Dear Students,
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.
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.
Yours truly
Pramod Maheshwari, B.Tech., IIT Delhi
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Editor : Pramod Maheshwari
Worry is a misuse of imagination.
Volume - 5 Issue - 9 March, 2010 (Monthly Magazine)
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Volume-5 Issue-9 March, 2010 (Monthly Magazine)
NEXT MONTHS ATTRACTIONS
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
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 made by sitting down.
• To succeed, we must first believe that we can.
• The secret of joy in work is contained in one word - excellence. To know how to do something well is to enjoy it.
• 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.
CONTENTS
INDEX PAGE
NEWS ARTICLE
4• Top medical honchos mull medical courses in IITs
• Promise of IITs, from Mamata
IITian ON THE PATH OF SUCCESS
7 Mr. R. MadhavanKNOW IIT-JEE
10Previous IIT-JEE Question
XTRAEDGE TEST SERIES
51Mock Test IIT-JEE Paper-1 & Paper-2 Mock Test AIEEE
Mock Test BIT SAT
SOLUTIONS
90Regulars ...
DYNAMIC PHYSICS
178-Challenging Problems [Set# 11] Students’ Forum
Physics Fundamentals
Matter Waves, Photo-electric Effect Thermal Expansion, Thermodynamics
CATALYST CHEMISTRY
32 Key ConceptPurification of Organic Compounds Boron & Carbon Family
Understanding: Inorganic Chemistry
DICEY MATHS
41Mathematical Challenges Students’ Forum Key Concept
Definite integrals & Area under curves Probability
Study Time...
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
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 of IITs, from Mamata
Mamata Banerjee today promised a deal with the Union human resource development ministry to build “schools, colleges and IITs” on railway land.
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 IIT-Kharagpur 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 Jhargram-Purulia and another Medinipur-Jhargram." 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.
"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
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 Indo-German 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
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
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 Under-graduate 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 under-graduate 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.
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 ‘world-class’ 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 conduc-ted 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
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 IIT-Kharagpur for placement at Singapore. IIT-Roorkee also achieved around.
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. . .
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.
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.
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!
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.
Mr. R. Madhavan
Mechanical Engineering, IIT Madras
Success Story
LEARNING FROM ISRAEL
In 1996, I visited Israel because I had heard that they are the best in water technology. Take the case of corn: they harvest 7 tonnes per acre whereas we produce less than a tonne. They harvest up to 200 tonnes of tomatoes, whereas here it is 6 tonnes, in similar area of land. I stayed in one of the kibbutz, which is a co-operative farm for 15 days. I understood what we do is quite primitive. It was an eye opener for me. They treat each plant as an industry. A plant producing one kilo of capsicum is an industry that has 1 kilo output. I learnt from them that we abuse water. Drip irrigation is not only for saving water but it enhances your plant productivity. We commonly practice flood irrigation where they just pump water. As per the 2005 statistics, instead of 1 litre, we use 750 litres of water.
DR. LAKSHMANAN
I met Dr Lakshmanan, a California-based NRI, who has been farming for the last 35 years on 50-60,000 acres of land. He taught me farming over the last one decade. Whatever little I have learnt, it is thanks to him. I knew a farm would give me much better returns in terms of money as well as happiness. Working for money and working for happiness are different. I work and get happiness. What more do you need?
NO GUIDANCE IN INDIA
I said at one platform that we have to change the curriculum of the agricultural universities. What they teach the students is not how to farm, but how to draw loans from a bank! What they learn cannot be transformed to reality or to the villages. The problem in the villages is not mentioned in the university. There is a wide gap and it is getting worse.
MAKING PROFITS
After burning my fingers for four years, from 1997 onwards, I started making profits. Even though it took me four years, I did not lose hope. I knew this was my path ven though I didn't have any guidance from anyone.
In those days, communication was slow. Today, I can get guidance from Dr Lakshmanan on Skype or Google Talk, or through e-mail. I send him the picture of my problem and ask his guidance. In those days, it took time to communicate. There was no Internet or connectivity. That was why it took me four years to learn farming. Today, I would not have taken more than six months or even less to learn the trick!
THE FARMING CYCLE
I started crop rotation after 1997. In August, I start with paddy and it is harvested in December. I plant vegetables in December itself and get the crops in February. After that, it is oil seeds like sesame and groundnut, which are drought-resistant, till May. During May, I go on trips to learn more about the craft. I come back in June-July and start preparations on the land to get ready for August. In 1999, I bought another four acres. My target is a net income of Rs 100,000 per annum per acre. I have achieved up to Rs 50,000.
I sell my produce on my own. I have a jeep and bring what I produce to my house and sell from there. People know that I sell at home. I don't go through any middle man. I take paddy to the mill, hull it and sell it on my own. In the future, I have plans to have a mill too. These days, people tell me in advance that they need rice from me. I have no problem selling my produce.
ENGINEERING HELPS IN FARMING
More than any other education, engineering helps in farming because toiling in the soil is only 20 per cent of the work. About 80 per cent of farming needs engineering skills. Science is a must for any farming. I have developed a number of simple, farmer-friendly tools for farming areas like seeding, weeding, etc. as we don't have any tools for 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 small farmers for agriculture. The village will prosper with food processing industries coming there. My yield will also be more with more land.
