XtraEdge_2010_03

Worry is a misuse of imagination.

Volume - 5 Issue - 9 March, 2010 (Monthly Magazine) Editorial / Mailing Office : 112-B, Shakti Nagar, Kota (Raj.) 324009

Editorial

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

Dear Students,

Pramod Maheshwari

All of us live and work within fixed patterns. These patterns and habits determine the quality of our life and the choices we make in life. There are a few vital things to know about ourselves. We should become aware of how much we influence others, how productive we are and what can help us to achieve our goals. It is important to create an environment which will promote our success. We should consciously create a system that would enable us to achieve our goals. Most of us live in systems which have come our way by an accident, circumstances or people we have met over a period of time. We are surrounded by our colleagues or subordinates who happened to be there by the fact of sheer recruitment earlier or later by the management. Our daily routines and schedules have been formed on the basis of convenience, coincidence, and the expectations of society and sometimes due to superstitions. The trick for success is to have an environment that helps in attaining our goals. Control your life. Make an effort to launch your day with a great start. A law of physics says that an object set in motion tends to remain in motion. It is the same thing with daily routine. To have a good start each morning will keep you upbeat during the day. If you begin the day stressed, you will tend to remain so that way. The best is to create a course of action or conditions where you are not hassled for being late for a meeting, worried about household affairs or distracted by happenings in the world.

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Aim to be highly successful. Control the direction of your life. Not only should you start the day on a cheerful note but also continue to do so during the day. Keep yourself stimulated and invigorated during the entire day. Start your day with a purpose. Have a daily direction and trajectory of action. It will keep you on your course all day long. Throughout the day reinforce your positive values and your choices. Anything that helps you in maintaining your highest values and your most important priorities should be welcome. Be in control of your life and work. Create and sustain a wonderful environment filled with beauty, peace, inspiration and hope. Plan your day in such a way that suits your plans objectives and makes you feel just right with the right amount of encouragement during the entire day. You should give a direction to your day and timing. Presenting forever positive ideas to your success.

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

Yours truly

Pramod Maheshwari, B.Tech., IIT Delhi

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MARCH 2010

XtraEdge for IIT-JEE

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Volume-5 Issue-9 March, 2010 (Monthly Magazine) NEXT MONTHS ATTRACTIONS

CONTENTS INDEX

Regulars ..........

Key Concepts & Problem Solving strategy for IIT-JEE. Know IIT-JEE With 15 Best Questions of IIT-JEE Challenging Problems in Physics, Chemistry & Maths Much more IIT-JEE News. Mock Test IIT-JEE Paper 1 & Paper II with Solution Mock Test AIEEE with Solution Mock Test BIT SAT with Solution

PAGE

NEWS ARTICLE

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IITian ON THE PATH OF SUCCESS

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• Top medical honchos mull medical courses in IITs • Promise of IITs, from Mamata Mr. R. Madhavan

KNOW IIT-JEE

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Previous IIT-JEE Question

Study Time........ DYNAMIC PHYSICS Success Tips for the Month • The difference between a successful person and others is not a lack of strength, not a lack of knowledge, but rather a lack of will.

• Footprints on the sands of time are not

8-Challenging Problems [Set# 11] Students’ Forum Physics Fundamentals Matter Waves, Photo-electric Effect Thermal Expansion, Thermodynamics

CATALYST CHEMISTRY

made by sitting down. can.

• The secret of joy in work is contained in

• Six essential qualities that are the key to success: Sincerity, personal integrity, humility, courtesy, wisdom, charity.

• Continuous efforts - not strength or intelligence - is the key to unlocking our potential.

• We can do anything we want to do if we stick to it long enough.

• The path to success is to take massive, determined action.

XtraEdge for IIT-JEE

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Key Concept Purification of Organic Compounds Boron & Carbon Family Understanding: Inorganic Chemistry

• To succeed, we must first believe that we

one word - excellence. To know how to do something well is to enjoy it.

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DICEY MATHS

41

Mathematical Challenges Students’ Forum Key Concept Definite integrals & Area under curves Probability

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

51

SOLUTIONS

90

Mock Test IIT-JEE Paper-1 & Paper-2 Mock Test AIEEE Mock Test BIT SAT

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MARCH 2010

Top medical honchos mull medical courses in IITs New Delhi: If the proposal of the Health secretary K Sujatha Rao comes into action, then the premier technology institutes of the country- the Indian Institutes of Technology (IITs) would soon see doctors passing out from the premier institute. Rao, in order to gain the consensus of various top honchos from top ministries, directors of medical institutes and chairmen of medical councils, who have been invited in a meeting by the health secretary today, will determine whether allowing institutes like IITs to teach medicine will "help medical education or dilute its quality." Those called to attend include directors of All India Institute of Medical Sciences (AIIMS), PGI (Chandigarh), Sanjay Gandhi Post Graduate Institute (Lucknow), JIPMER, NIMHANS, National Institute of Communicable Diseases, National Institute of Paramedical Sciences and principal of CMC Vellore. This will be the first major meeting to discuss the issue. Along with eminent doctors like Dr Ranjit Roychoudhury, Dr Devi Shetty, Dr Anupam Sibal and cardiologist Dr K Srinath Reddy, chairman of the Medical Council and the Nursing Council of India will also attend the meeting. The main pointers of the meeting would be to discuss major issues: if MBBS programme can be introduced at IITs, replacing the existing Medical Dental, Nursing and Pharma Councils of India to create the National Council for

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Human Resource in Health (NCHRH) - as the overarching regulatory body and will address the medical personnel crunch in India. An official from the health ministry said that, "We would like to know the experts views in order to understand the feasibility of the proposal, if starting medical courses in IITs will boost medical education in India. Once we have a consensus of all the experts we would send it to the Human Resource Development (HRD) Ministry for further consideration." In about three years, some IITs like Kharagpur and Hyderabad are working upon to start medical schools in about three years. According to the Ministry officials, a Memorandum of Understanding (MoU) has been signed between IIT Kharagpur and University of California, San Diego, to set up a hospital, which will offer graduate, postgraduate, and research programmes in medicine and bio-medical engineering. IIT Hyderabad has been expressing its interest to offer MD degrees in three years. HRD minister Kapil Sibal in a recent meeting with IIT directors, had asked them to expand their courses.

Promise Mamata

of

IITs,

from

Mamata Banerjee today promised a deal with the Union human resource development ministry to build “schools, colleges and IITs” on railway land. 4

The railway minister said she was in talks with the HRD ministry and a memorandum of understanding could be signed in a matter of days.

Railways to sign with IITKharagpur for research Kolkata : Indian Railways will sign a Memorandum of Understanding (MoU) with the Indian Institute of Technology, Kharagpur (IIT-K) to promote research and development in the organisation, Railway Minister Mamata Banerjee said here on Wednesday. "We are going to sign a MoU with the IIT-K on February 13, 2010, to promote research and development activities for the railways," Banerjee said after inaugurating a computerised reservation counter in southern part of Kolkata. "We may float scholarships for IIT-K students to carry out research activities," she said. The minister announced two new trains for West Bengal. "Two new trains would be flagged off on February 13. One from JhargramPurulia and another MedinipurJhargram." She said the railways would also provide its land for setting up 372 diagnostic centres, 44 first-class hospitals and 88 second-class hospitals across the country. MARCH 2010

"We'll provide land and the Union Health Ministry will develop the medical infrastructure on those lands. We have already signed a MoU with the Union Health Ministry to develop such health projects," Banerjee said. IANS Indo-German Computer Centre inaugurated at IIT Delhi An Indo-German Max Planck Centre on Computer Science (IMPECS) was inaugurated at the Indian Institute of Technology, Delhi (IIT-D) here on Wednesday by Mr. Prithviraj Chavan, Minister of Science & Technology and Earth Sciences along with H.E. Mr. Horst Koehler, President of the Federal Republic of Germany. IMPECS will be engaged in collaborative basic research in Computer Science between Indian and German scientists and serve as a bridge between computer scientific communities from both sides. The Centre will act as a center of excellence for faculty and students from both sides. The research areas envisaged under the Indo-German Computer Centre would be Algorithms and Complexity, Database and Information Retrieval; Graphics and Vision and Networking. It would also engage researchers from institutes including Tata Institute of Fundamental Research, Mumbai; Indian Institute of Technology, Kanpur (IIT-K); Indian Institute of Technology, Bombay (IIT-B) and Indian Institute of Technology, Madras (IIT-M) - from the Indian side and Max Planck Institute for informatics (MP-INF), Saarbruecken from the German side. The Centre is expected to benefit both countries. The major benefits for India would be further strengthening of the research base in Computer Science that would develop expertise of highest caliber needed by academia and industry. Germany would benefit through improved collaboration with leading

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Indian scientists and a highly visible outpost with highly professional environment and large pool of young talent. The Indo-German Centre on Computer Centre has been set up jointly by Department of Science & Technology (Govt. of India) and Max Planck Society of Germany under the over-all aegis of IndoGerman S&T Cooperation programme for an initial duration of 5 years at a total cost of approximately Rs.12 crores from Indian side by DST and approximately 2 million Euros by Max Planck Society from German side. The concept of setting-up the Indo-German Centre on Computer Science in India was discussed during the visit of German Chancellor to India in October 2007. Dr. T. Ramasami, Secretary, Department of Science & Technology, Prof Peter Gruss, President Max Planck Society, Germany and Prof Surendra Prasad, Director, Indian Institute of Technology, Delhi were the guest of honour during the inauguration function.

'Spectacular years ahead in space' - former ISRO chief New Delhi: "The last 50 years of space have been fantastic while the next 50 years will be spectacular," remarked Prof. U R Rao, former chairman ISRO, delivering his lecture on "Challenges in Space" in the 97th Indian Science Congress. The space age began with the launch of Sputnik-I, 52 years ago from the former Soviet Union. Since then, plenty of satellites have been launched. The Cosmic Background Radiation Explorer (COBE) launched in 1989, confirmed the prediction made by the Big Bang theory. The Wilkinson Microwave Anisotropy Probe and recently launched Herschel and Plank have 5

contributed a lot to the study of the universe. Dr. Rao, enumerated nine great challenges in space as food security, energy security, environmental security, resource security, space security, space transportation, search for life, exploration of the universe and colonisation of Mars. "The per capita food productivity of India which is currently about 1.7 ton/ ha should be increased to about 4 tons/ha by 2050 to meet the growing food requirements. This can be done by initiating a new green revolution that requires the application of space technology along with biological inputs," he observed. "Space technology can be used for better meteorological forecasting which would help mitigate the consequences of disasters," he added. Dr. Rao stressed the importance of energy security for industrial expansion, agriculture and infrastructure growth. He explained with figures that per capita energy usage of India is far lower than other developed and developing nations like U.S (15 times more), EU (7.5 times more) and China (2.3 times more). "Space technology can play a significant role in coping with India's energy deficit by the better utilisation of energy resources as well as by learning the effects of global warming, carbon dioxide emission and so on," he added. Three students bag Manmohan Singh scholarship at Cambridge New Delhi: Three students from Bangalore, Kolkata and Mumbai are set to receive the 2010 Manmohan Singh scholarship to fund their undergraduate studies at the University of Cambridge. For 2010, the scholarship will be given to Neal Duggal from Mallya Aditi International School, Bangalore, Jesika Haria from Dhirubhai Ambani International MARCH 2010

School, Mumbai, and Rudrajit Banerjee from The Cambridge School, Kolkata. Neal Duggal and Jesika Haria have received conditional offers of places at St John's College and Emmanuel College, Cambridge, to study for degrees in Economics and Engineering respectively, while Rudrajit Banerjee has received an unconditional offer to study Natural Sciences at Christ's College, Cambridge, the university said in a statement on Tuesday. The Manmohan Singh Undergraduate Scholarship programme was established in 2009 in honour of India's Prime Minister who graduated from the Cambridge varsity with a first class in Economics in the late 1950s. Singh was also awarded an honorary doctorate by the university in 2006. The scholarship is awarded to students who have received an offer of a place at the University of Cambridge. Two of the three places offered by University of Cambridge are conditional on these students achieving specific grades. The scholarship programme will provide full funding, covering fees and maintenance for undergraduate study at the Cambridge. IANS

Srikanth Jagabathula, President of India gold medal winner. Internet connectivity in rural areas at cheap rates? Well, this could be a reality if Srikanth's dream comes true. Meet Srikanth Jagabathula, IIT's pride, the President of India gold medal winner for 2005-06 for scoring the highest marks among all batches at Indian Institute of Technology-Bombay. After an enviable stint at the IIT, Srikanth is all set to fly to the United States to pursue his studies at the prestigious Massachusetts Institute of Technology. After five years he plans to come back to Indian to start his own communications company.

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As a kid he dreamt of becoming an engineer. Somehow he always thought an engineer's job would be very fascinating. He heard about the IIT when he was in the 7th standard. Since then IIT was his aim. After clearing his 10th class, he religiously worked towards cracking the IIT-Joint Entrance Examination. A rank of 38 at the IIT-JEE meant a smooth entry into IIT-Bombay. Srikanth, who hails from Hyderabad, was always a topper in school. Mathematics and physics were his favourite subjects, but he dreaded biology and chemistry.

World class university to be set up by Reliance To promote education and research in India, the Reliance Group plans to set up a ‘worldclass’ university in India. The university, modeled on the lines of American universities such as The University of Pennsylvania will be set up either in Delhi or Mumbai. “It will be international in scale and in best practices, but with an Indian soul.

IITs successfully conducted GATE online in two subjects New Delhi: Indian Institutes of Technology at Bombay, Delhi, Guwahati, Kanpur, Kharagpur, Madras, Roorkee and the Indian Institute of Science Bangalore successfully conducted online Graduate Aptitude Test in Engineering (GATE) 2010 for two out of 21 papers yesterday. Examinations in two subjects, namely Textile Engineering and Fiber Science (TF), and Mining Engineering (MN) were conducted using computers by these institutes. About 1700 candidates were registered for these examinations which were conducted simultaneously in eight cities over two shifts. This experiment was conducted in GATE 2010 this year for the first time and depending upon the experience, online examination 6

might be repeated on a larger scale in subsequent years. The offline version of the exam in other 19 papers shall be conducted all over the country on Feb 14, 2010. Graduate Aptitude test in Engineering (GATE) is an all India examination administered and conducted jointly by Indian Institute of Science and seven Indian Institutes of Technology on behalf of the National Coordination Board - GATE, Department of Higher Education, Ministry of Human Resource Development (MHRD). Admission to postgraduate programmes with MHRD and some other government scholarship/ assistanceship in engineering colleges/institutes is open to those who qualify through GATE. GATE qualified candidates with Bachelor's degree in Engineering / Technology / Architecture or master's degree in any branch of Science / Mathematics / Statistics / Computer Applications are eligible for admission to master's / doctoral program in Engineering / Technology / Architecture as well as for doctoral programs in relevant branches of Science with MHRD or other government scholarship/assistantship.

Placements in full swing at IITs In just over a month, around 70% of the students at IIT have been placed. The final placements began on Dec 1, 2009 and will continue till March 2010. However, because of the good response from IT companies, the IITs hope that the placements might be wrapped early. However, last year due to economic slowdown, the IITs were able to place only 75-80 per cent of their student pool. Many students had to opt for higher studies or jobs in teaching. Barclays Bank made the highest offer of Rs 22 lakh at IITKharagpur for placement at Singapore. IIT-Roorkee also achieved around. MARCH 2010

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

Mr. R. Madhavan Mechanical Engineering, IIT Madras

R. Madhavan IITians who preferred to be different, rather than get into a corporate rat race. One of the most interesting themes at this year's Pan-IIT event was the session on rural transformation. IITians who have chosen an offbeat career hogged the limelight at the event. The star at the event was R Madhavan, an alumnus of IIT-Madras. This is Madhavan's success story as a farmer. . .

I started my career at the Oil and Natural Gas Corporation (ONGC). My father refused to give me any money to start farming. So I asked the officials to let me work at the offshore sites, on the rigs. The advantage was that I could work on rigs for 14 days and then take 14 days off. I chose to work on the rigs for nine years, uninterrupted. After 4 years, I saved enough money to buy six acres of land. I bought land at Chengelpet near Chennai. I chose that land because the plot had access to road and water. Back in 1989, a man in a pair of trousers aroused curiosity among the farming community. That was not the image of a farmer!

PASSION FOR AGRICULTURE I had a passion for agriculture even when I was young. I don't know how my love for agriculture started. I only know that I have always been a nature lover. I used to have a garden even when I was a teenager. So, from a home garden, a kitchen garden, I gradually became a farmer! My mother used to be very happy with the vegetables I grew.

I became a full fledged farmer in 1993. It was tough in the beginning. Nobody taught me how to farm. There was no guidance from the gram sevaks or the University of Agriculture. I ran from pillar to post but couldn't find a single scientist who could help me. I burnt my fingers. My first crop was paddy and I produced 2 tonnes from the six acres of land, it was pathetic. When I lost all my money, my father said I was stupid. I told him, it didn't matter as I was learning. It was trial and error for me for three years. Until 1997, I was only experimenting by mingling various systems.

My family was against my ambition of becoming an agriculturist. So, I had to find a livelihood for myself. I wrote IIT-JEE and got selected to study at the Indian Institute of Technology, Madras. I enjoyed studying mechanical engineering. My intention was to transform what I study into what I love; mechanisation of farming. I felt the drudgery in farming is much more than in any other industry, and no one had looked into it.

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MARCH 2010

LEARNING FROM ISRAEL

In those days, communication was slow. Today, I can get guidance from Dr Lakshmanan on Skype or Google Talk,

In 1996, I visited Israel because I had heard that they are

or through e-mail. I send him the picture of my problem

the best in water technology. Take the case of corn: they

and ask his guidance. In those days, it took time to

harvest 7 tonnes per acre whereas we produce less than a

communicate. There was no Internet or connectivity.

tonne. They harvest up to 200 tonnes of tomatoes,

That was why it took me four years to learn farming.

whereas here it is 6 tonnes, in similar area of land. I

Today, I would not have taken more than six months or

stayed in one of the kibbutz, which is a co-operative farm

even less to learn the trick!

for 15 days. I understood what we do is quite primitive. It was an eye opener for me. They treat each plant as an

THE FARMING CYCLE

industry. A plant producing one kilo of capsicum is an

I started crop rotation after 1997. In August, I start with

industry that has 1 kilo output. I learnt from them that we

paddy and it is harvested in December. I plant vegetables

abuse water. Drip irrigation is not only for saving water

in December itself and get the crops in February. After

but it enhances your plant productivity. We commonly

that, it is oil seeds like sesame and groundnut, which are

practice flood irrigation where they just pump water. As

drought-resistant, till May. During May, I go on trips to

per the 2005 statistics, instead of 1 litre, we use 750 litres

learn more about the craft. I come back in June-July and

of water.

start preparations on the land to get ready for August. In

DR. LAKSHMANAN

1999, I bought another four acres. My target is a net income of Rs 100,000 per annum per acre. I have achieved

I met Dr Lakshmanan, a California-based NRI, who has

up to Rs 50,000.

been farming for the last 35 years on 50-60,000 acres of land. He taught me farming over the last one decade.

I sell my produce on my own. I have a jeep and bring what

Whatever little I have learnt, it is thanks to him. I knew a

I produce to my house and sell from there. People know

farm would give me much better returns in terms of

that I sell at home. I don't go through any middle man. I

money as well as happiness. Working for money and

take paddy to the mill, hull it and sell it on my own. In the

working for happiness are different. I work and get

future, I have plans to have a mill too. These days, people

happiness. What more do you need?

tell me in advance that they need rice from me. I have no problem selling my produce.

NO GUIDANCE IN INDIA I said at one platform that we have to change the

ENGINEERING HELPS IN FARMING

curriculum of the agricultural universities. What they

More than any other education, engineering helps in

teach the students is not how to farm, but how to draw

farming because toiling in the soil is only 20 per cent of

loans from a bank! What they learn cannot be

the work. About 80 per cent of farming needs engineering

transformed to reality or to the villages. The problem in

skills. Science is a must for any farming. I have developed a

the villages is not mentioned in the university. There is a

number of simple, farmer-friendly tools for farming areas

wide gap and it is getting worse.

like seeding, weeding, etc. as we don't have any tools for

MAKING PROFITS

small farmers. If I have 200 acres of land, I can go for food processing, etc. My next project is to lease land from the

After burning my fingers for four years, from 1997

small farmers for agriculture. The village will prosper with

onwards, I started making profits. Even though it took me

food processing industries coming there. My yield will also

four years, I did not lose hope. I knew this was my path

be more with more land.

ven though I didn't have any guidance from anyone.

Don't compare yourself with anyone in this world. If you do so, you are insulting yourself. – Alen Strike XtraEdge for IIT-JEE

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MARCH 2010

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MARCH 2010

KNOW IIT-JEE By Previous Exam Questions

PHYSICS

2mv 2 l Given that T = 0 when α = 90º 2mv 2 ∴ 0–0= ⇒v=0 l ∞ M

T – 2mg cos α =

1.

A cart is moving along + x direction with a velocity of 4 m/s. A person on the cart throws a stone with a velocity of 6 m/s relative to himself. In the frame of reference of the cart the stone is thrown in y-z plane making an angle of 30º with vertical z axis. At the highest point of its trajectory, the stone hits an object of equal mass hung vertically from the branch of a tree by means of a string of length L. A completely inelastic collision occurs, in which the stone gets embedded in the object. Determine: [IIT-1997] (i) The speed of the combined mass immediately after the collision with respect to an observer on the ground, (ii) The length L of the string such that the tension in the string becomes zero when the string becomes horizontal during the subsequent motion of the combined mass. Sol. θ = 30º, v = 6 m/s When the stone reaches the point Q, the component of velocity in the +Z direction (V cos θ) becomes zero due to the gravitational force in the –Z direction. +Z

+Y

V Vcosθ θ Vsinθ P t = 0 Vx=4m

Z

L

Q

Y

α Q 2mg sinα

2mg ⇒ Velocity is zero when α = 90º, i.e., in the horizontal position. Applying energy conservation from Q to M, we get 1 V2 (2.5) 2 2mV2 = 2mgl ⇒ l = = = 0.318 m 2 × 9.8 2 2g

2.

V´ Vx=4m/s

X t=t

Three particles A, B and C, each of mass m, are connected to each other by three massless rigid rods to form a rigid, equilateral triangular body of side l. This body is placed on a horizontal frictioness table (x-y plane) and is hinged to it at the point A so that it can move without friction about the vertical axis through A (see figure). The body is set into rotational motion on the table about A with a constant angular velocity ω. [IIT-2002] y A

The stone has two velocities at Q (i) Vx in the +X direction (4 m/s) (ii) V sin θ in the + Y direction (6 sin 30º = 3 m/s) The resultant velocity of the stone V´ =

α 2mg cosα

x ω

(Vx ) 2 + (V sin θ) 2 → F

= 4 2 + 32 = 5 m/s (i) Applying conservation of linear momentum at Q for collision with an mass of equal magnitude m × 5 = 2m × V [Since the collision is completely inelastic the two masses will stick together. V is the velocity of the two masses just after collision] ∴ V = 2.5 m/s (ii) When the string is undergoing circular motion, at any arbitrary position

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C B l (a) Find the magnitude of the horizontal force exerted by the hinge on the body. (b) At time T, when the side BC is parallel to the x-axis, a force F is applied on B along BC (as shown). Obtain the x-component and the y-component of the force exerted by the hinge on the body, immediately after time T.

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A large open top container of negligible mass and uniform cross-sectional area A has a small holes of cross-sectional area A/100 in its die wall near the bottom. The container is kept on a smooth horizontal floor and contains a liquid of density ρ and mass m0. Assuming that the liquid starts flowing out horizontally through the hole at t = 0, Calculate : [IIT-2000] (i) The acceleration of the container and (ii) The velocity when 75% of liquid drained out Sol. (i) Let at any instant of time during the flow, the height of liquid in the container is x. The velocity of flow of liquid through small hole in the orifice by Torricelli's theorem is

Sol. The mass B is moving in a circular path centred at A.

3.

2

The centripetal force (mlω ) required for this circular motion is provided by F′. Therefore a force F′ acts on A (the hinge) which is equal to mlω2. The same is the case for mass C. Therefore the net force on the hinge is Fnet = F'2 + F'2 +2F' F' cos 60º

Fnet = 2F' 2 +2F' 2 ×

1 = 3 F′ = 2

3 mlω2

Y A F′ l

X

60º F′

Fnet

The mass of liquid flowing per second through the orifice = ρ × volume of liquid flowing per second dm A = ρ × 2gx × …(ii) dt 100 Therefore the rate of change of momentum of the system in forward direction dm 2gx × A × ρ ×v= (From (i) and (ii)) = dt 100 The rate of change of momentum of the system in the backward direction = Force on backward direction = m × a where m is mass of liquid in the container at the instant t m = Vol. × density =A×x×ρ

l

F′

F′

C B l (b) The force F acting on B will provide a torque to the system. This torque is

l 3 = Iα 2 3l = (2ml2)α F× 2 3 F ⇒ α= 4 ml The total force acting on the system along x-direction is F + (Fnet)x This force is responsible for giving an acceleration ax to the system.

F×

c.m l

x

P

v ∴ The rate of change of momentum of the system in the backward direction = Axρ × a By conservation of linear momentum 2gxAρ Axρ × a = 100 g ⇒ a= 50

3 2

F Therefore F + (Fnet)x = 3m(ax) c.m. 3 F F F l = 3m Q ax = αr = × = 4 ml 4m 3 4 3F = 4 F ∴ (Fnet)x = 4 (Fnet)y remains the same as before = 3 mlω2.

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…(i)

v = 2gx

(ii) By toricell's theorem v′ = 2g × (0.25h )

where h is the initial height of the liquid in the container m0 is the initial mass m ∴ m0 = Ah × ρ ⇒ h = 0 Aρ

∴

11

v′ =

2g × 0.25 ×

m 0 gm 0 = Aρ 2Aρ

MARCH 2010

An ideal gas is enclosed in a vertical cylindrical container and supports a freely moving piston of mass M. The piston and the cylinder have equal crosssectional area A. Atmospheric pressure is P0, and when the piston is in equilibrium, the volume of the gas is V0. The piston is now displaced slightly from its equilibrium position. Assuming that the system is completely isolated from its surroundings, show that the piston executes simple harmonic motion and find [IIT-1981] the frequency of oscillation. Sol. Let the piston be displaced by a distance x. 4.

P2 = p0 +

Final pressure on the gas

Mg  γA 2 x  Ma =  p 0 +  A  V0 

⇒

Mg  γA 2 x  a = p0 +  A  V0 M 

Comparing it with a = ω2x we get Mg  γA 2 x  ω2 =  p 0 +  A  V0 M 

Mg  γA 2 x  ω =  p0 +  A  V0 M 

∴

Mg  γ  Mg   + p ( v 0 − Ax ) γ Then  p 0 + V0 =  p 0 + A A    

Q Initial pressure on the gas P1 = p0 +

⇒

If

Mg A

Mg is small as compared to p0 then A

ω=

Mg +p A

∴

P0 A V0

5.

x

where p is the extra pressure due to which the compression x takes place. Final volume of the gas V2 = V0 – Ax The above equation can be rearranged as

⇒

∴

⇒

F  Mg  γAx =  p0 +  A  A  V0

⇒

Mg  γA 2 x  F = p0 +  A  V0 

XtraEdge for IIT-JEE

p0γ V0 M

A hydrogen-like atom (described by the Bohr model) is observed to emit six wavelengths, originating from all possible transitions between a group of levels. These levels have energies between –0.85 eV and –0.544 eV (including both these values). [IIT-2002]

(Take hc = 1240 eV.nm. Ground state energy of hydrogen atom = – 13.6 eV) Sol. (a) If x is the difference in quantum number of the

states then x+1C2 = 6 ⇒ x = 3 n+3

   γAx   V   0   

n Smallest λ

γAx = Mg V0 p0 + A

∴

A 2π

(b) Calculate the smallest wavelength emitted in these transitions.

p

Mg  γAx  p =  p0 +  A  V0 

f=

(a) Find the atomic number of the atom.

