6Xtraedge 2010 June

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XtraEdge for IIT-JEE

1

JUNE 2010

XtraEdge for IIT-JEE

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JUNE 2010

To Determine Your Priorities, Examine Your Goals

Volume - 5 Issue - 12 June, 2010 (Monthly Magazine) Editorial / Mailing Office :

Editorial

112-B, Shakti Nagar, Kota (Raj.) 324009 Tel. : 0744-2500492, 2500692, 3040000 e-mail : [email protected] Editor :

Dear Students,

Everyone knows that setting goals will help you achieve more and adds excitement and meaning to life. But setting a goal is only the beginning. We often fail to follow through and our goals turn into unfulfilled daydreams. To eliminate that pitafall, here is a systematic, approach that will help you turn your goals into realities.

Pramod Maheshwari [B.Tech. IIT-Delhi] Cover Design Govind Saini, Om Gocher



Decide what you want to achieve. Determine exactly what you want. Be specific. Be sure your goal is measurable, so you can tell when you're making progress. Pick a target date for a achieving it. Be sure it is realistically achievable.



Ask yourself why it is important for your to achieve this goal. How you will benefit from reaching this goal ? Knowing why you want something raises your level of motivation. The higher your motivation level, the more likely you are to act on your goal.



• No Portion of the magazine can be published/ reproduced without the written permission of the publisher

Consider what obstacles, problems or personal shortcomings might block your progress. List every one you can think of some obstacles will be real, others may be only imaginary. You must conquer both.



Examine the obstacles one at a time, and think about how you might solve each problem.

• All disputes are subject to the exclusive jurisdiction of the Kota Courts only.



List the people or organizations who could help you achieve your goal. Decide specifically what you will ask them to do.

Every effort has been made to avoid errors or omission in this publication. In spite of this, errors are possible. Any mistake, error or discrepancy noted may be brought to our notice which shall be taken care of in the forthcoming edition, hence any suggestion is welcome. It is notified that neither the publisher nor the author or seller will be responsible for any damage or loss of action to any one, of any kind, in any manner, there from.



Consider what information you need that you don't have now. Where will you get it? What could you read? Who could you talk to? What seminars could you attend?



Write out a detailed action plan for achieving your goal. What are the priorities involved. ? Which tasks must be done first? When will different actions take place?

Layout Rajaram Gocher Circulation & Advertisement Ankesh Jain, Praveen Chandna Ph (0744)- 3040007, 9001799502 Subscription Suruchi Kabra Ph. 0744-2500492, 2500692 © Strictly reserved with the publishers

Unit Price Rs. 20/-

Setting a goal is a good step, but it is only the beginning. It takes all seven steps to make sure you actually follow through, and by so doing achieve your goal.

6 issues : Rs. 100 /- [One issue free ] 12 issues : Rs. 200 /- [Two issues free]

I guarantee that you will succeed and will secure a good rank in your exams if you make a habit of never to postpone your work.

Special Subscription Rates 24 issues : Rs. 400 /- [Four issues free]

Forever presenting positive ideas to your success. Yours truly

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

Editor : Pramod Maheshwari XtraEdge for IIT-JEE

Pramod Maheshwari, B.Tech., IIT Delhi

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JUNE 2010

XtraEdge for IIT-JEE

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JUNE 2010

Volume-5 Issue-12 June, 2010 (Monthly Magazine) NEXT MONTHS ATTRACTIONS

CONTENTS INDEX

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

PAGE

NEWS ARTICLE

4

IITian ON THE PATH OF SUCCESS

6

KNOW IIT-JEE

8

IIT Kharagpur set to devise high-speed trains IIT-Delhi to take in 850 more students in B-Tech this year Srikanth Jagabathula

Previous IIT-JEE Question

Study Time........ DYNAMIC PHYSICS S Success Tips for the Months • If you can't make a mistake, you can't make anything. • Sometimes a big step is safer; you can't cross a ditch in small jumps

8-Challenging Problems [Set# 2] Students’ Forum Physics Fundamentals Electrostatics - 2 Newton's Law of motion

CATALYSE CHEMISTRY

• Children focus on what they can’t do. Adults focus on what they can do.

• You never need to feel fear if you don't want to do anything. • You got to know when to hold ‘em and know when to fold ‘em…

DICEY MATHS

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

• To understand motivation, know the power of the Hunter.

XTRAEDGE TEST SERIES

51

Class XII – IIT-JEE 2011 Paper Class XI – IIT-JEE 2012 Paper AIEEE - 10 Examination paper with Solution

XtraEdge for IIT-JEE

39

Mathematical Challenges Students’ Forum Key Concept Inverse Trigonometric Function Quadratic Equation

• An ounce of success is worth a pound of positive thinking.

• Defeat is advance payment for victory.

31

Key Concept Nomenclature & Isomerism Electro Chemistry Understanding : Inorganic Chemistry

• Self-confidence grows not from what you can do, but what you know you can do.

• The secret of confidence is to know your resources.

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3

69

JUNE 2010

IIT Kharagpur set to devise high-speed trains As per a MoU signed recently between IIT Kharagpur and Railway Ministry, the former will carry out extensive research on developing cutting edge technologies , including highspeed trains and better security mechanisms for Indian Railways. As per media reports, 12 areas have been identified by the institute and the ministry in which research will be initiated. A Centre for Railway Research (CRR) has been set up by the ministry at the institute, which comprises faculty of at least 10 departments to initiate research in the chosen areas. The railway ministry will invest an initial amount of Rs 120 crore to get the centre working. The research would focus on areas such as heavy haul technology, vehicle dynamics, high-speed technologies, energy-efficient traction power supply systems, track research, use of artificial intelligence for predictive maintenance and management, material sciences for railway-related composites, including rubber, polymer and insulation materials, development of integrated/embedded processors for railway applications, applications for access control, security and safety, including biometrics, nonconventional drives and technology, including Maglev, LIM and remote sensing, and measurement of overhead equipment, tracks and signals, says a report published in The Times of India.

XtraEdge for IIT-JEE

Speaking to media, Siddhartha Mukherjee, a faculty member of the electrical engineering department who is also a spokesperson for CRR, said, “We are looking at high-speed trains. This doesn’t mean designing only the train and the engine, but also tracks that will support such trains. So, while a lot of stress would be laid on vehicle dynamics, an important portion of the research would focus on developing fracture-proof tracks and sensing equipment that would diagnose failures on time. Withstanding the load of high-speed trains is not easy and would mean extensive relaying of tracks.” Centre for Railway Research will offer PhD programmes in research areas related to the railways. It will also involve IIT BTech and MTech students in research projects and offer course electives related to railway technology.

IIT-Delhi to take in 850 more students in B-Tech this year In order to accommodate the quota for backward classes, the Indian Institute of Technology (IIT)-Delhi will increase its students intake by 27 percent this year. In all, the institute will admit 850 students at the B-Tech level. In an interview to IANS, IIT-Delhi director Surendra Prasad said, “We are adding 27 percent students more this time. With this, the government rule of increasing students’ intake by 54 percent to accommodate 27 percent OBC quota has been done.” 4

Authorities in IIT Delhi reveal that 300 more students have been enrolled in the premier institute in last three years to accommodate the OBC quota as mandated by the central government. As per the order issued by central government, all educational institutes aided by government have been ordered to implement the OBC quota.

Ropar in tier-I city league by IIT The faculty and the student community of IIT-Ropar are still at the budding stage. In the coming years they are all set to rule the academic affairs of this region. They also expect that the establishment of a esteemed institute like IIT will certainly bring Ropar in league of tier-I cities of the countries. The institute is expanding at a great pace and it is expected that it would certainly be profitable for other colleges because the expert faculty through interactive sessions will impart knowledge to them, revealed by IIT director MK Surrapa. Being the only IIT in the region IIT-Ropar will prove to be a boon for this region. He disclosed that the procedure for recruitment of expert in all the streams has already begun and by the end of March IIT-Ropar will have more than 50 regular faculty members. The total strength of the college is 209 out of which 190 are boys. British minister, Pat McFadden visited to the IIT for business, innovation and skills. He also interacted with the students and JUNE 2010

took a round of the institute. IIT being renowned institute, he said the collaboration of UK and IIT Ropar under research initiative will help in expansions of institutes and better research. He also added that the work between IIT and their universities will open great opportunities for the Indian students as well as British students. It’ll also include student exchange programmes for better development.

admission to students on the basis of these performance cards,” Baruah said.

Prof BK Dhindaw, dean academic and research, said, “We would be carrying forward the work that would be provided by varsities in UK,”

IIT JEE to transparent

The student’s view point is that after the establishment of IIT in Punjab, more multinational companies in the region would provide job opportunities.

IIT-JEE candidates to get performance cards now Students appearing for the next Joint Entrance Examination (JEE) for admission to IITs will get performance cards specifying marks and the ranks secured by them in the test. However, as per the new provision, they cannot seek regrading or re-totalling. For the first time, the JEE Board would issue performance cards which can be considered as certificates by many other institutions wanting to give admission to JEE candidates. The board will also put out the answers of the questions on its website to help students make assessment of their performance. IIT Guwahati Director Prof Gautam Baruah said the board had urged for issuing such performance cards which would serve as certificates for the students. “Many other institutes, which want to take JEE candidates, can give

XtraEdge for IIT-JEE

At present, the IITs are not issuing any scorecards to students. Certain institutes, which are giving admission to JEE candidates, have to get authenticated data from the JEE Board on the list of students and marks secured by them. The answers will be put in the JEE website two weeks after the exam.

be

more

The Central Information Commission (CIC) has taken a decision that the IIT Joint Entrance Examination process will be more transparent. The details of candidates such as registration number, name, parents’ name, category, and marks secured should be open to public scrutiny. Information Commissioner Shailesh Ghandhi overruled the objections by IIT Guwahati which organized the JEE 2009 said that these providing these details under the Right to Information Act does not constitute invasion of privacy. The decision is being viewed as an important step in making the process transparent as disclosure of 2006 data had shown that formulas for calculating subject cut-offs did not tally. Irregularities were alleged in marks scored by wards of faculty members.

Court refuses to IIT-JEE 2010 result

stay

New Delhi: The Delhi High Court declined to stay the declaration of this year's results for the Indian Institute Of Technology's (IIT) Joint Entrance Examination (JEE) but asked the institutes to explain as to why so many mistakes crept in during the all-India test. 5

A division bench of acting Chief Justice Madan B. Lokur and Justice Mukta Gupta declined to stay the declaration of IIT-JEE result but directed the IIT to explain by way of affidavit how the mistakes occurred. "Demonstrate to us the software with which you set these papers and also how the papers are scrutinized," the court said when the counsel for IIT claimed that their system is foolproof and is up to standard. The court directed the IIT to file an affidavit by June 2. The court was hearing a public interest litigation (PIL) of a nongovernment organisation (NGO) that has sought a stay on the declaration of result of the IIT entrance exam held on April 11. Raising the issue of errors in the instructions for examinees who took the IIT-JEE in Hindi, the NGO, Satya Foundation, filed the PIL. Chetan Upadhyaya, secretary and counsel of Satya Foundation, submitted before the court a list of serious blunders in the IIT-JEE 2010 and said that instead of accepting the faults and reconducting the examination, the Joint Admission Board was trying to cover up the issue with "corrective measures" which are "totally illogical and can't be digested by anybody". "The IIT-JEE board evolved corrective measures on May 2 to ensure that genuine candidates were not affected by the examination errors. It formulated a point-by-point remedial action and posted the same on the IIT-JEE website," Upadhyaya argued. However, after dismissal of Upadhyaya's petition, he said he will approach the Supreme Court. JUNE 2010

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

Srikanth Jagabathula President of India gold medal winner.

Internet connectivity in rural areas at cheap rates? Well,

From securing the first rank in the board exam in the 7th

this could be a reality if Srikanth's dream comes true.

standard, securing the 8th rank in the Andhra Pradesh State Science Talent Exam when he was in the 8th

Meet Srikanth Jagabathula, IIT's pride, the President of

standard to securing the President of India gold medal for

India gold medal winner for scoring the highest marks

topping across all the batches in 2005-06 in IIT-B,

among all batches at Indian Institute of TechnologyBombay.

Srikanth's success is an inspiring story.

After an enviable stint at the IIT, Srikanth is all set to fly to

Any IIT-ians in the family? "No," he says, "my father has retired from the technical education department in

the United States to pursue his studies at the prestigious

Andhra Pradesh and my mother is a homemaker. They

Massachusetts Institute of Technology. After five years he

are really happy to see me achieve my goals though they

plans to come back to Indian to start his own

are sad to see me go aboard for further studies."

communications company.

His role models: Apple Computer founder Steve Jobs,

As a kid he dreamt of becoming an engineer. Somehow he

Google founders Sergey Brin and Larry Page, Infosys

always thought an engineer's job would be very

founder N R Narayana Murthy and Nandan Nilekani.

fascinating. He heard about the IIT when he was in the 7th

Srikanth is excited about his next stint at MIT and takes us

standard. Since then IIT was his aim. After clearing his

through his journey from IIT to MIT.

10th class, he religiously worked towards cracking the IITJoint Entrance Examination. A rank of 38 at the IIT-JEE

My IIT experience

meant a smooth entry into IIT-Bombay. Srikanth, who

"IIT has been a dream come true for me. Initially, it was

hails from Hyderabad, was always a topper in school.

very difficult as I came from Hyderabad. It (Mumbai) was a

Mathematics and physics were his favourite subjects, but

new city and for the first time I was staying away from the

he dreaded biology and chemistry.

family. But from the second year, everything was perfect

A desire to top in everything he did from his schooldays

and I made good friends here.

helped him score high grades and he always lived up to his

It is really a great learning experience. IIT helps you excel

parents' expectations. Studies were always a priority for

as it provides you with the best facilities and faculty.

Srikanth and his efforts won him several awards.

XtraEdge for IIT-JEE

6

JUNE 2010

My mantra for success

Advice to IIT aspirants

I have a desire to excel. I am very organised, am always on

Don't get bogged down by pressure, many people are

time, and have deep love for the subjects I learn. I have

very talented, but it is important how you perform on the

always done things I like to do, which is very important.

entrance day. You have to prepare well for two years, you must learn to cope with the pressure.

Interests Reading books and participating in debates. I also like

On India

watching action-packed films and thrillers. On OBC quota.

Earlier foreign jobs offered much better prospects. But

The 50 per cent quota is not a good idea. The focus

now all the multinational corporations have set shop here. There are plenty of opportunities in India. So there is no

should be on improving basic education. Unless you get

need to actually go abroad. India is on a growth path and

the basics right, there is no point introducing the quota system at the highest level. There are many backward

in 5 to 6 years there will be big changes in India.

regions in the country, so the focus should be towards a

Future plans

region-based development than a caste-based reservation

I would like to come back to India and start my own

system.

company. It will basically be a communications company.

Next move

The idea is to develop products to provide cost-effective

"After I joined IIT, I heard more about MIT from my

Net connectivity and communication facilities in rural

seniors who studied there. I'm very happy that I got

India.

selected. I will be joining MIT soon to do my PhD. I'm

Now the cost is a big hindrance to Net penetration. My

very excited about it. I have got a scholarship which will

aim will be to remove the barriers and make

waive the entire tuition fees and I will get a stipend of

communication effective and increase its reach.

about $950 a month," he grins.

THE BRAND NEW EMOTIONAL ROBOT Most of the people always think, the robots are only the machines that didn’t have any feeling. It’s definitely impossible for the scientists to apply emotion on the robots. Well, please don’t be so sure about your thought yet, as the scientists at Georgia Tech had decided to test our ability to interpret a robot’s emotion. The research group discovered that older adults showed some unexpected differences in the way they read a robot’s face from the way younger adult did. Jenay Beer, a graduate student in Georgia Tech’s School of Psychology described that the home-based assistive robots have the potential to help older adults, as they can be used to keep the older adults independent longer. As a result, it reduces healthcare needs and provides everyday assistance to the elders. Based on the previous research, the robot found out that older adults are less accurate in recognizing anger, fear and happiness. Furthermore, the older adults have problem recognizing the happy robot compared with their success in recognizing happy people. Another interesting fact about the experiment was the researchers discovered that neither the young nor old could easily distinguish the emotion disgust on the virtual iCat. It might be due to the difficulty in programming a robot to show the emotion!

XtraEdge for IIT-JEE

7

JUNE 2010

KNOW IIT-JEE By Previous Exam Questions

⇒ m2x2 = V0tm1 + V0tm2 – V0tm1 + m1A(1 – cos ωt) ⇒ m2x2 = V0tm2 + m1A(1 – cos ωt) m ⇒ x2 = V0t + 1 A(1 – cos ωt) m2

PHYSICS 1.

A circular disc with a groove along its diameter is placed horizontally on a rough surface. A block of mass 1 kg is placed as shown. The co-efficient of friction between the block and all surfaces of groove and horizontal surface in contact is µ = 2/5. The disc has an acceleration of 25 m/s2 towards left. Find the acceleration of the block with respect to disc. Given 4 3 cos θ = , sin θ = . [IIT-2006] 5 5

(b) Given that s1 = V0t – A(1 – cos ωt) dx1 ∴ = V0 – Aω sin ωt dt

25 m/s2 θ

 m1   +1 A(1 – cos ωt) = l0  m2 

Sol. Applying pseudo force ma and resolving it. Applying Fnet = max for x-direction ma cos θ – (f1 + f2) = max ma cos θ – µN1 – µN2 = max ma cos θ – µma sin θ – µmg = max ⇒ ax = a cos θ – µa sin θ – µg 4 2 3 2 – × 25 × – × 10 = 10 m/s2 = 25 × 5 5 5 5

Also when

d 2 x1

=0 dt 2 cos ωt = 0 from (i) m  ∴ l0 =  1 +1 A  m2 

Two uniformly charged large plane sheets S1 and S2 having charge densities σ1 and σ2(σ1 > σ2) are placed at a distance d parallel to each other. A charge q0 is moved along a line of length a(a < d) at an angle 45º with the normal to S1. Calculate the work done by the electric field. [IIT-2004] σ1 σ2 Sol. E1 = ; E2 = ε0 ε0 3.

Two masses m1 and m2 connected by a light spring of natural length l0 is compressed completely and tied by a string. This system while moving with a velocity v0 along + ve x-axis pass through the origin at t = 0. At this position the string snaps. Position of mass m1 at time t is given by the equation x1(t) = v0t – A(1 – cos ωt). Calculate [IIT-2003] (a) position of the particle m2 as a function of time (b) l0 in terms of A. Sol. The string snaps and the spring force comes into play. The spring force being an internal force for the two mass-spring system will not be able to change the velocity of centre of mass. This means the location of center of mass at time t will be V0t m x + m2x 2 Now, xCM = 1 1 = V0t m1 + m 2 2.

E = E1 – E2 =

σ1

σ1 − σ 2 ε0 S1

S2 σ2 E1

E2

E a 45º a/ 2

⇒ m1[V0t – A(1 – cos ωt)] + m2x2 = V0t m1 + V0tm2

XtraEdge for IIT-JEE

d 2 x1

= – Aω2 cos ωt ...(i) dt 2 This is the acceleration of mass m1. When the spring comes to its natural length instantaneously, d 2 x1 = 0 and x2 – x1 = l0 then dt 2   m ∴ V0 t + 1 A(1 − cos ωt ) – [V0t – A(1 – cos ωt)] = l0 m2   ∴

d

8

JUNE 2010

W = q0 E × 4.

a

∴ W=

2

q 0 ( σ1 − σ 2 )a

sin i sin r n we get sin i = n sin r = n sin 30º = 2 The value of n at 600 nm is 10.8 × 10 4 n = 1.20 + = 1.50 (600) 2 From (1) and (2) 3 The angle of incidence is i = sin–1   4

Now using the expression n =

2ε 0

A prism of refractive index n1 and another prism of refractive index n2 are stuck together without a gap as shown in figure. The angles of the prisms are as shown. n1 and n2 depend on λ, the wavelength of 10.8 × 10 4 and light, according to n1 = 1.20 + λ2 1.80 × 10 4 n2 = 1.45 + , λ where is in nm. λ2 D

...(2)

A leaky parallel plate capacitor is filled completely with a material having dielectric constant k = 5 and electrical conductivity σ = 7.4 × 10–12 Ω–1m–1. If the charge on the plane at instant t = 0 is q = 8.85 mC, then calculate the leakage current at the instant [IIT-1997] t = 12 s. Sol. q0 = 8.85 × 10–6C at t = 0 q = q at t = 12 sec AV V = ...(i) Now, I = ρl R R = Resistance, V = Potential difference at t sec. S2 K=5 5.

70º

C

..(1)

n2 n1

20º

60º

40º

A B (a) Calculate the wavelength λ0 for which rays incident at any angle on the interface BC pass through without bending at that interface. (b) For light of wavelength λ0, find the angle of incidence i on the face AC such that the deviation produced by the combination of prisms is minimum. [IIT-1998] Sol. The following figure displays the given data.

σ =7.4×10–12

q

E

A

D C n2

n1 i

l V

60º r=30º

dq AV = ρl dt dq A q ⇒ – = ρl C dt

⇒ –

60º

A 60º B (a) The rays of wavelength λ0 incident at any angle on the interface BC will pass through without bending, provided the refractive indices n1 and n2 have the same value for the wavelength λ0. Equating the expressions of n1 and n2, we get 1.80 × 10 −4 1.80 × 10 −4 = 1.45 + 1.20 + λ20 λ20

1 λ20

(10.8 × 104 – 1.80 × 104) = 1.45 – 1.20 1/ 2

 9.0 × 10 4   = 600 nm or λ0 =   0.25    (b) For the wavelength 600 nm, the combination of prism acts as a single prism shaped like an isosceles triangle (ABE). At the minimum deviation, the ray inside the prism will be parallel to the base. Hence, the angle of refraction on the face AC will be r = 30º

XtraEdge for IIT-JEE

Aldt dq = q ρlKε 0 A

⇒ –

dq σ = dt dt Kε 0

Kε 0 A   Q C =  l    1 Q σ =  ρ  

7.4 × 10 −12 σ = = 0.1672 Kε 0 5 × 8.85 × 10 −12 dq ⇒ = – 0.1672 dt q On integrating t q dq = – 0.1672 dt 0 q0 q

(where λ0 is in nm) or

⇒ –

(Q q = CV)





q = – 0.1672 t ⇒ q0 When t = 12 sec

loge

9

q = q0e–0.1672 t

JUNE 2010

q=

q0

=

8.85 × 10 −6

=

PV = nRT PV or n= RT At room temperature, For NO, P = 1.053 atm, V = 250 ml = 0.250 L 1.053 × 0.250 ∴ Number of moles of NO = 0.0821× 300 = 0.01069 mol For O2, P = 0.789 atm, V = 100 ml = 0.1L 0.789 × 0.1 ∴ Number of moles of O2 = 0.0821× 300 = 0.00320 mol According to the given reaction, 2NO + O2 → 2NO2 → N2O4 Composition of gas after completion of reaction, Number of moles of O2 = 0 1 mol of O2 react with = 2 mol of NO ∴ 0.00320 mol of O2 react with = 2 × 0.00320 = 0.0064 mol of NO Number of moles of NO left = 0.01069 – 0.0064 = 0.00429 mol Also, 1 mol of O2 yields = 1 mol of N2O4 ∴ Number of moles of N2O4 formed = 0.00320 mol N2O4 condenses on cooling, ∴ 0.350 L (0.1 + 0.250) contains only 0.00429 mol of NO At T = 220 K, Pressure of the gas, nRT 0.00429 × 0.0821× 220 = = 0.221 atm P= V 0.350

8.85 × 10–6 7.439

e 0.1672 t e 0.1672 t ×1L = 1.1896 × 10–6C From (i) σA ql σ ∴ I= × = ×q l K ε 0 A Kε 0 = 0.1672 × 1.1896 × 10–6 I = 0.199 µA

CHEMISTRY How many grams of silver could be plated out on a serving tray by electrolysis of solution containing silver in +1 oxidation state for a period of 8.0 hours at a current of 8.46 amperes ? What is the area of the tray if the thickness of the silver plating is 0.00254 [IIT-1997] cm? Density of silver is 10.5 g cm–3. Sol. Given that, t = 8.0 hrs = 8 × 3600 s I = 8.46 A Thickness = 0.00254 cm Density = 10.5 g cm–3 M = 108 g mol–1 p =1 (for Ag) F = 96500 C mol–1 We know, according to Faraday's first law of electrolysis, 108 × 8.46 × 8 × 3600 MIt = m= 96500 × 1 Fp 6.

Also,



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

An organic compound A, C6H10O, on reaction with CH3MgBr followed by acid treatment gives compound B. The compound B on ozonolysis gives compound C, which in presence of a base gives 1acetyl cyclopentene D. The compound B on reaction with HBr gives compound E. Write the structures of A, B, C and E. Show how D is formed from C. [IIT-2000] Sol. The given reactions are as follows. CH3 OMgBr O CH3 Br CH3 8.

272.684 cm2 = 10224.37 cm2 10.5 × 0.00254

7.

At room temperature, the following reactions proceed nearly to completion : 2NO + O2 → 2NO2 → N2O4 The dimer, N2O4, solidified at 262 K. A 250 ml flask and a 100 ml flask are separated by a stopcock. At 300 K, the nitric oxide in the larger flask exerts a pressure of 1.053 atm and the smaller one contains oxygen at 0.789 atm. The gases are mixed by opening the stopcock and after the end of the reaction the flasks are cooled to 200 K. Neglecting the vapour pressure of the dimer, find out the pressure and composition of the gas remaining at 220 K. (Assume [IIT-1992] the gases to behave ideally) Sol. According to the gas equation,

XtraEdge for IIT-JEE

CH3MgBr

H+ –H2O

(A)

(B)

COCH3 Base (D)

HBr

COCH3 O

(E)

CH3 O O (C)

The conversion of C into D may involve the following mechanism. 10

JUNE 2010

CH2

COCH3

COCH3

COCH3 O

B+ –BH+

HC

O–

HC

O

(a) BH+ –B

Filter

Aqueous solution NaCl → White precipitate of A of B hot water

Residue

(C)

COCH3 –

COCH3 OH +B –BH+

COCH3

AlCl3

OH

C

PCl5 H2/Pd (BaSO4)

O

C

MATHEMATICS

C6H5

O H2NNH2 O

N

11. Find the centre and radius of the circle formed by all the points represented by z = x + iy satisfying the

N

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

O

C6H5



+

NH2

NH2 +

NH2

NH2 O



C6H5

O

relation

z−α =k(k ≠ 1), where α and β are constant z −β

complex numbers given by α = α1 + iα2, β = β1 + iβ2. [IIT-2004] Sol. As we know; |z|2 = z. z



N–H N–H



OH



C6H5

| z − α |2 2

| z −β|

= k2

(z – α)( z – α ) = k2(z – β)( z – β ) |z|2 – α z – α z + |α|2 = k2(|z|2 – β z – β z+ |β|2)

N N

or |z|2 (1 – k2) – (α – k2β) z – ( α – β k2) z + (|α|2 – k2|β|2) = 0

10. An aqueous solution of salt A gives a white crystalline precipitate B with NaCl solution. The filtrate gives a black precipitate C when H2S is passed through it. Compound B dissolves in hot water and the solution gives yellow precipitate D on treatment with potassium iodide and cooling. The compound A does not give any gas with dilute HCl but liberates a raddish brown gas on heating. Identify the compounds A to D giving the involved equation. [IIT-1976] Sol. The given information's are as follows.

