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XtraEdge for IIT-JEE

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JULY 2010

XtraEdge for IIT-JEE

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JULY 2010

Teachers open the door. You enter by yourself Volume - 6 Issue - 1 July, 2010 (Monthly Magazine) Editorial / Mailing Office : 112-B, Shakti Nagar, Kota (Raj.) 324009

Editorial

Tel. : 0744-2500492, 2500692, 3040000 e-mail : [email protected] Editor :

Dear Students,

Pramod Maheshwari

Is examination a common cause of stress? In most Asian cultures, the great emphasis on academic achievement and high expectations of success make it especially stressful for students. The strong negative stigma attached to failure also adds to the pressure. Like it or not, we have to accept that examinations are necessary in any educational system. Even though it is debatable whether they are accurate measures of actual ability, no better alternatives have been proposed. Examinations remain necessary to motivate students’ learning, measure their progress and ultimately, serve as evidence of attainment of certain skills, standards or qualifications. Success at examinations provides opportunities to proceed with higher education and improves employment prospects, underlining their importance. No matter how well prepared, many factors may influence one’s performance at the time of the examination and there is seemingly, no definite guarantee of success. Essentially, it is this vital importance attached to success at examinations coupled with the element of uncertainty that makes them so stressful. As with other sources of stress, the stress of examinations is not all bad. It is a strong incentive for students to study and poses a challenge for individual achievement. However, when stress becomes excessive, performance begins to suffer. There is thus a need to control levels of stress before it becomes overwhelming and detrimental. Reliase of stress is necessary for optimum performance, the means of which is relasing. Learn to relax The stress responses produces muscle tension, which you would commonly experience as backache, neck ache or tension headache at the end of the day. Often this is unconscious. So to relax these muscles, you need to consciously practice relaxation exercises. These could involve muscle relaxation, deep breathing exercises, body massage or guided imagery. Like any particular skill, you need to practice them regularly in order to reap the benefits. Another way to relax is to maintain a quiet time as part of the daily routine. Quiet time refers to a time for you with no interruption from external sources or distractions. This is a time where you may choose to just think of nothing and relax. Finally, you can always take up a hobby to help you relax. Do something you enjoy, be it listening to music. Ideally, the drive to study should be internally driven by a desire to achieve one’s own personal goals. Instead, many are driven more by the fear of failure, which is more stress-provoking and leads easily to discouragement. Attending school should not merely revolve around preparation for examinations. Interacting with teachers, socializing with friends, participating in sports or other extra-curricular activities are all valuable aspects of a ‘wellrounded’ education. Instead of wishing things would get easier, start looking at how you can get better...

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Editor : Pramod Maheshwari XtraEdge for IIT-JEE

Yours truly

Pramod Maheshwari, B.Tech., IIT Delhi

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JULY 2010

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Volume-6 Issue-1 July, 2010 (Monthly Magazine)

CONTENTS

NEXT MONTHS ATTRACTIONS

INDEX

PAGE

Regulars ..........

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- 2011 & 2012

NEWS ARTICLE

4

IITian ON THE PATH OF SUCCESS

6

KNOW IIT-JEE

7

Are Nanoparticles health hazard Nano is the new black Ms. Padmasree Warrior & Dr. Krishan K. Sabnani Previous IIT-JEE Question

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

• If you don’t notice when you win, you will only notice when you lose.

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

• It’s not bragging if you can do it.

CATALYSE CHEMISTRY

Success Tips for the Months

• Feel the power of yet. As in “I don't know how to do this yet.” • The difficult we do immediately. impossible takes a bit longer.

• To know what you are doing is an advantage. To look like you know what you are doing is essential. • First law of expertise: Never ask a barber if you need a haircut.

DICEY MATHS

42

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

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

• If you think you can, you are probably right. If you think you can't, you are certainly right.

XTRAEDGE TEST SERIES

• Don't do modesty unless you have earned it.

Class XII – IIT-JEE 2011 Paper Class XI – IIT-JEE 2012 Paper

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Key Concept Reaction Mechanism Energetics Understanding : Organic Chemistry

The

• Some look down the rapids and see the rocks. Hunters look down the rapids and see the flow around the rocks.

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3

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JULY 2010

Are

Nanoparticles

a

Health Hazard? Nanoparticles are a mega business opportunity for multinationals but they may pose a health hazard to users There is a new industrial revolution taking place all around us. The only problem is we can’t see it. The building blocks, being developed at the cost of billions of dollars by scientists, governments and multinational corporations, are just a few atoms or molecules thick — nanoparticles. Many are less than 100 nanometres (nm) — one-billionth of a metre — thick. A single human red blood cell in comparison is around 500 nm in diametre. It’s a pity though that our eyesight isn’t good enough at nanometre level, for if it were, we would see that nanoparticles of precious metals like gold, silver and titanium have already made the jump from research labs to our homes. Manufactured nanoparticles are today present in thousands of consumer products around the world — silver in washing machines and water purifiers to kill bacteria, zinc in cosmetics to protect against ultraviolet rays, carbon nano-tubes in tennis rackets to make them stronger and lighter, titanium in household paints to decompose dust and grime without human intervention.

cannot we write the entire 24 volumes of the Encyclopedia Britannica on the head of a pin?” he asked. “Because there isn’t much of a point, or money, in doing so!” is the answer he would have got today from nanotechnology researchers. Instead their time is mostly spent figuring out newer properties for nanoparticles which can then be embedded into commercial applications. Nanoparticles are highly reactive and prone to unusual properties. Describing gold, a metal that is normally inert to all other chemicals, Prof. C.N.R Rao, Honorary President and Linus Pauling Professor at the Jawaharlal Nehru Centre for Advanced Scientific Research (JNCASR) and the head of its Nanoscience centre, says “At 200-300 nm thickness, gold is not metallic, it does not shine — in fact it is not gold. And at 1.5-2 nm, it reacts like mad!” Gold that is not gold when shrunk to nanometer size might sound like an absurdity to many, but it’s exactly this change in physical properties that make nanoparticles popular. For example zinc oxide (ZnO) and titanium dioxide (TiO2) have been used as active ingredients in sunscreens for decades because of

Silver, an ornamental metal and a powerful bactericide, can be reduced to nanoparticle form to destroy disease-causing bacteria from all kinds of places — kitchen counters, contaminated water, dirty clothes and stinky underarms. Samsung claims its Silver Nano range of washing machines release hundreds of billions of silver nanoions with each wash to kill over 99 percent of the bacteria found in dirty clothes, while the same technology when lined on the doors of their refrigerators kill bacteria that could spoil stored food. Eureka Forbes’ water purifiers use nanosilver-coated filters, developed by Prof. Pradeep, head of IIT-Madras’ Nanoscience department, to destroy harmful bacteria from drinking water. Swach, the mass-market water filter introduced by the Tata Group, also uses nanosilver (coated on rice husk particles) to purify drinking water. During the last flu pandemic threat authorities in Hong-Kong sprayed subways with nanosilver to disinfect them. L’Oreal, the world’s largest cosmetics company, reportedly spends over $600 million each year researching and patenting nanoparticles. The head of its nanotechnology unit also sits on the management board.

Nano is the New Black “There’s Plenty of Room at the Bottom” So went the classic lecture by the Nobel prize-winning physicist Richard Feynman in 1959 that many nano-ficionados now consider the conceptual sun of the nanotechnology universe. “Why

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their ability to absorb ultraviolet rays and reflect back much of the other remaining sunlight. But they are both white — the reason many sunscreens leave a white residue on the face. When shrunk to nanometre size however, they become transparent without losing their light reflecting or absorbing abilities.

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Therefore a metal that is a poor second cousin to gold in a world where we value yellow over white, is the undisputed metal of choice in the nanoparticle world. In fact, silver is more popular than any other material, according to the database of consumer products using nanoparticles maintained by the Woodrow Wilson International Center for Scholars.

Obama admn nominates IIT alumnus for post of NSF Director IIT Madras alumnus Subra Suresh, popularly known as 'Bakthi Suresh' during his student days, has been nominated for the post of director of the National Science Foundation (NSF) by the Barrack Obama administration. An official relese from IIT Madras Alumni Association here said ''when confirmed by the Senate, Mr Suresh will become one of the highest ranking Indian-Americans ever to serve in an administration.'' An Indian-American technocrat, 53-year-old Subra Suresh completed his B.Tech Mechanical Engineering in 1977. Currently the dean of the MIT engineering school, he received the distinguished alumnus award in 1997. In a statement, President Obama said ''I am proud that such experienced and committed individuals have agreed to take on these important roles in my administration. I look forward to working with them in the coming months and years.'' The National Science Foundation is the funding source for nearly 20 per cent of all federally supported basic research and was an independent federal agency created by US Congress in 1950. Subra Suresh has been elected to the US National Academy of Engineering, the Indian National Academy of Engineering,

the American Academy of Arts and Sciences, the Indian Academy of Sciences in Bangalore, the German National Academy of Sciences, the Royal Spanish Academy of Sciences and the Academy of Sciences of the Developing World based in Trieste, Italy.

Major decision regarding Ganga at IIT-K KANPUR: A collaboration between the consortium of seven IITs (IIT-Kanpur, Madras, Bombay, Delhi, Kharagpur, Guwahati and Roorkee) and Ministry of Environment and Forest, Government of India, is being worked out for the purpose of cleaning the national river Ganga. The two are expected to sign an important memorandum of understanding (MoU) in this regard during Prime Minister Manmohan Singh's visit to the Indian Institute of Technology-Kanpur. Singh is scheduled to visit IIT-K on July 3 to take part in its convocation ceremony. It will be worth mentioning here that the Central government aims at meeting the formidable challenge of cleaning the Ganga. With this goal in mind, it had launched a new initiative and established the National Ganga River Basin Authority (NGRBA) last year. The Prime Minister, who is the Chairman of the NGRBA, asked the Ministry of Environment and Forest to involve IITs in the mega project. A joint meeting of all the seven IITs was convened on March 12, 2010 in which IITKanpur was represented by Prof Vinod Tare, also the convener of the mission. It is for the very first time that the Central government has involved the seven IITs together in one single project of such a large magnitude

Prof Tare further informed TOI that the 'zero discharge' of both treated and untreated sewage waste into the river had been proposed to the government under which the waste would not be allowed into the Ganga. "We will be doing this mega project in phases. The first phase will come to an end in 18 months wherein the concept of 'zero discharge' will be put into application. We also plan to apply the 'zero discharge' formula in four cities initially, viz. Hardwar, Rishikesh, Kanpur and Allahabad. If we are able to do so in these four cities, water of the river will become clean up to Allahabad and a major work will come to an end in the first phase." Meanwhile, Prof SG Dhande, director, IIT-Kanpur, and Prof Vinod Tare from IIT-Kanpur, Prof Devang Khakhar, director, IITBombay, took part in a meeting with the officials of the Ministry of Environment and Forest in New Delhi on May 19 and discussed all important aspects of the mega project. On the occasion, the team also handed over a set of proposals to the officials of the ministry. "A detailed project report will be given later as several social, legal aspects will have to be examined," said Prof Tare. The consortium of the seven IITs have been sanctioned Rs 16 crore by the government for formulating a proper plan of action for the purpose of cleaning the Ganga. The authority formed by the Central government has both regulatory and developmental functions. The authority will take measures for effective abatement of pollution and conservation of the Ganga in keeping with sustainable development needs.

Science Research : Conventional solar cell efficiency could be increased from the current limit of 30 percent to more than 60 percent, suggests new research on semiconductor nanocrystals, or quantum dots, led by chemist Xiaoyang Zhu at The University of Texas at Austin. The scientists have discovered a method to capture the higher energy sunlight that is lost as heat in conventional solar cells.

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Success Story This articles contains stories of person who have succeed after graduation from different IIT's

Ms. Padmasree Warrior

Dr. Krishan K. Sabnani

B.Tech, IIT Madras

B.Tech, IIT – Kanpur

Padmasree Warrior is senior vice president and chief technology officer for Motorola, with responsibility for Motorola Labs, the global software group and emerging early-stage businesses. Warrior's operational responsibilities include leading a global team of 4,600 technologists, prioritizing technology programs, creating value from intellectual property, guiding creative research from innovation through early-stage commercialization, and influencing standards and roadmaps. She also serves as a technology advisor to the office of the chairman and to the board's technology and design steering committee. Before assuming her current position in January 2003, Warrior was corporate vice president and general manager of Motorola's energy systems group, where she was responsible for profit and loss, sales, marketing, engineering and manufacturing. She also was general manager of Thought beam, Inc., a wholly owned subsidiary of Motorola, where she led the commercialization evaluation team related to compound semiconductor materials research. Prior to these assignments, Warrior was corporate vice president and chief technology officer for Motorola's Semiconductor Products Sector (SPS). A Motorola since 1984, she has been instrumental in driving innovative methods for technology commercialization realizing early "time to revenue" for the corporation. She has held many leadership positions within Motorola, was appointed vice president in 1999 and was elected a corporate officer in 2000. Warrior received a M.S. degree in chemical engineering from Cornell University, and a B.S. degree in chemical engineering from the Indian Institute of Technology (IIT) in New Delhi, India. Warrior served on the Texas Governor's Council for Digital Economy, and is a member of the Texas Higher Education Board review panel. She was one of six women nationwide selected to receive the "Women Elevating Science and Technology" award from Working Woman magazine in 2001. She also is a director of Ferro Corporation.

Krishan Sabnani is Senior Vice President of the Networking Research Laboratory at Bell Labs in New Jersey. For the past 23 years Krishan has been a member of Bell Labs Research. Krishan has conceived and launched several systems projects in the areas of Internetworking and wireless networking, led successful transfers of research ideas to products in Lucent and AT&T business units and conducted extensive personal research in data and wireless networking. He has built organizations known for technical excellence by recruiting and coaching the best people in the industry. Krishan has received the 2005 IEEE Eric E. Sumner Award and the 2005 IEEE W. Wallace McDowell Award - the only person ever to receive both awards. Krishan is a Bell Labs Fellow. He is also a fellow of the Institute of Electrical and Electronic Engineers (IEEE) and the Association of Computing Machinery (ACM). He received the Leonard G. Abraham Prize Paper Award from the IEEE Communications Society in 1991. Krishan will receive the 2005 Distinguished Alumni Award from Indian Institute of Technology (IIT), New Delhi, India. He has also won the 2005 Thomas Alva Edison Patent Award from the R&D Council of New Jersey. He holds 37 patents and has published more than 70 papers. In his personal research, Krishan has made major contributions to the communications protocols area. He has designed several protocols such as SNR, RMTP, and Airmail. He has also made significant contributions to conformance test generation, protocol validation, automated converter generation, and reverse engineering. Krishan received his Ph.D. in electrical engineering from Columbia University, New York, in 1981. He joined Bell Labs in 1981. Key Awards and Honors 1. 2005 IEEE Eric E. Sumner Award, received for seminal contributions to networking protocols 2. 2005 IEEE Computer Society W. Wallace McDowell Award 3. 2005 IIT Delhi Outstanding Alumni Award 4. 2005 Thomas Alva Edison Patent Award 5. 1991 Leonard G. Abraham Prize Paper Award 6. 1997 Bell Labs Fellow.

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KNOW IIT-JEE By Previous Exam Questions

2.

PHYSICS 1.

Masses M1, M2 and M3 are connected by strings of negligible mass which pass over massless and friction less pulleys P1 and P2 as shown in fig. The masses move such the portion of the string between P1 and P2 in parallel to the inclined plane and the portion of the string between P2 and M3 is horizontal. The masses M2 and M3 are 4.0 kg each and the coefficient of kinetic friction between the masses and the surfaces is 0.25. The inclined plane makes an angle of 37º with the horizontal. [IIT-1981] M P1 2 P2 M3 M1

A C B D Sol. K.E. of block = work against friction + P.E. of spring 1 1 mv2 = µk mg (2.14 + x) + kx2 2 2 1 1 2 × 0.5 × 3 = 0.2 × 0.5 × 9.8(2.14 + x) + 2 × x2 2 2 2.14+ x + x2 = 2.25 ∴ x2 + x – 0.11 = 0 11 On solving we get x = – 10 1 or x = = 0.1 (valid answer) 10 Here the body stops momentarily. Restoring force at y = kx = 2 × 0.1 = 0.2 N Frictional force at y = µs mg × x = 0.22 × 0.5 × 9.8 = 1.078 N Since friction force > Restoring force the body will stop here. ∴ The total distance travelled = AB + BD + DY = 2 + 2.14 + 0.1 = 4.24 m.

37º

If the mass M1 moves downwards with a uniform velocity, find (a) the mass of M1 (b) The tension in the horizontal portion of the string (g = 9.8 m/sec2, sin 37º ≈ 3/5) Sol. (a) Applying Fnet = ma on M1 we get ...(i) T – m1 . g = M1 × 0 = 0 ⇒ T = M1g Applying Fnet = Ma on M2 we get T – (T´ + M2g sin θ – f) = M2 × a T = T´ + M2g sin θ + f = T´ + M2g sin θ + µN [Q f = µN = µM2 g cos θ] ∴ T = T´ + M2g sin θ + µM2g cos θ ...(ii) P1

V

T

M1 T M1g

M2 θ

M2gcosθ

V

T´

M2g f

M2gsinθ

P2 T´

N

θ

A

M3g

2m

Applying Fnet = Ma for M3 we get T´ – f ´ = M3 × 0 ...(iii) ⇒ T´ = f ´ = µN´ = µM3g Putting the value of T and T´ from (i) and (iii) in (ii) we get M1g = µM3g + M2g sin θ – µ M2g cos θ M1 = 0.25 × 4 + 4 × sin 37º + 0.25 × 4 × cos 37º = 4.2 kg (b) The tension in the horizontal string will be T ´ = µM3g = 0.25 × 4 × 9.8 = 9.8 N

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

B D Rough L C 2.14m

x

Y

3.

7

A small sphere rolls down without slipping from the top of a track in a vertical plane. The track in a vertical plane. The track has an elevated section and a horizontal part, The horizontal part is 1.0 meter above the ground level and the top of the track is 2.4 metres above the ground. Find the distance on the ground with respect to the point B(which is vertically JULY 2010

speed of 0.001 ms–1. Calculate the current drawn from the battery during the process. (Dielectric constant of oil = 11, ε0 = 8.85 × 10–12C2N–1m–1) [IIT-1994] Sol. The adjacent figure is a case of parallel plate capacitor. The combined capacitance will be v

2.4 m

below the end of the track as shown in fig.) where the sphere lands. During its flight as a projectile, does the sphere continue to rotate about its centers of mass ? Explain. [IIT-1987]

A

+

1.0m B Sol. Applying law of conservation of energy at point D and point A P.E. at D = P.E. at A + (K.E.)T + (K.E.)R (K.E.)T = Translational K.E. 1 1 mg (2.4) = mg (1) + mv2 + Iω2 2 2 (K.E.)R = Rotational K.E. Since the case is of rolling without slipping D

2.4m

1–x x

d C = C1 + C2 kε 0 ( x × 1) ε [(1 − x ) × 1] = + 0 d d ε0 C= [kx + 1 – x] d After time dt, the dielectric rises by dx. The new equivalent capacitance will be C + dC = C1´ + C2´ kε 0 ε [1 − x − dx ) × 1] [(x + dx) × 1] + 0 = d d dC = Change of capacitance in time dt ε = 0 [kx + kdx + 1 – x – dx – kx – 1 + x] d ε = 0 (k – 1)dx d ε ε dC dx = 0 (k – 1) = 0 (k – 1)v ...(i) dt d dt d dx where v = dt We know that q = CV dq dC =V ...(ii) dt dt ε ⇒ I = V 0 (k – 1)v d From (i) and (ii) 500 × 8.85 × 10 −12 I= (11 – 1) × 0.001 0.01 = 4.425 × 10–9 Amp.

A 1m B

C

∴ v = rω v ∴ω= where r is the radius of the sphere Also r 2 I = mr2 5 ∴ mg(2.4) = mg(1) +

v2 1 1 2 mv2 + × mr2 × 2 2 2 5 r

⇒ v = 4.43 m/s After point A, the body takes a parabolic path. The vertical motion parameters of parabolic motion will be 1 uy = 0 S = ut + at2 2 1 = 4.9 ty2 Sy = 1m ay = 9.8 m/s2 1 ∴ ty = ? ty = = 0.45 sec 4.9 Applying this time in horizontal motion of parabolic path, BC = 4.43 × 0.45 = 2m During his flight as projectile, the sphere continues to rotate because of conservation of angular momentum. 4.

5.

Two square metal plates of side 1 m are kept 0.01 m apart like a parallel plate capacitor in air in such a way that one of their edges is perpendicular to an oil surface in a tank filled with an insulating oil. The plates are connected to a battery of emf 500 V. The plates are then lowered vertically into the oil at a

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1m

8

Two resistors, 400 ohms, and 800 ohms are connected in series with a 6-volt battery. It is desired to measure the current in the circuit. An ammeter of a 10 ohms resistance is used for this purpose. What will be the reading in the ammeter? Similarly, If a voltmeter of 10,000 ohms resistance is used to measure the potential difference across the 400-ohms resistor, What will be the reading in the voltmeter. [IIT-1982] JULY 2010

Sol. Applying Kirchoff's law moving in clockwise direction starting from battery we get 800Ω 400Ω 10Ω

Sol. (i) From the given data, it is evident that the t1/2 (half-life period)for the decomposition of X (g) is constant (100 minutes) therefore the order of reaction is one. 0.693 (ii) Rate constant, K = t1/ 2

A

0.693 = 6.93 × 10–3 min–1 100 (iii) Time taken for 75% completion of reaction = 2t1/2 = 2 × 100 = 200 minutes (iv) 2x → 2Y + 2Z Initial pressure 800 0 0 Ater time t (800 – 2p) 3P 2p when the pressure of X is 700 mm of Hg the, 800 – 2P = 700 2P = 100; P = 50 mm of Hg Total pressure = 800 – 2P + 3P + 2P = 800 + 150 = 950 mm of Hg.

=

6 volt + 6 – 10I – 400 I – 800 I = 0 ∴ 6 = 1210 I 6 ∴ I= = 4.96 × 10–3 A 1210 The voltmeter and 400 Ω resistor are in parallel and hence p.d. will be same ...(i) ∴ 10,000 I1 = 400 I2 Applying Kircoff's law in loop ABCDEA starting from A in clockwise direction. – 400 I2 – 800 I + 6 = 0 ∴ 6 = 400 I2 + 800 (I1 + I2) ∴ 6 = 400 I2 + 800(0.04 I2 + I2) From (i) putting the value of I1 ∴ 6 = 1232 I2 10,000Ω

F B I

V

G C

400Ω I 800Ω

A

A acid solution of Cu2+ salt containing 0.4 g of Cu2+ is electrolysed until all the copper is deposited. The electrolysis is continued for seven more minutes with the volume of solution kept at 100 ml. and the current at 1.2 amp. Calculate the volume of gases evolved at NTP during the entire electrolysis. [IIT-89] Sol. The chemical reactions taking place at the two electrodes are At cathode : Cu2+ + 2e– → Cu H2O H+ + OH– However, note that only Cu2+ ions will be discharged so as these are present in solution and H+ ions will be discharged only when all the cu2+ ions have been deposited. Atcathode : 2OH– → H2O + O + 2e– O + O → O2 Thus in first case, Cu2+ ion will be discharged at the cathode and O2 gas at the anode. Let us calculate the volume of gas (O2) discharged during electrolysis. According to Faraday's cecond law 31.75 g Cu ≡ 8 g of oxygen ≡ 5.6 litres of O2 at NTP 5 .6 0.4 g Cu = × 0.4 litres of O2 at NTP 31.75 = 0.07055 litres = 70.55 ml As explained earlier, when all the Cu2+ ion will be deposited at cathode, H+ ions will start going to cathode liberating hydrogen (H2) gas i.e. H+ + e– H H + H → H2 However, the anode reaction remains same as previous. Thus in the second (latter) case, amount of H2 collected at cathode should be calculated. 8 g of O2 = 1 g of H2 5.6 litres of O2 at NTP = 11.2 litres of hydrogen Quantity of electricity passed after 1st electrolysis, i.e. Q = i × t = 1.2 × 7 × 60 = 504 coulombs 7.

D

E

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

CHEMISTRY 6.

At constant temperature and volume, X decomposes as 2X(g) → 3Y(g) + 2Z(g); Px is the partial pressure of X.

Observation Time (in minute) Rx (in mm of Hg) No. 1 0 800 2 100 400 3 200 200 (i) What is the order of reaction with respect to X ? (ii) Find the rate constant. (iii) Find the time for 75% completion of the reaction. (iv) Find the total pressure when pressure of X is 700 mm of Hg. [IIT-2005]

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JULY 2010

Sol. Let us summarise the given facts.

5.6 × 504 = 29.24 ml 96500 of O2. Similarly, H2 liberated by 504 coulombs 504 = 11.2 × = 58.45 ml 96500 Total volume of O2 liberated = 70.55 + 29.24 = 99.79 ml vol. of H2 liberated = 58.48 ml. Cyclobutyl bromide on treatment with magnesium in dry ether forms an organometallic (A). The organometallic reacts with ethanal to give an alcohol (B) after mild acidification Prolonged treatment of alcohol (B) with an equivalent amount of HBr gives 1-bromo 1-methylcyclopentane (C). Write the structures of (A), (B) and explain how (C) is obtained from (B). [IIT-2001] MgBr Br

504 coulombs will liberate =

8.

112 ml of Basic colourless, (i) aq. HCl Nitrogen odourless (ii) NaNO2 0ºC Compound gas at S.T.P (0.295 g) + Residue

ether

Organic liquid (no N)

CH CHO , H O +

3 3 →

CH3

CH–CH3 OH (B)

1-Cyclobutylethanol (B)

CH–CH3 +

H→

+

CH–CH3

– H 2O +

OH2

→

⊕ Oxonium ion

(2º carbocation 4-membered ring)

H

exp ansion ring →

through 1, 2 − alkylshift

⊕

H CH3

CH3CH2CH2NH2

Hydride shift

H CH3 – ⊕ Br →

(3º carbocation)

9.

+

–

,I2 OH → No yellow ppt. (CH3)CHNH2 → (CH3)2CHOH + N2↑ Isopropylamine

H Br ⊕ CH3

→ (CH3)2CHOH

1-bromo-1-methyl Cyclopentane (C)

–

,I 2 OH →

( Haloform reaction )

Since the given reactions isopropylamine, the original isopropylamine, (CH3)2CHNH2

A basic, volatile nitrogen compound gave a foul smelling gas when treated with chloroform and alcoholic potash. A 0.295 g sample of the substance. Dissolved in aq. HCl and treated with NaNO2 solution at 0ºC, liberated a colorless, odourless gas whose volume corresponded to 112 ml at STP, After the evolution of the gas was complete, the aqueous solution was distilled to give an organic liquid which did not contain nitrogen and which on warming with alkali and iodine gave a yellow precipitate. Identify the original substance. Assume that it contains one N atom per molecule. [IIT-93]

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i ) HCl ( → CH3CH2CH2OH

(ii ) NaNO 2 / 0 º C

aq .sol .distill → CH3CH2CH2OH

(ring expansion) (2º carbocation in 5 membered ring

H

Yellow ppt.

N2↑ n-Propylamine

,2 – 1 →

H

OH–/I2

Reaction of the original compound with alcoholic potash and chloroform to give foul smelling gas indicates that it contains a primary –NH2 group. R–NH2 + CHCl3 + KOH → R–NC↑ (Basic compound) Carbylamine (foul smelling) Determination of mol. Weight of the amine. 112 ml. of gas is evolved at S.T.P. by 0.295 g of amine 0.295 22400 ml. of gas is evolved by = × 22400 = 59 112 Hence the mol. wt. of the amine = 59 ∴ Mol. wt. of the alkyl group = 59 – 16 = 43 Nature of alkyl gp. of mol. wt.= 43 = C3H7– Thus the amine may be either CH3 CH3CH2CH2NH2 or CHNH2 CH3 The reaction of amine with NaNO2 at 0ºC and all other reactions may thus be written as below.

Cyclobutymagnesium Bromide (A)

CH–OH

Foul smelling gas

Distilled aq. Sol.

dry →

Sol.