Mg     Ax  γ + p (V0 − Ax) γ  p  p0 +   1− + 1 A  Mg   V0  = 1= + p Mg  γ 0   A   p0 + V0    A  

 p p γAx  1=1+ – + Mg Mg V0  p 0 + p0 +  A A  Negligible as p and x are small

p 0 γA 2 = 2πf V0 M

2

Now, we have and

− z (13.6eV) n2

= – 0.85 eV

− z 2 (13.6eV) (n + 3) 2

…(i)

= – 0.544 eV …(ii)

Solving (i) and (ii) we get n = 12 and z = 3 (b) Smallest wavelength λ is given by hc = (0.85 – 0.544)eV λ Solving, we get λ ≈ 4052 nm.

12

MARCH 2010

CH3CH2CH2COOCH2CH3 (A)

CHEMISTRY 6.

LiAlH4

Compound A (C6H12O2) on reduction with LiAlH4 yields two compounds B and C. The compound B on oxidation gave D, which on treatment with aqueous alkali and subsequent heating furnished E. The latter on catalytic hydrogenation gave C. The compound D was oxidized further to give F which was found to be a monobasic acid (molar mass : 60.0 g mol–1). Deduce the structures of A, B, C, D and E.

CH3CH2OH CH3CH2CH2CH2OH (B) (C) [O] [O] CH3CHO → CH3COOH (D) (F) [H] aq KOH CH3CHCH2OH

[IIT-1990] Sol. The compound A is an ester. The equations involved in the given reactions are as follows. The compound F is a monobasic acid (molar mass = 60 g mol–1). This may be represented as RCOOH. From the molar mass of F, it is evident that the molar mass of R is 15 g mol–1 [= (60 – 45) g mol–1]. Hence, the compound F is CH3COOH (ethanoic acid). F is obtained by the oxidation of D. Hence, the compound D must be an aldehyde with the structure CH3CHO (ethanal). The compound D was obtained from the oxidation of B which must be an alcohol. Hence, the structure of B is CH3CH2OH (ethanol). D undergoes an aldol condensation (treatment with aqueous alkali) which subsequently gives E on heating. The reactions involved here are

OH

heating

CH3CH=CHCHO Alternatively, the compound A may be

CH3COOCH2CH2CH2CH3 (butylacetate) 7. A metallic element crystallizes into a lattice containing a sequence of layers of ABABAB ......... . Any packing of spheres leaves out voids in the lattice. What percentage by volume of this lattice is empty space ? [IIT-1996] Sol. A unit cell of hcp structure is a hexagonal cell, which is shown in figure (i) & (ii). Three such cells form one hcp unit. For hexagonal cell, a = b ≠ c; α = β = 90º and γ = 120º. It has 8 atoms at the corners and one inside, 8 Number of atoms per unit cell = +1=2 8 Area of the base = b × ON [From fig.(ii)] = b × a sin 60º

. KOH → CH 3CHCH 2 CHO 2CH3CHO aq | OH

=

heating

  → CH3CH=CHCHO

3 2 a 2

(Q b = a)

The reduction of E gives compound C. Hence, we have H] → CH 3CH 2 CH 2 CH 2 OH CH3CH=CHCHO [ ( C)

Finally, the structure of A can be obtained from the two alcohols (CH3CH2OH and CH3CH2CH2CH2OH) produced on treating A with LiAlH4. Thus, we have

c β αb a γ

 4 → CH 3CH 2 CH 2 COOCH 2 CH 3 LiAlH (A)

Figure (i) Volume of the hexagonal cell

CH 3CH 2 CH 2 CH 2 OH + CH 3CH 2 OH ( C) (B)

= Area of the base × height =

Thus, the reactions involved are as follows.

But c =

XtraEdge for IIT-JEE

13

2 2 3

3 2 a .c 2

a MARCH 2010

∴ Volume of the hexagonal cell 3 2 2 2 a . a = a3 2 = 2 3

or ∆H 0C 2 H 6 = – 20 kcal mol–1

From eq. (ii) ∆H 0reactant = 3 × ∆H 0CO 2 + 4 × ∆H 0H 2O – ∆H 0C3H8

a 2 Hence, fraction of total volume or atomic packing factor Volume of 2 atoms = Volume of the hexagonal cell O

and radius of the atom, r =

or – 530.0

= 3 × (–94.0) + 4 × (–68.0) – ∆H 0C3H8

or ∆H 0C3H8

= – 24 kcal mol–1

∆H 0reactant = 2 × [C(s) → C(g)] + 6 ×  1 H 2 → H  2

60º

– [1 × C – C + 6 × C – H] or – 20 = 2 × 172.0 + 3 × 104.0 – [1 × C – C + 6 × C – H] or [1 × C – C + 6 × C – H] = 676 kcal mol–1 From eq. (iv),

N b figure (ii) 3

4 a 4 2 × π  2 × πr 3 3 2 3 = = 3 a 2 a3 2 π = 0.74 = 74% = 3 2 ∴ The percentage of void space = 100 – 74 = 26%

2

...(v)



– [2 × C – C + 8 × C – H] – 24 = 3 × 172.0 + 4 × 104.0 – [2 × C – C + 8 × C – H] or [2 × C – C + 8 × C – H] = 956 kcal mol–1 ...(vi) Solving eq. (v) and eq. (vi), we get Bond energy of C – C bond = 82 kcal mol–1 Bond energy of C – H bond = 99 kcal mol–1 or

The electrode reactions for charging of a lead storage battery are PbSO4 + 2e– → Pb + SO42– PbSO4 + 2H2O → PbO2 + SO42– + 4H+ + 2e– The electrolyte in the battery is an aqueous solution of sulphuric acid. Before charging the specific gravity of the liquid was found to be 1.11 (15.7% H2SO4 by weight). After charging for 100 hours, the specific gravity of the liquid was found to be 1.28 (36.9% H2SO4 by weight). If the battery contained two litres of the liquid, calculate the average current used for charging the battery. Assume that the volume of the battery liquid remained constant during charging. [IIT-1972] 100 Sol. Volume of 100 g of 15.7% H2SO4 = = 90.9 ml 1.11 90.9 ml of 15.7% H2SO4 contains = 15.7 g H2SO4 ∴ 2.0 L of 15.7% H2SO4 contains 15.7 × 1000 × 2.0 = 90.9 = 345.43 g H2SO4 100 Volume of 100 g of 36.9% H2SO4 = = 78.125 ml 1.28 78.125 ml of 36.9% H2SO4 contains = 36.9 g H2SO4 ∴ 2.0 L of 36.9% H2SO4 contains 36.9 × 1000 × 2.0 = = 944.64 g H2SO4 78.125 9.

∆H 0Combustion (propane) = –530.0 ∆Hº for C (graphite) → C(g) = 172.0 Bond energy of H – H = 104.0 ∆H 0f of H2O (l) = – 68.0 ∆H 0f of CO2(g) = – 94.0 [IIT-1990] Sol. Given that, 7 C2H6(g) + O2(g) → 2CO2(g) + 3H2O(l) 2 ∆H = – 372.0 kcal mol–1 ...(i) C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ∆H = – 530.0 kcal mol–1 ...(ii) 0 ∆H for C(gr) → C(g) = 172.0 kcal mol–1 Bond energy of H – H = 104.0 kcal mol–1 ∆H 0f of H2O(l) = –68.0 kcal mol–1 ∆H 0f of CO2(g) = – 94.0 kcal mol–1 To find, 2C(g) + 6H(g) → C2H6(g) ∆H = ? ...(iii) 3C(g) + 8H(g) → C3H8(g) ∆H = ? ...(iv) and hence the bond energy of C – C and C – H bonds. We know that, ∆H 0reaction = ∆H 0product – ∆H 0reactant

From eq. (i), ∆H 0reactant = 2 × ∆H 0CO 2 + 3 × ∆H 0H 2O – ∆H 0C 2 H 6 = 2 × (–94.0) + 3 × (–68.0) – ∆H 0C 2 H 6

XtraEdge for IIT-JEE



∆H 0reactant = 3 × [C(s) → C(g)] + 8 ×  1 H 2 → H 

Using the data (all values are in kilocalories per mole at 25ºC) given below, calculate the bond energy of C – C and C – H bonds. ∆H 0Combustion (ethane) = – 372.0

or – 372.0

[for eq.(iv)]

∆H 0reactant = Sum of bond energies of reactants – Sum of bond energies of products From eq.(iii),

a

8.

[for eq.(iii)]

14

MARCH 2010

Amount of H2SO4 formed after charging = 944.64 – 345.43 = 599.21 g The overall reaction is Pb + PbO2 + 2H2SO4 → 2PbSO4 + 2H2O 98 g H2SO4 (2 × 1 + 32 + 4 × 16) required = 1 F electricity 599.21 F electricity ∴ 599.21 g H2SO4 requires = 98 599.21 × 96500 C electricity = 98 We know, Q = It Given that, t = 100 hrs = 100 × 3600 s Q 599.21× 96500 ∴ I= = A = 1.638 A t 98 × 100 × 3600 ∴ Current used for charging battery = 1.638 ampere

or

log K p 2 = – 1.507 = 2 .493 K p 2 = Antilog 2 .493 = 3.122 × 10–2

or

PH 2O = K p 2 = 3.122 × 10–2 atm

(c) CO(g) + H2O(l) Initial Pressure 1 Pressure at Equlibrium 1 – x Hence,

or x2 + 3124x – 3121 = 0 or x =

1

K p 3 = K p1 . K p 2 =

or 1.00 × 105 × 3.122 × 10–2 =

10. Given the following standard free energies of formation at 25ºC : CO(g) = – 32.807, CO2(g) = –94.260, H2O(g) = – 54.635, H2O (l) = – 56.69 kcal mol–1. (a) Find ∆Gº and the equilibrium constant Kp for the reaction CO(g) + H2O(g) CO2(g) + H2(g) at 25ºC. (b) Find the vapour pressure of H2O at 25ºC. (c) If CO, CO2 and H2 are mixed so that the partial pressure of each is 1.00 atm and the mixture is brought into contact with excess liquid H2O, what will be the partial pressure of each gas when equilibrium is attained at 25ºC. The volume available to the gases is constant. [IIT-1973] Sol. Given that, ∆G 0f (CO) = – 32.807 kcal mol–1

CO2(g) + H2(g)

1+x PCO 2 .PH 2

1 1+x

PCO

(1 + x )(1 + x ) 1− x

− b ± b 2 − 4ac 2a − 3124 ± (3124) 2 − [4 × 1× (−3121)

= 0.9987 2 ×1 ∴ PCO = 1 – x = 1 – 0.9987 = 1.3 × 10–3 atm PH 2 = PCO 2 = 1 + x = 1 + 0.9987 = 1.9987 atm =

MATHEMATICS 11. Complex numbers z1, z2, z3 are the vertices A, B, C respectively of an isosceles right angled triangle with right angle at C. Show that (z1 – z2)2 = 2(z1 – z3) (z3 – z2) [IIT-1986] Sol. Since, ∆is right angled isosceles ∆. ∴ Rotating z2 about z3 in anticlock wise direction through an angle of π/2, we get A(z1)

∆G 0f (CO2) = –94.260 kcal mol–1 ∆G 0f [H2O (g)] = – 54.635 kcal mol–1

∆G 0f [H2O(l) = – 56.69 kcal mol–1 (a) For the reaction, CO2(g) + H2(g) CO(g) + H2O(g) 0 0 0 ∆G reaction = ∆G f (CO2) – ∆G f [H2O(g)] – ∆G 0f (CO) = –94.260 – (– 54.635) – (–32.807) = – 6.818 kcal = – 6818 cal Also, ∆G0 = – 2.303 RT log K p1

B(z3)

z 2 − z3 | z − z 3 | iπ/2 = 2 e z1 − z 3 | z1 − z 3 | where, |z2 – z3| = |z1 – z3| ⇒ (z2 – z3) = i(z1 – z3) squarring both sides we get, (z2 – z3)2 = – (z1 – z3)2 ⇒ z22 + z32 – 2z2z3 = –z12 – z32 + 2z1z3 ⇒ z12 + z22 – 2z1z2 = 2z1z3 + 2z2z3 – 2z32 – 2z1z2 ⇒ (z1 – z2)2 = 2{(z1z3 – z32) + (z2z3 – z1z2)} ⇒ (z1 – z2)2 = 2(z1 – z3) (z3 – z2)

∴ –6818 = –2.303 × 1.987 × 298 × log K p1 or log K p1 = 5.00 or K p1 = Antilog 5.00 = 1.00 × 105 atm (b) For the reaction, H2O(l) ∆G 0reaction

H2O(g)

Also,

= [H2O(g) ] – ∆G 0f [H2O(l)] = – 54.635 – (– 56.69) = 2.055 kcal = 2055 cal ∆G0 = –2.303 RT log K p 2

∴

2055 = –2.303 × 1.987 × 298 log K p 2

∆G 0f

XtraEdge for IIT-JEE

C(z2)

12. The fourth power of the common difference of arithmetic progression with integer entries is added to the product of any four consecutive terms of it. Prove that resulting sum is the square of an integer. [IIT-2000]

15

MARCH 2010

Sol. Let four consecutive terms of the A.P. are a – 3d, a – d, a + d, a + 3d which are integers. Again P = (a – 3d) (a – d) (a + d) (a + 3d) + (2d)4 (by given condition) = (a2 – 9d2)(a2 – d2) + 16d4 = a4 – 10a2d2 + 9d4 + 16d4 = (a2 – 5d2)2 Now, a2 – 5d2 = a2 – 9d2 + 4d2 = (a – 3d)(a + 3d) + (2d)2 = I.I + I2 (given) = I 2 + I2 = I2 = I (where I is any integer) Therefore, P = (I)2 = Integer 13. Evaluate : Sol. I =

∫

∫

(cos 2x )1/ 2 dx sin x

=

2I =

∫

=–

∫ cos θ + cos

=–

∫ cos θ(1 + cos

=–

∫

(1 + cos 2 θ) − 2 cos 2 θ

∫

f(sin 2x) . (sin x + cos x) dx

π/ 4

0

f(sin 2x) . (sin x + cos x) dx π/ 4

∫

= 2 2 ∴ I=

Hence, =

∫

cos θ

dθ

π/ 4

∫

π/4

0

2

∫

∫

f(cos 2x) . cos x dx

0

π/2

0

2

f(cos 2x) . cos x dx

∫

f(sin(2x)) . sin x dx

π/ 4

0

f(cos 2x) . cos x dx

15. Find the area bounded by the curves, x2 + y2 = 25, 4y = |4 – x2| and x = 0 above the x-axis. [IIT-1987] Sol. Here, x2 + y2 = 25, 4y = |4 – x2| could be sketched as, whose point of intersection could be obtained by y 5

dθ dθ

∫ 1 + cos

= –log |sec θ + tan θ| + 2 = – log |sec θ + tan θ| +

2

= – log |cot x +

O –4

dθ

θ

cos θ

∫ 2 − sin dt

∫ 2−t 1

= – log |sec θ + tan θ| + 2.

2 2

2

2

θ

1

2 + sin θ

+c

2 − sin θ

∫

2

2 + 1 − tan x 2 − 1 − tan 2 x

+c

14. Show that f (sin 2x ) sin xdx =

2

∫

π/4 0

f (cos 2 x ) cos x dx

[IIT-1990]

XtraEdge for IIT-JEE

2

(4 − x ) = 25 16 2 2 ⇒ (x + 24) (x – 16) = 0 ⇒ x = ± 4 ∴ Required area  4 2 4 − x 2  4 2    dx −  x − 4  dx  25 − x 2 dx −  = 2   0 2  0  4    4  4  x 25 −1  x   2 25 − x + sin    = 2  2  5  0  2 2 4   1  x 3  1  x 3  25π   –  4x − −  − 4x   = 8+    4 3  4 3  4   2  0

x2 +

2

log

x

45

–5

, where sin θ = t log

2

–2

dθ

cot 2 x − 1 |

+

0

0

cos θ(1 + cos 2 θ)

= – sec θ dθ + 2

∫

π/2

∫

=–

π/ 2

∫

2

θ)

...(2)

π  f(sin 2x) sin  x +  dx 0 4  π/ 4  π π π  = 2 2 f  sin 2 − x   . sin  − x +  dx 0 4 4  4 

∫

2

 π π   f sin 2 − x  .sin  − x  dx 2 2   

f{sin 2x} . cos x dx

= 2 2

∫

θ 1 − cos 2 θ

0

0

...(1)

f (sin 2 x ) sin xdx

π/2

∫

π/2

∫

=2

[IIT-1987]

∫

3

0

adding (1) and (2), we get

cos 2 x − sin 2 x = dx = cot 2 x −1 dx sin x putting, cot x = sec θ ⇒ –cosec2x dx = sec θ tan θ dθ. We get, sec θ. tan θ I= sec 2 x −1 . dθ − (1 + sec 2 θ) 1 + sec 2 θ sin 2 θ

π/ 2

Then, I =

(cos 2x )1/ 2 dx sin x

sec θ. tan θ

∫

Sol. Let, I =

16

∫

∫

MARCH 2010

Physics Challenging Problems

Set #11

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

So lutions will b e p ub lished in nex t issue 4.

Passage # 1 (Q. No. 1 to 4) For the given phasor diagrams of AC circuit power factor seems to be same i.e. cos φ but the nature of the circuit is entirely different so to distinguish we use the following codes I V φ

Passage # 2 (Q. No. 5 to 7)

A circular platform, 5.0m in radius, has pair of 600Hz sirens, mounted on posts at opposite ends of a diameter. The platform rotates with an angular velocity of 0.80 rad/s. A stationary listener is located at a large distance from the platform. The speed of sound is 350 m/ sec.

φ

V I Phasor Diagram -1 Phasor Diagram -2 For circuit-1 Power factor cosφ (lagging)–Inductive nature For circuit-2 Power factor cosφ (leading)–Capacitive nature For a circuit shown in figure when V = 200 volt and f = 50 Hz then the voltmeter reading is zero. R = 10Ω

V/f

L

~

C

When source frequency get varied then the power 1 1 factor becomes (lagging) and freq. f2 and 2 2 (leading) at frequency f1 then. Which frequency relation(s) is/are correct (A) f > f1 > f2 (B) f < f1 < f2 (C) f1 < f < f2 (D) f2 < f < f1

2.

Value of ∆f = f2 – f1 is 1 R (A) (B) . 2π L 1 2R (D) (C) . 2π L

3.

1 R . 2π 2L 1 4R . 2π R

XtraEdge for IIT-JEE

In situation the longest wavelength reaching the listener from the sirens, in cm, is closest to (A) 58.3 (B) 59.6 (C) 59.0 (D) 57.7

6.

In situation, the highest siren frequency heard by the listener in S.I. units, is closest to (A) 605 (B) 607 (C) 611 (D) 609

7.

In situation, the listener mounts on a bicycle and rides directly away from the platform with a speed of 4.5 m/sec. The highest siren frequency heard by the listener, in SI units, is given by (A) 599 Hz (B) 585.6 Hz (C) 607 Hz (D) 614 Hz

8.

Figure shows a pulse travelling in the x-direction in a string stretched along x-axis. Then y

x 2 3 distance (in meter) Acceleration of particle at x = 1 m is in +ve ydirection Velocity of particle at x = 1m is in-ve y-direction Velocity of particle at x =1 m is zero None of these 1

(A) (B) (C) (D)

Value of watt less current when frequency is f1 (A) 5 2 Amp. (C) 10 Amp.

5.

V

1.

Value of charge on capacitor at frequency f (A) (5π)–1 cb (B) 10π cb (C) 10 cb (D) can not be calculated as the value of C is not known

(B) 10 2 Amp. (D) Zero Amp. 17

MARCH 2010

8

Solution

Physics Challenging Problems Qu e s tio ns we r e Pub lis he d in Feb rua ry I ssu e

[D] As the magnetic field and area vector of the coil are in same direction so the magnetic flux passing through the coil is

1.

4  = 4. − 0 = 8coulomb 2  As 96500 C = 1 Faraday 1 1 8C = × 8 Faraday As =x So 96500 96500 So induced charge = 8x faraday

→ →

φ = B . S . = Bs cos 0 φ = B( πr 2 )(1)

Induced emf. e − N

dφ at

n = 1 single turn coil

d dφ d = − (φ) = − (Bπr 2 ) dt dt dt dB e = −πr 2 As B = B0 + B1 t2 dt dB = 0 + 2B1 t e = − πr 2 .[2B.t ] So dt e = – πr2 2(–2)t = 4π2.t ... (i) e = 4πr2t e 4πr 2 .t as r = Induced current i = = R R i = 4t ... (ii)

[A] The forces working on metallic rod Fm = i L.B. Net force working on rod Fnet = Mg – i. LB vBL .LB = Mg − R vB 2 L2 = Mg − R Fm = i L.B.

5.

e=−

R π

2.

[B]

Induced current i = 4t

3.

[C]

RMS value of current T

1 2 i dt = T

∫

RMS

2

0

2

I =

RMS

Fg = Mg

vB2 L2 R a = acceleration of metallic rod ab vB2 L2 or acceleration = a = g − MR dv v − B 2 L2 =g− = g − kV or acceleration dt MR B 2 L2 Here k = MR dv dv = g − kv ⇒ = dt g − kv dt Integrating on both sides Fnet = Ma = Mg −

So at t = 0 i = 0

2

1 2 i dt ......(i) 2

∫ 0

2

2

4.

[A]

v

∫ 0

1 8 8 .16. = Amp. 2 3 3 2

0

XtraEdge for IIT-JEE

t

dv = dt g − kv

∫ 0

v

− 2

2

0

0

 t2  Induced charge q = idt = 4tdt = 4   2 

∫

b

i

t3   8 As ∫ i dt = ∫ (4t ) dt = 16  = 16. − 0 = 16(8 / 3)  3  3  0 0 0 2

X

a

e e 4πr 2 .t = = Induced electric field E = dist. 2πr 2πr E = 2rt ... (iii) Option (D) is correct, as induced electric field varies linearly with time for given value of radius.

I =

Set # 10

∫

1 1  1 ln(g − kv) = t ⇒ − ln(g − kv) −  −  lng = t k k  k 0

⇒

18

−

g − kv 1 g − kv ln =t⇒ = e − kt k g g

MARCH 2010

v v = e − kt ⇒ k = 1 − e − kt g g

⇒

1− k

and

g v = (1 – e–kt) k

acceleration a =

7.

[

]

8.

[A] Area traversed by rod = x.L

x distance travelled down and in time t

B 2 L2 where k = MR

So, Areal velocity =

[A] Acceleration a = ge–kt

When t = 1/k a = g.e

for t = 1/k

g (1 − e −1 ) = .63 (maximum velocity) k Time after which velocity is 63% of maximum velocity is t = 1/k = MR / B2L2

acceleration a = ge–kt

6.

g (1 − e − kt ) k

v=

dv g g = − e −kt (−k ) = e −kt k dt k

g v = (1 − e −kt ) k

[A] Velocity v =

− k.

1 k

=

= g.e −1 = .37g

dA dt d dx d ( A ) = ( xL) = L. = L.v dt dt dt

g = L. (1 − e − kt ) k

Acceleration a = 37 (maximum acceleration) So time after which acceleration is 37% of maximum

Areal velo.=

2

acceleration is t = 1/k = MR/B2L

gL (1 – e–kt) k

Why can’t the Sun melt Snow? There are some things in nature that have a great capacity to toss back or reflect a great deal of the sun’s light that falls on them. One of them is snow. Newly formed snow reflects about 90 per cent of the sunlight that falls upon it. This means that the sun is powerless to melt clean snow. And when snow does melt, it is not because of the sunlight. Snow does not melt on a spring day because of the sun’s heat. It melts because of the warm air from the sea. After snow becomes ice, a different problem arises. Clean ice absorbs about two-thirds of the sunlight that hits it - but ice is transparent enough for the light to penetrate quite a long way (10 metres or more) before the absorption takes place. It is remarkable what profound results follow from this simple property of transparency to sunlight. If, instead of penetrating deeply, the light were absorbed in a shallow surface layer of ice, the summer sun would quickly raise the temperature of the thin surface layer to melting point. And almost immediately, the water would run off. But when the sunlight penetrates a thick layer of ice before it can be absorbed, it cannot raise the temperature of the ice to melting point quickly enough. When the ice is very cold, the whole summer passes before any melting occurs at all. This is what happens today in the Antarctic, just as it must have happened in northern Europe during an Ice Age. Just imagine, if by magic, ice were suddenly made opaque to light, the glaciers that exist today would melt away in a few years, raising the sea level by 60 metres or more. It would flood at least half the world’s population. Simply amazing how so much depends on so simple a physical property! Clouds toss back about 50 per cent of the light that hits them. Ice and deserts reflect 35 per cent. Land areas are generally a good deal lower in reflectivity - usually 10 to 20 per cent, depending on the nature of vegetation. Oceans, which cover 71 per cent of the Earth’s surface, are least reflective of all - only about three per cent. That is why oceans appear dark in pictures of the Earth taken from the Moon or from artificial satellites. When all the sources of reflection are added together, our planet is found to turn back into space some 36 per cent of the solar radiation falling upon it.

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19

MARCH 2010

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

PHYSICS 1.

Two circular rings A and B, each of radius a = 30 cm, are placed coaxially with their axes vertical s shown in Fig. Distance between centres of these rings is h = 40 cm. Lower ring A has a positive charge of 10 µC, while upper ring B has a negative charge of 20 µC. A particle of mass m = 100gm carrying a positive charge of q = 10 µC is released from rest at the centre of the ring A.

Considering its free body diagram (Fig.),

(i) Calculate initial acceleration of the particle.

(i) to increase its gravitational potential energy by mgh and

ma = F – mg a = 47.6 ms–2

or

Ans.(i)

When particle is released, it starts moving upwards. During its motion, work is done by electric forces acting on it. That work is used in two ways –

(ii) Calculate velocity of particle when it reaches at the centre of upper ring B.

(ii) to provide kinetic energy

–2

(g = 10 ms )

1 mv2 to the particle. 2

But work done by electric field is W = q (V1 – V2) a

B

where V1 is potential at centre of ring A and

h

A

a

Sol. Since, Initially particle was at centre of lower ring A, therefore, no force acts on the particle due to charge of this ring. At initial moment, two forces act on the particle :

(i) its weight mg =0.1 × 10 = 1 Newton (downwards)

V2 is potential at centre of ring B.

V1 =

1 1 q1 + 4πε 0 a 4πε 0

V2 =

1 . 4πε0

q1 a2 + h2

q2 2

a +h

+

2

= – 6 × 104 volt

q 1 . 2 = – 42 × 104 volt 4πε0 a

∴

W = q (V1 – V2) = 3.6 joule

But

W = mgh +

∴

v = 8 ms–1.

1 mv2 2

Ans. (ii)

(ii) force F exerted by the charge on ring B. This force, F =

2.

q 2h 1 q 4πε 0 (a 2 + h 2 ) 3 / 2

where

q2 = – 20 µC (charge on ring B)

∴

F = 5.76 Newton (upwards)

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

Let acceleration of the particle be a. F

C2

C4 C3

ma A

mg

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B

20

C5 + –

MARCH 2010

Sol. In the given circuit capacitors C4 and C5 are in series with each other while capacitor C2 is in parallel with this series combination. Then capacitors C3, C1 and above said combination are in series with each other.

Current density in each plate is j per unit width. Calculate (i) magnetic induction in space between the plates and

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

(ii) force acting per unit area of each plate. Sol. If a large plate carries a current which is uniformly distributed over its width, then a uniform magnetic field is established around it.