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Black p recipitate (C)

From the information given in part (a), it may be concluded that the compound A is a lead salt. B is lead chloride as it is soluble in hot water. Yellow precipitate D is due to PbI2. Black precipitate C is due to PbS. Lead chloride being spraingly soluble in water, a little of it remains in the filtrate which is subsequently precipitated as PbS. From the information given in part (b), it may be concluded that anion associated with lead (II) is nitrate because lead nitrate dissociates on heating as 2Pb(NO3)2 → 2PbO + 4NO2 + O2 Brown coloured gas Hence, A is Pb(NO3)2, B is PbCl2, C is PbS and D is PbI2.

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

C6H5

Yellow precipitate (D)

 Saltheat (b) No gas ←  → Brown coloured gas

OH –OH–

9.

O

KI

HCl

(D)

O +

H2 S

Filtrate

Soluble



|z|2 –

(α − k 2β) (1 − k 2 )

+

z–

(α − β k 2 ) (1 − k 2 )

| α |2 −k 2 | β |2 (1 − k 2 )

z =0

...(i)

On comparing with equation of circle, |z|2 + a z + α z + b = 0 whose centre is (–a) and radius = | a | 2 − b

∴ centre for (i) 11

JUNE 2010

=

α − k 2β

1− k

2

dv = 2. sin x cos x dx dv 1 ⇒ = [2 sin x cos x] dx cos v 2 sin x cos x 2 sin x cos x ⇒ = = 1 − sin 2 v 1 − sin 4 x Put these values in equation (1)



and radius

 α − k 2β  α − k 2 β  α α − k 2β β −  =   1 − k 2  1 − k 2  1− k2   

radius =

k (α − β) 1− k2

⇒ At



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

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

Thus, the required equation of the normal is y – 1 = (– 1) (x – 0) i.e., y + x – 1 = 0. 14. Determine the equation of the curve passing through the origin in the from y = f(x), which satisfies the dy = sin (10x + 6y) [IIT-1996] differential equation dx dy Sol. = sin (10x + 6y) dx Let 10x + 6y = t (given) .....(1) dy  dt  ⇒ 10 + 6 =  dx  dx 

P(B ∩ C / E) 1 2 . P(B).P(C) 2 3 1 = = 2 P( E ) 2 3

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

dy 1  dt  −10  =  dx 6  dx  Now the given differential equation becomes



sin t =



1  dt   − 10  6  dx 

6sin t =

dt – 10 dx

dt = 6 sin t + 10 dx dt ⇒ = dx apply variable separable 6 sin t + 10 Integrating both the sides, we get



dy du  y  + log e (1 + x ) .....(2) =(1+x)y  + 1 x dx dx   Again, v = sin–1 sin2 x ⇒ sin v = sin2 x



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



⇒ P(E) = 1 – P( B ). P( C ) 1 2 2 =1– . = 2 3 3 Probability if A is hit by B and not by C.



.....(3)

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

12. A is targeting to B, B and C are targeting to A. Probability of hitting the target by A, B and C are 2 1 1 , and respectively. If A is hit, then find the 3 2 3 probability that B hits the target and C does not. [IIT-2003] Sol. Here, 2 P(A) = probability that A will hit B = 3 1 P(B) = probability that B will hit A = 2 1 P(C) = probability that C will hit A = 3 P(E) = probability that A will be hit



cos v

12

JUNE 2010

t    5 tan 2 + 3  1 –1 ⇒ tan  =x+c 4 4    

dt

∫ 6 sin t + 10 = ∫ dx ⇒

dt

1 2

∫ 3 sin t + 5 = x + c

.....(2)

t   5 tan + 3   2 ⇒ tan–1   = 4x + 4c 4    

dt 3 sin t + 5 Put tan t/2 = u



Let

I1 =



1 sec2 t/2 dt = du 2



dt =



dt =



dt =

Also,

I1 =

=

=



2du sec 2 t / 2 2du 1 + tan 2 t / 2

2du 1+ u2 dt

∫ 3 sin t + 5 = ∫

2 = 5

dt 2 tan t/2   3 +5 2  1 + tan t / 2 

(1 + tan 2 t / 2)dt t t (6 tan + 5 + 5 tan 2 ) 2 2



3 4 5x + 3y = tan–1  {tan( 4x + tan −1 3 / 4} −  5 5

2(1 + u 2 )du



3 4 3y = tan–1  {tan( 4x + tan −1 3 / 4} −  –5x 5 5



1 3  5x 4 y = tan–1  {tan(4x + tan −1 3 / 4} −  – 3 5 3 5

∫ (1 + u

2 = 5

∫u ∫

2

)(5u 2 + 6u + 5)

du 2

+ (6 / 5)u + 1

du

∫

=

2 5

∫

=

2 5  u + 3/5  . tan–1   5 4  4/5 

=

1  5u + 3  tan–1   2  4 

=

15. A tangent to the ellipse x2 + 4y2 = 4 meets the ellipse x2 + 2y2 = 6 at P and Q. Prove that the tangents at P and Q of the ellipse x2 + 2y2 = 6 arc at right angles. [IIT-1997] Sol. x2 + 4y2 = 4 (given)

du 6 9 9 u2 + u + − +1 5 25 25

2 5

=

1 [5 tan (5x + 3y) + 3] = tan (4x + 4c) 4 ⇒ 5tan (5x + 3y) + 3 = 4 tan (4x + 4c) When x = 0, y = 0 we get 5 tan 0 + 3 = 4 tan (4c) 3 ⇒ = tan 4c 4 3 ⇒ 4c = tan–1 4 Then, 5 tan (5x + 3y) + 3 = 4 tan (4x + tan–1 3/4) 4 3 ⇒ tan (5x + 3y) = tan (4x + tan–1 3/4) – 5 5



2

3  16 u +  + 5 25 

x 2 y2 + =1 .....(1) 4 1 Equation of any tangent to the ellipse on (1) can be written as x cos θ + ysinθ = 1 .....(2) 2 Equation of second ellipse A y



du 2

3  4 u +  +   5  5 

2

90º P

1  5 tan t / 2 + 3  tan–1   2 4  

– 6

1 I1 = x + c 2

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Q

1 –2 O

Putting this in (2) Now

3

2

x

6

–1 – 3

x2 + 2y2 = 6 13

(given) JUNE 2010



x 2 y2 + =1 6 3

.....(3)

Suppose the tangents at P and Q meet in A(h, k). h x ky + =1 6 3

.....(4)

But (4) and (2) represent the same straight line, so comparing (4) and (2) h/6 k /3 1 = = cos θ / 2 sin θ 1



h = 3cos θ

and

k = 3sinθ

Equation of the chord of contact of the tangents through A(h, k) is Therefore, coordinates of A are (3cosθ.3sinθ) Now, the joint equation of the tangents at A is given by T2 = SS1 2 2   h2 k2  y2  hx ky   x i.e.,  + − 1 =  + − 1  + − 1 .....(5) 3 3 3  6   6   6 

In equation (5) coefficient of x2 =

2.

A typical double mattress contains as many as two million house dust mites.

3.

The average human will grow 590 miles (949.5 km) of hair in their lifetime.

4.

About 51% of incoming solar radiation is absorbed by the earth's surface and 14% absorbed by the atmosphere.

5.

Only 2% of male red deer are seriously injured in their antler-rattling contests.

6.

The United States recycles 25 percent of its annual 180 million tons of household rubbish.

7.

The largest sapphire weighed 2,302 carats. It was found in Australia circa 1935, and was carved into the shape of the head of President Abraham Lincoln.

8.

When the pharaohs' tombs were opened in Egypt early in the nineteenth century, not only human mummies were found but also those of sacred animals such as cats and ibises

9.

Flies are one of the major success's of the insect world, and the 120 000+ species are divided into three sub-orders, the Nematocera, the Brachycera, and the Cyclorrhapha, and these in turn are divided into about 100 families.

1 k2 – 6 18

coefficient of y2 =

=

The storm cock or male mistlethrush sings as a thunderstorm approaches.

 h 2 1  h 2 k 2 + − 1 –  3 36 6  6 

h2 h2 k2 1 = – – + 36 36 18 6

=

1.

k2 1 – 9 3

  h2 k2    6 + 3 − 1  

k2 h2 k2 1 h2 1 – – + =– + 9 18 9 3 18 3

Again coefficient of x2 + coefficient of y2

10. The angle between Earth's equatorial and orbital planes varies by a few degrees every 40,000 years.

1 1 1 = – (h2 + k2) + + 18 6 3 =–

1 1 (9cos2 θ + 9sin2 θ) + 18 2

=–

9 1 + 18 2

11. The San Francisco earthquake and fire of 18th April 1906 caused the deaths of around 700 people, obliterated 500 city blocks and caused $500 million of damage.

1 1 + =0 2 2 which shows that two lines represent by (5) are at right angles to each other.

12. The slogan on New Hampshire licence plates is 'Live Free or Die'. These licence plates are manufactured by prisoners in the state prison in Concord.

=–

13. Venus is the hottest temperature of 480 °C.

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14

planet

with

a

JUNE 2010

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15

JUNE 2010

Physics Challenging Problems

Set # 2

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

So lutions wi ll b e p ub lish ed in n ex t is su e 1

Two particles having positive charges +Q and +2Q are fixed at equal distance x from centre of an conducting sphere having zero net charge and radius r as shown. Initially the switch S is open. After the switch S is closed, the net charge flowing out of sphere is +Q x r x +2Q

3

S

(A) 2

Qr x

(B)

2Qr x

(C)

3Qr x

(D)

(B)

6Qr x

(C) 4H

1K Ω



k

5H

(R) the current leads in phase to source voltage (S) the current lags in phase to source voltage (T) the magnitude of required phase

4

x q

A variable current flows through a 1Ω resistor for 2 seconds. Time dependence of the current is shown in the graph – I(A) 10

m

(A) maximum at extreme position and minimum at mean position (B) maximum at mean position and minimum at extreme position (C) uniform throughout the motion (D) both maximum and minimum at mean position

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3 µF

(D)

field E = E 0 i (E0 is a positive constant) is switched on at t = 0 sec. The block moves on horizontal surface without ever lifting off the surface. Then the normal reaction acting on the block is y B

(Q) the magnitude of phase difference is π/ 4

5H

A spring of spring constant ‘K’ is fixed at one end has a small block of mass m and charge q is attached at the other end. The block rests over a smooth horizontal surface. A uniform and constant magnetic field B exists normal to the plane of paper as shown in figure. An electric →

Four different circuit components are given in each situation of column – I and all the components are connected across an ac source of same angular frequency ω = 200 rad / sec . The information of phase difference between the current and source voltage in each situation of column – I is given in column – II. Column – I Column – II (A) (P) the magnitude of 10 Ω 500 µF required phase difference is π / 2

t(s) 2 (A) Total charge flown through the resistor is 10C (B) Average current through the resistor is 5A (C) Total heat produced in the resistor is 50J (D) Maximum power during the flow of current is 100W O

16

JUNE 2010

Q. 5

In the figure shown, ‘R’ is a fixed conducting fixed ring of negligible resistance and radius ‘a’. PQ is a uniform rod of resistance r. It is hinged at the centre of the ring and rotated about this point in clockwise direction with a uniform angular velocity ω . There is a uniform magnetic field of strength ‘B’ pointing inwards, ‘r’ is a stationary resistance, then-

⊗B R q

⊗B P

Q r

R

Q. 8

(A) Current through ‘r’ is zero 2Bωa 2 5r (C) Direction of current in external ‘r’ is from centre to circumference (D) Direction of current in external ‘r’ is from circumference to centre

(B) Current through ‘r’ is

Q. 6

O θ v0

P

(A)

qBR m sin θ

(B)

qBR 2m sin θ

(C)

2qBR m sin θ

(D)

3qBR 2m sin θ

A toroid having a rectangular cross section (a = 2.00 cm by b = 3.00 cm) and inner radius r = 4.00 cm consists of 500 turns of wire that carries a current I = I max sin ωt with I max = 50.0A and a ω = 60.0Hz. A coil that consists 2π of 20 turns of wire links with the toroid as shown in figure –

frequency f =

In column – I condition on velocity, force and acceleration of a particle is given. Resultant motion is described in column – II.

N = 500



u = instantaneous velocity – Column – I Column – II →

(A) u × F = 0 and

(P) path will be



b

circular path

→ →

(B) u . F = 0 and

(Q) speed will



F = constant

increase

→ →

(C) v . F = 0 all the →

time and | F | =

(A)

(R) path will be straight line

Time

constant and the particle always remains in one plane →





(D) u = 2 i − 3 j and acceleration at all →



(S) path will be parabolic

(B)



time a = 6 i − 9 j Q. 7

Time

A particle of charge q and mass m is projected with a velocity v0 towards a circular region having uniform magnetic field B perpendicular and into the plane of paper from point P as shown in the figure. R is the radius and O is the centre of the circular region. If the line OP makes angle θ with the direction of v0, then the value of v0 so that particle passes through O is

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N' = 20

Induced emf

F = constant

R

a

Induced emf



(C) Max value of induced emf is 0.422 V (D) Max value of induced emf is 0.122 V

17

JUNE 2010

1.

8

Solution

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

[D]

5.

After closing both switch potential difference across C1 and C2 will be different VC1 = V; VC 2 = 0

2.

6.

(B) → R

(C) → Q,R

(D) →P,T

[B] Vtotal = Vsolid + Vshell

 1 1 1 E 0 z 2 1 −  − E 0 z 2  −  = 3E 0  9 4 9

=

z=2

7.

 1 KE1 = E 0 1 −  − φ  9

λ2

KQ KQ − b a

Vinner surface of shell = 0

1 1 1 1 ∆V = KQ − ; W = KqQ −  a b a b  

= 8.5eV

[B,D] e AB

[C] Vsurface at inner solid sphere =

 1 KE 2 = E 0 z 2 1 −  − φ  4 1

KQ  2 a 2   − KQ   3a −  +  4   b  2a 3 

 11 1  = KQ  −   8a b 

λ1 =3 λ2

3.

(A) → T,R,S

[B,C]

KEα

Set # 1

8.

 dB  area of ∆AOB, using (E.M.I.) =   dt 

[B]

If outer ball is grounded. No change is charge distribution will occurs at

 1  3 = (4) ×  4 × × 2 × 2   2  2 

outer shell is already at zero potential

HOW HEAVY IS THE EARTH ?

Total e of loop   1 3 = 3×  4 × × 4 × × 2  × 2 = 2 × 24 3 = 48 3Volt   2 2  

4.

[1] C

C

Now despite science coming forward in leaps and bounds the simple question of how much the Earth weighs isn’t as straightforward as you might think. Certainly placing the earth onto a massive set of weighing scales isn’t an option. So how do we measure the total weight of the earth ?

2C C

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C

18

JUNE 2010

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

Find the gravitational force between a point like mass M and an infinitely long, thin rod, of mass density ρ, which is at a distance L from the mass M. Sol. Considering figure, let us divide the rod into very small mass elements, each one of magnitude dm, which can be expressed as : Ldθ rdθ dm = ρ ρ= cos θ cos 2 θ Note : We are using a polar co-ordinate system. r dθ on the rod. Hence, have dl = cos θ ρ

1.

L θ

m3

m1

collision process, or r r before Σ i m i v i r m3v0 v CM = = xˆ = v after CM Σi mi m1 + m 2 + m 3

M

Since this problem is unidimensional, we omit the vector notation from now on. (ii) Let us write the expressions for the kinetic energy: 1 Ebefore = E (km3 ) = m3 v 02 2 1 Eafter = E (km 2 + m3 ) = (m2 + m3) v 22+ 3 2 Notice that since the time that elapses when masses m2 and m3 stick together is short, mass m1 stays at rest during the process. We calculate v2+3 using the law of conservation of linear momentum. m3 m3v0 = (m2 + m3)v2+3 → v2+3 = v0 m 2 + m3





Substituting the value of v2+3 into Eafter, we find :

Two masses m1 ad m2 are attached to the two ends of a spring of force constant k. The system lies horizontally on a perfectly smooth surface. A third mass, m3, is thrown with a velocity of v0, horizontally onto the plane to hit mass m2. The two masses m2 and m3 stick together at the moment of collision. The sticking process occurs almost immediately, so that the length of the spring does not change. It is given 1 that m1 = 2m, m2 = m and m3 = m. 2

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k

What is the velocity of the center of mass before and after the collision ? (ii) How much kinetic energy is lost during the collision ? (iii) What are the velocities of the masses m1 and m2 + m3 immediately after the collision ? (iv) What is the maximum that the spring shrinks ? Sol. (i) There are no external forces acting on the system; therefore, its total linear momentum is conserved. r This means that v CM is conserved throughout the

Since the rod is symmetric, the components of the force element parallel to the rod will cancel each other, and the total force will be the sum of the force elements perpendicular to the rod. So, for a mass element dm at distance r from M, we have: GM(dm) MGρ F ⊥(dm) = cos θ = cos θ dθ 2 L  L     cos θ  Total force is : MGρ π / 2 cos θ dθ F= −π / 2 L 2MGρ π / 2 2MGρ cos θdθ = = 0 L L 2.

m2

(i)

r = L/cos θ

dm

v0

Eafter =

1 m 32 v 02 2 m 2 + m3

Therefore, the loss of kinetic energy, ∆E, is : ∆E = Ebefore – Eafter

19

=

m 32  2 1  m3 − v0 2  m 2 + m 3 

=

1 m 2m3 2 v0 2 m 2 + m3

JUNE 2010

(iii) Velocities in the centre of mass frame are determined by : v´ = v – vCM where v is the velocity in the laboratory frame. Hence, m1m 3 v '2+ 3 = v2+3 – vCM = (m 2 + m 3 )(m1 + m 2 + m 3 )

or, ∴

m3 v0 m 2 + m3



V 0 . A man of mass m rides the rear buggy. At a certain moment the man jumps into the front buggy →





...(1)



where V r = Velocity of rear buggy finally. The velocity of the man with respect to ground →









V man ground = V man buggy + V buggy ground = u + V r

(i)

Solving (1), we get →





mu Vr = V0 – M+m Conserving momentum for front buggy →









Vf = V0 +



Mm u

( M + m) 2



(ii)



where V f = Velocity of front buggy. 20

1 q1q 2 4πε 0 r

Where r is distance between their centres. Since the right sphere has a cavity, therefore it does not have uniformly distributed charge. This sphere can be assumed to be result of superposition of two solid spheres, one of radius R having a positive charge density ρ and a solid sphere of radius R/2 and having a charge density (– ρ). Required interaction energy will be the sum of the following two energies: Interaction energy U1 of left sphere and a solid sphere of radius R and having charge density ρ. Separation between centres of these spheres is r. ∴

M V 0 + m[ u + V r ] = (M + m) V f

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→ Mm u (M + m) V 0 + = (M + m) V f M+m

U=

with a velocity u relative to his buggy knowing that the mass of each buggy is equal to M. Find the velocity with which the buggies will move after that. Sol. Conserving momentum for the rear buggy. →





R/2 such that distance of centre of cavity is (r – R/2) from the centre of sphere A and R/2 from the centre of sphere B. Di-electric constant of material of each sphere is K = 1 and material of each sphere has a uniform charge density ρ per unit volume. Calculate interaction energy of the two spheres. Sol. Interaction energy between two spheres having uniformly distributed charges q1 and q2 is given by

Two identical buggies move one after the other due to inertia (without friction) with the same velocity







r

m1 k (m1 + m 3 )(m1 + m 2 + m 3 )

(M + m) V 0 = m[ u + V r ] + M V r



4. Distance between centres of two spheres A and B, each of radius R is r as shown in Figure. Sphere B has a spherical cavity of radius A B

1 Eafter = (m2 + m3) v 22+ 3 2 1 1 2 + k(∆x)2max E´after = (m1 + m2 + m3) v CM 2 2 to determine the maximal shrinking of the spring. E´after is the energy in the state of maximal shrinking. Note that in the state of maximal shrinking, the three masses move at the same velocity as the center of mass. Using the equality E´after = E´after, and substituting vCM and v2+3, we obtain :

3.



mu M+m

→ m2 u Hence, M V 0 + m u + m V 0 – = (M + m) V f M+m

(iv) We use the law of conservation of energy

(∆x)max = m3v0



Vr = V0 –



and since v1 = 0, we have v´1 = v1 – vCM = –



Finally as,

 4 3  4 3   πr ρ  πR ρ  3 6 1 3  3  = 4πρ R U1 = 4πε 0 r 9ε 0 r

Interaction energy U2 of left sphere and a solid sphere of radius R/2, having charge density (– ρ). Separation between centre of these spheres is (r – R/2) JUNE 2010

1 U2 = 4πε 0

3   4 3   4  R   πr ρ . π  (–ρ) 3   3  2  

R  r –  2  Hence, required interaction energy, U = U1 + U2

=

Learn How to Study

πρ 2 R 6 = 9ε 0 (2r – R )

Designate a specific time for studying. Cramming is not a reliable way to retain information.

πρ 2 R 6 (7 r – 4R ) ` 9ρ 0 r (2r – R )

• Take understandable lecture and textbook

Ans.

notes.

• Become a good listener in class and review

A particle of mass m having negative charge q moves along an ellipse around a fixed positive charge Q so that its maximum and minimum distances from fixed charge are equal to r1 and r2 respectively. Calculate angular momentum L of this particle. Sol. During movement, moving particle experiences an electrical force which is always directed towards fixed particle. Therefore, moment of this force about fixed particle remains always zero. Hence, angular momentum L of moving particle about fixed particle remains constant. v2 5.

B

r2 +Q

r1

your notes daily.

• Organize your class materials prior to each study session.

• Concentrate on the class material and keep in mind course objectives.

• Ask for help early if needed (such as your professor or teaching assistants).

• Tutoring assistance can provide a positive impact on your academic success.

• Form study groups and partners. Keep everyone committed to studying, not small

A

talk.

v1

At instant of maximum and minimum distance, velocity vectors of moving particle are perpendicular to respective radii vectors as shown in Figure. Let at these particle be υ1 and υ2 respectively. ∴ Angular momentum, L = mυ1r1 = mυ2r2 L L or υ1 = and υ2 = mr1 mr2

Roommate Tips • Communication is a must! Listen and learn from each other.

• Talk to each other early about problems that may arise. Don’t let problems build up.

• Living

Total energy of the system of two particles when moving particle is at A is E1 = kinetic energy of moving particle at A + electrical energy of the system 1 1 Q(– q) = mυ12 + 2 4πε 0 r1

with

someone

requires

increased

sensitivity to the needs of others.

• Respect each other’s privacy and personal space.

• Agree upon a set period of "quiet time" for study without distractions.

Similarly, total energy at B is E2 1 1 Q(– q) = mυ12 + 2 4πε 0 r2

If you have time to whine and complain about

But there is no external force on the system of two particles, therefore, total energy remains constant. ∴ Ε1 = Ε 2

Once our personal connection to what is wrong

2

or

1 Qq 1  L 1  L   – = m m 2  mr1  4πε 0 r1 2  mr2

or L =

mr1r2 Qq 2πε 0 (r1 + r2 )

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something then you have the time to do something about it. becomes clear, then we have to choose: we can go on as before, recognizing our dishonesty and

2

 1 Qq  – 4 πε 0 r2 

living with it the best we can, or we can begin the effort to change the way we think and live. Once you choose hope, anything's possible.

Ans.

21

JUNE 2010

P HYSICS F UNDAMENTAL F OR IIT-J EE

Electrostatics-2 KEY CONCEPTS & PROBLEM SOLVING STRATEGY

charge q between two fixed points having potential difference V is equal to, …(1) WAB = – UAB = q(VB – VA)qV

Electric Potential Energy: If a point charge q1 is present in an electric field where potential is V, by definition V = (U/q1) i.e., U = q1V And if the field is produced by a point charge q2 which is at a distance r12 from q1,  1 q2  1 q1q 2 …(1) U = q1   = 4πε 0 r12  4πε 0 r12  So in case of discrete distribution of charges qiq j  1  q1q 2 q 2 q 3 1 1  + + .......  = U=  2 4πε 0 i ≠ j rij 4πε 0  r12 r23 

And hence in moving a charged particle in an electric field work is always done unless the points are at same potential as shown in figure. [However, in magnetic field as force is always perpendicular to motion, work done is always zero.] A B

+q

(A)

(b) If the electric potential energy of a system in one configuration is U1 and in the other UF, work done in changing the configuration will be WIF = – UIF = – (U1 – UF) = UF – U1 And as potential energy at infinity is zero, work done in assembling or disassembling a given charge distribution will be respectively, W = UF [as U1 = 0] and W = – U1 [as UF = 0] (Assembling) (Disassembling)



(A) (c)

(B)

2qV m And if the field is uniform, i.e., E = (V/d)

υ=

υ=



(e)

(C)

2qEd m

…(3)

…(4)

In case of motion of a charged particle in a uniform electric field if force of gravity does not exist (or is balanced by some other force →



→ → F qE = = constant [as F = q E ] …(5) m m So equation of motion are valid. Now there are two possibilities: →

a =

(i) If the particle is initially at rest From Eq. υ = u + at, we get qE  qE  t as u = 0 and a = υ = at = m  m 

F = –q E

(B)

Electric field is conservative in an electric field work is path independent and work done in moving a point

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B

L

L

 2 2qV  …(2) u + m    And if the charged particle is initially at rest, i.e., u=0

As by definition of electric intensity E , F = q E , a point charge always experiences a force either at rest or in motion. (b) The direction of force is parallel to the field if the charge is positive and opposite to the field if charge is negative. E E + – → → → → F =qE

+Q

II +Q

υ=

or

Motion of a Charged Particle in an Electric Field In case of motion of a charged particle in an electric field : →

B

(d) When a charged particle is accelerated by an electric field (uniform or non-uniform) by Work energy theorem, i.e., ∆KE = W, we have 1 1 mυ 2 – mu 2 = qV [as from Eq. (1) W = qV] 2 2

Electrical potential energy is not localised but is distributed all over the field

(a)

A

A

Here it is worth noting

I

II

II



(a)

I

I

22

…(6)

JUNE 2010

Electric Dipole Definition: If two equal and opposite point charges are separated by a distance 2l such that the distance of field point r >>2l, the system is called a dipole. (a) Field of a Dipole Potential due to dipole at a point (r, θ) as shown in Figure will be

1 2 at 2 1 1 qE 2 s = at 2 = t …(7) 2 2 m i.e., the motion is accelerated translatory with

And from Eq. s = ut +

a ∝ t0 ; υ ∝ t and s ∝ t2 Further more in this situation : 1 1  qE  mυ2 = m  t  2 2 m 

W = ∆KE =

2

V = V 1 + V2 =

qEt   as fromEq.(6)υ = m   

=

1 q q   –  4πε 0  r1 r2  q  r1 – r2    4πε 0  r1r2  P

E + + +

– – –

+ + +

F

+q

r2

– – –

d PD=V

2lcosθ

+

+

+

+

+

Now as , r >> 2l r1 × r2 = r2 and r2 – r1 = 2l cosθ 1 q (2l cos θ) So, V= 4πε 0 r2

…(8)

V=

(b) Potential will be minimum when |cos θ| = min = 0, i.e., θ = 90º. So for broad on, equitorial or tan B position, potential is minimum and is zero, i.e., Vmin = 0 This all is shown Figure

–q

θ y

Y

A

+q O

B

End-on, Axial or tan A position (A)









L



D

2

…(11)

i.e., the path is a parabola. [However, under same conditions in magnetic field path is a circle.]