CHCl3 KOH

CHI3↓ (yellow) correspond compound

to is

10. Interpret the non-linear shape of H2S molecule and non-planar shape of PCl3 using valence shell electron pair repulsion (VSEPR) theory. (Atomic numbers : H = 1, P = 15, S = 16, Cl = 17.) [IIT-98] 1 Sol. In H2S, no. of hybrid orbitals = (6 + 2 – 0 + 0) = 4 2 Hence here sulphur is sp3 hybridised, so 2 2 6 2 2 1 1 16S = 1s , 2s 2p , 3s 3p x 3p y 3p z 144244 3 sp 3 hybridisation

10

JULY 2010

[Q O(0, 0) and C(2, 2) lie on the same side of AB 2 2 + – 1 < 0] Therefore, a b (2b + 2a – ab) ⇒ – =2 a 2 + b2

S S

or

H

H

H

H Due to repulsion between lp - lp; the geometry of H2S is distorted from tetrahedral to V-shape. 1 In PCl3, no. of hybrid orbitals = [5 + 3 – 0 + 0] = 4 2 Hence, here P shows sp3-hybridisation 2 1 1 1 2 2 6 15P = 1s , 2s 2p , 3s 3p x 3p y 3p z 144244 3

...(i) ⇒ 2a + 2b – ab + 2 a 2 + b 2 = 0 Let P(h, k) be the circumcentre of ∆OAB. Since ∆ OAB is a right angled triangle. So its circumcentre is the mid-point of AB. a b and k = ∴ h= 2 2 ⇒ a = 2h and b = 2k ...(ii) From (i) and (ii), we get

sp 3 hybridisation

P

P

4h + 4k – 4hk + 2 4h 2 + 4k 2 = 0 or

Cl

Cl

⇒ h + k – hk + h 2 + k 2 = 0 So, the locus of P(h, k) is

Cl

x + y – xy +

Cl Cl

Cl

x 2 + y2 = 0

But, the locus of the circumcentre is given to be

Thus due to repulsion between lp – bp, geometry is distorted from tetrahedral to pyramidal.

x + y – xy + k x 2 + y 2 = 0 Thus, the value of k is 1

MATHEMATICS

12. If A, B, C are the angles of a triangle ABC and the system of linear equations x sin A + y sin B + z sin C = 0 x sin B + y sin C + z sin A = 0 x sin C + y sin A + z sin B = 0 has a non trivial solution, prove that sin2A + sin2B + sin2C – (cos A + cos B + cos C + cos A cos B + cos B cos C + cos C cos A) = 0 [IIT-2002] Sol. The given system of linear equations has a non-trivial solution. Therefore, sin A sin B sin C sin B sin C sin A = 0 sin C sin A sin B

11. The circle x2 + y2 – 4x – 4y + 4 = 0 is inscribed in a triangle which has two of its sides along the coordinate axes. If the locus of the circumcentre of the triangle is

x + y – xy + k x 2 + y 2 = 0, [IIT-1987] find the value of k. Sol. Let OAB be the triangle in which the circle x2 + y2 – 4x – 4y + 4 = 0 is inscribed. Let the x y equation of AB be + = 1 a b y B(0,b)

C

sin A + sin B + sin C sin B sin C ⇒ sin B + sin C + sin A sin C sin A = 0 sin C + sin A + sin B sin A sin B

x y + =1 a b x+ y =1 a b

2 x´

O

x

(a, 0)A

Applying C1 → C1 + C2 + C3

y´

1 sin B sin C 2

2

⇒ (sin A + sin B + sin C) 1 sin C sin A = 0 1 sin A sin B

Since AB touches the circle x + y – 4x – 4y + 4 = 0. There fore, 2 2 2 2 + − 1 + −1 a b a b =2 ⇒– =2 1 1 1 1 + + a 2 b2 a 2 b2

XtraEdge for IIT-JEE

1 sin B sin C

⇒ 1 sin C sin A = 0 1 sin A sin B

11

Q sin A + sin B + sin C A B C = 4 cos 2 cos 2 cos 2 ≠ 0

JULY 2010

1

sin B

We claim that at least two y1, y2, and y3 are distinct. For if y1 = y2 = y3, then P, Q and R lie on a line parallel to x-axis and a line parallel to x-axis does not cross the circle in more than two points. Thus, we have either y1 ≠ y2 or, y1 ≠ y3 or, y2 ≠ y3. Subtracting (ii) from (iii) and (iv), we get

sin C

⇒ 0 sin C − sin B sin A − sin C = 0 0 sin A − sin B sin B − sin C Applying R2 → R2 – R1, R3 → R3 – R1 ⇒ –(sin B – sin C)2 – (sin A – sin C) (sin A – sin B) = 0 ⇒ sin2B + sin2C – 2 sin B sin C + sin2A – sin A sin B – sin C sin A + sin B sin C = 0 ⇒ sin2A + sin2B + sin2C – sin A sin B – sin B sin C – sin C sin A = 0 ⇒ sin2A + sin2B + sin2C – cos A cos B – cos B cos C – cos C cos A + cos (A + B) + cos (B + C) + cos (C + A) = 0 ⇒ sin2A + sin2B + sin2C – cos A cos B – cos B cos C – cos C cos A – cos A – cos B – cos C = 0

( x 22 + y 22 ) – ( x12 + y12 ) – 2 2 (y2 – y1) = 0

⇒ a1 – where,

a1 – ⇒

...(ii)

− 2 2 y2 = r – 2

...(iii)

x 32 + y 32 − 2 2 y3 = r2 – 2

...(iv)

+

y 22

XtraEdge for IIT-JEE

2

2 b1 = 0 a1 = b1

[From (v)]

2,

which is not

possible because

a1 is a rational b1

number and 2 is an irrational number. If b2 ≠ 0, then a a2 – 2 b2 = 0 ⇒ 2 = 2 , b2 which is not possible because

a2 b2

is a rational

number and 2 is an irrational number. Thus, in both the cases we arrive at a contradiction. This means that our supposition is wrong. Hence, there can be at most two rational points on circle C. 15. A rectangle PQRS has its side PQ parallel to the line y = mx and vertices P, Q and S lie on the lines y = a, x = b and x = –b, respectively. Find the locus of the vertex R. [IIT-1996] Sol. Let the coordinates of R be (h, k). It is given that P lies on y = a. So, let the coordinates of P be (x1, a). Since PQ is parallel to the line y = mx. Therefore, Slope of PQ = (Slope of y = mx) = m 1 And, Slope of PS = – (Slope of y = mx)

2 ) is

or, x2 + y2 – 2 2 y = r2 – 2 If possible, let P(x1, y1), Q(x2, y2) and R(x3, y3) be three distinct rational points on circle C. Then, x 22

...(v)

Clearly, a1, a2, b1, b2 are rational numbers as x1, x2, x3, y1, y2, y3 are rational numbers. Since either y1 ≠ y2 or, y1 ≠ y3 ∴ Either b1 ≠ 0 or, b2 ≠ 0 If b1 ≠ 0, then

2 )2 = r2, where r is any positive real

x12 + y12 − 2 2 y1 = r2 – 2

2 b2 = 0

a2 = ( x 32 + y 32 ) – ( x12 + y12 ) , b2 = 2(y3 – y1)

14. Let C be any circle with centre (0, 2 ). Prove that at most two rational points can be there on C. (A rational points is a point both of whose [IIT-1997] coordinates are rational numbers)

(x – 0)2 + (y – number.

2 b1 = 0 and a2 –

a1 = ( x 22 + y 22 ) – ( x12 + y12 ) , b1 = 2(y2 – y1)

13. Determine the name of the name of the curve described parametrically by the equations [IIT-1998] x = t2 + t + 1, y = t2 – t + 1 Sol. We have, x = t2 + t + 1 and, y = t2 – t + 1 ⇒ x + y = 2(t2 + 1) and, x – y = 2t 2 x − y ⇒ x + y = 2 + 1 2 2 ⇒ 2(x + y) = (x – y) + 4 ⇒ x2 + y2 – 2xy – 2x – 2y + 4 Comparing this equation with the equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, we get a = 1, b = 1, c = 4, h = –1, g = –1 and f = –1 ∴ abc + 2fgh – af2 – bg2 – ch2 = 4 – 2 – 1 – 1 – 4 ≠ 0 and , h2 – ab = 1 – 1 = 0 Thus, we have ∆ ≠ 0 and h2 = ab So, the given equations represent a parabola.

Sol. The equation of any circle C with centre (0, given by

( x 32 + y 32 ) – ( x12 + y12 ) – 2 2 (y3 – y1) = 0

and,

1 [∴ PS ⊥ PQ] m Now, equation of PQ is y – a = m(x – x1)

=–

12

...(i) JULY 2010

y (0, a)

⇒

y=0 P

x = –b

x=b Q

x´

S (0, – b)

O

(0, b)

⇒ x

BEWARE OF THE BATTERIES THAT YOU USE!

R y´

It is given that Q lies on x = b. So, Q is the point of intersection if (i) and x = b. Putting x = b in (i), we get y = a + m(b – x1) So, coordinates of Q are (b, a + m(b – x1)). 1 . Since PS passes through P(x1, a) and has slope – m 1 So, Equation of PS is y – a = – (x – x1) ...(ii) m It is given that S lies on x = – b. So, S is the point of intersection of (ii) and x = –b. 1 Solving (ii) and x = – b, we get y = a + (b + x1) m 1 So, coordinates of S are − b, a + (b + x 1 ) m 1 k − a − (b + x1 ) m =m Now, Slope of RS = h+b But RS is parallel to PQ. 1 k − a − (b + x1 ) m ∴ =m h+b ⇒ b + x1 = m(k – a) – m2(h + b) ...(iii) Similarly, k − a − m( b − x 1 ) Slope of RQ = h−b But, RQ is perpendicular to PQ whose slope is m. k − a − m( b − x1 ) 1 ∴ =– h−b m 1 1 ⇒ b – x1 = (k – a) + 2 (h – a) ...(iv) m m We have only one variable x1. To eliminate x1, add (iii) and (iv) to obtain 1 1 2b = (k – a) m + – m2(h + b) + 2 (h – b) m m ⇒

Have you ever noticed that the batteries are becoming smaller and smaller day after day? Many scientists and researchers have been finding the effective way to shrink the batteries into the smallest size as possible! In this case, Jae Kwon, an assistant professor of Electrical and computer engineering has recently developed a nuclear energy source, which is smaller, lighter and more efficient than the common batteries. Kwon’s described that the new discovered radioisotope battery can provide power density as much as six orders of magnitude higher than chemical batteries.

Kwon and his research team members have been cooperated and working on building a small nuclear battery. According to the information, the radioisotope batteries are having the size and thickness of a penny, but it’s powerful enough to power various micro or nanoelectromechanical systems. Even though the nuclear power sources have always been a safety concern, they’ve claimed to be safe, as the nuclear power sources have been used for powering many types of devices, including the pacemakers, space satellites and underwater systems. Kwon’s battery is in a liquid semiconductor rather than a solid semiconductor, as he believed that the liquid semiconductor can overcome the problem, where the lattice structure of the semiconductor being damaged, if it’s in the solid semiconductor form!

m2 +1 4 4 – h m +1 – b m +1 2b = (k – a) m m2 m2

m 2 + 1 h (m 2 − 1)(m 2 + 1) b(m 2 + 1) 2 – ⇒ (k–a) – =0 m m2 m2

XtraEdge for IIT-JEE

h (m 2 − 1) b(m 2 + 1) – =0 m m m(k – a) – h(m2 – 1) – b(m2 + 1) = 0 Hence, the locus of R(h, k) is m(y – a) – x(m2 – 1) – b(m2 + 1) = 0

(k – a) –

13

JULY 2010

Physics Challenging Problems

Set # 3

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

Passage # (Q. No. 4 to Q. No. 6) Resistance value of an unknown resistor is calculated using the formula R= V/I where V and I be the readings of the voltmeter and the ammeter respectively. Consider the circuits below. The internal resistances of the voltmeter and the ammeter (RV and RG respectively) are finite and non zero. V V A A R R E E r r

A circuit consisting of a constant e.m.f. ‘E’, a self induction ‘L’ and a resistance ‘R’ is closed at t = 0. The relation between the current I in the circuit and time t is as shown by curve ‘a’ in the figure. When one or more of parameters E, R and L are changed, the curve ‘b’ is obtained. The steady state current is same in both the cases. Then it is possible that I (a) (b)

Fig. (A) Fig. (B) Let RA and RB the calculated values in the two cases A and B respectively. 4. The relation between RA and the actual value R is (B) R < RA (A) R > RA (D) dependent upon E and r (C) R = RA 5. The relation between RB and the actual value R is (B) R > RB (A) R< RB (D) dependent upon E and R (C) R = RB 6. If the resistance of voltmeter is RV = 1 KΩ and that of ammeter is RG = 1Ω, the magnitude of the percentage error in the measurement of R (the value of R is nearly 10 Ω) is (A) zero in both cases (B) non-zero but equal in both cases (C) more in circuit A (D) more in circuit B Passage # (Q. No. 7 to Q. No. 8) The figure shows the interference pattern obtained in a double-slit experiment using light of wavelength 600 nm. 1,2,3,4 and 5 are marked on five fringes.

t (A) E and R are kept constant and L is increased (B) E and R are kept constant and L is decreased (C) E and R are both halved and L is kept constant (D) E and L are kept constant and R is decreased

2.

3.

Consider a resistor of uniform cross section area connected to a battery of internal resistance zero. If the length of the resistor is doubled by stretching it then (A) current will become four times (B) the electric field in the wire will become half (C) the thermal power produced by the resistor will become one fourth (D)the product of the current density and conductance will become half In front of an earthed conductor a point charge +q is placed as shown in figure +q

(A) On the surface of conductor the net charge is always negative (B) On the surface of conductor at the same points charges are negative and at some points charges may be positive distributed non uniformly (C) Inside the conductor electric field due to point charge is non-zero (D) None of these

XtraEdge for IIT-JEE

By : Dev Sharma Director Academics, Jodhpur Branch

7. 8.

14

The third order bright fringe is (A) 2 (B) 3 (C) 4 (D) 5 Which fringe results from a phase difference of 4π between the light waves incidenting from two slits (A) 2 (B) 3 (C) 4 (D) 5 JULY 2010

8

Solution

Physics Challenging Problems Qu e s tio ns we r e Pub lis he d in Jun e I s su e x −e x −0 4 =0 + r r 5x = 4e x = 4e/5 x 2Bωa 2 e 4e and i = = = i = r 5 r + r / 4 5r

Initially the potential at centre of sphere is 1 Q 1 2Q 1 3Q + = VC = 4πε0 x 4πε0 x 4πε0 x After the sphere grounded, potential at centre becomes zero. Let the net charge on sphere finally be q. 3Qr 1 q 1 3Q ∴ + = 0 or q = x 4πε0 r 4πε0 x

1.[C]

6.

→

3Qr x The velocity is maximum at mean position. Hence the magnetic force on block is maximum, at its mean position. The magnetic force on the block while it crosses the mean position towards right and left is as shown qvmaxB N1

→ →

(C) v . F = 0 means instantaneous velocity is always perpendicular to force. Hence the speed →

will remain constant. And also | F |= constant. Since the particle moves in one plane, the resulting motion has to be circular. →

∫ Idt = 5A ∫ dt Total heat produced = ∫ I 2 Rdt

Average current =

0

∧

∧

→

∧

∧

(D) u = 2 i − 3 j and a = 6 i − 9 j . Hence initial velocity is in same direction of constant acceleration, therefore particle moves in straight line with increasing speed.

3. (A) → Q,R, (B) → P,S, (C) → P,R, (D) → Q,S 4. [A,B,C] Total charge = ∫ Idt = area under curve = 10C

2

→

→

→ →

Mg + qvmaxB mg Case-1 Case-2 Hence normal reaction is maximum in case-1 and minimum in case-2. Hence correct option is D.

= ∫ (−5t + 10) 2 .1.dt =

→

(B) u . F = 0 and F = constant Initial velocity is perpendicular to constant force, hence the path will be parabolic with speed of particle increasing.

N2

vmaxvmax

(A) → R , (B) → Q,S, (C) → P, (D) → Q,R (A) F = constant and u × F = 0 Therefore initial velocity is either in direction of constant force or opposite to it. Hence the particle will move in straight line and speed may increase or decrease.

∴ The charge flowing out of sphere is

2. [D]

Set # 2

7. [B]

C r rθ R q

200 J 3

O

r=

v0

v0 =

mv 0 ; R = 2r sin θ qB

qBR 2m sin θ

Maximum power = I2R, when I is maximum current = 100 × 1 =100W.

x

ε r/2

r/2 ε

×× × ×× ×

r

x

r

φ total

x

Bωr 2 Bωa 2 = 2 2 By nodal equation, nodal

Induced emf e =

XtraEdge for IIT-JEE

dφ = B × hdr =

8. [B,C]

5. [B,D] Equivalent circuit

(∴ Radius = a)

15

h

µ 0 NIhdr 2πr

dr

R +b µ N×h = 0 log × I max × sin ωt 2 π R dφ e = total dt

JULY 2010

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

PHYSICS

A particle of mass m moves along a horizontal circle of radius R such that normal acceleration of particle varies with time as an = Kt2 , where K is a constant. Calculate (i) tangential force on particle at time t, (ii) total force on particle at time t, (iii) Power developed by total force at time t, and (iv) average power developed by total force over first t second. Sol. Since, the particle is moving along a circle, therefore, its normal acceleration is centripetal acceleration i.e. v2/R, where v is velocity of particle at time t. v2 ∴ = Kt2 or v = t. KR R …(1) Due to centripetal acceleration, particle follows a circular path but due to it velocity its magnitude does not change. Velocity magnitude increases due to tangential acceleration alone. d ∴ Tangential acceleration, at = . v = KR dt 1.

∴

Tangential force,

∴

Normal force, Resultant force on particle,

∴

∴ 2.

Ft = mat = m KR Ans. (i) Fn = man = mKt2 F= =m

C

Sol. Let the particle be displaced slightly through x along a line normal to plane of the figure. Then each spring is further elongated. Since, springs are identical, therefore, increase in tension of each spring will be the same. Let this increase be dE0. l l C A θ θ C (F0 + dF0) (F0 + dF0) P First considering forces exerted by spring AP and CP only as shown in Figure. Restoring force produced by these two springs = (F0 + dF0) 2 sin θ x Since x is very small, therefore, sin θ ≈ l Neglecting product of very small quantities, restoring force produced by these two springs

Ans. (ii)

Since, power developed by force F is given by →→

P = F v , therefore, power developed by normal force Fn is always zero because its direction is always perpendicular to the instantaneous direction of motion of the particle. Hence, power is developed by tangential force alone. Figure Ft

v

→ →

Ans.(iii) i.e. P = Ft v = mkRt Since, resultant force equals (mass × acceleration), therefore, resultant force is used to accelerate the

XtraEdge for IIT-JEE

P m

D

+ K 4t )

→

Fn

Figure Shows a particle of mass m = 100 gm, attached with four identical springs, each of length l = 10 cm. Initial tension in each spring is F0 = 25 newton. Neglecting gravity, calculate period of small oscillations of the particle along a line perpendicular to the plane of the figure B

A

Ft2 + Fn2 K (R

body. It means that velocity of the body increases due to resultant force. Hence power developed by resultant force is used to increase kinetic energy of the body. Average power developed by resultant force = Average rate of increase of KE Initial kinetic energy (at t = 0), E0 = 0 1 1 Kinetic energy at time t, E = mv2 = mKRt2 2 2 …(2) E – E0 1 Average power = = mKRt` Ans. (iv) t 2

16

JULY 2010

∴ ∴

∴

T = 2π

∫

∴

σ 2ε 0

∴

Sol.

r +y

0

r2 + y2

2

r +a

or,

V=

Put

∫

a

P

σ 2ε 0

a

ydy

0

r 2 + y2

∫

b (A)

C B A a

c b

(B)

1 q1 q 2 q 3 + + 4πε 0 a b c q1 q 2 q 3 + =0 + b c a Ans. or q1 = – 3µC Now, charges on different surface will be as shown in Figure(B) to calculate energy stored in the system, it can be as considered in three parts : (i) a spherical capacitor having radii a and b and having charge |q1| 3µC.

V= ∴

(r 2 + y 2 ) = P

XtraEdge for IIT-JEE

(+4µC) (–1µC) (+1µC) (–3µC) (+3µC)

c

q1 a

2πσ ydy . 2 4πε 0 r + y2

=

0

σ a2 q π . × = 2ε 0 2r π 4πε 0 r

C B A

q3 q2

dV

∫

1 a2 ≈ r 1 + 2 2 r

Three concentric, conducting spherical shells A,B and C have radii a = 10 cm, b = 20 cm and c = 30 cm respectively. The innermost shell A is earthed and charges q2 = 4µC and q3 = 3 µC are given to shells B and C respectively. Calculate charge q1 induced on shell A and energy stored in the system. Sol. System of three concentric shells is as shown in Figure(A) Since, Shell A is earthed, therefore its potential is zero. But its potential is

y =0

=

1/ 2

4.

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

σ 2 r + a 2 − r 2ε 0

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

r2 + y2

a

a σ 2 r + y2 0 ε 0

a 2 = r 1 + r

2

V=

e

r

=

r2 + y2

σ 2 r + a 2 − r 2ε 0 As a special case, if r >> a

Find the electric potential, at any point on the axis of a uniformly charged circular disc, whose surface charge density is σ, radius, a. Let us consider a small elemental thin ring of width dy. Area of the ring = 2πy dy Charge on this elemental ring = (2πy dy)σ Again, we can consider that this ring is divided into a large number of small elements. Each such element e is at the same distance from P. Hence, potential produced by this ring of width dy at the point P is given by dV, where 1 2πyσdy dV = 4πε 0 r 2 + y 2

O

pdp p

∫

∫ dp = p =

ydy

∫

=

∴V =

ml ml = π = 0.02π sec 4F0 F0

y

2

a

=

Ans. 3.

2

=

4F 2F x Resultant restoring force, F = 2 × 0 = 0 .x l l Restoring acceleration is directly proportional to displacement x, therefore, the particle executes SHM, displacement Its period T = 2π acceleration or

r2 + y2 = p2 2ydy = 2pdp ydy

∴ or

x l Similarly, restoring force produced by two remaining springs BP and DP will also be equal to 2F0 x l

= 2F0

17

JULY 2010

Its capacitance, C1 =

4πε 0 ab (b – a )

C1 + –

∴

5.

Applying Kirchhoff's voltage law on left mesh, of Figure (B) q1 q – q2 + 1 –E=0 …(1) C1 C2 For middle mesh, q2 q q – q2 + 2 – 1 =0 C3 C2 C2

Ans.

+ E –

C2

C3

C1 +E –

C2

…(2)

From equation (1) and (2), q1 = 50 µC and q2 = 20 µC Now consider the circuit when switch S is closed and steady state is reached. The circuit will be as shown in Figure (C)

In the circuit shown in Figure emf of each battery is E = 20 volts and capacitance is C1 = 5 µF, C2 = 3 µF and C3 = 6 µF. Calculate charge on capacitor C3 when switch S is closed and steady state is reached. Calculate also, heat generated in the circuit. C1

+ – E

S

(q + q 2 + q 3 ) 2 q12 (q + q 2 ) 2 + 1 + 1 2C1 2C 2 2C 3

= 0.45 joule

C1 + –

(B)

(iii) an isolated sphere of radius c and having charge (q3 + q2 + q1) = 4µC Energy stored in the system =

+ + C2 –(q1 –q2) C2– q2

+ E –

(ii) a spherical capacitor having radii b and c and having charge (q2 + q1) = 1 µC. 4πε 0 bc Its capacitance. C2 = (c – b)

C3 + –

+ E –

C3

C1 +q –

C1 – q+ + C2 –q

+ C2 –q

+E –

S

Sol.

When Switch S is closed and steady state is reached, the circuit becomes symmetric about the dotted line shown in Figure(A). C1 + E –

C2

C3 C2

(C)

Applying Kirchhoff's voltage law on left mesh of Figure(C) q q + –E=0 or q = 37.5 µC. C1 C2

C1 +E –

After shorting of switch S, increase in charge on capacitor C1 of Left mesh, ∆q1 = (q – q1) = – 12.5 µC That for capacitor C2 of left mesh, ∆q2 = q – (q1 –q2) = 7.5 µC That for capacitor C3 , ∆q3 = 0 – q2 = – 20µC That for capacitor C2 of right mesh, ∆q4 = (q –q2) = 17.5 µC That for capacitor C1 or right mesh, ∆q5 = q – 0 = 37.5 µC Since, heat generated in the circuit is given by

(A)

Right part of the circuit is exactly mirror image of the left part. Hence, charges on both plates of capacitor C3 should be identical. But charges on plates of a capacitor are always opposite to each other. It means one of the plates is always positively charged and the other is negatively charged. Both these conditions can be satisfied only if charge on capacitor C3 is zero. To calculate heat generated in the circuit initial and final charges on all the capacitors must be known. First analyse the circuit when switch S was open Charges on capacitors will be as shown in Figure (B)

( ∆q ) 2 2C

H=

∑

H=

(∆q 3 ) 2 (∆q1 ) 2 (∆q 2 ) 2 + + 2C1 2C 2 2C 3

+ = 250 × 10–6 joule

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18

(∆q 5 ) 2 ( ∆q 4 ) 2 + 2C 2 2C1

Ans. JULY 2010

P HYSICS F UNDAMENTAL F OR IIT-J EE

Capacitor-1 KEY CONCEPTS & PROBLEM SOLVING STRATEGY Capacitance : Whenever charge is given to a conductor of any shape its potential increases. The more the charge (Q) given to the conductor the more is its potential (V) i.e. Q∝V

(b) directly on the dielectric constant K of the medium between the conductors. (c) inversely on the distance of separation between the conductor. Principle of a condenser : Consider a conducting plate A which is given a charge Q such that its potential rises to V. Then C = Q/V Let us place another identical conducting plate B parallel to it such that charge is induced on plate B (as shown in figure). A

⇒ Q = CV where C is constant of proportionality called capacitance of the conductor C = Q/V, C = Q SI unit of capacitance is farad (F) and 1 F = 1 coulomb/volt (1CV–1) Energy stored in a charged capacitor : Q2 1 1 CV02 = = QV0 2C 2 2

W=

+ + + + + + + +

Capacitance of an isolated sphere : Let a conducting sphere of radius a acquire a potential V when a charge Q is given to it. The potential acquired by the sphere is

V=

Charge sharing Between two charged conductors : C2

V1

V2

V

V

q1 = C1V1

q2 = C2V2

q´1 = C1V

q´2 = C2V

(Initially)

V=

(Finally)

C´ =

C1V1 + C 2 V2 C1 + C 2

+ + + + + + + +

Q Q = V´ V + V+ − V−

⇒ C´ > C Further, if B is earthed from the outer side (see figure) then Vn = V – V– as the entire positive charge flows to the earth. So

(V1 – V2)2

Capacitor or Condenser : An arrangement which has capability of collecting (and storing) charge and whose capacitance can be varied is called a capacitor (or condenser) The capacitance of a capacitor depends. (a) directly on the size of the conductors of the capacitor.

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– – – – – – – –

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

There is always a loss in energy during the sharing process as some energy gets converted to heat. 1 C C Loss = – ∆U = 1 2 2 C1 + C 2

+ + + + + + + +

+ + + + + + + +

C2

C1

Q

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

Q Q ⇒C= = 4πε0a V 4πε 0 a

C1

+ + + + + + + +

C" =

Q V

n

=

Q ⇒ Cn >> C V − V−

So, if an identical earthed conductor is placed in the viscinty of a charged conductor then the capacitance of the charged conductor increases appreciably. This is the principle of a parallel plate capacitor. 19

JULY 2010

Parallel Plate Capacitor : B

A

A+

+σ

–σ

+ + + + + + + +

+ A = Area of plate + d = Separation + between the + + plates + +

– – – – – – – –

Special Case II : When the space between the parallel plate capacitor is partly filled by a conducting slab of thickness t( > S(liquid state) > (solid state) Expression of Entropy Function For a system which involves transferring infinitesimal heat at constant temperature, the entropy change of the system is given by dq rev dS = T

∆fusH = 6 kJ mol–1 Enthalpy of vaporization :

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∑

v g ,i ( products)

37

JULY 2010

here γ = Cp,m/Cv,m The symbols Cp,m and Cv,m represent molar heat capacities at constant pressure and volume conditions, respectively. For a monatomic ideal gas: Cv,m = (3/2)R; Cp,m = (5/2)R; and γ = 5/3 For a diatomic ideal gas: Cv,m = (5/2)R; Cp,m = (7/2)R; and γ = 7/5

For finite heat transferred at constant temperature, we have q ∆S = rev T For example, for a pure substance we have ∆ vap H ∆ H and ∆fusS = fus ∆vapS = Tb Tm where the subscripts vap and fus represent vaporization and fusion, respectively. Gibbs Function Gibbs function (or energy) or simply free energy is defined as G = H – TS For a process occurring at constant T and P, the change in Gibbs function is given by ∆G = ∆H – T ∆S For a process to be spontaneous, the value of ∆G is negative. For a nonspontaneous reaction, ∆G is positive. For a reaction at equilibrium, ∆G = 0 and temperature at which the system occurs at equilibrium is given by Teq = ∆H/∆S Pressure-Volume Work An ideal gas can undergo expansion of compression under isothermal or adiabatic conditions. The expansion ant compression may be carried out under reversible or irreversible conditions. We give below the expressions of p-V work under different conditions. Isothermal p–V Work In this case, temperature of the system remains constant, ie. ∆T = 0 For irreversible condition: w = – Pext (V2 – V1) For reversible condition: w = – nRT In (V2/V1) Adiabatic p–V Work In this case, heat can neither enter to or leave from the system, i.e. q = 0. From first law of thermodynamics, it follows that ∆U = w where ∆U is given by ∆U = Cv(T2 – T1) For a gas undergoing adiabatic irreversible volume change, the expression of work is given by w = – Pext (V2 –V1) For an ideal gas undergoing adiabatic reversible expansion/compression, we also have pVγ = constant pTγ(1–γ) = constant and TVγ–1 = constant

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CHEMISTRY JOKES If you didn't get the joke, you probably didn't understand the science behind it. If this is the case, it's a chance for you to learn a little chemistry.