If a section of plate, which is normal to the direction of flow of current, is considered then it will be as shown in Fig. P

C2

B

+ –

+ –

q1

– +

q1

+ –

l

Fig. (1) Let magnetic induction of the field induced due to current in one plate be B.

q2

+ –

C5

D

Considering a length l in the section as shown in Fig.(1) and applying Amperes's Circuital Law,

+ –

A

B. 2l = µ0 (lj) or

E

It is given that potential difference (VA – VB) is equal to 11 volt, or

Now consider an elemental width dx in the section of upper plate as shown in Fig.(2). This elemental width is similar to a long straight conductor carrying current di = jdx

(q1 − q 2 ) q 2 q 2 – – = 0 or q2 = 12.6 µC C2 C5 C 4

Potential difference across C5 = 3.

dx ×

q2 = 1.8 volt Ans. C5

System shown in Fig. consists of two large parallel metallic plates carrying current in opposite directions.

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×

×

×

Fig. (2) Magnetic induction at this conductor due to current in lower plate is B =

×

1 µ0j 2

∴ Resultant magnetic field induction between the plates = 2B = µ0j Ans. (i)

11C1C 3 q1 = = 30 µC C1 + C 3

Now applying Kirchhoff's voltage law on mesh BCDEB,

∴ +

B=

But there are two plates which carry equal current but in opposite directions. Therefore, magnetic fields due to these currents, in the space between the plates are unidirectional.

Since terminal A is connected with positive terminal of the battery, therefore, potential of A is higher than that of B.

q q ∴ + 1 + 1 = 11 C 3 C1

×

R

C

q2

E

×

B S

(q1 – q2) C4

C3

×

×

Hence, in steady state charges on different capacitors will be as shown in Fig. C1

Q B

×

21

1 µ0j (leftward) 2

MARCH 2010

∴ Volume (at atmospheric pressure) of air to be pumped in is

Hence, force on this conductor, dF = B di per unit length or

dF =

∆V = V1 – V0 = 10.5 × 10–3 m3

1 µ0j2 dx 2

Volume (at atmospheric pressure) of air pumped in each stroke is

per unit length

v = 500 cm3 = 0.5 × 10–3 m3

But area of unit length of the conductor considered = 1. dx = dx

∴ Number of strokes required

dF ∴ Force per unit area of upper plate = dx =

4.

=

1 µ0j2 Ans. (ii) 2

5.

Tyre of a bicycle has volume 2000 cm3. Initially, the tube is filled to 0.75 of its volume by air at atmospheric pressure of P0 = 105 Nm–2. Area of contact of tyre with road is A = 24 cm2 when total mass of bicycle and its rider is m = 120 kg. Calculate the number of strokes of pump, which delivers v = 500 cm3 volume of air in each stroke, required to inflate the tyre.

= 21

Ans.

y

x

O θ

Sol. Atmospheric pressure, P0 = 105 Nm–2

Air

Increase in pressure (due to weight of bicycle and its rider)

The medium has a variable index of refraction µ (y) given by µ = (0.25 + ky–2)1/2 where k = 1.0 m2. Calculate equation for trajectory of the ray in the medium.

120 × 10 mg = Nm–2 = 5 × 105 Nm–2 −4 A (24 × 10 )

∴ Final pressure of air in the tube,

Sol. If there are two transparent slabs having coefficients of refraction µ1 and µ2 and a ray is incident from air on first slab at angle i then it first refracts at interface of air and first slab as shown in fig.(1) and then at interface of two slabs. Let these angles of refraction be θ and r respectively.

P2 = 105 + (5 × 105) = 6 × 105 Nm–2 Final volume of air in the tube, V2 = 2000 cm3 = 2 × 10–3 m3 Let temperature be T. Finally, number of moles of air in the tube

(µ1)

PV (6 × 10 5 ) × (2 × 10 −3 ) n= 2 2 = RT RT

Volume of these moles at atmospheric pressure,

A

nRT (6 × 10 5 ) × (2 × 10 −3 ) 3 V1 = = m P0 10 5

i

= 12 × 10–3 m3

(µ2)

θ

r

θ

Fig. (1)

Initially, volume of air in tube (at atmospheric pressure) is V0 = 0.75 × 2000 cm3

Applying Snell's law at point A, sin i sin i = µ1 or sin θ = sin θ µ1

= 1.5 × 10–3 m3

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0.5 × 10 3

A ray of light travelling in air is incident at angle θ = 30º on a long rectangular slab of a transparent medium. The point of incidence is origin O of the coordinate system shown in Fig.

Asume that area of contact of tyre with road remains unchanged. (g = 10 ms–2)

=

10.5 × 10 3

22

… (i) MARCH 2010

Refractive index of second slab with respect to first

1 k  sec2 α = 4  + 2  4 y 

µ slab = 2 . µ1

or

Now applying Snell's law at point B,

∴

1 + tan2 α = 1 +

But

k = 1,

But

tan α =

∴

y dy = 2 dx

sin θ µ 2 = sin r µ1 Substituting value of sin θ from equation (1), sin i = µ2. sin r This relation shows that if there are several refracting surfaces parallel to each other then Snell's law can be applied at two points also. In that case i is angle of incidence at one point, r is angle of refraction at the other point and µ is refractive index of that medium in which angle r is measured with respect to that medium in which angle i is measured. Since, refractive index of given medium varies with y, therefore, it may be assumed that the given slab is composed of a large number of thin slabs having different refractive indices and refracting surfaces of all the slabs are normal to y-axis. Hence, angle of incidence and that of refraction are to be measured with y-axis.

∫ or

•

(90 – α)

α

• x

θ

•

Air

Fig. (2) Now consider a point P on trajectory of the refracted ray in medium as shown in fig. (2). Let inclination of tangent to the ray at this point with x-axis be α. Then angle of refraction is (90 – α). Applying Snell's law at points P and O,

•

sin θ µ = sin(90 − α ) 1

•

Substituting values of θ and µ, 1/ 2

1 k  1 =  + 2  2 cos α  4 y 

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therefore, tan α =

2 y

dy dx

y 0

y dy = 2

y2 = 4x

∫

x 0

dx

Ans.

SCIENCE TIPS

•

O

y2

Point O (x = 0, y = 0) and point P (x, y) lie on the trajectory of the ray. Hence, integrating above equation with these limits,

y

P

4k

•

23

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

MARCH 2010

P HYSICS F UNDAMENTAL F OR IIT-J EE

Matter Waves, Photo-electric Effect KEY CONCEPTS & PROBLEM SOLVING STRATEGY

behaves like a wave, while in other circumstances it behaves like a particle. The wave-particle is not the sole monopoly of e.m. waves. Even a material particle in motion according to de Broglie will have a wavelength. The de Broglie wavelength λ of the matter waves is also given by :

Matter Waves : Planck's quantum theory : Wave-particle duality Planck gave quantum theory while explaining the radiation spectrum of a black body. According to Planck's theory, energy is always exchanged in integral multiples of a quanta of light or photon. Each photon has an energy E that depends only on the frequency ν of electromagnetic radiation and is given by :

λ=

Here,

m=

Hence, λ =

hν

hν

v2 c2

hν

×c=

hν h h = = c c/ν λ

......(5) c The left hand side of the above equation involves the particle aspect of photons (momentum) while the right hand side involves the wave aspect (wavelength) and the Planck's constant is the bridge between the two sides. This shows that electromagnetic radiation exhibits a waveparticle duality. In certain circumstances, it 2

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h 2mqV

2mqV

=

12.34 V

Å

Photoelectric effect : When light of suitable frequency (electromagnetic radiation) is allowed to fall on a metal surface, electrons are emitted from the surface. These electrons are known as photoelectrons and the effect is known as photoelectric effect. Photoelectric effect, light energy is converted into electrical energy. Laws of photolectric effect : The kinetic energy of the emitted electron is independent of intensity of incident radiation. But the photoelectric current increases with the increase of intensity of incident radiation. The kinetic energy of the emitted electron depends on the frequency of the incident radiation. It increases with the increase of frequency of incident radiation. If the frequency of the incident radiation is less than a certain value, then photoelectric emission is not possible. This frequency is known as threshold frequency. This threshold frequency varies from emitter to emitter, i.e., depends on the material. There is no time lag between the arrival of light and the emission of photoelectrons, i.e., it is an instantaneous phenomenon.

and v = c c2 ....(4) Hence, m0 = 0 i.e., rest mass of photon is zero, i.e., energy of photon is totally kinetic. The momentum p of each photon is given by : p = mc =

2mK

1 mv2 = qV or mv = 2

.....(3) c2 where m represents the mass of a photon in motion. The velocity v of a photon is equal to that of light, i.e., v = c. According to theory of relativity, the rest mass m0 of a photon is given by : m0 = m 1 −

h

where K is the kinetic energy of the particle. If a particle of mass m kg and charge q coulomb is accelerated from rest through a potential difference of V volt. Then

E = hν .....(1) –34 joule-sec, is Planck's where h = 6.6 × 10 constant. In any interaction, the photon either gives up all of its energy or none of it. From Einstein's mass-energy equivalence principle, we have .....(2) E = mc2 Using equations (1) and (2), we get ; mc2 = hν or m =

h h = = mv p

24

MARCH 2010

(c) According to Einstein's equation, if the frequency of incident radiation is less than certain minimum value, the photoelectric emission is not possible. This frequency is known as threshold frequency. Hence, the frequency of incident radiation below which photoelectric emission is not possible is known as threshold frequency or cut-off frequency. It is given by : hν − (1 / 2)mv 2 ν0 = h On the other hand, if the wavelength of the incident radiation is more than certain critical value, then photoelectric emission is not possible. This wavelength is known as threshold wavelength of cut-off wavelength. It is given by : hc λ0 = [hν − (1 / 2)mv 2 ] (d) Since Einstein treated photoelectric effect as a collision between a photon and an atom, he explained the instantaneous nature of photoelectric effect. Some other important points : Stopping potential : The negative potential applied to the collector in order to prevent the electron from reaching the collector (i.e., to reduce the photoelectric current to zero) is known as stopping potential. 1 eV0 = mv 2max . = hν – W = h(ν – ν0) 2 Millikan measured K.E. of emitted electrons or stopping potentials for different frequencies of incident radiation for a given emitter. He plotted a graph with the frequency on x-axis and stopping potential on y-axis. The graph so obtained was a straight line as shown in figure.

Failure of wave theory :

Wave theory of light could not explain the laws of photoelectric effect. According to wave theory, the kinetic energy of the emitted electrons should increase with the increase of intensity of incident radiation. Kinetic energy of the emitted electron does not depend on the frequency of incident radiation according to wave theory. Wave theory failed to explain the existence of threshold frequency. According to wave theory there must be a time lag between the arrival of light and emission of photoelectrons. Einstein's theory of photoelectric effect :

Einstein explained the laws of photoelectric effect on the basis of Planck's quantum theory of radiation. Einstein treated photoelectric effect as a collision between a photon and an atom in which photon is absorbed by the atom and an electron is emitted. According to law of conservation of energy, hν = hν0 +

1 mv2 2

where hν is the energy of the incident photon; hv0 is the minimum energy required to detach the electron from the atom (work function or ionisation energy) and (1/2) mv2 is the kinetic energy of the emitted electron.

=

V0(stopping potential)

The above equation is known as Einstein's photoelectric equation. Kinetic energy of the emitted electron, 1 mv2 = h(ν – ν0) = hν – W 2

Explanation of laws of photoelectric effect :

(a) The KE of the emitted electron increases with the increase of frequency of incident radiation since W (work function) is constant for a given emitter. KE is directly proportional to (ν – ν0)

Millikan measured the slope of the straight line (=h/e) and calculated the value of Planck's constant. I

(b) Keeping the frequency of incident radiation constant if the intensity of incident light is increased, more photons collide with more atoms and more photoelectrons are emitted. The KE of the emitted electron remains constant since the same photon collides with the same atom (i.e., the nature of the collision does not change). With the increase in the intensity of incident light photoelectric current increases.

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ν0

Frequency of incident light

Full intensity 75% intensity 50% intensity 25% intensity

–

V0

+

Potential difference

25

MARCH 2010

The intercept of V0 versus ν graph on frequency axis is equal to threshold frequency (ν0). From this, the work function (hν0) can be calculated.

(ii) At stopping potential, if the wavelength of the incident light is kept at 4000 Å but the intensity of light is increased two times, will photoelectric current be obtained? Give reasons for your answer.

Graphs in photoelectric effect : (a) Photoelectric current versus potential difference graphs for varying intensity (keeping same metal plate and same frequency of incident light) : These graphs indicate that stopping potential is independent of the intensity and saturation current is directly proportional to the intensity of light. ν2>ν1

I

ν2

ν1

– (V0)2 (V0)1

and

hc = eV2 + W λ2

⇒

 1 1  −  = e(V2 – V1) hc  λ 2 λ1 

or h = +

Potential difference

e(V2 − V1 ) 1.6 × 10 −19 (1.85 − 0.82) =  1  1 1 1  8 − −  3 × 10  e  − 7 − 7 4 × 10   3 × 10  λ 2 λ1 

= 6.592 × 10–34 Js

(b) Photoelectric current versus potential difference graphs for varying frequency (keeping same metal plate and same intensity of incident light) : These graphs indicate that the stopping potential is constant for a given frequency. The stopping potential increases with increase of frequency. The KE of the emitted electrons is proportional to the frequency of incident light.

Stopping potential

hc = eV1 + W λ1

Sol. (i) We have

(ii) No, because the stopping potential depends only on the wavelength of light and not on its intensity. 2.

ν0 A1 A2 A3 Frequency

A small plate of a metal (work function = 1.17 eV) is plated at a distance of 2m from a monochromatic light source of wavelength 4.8 × 10–7 m and power 1.0 watt. The light falls normally on the plate. Find the number of photons striking the metal plate per square metre per second. If a constant magnetic field of strength 10–4 tesla is parallel to the metal surface, find the radius of the largest circular path followed by the emitted photoelectrons.

B1

Sol. Energy of one photon =

B2 B3

= 4.125 × 10–19 J

(c) Stopping potential versus frequency graphs for different metals : These graphs indicate that the stops is same for all metal, since they are parallel straight lines. The slope is a universal constant (=h/e). Further, the threshold frequency varies with emitter since the intercepts on frequency axis are different for different metals.

Number of photons emitted per second =

Solved Examples

4.125 × 10 −19

= 2.424 × 1018

2.424 × 1018 4 × 3.14 × (2)

2

= 4.82 × 1016

Maximum kinetic energy of photoelectrons emitted from the plate

(i) A stopping potential of 0.82 V is required to stop the emission of photoelectrons from the surface of a metal by light of wavelength 4000 Å. For light of wavelength 3000 Å, the stopping potential is 1.85 V. Find the value of Planck's constant.

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1.0

Number of photons striking the plate per square metre per second =

1.

hc 6.6 × 10 −34 × 3 × 108 = λ 4.8 × 10 −7

Emax =

hc –W λ

= 4.125 × 10–19 – 1.17 × 1.6 × 10–19 = 2.253 × 10–19 J 26

MARCH 2010

= 1.374 × 10–19 J = 0.8588 eV (ii) Energy of the photon emitted from a hydrogen atom

A monochromatic light source of frequency ν illuminates a metallic surface and ejects photoelectrons. The photoelectrons having maximum energy are just able to ionize the hydrogen atom in ground state. When the whole experiment is repeated with an incident radiation of frequency (5/6) ν, the photoelectrons so emitted are able to excite the hydrogen atom beam which then emits a radiation of wavelength 1215 Å. Find the work function of the metal and the frequency ν. Sol. In the first case, Emax = Ionization energy = 13.6 eV = 21.76 × 10–19 J 3.

So, hν = 21.76 × 10–19 J In the second case, E'max = =

λ=

....(1)

6.6 × 10 −34 × 3 × 108

5.

−10

5νh = 16.3 × 10–19 + W 6

...(2)

Dividing Eq.(1) by Eq.(2) 21.76 × 10 −19 + W 6 = 5 16.3 × 10 −19 + W Solving, we get W = 11.0 × 10 – 19 J = 6.875 eV

X-rays are produced in an X-ray tube by electrons accelerated through a potential difference of 50.0 kV. An electron makes three collisions in the target before coming to rest and loses half of its kinetic energy in each of the first two collisions. Determine the wavelengths of the resulting photons. Neglect the recoil of the heavy target atoms.

Sol. Initial kinetic energy of the electron = 50.0 keV

Kinetic energy after first collision = 25.0 keV Energy of the photon produced in the first collision, E1 = 50.0 – 25.0 = 25.0 keV Wavelength of this photon

21.76 × 10 −19 + 11.0 ×10 −19

From Eq.(1) ν =

6.62 × 10 −34 × 3 × 108

1.888 × 1.6 × 10 −19 = 6.572 × 10–7m = 6572 Å (iii) Work function of metal W = hν – Emax = 1.8888 – 0.8588 = 1.03 eV

hc λ

1215 × 10 =16.3×10–19 J

So,

1 hc 1 =  −  2 λ 32  2 = 1.888 eV Wavelength of radiation, hν =

6.6 × 10 −34 = 5 × 1015 Hz

λ1 =

6.6 × 10 −34 × 3 × 108 hc = E1 1.6 × 10 −19 × 25.0 × 103

= 0.495 × 10–10 m = 0.495 Å The radiation, emitted when an electron jumps from n = 3 to n = 2 orbit in a hydrogen atom, falls on a metal to produce photoelectrons. The electrons from the metal surface with maximum kinetic energy are made to move perpendicular to a magnetic field of 1/320 T in a radius of 10–3 m. Find (i) the kinetic energy of electrons, (ii) wavelength of radiation and (iii) the work function of metal. Sol. (i) Speed of an electron in the magnetic field, 4.

Kinetic energy of the electron after second collision = 12.5 eV Energy of the photon produced in the second collision, E2 = 25.0 – 12.5 = 12.5 keV Wavelength of this photon λ2 =

Ber m Kinetic energy of electrons

= 0.99 × 10–10 m

v=

= 0.99 Å Kinetic energy of the electron after third collision = 0

1 B2e 2 r 2 Emax = mv2 = 2 2m

Energy of the photon produced in the third collision, E3 = 12.5 – 0 = 12.5 keV

2

(1.6 × 10 −19 ) 2 × (10 −3 ) 2  1  =   ×  320  2 × 9.1× 10 −31

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hc 6.6 × 10 −34 × 3 × 108 = λ2 1.6 × 10 −19 × 25.0 × 103

This is same as E2. Therefore, wavelength of this photon, λ3 = λ2 = 0.99 Å.

27

MARCH 2010

P HYSICS F UNDAMENTAL F OR IIT-J EE

Thermal Expansion, Thermodynamics KEY CONCEPTS & PROBLEM SOLVING STRATEGY

Where ∆t is the loss or gain in time in a time interval t ∆T is change in temperature and d is coefficient of linear expansion. If a rod is heated or cooled but not allowed to expand or contract then the thermal stresses developed F = γα∆T. A If a scale is calibrated at a temperature T1 but used at a temperature T2, then the observed reading will be wrong. In this case the actual reading is given by R = R0(1 + α∆T) Where R0 is the observed reading, R is the actual reading. For difference between two rods to the same at all temperatures l 1α1 = l2α2. Thermodynamics According to first law of thermodynamics q = ∆U + W For an isothermal process (for a gaseous system) (a) The pressure volume relationship is ρV = constt. (b) ∆U = 0 (c) q = W (d) W = 2.303 nRT log10 Vf p = 2.303 nRT log10 i Vi pf

Thermal Expansion : .(a) When the temperature of a substance is increased, it expands. The heat energy which is supplied to the substance is gained by the constituent particles of the substance as its kinetic energy. Because of this the collisions between the constituents particles are accompanied with greater force which increase the distance between the constituent particles.

∆l = lα∆T ; ∆A = Aβ∆T ; ∆V = Vγ∆T or l' = l (1 + α∆T) ; A' = A(1 + β∆T) ; V' = V(1 + γ∆T) (b) Also ρ = ρ'(1 + γ∆T) where ρ' is the density at higher temperature clearly ρ' < ρ for substances which have positive value of γ * β = 2α and γ = 3α Water has negative value of γ for certain temperature range (0º to 4ºC). This means that for that temperature range the volume decreases with increase in temperature. In other words the density increases with increase in temperature. 30 ml 25 ml 20 ml 15 ml 10 ml 5 ml 0 ml If a liquid is kept in a container and the temperature of the system is increased then the volume of the liquid as well as the container increases. The apparent change in volume of the liquid as shown by the scale is ∆Vapp = V(γ – 3α) ∆T Where V is the volume of liquid at lower temperature ∆Vapp is the apparent change in volume γ is the coefficient of cubical expansion of liquid α is the coefficients of linear expansion of the container. Loss or gain in time by a pendulum clock with 1 change in temperature is ∆t = α(∆T) × t 2

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(e) Graphs T2 > T1 P P

V

T2 T1 V T T These lines are called isotherms (parameters at constant temperature) For an adiabatic process (for a gaseous system) (a) The pressure-volume relationship is PVγ = constt. (b) The pressure-volume-temperature relationship is PV = constt. T (c) From (a) and (b) TVγ–I = constt. (d) q = 0 (e) W = –∆U

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(f) ∆U = ncv∆T (g) W =

where cv =

Step 3: Execute the solution as follows: Solve for the target variables. Often you will be given two temperatures and asked to compute ∆T. Or you may be given an initial temperature T0 and asked to find a final temperature corresponding to a given length or volume change. In this case, plan to find ∆T first; then the final temperature is T0 + ∆T. Unit consistency is crucial, as always. L0 and ∆L (or V0 ∆V) must have the same units, and if you use a value or α or β in K–1 or (Cº)–1, then ∆T must be in kelvins or Celsius degrees (Cº). But you can use K and Cº interchangeably. Step 4: Evaluate your answer: Check whether your results make sense. Remember that the sizes of holes in a material expand with temperature just as the same way as any other linear dimension, and the volume of a hole (such as the volume of a container) expands the same way as the corresponding solid shape. Problem solving strategy : Thermodynamics Ist Law Step 1: Identify the relevant concepts : The first law of thermodynamics is the statement of the law of conservation of energy in its most general form. You can apply it to any situation in which you are concerned with changes in the internal energy of a system, with heat flow into or out of a system, and/or with work done by or on a system. Step 2: Set up the problem using the following steps Carefully define what the thermodynamics system is. The first law of thermodynamics focuses on systems that go through thermodynamic processes. Some problems involve processes with more than one step. so make sure that you identify the initial and final state for each step. Identify the known quantities and the target variables. Check whether you have enough equations. The first law, ∆U = Q – W, can be applied just once to each step in a thermodynamic process, so you will often need additional equations. These often

R γ −1

p i Vi − p f Vf nR (Ti − Tf ) = γ −1 γ −1

(h) Graphs P

P

V

T T V Please note that P-V graph line (isotherm) is steeper. For isochoric process (a) P ∝ T (b) W = 0 (c) q = ∆U R (d) ∆U = nCv∆T where Cv = γ −1

(e) Graphs P

P

V

V T For isobaric process (a) V ∝ T (b) W = P∆V = P(Vf – Vi) = nR(Tf – Ti) (c) ∆U = nCv∆T (d) q = nCp∆T (e) Graphs P P V

T

V T T For a cyclic process (a) ∆U = 0 ⇒ q = W (b) Work done is the area enclosed in p-V graph. For any process depicted by P-V diagram, area under the graph represents the word done. Kirchoff's law states that good absorbers are good emitters also. Problem solving Strategy : Thermal Expansion Step 1: Identify the relevant concepts: Decide whether the problem involves changes in length (linear thermal expansion) or in volume (volume thermal expansion) Step 2: Set up the problem using the following steps: Eq. ∆L = αL0∆T for linear expansion and Eq. ∆V = βV0∆T for volume expansion. Identify which quantities in Eq. ∆L = αL0∆T or ∆V = βV0∆T are known and which are the unknown target variables.

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V2

include Eq. W =

∫ p dV

for the work done in a

V1

volume change and the equation of state of the material that makes up the thermodynamic system (for an ideal gas, pV = nRT). Step 3: Execute the solution as follows : You shouldn't be surprised to be told that consistent units are essential. If p is a Pa and V in m3, then W is in joules. Otherwise, you may want to convert the pressure and volume units into units of Pa and m3. If a heat capacity is given in terms of calories, usually the simplest procedure is to convert it to joules. Be especially careful with moles. When you use n = mtot/M to convert 29

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between total mass and number of moles, remember that if mtot is in kilograms, M must be in kilograms per mole. The usual units for M are grams per mole; be careful !

V25 ρ (1 + 75γ ) 5 1 + 75γ = = × 100 4.9 1.0027 V25 × 1.0027 ρ100

or

The internal energy change ∆U in any thermodynamic process or series of processes in independent of the path, whether the substance is an ideal gas or not. This point is of the utmost importance in the problems in this topic. Sometimes you will be given enough information about one path between the given initial and final states to calculate ∆U for that path. Since ∆U is the same for every possible path between the same two states, you can then relate the various energy quantities for other paths. When a process consists of several distinct steps, it often helps to make a chart showing Q, W, and ∆U for each step. Put these quantities for each step on a different line, and arrange them so the Q's, W's, and ∆U's form columns. Then you can apply the first law to each line ; in addition, you can add each column and apply the first law to the sums. Do you see why ? Using above steps, solve for the target variables. Step 4: Evaluate your answer : Check your results for reasonableness. In particular, make sure that each of your answers has the correct algebraic sign. Remember that a positive Q means that heat flows into the system, and that a negative Q means that heat flows into the system, and that a negative Q means that heat flows out of the system. A positive W means that work is done by the system on its environment, while a negative W means that work is done on the system by its environment.

A one litre flask contains some mercury. It is found that at different temperature the volume of air inside the flask remains the same. What is the volume of mercury in flask ? Given that the coefficient of linear expansion of glass = 9 × 10–6(ºC)–1 and coefficient of volume expansion of mercury = 1.8 × 10–4 (ºC–1). Sol. Let V = Volume of the vessel V' = Volume of mercury For unoccupied volume to remain constant increase in volume of mercury should be equal to increase in volume of vessel. V × γg ∴ V' γm∆T = Vγg∆T or V' = γm 2.

∴

V' =

1000 × 27 × 10 −6 1.8 × 10

−4

= 150 cm3

A clock with a metallic pendulum gains 6 seconds each day when the temperature is 20ºC and loses 12 seconds each day when the temperature is 40ºC. Find the coefficient of linear expansion of the metal. Sol. Time taken for one oscillation of the pendulum is L L T = 2π or T2 = 4π2 × .....(1) g g 3.

Partially differentiating, we get ∆L .....(2) 2T∆t = 4π2 × g Dividing (2) by (1), we get 1 ∆T ∆L α L∆t = = = α∆t 2 T 2L 2L where ∆t is the change in temperature. Now, One day = 24 hours = 86400 sec Let t be the temperature at which the clock keeps correct time. At 20ºC, the gain in time is 1 6 = α × (t – 20) × 86400 ....(3) 2 At 40ºC, the loss in time is 1 ...(4) 12 = α× (40 – t) × 86400 2 Dividing (4) by (3), we have 12 40 − t = 6 t − 20 80 ºC. which gives t = 3 Using the value in equation(3), we have 1  80  6= × α ×  − 20  × 86400 2  3 

Solved Examples 1.

A metallic bob weighs 50 g in air. It it is immersed in a liquid at a temperature of 25ºC, it weighs 45 g. When the temperature of the liquid is raised to 100ºC, it weighs 45.1 g. Calculate the coefficient of cubical expansion of the liquid given that the coefficient of linear expansion of the metal is 2 × 10–6(ºC)–1. Sol. Loss in weight in liquid at 25ºC = (50 – 45) = 5 gm Weight of liquid displaced at 25ºC = V25ρ25g ∴ 5 = V25ρ25g ...(1) Similarly, V100ρ100g = 50 – 45.1 = 4.9 gm ...(2) From eq.(1) & (2) we get, V ρ 5 = 25 . 25 4.9 V100 ρ100 Now, V100 = V25(1 + γmetal × 75)= V25(1 + 3αmetal × 75) = V25(1 + 3 × 12 × 10–6 × 75) or V100 = V25(1 + 0.0027) = V25 × 1.0027 Also, ρ25 = ρ100(1 + γ × 75) where, γ = Required coefficient of expansion of the liquid

XtraEdge for IIT-JEE

γ = 3.1 × 10–4 (ºC)–1

which gives α = 2.1 × 10–5 perºC

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A piston can freely move inside a horizontal cylinder closed from both ends. Initially, the piston separates the inside space of the cylinder into two equal parts each of volume V0, in which an ideal gas is contained under the same pressure p0 and at the same temperature. What work has to be performed in order to increase isothermally the volume of one part of gas η times compared to that of the other by slowly moving the piston ? Sol. Let volume of chambers changes by ∆V. According to the problem, the final volume of left chamber is η times final volume of right chamber. ∴ V0 + ∆V = η(V0 – ∆V)  η −1   V0 or ∆V =   η +1 4.

