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Impatience never commanded success.

Volume - 7 Issue - 2 August, 2011 (Monthly Magazine) Editorial / Mailing Office : 112-B, Shakti Nagar, Kota (Raj.) Tel. : 0744-2500492, 2500692, 3040000 e-mail : xtraedge@gmail.com

Editorial

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Editor : Pramod Maheshwari XtraEdge for IIT-JEE

Dear Students, Life is like a buffet offering a variety of choices every day and still having a lot left and enough for everyone to have. But everything has a price tag. For instance to be fit we have to maintain a proper balance of vitamins, minerals proteins, carbohydrates, fats and fibres. The same is true of all our activities and responsibilities. if we undertake too many projects, we will find it difficult to handle all of them. Overseeing can create a long of health and self-image problems for us. If we do not maintain a proper balance of our activities to take care of our physical, mental, spiritual and social needs, we create stress, disharmony, unhappiness and eventually dysfunction in our lives. Organizing life will enable us to focus on things that are really important and for which we feel that we do not have time. This approach will also enable us to have more time3to relax and create time to do the things which we enjoy doing. Other benefits include making good note, being punctual and spending more time on really important things which require attention. The real key to success is to concentrate on what matters most to us it our lives. It also includes restricting the number of jobs we undertake at any given time or on any given day and concentrating on the most important ones. Focus is the most important things in life. Focusing on absolutely essential items and doing away with unimportant or not immediately relevant items is the best way to achieve success, health, happiness and our personal well-being. It is a fact that the definition of perfectionism varies from person to person and place to place. Their never were and there never will be perfect circumstances. It is futile to wait for them before starting to achieve your goal. Instead get started and create the right circumstances. It is the right self motivation which can fuel your enthusiasm and make you start working towards attaining your goal. Sometimes you can be inhibited by an external obstacle. The logical answer for this problem is to come up with solutions that can put you on the trail towards your goal again. All of us have faced the prospectus of temporary derailment of our goals. Never allows your courage to desert you. The only solution is to confront your fears. Isolate your fears and determine as to what you are afraid of imagine the worst that can happen in case any fear does have a concrete expression. Presenting forever positive ideas to your success. Yours truly

Pramod Maheshwari, B.Tech., IIT Delhi

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AUGUST 2011


Volume-7 Issue-2 August, 2011 (Monthly Magazine) NEXT MONTHS ATTRACTIONS Much more IIT-JEE News. Know IIT-JEE With 15 Best Questions of IIT-JEE Challenging Problems in Physics,, Chemistry & Maths Key Concepts & Problem Solving strategy for IIT-JEE. Xtra Edge Test Series for JEE- 2012 & 2013

CONTENTS INDEX

PAGE

Regulars .......... NEWS ARTICLE

3

• Krishna Pagilla Named Water Environment Federation Fellow • IIT-Gandhinagar all set to double up research facility

IITian ON THE PATH OF SUCCESS

5

Mr. Abhishek Kumar Bharti & Mr. Supreet Joshi

KNOW IIT-JEE

6

Previous IIT-JEE Question

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

S

Success Tips for the Months • What matters is not what you have, but what you can do. • It is not about what you can't do. It is about what you can do.

8-Challenging Problems [Set # 4] Students’ Forum Physics Fundamentals • Capacitor (Part-2) • Work, Power, Energy & conservation Law

CATALYSE CHEMISTRY

• If you want to be smart, find friends who are smarter than you are. • Don't be irreplaceable. If you can't be replaced, you can't be promoted. • Never test the depth of the water with both feet. • Many of life's failures are people who did not realize how close they were to success when they gave up.

XtraEdge for IIT-JEE

30

Key Concept • Reaction Mechanism • Solid State Understanding : Physical Chemistry

• Never mind what others do; do better than yourself, beat your own record from day to day, and you are a success. • If you don't know where you're going, who will follow you?

14

DICEY MATHS

39

Mathematical Challenges Students’ Forum Key Concept • Vector • Permutation & Combination

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

51

Class XII – IIT-JEE 2012 Paper Class XI – IIT-JEE 2013 Paper

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AUGUST 2011


Krishna Pagilla Named Water Environment Federation Fellow IIT Civil, Architectural & Environmental Engineering Professor Krishna Pagilla has been named a fellow of the Water Environment Federation (WEF), the organization announced this month. The WEF Fellows program "honors the professional achievements, stature and contributions of WEF members to the preservation and enhancement of the global water environment." A total of 17 WEF members were approved by the board of trustees to be honored during this year of the fellowship program.

"WEF takes a great deal of pride in the vast knowledge base of our members and in honor of that, felt it was very important to provide a distinguished recognition program," said WEF President Jeanette Brown. "We are very pleased to celebrate not only the personal achievements of this years' Fellows but also the invaluable role WEF has as a leading water quality organization." IIT-Gandhinagar all set to double up research facility The Indian Institute of Technology – Gandhinagar is all set to double up its research capacities. The institute is expanding its campus and adding another 2,000 sq mtrs of research facilities and laboratories. "We had developed 2,000 sq mtrs of research facilities ever since inception of IIT Gandhinagar. By August, we XtraEdge for IIT-JEE

will be adding another 2,000 sq mtrs of research building that will have advanced laboratories and other equipment," said Sudhir Jain, director of IIT Gandhinagar. "We have started to develop some very fine laboratories which comprise brand new equipment and technology as compared to other 30 year old laboratories," said Jain. The research facilities will house latest technology and equipment that can be used by chemical, civil, mechanical and other engineering streams alike.

PRADEEP

KUMAR, A L U M N U S , I S N E W CVC:

IIT

Central Vigilance Commissioner (CVC) to head India’s top anticorruption watchdog Kumar was selected by a three-member panel comprising Prime Minister Manmohan Singh, Home Minister P. Chidambaram and BJP leader Sushma Swaraj, Leader of Opposition in the Lok Sabha. Kumar’s selection came four months after the Supreme Court quashed the appointment of P.J. Thomas to the post as he faced corruption charge. Known to be a man of few words, Kumar boasts of an impressive resume that includes important postings in key ministries. He did his graduation in Electrical Engineering from the Indian Institute of Technology (IIT)-Delhi before joining the IAS in 1972. In 1986, he went to University of Wales, UK, on study leave to do a Masters in Economics and Social Sciences. Kumar, who turns 62 on Sep 29, has also worked as joint secretary in the department of heavy industry, additional secretary in the ministry of coal, chairman, National Highways Authority of India and secretary (Disinvestment) in the ministry of finance. 3

He has also served on the boards of a number of leading companies including Bharat Heavy Electricals Ltd., Maruti Udyog Ltd., Andrew Yule Ltd., Hindustan Paper Corporation Ltd., Coal India Ltd. and Neyveli Lignite Corporation Ltd.

IIT PROFESSOR BAGS G D BIRLA AWARD A professor of IIT-Mumbai was today selected for the prestigious G D Birla Award for Scientific Research for his contributions in the area of electronic communications. 48-year-old Subhasis Chaudhuri, who works in Department of Electrical Engineering, was conferred the award for his "important contributions" in the area of electronic communications. The award, constituted by K K Birla Foundation, carries a cash prize of Rs 1.50 lakh and is being given to scientists below the age of 50 years. Chaudhury joined IIT Mumbai in 1990 and is currently serving as an Institute Chair Professor and the Dean of International Relations. He works in the area of computer vision and image processing. "He has made original and substantive contributions in the specific areas of depth recovery from defocused observations and generation of super resolution images,"

Cracking IIT JEE The formula of JEE- has become comprehension based totally. One should clear every doubt and concept in each subject and topics. Making the effective notes while attending self study is the key of success to go depth knowledge of subject this gives more confidence to win over subject.

AUGUST 2011


All eyes on IIT-Delhi's Open House NEW DELHI: A low-cost Braille display for the visually-challenged, a page-turning machine for the physically-challenged, a body posture correction alarm and a voice-based complaint management system for municipal complaints are among the projects that will be showcased at Indian Institute of Technology-Delhi's Open House event -- I2 Tech 2011 The Open House will include display and showcasing of laboratories and research projects by students, staff and faculty along with industry interface with experts from major industries. “We are expecting over 1,000 people from industries, including 12 firms, to formally interact with our students during the Open House,” said IIT-D Director Prof. Surendra Prasad on Wednesday. Highlighting IIT's key achievements and progress during 2010-11, Prof. Prasad said the Institute was “promoting high-impact research”. Several departments at IIT-D are involved in research-based projects including smart materials like biotextiles (Textile Department). The research focuses on creating artificial tissue cells like blood vessels and bladder tissue that can prove useful in medical treatment. Naval and Army officers are also engaged in research to develop useful models for strategic use. The Centre for Applied Electronics is working on creating a self-sustained remote acoustic surveillance platform. The system is designed to detect lowpower waves and also act as a communication system using digital signal processing hardware. Another research project involving ultra-wide band communication would find widespread use in ground penetration radars, anti-vehicle collision systems in cars and in medical research. Speaking further about the Institute's achievements last year, Prof. Prasad said IIT-Delhi was in the process of setting up a centre for nanofabrication and nano-devices within the campus. The centre, being set up at a cost of nearly Rs.50 crore, would XtraEdge for IIT-JEE

focus on non-silicon devices, unlike similar centres at IIT-Bombay and Indian Institute of Science, Bangalore, that focus on silicon-based devices. IIT-Delhi has also taken up industrybased initiatives with companies like PepsiCo, Shell and Indus Towers. The Institute also approved 44 works last year for patents, according to Prof. Prasad. The I2Tech would include IIT's useful technological contributions including “Ferrite Phase Shifter” that are widely used in Defence radar operations, the “Fru wash” technology that prevents deterioration of unrefrigerated fresh vegetable and fruits by forming an organic coat over them and “axle counter for automatic signalling in railways” that is being used widely by Indian Railways.

IIT-Delhi & Aston setting up green plant to run fruit processing units Aston University, UK, and the Indian Institute of Technology (IIT), Delhi, are overseeing the construction of a combined heat and power plant (CHP) that will provide heat, steam and electricity to rice mill, fruit and vegetable processing plants and water distillation unit in a remote village in northern India. Fuelled by crop waste such as rice husks and prosopis wood, the 300kw biomass-solar plant is being built as part of a three-year project. The combination of solar and biomass power will reduce fuel consumption while allowing round-the-clock operation of the mill, processing plants and distillation unit. The plant will allow regional farmers and their families to access a cheap, renewable and reliable energy source that in-turn can help remote villages to generate an income and escape from a cycle of 'fuel poverty.' The research team wants to use this pioneering project to create a blueprint for renewably-powered combined heat and power boilers, capable of being replicated throughout India. "We want to develop a holistic business model which can be replicated to grow 4

wealth in rural communities. In India, over 70 per cent of people live in rural communities. We believe this project will empower and benefit the villagers involved, create sustainable rural development and encourage entrepreneurialism." "A reliable and readily available energy supply is critical for economic development. Bringing renewable and sustainable energy supplies to areas of rural India can ensure we can help people escape from a cycle of poverty. Most India farmers are small holders with limited technology for processing and preserving food. Reliable energy systems are needed to power such technologies and at the same time create employment. This research will create a wealth of ecological and economic benefits along the entire biomass chain, and will offer valuable new research in an evolving industry."

IIT-Guwahati search for science wizard among teens IIT-Guwahati is on the lookout for a science wizard among school and college students for which it is conducting an entrance exam in the 120 cities across India, including Nagpur. The entrance test is open only to students from Std IX-XII, registrations for which can be done. Students can register by logging on to www.techniche.org/technothlon.html The exams have been christened 'Technothlon International School Championship' and it is the only ticket which will allow students to participate in the big event. IITGuwahati's annual technomanagement festival called Techniche-2011 will be held at the institute campus later in the year. •

The exam, which will be held from 11am to 1pm on Sunday. Exam has to be given in teams consisting of two members only. There will be two categories for the exam, one will comprise students of Std IX-X and the other of students of Std XI-XII.

AUGUST 2011


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

achievement made by her son, Sangeeta Devi said, “We have been living without basic facilities at our end because we cannot afford to buy luxuries. Selection of my son has come as a gift to us from the God’s side. His hardwork has borne fruits.” Abhishek emerged as a fine example for others who wish to crack the entrance exam of various competitions. He not only lacked electricity at his house but it was too tough for him to study in a small room of 10-by-10 wherein his entire family lives. His father is the bread earner and many a times he used to work as a labourer for six to eight hours in a day to arrange for the two ends to meet. Abhishek used to assist his father in doing so. Mahesh Singh Chauhan, the teacher of Abhishek who provided the boy with free education was overwhelmed with the success.

Mr. Abhishek Kr. Bharti Cobbler’s son cracks IIT – Inspiring Success Story Abhishek Kumar Bharti, son of a poor shoemaker from Kanpur, whose family lives in a 10×10 house has cracked the entrance examination of the most premier institute of the country IIT. The son of a cobbler, Abhishek has managed to beat the odds and make it to the IIT with a rank of 154 in the SC/ ST category in the entrance examination. His life has been an endless struggle and it’s only his zeal to carry on that saw him reach thus far. A student of the UP board, Abhishek gathered 78 per cent marks in his class XII. Though he hails from a financially weak segment of the society but his deep inclination towards studies helped in meeting his goal – Goal to graduate from IIT. He helps his father, a cobbler, mend shoes in his spare time while his mother stitches rags to support the family, but financial hardships have not stopped Abhishek Kumar Bhartiya from coming out with flying colours in the IIT entrance exam. Abhishek has three young brothers and the family of six lives in a one room accommodation with no electricity. “We have just one small room where six of us live and that too without electricity. So, he used to study under the lantern for five-six hours in the night,” says his father Rajendra Prasad. Abhishek would work with his father as a shoeshine boy and at times would find a job as a labourer to earn some extra money. Talking about his elder son, Rajendra said, “He used to study in the night and help me in my work the whole day. His three brothers – Abhijit, Anshul and Aryan – are below 12 years of age and study in a municipal school. Their mother Sangeeta Devi repairs old clothes of poor people and earns about Rs 50 a day. “My husband gets around Rs 100 and I earn Rs 50 in a day. It is not enough for us. But we don’t want to beg. We want to live with our heads held high. My children know how to go ahead with their pride intact. Abhishek never demanded anything from us. The table of my sewing machine would turn into his study table at night. All I did was to ensure that there was enough kerosene in the lantern,” she said. Moved with the XtraEdge for IIT-JEE

Mr. Supreet Joshi M Tech dual degree in Electrical Engineering. Supreet is all smiles, for this is one of his happiest days. "It feels great to top my batch and win this award. The good thing is being here once more and sad part is leaving this place forever," says Supreet. Supreet, who hails from Dharwad in Karnataka, is excited about his new job at Texas Instruments. IIT experience IIT has been an amazing experience. The quality of life here has been outstanding. I made great friends here. We have good professors and the environment is very challenging. Here the environment is such that you can always excel. You don't have to make an effort, it come naturally. It also helps us develop entrepreneurial skills. My mantra for success Hard work comes first. Then good focus, concentration in everything I do. And I also have the strong will to excel. Advice to IIT aspirants There is no short cut to success, one has to be smart and sharp, dedicated and determined to face the most difficult tasks. Next move Have just joined Texas Instruments in Bangalore as design engineering. I'm enjoying my work. 5

AUGUST 2011


KNOW IIT-JEE By Previous Exam Questions Therefore (u cos θ) = gt – (u sin θ) …(iii) (The right hand side is written gt – u sin θ because the stone is in its downward motion. Therefore, gt > u sin θ. In upward motion u sin θ > gt). Multiplying equation (iii) with t we can write, (u cos θ) t + (u sin θ) t = 10t2 Now (iv)–(ii)– (i) gives 4.25 t2 – 4.25 = 0 or t = 1s Substituting t = 1s in (i) and (ii) we get, u sin θ = 6.25 m/s or uy = 6.25 m/s and u cos θ = 3.75 m/s

PHYSICS 1.

An object A is kept fixed at the point x = 3m and y = 1.25 m on a plank P raised above the ground. At time t = 0 the plank starts moving along the +x direction with an acceleration 1.5 m/s2. At the same instant a stone is projected from the origin with a → velocity u as shown. A stationary person on the ground observes the stone hitting the object during its downward motion at an angle of 45º to the horizontal. → All the motion are in the X-Y plane. Find u and the time after which the stone hits the object. Take [IIT-2000] g = 10 m/s2.

→ Or ux = 3.75 m/s therefore u = ux ˆi + uy ˆj

A

y

Or

P

1.25 m

2. u O

3.0 m

x

Sol. Let ‘t’ be the time after which the stone hits the object and θ be the angle which the velocity vector →

u makes with horizontal. According to question, we have following three conditions.

u sin θ

vx = u cos θ u

1.25m

45º

v vx=vy gt-usinθ= |vy|

θ ucosθ (i) Vertical displacement of stone is 1.25 m. 1 Therefore 1.25 = (u sin θ) t – gt2 2 where g = 10 m/s2 or (u sin θ)t = 1.25 + 5t2 (ii) Horizontal displacement of stone = 3 + displacement of object A 1 Therefore (u cos θ) t = 3 + at2 where a = 1.5 m/s2 2 2 or (u cos θ)t = 3 + 0.75 t …(ii) (iii) Horizontal component of velocity (of stone) = vertical component (because velocity vector is inclined) at 45º with horizontal).

XtraEdge for IIT-JEE

u = (3.75 ˆi + 6.25 ˆj ) m/s

Two blocks of mass m1 = 10 kg and m2 = 5 kg, connected to each other by a massless inextensible string of length 0.3 m are placed along a diameter of a turn table. The coefficient of friction between the table and m1 is 0.5 while there is no friction between m2 and the table. The table is rotating with an angular velocity of 10 rad/s about a vertical axis passing through its centre O. The masses are placed along the diameter of the table on either side of the centre O such that the mass m1 is at a distance of 0.124 m from O. The masses are observed to be at rest with respect to an observer on the turn table. (a) Calculate the frictional force on m1. (b) What should be the minimum angular speed of the turn table so that the masses will slip from this position? (c) How should the masses be placed with the string remaining taut, so that there is no frictional force acting on the mass m1? [IIT-1997]

Sol. ω

m1 T f

T

m2

0.176m 0.124m 0.3m

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(2) Heat lost by 100ºC water to change to 0ºC water = 0.05 × 4200 × 100 = 21,00 J (3) Heat required by 0.45 kg of ice to change its temperature from 253 K to 273 K = m × Sice × ∆T = 0.45 × 2100 × 20 = 18,900 J (4) Heat required by 0.45 kg ice at 273 K to convert into 0.45 kg water at 273 K = mLfusion = 0.45 × 336 × 1000 = 1,51,200 J From the above data it is clear that the amount of heat required by 0.45 kg of ice at 253 K to convert into 0.45 kg of water at 273 K (1,70,100 J) cannot be provided by heat lost by 0.05 kg of steam at 373 K to convert into water at 273 K. Therefore the final temperature will be 273 K or 0ºC.

(a) The tension in the string will remain the same The tension T acting on m2 is responsible in moving the mass m2 in circular motion ∴ Since m1 is also moving in a circular path, the centripetal force required by m1 will be = m1r1ω2 = 10 × 0.124 × 10 × 10 = 124 N Out of the 124 N required, 88 N will be provided by the tension T in the string The rest will be provided by frictional force Therefore frictional force acting on m1 = 124 – 88 = 36 N (b) Let ω' be the maximum angular speed for wich no slipping of masses occurs (or we may say that ω' is the minimum angular speed for which slipping occurs) The maximum frictional force that can act on mass m1 = µm1g = 0.5 × 10 × 9.8 = 49 N The equation for m1 to move in circular motion is T = f = m1r1ω2 when f = fmax then T = T' (say) and T' = m2r2ω'2 (for mass m2) and ω = ω' ∴ m1r2ω'2 + fmax = m1r1ω'2 ⇒ ω' = ⇒ ω' =

4.

A conducting sphere S1 of radius r is attached to an insulating handle. Another conducting sphere S2, of radius R is mounted on an insulating stand. S2, is initially uncharged. S1 is given a charge Q, brought into contact with S2 and removed. S1 is recharged such that the charge on it is again Q; and it is again brought into contact with S2 and removed. This procedure is repeated n times. [IIT-1998] (a) Find the electrostatic energy of S2 after n such contacts with S1. (b) What is the limiting value of this energy as n → ∞ Sol. When the spheres S1 and S2 come in contact, there is transference of charges till the potential of the two sphere becomes equal. During first contact V 1 = V2 [q1 charge shifts from S1 to S2]

f max m1r1 – m 2 r2 49 10 × 0.124 – 5 × 0.176

= 11.67 rad/s

(c) For no frictional force acting on mass m1 The tension should be sufficient to provide centripetal force for both the masses. Then T = m1r1ω2 and T = m1r2ω2 r m 5 1 = ...(i) ⇒ m1r1ω2 = m2r2ω2 ⇒ 1 = 2 = 10 r2 m1 2

During second contact again, V1 = V2 Kq 2 K[Q – (q2 – q1)] = R [(q1 – q2) charge shifts from S1 to S2] 2  R  R   ⇒ q2 = Q  +    R + r  R + r   On third contact again, V1 = V2 Kq 3 K[Q − (q 3 − q 2 )] = r R [(q3 – q2) charge shifts from S1 to S2]

But r1 + r2 = length of thread ⇒ r1 + r2 = 0.3 m Solving equation (i) and (ii) we get r1 = 0.1 m and r2 = 0.2 m. 3.

0.05 kg steam at 373 K and 0.45 kg of ice at 253 K are mixed in an insulated vessel. Find the equilibrium temperature of the mixture. Given, Lfusion = 80 cal/g = 336 J/g, Lvaporization = 540 cal/g = 2268 J/g, Sice = 2100 J/Kg K = 0.5 cal/gK and [IIT-2006] Swater = 4200 J/Kg K = 1 cal/gK Sol. (1) Heat lost by steam at 100ºC to change to 100ºC water mLvap = 0.05 × 2268 × 1000 = 1,13,400 J XtraEdge for IIT-JEE

Kq1 K (Q − q1 )  R  = ⇒ q1 = Q   r R R+r

2 3  R  R   R   + ⇒ q2 = Q   +    R + r  R + r   R + r   On nth contact by symmetry, V1 = V2 K qn K[Q − (q n − q n −1 )] = r R [(qn – qn–1) charge shifts from S1 to S2]

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AUGUST 2011


⇒ Current through R is 5A. Applying Ohm's law across R, we get 25 = 5 × R ⇒ R = 5 Ω

2 n  R  R   R   ⇒ qn = Q  +  + .... +     R + r    R + r  R + r 

 R  = Q  R+r

 {R /( R + r ) n }  1 −   1 − R /( R + r ) 

CHEMISTRY

n QR   R   = 1 −    r   R + r  

An ideal gas having initial pressure P, volume V and temperature T is allowed to expand adiabatically until its volume becomes 5.66 V, while its temperature falls to T/2. (a) How many degrees of freedom do the gas molecules have ? (b) Obtain the work done by the gas during the expansion as a function of the initial pressure P and volume V. [IIT-1990] Sol. (a) According to adiabatic gas equation, TVγ–1 = constant or T1V1γ–1 = T2V2γ–1 T2 = T/2 Here, T1 = T ; V1 = V and V2 = 5.66 V T × (5.66V)γ–1 Hence, TVγ–1 = 2 T × (5.66)γ–1 × Vγ–1 = 2 or (5.66)γ–1 = 2 ...(1) Taking log, (γ – 1)log 5.66 = log 2 0.3010 log 2 γ–1= = = 0.4 or 0.7528 log 5.66 6.

The electrostatic energy of S2 after n contact is  QR   R  n   1 q 2n 1 1 Un = = × × 1 −    2 C 2 4πε 0 R  r   R + r      (b) The limiting value

2

2  n 1 1  QR   R      . Lt U n = Lt 1 −     n →∝  2 4πε 0 R n →∞ r   R + r         

= 5.