Chemistry Joke 1: Outside his buckyball home, one molecule overheard another molecule saying, "I'm positive that a free electron once stripped me of an electron after he lepton me. You gotta keep your ion them."

Chemistry Joke 2: A chemistry professor couldn't resist interjecting a little philosophy into a class lecture. He interrupted his discussion on balancing chemical equations, saying, "Remember, if you're not part of the solution, you're part of the precipitate!"

Chemistry Joke 3: One day on the Tonight Show, Jay Leno showed a classified add that read: "Do you have mole problems? If so, call Avogadro at 602-1023."

Chemistry Joke 4: A student comes into his lab class right at the end of the hour. Fearing he'll get an "F", he asks a fellow student what she's been doing. "We've been observing water under the microscope. We're suppose to write up what we see." The page of her notebook is filled with little figures resembling circles and ellipses with hair on them. The panic-stricken student hears the bell go off, opens his notebook and writes, "During this laboratory, I examined water under the microscope and I saw twice as many H's as O's."

38

JULY 2010

UNDERSTANDING

U n d e r s t a n d i n g

Organic Chemistry

1.

An organic compound (A), C10H15N, undergoes carbylamine reaction but no diazotization. It reacts with HNO2 giving off N2 and a compound (B), C10H14O. (B) reacts with Lucas reagent immediately, but no colour in Victor meyer's test. (B) on heating with conc. H2SO4 eliminates water to give (C), C10H12, which decolourises Br2/CCl4 and cold dilute neutral KMnO4 solution. (C) on ozonolysis gives (D), C7H6O and (E), C3H6O. Compound (E) on heating with I2 and NaOH produced yellow precipitate and sodium acetate. Compound (D) reacts with conc. NaOH to give (F) and (G). Compound (G) on heating with sodalime gives benzene. Compound (F) gives a red colour with ceric ammonium nitrate, and on oxidation and heating the product with sodalime produced benzene. What are (A) to (G) ?

H C–CH OH

(A)

HNO2 –N2;–H2O

C10H14O

Conc. H2SO4

(B)

(B)

(A)

Two isomeric compounds (A) and (B) have the molecular formula C7H9N. (A) being soluble in water, the solution being alkaline to litmus It does not undergoes diazotization, but show carbylamine reaction and mustard oil reaction, it reacts with acetyl chloride and acetic anhydride. Its product with benzene sulphonyl chloride dissolves in KOH. (B) on the other hand, does not dissolve in water, but undergoes diazotization. Its product with C6H5SO2Cl dissolves in KOH. Its salt undergo hydrolysis in aqueous solution showing an acidic test. What are (A) and (B) ? Sol. As both (A) and (B) give products with C6H5SO2Cl, which are soluble in KOH, they contain –NH2 group. (B) can be diazotized so contains – NH2 in the nucleus. (A) cannot be diazotized, hence contains –NH2 in the side chain. The number of carbon and hydrogen atoms also indicates aromatic character. On the basis of above considerations we may show that (A) is benzylamine and (B) o–, m– or p-toludine. CH2NH2 NaNO2 + HCl CH2OH 2.

(I) O3

Conc. NaOH

(C)

C7H6O + C3H6O

(II) H2/Pd

C7H6O

(F) + (G)

(D)

(E)

Sodalime

C6H6

(D) [O]

C10H12

∆;–H2O

Product

Sodalime ∆

C6H6

I 2 + NaOH C 3 H 6 O 2 → CHI3 ↓ + CH3COONa (E)

+ 3NaI + 3H2O Since (C) decolourise Br2/CCl4 and KMnO4 colour, hence it has C=C bond. Its ozonolysis gives (D) and (E). Among these (D) undergoes Cannizaro's reaction, while (E) gives iodoform test, hence (D) is benzaldehyde and (E) acetone. Now joining (D) and (E), the structure of (C) can be determined. H C=O + O =C (D)

(E)

Benzylamine (A) CHCl3 + 3KOH

CH3 –2[O] CH3

H C=C (C)

CH2NC

CS2 + HgCl2

CH2NCS

C6H5SO2Cl

CH2NHSO2C6H5

– HCl

CH3

KOH

CH3

–H2O

CH2 N–SO2C6H5 K

Soluble

Since (C) is produced from (B), which is a t-alcohol, as it gives Lucas test immediately, hence (B) is.

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CH3

As (B) is obtained by the action of HNO2 on (A), hence (A) would be H CH3 C–CH CH3 NH2

Sol. The given data are : C10H15N

CH3

C6H5CH2NH2 + HOH → C6H5CH2N+H3OH–

39

JULY 2010

CH3 NaNO2 + HCl NH2

C6H4 (B)

C6H4

CH3

– CH.CH3

N2Cl (A)

C6H5SO2Cl –HCl

CH3

C6H4

KOH

NHSO2C6H5

Cl

[O] –H2O

NKSO2C6H5

C

68.32

H

6.4

Cl

25.26

Relative no. of atoms 68.32 = 5.59 12 6.4 = 6.40 1 25.26 = 0.71 35.5

(C)

(C)

Simplest ratio 5.59 =8 0.71 6.40 =9 0.71 0.71 =1 0.71

4.

Sol.

[O]

( B)

− H 2O

(C)

(iii) (C) reacts with C6H5NHNH2 to give C=O C8H8=N.NHC6H5, hence (C) contains a group. (iv) (C) on heating with I2 + NaOH gives CHI3, hence (C) contains –COCH3 group. Thus (C) is O

(C)

– C=N.NH.C6H5 CH3 Yellow ppt

O + 3H2O +

– COONa

An organic compound (A), C4H9Cl, on reacting with aqueous KOH gives (B) and on reaction with alcoholic KOH gives (C) which is also formed on passing vapours of (B) over heated copper. The compound (C) readily decolourises bromine water. Ozonolysis of (C) gives two compounds (D) and (E). Compound (D) reacts with NH2OH to gives (F) and the compound (E) reacts with NaOH to give an alcohol (G) and sodium salt (H) of an acid. (D) can also be prepared from propyne on treatment with water in presence of Hg++ and H2SO4. Identify (A) to (H) with proper reasoning. C4H9Cl

Alc. KOH ∆; –KCl

(A) (Alkyl halide) Aq.KOH ∆; –KCl

C4H8

(C) (Alkene) Cu C4H9OH ∆; –H2O (B) (Alcohol)

We know that p-alcohol on heating with Cu gives aldehyde while s-alcohol under similar conditions gives ketone. Thus, (B) is a t-alcohol because it, on heating with Cu gives an alkene (C). Since a talcohol is obtained by the hydrolysis of a t-alkyl halide, hence (A) is t-butyl chloride. Cl OH | | (A) = CH 3 − C − CH 3 and (B) = CH 3 − C − CH 3 | | CH 3 CH 3

– C – CH3

(v) Oxidation of (B) gives (C), hence (B) is a secondary alcohol, i.e., – CH – CH3 OH (vi) (B) is obtained by the hydrolysis of (A), hence it is : – CH.CH3

The alkene (C) on ozonolysis gives (D) and (E), hence (C) is not symmetrical alkene. In these compound (E) gives Cannizaro's reaction with NaOH. So, (E) is an aldehyde which does not contain α–H atom. Hence it is HCHO. Compound (D) can also be prepared by the hydration of propyne in the presence of acidic solution and Hg++.

Cl 1-Chloro-1-phenyl ethane

Now, different reactions are as follows :

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– COCH3

∆ – C – CH3 + 3I2 + 4NaOH → CHI3 + 3NaI

C8 H 9 Cl → C8 H 9 OH → C8 H 8 O (A)

CH3 –H2O

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

OH

(B)

– C = O + H2 N . NH . C6H5

An organic compound (A) of molecular weight 140.5, has 68.32% C, 6.4% H and 25.26% Cl. Hydrolysis of (A) with dilute acid gives compound (B), C8H10O. (B) can be oxidised under milder conditions to (C), C8H8O. (C) forms a phenyl hydrazone (D) with C6H5NHNH2 and gives positive iodoform test. What are (A) to (D) ? Sol. (i) Calculation of empirical formula of (A) 3.

%

– CH–CH3

CH3

C6H4

Soluble

Element

HOH/H+ –HCl

40

JULY 2010

A hydrocarbon (A) of the formula C8H10, on ozonolysis gives compound (B), C4H6O2, only. The compound (B) can also be obtained from the alkyl bromide, (C) (C3H4Br) upon treatment with Mg in dry ether, followed by treatment with CO2 and acidification. Identify (A), (B) and (C) and also equations for the reactions.

5.

++

CH3 – C ≡ CH + H2O Hg → CH 3 − C = CH 2 H+ | OH → CH 3 − C − CH 3 || O ( D)

Hence (D) is acetone and (E) is formaldehyde. Therefore, alkene (C) is 2-methyl propene. (CH3)2–C=CH2 (D) reacts with hydroxyl amine (NH2OH) to form oxime (F). CH3 –H2O CH3 C = O + H2 NOH C = NOH CH3 CH3 (D)

(ii ) H 2 O

If it was alkene its formula should be C8H16 (CnH2n), and if it was alkyne it should have the formula C8H14; it means it is neither a simple alkenen or simple alkyne. However it is definite that the compound has an unsaturated group, it appears that it is a cyclosubstituted ethyne. 2H H – C ≡ C – H − → C3H5 – C ≡ C – C3H5

+ C 6 H10

the C3H5 – corresponds to cyclopropyl (∆) radical, hence compund (A) is CH2 CH2 CH – C≡C – CH CH2 CH2

Reactions : OH Cl | | CH 3 − C − CH 3 CH 3 − C − CH 3 Aq.KOH → | | ∆ ;– KCl CH 3 CH 3

1,2-dicyclopropyl ethyne

The ozonolysis of above compound would give two moles of cyclopropane carboxylic acid (C4H6O2). CH2 CH2 (i) O3 CH – C≡C – CH CH2 CH2

( B)

300 º C Cu / → CH 3 − C = CH 2 + H 2 O − H 2O | CH 3

(A)

( C)

CH2

Alc.KOH / ∆

→ CH 3 − C = CH 2 − KCl; − H 2 O | CH 3

CH2 CH2

(C)

O

CH3

(I) O3

CH3

CH2

CH3

C = O + H2NOH (D)

O

CH3 CH3

CH – C – C – CH O

O

C = NOH

CH2 CH2

CH2

CH2 CH2

CH – Br

Mg ether

CH2 CH2

(H)

+ H2O2

CH – COOH (B)

O

CH . MgBr

C=O ∆

Cyclopropyl magnesium bromide

O

CH2

Hg + +

CH3 – C ≡ CH + H2O → CH3 – C – CH3 + H

CH2

H2O Warm

Compound (B) is prepared from cyclopropyl bromide as follows :

2HCHO + NaOH → CH 3OH + HCOONa (G )

CH2

O

2

(F) (E)

(A)

(D)

∆ –H2O

CH2

CH – C — C – CH

(E)

(C)

CH3

O

C=O+H–C–H

(II) H2O/Zn CH3

or a – C ≡ C – bond.

C=C

should have either

OH Cl | | (B) = CH 3 − C − CH 3 and (A) = CH 3 − C − CH 3 | | CH 3 CH 3

CH3 – C = CH2

( B)

Since compound (A) adds one mol of O3, hence it

(F)

(A)

(i ) O

A(C8H10) 3 → C 4 H 6 O 2

Sol.

CH2

(D)

CH .COOMgBr

HOH dil. HCl; –MgBrOH

CH2 CH2

CH–COOH

Addition compound

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41

JULY 2010

`tà{xÅtà|vtÄ V{tÄÄxÇzxá

Set

3

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.

2.

3.

Let y = f(x) be a curve satisfying dy – y ln 2 = 2sin x (cos x – 1). ln2, then dx (A) y is bounded when x → ∞ (B) f(x) = 2sin x + c . 2x, where c is an arbitrary constant (C) y = 2sinx, y is bounded when x → ∞ (D) f(x) = 2sinx does not have any solution if y is not bounded. In a right angled triangle the length of its hypotenuse is four times the length of the perpendicular drawn from its orthocentre on the hypotenuse. The acute angles of the triangle can be π π π 3π (A) , (B) , 6 3 8 8 π π π 5π (D) , (C) , 6 4 12 12

3, 2 − 3

(D) None of these

4.

If c1 is a fixed circle and c2 is a variable circle with fixed radius. The common transverse tangents to c1 and c2 are perpendicular to each other. The locus of the centre of variable circle is : (A) circle (B) ellipse (C) hyperbola (D) parabola

5.

The length of the latus rectum of the parabola 169 {(x – 1)2 + (y – 3)2} = (5x – 12y + 17)2 is – 14 56 28 (A) (B) (C) (D) None 13 13 13

6.

Evaluate :

∫

cos 5x + cos 4 x dx 1 − 2 cos 3x

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

8.

Given the base of a triangle and the sum of its sides prove that the locus of the centre of its incircle is an ellipse.

9.

A bag contains 7 tickets marked with the number 0, 1, 2, 3, 4, 5, 6 respectively. A ticket is drawn and replaced. Then the chance that after 4 drawings the sum of the numbers drawn is 8, is –

10. A polynomial in x of degree greater than 3 leaves remainders 2, 1 and – 1 when divided by (x – 1), (x + 2) and (x + 1) respectively. What would be the remainder if the polynomial is divided by (x2 – 1) (x + 2) ?

Let a, b ∈ R such that 0 < a < 1 and 0 < b < 1. The values of a and b such that the complex number z1 = –a + i, z2 = –1 + bi and z3 = 0 form an equilateral triangle are (A) 2 − 3 , 3 (B) 2 − 3 , 2 − 3

(C)

7.

42

•

William Bottke at Cornell University in the US has calculated that at least 900 asteroids of a kilometre or more across regularly sweep across Earth's path.

•

The Dutch astronomer Christiaan Huygens (1629 - 1695) drew Mars using an advanced telescope of his own design. He recorded a large, dark spot on Mars, probably Syrtis Major. He noticed that the spot returned to the same position at the same time the next day, and calculated that Mars has a 24 hour period. (It is actually 24 hours and 37 minutes)

•

Space debris travels through space at over 18,000 mph.

•

The nucleus of Comet Halley is approximately 16x8x8 kilometers. Contrary to prior expectations, Halley's nucleus is very dark: its albedo is only about 0.03 making it darker than coal and one of the darkest objects in the solar system. JULY 2010

MATHEMATICAL CHALLENGES SOLUTION FOR JUNE ISSUE (SET # 2) 1.

2.

3.

1st box can be filled in 4 ways. Next each box can be filled in 3 ways (except the ball of colour in previous box). Hence the required no. of ways = 4 × 35 = 972

Hence plane given by (3) is bisecting the acute angle between given two planes also. Hence the conclusion holds true. 5.

AA–1 = I ⇒ (AA–1)T = IT (A–1)TAT = I (as A is symmetric) (A–1)T A = I so by the definition of inverse A–1 = (A–1)T Hence A–1 is also symmetric. Given |A| ≠ 0;

The normal to hyperbola at the point P(a sec θ, b tan θ) is ax cos θ + by cot θ = a2 + b2 If it passes through (h, k) then a h cos θ + b k cot θ = a2 + b2

I2 =

Let ⇒

f–1(y) = x f(x) = y

((f −1 ( y)) 2 − a 2 ) dy

( x 2 − a 2 ) f´(x) dx

(

) ∫ b

I2 = ( x 2 − a 2 ) f ( x ) a – = (b2 – a2) f(b) – ...(1)

∫

=

Hence

6.

b

a

∫

=

Σ z1z2 = 0 = Σ e i ( θ1 + θ 2 ) = 0 Σ (cos (θ1 + θ2) + i sin (θ1 + θ2)) = 0 Hence Σ cos(θ1 + θ2) = Σ sin(θ1 + θ2) = 0

2x f(b) dx –

b

a

∫

b

a

∫

a

2x f(x) dx

2 x f(x) dx

b

a

b

2x f(x) dx

2 x (f(b) – f(x)) dx

I1 1 = I2 2

1 =2 y

y+

⇒ y=1 1 x+ = x

Planes are – x – 2y – 2z + 9 = 0 ....(1) and 4x – 3y + 12z + 13 = 0 ...(2) The plane bisecting the angle b/w these planes containing origin is − x − 2 y − 2z + 9 4 x − 3y + 12z + 13 =+ 3 13 i.e. 25x + 17y + 62z – 78 = 0 ...(3) If θ be the angle between (1) & (3) then 61 cos θ = 4758

⇒ x2 +

5+2

1 x2

= ( 5 + 2) – 2 =

x4 +

1 x4

⇒ x8 + ⇒ x16 +

5

=5–2 1 x8 1

=9–2

= 49 – 2 x16 ⇒ 47 + 1 + 1 = 49

1037 ∆Uac > ∆Uad > ∆Uac (B) ∆Uab = ∆Uac > ∆Uad > ∆Uac (C) ∆Uab = ∆Uac = ∆Uad = ∆Uac (D) ∆Uab < ∆Uac < ∆Uad < ∆Uac

x

(c)

The above graphs shows conduction of heat through materials A,B,C connected in series. Graph shows variations of temperature with distnace x-axis. Which of the above graph are not possible -

An ideal gas whose adiabatic exponent is γ is expanded so that the amount of heat transferred to the gas is equal to the decrease of its internal energy. Molar heat capacity of the gas for this process is -

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x

(a)

b c

2.

R γ –1 (D) R/2

(B)

3.

The figure shows two isotherms at temperatures T1 and T2. A gas is taken from one isotherm to another isotherm through different processes. Then change in internal energy ∆U has relation – P

R 1– γ (C) R

(A)

(A) a, b, c (C) a, c

52

(B) a, b (D) b, c

JULY 2010

4.

Figure shows a rectangular pulse and a triangular pulse approaching each other along x-axis. The pulse speed is 0.5 cm/s. What is the resultant displacement of medium particles due to superposition of waves x = 0.5 cm and t = 2 sec. y (cm) 0.5 cm/s 0.5 cm/s 2

9.

1 –2 –1 (A) 3.5 cm (C) 4 cm

5.

6.

0

Passage: II (Ques. 10 to 12)

x (cm)

2 1 3 (B) 2.5 cm (D) 3 cm

Two hydrogen like atoms A and equal number of protons and neutrons. The energy difference between the radiation corresponding to first Balmer lines emitted A and B is 5.667 eV. When the atoms A and B moving with the dame velocity, strikes a heavy target they rebound back with the same velocity. In this process the atom B imparts twice the momentum to the target than the A imparts.

Choose the correct statement (s) related to the photocurrent and the potential difference between the plate and the collector (A) Photocurrent always increase with the increase in potential difference (B) when the potential difference is zero, the photocurrent is also zero (C) Photocurrent attain a saturation value of some positive value of the potential difference (D) None of these

10. Ionization energy of Atom B is (A) 27.2 eV (B) 13.6 eV (C) 10.2 eV (D) 54.4 eV 11. Atomic number of atom A is (A) 1 (B) 2 (C) 3 (D) 4

Binding energy per nucleon of 1H2 and 2He4 are 1.1 MeV and 7.0 MeV respectively. Energy released in the process 1H2 + 1H2 = 2He4 is (A) 20.8 MeV (B) 16.6 MeV (C) 25.2 MeV (D) 23.6 MeV

12. Mass number of atom B & atom A (A) 2, 4 (B) 4, 2 (C) 2, 1 (D) 4, 1

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 : I (Ques. 7 to 9)

7.

(A) (C) 8.

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 DS, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

Two parallel plates in vacuum, separated by a small distance which is small compared with their linear dimensions, are at temperatures T1 & T2 respectively (T1 > T2). The plates are black bodies. Another plate (black body) at temperature T0 is a kept in between the two plates. Temperature of the plate kept i.e. T0 is T14

+ T24 2

= T04

T1T2 = T0

(B)

T14

– T24 2

P Q R S T A B C D

= T04

(D) 2T14 – T24 = T04

(C) σ

(T14 – T24 ) 2

XtraEdge for IIT-JEE

(B) σ

(T14 – T04 ) 2

(D) σ

(T04 – T24 ) 2

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.

Energy absorbed by the plate having temperature T0, per sec per unit area is (A) σ(T14 –T24)

If 'n' black Body plates are placed in between the two plates having temperature T1 & T2, then the net rate of emission of radiant energy by first plate is σ σ (A) (T14 – T24) (B) (T14 – T24) n n +1 n σ (T14 – T24) (D) (T14 – T24) (C) σ n –1 n +1

53

JULY 2010

13. Match the Column-I with Column-II : Column-I Column-II (A) An electron moves (P) Total Energy Potential Energy in an orbit in a = 2 Bohr atom (B) As a satellite moves (Q) Kinetic Energy = in a circular orbit Magnitude of Energy around total earth (C) In Rutherford's (R) Motion is under a α-scattering experiment central force as an α-particle moves in the electric field of a nucleus (D) As an object, released (S) Mechanical energy from some height above is conserved ground, falls towards earth, assuming negligible air resistance (T) None of these 14. Match the following : Column-I` Column-II (A) Steady state (P) A blackened platinum wire, when gradually heated appear first red and then blue. (B) Wein's displacement (Q) Radiated power is law proportional to fourth power of absolute temperature of body (C) Stefan's law (R) Energy absorbed is equal to energy emitted (D) Black body (S) Absorptive power of body is unity (T) None of these 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

0 1 2 3 4 5 6 7 8 9

XtraEdge for IIT-JEE

Y 0 1 2 3 4 5 6 7 8 9

Z 0 1 2 3 4 5 6 7 8 9

15. Four cylindrical rods of same material with length and radius (l,r), (2l,r), (2l,2r) and (l,2r) are connected between two reservoirs at 0ºC and 100ºC. Find the ratio of the maximum to minimum rate of conduction in them. 16. Using solar constant S = 20 kilo cal/mm –m2 and Joule's constant J = 4.2 J/cal, find the pressure exerted by sunlight ? (Ans. in …… × 10–6 N/m2) 17. The minimum intensity of audible sound is 10–12 W/m2 sec and density of air is 1.3 kg/m3. If the frequency of sound is 1000 Hz, find the amplitude (Ans. in …… × 10–11 m) of vibration ? [Speed of sound = 332 ms–1] 18. The size of a nucleus is of the order of –14 m. Calculate the velocity with which protons move inside the nucleus. The mass of a proton = 1.675 × 10–27 kg. [Ans. in …… × 107 ms–1] 19. Find the change in frequency of red light whose original frequency is 7.3 × 1014 Hz when it falls through 22.5 m losing gravitational potential energy. [Ans. in …… ×103]

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. The optically active tartaric acid is named as D–(+)– tartaric acid because it has a positive (A) optical rotation and is derived from D-glucose (B) pH in organic solvent (C) optical rotation and is derived from D–(+)– glyceraldehyde (D) optical rotation only when substituted by deuterium 2. Which of the following compounds is not coloured ? (A) Na2[CuCl4] (B) Na2[CdCl2] (D) K3[Fe(CN)6] (C) K4[Fe(CN)6] 3.

W 0 1 2 3 4 5 6 7 8 9

4.

54

The brown ring complex compound is formulated as [Fe(H2O)5(NO)]SO4. The oxidation state of iron is (A) 1 (B) 0 (C) 2 (D) 3 The following equilibrium is established when HCl is dissolved the acetic acid, Cl– + CH3COOH2+ HCl + CH3COOH the set that characterises the conjugate acid-base pairs is(A) (HCl, CH3COOH) and (Cl–, CH3COOH2+) (B) (HCl, CH3COOH2+) and (CH3COOH, Cl–) (C) (CH3COOH2+) HCl) and (Cl–, CH3COOH) (D) (HCl, Cl–) and (CH3COOH2+, CH3COOH) JULY 2010

5.

6.

Pure ammonia is placed in a vessel at a temperature where its dissociation constant(α) is appreciable. At equilibrium (A) kp does not change significantly with pressure (B) does not change with pressure (C) concentration of NH3 does not change with pressure (D) concentration of hydrogen is less than that of nitrogen

9.

Passage: II (Ques. 10 to 12)

The property of hydrides of p-block elements mostly depend on (i) Electronegatively difference between central atom and hydrogen (ii) Size of central atom (iii) Number of valence electrons in central atom. some undergo hydride in which central atom is less electronegative, react with OH– to given hydrogen while acidic property of hydride in a period depends on electronegativity of central atom i.e. more electronegative is the atom, more acidic is hydride. In a group, acidic property is proportional to size of central atom. Some electron deficient hydride behaves as lewis acid while only one hydride of an element in p-plock behaves as lewis base with lone pair of electrons. Hydrides in which central atom's electronegativity to close to hydrogen has no reaction with water.

Passage : I (Ques. 7 to 9)

The IUPAC has set guidelines for logical and methodical naming of organic compounds. The complex substituents are written in small brackets and their numbering is done separately. The bivalent radicals are named by adding 'idene' to the name of alkyl group. In polyfunctional compounds all lower priority groups are written in prefix. Now name the following compounds.

10. The hydride which do not react with water is (A) NH3 (B) PH3 (C) B2H6 (D) AsH3 11. Which one undergoes spontaneous combustion with exposure to air ? (A) PH3 (B) P2H4 (C) N2H4 (D) NH3

CHCHCH2OH is -

12. Which one is strongest base ? (A) OH– (B) HS– (C) HSe–

Br

(A) 3-(3'-isopropoxycarbonyl cyclopentylidene propane-1-ol (B)3-(2'-bromo-3'-hydroxypropylidene) cyclopentane carboxylate (C) Iso-propyl-3-(2'-bromo-3'-hydroxy propylidenyl) cyclopentane carboxylate (D) Iso-propyl-3-(2'-bromo-3' hydroxypropylidene) cyclopentane carboxylate

CH3CH2O

CH3CH2

is -

P Q R S T A B C D

(A) 2-(3'-Ethylphenyl)-1-(4'-ethoxyphenyl) ethane (B) 1-Ethyl-3-(2'-(4''-ethoxyphenyl) ethyl) benzene (C) 1-(3'-Ethylphenyl)-2-(4'-ethoxylphenyl) ethane (D) None of these

XtraEdge for IIT-JEE

(D) HTe–

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 DS, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

C2H5

8.

is -

O (A) 3-chlorocarbonyl-6-(N, N-diethylamino) hex-4ene-1-oic acid (B)4-chlorocarbonyl-3-(N, N-diethylamino) butanoic acid (C) 3-chlorocarbonyl-3-(3-N, N-diethylamino prop1'-enyl) butane-1-oic acid (D) 3-chlorocarbonylmethyl-6- (N, N-diethylamino) hex-4-en-1-oic acid

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.

(CH3)2CHOOC

N OH

O

At constant temperature, the equilibrium constant (kp) 2NO2 is for the decomposition reaction N2O4 expressed by kp = (4x2P)/(1–x2), where P = pressure, x = extent of decomposition. Which one of the following statement is true ? (A) kp increases with increase of P (B) kp increases with increase of x (C) kp increases with decrease of x (D) kp remains constant with change in p and x

7.