P0

P0

Sol. Let

A1 = Cross section of upper piston A2 = Cross section of lower piston T = Tension in the string P = Gas pressure m1 = Mass of upper piston m2 = Mass of lower piston Now, consider FBD of upper piston P0 A1

P0,v0,T0

P0,v0,T0

m1g PA1 From equilibrium consideration of upper piston we get, P0A1 + T + m1g = PA1 Similarly, consider FBD of lower piston T

As piston is moved slowly therefore, change in kinetic energy is zero. By work-energy theorem, we can write ext Wgas in right chamber + Wgas in left chamber + WAgent = ∆KE

PA2

ext WAgent = (Wgas(R) + Wgas(L))

We know that in isothermal process, work done is given by V  W = nRT ln  f   Vi  ∴ Work done by gas in left chamber (WL)  V + ∆V   2η   = P0V0 ln   = P0V0 ln  0  V  η +1 0  

P0 A2 m2g ∴ P0A2 + T = m2g + PA2 Eliminating T, we get (m1 + m 2 )g P = P0 + A1 − A 2 According to problem m = m1 + m2 and ∆S = A1 – A2 mg ∴ P = P0 + ∆S Now, PV = RT

Similarly, work done by gas in right chamber (WR)  V − ∆V   2η   = P0V0 ln   = P0V0 ln  0  V  η +1 0    2η  ext  – P0V0 ln WAgent = –P0V0 ln   η +1  η +1  = P0V0 ln   4η 

5.

 2η     η +1

2

XtraEdge for IIT-JEE

P∆V = R∆T

But

∆V = (A1 – A2)l = ∆S. l mg   ∆T =  P0 +  ∆S.l ∆S  

∴

A smooth vertical tube having two different sections is open from both ends equipped with two pistons of different areas figure. Each piston slides within a respective tube section. One mole of ideal gas is enclosed between the pistons tied with a nonstretchable thread. The cross-sectional area of the upper piston is ∆S greater than that of the lower one. The combined mass of the two pistons is equal to m. The outside air pressure is P0. By how many kelvins must the gas between the pistons be heated to shift the pistons through l.

or ∆T =

or

l

P∆V R

l

l

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KEY CONCEPT

Organic Chemistry Fundamentals

PURIFICATION OF ORGANIC COMPOUNDS

Qualitative Analysis : Qualitative analysis of an organic compound involves the detection of various elements present in it. The elements commonly present in organic compounds are carbon, hydrogen, oxygen, nitrogen, halogens, sulphur and sometimes phosphorus. Detection of Carbon and Hydrogen : This is done by heating the given organic compound with dry cupric oxide in a hard glass test tube when carbon present is oxidised to carbon dioxide and hydrogen is oxidised to water.

C + N + Na in organic compound

S + 2Na fusion → Na2S from organic compound sodium sulphide If nitrogen and sulphur both are present in any organic compound, sodium thiocyanate (NaSCN) is formed during fusion which in the presence of excess sodium, forms sodium cyanide and sodium sulphide. Na + C + N + S fusion → NaCNS in organic compound sodium thiocyanate Detection of Nitrogen : Take a small quantity of the sodium extract in a test tube. If not alkaline, make it alkaline by adding 2–3 drops of sodium hydroxide (NaOH) solution. To this solution, add 1 mL of freshly prepared solution of ferrous sulphate. Heat the mixture of the two solutions to boiling and then acidify it with dilute sulphuric acid. The appearance of prussion blue or green colouration or precipitate confirms the presence of nitrogen in the given organic compound. Chemistry of the test : The following reactions describe the chemistry of the tests of nitrogen. The carbon and nitrogen present in the organic compound on fusion with sodium metal give sodium cyanide (NaCN). NaCN being ionic salt dissolves in water. So, the sodium extract contains sodium cyanide. Sodium cyanide on reaction with ferrous sulphate gives sodium ferrocyanide. On heating, some of the ferrous salt is oxidised to the ferric salt and this reacts with sodium ferrocyanide to form ferric-ferrocyanide. 6 NaCN + FeSO4 → Na4[Fe(CN)6] + Na2SO4 sodium ferrocyanide 3Na4[Fe(CN)6] + 2Fe2(SO4)3 formed during boiling of the solution → Fe4[Fe(CN)6]3 + 6Na2SO4 prussian blue When nitrogen and sulphur both are present in any organic compound, sodium thiocyanate is formed during fusion. When extracted with water sodium thiocynate goes into the sodium extract and gives blood red colouration with ferric ions due to the formation of ferric thiocyanate

∆ H2O + Cu 2H + CuO → Carbon dioxide turns lime water milky. Ca(OH)2 + CO 2 → CaCO 3 + H2O ( Milky )

Water condenses on the cooler parts of the test tube and turns anhydrous copper sulphate blue. CuSO 4 + 5H2O → CuSO 4 .5H2O ( white )

( Blue)

Detection of Nitrogen, Sulphur and Halogens : Nitrogen, sulphur and halogens in any organic compound are detected by Lassaigne's test. Preparation of Lassaigne's Extract (or sodium extract): A small piece of sodium is gently heated in an ignition tube till it melts. The ignition tube is removed from the flame, about 50–60 mg of the organic compound added and the tube heated strongly for 2–3 minutes to fuse the material inside it. After cooling , the tube is carefully broken in a china dish containing about 20–30 mL of distilled water. The fused material along with the pieces of ignition tube are crushed with the help of a glass rod and the contents of the china dish are boiled for a few minutes. The sodium salts formed in the above reactions (i.e., NaCN, Na2S, NaX or NaSCN) dissolve in water. Excess of sodium, if any, reacts with water to give sodium hydroxide. This alkaline solution is called Lassaigne's extract or sodium extract. The solution is then filtered to remove the insoluble materials and the filtrate is used for making the tests for nitrogen, sulphur and halogens. Reactions : An organic compound containing C, H, N, S, halogens when fused with sodium metal gives the following reactions.

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NaCN sodium cyanide

X(Cl, Br, I) + Nafusion → NaX(X=Cl,Br, I) from organic compound sodium halide

∆ C + 2 CuO → CO2 + 2Cu

( from C )

fusion →

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Na + C +

N

+ S

NaCNS

(b)CS2 layer test for detecting bromine and iodine: Boil a small quantity of sodium extract with dilute HNO3 for 1–2 min and cool the solution. To this solution, add a few drops of carbon disulphide (CS2) and 1–2 mL fresh chlorine water, and shake. Appearance of orange colour in the CS2 layer confirms the presence of bromine, whereas that of a violet/purple colouration confirms the presence of iodine in the compound.

Sod. thiocyanate

from organic

→ Fe(CNS)3 + 3Na+ ferric thiocyanate (blood red) Note : (i) Some compounds like hydrazine (NH2NH2) although contain nitrogen, they do not respond Lassaigne's test because they do not have any carbon and hence NaCN in not formed. (ii) Diazonium salts do not show Lassaigne's test because they are unstable and lose nitrogen as N2 gas on heating. Hence during fusion, no NaCN is formed in Lassaigne's extract due the loss of nitrogen. Detection of Sulphur : The presence of sulphur in any organic compound is detected by using sodium extract as follows: (a) Lead acetate test : Acidify a small portion of sodium extract with acetic acid and add lead acetate solution to it. A black precipitate of lead sulphide indicates the presence of sulphur. 3NaCNS + Fe

3+

2NaBr(aq) + Cl2 CS  2 → Br2 + NaCl(aq) in sodium extract dissolves in CS2 to give orange colour. 2NaI(aq) + Cl2 CS  2 → I2 + 2NaCl(aq) in sodium extract dissolves in CS2 to give purple/violet colour Detection of Phosphorus : In order to detect phosphorus, the organic compound is fused with sodium peroxide, when phosphorus is converted into sodium phosphate.

+

5Na2O2 +

(CH3COO)2Pb + Na2S H→ PbS + 2CH3COONa lead acetate black ppt (b) Sodium nitroprusside test : To a small quantity of sodium extract taken in a test tube, add 2-3 drops of sodium nitroprusside solution. A violet colour indicates the presence of sulphur. This colour fades away slowly on standing.

Fuse  → 2 Na 3 PO 4 + 2Na2O Sod. phosphate

The fused mass is extracted with water and the water is extract is boiled with conc. HNO3. Upon cooling a few drops of ammonium molybdate solution are added. A yellow ppt. of ammonium phosphomolybdate indicates the presence of phosphorus in the organic compound. Na3PO4 + 3HNO3 → H3PO4 + 3NaNO3 H3PO4 + 12(NH4)2MoO4 + 21 HNO3 → ( NH 4 ) 3 PO 4 . 12 MoO3 + 21 NH4NO3 + 12H2O

Na2S + Na2[Fe(CN)5NO] → Na4[Fe(CN)5NOS] sodium nitroprusside violet or purple colour Detection of Halogens : The presence of halogens in any organic compound is detected by using sodium extract (Lassaigne's extract) by silver nitrate test. (a) Silver nitrate test: Sodium extract (or Lassaigne's extract) is boiled with dilute nitric acid to decompose sodium cyanide or sodium sulphide (if present) to hydrogen cyanide and hydrogen sulphide gases, respectively. This solution is cooled and silver nitrate solution added. A white precipitate soluble in ammonia shows chlorine, a yellowish precipitate sparingly soluble in ammonia indicates bromine, and a yellow precipitate insoluble in ammonia shows the presence of iodine in the given organic compound.

( yellow ppt .)

Quantitative Analysis : The quantitative analysis of an organic compound means the estimation of percentage composition of each element present in the organic compound. Estimation of Nitrogen : Duma's Method : Principle : A known mass of the organic substance is heated with excess of copper oxide in an atmosphere of CO2. Carbon, hydrogen and sulphur (if present) are oxidised to CO2, H2O and SO2 while nitrogen is set free. A small amount of nitrogen may be oxidised to oxides but they are reduced back to free nitrogen by passing over a hot reduced copper gauze.

NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3(aq) white precipitate (soluble in ammonia)

∆ Oxides of nitrogen + Cu → CuO + N2 The nitrogen thus formed is collected over conc. KOH solution taken in Schiff's nitrometer tube which absorbs all other gases i.e., CO2, H2O vapours, SO2 etc. The volume of nitrogen collected is converted to STP and from this the precentage of nitrogen can be calculated. % age of Nitrogen Vol. of N 2 at STP 28 = × 100 × 22400 Mass of Substance taken

NaBr(aq) + AgNO3(aq) → AgBr(s) + NaNO3(aq) light yellow ppt. (sparingly soluble in ammonia) NaI(aq) + AgNO3(aq) → AgI(s) + NaNO3(aq) yellow precipitate (insoluble in ammonia)

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2P

( Compound )

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KEY CONCEPT

BORON & CARBON FAMILY

Inorganic Chemistry Fundamentals

As a result of this back donation, the electron deficiency of boron gets compensated and its Lewis acid character decreases.

Boron Trihalides :

The trihalides of boron are electron deficient compounds having a planar structure as shown. They act as Lewis acids because of incomplete octet.

Now, the tendency for back donation is maximum in the case of fluorine due to its small size and more interelectronic repulsions, therefore, it is the least acidic. The tendency of back bonding falls as we move from BF3 to BCl3 and BCl3 to BBr3 due to increase in the size of halogen atoms consequently, the acidic character increase accordingly.

X 120º B

π

X Planar structure of X boron trihalides

BF3 Lewis acid BF3 Lewis acid

+ : NH 3 → Lewis base

+

:F

−

Lewis base

→

F3 B ← Addition product

F

NH3

F π

The acid strength of trihalides decreases as :

Empty 2p-orbital 2p-orbital with lone pair pπ-pπ back bonding

BF3 < BCl3 < BBr3 < BI3 Explanation :

This order of acid strength is reverse of what may normally be expected on the basis of electronegativity of halogens. Since F is most electronegative, hence BF3 should be most electron deficient and thus should be strongest acid. The anomalous behaviour is explained on the basis of tendency of halogen atom to back-donate its electrons to boron atom. For example, in BF3 one of the 2p-orbital of F atom having lone pair overlaps sidewise with the empty 2p-orbital of boron atom to form pπ-pπ back bonding. This is also known as back donation. Further, due to back-π donation of three surrounding fluorine atoms. BF3 can be represented as a resonance hybrid of following three structures. F F

B– = F+

+ F F

B– – F

F + F

Resonating forms of BF3

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F

B

BF4− Fluoroborate ion

B– — F ≡

F F

Acidic nature of H3BO3 or B(OH)3 :

Since B(OH)3 only partially reacts with water to form H3O+ and [B(OH)4]–, it behaves as a weak acid. Thus H3BO3 or (B(OH)3) cannot be titrated satisfactorily with NaOH, as a sharp end point is not obtained. If certain organic polyhydroxy compounds such as glycerol, mannitol or sugars are added to the titration mixture, then B(OH)3 behaves as a strong monobasic acid. It can now be titrated with NaOH, and the end point is detected using phenolphthalein as indicator (indicator changes at pH 8.3 – 10.0). 2B(OH)3 + 2NaOH Na[B(OH)4] + NaBO 2 + 2H 2 O sodium metaborate

–

B —F

The added compound must be a cis-diol, to enhance the acidic properties in this way. (This means that it has OH groups on adjacent carbon atoms in the cis configuration.) The cis-diol forms

Probable hybrid structure

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very stable complexes with the [B(OH)4]– formed by the forward reaction above, thus effectively removing it from solution. The reaction is reversible. Thus removal of one of the products at the right hand side of the equation upsets the balance, and the reaction proceeds completely to the right. Thus all the B(OH)3 reacts with NaOH : in effect it acts as a strong acid in the presence of the cis-diol. – C – OH HO +

OH B

– C – OH HO

–

OH

A convenient laboratory method for the preparation of diborane involves the oxidation of sodium borohydride with iodine. 2NaBH4 + I2 → B2H6 + 2NaI + H2 Diborane is produced on an industrial scale by the reaction of BF3 with sodium hydride. K 2BF3 + 6NaH 450  → B2H6 + 6NaF

– OH HO – C –

–C–O

–2H2O

4BF3 + 3 LiAlH4 → 2B2H6 + 3LiF + 3AlF3

B –C–O

Diborane is a colourless, highly toxic gas with a b.p. of 180 K. Diborane catches fire spontaneously upon exposure to air. It burns in oxygen releasing an enormous amount of energy.

+ OH HO – C –

B2H6 + 3O2 → B2O3 + 3H2O; –C–O –2H2O

O–C–

∆cHΘ = – 1976 kJ mol–1

–

Most of the higher boranes are also spontaneously flammable in air. Boranes are readily hydrolysed by water to give boric acid.

B –C–O

O–C–

B2H6 (g) + 6H2O(l) → 2B(OH)3 (aq) + 6H2 (g)

Borax :

Diborane undergoes cleavage reactions with Lewis bases (L) to give borane adducts, BH3.L

The most common metaborate is borax Na2[B4O5(OH)4] . 8H2O. It is a useful primary standard for titrating against acids.

B2H6 + 2 NMe3 → 2BH3.NMe3 B2H6 + 2 CO → 2BH3.CO

(Na2[B4O5(OH)4] . 8H2O) + 2HCl →

Reaction of ammonia with diborane gives initially B2H6. 2NH3 which is formulated as [BH2(NH3)2]+ [BH4]– ; further heating gives borazine, B3N3H6 known as "inorganic benzene" in view of its ring structure with alternate BH and NH groups.

2NaCl + 4H3BO3 + 5H2O One of the products H3BO3 is itself a weak acid. Thus the indicator used to detect the end point of this reaction must be one that is unaffected by H3BO3. Methyl orange is normally used, which changes in the pH range 3.1 – 4.4.

3B2H6 + 6NH3 → 3[BH2(NH3)2]+ [BH4]–

One mole of borax reacts with two moles of acid. This is because when borax is dissolved in water both B(OH)3 and [B(OH)4]– are formed, but only the [B(OH)4]– reacts with HCl. [B4O5(OH)4]2– + 5H2O

Heat  → 2B3N3H6 + 12H2

The structure of diborane is shown in Fig.(a). The four terminal hydrogen atoms and the two boron atoms lie in one plane. Above and below this plane, there are two bridging hydrogen atoms.

2B(OH)3 + 2[B(OH)4]–

2[B(OH)4]– + 2H3O+ → 2B(OH)3 + 4H2O The last reaction will titrate at pH 9.2, so the indicator must have pKa < 8. Borax is also used as a buffer since its aqueous solution contains equal amounts of weak acid and its salt.

H

97º 134pm

B 120º

B H

119pm

H

H

Fig. (a) The structure of diborane, B2H6

Diborane, B2H6 :

The four terminal B-H bonds are regular two centretwo electron bonds while the two bridge (B-H-B) bonds are different and can be described in terms of three centre-two electron bonds shown in Fig. (b).

The simplest boron hydride known, is diborane. It is prepared by treating boron trifluoride with LiAlH4 in diethyl ether.

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H

H

35

MARCH 2010

Boron also forms a series of hydridoborates; the most important one is the tetrahedral [BH4]– ion. Tetrahydridoborates of several metals are known. Lithium and sodium tetrahydridoborates, also known as borohydrides, are prepared by the reaction of metal hydrides with B2H6 in diethyl ether. 2MH + B2H6 → 2M+ [BH4]– H

H

B H

H

H

Polymerisation continues on both the ends and thus chain increases in length. RSiCl3 on hydrolysis gives a cross linked silicone. The formation can be explained in three steps : Cl

H

(i)

H

OH

R – Si – Cl Cl

Fig. (b) Bonding in diborane. Each B atom uses sp3 hybrids for bonding. Out of the four sp3 hybrids on each B atom, one is without an electron shown in broken lines. The terminal B-H bonds are normal 2-centre-2- electron bonds but the two bridge bonds are 3-centre-2-electron bonds. The 3-centre-2-electron bridge bonds are also referred to as banana bonds.

3H2O –3HCl

R

OH

OH

OH R

R

R

HO – Si – O – Si – O – Si – OH OH

OH

(iii) HO – Si – O – Si – O – Si – OH

These are organosilicon polymers containing Si – O – Si linkages. These are formed by the hydrolysis of alkyl or aryl substituded chlorosilanes and their subsequent polymerisation. The alkyl or aryl substitued chlorosilanes are prepared by the reaction of Grignard reagent and silicon tetrachloride.

OH

R

R

R

Silicones :

OH HO

OH HO

–3H2O

HO – Si – O – Si – O – Si – OH R

R

R R

R

R

– O – Si – O – Si – O – Si – O – O

O

O

– O – Si – O – Si – O – Si – O –

+ SiCl4 → R – SiCl3 + MgCl2

Grignard reagent

R R R Cross linked silicone

2RMgCl + SiCl4 → R2SiCl2 + 2MgCl2

Cyclic (ring) silicones are formed when water is eliminated from the terminal –OH group of linear silicones.

3RMgCl + SiCl4 → R3SiCl + 3MgCl2 R stands for – CH3, –C2H5 or –C6H5 groups Hydrolysis of substituted chlorosilanes yield corresponding silanols which udergo polymerisation.

R

R Si

R OH Si R OH Dialky silandiol

R

O

O

Si

Si

R

R O R3SiCl on hydrolysis forms only a dimer R

Polymerisation of dialkyl silandiol yields linear thermoplastic polymer.

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OH

(ii) HO – Si – OH + H O – Si – OH + H O – Si – OH

OH HO

R Cl H OH –2HCl Si + R Cl H OH

R – Si – OH

R

R

Both LiBH4 and NaBH4 are used as reducing agents in organic synthesis.

RMgCl

R

R

B H

R

R

HO – Si – O – Si – OH

H

H

R

R

H B

B

HO – Si – OH + H O – Si – OH

(M = Li or Na)

H

R

R

R3Si OH + OH Si R3

36

R3Si – O – Si R3

MARCH 2010

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37

MARCH 2010

UNDERSTANDING

Inorganic Chemistry

1.

For SO2(g) at 273 K and 1 atm pressure, the dielectric constant (or relative permittivity) is 1.00993. This molecule has a permanent dipole moment of 1.63 D. Assuming that SO2 behaves as an ideal gas, calculate per mol of (a) total, (b) orientation, (c) induced polarizations, and (d) distortion polarizability. Sol. We have ε εr = = 1.00993 ε0 p = 1.63 D = 1.63(3.3356 × 10–30 Cm) Vm = 22414 cm3 mol–1 at 1 atm and 273 K (a) Total polarization, ε −1 M Ptotal = r εr + 2 ρ

bond length in benzene is 0.14 nm, so that side of the box would be about 0.28 nm. Estimate wavelength for transition from ground state to first excited state of benzene, assuming that it is π-bonding electrons that are involved. Sol. For the one-dimensional box, h2 E= n2 8ml 2 Thus, the ground state energy E1 in a onedimensional box of length 0.14 nm is h2 E1 = 8m(0.14nm) 2 For the two-dimensional square box, h2 (n12 + n 22 ) E2 = 8ml 2 Now since l = 0.28 nm, we have E h2 (n12 + n 22 ) = 1 (n12 + n 22 ) E2 = 2 4 8m(2 × 0.14nm)

1.00993 − 1 × 22.414 cm3 mol–1 1.00993 + 2 = 73.95 cm3 mol–1 (b) Orientation polarization, N  p 2  P0 = A  3ε 0  3kT 

=

The various energy levels are as follows. n1 1 2 1 2

 {6.023 × 10 23 mol −1}  =    3(8.854 × 10 −12 C 2 N −1m − 2   (1.63 × 3.3356 × 10 −30 Cm) 2     3(1.38 × 10 − 23 J K −1 )(273 K )  = 59.31 × 10–6 m3 mol–1 = 59.31 cm3 mol–1 (c) Induced polarization, Pind = Ptotal – P0 = 73.95 cm3 mol–1 – 59.31 cm3 mol–1 = 14.64 cm3 mol–1 (d) Distortion polarizability, Pind αd = (1 / 3ε 0 ) N A

=

2.

E2 E1/2 (5/4)E1 (5/4)E1 2E1

}

degenerate

The first three energy levels will be doubly occupied in the ground state and hence the first excited state is obtained when the electron is promoted from n1 = 1, n2 = 2 state to n1 = 2, n2 = 2 state. Thus 5 3 ∆E = 2E1 – E1 = E1 4 4 Since the wavelength is inversely proportional to energy, the corresponding wavelength would be (4/3)λ, i.e. 4 × 70 nm = 93 nm 3

14.64 × 10 –6 m 3 mol −1 {1 /(3 × 8.854 × 10 −12 C 2 N −1m − 2 )}(6.023 × 10 23 mol −1 ) = 6.46 × 10–40 C2 N–1 m

3.

An electron confined to a one-dimensional box of length 0.14 nm has a ground-state energy corresponding to the radiation of wavelength about 70 nm. Benzene, as a rough approximation, may be considered to be a two-dimensional box that encompasses the regular hexagonal shape. The C—C

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n2 1 1 2 2

38

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.

MARCH 2010

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

(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

( B)

CH3 – CH – CH2 + 2NaNH2 Br

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 (E)

CH2OOCC17H35

CH2OH

CHOOCC17H35 + 3NaOH CH2OOCC17H35

∆

So lub le

Ag(NH3)2Cl + 2HNO3 → AgCl ↓ + 2NH4NO3

CH2OH + 3C17H35COONa (soap) CH2OH

white ppt . ( B)

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

(v) (F) is sodium peroxide as only peroxides gives H2O2 on reaction with dil. acids. K 2 Na 2 O 670  → Na 2 O 2 + 2 Na ( B)

∆

( F)

Re d gas

(A)

CrO2Cl2 + 4NaOH → Na 2 CrO 4

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

+ 2NaCl + 2H2O

Yellow solution ( C )

dil.

Na2CrO4 + (CH3COO)2Pb → PbCrO 4 + 2CH3COONa

(F) gives the following oxidations : Cr(OH)3 + 5OH– → CrO42– + 4H2O + 3e– Mn2+ + 8OH– → MnO4– + 4H2O + 5e– S2– + 8OH– → SO42– + 4H2O + 8e–

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Br

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 (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

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

( B)

CH3 – C ≡ CH Propyne

4.

NH3 + HCl → NH4Cl

( B)

∆

+ 2NaBr + 2NH3

( D)

( B)

(E)

(G )

Yellow ppt .

(ii) Na2B4O7 . 10H2O ∆

39

Na2B4O7 + 10H2O

MARCH 2010

Na2B4O7 ∆

Transparent bead

NiO + B2O3 ∆

B

Ni(BO 2 ) 2 [Oxidising flame]

Nickel meta borate ( Brown )

Ni(BO2)2 + C

∆

Hb

Ht

2 NaBO 2 + B 2 O 3 144 42444 3

97º 1.33Å

Ht B

121.5º

Ht 1.19Å

Hb Ht 1.77Å Thus, the diborane molecule has four two-centre-two electron bonds (2c-2e– bonds) also called usual bonds and two three-centre-two-electron bonds (3c-2e– bonds) also called banana bonds. Hydrogen attached to usual and banana bonds are called Ht (terminal H) and Hb (bridged H) respectively.

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)

TRUE OR FALSE

+ CO2 + H2O ∆ Ni 2 O 3 ↓ 2NiCO3 + 4NaOH + [O]

1.

Black ppt . ( D)

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

Green ppt . (E)

Ni(CN)2 + 2KCN → K 2 [ Ni(CN ) 4 ]

2.

( F)

NaOH + Br2 → NaOBr + HBr

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

3.

(D)

4.

Compound (X) on reduction with LiAlH4 gives a hydride (Y) containing 21.72% hydrogen along with other products. The compound (Y) reacts with air explosively resulting in boron trioxide. Identify (X) and (Y). Give balanced reactions involved in the formation of (Y) and its reaction with air. Draw the structure of (Y). Sol. Since B2O3 is formed by reaction of (Y) with air, (Y) therefore should be B2H6 in which % of hydrogen is 21.72. The compound (X) on reduction with LiAlH4 gives B2H6. Thus it is boron trihalide. The reactions are shown as: 4BX 3 + 3LiAlH4 → 2B 2 H 6 + 3LiX + 3AlX3

5.

(X)

5. 6.

(X = Cl or Br) B 2 H 6 + 3O2 → B2O3 + 3H2O + heat (Y)

Structure of B2H6 is as follows: Hb Ht B

B Ht

Ht

or

5 2 (sin 2πt + cos 2πt). Their amplitudes are in the ratio 1 : 2. When a dielectric is introduced between the plates of a capacitor at a constant potential difference, the charge on the plates remains unchanged. Heat can never be converted completely into work. The workings of a triode as an amplifier and a step-up transformer are same. A 60 dB sound has twice the intensity of a 30 dB sound. Two identical spherical air bubbles, one formed, inside water of a tank and the other outside the water, have equal pressure inside them.