Q2R 2(4πε 0 )r 2

A heater is designed to operate with a power of 1000 watts in a 100 volt line. It is connected in a combinations with a resistance of 10 ohms and a resistance R to a 100 volts mains as shown in the figure. What should be the value of R so that the heater operates with a power of 62.5 watts. [IIT-1997] Heater 10 Ω R

γ = 1.4 or If f, be the number of degrees of freedom, then 2 2 γ=1+ or 1.4 = 1 + f f 2 = 1.4 – 1 = 0.4 or f 2 =5 or f= 0.4 (b) According to adiabatic gas equation, P1V1γ = P2V2γ Here, P1 = P V1 = V V2 = 5.66 V Hence, PVγ = P2 × (5.66V)γ = P2 ×(5.66)γ × Vγ P P P or P2 = = = [using eq.(1)] γ 1.4 11 .32 (5.66) (5.66)

100V Sol. The resistance of the heater is

R=

100 × 100 V2 = = 10 Ω 1000 P

The power on which it operates is 62.5 Ω ∴

V=

The potential drop across AB = 75 V 75 V The current in AB = I = = = 7.5 A 10 R A B Heater 10 Ω R

R × P' = 10× 62.5 =

625 = 25

100V This current divides into two parts. Let I1 be the current that passes through the heater. Therefore 25 = I1 × 10 ⇒ I1 = 2.5 A

XtraEdge for IIT-JEE

Hence, work done by the gas during adiabatic expansion 8

AUGUST 2011


P V −P V = 1 1 2 2 = γ –1

PV −

P × 5.66V 11.32 1.4 – 1

PV 2 = PV = 1.25 PV 0.4 2 × 0.4

NH3

C≡N

N≡C 2+

Ni N≡C

C≡N

In [Ni(CO)4, nickel is present as Ni atom i.e. its oxidation number is zero and coordination number is four. 3d 4s 4p Ni in Complex sp3 hybridization

Its structure is as follows : CO

Ni OC

CO

CO (c) The transition metal is Cu2+. The compound is CuSO4.5H2O

 medium  → CuS ↓ + H2SO4 CuSO4 + H2S Acidic Black ppt

2CuSO4 + 4KI → Cu2I2 + I2 + 2K2SO4 (B) white – – I2 + I → I3 (yellow solution)

4p

The molar volume of liquid benzene (density = 0.877 g ml–1) increases by a factor of 2750 as it vaporizes at 20ºC and that of liquid toluene (density = 0.867 g ml–1) increases by a factor of 7720 at 20ºC. A solution of benzene and toluene at 20ºC has a vapour pressure of 46.0 torr. Find the mole fraction of benzene in vapour above the solution. [IIT-1996] Sol. Given that, Density of benzene = 0.877 g ml–1 Molecular mass of benzene (C6H6) = 6 × 12 + 6 × 1 = 78 78 ∴ Molar volume of benzene in liquid form = ml 0.877 78 1 × L = 244.58 L = 0.877 1000 8.

NH3

3+

H3N

or

Co H 3N

4p

Hence structure of [Ni(CN)4]2– is

d2sp3 hybridization

NH3

4s

dsp2 hybridization

Co3+ion in Complex ion

H3N

3d Ni ion in Complex ion

(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) 3+ (b) In [Co(NH3)6] 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 4s

4p

2+

7.

3d

4s

Ni ion =

PV −

=

3d 2+

NH3

NH3

NH3

Co3+

H3N

NH3

NH3

In [Ni(CN)42– nickel is present as Ni2+ ion and its coordination numbers is four Ni28 =1s2, 2s22p6, 3s23p63d8, 4s2 Ni2+ ion = 1s2, 2s22p6, 3s23p63d8 XtraEdge for IIT-JEE

9

AUGUST 2011


Both, Y and Z have the same molecular formula C6H12(CnH2n). Since, both Y and Z absorb one mol of H2 to give same alkane 2, 3-dimethyl butane, hence they should have the skeleton of this alkane.

And molar volume of benzene in vapour phse 78 2750 = × L = 244.58 L 0.877 1000 Density of toluene = 0.867 g ml–1 Molecular mass of toluene (C6H5CH3) = 6 × 12 + 5 × 1 + 1 × 12 + 3 × 1 = 92 ∴ Molar volume of toluene in liquid form 92 92 1 ml = × L = 0.867 0.867 1000 And molar volume of toluene in vapour phase 92 7720 = × L = 819.19 L 0.867 1000 Using the ideal gas equation, PV = nRT At T = 20ºC = 293 K nRT For benzene, P = PB0 = V =

2 CH3 – CH – CH – CH3 Y and Z (C6H12) H→

Ni

CH3 CH3 2,3-dimethyl butane

The above alkane can be prepared from two alkenes CH3 – C = C – CH3 and CH3 – CH – C = CH2

Ni

CH3 CH3

And

PM =

CH3 – C – CH – CH3

CH2 = C — CH – CH3 + CH3 – C = C – CH3 CH3 CH3

xT

(Z) 20%

Hence,

(Y) 80%

X, CH3 – C – CH – CH3 CH3 CH3

Y, CH3 – C = C – CH3 CH3 CH3

Z, CH3 – CH – C = CH2

An alkyl halide X, of formula C6H13Cl on treatment with potassium t-butoxide gives two isomeric alkenes Y and Z(C6H12). Both alkenes on hydrogenation give 2, 3-dimethyl butane. Predict the structures of X, Y and Z. [IIT-1996] Sol. The alkyl halide X, on dehydrohalogenation gives two isomeric alkenes. 9.

CH3 CH3

10. An ester A(C4H8O2), on treatment with excess methyl magnesium chloride followed by acidification, gives an alcohol B as the sole organic product. Alcohol B, on oxidation with NaOCl followed by acidification, gives acetic acid. Deduce the structures of A and B. Show the reactions involved. [IIT-1998]

K − t − butoxide C 6 H13Cl   → Y + Z

XtraEdge for IIT-JEE

CH3 CH3

Cl

(in vapour phase)

∆ ; – HCl

∆; –HCl

2-chloro-2,3-dimethyl butane (X)

or mole fraction of benzene in vapour form, P 74.48 × 0.457 x 'B = B = = 0.74 46.0 PM

X

K-t-butoxide

CH3 CH3

or 46.0 = 74.48 xB + 22.04 (1 – xB) Solving, xB = 0.457 According to Dalton's law, PB = PM x 'B

CH3 CH3

Both, Y and Z can be obtained from following alkyl halide : Cl

xT = 22.04 (1 – xB) xB +

CH3 – CH – CH – CH3

(Z)

PB = PB0 xB = 74.48 xB PT0

H2 Ni

CH3 – CH – C = CH2

= 22.04 torr (Q 1 atm = 760 torr) According to Raoult's law,

PB0

CH3 CH3

(Y)

1× 0.082 × 293 = 0.029 atm 819.19

PT =

(Z)

CH3 CH3

1× 0.082 × 293 = 0.098 atm 244.58

PT0

2,3-dimethyl

The hydrogenation of Y and Z is shown below : H2 CH3 – C = C – CH3 CH3 – CH – CH – CH3

= 74.48 torr (Q 1 atm = 760 torr) Similarly, for toluene, nRT P = PT0 = V =

CH3 CH3 butene-1

CH3 CH3 2,3-dimethyl butene-2 (Y)

C 6 H12

10

AUGUST 2011


Sol. The reactions of an ester with methyl magnesium chloride are as follows. O

OMgCl

O

R–C–OR´ CH3MgCl R–C–OR´ (A)

CH3

MATHEMATICS

+

H –HOMgCl

11. Tangent at a point P1 {other than (0, 0)} on the curve y = x3 meets the curve again at P2. The tangent at P2 meets the curve at P3, and so on. Show that the abscissae of P1, P2, P3 , ...., Pn, form a G.P. Also find the ratio [area (∆P1P2P3)]/[area (∆P2P3P4)] [IIT-1993] Sol. Let any point P1 on y = x3 be (h, h3) then tangent at P1 is y – h3 = 3h2(x – h) ...(i) It meets y = x3 at P2, Putting the value of y in (i) ∴ x3 – h3 = 3h2(x – h) ⇒ (x – h) (x2 + xh + h2) = 3h2(x – h) ⇒ x2 + xh + h2 = 3h2 or x = h 2 2 ⇒ x + xh – 2h = 0 ⇒ x2 + 2xh – xh – 2h2 = 0 ⇒ x(x + 2h) – h(x + 2h) = 0 ⇒ (x – h) (x + 2h) = 0 ⇒ x=h or x = – 2h therefore x = – 2h is the point P2. y = – 8h3 which implies hence, P2 ≡ – 2h, – 8h3 Again tangent at P2 is y + 8h3 = 3(2h)2(x + 2h) it meets y = x3 at P3 ⇒ x3 + 8h3 = 12h2(x + 2h) ⇒ x2 – 2hx – 8h2 = 0 ⇒ x2 – 4hx + 2hx – 8h2 = 0 ⇒ x(x – 4h) + 2h(x – 4h) = 0

R–C–CH3 + R´OH CH3MgCl OMgCl

OH H+ R–C–CH3 –HOMgCl

R–C–CH3

CH3

CH3

(B)

Since the given ester (C4H8O2) produces only one alcohol B, it follows that RC(CH3)2OH and R´OH must be identical. Thus, the alkyl group R´ must be RC(CH3)2 – and the given ester A is CH3

O

R – C – O – C – CH3 (molecular formula R2C4H6O2 ) R

From the molecular formula of A, we conclude that R must be H atom. Hence, the given ester is O H – C – O – CH – CH3

Isopropyl formate

CH3

The alcohol B is a secondary alcohol. Isopropyl alcohol

CH3 – CH – CH3 OH

⇒ (x + 2h) (x – 4h) = 0 Q x = 4h ⇒ y = 64h3 therefore, P3 ≡ (4h, 64h3) similarly, we get P4 ≡ (– 8h, – 83h3) hence the abscissae are h, – 2h, 4h, – 8h, .... which form a G.P. Let D´ = ∆P1P2P3 and D´´ = ∆ P2P3P4

The oxidation of alcohol B with NaOCl will give a ketone which further undergoes a haloform reaction. CH3 – CH – CH3 + NaOCl OH

CH3 – C – CH3 + NaCl + H2O O

CH3 – C – CH3 + 3NaOCl O

CH3 – C – CCl3 + 3NaOH

∆P1P2 P3 D´ = = D´´ ∆P2 P3 P4

O CH3 – C – CCl3 + NaOH O

CH3 – C – O–Na+ + CHCl3 O

The acidification of sodium acetate will produce acetic acid. =

XtraEdge for IIT-JEE

11

h h3 1 1 − 2 h − 8h 3 1 2 4h 64h 3 1 − 2 h − 8h 3 1 1 4h 64h 3 1 2 3 − 8h − 512h 1

h h3 1 1 3 − 2h − 8h 1 2 4h 64h 3 1 h h3 1 1 3 × −2 × −8 − 2h − 8h 1 2 4h 64h 3 1

AUGUST 2011


taking common –2 from C1 and –8 from C2 then D´ 1 = which is the required ratio. 16 D´´

A1 be the event exactly 4 white balls have been drawn. A2 be the event exactly 5 while balls have been drawn. A3 be the event exactly 6 white balls have been drawn B be the event exactly 1 white ball is drawn from two draws. 12 C 2 .6 C 4 10 C1.2 C1 12 C1.6 C 5 11 C1.1 C1 . 12 + 18 . 12 18 C6 C2 C6 C2 ∴ P(B/A) = 12 6 12 6 12 6 C. C C . C6 C2 . C4 + 181 5 + 180 18 C6 C6 C6

12. The sides of a triangle are three consecutive natural numbers and its largest angle is twice the smallest one. Determine the sides of the triangle. [IIT-1991] Sol. Let ABC be the triangle such that the lengths of its sides CA, AB and BC are x – 1, x and x + 1 respectively where x ∈ N and x > 1. Let ∠B = α be the smallest angle and ∠A = 2α be the largest angle. A 2α x B

=

x–1

α

Then by sine formula, we have sin 2α sin α = x −1 x +1 sin 2α x +1 x +1 ⇒ = ⇒ 2 cos α = sin α x −1 x −1 x +1 ∴ cos α = ...(1) 2( x − 1) Also cos α =

C 2 (12 C 2 .6 C 4 +12 C1.6 C 5 +12 C 0 .6 C 6 )

What is the ratio for the sides of the rectangle so that the window transmits the maximum light?[IIT-1991] Sol. Let '2b' be the diameter of the circular portion and 'a' be the lengths of the other sides of the rectangle.

Total perimeter = 2a + 4b + πb = K

x 2 + ( x + 1) 2 − ( x − 1) 2 , 2 x( x + 1)

using cosine law x+4 ⇒ cos α = 2( x + 1)

12

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

C

x+1

(12 C 2 .6 C 4 .10 C1.12 C1 ) + (12 C1.6 C 5 .11 C1.1 C1 )

Now, let the light transmission rate (per square metre) of the coloured glass be L and Q be the total amount of transmitted light. ...(2)

Coloured glass

from (1) and (2), we get x +1 x+4 = 2( x − 1) 2( x + 1)

a

⇒ (x + 1)2 = (x + 4))(x – 1) ⇒ x2 + 2x + 1 = x2 + 3x – 4 ⇒ x=5 Hence, the length of the sides of the triangle are 4, 5 and 6 units.

∑ P(A ).P(B / A )

1 2 πb (L) 2

L {πb2 + 12ab} 2

Q=

L {πb2 + 6b (K – 4b – πb)} 2

Q=

L {6Kb – 24b2 – 5πb2} 2

dQ L = {6K – 48b – 10πb} = 0 db 2

i

i =1

3

∑ P(A ) i

i =1

where A be the event at least 4 white balls have been drawn. XtraEdge for IIT-JEE

a

Q=

3

i

Clear glass

Then, Q = 2ab(3L) +

13. A bag contains 12 red balls and 6 white balls. Six balls are drawns one by one without replacement of which at least 4 balls are white. Find the probability that in the next two drawns exactly one white ball is drawn (leave the answer in nCr). [IIT-2004]

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

(say) ...(1)

12

b=

6K 48 +10π

...(2)

AUGUST 2011


and

d 2Q db

2

=

L {– 48 + 10π}La 2

TRUE OR FALSE

Thus, Q is maximum and from (1) and (2), 1.

(48 + 10π) b = 6K and K = 2a + 4b + πb ⇒ (48 + 10π) b = 6{2a + 4b + πb}

2.

2b 6 = a 6+π

Thus, the ratio =

3.

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

4. 5.

r3 r1 r2 r3 r1 r + 2 + = r − r1 r − r2 r − r3 (r − r1 )(r − r2 )(r − r3 )

[IIT-2000]

6.

Sol. The quadrilateral HEKJ is a square because all four angles are right angle and JK = JH.

7.

A

8.

A/2 A/2 K 90º 90º E

r1 J

F

r1 Circle

B

Circle

Therefore, HE = JK = r1 and IE = r ⇒

Sol. 1. [True] The number of moles of solute remains unchanged on dilution but molarity changes because No. of moles Molarity (M) = Volumes of solution (L) 2. [True] Since acetic acid is a weak acid as compared to sulphuric acid and hence the conjugate base of a weak acid will be stronger than that of strong acid. 3. [False] Zn + 2 HCl → ZnCl2 + H2 Pure HCl required = 14.6 gm 100 × 14.6 60% Cl required will be = = 24.33 gm 60 4. [False] Positively charged sol is formed as Fe(OH)3 + FeCl3 → [Fe(OH)3]Fe3+ : 3 Cl–

H

D 90º

C

(given)

IH = r – r1

Now, in right angled triangle IHJ, ∠JIH = π/2 – A/2 [Q ∠IEA = 90º, ∠IAE = A/2 and ∠JIH = ∠AIE] In triangle JIH tan(π/2 – A/2) = ⇒

cot A/2 =

r1 r − r1

r1 r − r1

Similarly, cot B/2 =

Positively charged sol

5.

r3 r2 and cot C/2 = r − r2 r − r3

adding above results, we obtain cot ⇒

6. 7.

A B C A B C + cot + cot = cot cot cot 2 2 2 2 2 2

r r1 r2 r3 r1 r + 2 + 3 = r − r1 r − r2 r − r3 (r − r1 )(r − r2 )(r − r3 )

XtraEdge for IIT-JEE

When a concentrated solution of a solute is diluted by adding more of solvent, the number of moles of solute remains unchanged. Acetate ion is stronger conjugated base than the sulphate ion. Mass of 60% HCl (by mass), required to completely react with 0.2 mol of zinc is 14.6 gm. Ferric hydroxide precipitate when agitated with dilute ferric chloride solution, gives negatively charged colloidal solution. Vapour pressures of ethanol and benzene at 293K are 43.9 mm and 74 mm respectively indicating stronger intermolecular forces in benzene as compared to ethanol. A mixture of molten zinc and lead is a heterogeneous system. The energy required to increase the surface area of a liquid by unit amount is called the surface tension of liquid. In aqueous solution cuprous ions dispro-portionate into cupric ions and metallic copper.

8.

13

[False] Higher the vapour pressure, weaker are the intermolecular forces and hence intermolecular forces in ethanol will be stronger than those in benzene. [True] [True] 2 Cu+ Cu2+ + Cu Since oxidation and reduction of Cu+ ion takes place simultaneously and hence it is known as the disproportion reaction. [False] Thomson through his experiment determined the charge to mass ratio of an electron and the value of 3/m is equal to 1.76 × 108 coulomb/gm. Hence one gm of electrons have charge 1.76 × 108 C. 1.60 × 10–19 coulomb is the charge on one electron. AUGUST 2011


Physics Challenging Problems

Set # 4

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

So lutions will b e p ub lished in nex t issue 1.

2.

If the energy of an electron in some excited state of H-atom is -2.4178×10-19 Joules then choose the correct options – (A) The excited state in n = 3 (B) The kinetic energy of electron is 1.51 eV (C) The potential energy of system is -4.835×10-19 J (D) The option (C) will be different if choice of zero potential level is changed

(A) The time constant of circuit is DC is RC (B) The time constant of circuit if source is DC is 4RC (C) The phase angle between voltage across capacitor and voltage across resister in series is tan −1 (ωRC) (D) The phase angle between net current and supply voltage is less than 45º if ω = 1 / RC

The decay of current in an L-R circuit is shown in figure. Select correct alternative(s)

4.

i(A) 10 5 1.369 0

r

r r

A

6.93

20

(A) Current in BC is zero (B) Current in AD is maximum (C) If we remove any of the resistance AB, AC, DB or DC, the change in current is same (D) Change in current is same if AD or BC is removed Passage # (Q. No. 5 to Q. No. 6) In the given circuits three inductors L1, L2 and L3 and a resistance R are connected to a V volt battery neglecting mutual inductance. At t = 0 switch is closed and L3 > L2 > L1. L2

A capacitor whose air capacitance is C is filled with a dielectric (constant K = 2) and electrical resistance R, is connected with another resistance R in series and connected across an ac source ( v 0 sin ωt ) . Tick the correct option(s) K

C

D

t(ms)

(A) The time constant of decay circuit will be 10ms (B) Determination of L and R is possible (C) The energy remaining on the inductor in first 13.86 ms is 6.25% (D) The fraction of energy lost by the inductor in first 6.93 sec is 3/4 3.

Choose the correct statement(s) regarding the circuit r B C r r

R

R

L3

L1

~ v0 sin ωt

R V

XtraEdge for IIT-JEE

14

AUGUST 2011


5.

6.

Final current through L1 will be (A)

L2 V R L1 + L 2 + L 3

(B)

L1 V R L1 + L 2 + L 3

(C)

L3 V R L1 + L 2 + L 3

Physics Facts Mechanics 1. Weight (force of gravity) decreases as you move away from the earth by distance squared. 2. Mass and inertia are the same thing.

(D) Can’t be determined

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

Time constant of the circuit is

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

L1 + L 2 + L 3 R 1 R (B)  1 1  L + L +L 1 2  3

(A)

(C)

5. Velocity, displacement [s], momentum, force and acceleration are vectors. 6. Speed, distance [d], time, and energy (joules) are scalar quantities.

   

7. The slope of acceleration.

1  1  1  R  +   L 2 L1 + L 3 

graph

is

9. Centripetal force and centripetal acceleration vectors are toward the center of the circle- while the velocity vector is tangent to the circle.

Passage # (Q. No. 7 to Q. No. 8) Hydrogen gas in the atomic state is excited to a energy of H-atom becomes – 1.7eV. Now a photoelectric plate having work function W = 2.3eV is exposed to the emission spectra of this gas. Assuming all transitions to be possible. Answer the following questions.

8.

velocity-time

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

(D) None of these

7.

the

10. An unbalanced force (object not in equilibrium) must produce acceleration. 11. The slope of the distance-tine graph is velocity. 12. The equilibrant force is equal in magnitude but opposite in direction to the resultant vector. 13. Momentum is conserved in all collision systems.

How many electric transition may result in an ejected photoelectron (A) 3 (B) 4 (C) 5 (D) 2

14. Magnitude is a term use to state how large a vector quantity is.

Energy 15. Mechanical energy is the sum of the potential and kinetic energy.

The minimum de-broglie wavelength of the ejected photo electron is (A) 13.43Å (B) 3.43Å (C) 2Å (D) None of these

16. Units: a = [m/sec2], F = [kg•m/sec2] (newton), work = pe= ke = [kg•m2/sec2] (joule) 17. An ev is an energy unit equal to 1.6 x 10-19 joules 18. Gravitational potential energy increases as height increases. 19. Kinetic energy changes only if velocity changes. 20. Mechanical energy (pe + ke) does not change for a free falling mass or a swinging pendulum. (when ignoring air friction) 21. The units for power are [joules/sec] or the rate of change of energy.

XtraEdge for IIT-JEE

15

AUGUST 2011


1.

8

Solution

Physics Challenging Problems Qu e s tio ns we r e Pub lis he d in Ju ly Is su e

[A,D]

Rate of heat flow, H = And for the inner layer,  l  700 − θ = H  i  KlA  ⇒ 700 − θ =

⇒ θ = 700 −

⇒ θ = 700 −

2.

3.

⇒ nαT0 e αV αV = C v αT0 e αV αV +

600 l li + 0 KlA K0A

⇒ α C = αC v +

⇒ C = Cv +

600 l li + 0 Kl K0A

 li    KlA

600 l li + 0 Kl K0A

 li    KlA

R V

R αV

[B] [100J /º C] × [0.2º C / sec] ⇒ 20 W

5.

[A] Let “x” be the rate of cooling when temperature of block becomes 40ºC From Newton’s law, 0.2 ∝ (50 − 30), x ∝ (40 − 30) 0.2 20 = ⇒ x = 0.1º C / sec x 10 So at this instant power radiated = 100 × 0.1 = 10W

.[A,C,D] TV-3 = k PV −3 ⇒ V =k nR For this polytrophic process, PV-2 = constant So x = - 2 nR (T2 − T1 ) nR (T2 − T1 ) 2 w= = = nRT0 1− x 3 3 3 ∆u = nC v ∆T = n R (3T0 − T0 ) = 3nRT0 2 11  2 ∆Q = ∆U = W =  + 3 nRT0 = nRT0 3  3 ∆Q = nC∆T ∆Q 11 nRT0 11 ⇒C= = = R n∆T 3 n (2T0 ) 6 So x = - 2

6.

[B] 200 × t = (100 × 20) + 10t [Power of heater = 200W] 190 t = 2000 T = 10.5 sec

7.

[B] E−

8.

σ =8 ∈0

[A] ε=

1 .B.ω.r 2 2 R1 R2 E

[B] From 1st law, ∆Q = ∆U + W nCαT + nC v αT + pαV

XtraEdge for IIT-JEE

nRT T0 e αV αV V

4.

600 l K 1+ i . l ll K0

⇒ nCαT = nC v αT +

Set # 3

nRT αV V

16

AUGUST 2011


Students' Forum PHYSICS 1.

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

Two identical blocks A and B each of mass 2 kg are hanging stationary by a light inextensible flexible string, passing over a light and frictionless pulley, as shown in figure. A shell C, of mass 1 kg moving vertically upwards with velocity 9 ms–1 collides with block B and gets stuck to it. Culculate (i) time after which block B starts moving downwards. (ii) maximum height reached by B and (iii) loss of mechanical energy upto that instant.

A

2 × 3 – J = 2v' ...(i) and for combined by J = 3v' ...(ii) From above equations v' = 1.2 ms–1 Now combined body starts to move upward with velocity v' = 1.2ms–1 and retardation a. 3–2 where a= .g = 2 ms–2 3+ 2 ∴ Combined body comes to an instantaneous rest after time t2 = v1/a = 0.6 sec. After this instant combined body starts moving down. Time interval between this instant and instant of collision = t1 + t2 = 0.3 + 0.6 = 0.9 sec. Ans.(i) Further height ascended by combined body during time t2 is 1 h2 = v't2 – at 22 = 0.36 m 2 ∴ Maximum height raised by combined body is h = h 1 + h2 = 0.81 m Ans. (ii) Loss of energy during collision of B and C is 1 1 × (1 + 2) v02 = 27 joule E1 = × 1 × 92 – 2 2 Loss of energy when impulse is developed in string is 1 1 E2 = × 2 × 32 – ( 2 + 3) (v')2 = 5.4 joule 2 2 ∴ Loss of mechanical energy till B reaches highest point is E = E1 + E2 = 27 + 5.4 = 32.4 joule Ans. (iii)

B

C Sol. When C strikes with B, combined body starts to rise vertically upward. According to law of conservation of momentum, velocity v0 of combined body (just after collision) is given by (2 + 1)v0 = 1 × 9 or v0 = 3 ms–1 . But A is at rest. Therefore, string becomes slack and bodies move under gravity, combined body upwards and A downwards till string again becomes taut. This happens when downward displacement of A becomes equal to upward displacement of combined body. Let it happen at instant t1.