Cl

55

P P P P

Q Q Q Q

R R R R

S S S S

T T T T

JULY 2010

16. Calculate the change in pressure (in atm) when 2 mole of NO and 16 gram O2 in a 6.25 litre originally at 27ºC react to produce the maximum quantity of NO2 possible according to the equation – 2NO(g) + O2(g) → 2NO2(g) 1 (Take R = ltr. Atm/mol-K) 12

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 extraction processes listed in Column-I with metals listed in Column-II : Column-I Column-II (A) Self reduction (P) Lead (B) Carbon reduction (Q) Silver (C) Complex formation (R) Copper and displacement by metal (D) Decomposition of (S) Boron iodide (T) Au

17. the number of isomers for the compound with molecular formula C2BrCIFI is. 18. In P4O10 each P atom is linked with ...........O atoms. 19. 0.15 mole of Pyridinium chloride has been added into 500 cm3 of 0.2 M pyridine solution. Calculate pH (Kb for pyridine = 1.5 × 10–9 M)

MATHEMATICS

14. Match statements in Column-I with appropriate solution in Column-II Column-I Column-II (A) A solution having (P) 5M HCl pH less then 7 (B) A solution having (Q) 1M NaCl pH more than 7 (C) A solution having (R) 0.1 M Na2CO3 pH almost equal to 7 (D) solution having (S) 0.1 M CaCl2 negative value of pH (T) 1M H2SO4

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.

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

n

2.

If

∑α

n

= an2 + + bn, where a, b are constants and

n =1

α1, α2, α3 ∈ {1, 2, 3........9} and 25α1, 37α2, 49α3 be α2 α3 α1 three digit number then 5 7 9 is 25α1 37α 2 49α 3 equal to (A) α1 + α2 + α3 (C) 7

15. A solution is prepared by mixing 50 mL 0.1 M HCl with 50 mL 2.9 M CH3CH2COOH and 100 mL 0.2 M CH3CH2COONa. Find pH of the resulting solution. Ka for CH3CH2COOH is 1 × 10–5.

XtraEdge for IIT-JEE

α 0 α If A = 2β β – β is an orthogonal matrix, then γ – γ γ the number of possible triplets (α,β,γ) is (A) 8 (B) 6 (C) 4 (D) 2

56

(B) α1 – α2 + α3 (D) 0

3.

Consider the matrix A, B, C, D with order 2 × 3, 3 × 4, 4 × 4, 4 × 2 respectively. Let x = (αABγC2D)3 where α & γ are scalars. Let |x| = k|ABC2D|3, then k is (A) αγ (B)α2γ2 4 4 (C) α γ (D) α6γ6

4.

Three numbers are selected at random from the set {1, 2, 3...........N}, one by one without replacement. If the first number is known to be smaller than second, then the probability that third selected number lies between the first two numbers is JULY 2010

(A)

1 2

(B)

1 3

(C)

1 6

(D)

Passage: II (Ques. 10 to 12)

1 8

consider the equation of two straight lines →

5.

6.

→

→

→

→

→

→

d

is

5 (B) 9

(A)

5 (C) 12

9.

^

^

^

^

2 5

1 (C) 2

→

^

^

^

→

^

^

^

(D)

^

^

^

^

^

^

(B) 2 29

29

(D) 4 29

(D) None of these

P Q R S T P Q R S T P Q R S T P Q R S 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. A B C D

1 4

13. Match the following : Column-I (A) Through (γ, γ + 1) there can not be more than one normal to parabola y2 = 4x if (B) If two cicles (x–1)2 + (y–1)2 = γ2 and x2 + y2 – 8x + 2y + 8 = 0 intersect at two points, then

2 (D) 3

Expectation of the number on the card is (A) 2 (B) 2.5 (C) 3 (D) 3.5

XtraEdge for IIT-JEE

^

P Q R S T

1 and then 10 a card is drawn. Let Ei represents the event that a card with number 'i' is drawn.

P(A3/E2) is equal to 1 1 (A) (B) 3 4

^

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 DS, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

the probability of selection of box Ai is

8.

^

(C) (–10, 53, –4)

There are four boxes A1, A2, A3 and A4. Box Ai has i cards and on each card a number is printed, the numbers are from 1 to i, A box is selected randomly,

(C)

^

12. The point of intersection of the lines is – 10 53 – 4 10 – 53 4 , , , (A) (B) , 3 3 3 3 3 3

5 (D) 18

Passage : I (Ques. 7 to 9)

P(E1) is equal to 1 1 (A) (B) 5 10

^

(C) 3 29

→ ^

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.

7.

^

11. The length of shortest distance between the lines is -

^

then c . i + c . j + c . k is equal to 5 (A) 3

^

^

(D) r = ( 4 i + 6 j + 8 k ) – δ( i + j + k )

→

→ → i j k – + and the angle between a and b is 30º, 6 3 3

→ ^

^

(C) r = ( 3 i + 5 j + 7 k ) – δ(4 i + 6 j + 8 k )

perpendicular to all of them. If ( a × b ) × ( c × d ) =

→ ^

^

(B) r = ( i + j + k ) + δ(3 i + 5 j + 7 k )

→

respectively are coplanar. A unit vector

^

^

(A) r = ( 3 i + 5 j + 7 k ) – δ( i + j + k )

Vectors a , b and c with magnitude 2, 3 & 4

^

^

x +1 y +1 z +1 = = . 7 –6 1 10. The equation of the line of shortest distance between the given lines is →

→

^

^

r = 3 i + 5 j + 7 k + λ( i – 2 j + k ) and

A bag 'A' contains 2 white and 3 red balls, another bag 'B' contains 4 white and 5 red balls. If one ball is drawn at random from one of the bag and it is found to be red, the probability that it was drawn from the bag B is 89 25 93 24 (A) (B) (C) (D) 245 52 256 663

57

Column-II (P) 2 < γ < 8

(Q) – 2 < γ < 2

JULY 2010

(C) Point (γ, γ + 1) lies inside the (R) γ < – 2 2 2 circle x + y = 1 for (D) Both equation x2 + y2 + 2γx + 4 = 0 (S) – 1 < γ < 0 and x2 + y2 – 4γy + 8 = 0 represent real circles if (T) γ > 8

15. There are N + 1 identical boxes each containing N wall clocks. The first box contains zero defective clocks. The second box contains one defective and (N – 1) effective clocks, in general rth box contains (r – 1) defective and (N – r + 1) effective clocks (1 ≤ r ≤ N + 1). Thus, the (N + 1)th box contains all defective clocks. A wall clock is selected and found an effective one. The probability that it is from kth γN – γK + γ find γ. box is N2 + N

3 – 4 1 / 2 3 14. Consider the matrix A = ;B= . 1 – 1 0 1 Let P be an orthogonal matrix and Q = PAPT, Rk = PTQk.P, S = PBPT & Tk = PTSkP where k ∈ N Column-I Column-II

16. If a determinant of order 3 × 3 is formed by using the numbers 1 or – 1then it minimum value of determinant is – γ find the value of γ.

5

(A)

∑a

k

, where ak represents the element

(P) –9

17. If equation of the plane through the straight line y+2 z x –1 = = and perpendicular to the plane –3 5 2 x – y + z + 2 = 0 ia ax – by + cz + 4 = 0, then find the 10 3 a + 10 2 b + 10c . value of 342

k =1

of first row & first column in matrix Rk 3

(B)

∑b

k

, where bk represents the element

(Q) 10

k =1

of second row & second column in matrix Rk

3 18. If A = 2 4 3 0 3 2 1 0 4 0 2

3 1 – 1 . Solve the system of equations – 3 2 x 8 2 y y = 1 + z , then find the value z 4 3y z y + of x + 3 2

∞

(C)

∑

X k , where Xk represents the element

(R) 35

k =1

of first row & first column in matrix Tk 10

(D)

∑y

k

, where yk represents the element

(S) 1

k =1

of second row and second column in Matrix TK (T) 15

19. Find the coefficient of x in the determinant

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 0 1 2 3 4 5 6 7 8 9

XtraEdge for IIT-JEE

Y 0 1 2 3 4 5 6 7 8 9

Z 0 1 2 3 4 5 6 7 8 9

–2

(1 + x ) a1b1

(1 + x ) a1b 2

(1 + x ) a1b3

(1 + x ) a 2 b1

(1 + x ) a 2 b 2

(1 + x ) a 2 b3 where ai, bj ∈N

(1 + x ) a 3b1

(1 + x ) a 3b 2

(1 + x ) a 3b3

How to Handle Difficult People A bully at your work is difficult for you to face. He is demanding you do part of his job without pay or credit. How do you handle it?

W 0 1 2 3 4 5 6 7 8 9

Your neighbors are constantly fighting. They wake you up in the middle of the night with their screams and curses. What do you say to them? Your father is unhappy about your career choice. He constantly criticizes your work and points out what he thinks you should do. How do you deal with him?

58

JULY 2010

Based on New Pattern

IIT-JEE 2012 XtraEdge Test Series # 3

Time : 3 Hours Syllabus : Physics : Calorimetry, K.T.G.,Thermodynamics, Heat Transfer, Thermal expansion, Transverse wave,

Sound wave, Doppler's effect. Chemistry : Chemical Equilibrium, Acid Base, Ionic Equilibrium, Classification & Nomenclature, Isomerism , Hydrogen Family, Boron Family & Carbon Family, S-block elements. Mathematics: Point, Straight line, Circle, Parabola, Ellipse, Hyperbola, Vector, 3-D

Instructions : Section - I • Question 1 to 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 Match type questions 8 marks will be awarded for correct answer and 0 mark for wrong answer. Section - IV • Question 15 to 19 are numerical response questions (with single digit Answer). 3 marks will be awarded for correct answer and 0 mark for wrong answer.

(A) 52 W (C) 512 W

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.

2.

3.

4.

In case of standing waves (A) At nodes particles displacement is time dependent (B) At antinodes displacement of particle may or may not be zero (C) Wave does not travel but energy is transmitted (D) Components waves traveling in same direction having same amplitude and same frequency are superimposed

5.

For an ideal gas graph is shown for three processes. Processes 1, 2 and 3 are respectively – Work done (magnitude)

The molar heat capacity for a process is : R α C= + , then process equation is 1– γ T (A) Ve–(α/RT)T = constant (B) Ve (α/R)T = constant (C) VT = constant (D) Veα/RT = constant

3

Three black bodies 1,2,3 have radius r1 < r2 < r3. Emissive powers of black bodies at Temperature T are E1, E2, E3, Then correct relation between them is (A) E1 < E2 < E3 (B) E1 > E2 > E3 (D) E1 > E2 < E3 (C) E1 = E2 = E3

2 1 Temperature change (A) Isochoric, isobaric, adiabatic (B) Isochoric, adiabatic, isobaric (C) isobaric, adiabatic, isochoric (D) Adiabatic, isobaric, isochoric

A taut string for which µ = 5 × 10–2 kg/m is under tension of 80 N. How much is the average rate of transport of potential energy if the frequency is 60 Hz and amplitude 6 cm - (Given 4π2 = 39.5)

XtraEdge for IIT-JEE

(B) 256 W (D) 215 W

59

JULY 2010

6.

Figure shows cyclic process. From c to b, 40 J is transferred as heat from b to a, 130 J is transferred as heat, and work done is 80 J from a to c, 400 J is transferred as heat then – P c

11. Change in internal energy in process CA (A) 900 R (B) 300 R (C) 1200 R (D) zero 12. Heat transferred in the process BC is (A) 1000 R (B) 500 R (C) 2000 R (D) 1500 R

b

a

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 DS, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

V (A) Work done in process a to c is 310 J (B) Net work done is cycle is 230 (C) Net change in internal energy in cycle is 130 J (D) None of these

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.

A B C D

Passage : I (Ques. 7 to 9)

Many waveforms are described in terms of combinations of travelling waves. Superposition principle is used to analyse such wave combinations. Two pulses travelling on same string are described by5 –5 y1 = , y2 = (3x – 4t ) 2 + 2 (3x + 4t – 6) 2 + 2 7.

The direction in which each pulse is travelling is (A) y1 is in positive x-axis, y2 is in positive x-axis (B) y1 is in negative x-axis, y2 is in negative x-axis (C) y1 is in positive x-axis, y2 is in negative x-axis (D) y1 is in negative x-axis, y2 is in positive x-axis

8.

The time when the two waves cancel everywhere (A) 1 sec (B) 0.5 sec (C) 0.25 sec (D) 0.75 sec

9.

The point where two waves always cancel(A) 0.25 m (B) 0.5 m (C) 0.75 m (D) 1 m

14.

C

P0 T

10. Work done in process AB is (A) 400 R (B) – 400 R (C) 200 R (D) – 300 R

XtraEdge for IIT-JEE

T T T T

13. Match the standing waves formed in column-II due to plane progressive waves in Column-I and also with conditions in Column-I Column-I Column-II (A) Incident wave is (P) y = 2A cos kx sin ωt y = A sin (kx – ωt) (B) Incident wave is (Q) y = 2A sin kx cos ωt y = A cos (kx – ωt) (C) x = 0 is rigid support (R) y = 2A sin kx cos ωt (D) x = 0 is flexible support (S) y = 2A cos kx cos ωt (T) None of these

One mole of monoatomic gas is taken through above cyclic process. TA = 300 K Process AB is defined as PT = constant P B

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.

Passage: II (Ques. 10 to 12)

3P0

P Q R S Q R S Q R S Q R S Q R S

P P P P

60

Column-I (A) Specific heat capacity S

Column-II (P) l1 – l2 = constant for l1α1 = l2α2 (B) Two metals (l1, α1) and (Q) Y is same (l2, α2) are heated uniformly (C) Thermal stress (R) S = ∞ for ∆T = 0 (D) Four wires of same (S) Y α ∆t material (T) None of these

JULY 2010

1.

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

(A) 2-Bromo-4-carbamoyl-5-chloroformyl-3-formyl hexanoic acid (B) 5-Bromo-3-carbamoyl-2-chloroformyl-4-formyl hexanoic (C) 4-Formyl-2-chloroformyl-3-carbamoyl-5-bromo hexanoic acid (D) 2-Chloroformyl-3-carbamoyl-4-formyl-5-bromo hexanoic acid 2.

Geometry in the given compound is CH3 H H

CH3

(A) cis (B) trans (C) cis as well as trans (D) no geometrical isomerism

15. If the volume of a block of metal changes by 0.12 % when heat is changed from 40ºC to 60ºC, find the linear expansion coefficient of the metal ? [Ans. in …… × 10–5/ºK]

3.

16. Calculate the pressure exerted by a mixture of 8 g of oxygen, 14 g of nitrogen and 22 g of carbon di-oxide in a container of 30 litres at a temperature of 27ºC. [Ans. in …… × 105 N/m2] 17. A sphere and a cube of same material and total surface area placed in an evacuated chamber turn by turn and heated to the same temperature. Calculate the ratio of the rate of cooling of spherical to cubical surface. [Ans. in …… × 10–1] π 18. Two oscillating waves have a phase difference of 2 is 25 oscillations. What is the percentage difference in their frequency ? 19. For a certain organ pipe, three successive resonance observed are 425, 595 and 765 Hz. Taking the speed of sound to be 340 ms–1 , find the length of the pipe, in metre.

CHEMISTRY

The structure of spiro [3,3] heptane is (A)

(B)

(C)

(D)

4.

The pH of a 10–8 molar solution of HCl in water is (A) 8 (B) – 8 (C) between 7 and 8 (D) between 6 and 7

5.

Consider the following equilibrium in a closed 2NO2(g). At a fixed container N2O4(g) temperature, the volume of the reaction container is halved. For this change, which of the following statements holds true regarding the equilibrium constant (kp) and degree of dissociation (α) ? (A) neither kp nor α changes (B) both kp and α change (C) kp changes, but α does not change (D) kp does not change, but α changes

6.

For H3PO3 and H3PO4 the correct choice is (A) H3PO3 is dibasic and reducing (B) H3PO3 is dibasic and non-reducing (C) H3PO4 is tribasic and reducing (D) H3PO3 is tribasic and non-reducing

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.

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.

XtraEdge for IIT-JEE

The IUPAC name of the given compound OHC CONH2 | | CH3–CH–CH–CH–CH–COOH is | | COCl Br

61

JULY 2010

10. Which one of the following is most stable conformer? Cl Cl CH3 Cl CH3 H (A) (B) H H CH3 CH3 H Cl Cl Cl Cl H Cl (C) (D) H H CH3 CH3 H CH3 CH3

Passage : I (Ques. 7 to 9)

In a reversible chemical reaction, the rate of forward reaction decreases and that of backward reaction increases with the passage of time; at equilibrium the rate of forward and backward reaction become same. Let us consider the formation of SO3(g) in the following reversible reaction : 2SO2(g) + O2(g) 2SO3 (g) Following graphs are plotted for this reactions

A. B. C.

Conc.

t1 t2 t3

9.

(A)

rf

(B)

rb

H

(C)

rb time

(D)

OH OH OH

H

OH

H

OH

(D)

CH3

H

CH3 OH

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 DS, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

rb rf time

Passage: II (Ques. 10 to 12)

A B C D

Different spatial arrangements of the atoms that result from restricted rotation about a single bond are conformers. n-Butane has four conformers eclipsed, fully eclipsed, gauche and anti. The stability order of these conformers are as follows: anti > gauche > partial eclipsed > fully eclipsed Although anti is more stable than gauche but in some cases gauche is more stable than anti.

XtraEdge for IIT-JEE

H

12. Number of possible conformers of n-butane is (A) 2 (B) 4 (C) 6 (D) infinite

time

rf

(B)

CH3

rb rf

rate of reaction

(C)

rate of reaction

time

OH CH3

Which of the following represent the rates of forward reaction (rf) and rates of backward reaction (rb) at equilibrium ? (A)

H CH3

In the above graph, the equilibrium state is attained at time (A) t1 (B) t2 (C) t3 (D) t4

rate of reaction

8.

t4

In the above graph, A,B & C respectively are (A) SO3, SO2 and O2 (B) SO3, O2 and SO2 (C) SO2, O2 and SO3 (D) O2, SO2 and SO3

rate of reaction

7.

time

11. Which one of the following is the most stable conformer ? CH3 CH3 HO H H CH3

P Q R S P Q R S P Q R S P Q R S P Q R S

T 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.

62

JULY 2010

13. Match the following : Column-I (A) N2(g) + 3H2(g) 2NH3(g) ; ∆H = –ve (B) N2(g) + O2(g) 2NO(g); ∆H = +ve (C) A(g) + B(g) 2C(g) + D(g); ∆H = +ve PCl3(g) + Cl2(g); ∆H = +ve (D) PCl5(g) Column-II (P) K increases with increase in temperature (Q) K decreases with increase in temperature (R) Pressure has no effect (S) Product concentration, increases due to addition of inert gas at constant pressure (T) Product concentration, increases due to addition of inert gas at constant volume

R(Gas constant) = Log e = 2.3}

16. Number of configurational dibromocinnamic acid is .

isomers

of

2,3-

17. Consider the reaction AB2(g) ABg + B(g). It the initial pressure of AB2 is 100 torr and equilibrium pressure is 120 torr. The equilibrium constant Kp in terms of torr is. 18. Dissociation of H3PO3 occurs in ......... stages. 19. The number of hydroxyl groups in pyrophosphoric acid is.

14. Match the following : Column-I Column-II 3+ + (A) Bi → (BiO) (P) Heat (B) [AlO2]– → Al(OH)3 (Q) Hydrolysis (C) [SiO4]4– → [Si2O7]6– (R) Acidification (D) [B4O7]2– → [B(OH)3] (S) Dilution by water (T) Basification

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

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

1.

A circle C1 of radius b touches the circle x2 + y2 = a2 externally and has its centre on the positive x-axis; another circle C2 of radius c touches the circle C1 externally and has its centre on the positive x-axis. Given a < b < c, then the three circles have a common tangent if a, b, c are in (A) A.P. (B) G.P. (C) H.P. (D) None of these

2.

P is a point on the axis of the parabola y2 = 4ax; Q and R are the extremities of its latus rectum, A is its vertex. If PQR is an equilateral triangle lying within the parabola and ∠AQP = θ, then cos θ = (A) (C)

3.

15. Calculate the pH at which the following conversion (reaction) will be at equilibrium in basic medium I–(aq.) + IO3– (aq.) I2 (s) When the equilibrium concentrations at 300 K are [I–] = 0.10 M and [IO3–] = 0.10 M. {Given → ∆Gfº (I–, aq.) = – 50 kJ./mol, ∆Gfº (IO3–, aq) = – 123.5 KJ/mol, ∆Gfº (H2O, l) = –233 KJ/mol ∆Gfº (OH–, aq.) = – 150 KJ/mol

XtraEdge for IIT-JEE

25 J/mol–K 3

2 5 5 –2 2 3

(B)

9 8 5

(D) None of these

x2 y2 + = 1, 25 9 perpendicular to the asymptote of the hyperbola x2 y2 – = 1 passing through the first and third 16 9 quadrants is : 100 150 (A) (B) 431 481

The length of the diameter of the ellipse

(C)

63

2– 3

25 3

(D) 11 2

JULY 2010

4.

5.

6.

→

→

→

→

→

9.

→

If a , b , c are such that [ a , b , c ] = 1, → → → → → → → 2π c = γ( a × b ), a ^ b < , and | a | = 2 , | b | 3 → → → 1 = 3,|c |= , then the angle between a and b 3 is π π π π (A) (C) (B) (D) 6 3 4 2

Passage: II (Ques. 10 to 12) →

→ → → → 3 | a || c | a c + → (A) → → → 3 | c | +2 | a | | a | | c | → → a c + → (B) → → → 3 | c | +2 | a | | a | | c | →

p1, p2 are lengths of perpendicular from foci on tangent to ellipse and p3, p4 are perpendiculars from extremities of major axis and p from centre of ellipse p p – p2 on same tangent, then 1 2 equals p 3p 4 – p 2

→ → → → 2 | a || c | a c + → (C) → → → 3 | c | +2 | a | | a | | c | (D) none of these

11. The position vector of point F is -

(B) e (D) None of these

1 | a |→ (A) a + → c 3 |c| 2| a |→ (C) a + → c |c|

XtraEdge for IIT-JEE

→

→

| a |→ (B) a + → c |c| →

→

| a |→ (D) a – → c |c|

→

12. The vector AF , is given by →

1| a |→ (A) c 3 → |c| →

| a |→ (C) 2 → c |c|

→

| a |→ (B) → c |c| →

| a |→ (D) – → c |c|

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 DS, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

If L represents the line joining the point P on C to its centre O, then equation of the tangent at M to the ellipse E is 5 =0

→

→

x2 y2 + = 1, L : y = 2x 9 4 P is a point on the circle C, the perpendicular PQ to the major axis of the ellipse E meets the ellipse at M, MQ is equal to then PQ (A) 1/3 (B) 2/3 (C) 1/2 (D) none of these

(C) x + 3y +

→

→

Let C : x2 + y2 = 9, E :

(A) x + 3y = 3 5

→

| a || c |

Passage : I (Ques. 7 to 9)

8.

→

10. The position vector of point P, is -

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.

7.

→

In a parallelogram OABC vector a , b , c are respectively the position vectors of vertices A, B, C with reference to O as origin. A point E is taken on the side BC which divides it in the ratio of 2 : 1. Also the line segment AE intersects the line besecting the angle O internally in point P. If CP when extended meets AB in point F, then

Equation of a plane which passes through the point of y–2 z–3 x –1 = and intersection of lines = 3 1 2 z–2 x –3 y –1 = = and at greatest distance 3 1 2 from the point (0, 0, 0) is (A) 4x + 3y + 5z = 25 (B) 4x + 3y + 5z = 50 (C) 3x + 4y + 5z = 49 (D) x + 7y – 5z = 2

(A) e (C) e2

Equation of the diameter of the ellipse E conjugate to the diameter represented by L is (A) 9x + 2y = 0 (B) 2x + 9y = 0 (C) 4x + 9y = 0 (D) 4x – 9y = 0

(B) 4x + 3y = 5 (D) 4x +3y + 5 = 0 64

JULY 2010

(D) The radius of the circular section

P Q R S T P Q R S T P Q R S T P Q R S T P Q R S 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 Column-II (A) If lines x + 2y – 1 = 0, ax + y + 3 = 0 (P) 4 and bx – y + 2 = 0 are concurrent, the least distance from origin to A B C D

(a, b) is S. The value of 58 . S is (B) A,B are two fixed points on a line L. Let locus of point P such that PA = 2PB be a curve cutting line 1 L at R and S. If slope of PR is – , 2 then slope of PS is (C) Let tangents at P and Q to curve y2 – 4x – 2y + 5 = 0 intersect at T. If S(2, 1) is a point such that SP.QS = 16 then the length ST is (D) Let the double ordinate PNP' of the hyperbola

x2 (π – 1)

2

–

of the sphere | r | = 5 by the plane →

X

0 1 2 3 4 5 6 7 8 9

(Q) 5

(R) 2

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. Two circle of radii 'a' and 'b' touching externally are

(S) π – 1

inscribed in area bounded by y = 1 – x 2 1 1 x-axis. If b = and a = , then k is ............... 2 k

and

16. If a circle S (x, y) = 0 touches at the point (2, 3) of

the line x + y = 5 and S (1, 2) = 0, then of such circle is.

Column-II (P) 1

(Q)

2 × radius

17. Consider two concentric circles C1 : x2 + y2 = 1, C2: x2 + y2 = 4. Tangents are drawn to C1 from any point P on C2. These tangents again meet circle C2 at A & B. It can be proved that locus of point of intersection of tangents drawn to C2 at A and B is a circle, what is the radius of that circle.

14.

18. In a regular tetrahedron let θ be the angle between any edge and a face not containing the edge. a If cos2θ = where a, b ∈ I+ also a and b are b 5 coprime, then find the value of (10a + b) 13

1 6

the lines

XtraEdge for IIT-JEE

^

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 :

(T) (π–1)2

x –1 z–3 y–2 = = 2 3 4 x–2 x –5 y–4 and = = is 3 5 4 (C) The points (0, –1, –1), (4, 5, 1), (3, 9, 4) and (–4, 4, k) are coplanar then k =

^

(T) 2

preduced both side to meet asymptotes in Q and Q'. The product PQ. PQ' is equal to

(B) The shortest distance between

^

r . ( i + j + k ) = 3 3 is

y2 = 1 is (π – 1)

Column-I (A) The distance of the point (1, – 2, 3) from the plane x – y + z = 5 measured parallel to the line 1 1 1 x = y = – z is 3 2 6

(S) 3

→

19. Let A (1, 2), B (3, 4) be two point and C (x,y) be a point such that (x – 1) (x – 3) + (y – 2) (y – 4) = 0. If area of ∆ABC is 1 sq, unit. Then maximum number of positions of C in xy plane is.