Sol. 1. [True] 2. [False] On introduction of dielectric the capacitance increases. Potential difference remaining constant, charge increases, because Q = CV. 3. [False] In an isothermal expansion of an ideal gas, heat can be completely converted into work. 4. [False] A step-up transformer increases the voltage but not the power. A triode increases both the voltage and the power. 5. [False] 6. [False]

(Y)

Ht

Two simple harmonic motion are represented by π  the equation x1 = 5 sin  2πt +  and x2 = 4 

Hb

XtraEdge for IIT-JEE

40

MARCH 2010

Set

`tà{xÅtà|vtÄ V{tÄÄxÇzxá 11 This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety of possible twists and turns of problems in mathematics that would be very helpful in facing IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and we hope that this section would prove a rich resource for practicing challenging problems and enhancing the preparation level of IIT JEE aspirants. By : Shailendra Maheshwari So lu t ion s wi l l b e p ub lished in nex t issue Joint Director Academics, Career Point, Kota 1.

For

complex

numbers

z1

=

x1

+ iy1

and

6.

z2 = x2 + iy2 we write z1 ∩ z2, if x1 ≤ x2 and

2.

3.

1− z ∩ 0, Justify the result. 1+ z

7.

AP and BQ are fixed parallel tangents to a circle, and a tangent at any point c cuts them at P and Q respectively. Show that CP.CQ is independent of the position of c on the circle.

|f(x)| ≤ 5/4 and |g(x)| ≤ 2 A straight line is drawn throguh the origin and parallel to the tangent to the curve  a + a 2 − y2 ln   y 

a

  at an arbitrary point M. Show that  

n

Show that

∑

(−2) r

r =0

n

Cr Cr

tan −1 x dx , then expression In in terms

2 + f ( x ) + f ( y) x+y If f   = 3  3 

g(f(x)) is (A) Many one into (B) Many one onto (C) One one on to (D) One one into

9.

f(g(x)) = p has (A) no real root (B) at least one real root (C) infinity many roots (D) exactly one real root

10. g(f(x)) is – (A) non periodic function (B) odd function (C) even function (D) None of these

r+2

 1 , If n is even  =  n +1 1  , If n is odd n + 2

XtraEdge for IIT-JEE

n

8.

=

the locus of the points P of intersection of this straight line and the straight line parallel to the x-axis and passing through the point M is a circle. 5.

0

Passage : Let Z denotes the set of integers. Let p be a prime number and let z1 ≡ {0, 1}. Let f : z → z and g : z → z1 are two functions defined as follows : f(n) = pn; if n ∈ z and g(n) = 1; if n is a perfect square = 0, otherwise.

Let f(x) = ax2 + bx + c & g(x) = cx2 + bx + a, such

x + a 2 − y2

∫x

for all real x and y. If f ´(2) = 2, then f(2) is -

that |f(0)| ≤ 1, |f(1)| ≤ 1 and |f(–1)| ≤ 1. prove that

4.

1

of In–2.

y1 ≤ y2. The for all complex numbers z with 1 ∩ z, we have

Let In =

41

MARCH 2010

MATHEMATICAL CHALLENGES SOLUTION FOR FEBRUARY ISSUE (SET # 10)

1.

g(x) = sin x ; 0 ≤ x < π/2 1 ; π/2 ≤ x ≤ π sin2 x/2 ; π θ θ2 so f(θ) ↓ so f(x) < f(sinx) as sin x < x

use it in line AD .  b c  b | c | +c | b | | b || c | = pt D : + .  | b | + | c |  | b | | c |  |b|+|c| which divides BC in ratio of |c| : |b| similary use eq. of external angle bisector line AE  b c   − ⇒ r = p   | b| | c| solve it with BC to find pt. E.

6.5 = 15 2 (ii) coeff. of x4 in (1 – x)–6 9.8.7.6 = 126 = 4 + 6 – 1C6 – 1 = 9C5 = 4.3.2 (iii) select 3 different flavours : 6C3 ways choose (at least one from each) 4 cones : 4–1 C3 – 1 = 3C2 = 3 ways 6.5.4 so required ways = 6C3 × 3 = × 3 = 60 3.2 (iv) Select 2 different flavours : 6C2 ways choose (at least one from each) 4 cones ; 4–1 C2 – 1 = 3C1 = 3 so required ways (either 2 or 3 different flavours) 6.5 × 3 = 105 = 60 + 6C2 3 = 60 + 2 Let A at origin & P.V. of B & C are b & c.  b c   + So line AD ⇒ r = t  | b | | c |   & line BC ⇒ r = b + ∆ ( b – c ) solve them together to find pt. D  b c   = b +s(b – c) t  +  | b| | c|

(i) 6C4 =

XtraEdge for IIT-JEE

B

5.

Consider eix(1 + eix)n = eix [1 + nC1eix + nC2ei2x + .... + nCneinx]  n+2  i x 2 

e n

. 2cosn

x = eix + nC1ei2x + nC2ei3x +... + 2

Cnei(n+ 1)x Compare real parts & get (a) Compare imag. parts & get (b) 6.

Let Ei = the event that originator will not receive a letter in the ith stage. Originator sands letters to two persons so in 1st stage he will not get letter. Prob. that letter sent by 1st received is not received n −2 C (n − 2)(n − 3) n − 3 = by originator is n −1 2 = (n − 1)(n − 2) n −1 C1 similarly prob. that letter sent by 2nd receipiant is not n −3 received by originator is n −1 so P(E2) = prob. that originator not received letter in 2

 n −3 2nd stage is =   .  n −1 

42

MARCH 2010

similarly P(E3) = prob. that originator not receive letter sent by the four person getting letters from two recipients is 4

 n −3  n −3  n −3  n −3  n −3  n −3    .  .  .  =  =  n −1   n −1   n −1   n −1   n −1   n −1  8

 n −3  n −3 P(E4) =   =    n −1   n −1 

Similarly,

π/ 2 1 1 1 – + – sin 2θ sec θdθ 0 7 5 3 29 181 + 2(cos θ) 0π / 2 – 0 dθ = – = 105 105

∫

∫

22

9.

23

2 k –1

 n −3 Similarly, P(Ek) =    n −1  So the required prob. is P(E) = prob. the originator not receive letter in 1st k stages = P(E1) . P(E2) . ........ P(Ek)  n −3 =    n −1   n −3 =    n −1 

7.

y = f(x) = y´ =

∫

x

0

∫

x

0

2 + 2 2 + 2 3 + ....2 k −1

2.

2 k −1 −1 2 −1

 n −3 =   n −1 

2

e zx − z dz =

∫

∫

x

0

2

 4 x + 3y + m   3x − 4 y + l   = a   5 5     2 (3x – 4y + l) – 5a(4x + 3y + m) = 0 9x2 – 24xy + 16y2 + (6l – 20a)x + (–8l – 15a)y + (l2 – 5am) = 0 comp. it with given equation. 6l – 20a = –18 ...(1) ⇒ 24l – 80a = –72 –8l –15a = –101 ...(2) ⇒ –24l – 45a = –303 From (1) & (2) ⇒ 125a = –375 ⇒ a=3

( 2 k − 2)

2

e zx .e − z dz

1 2

e zx .e − z dz + 1 = –

1 2 = – (e z .e zx ) 0x − 2 dy 1 – xy = 1 dx 2

x

0

2

∫

x

0

2

e zx (−2ze − z ) dz + 1

1  xe − z .e zx dz  + 1 = xy + 1 2  2

− x / 2 dx 2 = e−x / 4 I.F. = e ∫

Sol is y . e y = ex 8.

2

/4

∫

x

0

−x2 / 4

e −z

2

=

/4

∫e

−x2 / 4

∫

dx =

x

0

e

−z2 / 4

10. circle : (x – 1)2 + (y – 1)2 = 1 ⇒ x2 + y2 – 2x – 2y + 1 = 0

dz

dz. (0,1)B

∫ sin n θ sec θ dθ = ∫ sin (n –1 + 1) θ sec θ dθ = ∫ sin (n – 1)θ + cos (n – 1)θ sin θ sec θ ) dθ = ∫ sin (n – 1)θ + [ sin (n – 1)θ cos θ – sin (n – 2)θ sec θ ] dθ = ∫ (2 sin (n – 1)θ – sin (n – 2)θ sec θ ) dθ 2 cos(n − 1)θ =– – ∫ sin (n – 2)θ secθ dθ n −1 1 π 2 sin 8θ − sin 2θ = dθ 2 0 cos θ 1  2   − cos 7θ  − 2  7 0 π/2

∫

π2

0

A(1,0)

Let the line be y = mx 1 Altitude of ∆ = 1+ m2 For DE length : solve line with circle. x2 + m2x2 – 2x – 2mx + 1 = 0 (1 + m2)x2 – 2(1 + m)x + 1 = 0

sin 6θ sec θdθ

|x1 – x2| = −

1 2 2 2 − − (cos 3θ) 0π / 2 −  2 7 5 3

∫

π/2

0

∫

π2

0

 sin 2θ sec θdθ 

=

sin 2θ sec θdθ −

XtraEdge for IIT-JEE

E

D

∫

=

9x2 – 24xy + 16y2 – 18x – 101y + 19 = 0] (3x – 4y)2 = 18x + 101y – 19. Let the vertex of the parabola be A(α, β). Shift origin to A and y-axis along the tangent at vertex (3x – 4y + l) . So the axis of parabola be 4x + 3y + m = 0 (along x axis) If L.R. of parabola be a then it’s equation is

∫

π2

0

|DE| =

 sin 2θ sec θdθ 

43

( x1 + x 2 ) 2 − 4x1x 2

4(1 + m) 2 2 1 −4 = 2 2 2 (1 + m ) 1+ m2 1+ m x 12 + 1 |x1 – x2| = 2

2m

2m 1+ m2

MARCH 2010

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

MATHS 1.

A ray of light is coming along the line L = 0 and strikes the plane mirror kept along the plane p = 0 at x−2 z−6 y −1 = = and B. It is given that L = 0 is 3 5 4 p = 0 is x + y – 2z = 3, then find the co-ordinates of B and the equation of line along reflected ray. x−2 z−6 y −1 Sol. let = = =λ 3 5 4 ⇒ x = 2 + 3λ, y = 1 + 4λ, z = 6 + 5λ lies on plane x + y – 2z = 3 (2 + 3λ) + (1 + 4λ) – 2(6 + 5λ) = 3 ⇒ λ = –4 point B ≡ (–10, –15, –14) Let equation of reflected ray L1 = 0 is line joining Q(x2, y2, z2) and B(–10, –15, –14) x + 10 y + 15 z + 14 = = i.e. 16 20 12 2.

Let f be a polynomial function such that f(x) . f(y) + 2 = f(x) + f(y) + f(xy) ∀ x ∈ R+, y ∈ R+ ∪ {0} and f(x) is one one ∀ x ∈ R+ with f(0) = 1, f´(1) = 2, then find the area bounded between the curve y = x2 and y = g(x) where g(x) = 2 and x-axis and also find the no. of points of f (x) nondifferentiability of h(x) = min {g(x), x2, |1 – |x||} Sol. Let f(x) . f(y) + 2 = f(x) + f(y) + f(xy) ....(1) putting x = y = 1 f(1) . f(1) + 2 = 3f(1) ⇒ f(1) = 2, 1 f is given one-one and f(0) = 1 ⇒ f(1) = 2 1 in (1) than replacing y by x 1 1 f(x) . f   + 2 = f(x) + f   + 2 x x 3.

⇒ f(x) = 1 + xn also f´(1) = 2 ⇒ n = 2 ⇒ f(x) = 1 + x2 Now to find the area, 1 2 2   2π 1  − x 2  dx = 2  −  =π– Area = 2  01+ x2 3 4 3    clearly by graph you can find there is 6 points of non differentiability.

Six points (xi. yi), i = 1, 2, ...6 on the circle x2 + y2 = 4 6

such that

∑

6

x i = 8 and

i =1

∑y

i

= 4. The line segment

∫

i =1

joining orthocentre of a ∆ made by these points and the centroid of the ∆ made by other three points passes through a fixed point, find that point. 6

Sol. let

∑

6

x i = α and

i =1

∑y

i

If f(x) = x3 + ax2 + bx + c = 0 has three distinct integral roots and (x2 + 2x + 2)3 + a(x2 + 2x + 2)2 + b(x2 + 2x + 2) + c = 0 has no real roots then find the values of a, b and c Sol. x3 + ax2 + bx + c = 0 has three distinct integral roots and (f(x))3 + a(f(x))2 + b(f(x)) + c = 0 has no real roots, where f(x) = x2 + 2x + 2 Let the roots of x3 + ax2 + bx + c = 0 be x1 > x2 > x3 respectively. Since f(x) can take all values from [1, ∞]. ⇒ x1 ≤ 0 x2 ≤ –1 x3 ≤ –2 ⇒ a = –(x1 + x2 + x3) ≥ 3 b = x1x2 + x2x3 + x3x4 ⇒ b≥2 and c = –(x1x2x3) ⇒ c≥0 4.

=β

i =1

let θ be the ortho centre of the ∆ made by (xi, yi), i = 1, 2, 3 ⇒ 0 is (x1 + x2 + x3, y1 + y2 + y3) and G be the centroid of the ∆ made by (xi, yi), i = 4, 5, 6  x + x 5 + x 6 y 4 + y5 + y6  ⇒ G is  4 ,  3 3    α − α1 β − β1  ⇒ G is  ,  3   3 α β Here you can say the point  ,  divides to the OG  4 4 in 3 : 1.

XtraEdge for IIT-JEE

44

MARCH 2010

5.

Let A, B, and C are points represented by complex No. Z1, Z2, Z3. If the circum centre of the ∆ABC is at the origin and the altitude AD of the triangle meets the circumcircle again at P, then prove that P − Z 2 Z3 . represents the complex number Z1 Sol. Let circumcentre is point O. A Z1

x  x   x  r. n C r  1 +   =n 1− x  1− x   1− x  r =0

∑ n

⇒

∑r . C y 2 n

O

∑

n

C(Z3)

Do you know

C r (r – nx)2 . xr(1 – x)n – r;

x ≠ 0, 1, n ∈ N > 2 Sol.

∑

n

2

r

C r (r – nx) . x (1 – x)

n

∑

r

n

r =0

 x  n C r (r2 + n2x2 – 2nxr)   (1 – x) 1− x 

r r n  n  x   x  2 2 n = (1 – x)n  r 2 n C r  Cr   +n x   1− x   1− x   r =0 r =0 r n  x   − 2nx r. n C r     1 − x   r =0

∑

∑

∑

n

we know that

∑ r.

n

C r y r = (1 + y)r

...(1)

r =0

n

r

∑

∑ r.

n

∴

The human eye blinks an average of 4,200,000 times a year.

•

Skylab, the first American space station, fell to the earth in thousands of pieces in 1979. Thankfully most over the ocean.

•

It takes approximately 12 hours for food to entirely digest.

•

Human jaw muscles can generate a force of 200 pounds (90.8 kilograms) on the molars.

•

The Skylab astronauts grew 1.5 - 2.25 inches (3.8 - 5.7 centimeters) due to spinal lengthening and straightening as a result of zero gravity.

•

An inch (2.5 centimeters) of rain water is equivalent to 15 inches (38.1 centimeters) of dry, powdery snow.

•

Tremendous erosion at the base of Niagara Falls (USA) undermines the shale cliffs and as a result the falls have receded approximately 7 miles over the last 10,000 years.

•

40 to 50 percent of body heat can be lost through the head (no hat) as a result of its extensive circulatory network.

•

A large swarm of desert locusts (Schistocerca gregaria) can consume 20,000 tons (18,160,000 kilograms) of vegetation a day.

C r y r −1 = n(1 + y)n – 1

r =0 n

•

n

n

⇒

The largest meteorite crater in the world is in Winslow, Arizona. It is 4,150 feet across and 150 feet deep.

n

x    x  –n Cr   = 1 +  = (1 – x) 1 x 1 − x −     r =0 Differentiating (1) w.r.t. y we get

⇒

•

n–r

r =0

=

∑ r.

n

r

C r y = ny (1 + y)

n–1

....(2)

r =0

XtraEdge for IIT-JEE

∑

r

x  x    nx + 1 1 +   1− x  1− x  1− x  = nx(nx + 1 – x) (1 – x)–n given sum is equal to (1 – x)n {nx(nx + 1 – x) (1 – x)–n + n2x2(1 – x)–n – 2nx . nx (1 – x)–n} 2 2 2 2 = nx(nx + 1 – x) + n x – 2n x = nx(1 – x)

r =0

n

 x  r 2 .n C r    1− x  r =0

n −2

D

n

= n(1 + y)n–2{y(n – 1) + (1 + y)}

=n.

P (Z) ∠POC = π – 2C and ∠BOA = 2C Now applying coni's method ...(1) Z3 = Zei(π –2C) ....(2) Z2 = Z1 ei 2 C Multiplying (1) and (2) Z3Z2 = ZZ1 eiπ = –ZZ1 −Z 2 Z3 ⇒ Z=– Z1

Evaluate :

r −1

r =0

π–2C

6.

r

= ny(1 + y)n–2(ny + 1) n

B(Z2)

n −1

= nx(1 – x)–n Differentiating (2) w.r.t. y we get

⇒ 2C

r

n

⇒

45

MARCH 2010

MATHS

DEFINITE INTEGRALS & AREA UNDER CURVES Mathematics Fundamentals

Properties 1 :

Every continuous function defined on [a, b] is integrable over [a, b]. Every monotonic function defined on [a, b] is integrable over [a, b] If f(x) is a continuous function defined on [a, b], then there exists c ∈ (a, b)such that

∫ f (x) dx = F(x), then

If

∫

b

a

f ( x ) dx = F(b) – F(a), b ≥ a

Where F(x) is one of the antiderivatives of the function f(x), i.e. F´(x) = f(x) (a ≤ x ≤ b). Remark : When evaluating integrals with the help of the above formula, the students should keep in mind the condition for its legitimate use. This formula is used to compute the definite integral of a function continuous on the interval [a, b] only when the equality F´(x) = f(x) is fulfilled in the whole interval [a, b], where F(x) is antiderivative of the function f(x). In particular, the antiderivative must be a function continuous on the whole interval [a, b]. A discontinuous function used as an antiderivative will lead to wrong result. If F(x) =

∫

x

a

∫

a

b 1 f ( x ) dx is called the (b − a ) a mean value of the function f(x) on the interval [a, b]. If f is continous on [a, b], then the integral function g

∫

b

a

f ( x ) dx =

∫

b

a

b

b

a

∫

b

a

f ( x ) dx =

∫

m(b – a) ≤

a

a

0

0

d   dx 

∫

b

c

∫

b

a

f ( x ) dx =

∫

b

a

∫ ∫

2a

0

∫

b

a

 2 f ( x ) dx =   0

∫

a

XtraEdge for IIT-JEE

∫

if f(–x) = – f ( x )

a

∫

b

a

f ( x ) dx ≤ M(b – a)

∫

ψ( x)

φ( x )

 f ( t ) dt  = f(ψ(x)) ψ´(x) – f(φ(x)) φ´(x) 

f ( x ) dx ≤

∫

b

a

| f ( x ) | dx

 f ( x )g ( x ) dx ≤  

∫

b

a

1/ 2

 f 2 ( x ) dx  

  

∫

b

a

1/ 2

 g 2 ( x ) dx  

Change of variables : If the function f(x) is continuous on [a, b] and the function x = φ(t) is continuously differentiable on the interval [t1, t2] and a = φ(t1), b = φ(t2), then

f ( x ) dx if f(–x) = f(x)

b

b

∫

f (a + b − x ) dx

 2 f ( x ) dx =  −a  0 a

f ( t ) dt for x ∈ [a, b] is

If f2(x) and g2(x) are integrable on [a, b], then

f ( x ) dx, a < c < b

∫ f (x) dx = ∫ f (a − x) dx or

b

∫

a

f ( x ) dx +

a

If the function φ(x) and ψ(x) are defined on [a, b] and differentiable at a point x ∈ (a, b) and f(t) is continuous for φ(a) ≤ t ≤ ψ(b), then

f ( x ) dx ≥ 0

f ( x ) dx

c

a

a

∫

x

derivable on [a, b] and g´(x) = f(x) for all x ∈ [a, b]. If m and M are the smallest and greatest values of a function f(x) on an interval [a, b], then

f ( t ) dt

a

∫ f (x) dx = – ∫

∫

∫

defined by g(x) =

Properties of Definite Integrals : b

f ( x ) dx = f(c) . (b – a)

The number f(c) =

f ( t ) dt, t ≥ a, then F´(x) = f(x)

If f(x) ≥ 0 on the interval [a, b], then

b

b

a

f ( x ) dx =

∫

t2

t1

f (φ( t )) φ´(t) dt

Let a function f(x, α) be continuous for a ≤ x ≤ b and c ≤ α ≤ d. Then for any α ∈ [c, d], if

f ( x ) dx if f(2a – x) = f(x) if f(2a – x) = – f ( x )

I(α) = 46

∫

b

a

f ( x , α) dx, then I´(α) =

∫

b

a

f ´(x , α) dx,

MARCH 2010

Where I´(α) is the derivative of I(α) w.r.t. α and f´(x, α) is the derivative of f(x, α) w.r.t. α, kepping x constant.

2  n −1 n − 3 n − 5  n . n − 2 . n − 4 ..... 3 In =  n −1 n − 3 n − 5 1 π  ..... . . . 2 2  n n−2 n−4

Integrals with Infinite Limits :

If a function f(x) is continuous for a ≤ x < ∞, then by definition

∫

∞

a

f ( x ) dx = lim

b

∫

f ( x ) dx

b →∞ a

If In =

∫ ∫

∞

−∞

−∞

∫

b

a →−∞ a

f ( x ) dx =

∫

a

−∞

d dx

f ( x ) dx and

f ( x ) dx +

∫

∞

a

a

0

0

f ( x ) dx

=

1 a 2

a

∫ x f (x) 0

log sin x dx =

∫

π/ 2

∫

∞

0

e −ax cos bx dx = e −ax sin bx dx =

e −ax xndx =

If In =

0

r =0

0

1   1  cosα + (n − 1)β sin  nβ  2   2  cos(α + rβ) = 1  r =0 sin  β  2 

log cos x dx

n −1

∑ 1

π

∫

π/ 2

0

1

–

12 1

+

22

+

2

1

+

2

1 32 1 2

– .... =

π2 12

+ .... =

π2 6

1 2 3 Area under Curves : Area bounded by the curve y = f(x), the x-axis and the ordinates x = a, x = b

=

Reduction Formulae of some Define Integrals :

0

1

∑

a

∫

∞

r 1

∑ f  n  . n = ∫ f (x) dx

1   1  sin α + (n − 1)β sin  nβ  2   2  sin(α + rβ) = 1  r =0 sin  β  2 

 m +1  n +1 Γ  Γ  π/ 2 2   2  sin m x cos n x dx =  0 m+n+2 2Γ  2  

∫

d d v(x) – f{u(x) u(x) dx dx

n −1

If m and n are non-negative integers, then

0

u(x)

n →∞

if f(a – x) = f(x)

1 Γ(n + 1) = n Γ (n), Γ(1) = 1, Γ   = 2

∞

f ( t ) dt = f{v(x)}

n −1

π π 1 = – log 2 = log 2 2 2

∫

v( x )

∫

lim

f (x) a and dx = 0 f ( x ) + f (a − x ) 2

∫

( when n is even)

Some Important Results :

∫ x f (x) dx π/ 2

( when n is odd )

Summation of Series by Integration :

properties :

∫

cos n x dx , then

Leibnitz's Rule : If f(x) is continuous and u(x), v(x) are differentiable functions in the interval [a, b], then

Geometrically, the improper integral (i) for f(x) > 0, is the area of the figure bounded by the graph of the function y = f(x), the straight line x = a and the xaxis. Similarly, f ( x ) dx = lim

0

( when n is even)

2  n −1 n − 3 n − 5  n . n − 2 . n − 4 ..... 3 Im =  1 π n −1 n − 3 n − 5  ..... . . . 2 2  n n−2 n−4

....(i)

If there exists a finite limit on the right hand side of (i), then the improper integrals is said to be convergent; otherwise it is divergent.

b

π/ 2

∫

( when n is odd )

a

∫

b

a

y dx =

∫

b

a

f ( x ) dx

Y

2

a + b2

y = f(x)

b y

2

a + b2

x=b

n! a n +1

O

n

sin x dx , then

XtraEdge for IIT-JEE

δx

X

Area bounded by the curve x = f(y), the y-axis and the abscissae y = a, y = b 47

MARCH 2010

=

∫

b

a

x dy =

∫

b

Y

ENERGY

f ( y) dy

a

y=b x

δy

x = f(y)

y=a

•

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

•

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

•

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

•

Gravitational potential energy increases as height increases.

•

Kinetic energy changes only if velocity changes.

•

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

•

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

O

X The area of the region bounded by y1 = f1(x), y2 = f2(x) and the ordinates x = a and x = b is given by

=

∫

b

a

∫

f 2 ( x ) dx –

b

a

f1 ( x ) dx

Y B

x=a

x=b

A

ELECTRICITY

O

X where f2(x) is y2 of the upper curve and f1(x) is y1 of the lower curve, i.e. the required area

=

∫

b

a

[f 2 ( x ) − f1 ( x )] dx =

∫

b

a

( y 2 − y1 ) dx

f(x) ≤ 0 for all x in a ≤ x ≤ b, then area bounded by xaxis, the curve y = f(x) and the ordinates x = a, x = b is given by =–

∫

b

a

f ( x ) dx Y

C X

D

O

•

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

•

Short fat cold wires make the best conductors.

•

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

•

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

•

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

•

All resistors in series have equal current (I).

•

All resistors in parallel have equal voltage (V).

•

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

•

Insulators contain no free electrons.

•

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

•

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

•

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

•

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

•

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

B A

If f(x) ≥ 0 for a ≤ x ≤ c and f(x) ≤ 0 for c ≤ x ≤ b, then area bounded by y = f(x), x-axis and the ordinates x = a, x = b is given by b

a

c

∫ f (x) dx + ∫

− f ( x ) dx =

x=a

A f(x)≥0

O

M

C

f(x)≤0

c

b

a

c

∫ f (x) dx – ∫

f ( x ) dx

N x=b

=

c

B

XtraEdge for IIT-JEE

48

MARCH 2010

MATHS

PROBABILITY Mathematics Fundamentals Probability : In a random experiment, let S be the sample space and E ⊆ S, then E is an event. The probability of occurrence of event E is defined as

Some Definitions : Experiment : A operation which can produce some well defined outcomes is known as an experiment. Random experiment : If in each trail of an experiment conducted under identical conditions, the outcome is not unique, then such an experiment is called a random experiment. Sample space : The set of all possible outcomes in an experiment is called a sample space. For example, in a throw of dice, the sample space is {1, 2, 3, 4, 5, 6}. Each element of a sample space is called a sample point. Event : An event is a subset of a sample space. Simple event : An event containing only a single sample point is called an elementary or simple event. Events other than elementary are called composite or compound or mixed events. For example, in a single toss of coin, the event of getting a head is a simple event. Here S = {H, T} and E = {H} In a simultaneous toss of two coins, the event of getting at least one head is a compound event. Here S = {HH, HT, TH, TT} and E = {HH, HT, TH} Equally likely events : The given events are said to be equally likely, if none of them is expected to occur in preference to the other. Mutually exclusive events : If two or more events have no point in common, the events are said to be mutually exclusive. Thus E1 and E2 are mutually exclusive in E1 ∩ E2 = φ. The events which are not mutually exclusive are known as compatible events. Exhaustive events : A set of events is said to be totally exhaustive (simply exhaustive), if no event out side this set occurs and at least one of these event must happen as a result of an experiment. Independent and dependent events : If there are events in which the occurrence of one does not depend upon the occurrence of the other, such events are known as independent events. On the other hand, if occurrence of one depend upon other, such events are known as dependent events.