2.

1 2 1 gt1 = 3t1 – gt12 or t1 = 0.3 sec 2 2 ∴ Displacement of each block, at that instant, 1 h1 = gt12 = 0.45 m and velocity of combined body 2 = 3 – gt1 = 0 and velocity of A gt1 = 3 ms–1 (downward) Since, at this instant, velocities are different, therefore, an impulse is developed in string and magnitude of velocities of bodies becomes equal. Let that velocity magnitude be v' and impulse developed be J. This impulse acts upwards on both the bodies. Hence, for A

Then,

XtraEdge for IIT-JEE

A right angled wedge ABC of mass M = 4 kg and base angle α = 53º is resting over a smooth horizontal plane. A shell of mass m = 0.5 kg moving horizontally with velocity v0 = 40 ms–1 , collides with the wedge, just above point A. As a consequence, wedge starts to move towards left with velocity v = 5 ms–1. C

v0

53º B

17

A

AUGUST 2011


Calculate (i) heat generated during collision, (ii) maximum height reached by the shell, and (iii) distance of point A of wedge from the shell when shell strikes the plane. Sol. When shell strikes plane AC of the wedge, wedge exerts a normal force on the shell. Let impulse of that normal force be J. Impulse exerted by the shell on wedge is also J. Due to horizontal component J.cos 37º of J, the wedge acquires horizontal velocity and vertical component of J is balanced by impulsive reaction of the floor as shown in figure. C

A ball of radius R = 20 cm has mass m = 0.75 kg and moment of inertia (about its diameter) I = 0.0125 kg m2. The ball rolls without sliding over a rough horizontal floor with velocity v0 = 10 ms–1 towards a smooth vertical wall. If coefficient of restitution between the wall and the ball is e= 0.7, calculate velocity v of the ball long after the collision. (g = 10 ms–2) Sol. Since the ball is rolling without sliding. therefore, its angular velocity (ω0), just before collision with wall is v0/R = 10/0.2 = 50 rad s–1. (Figure(i)) Translational velocity of centre of ball, just after collision = ev0 = 0.7 × 10 = 7 ms–1 (Rightwards). ω0 3.

J 37º

v0

37º 53º

B M.v

Fig.(i) Since, the wall is smooth, therefore not tangential force is applied by the wall. Hence, angular velocity ω0 of ball remains unchanged during collision. Now surface of ball slides on floor to the right as shown in figure (ii). Therefore, friction offered by the floor acts on the ball towards left. ω

A

J'

J. cos37º = M.v or J = 25 kg ms–1 Let horizontal and vertical components of velocity of shell (just after collision) be vx (leftwards) and vy (upward) respectively. Then, mvx = mv0 – J cos 37º or vx = 0 and mvy = J sin 37º or vy = 30 ms–1 heat generated during collision = loss of kinetic energy 1 1  1 = mv 02 –  mv 2 – m( v 2x + v 2y ) 2 2 2  

Fig.(ii) Let co-efficient of friction between ball and the floor be µ. Considering free body diagram (Figure(iii)) of ball while sliding, mg

Ans. (i) = 125 joule Let maximum height reached by the shell be H. Considering its vertically upward motion from instant of collision to the highest point of the shell, u = 30 ms–1, acceleration = – g = – 10 ms–2, u = 0, T s = H = ?, t = = ? 2 where, T is total time of flight of the shell. Using, v2 = u2 + 2as, H = 45 m T Now using v = u + at, t = = 3 second or time of 2 flight of shell is T = 6 second. During this interval distance moved by the wedge = vT = 30 m And horizontal displacement of shell is zero because vx is equal to zero. ∴ Distance between the wedge and the shell, when shell strikes the ground = 30 m Ans. (iii)

XtraEdge for IIT-JEE

lα ma µN N Fig.(iii) For vertical forces, N = mg For horizontal forces, µN = ma. or a = µg = 10 µ. Taking moments (about O) of forces acting on the ball, µNR = Iα or α = 120 µ. Long after the collision, there will be no sliding or it will be pure rolling. Let sliding stop after a time t after collision, then final translational velocity,

18

AUGUST 2011


v = (7 – at) or v = 7 – 10 µt and final angular velocity, ω = (– ω0) + αt or ω = (120µt – 50) rad s–1 (clockwise) But at that instant v = Rω ∴ (7 – 10µt) = 0.2 (120µt – 50) ∴ µt = 0.5 or v = 2 ms–1 4.

Now applying Kirchhoff's voltage law on mesh BCDEB, (q − q 2 ) q 2 q 2 – – = 0 or q2 = 12.6 µC ∴+ 1 C2 C5 C 4

In the circuit shown in Fig., C1 = 5 µF, C2 = 2.9 µF, C3 = 6 µF, C4 = 3 µF and C5 = 7 µF. If in steady state potential difference between points A and B is 11 volt, calculate potential difference across C5. C1

5.

C2

B

2Ω

C3

2Ω 1Ω + 1µF 10V – Sol. Since, in steady state no current flows through capacitors, therefore current through 1Ω resistor becomes zero.

C5 + –

Sol. In the given circuit capacitors C4 and C5 are in series with each other while capacitor C2 is in parallel with this series combination. Then capacitors C3, C1 and above said combination are in series with each other. When steady state is reached, no current flows through the circuit. To analyse the given circuit, it may be assumed that a charge q1 leaves the battery from its positive terminal and flows from A to B. This charge first reaches lower plate of capacitor C3. Hence, this plate becomes positively charged and upper plate negatively charged. Now charge q1 reaches the capacitor C1. Its left plate becomes positively charged and right plate negatively charged. Now charge q1 reaches the junction B. From where it gets divided into two parts. Let a charge q2 flow through series combination of capacitors C4 and C5. Then a charge (q1 – q2) flows through capacitor C2. Hence, in steady state charges on different capacitors will be as shown in Fig. C1

C2

B

+ –

+ –

q1

C3

A

+

E

q1

+ –

q2

F 2Ω I 10V

C5

+ – E q1

+ – D q2

I

H

2Ω

+ q – 3 1µF L K

J

I

Current through resistors and charges on capacitors will be as shown in figure. Applying KVL on mesh ABCHJMGA, 3I + 3I + 2I – 10 + 2I = 0 ∴ I = 1A  q1  Mesh GFELMG,+   + (1 × 0) – 10 + 2I = 0 –6  2 × 10  q1 = 16 µC  q2  Mesh EDHJLE, +   + 2I + (1 × 0) = 0 –6  2 × 10  q2 = – 4µC q3 =0 Mesh BCHJKB, 3I + 2I – (1× 10 – 6 )

C

D

q3 = 5 µC Energy stored in capacitors, U =

+ –

E

=

Since terminal A is connected with positive terminal of the battery, therefore, potential of A is higher than that of B. It is given that potential difference (VA – VB) is equal to 11 volt, 11C1C 3 q q or q1 = = 30 µC ∴ + 1 + 1 = 11 C 3 C1 C1 + C 3 XtraEdge for IIT-JEE

2µF

+ – I

I B I

2µF

1Ω

M

q2

+ –

I

G

3Ω

3Ω

A

(q1 – q2) C4

2µF

2µF

C4

A

q2 = 1.8 volt Ans. C5 In steady state, calculate energy stored in capacitors shown in Figure and the rate at which battery supplies energy. 3Ω 3Ω

Potential difference across C5 =

q12

q 22

∑ 2C +

q 33

2 × (2 × 10 – 6 ) 2 × (1× 10 – 6 ) = 80.5 × 10–6 J Ans. Rate of supply of energy by battery is p = EI = 10 × 1 watt = 10 W Ans.

19

2 × (2 × 10 – 6 )

+

q

2

AUGUST 2011


P HYSICS F UNDAMENTAL F OR IIT-J EE

Capacitor (Part-2) KEY CONCEPTS & PROBLEM SOLVING STRATEGY Capacitors in Series : +Q –Q

+Q –Q

A C1

C2

V1

If CpV is the net capacitance for the parallel combination of capacitors then

+Q –Q C3

V2

CpV = C1V + C2V + C3V ⇒ Cp = C1 + C2 + C3 Important terms : (a) If C1, C2, C3 .... are capacitors connected in series and if total potential across all is V, then potential across each capacitor is

B

V3

V

In this arrangement of capacitor the charge has no alternative path(s) to flow. (a) The charges on each capacitor are equal i.e. Q = C1V1 = C2V2 = C3V3 ...(1) (b) The total potential difference across AB is shared by the capacitors in the inverse ratio of the capacitances. ...(2) V = V 1 + V2 + V 3 If Cs is the net capacitance of the series combination, then

  V1 =    

C Q1 =  1  Cp 

U=

Capacitors in Parallel :

A

–Q1 C1

+Q2

–Q2 C2

+Q3

–Q3 C3

B

⇒ U=

Q Q1 Q = 2 = 3 C1 C2 C3

  V3 =    

1 C3 1 Cs

  V   

1 1 1 1 1 = + + + .... + Cn C1 C 2 C 3 Cs

  Q;  

C Q2 =  2  Cp 

  Q;  

C Q3 =  3  Cp 

 Q  

 ε0A σ and V = Ed where E =  d ε0  

1 ε0A 2 2 Ed 2 d

 1 ⇒ U =  ε 0 E 2  (Ad)  2

In such an arrangement of capacitors the charge has an alternative path(s) to flow (a) The potential difference across each capacitor is same and equals the total potential applied. ...(1) i.e. V = V 1 = V2 = V 3 V=

   V;   

1 CV2 2

where C =

V

1 C2 1 Cs

and so on, where Cp = C1 + C2 + C3 + ... + Cn Energy Density : For a parallel plate capacitor

Q Q and V = C1 Cs

+Q1

  V2 =    

(b) If C1, C2, C3 ... are capacitors connected in parallel and if Q is total charge on the combination, then charge on each capacitor is

1 1 1 1 + + = Cs C1 C 2 C 3

Further V1 =

   V;   

and so on, where

Q Q Q Q + + = Cs C1 C 2 C 3

1 C1 1 Cs

U=

1 ε0E2τ 2

where τ is volume of the capacitor ⇒

U Electrostatic Energy = Ue = τ Volume = Electrostatic Pressure

...(2)

=

(b) The total charge Q is shared by each capacitor in the direct ratio of the capacitances.

σ2 1 ε0E2 = 2 2ε 0

 σ Q E =  ε0  

⇒ Q = Q 1 + Q2 + Q 3 XtraEdge for IIT-JEE

20

AUGUST 2011


Energy for series and parallel combinations : Series Combination : For a series combination of capacitor Q = constant and

F=

 σ2 Q2 =  2ε 2ε 0 A  0

 σ Q Q = σA, E =  ε0  

1 1 1 1 = + + + ... C1 C 2 C 3 Cs

Kirochhoff's laws for capacitor circuits : Kirchhoff's first law or junction law : Charge can never accumulate at a junction i.e. at the junction

Q2 Q2 Q2 Q2 = + + + .... 2C s 2C1 2C 2 2C 3

⇒ Us = U1 + U2 + U3 + ...... Parallel Combination : For a parallel combination of capacitors V = constant and Cp = C1 + C2 + C3 + .... ⇒

∑q = 0 Important terms : This law is helpful in determining the nature of charge on an unknown capacitor plate. Charge on capacitor C can be determined by using this rule. As no charge must accumulate at the junction O, so if x is charge on plate 1 of C, then –q1 + q2 + x = 0

1 1 1 1 CPV2 = C1V2 + C2V2 + C3V2 + ... 2 2 2 2

⇒ Up = U1 + U2 + U3 + .... Electrostatic force between the plates of a parallel plate capacitor : The plates of the capacitor each carry equal and opposite charges, hence they must attract each other with a force, say F. +Q –Q + – + – + – + – + – – +

⇒ x = q1 – q2 +

Also

U=

 Q  x   2ε A   0 

Q2 Q2 = (x – dx) 2C´ 2ε 0 A

If dU is the change in potential energy, then dU = Uf – Ui Q2 Q2 (x – dx) – x 2ε 0 A 2ε 0 A

dU =

dU = –

Q2 dx 2ε 0 A

dU dx

XtraEdge for IIT-JEE

+

C

∑V = 0

(d) In a loop, across a capacitor, if we go from positive plate to the negative plate of the capacitor then there is a potential fall and a –ve sign is to be taken with the potential difference across the q capacitor i.e. ∆V = – . C

Further since F=–

Conventions followed to apply loop law : (a) In a loop, across a battery, if we travel from negative terminal of battery to the positive terminal then there is a potential rise and a +ve sign is applied with voltage of the battery. (b) In a loop, across a battery, if we travel from positive terminal of the battery to the negative terminal then there is a potential fall and a –ve sign is applied with voltage of the battery. (c) In a loop, across a capacitor, if we go from negative plate to the positive plate of the capacitor then there is a potential rise and a +ve sign is to be taken with potential difference across the q capacitor i.e. ∆V = + . C

2

Let the plates be moved towards each other through dx, such that the new separation between the plates is (x – dx). If Uf is the final potential energy, then Uf =

B

i.e. plate 1 has a charge (q1 – q2) and plate 2 has a charge –(q1 – q2). Kirchhoffs second law or loop law : In a closed loop (a closed loop is the one which starts and ends at the same point), the algebraic sum of potential differences across each element of a closed circuit is zero. ⇒

Q ⇒U= 2C

+q1 –q1 +q2 –q2

1 2

ε0A x 2

A

At any instant let the plate separation be x, then C=

  A =  1 ε 0 E 2  A  2  

21

AUGUST 2011


2.

Finding net capacitance of circuits : A. Simple Circuits : Analyse the circuit carefully to conclude which pair of capacitors are in series and which are in parallel (This all should be done keeping in mind the points across which net capacitance has to be calculated). Find their net capacitance and again draw an equivalent diagram to apply the above specified technique repeatedly so as to get the total capacitance between the specified points. B. Concept of line of symmetry : Line of symmetry (L.O.S.) is an imagination of our mind to divide a highly symmetric circuit into two equal halves such that the points of the circuit through which LOS passes are at equal potential.

Find the equivalent capacitance between the point A and B in figure. C1

C3

A

C2

C

Let a charge Q1 goes to the plate a and the rest Q – Q1 goes to the plate e. The charge –Q supplied by the negative terminal is divided between plates d and h. Using the symmetry of the figure, charge –Q1 goes to the plate h (as it has a capacitance C1) and –(Q – Q1) to the plate d (as it has a capacitance C2). This is because if we look into the circuit from A or from B, the circuit looks identical. The division of charge at A and at B should, therefore, be similar. The charges on the other plates may be written easily. The charge on the plate i is 2Q1 – Q which ensures that the total charge on plates b, c and i remains zero as these three plates form an isolated system. We have VA – VB = (VA – VD) + (VD – VB)

C C B C

C

Sol. This circuit is highly symmetric and so we can consider the line of symmetry to pass through the circuit to divide it into two equal (identical) halves. If line of symmetry passes through a branch possessing a capacitor, then on each side of line of Symmetry the capacitance will become 2C (2C and 2C in series will gives C), as shown. 1 C 3

A

2C

2C

C

4

C

or VA – VB =

C

C

Q1 Q − Q1 + C1 C2

...(1)

Also, VA – VB = (VA – VD) + (VD – VE) + (VE – VB)

P

P

C

B

or

LOS

Now, the concept of line of Symmetry makes our job easy to calculate capacitance across AP. (1) and (2) are in parallel further in series with (3), whose resultant capacitance is in parallel with (4). Resultant of (1) and (2) is 3C 3C Resultant of 3C and (3) is 4 3C 7C and (4) is Resultant of 4 4 So total capacitance across AB is C 7C CAB = AP ⇒ CAB = 8 2 XtraEdge for IIT-JEE

B

e f E g h Q–Q1 –(Q–Q1) Q1 –Q1 C2 C1

C

A

C1

C2 C1 Q1 –Q1 Q–Q1 –(Q–Q1) a b D (2Q1–Q) i –(2Q1–Q) C3 j

A

Find the net capacitance of the circuit shown between the points A and B.

C

B

Sol. Let us connect a battery between the points A and B. The charge distribution is shown in figure. Suppose the positive terminal of the battery supplies a charge +Q and the negative terminal a charge –Q. The charge Q is divided between plates a and e.

Solved Examples 1.

C2

VA – VB =

Q1 2Q1 − Q Q + + 1 C1 C3 C1

...(2)

We have to eliminate Q1 from these equation to get Q the equivalent capacitance . (VA − VB ) The first equation may be written as  1 1  Q  + VA – VB = Q1  − C C C 2  2  1

or

C1C 2 C1 (VA – VB) = Q1 + Q C 2 − C1 C 2 − C1

...(3)

The second equation may be written as 22

AUGUST 2011


The capacitors C12 and C32 are in parallel and hence their capacity is C0 + C0 = 2C0. The capacitor C54 is in series with effective capacitor of capacity 2C0. Hence the resultant capacity will be

 1 1  Q  – + VA – VB = 2Q1    C1 C 3  C 3

or

C1C 2 C1 (VA – VB) = Q1 – Q ...(4) 2(C1 + C 3 ) 2(C1 + C 3 )

C 0 × 2C 0 C 0 + 2C 0

Subtracting (4) from (3)

Further C34 is again in parallel. Hence the effective capacity

 CC C1C 3  (VA – VB)  1 2 −   C 2 − C1 2(C1 + C 3 ) 

= C0 +

 C1  C1 =  + Q  C 2 − C1 2(C1 + C 3 ) 

(ii) Charge on the plate 5 = charge on the uper half of parallel combination

or (VA – VB)[2C1C2(C1 + C3) – C1C3(C2 – C1)] = C1[2(C1 + C3) + (C2 – C1)]Q or C =

2 Kε 0 AV0 2  ∴Q5 = V0  C 0  = 3 d 3 

2C1C 2 + C 2 C 3 + C 3C1 Q = VA − VB C1 + C 2 + 2C 3

Charge on plate 3 on the surface facing 4 ∴ V0C0 =

Five identical conducting plates 1, 2, 3, 4 and 5 are fixed parallel to and equidistant from each other as shown in fig. Plates 2 and 5 are connected by a conductor while 1 and 3 are joined by another conductor. The junction of 1 and 3 the plate 4 are connected to a source of constant e.m.f. V0. Find (i) The effective capacity of the system between the terminals of the source (ii) the charge on plates 3 and 5. Given d = distance between any two successive plates and A = are of either face of each plate. Sol. (i) The equivalent circuits is shown in fig. The system consists of four capacitors. 5 3.

kε 0 AV0 d

Charge on plate 3 on the surface facing 2 = [potential difference across (3 – 2)]C0 = V0

AV0 C0 C0 = Kε0 3d C 0 + 2C 0

∴ Q3 =

= 4.

Kε 0 AV0 AV0 + Kε0 d 3d Kε 0 AV0 d

3µF

(+)

3µF

Q

1µF

1µF

2

20Ω

(Q2/2) (Q2/2)

10Ω

100 V

C

A

3

2

5

Sol. The circuit is redrawn in fig (a, b, c)

4

Q2

3µF

(–)

(+) 3

1µF B

4

Q1

3µF

(b)

i.e., C12, C32, C34 and C54. The capacity of each  Kε A  capacitor is  0  = C0. The effective capacity  d  across the source can be calculated as follows : XtraEdge for IIT-JEE

1µF

B

(a) 1

4 A  1 1 + 3  = 3 Kε0 d V0  

In diagram find the potential difference between the points A and B and between the points B and C in the steady state.

(–)

4 3 2 1

C 0 × 2C 0 5 5 A = C0 = Kε0 . 3 3 d C 0 + 2C 0

20Ω A

23

1µF 1µF

10Ω C

100 V Fig.(a)

AUGUST 2011


6µF

B

2µF

3/2 µF

R

1µF

∴ qB = CV = C1V1 or CV = 3CV1

1µF

P

Q

P

10Ω

20Ω

C

A

Let the potential across capacitor B is V1

S

100 V

Q

∴ V1 =

10Ω

20Ω

C

A

Initial energy of capacitor A =

100 V Fig.(c)

Fig.(b)

From fig. (c). potential difference between P and Q = Potential difference between R and S = 100 volt

energy of capacitor B =

1 3 × (3C)V2 = CV2 2 2 Final energy of capacitor B

=

2

=

Ef =

−6

5.

150 × 10 −6 2 × 10 −6

= 25 volt. ∴

C

B

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

C

Sol. Initially the charge on either capacitor, i.e. qA or qB is CV coulomb. When dielectric is introduced, the new capacitance of either capacitor K1 C = 3C. K After the opening of switch S, the potential across capacitor A is volt.

C1 =

XtraEdge for IIT-JEE

Ei 3 CV 2 = = 5 Ef (5 / 3)CV 2

Do you know

S

A

CV 2 5 3 CV2 + = CV2 6 3 2

= 75 volt

Fig. shows two identical parallel plate capacitors connected to a battery with switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant (or relative permittivity) 3. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.

V

CV 2 1 V × (3C)   = 6 2 3

∴ Total final energy

Again potential difference between C and D = potential difference across the two ends of condenser of capacity 2µF V2 =

1 CV2 2

1 1 CV2 + CV2 = CV2 2 2 Final energy of capacitor A

3 × 10–6 × 100 2 = 150 × 10–6 coulomb Now according to fig.(b), the charge flowing through capacitors of capacity 6 µF and 2 µF is 150 × 10–6 coulomb because they are connected in series. Potential difference between A and B = Potential difference across the two ends of condenser of capacity 6 µF. Q 150 × 10 = capacity 6 × 10 −6

1 CV2 2

∴ Total energy Ei =

∴ Q = capacity × volt =

∴ V1 =

V volt 3

24

AUGUST 2011


P HYSICS F UNDAMENTAL F OR IIT-J EE

Work, Power, Energy & Conservation Law KEY CONCEPTS & PROBLEM SOLVING STRATEGY Work, Energy and Power :

The potential energy of particle in the gravitational field is given by U = U0 + mgh

Work is done when a force (F) is displaced. dr

where U0 = potential energy of the body at the ground level. This is true only for objects near the surface of the earth because g is uniform only near the surface of the earth. The strain potential energy of a spring is given by U 1 = kx2, where k is the force constant of the spring 2 and x is the charge in length of the spring. This change in length may be either a compression or on extension.

θ F

The work done is dW = F dr cos θ Using vector notation rr dW = F.dr When the force and the displacement are in the same direction, θ = 0, cos θ = +1, work done is positive. When the force and the displacement are in opposite directions, θ = 180º, cos θ = –1, work done is negative. When the displacement is perpendicular to the direction of the force, θ = 90º, cos θ = 0, no work is done. r r For a system of particles the quantity F.dx cm is

Potential Energy and force

Fx = –

Principle of Conservation of energy : Conservative and Non-conservative Force : If the work done by a force in moving a body from one point to another depends only on the positions of the body and not on the process or the path taken, the force is said to be conservative. Gravitational force, spring force, elastic forces, electric and magnetic forces are examples of conservative forces. If the work done depends on the paths taken, the force is said to be non-conservative. Frictional force is a nonconservative force.

called pseudo work. At times actual work may be zero but not pseudo work. Work is a scalar quantity. Its unit is joule. Power is the rate of doing work. Thus Power =

work done time taken

Work-energy Theorem : The work by external forces on a body is equal to the change of kinetic energy of the body. This is true for both constant forces and variable forces (variable in both magnitude and direction).

The unit of power is the watt (= joules/second). rr The power of an agent is given by P = F.v where F is the force applied by the agent and v is the velocity of the body on which the agent applies the force.

For a particle W = ∆K. For a system of particles Wnet = Wreal + Wpseudo = ∆Kcm Principle of Conservation of Energy : Energy can neither be created nor destroyed by any process. For a particle K + U = a constant. For a system of particles Kcm + Uext + Eint = a constant. However, energy can be transformed from one form into another.