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P

Vmax =

1 P 4πε0 r 3

–q P

+q A

O

B

Broad on, Equitorial or tan B position (B)

(c) Now as electric field, → dV → E =– n dx So component of electric intensity in the direction of r : d  1 p cos θ  1 2p cos θ …(i) Er = –   = dr  4πε 0 r 2  4πε 0 r3 And perpendicular to r,



So eliminating t between equation for x and y, we have qE  x  qE y=   = 2m  υ 0  2mυ 02

Vmax = 0

1 P 4πε0 r 2 Vmax = 1 2P 4πε 0 r 3 Vmax =

V0 –

1 P cos θ . 2 [as p = q × 2l] ] …(1) 4πε 0 r

r

+

–q

+q

l

l

(ii) If the particle is projected perpendicular to the field with an initial velocity υ0 1 From Eq. υ = u + at and s = ut + at2 respectively 2 for motion along x-axis as u = υ0 and a = 0, υx = υ0 = constt. and x = υ0t …(9) While for motion along y-axis as u = 0 and a = (qE/m), 1  qE   qE  υy =   t and y =   t 2 …(10) 2m m +

θ

–q

Which in the light of Eq. (7) with s = d, gives W = qEd = qV [as E = V/d]

+

r1

r

23

JUNE 2010

Eθ = –

(h) The field in a cavity inside a conductor is zero resulting in 'electrostatic shielding.'

d  1 p cos θ  1 p sin θ   =   2 rdθ  4πε 0 r 4πε 0 r 3 

dF σ 2 = ds 2ε 0

…(ii)

+ +

E 1 So that, tanφ = θ = tan θ Er 2 E 2r + E θ2 =

E0 = 0

(B)

Solved Examples 1.

Three point charges q, 2q and 8q are to be placed on a 9 cm long straight line. Find the position where the charges should be placed such that the potential energy of this system is minimum. In this situation, what is the electric field at the position of the charge q due to the other two charges ? Sol. As potential energy of two point charges separated by a distance r is given by U(=q1q2/4πε0r), so to have minimum potential energy the charges of greater value should be farthest, i.e., q must be between 2q and 8q. Let q be at a distance x from 2q, then potential energy of the system will be 1  2qq 8qq 8q × 2q  U= + +   4πε 0  x (d – x ) d 

(3) E will be minimum when cos2 θ = min = 0, i.e., θ = 90º, i.e., for broad on, equitorial or tan B position E is minimum and is, 1 p Emin = 4πε 0 r 3 Conductor in Electrostatics The substances such as metals which allow the charge to flow freely through them are called conductors. In metals conduction envolves the movement of free electrons. In case of conductors in electrostatics, it is worth noting that : (a) In charging a conductor electrons are removed, conductor becomes positively charged and its potential increases and if added, it becomes negatively charged and its potential decreases. (b) When a conductor is charges by induction, induced charge (Which is free to move) is equal and opposite to the inducing charge, i.e.,q' = –q (c) Charge resides on the outer surface of a conductor. However, distribution of charge on the surface is generally not uniform and surface density of charge varies inversly as the radius of curvature of that part of the conductor, i.e., σ ∝ (1/R) (d) The dielectric constant of conductors in electrostatics is infinite, i.e., K = ∞ (e) Electric intensity inside a conductor is zero while outside (near its surface) is (σ/ε0), i.e., Ein = 0 and Eout = (σ/ε0) (f) Conductor is an equipotential surface, i.e., potential at its surface or inside everywhere is same, i.e., for a conductor V = constt. (g) Electric field and hence lines of force are normal to the surface of a conductor.

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σ ε0

Ein = 0

Vs = vin = constt. ++ + + ++ + + + + σ1(>σ2) σ2 (A)

1 P (1 + 3 cos 2 θ) 4πε 0 r 3 …(2) From this it is clear that : (1) Intensity due to a dipole varies as (1/r3) and can never be zero unless r → ∞ or p → 0. (2) E will be maximum when cos2θ = max = 1, i.e., θ = 0º, i.e., for end on, axial or tan A position E is maximum and is, 1 2p Emax = 4πε 0 r 3

and, E =

E out =

For U to be minimum (dU/dx) = 0 ] 2q 2 8q 2 i.e., – 2 + =0 x (d – x ) 2 q 2q 8q x

(d–x) d i.e., 2x = (d – x) or x = (d/3) = (9/3) = 3 cm So to have minimum potential energy, the charge q must be placed at a distance of 3 cm from 2q between 2q and 8q on the line joining the charges. In this situation, Field at q 1  2q 8q  E= Ans. =0  2 – 4πε 0  (3) (6) 2 

2.

Three point charges 1C, 2C and 3C are placed at the corners of an equilateral triangle of side 1m. Calculate the work required to move these charges to the corners of a smaller equilateral triangle of side 0.5 m as shown in Figure (A) A 1 A

C

B B

24

(A)

C

3

JUNE 2010

Sol. As potential energy of two charges separated by a distance r is given by U = [q1q2/4πε0r], the initial and final potential energy of the system will be 1 1 × 2 2 × 3 3 × 1  + + (US)I = 4πε 0  1 1 1  = 9 × 109 × 11 = 9.9 × 1010 J 1 1 × 2 2 × 3 3 × 1  + + (US)F = 4πε 0  0.5 0.5 0.5  = 9 × 109 × 22 = 19.8 × 1010J So work done in changing the configuration of the system: W = (US)F – (US)I = (19.8 – 9.9) × 1010 Ans. = 9.9 × 1010J

Sol. Keeping in mind that here both electric and gravitational potential energy are changing and for external point a charged sphere behaves as whole of its charge were concentrated at its applying conservation of energy between initial and final position, we have 1 qq 1 q2 + mg × 9 = + mg × 1 4πε 0 1 4πε 0 9 10 9 q = 28 µC

Ans.

The distance between the two plates of a cathode-ray oscillograph is 1 cm and potential difference between them is 1200 volt. If an electron of energy 2000 eV enters at right-angles to the field, what will be its deflection if the plates be 1.5 cm long ? Sol. As distance between the plates is 1 cm and potential difference 1200 V, the field between the plates 1200 V V E= …(1) = 1.2 × 105 = –2 d m 1× 10

A particle of mass 40 µg and carrying a charge 5 × 10–9 C is moving directly towards fixed positive point charge of magnitude 10–8 C. When it is at a distance of 10 cm from the fixed point charge it has a velocity of 50 cm/s. At what distance from the fixed point charge will the particle come momentarily to rest ? Is the acceleration constant during motion ? Sol. If the particle comes to rest momentarily at a distance r from the fixed charge, then from 'conservation of energy' we have 1 1 Qq 1 Qq + = 2 4πε 0 a 4πε 0 r 2mu Substituting the given date, we get : 1 1 1 1  × 40 × 10–6 × × = 9 × 109 × 10–8 × 5 × 10–9  – 10 r 2 2 2  

y +

+

+

+

+

+ y

d

a

–e

V0 –





x

E –





L

5 × 10 –6 1 190 – 10 = = –8 9 r 9 × 5 × 10 100 100 1 or = + 10 = m 9 9 r i.e., r = 4.7 × 10–2 m F 1 1 qQ So acc. = ∝ 2 As here, F= m 4πε 0 r 2 r i.e., acceleration is not constant during motion.

or

So the electron will experience a force Fe = eE opposite to the field as shown in Figure and hence acceleration of electron along y-axis: F eE = a= m m = constt. ...(2) So from equation of motion, 1 s = ut + at 2 2 Along x-axis, L = υ0t [as a = 0] …(3) 1 2 …(4) and along y-axis, y = at [as u = 0] 2 Eliminating t, between Eqs. (3) and (4)

A very small sphere of mass 80 g having a charge q is held at a height 9 m vertically above the centre of a fixed conducting sphere of radius 1m, carrying an equal charge q. When released it falls until it is repelled just before it comes in contact with the sphere. Calculate the charge q. [g = 9.8 m/s2] A

80 × 10 –3 × 9.8

5.

3.

4.

q2 =

or

2

y=

+q

1 L 1 eEL2  eE  as from Eq. (2) a =  a  = 2  u0  2 mu 02  m

1 eE 2 1  2 L as K = 2 mυ 0  4 K   Substituting the given data and value of E from Eq. (1), 1 (1.6 × 10 –19 )(1.2 × 10 5 ) × (1.5 × 10 –2 ) 2 y= 4 200 × 1.6 × 10 –19 Ans. = 3.375 mm

or y =

9m

1m

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+ + +

+

+

B +

+ + +

+

+ + + +

25

JUNE 2010

P HYSICS F UNDAMENTAL F OR IIT-J EE

Newton's Law of motion KEY CONCEPTS & PROBLEM SOLVING STRATEGY Newton's First Law (or Law of Inertia) : A body continues to maintain its state of equilibrium till disturbed by an unbalanced force i.e. it continues to maintain its state of rest or of uniform motion till an unbalanced external force disturbs it. This law is also called Galileo Law or Law of Inertia. Newton's second Law : The rate of change of momentum is equal to the force applied on the body and this change takes place in the r → dp direction of force applied i.e. F = dt

defined as the product of extremely large force with the very small time value. Impulse is a vector quantity having direction along the force. Equilibrium : A body is said to be in the equilibrium state when (a) no net force acts on the body r r F = 0 (Condition for translational equilibrium)





∑ ⇒ ∑τ



Mathematically F AB = F BA

p2

t2

r F dt = p 2 – p1





∑ ∑

y

=0

∑F

z

=0

x

=0

∑τ

y

=0

∑τ

z

=0

∑ ∑

Frames of Reference : The system/co-ordinate system/a platform w.r.t. which the position or the motion of a body is determined is called a frame of reference. The simplest frame of reference having all the properties of a frame is the Cartesian co-ordinate frame/system. Frame of reference are of two types :

t1

S.N. Inertial frame Non-Inertial frame 1. Newton's laws are valid Newton's Laws are not in the inertial frames. valid in the non-inertial frame. They are to be modified by introducing the concept of pseudo force. 2. All non-accelerated All accelerated frames frames (frames at rest are non-inertial frames. or frames moving with uniform velocity) are inertial frames.

t1

So, Impulse = total change in momentum Also called Impulse-Momentum theorem Important : Newton's second law is the real law of motion as the First law and Third law can be derived form it. The concept of impulse must be applied at those places where an extremely large force acts on a body for a very small time interval. Then, impulse is just

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∑F

Important : For Rotational Equilibrium  total clockwise   total anticlockwise    =   momentus  momentus   

r F dt



p1

r ⇒ I=

t2

r dp =

=0

This statement is none other than law of conservation of moments according to which the above condition can be restated as  total clockwise   total anticlockwise    =   momentus  momentus   

or mAaA = mBaB (in magnitude) i.e. for the same force acting on two bodies the massive body has less acceleration than a light body. Impulse : If two bodies moving along a straight line collide, then the collision is small and the force experienced during collision on any of the two bodies varies with time and has a large value. In such cases the net effect of force can be measured with the help of a physical quantity called Impulse. r r → r → dp ⇒ dp = F dt = d I since F = dt r ⇒ I=

x

(b) no net torque acts on the body : r r p = 0 (condition for rotational equilibrium)

Newton's third law (Action-reaction law) : To every action there is equal and opposite reaction and both act on two different bodies. →

∑F

26

JUNE 2010

3.

A particle moves with uniform velocity in the absence of an external force. A frame of reference moving with constant velocity with respect to an inertial frame is also inertial.

4.

include the magnitude of one of the forces, the components of a force, or the direction of a force. Step 2: Set up the problem using the following steps: Draw a very simple sketch of the physical situation, showing dimensions and angles. You don't have to be an artist! For each body that is in equilibrium, draw a freebody diagram of this body. For the present, we consider the body as a particle, so a large dot will do to represent it. In your free-body diagram, do not include the other bodies that interact with it, such as surface it may be resting on, or a rope pulling on it. Now ask yourself what is interacting with the body by touching it or in any other way. On your free-body diagram, draw a force vector for each interaction. If you know the angle at which a force is directed, draw the angle accurately and label it. A surface in contact with the body exerts a normal force perpendicular to the surface and possibly a friction force parallel to the surface. Remember that a rope or chain can't push on a body, but can only pull in a direction along its length. Be sure to include the body's weight, except in case where the body has negligible mass (and hence negligible weight). If the mass is given, use w = mg to find the weight. Label each force with a symbol representing the magnitude of the force. Do not show in the free-body diagram any forces exerted by the body on any other body. The sums r in Eq. F = 0 (particle in equilibrium, vector

In this frame of reference the particle doesn't move with uniform velocity. A rotating frame of reference is a noninertial frames and an example for this is the earth.

Concept of Pseudo force : If a body of mass m is placed in a non-inertial frame r having acceleration a 0 then it experiences a pseudo r force m a 0 acting in a direction opposite to the r direction of a 0 (the acceleration of non-inertial frame). So, r r Fpseudo = – m a 0

where, negative sign indicates the pseudo force is always directed in a direction opposite to the direction of the acceleration of the frame. While drawing free Body Diagrams (FBDs) in which pseudo force is involved, we must first see the acceleration of the non internal frame and then in the FBD, plot the pseudo force with a value ma in a direction opposite to the acceleration of non-inertial frame. y



a0 m

Fpseudo = ma0

form) and

x

= 0;

∑F

y

= 0 (particle in

equilibrium, component form) include only forces that act on the body. Make sure you can answer the question "What other body causes that force?" for each force. If you can't answer that question, you may be imagining a force that isn't there. Choose a set of coordinate axes and include them in your free-body diagram. (If there is more than one body in the problem, you'll need to choose axes for each body separately.) Make sure you label the positive direction for each axis. This will be crucially important when you take components of the force vectors as part of your solution. Often you can simplify the problem by your choice of coordinate axes. For example, when a body rests or sides on a plane surface, it's usually simplest to take the axes in the directions parallel and perpendicular to this surface, even when the plane is tilted. Step 3 : Execute the solution as follows : Find the components of each force along of the body's coordinate axes. Draw a wiggly line through each force vector that has been replaced

x

"No negative sign has then to be applied to the value of pseudo force as its direction in the indicated correctly". Problem Solving strategy Newton's First Law : Equilibrium of a Particle Step 1 : Identify the relevant concepts : You must use Newton's first law for any problem that involves forces acting on a body in equilibrium. Remember that "equilibrium" means that the body either remains at rest or moves with constant velocity. For example, a car is in equilibrium when it's parked, but also when it's driving down a straight road at a steady speed. If the problem involves more than one body, and the bodies interact with each other, you'll also need to use Newton's third law. This law allows you to relate the force that one body exerts on a second body to the force that the second body exerts on the first one. Be certain that you identify the target variable(s). Common target variables in equilibrium problems

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∑F

27

JUNE 2010

by its components, so you don't count it twice. Keep in mind that while the magnitude of a force is always positive, the component of a force along a particular direction may be positive or negative. Set the algebraic sum of all x-components of force equal to zero. In a separate equation, set the algebraic sum of all y-components equal to zero. (Never add x-and y-components in a single equations.) You can then solve these equations for up to two unknown quantities, which may be force magnitudes, components, or angles. If there are two or more bodies, repeat all of the above steps for each body. If the bodies interact with each other, use Newton's third law to relate the forces they exert on each other. Make sure that you have as many independent equations as the number of unknown quantities. Then solve these equations to obtain the target variables. This part is algebra, not physics, but it's an essential step. Step 4 : Evaluate your answer : Look at your results and ask whether they make sense. When the result is a symbolic expression or formula, try to think of special cases (particular values or extreme cases for the various quantities) for which you can guess what the results ought to be. Check to see that your formula works in these particular cases. Newton's Second Law : Dynamics of Particles Step 1: Identify the relevant concepts : You have to use Newton's second law for any problem that involves forces acting on an accelerating body. As with any problem, identify the target variable – usually an acceleration or a force. If the target variable is something else, you'll need to identify another concept to use. For example, suppose the problem asks you to find how fast a sled is moving when it reaches the bottom of a hill. This means your target variable is the sled's final velocity. To find this, you'll first need to use Newton's second law to find the sled's acceleration. Then you'll also have to use the constant –acceleration relationships to find velocity from acceleration. Step 2: Set up the problem using the following steps: Draw a simple sketch of the situation. Identify one or more moving bodies to which you'll apply Newton's second law. For each body you identified, draw a free-body diagram that shows all the forces acting on the body. (Don't try to be fancy–just represent the object by a point.) Be careful not to include any forces exerted by the object on some other object. Remember, the acceleration of a body is determined by the forces that act on it, not by the forces that it exerts on anything else. Make sure you can answer the question "What other body is applying this force ?" for each force in your

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diagram. Furthermore, never include the quantity r ma in your free-body diagram; it's not a force!

Label each force with an algebraic symbol for the force's magnitude, as well as a numerical value of the magnitude if it's given in the problem. (Remember that magnitudes are always positive. Minus signs show up later when you take components of the forces.) Usually, one of the forces will be the body's weight; it's usually best to label this as w = mg. If a numerical value of mass is given, you can compute the corresponding weight. Choose your x-and y-coordinate axes for each object, and show them explicitly in each free body diagram. Be sure to indicate which is the positive direction for each axis. If you know the direction of the acceleration, it usually simplifies things to take one positive axis along that direction. Note that if your problem involves more than one object and the objects accelerate in different directions, you can use a different set of axes for each object. r r In addition to Newton's second law, F = ma ,



identify any other equations you might need. (You need one equation for each target variable.) For example, you might need one or more of the equations for motion with constant acceleration. If more than one body is involved, there may be relationships among their motions; for example, they may be connected by a rope. Express any such relationships as equations relating the accelerations of the various bodies. Step 3 : Execute the solution as follows : For each object, determine the components of the forces along each of the object's coordinate axes. When you represent a force in terms of its components, draw a wiggly line through the original force vector to remind you not to include it twice. For each object, write a separate equation for each component of Newton's second law, as in Eq. ΣFx = max; ΣFy = may (Newton's second law, component form) Make a list of all the known and unknown quantities. In your list, identify the target variable or variables. Check that you have as many equations as there are unknowns. If you have too few equations, go back to step 5 of "Set up the problem." If you have too many equations, perhaps there is an unknown quantity that you haven't identified as such. Do the easy part – the math! Solve the equations to find the target variable(s).

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figure. What horizontal force F must be applied on M so that m1 and m2 do not move relative to it ? Sol. Since m1 and m2 are in accelerating frame, we can assume that inertial force m1a and m2a act on them, respectively, a being the acceleration of the system. Clearly,

Step 4 : Evaluate your answer : Does your answer have the correct units ? (When appropriate, use the conversion 1N = 1kg . m/s2) Does it have the correct algebraic sign ? (If the problem is about a sled sliding downhill, you probably took the positive x-axis to point down the hill. If you then find that the sled has a negative acceleration – that is, the acceleration is uphill – then something went wrong in your calculations.) When possible, consider particular values or extreme cases of quantities and compare the results with your intuitive expectations. Ask, "Does this result make sense ?"

a=

F M + m1 + m 2 N1

m1a

m1

M

Solved Examples

m2

m2a

N2

The forces acting on m1 and m2 are shown in fig. We have T = m1a For m1 : T = m2g For m2 : ⇒ m1a = m2g

A balloon is descending with a constant acceleration a. The mass of the balloon and its contents is M. What mass m of its contents should be released so that the balloon starts ascending with the same acceleration a ? Assume that the volume of the balloon remains the same when the mass m is released. Sol. The forces acting on the balloon are its weight and the upthrust U due to air. Since the volume of the balloon remains the same the upthrust is the same in both the cases. We have, According to Newton's second law, Mg – U = Ma and U – (M – m)g = (M – m)a Solving these, we get

or

a=

m2 g m1

...(2)

Eqs. (1) & (2) give F = (M + m1 + m2)

m2 g m1

The total mass of an elevator with a 80 kg man in it is 1000 kg. This elevator moving upward with a speed of 8 m/sec, is brought to rest over a distance of 16m. Calculate (a) the tension T in the cables supporting the elevator and (b) the force exerted on the man by the elevator floor. Sol. (a) The elevator having an initial upward speed of 8 m/sec is brought to rest within a distance of 16 m. Hence, 3.

2a M a+g U

a

T

m2g

1.

U

T

m1g

F

m=

...(1)

0 = (8)2 + 2a(16)

a

(Q v2 = u2 + 2as)

8×8 = – 2 m/sec2 2 × 16 Resultant upward force on elevator = T – mg. According to Newton's law, T – mg = ma or T = mg + ma = m(g + a) = 1000 (9.8 – 2) = 7800 N. (b) Let P be the upward force exerted on the man by the elevator floor. If m´ be the mass of the man, then Weight of the man acting downward = m´g Upward force on the man = P – m´g. a=–

2.

Mg

(M – m)g

(a)

(b)

A frictionless cart of mass M carries two other frictionless carts having masses m1 and m2 connected by a string passing over a pulley, as shown in the

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29

JUNE 2010

According to Newton's law, P – m´g = m´a or P = m´(a + g) or = 80 (– 2 + 9.8) = 624 N

A

θ

x

What is the least horizontal force needed to pull a cylinder of radius a and weight W over an obstacle of height b ? Sol. The situation is shown in fig. The different forces acting on the cylinder are shown in fig. The weight W of the cylinder acts downwards. The applied force F is horizontal towards the left.

T

4.

O

a

A

b

Sol. The force acting on A towards B is given by T cos θ. From figure, F = 2 T sin θ

F 2 sin θ Now, acceleration of A towards B, a1 = T cos θ/m and acceleration of B towards A, a2 = T cos θ/m Substituting the value of T, from eq. and (3), we get or T =

b B W (b)

W (a) As the cylinder is pulled up and not rolled up, the algebraic sum of the moments about a point A, which is in contact with the obstacle must be zero, hence F × OB = W × AB From fig. (b) , OB = (a – b) and

AB =

[(OA) 2 − (OB) 2 ] =

=

(a 2 − a 2 + 2ab − b 2 )

=

[b(2a − b)]

a1 = a2 =

5.

F=

=

[a 2 − (a − b) 2 ]

(a − b )

connected by a light string of length 2l as shown in fig. A constant force F is applied continuously at the mid-point of the string, always along the perpendicular bisector of the straight line joining the two particles. Show that when the distance between the two particles is x, the acceleration of the particle x F . m (l 2 − x 2 )1/ 2 2l

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F (a)

...(3) (1) in eqs. (2)

F cot θ 2m

x F . 2 m (l − x 2 )1/ 2

Lets pause briefly to discuss a situation that illustrates the concept of electric potential difference. Consider the common 12-V automobile battery. Such a battery maintains a potential differences across its terminals, where the positive terminal is 12 V higher in potential than the negative terminal. In practice, the negative terminal is usually connected to the metal body of the car, which can be considered at a potential of zero volts. The battery becomes a useful device when it is connected by conducting wires to such things as headlight, a radio, power windows, motors, and so forth. Now consider a charge of +1C, to be moved around a circuit that contains the battery connected to some of these external devices. As the charge is moved inside the battery from the negative terminal (at 0V) to the positive terminal (at 12 V), the work done by the battery on the charge is 12 J. Thus, every coulomb of positive charge that leaves the positive terminal of the battery carries energy of 12 J. As this charge moves through the external circuit towards the negative terminal, it gives up its 12 J of electrical energy to the external devices. When the charge reaches the negative terminal, its electrical energy is zero.

W [b(2a − b)]

m

...(2)

AUTOMOBILE BATTERY

Two particles of equals masses m and m are

is a =

...(1)

F cos θ F cot θ = 2m sin θ 2m

∴ Acceleration of approach = 2 ×

∴ F × (a – b) = W [b(2a − b)]

or

F (b)

O (a – b)

F

B

θ

l2 − x 2

T l

F

T cosθ

T cosθ

m

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KEY CONCEPT

NOMENCLATURE & ISOMERISM

Organic Chemistry Fundamentals

objects are chiral, while other objects can be shown to be chiral only by applying the universal test for chirality–the nonsuperposability of the object and its mirror image. (hydrogen)

Alicyclic compounds : In addition to the simple monocyclic compounds, there are more complicated compounds with bridges linked across the ring e.g., CH CH–CH2 H2C CH2 CH2 CH2 CH2 H2C CH2 CH–CH

H 1

bicyclo[2,2,1] heptatne bicyclo[3,1,0] hexane According to the I.U.P.A.C system, cycloalkanes consisting of two rings only and having two or more atoms in common, take the prefix bicyclo followed by the name of the alkane containing the same total number of carbon atoms. The number of carbon atoms in each of the three bridges connecting the two tertiary carbon atoms is indicated in brackets in descending order. Numbering begins with one of the bridgeheads and proceeds by the longest possible path to the second bridgehead; numbering is then continued from this atom by the longer unnumbered path back to the first bridgehead and is completed by the shortest path e.g., CH3 2

1 CH2––C———CHC 2H5 6

3

CHCH3 CH2

5

4

CHCl–CH——CH2

6-chloro-2-ethyl-1, 8-dimethylbicyclo[3,2,1]octane

N.B.A bridged system is considered to have a number of rings equal to the number of scissions required to convert the system into an acyclic compound. Enantiomers and chiral molecules : Enantiomers occur only with compounds whose molecules are chiral. A chiral molecule is defined as one that is not superposable on its mirror image. Alkene stereoisomers are not chiral, whereas the trans-1, 2-dimethylcyclopentane isomers are chiral. A chiral molecule and its mirror image are called a pair of enantiomers. The relationship between them is defined as enantiomeric. Molecules (and objects) that are superposable on their image are achiral (meaning not chiral). Most socks, for example, are achiral, whereas shoes are chiral. Many familiar

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4

(ethyl)

OH (hydroxy) An important property of enantiomers such as these is that interchanging any two groups at the tetrahedral atom that bears four different groups converts one enantiomer into the other. In the figure, it is easy to see that interchanging the OH group and H-atom convert one enantiomer into the other. Because interchanging two groups at C2 converts one stereoisomer into another, C2 is an example of what is called a stereogenic carbon. A stereogenic carbon is defined as a carbon atom bearing groups of such nature that an interchange of any two groups will produce a stereoisomer. Carbon-2 of butanol is an example of a tetrahedral stereogenic carbon. Not all stereogenic cabons are tetrahedral, however. The carbon atoms of cis- and trans-1, 2-dichloroethene are examples of trigonal planar stereogenic carbons because an interchange of groups at either atom also produces a stereoisomer (a diastereomer). In general, any location where an interchange of groups leads to a stereoisomer is called a stereogenic centre. When we discuss interchanging groups like this, we must take care to notice that what we are describing is something we do to a molecular model or something we do on paper. An interchange of groups in a real molecule, if it can be done, requires breaking covalent bonds, and this is something that requires a large input of energy. This means that enantiomers such as the 2-butanol enantiomers do not interconvert spontaneously. Tests for chirality : Plane of symmetry The ultimate way to test for molecular chirality is to construct models of the molecule and its mirror image and then determine whether they are superposable. If the two models are superposable, the molecule that they represent is achiral. If the models are not superposable, then the molecules that they represent are chiral. We can apply this test with

CH

8

3

(methyl) CH3 – C – CH2CH3

2

7

2

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quantities as well. A specific rotation might be given as follows :

actual models, as we have just described, or we can apply it by drawing three-dimensional structures and attempting to superpose them in our minds. There are other aids, however, that will assist us in a recognizing chiral molecules. We have mentioned one already : the presence of a single tetrahedral stereogenic carbon. The other aids are based on the absence in the molecule of certain symmetry elements. A molecule will not be chiral, for example, if it possesses a plane of symmetry. A plane of symmetry (also called a mirror plane) is defined as an imaginary plane that bisects a molecule in such a way that the two halves of the molecule are mirror images of each other. The plane may pass through atoms, between atoms, or both. For example, 2-chloropropane has a plane of symmetry fig.(a), whereas 2-chlorobutane does not fig(b). All molecules with a plane of symmetry are achiral.