XtraEdge for IIT-JEE

P(E) = =

number of distinct elements in E n(E) = number of distinct element in S n(S)

number of outocomes favourable to occurrence of E number of all possible outcomes

Notations : Let A and B be two events, then

A ∪ B or A + B stands for the occurrence of at least one of A and B. A ∩ B or AB stands for the simultaneous occurrence of A and B. A´ ∩ B´ stands for the non-occurrence of both A and B. A ⊆ B stands for "the occurrence of A implies occurrence of B". Random variable : A random variable is a real valued function whose domain is the sample space of a random experiment. Bay’s rule : Let (Hj) be mutually exclusive events such that n

P(Hj) > 0 for j = 1, 2, ..... n and S = U H j . Let A be j=1

an events with P(A) > 0, then for j = 1, 2, .... , n P( H j ) P(A / H j )  Hj   = P  n   A  ∑ P(H k ) P( A / H k ) k =1

Binomial Distribution : If the probability of happening of an event in a single trial of an experiment be p, then the probability of happening of that event r times in n trials will be nCr pr (1 – p)n – r. Some important results : (A)

P(A) = =

49

Number of cases favourable to event A Total number of cases n(A) n(S)

MARCH 2010

P(A) =

(i) Probability of happening none of them

Number of cases not favourable to event A Total number of cases

=

= (1 – p1) (1 – p2) ........ (1 – pn) (ii) Probability of happening at least one of them

n(A) n(S)

= 1 – (1 – p1) (1 – p2) ....... (1 – pn) (iii) Probability of happening of first event and not happening of the remaining

(B) Odd in favour and odds against an event : As a result of an experiment if “a” of the outcomes are favourable to an event E and b of the outcomes are against it, then we say that odds are a to b in favour of E or odds are b to a against E. Thus odds in favour of an event E

=

= p1(1 – p2) (1 – p3) ....... (1 – pn) If A and B are any two events, then B P(A ∩ B) = P(A) . P   or A

Number of favourable cases a = Number of unfavourable cases b

B P(AB) = P(A) . P   A

Similarly, odds against an event E =

Number of unfavourable cases b = Number of favorable cases a

B Where P   is known as conditional probability A means probability of B when A has occurred.

Note : If odds in favour of an event are a : b, then the probability of the occurrence of that event is a and the probability of non-occurrence of a+b b that event is . a+b

Difference between mutually exclusiveness and independence : Mutually exclusiveness is used when the events are taken from the same experiment and independence is used when the events are taken from the same experiments. (E)

If odds against an event are a : b, then the probability of the occurrence of that event is b and the probability of non-occurrence of a+b a that event is . a+b (C)

P(AB) + P( AB ) = 1 P( A B) = P(B) – P(AB) P(A B ) = P(A) – P(AB) P(A + B) = P(A B ) + P( A B) + P(AB) Some important remark about coins, dice and playing cards :

P(A) + P( A ) = 1 0 ≤ P(A) ≤ 1

(D)

Coins : A coin has a head side and a tail side. If an experiment consists of more than a coin, then coins are considered to be distinct if not otherwise stated.

P(φ) = 0 P(S) = 1 If S = {A1, A2, ..... An}, then P(A1) + P(A2) + .... + P(An) = 1 If the probability of happening of an event in one trial be p, then the probability of successive happening of that event in r trials is pr. If A and B are mutually exclusive events, then P(A ∪ B) = P(A) + P(B) or P(A + B) = P(A) + P(B) If A and B are any two events, then

Dice : A die (cubical) has six faces marked 1, 2, 3, 4, 5, 6. We may have tetrahedral (having four faces 1, 2, 3, 4,) or pentagonal (having five faces 1, 2, 3, 4, 5) die. As in the case of coins, If we have more than one die, then all dice are considered to be distinct if not otherwise stated. Playing cards : A pack of playing cards usually has 52 cards. There are 4 suits (Spade, Heart, Diamond and Club) each having 13 cards. There are two colours red (Heart and Diamond) and black (Spade and Club) each having 26 cards.

P(A ∪ B) = P(A) + P(B) – P(A ∩ B) or P(A + B) = P(A) + P(B) – P(AB) If A and B are two independent events, then

In thirteen cards of each suit, there are 3 face cards or coart card namely king, queen and jack. So there are in all 12 face cards (4 kings, 4 queens and 4 jacks). Also there are 16 honour cards, 4 of each suit namely ace, king, queen and jack.

P(A ∩ B) = P(A) . P(B) or P(AB) = P(A) . P(B) If the probabilities of happening of n independent events be p1, p2, ...... , pn respectively, then

XtraEdge for IIT-JEE

P(A A ) = 0

50

MARCH 2010

MOCK TEST FOR IIT-JEE PAPER - I

Time : 3 Hours

Total Marks : 240

Instructions :

• This question paper contains 60 questions in Chemistry (20), Mathematics (20) & Physics (20). • In section -I (8 Ques) of each paper +3 marks will be given for correct answer & –1 mark for wrong answer. • In section -II (4 Ques) of each paper +4 marks will be given for correct answer –1 mark for wrong answer. • In section -III contains 2 groups of questions (2 × 3 = 6 Ques.) of each paper +4 marks will be given for each correct answer & –1 mark for wrong answer. • In section -IV (2 Ques.) of each paper +8(2×4) marks will be given for correct answer & No Negative marking for wrong answer.

CHEMISTRY

5.

Calculate the pH of a solution of 0.1 M Fe(NO3)3 if acid dissociation constant for the given reaction is 1.0 × 10–3 H3O+ (aq) + [Fe(H2O)5OH]2+ [Fe(H2O)6] 3+ + H2O(l) (A) 1.5 (B) 2.02 (C) 2.64 (D) 3

6.

A real gas of molar mass 60 g mol–1 has density at critical point equal to 0.80 g/cm3 and its critical

SECTION – I Straight Objective Type This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1.

2.

4 × 105 K. Then the 821 van der Waal's constant 'a' (atm L2 mol–2) will be (A) 0.025 (B) 0.325 (C) 3.375 (D) 33.750

If the wavelength of series limit of Lyman series for He+ ion is x Å then what will be the wavelength of series limit of Balmer series for Li+2 ion ? 9x 4x Å (B) Å (A) 4 7 5x 16 x Å (D) Å (C) 4 9

temperature is given by Tc =

7.

The molecular size of ICl and Br2 is approximately same, but boiling point of ICl is about 40° higher than that of Br2 this difference in boiling point is observed because(A) ICl bond is stronger than Br–Br bond (B) I.E. of iodine < I.E. of bromine (C) Iodine is larger than bromine (D) ICl is polar while Br2 is non polar

8.

Which of the following reaction leads to formation of pair of enantiomers ?

Select the correct order of decreasing boiling point of the following compounds

| OH (I)

N | Me (II)

(A) I > II > III > IV (C) IV > III > II > I

N | H (III)

O (IV)

(B) I > III > IV > II (D) IV > III > I > II

3.

D-glucose and D-fructose can be distinguished by (A) Fehling's solution (B) Tollene's reagent (D) Benedict's test (C) Br2/H2O

4.

Which of the following is an organo silicon polymer? (A) Silica (B) Silicon (C) Silicic Acid (D) Silicon carbide

XtraEdge for IIT-JEE

CH OH

3 →

(A)

I

H

I

H

(B)

H

(C)

I

51

H O

2  →

–

CN  → DMSO

MARCH 2010

Me

I

(D)

H

T

OH →  DMF

(D)

(C)

SECTION – II Multiple Correct Answers Type

v

n2

1/n

This section contains 4 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct. 9.

SECTION – III

Which of the following reaction occur in Bessemer's converter ? (A) 2Cu2S + 5O2 → 2CuSO4 + 2CuO (B) 2Cu2S + 3O2 → 2Cu2O + 2SO2 ↑ (C) 2CuFeS2 + O2 → Cu2S + 2FeS + SO2 (D) FeO + SiO2 → FeSiO3

Comprehension Type This section contains 2 groups of questions. Each group has 3 multiple choice questions based on a paragraph. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct. Paragraph # 1 (Ques. 13 to 15) Generally indicators used in acid-base titration reactions are either weak acid or weak bases. Their equilibria may be presented as H+ + In– HIn Where HIn is its acidic form and In– is its basic form. If KIn is the indicator ionisation constant then [H+] of indicator can be expressed in the following way [HIn] [H+] = KIn . [In − ]

10. Identify the compounds which do not give positive iodoform test in the following sequence of the reaction (i) Hydrolysis (ii) Heating (iii) I2 + NaOH CO2Et O

(A) CH3 – CH – C – Et (B)

O

O

CO2Et O

O–Et

(C)

(D) Me – CH – CO2Et

12

CO2Et 11. A sample of H2O2 solution labelled as "28V" has density of 265 gL–1. Identify the correct statement (s) w (A) M H 2O2 = 2.5 (B) % = 8.5 v (C) m H 2O2 = 13.88

8 4 3 2

(D) Mole fraction of H2O2 = 0.2

The indicator used in a particular acid base titration depends on the nature of acid or base. One such indicator diagram is given. 13. The pH range of an indicator is 4-6. If it is 50% ionised in a given solution then its ionisation constant would be (A) 10–4 (B) 10–5 –6 (C) 10 (D) None

r U

(B)

HB Vol.of NaOH (in mL)

12. Select the correctly presented graph if v = velocity of e in Bohr's orbit r = radius of Bohr's orbit U = potential energy of e– in Bohr's orbit T = kinetic energy of e– in Bohr's orbit

(A)

HA

pH

1/n2

14. Calculate the pH at equivalence point when 5 milli mol of HB is titrated with 0.1 M NaOH. (A) 8.5 (B) 8.75 (C) 8.85 (D) 9.0

n2

15. Which of the following indicator is most suitable for titration of HB with strong base

XtraEdge for IIT-JEE

52

MARCH 2010

(A) Phenolphthalein (8.3 – 10) (B) Bromothymol blue (6 – 7.6) (C) Methyl Red (4.2 - 6.3) (D) Malachite green (11.4–13)

labeled p, q, r, s and t. any given statement in Column I can have correct matching with ONE OR MORE statements (s) in column II. The appropriate bubbled corresponding to the answers to these questions have to be darkened as illustrated in the following example : If the correct matches are A – p, s and t; B – q and r; C – p and q; and D – s and t; then the correct darkening of bubbles will look like the following. p q r s t q r A

Paragraph # 2 (Ques. 16 to 18) A useful method to convert oxime to substituted amide is Beckmann rearrangement which occurs through following steps,

Ph

Ph

C=N

+

H→

C=N CH3

( I)

OH

Me

B

+ OH2

D

⊕

( III)

(IV)

CH3 –C–NH–Ph

OH 16. Rate determining step in Beckmann rearrangement is (A) I (B) II (C) III (D) IV

Me C=N when treated Ph OH with H2SO4 and hydrolysed the products formed are (A) CH3COOH and PhNH2 (B) CH3NH2 and PhCOOH (C) PhCH2NH2 and CH3COOH (D) PhCH2COOH and CH3NH2

17. The compound

NH OH

→ I PCl  5 → P –CH3  2 pH = 4−6

∆

the product P may be (A) PhCOOH O (B) CH3–

q

s

t

s

t

r

Column II (Isomers) (p) Total stereoisomers are 10 (q) Total meso isomers are 2 (r) Total optical isomers are 4 (s) Total stereoisomers are 2 (t) Meso isomers zero

20. Match species of column-I with those species of column-II which has same hybridization. Column I Column II (A) B3N3H6 (p) ClO– (B) S2Cl2 (q) IF7 (C) XeF5– (r) CO32– (D) ICl4– (s) S8 (t) XeO2F4

18. In the following sequence of reaction O Ph – C –

p

19. Match the column Column I (Compounds) (A) 2,3-Dihydroxy butanoic acid (B) 1,3-Dichloro-1methyl cyclobutane (C) 2,3,4,5-Tetrahydroxy hexane-1, 6-diol (D) 3-Chloro butane-2-ol

O CH3 –C = N–Ph

r

C

H 2O 2O − → CH 3 − C = N − Ph H  → ( II)

p

MATHEMATICS

–C–NH2

SECTION – I

O (C) Ph – C – NH –

Straight Objective Type

– CH3

This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

O (D) Ph – C – NH2

1.

SECTION – IV Matrix – Match Type This section contains 2 questions. Each question contains statements given in two columns, which have to be matched. The statements in Column I are labeled A, B, C and D, while the statements in Column II are

XtraEdge for IIT-JEE

53

Given the function f(x) = 1 /(1 – x), the points of discontinuity of the composite function y = f 3n (x), where f n(x) = fof …. of (n times) are (n ∈ N) (A) 0, 1 (B) 2n (C) 3n (D) 2n + 1

MARCH 2010

2.

3.

r The lines r = i – j + λ(2i + k) and r r = (2i – j) + µ(i + j – k) intersect for (A) λ = 1, µ = 1 (B) λ = 2, µ = 3 (C) all values of λ and µ (D) no value of λ and µ

The tangent at the point P(x1, y1) to the parabola y2 = 4ax meets the parabola y2 = 4a(x + b) at Q and R, the coordinates of the mid-point of QR are (A) (x1 – a, y1 + b) (B) (x1, y1) (D) (x1 – b, y1 – b) (C) (x1 + b, y1 + a)

5.

Equation of the line which is parallel to the line common to the pair of lines given by 6x2 – xy – 12y2 = 0 and 15x2 + 14xy – 8y2 = 0 and the sum of whose intercepts on the axes is 7, is (A) 2x – 3y = 42 (B) 3x + 4y = 12 (C) 5x – 2y = 10 (D) None of these

6.

If 2 sin2 ((π/2) cos2x) = 1 – cos (π sin 2x), x ≠ (2n + 1) π/2, n ∈ I, then cos 2x is equal to (A) 1/5 (B) 3/5 (C) 4/5 (D) 1

7.

The value of tan 3 α cot α cannot lie in (A) ] 0, 2/3 [ (B) ] 1/3, 3 [ (C) ] 4/3, 4 [ (D) ] 2, 10/3 [

n

∫ | x − 3 | dx = 2A + B then 1

(A) A = 3/2, B = 4 (C) A = 2, B = –3/2

(B) A = 1, B = 1/2 (D) A = 1/2, B = 3/2

3x 2 + 12 x − 1 , − 1 ≤ x ≤ 2 11. If f(x) =  . Then 37 − x , 2 0 and satisfies f(x2) = x3 for all x > 0. Then the value of f ′(4) is ___.

This section contains 8 questions. 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 ORS 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

1 1 1 , , , ……. 2 4 8

19. The value of y

( 8 )− 13 if (1 + x ) 2

dy = x(1 – y), dx

y(0) = 4/3 is _____.

PHYSICS

W 0 1 2 3 4 5 6 7 8 9

SECTION – I Straight Objective Type This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1.

68

A pendulum of length 10 cm is hanged by wall making an angle 3° with vertical. It is swinged to position B. Time period of pendulum will be –

MARCH 2010

between block and the plane 'µ' is insufficient to stop slipping, then acceleration of block is –

3° 6° B

B

A

(A) π/5 sec

(B)

θ

2π sec 15

(A) g{sin α – µ cos α}

(C) π/6 sec (D) Subsequent motion will not be periodic 2.

 cos α  (B) g sin α − µ  sin( θ / 2)  

A rope of mass M is hanged from two support 'A' & 'B' as shown in figure. Maximum and minimum tension in the rope is –

(C) g {sin α – 2 µ cos α cos θ/2} (D) g {sin α – µ cos α. sin (θ/2)}

A

SECTION – II

θ1

Multiple Correct Answers Type This section contains 5 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct.

B θ2

(A)

Mg cos θ 2 Mg cos θ1 , sin(θ1 + θ 2 ) sin (θ1 + θ 2 )

5.

(B) Mg, Mg cos θ1

3.

λ where λ is wavelength of light 4 and S1, S2 are slits separated by distance 2λ. Then value/s of θ for which a maxima is obtained will be –

Mg cos θ1 Mg cos θ2 , sin (θ1 + θ 2 ) cos (θ2 − θ1 )

S1

Two ring of mass m and 2 m are connected with a mass less spring and can slips over two frictionless parallel horizontal rails as shown in figure. Ring of mass m is given velocity 'v0' in the direction shown. Maximum stretch in spring will be – m v0

O

–1

(A) sin (1/8) (C) sin–1 (5/6)

k

6. 2m

4.

(A)

m v0 k

(B)

3m v0 k

(C)

2m v0 3k

(D)

2m v0 k

θ

S2 (B) sin–1 (–1/4) (D) sin–1 (–7/8)

A particle of charge q and mass m moves rectilinearly under the action of an electric field E = α – βx. Here α and β are positive constants and x is the distance from the point where the particle was initially at rest then(A) motion of particle is oscillatory (B) amplitude of the particle is α/β (C) mean position of the particle is at x =

A block 'B' is just fitting between two plane inclined at an angle 'θ'. The combination of plane is inclined at angle 'α' with horizontal. If coefficient of friction

XtraEdge for IIT-JEE

In a modified YDSE experiment if point source of monochromatic light O is placed in such a manner that OS1– OS2 =

Mg cos θ 2 Mg cos θ1 cos θ2 (C) , sin (θ1 + θ 2 ) sin (θ1 + θ2 )

(D)

α

α β

(D) the maximum acceleration of the particle is

69

qα m

MARCH 2010

7.

8.

Speed of a body moving in a circular path changes with time as v = 2t, then – (A) Magnitude of acceleration remains constant (B) Magnitude of acceleration increases (C) Angle between velocity and acceleration remains constant (D) Angle between velocity and acceleration increases

p A B

D

Radius of rod changes linearly from a to b (a > b).Temperature of the two ends are maintained at θ1 and θ2 (θ1 > θ2) respectively. Let 'i' be the heat passing per unit cross sectional area of rod and 'θ' be temperature at a distance 'r' from one end (having cross section radius 'a'), then which of the following graphs is/are correct – θ1 θ2

(B) θ r

(C)

θ2 r

i

b

(D) i r

b

r

b

SECTION – III Matrix - Match Type This section contains 2 questions. Each question contains statements given in two columns, which have to be matched. The statements in Column I are labeled A, B, C and D, while the statements in Column II are labeled p, q, r, s and t. any given statement in Column I can have correct matching with ONE OR MORE statements (s) in column II. The appropriate bubbled corresponding to the answers to these questions have to be darkened as illustrated in the following example : If the correct matches are A – p, s and t; B – q and r; C – p and q; and D – s and t; then the correct darkening of bubbles will look like the following.

XtraEdge for IIT-JEE

s

t

s

t

r

11. Column-I contains different processes undergone by a diatomic ideal gas. Column-II change in different parameter of ideal gas. Column I Column II –1 (A) PV = constant and (p) Heat is given to volume is increased gas twice (B) P2V = constant and (q) Heat is rejected by pressure is increased gas twice (C) PV6/5 = constant and (r) Work done by gas volume is reduced is negative to half the initial volume (D) PV2 = constant (s) Internal energy and pressure is increase increased 3 times (t) None of these

θ1

b

q

t

r r

p

s

10. Column-I contains the process of emission of electrons while column-II contains the method to achieve emission. Match column I and II. Column I Column II (A) Thermionic emission (p) By irradiating with light (B) Photoelectric (q) By applied strong emission electric field (C) Field emission (r) By colliding accelerated electrons on metals (D) Secondary emission (s) By heating (t) None of these

A wave disturbance in medium is given by y(x,t) = 0.04 cos (25πt + π/2) cos (5πx), where x and y are in meter and t is in second – (A) An antinode occurs at x = 0 (B) Speed of wave is 5 m/s (C) A node occurs at x = 20 cm

(A) θ

r

p

C

(D) Maximum velocity of medium particle is π m/s 9.

q q

SECTION – IV Integer answer type This section contains 8 questions. 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 ORS 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 :

70

MARCH 2010

X 0 1 2 3 4 5 6 7 8 9

Y 0 1 2 3 4 5 6 7 8 9

Z 0 1 2 3 4 5 6 7 8 9

calculate the resultant intensity at P. (in W/m2). (Ans.. ×101)

W 0 1 2 3 4 5 6 7 8 9

d

S1

P

S2 d

Screen

S3 D>>d

16. A cubical block of mass 6 kg and side 16.1 cm is placed on frictionless horizontal surface. It is hit by a cue at the top as to impart-impulse in horizontal direction. Minimum impulse imparted to topple the block must be greater than.

12. A small body is projected with a velocity just sufficient to make it reach from the surface of a planet (of radius 2R and mass 4 M) to the surface of another planet (of radius R = 1000 km and mass M). The distance between the centres of the two spherical planets is 6 R. Find the distance of the body from the centre of bigger planet in kilometers where the speed of the body is minimum. Assume motion of body along the line joining centres of planets. (Ans... ×103)

17. An unstable element is produced in a nuclear reactor at a constant rate. If its half life is 100 years, how much time in years is required to produce 50% of the equilibrium quantity ? (Ans.... ×102)

13. Light from sun is found to be maximum intensity near 470 nm. Distance of sun from earth is 1.5 × 1011 and radius of sun is 7 × 108 m. Treating sun as a black body, calculate the intensity of radiation from sun at the surface of earth in watt/m2. (Ans. ... ×103)

18. The wavelength of light incident on a metal surface is reduced from 300 nm to 200 nm (both are less than threshold wavelength). What is the change in the stopping potential for photoelectrons emitted from the surface. (Take h = 6.6 × 10– 34 J-sec)

14. 1 gm of helium gas undergoes process ABCA as shown in figure. Calculate the maximum temperature of gas in degree centigrade. (Ans. ... ×102)

19. A mercury pallet is trapped in a tube as shown in figure. The tube is slowly heated to expel all mercury inside it (Isothermal condition). Calculate the heat

P

225

(in kPa) 200

given to the tube in J. (ρHg = 13.6 gm/cc, Atmospheric pressure = 105 Pa, cross–section area of tube = 2 cm2)

B

A

C

10 cm

4

3

5 cm

V (in liters)

10 cm

15. Consider the interference at P between waves emanating from three coherent sources in same phase located at S1, S2 and S3. If intensity due to each

source is I0 = 12 W/m2 at P and

XtraEdge for IIT-JEE

λ d2 = then 3 2D

71

MARCH 2010

MOCK TEST - AIEEE PATTERN SYLLABUS : Physics : Full syllabus

Chemistry : Full syllabus

Mathematics : Full syllabus

Time : 3 Hours

Total Marks : 432

Instructions : • Part A – Physics (144 Marks) – Questions No. 1 to 2 and 9 to 30 consist FOUR (4) marks each and Question No. 3 to 8 consist EIGHT (8) marks each for each correct response. Part B – Chemistry (144 Marks) – Questions No. 31 to 39 and 46 to 60 consist FOUR (4) marks each and Question No. 40 to 45 consist EIGHT (8) marks each for each correct response. Part C – Mathematics (144 Marks) – Questions No.61 to 82 and 89 to 90 consist FOUR (4) marks each and Question No. 83 to 88 consist EIGHT (8) marks each for each correct response • For each incorrect response, ¼ (one fourth) of the weightage marks allotted of the would be deducted.

PHYSICS 1.

An elevator, in which a man is standing, is moving upwards with a speed of 10 m/s. If the man drops the coin from a height of 2.45 m, it reaches the floor of the elevator after a time 1 (A) sec (B) 2 sec 2 1 (C) 2 sec (D) sec 2

2.

Graph between the mass of liquid inside the capillary and radius of capillary is –

(A)

3.

4.

If input in a full-wave rectifier is e = 50 sin 314t volt, diode resistance is 100 Ω and load resistance is 1K Ω then. (1) Pulse frequency output voltage is 100. (2) Input power is 1136 mw (3) Output power is 827 mw (4) Efficiency is 81.2 % (A) 1, 3 (B) 1, 2 (C) 1, 2, 3 (D) 1,2,3,4

6.

An n-p-n transistor circuit is arranged as shown, it is a–

m

m

(B)

V

r

(C)

5.

m

R = 10 K

V

r

(D)

(A) Common base amplifier circuit (B) Common-emitter amplifier circuit (C) Common-collector amplifier circuit (D) None

m

r r If the De –Broglie wavelength of an electron in first Bohr's orbit be λ then the minimum radial distance between the electrons in the first and second Bohr's orbit is – λ λ (A) λ (B) (C) 2λ (D) 2π 2

Photoelectron are emitted with maximum kinetic energy E from a metal surface when light of frequency υ falls on it when light of frequency υ' falls on the same metal, the max. KE. of emitted Photoelectrons is found to be 2E then υ' is (A) υ´ = υ (B) υ´ = 2υ (C) υ´ > 2υ (D) υ´ < 2υ

XtraEdge for IIT-JEE

N P N

72

7.

At what temperature will wood and iron appear equally hot or equally cold (A) 37° C (B) 98.6° (C) Temperature of human body (D) all of the above

8.

In an Experiment to find loss of energy w.r.t time in case of swinging simple pendulum mark graph between (amplitude)2 and time is –

MARCH 2010

a2

(A)

(B) t

t

a2

a2

(C)

(D)

15. A magnet is suspended horizontally in the earth's magnetic field. When it is displaced and released, it oscillates in a horizontal plane with a period T. If a piece of wood of same M.I as the magnet is attached to the magnet is attached to the magnet, the new period of oscillation of the system would be T T T (A) (B) (C) (D) 2T 3 2 2

t

t

9.

14. A cyclotron is accelerating proton, where the applied magnetic field is 2T and the potential gap is 100 keV. To acquire a kinetic energy of 20 MeV, the number of turns, the proton has to move between the dees is (A) 200 (B) 300 (C) 150 (D) 100

a2

A body weights 24.2g when placed is one pan of a balance and 20g when placed in other. What is the true mass of the body if the arms have un equal length (A) 24.2g (B) 20g (C) 22.1g (D) 22g

10. In the YDSE apparatus shown in the fig ∆x is the path difference between S2P and S1P. Now a glass slab is introduced in front of S2 then the number of fring between O and P will –

16. A square loop of side 1m is placed in a perpendicular magnetic field. Half of the area of the loop lies inside the magnetic field. A battery of emf 10V and negligible internal resistance is connected in the loop. The magnetic field change with time according to the relation B = (0.01 –2t) tesla. The total emf of the battery will be -

P S1

O S2

(A) Increase (B) decrease (C) Many increase or decrease depends upon ∆x (D) remains constant

X

X

X

X

X

X

X

X

X

X

X

X

X

X

– 10V

(A) 11V (C) 12 V

11. A capacitor is charged until its stored energy is 3 J and the charging battery is removed. Now another uncharged capacitor is connected across it and it is found that charge distributes equally. The final value of total energy stored in the electric fields is (A) 1.5 J (B) 3 J (C) 2.5 J (D) 2 J

+

(B) 9 V (D) 6 V

17. A Smooth ring of Mass 'M' is threaded on a string as shown in the figure. Various portions of strings are vertical. What is the condition if the ring alone is to remain at rest –

12. A potential difference of 30 V is applied between the ends of a conductor of length 100 m and resistance 0.5 Ω and uniform area of cross-section. The total linear momentum of free electrons is (A) 3.4 × 10–6 kg/s (B) 4.3 × 10–6 kg/s –8 (C) 3.4 × 10 kg/s (D) 4.3 × 10–8 kg/s

m'

m

13. A certain unknown resistance is connected in the left gap and a resistance box in the right gap of a metre bridge. By introducing a resistance of 10 Ω with the help of resistance box, the balance point is determined. If the balance point shift by 20 cm on increasing the resistance from the resistance box by 12.5 Ω ,then value of unknown resistance is (A) 15 Ω (B) 25 Ω (C) 10 Ω (D) 20 Ω

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X

M

73

(A)

1 4 1 = + M m m'

(B)

1 2 1 = + M m m'

(C)

1 1 1 = + M m m'

(D)

1 3 1 = + M m m'

MARCH 2010

18. A Particle of mass 100 gm moves in a potential well given by U = 8x2 – 4x + 400 Joule. Find its acceleration at a distance of 25 cm from equilibrium in positive direction (A) 0 (B) 40 m/s2 2 (D) 20 m/s2 (C) –40 m/s

Statement based questions: Each of the questions (Q.24 to 25) given below consist of Statement-I and Statement-II. Use the following Key to choose the appropriate answer. (A) If both Statement-I and Statement-II are true, and Statement-II is the correct explanation of Statement-I. (B) If both Statement - I and Statement-II are true but Statement-II is not the correct explanation of Statement-I. (C) If Statement-I is true but Statement-II is false. (D) If Statement-I is false but Statement-II is true.