The energy of a system is its capacity of doing work. Mechanical energy may be of two types : (i) kinetic energy and (ii) potential energy. The kinetic energy of a particle is T =

1 mv2. 2

The kinetic energy of a system is T =

1 2 Mv cm 2

XtraEdge for IIT-JEE

∂U ∂x

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AUGUST 2011


Collision of Bodies : Elastic Collision : When two bodies meet a with certain relative speed they are said to collide with each other. In a collision, kinetic energy is transferred, from one body to another. When the transfer of kinetic energy takes place in such a way that the total kinetic energy is conserved, the collision is said to be perfectly elastic, or simply elastic. When kinetic energy is not conserved the collision is said to be inelastic. Further, in a collision, if one body gets embedded in the other and kinetic energy is not conserved, it is a completely inelastic collision. In inelastic and completely inelastic collisions there is always a loss of kinetic energy and this energy is converted into other forms of energy, mostly heat. A collision is said to be direct or head-on if the relative motion before and after the collision is in the same direction; if not it an oblique collision. Remember the following points while solving problems on the collision of bodies. (i) Apply the principle of conservation of momentum. In one-dimensional direct collisions, one equation is obtained by equating momenta before and after collision in the direction of motion. In two-dimensional collisions, select the line of impact as the X-axis and the line perpendicular to it as the Y-axis and obtain two equations by equation by equating momenta before and after the collision along the X- and Yaxes. Remember that momentum is a vector quantity. It may be positive or negative depending on the direction. Choose any one direction as positive; the opposite will be negative. (ii) If it is an elastic collision, apply the principle of conservation of kinetic energy. For inelastic collisions, apply the principle of conservation of energy to obtain an additional equation. (iii) Remember there is no change in momentum along the common tangent to the colliding bodies. Coefficient of restitution : According to Newton, the relative velocity of a body after collision is proportional to its relative velocity in the same direction before collision, with a reversal of sign. Here, relative velocity means the velocity of any one of the colliding bodies (say A) with respect to the other colliding body (say B). The constant of proportionality is called the coefficient of restitution (e). That is VAB (after collision) = –e × V´AB (before collision) This is Newton's law of collision. For elastic collisions, e = 1. For inelastic collisions, e<1 XtraEdge for IIT-JEE

(A) Problem solving strategy : Work and Kinetic Energy : Step 1 : Identify the relevant concepts : The workenergy theorem is extremely useful in situations where you want to relate a body’s speed v1 at one point in its motion to its speed v2 at a different point. This approach is less useful for problems that involve time, such as finding the time it takes a body to go from point 1 to point 2. The reason is that the workenergy theorem. Wtot = K2 – K1, doesn’t involve time at all. For problems that involve time, it’s usually best to use the relationships among time, position, velocity, and acceleration Step 2 : Set up the problem using the following steps : Choose the initial and final positions of the body, and draw a free-body diagram showing all the forces that act on the body. Choose a coordinate system. (If the motion is along a straight line, it’s usually easiest to have both the initial and final positions lie along the xaxis.) List the unknown and known quantities, and decide which unknowns are your target variables. In some cases the target variable will be the body’s initial or final speed; in other cases it will be the magnitude of one of the forces acting on the body. Step 3 : Execute the solution : Calculate the work done by each force. If the force is constant and the displacement is a straight line, you can use Eq. rr W = Fs cos φ or W = F.S . (Latter in this chapter we’ll see how to handle varying forces and curved trajectories.) Be sure to check the sign of the work for each force; it must be positive if the force has a component in the direction of the displacement, negative if it has a component opposite the displacement, and zero if the force and displacement are perpendicular. Add the amounts of work done by each force to find the total work Wtot. Be careful with signs! Sometimes it may be easier to calculate the vector sum of the forces (the net force), then find the work done by the net force; this value is also equal to Wtot. Write expressions for the initial and final kinetic energies, K1 and K2. Note that kinetic energy involves mass, not weight; if you are given the body’s weight, you’ll need to use the relationship W = mg to find the mass. Finally, use the relationship Wtot = K2 – K1 to solve for the target variable. Remember that the right-hand side of this equation is the final kinetic energy minus the initial kinetic energy, never the other way around. Step 4 : Evaluate your answer : Check whether your answer makes physical sense. A key item to

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Step 4 : Evaluate your answer : Check whether your answer makes physical sense. Keep in mind, here and in later sections, that the work done by each force must be represented either in U1 – U2 = –∆U, so make sure you did not include in ∆U, so make sure you did not include it again in Wother. (C) Problem solving strategy : Conservation of Momentum : Step 1 : Identify the relevant concepts : Before applying conservation of momentum to a problem, you must first decide whether momentum is conserved. This will be true only if the vector sum of the external forces acting on the system of particles is zero. If this is not the case, you can’t use conservation of momentum. Step 2 : Set up the problem using the following steps : Define a coordinate system. Make a sketch showing the coordinate axes, including the positive direction for each. Often it is easiest to choose the x-axis to have the direction of one of the initial velocities. Make sure you are using an inertial frame of reference. Most of the problems in this chapter deal with two-dimensional situations, in which the vectors have only x- and y-components; all of the following statements can be generalized to includes-components when necessary. Treat each body as a particle. Draw “before” and “after” sketches, and include vectors on each to represent all known velocities. Label the vectors with magnitudes, angles, components, or whatever information is given, and give each unknown magnitude, angle, or component an algebraic symbol. You may find it helpful to use the subscripts 1 and 2 for velocities before and after the interaction, respectively; if you use these subscripts, use letters (not numbers) to label each paritcle. As always, identify the target variable(s) from among the unknowns. Step 3 : Execute the solution as follows : Write an equation in terms of symbols equating the total initial x-component of momentum (that is, before the interaction) to the total final xcomponent of momentum (that is, after the interaction), using px = mvx for each particle. Write another equation for the y-components, using py = mvy for each particle. Remember that the x- and y-components of velocity or momentum are never added together in the same equation ! Even when all the velocities lie along a line (such as the x-axis), the components of velocity along this line can be positive or negative; be careful with signs !

1 mv2 can never 2 be negative. If you come up with a negative value of K, you’ve made a mistake. Perhaps you interchanged the initial and final kinetic energies in Wtot = K2 – K1 or made a sign error in one of the work calculations. (B) Problems using Mechanical Energy : Step 1 : Identify the relevant concepts : First decide whether the problem should be solved by energy r r methods, by using F = ma directly, or by a

remember is that kinetic energy K =

combination of these strategies. The energy approach is particularly useful when the problem involves motion with varying forces, along a curved path (discussed later in this section), or both. But if the problem involves elapsed time, the energy approach is usually not the best choice, because this approach doesn’t involve time directly. Step 2 : Set up the problem using the following steps : When using the energy approach, first decide what the initial and final states (the positions and velocities) of the system are Use the subscript 1 for the initial state and the subscript 2 for the final state. It helps to draw sketches showing the initial and final states. Define your coordinate system, particularly the level at which y = 0. You will use it to compute gravitational potential energies. Equation U = mgy (gravitational potential energy) assumes that the positive direction for y is upward; we suggest that you use this choice consistently. Identify all no gravitational forces that do work. A free body diagram is always helpful. It some of the quantities you need are unknown, represent them by algebraic symbols. List the unknown and known quantities, including the coordinates and velocities at each point. Decide which unknowns are your target variables. Step 3 : Execute the solution : Write expressions for the initial and final kinetic and potential energies – that is K1, K2, U1, and U2. In general some of these quantities will be known and some will be unknown. Then relate the kinetic and potential energies and the no gravitation work Wother using eq. K1 + U1 + Wother = K2 + U2 (you will have to calculate Wother in terms of the nongravitational forces.) If there is no nongravitational work, this expression becomes eq. 1 1 mv12 + mgy1 = mv22 + mgy2 2 2 It’s helpful to draw bar graphs showing the initial and final values of K, U, and E = K + U. Then solve to find whatever unknown quantity is required.

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∴ Total work done 20  x W=  20 −  g dx = g 0  2

Solve these equations to determine whatever results are required. In some problems you will have to convert from the x-and y-components of a velocity to its magnitude and direction, or the reverse. In some problems, energy considerations give additional relationships among the various velocities. Step 4 : Evaluate your answer: Does your answer make physical sense ? If your target variable is a certain body’s momentum, check that the direction of the momentum is reasonable.

20 

1  20dx − 2 xdx   

0

20 

  x 2  = g {20 x}020 −      4  0    = g[400 – 100] = 300g = 300 × 10 = 3000J

A body of mass m is thrown at an angle α to the horizontal with an initial velocity v0. Find the mean power developed by gravity over the whole time of motion of the body and the instantaneous power of gravity as a function of time. Sol. We know that Pinstan = F.v The velocity of the particle after time t is given by v = v0 cos α i + (v0 sin α – g t)j) and F = – mg j ∴ Pinstan. = (–mg j) . {v0 cos α i + (v0 sin α – g t)j} = – mg(v0 sin α – g t) The average power is given by 3.

Solved Examples 1.

A bus of mass 1000 kg has an engine which produces a constant power of 50 kW. If the resistance to motion, assumed constant is 1000 N, find the maximum speed at which the bus can travel on level road and the acceleration when it is traveling at 25 m/s. Sol. At maximum speed all the power is used to overcome the resistance to motion. Hence if the maximum speed is v, then 50,000 = 1000 × v or v = 50 m/s The maximum speed = 50 m/s At 25 m/s, let the pull of the engine be P. Then the power 50,000 = P × 25 50000 = 2000 N or P = 25 1000 N 2000 N

(P) =

T

0

− mg

P( t ) dt

T

0

= dt

T

0

( v 0 sin α − gt )dt T

− mg[ v 0 sin αT − gT 2 / 2] T Here T is total time of flight T = (2 v0 sin α)/g Substituting this value, we have  v sin α(2v 0 sin α) / g − g (4 v 02 sin 2 α / 2g 2 )  (P) = – mg  0  (2 v 0 sin α) / g   Solving we get (P) = 0

=

a Now resultant force = 2000 – 1000 = 1000 N Applying Newton's law, F = ma, we have 1000 = 1000 a or a = 1.0 m/s2 2. A man is drawing water from a well with a bucket which leaks uniformly. The bucket when full weighs 20 kg and when it arrives the top only half the water remains. The depth of the water is 20 metres. If g = 10 m/sec2, what is the work done ? Sol. When the bucket arrives at the top, the mass is 10 kg. Hence loss in mass = 20 – 10 = 10 kg. The depth of the well is 20 metres. 10 1 ∴ mass lost per unit distance = = kg 20 2 Consider a point at a height x from the bottom of the well. At height x from the bottom, the bucket weighs x  =  20 −  kg. The work done against the force 2 

Two blocks of masses m1 = 2 kg and m2 = 5 kg are moving in the same direction along a frictionless surface with speeds 10 m/s and 3m/s respectively, m2 being ahead of m1. An ideal spring with k = 1120 nt/m is attached to the back side of m2. Find the maximum compression of the spring when the blocks collide. Sol. The situation is shown in fig. u1 u2 4.

m1

m2

Let v be the speed of the system after collision. Applying the law of conservation of energy, we have m1u1 + m2u2 = (m1 + m2)v Substituting the given values (2 × 10) + (5 × 3) = (2 + 5)v

x  during elementary displacement dx =  20 −  dx.g 2 

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v=

Here we have taken the velocity positive in downward direction.

20 + 15 = 5 m/s 7

9.8 − 90.2 = – 40.2 m/s 2 The negative sign shows that the velocity of combined mass is in the upward direction. The height to which the combined mass rises after impact

Now applying the law of conservation of kinetic energy, we get

v=

1 1 1 1 m1u12 + m 2 u 22 = (m1 + m2)v2 + kx2 2 2 2 2 2 2 2 2 or m1u1 + m2u2 = (m1 + m2)v + kx (2 × 100) + (5 × 9) = (7 × 25) + (1120 × x2) 200 + 45 = 175 + 1120 x2

x2 = x=

=

200 + 45 − 175 1 = 1120 16

(40.2) 2 v2 = = 82.45 m 2 × 9.8 2g

The height to which it rises above the cliff is = 82.45 – 4.90 = 77.5 m

1 = 0.25 m 4

A wooden block of mass 10g is dropped from the top of a cliff 100 metres high. simultaneously, a bullet of mass 10 g is fired from the foot of the cliff vertically upwards with a velocity of 100 m/sec. (i) Where and after what time will they meet ? (ii) If the bullet, after striking the block, gets embedded in it, how high will it rise above the cliff before it starts falling ? Sol. (i) Let the wooden block and bullet meet after a time t seconds. The distance s1 moved by the block is given by 5.

Chemistry Facts

1 s1 = gt2 ...(1) 2 The distance s2 moved by the bullet in time t second is given by 1 2 1 gt = 100t – gt2 2 2 Adding eqs. (1) and (2) s1 + s2 = 100 t

s2 = ut –

or

100 = 100t

t = 1 sec.

1.

The element with the lowest boiling point is also helium at -452.07 degrees Fahrenheit (268.93 degrees Celsius.

2.

The word "atom" comes from the Greek word atomos, meaning "uncut."

3.

In 1964, scientists in Russia discovered element 104, and suggested the name Kurchatovium and symbol Ku in honor of Igor Vasilevich Kurchatov. Then in 1969, scientists in the U.S. also found element 104, and propsed the name Rutherfordium (symbol Rf), in the honor of the New Zealand physicist Ernest R. Rutherford. To get the names past the I.U.P.A.C., it won with rutherfordium.

4.

The first and relatively pure atom of tantalum was produced by von Bolton in 1907.

5.

Andres Manual del Rio discovered what we call today vanadium. He called it panchromium, and then changed it to erythronium (red), after noting that upon heating it turned red. In 1831, Nils Gabriel Sefström (a Swedish chemist) was working with some iron ores and this matter was lead to honor the Northern Germanic tribes' goddess Vanadis due to its inspiration in multi-colors. In the same year, Friedrich Wöhler came into posession of del Rio's erythronium, and confirmed it to be vanadium, after Vanadis. The name Vanadium is now being used instead of del Rio's erythronium.

6.

Hafnium was named Copenhagen, Denmark.

7.

The heaviest type of lepton is the tau.

...(2)

(Q s1 + s2 = 100 m)

1 2 1 gt = × 9.8 × 1 = 4.9 m 2 2 Thus the two meet after 1 sec. at distance of 4.9 m from the top of the cliff. (ii) The velocity of the block before impact u1 = 0 + gt = 9.8 m/s The velocity of the bullet before impact u2 = u – gt

Now

s1 =

∴ u2 = 100 – (9.8 × 1) = 90.2 m/s Let after the impact. v be the velocity of combined mass. Applying the law of conservation of linear momentum, we have m1u1 + m2u2 = (m1 + m2)v 10 × 9.8 + 10 × (– 90.2) = (10 + 10)v XtraEdge for IIT-JEE

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after

the

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REACTION

Organic Chemistry Fundamentals

MECHANISM

Mechanism for the E2 Reaction : When isopropyl bromide is heated with sodium ethoxide in ethanol to form propene, the reaction rate depends on the concentration of isopropyl bromide and the concentration of ethoxide ion. The rate equation is first order in each reactant and second order overall :

substitution products in 83% yield and an elimination product (2-methyl propene) in 17% yield : CH3 CH3 SN1

CH3 CH3C—Cl

Rate ∝ [CH3CHBrCH3][C2H5O–] Rate = k[CH3CHBrCH3][C2H5O–] From this we infer that the transition state for ratedetermining step must involve both the alky halide and the alkoxide ion. The reaction must be bimolecular. Considerable experimental evidence indicates that the reaction takes place in the following way Reaction :

CH3

C2H5 – O

H

–δ C2H5 – O ... H

CH3

H C– C βα H Br H (I)

H H

C=C

H

CH3 C— C βα H

H Br

H

+

CH3

CH3 tert-Butyl ethyl ether

tert-Butyl alcohol

E1

(83%)

CH3 CH2= C

CH3

The initial step for both reactions is the formation of a tert-butyl cation. This is also the rate-determing step for both reactions; thus, reactions are unimolecular: CH3 CH3 – slow CH C—Cl CH C+ + Cl – 3

3

CH3

CH3

(solvated)

(solvated)

Whether substitution or elimination takes place depends on the next step (the fast step). If a solvent molecule reacts as a nucleophile at the positive carbon atom of the tert-butyl cation, the product is tert-butyl alcohol or tert-butyl ethyl ether and the reaction is SN1 : CH3 Sol CH3 + fast + CH C—O CH C Sol—OH

+ +

δ–

C2H5–OH + Br–

3

3

CH3

(III)

CH3

(Sol=H– or CH3CH2–)

(I)

The basic ethoxide ion begins to remove a proton from the β carbon using its electron pair to form a bond to it. At the same time, the electron pair of the β C–H bond begins to move in to become the π bond of a double bond, and the bromine begins to depart with the electrons that bonded it to the α carbon.

H H–O–Sol

H CH3C – O – Sol + H – O+– Sol CH3

CH3

(II)

Partial bonds in the transition state extend from the oxygen atom that is removing the β hydrogen, through the carbon skeleton of the developing double bond, to the departing leaving group. The flow of electron density is from the base toward the leaving group as an electron pair fills the π bonding orbital to the alkene.

SN1 reaction

If, however, a solvent molecule acts as a base and removes one of the β hydrogen atoms as a proton, the product is 2-methylpropene and the reaction is E1. E1 reactions almost always accompany SN1 react CH3 CH3 + + fast Sol–O –H + CH2=C Sol–O H–CH2–C CH3 CH 3 H H 2-methylpropene

(III)

EI reacton

At completion of the reaction, the double bond is fully formed and the alkene has a trigonal planar geometry at each carbon atom. The other products are a molecule of ethanol and a bromide ion.

Mechanism for the E1 Reaction Eliminations may take a different pathway from E2 reaction. Treating ter-tbutyl chloride with 80% aqueous ethanol at 25ºC, for example, gives XtraEdge for IIT-JEE

CH3C—OH + CH3C—OCH2CH3

2-Methylpropene (17%)

(II) Transition state

CH3

20% H2O 25ºC

tert-Butyl chloride

C2H5O– + CH3CHBrCH3 → CH2 = CHCH3 + C2H5OH + Br– Mechanism : –

80% C2H5OH

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Reaction : +

2 H 5 OH  → CH3CH2O–Na+ + CH3CH2Br C

55 º C ( − NaBr )

(CH3)3CCl + H2O → CH2 = C(CH3)2 + H2O + Cl Mechanism : Step 1 : CH3 CH3 slow H3C—C—Cl H3C—C+ + Cl – H2O CH3 CH3

CH3CH2OCH2CH3 + CH2 = CH2 E2 (10%) SN2 (90%) Secondary Substrate : With secondary halides, however, a strong base favors elimination because steric hindrance in the substrate makes substitution more difficult : CH3CH2O–Na+ + CH3CHCH3 C2H5OH 55ºC (–NaBr) Br

(II)

(I)

(I) Aided by the polar solvent, a chlorine departs with the electron pair that bonded it to the carbon. (II) This slow step produces the relatively stable 3º carbocation and a chloride ion. The ions are solvated (and stabillized) by surrounding water molecules.

CH3CHCH3 + CH2 = CHCH3 OCH2CH3 SN2 (21%)

Step 2 : H CH3 H–O + H–C–C+ βα H H CH3 (I)

CH3 H + H–O–H + C = C CH3 H H

Tertiary Substrate : With tertiary halides, steric hindrance in the substrate is severe and an SN2 reaction cannot take place. Elimination is highly favored, especially when the reaction is carried out at higher temperatures. Any substitution that occurs must take place through an SN1 mechanism:

(II)

(I) A molecule of water removes one of the hydrogens from the β carbon of the carbocation. These hydrogens are acidic due to the adjacent positive charge. At the same time an electron pair moves in to form a double bond between the α and β carbon atoms.

CH3

CH3 CH3 CH3CH2O Na + CH3CCH3 C2H5OH CH3CCH3 + CH2 = CCH3 55ºC Br (–NaBr) OCH2CH3 –

(II) This step produces the alkene and a hydronium ion.

Substitution versus Elimination : All nucleophiles are potential bases and all bases are potential nucleophiles. This is because the reactive part of both nucleophiles and bases is an unshared electron pair. It should not be surprising, then, that nucleophilic substitution reactions and elimination reactions often compete with each other. SN2 versus E2 SN2 and E2 reactions are both favored by a high concentration of a strong nucleophile or base. When the nucleophile (base) attacks a β hydrogen atom, elimination occurs. When the nucleophile attacks the carbon atom bearing the leaving group, substitution results : (a) Θ

Nu (b)

H–C–

(a) elimination E2

C C

Mainly E2 (91%)

CH3

CH3 C H OH 2 5 CH3CH2O Na + CH3CCH3 CH2=CCH3 + CH3CH2OH 55ºC Br (–NaBr) E2 + E1 (100%) –

+

Temperature : Increasing the reaction temperature favors elimination (E1 and E2) over substitution. Elimination reactions have greater free energies of activation than substitution reactions because more bonding changes occur during elimination. When higher temperature is used, the proportion of molecules able to surmount the energy of activation barrier for elimination increases more than the proportion of molecules able to undergo substitution, although the rate of both substitution and elimination will be increased. Further more, elimination reactions are entropically favored over substitution because the products of an elimination reaction are greater in number than the reactants. Additionally, because temperature is the coefficient of the entropy term in the Gibbs freeenergy equation ∆Gº = ∆Hº – T ∆Sº, an increase in temperature further enhances the entropy effect. Size of the Base/Nucleophile : Increasing the reaction temperature is one way of favorably influencing an elimination reaction of an alkyl halide. Another way is to use a strong sterically hindered base such as the tert-butoxide ion. The

+ Nu – H + X–

H – C – + X:– Nu– C –

Consider the following examples with small (unhindered) nucleophiles and alkyl halides of different classes. Primary Substrate : When the substrate is a primary halide and the base is unhindered, like ethoxide ion, substitution is highly favored because the base can easily approach the carbon bearing the leaving group:

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+

SN1 (9%)

–C–X (b) substitution SN2

E2(79%)

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bulky methyl groups of the tert-butoxide ion inhibit its reacting by substitution, allowing elimination reactions to take precedence. We can see an example of this effect in the following two reactions. The relatively unhindered methoxide ion reacts with octadecyl bromide primarily by substitution; the bulky tert-butoxide ion gives mainly elimination. Unhindered (small) Base/Nucleophile

It is usually difficult to influence the relative partition between SN1 and E1 products because the free energy of activation for either reaction proceeding from the carbocation (loss of a proton of combination with a molecule of the solvent) is very small. In most unimolecular reactions the SN1 reaction is favored over the E1 reaction, especially at lower temperatures. In general, however, substitution reactions of tertiary halides do not find wide use as synthetic methods. Such halides undergo eliminations much too easily. Increasing the temperature of the reaction favors reaction by the E1 mechanism at the expense of the SN1 mechanism. If the elimination product is desired, however, it is more convenient to add a strong base and force an E2 reaction to take place instead.

OH CH3O– + CH3(CH2)15CH2CH2 – Br CH 3 → 65 º C

CH3(CH2)15CH = CH2 + CH3(CH2)15CH2CH2OCH3 E2 (1%) SN2(99%) Hindered Base/Nucleophile CH 3 | CH 3 ) 3 COH (  → CH 3 − C − O − +CH3(CH2)15CH2CH2 – Br 40 º C | CH 3

CH 3 | CH3(CH2)15CH=CH2+CH3(CH2)15CH2CH2– O − C − CH 3 | CH 3

SCIENCE TIPS • A porcelain funnel used for filtration by suction is known as ® Bucher Funnel

E2 (85%) SN2 (15%) Basicity and polarizability : Another factor that affects the relative rates of E2 and SN2 reactions is the relative basicity and polarizability of the base/nucleophile. Use of a strong, slightly polarizable base such as amide ion (NH2–) or alkoxide ion (especially a hindered one) tends to increase the likelihood of elimination (E2). Use of a weakly basic ion such as a chloride ion (Cl–) or an acetate ion (CH3CO2–) or a weakly basic and highly polarizable one such as Br–, I–, or RS– increases the likelihood of substitution (SN2). Acetate ion, for example, reacts with isopropyl bromide almost exclusively by the SN2 path : O CH3 CH3C – O– + CH3CH – Br

• What is diazomethane ? +

• A drying chamber, containing chemicals such as concentrated sulphuric acid or silica gel is known as ® Desiccator • Reforming of a gasoline fraction to increase branching in presence of AlCl3 is known as ® Isomerization

• A condenser consisting of glass tube surrounded by another glass tube through which cooling water flows is known as ® Liebig condenser • For wattles current what should be the value of the power factor of the circuit ? ® Zero

SN2 (~ 100%)

O

• For which colour is the critical angle of light, pasing from glass to air, minimum ? ® Violet

CH3

• Give an example of application of mutual induction ® Transformer in any device.

