maximum current drawn from the generator. (ii) What will be the orientation of the coil w.r.t. the magnetic field to have (a) maximum (b) zero magnetic flux. (iii) Would the generator work if the coil were stationary and instead the poles were rotated with same speed as above.

20. If focal lengths of objective and eye-piece lens of a compound microscope is 2cm and 3cm respectively and distance between both the lenses is 15cm then calculate distance of object from objective lens if final image forms at infinity. 21. If a gas is at temp T, then derive an expression for debroglie wavelength of its molecule of mass m. 22. For a common emitter transistor amplifier current gain is 72. Calculate the base current for which emitter current is 8.9mA.

30. Calculate the electric field intensity for following points due to a uniformly charged non-conducting sphere. Represent the results by graph – (A) at any point outside the sphere (B) at any point on the surface of sphere (C) at any point inside the sphere. OR What is a Capacitor ? Explain its principles. Derive the relations for equivalent capacity of series and parallel combinations of capacitors.

23. Give one use of each of the following (i) microwaves (ii) infra-red waves (iii) ultraviolet radiation (iv) gamma rays.

State the principle of potentiometer with the help of circuit diagram, describe a method to find the internal resistance of a primary cell. r 25. A proton is shot into the magnetic field B = 0.8ˆjT with a velocity (2 × 106 ˆi + 3 × 106 ˆj) ms–1. Calculate

24.

CHEMISTRY

the radius and pitch of the helix path followed by proton.

26. Derive an expression for the torque on a rectangular coil of area A, carrying a current I placed in a magnetic field B. The angle between the direction of B and vector perpendicular to plane of coil is θ.

OH COOC2H5

27. Using Kirchhoff's laws in the given network, calculate the values of I1, I2 and I3. A B C 5Ω

2Ω

3Ω

12V

6V F

28. With the help of ray diagram, explain the phenomenon of total interval reflection. Obtain the relation between critical angle and refractive index of the medium. OR XtraEdge for IIT-JEE

Write the IUPAC names of the following : Br

How would ethylamine ?

you

convert

methylamine

into

What are essential amino acids ? Give two example.

What is meant by the term peptization ?

Identify the reaction order if the unit of rate constant is sec–1.

In an alloy of gold and cadmium if gold crystallizes in cubic structure occupying the corners only and cadmium fits into edge centre voids, what is the formula of the alloy ?

DECEMBER 2011

Which oxide of nitrogen has oxidation number of N same as that in nitric acid.

What is froth floatation for which ores it is used ?

Explain as to why haloarenes are much less reactive than haloalkanes towards nucleophilic substitution reactions.

19. Name the reagents which are used in the following conversions : (i) A primary alcohol to an aldehyde (ii) Butan-2-one to butan-2-ol (iii) Phenol to 2,4,6-tribromopheno 20. Give the structures of following polymers. (A) Perlon (B) Orlon (C) Neoprene 21. What are tranquilizers ? Explain with example tranquilizers are neurologically active drugs which are used to reduce strain or anxiety.

10. Assign a reason for each of the following statements : (i) Alkylamines are stronger bases than arylamines (ii) Acetonitrile is preferred as solvent for carrying out several organic reactions.

22. Name any three fat soluble vitamins & their deficiency disease.

11. Write one chemical equation each to exemplify the following reactions : (i) Carbylamine reaction (ii) Hofmann bromamide reaction

23. (a) State the products of electrolysis obtained on the cathode and the anode in the following cases : [2 + 1] (i) A dilute solution of H2SO4 with platinum electrodes. (ii) An aqueous solution of AgNO3 with silver electrodes (b) Write the cell formulation and calculate the standard cell potential of the galvanic cell in which the following reaction takes place : Fe2+ (aq) + Ag+ (aq) → Fe3+ (aq) + Ag (s) Calculate ∆rGº for the above reaction. [Given : E oAg + / Ag = + 0.80 V ;

12. Identify A and B in the following : CH2Br (i) (ii) R2CO

CN–

NH3

A A

LiAlH4

Ni/H2

B B

13. Write any two feature which physisorption from Chemisorption.

distinguish

14. The decomposition of a compound is found to follow a first order rate law. If it takes 15 minutes for 20% of the original material to react, calculate(i) Specific rate constant. (ii) the time at which 10% of the original material remains unreacted.

E oFe3+ / Fe 2 + = + 0.77 V ; 1 F = 96500 C mol–1]

24. (a) Explain each of the following with a suitable example : (i) Paramagnetism (ii) Frenkel defect in crystals (b) An element occurs in bcc structure with cell edge 300 pm. The density of the element is 5.2 g cm–3. How many atoms of the element does 200 g of the element contain ?

15. Prove that the time required for the completion of 3/4th of the reaction of first order is twice the time required for the completion of half of the reaction.