(R) 4

65

JULY 2010

XtraEdge Test Series ANSWER KEY IIT- JEE 2011 (July issue) PHYSICS Ques Ans Column Match Numerical Response

1 C 13 14 Ques Ans

2 3 A B A → P, Q, R, S A→R 15 16 8 5

Ques Ans Column Match Numerical Response

1 C 13 14 Ques Ans

2 3 B C A → P, R A → P, S, T 15 16 4 2

Ques Ans Column Match Numerical Response

1 A 13 14 Ques Ans

2 3 D D A → Q, R, S A→R 15 16 2 4

4 D

17 1

5 6 C D B → P, Q, R, S B→P 18 19 4 2

7 A

8 C C → R, S C→Q

9 B

10 D

11 A D → R, S D → P, S

12 B

7 D

8 B C → Q, T C→Q

9 D

10 D

11 B D→S D→P

12 A

8 B C→S C→S

9 A

10 C

11 12 B D D → P, R, T D→Q

7 C

8 D C → Q, R C→S

9 D

10 B

11 A D → P, S D→Q

12 C

7 A

8 C C → P, S C→P

9 A

10 A

11 C D → P, S D→R

12 D

8 A C→P C→R

9 B

10 A

11 A D→S D→R

12 A

C HE M ISTR Y 4 A

17 6

5 A B → P, R B → R, S 18 4

6 D

19 5

MATHEMATICS 4 C

17 5

5 B B→P B→P 18 3

6 D

7 C

19 0

IIT- JEE 2012 (July issue) PHYSICS Ques Ans Column Match Numerical Response

1 D 13 14 Ques Ans

2 C A → P, R A→R 15 2

Ques Ans Column Match Numerical Response

1 B 13 14 Ques Ans

2 B A→Q A→Q 15 8

3 B

4 B

16 3

17 7

3 D

4 D

16 4

17 5

5 A B → Q, S B→P 18 5

6 A

19 7

C HE M ISTR Y 5 D B → P, R B→S 18 2

6 A

19 4

MATHEMATICS Ques Ans Column Match Numerical Response

1 B 13 14 Ques Ans

2 A A→Q A→P 15 4

XtraEdge for IIT-JEE

3 B

4 B

16 1

17 4

5 B B→R B→Q 18 5

6 C

7 B

19 4

66

JULY 2010

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Teachers open the door. You enter by yourself Volume - 6 Issue - 1 July, 2010 (Monthly Magazine) Editorial / Mailing Office : 112-B, Shakti Nagar, Kota (Raj.) 324009

Editorial

Tel. : 0744-2500492, 2500692, 3040000 e-mail : [email protected] Editor :

Dear Students,

Pramod Maheshwari

Is examination a common cause of stress? In most Asian cultures, the great emphasis on academic achievement and high expectations of success make it especially stressful for students. The strong negative stigma attached to failure also adds to the pressure. Like it or not, we have to accept that examinations are necessary in any educational system. Even though it is debatable whether they are accurate measures of actual ability, no better alternatives have been proposed. Examinations remain necessary to motivate students’ learning, measure their progress and ultimately, serve as evidence of attainment of certain skills, standards or qualifications. Success at examinations provides opportunities to proceed with higher education and improves employment prospects, underlining their importance. No matter how well prepared, many factors may influence one’s performance at the time of the examination and there is seemingly, no definite guarantee of success. Essentially, it is this vital importance attached to success at examinations coupled with the element of uncertainty that makes them so stressful. As with other sources of stress, the stress of examinations is not all bad. It is a strong incentive for students to study and poses a challenge for individual achievement. However, when stress becomes excessive, performance begins to suffer. There is thus a need to control levels of stress before it becomes overwhelming and detrimental. Reliase of stress is necessary for optimum performance, the means of which is relasing. Learn to relax The stress responses produces muscle tension, which you would commonly experience as backache, neck ache or tension headache at the end of the day. Often this is unconscious. So to relax these muscles, you need to consciously practice relaxation exercises. These could involve muscle relaxation, deep breathing exercises, body massage or guided imagery. Like any particular skill, you need to practice them regularly in order to reap the benefits. Another way to relax is to maintain a quiet time as part of the daily routine. Quiet time refers to a time for you with no interruption from external sources or distractions. This is a time where you may choose to just think of nothing and relax. Finally, you can always take up a hobby to help you relax. Do something you enjoy, be it listening to music. Ideally, the drive to study should be internally driven by a desire to achieve one’s own personal goals. Instead, many are driven more by the fear of failure, which is more stress-provoking and leads easily to discouragement. Attending school should not merely revolve around preparation for examinations. Interacting with teachers, socializing with friends, participating in sports or other extra-curricular activities are all valuable aspects of a ‘wellrounded’ education. Instead of wishing things would get easier, start looking at how you can get better...

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Editor : Pramod Maheshwari XtraEdge for IIT-JEE

Yours truly

Pramod Maheshwari, B.Tech., IIT Delhi

1

JULY 2010

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JULY 2010

Volume-6 Issue-1 July, 2010 (Monthly Magazine)

CONTENTS

NEXT MONTHS ATTRACTIONS

INDEX

PAGE

Regulars ..........

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- 2011 & 2012

NEWS ARTICLE

4

IITian ON THE PATH OF SUCCESS

6

KNOW IIT-JEE

7

Are Nanoparticles health hazard Nano is the new black Ms. Padmasree Warrior & Dr. Krishan K. Sabnani Previous IIT-JEE Question

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

• If you don’t notice when you win, you will only notice when you lose.

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

• It’s not bragging if you can do it.

CATALYSE CHEMISTRY

Success Tips for the Months

• Feel the power of yet. As in “I don't know how to do this yet.” • The difficult we do immediately. impossible takes a bit longer.

• To know what you are doing is an advantage. To look like you know what you are doing is essential. • First law of expertise: Never ask a barber if you need a haircut.

DICEY MATHS

42

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

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

• If you think you can, you are probably right. If you think you can't, you are certainly right.

XTRAEDGE TEST SERIES

• Don't do modesty unless you have earned it.

Class XII – IIT-JEE 2011 Paper Class XI – IIT-JEE 2012 Paper

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29

Key Concept Reaction Mechanism Energetics Understanding : Organic Chemistry

The

• Some look down the rapids and see the rocks. Hunters look down the rapids and see the flow around the rocks.

14

3

52

JULY 2010

Are

Nanoparticles

a

Health Hazard? Nanoparticles are a mega business opportunity for multinationals but they may pose a health hazard to users There is a new industrial revolution taking place all around us. The only problem is we can’t see it. The building blocks, being developed at the cost of billions of dollars by scientists, governments and multinational corporations, are just a few atoms or molecules thick — nanoparticles. Many are less than 100 nanometres (nm) — one-billionth of a metre — thick. A single human red blood cell in comparison is around 500 nm in diametre. It’s a pity though that our eyesight isn’t good enough at nanometre level, for if it were, we would see that nanoparticles of precious metals like gold, silver and titanium have already made the jump from research labs to our homes. Manufactured nanoparticles are today present in thousands of consumer products around the world — silver in washing machines and water purifiers to kill bacteria, zinc in cosmetics to protect against ultraviolet rays, carbon nano-tubes in tennis rackets to make them stronger and lighter, titanium in household paints to decompose dust and grime without human intervention.

cannot we write the entire 24 volumes of the Encyclopedia Britannica on the head of a pin?” he asked. “Because there isn’t much of a point, or money, in doing so!” is the answer he would have got today from nanotechnology researchers. Instead their time is mostly spent figuring out newer properties for nanoparticles which can then be embedded into commercial applications. Nanoparticles are highly reactive and prone to unusual properties. Describing gold, a metal that is normally inert to all other chemicals, Prof. C.N.R Rao, Honorary President and Linus Pauling Professor at the Jawaharlal Nehru Centre for Advanced Scientific Research (JNCASR) and the head of its Nanoscience centre, says “At 200-300 nm thickness, gold is not metallic, it does not shine — in fact it is not gold. And at 1.5-2 nm, it reacts like mad!” Gold that is not gold when shrunk to nanometer size might sound like an absurdity to many, but it’s exactly this change in physical properties that make nanoparticles popular. For example zinc oxide (ZnO) and titanium dioxide (TiO2) have been used as active ingredients in sunscreens for decades because of

Silver, an ornamental metal and a powerful bactericide, can be reduced to nanoparticle form to destroy disease-causing bacteria from all kinds of places — kitchen counters, contaminated water, dirty clothes and stinky underarms. Samsung claims its Silver Nano range of washing machines release hundreds of billions of silver nanoions with each wash to kill over 99 percent of the bacteria found in dirty clothes, while the same technology when lined on the doors of their refrigerators kill bacteria that could spoil stored food. Eureka Forbes’ water purifiers use nanosilver-coated filters, developed by Prof. Pradeep, head of IIT-Madras’ Nanoscience department, to destroy harmful bacteria from drinking water. Swach, the mass-market water filter introduced by the Tata Group, also uses nanosilver (coated on rice husk particles) to purify drinking water. During the last flu pandemic threat authorities in Hong-Kong sprayed subways with nanosilver to disinfect them. L’Oreal, the world’s largest cosmetics company, reportedly spends over $600 million each year researching and patenting nanoparticles. The head of its nanotechnology unit also sits on the management board.

Nano is the New Black “There’s Plenty of Room at the Bottom” So went the classic lecture by the Nobel prize-winning physicist Richard Feynman in 1959 that many nano-ficionados now consider the conceptual sun of the nanotechnology universe. “Why

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their ability to absorb ultraviolet rays and reflect back much of the other remaining sunlight. But they are both white — the reason many sunscreens leave a white residue on the face. When shrunk to nanometre size however, they become transparent without losing their light reflecting or absorbing abilities.

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Therefore a metal that is a poor second cousin to gold in a world where we value yellow over white, is the undisputed metal of choice in the nanoparticle world. In fact, silver is more popular than any other material, according to the database of consumer products using nanoparticles maintained by the Woodrow Wilson International Center for Scholars.

Obama admn nominates IIT alumnus for post of NSF Director IIT Madras alumnus Subra Suresh, popularly known as 'Bakthi Suresh' during his student days, has been nominated for the post of director of the National Science Foundation (NSF) by the Barrack Obama administration. An official relese from IIT Madras Alumni Association here said ''when confirmed by the Senate, Mr Suresh will become one of the highest ranking Indian-Americans ever to serve in an administration.'' An Indian-American technocrat, 53-year-old Subra Suresh completed his B.Tech Mechanical Engineering in 1977. Currently the dean of the MIT engineering school, he received the distinguished alumnus award in 1997. In a statement, President Obama said ''I am proud that such experienced and committed individuals have agreed to take on these important roles in my administration. I look forward to working with them in the coming months and years.'' The National Science Foundation is the funding source for nearly 20 per cent of all federally supported basic research and was an independent federal agency created by US Congress in 1950. Subra Suresh has been elected to the US National Academy of Engineering, the Indian National Academy of Engineering,

the American Academy of Arts and Sciences, the Indian Academy of Sciences in Bangalore, the German National Academy of Sciences, the Royal Spanish Academy of Sciences and the Academy of Sciences of the Developing World based in Trieste, Italy.

Major decision regarding Ganga at IIT-K KANPUR: A collaboration between the consortium of seven IITs (IIT-Kanpur, Madras, Bombay, Delhi, Kharagpur, Guwahati and Roorkee) and Ministry of Environment and Forest, Government of India, is being worked out for the purpose of cleaning the national river Ganga. The two are expected to sign an important memorandum of understanding (MoU) in this regard during Prime Minister Manmohan Singh's visit to the Indian Institute of Technology-Kanpur. Singh is scheduled to visit IIT-K on July 3 to take part in its convocation ceremony. It will be worth mentioning here that the Central government aims at meeting the formidable challenge of cleaning the Ganga. With this goal in mind, it had launched a new initiative and established the National Ganga River Basin Authority (NGRBA) last year. The Prime Minister, who is the Chairman of the NGRBA, asked the Ministry of Environment and Forest to involve IITs in the mega project. A joint meeting of all the seven IITs was convened on March 12, 2010 in which IITKanpur was represented by Prof Vinod Tare, also the convener of the mission. It is for the very first time that the Central government has involved the seven IITs together in one single project of such a large magnitude

Prof Tare further informed TOI that the 'zero discharge' of both treated and untreated sewage waste into the river had been proposed to the government under which the waste would not be allowed into the Ganga. "We will be doing this mega project in phases. The first phase will come to an end in 18 months wherein the concept of 'zero discharge' will be put into application. We also plan to apply the 'zero discharge' formula in four cities initially, viz. Hardwar, Rishikesh, Kanpur and Allahabad. If we are able to do so in these four cities, water of the river will become clean up to Allahabad and a major work will come to an end in the first phase." Meanwhile, Prof SG Dhande, director, IIT-Kanpur, and Prof Vinod Tare from IIT-Kanpur, Prof Devang Khakhar, director, IITBombay, took part in a meeting with the officials of the Ministry of Environment and Forest in New Delhi on May 19 and discussed all important aspects of the mega project. On the occasion, the team also handed over a set of proposals to the officials of the ministry. "A detailed project report will be given later as several social, legal aspects will have to be examined," said Prof Tare. The consortium of the seven IITs have been sanctioned Rs 16 crore by the government for formulating a proper plan of action for the purpose of cleaning the Ganga. The authority formed by the Central government has both regulatory and developmental functions. The authority will take measures for effective abatement of pollution and conservation of the Ganga in keeping with sustainable development needs.

Science Research : Conventional solar cell efficiency could be increased from the current limit of 30 percent to more than 60 percent, suggests new research on semiconductor nanocrystals, or quantum dots, led by chemist Xiaoyang Zhu at The University of Texas at Austin. The scientists have discovered a method to capture the higher energy sunlight that is lost as heat in conventional solar cells.

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JULY 2010

Success Story This articles contains stories of person who have succeed after graduation from different IIT's

Ms. Padmasree Warrior

Dr. Krishan K. Sabnani

B.Tech, IIT Madras

B.Tech, IIT – Kanpur

Padmasree Warrior is senior vice president and chief technology officer for Motorola, with responsibility for Motorola Labs, the global software group and emerging early-stage businesses. Warrior's operational responsibilities include leading a global team of 4,600 technologists, prioritizing technology programs, creating value from intellectual property, guiding creative research from innovation through early-stage commercialization, and influencing standards and roadmaps. She also serves as a technology advisor to the office of the chairman and to the board's technology and design steering committee. Before assuming her current position in January 2003, Warrior was corporate vice president and general manager of Motorola's energy systems group, where she was responsible for profit and loss, sales, marketing, engineering and manufacturing. She also was general manager of Thought beam, Inc., a wholly owned subsidiary of Motorola, where she led the commercialization evaluation team related to compound semiconductor materials research. Prior to these assignments, Warrior was corporate vice president and chief technology officer for Motorola's Semiconductor Products Sector (SPS). A Motorola since 1984, she has been instrumental in driving innovative methods for technology commercialization realizing early "time to revenue" for the corporation. She has held many leadership positions within Motorola, was appointed vice president in 1999 and was elected a corporate officer in 2000. Warrior received a M.S. degree in chemical engineering from Cornell University, and a B.S. degree in chemical engineering from the Indian Institute of Technology (IIT) in New Delhi, India. Warrior served on the Texas Governor's Council for Digital Economy, and is a member of the Texas Higher Education Board review panel. She was one of six women nationwide selected to receive the "Women Elevating Science and Technology" award from Working Woman magazine in 2001. She also is a director of Ferro Corporation.

Krishan Sabnani is Senior Vice President of the Networking Research Laboratory at Bell Labs in New Jersey. For the past 23 years Krishan has been a member of Bell Labs Research. Krishan has conceived and launched several systems projects in the areas of Internetworking and wireless networking, led successful transfers of research ideas to products in Lucent and AT&T business units and conducted extensive personal research in data and wireless networking. He has built organizations known for technical excellence by recruiting and coaching the best people in the industry. Krishan has received the 2005 IEEE Eric E. Sumner Award and the 2005 IEEE W. Wallace McDowell Award - the only person ever to receive both awards. Krishan is a Bell Labs Fellow. He is also a fellow of the Institute of Electrical and Electronic Engineers (IEEE) and the Association of Computing Machinery (ACM). He received the Leonard G. Abraham Prize Paper Award from the IEEE Communications Society in 1991. Krishan will receive the 2005 Distinguished Alumni Award from Indian Institute of Technology (IIT), New Delhi, India. He has also won the 2005 Thomas Alva Edison Patent Award from the R&D Council of New Jersey. He holds 37 patents and has published more than 70 papers. In his personal research, Krishan has made major contributions to the communications protocols area. He has designed several protocols such as SNR, RMTP, and Airmail. He has also made significant contributions to conformance test generation, protocol validation, automated converter generation, and reverse engineering. Krishan received his Ph.D. in electrical engineering from Columbia University, New York, in 1981. He joined Bell Labs in 1981. Key Awards and Honors 1. 2005 IEEE Eric E. Sumner Award, received for seminal contributions to networking protocols 2. 2005 IEEE Computer Society W. Wallace McDowell Award 3. 2005 IIT Delhi Outstanding Alumni Award 4. 2005 Thomas Alva Edison Patent Award 5. 1991 Leonard G. Abraham Prize Paper Award 6. 1997 Bell Labs Fellow.

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JULY 2010

KNOW IIT-JEE By Previous Exam Questions

2.

PHYSICS 1.

Masses M1, M2 and M3 are connected by strings of negligible mass which pass over massless and friction less pulleys P1 and P2 as shown in fig. The masses move such the portion of the string between P1 and P2 in parallel to the inclined plane and the portion of the string between P2 and M3 is horizontal. The masses M2 and M3 are 4.0 kg each and the coefficient of kinetic friction between the masses and the surfaces is 0.25. The inclined plane makes an angle of 37º with the horizontal. [IIT-1981] M P1 2 P2 M3 M1

A C B D Sol. K.E. of block = work against friction + P.E. of spring 1 1 mv2 = µk mg (2.14 + x) + kx2 2 2 1 1 2 × 0.5 × 3 = 0.2 × 0.5 × 9.8(2.14 + x) + 2 × x2 2 2 2.14+ x + x2 = 2.25 ∴ x2 + x – 0.11 = 0 11 On solving we get x = – 10 1 or x = = 0.1 (valid answer) 10 Here the body stops momentarily. Restoring force at y = kx = 2 × 0.1 = 0.2 N Frictional force at y = µs mg × x = 0.22 × 0.5 × 9.8 = 1.078 N Since friction force > Restoring force the body will stop here. ∴ The total distance travelled = AB + BD + DY = 2 + 2.14 + 0.1 = 4.24 m.

37º

If the mass M1 moves downwards with a uniform velocity, find (a) the mass of M1 (b) The tension in the horizontal portion of the string (g = 9.8 m/sec2, sin 37º ≈ 3/5) Sol. (a) Applying Fnet = ma on M1 we get ...(i) T – m1 . g = M1 × 0 = 0 ⇒ T = M1g Applying Fnet = Ma on M2 we get T – (T´ + M2g sin θ – f) = M2 × a T = T´ + M2g sin θ + f = T´ + M2g sin θ + µN [Q f = µN = µM2 g cos θ] ∴ T = T´ + M2g sin θ + µM2g cos θ ...(ii) P1

V

T

M1 T M1g

M2 θ

M2gcosθ

V

T´

M2g f

M2gsinθ

P2 T´

N

θ

A

M3g

2m

Applying Fnet = Ma for M3 we get T´ – f ´ = M3 × 0 ...(iii) ⇒ T´ = f ´ = µN´ = µM3g Putting the value of T and T´ from (i) and (iii) in (ii) we get M1g = µM3g + M2g sin θ – µ M2g cos θ M1 = 0.25 × 4 + 4 × sin 37º + 0.25 × 4 × cos 37º = 4.2 kg (b) The tension in the horizontal string will be T ´ = µM3g = 0.25 × 4 × 9.8 = 9.8 N

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

B D Rough L C 2.14m

x

Y

3.

7

A small sphere rolls down without slipping from the top of a track in a vertical plane. The track in a vertical plane. The track has an elevated section and a horizontal part, The horizontal part is 1.0 meter above the ground level and the top of the track is 2.4 metres above the ground. Find the distance on the ground with respect to the point B(which is vertically JULY 2010

speed of 0.001 ms–1. Calculate the current drawn from the battery during the process. (Dielectric constant of oil = 11, ε0 = 8.85 × 10–12C2N–1m–1) [IIT-1994] Sol. The adjacent figure is a case of parallel plate capacitor. The combined capacitance will be v

2.4 m

below the end of the track as shown in fig.) where the sphere lands. During its flight as a projectile, does the sphere continue to rotate about its centers of mass ? Explain. [IIT-1987]

A

+

1.0m B Sol. Applying law of conservation of energy at point D and point A P.E. at D = P.E. at A + (K.E.)T + (K.E.)R (K.E.)T = Translational K.E. 1 1 mg (2.4) = mg (1) + mv2 + Iω2 2 2 (K.E.)R = Rotational K.E. Since the case is of rolling without slipping D

2.4m

1–x x

d C = C1 + C2 kε 0 ( x × 1) ε [(1 − x ) × 1] = + 0 d d ε0 C= [kx + 1 – x] d After time dt, the dielectric rises by dx. The new equivalent capacitance will be C + dC = C1´ + C2´ kε 0 ε [1 − x − dx ) × 1] [(x + dx) × 1] + 0 = d d dC = Change of capacitance in time dt ε = 0 [kx + kdx + 1 – x – dx – kx – 1 + x] d ε = 0 (k – 1)dx d ε ε dC dx = 0 (k – 1) = 0 (k – 1)v ...(i) dt d dt d dx where v = dt We know that q = CV dq dC =V ...(ii) dt dt ε ⇒ I = V 0 (k – 1)v d From (i) and (ii) 500 × 8.85 × 10 −12 I= (11 – 1) × 0.001 0.01 = 4.425 × 10–9 Amp.

A 1m B

C

∴ v = rω v ∴ω= where r is the radius of the sphere Also r 2 I = mr2 5 ∴ mg(2.4) = mg(1) +

v2 1 1 2 mv2 + × mr2 × 2 2 2 5 r

⇒ v = 4.43 m/s After point A, the body takes a parabolic path. The vertical motion parameters of parabolic motion will be 1 uy = 0 S = ut + at2 2 1 = 4.9 ty2 Sy = 1m ay = 9.8 m/s2 1 ∴ ty = ? ty = = 0.45 sec 4.9 Applying this time in horizontal motion of parabolic path, BC = 4.43 × 0.45 = 2m During his flight as projectile, the sphere continues to rotate because of conservation of angular momentum. 4.

5.

Two square metal plates of side 1 m are kept 0.01 m apart like a parallel plate capacitor in air in such a way that one of their edges is perpendicular to an oil surface in a tank filled with an insulating oil. The plates are connected to a battery of emf 500 V. The plates are then lowered vertically into the oil at a

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1m

8

Two resistors, 400 ohms, and 800 ohms are connected in series with a 6-volt battery. It is desired to measure the current in the circuit. An ammeter of a 10 ohms resistance is used for this purpose. What will be the reading in the ammeter? Similarly, If a voltmeter of 10,000 ohms resistance is used to measure the potential difference across the 400-ohms resistor, What will be the reading in the voltmeter. [IIT-1982] JULY 2010

Sol. Applying Kirchoff's law moving in clockwise direction starting from battery we get 800Ω 400Ω 10Ω

Sol. (i) From the given data, it is evident that the t1/2 (half-life period)for the decomposition of X (g) is constant (100 minutes) therefore the order of reaction is one. 0.693 (ii) Rate constant, K = t1/ 2

A

0.693 = 6.93 × 10–3 min–1 100 (iii) Time taken for 75% completion of reaction = 2t1/2 = 2 × 100 = 200 minutes (iv) 2x → 2Y + 2Z Initial pressure 800 0 0 Ater time t (800 – 2p) 3P 2p when the pressure of X is 700 mm of Hg the, 800 – 2P = 700 2P = 100; P = 50 mm of Hg Total pressure = 800 – 2P + 3P + 2P = 800 + 150 = 950 mm of Hg.

=

6 volt + 6 – 10I – 400 I – 800 I = 0 ∴ 6 = 1210 I 6 ∴ I= = 4.96 × 10–3 A 1210 The voltmeter and 400 Ω resistor are in parallel and hence p.d. will be same ...(i) ∴ 10,000 I1 = 400 I2 Applying Kircoff's law in loop ABCDEA starting from A in clockwise direction. – 400 I2 – 800 I + 6 = 0 ∴ 6 = 400 I2 + 800 (I1 + I2) ∴ 6 = 400 I2 + 800(0.04 I2 + I2) From (i) putting the value of I1 ∴ 6 = 1232 I2 10,000Ω

F B I

V

G C

400Ω I 800Ω

A

A acid solution of Cu2+ salt containing 0.4 g of Cu2+ is electrolysed until all the copper is deposited. The electrolysis is continued for seven more minutes with the volume of solution kept at 100 ml. and the current at 1.2 amp. Calculate the volume of gases evolved at NTP during the entire electrolysis. [IIT-89] Sol. The chemical reactions taking place at the two electrodes are At cathode : Cu2+ + 2e– → Cu H2O H+ + OH– However, note that only Cu2+ ions will be discharged so as these are present in solution and H+ ions will be discharged only when all the cu2+ ions have been deposited. Atcathode : 2OH– → H2O + O + 2e– O + O → O2 Thus in first case, Cu2+ ion will be discharged at the cathode and O2 gas at the anode. Let us calculate the volume of gas (O2) discharged during electrolysis. According to Faraday's cecond law 31.75 g Cu ≡ 8 g of oxygen ≡ 5.6 litres of O2 at NTP 5 .6 0.4 g Cu = × 0.4 litres of O2 at NTP 31.75 = 0.07055 litres = 70.55 ml As explained earlier, when all the Cu2+ ion will be deposited at cathode, H+ ions will start going to cathode liberating hydrogen (H2) gas i.e. H+ + e– H H + H → H2 However, the anode reaction remains same as previous. Thus in the second (latter) case, amount of H2 collected at cathode should be calculated. 8 g of O2 = 1 g of H2 5.6 litres of O2 at NTP = 11.2 litres of hydrogen Quantity of electricity passed after 1st electrolysis, i.e. Q = i × t = 1.2 × 7 × 60 = 504 coulombs 7.

D

E

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

CHEMISTRY 6.

At constant temperature and volume, X decomposes as 2X(g) → 3Y(g) + 2Z(g); Px is the partial pressure of X.

Observation Time (in minute) Rx (in mm of Hg) No. 1 0 800 2 100 400 3 200 200 (i) What is the order of reaction with respect to X ? (ii) Find the rate constant. (iii) Find the time for 75% completion of the reaction. (iv) Find the total pressure when pressure of X is 700 mm of Hg. [IIT-2005]

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9

JULY 2010

Sol. Let us summarise the given facts.

5.6 × 504 = 29.24 ml 96500 of O2. Similarly, H2 liberated by 504 coulombs 504 = 11.2 × = 58.45 ml 96500 Total volume of O2 liberated = 70.55 + 29.24 = 99.79 ml vol. of H2 liberated = 58.48 ml. Cyclobutyl bromide on treatment with magnesium in dry ether forms an organometallic (A). The organometallic reacts with ethanal to give an alcohol (B) after mild acidification Prolonged treatment of alcohol (B) with an equivalent amount of HBr gives 1-bromo 1-methylcyclopentane (C). Write the structures of (A), (B) and explain how (C) is obtained from (B). [IIT-2001] MgBr Br

504 coulombs will liberate =

8.

112 ml of Basic colourless, (i) aq. HCl Nitrogen odourless (ii) NaNO2 0ºC Compound gas at S.T.P (0.295 g) + Residue

ether

Organic liquid (no N)

CH CHO , H O +

3 3 →

CH3

CH–CH3 OH (B)

1-Cyclobutylethanol (B)

CH–CH3 +

H→

+

CH–CH3

– H 2O +

OH2

→

⊕ Oxonium ion

(2º carbocation 4-membered ring)

H

exp ansion ring →

through 1, 2 − alkylshift

⊕

H CH3

CH3CH2CH2NH2

Hydride shift

H CH3 – ⊕ Br →

(3º carbocation)

9.

+

–

,I2 OH → No yellow ppt. (CH3)CHNH2 → (CH3)2CHOH + N2↑ Isopropylamine

H Br ⊕ CH3

→ (CH3)2CHOH

1-bromo-1-methyl Cyclopentane (C)

–

,I 2 OH →

( Haloform reaction )

Since the given reactions isopropylamine, the original isopropylamine, (CH3)2CHNH2

A basic, volatile nitrogen compound gave a foul smelling gas when treated with chloroform and alcoholic potash. A 0.295 g sample of the substance. Dissolved in aq. HCl and treated with NaNO2 solution at 0ºC, liberated a colorless, odourless gas whose volume corresponded to 112 ml at STP, After the evolution of the gas was complete, the aqueous solution was distilled to give an organic liquid which did not contain nitrogen and which on warming with alkali and iodine gave a yellow precipitate. Identify the original substance. Assume that it contains one N atom per molecule. [IIT-93]

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i ) HCl ( → CH3CH2CH2OH

(ii ) NaNO 2 / 0 º C

aq .sol .distill → CH3CH2CH2OH

(ring expansion) (2º carbocation in 5 membered ring

H

Yellow ppt.

N2↑ n-Propylamine

,2 – 1 →

H

OH–/I2

Reaction of the original compound with alcoholic potash and chloroform to give foul smelling gas indicates that it contains a primary –NH2 group. R–NH2 + CHCl3 + KOH → R–NC↑ (Basic compound) Carbylamine (foul smelling) Determination of mol. Weight of the amine. 112 ml. of gas is evolved at S.T.P. by 0.295 g of amine 0.295 22400 ml. of gas is evolved by = × 22400 = 59 112 Hence the mol. wt. of the amine = 59 ∴ Mol. wt. of the alkyl group = 59 – 16 = 43 Nature of alkyl gp. of mol. wt.= 43 = C3H7– Thus the amine may be either CH3 CH3CH2CH2NH2 or CHNH2 CH3 The reaction of amine with NaNO2 at 0ºC and all other reactions may thus be written as below.

Cyclobutymagnesium Bromide (A)

CH–OH

Foul smelling gas

Distilled aq. Sol.

dry →

Sol.