19. Side of a cube is measured with a standard vernier callipers. The main scale reads 10 mm and first division of vernier scale coincides with that of main scale. Measured value of side of cube is (A) 1.1 cm (B) 1.01 cm (C) 1.001 cm (D) 1.02 cm

24. Statement-I : Heat supplied to a gas in a process is 100 J and work done by the gas in the same process is 120 J, then pressure of the gas in the process should increase. Statement-II : Work done by the gas is greater than the heat supplied to the gas. Hence, internal energy of the gas should decrease.

20. While measuring the speed of sound by performing a resonance column experiment a student gets the first resonance condition at a column length of 18 cm during winter. Repeating the same experiment during summer, she measures the column length to be x cm for the second resonance then (A) x > 54 (B) 54 > x > 36 (C) 36 > x > 18 (D) 18 > x

25. Statement-I : In the circuit shown in figure current I through the battery rises instantly to its steady state value V/R when the switch is closed, provided

R = L/C .

21. An ideal gas is taken through the cycle A → B → C → A as shown in figure. If the net heat supplied to the gas in the cycle is 5 J. The work done by the gas in the process C → A is – 2

C

R

L

R

C

I

B

S

V

V(m3)

Statement-II : At R = L / C , τL = τC. A

1 P (N/m3)

(A) –5J (C) –15J

Passage Based questions (Q. 26 to 27) : When two concentric shells are connected by a thin conducting wire, whole of the charge of inner shell transfers to the outer shell and potential difference between them becomes zero.

10

(B) –10J (D) –20 J

–σ

22. The apparent frequency of the whistle of an engine changes by the ratio 5/3 as the engine passes a stationary observer. If the velocity of sound is 340 m/s, then the velocity of the engine is (A) 340 m/s (B) 170 m/s (C) 85 m/s (D) 42.5 m/s

+σ

R

Questions: Surface charge densities of two thin concentric spherical shells are σ and – σ respectively. They radii are R and 2R. Now they are connected by a thin wire.

23. A ray incident at sphere an angle of incidence 60°

enters a glass sphere of R.I µ = 3 . This ray is reflected and refracted at the farther surface of the sphere. The angle between reflected and refracted rays in this surface is (A) 90° (B) 60° (C) 70° (D) 40°

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2R

26. Potential on either of the shells will be 3σR 2σR (A) – (B) – 2ε 0 ε0

(C) –

74

σR 2ε 0

(D) zero

MARCH 2010

connecting the two shells, (A) zero

(B) 1

E2 then is E1

(C) 2

34. Which of the following has minimum flocculating power(C) Sr+2 (D) Na+ (A) Pb+2 (B) Pb+4

(D) 1/2

Passage Based questions (Q. 28 to 30) : In perfectly inelastic collision between two bodies momentum remains constant and the bodies stick together. Angular frequency of a spring block system

35. The conjugate base of H3BO3 is(B) H2BO3– (A) B(OH)4– – (D) H4BO3+ (C) HBO3 36. Calculate equivalent weight of C6H12O6 in given redox changeC6H12O6 →CO2 (A) M/2 (B) M/4 (C) M/24 (D) M/6

K and maximum speed of particle is SHM is m ωA, where A is the amplitude. Question: Two identical blocks P and Q have mass m each. They are attached to two identical springs initially unstretched. Now the left spring (along with P) is compressed by A/2 and the right spring (along with Q) is compressed by A. Both the blocks are released simultaneously. They collide perfectly inelastically. Initially time period of both the block was T.

is ω =

A 2

CHO

37. Give correct IUPAC name for

(A) 3-Aldo pentane-1,5-dial (B) 3-Formyl-1,5-pentanedial (C) Propane-1,2,3-tricarbaldehyde (D) Propane-1,2,3- trial

A P Q

38. Which of the following compounds would be hydrolysed most easily(B) H3C-Br (A) C2H5Br (D) H2C=CH-CH2Br (C) H2C=CH-Br 39. Which of the following alcohols is most soluble in H2O (A) n-Butyl alcohol (B) iso-Butyl alcohol (C) sec-Butyl alcohol (D) tert-Butyl alcohol

29. The amplitude of combined mass is A 4

(B)

A 2

(C)

2A 3

(D)

3A 4

30. What is energy of oscillation of the combined mass ? 1 kA2 2 1 (C) kA2 8

(A)

40. Which of the following reaction does not give amine(A) R-X + NH3 →

1 kA2 4 1 (D) kA2 16

(B)

Na / C H OH

(B) R-CH=NOH  2 5 → +

/H (C) R-CN H2O → 4 (D) R-CONH2 LiAlH   →

CHEMISTRY

41. Aniline reacts with conc. HNO3 to give -

(A) H2N

31. Which of the following oxides of Chromium is amphoteric ? (A) CrO (B) Cr2O3 (C) CrO3 (D) CrO5

NH2

NH2

NH2 NO2

(B)

32. Which of the following gives foul smelling gas with smell of rotten eggs with dil H2SO4? (D) NO2– (A) CO32– (B) SO32– (C) S2–

and NO2

O

(D)

(C)

33. The Nessler's reagent is used for the detection of ammonia the active species involved in this is-

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CHO CHO

28. The time period of oscillation of combined mass is T T (B) 2T (C) T (D) (A) 2 2

(A)

(B) HgCl42+ (D) Hg2I2

(A) HgCl2 (C) HgI42–

27. Suppose electric field at a distance r (> 2R) was E1 before connecting the two shells and E2 after

NO2

O

75

MARCH 2010

42. Benzaldehyde reacts with ammonia to form(A) Benzal amine (B) Urotropine (C) Hydrobenzamide (D) Benzaldehyde ammonia

50. Find the t2/3of the first order reaction in which K1 = 5.48×10–14sec–1. (B) 2.01×1010sec (A) 2.01×1013sec 16 (D) 2.01×105sec (C) 2.01×10 sec 51. The electrode potential of Mg+2/Mg electrode in which concentration of Mg+2 is 0.01 M is, Given that

43. Several blocks of magnesium are fixed to the bottom of a ship to(A) Prevent puncturing by under sea rocks. (B) Keep away the sharks (C) Make a ship lighter (D) Prevent action of water & salt

E ºMg 2 + / Mg = –2.36 volt.

(A) 2.36 volt (C) 2.42 volt

52. 50ml of 2N CH3COOH mixed with 10ml of 1N CH3COONa solution will have pH of.......(Ka= 10–5) (A) 4 (B) 5 (C) 6 (D) 7

44. Which of the following metal reacts with hot solution of NaOH and liberates H2 gas(A) Tin (B) Lead (C) Zinc (D) None of These

53. An aqueous solution containing 2gm of solute dissolved in 100 gm of water freezes at –0.5°C. What is the molecular wt. of solute ? Molar heat of fusion of ice at 0°C is 1.44 KCals. & R = 2cals(A) 74.4 (B) 84.6 (C) 48.6 (D) 90.2

45. Which of the following compound is aspirin(A) Methyl salicylate (B) Acetyl salicylic Acid (C) Phenyl salicylate (D) Salicylic acid 46. A coloured ppt. is obtained when H2S gas is passed through an aqueous solution of the salt in presence of NH4OH. The ppt. dissolves in dil. HCl and reacts with NaOH to give white ppt which on standing turns into brown/black mass. The cation present in the salt is (B) Mg2+ (A) Cu2+ 2+ (D) Mn2+ (C) Ni

54. Calculate the maximum no. of possible e– for which 4 0, b > 0, c > 0 then (a + b) (b + c) (c + a) is greater than (A) 2(a + b + c) (B) 6abc (C) 3(a + b + c) (D) 8abc

60. Silver ore dissolved in dilute soln of NaCN in the presence of air to form(A) AgCN (B) [Ag(CN)2]– (C) AgCNO (D) [Ag(CN)4]3–

69. The order of the differential eqn of all conics whose axes coincide with the axes of coordinates is (A) 2 (B) 3 (C) 4 (D) 1

MATHEMATICS

70. The area bounded by y = [x] and the two ordinates x = 1 and x = 1.7 is -

(A)

61. If sinx + cos x = then tan2x is 25 7 25 (A) (B) (C) 17 25 7

(D)

24 7

f(x) =

(

)

(

)

(B)

n n 2 −1 12

(C)

n 2 +1 12

(D)

n n2 +1 12

17 5

(D)

7 10

x 4 +1

is (B) [–1, 1] (D) None of these

73. The value of k for which points A(1, 0, 3), B(–1, 3, 4), C(1, 2, 1) & D(k, 2, 5) are coplaner, is (A) 1 (B) 2 (C) 0 (D) –1

64. The negation of statement (p ∧ q ) → (q ∨ ~ r) will be. (A) (~ p ∨ ~ q) → (~ q ∧ r) (B) (p ∧ q) ∨ (~ q ∧ r) (C) (p ∧ q) ∧ (~ q ∧ r) (D) (p ∧ q) ∧ (q ∨ ~ r)

74. Area of triangle formed by the positive x-axis, the normal & the tangent to the circle x2 + y2 = 4 at point

(1, 3 ) is -

65. The angle between the lines given by the equation ay2 – (1 + λ2)xy – ax2 = 0 is same as the angle between the lines (A) 5x2 +2xy–3y2 = 0 (B) x2 – y2 = 100 (C) xy = 0 (D) B & C both

(A)

3 sq. unit 2

(C) 2 3 sq. unit

(B) 3 sq. unit (D) 6 sq. unit

75. A & B are two candidates seeking admission in AIEEE. The probability that A is selected is 0.5 and probability that both A & B are selected is at most 0.3. The probability of B getting selected can not exceed (A) 0.6 (B) 0.7 (C) 0.8 (D) 0.9

66. The vector ((i – j) × (j – k)) × (i + 5k) is equal to (A) 5i – 4j – k (B) 3i – 2j + 5k (C) 4i – 5j – k (D) 5i + 4j – k

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(C)

72. If A and B are square matrices of the same order and AB = 3I then A–1 is equal to 1 (A) 3B (B) B 3 –1 (D) None of these (C) 3B

63. The variance of the first n natural numbers is n 2 −1 12

sin( π[ x 2 + 1])

(A) [0, 1] (C) {0}

1 (D) tan–1    18 

(A)

(B) b = 1

71. The range of the function

62. If tan(x + y) = 33 and x = tan–13 then y will be (A) 0.3 (B) tan–1(1.3)

(C) tan–1 (0.3)

17 10

77

MARCH 2010

76. If a plane meets the co-ordinate axes in A, B & C such that the centroid of triangle ABC is a point (1, 2r, 3r2) then equation of plane is z y z y (A) x + + 2 = 9 (B) x + + 2 = 3 2r 3r r r

(C) 6x +

a1 ∆ = a2 a3

3 y 2z + 2 =18 (D) None of these r r

b1

c1

b2

c2

b3

c3

A1

B1

C1

A2 A3

B2 B3

C2 C3

84. The value of

78. If shortest distance of a line x – y = 3a from a point (a, b) is |a| then b/a must be root of equation (a, b, ∈ Ro) (B) x2 + 4x + 2 = 0 (A) x2 + 4x – 2 = 0 2 (D) None of these (C) x – x – 1 = 0

lim + x→0

16

r −q q−p

85.

2rπ 

(B) i (D) – 1 x sin −1 −1  1− x

2x 1+ x2

 −1 3x − x 3  tan cos  2  1 − 3x 2 1+ x 

(A) 1/2 (C) 1/4

2

=

(B) 1/3 (D) 1/6

a sin x + b cos x decreases for all x if c sin x + d cos x (A) ad – bc < 0 (B) ad – bc > 0 (C) ab – cd > 0 (D) ab – cd < 0

86. f(x) =

87. The minimum value of 27cos2x.81sin2x is 1 (B) – 5 (A) 243 (C) 1/5 (D) None of these

p−q p−r

(D) None of these

88. The value of

81. There are four balls of different colours and four boxes of coluers same as those of the balls. The number of ways in which the balls, one in each box, could be placed such that a ball does not go to box of its own colour is : (A) 8 (B) 7 (C) 9 (D) None of these

(A)

3 2 aπ 8

(C) 3π/8

∫

a+π / 2 a

(sin 4 x + cos 4 x ) dx is -

(B) a(π/2)2 (D) None of these

89. The number of products that can be formed with 10 prime number taken two or more at a time is (B) 210–1 (A) 210 (D) 210–10 (C) 210–11

82. If T0, T1, T2......are the terms in expansion of (x + a)n then value of (T0 –T2 + T4 – ...........)2 + (T1 – T3 + T5–........)2 is (B) (x2 + a2)n (A) (x2 + a2) (D) (x2 + a2)–1/n (C) (x2 + a2)1/n

90. If A ≡ (3, 4) & B is a variable point on line |x| = 6 if AB ≤ 4 then no of position of point B with integral co-ordinates is (A) 5 (B) 6 (C) 10 (D) 12

83. If the capital latters denote the cofactors of the corresponding small letters in the determinant

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2rπ



(D) 0

∑  sin 17 + i cos 17  is -

(A) 1 (C) –i

80. If pth, qth and rth term of a A.P. are three consecutive terms of G.P. find common ratio of the G.P.

(C)

(C) 2∆

r =1

79. If both the roots of equation 2x2 + 3 2 x + 6 = 0 are real and equal then both the roots of equation x2 – bx + 1 = 0 are (A) Imaginary (B) both are –ve (C) One is +ve other is –ve (D) both are +ve

(B)

is -

(B) ∆2

(A) ∆

77. If the equation of an ellipse whose focus is (–1, 1) & eccentricity is 1/2 and directrix is x – y + 3 = 0 is ax2 + by2 + 2hxy + 2gx + 2fy + c = 0 then (B) a + b = c (A) h2 – abc = 0 (C) a + b = h (D) a + b = 2ch

(A) 1

then the value of the determinant

78

MARCH 2010

MOCK TEST – BIT-SAT Time : 3 Hours

Total Marks : 450

Instructions :

•

This question paper contains 150 questions in Physics (40) Chemistry (40), Mathematics (45), Logical Reasoning (10) & English (15). There is Negative Marking

•

Each question has four option & out of them, ONLY ONE is the correct answer. There is – ve marking.

•

+3 Marks for each correct & – 1 Mark for the incorrect answer.

4.

PHYSICS 1.

A particle is given an initial speed u inside a smooth spherical shell of radius R = 1 m that it is just able to complete the circle. Acceleration of the particle when its velocity is vertical is -

A system is shown in the figure. The time period for small oscillations of the two blocks will be 2k k m m (A) 2π

3m k

(B) 2π

3m 4k

(C) 2π

3m 8k

(D) 2π

3m 2k

R

5.

u

2.

(A) g 10

(B) g

(C) g

(D) 3g

2

For what value of ω energy of both the particles is same ? (A) 16 unit (B) 6 unit (C) 4 unit (D) 8 unit

A rigid rod leans against a vertical wall (y-axis) as shown in figure. The other end of the rod is on the horizontal floor. Point A is pushed downwards with constant velocity. Path of the centre of the rod is – y

6.

A

x B (A) a straight line passing through origin (B) a straight line not passing through origin (C) a circle of radius l/2 and centre at origin (D) a circle of radius l/2 but centre not at origin

3.

XtraEdge for IIT-JEE

(D)

A solid sphere of mass M and radius R is placed on a smooth horizontal surface. It is given a horizontal impulse J at a height h above the centre of mass and sphere starts rolling then, the value of h and speed of centre of mass are – J h M C

(A) h =

The height at which the acceleration due to gravity g (where g = the acceleration due to becomes 9 gravity on the surface of the earth) in terms of R, the radius of the earth, is – R (A) 2R (B) 2 (C) R / 2

The displacement of two identical particles executing SHM are represented by equations π  x1 = 4 sin 10 t +  and x2 = 5 cos ωt 6 

(B) h = (C) h = (D) h =

2R 79

R

µ=0 2 J R and v = 5 M 2 2 J R and v = 5 5 M 7 7 J R and v = 5 5 M 7 J R and v = 5 M

MARCH 2010

7.

through a consumer of unknown resistance, what energy does the consumer give out to its surrounding? Assume d1 = d2 = d A

As shown in figure, wheel A of radius rA = 10 cm is coupled by belt B to wheel C of radius rC = 25 cm. The angular speed of wheel A is increased from rest at a constant rate of 1.6 rad/s2. Time after which wheel C reaches a rotational speed of 100 rpm, assuming the belt does not slip, is nearlyB A

C

(A) 4 sec (C) 12 sec

(B) 8 sec (D) 16 sec

B

+q

C

–q

K 2

8.

r be the charge density distribution πR 4 for a solid sphere of radius R and total charge Q. For a point ‘p’ inside the sphere at distance r1 from the centre of the sphere, the magnitude of electric field isQ (A) 0 (B) 4π ∈0 r12 Q r12 4π ∈0 R 4

(D)

rT ∈0

q d q 2d (B) 4ε 0 A ε0A

q 2d 2ε 0 A

(D)

2q 2d ε0A

Q r12 3π ∈0 R 4

B

(A) (5/3) R (C) (5/12) R

(D) 8 π r

2rT ∈0

2 amp D 3Ω

(A) +2 (C) –1 4.0 Volt

t(s) 4.0 sec The type of the circuit element is : (A) capacitance of 2 F (B) resistance of 2Ω (C) capacitance of 1 F (D) a voltage source of e.m.f 1 V

4.0 sec

(B) (5/6) R (D) None of these

13. A current of 2 ampere flows in a system of conductors as shown in the following figure. The potential difference (VA – VB) will be - (in volt) A 3Ω 2Ω

Current versus time and voltage versus time graphs of a circuit element are shown in figure. V(Volt) I(A)

1.0 amp

(C)

12. Consider the network of equal resistances (each R) shown in Figure. Then the effective resistance between points A an B is – A

An isolated and charged spherical soap bubble has a radius 'r' and the pressure inside is atmospheric. If 'T' is the surface tension of soap solution, then charge on drop is 2rT (A) 2 (B) 8 π r 2rT ∈0 ∈0 (C) 8 π r

10.

Q

Let P(r ) =

(C) 9.

(A)

C

B

2Ω

(B) +1 (D) –2

14. Consider a toroid of circular cross-section of radius b, major radius R much greater than minor radius b, (see diagram) find the total energy stored in magnetic field of toroid –

t(s)

11. Three identical metal plates of area 'A' are at distance d1 & d2 from each other. Metal plate A is uncharged, while plate B & C have respective charges +q & – q. If metal plates A &C are connected by switch K

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80

MARCH 2010

(A)

B2 π2b 2R 2µ 0

(B)

B2 π2b 2R 4µ 0

(C)

B2 π2b 2R 8µ 0

(D)

B2 π2b 2R µ0

18. A step down transformer reduces 220 V to 110 V. The primary draws 5 ampere of current and secondary supplies 9 ampere. The efficiency of transformer is (A) 20% (B) 44% (C) 90% (D)100%

15. AB and CD are smooth parallel rails, separated by a distance L and inclined to the horizontal at an angle θ. A uniform magnetic field of magnitude B, directed vertically upwards, exists in the region. EF is a conductor of mass m, carrying a current I. For EF to be in equilibrium: D F θ C

19. Of the following transitions in hydrogen atom, the one which gives emission line of minimum frequency is (A) n = 1 to n = 2 (B) n = 3 to n = 10 (C) n = 10 to n = 3 (D) n = 2 to n = 1 20. In uranium (Z = 92) the K absorption edge is 0.107 Å and the Kα line is 0.126 Å the, wavelength of the L absorption edge is (A) 0.7 Å (B) 1 Å (C) 2 Å (D) 3.2 Å

B

L E

21. A material whose K absorption edge is 0.15 Å is irradiated with 0.1 Å X-rays. The maximum kinetic energy of photoelectrons that are emitted from Kshell is(A) 41 KeV (B) 51 KeV (C) 61 KeV (D) 71 KeV

θ

A (A) I must flow from E to F (B) BIL = mg cos θ (C) BIL = mg sin θ (D) BIL = mg

22. The element which has Kα X-ray line whose wavelength is 0.18 nm is – (A) Iron (B) Cobalt (C) Nickel (D) Copper

16. In the circuit shown the cell is ideal. The coil has an inductance of 4H and zero resistance. F is a fuse of zero resistance and will blow when the current through it reaches 5A.The switch is closed at t = 0. The fuse will blow -

+ 2V – (A) after 5 sec (C) after 10 sec

F S

23. The momentum of a photon having energy equal to the rest energy of an electron is: (A) zero (B) 2.73 × 10–22 kg ms–1 (C) 1.99 × 10–24 kg ms–1 (D) infinite

L=4H

(B) after 2 sec (D) almost at once

24. A parallel beam of uniform, monochromatic light of wavelength 2640 Å has an intensity of 100 W/m2. The number of photons in 1 mm3 of this radiation are – (A) 222 (B) 335 (C) 442 (D) 555

17. In the circuit shown X is joined to Y for a long time and then X is joined to Z. The total heat produced in R2 is – R2 Z X Y

L

E

(A)

LE 2 2R 12

LE 2 (C) 2 R 1R 2

25. The figure shows the variation of photo current with anode potential for a photo-sensitive surface for three different radiations. Let Ia, Ib and Ic be the intensities and fa, fb and fc be the frequencies for the curves a, b and c respectively Photo current

R1 Fig.

(B) (D)

LE 2 2R 22

c

LE 2 R 2

b

a

2R 13 O Anode potential

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81

MARCH 2010

(A) fa = fb and Ia ≠ Ib (B) fa = fc and Ia = Ic (C) fa = fb and Ia = Ib (D) fb = fc and Ib = Ic

(A) N + 0.01 n (B) N + 0.001 n (C) 0.5 N + 0.001 n (D) 5(0.1 N+0.0001 n)

26. The internal resistance of a cell is determined by using a potentiometer. In an experiment, an internal resistance of 100 Ω is used across the given cell. When the key K2 is closed, the balance length on the potentiometer decreases from 90 cm to 72 cm. Calculate the internal resistance of the cell (A) 100Ω (B) 75Ω (C) 50Ω (D) 25Ω

30. When 36 Li is bombarded with 4 MeV deutrons, one reaction that is observed is the formation of two α-particles, each with 13.2 MeV of energy. The Q-value for this reaction is (A) 13.2 MeV (B) 26.4 MeV (C) 22.4 MeV (D) 4 MeV 31. In a radioactive decay, let N represent the number of residual active nuclei, D the number of daughter nuclei, and R the rate of decay at any time t. Three curves are shown in Fig. The correct ones are –

27. In the potentiometer arrangement shown, the driving cell D has e.m.f. E and internal resistance r. The cell C whose e.m.f. is to be measured has e.m.f. E/2 and internal resistance 2r. The potentiometer wire is 100 cm long. If the balance is obtained the length AP = l, thenD(E,r)

P

A

N

t (1) (A) 1 and 3 (C) 1 and 2

(A) l = 50 cm (B) l > 50 cm (C) l < 50 cm (D) Balance will not obtained

34. A concave mirror of focal length 15 cm forms an image having twice the linear dimensions of the object. The position of the object when the image is virtual will be(A) 22.5 cm (B) 7.5 cm (C) 30 cm (D) 45 cm

R J

B

If R is now made 8 Ω, through what distance will J have to be moved to obtain balance? (A) 10 cm (B) 20 cm (C) 30 cm (D) 40 cm

35. A telescope has focal length of objective and eye-piece as 200 cm and 5 cm. What is the magnification of telescope ? (A) 40 (B) 80 (C) 50 (D) 101

29. The pitch of a screw gauge is 0.1 cm. The number of divisions on its circular scale is 100. In the measurement of diameter of a wire with this screw gauge the linear scale reading is 'N' cm and the number of division on the reference line is n. Then the radius of the wire in cm will be -

XtraEdge for IIT-JEE

t (3) (B) 2 and 3 (D) all three

33. A particle moves in a circle of diameter 1 cm with a constant angular velocity. A concave mirror of focal length 10 cm is placed with its principal axis passing through the centre of the circle and perpendicular to its plane. The distance between the pole of the mirror and the centre of the circle is 30 cm. The ratio of acceleration of image to that of object is 1 1 (A) (B) (C) 2 (D) 4 2 4

28. The figure shows a metre-bridge circuit, with AB = 100 cm, X = 12 Ω and R = 18 Ω, and the jockey J in the position of balance. – +

A

t (2)

32. Young's double slit experiment is made in a liquid. The 10th bright fringe in liquid lies where 6th dark fringe lies in vacuum. The refractive index of the liquid is approximately(A) 1.8 (B) 1.54 (C) 1.67 (D) 1.2

B

G C (E/2, 2r)

X

R N

D

36. A compound microscope has magnifying power as 32 and magnifying power of eye-piece is 4, then the magnifying power of objective is (A) 8 (B) 10 (C) 6 (D) 12

82

MARCH 2010

37. Two blocks are connected by a massless string through an ideal pulley as shown. A force of 22N is applied on block B when initially the blocks are at rest. Then speed of centre of mass of block A and block B, 2 sec, after the application of force is (masses of A and B are 4 kg and 6 kg respectively and surfaces are smooth) –

B 6kg

A 4kg

(A) 1.4 m/s2 (C) 2 m/s2

O

 (A) 10.2 N upwards (B) 4.2 N downwards (C) 8.3 N downwards (D) 6.2 N upwards

CHEMISTRY

F = 22 N

1.

According to Bohr’s theory, angular momentum of an electron in fourth orbit is h 2h 4h h (B) (C) (D) (A) 2π 4π π π

2.

1.25g of a solid dibasic acid is completely neutralized by 25 ml. of 0.25 molar Ba(OH)2 solution. Molecular mass of the acid is (A) 100 (B) 150 (C) 120 (D) 200

3.

Rates of effusion of hydrogen and deuterium under similar conditions are in the ratio -

(B) 1 m/s2 (D) None of these

38. A chain of length 1.5 πR and mass ‘m’ is put on a mounted half cylinder as shown in figure. Chain is pulled by vertically downward force 2 mg. Assuming surfaces to be friction less, acceleration of chain is –

R

(A) 1 : 1 F = 2mg

(A) 2g

(B)

2g 3

(C)

g 2

(D)

XtraEdge for IIT-JEE

(D) 1 : 4

5.

Given that H2O (l) → H2O(g) ; ∆H = + 43.7 kJ H2O (s) → H2O (l) ; ∆H = + 6.05 kJ ∆Hsublimation of ice is (B) 37.65 kJ mol–1 (A) 49.75 kJ mol–1 –1 (D) – 43.67 kJ mol–1 (C) 43.7 kJ mol

6.

Which of the following is a Lewis base ? (B) BF3 (A) CO2 (C) Al3+ (D) CH3NH2

7.

The solubility product Ksp of sparingly soluble salt Ag2CrO4 is 4 × 10–12. The solubility of the salt is (B) 2 × 10–6 M (A) 1 × 10–12 M –6 (D) 1 × 10–4 M (C) 1 × 10 M

8.

Which of the following chemical reactions depicts the oxidising behaviour of H2SO4 ? (A) 2HI + H2SO4 → I2 + SO2 + 2H2O (B) Ca(OH)2 + H2SO4 → CaSO4 + 2H2O (C) NaCl + H2SO4 → NaHSO4 + HCl (D) 2PCl5 + H2SO4 → 2POCl3 + 2HCl + SO2Cl2

(B) Mg/2 (D) Mg/8

83

2 : 1 (C) 2 : 1

NH3(g) + H2S(g) For equilibrium NH4HS(s) KC = 1.8 × 10–4 at 298 K. The value of Kp at 298 K is(A) 0.108 (B) 4.4 × 10–3 (C) 1.8 × 10–4 (D) 4.4 × 10–4

M

40. A uniform rod of length 2.0 m specific gravity 0.5 and mass 2 kg is hinged atone end to the bottom of a tank of water (specific gravity = 1.0) filled upto a height of 1.0 m as shown in figure. Taking the case θ ≠ 0º the force exerted by the hinge on the rod is : (g = 10 m/s2) –

(B)

4. 5g 3

39. In hydraulic press radii of connecting pipes r1 and r2 are in ratio 1 : 2. In order to lift a heavy mass M on larger piston, the small piston must be pressed through a minimum force f equal to f

(A) Mg (C) Mg/4

1.0 m

θ

MARCH 2010

9.