CH3C –O – CHCH3 + Br–

The more strongly basic ethoxide ion reacts with the same compound mainly by an E2 mechanism. Tertiary Halides : SN1 versus E1 Because E1 and SN1 reactions proceed through the formation of a common intermediate, the two types respond in similar ways to factors affecting reactivities. E1 reactions are favored with substrates that can form stable carbocations (i.e., tertiary halides); they are also favored by the use of poor nucleophiles (weak bases) and they are generally favored by the use of polar solvents. XtraEdge for IIT-JEE

® [CH 2 = N = N or CH 2 N 2 ]

• What is the correct sequence of the semiconductors silicon, tellunium and germanium in the increasing order of their energy gap ? ® Tellurium, germanium, silicon

• Which ammeter is used to measure alternating ® Hot wire ammeter current ? • What quantity has the ampere-second as its unit ? ® Quantity of electricity

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KEY CONCEPT

Physical Chemistry Fundamentals

SOLID STATE In this structure, each sphere is in contact with six nearest neighbours (four in the same base, one above and one below). The percentage of occupied volume in this structure can be calculate as follows: The edge length a of the cube will be twice the radius of the sphere, i.e. a = 2r. Since in the primitive cubic lattice, there is only one sphere present in the unit lattice, the volume occupied by the sphere is

Density of Cubic Crystals : The density based on the structure can be calculated from the mass contained in a unit cell and its volume. If N is the number of molecules per unit cubic cell of edge length a, then the mass and volume per unit cell are  M Mass =   NA

  N 

Therefore, Density =

Volume = a3 mass NM = 3 volume a NA

V=

The value of N for the three cubic cells can be calculated as follows : Primitive cubic cell : In a primitive cubic cell, atoms are present at the corners of the cube. There are eight corners of a cube and thus eight atoms are present at these corners. Now, any particular corner of the cube is actually shared amongst eight such cubic unit cells placed adjacent to one another. Thus, the contribution of the atom placed at one of the corners to the single cubic unit cell is 1/8. Since there are eight corners of a cube, the number of atoms associated with a single primitive unit cell is 8/8 = 1. Body-centred cubic cell : In a body-centred cubic unit cell, besides atoms being present at the corners, there is one atom in the centre of the cube which belongs exclusively to this cubic unit cell. Therefore, number of atoms per unit cell are two. Face-centred cubic cell : Here, atoms, besides being at the corners, are also present at the centre of the six faces. Each of these atoms is shared between two such unit cells. Thus, their contribution to the unit cell is 6/2 = 3 atoms, making a total of 4 atoms per cubic unit cell Packing in a simple Cubic Lattice : In a lattice of this type, the spheres are packed in the form of a square array by laying down a base of spheres and then piling upon the base other layers in such a way that each sphere is immediately above the other sphere, as shown in fig.

4 3 πr 3

or

V=

4 a π  3 2

3

The fraction of the total volume occupied by the sphere is 3

4 a π  π 3 2 = = 0.5236 φ= 3 6 a or 52.36 percent Thus, the structure is relatively open since only 52.36% (π/6) of the total volume is occupied by the spheres. The remainder, i.e. 0.4764 of the total volume is empty space or void volume. No crystalline element has been found to have this structure. Closest Packing : In closest packing arrangements, each sphere is in contact with the maximum possible number of nearest neighbours. Fig. shows a closest packed layer of spheres. Each sphere is surrounded by six nearest neighbours lying in the plane, three spheres Just above it and three below it, thus making the total number of nearest neighbours equal to twelve. If the spheres are packed in the same plane, then just above these spheres

A B C A

A B C

A B A

B C

B

B

B C

A

C A

A

C

A

A

C A

A C

A

A

Packing in a simple cubic lattice

Fig. (a) Closest packed layers of spheres

XtraEdge for IIT-JEE

34

AUGUST 2011


determined by reference to a face centred cubic lattice with unit cell length a. In such a lattice, the distance between closest-packed layers (Miller indices 111)is one –third of the body diagonal, i.e.

B

A B

A

A

C B

B

3 a./3.

A

A B A

Thus,

C B A

C = 2

3a 3 Layer A

Fig. (b) Two types of packing

Layer B

there exist two different types of voids, pointing in different directions as shown in fig. (a). Thus, we can have three different types of locations as shown by A, B and C in fig. (a). Location A is occupied by the spheres while B and C are the two different types of voids. But because of the size of the spheres, both types of voids cannot be occupied simultaneously. The third layer of closest-packed sphere can be formed in two different ways. If, for example, we choose to place the spheres of the second layer in B sites, one of the available sets of voids for the third layer will be directly above the spheres in the original layer. These are A sites. The other set of voids will be directly above the voids designated by C in the original layer. Types of Packing : Thus, two types of packing (fig. b) are possible ABABA.... or ABCABC .... We can have many other varieties of patterns such as ABCACB....., ABAC .... etc. But for many of the common substances that form closest-packed structures, one of the above two symmetrical arrangements is observed. Hexagonal Closest Packed Structure : The packing ABAB.... is known as a hexagonal closest-packed structure (HCP). The unit cell of shown in figure.

C/2 Layer A a 2r Hexagonal closest-packed structure

Now, in the face-centred lattice spheres touch one another along the face diagonal. Thus, we have 4r = 2 a

4 2

r

 3   3 4r   = 8 r a = 2 C = 2  3   3 2 6    

The hexagonal base consists of six equilateral triangles, each with side 2r and with an altitude of 2r sin 60º, i.e.

3 r. Therefore,

1 Area of the base = 6  2

( 3r )(2r) (

= 6 3 r2

)

 8  r  = 24 2 r3 Volume of the prism = 6 3r 2   6 

Number of spheres belonging to this prism 3 spheres in B layers exclusively belong to this prism. 1 from the centre of the base. There are two spheres of this type and each is shared by two prisms. 2 from the corners. There are twelve such spheres and each is shared amongst six prisms of this type. Thus, the total number of spheres is 6. The fraction of volume of the prism actually occupied by the spheres is

B A Hexagonal closest -packed

Unit cell formed by ABA packing

The fraction of the volume occupied in HCP can be calculated as described in the following. The distance C/2 (in figure) is the distance between the layers A and B. This distance will be from the centre of a sphere to the plane of the three spheres that are in contact with it. This distance can be XtraEdge for IIT-JEE

a=

With this, the distance C becomes

A

Exploded view

or

4  6 πr 3  3  = 24 2 r 3

2π = 0.7405 6

or 70.05 percent Example of HCP are Ca, Cd, Cr, Mg and Zn. 35

AUGUST 2011


this arrangement, spheres are touching one another along the cross diagonal of the cube, making its

Cubical Closest-Packed Structure The packing ABCABC, .... is a cubical closestpacking (CCP) or face-centred cubic packing. The fraction of volume occupied in CCP can be calculated as follows : The radius of the sphere in terms of the unit length of the face-centred cube is given by

distance equal to 4r. This must be equal to Thus, 4r =

3 a,

3 a 4 Volume of the cube = a3

i.e.,

2a 4 since the sphere will be touching each other along the diagonal of the face of the cube. In the face-centred cubic lattice, there are four spheres per unit cell. Therefore, fraction of volume occupied by the spheres is

r=

3     4  2a   4 π     3  4   = a3

3 a.

r=

4 4  3  a Volume of one sphere = πr3 = π  3 3  4 

3

Since there are two spheres in each unit cell, the total volume occupied will be 3     4  3   2 π a     3  4  

2π = 0.7405 6

The fraction of the volume occupied by the spheres 3     4  3   2 π a     3  4   = φ = a3

or 74.05 percent A C

3π = 0.6802 8

or 68.02 percent In this arrangement each sphere has eight nearest neighbours.

B A Exploded view

SCIENCE TIPS

Cubical closest-packed stricture

Face-cented cubic unit cell formed by ABCA packing

• A porcelain funnel used for filtration by suction is known as ® Bucher Funnel

out of all these packings, HCP and CCP are more common for uniform spheres. In general, the packing fraction, i.e. fraction of volume occupied, is independent of the radius of the sphere and depends only on the nature of packing. From the values of packing fractions, it follows that the density of a substance in HCP and CCP structures will be more than in the other two packings.

• What is diazomethane ? +

® [CH 2 = N = N or CH 2 N 2 ]

• A drying chamber, containing chemicals such as concentrated sulphuric acid or silica gel is known as ® Desiccator

Packing in a Body Centred Cubic Lattice : Here the packing consists of a base of spheres, followed by a second layer where each sphere rests in the hollow at the junction of four spheres below it, as shown in figure.

• Reforming of a gasoline fraction to increase branching in presence of AlCl3 is known as ® Isomerization

• A condenser consisting of glass tube surrounded by another glass tube through which cooling water flows is known as ® Liebig condenser • What is • ® Hot wire ammeter

Packing in a body-centred cubic lattice

• What quantity has the

The third layer then rests on these in arrangement which corresponds exactly to that in the first layer. In XtraEdge for IIT-JEE

36

AUGUST 2011


UNDERSTANDING

Physical Chemistry

1.

At 298 K, the emf of the cell -3

–3

Sol. (a) Amount of alum =

-3

Hg 0.01 mol dm KCl 1 mol dm KNO3 0.01 mol dm KOH Hg sat. with Hg 2 Cl 2

sat. with HgO

is found to be 0.1634 V and the temperature coefficient of the emf to be 0.000837 VK–1. Calculate the enthalpy and entropy changes of the reaction. What is the reaction that occurs in the cell ? Sol. For the given cell, we have Electrode Reduction reaction Right HgO(s) + H2O(1) + 2e– = Hg(1) + 2OH–(aq) ...(i) – – = 2Hg(1) + 2Cl (aq) Left Hg2Cl2(s) + 2e ...(ii) Subtracting Eq. (ii) from Eq. (i), we get HgO(s) + H2O(1) + Hg(1) + 2Cl–(aq) = Hg2Cl2(s) + 2OH–(aq) The number of electrons involved in the electrode reactions is 2. Thus ∆G = –nFE = –2(96500 C mol–1)(0.1634 V) = 31536.2 J mol–1

= 0.024 mol

0.024 mol 0.1 dm 3

= 0.24 M Hydrolysis of Al3+ is Al3+ + 2H2O Kh =

Al(OH)2+ + H3O+

[Al(OH) 2 + ][H 3O + ] [Al3+ ]

If x is the concentration of Al3+ that has hydrolyzed, we have ( x )( x ) = 1.4 × 10–5 M Kh = 0.24 M − x Solving for x, we get [H3O+] = x = 1.82 × 10–3 M (b) We will have to consider the following equilibria. Al(OH)2+ + H3O+ Al3+ + 2H2O + 2– H3O + SO4 HSO4– + H2O Let z be the concentration of SO42– that combines with H3O+ and y be the net concentration of H3O+ that is present in the solution. Since the concentration z of SO42– combines with the concentration z of H3O+, it is obvious that the net concentration of H3O+ produced in the hydrolysis reaction of Al3+ is (y + z). Thus, the concentration (y + z) of Al3+ out of 0.24 M hydrolyzes in the solution. With these, the concentrations of various species in the solution are

= –2(96500 C mol–1)[(0.1634V) – (298 K × 0.000837VK–1)] = 16603 J mol–1  ∂E  ∆S = nF   = 2(96500 C mol–1)(0.000837 V K–1)  ∂T  p

= 161.54 JK–1mol–1

Al3+

0.24 M − y − z

Potassium alum is KA1(SO4)2.12H2O. As a strong electrolyte, it is considered to be 100% dissociated into K+, Al3+, and SO42–. The solution is acidic because of the hydrolysis of Al3+, but not so acidic as might be expected, because the SO42– can sponge up some of the H3O+ by forming HSO4–. Given a solution made by dissolving 11.4 g of KA1(SO4)2.12H2O in enough water to make 0.10 dm3 of solution, calculate its [H3O+] : (a) Considering the hydrolysis Al(OH)2+ + H3O+ Al3+ + 2H2O with Kh = 1.4 × 10–5 M (b) Allowing also for the equilibrium H3O+ + SO42– HSO4– + H2O –2 with K2 = 1.26 × 10 M

XtraEdge for IIT-JEE

474.38 g mol −1

Molarity of the prepared solution =

  ∂E   ∆H = – nF E − T    ∂T  p  

2.

11.4 g

+ 2H2O

H 3O + + SO 24 − y

0.48 M − z

Thus, Kh = K2 =

Al(OH) 2+ + H 3O + y+ z

y

HSO −4 + H2O z

( y + z)( y) = 1.4 × 10–5 M (0.24M − y − z)

...(i)

z 1 = y(0.48M − z) 1.26 × 10 − 2 M

...(ii)

From Eq. (ii), we get (0.48 M ) y z= (1.26 × 10 − 2 M ) + y Substituting this in Eq. (i), we get

37

AUGUST 2011


Substituting the values, we have

  (0.48M) y y+ y −2  (1.26 × 10 M ) + y   = 1.4 × 10–5   ( 0 . 48 M ) y  0.24 − y −   (1.26 × 10 − 2 M ) + y   –2

Making an assumption that y <<1.26 × 10 then solving for y, we get [H3O+] = y = 2.932 × 10–4 M

The concentration of hydrogen ions is given as [H3O+] = cα = (0.5 M) (6 × 10–3) = 3 × 10–3 M Hence, pH = – log {[H3O+]/M} = – log (3 × 10–3) = 2.52 (a) To double the pH Thus pH = 5.04 Since pH = – log {[H3O+]/M} therefore [H3O+]/M= 10–pH . Substituting the value of pH , we have [H3O+]/M = 10–5.04 = 0.912 × 10–5 = 9.12 × 10–6 Thus, c1α = 9.12 × 10–6 M In dilution, α will increase, and its value will not be negligible in comparison to one. Thus, we shall have to use the expression

M, and

3.

What is the solubility of AgCl in 0.20 M NH3 ? Given : Ksp(AgCl) = 1.7 × 10–10 M2 K1 = [Ag(NH3)+] / [Ag+] [NH3] = 2.33 × 103 M–1 and K2 = [Ag(NH3)2+]/[Ag(NH3)+][NH3] = 7.14 × 103 M–1 Sol. If x be the concentration of AgCl in the solution, then [Cl–] = x From the Ksp for AgCl, we derive [Ag+] =

K sp [Cl − ]

=

1.7 × 10 −10 M 2 x

If we assume that the majority of the dissolved Ag+ goes into solution as Ag(NH3)2+ then [Ag(NH3)2+] = x Since two molecules of NH3 are required for every Ag(NH3)2+ ion formed, we have [NH3] = 0.20 M – 2x Therefore,

K a=

= 0.6637 27.12 × 10 – 6 M Since c1α = 9.12 × 10–6 M and α = 0.6637 Therefore,

= 6.0 × 10–8 M2 From which we derive (0.20M − 2 x )

−8

6.0 × 10 M

9.12 × 10 –6 M = 1.374 × 10–5 M 0.6637 Volume to which the solution should be diluted

c1 =

2

= 3.5 × 102 x 1.7 × 10 −10 M 2 which gives x = [Ag(NH3)2+] = 9.6 × 10–3 M, which is the solubility of AgCl in 0.20 M NH3 2

=

=

[OH–] =

(1.0 × 10 –14 M 2 ) (3 × 10 – 3 M)

In the present case, the concentration of hydroxyl becomes [OH–] =

2(1.0 × 10 –14 M 2 ) (3 × 10 – 3 M )

which gives (3 × 10 –3 M) = 1.5 × 10–3 M 2 For the concentration, we can use Ka = c2α2 = (c2α) (α)

[CH 3COO – ][H 3O + ] (cα)(cα ) = = cα2 Ka = [CH 3COOH] c(1 – α)

[H3O+] =

Ka c

XtraEdge for IIT-JEE

(0.5 M ) (1 dm 3 ) cV = = 3.369 × 104 dm3 c1 (1.374 × 10 – 5 M)

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

Given a solution that is 0.5 M CH3COOH. To what volume at 25ºC must one dm3 of this solution be diluted to (a) double the pH; (b) double the hydroxide-ion concentration ? Given that Ka = 1.8 × 10–5 M. Sol. If α is the degree of dissociation of acetic acid of concentration c then the concentrations of various species in the solution are CH3COOH + H2O CH3COO– + H2O+ c(1 – α) cα cα With these concentrations, the equilibrium constant becomes 4.

or = α =

1.8 × 10 –5 M

or a =

2

2

c α2 (c1α ) 2 (c α )α (9.12 × 10 –6 M)α = 1 = 1 = c1 (1 – α) 1 – α 1– α 1– α

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

 1.7 × 10 −10 M 2   (0.20M − 2x ) 2   x [Ag ][ NH 3 ]   = Kinst = x [Ag( NH 3 ) +2 ] +

(1.8 × 10 –5 M ) = 6 × 10–3 (0.5 M)

α=

38

AUGUST 2011


or α =

263.15 K

Ka (1.8 × 10 –5 M ) = = 1.2 × 10–2 (c 2 α ) (1.5 × 10 – 3 M) (1.5 × 10 –3 M )

Thus, c2 =

∆rS3 =

273.15 K

cV (0.5M )(1 dm 3 ) = = 4 dm3 c2 (0.125 M )

It is possible to supercool water without freezing. 18 g of water are supercooled to 263.15 K(–10ºC) in a thermostat held at this temperature, and then crystallization takes place. Calculate ∆rG for this process. Given: Cp(H2O,1) = 75.312 J K–1 mol–1 Cp (H2O,s) = 36.400 J K–1 mol–1 ∆fusH (at 0ºC) = 6.008 kJ mol–1 Sol. The process of crystallization at 0ºC and at 101.325 kPa pressure is an equilibrium process, for which ∆G = 0. The crystallization of supercooled water is a spontaneous phase transformation, for which ∆G must be less than zero. Its value for this process can be calculated as shown below. The given process H2O(1, – 10ºC) → H2O(s, –10ºC) is replaced by the following reversible steps. (a) H2O(1, – 10ºC) → H2O(1, 0ºC) ...(1) 5.

∫C

p , m (1) dT

263.15 K

= (75.312 J K–1 mol–1 ) (10 K) = 753.12 J mol–1 273.15K

C p, m (1)

∆rS1 =

263.15K

dT

R

 273.15 K   = (75.312 J K–1mol–1) × ln   263.15 K 

= 2.809 J K–1 mol–1 (b) H2O(1, 0ºC) → H2O(s, 0ºC)

...(2)

–1

∆rH2 = – 6.008 kJ mol ∆rS2 = –

(6008 J mol –1 ) = – 21.995 J K–1 mol–1 (273.15 K )

(c) H2O(s, 0ºC) → H2O(s, –10ºC)

...(3) •

263.15 K

∆rH3 =

∫C

p , m (s)

dT

273.15 K

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

XtraEdge for IIT-JEE

T

dT

SCIENCE TIPS

273.15 K

∆rH1 =

C p, m (s)

 263.15 K   = (36.400 J K–1 mol–1) ×ln   273.15 K  = – 1.358 J K–1 mol–1 The overall process is obtained by adding Eqs. (1), (2) and (3), i.e. H2O(1, –10ºC) → H2O(s, –10ºC) The total changes in ∆rH and ∆rS are given by ∆rH = ∆rH1 + ∆rH2 + ∆rH3 =(753.12 – 6008 – 364.0) J mol–1 = – 5618.88 J mol–1 ∆rS = ∆rS1 + ∆rS2 + ∆rS3 = (2.809 – 21.995 – 1.358) J K–1 mol–1 = – 20.544 J K–1 mol–1 Now ∆rG of this process is given by ∆rG = ∆rH – T∆rS = – 5618.88 J mol–1 – (263.15 K)( –20.544 J K–1 mol–1 ) = – 212.726 J mol–1

–1

= 1.25 × 10 M = 0.125 M 1.2 × 10 – 2 Volume to which the solution should be diluted =

39

Why is the cooling inside a refrigerator not proper when a thick layer of ice deposits on the freezer? ® Because ice is a bad conductor of heat Which type of computer is often found in small business and in homes and classrooms ? ® The micro computer. It is the smallest and the least costly type of computer Out of joule, calorie, kilowatt and electron-volt which one is not the unit of energy ? ® kilowatt How does the atmospheric pressure vary with height ? ® Atmospheric pressure P decreases with height h above sea level. For an 'ideal' atmosphere at constant temperature P = P0 e–kh where k is a constant and P0 is the pressure at the surface How is r.m.s. velocity of gas molecules related to absolute temperature of the gas ? ® vrms ∝ T What are transducers ? ® Devices which change signals from one form to another (e.g. sound to electrical) are called transducers Is polarization the property of all types of waves? ® No, it is property of only transverse waves AUGUST 2011


`tà{xÅtà|vtÄ V{tÄÄxÇzxá

Set

4

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 So lu t ion s wi l l b e p ub lished in nex t issue Joint Director Academics, Career Point, Kota 1.

If two circles cut orthogonally, prove that the polar of any point P on the first circle with respect to the second passes through the other end of diameter of the first circle which goes through P.

2.

Let ABCD be a tetrahedron. If perpendiculars from B and C to the opposite faces intersect, then show that BC is perpendicular to AD and the perpendiculars from A and D to the opposite faces will also intersect.

3.

8.

9.

For a real number u ; π

I(u) = ln (1 − 2u cos x + u 2 ) dx;

Generalize the result as

2

n

n

I (u 2 ) .

4.

Let f (x) = x2 + ax + b be a quadratic polynomial where a and b are integers. Let n be an integer, show that f (n). f (n + 1) = f (m) for some integer m.

5.

Show that the straight lines joining any two fixed points on a rectangular hyperbola to any variable point on it intercept a constant length on either asymptote.

6.

A constant function and e^x are walking on Broadway. Then suddenly the constant function sees a differential operator approaching and runs away. so e to-the x follows him and asks why the hurry.

Solve : 1 y2  dx +  − 2  x ( x − y ) 

7.

1 sin 9 x sin x sin 3 x + + = [tan 27x – tan x] cos 3 x cos 9 x cos 27 x 2

1 2 I(u ). 2

1

"Well, you see, there's this diff.operator coming this way, and when we meet, he'll differentiate me and nothing will be left of me...!"

 x2 1 −  dy = 0  2 y   ( x − y )

"Ah," says e^x, "he won't bother ME, I'm e to-the x!" and he walks on. Of course he meets the differential operator after a short distance.

Through a focus of an ellipse two chords are drawn and a conic is described to pass through their extremities, and also through the center of the ellipse. Prove that it cuts the major axis in another fixed point.

XtraEdge for IIT-JEE

Let there be n straight lines in a plane, no two of which being parallel or coincident and no three of them meet at a point, then show that they divide the 1 plane in (n2 + n + 2) parts. 2

10. Prove that

0

prove that I (u) = I (– u) =

 xy  f ( x). f ( y ) for all real x & y. If Let f   = 2  2  f ′(1) = f (1) ≠ 0 then show that f (x) is differentiable ∀ x ∈ R except zero. Also find f (x) for all x ≠ 0.

e^x : "Hi, I'm e^x" diff.op. : "Hi, I'm d/dy"

40

AUGUST 2011


MATHEMATICAL CHALLENGES SOLUTION FOR JULY ISSUE (SET # 3)

1.

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

 1  3  3 1 = − a − + i − a+   2  2 2  2   

− ln 2 dx I.F. = e ∫ = 2–x –x

so y2 =

∫2

sin x −x

–1 = –

. (cos x – 1). ln 2 dx

⇒ a=2– 3

y2–x = 2sin x – x + c Ans (B) y = 2sin x + c.2x y . 2–x = 2sin x – x + c Now if x → ∞ and y is bounded (finite) then c = 0 Ans (C) sin x For option (D) : f (x) = 2 c.2x = 0, so c = 0 but in this case y is bounded so this is not correct option. Hence correct answers are (B, C) 2.

1 3 a– 2 2

3 1 a+ 2 2 3 1 ⇒ b = – 3 + + = 12 – 3 2 2 Hence correct answer is (B)

b=–

4.

c1 ≡ x2 + y2 = a2 c2 ≡ (x – h)2 + (y – k)2 = r2

AC = 4p A

P α

p

B

tan α =

D α

c1

Director circle x2 + y2 = 2a2  ah + r.0 ak + r.0  Pt. P  ;  a+r   a+r it lies on eqn. (1) a2h2+ a2k2 = (a + r)2 2a2 h2 + k2 = 2(a + r)2 x3 + y3 = 2(a + r)2 Hence correct answer is (A)

C

p AD =4 = p DC

cot α + tan α =

DC AD + =4 p p

tan2α – 4 tan α + 1 = 0 4 ± 16 − 4 =2± 3 2 α = 15º & 75º Hence correct answer is (D)

tan α =

3.

c2

5.