CHCl3 KOH

CHI3↓ (yellow) correspond compound

to is

10. Interpret the non-linear shape of H2S molecule and non-planar shape of PCl3 using valence shell electron pair repulsion (VSEPR) theory. (Atomic numbers : H = 1, P = 15, S = 16, Cl = 17.) [IIT-98] 1 Sol. In H2S, no. of hybrid orbitals = (6 + 2 – 0 + 0) = 4 2 Hence here sulphur is sp3 hybridised, so 2 2 6 2 2 1 1 16S = 1s , 2s 2p , 3s 3p x 3p y 3p z 144244 3 sp 3 hybridisation

10

JULY 2010

[Q O(0, 0) and C(2, 2) lie on the same side of AB 2 2 + – 1 < 0] Therefore, a b (2b + 2a – ab) ⇒ – =2 a 2 + b2

S S

or

H

H

H

H Due to repulsion between lp - lp; the geometry of H2S is distorted from tetrahedral to V-shape. 1 In PCl3, no. of hybrid orbitals = [5 + 3 – 0 + 0] = 4 2 Hence, here P shows sp3-hybridisation 2 1 1 1 2 2 6 15P = 1s , 2s 2p , 3s 3p x 3p y 3p z 144244 3

...(i) ⇒ 2a + 2b – ab + 2 a 2 + b 2 = 0 Let P(h, k) be the circumcentre of ∆OAB. Since ∆ OAB is a right angled triangle. So its circumcentre is the mid-point of AB. a b and k = ∴ h= 2 2 ⇒ a = 2h and b = 2k ...(ii) From (i) and (ii), we get

sp 3 hybridisation

P

P

4h + 4k – 4hk + 2 4h 2 + 4k 2 = 0 or

Cl

Cl

⇒ h + k – hk + h 2 + k 2 = 0 So, the locus of P(h, k) is

Cl

x + y – xy +

Cl Cl

Cl

x 2 + y2 = 0

But, the locus of the circumcentre is given to be

Thus due to repulsion between lp – bp, geometry is distorted from tetrahedral to pyramidal.

x + y – xy + k x 2 + y 2 = 0 Thus, the value of k is 1

MATHEMATICS

12. If A, B, C are the angles of a triangle ABC and the system of linear equations x sin A + y sin B + z sin C = 0 x sin B + y sin C + z sin A = 0 x sin C + y sin A + z sin B = 0 has a non trivial solution, prove that sin2A + sin2B + sin2C – (cos A + cos B + cos C + cos A cos B + cos B cos C + cos C cos A) = 0 [IIT-2002] Sol. The given system of linear equations has a non-trivial solution. Therefore, sin A sin B sin C sin B sin C sin A = 0 sin C sin A sin B

11. The circle x2 + y2 – 4x – 4y + 4 = 0 is inscribed in a triangle which has two of its sides along the coordinate axes. If the locus of the circumcentre of the triangle is

x + y – xy + k x 2 + y 2 = 0, [IIT-1987] find the value of k. Sol. Let OAB be the triangle in which the circle x2 + y2 – 4x – 4y + 4 = 0 is inscribed. Let the x y equation of AB be + = 1 a b y B(0,b)

C

sin A + sin B + sin C sin B sin C ⇒ sin B + sin C + sin A sin C sin A = 0 sin C + sin A + sin B sin A sin B

x y + =1 a b x+ y =1 a b

2 x´

O

x

(a, 0)A

Applying C1 → C1 + C2 + C3

y´

1 sin B sin C 2

2

⇒ (sin A + sin B + sin C) 1 sin C sin A = 0 1 sin A sin B

Since AB touches the circle x + y – 4x – 4y + 4 = 0. There fore, 2 2 2 2 + − 1 + −1 a b a b =2 ⇒– =2 1 1 1 1 + + a 2 b2 a 2 b2

XtraEdge for IIT-JEE

1 sin B sin C

⇒ 1 sin C sin A = 0 1 sin A sin B

11

Q sin A + sin B + sin C A B C = 4 cos 2 cos 2 cos 2 ≠ 0

JULY 2010

1

sin B

We claim that at least two y1, y2, and y3 are distinct. For if y1 = y2 = y3, then P, Q and R lie on a line parallel to x-axis and a line parallel to x-axis does not cross the circle in more than two points. Thus, we have either y1 ≠ y2 or, y1 ≠ y3 or, y2 ≠ y3. Subtracting (ii) from (iii) and (iv), we get

sin C

⇒ 0 sin C − sin B sin A − sin C = 0 0 sin A − sin B sin B − sin C Applying R2 → R2 – R1, R3 → R3 – R1 ⇒ –(sin B – sin C)2 – (sin A – sin C) (sin A – sin B) = 0 ⇒ sin2B + sin2C – 2 sin B sin C + sin2A – sin A sin B – sin C sin A + sin B sin C = 0 ⇒ sin2A + sin2B + sin2C – sin A sin B – sin B sin C – sin C sin A = 0 ⇒ sin2A + sin2B + sin2C – cos A cos B – cos B cos C – cos C cos A + cos (A + B) + cos (B + C) + cos (C + A) = 0 ⇒ sin2A + sin2B + sin2C – cos A cos B – cos B cos C – cos C cos A – cos A – cos B – cos C = 0

( x 22 + y 22 ) – ( x12 + y12 ) – 2 2 (y2 – y1) = 0

⇒ a1 – where,

a1 – ⇒

...(ii)

− 2 2 y2 = r – 2

...(iii)

x 32 + y 32 − 2 2 y3 = r2 – 2

...(iv)

+

y 22

XtraEdge for IIT-JEE

2

2 b1 = 0 a1 = b1

[From (v)]

2,

which is not

possible because

a1 is a rational b1

number and 2 is an irrational number. If b2 ≠ 0, then a a2 – 2 b2 = 0 ⇒ 2 = 2 , b2 which is not possible because

a2 b2

is a rational

number and 2 is an irrational number. Thus, in both the cases we arrive at a contradiction. This means that our supposition is wrong. Hence, there can be at most two rational points on circle C. 15. A rectangle PQRS has its side PQ parallel to the line y = mx and vertices P, Q and S lie on the lines y = a, x = b and x = –b, respectively. Find the locus of the vertex R. [IIT-1996] Sol. Let the coordinates of R be (h, k). It is given that P lies on y = a. So, let the coordinates of P be (x1, a). Since PQ is parallel to the line y = mx. Therefore, Slope of PQ = (Slope of y = mx) = m 1 And, Slope of PS = – (Slope of y = mx)

2 ) is

or, x2 + y2 – 2 2 y = r2 – 2 If possible, let P(x1, y1), Q(x2, y2) and R(x3, y3) be three distinct rational points on circle C. Then, x 22

...(v)

Clearly, a1, a2, b1, b2 are rational numbers as x1, x2, x3, y1, y2, y3 are rational numbers. Since either y1 ≠ y2 or, y1 ≠ y3 ∴ Either b1 ≠ 0 or, b2 ≠ 0 If b1 ≠ 0, then

2 )2 = r2, where r is any positive real

x12 + y12 − 2 2 y1 = r2 – 2

2 b2 = 0

a2 = ( x 32 + y 32 ) – ( x12 + y12 ) , b2 = 2(y3 – y1)

14. Let C be any circle with centre (0, 2 ). Prove that at most two rational points can be there on C. (A rational points is a point both of whose [IIT-1997] coordinates are rational numbers)

(x – 0)2 + (y – number.

2 b1 = 0 and a2 –

a1 = ( x 22 + y 22 ) – ( x12 + y12 ) , b1 = 2(y2 – y1)

13. Determine the name of the name of the curve described parametrically by the equations [IIT-1998] x = t2 + t + 1, y = t2 – t + 1 Sol. We have, x = t2 + t + 1 and, y = t2 – t + 1 ⇒ x + y = 2(t2 + 1) and, x – y = 2t 2 x − y ⇒ x + y = 2 + 1 2 2 ⇒ 2(x + y) = (x – y) + 4 ⇒ x2 + y2 – 2xy – 2x – 2y + 4 Comparing this equation with the equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, we get a = 1, b = 1, c = 4, h = –1, g = –1 and f = –1 ∴ abc + 2fgh – af2 – bg2 – ch2 = 4 – 2 – 1 – 1 – 4 ≠ 0 and , h2 – ab = 1 – 1 = 0 Thus, we have ∆ ≠ 0 and h2 = ab So, the given equations represent a parabola.

Sol. The equation of any circle C with centre (0, given by

( x 32 + y 32 ) – ( x12 + y12 ) – 2 2 (y3 – y1) = 0

and,

1 [∴ PS ⊥ PQ] m Now, equation of PQ is y – a = m(x – x1)

=–

12

...(i) JULY 2010

y (0, a)

⇒

y=0 P

x = –b

x=b Q

x´

S (0, – b)

O

(0, b)

⇒ x

BEWARE OF THE BATTERIES THAT YOU USE!

R y´

It is given that Q lies on x = b. So, Q is the point of intersection if (i) and x = b. Putting x = b in (i), we get y = a + m(b – x1) So, coordinates of Q are (b, a + m(b – x1)). 1 . Since PS passes through P(x1, a) and has slope – m 1 So, Equation of PS is y – a = – (x – x1) ...(ii) m It is given that S lies on x = – b. So, S is the point of intersection of (ii) and x = –b. 1 Solving (ii) and x = – b, we get y = a + (b + x1) m 1 So, coordinates of S are − b, a + (b + x 1 ) m 1 k − a − (b + x1 ) m =m Now, Slope of RS = h+b But RS is parallel to PQ. 1 k − a − (b + x1 ) m ∴ =m h+b ⇒ b + x1 = m(k – a) – m2(h + b) ...(iii) Similarly, k − a − m( b − x 1 ) Slope of RQ = h−b But, RQ is perpendicular to PQ whose slope is m. k − a − m( b − x1 ) 1 ∴ =– h−b m 1 1 ⇒ b – x1 = (k – a) + 2 (h – a) ...(iv) m m We have only one variable x1. To eliminate x1, add (iii) and (iv) to obtain 1 1 2b = (k – a) m + – m2(h + b) + 2 (h – b) m m ⇒

Have you ever noticed that the batteries are becoming smaller and smaller day after day? Many scientists and researchers have been finding the effective way to shrink the batteries into the smallest size as possible! In this case, Jae Kwon, an assistant professor of Electrical and computer engineering has recently developed a nuclear energy source, which is smaller, lighter and more efficient than the common batteries. Kwon’s described that the new discovered radioisotope battery can provide power density as much as six orders of magnitude higher than chemical batteries.

Kwon and his research team members have been cooperated and working on building a small nuclear battery. According to the information, the radioisotope batteries are having the size and thickness of a penny, but it’s powerful enough to power various micro or nanoelectromechanical systems. Even though the nuclear power sources have always been a safety concern, they’ve claimed to be safe, as the nuclear power sources have been used for powering many types of devices, including the pacemakers, space satellites and underwater systems. Kwon’s battery is in a liquid semiconductor rather than a solid semiconductor, as he believed that the liquid semiconductor can overcome the problem, where the lattice structure of the semiconductor being damaged, if it’s in the solid semiconductor form!

m2 +1 4 4 – h m +1 – b m +1 2b = (k – a) m m2 m2

m 2 + 1 h (m 2 − 1)(m 2 + 1) b(m 2 + 1) 2 – ⇒ (k–a) – =0 m m2 m2

XtraEdge for IIT-JEE

h (m 2 − 1) b(m 2 + 1) – =0 m m m(k – a) – h(m2 – 1) – b(m2 + 1) = 0 Hence, the locus of R(h, k) is m(y – a) – x(m2 – 1) – b(m2 + 1) = 0

(k – a) –

13

JULY 2010

Physics Challenging Problems

Set # 3

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

Passage # (Q. No. 4 to Q. No. 6) Resistance value of an unknown resistor is calculated using the formula R= V/I where V and I be the readings of the voltmeter and the ammeter respectively. Consider the circuits below. The internal resistances of the voltmeter and the ammeter (RV and RG respectively) are finite and non zero. V V A A R R E E r r

A circuit consisting of a constant e.m.f. ‘E’, a self induction ‘L’ and a resistance ‘R’ is closed at t = 0. The relation between the current I in the circuit and time t is as shown by curve ‘a’ in the figure. When one or more of parameters E, R and L are changed, the curve ‘b’ is obtained. The steady state current is same in both the cases. Then it is possible that I (a) (b)

Fig. (A) Fig. (B) Let RA and RB the calculated values in the two cases A and B respectively. 4. The relation between RA and the actual value R is (B) R < RA (A) R > RA (D) dependent upon E and r (C) R = RA 5. The relation between RB and the actual value R is (B) R > RB (A) R< RB (D) dependent upon E and R (C) R = RB 6. If the resistance of voltmeter is RV = 1 KΩ and that of ammeter is RG = 1Ω, the magnitude of the percentage error in the measurement of R (the value of R is nearly 10 Ω) is (A) zero in both cases (B) non-zero but equal in both cases (C) more in circuit A (D) more in circuit B Passage # (Q. No. 7 to Q. No. 8) The figure shows the interference pattern obtained in a double-slit experiment using light of wavelength 600 nm. 1,2,3,4 and 5 are marked on five fringes.

t (A) E and R are kept constant and L is increased (B) E and R are kept constant and L is decreased (C) E and R are both halved and L is kept constant (D) E and L are kept constant and R is decreased

2.

3.

Consider a resistor of uniform cross section area connected to a battery of internal resistance zero. If the length of the resistor is doubled by stretching it then (A) current will become four times (B) the electric field in the wire will become half (C) the thermal power produced by the resistor will become one fourth (D)the product of the current density and conductance will become half In front of an earthed conductor a point charge +q is placed as shown in figure +q

(A) On the surface of conductor the net charge is always negative (B) On the surface of conductor at the same points charges are negative and at some points charges may be positive distributed non uniformly (C) Inside the conductor electric field due to point charge is non-zero (D) None of these

XtraEdge for IIT-JEE

By : Dev Sharma Director Academics, Jodhpur Branch

7. 8.

14

The third order bright fringe is (A) 2 (B) 3 (C) 4 (D) 5 Which fringe results from a phase difference of 4π between the light waves incidenting from two slits (A) 2 (B) 3 (C) 4 (D) 5 JULY 2010

8

Solution

Physics Challenging Problems Qu e s tio ns we r e Pub lis he d in Jun e I s su e x −e x −0 4 =0 + r r 5x = 4e x = 4e/5 x 2Bωa 2 e 4e and i = = = i = r 5 r + r / 4 5r

Initially the potential at centre of sphere is 1 Q 1 2Q 1 3Q + = VC = 4πε0 x 4πε0 x 4πε0 x After the sphere grounded, potential at centre becomes zero. Let the net charge on sphere finally be q. 3Qr 1 q 1 3Q ∴ + = 0 or q = x 4πε0 r 4πε0 x

1.[C]

6.

→

3Qr x The velocity is maximum at mean position. Hence the magnetic force on block is maximum, at its mean position. The magnetic force on the block while it crosses the mean position towards right and left is as shown qvmaxB N1

→ →

(C) v . F = 0 means instantaneous velocity is always perpendicular to force. Hence the speed →

will remain constant. And also | F |= constant. Since the particle moves in one plane, the resulting motion has to be circular. →

∫ Idt = 5A ∫ dt Total heat produced = ∫ I 2 Rdt

Average current =

0

∧

∧

→

∧

∧

(D) u = 2 i − 3 j and a = 6 i − 9 j . Hence initial velocity is in same direction of constant acceleration, therefore particle moves in straight line with increasing speed.

3. (A) → Q,R, (B) → P,S, (C) → P,R, (D) → Q,S 4. [A,B,C] Total charge = ∫ Idt = area under curve = 10C

2

→

→

→ →

Mg + qvmaxB mg Case-1 Case-2 Hence normal reaction is maximum in case-1 and minimum in case-2. Hence correct option is D.

= ∫ (−5t + 10) 2 .1.dt =

→

(B) u . F = 0 and F = constant Initial velocity is perpendicular to constant force, hence the path will be parabolic with speed of particle increasing.

N2

vmaxvmax

(A) → R , (B) → Q,S, (C) → P, (D) → Q,R (A) F = constant and u × F = 0 Therefore initial velocity is either in direction of constant force or opposite to it. Hence the particle will move in straight line and speed may increase or decrease.

∴ The charge flowing out of sphere is

2. [D]

Set # 2

7. [B]

C r rθ R q

200 J 3

O

r=

v0

v0 =

mv 0 ; R = 2r sin θ qB

qBR 2m sin θ

Maximum power = I2R, when I is maximum current = 100 × 1 =100W.

x

ε r/2

r/2 ε

×× × ×× ×

r

x

r

φ total

x

Bωr 2 Bωa 2 = 2 2 By nodal equation, nodal

Induced emf e =

XtraEdge for IIT-JEE

dφ = B × hdr =

8. [B,C]

5. [B,D] Equivalent circuit

(∴ Radius = a)

15

h

µ 0 NIhdr 2πr

dr

R +b µ N×h = 0 log × I max × sin ωt 2 π R dφ e = total dt

JULY 2010

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

PHYSICS

A particle of mass m moves along a horizontal circle of radius R such that normal acceleration of particle varies with time as an = Kt2 , where K is a constant. Calculate (i) tangential force on particle at time t, (ii) total force on particle at time t, (iii) Power developed by total force at time t, and (iv) average power developed by total force over first t second. Sol. Since, the particle is moving along a circle, therefore, its normal acceleration is centripetal acceleration i.e. v2/R, where v is velocity of particle at time t. v2 ∴ = Kt2 or v = t. KR R …(1) Due to centripetal acceleration, particle follows a circular path but due to it velocity its magnitude does not change. Velocity magnitude increases due to tangential acceleration alone. d ∴ Tangential acceleration, at = . v = KR dt 1.

∴

Tangential force,

∴

Normal force, Resultant force on particle,

∴

∴ 2.

Ft = mat = m KR Ans. (i) Fn = man = mKt2 F= =m

C

Sol. Let the particle be displaced slightly through x along a line normal to plane of the figure. Then each spring is further elongated. Since, springs are identical, therefore, increase in tension of each spring will be the same. Let this increase be dE0. l l C A θ θ C (F0 + dF0) (F0 + dF0) P First considering forces exerted by spring AP and CP only as shown in Figure. Restoring force produced by these two springs = (F0 + dF0) 2 sin θ x Since x is very small, therefore, sin θ ≈ l Neglecting product of very small quantities, restoring force produced by these two springs

Ans. (ii)

Since, power developed by force F is given by →→

P = F v , therefore, power developed by normal force Fn is always zero because its direction is always perpendicular to the instantaneous direction of motion of the particle. Hence, power is developed by tangential force alone. Figure Ft

v

→ →

Ans.(iii) i.e. P = Ft v = mkRt Since, resultant force equals (mass × acceleration), therefore, resultant force is used to accelerate the

XtraEdge for IIT-JEE

P m

D

+ K 4t )

→

Fn

Figure Shows a particle of mass m = 100 gm, attached with four identical springs, each of length l = 10 cm. Initial tension in each spring is F0 = 25 newton. Neglecting gravity, calculate period of small oscillations of the particle along a line perpendicular to the plane of the figure B

A

Ft2 + Fn2 K (R

body. It means that velocity of the body increases due to resultant force. Hence power developed by resultant force is used to increase kinetic energy of the body. Average power developed by resultant force = Average rate of increase of KE Initial kinetic energy (at t = 0), E0 = 0 1 1 Kinetic energy at time t, E = mv2 = mKRt2 2 2 …(2) E – E0 1 Average power = = mKRt` Ans. (iv) t 2

16

JULY 2010

∴ ∴

∴

T = 2π

∫

∴

σ 2ε 0

∴

Sol.

r +y

0

r2 + y2

2

r +a

or,

V=

Put

∫

a

P

σ 2ε 0

a

ydy

0

r 2 + y2

∫

b (A)

C B A a

c b

(B)

1 q1 q 2 q 3 + + 4πε 0 a b c q1 q 2 q 3 + =0 + b c a Ans. or q1 = – 3µC Now, charges on different surface will be as shown in Figure(B) to calculate energy stored in the system, it can be as considered in three parts : (i) a spherical capacitor having radii a and b and having charge |q1| 3µC.

V= ∴

(r 2 + y 2 ) = P

XtraEdge for IIT-JEE

(+4µC) (–1µC) (+1µC) (–3µC) (+3µC)

c

q1 a

2πσ ydy . 2 4πε 0 r + y2

=

0

σ a2 q π . × = 2ε 0 2r π 4πε 0 r

C B A

q3 q2

dV

∫

1 a2 ≈ r 1 + 2 2 r

Three concentric, conducting spherical shells A,B and C have radii a = 10 cm, b = 20 cm and c = 30 cm respectively. The innermost shell A is earthed and charges q2 = 4µC and q3 = 3 µC are given to shells B and C respectively. Calculate charge q1 induced on shell A and energy stored in the system. Sol. System of three concentric shells is as shown in Figure(A) Since, Shell A is earthed, therefore its potential is zero. But its potential is

y =0

=

1/ 2

4.

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

σ 2 r + a 2 − r 2ε 0

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

r2 + y2

a

a σ 2 r + y2 0 ε 0

a 2 = r 1 + r

2

V=

e

r

=

r2 + y2

σ 2 r + a 2 − r 2ε 0 As a special case, if r >> a

Find the electric potential, at any point on the axis of a uniformly charged circular disc, whose surface charge density is σ, radius, a. Let us consider a small elemental thin ring of width dy. Area of the ring = 2πy dy Charge on this elemental ring = (2πy dy)σ Again, we can consider that this ring is divided into a large number of small elements. Each such element e is at the same distance from P. Hence, potential produced by this ring of width dy at the point P is given by dV, where 1 2πyσdy dV = 4πε 0 r 2 + y 2

O

pdp p

∫

∫ dp = p =

ydy

∫

=

∴V =

ml ml = π = 0.02π sec 4F0 F0

y

2

a

=

Ans. 3.

2

=

4F 2F x Resultant restoring force, F = 2 × 0 = 0 .x l l Restoring acceleration is directly proportional to displacement x, therefore, the particle executes SHM, displacement Its period T = 2π acceleration or

r2 + y2 = p2 2ydy = 2pdp ydy

∴ or

x l Similarly, restoring force produced by two remaining springs BP and DP will also be equal to 2F0 x l

= 2F0

17

JULY 2010

Its capacitance, C1 =

4πε 0 ab (b – a )

C1 + –

∴

5.

Applying Kirchhoff's voltage law on left mesh, of Figure (B) q1 q – q2 + 1 –E=0 …(1) C1 C2 For middle mesh, q2 q q – q2 + 2 – 1 =0 C3 C2 C2

Ans.

+ E –

C2

C3

C1 +E –

C2

…(2)

From equation (1) and (2), q1 = 50 µC and q2 = 20 µC Now consider the circuit when switch S is closed and steady state is reached. The circuit will be as shown in Figure (C)

In the circuit shown in Figure emf of each battery is E = 20 volts and capacitance is C1 = 5 µF, C2 = 3 µF and C3 = 6 µF. Calculate charge on capacitor C3 when switch S is closed and steady state is reached. Calculate also, heat generated in the circuit. C1

+ – E

S

(q + q 2 + q 3 ) 2 q12 (q + q 2 ) 2 + 1 + 1 2C1 2C 2 2C 3

= 0.45 joule

C1 + –

(B)

(iii) an isolated sphere of radius c and having charge (q3 + q2 + q1) = 4µC Energy stored in the system =

+ + C2 –(q1 –q2) C2– q2

+ E –

(ii) a spherical capacitor having radii b and c and having charge (q2 + q1) = 1 µC. 4πε 0 bc Its capacitance. C2 = (c – b)

C3 + –

+ E –

C3

C1 +q –

C1 – q+ + C2 –q

+ C2 –q

+E –

S

Sol.

When Switch S is closed and steady state is reached, the circuit becomes symmetric about the dotted line shown in Figure(A). C1 + E –

C2

C3 C2

(C)

Applying Kirchhoff's voltage law on left mesh of Figure(C) q q + –E=0 or q = 37.5 µC. C1 C2

C1 +E –

After shorting of switch S, increase in charge on capacitor C1 of Left mesh, ∆q1 = (q – q1) = – 12.5 µC That for capacitor C2 of left mesh, ∆q2 = q – (q1 –q2) = 7.5 µC That for capacitor C3 , ∆q3 = 0 – q2 = – 20µC That for capacitor C2 of right mesh, ∆q4 = (q –q2) = 17.5 µC That for capacitor C1 or right mesh, ∆q5 = q – 0 = 37.5 µC Since, heat generated in the circuit is given by

(A)

Right part of the circuit is exactly mirror image of the left part. Hence, charges on both plates of capacitor C3 should be identical. But charges on plates of a capacitor are always opposite to each other. It means one of the plates is always positively charged and the other is negatively charged. Both these conditions can be satisfied only if charge on capacitor C3 is zero. To calculate heat generated in the circuit initial and final charges on all the capacitors must be known. First analyse the circuit when switch S was open Charges on capacitors will be as shown in Figure (B)

( ∆q ) 2 2C

H=

∑

H=

(∆q 3 ) 2 (∆q1 ) 2 (∆q 2 ) 2 + + 2C1 2C 2 2C 3

+ = 250 × 10–6 joule

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18

(∆q 5 ) 2 ( ∆q 4 ) 2 + 2C 2 2C1

Ans. JULY 2010

P HYSICS F UNDAMENTAL F OR IIT-J EE

Capacitor-1 KEY CONCEPTS & PROBLEM SOLVING STRATEGY Capacitance : Whenever charge is given to a conductor of any shape its potential increases. The more the charge (Q) given to the conductor the more is its potential (V) i.e. Q∝V

(b) directly on the dielectric constant K of the medium between the conductors. (c) inversely on the distance of separation between the conductor. Principle of a condenser : Consider a conducting plate A which is given a charge Q such that its potential rises to V. Then C = Q/V Let us place another identical conducting plate B parallel to it such that charge is induced on plate B (as shown in figure). A

⇒ Q = CV where C is constant of proportionality called capacitance of the conductor C = Q/V, C = Q SI unit of capacitance is farad (F) and 1 F = 1 coulomb/volt (1CV–1) Energy stored in a charged capacitor : Q2 1 1 CV02 = = QV0 2C 2 2

W=

+ + + + + + + +

Capacitance of an isolated sphere : Let a conducting sphere of radius a acquire a potential V when a charge Q is given to it. The potential acquired by the sphere is

V=

Charge sharing Between two charged conductors : C2

V1

V2

V

V

q1 = C1V1

q2 = C2V2

q´1 = C1V

q´2 = C2V

(Initially)

V=

(Finally)

C´ =

C1V1 + C 2 V2 C1 + C 2

+ + + + + + + +

Q Q = V´ V + V+ − V−

⇒ C´ > C Further, if B is earthed from the outer side (see figure) then Vn = V – V– as the entire positive charge flows to the earth. So

(V1 – V2)2

Capacitor or Condenser : An arrangement which has capability of collecting (and storing) charge and whose capacitance can be varied is called a capacitor (or condenser) The capacitance of a capacitor depends. (a) directly on the size of the conductors of the capacitor.

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– – – – – – – –

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

There is always a loss in energy during the sharing process as some energy gets converted to heat. 1 C C Loss = – ∆U = 1 2 2 C1 + C 2

+ + + + + + + +

+ + + + + + + +

C2

C1

Q

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

Q Q ⇒C= = 4πε0a V 4πε 0 a

C1

+ + + + + + + +

C" =

Q V

n

=

Q ⇒ Cn >> C V − V−

So, if an identical earthed conductor is placed in the viscinty of a charged conductor then the capacitance of the charged conductor increases appreciably. This is the principle of a parallel plate capacitor. 19

JULY 2010

Parallel Plate Capacitor : B

A

A+

+σ

–σ

+ + + + + + + +

+ A = Area of plate + d = Separation + between the + + plates + +

– – – – – – – –

Special Case II : When the space between the parallel plate capacitor is partly filled by a conducting slab of thickness t( > S(liquid state) > (solid state) Expression of Entropy Function For a system which involves transferring infinitesimal heat at constant temperature, the entropy change of the system is given by dq rev dS = T

∆fusH = 6 kJ mol–1 Enthalpy of vaporization :

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∑

v g ,i ( products)

37

JULY 2010

here γ = Cp,m/Cv,m The symbols Cp,m and Cv,m represent molar heat capacities at constant pressure and volume conditions, respectively. For a monatomic ideal gas: Cv,m = (3/2)R; Cp,m = (5/2)R; and γ = 5/3 For a diatomic ideal gas: Cv,m = (5/2)R; Cp,m = (7/2)R; and γ = 7/5

For finite heat transferred at constant temperature, we have q ∆S = rev T For example, for a pure substance we have ∆ vap H ∆ H and ∆fusS = fus ∆vapS = Tb Tm where the subscripts vap and fus represent vaporization and fusion, respectively. Gibbs Function Gibbs function (or energy) or simply free energy is defined as G = H – TS For a process occurring at constant T and P, the change in Gibbs function is given by ∆G = ∆H – T ∆S For a process to be spontaneous, the value of ∆G is negative. For a nonspontaneous reaction, ∆G is positive. For a reaction at equilibrium, ∆G = 0 and temperature at which the system occurs at equilibrium is given by Teq = ∆H/∆S Pressure-Volume Work An ideal gas can undergo expansion of compression under isothermal or adiabatic conditions. The expansion ant compression may be carried out under reversible or irreversible conditions. We give below the expressions of p-V work under different conditions. Isothermal p–V Work In this case, temperature of the system remains constant, ie. ∆T = 0 For irreversible condition: w = – Pext (V2 – V1) For reversible condition: w = – nRT In (V2/V1) Adiabatic p–V Work In this case, heat can neither enter to or leave from the system, i.e. q = 0. From first law of thermodynamics, it follows that ∆U = w where ∆U is given by ∆U = Cv(T2 – T1) For a gas undergoing adiabatic irreversible volume change, the expression of work is given by w = – Pext (V2 –V1) For an ideal gas undergoing adiabatic reversible expansion/compression, we also have pVγ = constant pTγ(1–γ) = constant and TVγ–1 = constant

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CHEMISTRY JOKES If you didn't get the joke, you probably didn't understand the science behind it. If this is the case, it's a chance for you to learn a little chemistry.