CH3 H

This means that the D line of a sodium lamp (λ = 589.6nm) was used for the light, that a temperature of 25ºC was maintained, and that a sample containing 1.00 g mL–1 of the optically active substance, in a 1-dm tube, produced a rotation of 3.12º in a clockwise direction.The specific rotations of (R)-2butanol and (S)-2-butanol are given here : CH3 CH3 HO

CH3

H

C

H

CH3 (R)-2-Butanol

CH3 (S)-2-Butanol

[α]1025 = –13.52º [α]1025 = +13.52º The direction of rotation of plane-polarized light is often incorporated into the names of optically active compounds. The following two sets of enantiomers show how this is done : CH3 CH3

C2H5

HOCH2

H

H

C

H

C2H5

(R)-(+)-2-Methyl-1-Butanol [α]1025 = +5.756º

(a) (b) (a) 2-Chloropropane has plane of symmetry & achiral. (b) 2-Chlorobutane does not possess a plane of symmetry and is chiral. Specific Rotation : The number of degrees that the plane of polarization is rotated as the light passes through a solution of an enantiomer depends on the number of chiral molecules that it encounters. This, of course, depends on the length of the tube and the concentration of the enantiomer. In order to place measured rotations on a standard basis, chemists calculate a quantity called the specific rotation, [α], by the following equation: α [α] = c.l

(S)-(–)-2-Methyl-1-Butanol [α]1025 = –5.756º

CH3 ClCH2

C

CH2OH

C

C2H5

CH3 H

H

C2H5

C

CH2Cl

C2H5

(R)-(–)-1-Chloro-2-methylbutane [α]1025 = –1.64º

(S)-(+)-1-Chloro-2-methylbutane [α]1025 = +1.64º

The previous compounds also illustrate an important principle : No obvious correlation exists between the configurations of enantiomers and the direction [(+) or (–)] in which they rotate plane-polarized light. (R)-(+)-2-Methyl-1-butanol and (R)-(–)-2-Methyl-1butanol chloro-2-methyl butane have the same configuration; that is, they have the same general arrangement of their atoms in space. They have, however, an opposite effect on the direction of rotation of the plane of plane-polarized light : CH3 CH3

where [α] = the specific rotation α = the observed rotation c = the concentration of the solution in grams per milliliter of solution (or density in g mL–1 for neat liquids) l = the length of the tube in decimeters (1 dm = 10 cm) The specific rotation also depends on the temperature and the wavelength of light that is employed. Specific rotations are reported so as to incorporate these

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OH

C CH2

CH2

Cl

Cl

CH3

[α]25 D = + 3.12º

HOCH2

C

H

C2H5 (R)-(+)-2-Methyl-1-butanol

ClCH2

C

H

C2H5 (R)-(–)-1-Chloro-2-methylbutane

These same compounds also illustrate a second important principle: No necessary correlation exist 32

JUNE 2010

between the (R) and (S) designation and the direction of rotation of plane-polarized light. (R)-2-Methyl-1butanol is dextrorotatory (+), and (R)-1-chloro-2methyl butane is levorotatory (–)

=

The enantiomeric excess can be calculated from optical rotations : % Enantiomeric excess observed specific rotation ×100 = specific rotation of the pure enantiomer

Racemic Forms

A sample that consists exclusively or predominantly of one enantiomer causes a net rotation of plane – polarized light. A plane of polarized light as it encounters a molecule of (R)-2-butanol, causing the plane of polarization to rotate slightly in one direction. (for the remaining purposes of our discussion we shall limit our description of polarized light to the resultant plane, neglecting consideration of the circularly polarized components from which plane-polarized light arises.) Each additional molecule of (R)-2-butanol that the beam encounters would cause further rotation in the same direction. If, on the other hand, the mixture contained molecules of (S)-2-butanol, each molecule of this enantiomer would cause the plane of polarization to rotate in the opposite direction. If the (R) and (S) enantiomers were present in equal amounts, there would be no net rotation of the plane of polarized light.

Meso Compounds : A structure with two stereogenic carbons does not always have four possible stereoisomers. Sometimes there are only three. This happens because some molecules are achiral even though they contain stereogenic carbons. To understand this, let us write stereochemical formulas for 2,3-dibromobutane shown here: CH3 * CHBr * CHBr CH3 2,3-Dibromobutane

We begin in the same way as we did before. We write the formula for one stereoisomer and for its mirror image: CH3 CH3

An equimolar mixtrure of two enantiomers such as the example above is called a racemic mixture (or racemate or racemic form). A racemic mixture causes no net rotation of plane-polarized light. In a racemic mixture the effect of each molecule of one enantiomer on the in circularly polarized beam cancels the effect of molecules of the other enantiomers, resulting in no net optical activity.

Br

C C

H

H

H

Br

Br

C C

Br

H

CH3 CH3 (A) (B) Structures A and B are nonsuperposable and represent a pair of enantiomers. When we write structure C and its mirror image D, however, the situation is different. The two structures are superposable. This means that C and D do not represent a pair of enantiomers. Formulas C and D represent two different orientations of the same compounds :

The racemic form of a sample is often designated as being (±).A racemic mixture of (R)-(–)-2-butanol and (S)-(+)-2-butanol might be indicated as (±)-2-butanol or as (±)- CH3CH2CHOHCH3 Racemic forms and enantiomeric excess :

A sample of an optically active substance that consists of a single enantiomer is said to be enantiomerically pure or to have an enantiomeric excess of 100%. An enantiomerically pure sample of (S)-(+)-2-butanol shows a specific rotation of +13.52º

CH3 H

( [α]25 D = + 13.52º). On the other hand, a sample of (S)-(+)-2-butanol that contains less than an equimolar amount of (R)-(–)-2-butanol that contains less than an equimolar amount of (R)-(–)-2-butanol will show a specific rotation that is less than + 13.52º but greater than 0º. Such a sample is said to have an enantiomeric excess less than 100%. The enantiomeric excess (ee) is defined as follows :

H

C C CH3 (C)

CH3 Br

Br

Br

Br

C C CH3 (D)

H

H

This structure when turned by 180º in the plane of the page can be superposed on C.

The molecule represented by structure C (or D) is not chiral even though it contains tetrahedral atoms with four different attached groups. Such molecules are called meso compounds. Meso compounds, because they are achiral, are optically inactive.

% Enantiomeric excess

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moles of one enantiomer - moles of other enantiomer ×100 total moles of both enantiomers

33

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KEY CONCEPT

Physical Chemistry Fundamentals

ELECTRO CHEMISTRY contain between them the whole solution. Thus, it gives the conducting power of the ions produced by 1 mole of an electrolyte at any particular concentration. This can be calculated using the expressions analogous to eqs (ii) and (iii). Λm = κV = κ(1/c) i.e. Λm = κ/c where V is the volume of the solution containing one mole of the substance and c is the resultant molar concentration. Note V carries the unit of m3 mol–1 The unit of Λm will be Λm = (Ω–1m–1)(m3 mol–1) = Ω–1 m2 mol–1 ≡ Sm2 mol–1 Variation of conductivity and molar conductivity with concentration : Both the conductivity and molar conductivity of a solution vary with concentration. The conductivity increases with increase in concentration whereas the molar conductivity increases on dilution (i.e. decrease in concentration). For strong electrolytes, conductivity increases sharply with increase in concentration while for weak electrolytes it starts at lower value in dilute solutions and increases much more gradually. In both the cases, this increase is due to the increase in the number of ions per unit volume of the solution. For strong electrolytes, the number of ions per unit volume increases in proportion to the concentration and that is why the increase in conductivity is very rapid. In weak electrolytes, however, the increase in the number of ions is basically due to the change in the partial ionization of the solute, and consequently, the conductivity increases very gradually. As stated above, molar conductivity Λm of both strong and weak electrolytes increases on dilution. The basic reason for this is that the decreases in conductivity is more than compensated by increase in the value of 1/c on dilution. The variation of molar conductivity on dilution for strong and weak electrolytes shows altogether different behavior as can be seen from figure where

Equivalent and Molar conductivities : Since the charges of solute ions are critical in determining the conductance of a solution, the comparison of conductance data is made between values for solutions corresponding to a total of unit charge on each ion of the solute. It is because of this that the equivalent conductivity of the solution is employed for comparison purpose. Suppose 1 equivalent mass of an electrolyte is dissolved in volume V of the solution. Let this whole solution be placed in a conductivity cell. Multiplying and dividing the right side of Eq. (i) by the distance l between the two electrodes of the cell, we get eq. (ii) κV A×l l l 1 κ=  G=   ; ...(i) G = κ 2 = 2 A A R l l    

...(ii) or Gl2 = κV Note : Both V and l2 in Eq. (ii) carry the units of m3 eq–1 and m2 eq–1, respectively. The term Gl2 is known as equivalent conductivity, abbreviated as Λeq. Thus, we have , Λeq = Gl2 = κV The equivalent conductivity of an electrolyte may be defined as the conductance of a volume of solution containing one equivalent mass of a dissolved substance when placed between two parallel electrodes which are at a unit distance apart, and large enough to contain between them the whole solution. The equivalent conductivity thus gives the conducting power of the ions produced by 1 equivalent mass (i.e., mass corresponding to a total of unit charge on each ion) of an electrolyte at any particular concentration. The unit of Λeq in CGS units are : Λeq = (Ω–1 cm–1) (cm3 eq–1) = Ω–1 cm2 eq–1 ≡ S cm2 eq–1 If c is the concentration of the solution (in equivalent per unit volume), then V (which carry a unit of m3 eq–1) will be equal to 1/c. Hence, Eq. (ii) may be written as Λeq = κ(1/c) i.e. Λeq = κ/c ...(iii) In SI units, c is expressed as the amount per unit volume instead of equivalent per unit volume, and thus one uses the term molar conductivity as defined below. The molar conductivity, Λm, of an electrolyte may be defined as the conductance of a volume of solution containing one mole of a dissolved substance when placed between two parallel electrodes which are at a unit distance apart, and large enough to

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Λm has been plotted against c . For strong electrolytes, the variation is almost linear in dilute solutions while that for weak electrolytes, the variation is very rapid. As the molar conductivity is a measure of the conducting power of all the ions that are available in 1 mole of the substance, it is, therefore, obvious that the number of ions that are available for conductance increases on dilution. For weak electrolytes, the increase in the number of ions

34

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ions), each ion migrates independently of its co-ion and contributes to the total molar conductivity of an electrolyte a definite share which depends only on its own nature and not at all on the ion with which it is associated. Thus, Λ∞m of the electrolyte must be equal to the sum of the molar conductivities of the ions composing it. Thus

has been explained on the basis of Arrhenius theory of electrolytic dissociation whereas that for strong electrolytes has been explained on the basis of Debye-Huckel-Onsager theory. In brief, the increase in the number of ions in case of weak electrolytes is due to the increase in the degree of ionization of the electrolyte on dilution, whereas in the case of strong electrolytes, this increase is due to the weakening of the ion-ion interactions on dilution. A brief account of the above two theories is given in the following.

Λ∞m (AB) = λ∞(A+) + λ∞(B–)

Λ

(i

(ii √ Variation of molar conductivity on dilution (i) for strong electrolyte and (ii) weak electrolyte Kohlrausch's law of independent migration of ions: For a strong electrolyte, the value of Λm in a very dilute solution, is very close to the limiting value of the conductivity Λ∞m at infinite dilution (or at zero concentration obtained by extrapolation). On the other hand, the corresponding value for a weak electrolyte is very far away from the limiting value at zero concentration. For example, at 25 ºC for 0.001 M sodium chloride solution Λm is 123.7 ohm–1 cm2 mol–1 as against Λ∞m of 126.5 ohm–1 cm2 mol–1. At the same concentration and temperature, the value for acetic acid is 49.2 ohm–1 cm2 mol–1 as compared to 390.7 ohm–1 cm2 mol–1 for the value of Λ∞m .



H

Λm = Λ∞m – b c

H

H

+

H – O H – O H – O+– H O – H O – H H H H H H H–O H–O–H O–H O–H O–H + H H H H H H–O–H O–H O–H O–H O–H + H H H H H O–H O–H O–H O–H O–H –

H

H

H

H

H

+

H–O H–O H–O H–O O – H H H H H

...(i)

where b is constant. The value of Λ∞m can, thus, be obtained by extrapolating the above curve to a value

H–O H–O H–O O–H O –

of c = 0. Statement of Kohlrausch's Law : The extrapolation method cannot be employed for a weak electrolyte as

H

H

H

H

H

H–O H–O O–H O–H O – H

Λm versus c curve does not approach linearity in solutions as dilute as 0.0001 M. In fact, the variation of Λm with dilution is very rapid. However, the value of Λ∞m for a weak electrolyte can be determined by the application of Kohlrausch´s law of independent migration of ions. This law states that at infinite dilution, where dissociation for all electrolytes is complete (including weak electrolytes since α → 1 as c → 0; Ostwald dilution law) and where all interionic effects disappear (because of larger distance between

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H

H–O H–O H–O H–O–H O–H + H H H H H

c , the curve approaches linearity in dilute

solutions, i.e.

H

H–O H–O H–O H–O H–O–H + H H H H H

Kohlrausch was the first to point out that when Λm for a uni-univalent strong electrolyte is plotted against

...(ii)

Abnormally high conductivities of H+ and OH–The molar conductivities of the hydrogen ion and the hydroxyl ion are much larger than those of other ions. This was first explained by von Grotthus and hence is known as Grotthus conductance. It is explained on the basis of a proton jump from one water molecule to another. The process of proton transfer results in a more rapid transfer of positive charge from one region of the solution to another, than would be possible if the ion H3O+ has to push its way through the solution as other ions do. The mechanisms of conduction of H+ ion and OH– ion are shown below.

H

H

H

H

H–O O–H O–H O–H O – H

H

H

H

H

O – H O – H O – H O – H O–

This type of mechanism also prevails in any other solvent. Thus, in a given solvent (for example, liquor ammonia) the molar conductivities of its characteristic cation and anion (namely, NH4+ and NH2–) will have unusually high values than any other cations and anions. 35

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UNDERSTANDING

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

Inorganic Chemistry

A metal (A) gives the following observations : (i) It gives golden yellow flame. (ii) It is highly reactive and used in photoelectric cells as well as used in the preparation of Lassaigane solution. (iii) (A) on fusion with NaN3 and NaNO3 separately, yields an alkaline oxide (B) and an inert gas (C). The gas (C) when mixed with H2 in Haber's process gives another gas (D). (D) turns red litmus blue and gives white dense fumes with HCl. (iv) Compound (B) react with water forming on alkaline solution (E). (E) is used for the saponification of oils and fats to give glycerol and a hard soap. (v) (B) on heating at 670 K give (F) and (A). The compound (F) liberates H2O2 on reaction with dil. mineral acids. It is an oxidising agent and oxidises Cr(OH)3 to chromate, manganous salt to manganate, sulphides to sulphates. (vi) (B) reacts with liquid ammonia to give (G) and (E). (G) is used for the conversion of 1, 2 dihaloalkanes into alkynes. What are (A) to (G)? Explain the reactions involved. Sol. (i) (A) appears to be Na as it gives the golden yellow flame. It is also used in the preparation of Lassaigane solution which is sodium extract of organic compounds. Na + C + N → NaCN Na + Cl → NaCl 2Na + S → Na2S (ii) Compound (B) is Na2O and (C) is N2 while (D) is NH3, as (D) is alkaline and turns red litmus blue and gives white fumes with HCl (C) + H2 → NH3 N2 + 3H2 2 NH 3

1.

CH2OOCC17H35 CHOOCC17H35 + 3NaOH CH2OOCC17H35

(A)

( F)

Na 2 O 2 + H 2SO 4 → H2O2 + Na2SO4 ( F)

dil.

(F) gives the following oxidations : Cr(OH)3 + 5OH– → CrO42– + 4H2O + 3e– Mn2+ + 8OH– → MnO4– + 4H2O + 5e– S2– + 8OH– → SO42– + 4H2O + 8e– The reduction equation of (F) is O22– + 2H2O + 2e– → 4OH– (vi) (G) is sodamide because it is used in the dehydrohalogenation reactions. Na 2 O + NH3(l) → NaNH 2 + NaOH ( B)

Br

(E)

(G )

CH3 – CH – CH2 + 2NaNH2 Br



CH3 – C ≡ CH Propyne + 2NaBr + 2NH3

2.

White fumes

(C)

3NaN3 + NaNO2 → 2 Na 2O + N 2 (C)

(iv) Compound (E) is NaOH as it is used in the preparation of soaps. Na 2O + H2O → 2 NaOH

XtraEdge for IIT-JEE



( B)

(iii) is prepared from Na as follows. 2NaNO3 + 10 Na → 6 Na 2 O + N 2

( B)

CH2OH + 3C17H35COONa (soap) CH2OH

K 2 Na 2 O 670  → Na 2 O 2 + 2 Na

NH3 + HCl → NH4Cl

( B)



(v) (F) is sodium peroxide as only peroxides gives H2O2 on reaction with dil. acids.

( D)

( B)

CH2OH

(E)

36

A green coloured compound (A) gave the following reactions : (i) (A) dissolves in water to give a green solution. The solution on reaction with AgNO3 gives a white ppt. (B) which dissolves in NH4OH solution and reappears on addition of dil. HNO3. It on heating with K2Cr2O7 and conc. H2SO4 produced a red gas which dissolves in NaOH to give yellow solution (C). Addition of lead acetate solution to (C) gives a yellow ppt. which is used as a paint. (ii) The hydroxide of cation of (A) in borax bead test gives brown colour in oxidising flame and grey colour in reducing flame. (iii) Aqueous solution of (A) gives a black ppt. on passing H2S gas. The black ppt. dissolves in aquaregia and gives back (A). (iv) (A) on boiling with NaHCO3 and Br2 water gives a black ppt. (D) (v) (A) on treatment with KCN gives a light green ppt. (E) which dissolves in excess of KCN to give JUNE 2010

(F). (F) on heating with alkaline bromine water gives the same black ppt. as (D). Identify compounds (A) to (F) and give balanced equations of the reactions. Sol. Reaction (i) indicates that (A) contains Cl– ions because, it gives white ppt. soluble in NH4OH. It is again confirmed because it gives chromyl chloride test. The colour of oxidising and reducing flames indicate that (A) also contains Ni2+ ions. Hence, (A) is NiCl2. The different reactions are : (i) NiCl2 + 2AgNO3 → 2AgCl + Ni(NO3)2 AgCl + 2NH3 → [Ag( NH 3 ) 2 ]Cl

Calculate mol of Ca(OH)2 required to carry out following conversion taking one mol in each case : COO COOH into Ca (a) COO COOH (b) H3PO4 into CaHPO4 (c) NH4Cl into NH3 (d) NaHCO3 into CaCO3 COOH Sol. (a) is a dibasic acid COOH COOH COO + Ca(OH)2 Ca COOH COO 1 mo l 1 mo l Ca(OH)2 required = 1 mol (b) H3PO4 + Ca(OH)2 → CaHPO4 + 2H2O 1 mol of H3PO4 ≡ 2H+ neutralised by 1 mol of Ca(OH)2 Ca(OH)2 required = 1 mol (c) 2NH4Cl + Ca(OH)2 → CaCl2 + 2NH3 + 2H2O 2 mol NH4Cl ≡ 1 mol Ca(OH)2 1 mol NH4Cl ≡ 0.5 mol Ca(OH)2 (d) 2NaHCO3 + Ca(OH)2 → Na2CO3 + CaCO3 + 2H2O 2 mol NaHCO3 ≡ 1 mol Ca(OH)2 1 mol NaHCO3 ≡ 0.5 mol Ca(OH)2 3.

So lub le

Ag(NH3)2Cl + 2HNO3 → AgCl ↓ + 2NH4NO3 white ppt . ( B)

The equations of chromyl chloride tests are : NiCl2 + Na2CO3 → 2NaCl + NiCO3 4NaCl + K2Cr2O7 + 6H2SO4 → 4NaHSO4 + 2KHSO4 + 3H2O + 2CrO 2 Cl 2 Re d gas

CrO2Cl2 + 4NaOH → Na 2 CrO 4

+ 2NaCl + 2H2O

Yellow solution ( C )

Na2CrO4 + (CH3COO)2Pb → PbCrO 4 + 2CH3COONa Yellow ppt .

(ii) Na2B4O7 . 10H2O ∆ Na2B4O7 + 10H2O Na2B4O7 ∆ 2 NaBO 2 + B 2 O 3 144 42444 3

4.

A compound Co(en)2 (NO2)2Cl has been prepared in a number of isomeric forms. One form undergoes no reaction with AgNO3 or (en) and is optically inactive. A second form reacts with AgNO3 but not with (en) and is optically inactive. A third form is optically active and reacts with both AgNO3 and (en). Identify each of these isomeric forms. Name and sketch each of the structures. Sol. Based on reactions with AgNO3 and (en), and optical activity, isomers can be identified. First form : There is no reaction with AgNO3, hence no Cl– ions outside coordination sphere. Also there is no reaction with bidentate (en), hence these ligands are trans to each other. Optical inactivity is also due to trans structure. Thus, it may have structure : + Cl

Transparent bead

NiO + B2O3 ∆

Ni(BO 2 ) 2 [Oxidising flame]

Nickel meta borate ( Brown )

Ni(BO2)2 + C



Ni + B2O3 + CO

Grey

[Reducing flame] (iii) NiCl2 + H2S → 2HCl + NiS ↓ Black ppt .

NiS + 2HCl + [O] → NiCl 2 + H2S ↑ (A)

(iv) NiCl 2 + 2NaHCO3 → NiCO3 + 2NaCl (A)

+ CO2 + H2O 2NiCO3 + 4NaOH + [O] ∆ Ni 2 O 3 ↓ Black ppt . ( D)

+ 2Na2CO3 + H2O (v) NiCl 2 + 2KCN → Ni(CN) 2 + 2KCl (A)

en

Ni(CN)2 + 2KCN → K 2 [ Ni(CN ) 4 ]

en NO2–

NO2

( F)

trans-chloronitrobis (ethylenediamine) cobalt (III) nitrite.

NaOH + Br2 → NaOBr + HBr

2K2[Ni(CN)4] + 4NaOH + 9NaOBr ∆ Ni 2 O 3 ↓ + 4KCNO + 9NaBr + 4NaCNO

Secon form : In this Cl– is outside coordination sphere since it reacts with AgNO3. As in the first

(D)

XtraEdge for IIT-JEE

Co

Green ppt . (E)

37

JUNE 2010

form NO2– ligands are trans to each other being optically inactive. This is represented as, + NO

∆ (NH4)2C2O4 + 2NaOH→ Na2C2O4 + 2NH3 + 2H2O (B) Na2C2O4 + CaCl2 → CaC 2 O 4 ↓ + 2NaCl

2

white ppt ( E )

en

Co

3NaOH + NH3 + 2K2HgI4 → Hg

en Cl–

NH2I ↓ + 4KI + 3NaI + 2H2O Hg brown ppt (Iodide of Millon's base) O

NO2

trans-bis (ethylenediamine) dinitrocobalt (III) chloride Third form : In this case also, Cl– is outside coordination sphere. Also it shows reaction with (en) hence monodentate ligands are cis to each other. Being optically active, mirror image should not superimpose. Thus, it can have structure : + en NO2 Co Cl– en

2MnO 4− + 16H+ + 5C2O42– violet

→ 10 CO2 + 2Mn 2+ + 8H2O colourless

NO2

cis-bis (ethylenediamine) dinitrocobalt (III) chloride A colourless salt (A), soluble in water, gives a mixture of three gases (B), (C) and (D) along with water vapours. (B) is blue, (C) is red and (D) is neutral towards litmus paper. Gas (B) is also obtained when (A) is heated with NaOH and gives brown ppt with K2HgI4. Solution thus obtained gives white ppt (E) with CaCl2 solution in presence of CH3COOH. (E) decolorises MnO4–/H+. Gas (C) turns lime water milky while gas (D) burns with blue flame and is fatal when inhaled. Identify (A) to (D) and explain reactions. Sol. Gas (B) gives brown ppt with K2HgI4 ⇒ gas (B) is NH3 ⇒ (A) has NH4+ ion Gas (C) turns lime water milky ⇒ gas (C) can be SO2 or CO2 Gas (D) is also obtained along with (C). Gas (D) burns with blue flame and is fatal when inhaled ⇒ gas (D) is CO ⇒ gas (C) is CO2 ⇒ (A) has C2O42– ion It is confirmed by the fact that CaCl2 gives white ppt CaC2O4 (E) which decolorises MnO4–/H+ ⇒ (A) is (NH4)2C2O4 Explanation : 5.

There is clear geological evidence that there has been water on Mars.

2.

George Ellery Hale was the 20th century's most important builder of telescopes. In 1897, Hale built a 40 inch wide telescope, the largest ever built at that time. Later, during the building of his 100 inch lens Hale spent time in a sanatorium and would only discuss his plans for the telescope with 'a sympathetic green elf'.