4a = 2(2a) = 2 .

...(1)

5 − 36 + 17 13

14 28 = 13 13 Hence correct answer is (C) =2.

z2 = z1iπ/3 z1(–a + i) z2 (–1+bi)

6.

O z3

1 3  –1 + bi = (– a + i)  + i 2 2  

XtraEdge for IIT-JEE

∫ =

41

9x x cos . sin 3x 2 2 sin 3x − sin 6 x x 9x 2 cos cos . sin 3x 2 2 9x 3x − 2 cos sin 2 2

2 cos

AUGUST 2011


x

A

3x

∫ 2 cos 2 dx = – ∫ (cos 2 x + cos x ) dx = – 2 cos

=– 7.

R

sin 2 x – sin x + c 2

r B

(x – 1)2 = a2 a = 1 ± |a| f (x) = x2 – 2(a + 1)x + a(a – 1) = 0

2

I(h,k)

M

N

h = ON =

& r=k

1

so r = k =

2sk = D = 4(a + 1)2 – 4a (a – 1) > 0 a2 + 2a + 1 – a2 + a > 0 1 a>– ...(1) not essential 3 f (1 ± |a|) < 0 ⇒ (1 ± |a|)2 – 2(a + 1) (1 ± |a|) + a(a – 1) < 0 ⇒ 1 + a2 ± 2|a| – 2a m 2a|a| – 2 m 2|a| + a2 – a < 0 ⇒ 2a2 m 2a|a| – 3a – 1 < 0 if a ≥ 0 2a2 – 2a2 – 3a – 1 < 0 1 ⇒ a>– ⇒a≥0 3 2a2 + 2a2 – 3a – 1 < 0 ⇒ 4a2 – 3a – 1 < 0 (4a + 1) (a – 1) < 0 1 ⇒ – <a<1 4 ⇒ 0≤a<1 So, similarly if a < 0 then ⇒ f (1 m a) < 0 f (1 ± |a|) < 0 1 so f (1 ± |a|) < 0 ⇒ – < a < 1 ...(2) 4 1 eqn. (1) ∩ (2) – < a < 1 4

= 2sk =

∆ = s

s ( s − a )( s − b)( s − c) s

s ( s − a )( s − b)( s − c) s

s ( s − a )(a − b + c)(a + b − c)

s ( s − a )(a − 2 x)(a + 2 x) s ( s − a)(a 2 − 4h 2 )

required lows is 4s2y2 = A(a2 – 4x2) ⇒ s2y2 + A x2 =

Aa 2 4

where A is = s(s – a) here h2 < ab so it is an ellipse. 9.

Total no of ways of drawing = 74 = 2401 favorable no of drawings = coeffi. of x8 in the expansion of (x0 + x1 + x2 + ...... x6)4 = coeffi. of x8 is (1 – x7)4 (1 – x)–4 = coeffi of x8 in (1 – 4x7 + ....) (1 + 4x + .... + 165x8 +...) = 165 – 16 = 149 149 so required probability = 2401

10. f (x) = (x – α)Q(x) + R f (+1) = 2; f (–2) = 1, f (–1) = –1. If f (x) is divided by (x2 – 1) (x + 2) which is a cubic expression then remainder should be a quadratic expression. f (x) = (x2 – 1) (x + 2) Q(x) + (ax2 + bx + c) f (1) = a + b + c = 2 f (–2) = 4a – 2b + c = 1 f (–1) = a – b + c = –1 3 7 2 Solving these eqn, a = , b = , c = – 6 3 2 3 7 2 so the remainder is x2 + x – 6 3 2

2r = a + b + c ON = – BN + BO Let BN = x 2BN + 2CN + 2AR = 2s x + (a – x) + (b – a + x) = s x=s–b

XtraEdge for IIT-JEE

C

a – (s – b) 2 b−c −2s + a + 2b = = 2 2

so

r=k=

8.

O

42

AUGUST 2011


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

MATHS 1.

Prove that for a differentiable function f(x) n n 1 1  f ´(x)[ x] − x +  dx = f ´(x) dx – f (0) + 0 0 2 2 

1 f(n) – 2

Sol. L.H.S. =

∑ f (r ) , [.] denotes greatest integer. r =0

0

f ´(x)[ x] dx –

n

0

n

=

r =1

r

r

n

=

1 2

 f ´(x)[ x] dx–  xf ( x) | 0n − r −1 

∑∫

(r − 1) f ´(x)dx – nf(n)+

r =1

r −1

AQ =

x. f ´(x) dx

+

n 0

n

n

 1 f ( x)  + f (x) | 0n  2

0

0

b 3

f ( x)dx +

r =1

n 0

θ

n

=–

f (r ) +

r =0

2.

1 1 f (n) – f (0) + 2 2

n 0

0

a2 + b2 + c2 2 3

R

B

 b  (sin A − 3 cos A) + C cos θ =   3 

+

1 3

[b(cos A +

3 sin A) + C

]

We know a cos θ + b sin θ lies between f ( x) dx

a2 + b2 &

a2 + b2 .

Using this

f ( x) dx

(PQ)2 maxm =

2 2 8∆ (a + b2 + c2) + 3 3

1 b2 + c2 − a2 bc sin A = ∆, cos A = ) greatest 2bc 2 area of equilateral

(Q

.

∆PQR =

Sol. Let PQR be an equilateral ∆ circumscribing the ∆ABC. Let ∠ PAB = θ Apply sine rule in ∆PAB

XtraEdge for IIT-JEE

a

From fig : PQ = AP + AQ

A, B and C are three non collinear points in a plane. Show that the area of the greatest equilateral triangle, which can be drawn to circumscribe ∆ ABC is : 2∆ +

C

c

P

f ( x) dx

b

A

A

A+θ – 60º

180 – (A+θ)

1 (f (n) – f (0)) 2 = (f (2) – f (1)) + 2{f (3) – f (2)} + .... + (n – 1) (f (n) n 1 1  – f (n –1)) –  n −  f (n) – f (0) + f ( x) dx 0 2 2  n

3 sin A) sin θ]

Q

+

= –f (1) – f (2) – f (3) .... f (n – 1) – f(n) 1 1  +  n − n +  f (n) – f (0) + 2 2 

3 cos A) cos θ

+ (cos A +

1 (f (n) – f (0)) 2

∑ (r − 1) {f(r) – f (r – 1)} – n f(n) + ∫

( 3 cos θ + sin θ)

[(sin A –

f ´(x) dx

n

=

3

Apply sine rule in ∆ QAC 2b ⇒ AQ = sin (A +θ – 60º) 3

n

n

C

⇒ AP =

3 (PQ maxm)2 4

= 2∆ +

43

a2 + b2 + c2

2 3

AUGUST 2011


3.

Prove that if sin2x + sin2y < 1 for all x ∈R, y ∈R, then  π π (sin–1(tan x . tan y)) ∈  − ,  .  2 2 sin 2 x. sin 2 y

Sol. Consider (tan x . tan y)2 = 2

2

cos 2 x. cos 2 y

(

⇒ 9

2

cos x. cos y

2

9– a ≥ 3 ⇒ a ≤ – 6.

Sol. (i) Ist method : Let z = |z| (cos θ + i sin θ) or |z|eiθ

where θ = arg (z) L.H.S. = |cos θ + i sin θ – 1| = − 2 sin 2

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

Now P(A) =

= 2 sin

0≤i≤k

Ck

P(A/Ei) =

and

∑ j=0

j × 20

6

θ θ θ + i 2 sin . cos 2 2 2

θ θ θ θ = θ = arg (z) sin − i cos ≤ 2 2 2 2 2

(Q sin x ≤ x for x > 0) z complex no. lies on a circle with |z| unit radius i.e. x2 + y2 = 1

14 − i 20 − k

k 14 C

z − 1 ≤ arg (z) |z|

(ii) |z – 1| ≤ ||z| – 1| + |z| |arg (z)|

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

Ci × 6 C k −i

1 2

For any complex no. z, Prove that following inequalities. (i)

 π π = sin (tan x . tan y) ∈  − ,   2 2 –1

20

+3≥0

⇒ t ≥ 3 is possible as t > 1.

+1

= tan x tan y ∈ (–1, 1)

14

IInd method :

C k −i

Ck

+

14 − j 20 − k

P  z    | z |

Also (1 + x)14 – 1 (1 + x)6

θ

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

∑(

14 −1

≥0

(t – 1) (t – 3) ≥ 0 ⇒ t ≤ 1 or t ≥ 3

= tan2x . tan2y < 1

P(Ei) =

– 4.9

–a

0 a

where t = 9– a and t ∈(1, ∞)

+1

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

4.

–2a

)

⇒ t – 4t + 3 ≥ 0

2

2

)

− 2.9 −t dt ≥ 0

⇒ − 9 − 2t + 4(9) −t

+1

cos x. cos y sin x. sin y – 1

− 2t

a

2

2

=

0

0

∀ x, y ∈ R

sin 2 x. sin 2 y – 1 + sin 2 x + sin 2 y – sin 2 x. sin 2 y 2

∫ (9

 9 − 2t 2.9 −t  ⇒  ≥0 −  − 2 log 9 − log 9  a 

2

sin x. sin y − cos x cos y

=

=

2

cos 2 x. cos 2 y

Sol. Here,

Q (1, 0)

C j + 6 C k − j ) = co-efficient of xk

j=0

5.

z –1 |z|

14 14 P(E) = = 6 + 14 20

QP =

20P(Ε) = 14

| QP | ≤ length of the arc QP z − 1 ≤ 1 arg (z) |z|

Find all possible negative real values of 'a' such that

∫ (9 0

a

−2t

)

− 2.9 − t dt ≥ 0

XtraEdge for IIT-JEE

Hence proved 44

AUGUST 2011


(ii) Ist method :

MEMORABLE POINTS

L.H.S. = |z – 1| = |z – |z| + |z| – 1| By using ∆ inequality |z – 1| ≤ |z – |z|| + | |z| – 1|

• Can a current be measured by a voltameter ? ® Yes, direct current can be measured by a voltameter

z ≤ | |z| – 1| + |z| −1 |z|

≤ | |z| – 1| + |z| arg (z) [by using (i)]

• The equivalent resistance of n resistances each equal to r and connected in parallel is given by ® r/n

IInd method : Let P(z) is any point on the circle with radius |z|.

• Repeated use of which digital gate or gates can produce all the three basic gates (OR, AND and NOT) ® NAND gate and NOR gate

P(z) θ O

• What is Turnbull's blue ?

R Q (1, 0)

® Fe4[Fe(CN)6]3

• An hypothesis tested by experiments is known as ® Theory • What is magnesia alba ? ® Mg(OH)2.MgCO3.3 H2O • The humidity of air is measured by

In ∆PQR, use ∆ PQR, use ∆ inequality

• What is oleum ?

PQ ≤ PR + QR

® Hygrometer ® H2S2O7

Also known as fuming sulphuric acid

|z – 1| ≤ |z| arg (z) + ||z| – 1| Note : you can take Q point out side the circle, too.

• Vulcanised rubber was invented by ® Charies Goodyear (1839)

The co-ordination of the ion

• An amino acid which does not contain a chiral centre. ® Glycine [NH2–CH2–COOH] • Scientist who perfected the technique for converting pig iron into steel. ® Hynry Besemer (1856)

Splitting is greater if the ion is octahedral than if it is tetrahedral, and therefore the colour will change with a change of co-ordination. Unfortunately, I can't think of a single example to illustrate this with! The problem is that an ion will only change co-ordination if you change the ligand - and changing the ligand will change the colour as well. You can't isolate out the effect of the co-ordination change. For example, a commonly quoted case comes from cobalt(II) chemistry, with the ions [Co(H2O)6]2+ and [CoCl4]2-.

• When the pH of the blood is lower than the normal value, this condition is known as ® Acidosis • The electrolytic method of obtaining aluminium from bauxite was first developed by ® Charles Hall (1886)

• Which compound possesses characteristic smell like that of mustard oil ? ® Ethyl isothiocyanate [C2H5N = C = S] • First solar battery was developed in the ® Bell Telephone Laboratory (1954) • What is Wilkinson's catalyst ? ® tris (triphenylphosphine) chlororhodlum (I) • In 1836 the galvanised iron was introduced first in ® France • What is caro's acid ? ® Permonosulphuric acid [H2SO5]

The difference in the colours is going to be a combination of the effect of the change of ligand, and the change of the number of ligands. XtraEdge for IIT-JEE

45

AUGUST 2011


MATHS

VECTOR Mathematics Fundamentals Properties of vectors : (i) Addition of vectors

Representation of vectors : Geometrically a vector is represent by a line segment.

Triangle law of addition : If in a ∆ABC,

For example, a = AB . Here A is called the initial point and B, the terminal point or tip. Magnitude or modulus of a is expressed as

AB = a, BC = b and AC = c, then

AB + BC = AC i.e., a + b = c

|a| = | AB | = AB.

C B

c=a+b

a

B A a Parallelogram law of addition : If in a

A Types of Vector: Zero or null vector : A vector whose magnitude is zero is called zero or null vector and it is

parallelogram OACB, OA = a, OB = b and OC = c B

represented by O . Unit vector : A vector whose modulus is unity, is called a unit vector. The unit vector in the direction of a vector a is denoted by aˆ , read as “a cap”. Thus, | aˆ | = 1. aˆ =

C c=a+b

b

Vector a a = |a| Magnitude of a

O

a

A

Then OA + OB = OC i.e., a + b = c, where OC is a diagonal of the parallelogram OABC. Addition in component form : If the vectors are defined in terms of i, j, and k, i.e., if a = a1i + a2j + a3k and b = b1i + b2j + b3k, then their sum is defined as a + b = (a1 + b1)i + (a2 + b2)j + (a3 + b3)k. Properties of vector addition : Vector addition has the following properties. Binary operation : The sum of two vectors is always a vector. Commutativity : For any two vectors a and b, a + b = b + a. Associativity : For any three vectors a, b and c, a + (b + c) = (a + b) + c Identity : Zero vector is the identity for addition. For any vector a, 0 + a = a = a + 0 Additive inverse : For every vector a its negative vector – a exits such that a + (– a) = (– a) + a = 0 i.e., (– a) is the additive inverse of the vector a.

Like and unlike vectors : Vectors are said to be like when they have the same sense of direction and unlike when they have opposite directions. Collinear or parallel vectors : Vectors having the same or parallel supports are called collinear or parallel vectors. Co-initial vectors : Vectors having the same initial point are called co-initial vectors. Coplanar vectors : A system of vectors is said to be coplanar, if their supports are parallel to the same plane. Two vectors having the same initial point are always coplanar but such three or more vectors may or may not be coplanar. Negative of a vector : The vector which has the same magnitude as the vector a but opposite direction, is called the negative of a and is

denoted by – a. Thus, if PQ = a, then QP = – a.

XtraEdge for IIT-JEE

b

46

AUGUST 2011


Subtraction of vectors : If a and b are two vectors, then their subtraction a – b is defined as a – b = a + (– b) where – b is the negative of b having magnitude equal to that of b and direction opposite to b. If a = a1i + a2j + a3k, b = b1i + b2j + b3k Then a – b = (a1 – b1)i + (a2 – b2)j + (a3 – b3)k. B a+b O

Scalar or Dot product Scalar or Dot product of two vectors : If a and b are two non-zero vectors and θ be the angle between them, then their scalar product (or dot product) is denoted by a . b and is defined as the scalar |a| |b| cosθ, where |a| and |b| are modulii of a and b respectively and 0 ≤ θ ≤ π. Dot product of two vectors is a scalar quantity. B

b a

b

A

θ A O a Angle between two vectors : If a, b be two vectors inclined at an angle θ, then a . b = |a| |b| cos θ a.b ⇒ cos θ = | a || b |

–b

a + (–b) = a – b

B´ Properties of vector subtraction :

(i) a – b ≠ b – a (ii) (a – b) – c ≠ a – (b – c) (iii) Since any one side of a triangle is less than the sum and greater than the difference of the other two sides, so for any two vectors a and b, we have

(a) |a + b| ≤ |a| + |b|

If a = a1i + a2j + a3k and b = b1i + b2j + b3k; then

(b) |a + b| ≥ |a| – |b|

 a1b1 + a2 b2 + a3b3  θ = cos–1   a2 + a2 + a2 b2 + b2 + b2 2 3 1 2 3  1

(c) |a – b| ≤ |a| + |b| (d) |a – b| ≥ |a| – |b| Multiplication of a vector by a scalar : If a is a vector and m is a scalar (i.e., a real number) then ma is a vector whose magnitude is m times that of a and whose direction is the same as that of a, if m is positive and opposite to that of a, if m is negative. Properties of Multiplication of vector by a scalar : The following are properties of multiplication of vectors by scalars, for vector a, b and scalars m, n. (i) m(– a) = (– m)a = –(ma) (ii) (– m) (– a) = ma (iii) m(na) = (mn)a = n(ma) (iv) (m + n)a = ma + na (v) m(a + b) = ma + mb Position vector :

   

Properties of scalar product Commutativity : The scalar product of two vector is commutative i.e., a . b = b . a Distributivity of scalar product over vector addition : The scalar product of vectors is distributive over vector addition i.e., (a) a . (b + c) = a . b + a . c, (Left distributivity) (b) (b + c) . a = b . a + c . a, (Right distributivity) Let a and b be two non-zero vectors a . b = 0

⇔ a ⊥ b. As i, j, k are mutually perpendicular unit vectors along the coordinate axes, therefore, i . j = j . i = 0; j . k = k . j = 0; k . i = i . k = 0. For any vector a . a . a = |a|2. As i. j. k are unit vectors along the co-ordinate axes, therefore i . i = |i|2 = 1, j . j = |j|2 = 1 and k . k = |k|2 = 1 If m, n are scalars and a . b be two vectors, then ma . nb = mn(a . b) = (mna) . b = a .(mnb) For any vectors a and b, we have (a) a . (– b) = – (a . b) = (– a). b (b) (– a) . (– b) = a . b For any two vectors a and b, we have (a) |a + b|2 = |a|2 + |b|2 + 2a.b (b) |a – b|2 = |a|2 + |b|2 – 2a.b

AB in terms of the position vectors of points A and B : If a and b are position vectors of points A

and B respectively. Then, OA = a, OB = b

∴ AB = (Position vector of B)– (Position vector of A) = OB – OA = b – a Position vector of a dividing point : The position vectors of the points dividing the line AB in the ratio m : n internally or externally are mb + na mb − na or . m−n m+n

XtraEdge for IIT-JEE

 a.b   θ = cos–1   | a || b | 

47

AUGUST 2011


(c) (a + b) . (a – b) = |a|2 – |b|2

Right handed system of vectors : Three mutually perpendicular vectors a, b, c form a right handed system of vector iff a × b = c, b × c = a, c × a = b Left handed system of vectors : The vectors a, b, c mutually perpendicular to one another form a left handed system of vector iff c × b = a, a × c = b, b × a = c. Area of parallelogram and triangle : The area of a parallelogram with adjacent sides a and b is |a × b|. The area of a plane quadrilateral ABCD is 1 | AC × BD | , where AC and BC are its 2 diagonals.

(d) |a + b| = |a| + |b| ⇒ a | | b (e) |a + b|2 = |a|2 + |b|2 ⇒ a ⊥ b (f) |a + b| = |a – b| ⇒ a ⊥ b Vector or Cross product Vector product of two vectors : Let a, b be two non-zero, non-parallel vectors.

b θ O

a

The area of a triangle ABC is

Then a × b = |a| |b| sin θ ηˆ , and |a × b| = |a| |b| sin θ, where θ is the angle between a and b, ηˆ is a unit vector perpendicular to the plane of a and b such that a, b, ηˆ form a right-handed system.

1 1 | BC × BA | or | CB × CA | 2 2 Scalar triple product Scalar triple product of three vectors : If a, b, c are three vectors, then scalar triple product is defined as the dot product of two vectors a and b × c. It is generally denoted by a . (b × c) or [a b c]. Properties of scalar triple product : If a, b, c are cyclically permuted, the value of scalar triple product remains the same. i.e., (a × b) . c = (b × c) . a = (c × a). b or [a b c] = [b a c] = [c a b] Vector triple product Let a, b, c be any three vectors, then the vectors a × (b × c) and (a × b) × c are called vector triple product of a, b, c. Thus, a × (b × c) = (a . c) b – (a . b)c Properties of vector triple product : The vector triple product a × (b × c) is a linear combination of those two vectors which are within brackets. The vector r = a × (b × c) is perpendicular to a and lies in the plane of b and c. The formula a × (b × c) = (a . c)b – (a . b)c is true only when the vector outside the bracket is on the left most side. If it is not, we first shift on left by using the properties of cross product and then apply the same formula. Thus, (b × c) × a = – {a × (b × c)} = {(a . c)b – (a . b)c} = (a . b)c – (a . c)b Vector triple product is a vector quantity.

Properties of vector product : Vector product is not commutative i.e., if a and b are any two vectors, then a × b ≠ b × a, however, a × b = – (b × a) If a, b are two vectors and m, n are scalars, then ma × nb = mn(a × b) = m(a × nb) = n(ma × b). Distributivity of vector product over vector addition. Let a, b, c be any three vectors. Then (a) a × (b + c) = a × b + a × c (left distributivity) (b) (b + c) × a = b × a + c × a (Right disributivity) For any three vectors a, b, c we have a × (b – c) = a × b – a × c. The vector product of two non-zero vectors is zero vector iff they are parallel (Collinear) i.e., a × b = 0 ⇔ a| | b, a, b are non-zero vectors. It follows from the above property that a × a = 0 for every non-zero vector a, which in turn implies that i × i = j × j = k × k = 0. Vector product of orthonormal triad of unit vectors i, j, k using the definition of the vector product, we obtain i × j = k, j × k = i, k × i = j, j × i = – k, k × j = – i, i × k = – j. Vector product in terms of components :

If a = a1i + a2j + a3k and b = b1i + b2j + b3k. i

j

k

Then, a × b = a1 b1

a2

a3

b2

b3

Angle between two vectors :

If θ is the angle between a and b then sin θ =

XtraEdge for IIT-JEE

1 | AB × AC | or 2

a × (b × c) ≠ (a × b) × c

| a×b | | a || b |

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AUGUST 2011


MATHS

PERMUTATION & COMBINATION Mathematics Fundamentals

Permutation : Definition : The ways of arranging or selecting a smaller or an equal number of persons or objects at a time from a given group of person or objects with due regard being paid to the order of arrangement or selection are called the (different) permutations. Number of permutations without repetition : Arranging n objects, taken r at a time equivalent to filling r places from n things. r-places :

1

2

3

4

Condition permutations : Number of permutations of n dissimilar things taken r at a time when p particular things always occur = n – pCr – p r !. Number of permutations of n dissimilar things taken r at a time when p particular things never occur = n – pCr r !. The total number of permutations of n different things taken not more than r at a time, when each thing may be repeated any number of times, is n(n r − 1) . n −1 Number of permutations of n different things, taken all at a time, when m specified things always come together is m ! × (n – m + 1) !. Number of permutation of n different things, taken all at a time, when m specified things never come together is n ! – m! × (n – m + 1) !. Let there be n objects, of which m objects are alike of one kind, and the remaining (n – m) objects are alike of another kind. Then, the total number of mutually distinguishable permutations that can be formed from these objects is n! . (m!) × (n − m)!

r

Number of choice n (n–1)(n–2) (n–3)

n–(r – 1)

The number of ways of arranging = The number of ways of filling r places. = n(n – 1) (n – 2) ....... (n – r + 1) =

n! n(n − 1)(n − 2)....(n − r + 1)((n − r )!) = (n − r )! (n − r ) !

= nPr The number of arrangements of n different objects taken all at a time = nPn = n ! (i)

n

P0 =

n! = 1; nPr = n. n–1Pr – 1 n!

(ii) 0 ! = 1;

1 = 0 or (– r) ! = ∞ (r ∈ N) (– r )!

The above theorem can be extended further i.e., if there are n objects, of which p1 are alike of one kind; p2 are alike of another kind; p3 are alike of 3rd kind; .....; pr are alike of rth kind such that p1 + p2 + .... + pr = n; then the number of permutations of these n n! objects is . ( p1 !) × ( p 2 !) × .... × ( p r !)