Chemistry Joke 1: Outside his buckyball home, one molecule overheard another molecule saying, "I'm positive that a free electron once stripped me of an electron after he lepton me. You gotta keep your ion them."

Chemistry Joke 2: A chemistry professor couldn't resist interjecting a little philosophy into a class lecture. He interrupted his discussion on balancing chemical equations, saying, "Remember, if you're not part of the solution, you're part of the precipitate!"

Chemistry Joke 3: One day on the Tonight Show, Jay Leno showed a classified add that read: "Do you have mole problems? If so, call Avogadro at 602-1023."

Chemistry Joke 4: A student comes into his lab class right at the end of the hour. Fearing he'll get an "F", he asks a fellow student what she's been doing. "We've been observing water under the microscope. We're suppose to write up what we see." The page of her notebook is filled with little figures resembling circles and ellipses with hair on them. The panic-stricken student hears the bell go off, opens his notebook and writes, "During this laboratory, I examined water under the microscope and I saw twice as many H's as O's."

38

JULY 2010

UNDERSTANDING

U n d e r s t a n d i n g

Organic Chemistry

1.

An organic compound (A), C10H15N, undergoes carbylamine reaction but no diazotization. It reacts with HNO2 giving off N2 and a compound (B), C10H14O. (B) reacts with Lucas reagent immediately, but no colour in Victor meyer's test. (B) on heating with conc. H2SO4 eliminates water to give (C), C10H12, which decolourises Br2/CCl4 and cold dilute neutral KMnO4 solution. (C) on ozonolysis gives (D), C7H6O and (E), C3H6O. Compound (E) on heating with I2 and NaOH produced yellow precipitate and sodium acetate. Compound (D) reacts with conc. NaOH to give (F) and (G). Compound (G) on heating with sodalime gives benzene. Compound (F) gives a red colour with ceric ammonium nitrate, and on oxidation and heating the product with sodalime produced benzene. What are (A) to (G) ?

H C–CH OH

(A)

HNO2 –N2;–H2O

C10H14O

Conc. H2SO4

(B)

(B)

(A)

Two isomeric compounds (A) and (B) have the molecular formula C7H9N. (A) being soluble in water, the solution being alkaline to litmus It does not undergoes diazotization, but show carbylamine reaction and mustard oil reaction, it reacts with acetyl chloride and acetic anhydride. Its product with benzene sulphonyl chloride dissolves in KOH. (B) on the other hand, does not dissolve in water, but undergoes diazotization. Its product with C6H5SO2Cl dissolves in KOH. Its salt undergo hydrolysis in aqueous solution showing an acidic test. What are (A) and (B) ? Sol. As both (A) and (B) give products with C6H5SO2Cl, which are soluble in KOH, they contain –NH2 group. (B) can be diazotized so contains – NH2 in the nucleus. (A) cannot be diazotized, hence contains –NH2 in the side chain. The number of carbon and hydrogen atoms also indicates aromatic character. On the basis of above considerations we may show that (A) is benzylamine and (B) o–, m– or p-toludine. CH2NH2 NaNO2 + HCl CH2OH 2.

(I) O3

Conc. NaOH

(C)

C7H6O + C3H6O

(II) H2/Pd

C7H6O

(F) + (G)

(D)

(E)

Sodalime

C6H6

(D) [O]

C10H12

∆;–H2O

Product

Sodalime ∆

C6H6

I 2 + NaOH C 3 H 6 O 2 → CHI3 ↓ + CH3COONa (E)

+ 3NaI + 3H2O Since (C) decolourise Br2/CCl4 and KMnO4 colour, hence it has C=C bond. Its ozonolysis gives (D) and (E). Among these (D) undergoes Cannizaro's reaction, while (E) gives iodoform test, hence (D) is benzaldehyde and (E) acetone. Now joining (D) and (E), the structure of (C) can be determined. H C=O + O =C (D)

(E)

Benzylamine (A) CHCl3 + 3KOH

CH3 –2[O] CH3

H C=C (C)

CH2NC

CS2 + HgCl2

CH2NCS

C6H5SO2Cl

CH2NHSO2C6H5

– HCl

CH3

KOH

CH3

–H2O

CH2 N–SO2C6H5 K

Soluble

Since (C) is produced from (B), which is a t-alcohol, as it gives Lucas test immediately, hence (B) is.

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CH3

As (B) is obtained by the action of HNO2 on (A), hence (A) would be H CH3 C–CH CH3 NH2

Sol. The given data are : C10H15N

CH3

C6H5CH2NH2 + HOH → C6H5CH2N+H3OH–

39

JULY 2010

CH3 NaNO2 + HCl NH2

C6H4 (B)

C6H4

CH3

– CH.CH3

N2Cl (A)

C6H5SO2Cl –HCl

CH3

C6H4

KOH

NHSO2C6H5

Cl

[O] –H2O

NKSO2C6H5

C

68.32

H

6.4

Cl

25.26

Relative no. of atoms 68.32 = 5.59 12 6.4 = 6.40 1 25.26 = 0.71 35.5

(C)

(C)

Simplest ratio 5.59 =8 0.71 6.40 =9 0.71 0.71 =1 0.71

4.

Sol.

[O]

( B)

− H 2O

(C)

(iii) (C) reacts with C6H5NHNH2 to give C=O C8H8=N.NHC6H5, hence (C) contains a group. (iv) (C) on heating with I2 + NaOH gives CHI3, hence (C) contains –COCH3 group. Thus (C) is O

(C)

– C=N.NH.C6H5 CH3 Yellow ppt

O + 3H2O +

– COONa

An organic compound (A), C4H9Cl, on reacting with aqueous KOH gives (B) and on reaction with alcoholic KOH gives (C) which is also formed on passing vapours of (B) over heated copper. The compound (C) readily decolourises bromine water. Ozonolysis of (C) gives two compounds (D) and (E). Compound (D) reacts with NH2OH to gives (F) and the compound (E) reacts with NaOH to give an alcohol (G) and sodium salt (H) of an acid. (D) can also be prepared from propyne on treatment with water in presence of Hg++ and H2SO4. Identify (A) to (H) with proper reasoning. C4H9Cl

Alc. KOH ∆; –KCl

(A) (Alkyl halide) Aq.KOH ∆; –KCl

C4H8

(C) (Alkene) Cu C4H9OH ∆; –H2O (B) (Alcohol)

We know that p-alcohol on heating with Cu gives aldehyde while s-alcohol under similar conditions gives ketone. Thus, (B) is a t-alcohol because it, on heating with Cu gives an alkene (C). Since a talcohol is obtained by the hydrolysis of a t-alkyl halide, hence (A) is t-butyl chloride. Cl OH | | (A) = CH 3 − C − CH 3 and (B) = CH 3 − C − CH 3 | | CH 3 CH 3

– C – CH3

(v) Oxidation of (B) gives (C), hence (B) is a secondary alcohol, i.e., – CH – CH3 OH (vi) (B) is obtained by the hydrolysis of (A), hence it is : – CH.CH3

The alkene (C) on ozonolysis gives (D) and (E), hence (C) is not symmetrical alkene. In these compound (E) gives Cannizaro's reaction with NaOH. So, (E) is an aldehyde which does not contain α–H atom. Hence it is HCHO. Compound (D) can also be prepared by the hydration of propyne in the presence of acidic solution and Hg++.

Cl 1-Chloro-1-phenyl ethane

Now, different reactions are as follows :

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– COCH3

∆ – C – CH3 + 3I2 + 4NaOH → CHI3 + 3NaI

C8 H 9 Cl → C8 H 9 OH → C8 H 8 O (A)

CH3 –H2O

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

OH

(B)

– C = O + H2 N . NH . C6H5

An organic compound (A) of molecular weight 140.5, has 68.32% C, 6.4% H and 25.26% Cl. Hydrolysis of (A) with dilute acid gives compound (B), C8H10O. (B) can be oxidised under milder conditions to (C), C8H8O. (C) forms a phenyl hydrazone (D) with C6H5NHNH2 and gives positive iodoform test. What are (A) to (D) ? Sol. (i) Calculation of empirical formula of (A) 3.

%

– CH–CH3

CH3

C6H4

Soluble

Element

HOH/H+ –HCl

40

JULY 2010

A hydrocarbon (A) of the formula C8H10, on ozonolysis gives compound (B), C4H6O2, only. The compound (B) can also be obtained from the alkyl bromide, (C) (C3H4Br) upon treatment with Mg in dry ether, followed by treatment with CO2 and acidification. Identify (A), (B) and (C) and also equations for the reactions.

5.

++

CH3 – C ≡ CH + H2O Hg → CH 3 − C = CH 2 H+ | OH → CH 3 − C − CH 3 || O ( D)

Hence (D) is acetone and (E) is formaldehyde. Therefore, alkene (C) is 2-methyl propene. (CH3)2–C=CH2 (D) reacts with hydroxyl amine (NH2OH) to form oxime (F). CH3 –H2O CH3 C = O + H2 NOH C = NOH CH3 CH3 (D)

(ii ) H 2 O

If it was alkene its formula should be C8H16 (CnH2n), and if it was alkyne it should have the formula C8H14; it means it is neither a simple alkenen or simple alkyne. However it is definite that the compound has an unsaturated group, it appears that it is a cyclosubstituted ethyne. 2H H – C ≡ C – H − → C3H5 – C ≡ C – C3H5

+ C 6 H10

the C3H5 – corresponds to cyclopropyl (∆) radical, hence compund (A) is CH2 CH2 CH – C≡C – CH CH2 CH2

Reactions : OH Cl | | CH 3 − C − CH 3 CH 3 − C − CH 3 Aq.KOH → | | ∆ ;– KCl CH 3 CH 3

1,2-dicyclopropyl ethyne

The ozonolysis of above compound would give two moles of cyclopropane carboxylic acid (C4H6O2). CH2 CH2 (i) O3 CH – C≡C – CH CH2 CH2

( B)

300 º C Cu / → CH 3 − C = CH 2 + H 2 O − H 2O | CH 3

(A)

( C)

CH2

Alc.KOH / ∆

→ CH 3 − C = CH 2 − KCl; − H 2 O | CH 3

CH2 CH2

(C)

O

CH3

(I) O3

CH3

CH2

CH3

C = O + H2NOH (D)

O

CH3 CH3

CH – C – C – CH O

O

C = NOH

CH2 CH2

CH2

CH2 CH2

CH – Br

Mg ether

CH2 CH2

(H)

+ H2O2

CH – COOH (B)

O

CH . MgBr

C=O ∆

Cyclopropyl magnesium bromide

O

CH2

Hg + +

CH3 – C ≡ CH + H2O → CH3 – C – CH3 + H

CH2

H2O Warm

Compound (B) is prepared from cyclopropyl bromide as follows :

2HCHO + NaOH → CH 3OH + HCOONa (G )

CH2

O

2

(F) (E)

(A)

(D)

∆ –H2O

CH2

CH – C — C – CH

(E)

(C)

CH3

O

C=O+H–C–H

(II) H2O/Zn CH3

or a – C ≡ C – bond.

C=C

should have either

OH Cl | | (B) = CH 3 − C − CH 3 and (A) = CH 3 − C − CH 3 | | CH 3 CH 3

CH3 – C = CH2

( B)

Since compound (A) adds one mol of O3, hence it

(F)

(A)

(i ) O

A(C8H10) 3 → C 4 H 6 O 2

Sol.

CH2

(D)

CH .COOMgBr

HOH dil. HCl; –MgBrOH

CH2 CH2

CH–COOH

Addition compound

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41

JULY 2010

`tà{xÅtà|vtÄ V{tÄÄxÇzxá

Set

3

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.

2.

3.

Let y = f(x) be a curve satisfying dy – y ln 2 = 2sin x (cos x – 1). ln2, then dx (A) y is bounded when x → ∞ (B) f(x) = 2sin x + c . 2x, where c is an arbitrary constant (C) y = 2sinx, y is bounded when x → ∞ (D) f(x) = 2sinx does not have any solution if y is not bounded. In a right angled triangle the length of its hypotenuse is four times the length of the perpendicular drawn from its orthocentre on the hypotenuse. The acute angles of the triangle can be π π π 3π (A) , (B) , 6 3 8 8 π π π 5π (D) , (C) , 6 4 12 12

3, 2 − 3

(D) None of these

4.

If c1 is a fixed circle and c2 is a variable circle with fixed radius. The common transverse tangents to c1 and c2 are perpendicular to each other. The locus of the centre of variable circle is : (A) circle (B) ellipse (C) hyperbola (D) parabola

5.

The length of the latus rectum of the parabola 169 {(x – 1)2 + (y – 3)2} = (5x – 12y + 17)2 is – 14 56 28 (A) (B) (C) (D) None 13 13 13

6.

Evaluate :

∫

cos 5x + cos 4 x dx 1 − 2 cos 3x

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

8.

Given the base of a triangle and the sum of its sides prove that the locus of the centre of its incircle is an ellipse.

9.

A bag contains 7 tickets marked with the number 0, 1, 2, 3, 4, 5, 6 respectively. A ticket is drawn and replaced. Then the chance that after 4 drawings the sum of the numbers drawn is 8, is –

10. A polynomial in x of degree greater than 3 leaves remainders 2, 1 and – 1 when divided by (x – 1), (x + 2) and (x + 1) respectively. What would be the remainder if the polynomial is divided by (x2 – 1) (x + 2) ?

Let a, b ∈ R such that 0 < a < 1 and 0 < b < 1. The values of a and b such that the complex number z1 = –a + i, z2 = –1 + bi and z3 = 0 form an equilateral triangle are (A) 2 − 3 , 3 (B) 2 − 3 , 2 − 3

(C)

7.

42

•

William Bottke at Cornell University in the US has calculated that at least 900 asteroids of a kilometre or more across regularly sweep across Earth's path.

•

The Dutch astronomer Christiaan Huygens (1629 - 1695) drew Mars using an advanced telescope of his own design. He recorded a large, dark spot on Mars, probably Syrtis Major. He noticed that the spot returned to the same position at the same time the next day, and calculated that Mars has a 24 hour period. (It is actually 24 hours and 37 minutes)

•

Space debris travels through space at over 18,000 mph.

•

The nucleus of Comet Halley is approximately 16x8x8 kilometers. Contrary to prior expectations, Halley's nucleus is very dark: its albedo is only about 0.03 making it darker than coal and one of the darkest objects in the solar system. JULY 2010

MATHEMATICAL CHALLENGES SOLUTION FOR JUNE ISSUE (SET # 2) 1.

2.

3.

1st box can be filled in 4 ways. Next each box can be filled in 3 ways (except the ball of colour in previous box). Hence the required no. of ways = 4 × 35 = 972

Hence plane given by (3) is bisecting the acute angle between given two planes also. Hence the conclusion holds true. 5.

AA–1 = I ⇒ (AA–1)T = IT (A–1)TAT = I (as A is symmetric) (A–1)T A = I so by the definition of inverse A–1 = (A–1)T Hence A–1 is also symmetric. Given |A| ≠ 0;

The normal to hyperbola at the point P(a sec θ, b tan θ) is ax cos θ + by cot θ = a2 + b2 If it passes through (h, k) then a h cos θ + b k cot θ = a2 + b2

I2 =

Let ⇒

f–1(y) = x f(x) = y

((f −1 ( y)) 2 − a 2 ) dy

( x 2 − a 2 ) f´(x) dx

(

) ∫ b

I2 = ( x 2 − a 2 ) f ( x ) a – = (b2 – a2) f(b) – ...(1)

∫

=

Hence

6.

b

a

∫

=

Σ z1z2 = 0 = Σ e i ( θ1 + θ 2 ) = 0 Σ (cos (θ1 + θ2) + i sin (θ1 + θ2)) = 0 Hence Σ cos(θ1 + θ2) = Σ sin(θ1 + θ2) = 0

2x f(b) dx –

b

a

∫

b

a

∫

a

2x f(x) dx

2 x f(x) dx

b

a

b

2x f(x) dx

2 x (f(b) – f(x)) dx

I1 1 = I2 2

1 =2 y

y+

⇒ y=1 1 x+ = x

Planes are – x – 2y – 2z + 9 = 0 ....(1) and 4x – 3y + 12z + 13 = 0 ...(2) The plane bisecting the angle b/w these planes containing origin is − x − 2 y − 2z + 9 4 x − 3y + 12z + 13 =+ 3 13 i.e. 25x + 17y + 62z – 78 = 0 ...(3) If θ be the angle between (1) & (3) then 61 cos θ = 4758

⇒ x2 +

5+2

1 x2

= ( 5 + 2) – 2 =

x4 +

1 x4

⇒ x8 + ⇒ x16 +

5

=5–2 1 x8 1

=9–2

= 49 – 2 x16 ⇒ 47 + 1 + 1 = 49

1037 ∆Uac > ∆Uad > ∆Uac (B) ∆Uab = ∆Uac > ∆Uad > ∆Uac (C) ∆Uab = ∆Uac = ∆Uad = ∆Uac (D) ∆Uab < ∆Uac < ∆Uad < ∆Uac

x

(c)

The above graphs shows conduction of heat through materials A,B,C connected in series. Graph shows variations of temperature with distnace x-axis. Which of the above graph are not possible -

An ideal gas whose adiabatic exponent is γ is expanded so that the amount of heat transferred to the gas is equal to the decrease of its internal energy. Molar heat capacity of the gas for this process is -

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x

(a)

b c

2.

R γ –1 (D) R/2

(B)

3.

The figure shows two isotherms at temperatures T1 and T2. A gas is taken from one isotherm to another isotherm through different processes. Then change in internal energy ∆U has relation – P

R 1– γ (C) R

(A)

(A) a, b, c (C) a, c

52

(B) a, b (D) b, c

JULY 2010

4.

Figure shows a rectangular pulse and a triangular pulse approaching each other along x-axis. The pulse speed is 0.5 cm/s. What is the resultant displacement of medium particles due to superposition of waves x = 0.5 cm and t = 2 sec. y (cm) 0.5 cm/s 0.5 cm/s 2

9.

1 –2 –1 (A) 3.5 cm (C) 4 cm

5.

6.

0

Passage: II (Ques. 10 to 12)

x (cm)

2 1 3 (B) 2.5 cm (D) 3 cm

Two hydrogen like atoms A and equal number of protons and neutrons. The energy difference between the radiation corresponding to first Balmer lines emitted A and B is 5.667 eV. When the atoms A and B moving with the dame velocity, strikes a heavy target they rebound back with the same velocity. In this process the atom B imparts twice the momentum to the target than the A imparts.

Choose the correct statement (s) related to the photocurrent and the potential difference between the plate and the collector (A) Photocurrent always increase with the increase in potential difference (B) when the potential difference is zero, the photocurrent is also zero (C) Photocurrent attain a saturation value of some positive value of the potential difference (D) None of these

10. Ionization energy of Atom B is (A) 27.2 eV (B) 13.6 eV (C) 10.2 eV (D) 54.4 eV 11. Atomic number of atom A is (A) 1 (B) 2 (C) 3 (D) 4

Binding energy per nucleon of 1H2 and 2He4 are 1.1 MeV and 7.0 MeV respectively. Energy released in the process 1H2 + 1H2 = 2He4 is (A) 20.8 MeV (B) 16.6 MeV (C) 25.2 MeV (D) 23.6 MeV

12. Mass number of atom B & atom A (A) 2, 4 (B) 4, 2 (C) 2, 1 (D) 4, 1

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 : I (Ques. 7 to 9)

7.

(A) (C) 8.

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 DS, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

Two parallel plates in vacuum, separated by a small distance which is small compared with their linear dimensions, are at temperatures T1 & T2 respectively (T1 > T2). The plates are black bodies. Another plate (black body) at temperature T0 is a kept in between the two plates. Temperature of the plate kept i.e. T0 is T14

+ T24 2

= T04

T1T2 = T0

(B)

T14

– T24 2

P Q R S T A B C D

= T04

(D) 2T14 – T24 = T04

(C) σ

(T14 – T24 ) 2

XtraEdge for IIT-JEE

(B) σ

(T14 – T04 ) 2

(D) σ

(T04 – T24 ) 2

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.

Energy absorbed by the plate having temperature T0, per sec per unit area is (A) σ(T14 –T24)

If 'n' black Body plates are placed in between the two plates having temperature T1 & T2, then the net rate of emission of radiant energy by first plate is σ σ (A) (T14 – T24) (B) (T14 – T24) n n +1 n σ (T14 – T24) (D) (T14 – T24) (C) σ n –1 n +1

53

JULY 2010

13. Match the Column-I with Column-II : Column-I Column-II (A) An electron moves (P) Total Energy Potential Energy in an orbit in a = 2 Bohr atom (B) As a satellite moves (Q) Kinetic Energy = in a circular orbit Magnitude of Energy around total earth (C) In Rutherford's (R) Motion is under a α-scattering experiment central force as an α-particle moves in the electric field of a nucleus (D) As an object, released (S) Mechanical energy from some height above is conserved ground, falls towards earth, assuming negligible air resistance (T) None of these 14. Match the following : Column-I` Column-II (A) Steady state (P) A blackened platinum wire, when gradually heated appear first red and then blue. (B) Wein's displacement (Q) Radiated power is law proportional to fourth power of absolute temperature of body (C) Stefan's law (R) Energy absorbed is equal to energy emitted (D) Black body (S) Absorptive power of body is unity (T) None of these 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

0 1 2 3 4 5 6 7 8 9

XtraEdge for IIT-JEE

Y 0 1 2 3 4 5 6 7 8 9

Z 0 1 2 3 4 5 6 7 8 9

15. Four cylindrical rods of same material with length and radius (l,r), (2l,r), (2l,2r) and (l,2r) are connected between two reservoirs at 0ºC and 100ºC. Find the ratio of the maximum to minimum rate of conduction in them. 16. Using solar constant S = 20 kilo cal/mm –m2 and Joule's constant J = 4.2 J/cal, find the pressure exerted by sunlight ? (Ans. in …… × 10–6 N/m2) 17. The minimum intensity of audible sound is 10–12 W/m2 sec and density of air is 1.3 kg/m3. If the frequency of sound is 1000 Hz, find the amplitude (Ans. in …… × 10–11 m) of vibration ? [Speed of sound = 332 ms–1] 18. The size of a nucleus is of the order of –14 m. Calculate the velocity with which protons move inside the nucleus. The mass of a proton = 1.675 × 10–27 kg. [Ans. in …… × 107 ms–1] 19. Find the change in frequency of red light whose original frequency is 7.3 × 1014 Hz when it falls through 22.5 m losing gravitational potential energy. [Ans. in …… ×103]

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. The optically active tartaric acid is named as D–(+)– tartaric acid because it has a positive (A) optical rotation and is derived from D-glucose (B) pH in organic solvent (C) optical rotation and is derived from D–(+)– glyceraldehyde (D) optical rotation only when substituted by deuterium 2. Which of the following compounds is not coloured ? (A) Na2[CuCl4] (B) Na2[CdCl2] (D) K3[Fe(CN)6] (C) K4[Fe(CN)6] 3.

W 0 1 2 3 4 5 6 7 8 9

4.

54

The brown ring complex compound is formulated as [Fe(H2O)5(NO)]SO4. The oxidation state of iron is (A) 1 (B) 0 (C) 2 (D) 3 The following equilibrium is established when HCl is dissolved the acetic acid, Cl– + CH3COOH2+ HCl + CH3COOH the set that characterises the conjugate acid-base pairs is(A) (HCl, CH3COOH) and (Cl–, CH3COOH2+) (B) (HCl, CH3COOH2+) and (CH3COOH, Cl–) (C) (CH3COOH2+) HCl) and (Cl–, CH3COOH) (D) (HCl, Cl–) and (CH3COOH2+, CH3COOH) JULY 2010

5.

6.

Pure ammonia is placed in a vessel at a temperature where its dissociation constant(α) is appreciable. At equilibrium (A) kp does not change significantly with pressure (B) does not change with pressure (C) concentration of NH3 does not change with pressure (D) concentration of hydrogen is less than that of nitrogen

9.

Passage: II (Ques. 10 to 12)

The property of hydrides of p-block elements mostly depend on (i) Electronegatively difference between central atom and hydrogen (ii) Size of central atom (iii) Number of valence electrons in central atom. some undergo hydride in which central atom is less electronegative, react with OH– to given hydrogen while acidic property of hydride in a period depends on electronegativity of central atom i.e. more electronegative is the atom, more acidic is hydride. In a group, acidic property is proportional to size of central atom. Some electron deficient hydride behaves as lewis acid while only one hydride of an element in p-plock behaves as lewis base with lone pair of electrons. Hydrides in which central atom's electronegativity to close to hydrogen has no reaction with water.

Passage : I (Ques. 7 to 9)

The IUPAC has set guidelines for logical and methodical naming of organic compounds. The complex substituents are written in small brackets and their numbering is done separately. The bivalent radicals are named by adding 'idene' to the name of alkyl group. In polyfunctional compounds all lower priority groups are written in prefix. Now name the following compounds.

10. The hydride which do not react with water is (A) NH3 (B) PH3 (C) B2H6 (D) AsH3 11. Which one undergoes spontaneous combustion with exposure to air ? (A) PH3 (B) P2H4 (C) N2H4 (D) NH3

CHCHCH2OH is -

12. Which one is strongest base ? (A) OH– (B) HS– (C) HSe–

Br

(A) 3-(3'-isopropoxycarbonyl cyclopentylidene propane-1-ol (B)3-(2'-bromo-3'-hydroxypropylidene) cyclopentane carboxylate (C) Iso-propyl-3-(2'-bromo-3'-hydroxy propylidenyl) cyclopentane carboxylate (D) Iso-propyl-3-(2'-bromo-3' hydroxypropylidene) cyclopentane carboxylate

CH3CH2O

CH3CH2

is -

P Q R S T A B C D

(A) 2-(3'-Ethylphenyl)-1-(4'-ethoxyphenyl) ethane (B) 1-Ethyl-3-(2'-(4''-ethoxyphenyl) ethyl) benzene (C) 1-(3'-Ethylphenyl)-2-(4'-ethoxylphenyl) ethane (D) None of these

XtraEdge for IIT-JEE

(D) HTe–

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 DS, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

C2H5

8.

is -

O (A) 3-chlorocarbonyl-6-(N, N-diethylamino) hex-4ene-1-oic acid (B)4-chlorocarbonyl-3-(N, N-diethylamino) butanoic acid (C) 3-chlorocarbonyl-3-(3-N, N-diethylamino prop1'-enyl) butane-1-oic acid (D) 3-chlorocarbonylmethyl-6- (N, N-diethylamino) hex-4-en-1-oic acid

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.

(CH3)2CHOOC

N OH

O

At constant temperature, the equilibrium constant (kp) 2NO2 is for the decomposition reaction N2O4 expressed by kp = (4x2P)/(1–x2), where P = pressure, x = extent of decomposition. Which one of the following statement is true ? (A) kp increases with increase of P (B) kp increases with increase of x (C) kp increases with decrease of x (D) kp remains constant with change in p and x

7.