25. (a) Describe the preparation of KMnO4 from pyrolusite ore. (b) Among ionic species, Sc+3, Ce+4 and Eu+2, which one is good oxidizing agent ? (Atomic numbers : Sc = 21, Ce = 58, Eu = 63 )

16. When a certain conductivity cell was filled with 0.1 M KCl, it has a resistance of 85 Ω at 25ºC. When the same cell was filled with an aqueous solution of 0.052 M unknown electrolyte the resistance was 96 Ω. Calculate the molar conductivity of the unknown electrolyte at this concentration. (Specific conductivity of 0.1 M KCl = 129 × 10–2 ohm–1cm–1).

26. (a) Assign a reason for each of the following statements : (i) Di-methyl amine is stronger base than tri-methyl amine in aq. solution. (ii) Explain why ? Benzamide is more basic in comparision to acetamide.

17. Give methods of preparation of XeO3 and XeOF4.

(b) How would you convert aryl amine into cynobenzene.

18. Explain giving reason : (i) Copper (I) is diamagnetic whereas copper (II) is paramagnetic (ii) K2PtCl6 is a well known compound whereas the corresponding Ni compound does not exist. XtraEdge for IIT-JEE

DECEMBER 2011

27. (a) Why chelated complexes are more stable than unchelated complexes ? (b) Write IUPAC names of : (i) K3 [Al(C2O4)3] (ii) [Mn (H2O)6] S

MATHEMATICS Section A

28. You are provided with four reagents : LiAlH4, I2/NaOH, NaHSO3 and Schiff's reagent (a) Write which two reagents can be used to distinguish between the compounds in each of the following pairs : [3+2=5] (i) CH3CHO and CH3COCH3 (ii) CH3CHO and C6H5CHO (iii) C6H5COCH3 and C6H5COC6H5 (b) Account for the following : (i) The order of reactivity of halogen acids with ether is HI > HBr > HCl. (ii) The pKa value of chloroacetic acid is lower than the pKa value of acetic acid. Or (a) An organic compound contains 69.77% carbon, 11.63% hydrogen and the rest is oxygen. The molecular mass of the compound is 86. It does not reduce Tollens' reagent but forms an addition compound with sodium hydrogen sulphite and gives a positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acids. Deduce the possible structure of the organic compound. (b) State reasons for the following : (i) Monochloroethanoic acid has a higher pKa value than dichloroethanoic acid. (ii) Ethanoic acid is a weaker acid than benzoic acid.

Find order and degree of diff. equation 2

1  dy  =2   + dy  dx  dx

∫ tan

−1

Evaluate :

x dx .

Find unit vector in the direction of vector →

a = 2 iˆ + 3 ˆj + kˆ .

29. (a) Density of 0.8 M aqueous solution of H2SO4 is 1.06 g mL–1. Calculate the concentration of solution in (i) mol fraction (ii) molality (molar mass of H2SO4 = 98 g mol–1). (b) Heptane and octane form ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa, respectively. What will be the vapour pressure, in bar, of a mixture of 25.0 g of heptane and 35.0 g of octane ?

→

If a = iˆ + ˆj – 3 kˆ and b = ˆj + 2 kˆ . Find |2 b × a |

Find the angle between the 3x – 6y – 2z = 7 and 2x + 2y – 2z = 5

Construct a matrix of order 3 × 3, whose element aij is given by rule aij = (i + j)2.

Find a matrix X such that A + 2B + X = 0, where 2 − 1 − 1 1 A=   ; B =  0 2 . 3 5    

Find value of

Let f = {(1, 3), (2, 1), (2, 1), (3, 2)} and g = {(1, 2), (2, 3), (3, 1)} then find gof (1).

two

planes

sin 10º − cos 10º . sin 80º cos 80º

1 − x 2 ) w.r.to x.

10. Find derivative of sin–1 (2 x

Section B

30. (a) Explain the following giving reasons : (i) H3PO3 is diprotic. (ii) Nitrogen does not form pentahalides (iii) SF6 is well known but SH6 is not known (b) Complete the equations : (i) XeF6 + PF5 → (ii) AlN + H2O →

11. A and B are mutually exclusive events of an experiment. If P (Not A) = 0.65, P (A ∪ B) = 0.65 and P (B) = p. Find the value of p. x 1 + sin

x cos x   dx. 

12. Evaluate

∫e

13. Evaluate

∫ (2 + sin θ)(3 + sin θ) dθ

 

cos 2 x cos θ

OR XtraEdge for IIT-JEE

DECEMBER 2011

Evaluate :

∫ 2x

3x + 1 2

− 2x + 3

Section C

dx .

23. The bag A contains 5 red and 3 green balls and bag B contains 3 red and 5 green balls. One ball is drawn from bag A and two from bag B. Find the Probability that of the three balls drawn two are red and one is green.