Number of permutations with repetition : The number of permutations (arrangements) of n different objects, taken r at a time, when each object may occur once, twice, thrice, ....... upto r times in any arrangement = The number of ways of filling r places where each place can be filled by any one of n objects. r-places :

1

2

3

4

Number of choices : n (n) (n) (n)

Circular permutations : Difference between clockwise and anti-clockwise arrangement : If anti-clockwise and clockwise order of arrangement are not distinct e.g., arrangement of beads in a necklace, arrangement of flowers in garland etc. then the number of circular permutations (n − 1)! of n distinct items is . 2 Number of circular permutations of n different things, taken r at a time, when clockwise and n pr anticlockwise orders are taken as different is . r

r n

The number of permutations = The number of ways of filling r places = (n)r. The number of arrangements that can be formed using n objects out of which p are identical (and of one kind) q are identical (and of another kind), r are identical (and of another kind) and the rest n! are distinct is . p ! q !r ! XtraEdge for IIT-JEE

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Number or circular permutations of n different things, taken r at a time, when clockwise and n Pr anticlockwise orders are not different is . 2r Theorems on circular permutations : Theorem (i) : The number of circular permutations on n different objects is (n – 1) !. Theorem (ii) : Then number of ways in which n persons can be seated round a table is (n – 1) !. Theorem (iii) : The number of ways in which n different beads can be arranged to form a 1 necklace, is (n − 1) ! . 2 Combinations : Definition : Each of the different groups or selection which can be formed by taking some or all of a number of objects, irrespective of their arrangements, called a combination. Notation : The number of all combinations of n things, taken r at a time is denoted by C(n, r) or nCr or  n  . nCr is always a natural number. r  Difference between a permutation and combination : In a combination only selection is made whereas in a permutation not only a selection is made but also an arrangement in a definite order is considered. Each combination corresponds to many permutations. For example, the six permutations ABC, ACB, BCA, CBA and CAB correspond to the same combination ABC. Number of combinations without repetition The number of combinations (selections or groups) that can be formed from n different objects taken n! r (0 ≤ r ≤ n) at a time is nCr = . Also r !(n − r )! n Cr = nCn – r. Let the total number of selections (or groups) = x. Each group contains r objects, which can be arranged in r ! ways. Hence the number of arrangements of r objects = x × (r!). But the number of arrangements = npr. n! ⇒ x(r !) = nPr ⇒ x = = nCr r !(n − r )!

The total number of ways in which it is possible to make groups by taking some or all out of n = (n1 + n2 + .....) things, when n1 are alike of one kind, n2 are alike of second kind, and so on is {(n1 + 1) (n2 + 1) .....} – 1. The number of selections taking at least one out of a1 + a2 + a3 + .... + an + k objects, where a1 are alike (of one kind), a2 are alike (of second kind) and so on ............... an are alike (of nth kind) and k are distinct = [(a1 + 1) (a2 + 1) (a3 + 1) ......... (an + 1)]2k – 1 Conditional combinations : The number of ways in which r objects can be selected from n different objects if k particular objects are (i) Always included = n – kCr–k (ii) Never included = n – kCr The number of combinations of n objects, of which p are identical, taken r at a time is

Number of combinations with repetition and all possible selections : The number of combinations of n distinct objects taken r at a time when any object may be repeated any number of times. = Coefficient of xr in (1 + x + x2 + ...... + xr)n = Coefficient of xr in (1 – x)– n = n + r – 1Cr The total number of ways in which it is possible to form groups by taking some or all of n things at a time is nC1 + nC2 + .... + nCn = 2n – 1.

Derangement : Any change in the given order of the things is called a derangement. If n things form an arrangement in a row, the number of ways in which they can be deranged so that no one of them occupies its original place is

XtraEdge for IIT-JEE

n–p

Cr + n – pCr – 1 +...........+ n – pC0, if r ≤ p and n–p Cr + n – pCr – 1 + ........... + n – pCr – p, if r > p. Division into groups The number of ways in which n different things can be arranged into r different groups is n + r – 1Pn or n ! n –1Cr – 1 according as blank group are or are not admissible. Number of ways in which m × n different objects can be distributed equally among n persons (or numbered groups) = (number of ways of dividing into groups) × (number of groups)! =

(mn)!n! (mn)! = (m!) n n! (m!) n

If order of group is not important: The number of ways in which mn different things can be divided (mn)! equally into m groups is . (m!) m m! If order of groups is important : The number of ways in which mn different things can be divided equally into m distinct groups is (mn)! (mn)! ×m!= m (n!) m! (n!) m

 1 1 1 1 n ! 1 − + − + ... + (−1) n .  n!  1! 2 ! 3!

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AUGUST 2011


Based on New Pattern

IIT-JEE 2012 XtraEdge Test Series # 3

Time : 3 Hours Syllabus : Physics: Essential Mathematics, Vector, Units & Dimension, Motion in One dimension, Projectile motion, Circular motion, Electrostatics & Gauss's Law, Capacitance, Current electricity, Alternating Current, Magnetic Field, E.M.I. Chemistry : Mole Concept, Chemical Bonding, Atomic Structure, Periodic Table, Chemical Kinetics, Electro Chemistry, Solid state, Solutions, Surface Chemistry, Nuclear Chemistry. Mathematics: Trigonometric Ratios, Trigonmetrical Equation, Inverse Trigonmetrical Functions, Properties of Triangle, Radii of Circle, Function, Limit, Continiuty, Differentiation, Application of Differentiation (Tangent & Normal, Monotonicity, Maxima & Minima)

Instructions : [Each subject contain] Section – I :

Question 1 to 6 are multiple choice questions with only one correct answer. +5 marks will be awarded for correct answer and -2 mark for wrong answer.

Section – II :

Question 7 to 12 are passage based single correct type questions. +3 marks will be awarded for correct answer and -1 mark for wrong answer

Section – III : Question 13 to 14 are Column Matching type questions. +8 marks will be awarded for the complete correctly matched answer (i.e. +2 marks for each correct row) and –2 mark for wrong answer.. Section – IV : Question 16 to 19 are Numerical Response Question (single digit Ans. type) +3 marks will be awarded for correct answer and No Negative marks for wrong answer. z × × × × × ×× × × ×× × × × × ×× × × × × R × × × × × × × ×

PHYSICS Questions 1 to 6 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 5 marks will be given for each correct answer and – 2 mark for each wrong answer. 1. Find equivalent resistance between A and B. A

4Ω

4Ω 20Ω

4Ω

4Ω

4Ω 2Ω

B

(A) 6Ω 2.

(B) 8Ω

3.

8Ω 8Ω 3Ω

(C) 12Ω

6Ω

− 2Bπa 2 r ˆ k mR

(B)

− Bπa 2 r ˆ k 3mR

(C)

− Bπa 2 λ ˆ k 2mR

(D)

− Bπa 2 λ ˆ k mR

A disc, having plane parallel to the horizontal is moving such that velocity of point P with respect to ground on its periphery is 2m/s ˆj as shown in the figure. If radius of disc is R = 1m and angular speed of disc about vertical axis passing through disc is ω = 2 rad/s, the velocity of centre of disc in m/s is -

(D) None

2ˆj

A line charge λ per unit length is pasted uniformly on to the rim of a wheel of mass m and radius R. The wheel has light non-conducting spokes and is free to rotate about a vertical axis as shown in figure. A uniform magnetic field extends over a radial region given by r B = – B 0 k (r ≤ a) and B = 0 for r > a What is the angular velocity of the wheel when this field is suddenly switched off ?

XtraEdge for IIT-JEE

(A)

P C

51

ˆj ˆi

ω

(A) 2 ˆj

(B) 2 ˆi + 2 ˆj

(C) –2 ˆi + 2 ˆj

(D) none of these AUGUST 2011


4.

A large box is moving on a horizontal floor with constant acceleration a = g. A particle is projected inside box with velocity u and angle θ with horizontal from box frame. For the given u, the value of θ for which horizontal range inside the box will be maximum if -

Passage # 1 (Ques. 7 to 9) A particle is always moving along the sides of a regular hexagon starting from A with rest. It moves such that its speed increase at constant rate of 3m/s2. Assume there is no change of speed at corners. Given that each side of hexagon is 4m. Answer the following question. D E

g u

F

C

θ

(A) θ = π/4 (C) θ = 3π/8 5.

6.

(B) θ = π/8 (D) θ = π/3

7.

Which of the following statements is incorrect? (A) A capacitor acts as an infinite resistance for AC as well as DC current (B) An electric transformer can step up or down the AC as well as DC voltage. (C) A hot wire ammeter and voltmeter can measure current and voltage respectively for an AC and DC voltage both. (D) We use a choke coil in series with a tube light to reduce voltage across tube light, without losing electrical energy in the form of heat as an ideal inductor does not consume power in a circuit

8.

9.

O'

(A) 12 sec

(B) 2 6 sec

(C) 2 3 sec

(D) None of these

The magnitude and direction of change in velocity at B (A) 0 m/s

Passage # 2 (Ques. 10 to 12)

In the circuit shown switch S1 is initially closed and S2 is open.

Q

f

1Ω

b

P

5Ω

d

10µC

O

e

(A) If all slopes are frictionless then bead released from R will reach O first (B) If the slopes are frictionless then all beads will take same time to reach O (C) If there is friction, bead released from Q will be the first to reach O (D) If there is friction, then bead released from P will be the first to reach O

S2 3Ω

3Ω a

c

S1 24 V

10. What is the value of Va – Vb = (A) 10 V (B) 12 V (C) 6 V

(D) 8 V

11. After S2 is closed what is Va – Vb just after closing of switch S2 (A) 10 V (B) 8 V (C) 6 V (D) None

This section contains 2 paragraphs, each has 3 multiple choice questions. (Questions 7 to 12) 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. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer. XtraEdge for IIT-JEE

Time taken to complete one round trip -

(B) 4 2 m/s towards centre (C) 4 m/s towards centre (D) None of these

As shown in the figure, a vertical fixed ring has 3 spokes OP, OQ and OR. 3 beads each of mass m are released from point P, Q and R simultaneously. R

A B The time particle takes from B to C (A) 2 sec (B) 0.8284 sec (C) 1.464 sec (D) None of these

12. S1 is opened and S2 is closed. what is the time constant for the capacitor discharging (A) 368 µs (B) 26.7 µs (C) 16.7 µs (D) None of these

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AUGUST 2011


(C) Maximum current (R) R, C in the circuit depends on (D) Time to achieve (S) R 50% charging of the capacitor depends on (T) None

This section contains 2 questions (Questions 13, 14). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows : P Q R S T A B C D

P P P P

Q Q Q Q

S S S S

R R R R

This section contains 5 questions (Q.15 to 19). +3 marks will be given for each correct answer and no negative marking. 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 has 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

T T T T

Mark your response in OMR sheet against the question number of that question in section-II. + 8 marks will be given for complete correct answer (i.e. +2 marks for each correct row) and NO NEGATIVE MARKING for wrong answer. 13. Match the following : Column-I (A) Constant positive acceleration (B) Constant negative acceleration (C) Angle between acceleration and velocity is 90º (D) Angle between acceleration and velocity is less than 90º

Column-II (P) Speed may decrease

(Q) Speed must increase (R) Speed is constant

(S) None

15. A parallel plate capacitor is maintained at a certain potential difference. When a 3mm thick slab is introduced between the plates, in order to maintain the same potential difference the distance between the plates is increased by 2.4 mm. Find the dielectric constant of slab.

(T) None 14. In the circuit show in figure the capacitor is initially uncharged. At t = 0, the switch is closed at position (1) and remain closed fro long time. Then at t = t', switch is shifted to position (2).

16. A charged particle is accelerated through a potential difference of 12 kV and acquires a speed of 106 ms–1 . It is projected perpendicularly into the magnetic field of strength 0.2 T. The radius of circle described is ....................× 10 cm.

C

L

2 S R

Column-I (A) As the capacitor charges from t = 0 to t = t' (B) As the capacitor discharge i.e. for t > t' XtraEdge for IIT-JEE

1

17.

Column-II (P) Current in the circuit falls exponentially (Q) Current in the circuit grows exponentially

C V ~ AC source Current in the inductance is 0.8 A while in the capacitance is 0.6 A. The current drawn from the sources ........×10–1 Amp.

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AUGUST 2011


18.

19.

Figure shows a square loop. 20 cm on each side in the x-y plane with its centre at the origin. The loop carries a current of z 7A. Above it at y = 0, z = 12 cm is an D d y infinitely long wire parallel to the x axis A C carrying a current of 10 A. The net force x on the loop is ..... B ......×10–4N.

3.

1Ω

Cl Cl

A

B

1Ω

C

Suppose that a hypothetical H-like atom gives a red Green, blue & violet line spectrum. Which jump according to figure would give off the red spectral line. n=4 n=3 n=2 n=1

2.

(B) 2 → 1

(C) 4 – 1

Which one of the following sets of ions represents the collection of isoelectronic species ? (A) K+, Cl–, Mg2+, Sc3+ (B) Na+, Mg+2, Al3+, Cl (C) K+, Ca2+, Sc3+, Cl– (D) Na+, Mg2+, Cl–, Al3+

5.

A conductance cell was filled with a 0.02 M KCl solution which has a specific conductance of 2.768 × 10–3 ohm–1 cm–1. If its resistance is 82.4 ohm at 25ºC, the cell constant is(A) 0.2182 cm–1 (B) 0.2281 cm–1 –1 (C) 0.2821 cm (D) 0.2381 cm–1

6.

Choose the incorrect option(A) Among α, β and γ rays, γ ray has highest penetration power (B) γ ray has highest velocity (C) γ ray has highest ionization power (D) α particle is also called helium nucleus

This section contains 2 paragraphs, each has 3 multiple choice questions. (Questions 7 to 12) 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. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer. Passage # 1 (Ques. 7 to 9) Vapour pressure of a solvent is the pressure exerted by vapours when they are in equilibrium with its solvent at that temperature. The vapour pressure of solvent is dependent on nature of solvent, temperature, addition of non-volatile solute as well as nature of solute to dissociate or associate. The vapour pressure of a mixture obtained by mixing two volatile liquids is given by PM = PºA . XA + PºB . XB where PºA and PºB are vapour pressures of pure components A and B and XA, XB are their mole fraction in mixture. For solute-solvent system, the relation becomes PM = PºA . XA where B is non-volatile solute.

(D) 3 → 2)

10 ml of a solution containing Na2CO3 and NaHCO3 is titrated by HCl using phenolphthalein and then methyl orange (added after first end point). The first and second end points were found after N adding 10 ml and 15 ml of   HCl respectively.  10  The ratio of milliequivalents of Na2CO3 and NaHCO3 in the solution is 1 2 1 5 (B) (C) (D) ) (A) 5 1 2 1

XtraEdge for IIT-JEE

Cl

4.

E

Questions 1 to 6 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 5 marks will be given for each correct answer and – 2 mark for each wrong answer.

(A) 3 – 1

(D)

Cl

CHEMISTRY

1.

Cl

(C)

1Ω

D

Cl Cl

Cl

14 A

1Ω

(B)

(A)

A milliammeter of range 1 A resistance 14 Ω is joined in a circuit as shown in figure. The meter gives full scale deflection when current in the main circuit is 'I' and 'A' and 'D' are used as terminals. Find the value of 'I' in Ampere. 1A

Which of the following species have maximum dipole moment ? Cl Cl Cl

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AUGUST 2011


7.

The vapour pressure of benzene and its solution with a non-electrolyte are 640 and 600 mm respectively. The molality of the solution is (A) 0.80 (B) 0.86 (C) 0.90 (D) 0.95

8.

A mixture of two volatile liquids A and B for 1 and 3 moles respectivley has a V.P. of 300 mm at 27ºC. If one mole of A is further added to this solution, the vapour pressure becomes 290 mm at 27ºC. The vapour pressure of pure A is (A) 250 mm (B) 316 mm (C) 220 mm (D) 270 mm

9.

The amount of solute (mol. wt. 60) required to dissolve in 180 g of water to reduce the vapour pressure to 4/5 of the pure water (A) 120 g (B) 150 g (C) 200 g (D) 60 g

This section contains 2 questions (Questions 13, 14). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows : P Q R S T A B C D

Passage # 2 (Ques. 10 to 12) For the following reaction at 1000K the standard equilibrium constant is represented by K°p. Given at 1000K, lnK°p = 1 , ln2 = 0.693 1 A(g) + B2(g) AB(g) 2 –1 Given : R = 8.314 J K mol–1 10. What is the standard free energy change of the 2AB(g) at 1000K? reaction 2A(g) + B2(g) (A) –8.314 KJ (B) –16.628 KJ (C) +8.314 KJ (D) +16.628 KJ 11. At any instant if the partial pressures of A(g), B2(g) and AB(g) are also equal to 2 bar at 1000K, then for the reaction 1 A(g) + B2(g) AB(g) 2 (A) The equilibrium condition is reached. (B) The reaction become non spontaneous (C) At the standard condition the above reaction was non spontaneous but at the given condition the reaction become spontaneous (D) At the standard condition the above reaction was spontaneous and at the given condition, the given reaction also remain spontaneous 12. Egg albumin is protein whose structure is maintained by an immense number of hydrogen bonds. Therefore (A) The process of boiling of egg albumin is endothermic and non spontaneous (B) The process of boiling of egg albumin is non spontaneous because order of protein molecule is lost (C) The process of boiling of egg albumin is spontaneous because order of protein molecule is lost (D) The process of boiling of egg albumin is spontaneous because collapse of the structure of protein molecule at higher temperature, increases T∆S more than enough to counteract the large endothemicity required for breaking hydrogen bonds XtraEdge for IIT-JEE

P P P P

Q Q Q Q

R R R R

S S S S

T T T T

Mark your response in OMR sheet against the question number of that question in section-II. + 8 marks will be given for complete correct answer (i.e. +2 marks for each correct row) and NO NEGATIVE MARKING for wrong answer. 13. Column-I (A) Orbitals having equal energy (B) Orbitals having zero orbital angular momentum (C) Orbitals with only one spherical node (D) directional character 14. Column-I

(A) Ecell when salt bridge removed

Column-II (P) 3p, 3d

(Q) 2s and 3s

(R) Degenerate orbitals (S) 2s and 3p (T) 3d and 4s

Column-II –

(P) Reversible ion –Cl

+

(B) Junction potential (Q) Reversible ion – H in a cell with no salt bridge (C) Calomel electrode (R) Zero volt (D) Quinhydrone (S) Non-zero, opposing electrode

Ecell (T) 1 volt

This section contains 5 questions (Q.15 to 19). +3 marks will be given for each correct answer and no negative marking. 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 has 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 :

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AUGUST 2011


X 0 1 2 3 4 5 6 7 8 9

Y 0 1 2 3 4 5 6 7 8 9

Z 0 1 2 3 4 5 6 7 8 9

W 0 1 2 3 4 5 6 7 8 9

15. In an ionic solid r(+) = 1.6Å and r(-) = 1.864 Å. Use the radius ratio rule to determine the edge length of the cubic unit cell in Å. 16. A porous catalyst for chemical reactions has an internal surface area of 800 m2 per cubic centimeter of bulk material. Ten percent of the bulk volume consists of the pores (holes), while the other 90% of the volume is made up of the solid substance. Assume that the pores are all cyllindrical tubules of uniform diameter and length l, and that the measured internal surface area is the total area of the curved surfaces of the tubules. What is the diameter (in Å) of each pore ? 17. The number of moles present in 2 litre of 0.5 M NaOH is-

2.

Let the sequence < bn> of real numbers satisfy the recurrence relation : 1 125  bn +1 =  2bn + 2  , bn ≠ 0, then lim bn is equal  n→∞ 3 bn  to 2 (C) 5 (D) (A) 0 (B) ∞ 3

3.

Let f : R → R be a one one onto differentiable function, such that f (2) = 1 and f ' (2) = 3, then the d  value of  ( f −1 ( x))  is  dx  x =1 1 1 1 (A) (B) (C) (D) 6 3 6 2

4.

O is the circumcentre of the triangle ABC and R1, R2, R3 are the radii of the circumcircles of the triangle OBC, OCA and OAB respectively. Then a b c + + is equal to R1 R2 R3

19.

NiO adopts a rock-salt structure. coordination number of the Ni2+ ion is -

The

MATHEMATICS

(B)

(C)

a+b+c R

(D)

If A > 0, c, d, u, v are non-zero constants, and the graphs of f (x) = |Ax + c| + d and g(x) = – |Ax + u | + v intersect exactly at 2 points u+c equals to (1, 4) and (3, 1) then the value of A (A) 4 (B) –4 (C) 2 (D) –2

R3 a2 + b2 + c2 R2

The minimum value of the expression sin α + sin β + sin γ, where α, β, γ are real numbers satisfying α + β + γ = π, is (A) Positive (B) Zero (C) Negative (D) None of these

6.

Which of the following numbers is the maximum value of the function 5 sin 3 x cos x f (x) = , ∀ x∈R? tan 2 x + 1 (A) 5/8 (B) 3/4 (C) 1 (D) 5/ 2

and g(x) = XtraEdge for IIT-JEE

abc

This section contains 2 paragraphs, each has 3 multiple choice questions. (Questions 7 to 12) 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. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer. Passage # 1 (Ques. 7 to 9) Consider the following functions π  −1  sin x + 2 ; − 1 ≤ x ≤ 0  f(x) = ln x + 1 − π ; 0 < x < π, x ≠ π − 1  sin x ; π ≤ x ≤ 4π  

Questions 1 to 6 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 5 marks will be given for each correct answer and – 2 mark for each wrong answer. 1.

abc R

5.

18. How many of the following are paramagnetic –

C2, B2, O 2–2 , BN, Cl +2 & NO+.

(A)

56

8 f (| x |) AUGUST 2011


7.

8.

Number of points where f (x) is non differentiable is (A) 4 (B) 5 (C) 3 (D) 6

13. Match the column : Column-I

Domain of g (x) is (A) [–2π, 1 – π] ∪ [π – 1, π] (B) [π – 1, 2π] (C) [– 4π, – 2π] ∪ [– π,1–π] ∪ [π – 1, π] ∪ [2π, 4π]

π , then k is 4 (B) Number of integral values of 'a' for (Q) 3 which the function f (x) = x3 + (a + 2)x2 + 3ax + 5 is monotonic in R is (C) The value of 3 + (cot 14°– 1) (cot31º–1) is (R) 4 (D) If M is the maximum value of the function (S) 5

is equal to k.

(D) [– 4π, – π] ∪ [2 – π, π – 2] ∪ [π, 4π] 9.

Number of solutions of the equation g2(x) = 1 will be (A) 10 (B) 14 (C) 16 (D) None of these

1 f (x) = x2 e–2x, x > 0, then   2

14. Match the column: Column-I (A) Period of cos 2πx + [2x] + [–2x] is 2 4 18  tan  cos –1 + tan −1  is (B) 5 3 17  equal to

Consider the system of equation x cos3 y + 3x cos y sin2 y = 14 x sin3 y + 3x cos2y sin y = 13. The values of x are (A) ± 5 5 (C) ±

(B) ± 5

1

10

(C) 4 + ∑ [sin–1sin r] is equal to

(D) none

5

r =1

11.

The number of values of y ∈ [0, 6π] are (A) 5 (B) 3 (C) 4 (D) 16

12.

The value of sin2y + 2 cos2y is 4 9 (A) (B) (C) 2 5 5

3  12  (D) If tan–1  x  = 2 sin–1 then 5 7  x is (where [⋅] denotes greatest integer function)

P P P P

Q Q Q Q

R R R R

S S S S

Column-II (P) 4

(Q) 3

(R) 2 (S) 1

(D) none This section contains 5 questions (Q.15 to 19). +3 marks will be given for each correct answer and no negative marking. 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 has 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

T T T T

Mark your response in OMR sheet against the question number of that question in section-II. + 8 marks will be given for complete correct answer (i.e. +2 marks for each correct row) and NO NEGATIVE MARKING for wrong answer. XtraEdge for IIT-JEE

is

(T) 0

This section contains 2 questions (Questions 13, 14). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows : P Q R S T A B C D

lnM

(T) 1

Passage # 2 (Ques. 10 to 12)

10.

Column-II

 1  (P) 2 1 1 (A) If lim  + + + ..... n →∞  n   (2n – 1) 2 ( 4 n – 2) 2  

57

AUGUST 2011


15.