3.

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

4.

Every autumn, monarch butterflies fly 3500 kilometres from the Northern US to Mexico. It has been hypothesised that they have some kind of sun compass linked to their body clock which tells them which way to go. 'The Adventures of Tom Sawyer', written by Mark Twain, was the first novel ever to be written on a typewriter. A mature yew (if stripped) will yield between five and 20 pounds of bark.

5.

6. 7.



(NH4)2C2O4 → 2NH3 + CO2 + CO + H2O (A) (B) (C) (D) (B) is blue towards litmus (basic) (C) is red towards litmus (acidic) (D) is neutral

XtraEdge for IIT-JEE

1.

38

Since about 1990, GHB (gammahydroxybutyrate) has been abused in the U.S. for euphoric, sedative, and anabolic (bodybuilding) effects. GHB use associated with sexual assault has surpassed Rohypnol use associated with sexual assault. JUNE 2010

Set

2

`tà{xÅtà|vtÄ V{tÄÄxÇzxá

This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety of possible twists and turns of problems in mathematics that would be very helpful in facing IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and we hope that this section would prove a rich resource for practicing challenging problems and enhancing the preparation level of IIT JEE aspirants. By : Shailendra Maheshwari So lu t ion s wi l l b e p ub lished in nex t issue Joint Director Academics, Career Point, Kota Six different boxes are placed in a row. A ball is to be put in each box. If unlimited balls of red, blue, green and white colour are available, then find the number of ways in which all the boxes can be filled so that no two adjacent boxes have balls of the same colour in them.

7.

2.

If the non singular matrix A is symmetric, then justify that A–1 is also symmetric.

8.

3.

If θ1, θ2, θ3 and θ4 are eccentric angles of four

1.

x2



4.

Show that the origin lies in the acute angle between the planes x + 2y + 2z = 9 and 4x – 3y + 12z + 13 = 0.

5.

Let f(x) be a differentiable function on the interval a ≤ x ≤ b, with a > 0 and suppose that f(x) has a differentiable inverse, f–1(x). Evaluate :





b

a f (b)

f (a )

6.

x16 +

1 x

16

5 + 2 and y +

+ y20 +

1 y 20

9.

1 = 2, then y

is

(A) 49

(B) 51

(C) 2209

(D) 102

XtraEdge for IIT-JEE

2 cm

(C) 12 – 6 2 cm

(D) 18 – 12 2 cm

Find the sum of all the natural number 1 to 120 which are divisible neither by 2 nor 7. (A) 3042 (B) 3084 (C) 2529 (D) 3033

Given below are the definitions of functions F(x), f(x), g(x) and h(x) F(x) = x3 ; 0 ≤ x ≤ 1 = –x3 ; – 1 ≤ x < 0 = 1 ; otherwise f(x) = F(–x); x ∈ R g(x) = –F(x); x ∈ R h(x) = – F(–x) ; x ∈ R Answer the following questions based on these information’s.

((f −1 ( y)) 2 − a 2 ) dy

1 = x

(B)

Passage :

x (f (b) − f ( x )) dx

If x +

(A) 2cm

y2

= 1, the a2 b2 normals at which are concurrent, then prove that Σcos(θ1 + θ2) = 0. conormal points on the hyperbola

A circle S1 of area 36π cm2 touches the coordinate axes. Another circle S2 smaller than S1 also touches the coordinate axes as well as S1 also, then the radius of S2 is –

How many of the following expressions are necessarily zero for every real value of x : f(x) + h(x); g(x) – h(x); F(x) + f(x); f(x) – g(x) ? (A) 1 (B) 2 (C) 3 (D) 4

10. Which of the following relations is necessarily true ? (A) h(x) = f(–x); x ∈ R (B) F(x) = –f(–x); x ∈ R (C) g(x) + f(–x) = 0, x ∈ R (D) h(x) – F(–x) = 0; x ∈ R

39

JUNE 2010

MATHEMATICAL CHALLENGES SOLUTION FOR MAY ISSUE (SET # 1) 1.

[A] Let Y draws any card, the probability that X draws the same card = 1/n

2.

[B] Let ‘X’ draws a card marked with r, then y can draw any card marked 1, 2, 3, .... r –1. Hence the n

required probability =

x ⇒ x2 = 20 y 20 normal is x = my – 10m – 5m3 it passes through (15, 0) 15 + 10m + 5m3 = 0 m3 + 2m + 3 = 0 (m + 1) (m2 – m + 3) = 0 ⇒ m = –1 so point on parabola (–2am, am2) = (+10, 5) Required minimum distance is

y=

[B] Let ‘Y’ draws rth card, then ‘X’ draws any card marked 1, 2, 3 ...... (r – 1), n

Im, n =



1



1

0

=2

0

xm(1 – x)n dx,

1  r −1  n −1  = n  2n

∑ n  r =1

let x = sin2θ

(15 − 10) 2 + (5) 2

d=

sin2m + 1 θ . cos2n + 1 θ dθ

25 + 25 = 5 2 – 2

=

m +1 n +1

=

dmin2

so

m+n+2

I 24, 49

=

77 25 50

=

25 50 . 74 76 24 49

=

25 × 50 76 × 75

6.

5.

I 26 , 51 I 25, 50

= 50

 x2  d = (x1 – x2) +  1 − (17 − x 2 ) ( x 2 − 13)   20    2

2

2

Lt

x →0

2

(17 − x 2 )( x 2 − 13)

 π π for ∀ x ∈  − ,   2 2

f ( x ) − f ( 0) tan x ≤ Lt x →0 x x

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

 x2  it is the sq. of distance b/w two points  x1 , 1  &  20  

(x ,

as | f(x) | ≤ | tan x |

so f(0) = 0 so | f(x) – f(0) | ≤ | tan x | divide both sides by | x | f ( x ) − f (0) tan x ≤ x x

75

228

= ( 5 2 – 2)2 = 50 + 4 – 20 2 = 54 – 20 2

26 . 51 I 25, 50

225 − 221 = 2

r=

2

1 (r − 1) n −1 = n 2n

Hence required probability =

4.

centre (15, 0),

∑n r =1

3.

x12 20 2nd point lies on y2 + x2 – 30x + 221 = 0

First point lies on y =

)

n

ai

∑i

≤1

i =1

7.

(15, 0)

b 2 − 4ac ≤ |b| 1 −  2ac ≤ |b| 1 + 2 b 

XtraEdge for IIT-JEE

40

4ac b

2

≤ |b| 1 +

4ac b2

  

JUNE 2010

b 2 − 4ac ≤ |b| +

so

so that −

2ac b

Now,

= c.d – c.b – a.d + a.b using (1) & (2) in it AC . BD = 0 so AC ⊥ BD x2 y2 10. Let 2 + 2 = 1 be the ellipse and a b y = m1(x – ae) and y = m2(x – ae) are two chords through its focus (ae, 0). Any conic through the extrimities of these chords can be defined as {y – m1(x – ae)} {y – m2(x – ae)}   x 2 y2 ...(1) + λ  2 + 2 − 1 = 0 b  a If it passes through origin, then ...(2) m1m2a2e2 – λ = 0 solving (1) with x axis  x2  m1m2 (x – ae)2 + λ  2 − 1 = 0  a   using (2) in it (x – ae)2 + e2 (x2 – a2) = 0 (1 + e2)x2 – 2aex = 0 2ae x=0 & x= 1 + e2 so other point on x-axis through which this conic  2ae  passes is  , 0  which is a fixed point. 2 1+ e  Hence proved.

2

b b − 4ac b b c ± ≤ + + 2a 2a 2a 2a b

b c + a b Hence the solutions of az2 + bz + c = 0 satisfy the condition b c + |z| ≤ + a b =

8.

f(x) = (x – 1)3 g(x) + 1 ⇒ f ′(x) is divisible by (x – 1)2 & f(x) = (x + 1)3 g(x) – 1 ⇒ f ′(x) is divisible by (x + 1)2 so f ′(x) = a(x2 – 1)2 = a(x4 – 2x2 + 1) ax 5 2x 3 Now, f(x) = – a + ax + c 5 3 a 2a – +a+c=1 f(+1) = 5 3 ⇒ 8a + 15 c = 15 ...(1) a 2a & f(–1) = – + – a + c = –1 5 3 ⇒ –8a + 15 c = –15 ...(2) 15 so c = 0, so a = 8 1 so f(x) = (3x5 – 10x3 + 15x) 8

9.

Let the centre be O and P.V. of the pt. are respectively a , b , c , d D C O A

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

B

so | a | = | b | = | c | = | d | = R. given AB2 + CD2 = 4R2 | b − a |2 + | d − c |2 = 4R2

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

| b | 2 + | a |2 – 2a.b + | d | 2 + | c |2 – 2c.d = 4R2

4R2 – 2a.b – 2c.d = 4R2 so a.b + c.d = 0 cos (∠AOB) + cos (∠COD) = 0 cos ∠AOB = cos (π – ∠COD) so (∠AOB) + ∠COD = π so ∠AOD + ∠BOC = π ⇒ cos(∠AOD) + cos(∠BOC) = 0 so a.d + c.b = 0

XtraEdge for IIT-JEE

AC . BD = ( c − a ) . ( d − b )

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

...(1)

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

...(2) 41

JUNE 2010

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

MATHS 1.

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

also

0

∫ (9

−2t

− 2.9 − t ) dt ≥ 0



0

∫ (9

−2t

− 2.9 ) dt ≥ 0 0

(

| a× b | 1 + λ1 + λ 2 + λ1λ 2 2 λ + λ2 where, 1 ≥ λ1λ 2 2

)

0 a

≥0

→ →



→ →

= ∴



3.



D with respect to E be a , b , – λ1. a and –λ2 b ; where λ, λ2 ∈ R+ → λλ → → 1 → Now, ∆1 = | E C × E D | = 1 2 | a × b | 2 2

E( 0)

B( b )

A(a )

and

1 → → | a× b | | a × b || λ1λ 2 | = 2 2 → 1 → 1 → → ∆2 = | E B × EA | = | a × b | 2 2

∆1 =

λ1λ 2 ...(i)

→ →



∆2 =

XtraEdge for IIT-JEE

| a× b | 2

∆ ≥

1 → → | a× b | + 2

(1 + λ1λ 2 ) 2

→ →

| a × b | | λ1λ 2 | 2

∆2 +

∆1 {using (i) and (ii)

Let S be the coefficients of x49 in given expression f(x) and if P be product of roots of the equation S f(x) = 0, then find the value of , given that : P 1 x 1 x   f(x) = (x – 1)2  − 2   x −   − 3   x −  , 2  3 3 2  

 1  1  1   × ( x − 1) x −  x − ... x −  2  3  25    Now roots of f(x) = 0 are; 1 1 1 12, 22, 32, ..... , 252 and 1, , , ....., 25 2 3 Now f(x) is the polynomial of degree 50, So coefficient of x49 will be : S = – (sum of roots)

→ →



| a× b | + 2

1 + 2 λ1 λ 2 + λ1λ 2

1   x  .........  − 25   x −  25   25  Sol. Here we can write f(x) as :  x  x   x  f(x) = ( x − 1) − 2  − 3 ... − 25  2  3   25  

C( − λ 1 , a )

D( − λ 2 , b )

| a× b | 2

It is clear that equality holds if λ1 = λ2 and in this case side AB and DC will become parallel.

quadrilateral ABCD. Also discuss the case when the equality holds. Sol. Let the position vector of the points A, B, C and and →

∆ =



∆ 2 , where ∆ is the area of the



(1 + λ1 )(1 + λ 2 )

∆ =

Let ABCD be any arbitrary plane quadrilateral in the space having E as the point of intersection of its diagonals. If ∆1 and ∆2 be the areas of triangles DEC and AEB, using vector method prove that ∆1 +

| a× b | 2 → →

 9 −2t 2.9 − t  ⇒ ≥ 0 ⇒ − 9 − 2 t + (49) − t −  − 2 log 9 − log 9   a ⇒ 9–2a – 4.9–a + 3 ≥ 0 ⇒ t2 – 4t + 3 ≥ 0; where t = 9–a and t ∈(1, ∞ ) ⇒ (t – 1) (t – 3) ≥ 0 ⇒ t ≤ 1 or t ≥ 3 is possible as t > 1. 1 ∴ 9–a ≥ 3 ⇒ a ≤ – ; |3–a ≥ 3 ⇒ a ≤ –1 2

∆ ≥

∆ =

−t

a

2.

→ (1 + λ1 )(1 + λ 2 ) → → 1 → | A C× BD | = | a× b | 2 2 → →

a

Sol. Here

∆=

...(ii)

42

JUNE 2010

XtraEdge for IIT-JEE

43

JUNE 2010

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44

JUNE 2010

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45

JUNE 2010

XtraEdge for IIT-JEE

46

JUNE 2010

y

1   1 1 = – (12 + 22 + ... + 252) – 1 + + + .... +  2 3 25  

O

25

1  25 × 26 × 51  =–  + K  where, K = n 6   n =1 ⇒ S = –(K + 5525). Product of roots : 1 1 1 . .... = 1 . 2 . 3 ...25 12 . 22 . 32 .... 252 . 1 . 25 2 3 ∴ P = 25 !



Hence

S −(K + 5525) = , where K = P 25!

25

C

x

PA.PD = PB.PC Equation of any line through point 'P' is : x−α y−2 = =r cos θ sin θ or x = α + r cos θ, y = 2 + r sin θ Putting this point in the equation of given ellipse, we get 4(r cos θ + α)2 + 9(2 + r sin θ)2 = 36 ⇒ r2 (4 cos2 θ + 9 sin2θ) + 4r (9 sin θ + 2 α cos θ) + 4 α2 = 0 Since PA and PD are the roots of this quadratic in r, we get 4α 2 PA.PD = ...(i) (4 cos 2 θ + 9 sin 2 θ)

1

∑n n =1

Find the point inside a triangle from which the sum of the squares of distance to the three side is minimum. Find also the minimum value of the sum of squares of distance. Sol. If a, b, c are the lengths of the sides of the ∆ and x, y, z are length of perpendicular from the points on the sides BC, CA and AB respectively, we have to minimise : ∆ = x2 + y2 + z2 1 1 1 we have, ax + by + cz = ∆ 2 2 2 ⇒ ax + by + cz = 2∆ A

Similarly, putting x = r cos θ + α, y = r sin θ + 2 in the equation of coordinate axis i.e. xy = 0 (r cos θ + α). (r sin θ + 2) = 0 ⇒ r2 sin θ cos θ + r (2 cos θ + α sin θ) + 2α = 0 Since PB and PC and the roots of this quadratic in 'r', 2α 4α = ...(ii) we get, PB.PC = sin θ cos θ sin 2α Thus, we get

y

4α 4α 2 = {from (i) and (ii)} sin 2θ 4 cos 2 θ + 9 sin 2 θ 8α 2 4α ⇒ = sin 2θ 4(1 + cos 2θ) + 9(1 − cos 2θ)

x B C where ∆ is the area of ∆ABC. We have the identity : ⇒ (x2 + y2 + z2) (a2 + b2 + c2) – (ax + by + cz)2 = (ax – by)2 + (by – cz)2 + (cz – ax)2 2 2 ⇒ (x + y + z2)(a2 + b2 + c2) ≥ (ax + by + cz)2 ⇒ (x2 + y2 + z2) (a2 + b2 + c2 ≥ 4∆2 4∆2 ⇒ x2 + y2 + z-2 ≥ 2 a + b2 + c2 Equality holds only when x y z ax + by + cz 2∆ = = = 2 = 2 2 2 a b c a +b +c a + b2 + c2 ∴ The minimum value of ∆ is ; 4∆2 4(s − a )(s − b)(s − c)s = 2 2 2 a +b +c a 2 + b2 + c2

1 2α = sin 2θ 13 − 5 cos 2θ ⇒ 13 = 5 cos 2θ + 2α sin 2θ



25 + 4α 2 169 − 25 ∴ 4α 2 + 25 ≥ 13 ⇒ α2 ≥ = 36 4 ⇒ α ∈ (–∞, – 6] ∪ [6, ∞)

where; 5 cos 2θ + 2α sin 2θ ≤

Let f(x) be a polynomial with integral coefficients suppose that both f(1) and f(2) are odd. Then, prove that for any integer n, f(n) ≠ 0. Sol. Suppose f(x) = 0 for some integer n. Then (x – n) divides f(x) So; f(x) = (x – n) g (x) Now, f(1) = (1 – n) . g(1) and f(2) = (2 – n) g (2) Now g (1) and g(2) are both integers, and one of (1 – n) or (2 – n) is even. So one of f(1) or f(2) is even, which is contradictory, so there is no integer n, for which f(n) = 0 6.

From an external point P(α, 2) a variable line is x 2 y2 drawn to meet the ellipse + = 1 at the points 9 4 A and D. Same line meets the x-axis and y-axis at the points B and C respectively. Find the range of values of 'α' such that PA. PD = PB.PC. Sol. We have been given, 5.

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B

D

4.

z

P(α,2) A

47

JUNE 2010

MATHS

INVERSE TRIGONOMETRIC FUNCTION Mathematics Fundamentals (d). 1. sin–1x + cos–1x = π/2 2. tan–1x + cot–1x = π/2 3. sec–1x + cosec–1x = π/2 Formulae for Sum and Difference of Inverse Function –

Meaning of inverse function : 1. sin θ = x ⇔ sin–1 x = θ 2. cos θ = x ⇔ cos–1x = θ 3. tan θ = x ⇔ tan–1x = θ 4. cot θ = x ⇔ cot–1x = θ

 −1 x + y  tan 1 − xy where xy < 1 1. tan–1x + tan–1y =  x+y π + tan −1 when xy > 1  1 − xy

5. sec θ = x ⇔ sec—1x = θ 6. cosec θ = x ⇔ cosec–1x = θ Domains and Range of Functions : Function sin–1x

–1≤ x ≤ 1

Domain

Range

cos–1x

–1≤ x ≤ 1

0≤θ≤π

tan–1x

–∞ < x < ∞, i.e. x ∈ R



cosec–1x

x ≤ –1, x ≥ 1

sec–1x

x ≤ –1, x ≥ 1

cot–1x

–∞ < x < ∞ i.e. x ∈ R



2. tan–1x – tan–1y = tan–1

π π ≤θ≤ 2 2

3. sin–1x ± sin–1y = sin–1 x 1 − y 2 ± y 1 − x 2   

π π vb > Vc (B) va = vb = vc (C) va < vb = vc (D) va < vb < vc

6.

Light is incident from a medium X at angle of incidence i and refracted into a medium Y at angle of refraction r. The graph is sin i versus sin r is shown in the figure. Which of he following conclusions would fit the situation ?

9.

sin r

(B)

2

(C)

1

(D) 0

2

A particle of mass 5 kg moving in the x - y plane has its potential energy given by U = (–7x + 24y) J, Where x and y are in metre. The particle is initially at →

^

^

(A) The particle has a speed of 25 ms–1 at t = 4 s (B) The particle has an acceleration of 5 ms–2 (C) The acceleration of the particle is perpendicular to its initial velocity (D) None of the above is correct 10. A wire of length L, cross-sectional area A is made of a material of Young's modulus Y. The wire is stretched by an amount. x, which lies well within the elastic limit. If k be the equivalent force constant of the wire and W be the work done then YA YA (A) k = (B) k = 2L L

0.4 sin i

3 times that in

1 YAx 2 YAx 2 (D) W = 2 L L 11. A body of mass m falls from a height h onto a pan (of negligible mass) of a spring balance as shown. The spring also possesses negligible mass and has spring constant k. Just after striking the pan, the body starts oscillatory motion on vertical direction of amplitude A and energy E

(C) W =

(b) Speed of light in medium Y is 1/ 3 times that in medium X (c) Total internal reflection will occur above a certain i value (A) b and c (B) a and c (C) b only (D) c only Two transparent slabs have the same thickness as shown. One is made of material A of refractive index 1.5. The other is made of two materials B and C with thickness in the ratio 1 : 2. The refractive index of C is 1.6 If a monochromatic parallel beam passing in the slabs has the same number of wavelengths inside both, the refractive index of B is

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origin and has a velocity u = (14.4 i + 4.2 j ) ms–1 .

30º

7.

(D) 1.4

Questions 9 to 13 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) 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 no negative marks.

0.1

(a) Speed of light in medium Y is medium X

(C) 1.3

The potential energy for a force field F is given by U (x, y) = sin (x + y). The force acting on the particle  π of mass m at  0,  is  4 (A) 1

0.2

0.2

(B) 1.2

h

52

JUNE 2010

(A) A =

mg k

(C) E = mgh +

Passage : I (Ques. 14 to 15)

mg 2kh 1+ k mg

(B) A =

1  mg  kA 2 (D) E = mgh +   2  2k 

A rod AB of length 2 m and mass 2 kg is lying on smooth horizontal x – y plane with its centre at origin O as shown in figure. An impulse J of magnitude 10 Ns applied perpendicular to AB at A.

2

J

12. Each of the three balls has a mass m and is welded to the rigid equiangular from of negligible mass. The assembly rests on a smooth horizontal surface. A force F is suddenly applied to one bar shown.

x B

14. The distance of point P from centre of the rod which is at rest just after the impact is 2 1 (A) m (B) m 3 3

y m →

r

F

b

(C)

O 120º r

120º

x

r





1 F^ (A) Acceleration of point O is a = i 3m →



F^ i m (C) The magnitude of angular acceleration of the 1 Fb frame is α = 3 mr 2 (D) The magnitude of angular acceleration of the Fb frame is α = mr 2

(B) Acceleration of point O is a =

2 and medium 2 with z < 0 has a

refractive index

3 . A ray of light in medium 1 →

^

^

1 m 4

Collision is the transfer of momentum due to only the internal forces between the particles taking part in collision. When exchange of a momentum takes place between two physics bodies due to their mutual interactive force, it is defined as collision between two bodies. Two bodies move in different directions interact each other at the point of intersection of their line of motion and the reaction due to their physical contact is the interaction force which is the cause of transfer of momentum from one body to another. Collision may be either elastic or inelastic. In case of elastic collision, momentum and K.E. are both conserved but in case of inelastic collision only momentum is conserved and K.E. is not conserved.

^

given by the vector A = 6 3 i + 8 3 j – 10 k is incident on the plane of separation. The refracted ray makes angle r with +z axis and incident ray makes an angle i with –z axis. Then, (A) i = 120º (B) i = 60º (C) r = 45º (D) r = 135º

16. A simple pendulum is suspended from a peg on a vertical wall. The pendulum is pulled away from the wall to a horizontal position and released. The ball hits the wall, the coefficient of restitution being

This section contains 2 paragraphs; passage- I has 2 multiple choice questions (No. 14 & 15) and passage- II has 3 multiple (No. 16 to 18). 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.

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(D)

Passage: II (Ques. 16 to 18)

13. The x – y plane is the boundary between tow transparent media. Medium 1 with z ≥ O has a

refractive index

1 m 2

15. If the rod is hinged at B and instead of impulse J, a point mass of 1 kg moving with a speed 10 m/s collide at A and sticks to it. Find the angular velocity of rod just after the collision (A) 1 rad/s (B) 2 rad/s (C) 3 rad/s (D) 4 rad/s

m

m

A

 2   e =  . The number of collision after which the 5  amplitude of oscillation becomes less than 60º is : (A) n = 3 (B) n = 2 (C) n = 5 (D) n = 4

53

JUNE 2010

20. A plate of area 100 cm2 is placed on the upper surface of castor oil, 2mm thick. Taking the coefficient of viscosity to be 15.5 poise, calculate the horizontal force necessary to move the plate with a velocity 3 cm/sec. [in 10–5N].

→  ^ ^ 17. Two particles having the position r1 =  3 i + 5 j  m   → →  ^ ^ and r2 =  – 5 i – 3 j  m move with velocities v1 =   →  ^ ^  ^ ^  4 i + 3 j  m/s and v 2 =  a i + 7 j  m/s. If the     particles collide, then value of 'a' must be : (A) 8 (B) 6 (C) 4 (D) 2 18. A block of mass m is moved towards a movable wedge of mass M kg and height h with velocity u (All surfaces are smooth). If the block just reaches the top of the wedge, the value of u is :

21. A ball of 1 kg attached to a string of 0.5 m length is whirled round in a horizontal circle. If the breaking stress is given by 1.5 N find the maximum permissible angular velocity. [in rad-s–1] 22. Distance between two consecutive dark bands is 0.4 mm when yellow light of 6000 Å wavelength is used. Find the distance between two consecutive bright bands with a wavelength of 4500 Å [in 10–4 m] 23. A 4 cm object is placed perpendicular to the principal axis of a convex mirror of focal length 7.5 cm. Find distance of the image from the mirror if it is 0.6 cm is size. 24. A spot light rotates in a horizontal plane with a constant angular velocity of 0.1 rad/s. The spot of light P moves along the wall at a distance 3 m. What is the velocity of the spot P when θ = 45º ? (Ans. ……× 10–1)

h m

u

M kg

(A)

2gh

(B)

2ghK 1+ K

(C)

2gh (1 + K ) K

(D)

1  2gh 1 –  K  

Wall

P θ

x

This section contains 9 questions (Q.19 to 27). +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 have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following : X Y Z W 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9

S

O

25. A ball moving with a speed of 9 m/s strikes an identical stationary ball such that after the collision the direction of each ball makes an angle of 30º with the original line of motion. Find the speed of each ball after the collision. 9 m/s

V1

30º 30º

V2

26. The 10 kg block is resting on the horizontal surface when the force F is applied to it for 7 sec. What is the maximum velocity gained by the block. 10 kg µ5 = µx =

19. Find the mass of water flowing in 10 minutes through a tube 0.1 cm in diameter 40 cm long if there is a constant pressure head of 20 cm of water. Given η for water = 0.008 C.G.S. units and g = 1000 cm/s2. [in gram]

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3 cm

F

1 2

100 N F

40 4 sec. 7 sec. t sec.

54

JUNE 2010

27. Moment of inertia of a semicircular ring of mass π kg and radius 2 cm about the axis passing through point P and perpendicular to its plane is : [in 10–5 kg m2]

4.

P R = 2 cm

CHEMISTRY

CH3

5.

Questions 1 to 8 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. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer. 1.

(C)

(B)

(C)

3.

6.

Given : HCO3–(aq) + OH–(aq) → H2O(1) + CO32–(aq); ∆H1 = – 41.84 kJ mol–1 H+(aq) + OH–(aq) → H2O(1); ∆H2 = – 57.32 kJ mol–1 The enthalpy change for the reaction HCO3–(aq) → H+(aq) + CO32–(aq) would be (A) (–41.84 – 57.32) kJ mol–1 (B) (– 41.84 + 57.32) kJ mol–1 (C) (41.84 + 57.32) kJ mol–1 (D) (41.84 – 57.32) kJ mol–1

7.