Cl

55

P P P P

Q Q Q Q

R R R R

S S S S

T T T T

JULY 2010

16. Calculate the change in pressure (in atm) when 2 mole of NO and 16 gram O2 in a 6.25 litre originally at 27ºC react to produce the maximum quantity of NO2 possible according to the equation – 2NO(g) + O2(g) → 2NO2(g) 1 (Take R = ltr. Atm/mol-K) 12

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 extraction processes listed in Column-I with metals listed in Column-II : Column-I Column-II (A) Self reduction (P) Lead (B) Carbon reduction (Q) Silver (C) Complex formation (R) Copper and displacement by metal (D) Decomposition of (S) Boron iodide (T) Au

17. the number of isomers for the compound with molecular formula C2BrCIFI is. 18. In P4O10 each P atom is linked with ...........O atoms. 19. 0.15 mole of Pyridinium chloride has been added into 500 cm3 of 0.2 M pyridine solution. Calculate pH (Kb for pyridine = 1.5 × 10–9 M)

MATHEMATICS

14. Match statements in Column-I with appropriate solution in Column-II Column-I Column-II (A) A solution having (P) 5M HCl pH less then 7 (B) A solution having (Q) 1M NaCl pH more than 7 (C) A solution having (R) 0.1 M Na2CO3 pH almost equal to 7 (D) solution having (S) 0.1 M CaCl2 negative value of pH (T) 1M H2SO4

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.

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

n

2.

If

∑α

n

= an2 + + bn, where a, b are constants and

n =1

α1, α2, α3 ∈ {1, 2, 3........9} and 25α1, 37α2, 49α3 be α2 α3 α1 three digit number then 5 7 9 is 25α1 37α 2 49α 3 equal to (A) α1 + α2 + α3 (C) 7

15. A solution is prepared by mixing 50 mL 0.1 M HCl with 50 mL 2.9 M CH3CH2COOH and 100 mL 0.2 M CH3CH2COONa. Find pH of the resulting solution. Ka for CH3CH2COOH is 1 × 10–5.

XtraEdge for IIT-JEE

α 0 α If A = 2β β – β is an orthogonal matrix, then γ – γ γ the number of possible triplets (α,β,γ) is (A) 8 (B) 6 (C) 4 (D) 2

56

(B) α1 – α2 + α3 (D) 0

3.

Consider the matrix A, B, C, D with order 2 × 3, 3 × 4, 4 × 4, 4 × 2 respectively. Let x = (αABγC2D)3 where α & γ are scalars. Let |x| = k|ABC2D|3, then k is (A) αγ (B)α2γ2 4 4 (C) α γ (D) α6γ6

4.

Three numbers are selected at random from the set {1, 2, 3...........N}, one by one without replacement. If the first number is known to be smaller than second, then the probability that third selected number lies between the first two numbers is JULY 2010

(A)

1 2

(B)

1 3

(C)

1 6

(D)

Passage: II (Ques. 10 to 12)

1 8

consider the equation of two straight lines →

5.

6.

→

→

→

→

→

→

d

is

5 (B) 9

(A)

5 (C) 12

9.

^

^

^

^

2 5

1 (C) 2

→

^

^

^

→

^

^

^

(D)

^

^

^

^

^

^

(B) 2 29

29

(D) 4 29

(D) None of these

P Q R S T P Q R S T P Q R S T P Q R S 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. A B C D

1 4

13. Match the following : Column-I (A) Through (γ, γ + 1) there can not be more than one normal to parabola y2 = 4x if (B) If two cicles (x–1)2 + (y–1)2 = γ2 and x2 + y2 – 8x + 2y + 8 = 0 intersect at two points, then

2 (D) 3

Expectation of the number on the card is (A) 2 (B) 2.5 (C) 3 (D) 3.5

XtraEdge for IIT-JEE

^

P Q R S T

1 and then 10 a card is drawn. Let Ei represents the event that a card with number 'i' is drawn.

P(A3/E2) is equal to 1 1 (A) (B) 3 4

^

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 DS, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

the probability of selection of box Ai is

8.

^

(C) (–10, 53, –4)

There are four boxes A1, A2, A3 and A4. Box Ai has i cards and on each card a number is printed, the numbers are from 1 to i, A box is selected randomly,

(C)

^

12. The point of intersection of the lines is – 10 53 – 4 10 – 53 4 , , , (A) (B) , 3 3 3 3 3 3

5 (D) 18

Passage : I (Ques. 7 to 9)

P(E1) is equal to 1 1 (A) (B) 5 10

^

(C) 3 29

→ ^

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.

7.

^

11. The length of shortest distance between the lines is -

^

then c . i + c . j + c . k is equal to 5 (A) 3

^

^

(D) r = ( 4 i + 6 j + 8 k ) – δ( i + j + k )

→

→ → i j k – + and the angle between a and b is 30º, 6 3 3

→ ^

^

(C) r = ( 3 i + 5 j + 7 k ) – δ(4 i + 6 j + 8 k )

perpendicular to all of them. If ( a × b ) × ( c × d ) =

→ ^

^

(B) r = ( i + j + k ) + δ(3 i + 5 j + 7 k )

→

respectively are coplanar. A unit vector

^

^

(A) r = ( 3 i + 5 j + 7 k ) – δ( i + j + k )

Vectors a , b and c with magnitude 2, 3 & 4

^

^

x +1 y +1 z +1 = = . 7 –6 1 10. The equation of the line of shortest distance between the given lines is →

→

^

^

r = 3 i + 5 j + 7 k + λ( i – 2 j + k ) and

A bag 'A' contains 2 white and 3 red balls, another bag 'B' contains 4 white and 5 red balls. If one ball is drawn at random from one of the bag and it is found to be red, the probability that it was drawn from the bag B is 89 25 93 24 (A) (B) (C) (D) 245 52 256 663

57

Column-II (P) 2 < γ < 8

(Q) – 2 < γ < 2

JULY 2010

(C) Point (γ, γ + 1) lies inside the (R) γ < – 2 2 2 circle x + y = 1 for (D) Both equation x2 + y2 + 2γx + 4 = 0 (S) – 1 < γ < 0 and x2 + y2 – 4γy + 8 = 0 represent real circles if (T) γ > 8

15. There are N + 1 identical boxes each containing N wall clocks. The first box contains zero defective clocks. The second box contains one defective and (N – 1) effective clocks, in general rth box contains (r – 1) defective and (N – r + 1) effective clocks (1 ≤ r ≤ N + 1). Thus, the (N + 1)th box contains all defective clocks. A wall clock is selected and found an effective one. The probability that it is from kth γN – γK + γ find γ. box is N2 + N

3 – 4 1 / 2 3 14. Consider the matrix A = ;B= . 1 – 1 0 1 Let P be an orthogonal matrix and Q = PAPT, Rk = PTQk.P, S = PBPT & Tk = PTSkP where k ∈ N Column-I Column-II

16. If a determinant of order 3 × 3 is formed by using the numbers 1 or – 1then it minimum value of determinant is – γ find the value of γ.

5

(A)

∑a

k

, where ak represents the element

(P) –9

17. If equation of the plane through the straight line y+2 z x –1 = = and perpendicular to the plane –3 5 2 x – y + z + 2 = 0 ia ax – by + cz + 4 = 0, then find the 10 3 a + 10 2 b + 10c . value of 342

k =1

of first row & first column in matrix Rk 3

(B)

∑b

k

, where bk represents the element

(Q) 10

k =1

of second row & second column in matrix Rk

3 18. If A = 2 4 3 0 3 2 1 0 4 0 2

3 1 – 1 . Solve the system of equations – 3 2 x 8 2 y y = 1 + z , then find the value z 4 3y z y + of x + 3 2

∞

(C)

∑

X k , where Xk represents the element

(R) 35

k =1

of first row & first column in matrix Tk 10

(D)

∑y

k

, where yk represents the element

(S) 1

k =1

of second row and second column in Matrix TK (T) 15

19. Find the coefficient of x in the determinant

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 0 1 2 3 4 5 6 7 8 9

XtraEdge for IIT-JEE

Y 0 1 2 3 4 5 6 7 8 9

Z 0 1 2 3 4 5 6 7 8 9

–2

(1 + x ) a1b1

(1 + x ) a1b 2

(1 + x ) a1b3

(1 + x ) a 2 b1

(1 + x ) a 2 b 2

(1 + x ) a 2 b3 where ai, bj ∈N

(1 + x ) a 3b1

(1 + x ) a 3b 2

(1 + x ) a 3b3

How to Handle Difficult People A bully at your work is difficult for you to face. He is demanding you do part of his job without pay or credit. How do you handle it?

W 0 1 2 3 4 5 6 7 8 9

Your neighbors are constantly fighting. They wake you up in the middle of the night with their screams and curses. What do you say to them? Your father is unhappy about your career choice. He constantly criticizes your work and points out what he thinks you should do. How do you deal with him?

58

JULY 2010

Based on New Pattern

IIT-JEE 2012 XtraEdge Test Series # 3

Time : 3 Hours Syllabus : Physics : Calorimetry, K.T.G.,Thermodynamics, Heat Transfer, Thermal expansion, Transverse wave,

Sound wave, Doppler's effect. Chemistry : Chemical Equilibrium, Acid Base, Ionic Equilibrium, Classification & Nomenclature, Isomerism , Hydrogen Family, Boron Family & Carbon Family, S-block elements. Mathematics: Point, Straight line, Circle, Parabola, Ellipse, Hyperbola, Vector, 3-D

Instructions : Section - I • Question 1 to 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 Match type questions 8 marks will be awarded for correct answer and 0 mark for wrong answer. Section - IV • Question 15 to 19 are numerical response questions (with single digit Answer). 3 marks will be awarded for correct answer and 0 mark for wrong answer.

(A) 52 W (C) 512 W

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.

2.

3.

4.

In case of standing waves (A) At nodes particles displacement is time dependent (B) At antinodes displacement of particle may or may not be zero (C) Wave does not travel but energy is transmitted (D) Components waves traveling in same direction having same amplitude and same frequency are superimposed

5.

For an ideal gas graph is shown for three processes. Processes 1, 2 and 3 are respectively – Work done (magnitude)

The molar heat capacity for a process is : R α C= + , then process equation is 1– γ T (A) Ve–(α/RT)T = constant (B) Ve (α/R)T = constant (C) VT = constant (D) Veα/RT = constant

3

Three black bodies 1,2,3 have radius r1 < r2 < r3. Emissive powers of black bodies at Temperature T are E1, E2, E3, Then correct relation between them is (A) E1 < E2 < E3 (B) E1 > E2 > E3 (D) E1 > E2 < E3 (C) E1 = E2 = E3

2 1 Temperature change (A) Isochoric, isobaric, adiabatic (B) Isochoric, adiabatic, isobaric (C) isobaric, adiabatic, isochoric (D) Adiabatic, isobaric, isochoric

A taut string for which µ = 5 × 10–2 kg/m is under tension of 80 N. How much is the average rate of transport of potential energy if the frequency is 60 Hz and amplitude 6 cm - (Given 4π2 = 39.5)

XtraEdge for IIT-JEE

(B) 256 W (D) 215 W

59

JULY 2010

6.

Figure shows cyclic process. From c to b, 40 J is transferred as heat from b to a, 130 J is transferred as heat, and work done is 80 J from a to c, 400 J is transferred as heat then – P c

11. Change in internal energy in process CA (A) 900 R (B) 300 R (C) 1200 R (D) zero 12. Heat transferred in the process BC is (A) 1000 R (B) 500 R (C) 2000 R (D) 1500 R

b

a

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 DS, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

V (A) Work done in process a to c is 310 J (B) Net work done is cycle is 230 (C) Net change in internal energy in cycle is 130 J (D) None of these

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.

A B C D

Passage : I (Ques. 7 to 9)

Many waveforms are described in terms of combinations of travelling waves. Superposition principle is used to analyse such wave combinations. Two pulses travelling on same string are described by5 –5 y1 = , y2 = (3x – 4t ) 2 + 2 (3x + 4t – 6) 2 + 2 7.

The direction in which each pulse is travelling is (A) y1 is in positive x-axis, y2 is in positive x-axis (B) y1 is in negative x-axis, y2 is in negative x-axis (C) y1 is in positive x-axis, y2 is in negative x-axis (D) y1 is in negative x-axis, y2 is in positive x-axis

8.

The time when the two waves cancel everywhere (A) 1 sec (B) 0.5 sec (C) 0.25 sec (D) 0.75 sec

9.

The point where two waves always cancel(A) 0.25 m (B) 0.5 m (C) 0.75 m (D) 1 m

14.

C

P0 T

10. Work done in process AB is (A) 400 R (B) – 400 R (C) 200 R (D) – 300 R

XtraEdge for IIT-JEE

T T T T

13. Match the standing waves formed in column-II due to plane progressive waves in Column-I and also with conditions in Column-I Column-I Column-II (A) Incident wave is (P) y = 2A cos kx sin ωt y = A sin (kx – ωt) (B) Incident wave is (Q) y = 2A sin kx cos ωt y = A cos (kx – ωt) (C) x = 0 is rigid support (R) y = 2A sin kx cos ωt (D) x = 0 is flexible support (S) y = 2A cos kx cos ωt (T) None of these

One mole of monoatomic gas is taken through above cyclic process. TA = 300 K Process AB is defined as PT = constant P B

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.

Passage: II (Ques. 10 to 12)

3P0

P Q R S Q R S Q R S Q R S Q R S

P P P P

60

Column-I (A) Specific heat capacity S

Column-II (P) l1 – l2 = constant for l1α1 = l2α2 (B) Two metals (l1, α1) and (Q) Y is same (l2, α2) are heated uniformly (C) Thermal stress (R) S = ∞ for ∆T = 0 (D) Four wires of same (S) Y α ∆t material (T) None of these

JULY 2010

1.

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

(A) 2-Bromo-4-carbamoyl-5-chloroformyl-3-formyl hexanoic acid (B) 5-Bromo-3-carbamoyl-2-chloroformyl-4-formyl hexanoic (C) 4-Formyl-2-chloroformyl-3-carbamoyl-5-bromo hexanoic acid (D) 2-Chloroformyl-3-carbamoyl-4-formyl-5-bromo hexanoic acid 2.

Geometry in the given compound is CH3 H H

CH3

(A) cis (B) trans (C) cis as well as trans (D) no geometrical isomerism

15. If the volume of a block of metal changes by 0.12 % when heat is changed from 40ºC to 60ºC, find the linear expansion coefficient of the metal ? [Ans. in …… × 10–5/ºK]

3.

16. Calculate the pressure exerted by a mixture of 8 g of oxygen, 14 g of nitrogen and 22 g of carbon di-oxide in a container of 30 litres at a temperature of 27ºC. [Ans. in …… × 105 N/m2] 17. A sphere and a cube of same material and total surface area placed in an evacuated chamber turn by turn and heated to the same temperature. Calculate the ratio of the rate of cooling of spherical to cubical surface. [Ans. in …… × 10–1] π 18. Two oscillating waves have a phase difference of 2 is 25 oscillations. What is the percentage difference in their frequency ? 19. For a certain organ pipe, three successive resonance observed are 425, 595 and 765 Hz. Taking the speed of sound to be 340 ms–1 , find the length of the pipe, in metre.

CHEMISTRY

The structure of spiro [3,3] heptane is (A)

(B)

(C)

(D)

4.

The pH of a 10–8 molar solution of HCl in water is (A) 8 (B) – 8 (C) between 7 and 8 (D) between 6 and 7

5.

Consider the following equilibrium in a closed 2NO2(g). At a fixed container N2O4(g) temperature, the volume of the reaction container is halved. For this change, which of the following statements holds true regarding the equilibrium constant (kp) and degree of dissociation (α) ? (A) neither kp nor α changes (B) both kp and α change (C) kp changes, but α does not change (D) kp does not change, but α changes

6.

For H3PO3 and H3PO4 the correct choice is (A) H3PO3 is dibasic and reducing (B) H3PO3 is dibasic and non-reducing (C) H3PO4 is tribasic and reducing (D) H3PO3 is tribasic and non-reducing

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.

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.

XtraEdge for IIT-JEE

The IUPAC name of the given compound OHC CONH2 | | CH3–CH–CH–CH–CH–COOH is | | COCl Br

61

JULY 2010

10. Which one of the following is most stable conformer? Cl Cl CH3 Cl CH3 H (A) (B) H H CH3 CH3 H Cl Cl Cl Cl H Cl (C) (D) H H CH3 CH3 H CH3 CH3

Passage : I (Ques. 7 to 9)

In a reversible chemical reaction, the rate of forward reaction decreases and that of backward reaction increases with the passage of time; at equilibrium the rate of forward and backward reaction become same. Let us consider the formation of SO3(g) in the following reversible reaction : 2SO2(g) + O2(g) 2SO3 (g) Following graphs are plotted for this reactions

A. B. C.

Conc.

t1 t2 t3

9.

(A)

rf

(B)

rb

H

(C)

rb time

(D)

OH OH OH

H

OH

H

OH

(D)

CH3

H

CH3 OH

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 DS, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

rb rf time

Passage: II (Ques. 10 to 12)

A B C D

Different spatial arrangements of the atoms that result from restricted rotation about a single bond are conformers. n-Butane has four conformers eclipsed, fully eclipsed, gauche and anti. The stability order of these conformers are as follows: anti > gauche > partial eclipsed > fully eclipsed Although anti is more stable than gauche but in some cases gauche is more stable than anti.

XtraEdge for IIT-JEE

H

12. Number of possible conformers of n-butane is (A) 2 (B) 4 (C) 6 (D) infinite

time

rf

(B)

CH3

rb rf

rate of reaction

(C)

rate of reaction

time

OH CH3

Which of the following represent the rates of forward reaction (rf) and rates of backward reaction (rb) at equilibrium ? (A)

H CH3

In the above graph, the equilibrium state is attained at time (A) t1 (B) t2 (C) t3 (D) t4

rate of reaction

8.

t4

In the above graph, A,B & C respectively are (A) SO3, SO2 and O2 (B) SO3, O2 and SO2 (C) SO2, O2 and SO3 (D) O2, SO2 and SO3

rate of reaction

7.

time

11. Which one of the following is the most stable conformer ? CH3 CH3 HO H H CH3

P Q R S P Q R S P Q R S P Q R S P Q R S

T 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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JULY 2010

13. Match the following : Column-I (A) N2(g) + 3H2(g) 2NH3(g) ; ∆H = –ve (B) N2(g) + O2(g) 2NO(g); ∆H = +ve (C) A(g) + B(g) 2C(g) + D(g); ∆H = +ve PCl3(g) + Cl2(g); ∆H = +ve (D) PCl5(g) Column-II (P) K increases with increase in temperature (Q) K decreases with increase in temperature (R) Pressure has no effect (S) Product concentration, increases due to addition of inert gas at constant pressure (T) Product concentration, increases due to addition of inert gas at constant volume

R(Gas constant) = Log e = 2.3}

16. Number of configurational dibromocinnamic acid is .

isomers

of

2,3-

17. Consider the reaction AB2(g) ABg + B(g). It the initial pressure of AB2 is 100 torr and equilibrium pressure is 120 torr. The equilibrium constant Kp in terms of torr is. 18. Dissociation of H3PO3 occurs in ......... stages. 19. The number of hydroxyl groups in pyrophosphoric acid is.

14. Match the following : Column-I Column-II 3+ + (A) Bi → (BiO) (P) Heat (B) [AlO2]– → Al(OH)3 (Q) Hydrolysis (C) [SiO4]4– → [Si2O7]6– (R) Acidification (D) [B4O7]2– → [B(OH)3] (S) Dilution by water (T) Basification

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

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

1.

A circle C1 of radius b touches the circle x2 + y2 = a2 externally and has its centre on the positive x-axis; another circle C2 of radius c touches the circle C1 externally and has its centre on the positive x-axis. Given a < b < c, then the three circles have a common tangent if a, b, c are in (A) A.P. (B) G.P. (C) H.P. (D) None of these

2.

P is a point on the axis of the parabola y2 = 4ax; Q and R are the extremities of its latus rectum, A is its vertex. If PQR is an equilateral triangle lying within the parabola and ∠AQP = θ, then cos θ = (A) (C)

3.

15. Calculate the pH at which the following conversion (reaction) will be at equilibrium in basic medium I–(aq.) + IO3– (aq.) I2 (s) When the equilibrium concentrations at 300 K are [I–] = 0.10 M and [IO3–] = 0.10 M. {Given → ∆Gfº (I–, aq.) = – 50 kJ./mol, ∆Gfº (IO3–, aq) = – 123.5 KJ/mol, ∆Gfº (H2O, l) = –233 KJ/mol ∆Gfº (OH–, aq.) = – 150 KJ/mol

XtraEdge for IIT-JEE

25 J/mol–K 3

2 5 5 –2 2 3

(B)

9 8 5

(D) None of these

x2 y2 + = 1, 25 9 perpendicular to the asymptote of the hyperbola x2 y2 – = 1 passing through the first and third 16 9 quadrants is : 100 150 (A) (B) 431 481

The length of the diameter of the ellipse

(C)

63

2– 3

25 3

(D) 11 2

JULY 2010

4.

5.

6.

→

→

→

→

→

9.

→

If a , b , c are such that [ a , b , c ] = 1, → → → → → → → 2π c = γ( a × b ), a ^ b < , and | a | = 2 , | b | 3 → → → 1 = 3,|c |= , then the angle between a and b 3 is π π π π (A) (C) (B) (D) 6 3 4 2

Passage: II (Ques. 10 to 12) →

→ → → → 3 | a || c | a c + → (A) → → → 3 | c | +2 | a | | a | | c | → → a c + → (B) → → → 3 | c | +2 | a | | a | | c | →

p1, p2 are lengths of perpendicular from foci on tangent to ellipse and p3, p4 are perpendiculars from extremities of major axis and p from centre of ellipse p p – p2 on same tangent, then 1 2 equals p 3p 4 – p 2

→ → → → 2 | a || c | a c + → (C) → → → 3 | c | +2 | a | | a | | c | (D) none of these

11. The position vector of point F is -

(B) e (D) None of these

1 | a |→ (A) a + → c 3 |c| 2| a |→ (C) a + → c |c|

XtraEdge for IIT-JEE

→

→

| a |→ (B) a + → c |c| →

→

| a |→ (D) a – → c |c|

→

12. The vector AF , is given by →

1| a |→ (A) c 3 → |c| →

| a |→ (C) 2 → c |c|

→

| a |→ (B) → c |c| →

| a |→ (D) – → c |c|

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 DS, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

If L represents the line joining the point P on C to its centre O, then equation of the tangent at M to the ellipse E is 5 =0

→

→

x2 y2 + = 1, L : y = 2x 9 4 P is a point on the circle C, the perpendicular PQ to the major axis of the ellipse E meets the ellipse at M, MQ is equal to then PQ (A) 1/3 (B) 2/3 (C) 1/2 (D) none of these

(C) x + 3y +

→

→

Let C : x2 + y2 = 9, E :

(A) x + 3y = 3 5

→

| a || c |

Passage : I (Ques. 7 to 9)

8.

→

10. The position vector of point P, is -

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.

7.

→

In a parallelogram OABC vector a , b , c are respectively the position vectors of vertices A, B, C with reference to O as origin. A point E is taken on the side BC which divides it in the ratio of 2 : 1. Also the line segment AE intersects the line besecting the angle O internally in point P. If CP when extended meets AB in point F, then

Equation of a plane which passes through the point of y–2 z–3 x –1 = and intersection of lines = 3 1 2 z–2 x –3 y –1 = = and at greatest distance 3 1 2 from the point (0, 0, 0) is (A) 4x + 3y + 5z = 25 (B) 4x + 3y + 5z = 50 (C) 3x + 4y + 5z = 49 (D) x + 7y – 5z = 2

(A) e (C) e2

Equation of the diameter of the ellipse E conjugate to the diameter represented by L is (A) 9x + 2y = 0 (B) 2x + 9y = 0 (C) 4x + 9y = 0 (D) 4x – 9y = 0

(B) 4x + 3y = 5 (D) 4x +3y + 5 = 0 64

JULY 2010

(D) The radius of the circular section

P Q R S T P Q R S T P Q R S T P Q R S T P Q R S 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 Column-II (A) If lines x + 2y – 1 = 0, ax + y + 3 = 0 (P) 4 and bx – y + 2 = 0 are concurrent, the least distance from origin to A B C D

(a, b) is S. The value of 58 . S is (B) A,B are two fixed points on a line L. Let locus of point P such that PA = 2PB be a curve cutting line 1 L at R and S. If slope of PR is – , 2 then slope of PS is (C) Let tangents at P and Q to curve y2 – 4x – 2y + 5 = 0 intersect at T. If S(2, 1) is a point such that SP.QS = 16 then the length ST is (D) Let the double ordinate PNP' of the hyperbola

x2 (π – 1)

2

–

of the sphere | r | = 5 by the plane →

X

0 1 2 3 4 5 6 7 8 9

(Q) 5

(R) 2

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. Two circle of radii 'a' and 'b' touching externally are

(S) π – 1

inscribed in area bounded by y = 1 – x 2 1 1 x-axis. If b = and a = , then k is ............... 2 k

and

16. If a circle S (x, y) = 0 touches at the point (2, 3) of

the line x + y = 5 and S (1, 2) = 0, then of such circle is.

Column-II (P) 1

(Q)

2 × radius

17. Consider two concentric circles C1 : x2 + y2 = 1, C2: x2 + y2 = 4. Tangents are drawn to C1 from any point P on C2. These tangents again meet circle C2 at A & B. It can be proved that locus of point of intersection of tangents drawn to C2 at A and B is a circle, what is the radius of that circle.

14.

18. In a regular tetrahedron let θ be the angle between any edge and a face not containing the edge. a If cos2θ = where a, b ∈ I+ also a and b are b 5 coprime, then find the value of (10a + b) 13

1 6

the lines

XtraEdge for IIT-JEE

^

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 :

(T) (π–1)2

x –1 z–3 y–2 = = 2 3 4 x–2 x –5 y–4 and = = is 3 5 4 (C) The points (0, –1, –1), (4, 5, 1), (3, 9, 4) and (–4, 4, k) are coplanar then k =

^

(T) 2

preduced both side to meet asymptotes in Q and Q'. The product PQ. PQ' is equal to

(B) The shortest distance between

^

r . ( i + j + k ) = 3 3 is

y2 = 1 is (π – 1)

Column-I (A) The distance of the point (1, – 2, 3) from the plane x – y + z = 5 measured parallel to the line 1 1 1 x = y = – z is 3 2 6

(S) 3

→

19. Let A (1, 2), B (3, 4) be two point and C (x,y) be a point such that (x – 1) (x – 3) + (y – 2) (y – 4) = 0. If area of ∆ABC is 1 sq, unit. Then maximum number of positions of C in xy plane is.

(R) 4

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JULY 2010

XtraEdge Test Series ANSWER KEY IIT- JEE 2011 (July issue) PHYSICS Ques Ans Column Match Numerical Response

1 C 13 14 Ques Ans

2 3 A B A → P, Q, R, S A→R 15 16 8 5

Ques Ans Column Match Numerical Response

1 C 13 14 Ques Ans

2 3 B C A → P, R A → P, S, T 15 16 4 2

Ques Ans Column Match Numerical Response

1 A 13 14 Ques Ans

2 3 D D A → Q, R, S A→R 15 16 2 4

4 D

17 1

5 6 C D B → P, Q, R, S B→P 18 19 4 2

7 A

8 C C → R, S C→Q

9 B

10 D

11 A D → R, S D → P, S

12 B

7 D

8 B C → Q, T C→Q

9 D

10 D

11 B D→S D→P

12 A

8 B C→S C→S

9 A

10 C

11 12 B D D → P, R, T D→Q

7 C

8 D C → Q, R C→S

9 D

10 B

11 A D → P, S D→Q

12 C

7 A

8 C C → P, S C→P

9 A

10 A

11 C D → P, S D→R

12 D

8 A C→P C→R

9 B

10 A

11 A D→S D→R

12 A

C HE M ISTR Y 4 A

17 6

5 A B → P, R B → R, S 18 4

6 D

19 5

MATHEMATICS 4 C

17 5

5 B B→P B→P 18 3

6 D

7 C

19 0

IIT- JEE 2012 (July issue) PHYSICS Ques Ans Column Match Numerical Response

1 D 13 14 Ques Ans

2 C A → P, R A→R 15 2

Ques Ans Column Match Numerical Response

1 B 13 14 Ques Ans

2 B A→Q A→Q 15 8

3 B

4 B

16 3

17 7

3 D

4 D

16 4

17 5

5 A B → Q, S B→P 18 5

6 A

19 7

C HE M ISTR Y 5 D B → P, R B→S 18 2

6 A

19 4

MATHEMATICS Ques Ans Column Match Numerical Response

1 B 13 14 Ques Ans

2 A A→Q A→P 15 4

XtraEdge for IIT-JEE

3 B

4 B

16 1

17 4

5 B B→R B→Q 18 5

6 C

7 B

19 4

66

JULY 2010

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