If p1, p2 and p3 are the altitudes of a triangle from vertices A, B and C respectively, and ∆ is the area of the triangle. λ ab 1 1 1 C If + – = cos2 . find λ p1 p 2 p3 (a + b + c)∆ 2

Why can’t the Sun melt Snow? There are some things in nature that have a great capacity to toss back or reflect a great deal of the sun’s light that falls on them. One of them is snow. Newly formed snow reflects about 90 per cent of the sunlight that falls upon it. This means that the sun is powerless to melt clean snow. And when snow does melt, it is not because of the sunlight. Snow does not melt on a spring day because of the sun’s heat. It melts because of the warm air from the sea.

16. If y = (sinx + cosecx)2 + (cosx + secx)2 then minimum value of y ∀ x ∈ R is 17.

From a point on the curve y = sin–1 x + cos–1 − x a π tangent is drawn to the curve g(x) = x + cosx + – 1. 2 Then sum of its intercepts on coordinate axes is

18.

If f (x) = x – 2sinx, 0 ≤ x ≤ 2π is increasing in the interval [aπ, bπ], then a + b is equal to

19.

After snow becomes ice, a different problem arises. Clean ice absorbs about two-thirds of the sunlight that hits it - but ice is transparent enough for the light to penetrate quite a long way (10 metres or more) before the absorption takes place.

 2[ x] − {x} , If x < 0 Let f (x) =  , where {x} and [x] + 3{x} , If x ≥ 0 [x] are the fractional part and greatest integer of x respectively. The number of solutions of the equation f (x) – x = {x} ∀ x ∈ [–8, 8] are.

It is remarkable what profound results follow from this simple property of transparency to sunlight. If, instead of penetrating deeply, the light were absorbed in a shallow surface layer of ice, the summer sun would quickly raise the temperature of the thin surface layer to melting point. And almost immediately, the water would run off. But when the sunlight penetrates a thick layer of ice before it can be absorbed, it cannot raise the temperature of the ice to melting point quickly enough. When the ice is very cold, the whole summer passes before any melting occurs at all. This is what happens today in the Antarctic, just as it must have happened in northern Europe during an Ice Age.

MEMORABLE POINTS MECHANICS 1.

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

2.

Mass and inertia are the same thing.

3.

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

4.

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

5.

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

6.

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

7.

The slope of the velocity-time graph is acceleration.

8.

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

9.

Centripetal force and centripetal acceleration vectors are toward the center of the circlewhile the velocity vector is tangent to the circle.

Just imagine, if by magic, ice were suddenly made opaque to light, the glaciers that exist today would melt away in a few years, raising the sea level by 60 metres or more. It would flood at least half the world’s population. Simply amazing how so much depends on so simple a physical property! Clouds toss back about 50 per cent of the light that hits them. Ice and deserts reflect 35 per cent. Land areas are generally a good deal lower in reflectivity - usually 10 to 20 per cent, depending on the nature of vegetation. Oceans, which cover 71 per cent of the Earth’s surface, are least reflective of all - only about three per cent. That is why oceans appear dark in pictures of the Earth taken from the Moon or from artificial satellites. When all the sources of reflection are added together, our planet is found to turn back into space some 36 per cent of the solar radiation falling upon it.

10. An unbalanced force (object not in equilibrium) must produce acceleration.

XtraEdge for IIT-JEE

58

AUGUST 2011


Based on New Pattern

IIT-JEE 2013 XtraEdge Test Series # 4

Time : 3 Hours Syllabus : Physics : Essential Mathematics, Vector, Units & Dimension, Motion in One dimension, Projectile motion, Circular motion. Chemistry : Mole Concept, Chemical Bonding, Atomic Structure, Periodic Table. Mathematics: Trigonometric Ratios, Trigonmetrical Equation, Inverse Trigonmetrical Functions, Properties of Triangle, Radii of Circle.

Instructions : [Each subject contain] Section – I :

Question 1 to 6 are multiple choice questions with only one correct answer. +5 marks will be awarded for correct answer and -2 mark for wrong answer.

Section – II :

Question 7 to 12 are passage based single correct type questions. +3 marks will be awarded for correct answer and -1 mark for wrong answer

Section – III : Question 13 to 14 are Column Matching type questions. +8 marks will be awarded for the complete correctly matched answer (i.e. +2 marks for each correct row) and –2 mark for wrong answer.. Section – IV : Question 16 to 19 are Numerical Response Question (single digit Ans. type) +3 marks will be awarded for correct answer and No Negative marks for wrong answer.

3.

PHYSICS Questions 1 to 6 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 5 marks will be given for each correct answer and – 2 mark for each wrong answer. r r 1. Two vectors A and B have magnitudes 2 and r r r r 2 2 respectively. It is found that A.B = | A × B | , r r | A + B| then the value of r r will be | A −B|

(A) 5 (C)

2.

(B) 2 +1 2 −1

(D)

A

(A) 4.

5

v0

θ2 kv0

B

cos θ 2 cos θ1

Point of collision

θ1 d

(B)

tan θ 2 cot θ 2 sin θ 2 (C) (D) cos θ1 tan θ1 cot θ1

Position vector of a particle moving in x-y plane at →

time t is r = a (1 – cos ωt) ˆi + a sin ωt ˆj . The path of the particle is : (A) a circle of radius a and centre at (a, 0) (B) a circle of radius a and centre at (0, 0) (C) an ellipse (D) neither a circle nor an ellipse

2 −1 2 +1

Two bodies P and Q have to move equal distances starting form rest. P is accelerated with acceleration of 2a for first half distance then its acceleration becomes a for last half distance whereas Q has acceleration a for first half distance and acceleration 2a for last half distance then – (A) both take same time. (B) P will take less time (C) Q will take less time (D) None of these

XtraEdge for IIT-JEE

At the same instant two boys throw balls A and B from the positions shown with a speed v0 and kv0 respectively, where k is a constant. For what value of k, balls will collide ? Relevant data are available in figure –

5.

Figure shows the variation of y vs x, the rate dy/dx when x increases y

x

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AUGUST 2011


(A) increases (B) decreases (C) first increases then decreases (D) first decreases then increases 6.

Passage # 2 (Ques. 10 to 12)

A particle is moving along x-axis and its initial velocity is 27 m/s. Acceleration of particle is a = (– 6t) m/s2, where t is in seconds. At t = 0 particle is at x = 0. 10. The velocity of particle when it travels 26 m (A) 21 m/s (B) 15 m/s (C) 24 m/s (D) 18 m/s

[M–1L3T–2] is the dimensional formula of (A) Gravitational constant (B) Planck's constant (C) Surface tension (D) Modulus of rigidity

11. Maximum value of velocity along positive x-direction is (A) 35 m/s (B) 33 m/s (C) 27 m/s (D) 30 m/s 12. Maximum value of displacement along positive x-direction is (A) 54 m (B) 27 m (C) 120 m (D) None

This section contains 2 paragraphs, each has 3 multiple choice questions. (Questions 7 to 12) 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. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer. Passage # 1 (Ques. 7 to 9) An inclined plane makes an angle θ = 45º with horizontal. A stone is projected normally from the inclined plane, with speed 4 m/sec at t = 0. x and y-axis are drawn from point of projection along and normal to inclined plane as shown. The length of inclined plane is sufficient for stone to land on it and neglect air friction. y

This section contains 2 questions (Questions 13, 14). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows : P Q R S T A B C D

u

x

8.

2 2u g

(B)

2u g

(C)

2u g

(D)

u 2g

The instant of time at which velocity of stone makes an angle θ = 45º with positive x-axis : (A)

9.

13.

The time instant at which velocity of stone is parallel to x-axis (A)

2 2u g

(B)

2u g

(C)

2u g

(D)

u 2g

The time instant till which (starting t = 0) component of displacement along x-axis is half the range on inclined plane is 2 2u (A) g

2u (B) g

XtraEdge for IIT-JEE

(C)

Q Q Q Q

R R R R

S S S S

T T T T

Mark your response in OMR sheet against the question number of that question in section-II. + 8 marks will be given for complete correct answer (i.e. +2 marks for each correct row) and NO NEGATIVE MARKING for wrong answer.

θ = 45º

7.

P P P P

2u g

(D)

In column-I condition on velocity, force and acceleration of a particle is given. Resultant motion r is described in column-II. u = instantaneous velocity Column-I Column-II r r (P) path will be (A) u × F = 0 and r F = constant circular path r r (B) u . F = 0 and (Q) speed will increase r F = constant rr (R) path will be (C) v.F = 0 all the time r straight line and | F | = constant and the particle always remains in one plane r (D) u = 2 ˆi – 3 ˆj and (S) path will be acceleration at all r time a = 6ˆi − 9ˆj

u 2g

parabolic (T) None

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14.

Two bodies are projected from ground with same speed at angles 30º and 60º. If R1 is range of first and R2 is range of second similarly H1 and H2 are their maximum heights and T1 and T2 are time of flights. Column-I Column-II R1 1 (A) (P) 3 R2 (B)

H1 H2

T (C) 2 T1

(D)

T1H1R 1 T2 H 2 R 2

18.

v = 10 2 m/s as shown. After elastic collision with the inclined plane the particle rebounds normally with the plane and retraces its path to come back at its point of projection. Then find the time in seconds in which particle returns to the point of projection. (g = 10 m/s2)

(Q) 1 v

(R) (S)

 1   α = sin –1   3

3 1

19.

3 3 (T) None

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 -

This section contains 5 questions (Q.15 to 19). +3 marks will be given for each correct answer and no negative marking. 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 has 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

O l v = kgl

CHEMISTRY Questions 1 to 6 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 5 marks will be given for each correct answer and – 2 mark for each wrong answer. 1. If Auf-Bau principle is violated then K(Z = 19) is found in which block is modern periodic table ? (A) s-block (B) p-block (C) d-block (D) f block

15. The length of a cylinder is measured with a metre rod having least count 0.1 cm. Its diameter is measured with vernier calipers having least count 0.01 cm. Given that length is 5.0 cm and radius is 2.0 cm. The percentage error in the calculated value of the volume will be -

2.

16. Two motor vehicles run at constant speeds 5 m/s each along highways intersecting at an angle 60º. In what time after they meet at the intersection

H, D and He+ are all one electron species. The wavelength of radiations emitted for their downward transitions from Ist excited state to ground state are λ1, λ2 and λ3 respectively. Then approximately (A) 4λ1 = 2λ2 = 2λ3 (C) λ1 = λ2 = 2λ3

will the distance between the vehicles be 10 3 m. 3.

17. A police jeep is chasing a culprit going on a motor bike. The motor bike crosses a turning at a speed of 72 km/h. The jeep follows it a speed of 90 km/h crossing the turning ten seconds later than the bike. Assuming that they travel at constant speeds, how far from the turning will the jeep catch up with the bike ? (in km) XtraEdge for IIT-JEE

A particle is projected with initial velocity

61

(B) λ1 = 2λ2 = 2 2λ 3 (D) λ1 = λ2 = 4λ3

Arrange the following gases in the increasing order of their dipole moment (CO2, H2O, H2,HF) (A) H2 = CO2 < H2O < HF (B) H2O < CO2 < H2 < HF (C) CO2 < H2O < H2 < HF (D) H2O < H2 < CO2< HF AUGUST 2011


4.

Which of the following are isoelectronic and isostructural ?

Passage # 2 (Ques. 10 to 12)

Nature of bond can be predicted on the basis of electronegativity of bonded atoms, greater difference in electronegativity (X), more will be the polarity of bond, and polar bond are easily broken in polar solvent like water. For hydroxyl acids XO – XA difference predict the nature of oxide formed by the element A. |XO – XA| > |XO – XH| then A – O – H show basic nature (NaOH) |XO – XA| < |XO – XH| then A – O – H show acidic nature (H–O–Cl) With the help of EN values [ENA = 1.8, ENB = 2.6, ENC = 1.6, END = 2.8] answer the following questions for the compounds HAO, HBO, HCO, HDO. 10. Compounds whose aqueous solution is acidic and order of their acidic strength (A) AOH, COH ; AOH < COH (B) HDO, HBO ; HDO > HBO (C) AOH, COH ; AOH > COH (D) HDO, HBO ; HDO < HBO

NO 3− , CO 32− , ClO 3− , SO3

5.

(A) NO 3− , CO 32−

(B) SO3, NO 3−

(C) ClO 3− , CO 32−

(D) CO 32− , ClO 3−

The number of moles of KMnO4 that will be required to react with one mole SO32– ion in acidic medium is (A)

6.

2 5

(B)

3 5

(C)

4 5

(D) 1

The orbital diagram in which both Pauli's exclusion principle and Hund's rule are violated is 2s

2p

(A) ↑↓

↑↑ ↑

(B) ↑↓

↑↓ ↑↓

(C) ↑↓

↓ ↓ ↓

(D) ↑↓

↑↓ ↑↓ ↑

11. Compounds whose aqueous solution is basic and order of their basic strength (A) AOH, COH ; AOH < COH (B) HDO, HBO ; HDO > HBO (C) AOH, COH ; AOH > COH (D) HDO, HBO ; HDO < HBO

This section contains 2 paragraphs, each has 3 multiple choice questions. (Questions 7 to 12) 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. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer. Passage # 1 (Ques. 7 to 9) Hydrogen can be generated by the reaction of calcium hydride with water and this method is used as a source of hydrogen in portable hydrogen generators. Reaction of CaH2 (calcium hydride) with water is given as : CaH2 + 2H2O → Ca(OH)2 + 2H2 7.

12. Percentage ionic character of compound AB is (A) 42.42% (B) 24.24% (C) 15.04% (D) None of these This section contains 2 questions (Questions 13, 14). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

Weight of calcium hydride required to produce one kg of hydrogen is (atomic wt. of Ca = 40, O = 16, H = 1) (A) 5.25 kg (B) 10.5 kg (C) 15.75 kg (D) 21 kg

8.

The amount of calcium hydroxide formed from 2.1 kg of calcium hydride is (A) 8.4 kg (B) 7.4 kg (C) 4.2 kg (D) 3.7 kg

9.

If the amount of Ca(OH)2 obtained in question 18 is neutralized by HCl solution of 0.5 M concentration, then volume of HCl consumed is (A) 10 Litre (B) 100 Litre (C) 10 mL (D) 100 mL

XtraEdge for IIT-JEE

P Q R S T A B C D

P P P P

Q Q Q Q

R R R R

S S S S

T T T T

Mark your response in OMR sheet against the question number of that question in section-II. + 8 marks will be given for complete correct answer (i.e. +2 marks for each correct row) and NO NEGATIVE MARKING for wrong answer.

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AUGUST 2011


13. Match the column : Column-A (A)

17. Out of the following elements, the number of elements forming basic hydroxides ? Na, K, Cl, Br, Ca, Cs

Column-B

Be(OH)2 < Mg(OH)2 <

(P) Solubility

Ca(OH)2 < Ba(OH)2

18.

No. of unpaired electrons present in Fe3+ is ?

19.

The maximum value of l for an electron in fifth energy level is.

(B) Al2O3 < MgO < Na2O < K2O (Q) Acidic character (C) NH3 < PH3 < AsH3 < SbH3

(R) Bond moment

(D) PH3 < AsH3 < SbH3 < NH3

(S) Basic character

MATHEMATICS

(T) Bond length

14. Match the column : Column-I (A) 4 (B) 5

(C) 3 (D) 2

Questions 1 to 6 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 5 marks will be given for each correct answer and – 2 mark for each wrong answer.

Column-II (P) No. of nodes in 3s (Q) No. of sub-shell in third energy level (R) No. of unpaired electron in Fe2+

1.

If sin 2θ = k, then

1 + tan 2 θ

1− k 2 k (C) 1 + k2

(S) No. of e– with ml = 0 & ms = +1/2 in an atom of phosphorus (T) No. of nodes in 5s

2.

+

cot 3 θ 1 + cot 2 θ

=

2−k2 k (D) 2 – k2

(A)

(B)

The most general values of x for which

3 sin x – cos x = min {2, e2, π, λ2– 4λ + 7} are

This section contains 5 questions (Q.15 to 19). +3 marks will be given for each correct answer and no negative marking. 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 has 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

λεR

given by (A) 2nπ (C) nπ + (–1)n

15. In (HF)4 the number of H bonds is ………

(B) 2nπ + π π + 4 6

2π 3

(D) nπ + (–1)n+1

π π – 4 3

3.

Which of the following is negative (A) cos (tan–1 (tan 4)) (B) sin (cot–1 (cot 4)) –1 (C) tan (cos (cos 5)) (D) cot (sin–1 (sin 4))

4.

A circle is inscribed in a triangle ABC touching the side AB at D such that AD = 5, BD = 3. If ∠A = 60° then length BC equals 120 (A) 9 (B) (C) 13 (D) 12 13

5.

If a right-angled ∆ABC of maximum area is inscribed within a circle of radius R, then(A) ∆ = 2R2 (B)

1 1 1 + + = r3 r1 r2

2 +1 R

(C) r = ( 2 – 1) R

16. The number of resonating structures exist for the azide ion, N 3– are ……………

XtraEdge for IIT-JEE

tan 3 θ

(D) s = (1 +

63

2 )R

AUGUST 2011


6.

Sum of the roots of the equation tan233x = cos 2x–1 lying in the interval [0, 314] is(A) 4950π (B) 5050π (C) 4851π (D) None of these

12.

This section contains 2 paragraphs, each has 3 multiple choice questions. (Questions 7 to 12) 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. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer. Passage # 1 (Ques. 7 to 9) We know Inverse trigonometric function have defined domain & range & [x] denotes greatest integer function less than or equals to x. Answer the following If [sin–1 cos–1 sin–1 tan–1 x] = 1 then x belong to the 7. interval (A) [–1, 1] (B) [tan sin cos 1, tan sin cos sin 1] (C) [tan sin cos2, tan sin cos sin 1] (D) None of these 8.

(C) 9.

This section contains 2 questions (Questions 13, 14). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows : P Q R S T A B C D

(B) 2

3 2

The sum of series 10 + sec–1 3

sec–1 2 + sec–1 …+ sec–1

50 +..... 7

(n 2 + 1) (n 2 − 2n + 2) (n 2 − n + 1) 2

(A) tan–1 1 (C) tan–1(n+1)

is

(B) tan–1 n (D) tan–1 (n – 1)

Passage # 2 (Ques. 10 to 12)

10.

If R be the circumradius, r the inradius and r1, r2, r3 be the radii of the three escribed circles of a triangle, s be the semi-perimeter of the triangle. The value of r1 + r2 + r3 is (A) 4R – r (B) 4R + r (C) R + r (D) R – r

11.

∑r r

1 2

(A) s (C) r . s XtraEdge for IIT-JEE

R R R R

S S S S

14. Match the column: Column-I 3π 5π π + cos (A) If cos + cos 7 7 7 2π 4π π = k cos cos cos 7 7 7 then value of k is (B) Value of expression sin 20° (4 cos 20° + 1) is cos 20°. cos 30°

is -

2

Q Q Q Q

T T T T

13. Match the column : Column-I Column-II (A) Number of solution (P) 1 of the equation sin x = – 1/6 in the interval [–7π, 5π] is (B) (cos nπ)12, n ∈ I is equal to (Q) 12 (C) If the equation (R) 0 sec2θ = –x2 + 14x – 23 is possible, then the greatest integral value of x, is (D) Solution(s) of the equation (S) –1 6.91/x – 13.61/x + 6.41/x = 0 is (T) 4

(D) None of these

3

P P P P

Mark your response in OMR sheet against the question number of that question in section-II. + 8 marks will be given for complete correct answer (i.e. +2 marks for each correct row) and NO NEGATIVE MARKING for wrong answer.

Positive value of x for cos (2 sin–1 (cot ( tan–1 ( sec (6 cosec–1 x))))) + 1 = 0 is (A) 3

The cubic equation formed by r1, r2, r3 as roots of the equation, is – (A) x3 – (4R + r) x2 + s2x – r . s2 = 0 (B) x3 + (4R – r) x2 + r . s. x + r2x = 0 (C) x3 + (R + r)x2 + r. s x + r2 . s = 0 (D) None of these

(B) s (D) ∆ 64

Column-II

(P) 3

(Q) – 4

AUGUST 2011


(C) value of

b−c c−a a −b + + r1 r2 r3

ENERGY

(R) 2

is where r1, r2, r3 are exradii of ∆. corresponding to ∠A, ∠B & ∠C respectively (D) In any ∆ABC, minimum value of (S) 0 sin 2 A + sin A + 1 is sin A (T) 4 This section contains 5 questions (Q.15 to 19). +3 marks will be given for each correct answer and no negative marking. 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 has 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

Mechanical energy is the sum of the potential and kinetic energy.

Units: a = [m/sec2], F = [kg•m/sec2] (newton), work = pe= ke = [kg•m2/sec2] (joule)

An ev is an energy unit equal to 1.6 × 10–19 joules

Gravitational potential energy increases as height increases.

Kinetic energy changes only if velocity changes.

Mechanical energy (pe + ke) does not change for a free falling mass or a swinging pendulum. (when ignoring air friction)

The units for power are [joules/sec] or the rate of change of energy.

ELECTRICITY •

A coulomb is charge, an amp is current [coulomb/sec] and a volt is potential difference [joule/coulomb].

Short fat cold wires make the best conductors.

Electrons and protons have equal amounts of charge (1.6 x 10-19 coulombs each).

Adding a resistor in parallel decreases the total resistance of a circuit.

Adding a resistor in series increases the total resistance of a circuit.

All resistors in series have equal current (I).

All resistors in parallel have equal voltage (V).

If two charged spheres touch each other add the charges and divide by two to find the final charge on each sphere.

Insulators contain no free electrons.

Ionized gases conduct electric current using positive ions, negative ions and electrons.

Electric fields all point in the direction of the force on a positive test charge.

18. If A + B + C = π, then the greatest value of k cos A + cos B + cos C is . Then k is. 2

Electric fields between two parallel plates are uniform in strength except at the edges.

Millikan determined the charge on a single electron using his famous oil-drop experiment.

19. Find no. of solution of x for excot x = 1 where x ∈ (0, 2π).

All charge changes result from the movement of electrons not protons (an object becomes positive by losing electrons).

15. If distance between incentre & one of the excentre of equilateral triangle is 4 unit. Then inradius of triangle is 16. Number of solution(s) of the equation π  | x −1 | + | x − 2 |  –1  | x − 3 |  sin–1  is.  +cos   = 2 2    2  2009

 rπ  17. The sum cos  equals  6  r =1

XtraEdge for IIT-JEE

65

AUGUST 2011


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

Column Match Numerical Response

Ques Ans

Column Match Numerical Response

1 A

2 D

3 C

4 A

PHYSICS

5 A

6 B

(A) → Q, (B) → P, (C) → R, (D) → Q

13.

7

8

9

10

11

12

C

B

C

D

D

B

(A) → P, (B) → P, (C) → S, (D) → R,

14.

15

16

17

18

19

5

1

3

6

6

1 D

2 A

3 A

4 C

CHEMISTRY 5 B

6 C

(A) → R, (B) → Q, (C) → S, (D) → P

13.

7

8

9

10

11

12

B

A

B

B

D

C

(A) → R, (B) → S, (C) → P, (D) → Q,

14.

15

16

17

18

19

4

5

1

2

6

MATHEMATICS Ques Ans

Column Match Numerical Response

1 B

2 C

3 B

4 B

5 C

6 A

(A) → P, (B) → R, (C) → S, (D) → R

13.

7

8

9

10

11

12

A

D

B

A

D

B

(A) → S, (B) → Q, (C) → R, (D) → R,

14.

15

16

17

18

19

2

9

0

2

9

IIT- JEE 2013 (August issue) Ques Ans

Column Match Numerical Response

Ques Ans

Column Match Numerical Response Ques Ans

Column Match Numerical Response

1 B

2 B

3 D

4 A

PHYSICS

5 D

6 A

(A) → R, (B) → Q,S (C) → P, (D) → Q,R

13.

7

8

9

10

11

12

C

D

B

C

C

A

(A) → Q, (B) → P, (C) → R, (D) → S,

14.

15

16

17

18

19

1

2

3

4

4

1 C

2 D

3 A

4 A

CHEMISTRY 5 A

6 A

(A) → P,S, (B) → P,S, (C) → Q, (D) → R

13.

7

8

9

10

11

12

B

D

B

B

A

C

(A) → R, (B) → S, (C) → Q, (D) → P,

14.

15

16

17

18

19

3

3

4

5

4

1 B

2 B

3 D

4 C

MATHEMATICS 5 B

6 A

(A) → Q, (B) → P, (C) → Q, (D) → P,S

13.

7

8

9

10

11

12

B

C

B

B

A

A

(A) → Q, (B) → R, (C) → S, (D) → P,

14.

15

16

17

18

19

1

1

0

3

2

XtraEdge for IIT-JEE

66

AUGUST 2011


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