COONa

Potassium has a bcc structure with nearest neighbour distance of 4.52 Å. If atomic mass of potassium is 3a, its density is (B) 804 kg m–3 (A) 454 kg m–3 –3 (C) 852 kg m (D) 900 kg m–3

(B) CH3

0 10. If E 0Zn 2 + / Zn = – 0.763 V and E Cd = – 0.403 V, 2+ / Cd

; SO3

(C)

the emf of the cell Zn | Zn2+ ||Cd2+|Cd (a = 0.004), (a = 0.2) will be given by 0.059 0.004 (A) E = – 0.36 + log 2 2 0.059 0.04 log (B) E = + 0.36 + 2 2 0.059 0.2 log (C) E = – 0.36 + 0.004 2 0.059 0.2 log (D) E = + 0.36 + 2 0.004

Br SO2 – O – C – CH3 O

(D)

; NaOH CH3

17. The product(s) obtained via oxymercuration (HgSO4 + H2SO4) of 1-butyne would be – (A) CH3CH2COCH3 (B) CH3CH2CH2CHO (C) CH3CH2CHO + HCHO (D) CH3CH2COOH + HCOOH

11. The value of P° for benzene of certain temperature is 640 mm of Hg. The vapour pressure of solution containing 2.5 g of a certain substance ‘A’ in 39.0 g of benzene is 600 mm of Hg. The molecular mass of A is (A) 65.25 (B) 130 (C) 40 (D) 80

18. Acetophenone is prepared by the reaction of which of the following in the presence of AlCl3 catalyst – (A) Phenol and acetic acid (B) Benzene and acetone (C) Benzene and acetyl chloride (D) Phenol and acetone OCH3

12. For adsorption, ∆H is (A) + ve (B) – ve (C) zero (D) may + ve or –ve

.Br2 / NaOH 1 →

19.

2 − heat

CH3

13. A reaction which is of first order w.r.t. reactant A, has a rate constant 6 min–1. If we start with [A] = 0.5 mol L–1, when would [A] reach the value of 0.05 mol L–1 ? (A) 0.384 min (B) 0.15 min (C) 3 min (D) 3.84 min

OCH3

OCH3 Br

(A)

(B)

CH3 OCH3

14. The number of molecules present in 1 cm3 of water is (density of H2O = 1 g cm–3) (B) 3.3 × 1022 (A) 2.7 × 1018 20 (C) 6.02 × 10 (D) 1000

(C)

Br

CH3 Br OCH3

(D) Br

15. CH3NH2 + CHCl3 + KOH → Nitrogen containing compound + KCl + H2O Nitrogen containing compound is – (A) CH3 – C ≡ N (B) CH3 – NH – CH3 (C) CH3 – Ν ≡ C+ (D) CH3 – N+ ≡ C–

CH3

CH3

/ H 2SO 4 H 2O 20. Phenol NaNO  2  → B  → C NaOH → D Name of the above reaction is – (A) Libermann's reaction (B) Phthalein fusion test (C) Reimer-Tiemann reaction (D) Schotten-Baumann reaction

16. 4-methyl benzene sulphonic acid react with sodium acetate to give – CH3

(A)

; SO3

CCl3

21.

; CH3COOH

1 eqv. of Br / Fe

  2 → A. Compound A is -

SO3Na

XtraEdge for IIT-JEE

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MARCH 2010

CCl3

CCl3

(A)

27. The IUPAC name of the compound Br

(B)

(A) (2E, 4E)-2, 4-hexadiene (B) (2Z, 4Z)-2, 4-hexadiene (C) (2Z, 4E)-2, 4-hexadiene (D) (2E, 4Z)-4, 2-hexadiene

Br CCl3

CCl3

(C)

28. The brown ring test for NO −2 and NO 3− is due to the formation of complex ion with the formula – (B) [Fe(NO)(CN)5]2+ (A) [Fe(H2O)6]2+ 2+ (D) [Fe(H2O) (NO)5]2+ (C) [Fe(H2O)5NO]

(D) Br

Br

Br

22. In a reaction

29. The correct order for the wavelength of absorption in the visible region is – (A) [Ni (NO2)6]4– < [Ni(NH3)6]2+ < [Ni(H2O)6]2+ (B) [Ni (NO2)6]4– < [Ni(H2O)6]2+ < [Ni(NH3)6]2+ (C) [Ni(H2O)6]2+ < [Ni(NH3)6]2+ < [Ni (NO2)6]4– (D) [Ni(NH3)6]2+ < [Ni(H2O)6]2+ < [Ni (NO2)6]4–

Hypochlorous R CH2 – OH → M → CH2 = CH2  acid CH2 – OH where M = molecule R = Reagent M and R are (A) CH3CH2Cl and NaOH (B) CH2Cl – CH2OH and aq. NaHCO3 (C) CH3CH2OH and HCl (D) CH2 = CH2 and heat

30. In nitroprusside ion,, the iron and NO exists as Fe (II) and NO+ rather than Fe(III) and NO these forms can be differentiated by – (A) Estimating the concentration of iron (B) Measuring the concentration of CN– (C) Measuring the solid state magnetic moment (D) Thermally decomposing the compound

23. Which of the following will have least hindered rotation about carbon-carbon bond – (A) Ethane (B) Ethylene (C) Acetylene (D) Hexachloroethane 24. Which is least reactive substitution (SN2) (A) CH2 = CH2 – CH2 – Cl CH3 (B) CH3 – C – Cl

(C)

towards

31. Four reactions are given below I 2Li + 2H2O → 2LiOH + H2 II 2Na + 2H2O → 2NaOH + H2

nucleophilic

III

2LiNO3 heat  → 2LiNO2 + O2

IV 2NaNO3 heat  → 2NaNO2 + O2 Which of the above if any is wrong (A) IV (B) III (C) I (D) None of these

CH3 Cl

32. Name of the structure of silicates in which three oxygen atoms of [SiO4]4– are shared is – (A) Pyrosilicate (B) Sheet silicate (C) Linear chain silicate (D) Three dimensional silicate 33. The metallic lusture exhibited by sodium is explained by – (A) Diffusion of sodium ions (B) Oscillation of loose electron (C) Excitation of free protons (D) Existence of body centred cubic lattice

(D) CH3 – CH – CH3 Cl

-

25. Among the following the least stable reasonance structure is – - - O O (B) (A) N N OO- O - O (C) (D) N N O O-

-

-

-

-

34. Hydrogen is evolved by the action of cold dil. HNO3 on – (A) Fe (B) Mn (C) Cu (D) Al

26. Homolytic fission of C–C bond in ethane gives an intermediate in which carbon is (A) sp3 hybridised (B) sp2 hybridised (C) sp hybridised (D) sp2d hybridised

XtraEdge for IIT-JEE

is –

85

MARCH 2010

35. 'Lapis-Lazuli' is a blue coloured precious stone. It is mineral of the class – (A) Sodium alumino silicate (B) Zinc-cobaltate (C) Basic copper carbonate (D) Prussian blue 36. In which of the following arrangements the order is not according to the property indicating against it – (A) Al3+ < Mg2+ < Na+ < F– (increasing ionic size) (B) B < C < N < O (increasing first I.E.) (C) I < Br < F < Cl (increasing electron gain enthalpy (–ve)) (D) Li < Na < K < Rb (increasing metallic radius)

3.

If the radius of a spherical balloon is measured with in 1 % the error (in percent) in the volume is – (A) 4 π r2 % (B) 3 %  88  (D) None (C)   % 7

4.

The relation R defined on the set A = {1, 2, 3} is given by R = {(1, 1) (2, 2)} then number of correct choices from the following is (i) reflexive (ii) symmetric (iii) Transitive (iv) anti symmetric (A) 1 (B) 2 (C) 3 (D) 4

5.

Let U be the universal set and A ∪ B ∪ C = U then {(A – B) ∪ (B – C) ∪ (C – A)}c = (A) A ∩ (B ∩ C) (B) A ∩ (B ∪ C) (C) (A ∩ B ∩ C) (D) None of these

6.

If A and B are square matrices of same size and | B | ≠ 0 then (B–1 AB)4 = (B) BA4B–1 (A) (B4)–1 AB4 –1 4 (D) None of these (C) B A B

37. Which set of hybridisation is correct for the following compound

NO2, SF4, PF6− (A) sp, sp2, sp3 (C) sp2, sp3, d2sp3

(B) sp, sp3d, sp3d2 (D) sp3, sp3d2, sp3d2

38. The increasing order of atomic radius for the elements Na, Rb, K and Mg is – (A) Mg < Na < K < Rb (B) K < Na < Mg < Rb (C) Na < Mg < K < Rb (D) Rb < K < Mg < Na

7.

(A) 0

8.

2d

e

2a

b

e

If ∆1 = 2z 4x 2 y and ∆2 = 2d e f then e 2a b 4 x 2 y 2z ∆1/∆2 =

(A) 1

MATHEMATICS

9.

y = 2x2 – log | x | passes (A) two minima & one maxima (B) Two maxima and one minima (C) Only two minima (D) Only two maxima

(B) 2

(C)

1 2

(D) None

The number of ways in which 20 one rupee coin can be distributed among 5 people such that each person gets at least 3 rupee is – (A) 26 (B) 63 (C) 125 (D) None

10. The total number of six digit number x1 x2 x3 x4 x5 x6 have the property that x1 < x2 ≤ x3 < x4 < x5 ≤ x6 is equal to – (A) 10C6 (B) 12C6 (C) 11C6 (D) None

The function f(x) = 1 + x sin x [cos x], π 0 28.[B]

| m |=

l0 E 50 cm 2

ωo = ωI a r 1 ∴ I = I = a o ro 2

X l = for balance R 100 − l l' 12 12 l Initially, = , finally = 18 100 − l 8 100 − l' or JJ′ = l′ – l = 20 cm

34.[B]

0.1 100 0.1 ) 2r = N + n ( 100 N n r= + (0.001) 2 2 r = 0.5 N + n (0.0005) r = 5 (0.1 N + 0.0001 n)

29.[D] Least count =

30.[C]

6 2 3 Li +1

f = – 15cm for virtual & 2 times large image m=+2 −15 f or + 2 = m= f −u − 15 – u – 30 – 2u = – 15 – 2u = 15 u = – 7.5 cm

35.[A] MP =

f 0 200 = = 40 fe 5

36.[A] M.P. = m0 × me 32 = m0 × 4 ∴ m0 = 8

H → 42 He + 42 He

Q = (K (K α1 + K α 2 ) – Kd = (13.2 + 13.2) MeV – 4MeV = 22.4 MeV

37.[A] F – 2T = 6a and T = 4 × 2a ∴ F – 16 a = 6a F ⇒ a= ⇒ a = 1m/s2 22 6 ×1 + 4 × 2 ∴ aCM = = 1.4 m/s2 10

31.[D] N = N0e–λt, D = N0 (1 – e–λt) R R 0 e – λt R 0 R = R0e–λt, = = =λ N N 0 e – λt N 0

= const.

38.[D] Acceleration of chain is given by 2mg − mg / 3 5g = a= 3 m

32.[C]

10 β1 = 10 ×

f 1 − 10 = = f − u − 10 + 30 2

λD µd

39.[C] According to Pascal’s principle

in liquid λD β2 = d 6β2 = 10β1 6λD 10λD = µd d 10 = 1.67 µ= 6

r2 f1 A 1 = 1 = 1 = 2 f2 A2 4 r2

f1 =

1 Mg 4

40.[C] Length of rod inside the water = 1.0 secθ = secθ F

33.[A] f = 10 cm

θ

1 cm

W

1.0 m

O 2 Upthrust F =   (sec θ) 2

30 cm

XtraEdge for IIT-JEE

115

 1    (1000) (10)  500 

MARCH 2010

or F = 20 sec θ Weight of rod W = 2 × 10 = 20 N For rotational equilibrium of rod net torque about O should be zero.  secθ  ∴F   (sin θ) = W = (1.0 sin θ)  2 

11.[D]

20 sec2θ = 20 2 or θ = 45º ∴ F = 20 sec 45º

or

12.[B]

= 20 2 N

2.[D]

Meq. of Acid = Meq. of Ba(OH)2 1.25 ⇒ × 1000 = (0.25 × 2) × 25 M/2 ⇒ M = 200

3.[B]

4.[A]

r( H 2 ) r( D 2 )

M ( D2 )

=

M ( H2 )

14.[B]

=

∆Hsublimation = ∆Hfusion + ∆Hvap

6.[D]

In CH3NH2, N has one lone pair of electrons.

9.[D]

a3 × NA

–12

=

Ecell = E 0cell +

1× 6.023 × 1023 18

CH3 – CH2 – C ≡ CH + H2O HgSO   4 → But–1–yne H SO 2

4

CH3 – CH2 – C = CH2 OH − enol tautomeris m Keto   → CH3 – CH2 – C – CH3 O

–4

18.[C]

Anhyd . AlCl

+ CH3COCl   3 → COCH3

2 × (3a × 10 –3 ) 3

+ HCl

 4 × 4.52 × 10 – 3    × 6.02 × 10 23   2 3  

= 900 kg m–3 10.[B]

1 cm3 H2O = 1 g H2O

16.[A] 4-methyl benzene sulphonic acid is stronger than acetic acid thus it will release acetic acid from sodium acetate.

Ksp = 4s = 4 × 10 ⇒ s = 10 M Oxidant is the one whose O.N. decreases during the reaction. H2SO4 (O.N. of S = + 6) changes to SO2 (O.N. of S = + 4) d=

Adsorption is exothermic process due to attraction between adsorbate and adsorbent.

15.[D] Isocyanide test also known as carbylamine test.

2 4 = 2 1

5.[A]

Z× M

2.5 × 78 × 640 = 80 39 × 40

= 3.3 × 1022

17.[A]

7.[D] 8.[A]

⇒ MB =

No. of molecules in 1 g H2O =

From Kp = Kc (RT)∆ng = 1.8 × 10–4 × (0.082 × 298)2 = 0.108

3

WBM A WA (∆P / P°)

13.[A] t =

h 2h nh =4× = 2π 2π π

mvr =

or MB =

2.303 a log a−x k 2.303 0 .5 = log 6 0.05 = 0.384 min

CHEMISTRY 1.[C]

∆P WB M A = M B WA P°

OCH3

[Cathode] 0.059 log [Anode] n

19.[B] Br

0.059 0.004 log = [– 0.0403 – (– 0.763)] + 0.2 2 0.059 0.04 log = + 0.36 + 2 2

XtraEdge for IIT-JEE

CH3

20.[A] Libermann's reaction

116

MARCH 2010

CCl3

22.[B]

Br

1 eqv. at Br / Fe

  2 →

21.[A]

30.[C] The existance of Fe2+ and NO+ in nitroprusside ion [Fe(CN)5NO]2– can be established by measuring the magnetic moment of the solid compound which should correspond to Fe2+ = 3d6 four unpaired electron.

CCl3

CH2 = CH2 HOCl →

31.[B]

CH2 – CH2   → CH2 – OH (glycol) Cl CH2 – OH OH

32.[D] Three dimensional sheet structure are formed when three oxygen atoms of each [SiO4]4– tetrahedral are shared.

23.[A] Free rotation around carbon-carbon bond takes place easily in alkanes. Now ethane and hexachloroethane both are alkanes, but in hexachloroethane bulky chlorine atom is present while ethane is least hindered. 24.[C] Due to the presence of –Cl group which is a +M group.

33.[B]

Due to oscillation of free electron Na metal shows metallic lusture.

34.[B]

Mn + 2HNO3 → Mn (NO3)2 + H2

35.[A] 'Lapis Lazuli' is the aluminium silicate present in the earth rocks as blue stone.

-

25.[A] Due to similar charges on adjacent atom the structure is least stable. - O N O-

36.[B]

B < C < N < O when we move from B to O in a periodic table the first ionization enthalpy increase due to the attraction of nucleus towards the outer most of electron and IE of N > O.

37.[B]

NO2 → sp SF4 → sp3d

-

26.[B]

•

•

CH3 – CH3 Homolytic  → C H 3 + C H 3 bond fission

methyl free radical

Free radical is formed which is sp2 hybridised H H

LiNO3 on heating gives ∆ 2Li2O(s) + 4NO2 + O2 4LiNO3 →

aq . NaHCO 3

PF6− → sp3d2

C—H

38.[A] Mg belongs to group 2. Therefore its size is less than that of Na.

27.[A] If atom or group of higher priority are on opposite direction at the double bond of each carbon atom then the configuration is known as E and if they are in same direction then the configuration is known as Z-configuration.

39.[B]

Alkali metal hydroxide KOH is highly soluble in water.

40.[B]

Na2CO3 + H2O + CO2 → 2NaHCO3

MATHEMATICS

(2E, 4E) – 2, 4-hexadiene 28.[C] The brown ring test for NO −2 and NO 3− is due to formation of [Fe(H2O)5NO]2+

1.[C]

29.[A] The absorption of energy or observation of color in a complex transition compounds depend upon the charge of the metal ion and the nature of the ligand attached. The same metal ion with different ligands shows different absorption depending upon the type of ligand, the presence of weak field ligand make the central metal ion to absorb low energies i.e. of higher wavelength.

XtraEdge for IIT-JEE

y = 2x2 – log | x | 1 |x| 1 dy = 4x – × = 4x – dx |x| x x dy = dx

1 1 4( x + )( x − ) 2 2 x – – + 0 –1/2

+ 1/2

1 1 and x = but x = 0 2 2 is not point of maxima as x = 0 is not in the domain.

∴ y has minima at x = –

117

MARCH 2010

2.[A]

3.[B]

f(x) = 1 + x sin x [cos x] π Q0 cos2 x ∴

dy ⇒ < cot–1 < π, n∈ N π 6 6 π

{ Q cot–1x ∈ 0 (a, π)} n ⇒–∞< < 3 π

2

– ∞0 dx

 π in  0,  at which f(x) has local maxima.  2

1− 2 sin 2 x − cos 2 x + 1 ⇒ ≤ ≤ 2 2

35.[B]

∴

∴ at x1 slope change from + ve to – ve ∴ There is only one critical point

3K.7 K.8K 7 7 R = ⇒ = 4.6K 2K K 2 r 2

1− 2 ≤K≤ 2

π ) say x, at 2

dy =0 dx

at x1 – h ⇒ x2 < cos2x

34.[D] sinx (sinx + cosx) = K ⇒ sin2 x + sinx cos x = K 1 − cos 2 x sin 2 x ⇒ + =K 2 2 1 ⇒ (sin2x – cos2x + 1) = K 2

⇒

π

π/2

There is only one point in (0,

Let a = 3K, b = 7 K and c = 8K a+b+c ∴s= = 9K 2 abcs R abc s = . = there r 4∆ ∆ 4s(s − a )(s − b)(s − c)

Q – 2 ≤ sin 2x – cos 2x ≤

y = cos2x

dy = 0 ⇒ x2 = cos2x dx

32.[D] Triangle is right angled at O(0, 0). Therefore orthocentre is O(0, 0) and circumcentre is mid a b point of hypotense i.e.  ,  2 2

=

(1 + x tan x ) 2

O

Clearly a > 0 Also P lies on that side of line x + y = 2 Where origin lies ∴ a + a2 – 2 < 0 ⇒ (a – 1) (a + 2) < 0 ⇒ – 2 < a < 1 but a > 0 ∴0 0 and [x] ≠ 1 ⇒ x ≥ 2 ∴ x ∈ [2, ∞)

For domain

⇒

⇒

x →a

x →0

45.[B]

log e {1 + 6f ( x )} =2×1=2 6f ( x )

1 x2f(x) + f   = 2 x 3

∫ f (x )dx

I=

1/ 3

⇒I= –

h →0

42.[B]

3

⇒ 2I =

=

(1 + h ) − 1

= lim

h →0

1−1 2h + h 2

=0

0 −1 = lim =∞ f(1 – h) = lim 2 h →0 (1 − h ) − 1 h →0 − 2h + h 2

43.[A] (a + bx)ey/x = x ... (1) Differentiating, w.r.t. x we get

be

+ (a + bx)e

 x. y − y  .  12  = 1   x

∫

3

2

∫x

2

dx

1/ 3

8 3

The pattern is x2 +1, x2 + 2, . . . . Missing number = 28 × 2 + 3 = 59

2. [A]

A car runs on petrol and a television works by electricity.

3.[A]

All except Titans are planets of the solar system.

4. [C]

 xy − y  ⇒ bey/x + x.  1 2  = 1 {Q (a + bx)ey/x = x}  x 

5. [B]

⇒ bxey/x + xy1 – y = x ⇒ xy1 – y = x – bxey/x ⇒ xy1 – y = aey/x ... (2) (from (1))

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 1  f   dx  x 

1. [D]

⇒ f(x) is discontinuous at x = 1

y/x

dx

LOGICAL REASONING

[(1 – h ) 2 ] – 1

y/x

2

2

1/ 3

 2  1  1  x f ( x ) + f   2 dx =  x  x 1/ 3 

⇒I=

h →0

1

1 1

∫ f  x . x

1  1  16 = − 2  = – 2  − 3 = 3  x 1 / 3 3 

f(1) = 0 2



3

3

∴P=0

[(1 + h ) 2 ] − 1

1 1 f  . 2 dt = t t

∫  f (x) + x

1/ 3

1 1 , dx = − 2 dt t t

3

2

f(1 + 0) = lim

∫ 3

2

h2 + h =0 1− h

put x =

1/ 3

x − 9 x + 20 (4 − h ) − 9(4 − h ) + 20 lim– = lim h →0 4 − h − [4 − h ] x →4 x − [x]

= lim

2

1

2

∫ f (x )dx

= (–2)3 + (–1)3 + 03 + 13 + 23 = 0

h +h (5 + h ) − 9(5 + h ) + 20 = lim =0 h →0 h 5 + h − [5 + h ]

h →0

0

3

+ f ( x )dx +

2

= lim

∫

∫

π 2

x − 9 x + 20 x − [x]

lim

x →5 +

∫

1

f ( x )dx + f ( x )dx

−1

−2

2

41.[D]

∫

0

f ( x )dxt 2

log e {1 + x} =1 x

Q lim

∫

f ( x )dx =

−2

40.[C] ∴ f(a) = 0 log e {1 + 6f ( x )}  0  ∴ lim   form x →a 3f ( x ) 0

⇒ lim 2 ×

−1

3

44.[B]

|x| | x | = 1 then log[x]   =0 x  x 

f(x) = cos–10 =

1 (xy1 – y)2 = x3 y2

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Delicious : Irrelevant as it means 'something very tasty.' Comforting : 'Irrelevant' as it means 'giving necessary comforts', whereas 'Plush' means more than comforts. Tasty : (Irrelevant) It means 'delicious'

6. [D]

7.[B]

The third figure in each row comprises of parts which are not common to the first two figure.

8. [A]

9. [C]

4.[A]

Lively : Correct synonym to 'sprightly' as both means, 'someone dashing/energetic/enthusiastic'. Beautiful : (Irrelevant) Sportive : (Irrelevant) Intelligent : (Irrelevant)

5.[D]

Wicked : It is almost a synonym to 'Astute' Impolite : Irrelevant because it is the antonym of 'polite'. Cowardly : Irrelevant as it is the opposite of 'bravely'. Foolish : (It's the correct antonym of 'Astute' which itself means 'clever, shrewd'.

6.[D]

Deadly : It means 'Fatal'. Hence, this is not a proper antonym to 'innocuous'. Ferocious : It means 'horrible' Hence, irrelevant to the opposite of 'innocuous'. Poisonous : It means 'venomous'. Hence, an irrelevant 'antonym'. Harmful : It is a perfect antonym of innocuous which itself means 'harmless'.

7.[D]

Corruption : Irrelevant Worldliness : Irrelevant Favouritism : Irrelevant Nepotism : (Correct Answer) because It's a kind of corruption in which the authority in power takes the advantage of giving opportunity to their relatives in their self interest.

8.[B]

Cross : (to pass by, to intersect) It means different Hence, irrelevant. Shuttle : (Proper answer)

10.[A]

ENGLISH 1.[B]

Geraff : Incorrect spelling.

• 'e' should be replaced with 'i' • The word should end with 'e' after 'ff' Giraffe : Correct spelling. Giraf : 'fe' is to be added in the end. Gerraffe : • 'Ge' is to be replaced with 'Gi' to make the correct spelling. 2.[B]

3.[A]

Puncture : No error. It makes the tyre flat. Puntuation : Error of spelling Correct spelling is 'Punctuation' Hence 'c' is missing. Pudding : No error It is used as 'Dessert' Pungent : No Error It is some what 'sharp' and 'shrill'. Luxurious : (Plush) Something full of all 'amenities' making life 'cozy' and 'snug'.

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123

MARCH 2010

It's a kind of "regular beats" of an air flight or bus service between the two stations. Travel : It means to journey. Hence, irrelavent. Run : (to move regularly) Hence, irrelevant. 9.[D]

12.[A] "Meatless days" This is the name of a novel. Hence, no error is there. Have been made : (Erroneous) Because 'have' should be replaced with 'has' because 'meatless days' is a singular noun. Into a film : No error in this part of the sentence. No error : Incorrect option because there is an error in the sentence.

Only 1 is correct : Inappropriate answer because sentence 1 can't be correct using 'practise' as it is a verb, whereas the required word should be a noun. Only 2 is correct : Sentence 2 is also wrong because the word 'practice' is wrongly used as a verb. It should be a verb like 'practise'. Hence, incorrect answer. Both the sentences 1 and 2 are correct. This is not relevant. Both the sentences 1 and 2 are not correct. Correct option, if both the words, i.e. 'practice' and 'practise' are interchanged respectively, it really makes a meaningful sentence.

13.[C] Looking forward : (No error) This is a phrase. 'to' (no error) This is a preposition. 'Meet you here' (erroneous) Because 'meet will be replaced with 'meeting' Phrase 'looking forward to' is followed by present participle (V. I + ing) form of the Verb. No error : (incorrect option) Part 'C' is erroneous. 14.[C] Good and Evil This is a wrong interpretation. Former and Latter : Wrong interpretation. For and against a thing. Appropriate option as it really suits the Idiom ins and outs. Foul and Fair : (by hook or by crook) This is an inappropriate option.

10.[C] Sentence 1 is correct : This option is wrong because the word 'ingenuous' means 'frank and simple' which is inappropriate. Sentence 2 is correct : This option is also wrong because the word 'ingenious' means 'clever or prudent' and this is inappropriate. Both the words, i.e. 'ingenuous' and 'ingenious' if interchanged together respectively, it really makes both the sentences meaningful. Hence, appropriate option. Both the sentences can't be interchanged. This is an incorrect option because words have been misinterpreted together. Incorrect option.

15.[A] Broke out : (to start suddenly) 'Correct and relevant' option because it is used for 'wars' and 'diseases' e.g. cholera broke out in Surat in 1985. Set out : (to start) it is different because it is used when one leaves for somewhere e.g. He set out on his long voyage to Achilese. took out : (incorrect use) Because it means differently. e.g. He took out a one rupee coin to give to the beggar. Went out : (Incorrect use) Because meaning is different e.g. : The light went out when I was preparing for my Board Exams.

11.[C] Far off : It can't be used in place of 'aloof' as far off' means long-long ago. Hence, incorrect alternative . Introvert : It means 'self-centred', Hence, It is an incorrect alternative. distance : This is an appropriate word because one of the meaning of 'aloof' is distant also while keeping distance between two nouns. Depressed : (it means 'hopeless') Hence, quite irrelevant.

Hence,

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inappropriate

MARCH 2010

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125

MARCH 2010