Which of the following reactions is a redox reaction? (A) Cr2O3 + 6HCl → 2CrCl3 + 3H2O (B) CrO3 + 2NaOH → Na2CrO4 + H2O Cr2O72– + OH– (C) 2CrO42– + H+ – + 2– (D) Cr2O7 + 6I + 14H 2Cr3+ + 3I2 + 7H2O

8.

The propagation steps involved in the free radical addition of HX across a double bond are

CH3

CH=CH

–2

10 moles of Fe3O4 is treated with excess of KI solution in presence of dilute H2SO4, the products are 2+ Fe and I2 (g). What volume of 0.1 (M) Na2S2O3 will be needed to reduce the liberated I2 (g)? (A) 50 ml (B) 100 ml (C) 200 ml (D) 400 ml Most stable free radical is CH3

(C)

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CH2 – CH = C – CH3 CH3

CH2–CH=CH–CH2

(A)

CH = CH – CH – CH3

(D) None of these

CH3

2.

CH2 – CH2 – C = CH2

CH3

CH3

(D)

A,

CH3

C=C CH3

(CH3)3COΘ

CH2 – CH – CH – CH3 t-BuOH OTs pre dominant product A is

(A)

An alkene (A) C16H16 on ozonolysis gives only one product (B) (C8H8O). Compound (B) on reaction with NH2OH/H2SO4,∆ gives N-methyl benzamide the compound 'A' is H C=C (A) CH3 CH3 H (B)

A gas is present in a cylinder fitted with movable piston. Above and below of the piston there is equal number of moles of gas. The volume above is two times the volume below at a temperature of 300K. At what temperature will the volume above be four times the volume below(A) 600 K (B) 400 K (C) 200 K (D) 120 K

Step 1

X +

C=C

Step 2 X – C – C + HX

(B)

X–C–C

X–C–C–H+X

HCl does not follow free-radical addition because (A) Step 1 is exothermic and step 2 is endothermic (B) Step 1 is endothermic and step 2 is exothermic (C) Steps 1 and 2 are exothermic (D) Steps 1 and 2 are endothermic

(D)

55

JUNE 2010

12. Which of the following statements is incorrect (A) Alkynes are more reactive than alkenes towards halogen additions (B) Alkynes are less reactive than alkenes towards halogen additions (C) Both alkynes and alkenes are equally reactive towards halogen additions

Questions 9 to 13 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) 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 no negative marks. 9.

Consider the reaction O C – OH

+

(D) Primary vinylic cation (RCH = C H) is more reactive than secondary vinylic cation

Na , NH (l )

O , Me S

EtOH

CH 2Cl 2

+

2 3    → A 3  → B+C

( R C = CH 2 ) 13. When nitrobenzene is treated with Br2 in presence of FeBr3, the major product formed is mbromonitrobenzene. Statements which are related to obtain the m-isomer are : (A) The electron density on meta carbon is more than that on ortho and para positions (B) The intermediate carbonium ion formed after initial attack of Br+ at the meta position is least destabilised (C) Loss of aromaticity when Br+ attacks at the ortho and para positions and not at meta position (D) Easier loss of H+ to regain aromaticity from the meta position than from ortho and para positions

D

∆ Identify the correct representation of structure of the products COOH

(A) A is (B) The intermediate formed in the conversion of B to D is enol (C) The structure of C is O O (D) A can also be formed from the reaction + COOH

This section contains 2 paragraphs; passage- I has 2 multiple choice questions (No. 14 & 15) and passage- II has 3 multiple (No. 16 to 18). 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.

HBr / Boiling

OH   → product.

10.

Which of the following are possible products (in significant amount) (A)

(C) 11.

(B)

Br

Passage : I (Ques. 14 to 15) OH

Br

CHCl3 + NaOH CH3

(D)

H⊕ Isomerisation CH3 CHCl2 OHΘ Q

Which of the following statements is/are correct ? (A) At 300 ºC, with reaction in the vapour phase, the relative rates of substitution of primary, secondary and tertiary hydrogen atom by chlorine are 1.00 : 3.25 : 4.43 (B) As the temperature increases, the relative rates of substitution of primary, secondary and tertiary hydrogen atom by chlorine become equal, i.e. 1:1:1 (C) Increasing pressure causes a decrease in the relative rates of primary hydrogen substitution by chlorine (D) Bromination of alkene is more selective than chlorination

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P HCN/NH4Cl H3O+/∆ (R) α-amino acid (zwitter ion) O

14. Product P is OH

OH

(A)

CHO

(B) CHO

OHC

CHO

(C) CH3

56

OH

OH

CHO

(D) CH3

JUNE 2010

15. Product Q is OH

(A)

(C) For H2O if reduced pressure, reduced volume and reduced temperature are 20, 0.6 and 2 respectively then intermolecular force of repulsion predominate over intermolecular H-bonding among H2O molecules. (D) All of the above are correct.

OH

(B)

CHO

CHCl2 CH3

CH3 OH

O

(C)

This section contains 9 questions (Q.19 to 27). +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 have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following : X Y Z W 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9

(D)

CH3

CH3 CHO

CHO

Passage: II (Ques. 16 to 18) J C

218 Pressure 85 in atm

G

673 K 647 K 573 K

15

F

1

H

A

56 Vm (in cm3/mol)

B

E

473 K

In the given figure P-Vm isotherm of H2O is shown. The line (……) represent, the vanderwaals plot for H2O at 473 K. The vanderwaals constant of H2O is represented by a and b. 16

What is the equation of the dotted line (- - - ) AICFB?  d 2P   dP   =0  = 0 and  (A)  2   dV dV  m T  m T (B) P =

a  2b 1 + 2  Vm  Vm

(C) P =

a  2b  1 –  2  V Vm  m 

 d 2P (D)   dV 2  m

19. How many following compounds give positive test with tollen's reagent? [Glucose, Acetaldehyde, Sucrose, α-Hydroxy Ketone, fructose, Acetone, Maltose, Formic acid, Starch, Acetic acid, Lactose, Cellulose, Glycogen, Benzaldehyde.]

  

20.

  =0  T

21. pH of the half equivalence point of the titration of valine (CH3)2 CH CH NH2 COOH, is 2.284 with HCl and 9.716 with NaOH. Calculate the isoelectric point of valine that is, the pH at which the dipole ion does not migrate in an electric field.

As per the vanderwaals line I H G F (- - - -) which of the following section against the behaviour of gas(A) I H (B) H G (C) G F (D) All of the given 18. For H2O which of the following is / are correct(A) For H2O, compressibility factor (Zc) is equal to 0.23. (B) For H2O, compressibility factor (Zc) is lesser than 0.375 because of stronger intermolecular attraction among H2O molecules. 17

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1.245 g of CuSO4 . x H2O was dissolved in water and H2S was passed into it till CuS was completely precipitated. The H2SO4 produced in the filtrate required 10 mL of 1 N NaOH solution. Calculate x.

22. 80 Volumes of ammonia gas effuse out from a vessel in 8 minutes at –5ºC and 500 mm Hg pressure. How long (in minutes) will 40 volumes of a certain gas (X) with molecular mass 68 take to effuse out from the same vessel at the same temperature & pressure ?

57

JUNE 2010

23. Kinetic energy (in atm × litre) of 0.30 moles of He gas in a container of maximum capacity of 4 litres at 5 atm. must be : (R = 0.0821 atm litre mol–1 K–1)

(A) ln

 x 2 +1   +C (B) ln sec   2   

24. Maltose molecule reacts with 'X' number of phenyl hydrazine molecules to yield monophenylosa zone. The value of 'X' is.

(C)

1 ln | sec(x2 + 1) | + C 2

(D)

1 ln 2

25. For the reaction 1 O2 if one mole of 2 MnO4– oxidises 1.67 moles of Mx+ to MO3–, then the value of x in the reaction is :

Mx+ + MnO4– → MO3– + Mn2+ +

4.

26. 2 moles of an ideal gas (Cv = 5/2 R) was compressed adiabatically against constant pressure of 2 atm. Which was initially at 350 K and 1 atm pressure. The work involve (in term R) in the process is equal to : Give your answer after divide actual answer by 100 R.

s

n

∑∑

3.

n

C s s C r is

(A) (B) 3n + 1 (D) 3 (3n – 1)

3n n +1

(B)

6n n +1

(C)

9n n +1

(D)

12n n +1

Let x2 ≠ nπ – 1, n ∈ N then 2 sin ( x 2 + 1) – sin 2 ( x 2 + 1) 2 sin ( x 2 + 1) + sin 2 ( x 2 + 1)

XtraEdge for IIT-JEE

2

2

2

2

2

+ a ) (x + b ) (x + c )

=

π 2(a + b)(b + c)(c + a)

∫ (x

dx 2

0

(A)



2

x2 dx

then

π 60

(B)

π 20

(C)

π 40

+ 4)( x 2 + 9)

(D)

=

π 80

6.

The positive value of parameter ‘a’ for which the area of figure bounded by y = sin ax, y = 0, π π and x = is 3 unit, is x= a 3a 1 (A) 1 (B) 3 1 1 (D) (C) 2 4

7.

Solution of (1 + ex/y ) dx + ex/y (1 – x/y) dy = 0 is (A) xex/y + x = c (B) yex/y – x = c (D) yex/y + x = c (C) yex/y + y = c

The sum to n terms of the series 3 5 7 + 2 + 2 + …… is 2 2 1 1 +2 1 + 2 2 + 32

x

1 +1 + c x



r ≤s

2.

∫ (x 0

1 +1 + λ x

1 1 tan −1 x + + 1 + c 2 x



r = 0 s =1

(A) 3n – 1 (C) 3n

1 +1 + c x

2 tan −1 x +

(D)

5.

is

x ( x 2 + x + 1)

(C) 2 tan –1 x +

Questions 1 to 8 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. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer.

+C

sec ( x 2 + 1)

( x − 1)dx

∫ (x + 1)

(B)

MATHEMATICS

The value of

2

(A) tan −1 x +

27. 1.8 g of a carbohydrate (molar mass 180 g mol–1) requires 3.925 g acetyl chloride (molar mass 78.5 g mol–1 ) in acetylation of hydroxyl group. Calculate number of hydroxyl groups per unit of carbohydrate.

1.

1 sec ( x 2 + 1) + C 2

dx equals 58

JUNE 2010

π/2

8.



If In =

13

x n sinx dx then I7 + 42 I5 =

0 7

π (A)   2

π (C) 7   2

π (B)   2 6

6

π (D) 7   2

7

This section contains 2 paragraphs; passage- I has 2 multiple choice questions (No. 14 & 15) and passage- II has 3 multiple (No. 16 to 18). 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 9 to 13 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) 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 no negative marks. 1/ 2

9.

Value of

∫ 0

dx 1 − x 2n

; (n∈N) is

Passage : I (Ques. 14 to 15) Consider the differential equation ex (ydx – dy) = e–x (ydx + dy). Let y = f(x) be a particular solution to this differential equation which passes through the point (0, 2). Let 1 1  C ≡ y = log1/4  x –  + log4 (16x2 – 8x + 1), be 4 2   another curve

(A) less than or equal to π/6 (B) greater than or equal to 1/2 (C) non-negative (D) greater than 1 10. If

∫ x ln (1 + x ) dx = φ(x) ln(1 + x ) + ψ(x) + c then 2

2

(A) φ(x) =

1+ x2 2

(B) ψ(x) =

(C) ψ(x) =

−1+ x2 2

(D) None of these

If Ai is the area bounded by |x –ai| + |y| = bi, i ∈ N, b 3 where ai +1= ai + bi and bi + 1 = i , a1 = 0, b1= 32, 2 2 then: (A) A3 = 128 (B)A3 = 256 n n 8 4 (C) lim ∑ A i = (32)2 (D) lim ∑ A i = (16)2 n →∞ n →∞ 3 3 i =1 i =1

14. The area bounded by the curve C, parabola x = y2 + 1/4 and the line x = 1/4 is (A) 1 (B) 3 2 1 (D) (C) 3 3

1+ x2 2

11. Solution of equation  dy  x cos x   + y (x sin x + cos x) = 1 is  dx  (A) xy = sin x + c cosx (B) xy sec x = tan x + c (C) xy + sin x + c cos x = 0 (D) xy = tan x + c

15. If the area bounded by the curve y = f(x), curve C, ordinate x = 1/4 & the ordinate x = a is 1 4 – ln 4 + 1/ 4 – e1/4, then value of a is e (B) ln 4 (A) ln 6 (C) 4 (D) ln 12

12. If f(2 –x) = f(2 + x) and f(4 –x) = f (4 + x) for all x

Passage: II (Ques. 16 to 18) By using by parts it is possible to reduce an integral dependent on the integer n (n > 0), to an integral of the same type with smaller value of n.

2

and f(x) is a function for which

∫ f (x ) dx = 5, then 0

50



16. If In =

f ( x ) dx is equal to

0

(A) 125

(B)

51

(C)



f ( x ) dx

1

XtraEdge for IIT-JEE

46

(A)

−4

(B)

∫ f (x) dx

52

(D)



f ( x ) dx

(C)

2

∫ (x

dx 2

+ a 2 )n

then I5 –

7 8a 2

I4 is equal to

x 2

(x + a 2 ) 4 1 8a 2 ( x 2 + a 2 ) 3 x

8a 2 ( x 2 + a 2 ) 4 (D) None of these

59

JUNE 2010

π/ 4

17.

sin n x

∫ cos

If In,–m =

m

0

x

dx then In,– m +

22

n −1 In – 2, 2–m is m −1

n −m

(B)

1  1    (C) n − 1  2 

n −m

1  1    m − 1  2 

n −1  1    (D) m − 1  2 

the solution of differential equation dy d y x2 2 + 2x = 12y is y = Axm + Bx–n then find the dx dx value of m + n, if m & n ∈ N. 2

equal to  1   (A)   2

If

n −m

23

Let f(x) = min.{x – [x], – x – [– x]}. 2

n −m

Then find value of

−2

(Here [⋅] stands for G. I. F)

1

18. If Im, n =

∫x

m

24

n

(1 – x) dx then value of I3, 4 is equal to

0

1 280 2 (C) 173

1 140 3 (D) 155

25

This section contains 9 questions (Q.19 to 27). +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 have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following : X Y Z W 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9 19

26

27

Determine a positive integer n such that

∫ 0

21

x n sin x dx =

n

∑ r =1

2k  rπ  sin2k  2n  = ( x ) k (k!) 2 . Find the value

y =

f(x) passes through (1, 2) and satisfies the 1 relation y(1 + xy) dx – xdy= 0. Find value of 7f   . 2



2 15

(x + 1 + x )

x + 1 + x 2      dx = b

1+ x 2 then find the value of (a + b)/ 6

a



Area bounded between maxima and minima of function y = x3 – 3x + 4 with curve and X-axis is A. Find number of even divisors of 3A.

Trees live much longer than any other type of plant or animal. In fact it’s possible to know the age of a tree by counting the rings in its trunk (One ring generally equates to one year).

If ∫ xe x cos x dx = aex (b(1 – x) sin x + cx cos x) + d,

π/ 2

n →∞

1 n

THE OLDEST LIVING THING ON EARTH

then find the value of 2a + b + c. 20

Lim

of x.

(B)

(A)

∫ f (x) dx.

The oldest living tree and hence one of the oldest known living things on Earth is a bristlecone pine tree located in California, North America, it’s actually

3 2 (π − 8) 4

Find the area enclosed between the curves

over 4600 years old, now thats old.

1 y = loge (x + e), x = loge   and the x-axis. y

XtraEdge for IIT-JEE

60

JUNE 2010

Based on New Pattern

IIT-JEE 2012 XtraEdge Test Series # 2

Time : 3 Hours Syllabus : Physics : Laws of motion, Friction, Work Power Energy, Gravitation, S.H.M., Laws of Conservations of Momentum, Rotational Motion (Rigid Body), Elasticity, Fluid Mechanics, Surface Tension, Viscosity. Chemistry : Gaseous state, Chemical Energetics, Oxidation-Reduction, Equivalent Concept, Volumetric Analysis. Mathematics: Logarithm & Modulus Function, Quadratic Equation, Progressions, Binomial Theorem, Permutation & Combination, Complex Number

Instructions : Section - I • Question 1 to 8 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct answer and -1 mark for wrong answer. Section - II • Question 9 to 13 are multiple choice questions with multiple correct answer. +3 marks will be awarded for correct answer and 0 mark for wrong answer. Section - III • Question 14 to 18 are passage based single correct type questions. +3 marks will be awarded for correct answer and -1 mark for wrong answer. Section - IV • Question 19 to 27 are numerical response questions (with single digit Answer). +3 marks will be awarded for correct answer and 0 mark for wrong answer.

PHYSICS Questions 1 to 8 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. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer. 1.

2.

3.

A body at rest on a rough horizontal surface starts with an acceleration when a force 'F' is applied. When the force is doubled, the acceleration increases to 4 times. The force necessary just to start the body will be – (A) F (B) 2F 2F F (D) (C) 3 2

3 2

(B) µ ≤

3 2

(C) µ ≥

3

(D) µ ≤

3

Power supplied to a particle of mass 2 kg varies with 3t 2 watt. Here t is in seconds. If velocity time at P = 2 of particle at t = 0 is v = 0, the velocity of particle at time t = 2 sec will be (A) 1 m/s (B) 4 m/s (C) 3 m/s

4.

A block of mass M rests on a rough horizontal surface. The coefficient of friction between the block and the surface is µ. A force F = mg acting at an angle φ = 60º with the vertical side of block pulls it. In which of the following case, the block can be pulled along the surface.

XtraEdge for IIT-JEE

(A) µ ≥

(D) 2 2 m/s

A soap bubble is blown slowly at the end of a tube by a pump supplying air at a constant rate. Which one of the following graphs represent the correct variation of the excess of pressure inside the bubble with time (A)

(B)

p

p

t

(C)

(D)

p t

61

t

p t

JUNE 2010

5.

2mu (A) 2π MA 1 2 mu (C) 2π 3 MA

6.

9.

A bob of mass M is hung using a string of length l. A mss m moving with a velocity u pierces through the u bob and emerges out with a velocity . The 3 frequency of oscillation of the bob considering an amplitude A is -

(A) T2 =

1 2m (B) 2π 3MA

π2 (rA + rP)3 2Gm (C) vArA = vPrP (D) vA < vP ; rA > rP

(D) cannot be found

A

10. A student holds the axle of a spinning bicycle wheel while seated on a pivoted stool. The student and the stool are initially at rest while the wheel is spinning

F3



O B

in a horizontal plane with an angular momentum L 0 pointing upward. The wheel is inverted about its centre by 180º (A) The angular momentum of the system is conserved (B) The angular momentum of the system is not conserved (C) The final angular momentum of "student plus

F2

C F1

(C) F1 – F2

π2 (rA + rP)3 2Gm

(B) T2 =

What should be the value of F3. So that torque about O is zero, in the given situation :

(A) 2 (F1 + F2)

A planet is revolving round the sun. Its distance from the sun at Apogee is rA and that at perigee is rp. The mass of planet and sun is m and M respectively vA and vp is the velocity of planet at Apogee and Perigee respectively and T is the time period of revolution of planet round the sun -

F1 + F2 2 (D) F1 + F2

(B)



7.

stool" will be 2 L 0 (D) The final angular momentum of "student plus stool" will be zero

A bar of mass m resting on a smooth horizontal plane mg of constant starts moving due to a force F = 3 magnitude. In the process of its rectilinear motion, the angle θ between the direction of this force and the horizontal varies as θ = ks, where k is a constant. The velocity of bar as a function of θ is (A) υ =

2g sin θ 3

(B) υ =

2g cos θ 3k

(C) υ =

2g sin θ 3k

(D) υ =

2g cos θ 3



11. The potential energy U for a force field F is such that U = – kxy, where k is a constant →

^

^



^

^

(A) F = ky i + kx j (B) F = kx i + ky j →

(C) The force F is a conservative force →

(D) The force F is a non-conservative force 8.

A planet of mass m is revolving round the sun →

12. The spheres A and B as shown have mass M each. The strings SA and AB are light and inextensible with tensions T1 and T2 respectively. A constant horizontal force F = Mg is acting on B. For the system to be in equilibrium we have

(of mass Ms) in an elliptical orbit. If v is the velocity of the planet when its position vector from the sun is →

r , then areal velocity of the planet is → →

(A) v × r (C)

1 → →  v × r  2  

→ →

(B) r × v (D)

S

1 → →  r × v  2  

θ

T2

A

Questions 9 to 13 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) 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 no negative marks.

XtraEdge for IIT-JEE

T1

φ

62

B

F = mg

(A) tan φ = 1

(B) tan θ = 0.5

(C) T2 =

(D) T1 =

2 mg

5 mg

JUNE 2010

13. Which of the following statement(s) is (are) correct about friction ? (A) The coefficient of friction between two bodies is largely independent of area of contact (B) The frictional force can never exceed the reaction force on a body from the supporting surface (C) Rolling friction is generally smaller than sliding friction (D) Friction is due to irregularities of the surfaces in contact.

16. The ration of the time taken to reduce the height to H 3H and then from H is (liquid flowing through a 4 2 hole at its base) :

(C) 15.

2gH

(B)

5 2gH 3

(D) none of these

The height of the bob of mass rise after the collision(A)

25 H 9

(B)

(C)

16H 9

(D) none of these

H 9

Passage: II (No. 16 to 18)

(D) (2 –

2)

(H – h ) R2

(B) R1 =

(C) R1 = 4R2

(D) hR1 = (H – h)R2

K1 = 1500 N/m

`m1

2x . Rate of flow will obviously change as the g

2 kg

K2 = 500 N/m

position of the hole is altered and also when the liquid level drops.

XtraEdge for IIT-JEE

3):( 3 –

(A) R1 = R2

2g (h – x ) . For streamlined liquid flow the time

depends on the height at which the hole is made. Range-the horizontal distance at which the water touches the ground is given by R = vt where t=

3H : 4

H2

19. Find the potential energy stored in the springs in equilibrium. (Ans. ………..× 10–1)

If a hole is made at a height x from the ground for a cylinder holding water to height 'h', the velocity of efflux is given by toricellian theorem as v=

(C)

H 2

H1 :

This section contains 9 questions (Q.19 to 27). +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 have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following : X Y Z W 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9

14. Velocity of the bob of mass m is -

3

(B)

18. The principle used to arrive at the velocity of low can be derived from : (A) Pascal's law (B) Bernoulli's theorem (C) Stoke's law (D) All of these

Passage : I (Ques. 14 to 15) Two pendulum bobs of mass m and 2m collide elastically at the lowest point in their motion. If both the balls are released from height H above the lowest point .

2gH

H1 H 2 : 1

17. For two holes made at distance h from the top and bottom, the range are R1 and R2. Then :

This section contains 2 paragraphs; passage- I has 2 multiple choice questions (No. 14 & 15) and passage- II has 3 multiple (No. 16 to 18). 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)

(A)

m2

63

1 kg

JUNE 2010

20. A wire of length 1 m and area of cross section 2 × 10–6 m2 is suspended from the top of a roof at one end and a load of 20 N is applied at the other end. If the length of the wire is increased by 0.5 × 10–4 m. Calculate its Young's modulus ? [in 1011 N/m2] 21. A body of mass 2 kg is attached to a spring. Now a mass of 300 g is also attached to it and its length is increased by 2 cm. What will be the time period of 2 kg mass, if the second body is now removed ? (Take g = 10 ms–2)

1.

Adiabatic reversible expansion of a monoatomic gas (M) and a diatomic gas (D) at an initial temperatrue Ti has been carried out independently from initial volume V1 to final volume V2. The final temperature (TM for monoatomic gas and TD for diatomic gas) attained will be : (A) TM = TD > Ti (B) TM < TD < Ti (C) TM > TD > Ti (D) TM = TD = Ti

2.

For a given mixture of NaHCO3 and Na2CO3, volume of a given HCl required is x mL with phenolphthalein indicator and further y mL required with methyl orange indicator. Hence, volume of HCl for complete reaction of NaHCO3 is : (A) 2x (B) x/2 (C) y (D) (y – x)

3.

Helium atom is two times heavier than a hydrogen molecule. At 298 K, the average kinetic energy of a helium atom is (A) two times that of a hydrogen molecule (B) same as that of a hydrogen molecule (C) four times that of a hydrogen molecule (D) half that of hydrogen molecule

4.

In van der Waals equation of state for a nonideal gas the term that accounts for intermolecular forces is a   (A) (V – b) (B)  p + 2  V   –1 (C) RT (D) (RT)

5.

White phosphorus reacts with caustic soda. The products are PH3 and NaH2PO2. This reaction is an example of (A) oxidation (B) reduction (C) oxidation and reduction (D) neutralization

6.

Which of the following reactions is a redox reaction? (A) Cr2O3 + 6HCl → 2CrCl3 + 3H2O (B) CrO3 + 2NaOH → Na2CrO4 + H2O Cr2O72– + OH– (C) 2CrO42– + H+

22. The gravitational potential in a region is given by V = 20 (x + y) J/kg. Find the magnitude of gravitational force on the particle of mass 0.5 kg placed at the origin. 23. A cylinder of radius 10 cm rides between two horizontal bars moving in opposite direction as shown in the figure. Where is the instantaneous axis of rotation of roller ? [Neglect slipping at P and Q] [in cm] P

15 m/s

Q 10 m/s

24. An object of mass 1 kg moves under the action of a force F in a straight line with its velocity V, changing

with displacement x as v = 10 x . Find the work done in a displacement from x = 0 to x = 2. [in Joule] (Ans. ………..× 102) 25. An object falling from height of 80 m explodes into two parts of mass ratio 1 : 2 after 2 second of free fall. Velocity of the bigger part in the frame of reference of C.O.M. is 20 m/s. What is velocity of each part w.r.t. ground ? (Ans. ………..× 10–1) 26. A block of mass 2 kg is pulled by a constant power 100 W is placed on a rough horizontal plane. The 1 frictional coefficient between block and surface is . 4 Find the maximum velocity of block. (Ans. …….× 101)

(D) Cr2O72– + 6I– + 14H+

27. A solid sphere (a) rolls (b) slides, from rest down an inclined plane. Find the ratio of their accelerations.

7.

CHEMISTRY Questions 1 to 8 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. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer.

XtraEdge for IIT-JEE

64

2Cr3+ + 3I2 + 7H2O

The reaction of cyanamide, NH2CN(s), with oxygen was run in a bomb calorimeter and ∆U at 300 K was found to be –743 kJ mol–1. The value of ∆H at 300K for the combustion reaction NH2CN(s) + (3/2) O2(g) → N2(g) + CO2(g) + H2O (1) would be (A) – 741.75 kJ mol–1 (B) –743 kJ mol–1 (C) – 744.25 kJ mol–1 (D) – 740.5 kJ mol–1

JUNE 2010

8.