14. Find the diff. equation of the family of curves y = a sin (bx + c), a and c being parameter. OR dy Solve the differential equation : x − y − 2x 3 = 0 . dx →

→

24. Find area between x2 + y2 = 4 and line x = 3 y in first quadrant OR Find the area of the region [(x, y) : y2 ≤ 4x, 4x2 + 4y2 ≤ 9]

→

15. If a = iˆ – ˆj + 2 kˆ , b =2 iˆ + ˆj – 3 kˆ , c = iˆ +2 ˆj – kˆ , →

→

→ →

→

→ →

→

verify that a × ( b × c ) = ( a . c ) b –( a . b ) c . 16. Find the vector equation of the plane passing through →

25. Find the image of the point (1, 6, 3) in the line x y −1 z − 2 = = . 1 2 3

the intersection of the planes r .(2 iˆ –7 ˆj + 4 kˆ ) =3 →

and r .(3 iˆ – 5 ˆj + 4 kˆ ) + 11=0 and passing through the point (–2, 1, 3). OR Find the distance of the point (2, 3, 4) from the plane 3x + 2y + 2z + 5 = 0, measured parallel to the line x+3 y−2 z = = 3 6 2

26. A dealer wishes to purchase a number of fans and radios. He has only Rs 5,760 to invest and has a space for at most 20 items. A fan costs him Rs. 360 and a radio Rs. 240. His expectation is that he can sell a fan at a profit of Rs. 22 and a radio at a profit of Rs. 18. Assuming that he can sell all the items he buys, how should he invest his money for maximum profit ? Translate the problem as LPP and solve it graphically.

17. Using properties of determinats, prove that 1+ a 1 1  1 1 1 1 1+ b 1 = abc 1 + + +   a b c 1 1 1+ c

27. Evaluate the following integrals as limit of sums 2

∫ (x

OR 28.

18. If f (x) = cos x, g(x) = x2, then show that fog (x) ≠ gof (x) 19. Show that curve xy = a2 and x2 + y2 = 2a2 touch each other. 20. Find the derivative of xex from first principle. 21. If cos y = x cos (a + y), then prove that dy cos 2 (a + y ) = . dx sin a

XtraEdge for IIT-JEE

function f (x) is defined by    

+ 3) dx .

4 − 5 − 11 If A = 1 − 3 1  find A–1 2 3 − 7 

22. Determine f (0) so that (4 x − 1) 3 f (x) = x2  x  sin   log1 + 3 4 

1 − 1 0  Given that A = 2 3 4 and 0 1 2 2 − 4  2  B = − 4 2 − 4 find A.B. Use this to solve the  2 − 1 5  following system of equations x–y=3 2x + 3y + 4z = 17 y + 2z = 7 OR Using matrix method solve the following system of linear equations 2x – 3y + 5z = 11 3x + 2y – 4z = –5 x + y – 2z = –3

29. If the sum of the lengths of the hypotenuse and a side of right angled triangle is given, show that the area of triangle is maximum when the angle between them is π/3.

becomes continuous at x = 0.

DECEMBER 2011

XtraEdge Test Series ANSWER KEY IIT- JEE 2012 (December issue) PHYSICS Ques Ans Ques Ans

1 A,C 10 3

2 C 11 4

3 A 12 6

4 D 13 4

5 A 14 8

6 A 15 7

7 A 16 0012

8 D 17 0049

9 D 18 0005

6 A,B,C 15 6

7 A 16 0108

8 A 17 0001

9 D 18 0041

6 B,D 15 3

7 D 16 0012

8 C 17 1440

9 D 18 0000

6 D 15 4

7 B 16 0200

8 D 17 0004

9 D 18 0008

6 D 15 4

7 C 16 0002

8 A 17 0094

9 B 18 0496

6 B,C 15 6

7 B 16 0766

8 C 17 0010

9 A 18 1331

C HE M ISTR Y Ques Ans Ques Ans

1 A 10 4

2 B 11 4

3 A 12 3

4 C 13 1

5 A,B 14 4

MATHEMATICS Ques Ans Ques Ans

1 B,C,D 10 6

2 A 11 3

3 C,D 12 5

4 C 13 9

5 C 14 6

IIT- JEE 2013 (December issue) PHYSICS Ques Ans Ques Ans

1 A 10 6

2 A 11 7

3 B 12 1

4 C 13 4

5 B 14 5

C HE M ISTR Y Ques Ans Ques Ans

1 B,C,D 10 4

2 B 11 1

3 A 12 1

4 A 13 6

5 B,C,D 14 1

MATHEMATICS Ques Ans Ques Ans

1 A,B,C 10 6

XtraEdge for IIT-JEE

2 B 11 0

3 C,D 12 3

4 A 13 5

5 C 14 6

DECEMBER 2011