The combustion reaction occurring in an automobile is 2C8H18(s) + 5O2(g) → 16CO2(g) + 18H2O(1). This reaction is accompanied with (A) ∆H = –ve, ∆S = + ve, ∆G = + ve (B) ∆H = + ve, ∆S = –ve, ∆G = + ve (C) ∆H = – ve, ∆S = +ve, ∆G = – ve (D) ∆Η = +ve, ∆S = +ve, ∆G = – ve

This section contains 2 paragraphs; passage- I has 2 multiple choice questions (No. 14 & 15) and passage- II has 3 multiple (No. 16 to 18). 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 9 to 13 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) 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 no negative marks. 9.

Passage : I (Ques. 14 to 15) The kinetic-molecular theory accounts for the properties of gases by assuming that gas particles act independently of one another. Because the attractive forces between them are so weak, the particles in gases are free to move about at random and occupy whatever space is available. The same is not true in liquids and solids, however liquids and solids are distinguished from gases by the presence of strong attractive forces between particles. In liquids, these attractive forces are strong enough to hold the particles in close contact while still letting them slip and slide over one another. In solids, the forces are strong that they hold the particles rigidly in place and prevent their movement. 14. Which pair of molecules have the strongest dipoledipole interactions ? (A) NH3 and CH4 (B) NH3 and NH3 (C) CH4 and CH4 (D) CO2 and CO2

Which of the following are disproportionation reactions ? (A) 2

CHO

Al(OC2H5)3

COOCH2 ∆ (B) 4H3PO3 → 3H3PO4 + PH3

∆ N2O + 2H2O (C) NH4NO3 → ∆ PCl3 + Cl2 (D) PCl5 →

10. 10 volume H2O2 solution is present, then it means (A) 10 ml of H2O2 solution liberates 1 ml of oxygen at STP (B) 1 ml of H2O2 solution liberates 10 ml of oxygen at STP (C) 0.0303 g of H2O2 in 10 ml of solution liberates 10 ml O2 at STP (D) 0.0303 g of H2O2 in 1 ml of the solution liberates 10 ml O2 at STP

 ∂P  15. For one mole of an ideal gas    ∂T  V (A) –1

(B) –

R2 P

2

(C)

R2 P2

 ∂V     ∂T  P

 ∂V    =  ∂P  T

(D) +1

11. Silver metal in ore is dissolved by potassium cyanide solution in the presence of air by the reaction 4Ag + 8KCN + O2 + 2H2O → 4KAg (CN)2 + 4KOH (A) The amount of KCN required to dissolve 100 g of pure Ag is 120 g. (B) The amount of oxygen used in this process is 0.742 g (C) The amount of oxygen used in this process is 7.40g (D) The volume of oxygen used at STP is 5.20 litres.

Passage: II (Ques. 16 to 18) Redox reactions are those in which oxidation and reduction take place simultaneously. Oxidising agent can gain electron whereas reducing agent can lose electron easily. The oxidation state of any element can never be in fraction. If oxidation number of any element comes out be in fraction, it is average oxidation number of that element which is present in different oxidation states.

12. Identify the intensive properties among the following: (A) Enthalpy (B) Temperature (C) Volume (D) Refractive index

16.

1

N

3

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

2

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

13. During the Joule Thomson effect (A) A gas is allowed to expand adiabatically from a high pressure region to a low pressure region (B) A gas is allowed to expand adiabatically at constant pressure (C) ∆Η = 0 for ideal gas (D) ∆U = 0 for ideal gas

XtraEdge for IIT-JEE

N

17. Equivalant weight of chlorine molecule in the equation 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O

(A) 42.6 65

(B) 35.5

(C) 59.1

(D) 71 JUNE 2010

18. The oxidation number of sulphur in K2S2O8 is (A) + 2 (B) + 4 (C) + 7 (D) +

24. A → B, ∆H = +ve. Graph between log10P and

straight line of slope

This section contains 9 questions (Q.19 to 27). +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 have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following : X Y Z W 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9

1 . Hence, ∆H (in cal) is : 4.606

25. 9.8 g of FeSO4 (NH4)2SO4.xH2O was dissolved in 250 mL of its solution. 20 mL of this solution required 20 ml of KMnO4 solution containing 3.53 g of 90% pure KMnO4 dissolved per litre. Calculate x. 26. Two moles of a perfect gas undergo the following process: A B 1 P(atm)

C

0.5

40 V(L) What is the value of ∆S for the overall process ? 20

27. Heat of formation of ethylene from the following data at 20ºC is given by 2n kcal. What is the value of n? H2(g) + 1/2O2(g) → H2O (l) ; ∆H = – 65 kcal C(s) + O2(g) → CO2 (g) ; ∆H = – 97 kcal C2H4 (g) + 3O2(g) → 2CO2 (g) + 2H2O (l) ; ∆H = – 340 kcal

19. A weather balloon is inflated with helium. The balloon has a volume of 100 m3 and it must be inflated to a pressure of 0.10 atm. If 50 L gas cylinders of helium at a pressure of 100 atm are used, how many cylinders are needed ? Assume that the temperature is constant.

MATHEMATICS Questions 1 to 8 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. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer.

20. The compressibility factor for nitrogen at 330 K and 800 atm is 1.90 and at 570 K and 200 atm is 1.10. A certain mass of N2 occupies a volume of 1dm3 at 300 K and 800 atm. Calculate volume occupied (in litre) by same quantity of N2 gas at 750 K and 200 atm. 21. The ion An+ oxidised to AO3– by MnO4–, changing to Mn2+ in acidic solution. Given that 2.68 × 10–3 mol of An+ requires 1.61 × 10–3 mole of MnO4– . What is the value of n ?

1.

22. 100 mL of 1.44% solution of pure FeC2O4 in dil. HCl is oxidised by 0.01 M KMnO4. Then volume (in mL) of KMnO4 required is : Given your answer after divide actual answer by 100.

2.

23. For an isomerisation reaction A B, the temperature dependence of equilibrium constant is given by 2000 loge K = 4.0 – T The value of ∆Sº at 300 K (in term ……….× R) is :

XtraEdge for IIT-JEE

1 is a T

3

Let a,b,c be the three sides of a triangle, then the quadratic equation b2x2 +(b2 + c2 – a2) x + c2 = 0 has (A) both roots positive (B) both roots negative (C) both roots imaginary (D) None of these Coefficient of abc3de2 in expansion of (a + b + c + d + e)8 is (A) 3630 (B) 3600 (C) 3360 (D) None Number of rectangles in fig. shown which are not squares is

(A) 159 66

(B) 160

(C) 161

(D) None JUNE 2010

4

Let f(x) = x3 – 3x2 + 2x. If the equation f(x) = k has exactly one positive & one negative solution then k equals 2 3 −2 (A) (B) 9 9 −2 3 (C) 9

5

If an =

∑ k =0

2 (D) 9

8

11



r n

Cr

n

=

∑ r =0

n 2 − 3n + 3 2.n C r

, then

(B) n = 2 (D) None of these

This section contains 2 paragraphs; passage- I has 2 multiple choice questions (No. 14 & 15) and passage- II has 3 multiple (No. 16 to 18). 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.

(log e 10) n for n ≥ 0, then k!(n − k )!

Passage : I (Ques. 14 to 15) Given three equations in the complex plane : …...(1) iz + z + 1 + i = 0 (2 – i) z = (2 + i) z ......(2) (2 + i) z + (i – 2) z – 4i = 0 ......(3)

If a1, a2, a3 ……an are in H.P. &  n  f(k)=  ∑ a r  – ak then    r =1  a a1 a a , 2 , 3 ,…….., n are in f (1) f (2) f (3) f (n ) (A) A.P. (B) G.P. (C) H.P. (D) None of these

14. If the intersection point of the lines (2) and (3) is denoted by z1, then z1 equals 1  (A)  + i  (B) (1 + i) 2  i  (C) 1 +  (D) none of these  2

If ‘z’ is complex number then the locus of ‘z’ satisfying the condition |2z – 1| = |z – 1| is (A) perpendicular bisector of line segment joining 1 and 1 2 (B) circle (C) parabola (D) none of the above curves

15

The distance between the point z1 and the line (1) is 3 (A) (B) 3 2 2 3 (C) (D) none of these 2 2

Passage: II (Ques. 16 to 18)

The integer a,b,c are selected from 3n consecutive integer (1,2,3,......3n), then in how many ways can these integers be selected such that

If sec2θ & cosec2θ are roots of ax2 + bx + c = 0 (a > 0) then (A) b + c = 0 (B) b2 – 4ac ≥ 0 (C) c ≥ 4a (D) b + 4a ≥ 0 If y = log7–a (2x2 + 2x + a + 3) is defined ∀x∈R then possible integer value of a is/are (A) 4 (B) – 3 (C) – 2 (D) 5

XtraEdge for IIT-JEE

If

(A) n = 1 (C) n = 3

Questions 9 to 13 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) 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 no negative marks. 9. If z1 lies on |z| = 1 and z2 lies on |z| = 2, then (A) 3 ≤ |z1 – 2z2| ≤ 5 (B) 1 ≤ |z1 + z2| ≤ 3 (C) |z1 – 3z2| ≥ 5 (D) |z1 – z2| ≥ 1 10

13

r =0

a0 + a1 + a2 + a3 …………infinite equals to (A) 10 (B) 100 (C) 1000 (D) None of these 7

If 12! = 2a3b5c7d11e then (A) a = 5c (B) a = 10d (C) b = 5e (D) None of these n

The number of solutions of equation log6(x + 3) = 7 – x is (A) 0 (B) 2 (C) 1 (D) None of these n

6

12

67

16

Their sum is divisible by 3 is n (A) (3n2 – 3n + 2) (B) 3n2 – 3n + 2 2 n (C) (D) None 2

17

(a2 – b2) is divisible by 3 is n (n − 1) n (n + 1) (B) n2 + (A) n2 + 2 2 3n (n − 1) 2 (C) n + (D) None 2

JUNE 2010

18

(a3 + b3) is divisible by 3 is 3n 2 − n 3n 2 + n (B) (A) 2 2 n (n + 1) (C) (D) None 2

25

( 3 + 7 )17 is equal to 3k + k2 then k equals

This section contains 9 questions (Q.19 to 27). +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 have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following : X Y Z W 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9 19 20

21

22

1 1 + y lnz lnx

24

1 1 + lnx lny

x . . z is e – k then k equals ……… There are 2n white & 2n red counters. Counters are all alike except for the colour. If the number of ways in which they can be arranged in a line so that they are symmetric w.r.t. a central mark is 70 then n equals ……..

If the fraction a quotient equals………

23

x 3 + (a − 10) x 2 − x + a − 6

sum log10 3 2

of

27

The value of ‘‘a’’ for which all roots of quadratic equation, f(x) = (a – 2)x2 + 2ax + a + 3 = 0 1  lies in (– 2, 1) blongs to  − ∞,−  ∪ (m, n] then 4  value of n – m is

Element

Approximate % by weight

Oxygen

46.6

Silicon

27.7

Aluminum

8.1

Iron

5.0

Calcium

3.6

Sodium

2.8

Potassium

2.6

Magnesium

2.1

All others

1.5

x 3 + (a − 6) x 2 − x + a − 10 of two linear functions

Given the abundance of oxygen and silicon in the crust, it should not be surprising that the most abundant minerals in the earth's crust are the silicates. Although the Earth's material must have had the same composition as the Sun originally, the present composition of the Sun is quite different. The elemental composition of the human body and life in general is quite different.

reduces to then

a

 x2  If log  3  = 1 & log(x 2 y 3 ) = 7 then log |xy| is y  equal to……..

If

Number of zeroes at the end in product of 56 × 66 × 76 ……316 is 8k2 + 2k + 1 then k equals

3 is ………

Suppose x, y, z > 0 & different from one and ln x + ln y + ln z = 0, then value of 1 1 + lny lnz

26

Abundances of the Elements in the Earth's Crust

The value of expression x4 – 8x3 + 18x2 – 8x + 2 when x = 2 +

Number of irrational terms in expansion of

all log10 x

solutions

of

the

equation

These

log b c

where a,b,c∈N (x ) − (3 ) – 2 = 0 is a & a, b are prime numbers then a + b equals

XtraEdge for IIT-JEE

general

element

abundances

are

reflected in the composition of igneous rocks

68

JUNE 2010

AIEEE PAPER 2010 (PAPER & SOLUTION)

Time : 3 Hours

Total Marks : 432

Instructions : •

Part A – Physics (144 Marks) –Questions No. 1 to 20 and 23 to 26 consist of FOUR (4) marks each and Questions No. 21 to 22 and 27 to 30 consist of EIGHT (8) marks each for each correct response. Part B – Chemistry (144 Marks) – Questions No. 31 to 39 and 43 to 57 consist of FOUR (4) marks each and Questions No.40 to 42 and 58 to 60 consist of EIGHT (8) marks each for each correct response. Part C – Mathematics (144 Marks) – Questions No.61 to 66, 70 to 83 and 87 to 90 consist of FOUR (4) marks each and Questions No. 67 to 69 and 84 to 86 consist of EIGHT (8) marks each for each correct response • Candidates will be awarded marks as stated above in instructions No. 5 for correct response of each question. ¼ (one fourth) marks will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer sheet.

PHYSICS

4

Directions: Questions number 1 – 3 are based on the following paragraph. An initially parallel cylindrical beam travels in a medium of refractive index µ (I) = µ0 + µ2I, where µ0 and µ2 are positive constants and I is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius.

(3) c Sol.

2 ∆m M

∆mc 2 = 2 ×

(4) c

∆m M

1 M 2 ×  v 2  2

2∆mc 2 2 ∆m , v=c M M Ans. (3) v2 =

The initial shape of the wavefront of the beam is (1) planar (2) convex (3) concave (4) convex near the axis and concave near the periphery Ans.[3] 1.

The binding energy per nucleon for the parent nucleus is E1 and that for the daughter nuclei is E2. Then – (1) E1 = 2E2 (2) E2 = 2E1 (3) E1 > E2 (4) E2 > E1 Ans. [4] 5.

The speed of light in the medium is – (1) maximum on the axis of the beam (2) minimum on the axis of the beam (3) the same everywhere in the beam (4) directly proportional to the intensity I Ans. [2] 3. As the beam enters the medium, it will (1) travel as a cylindrical beam (2) diverge (3) converge (4) diverge near the axis and converge near the periphery Ans. [3] 2.

Directions: Questions number 6 – 7 contain Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements. 6.

Directions: Questions number 4 – 5 are based on the following paragraph. A nucleus of mass M + ∆m is at rest and decays into M each. Speed of two daughter nuclei of equal mass 2 light is c.

XtraEdge for IIT-JEE

The speed of daughter nuclei is – ∆m ∆m (1) c (2) c M + ∆m M + ∆m

69

Statement- 1: When ultraviolet light is incident on a photocell, its stopping potential is V0 and the maximum kinetic energy of the photoelectrons is Kmax. When the ultraviolet light is replaced by Xrays, both V0 and Kmax increase – Statement – 2 : Photoelectrons are emitted with speeds ranging from zero to a maximum value because the range of frequencies present in the incident light. (1) Statement-1 is true, Statement-2 is false. (2) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1 JUNE 2010

(3)

Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1. (4) Statement-1 is false, Statement-2 is true. Ans. [1]

7.

Sol. 8.

(3)

Statement- 1: Two particles moving in the same direction do not lose all their energy in a completely inelastic collision. Statement- 2: Principle of conservation of momentum holds true for all kinds of collisions. (1) Statement-1 is true, Statement-2 is false. (2) Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1 (3) Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1. (4) Statement-1 is false, Statement-2 is true. Both are true but not explain the Ist. Ans. (3)

(4) Ans. [2] 10.

A ball is made of a material of density ρ where ρoil < ρ < ρwater with ρoil and ρwater representing the densities of oil and water, respectively. The oil and water are immiscible. If the above ball is in equilibrium in a mixture of this oil and water, which of the following pictures represents its equilibrium position?

(1)

(2)

(3)

(4)

The figure shows the position – time (x – t) graph of one-dimensional motion of the body of mass 0.4 kg. The magnitude of each impulse is –

Sol.

(1) 0.2 Ns (2) 0.4 Ns (3) 0.8 Ns (4) 1.6 Ns Sol. From graph, v1 = 1 ms–1 , v2 = –1 ms–1



11.



∴ J = F dt = dP = m ∆V

ρoil < ρ < ρwater, so ball will not sink in water but sink in oil. Ans. (3) A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net r field E at the centre O is – Q

= 0.4 × 2 = 0.8 N.s. Ans. (3) 9.

Two long parallel wires are at a distance 2d apart. They carry steady equal currents flowing out of the plane of the paper as shown. The variation of the magnetic field B along the line XX’ is given by -

(1)

q 2π 2 ε 0 r 2

(3) −

ˆj

q 2

4π ε 0 r

2

(2) ˆj

q 4π 2 ε 0 r 2

(4) −

ˆj

q 2

2π ε 0 r 2

ˆj

(1)

Sol.

(2)

XtraEdge for IIT-JEE

dE

70

JUNE 2010

π/2

E=



π/2

dE cos θ = 2

−π / 2

r E=

r E=

0

Sol.

R2

cos θ

Sol.

qR 2 [sin θ ]π0 / 2 = 2 q 2 [ Sin90 − sin 0](− ˆj ) 4πε 0 πR R 2 2π ε 0 R 16.

q

2π ε 0 R 2 Ans. (4) 12.



kλRdθ

2

(− ˆj )

A diatomic ideal gas is used in a Carnot engine as the working substance. If during the adiabatic expansion part of the cycle the volume of the gas increases from V to 32 V, the efficiency of the engine is(1) 0.25 (2) 0.5 (3) 0.75 (4) 0.99  T  η = 1 − 2  T1  

Sol.

γ −1

Sol.

14.

A radioactive nucleus (initial mass number A and atomic number Z) emits 3 α-particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be – A−Z−4 A − Z−8 (1) (2) Z−2 Z−4 A−Z−4 A − Z − 12 (4) (3) Z−8 Z−4 A ZX

A −3×4

3α + 2 positron ) (   →Z−3×2− 2 × 1 X =

17.

γ −1

The respective number of significant figures for the numbers 23.023, 0.0003 and 2.1 × 10–3 are (1) 4, 4, 2 (2) 5, 1, 2 (3) 5, 1, 5 (4) 5, 5, 2 23.023 significant fig. 5 0.0003 significant fig. 1 2.1 × 10–3 significant fig. 2 Ans. (2)

Sol.

Let there be a spherically symmetric charge distribution with charge density varying as 5 r  ρ(r ) = ρ 0  −  upto r = R, and ρ(r) = 0 for 4 R r > R, where r is the distance from the origin. The electric field at a distance r (r < R) from the origin is given by – ρ r 5 r  4πρ 0 r  5 r  (1) 0  −  (2)  −  3ε 0  4 R  3ε 0  3 R  ρ r 5 r  4ρ 0 r  5 r  (4) (3) 0  −   −  4ε 0  3 R  3ε 0  4 R  r F − > Na + > Mg +2 > Al +3

and

Ans. (1)

K 2 = 4.8 x 10 −11 Selection the correct statement 34.

for a saturated 0.034 M solution of the carbonic acid. (1) The concentration of H + is double that of CO32– (2) The concentration of CO32– is 0.034 M. (3) The concentration of CO32– is greater than that of H CO3–

H 2 CO 3

H + + HCO 3−

0.034 M

x

HCO 3−

NaNO

HCl, 278 K

The compounds 'A' and 'B' respectively are (1) nitrobenzene and chlorobenzene (2) nitrobenzene and flurobenzene (3) phenol and benzene (4) benzene diazonium chloride and flurobenzene

k 1 = 4.2 × 10 −7

Sol.

H +

CO 3−2

k1 = 4.8 × 10

−11

and hence (H ) ≅

[HCO 3−

HBF4

+ HCl NaNO  2 →

x– y x +5 As k 2 Mg 2 > Na + > F − > O 2− +

+



(3) Na + > Mg 2 > Al3 > O 2 > F − +



(4) Na + > F − > Mg 2 O 2 Al 3

XtraEdge for IIT-JEE

(2) 5.56 x 10 −3 mol

(3) 1.53 x10 −2 mol

(4) 4346 x 10 −2 mol

PV = nRT N 3170 2 × 10 −3 m 3 = n × 8.314 × 300 m 3170 × 10 −3

24.93 × 10 −2 Ans. (1)

The correct sequence which shows decreasing order of the ionic radii of the elements is (1) O 2− > F − > Na + > Al

(1) 1.27 x 10 −3 mol

n=

Ans. (3) 33.

If 10 −4 dm −3 of water is introduced into a 1.0

+

75

=

31.7 × 10 −1 24.93 × 10 −2

= 1.287 × 10 −1 m

From amongst the following alcohols the one that would react fastest with conc, HCl and anhydrous ZnCl2 is (1) 1- Butanol (2) 2- Butanol (3) 2- Methylpropan -2-ol (4) 2- Methylpropanol JUNE 2010

Sol.

Reactivity of Lucas reagent for alcohol = 3° > 2 ° > 1° > CH 3

Sol.

CH3 CH3

C



H → C6H5–CH2–CH–CH–CH3  | | OH CH3

⊕ C6H5–CH2–CH–CH–CH3 | CH3

CH3

OH

2-methyl. 2-propanol (3° -alcohol) highest reactivity Ans. (3)

HΘShift

If sodium sulphate is considered to be completely dissociated into cations and anions in equeous solution, the change in freezing point of water ( ∆Tf ), When 0.01 mol of sodium sulphate is dissolved in 1 kg of water, is ( K f = 1.86K Kg mol −1 ) (1) 0.0186 K (2) 0.0372 K (3) 0.0558 K (4) 0.0744 K Sol. Na 2SO 4 2 Na 2 + SO 4−2 i=3 ∆Tf = iK f .m. = 3 × 1.86 × 0.0 = 0.0558 K Ans. (3) 37.

38.

⊕ C6H5–CH–CH2–CH–CH3 | CH3

Benzyl Carbon cation stable by Resonance Removal of H ⊕ C6H5 H C=C CH(CH H 3) 2 Trans is more stable than cis Alkene

Ans. (2)

Three reactions involving H2PO4– are given below:

40.

(1) H 3 PO 4 + H 2 O → H 3 O + + H 2 PO 4 – –

(2) H 2 PO 4 + H 2 O → HPO 4

+ H 3O

bonds in Cl 2 is 242 kJ mol −1 . The longest

+

wavelength of light capable of breaking a single Cl–Cl bond is



(3) H 2 PO 4 + OH → H 3 PO 4 + O 2− In which of the above does H2PO4– act as an acid? (1) (i) Only (2) (ii) Only (3) (iii) and (ii) (4) (iii) only (ii) only –

Sol.

2–

(C = 3 x 10 8 ms −1 and N A = 6.02 x 10 23 mol −1 )

H 2 PO 4− + H 2 O  → HPO 4−2 + H 3 O + acid base Ans. (2)

39.

Sol.

The main product of the following reaction is .H 2SO 4 C 6 H 5 CH 2 CH (OH )CH (CH 3 ) 2 conc  → ? H5C6CH2CH2 (1) C = CH 2

H 3C

(2)

H5C6

H C=C

H

(3)

C6H5CH2

CH(CH3)2

CH3

H

(4)

C6H5

C=C

H

XtraEdge for IIT-JEE

41.

CH3

C=C

The energy required to break one mole of Cl–Cl

CH(CH3)2 H

76

(1) 494 nm (2) 594 (3) 640 nm (4) 700 nm hc E= λ 242 × 10 +3 6.62 × 10 −34 × 3 × 108 = = λ 6.02 × 10 23 −26 6.62 × 10 × 3 × 6.02 × 10 20 λ = 242 6.62 × 18.06 × 10 −6 = 242 = 0.494 × 10 −6 = 4.94 × 10 −7 m = 494 nm Ans. (1) 29.5 mg of an organic compound containing nitrogen was digested according to Kjeldahl's method and the evolved ammonia was absorbed in 20 mL of 0.1 M HCl solution. The excess of the acid required 15 mL of 0.1 M NaOH solution for complete neutralization. The percentage of nitrogen in the compound is (1) 29.5 (2) 59.0 (3) 47.4 (4) 23.7 JUNE 2010

Sol.

42.

45.

Equivalent of NH 3 = (0.1 × 20) − (0.1 × 15) = 0.5 wt. of NH 3 = 0.5 × 17 = 8.5mg wt. of 'N' 14 = × 8.5mg = 7mg 17 7 % of 'N' = × 100 = 23.7 29.5 Ans. (4)

Sol.



Ionisation energy of He + is 19.6 x 10 −18 J atom −1 . The energy of the first stationary state

(A) Ans. (2)

+

46.

(2) 4.41 x 10 −16 J atom −1 (3) − 4.41 x 10 −17 J atom −1 (4) -2.2 x 10 −15 J atom −1 I.E1 Z12 = I.E 2 Z 22

Sol.

19.6 × 10 4 = x 9 9 x = × 19.6 × 10 −18 = 44.1 × 10 −18 J / atm. 4 = 4.41 × 10 −17 J / atm

=

47.

Ans. (3)

Sol.

Sol.

Sol.

XtraEdge for IIT-JEE

Consider the reaction : Cl 2 (aq) + H 2 S(aq) → S(s) + 2H + (aq) + 2 Cl − (aq) The rate equation for this reaction is Rate = k [Cl 2 ][H 2 S]

H 2S ⇔ H + + HS − (fast equilibrium)

Cl 2 + HS − → 2Cl − + H + + S (slow ) (1) A only (2) B only (3) Both A and B (4) Neither A nor B r = k[Cl 2 ][H 2S]

∴ According to A → r = k[ H 2 S ][Cl2 ] ∴ According to B → r = k[Cl 2 ][HS]

or K eq =

[H + ][HS] [H 2S]

[HS] = K eq

Which one of the following has an optical isomer ? (1) [ Zn (en ) 2 ] 2+ (2) [ Zn (en )( NH 3 ) 2 ] 2+ (4) [Co(H 2 O) 4 (en )] (3) [Co(en ) 3 ] [M (AA) 3 ] type of compound ∴ optically active Ans. (3)

One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of 44 u. The alkene is (1) ethene (2) propene (3) 1-butene (4) 2-butene Molecular weight = 44 ∴ [CH 3 − CHO]

(B)

0.3 0.25 + 105 × 25 0.55 0.55 = 45 × 0.545 + 105 × 0.454 = 72.25 kPa. Ans. (2)

3+

(C)

Which of these mechanisms is/are consistent with this rate equation ? Cl 2 + H 2 S → H + + Cl − + Cl + + HS − (slow ) (A) Cl + + HS − → H + + Cl − + S (fast )

On mixing, heptane and octane form an ideal solution At 373 K, the vapour pressures of the two liquid components (Heptane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure of the solution obtained by mixing 25.0 g of heptane and 35 g of octane will be (molar mass of heptane = 100 g mol −1 and of octane =114 g mol −1 ) (1) 144.5 kPa (2) 72.0 kPa (4) 96.2 kPa (3) 36.1 kPa PT = P0 x 0 + Phep x hep = 45 ×

44.

⊕ 2° Carbocation

CH 3 − CH = O + O = CH − CH 3  → CH 3 − CH = CH − CH 3 2-butene Ans. (4)

−18

43.

⊕ stable by resonance (B)

1° carbocation

(n = 1) of Li 2 is (1) 8.82 x 10 −17 J atom −1

Sol.

Consider the following bromides Me Me Me Me Br Br Br (A) (B) (C) The correct sorder of S N 1 reactivity is (1) A > B > C (2) B > C > A (3) B > A > C (4) C > B > A Reactivity for SN ' ∝ Stability of carbocation

[H 2S] H+

r = k[Cl 2 ] K eq

3+

= K'

[H 2S] [H + ]

[Cl 2 ][H 2S]

[H + ] ∴ (A) Only Ans. (1)

77

JUNE 2010

48.

Sol.

49.

The Gibbs energy for the decomposition of Al 2 O 3 at 500 °C is as follows: 2 4 Al 2 O 3 → Al + O 2 , ∆ r G = +966kJ mol −1 3 3 The potential difference needed for electrolytic reduction of Al 2 O 3 at 500 °C is at least (1) 5.0 V (2) 4.5 V (3) 3.0 V (4) 2.5 V ∆G = −nFE n=4 966 × 10 3 = −4 × 96500 × E = 2.5 V Ans. (4)

52.

Sol.

53.

The correct order of increasing basicity of the given conjugate bases (R = CH 3 ) is (1) RCOO < HC ≡ C < NH 2 < R Sol.

(2) RCOO < HC ≡ C < R < NH 2 (3) R < HC ≡ C < RCOO < NH 2 (4) RCOO < NH 2 < HC ≡ C < R Sol. Conjugated acid RCOOH CH ≡ CH NH 3 R−H Order to A.S. ⇒ RCOOH > CH ≡ CH > NH 3 > R − H Order to B.S.

54.

Sol.

⇒ RCOO − < CH ≡ C − < NH 2 – < R − Ans. (1)

50.

Sol.

55.

The edge length of a face centered cubic cell of an ionic substance is 508 pm. If the radius of the cation is 110 pm, the radius of the anion is (1) 144 pm (1) 288 pm (3) 398 pm (4) 618 pm a r⊕ + r( − ) = 2 508 110 + r( − ) = 2

Sol.

The polymer containing strong intermolecular forces e.g. hydrogen bonding, is (1) natural rubber (2) teflon (3) nylon 6,6 (4) polystyrene Fact Ans. (3) At 25°C, the solubility product of Mg (OH) 2 is

solution of 0.001 M Mg 2+ ions ? (1) 8 (2) 9 (3) 10 (4) 11 K sp = [ Mg +2 ][OH − ]2 1 × 10 −11 = [0.001][OH − ]2 [OH − ] = 10 −4 pOH = 4; pH = 10 Ans. (3)

= 144 nm Ans. (1)

56.

The correct order of E °M 2 + / M values with negative sign for the four successive elements Cr, Mn, Fe and Co is (1) Cr > Mn > Fe > Co (2) Mn > Cr > Fe > Co (3) Cr > Fe > Mn > Co (4) Fe > Mn > Cr > Co

Out of the following the alkene that exhibits optical isomerism is (1) 2-methyl-2-pentene (2) 3-methyl-2-pentene (3) 4-methyl-pentene (4) 3-methyl-1-pentene

Sol.

Percentages of free space in cubic close packed structure and in body centered packed structure are respectively (1) 48% and 26% (2) 30% and 26% (3) 26% and 32% (4) 32% and 48% ccp : p.f. = 74%; 100 – 74 = 26% bcc : p.f. = 68%; 100 – 68 = 32% Ans. (3)

1.0 × 10 −11 . At which pH, will Mg 2+ ions start precipitating in the form of Mg (OH) 2 from a

r( − ) = 254 – 110

51

For a particular reversible reaction at temperature T, ∆H and ∆S were found to be both +ve. If Te is the temperature at equilibrium, the reaction would be spontaneous when (2) Te > T (1) T = Te (4) Te is 5 times T (3) T > Te ∆G = ∆H − T∆S +ve +ve T > Te for ∆G = − ve Ans. (3)

Sol. E 0M + 2 / M

CH3 CH3–CH2

C

Ti

Cr Mn Fe Co V − 1.67 − 1.18 − 0.91 − 1.18 − 0.44 − 0.28 Ni

CH= CH2

Zn

− 0.24 + o.34 − 0.76

Ans. (2)

H

Due to presence of chiral carbon atom 4 is show optical isomerism. Ans. (4)

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Cu

57.

78

Biuret test is not given by (1) proteins (2) carbohydrates (3) polypeptides (4) urea JUNE 2010

Sol.

Carbohydrate does not give biuret test. Due to absence of amide group. Ans. (2)

58.

The time for half life period of a certain reaction A → Products is 1 hour. When the initial

MATHEMATICS 61.

concentration of the reactant 'A', is 2.0 mol L−1 , how much time does it take for its concentration come from 0.50 to 0.25 mol L−1 if it is a zero order reaction ? (1) 1h (2) 4 h (3) 0.5 h (4) 0.25 h Ans.[4] 59.

A

solution

containing

2.675

g

CoCl 3 ⋅ 6 NH 3 (molar mass = 267.5 g mol

−1

of

Sol.

) is

passed through a cation exchanger . The chloride ions obtained in solution were treated with excess of AgNO 3 to give 4.78 g of AgCl (molar mass =143.5 g mol −1 ). The formula of the complex is (At. mass of Ag = 108 u) (1) [CoCl( NH 3 ) 5 ]Cl 2 (2) [Co( NH 3 ) 6 ]Cl 3 (3) [CoCl 2 ( NH 3 ) 4 ]Cl Sol.

Reflexive

4.78 = 0.03 moles 143.5

Ans. (2) 60.

p m S → pn = mq q n hence function symmetric m p S → mq = pn (1) n q p r S → ps = qr (2) q s eqn. (1)/(2) m r m r = → S n s n s hence transitive So S is equivalence relation Ans. (3)

Then

The standard enthalpy of formation of NH 3 is − 46.0 kJ mol −1 . If the enthalpy of formation of H 2 from its atoms is − 436 kJ mol −1 and that of N 2 is − 712 kJ mol −1 , the average bond enthalpy

of N − H bond in NH 3 ]

Sol.

(1) − 1102 kJ mol −1

(2) − 964 kJ mol −1

(3) + 352 kJ mol −1

(4) + 1056 kJ mol −1

1 3 N 2 (g ) + H ( g )  → NH 3 2 2 1 3 ∆H f = B − E N − N + BE H −H − 3.B.E N −H 2 2 1 3 –46 = × (−712) + × (−436) − 3 × x 2 2 1056 = 352 kJ/ml. x= 3 Ans. (3)

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n, q ≠ 0 and qm = pn}. Then (1) R is an equivalence relation but S is not an equivalence relation (2) Neither R nor S is an equivalence relation (3) S is an equivalence relation but R is not an equivalence relation (4) R and S both are equivalence relations Probable part of R is {(0, 1), (0, 2)} But (1, 0) ∉ R as 1 = (w) 0 So not symmetric ie. not equivalence Relation m p S → qm = pn n q m m S → mm = mn n n hence function reflexive . m p S → qm = pn Let n q

(4) [CoCl 3 ( NH 3 ) 3 ]

CoCl 3 .6 NH 3  → AgCl

4.78 g or

Consider the following relations R = {(x, y) | x, y are real numbers and x = wy for some rational number w}; m p S = {( , ) | m, n, p and q are integers such that n q

62.

Sol.

79

The number of complex numbers z such that |z –1 | = | z + 1| = | z – i| equals (1) 0 (2) 1 (3) 2 (4) ∞ |z – 1| = |z + 1| = |z – i| The point z is equidistance from (–1, 0), (1, 0) and (0, 1) is only (0, 0) hence z is only point (0, 0) Ans. (2) JUNE 2010

If α and β are the roots of the equation x2 – x + 1 = 0, then α2009 + β2009 = (1) – 2 (2) – 1 (3) 1 (4) 2 Sol . Here roots of x2 – x + 1 = 0 are – ω and – ω2. (– ω)2009 + (– ω2)2009 = (– ω)2007 × (– ω)2 + (– ω2)2007 × (– ω2)2 –( ω2 + ω) = 1. Ans. (3) 63.

64.

67.

Let f : R → R be a positive increasing function f (3x) f (2 x) = 1. Then lim = with lim x→∞ f ( x ) x→∞ f ( x ) 2 3 (1) 1 (2) (3) (4) 3 3 2

Sol.

Function (↑) f(x) < f(2x) < f(3x) f (2 x) f (3x) ≤ 1≤ f ( x) f ( x) f (3 x) =1 given that f ( x) f (2 x) ≤1 hence 1 ≤ f ( x) f (2 x) = 1 (by sandwich theorem) hence lim x → ∞ f ( x) Ans. (1)

68.

Let p(x) be a function defined on R such that p'(x) = p'(1 – x), for all x ∈ [0, 1], p(0) = 1 and

Consider the system of linear equations : x1 + 2 x2 + x3 = 3 2 x1 + 3 x2 + x3 = 3 3x1 + 5 x2 + 2 x3 = 1

The system has (1) Infinite number of solutions (2) Exactly 3 solutions (3) a unique solution (4) No solution 1 2 1

Sol.

Here ∆ = 2 3 1 = 1 (1) – 2(1) + 1 (1) = 0 3 5 2

1

p(1) = 41. Then

0

3 2 1

(1) 41

∆1 = 3 3 1 = 3 (1) – 2 (5) + 1 (12) = 5 1 5 2

Sol.

∆1 ≠ 0 When ∆ = 0 and if ∆1, ∆2, ∆3, are not zero then no solution Ans. (4) 65.

Sol.

There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urn two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is (1) 3 (2) 36 (3) 66 (4) 108

(4) 42

P'(x) = P' (1 – x) integrate P(x) = – P (1 – x) + k ------- (1) put x = 1 P(1) = – P(0) + k 41 = – 1 + k K = 42 Put in (1) P(x) = –P(1 – x) + 42 ------(2)



0 1



also I = P(1 − x)dx 0

1

9

Total no. of ways C2 × C2 = 108



2 I = ( P ( x) + P(1 − x))dx 0

1



Let f : (–1, 1) → R be a differentiable function with f(0) = – 1 and f '(0) = 1. Let g(x) = [f(2f(x) + 2)]2,. Then g ' (0) = (1) 4 (2) –4 (3) 0 (4) –2

using (2) 2 I = 42 dx

= 42( x)10

0

I = 21 Ans. (2)

g'(x) = 2 [f(2f(x) + 2)]. f '(2 f(x) + 2) .2f ' (x) g'(0) = 2 [f(2.f(0)+2)] . f ' (2. f(0) + 2). 2f '(0) = 2[f(0)]. f '(0) .2 = 2(–1) .(1).2 = –4 Ans. (2)

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(3) 41

1

Ans. (4)

Sol.

(2) 21

Now I = P ( x)dx

By 3C2 way we can select 2 balls from A and By 9 C2 ways we can select 2 balls from B 3

66.

∫ p( x)dx equals -

69.

80

A person is to count 4500 currency notes. Let an denote the number of notes he counts in the nth minute. If a1 = a2 = .... = a10 = 150 and a10, a11,... are in an AP with common difference – 2, then the time taken by him to count all notes is (1) 24 minutes (2) 34 minutes (3) 125 minutes (4) 135 minutes JUNE 2010

Sol.

π

a1 = a2 = a3 ..... a9 = 150 a1 + a2 + a3 + .... + a9 = 1350 a10 + a11 + ......... + an = 4500 – 1350 = 3150 n [2 × 150 + (n − 1)(−2)] = 3150 2 150 n – n2 + n = 3150 n2 – 151n + 3150 = 0 n = 25 min hence total time = 25 + 9 = 34 min Ans. (2)

4

(3) 4 2 − 1

(4) 4 2 + 1

π

Sol. π/4

4

  1 1   + − 1 + 0 −  − − 2 2   

= 2 −1+

4 2

−1+ 2 = 4 2 − 2

Ans. (1) 72.

Solution of the differential equation cos x dy = y (sin x – y) dx, 0 < x <

π

is 2 (1) sec x = (tan x + c) y (2) y sec x = tan x + c (3) y tan x = sec x + c (4) tan x = (sec x + c) y Sol.

dy = y sin x − y 2 dx dy cos x − sin x. y = − y 2 dx 1 dy 1 − tan x = sec x y 2 dx y

cos x



1 =z y

1 dy dz = y 2 dx dx dz + tan x.z = sec x dx

∫ tan x dx

I.F. = e Solution of above differential equation is z. sec x =

73. y=0

∫ sec

2

x dx

sec x = tan x + c y sec x = y (tan x + c) Ans. (1) r r r Let a = ˆj − kˆ and c = iˆ − ˆj − kˆ. Then the vector b r r r r r r satisfying a × b + c = 0 and a .b = 3 is (1) − iˆ + ˆj − 2kˆ (2) 2iˆ − ˆj + 2kˆ

(3) iˆ − ˆj − 2kˆ

Area

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  1  1 1 1 1  1 − − − = + − (0 + 1) − −  2 2 2 2 2    2

sin x 5π/4 3π/2



4

X=0

cos x

5 4

= [sin x + cos x ] 04 + [− cos x − sin x ] π4 + [sin x + cos x ] 52π

The area bounded by the curves y = cos x and y = 3π is sin x between the ordinates x = 0 and x = 2 (2) 4 2 + 2

4

0

The equation of the tangent to the curve 4 y = x + 2 , that is parallel to the x – axis, is x (1) y = 0 (2) y = 1 (3) y = 2 (4) y = 3 4 Sol. y = x + 2 x dy 8 = 1− 3 = 0 dx x 8 1= 3 x x3 = 8 x=2 4 at x = 2, y = x + 2 x 4 = 2+ =3 4 tangent y – 3 = 0 (x – 2) y=3 Ans. (4)

(1) 4 2 − 2

3π 2

∫ (cos x − sin x)dx + π∫ (sin x − cos x)dx + ∫π (cos x − sin x)dx

70.

71.

5π 4

Sol.

81

(4) iˆ + ˆj − 2kˆ

Let b = b1iˆ + b2 ˆj + b3kˆ r r given a . b = 3 b2 – b3 = 3 (1) JUNE 2010

r r r and a × b + c = 0 r r r a × b = −c i j k 0 1 − 1 = −iˆ + ˆj + kˆ b1 b2 b3

Sol.

b3 + b2 = – 1 (2) – b1 =1 ----- (3) – b1 = 1 ------(4) b1 = – 1 from (1) and (2) b2 = 1 b3 = – 2 r b = −iˆ + ˆj − 2kˆ Ans. (1) 74.

Sol.

75.

Sol.

76.

r r If the vectors a = iˆ − ˆj + 2kˆ , b = 2iˆ + 4 ˆj + kˆ and r c = λiˆ + ˆj + µkˆ are mutually orthogonal, then (λ, µ) = (1) (–3, 2) (2) (2, –3) (3) (–2, 3) (4) (3, –2) r r r r a ⊥b ∴a .b =0 r r r r b ⊥ c ∴b .c = 0 2λ + 4 + µ = 0 .....(1) r r rr a ⊥ c ∴ a.c = 0 λ – 1 + 2µ = 0 .....(2) solving (1) and (2), we get λ=–3 µ=2 Ans. (1)

If two tangents drawn from a point P to the parabola y2 = 4x are at right angles, then the locus of P is (1) x = 1 (2) 2x + 1 = 0 (3) x = –1 (4) 2x – 1 = 0 y2 = 4x comparing with y2 = 4ax a=1 Locus of point P will be directrix of given parabola as tangents drawn from P are at right angles, therefore required locus is x = – a x=–1 Ans. (3) x y The line L given by + = 1 passes through the 5 b point (13, 32). The line K is parallel to L and has x y the equation + = 1 . Then the distance between c 3 L and K is 23 17 23 (1) (2) 17 (3) (4) 15 15 17

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82

x y + =1 (1) Passess through (13, 32) 5 b 13 32 + = 1 ⇒ 13b + 160 = 5b ⇒ b = – 20 5 b so line is –20x + 5y = – 100 (1) second line x y + =1 c 3 3x + cy = 3c (2) (1) and (2) are parallel c −3 3 = ⇒ c= − 20 5 4 3 9 Line 3x − y = − 4 4 12x –3y = – 9  5 − 20 x + 5 y = −9 ×  −   3 –20x + 5y = 15 ......(2) Distance between (1) and (2) | −100 − 15 | 115 115 23 = = = = 400 + 25 425 5 17 17 Ans. (4)

77.

A line AB in three dimensional space makes angles 45° and 120° with the positive x – axis and the positive y – axis respectively. If AB makes an acute angle θ with the positive z – axis, then θ equals (1) 30° (2) 45° (3) 60° (4) 75°

Sol.

cos2 α + cos2 β + cos2 γ = 1 cos2 45° + cos2 120° + cos2 γ = 1 1 1 + + cos2y = 1 2 4 1 Cos2y = 4 1 cos y = ± 2 y = 60º Ans. (3)

78.

Let S be a non- empty subset of R. Consider the following statement : P : There is a rational number x ∈ S such that x>0 Which of the following statements is the negation of the statement P ? (1) There is a rational number x ∈ S such that x < 0. (2) There is no rational number x ∈ S such that x < 0. (3) Every rational number x ∈ S satisfies x < 0. (4) x ∈ S and x < 0 ⇒ x is not rational Ans. [3] JUNE 2010

79.

cos(α + β ) =

Let

where 0 ≤ α , β ≤

4 and 5

π 4

let

sin(α − β ) =

5 , 13

5=

∑x (as variance =

25 16 19 (3) 12

56 33 20 (4) 7

(2)

tan(α + β ) + tan(α − β ) 1 − tan(α + β ) tan(α − β ) as cos(α + β) = 4/5, sin (α – β) = 5/13 3 5 9+5 + 56 4 12 tan 2α = = 12 = 3 5 16 − 5 33 1− . 4 12 16 Ans. (2)

tan 2α = tan[(α + β ) + (α − β )] =

Sol.

The circle x2 + y2 = 4x + 8y + 5 intersects the line 3x –4y = m at two distinct points if (1) – 85 < m < – 35 (2) – 35 < m < 15 (3) 15 < m < 65 (4) 35 < m < 85 x2 + y2 –4x –8y –5 = 0 centre = (2, 4) and radius = 5 P < r for if line is intersecting the circle at two points P=

3(2) − 4(4) − m 32 + 4 2

82.

An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colours is 1 2 1 2 (2) (3) (4) (1) 3 7 21 23

Sol.

Total balls = 3 red balls + 4 blue balls + 2 green balls = 9 balls 3

required probability =

83.

−1 If f has a local minimum at x = –1, then a possible value of k is (1) 1 (2) 0 1 (4) –1 (3) − 2 Let

f

f:R → R

K – 2x

lim

x < − 1  x >1 

10

87.

Let

10

S3 =

O This is true where k = –1



j ( j − 1) 10C j , S 2 =

∑j

2 10

10

∑j

10

Cj

and

j =1

C j.

j =1

Statement – 1 : S3 = 55 × 29. Statement – 2 : S1 = 90 × 28 and S2 = 10 × 28. (1) Statement -1 is true, Statement -2 is true; Statement -2 is a correct explanation for Statement -1 (2) Statement -1 is true, Statement -2 is true; Statement -2 is not a correct explanation for Statement -1. (3) Statement -1 is true, Statement -2 is false. (4) Statement -1 is false, Statement -2 is ture.

k − 2 x = +1 k = −1 Ans. (4) Directions : Questions number 86 to 90 are Assertion – Reason type questions. Each of these questions contains two statements: Statement – 1 (Assertion) and Statement – 2 (Reason).

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S1 =

j =1

1

F(x)≤–1

57 4! 3.4.3.2.1 = 20.19.18.17 20.18.17. 1 = 85

S : 1 is true. S : 2 possible cases of common difference are [± 1, ± 2,± 3, ± 4,±5,±6] S:2 is false Ans. (3)

2x + 3

–1 x → –1–

=

f(x) =

− 2 k − 2x x ≤ −1   f ' (x) =   2x + 3 x > −1 2

(17 + 14 + 11 + 8 + 5 + 2)4! 20 C4 4!

required arability =

84

JUNE 2010

Sol.

Sol. 10

S1 =

10!

∑ j( j − 1) j( j − 1)( j − 2)!(10 − j)! j=1

10

= 90

8!

∑ ( j − 2)!(8 − ( j − 2))! = 90 × 2

AM > GM 2 ex + x e ≥ (e x ) 2  2  ex 

8

j= 2

10

S2 =

∑ j=1

ex +

10! j j( j − 1)!(9 − ( j − 1))! 10

= 10

9!

∑ ( j − 1)!(9 − ( j − 1))! = 10 × 2

2 ≥2 2 ex

Q ex > 0 ⇒ ex +

9

j=1

10

10! = S3 = [ j( j − 1) + j] j ( 10 − j)! j=1

∑ 10

=

∑ ( j)

10

10

∑ ( j − 1)

10

Cj

89.

j=1

C j =90.2 8 + 10.2 9 = 110.2 8 = 55.2 9

90.

Statement – 1 : The point A(3, 1, 6) is the mirror image of the point B(1, 3, 4) in the plane x – y + z = 5. Statement – 2 : The plane x – y + z = 5 bisects the line segment joining A(3, 1, 6) and B(1, 3, 4). (1) Statement -1 is true, Statement -2 is true; Statement -2 is a correct explanation for Statement -1 (2) Statement -1 is true, Statement -2 is true; Statement -2 is not a correct explanation for Statement -1. (3) Statement -1 is true, Statement -2 is false. (4) Statement -1 is false, Statement -2 is true. Mid point of A(3, 1, 6) and B(1, 3, 4) should lie in the plane mid point : (2, 2, 5) it satisfies the plane x – y + z = 5. Also AB ⊥ to plane. Hence Dr's of AB are < 1, – 1, 1 > statement 1 and 2 are true Ans. (1)

Sol.

x

Let A be a 2 × 2 matrix with non zero entries and let A2 = I, where I is 2 × 2 identity matrix. Define Tr(A) = sum of diagonal elements of A and |A| = determinant of matrix A. Statement – 1 : Tr(A) = 0 Statement – 2 : |A| = 1 (1) Statement -1 is true, Statement -2 is true; Statement -2 is a correct explanation for Statement -1 (2) Statement -1 is true, Statement -2 is true; Statement -2 is not a correct explanation for Statement -1. (3) Statement -1 is true, Statement -2 is false. (4) Statement -1 is false, Statement -2 is ture. a b  let   c d A2 =I a b  a b 1 0  c d   c d  = 0 1      

Let f : R → R be a continuous function defined by 1 f ( x) = x e + 2e − x 1 Statement – 1 : f (c) = , for some c ∈ R. 3 1 Statement – 2 : 0 < f ( x) ≤ , for all x ∈ R. 2 2 (1) Statement -1 is true, Statement -2 is true; Statement -2 is a correct explanation for Statement -1 (2) Statement -1 is true, Statement -2 is true; Statement -2 is not a correct explanation for Statement -1. (3) Statement -1 is true, Statement -2 is false. (4) Statement -1 is false, Statement -2 is ture.

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(2)

e +

j=1

Sol.

2 >0 ex 1

< 2 2 2 x e also f(c) = 1/3 for c = 0 so statement 1 : is true statement 2 : is also true with correct explanation Ans. (1) 0<

Hence statement 1 is true, statement 2 is false Ans. (3) 88.

1

(1)

a 2 + bc ab + bd  1 0 =   2  ac + dc bc + d  0 1

ab + bd = 0 b (a + d) =0 b≠0 so, a = – d a b  A=  c d 

a+d=0

Tr (A) = 0

But |A| ≠ 1. So, statement I is true and statement 2 is false. Ans. (3) 85

JUNE 2010

XtraEdge Test Series ANSWER KEY IIT- JEE 2011 (June issue) Ques Ans Ques Ans Numerical Response

1 A 11 B,D Ques Ans

2 C 12 A,D 19 2

3 C 13 B,C 20 2

4 A 14 B 21 2

Ques Ans Ques Ans Numerical Response

1 B 11 A,B,D Ques Ans

2 C 12 A,C,D 19 8

3 C 13 A,B 20 5

4 D 14 D 21 6

Ques Ans Ques Ans Numerical Response

1 A 11 A,B Ques Ans

2 B 12 A,B,D 19 1

3 B 13 A,C 20 3

4 C 14 D 21 2

PHYSICS 5 D 15 C 22 3

C HE M ISTR Y 5 B 15 C 22 8

6 C 16 D 23 6

7 C 17 A 24 6

8 B 18 C 25 5

9 B,C

10 B,C

26 5

27 2

6 B 16 C 23 9

7 D 17 B 24 2

8 A 18 D 25 2

9 B,C,D

10 B,C

26 5

27 5

7 D 17 B 24 4

8 C 18 A 25 4

9 A,B,C

10 A,C

26 5

27 6

6 D 16 D 23 8

7 C 17 A 24 1

8 D 18 B 25 3

9 B,C,D

10 A,C

26 2

27 7

6 D 16 B 23 4

7 A 17 A 24 1

8 C 18 D 25 6

9 B,C

10 B,D

26 0

27 4

7 C 17 C 24 5

8 B 18 A 25 3

9 A,B,C,D

10 A,B,C

26 4

27 1

MATHEMATICS 5 A 15 B 22 7

6 C 16 C 23 1

IIT- JEE 2012 (June issue) Ques Ans Ques Ans Numerical Response

1 D 11 A,C Ques Ans

2 D 12 A,B,C,D 19 4

3 C 13 A,D 20 2

4 B 14 B 21 2

Ques Ans Ques Ans Numerical Response

1 B 11 A,C,D Ques Ans

2 D 12 B,D 19 2

3 B 13 A,C,D 20 4

4 B 14 B 21 2

Ques Ans Ques Ans Numerical Response

1 C 11 A,C,D Ques Ans

2 C 12 A,B,C 19 1

3 B 13 A,C 20 3

4 C 14 C 21 4

XtraEdge for IIT-JEE

PHYSICS 5 C 15 B 22 1

C HE M ISTR Y 5 C 15 B 22 6

MATHEMATICS 5 C 15 C 22 8

86

6 B 16 A 23 3

JUNE 2010

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"XtraEdge for IIT JEE" magazine makes sure you're updated & at the forefront. Every month get the XtraEdge Advantage at your door step. Magazine content is prepared by highly experienced faculty members on the latest trend of the IIT JEE. Predict future paper trends with XtraEdge Test Series every month to give students practice, practice & more practice. Take advantage of experts' articles on concepts development and problem solving skills Stay informed about latest exam dates, syllabus, new study techniques, time management skills and much more XtraFunda. ✓ Confidence building exercises with Self Tests and success stories of IITians ✓ Elevate you to the international arena with international Olympiad/ Contests problems and Challenging Questions.

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