Don't compare yourself with anyone in this world.
If you do so, you are insulting yourself. – Alen Strike
PHYSICS
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 Q Y Vx=4m/s V´ X +Y Vsinθ Vx=4m t = 0 t = t Vcosθ +Z P V θ L
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´ = 2 2
x) (Vsin )
V
( + θ
= 42+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
T – 2mg cos α =
l
2
mv 2
Given that T = 0 when α = 90º ∴ 0 – 0 = l 2 mv 2 ⇒ v = 0 2mg cosα ∞ M α α 2mg sinα 2mg Q
⇒ Velocity is zero when α = 90º, i.e., in the horizontal position.
Applying energy conservation from Q to M, we get
2 1 2mV2 = 2mgl ⇒ l = g 2 V2 = 8 . 9 2 ) 5 . 2 ( 2 × = 0.318 m
2. 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] F → x l ω B C A y
(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.
KNOW IIT-JEE
By Previous Exam QuestionsSol. The mass B is moving in a circular path centred at A. The centripetal force (mlω2) 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'cos60º Fnet = 2 1 ' F 2 ' F 2 2+ 2× = 3F′ = 3mlω2 X l l B C A Y 60º Fnet F′ F′ l F′ F′
(b) The force F acting on B will provide a torque to the system. This torque is
F × 2 3 l = Iα F × 2 3l = (2ml2)α ⇒ α = l m F 4 3
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. l 2 3 F c.m Therefore F + (Fnet)x = 3m(ax) c.m. = 3m m 4 F Q ax = αr = l m F 4 3 × 3 l = 4 F = 4 F 3 ∴ (Fnet)x = 4 F
(Fnet)y remains the same as before = 3mlω2.
3. 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
v = 2gx …(i) The mass of liquid flowing per second through the orifice
= ρ × volume of liquid flowing per second dt
dm = ρ × 2gx× 100
A …(ii)
Therefore the rate of change of momentum of the system in forward direction
= dt dm × v = 100 A gx 2 × ×ρ
(From (i) and (ii)) 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 P v x
∴ The rate of change of momentum of the system in the backward direction
= Axρ × a
By conservation of linear momentum
Axρ × a = 100 gxA 2 ρ ⇒ a = 50 g
(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
∴ m0 = Ah × ρ ⇒ h = ρ A m0 ∴ v′ = ρ = ρ × × A 2 gm A m 25 . 0 g 2 0 0
4. 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 cross-sectional 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 the frequency of oscillation. [IIT-1981] Sol. Let the piston be displaced by a distance x.
Then γ − γ + + = + ) Ax v ( p A Mg p V A Mg p0 0 0 0
Q Initial pressure on the gas P1 = p0 +
A Mg
Final pressure on the gas P2 = p0 +
A Mg+ p
V0 x
A P0
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
1 = γ γ + − + + 0 0 0 0 V A Mg p ) Ax V ( p A Mg p = γ − + + 0 0 V Ax 1 A Mg p p 1 ⇒ 1 = 1 + A Mg p p 0+ – 0 V Ax γ + γ + 0 0 V Ax A Mg p p
Negligible as p and x are small ∴ A Mg p p 0+ = 0 V Ax γ ∴ p = 0 0 V Ax A Mg p γ + ⇒ 0 0 MgA VAx p A F γ + = ⇒ F = 0 2 0 V x A A Mg p γ + ⇒ Ma = 0 2 0 MgA AV x p γ + ⇒ a = M V x A A Mg p 0 2 0 γ +
Comparing it with a = ω2x we get ω2 = M V x A A Mg p 0 2 0 γ + ∴ ω = M V x A A Mg p 0 2 0 γ + If A Mg
is small as compared to p0 then
ω = M V A p 0 2 0γ = 2πf ∴ f = π 2 A M V p 0 0γ
5. 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] (a) Find the atomic number of the atom.
(b) Calculate the smallest wavelength emitted in these transitions.
(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 n Smallest λ Now, we have 2 2 n ) eV 6 . 13 ( z − = – 0.85 eV …(i) and 2 2 ) 3 n ( ) eV 6 . 13 ( z + − = – 0.544 eV …(ii) Solving (i) and (ii) we get n = 12 and z = 3 (b) Smallest wavelength λ is given by
λ
hc= (0.85 – 0.544)eV
CHEMISTRY
6. 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.
[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
2CH3CHO aq.KOH→ OH| CHO CHCH CH3 2 → heating CH3CH=CHCHO
The reduction of E gives compound C. Hence, we have CH3CH=CHCHO [→H] ) C ( OH CH CH CH CH3 2 2 2
Finally, the structure of A can be obtained from the two alcohols (CH3CH2OH and CH3CH2CH2CH2OH) produced on treating A with LiAlH4. Thus, we have
) A ( CH COOCH CH CH CH3 2 2 2 3 →LiAlH4 ) C ( OH CH CH CH CH3 2 2 2 + ) B ( OH CH CH3 2
Thus, the reactions involved are as follows.
CH3CH2CH2COOCH2CH3 (A) LiAlH4 CH3CH2OH (B) CH3CH2CH2CH2OH (C) [O] CH3CHO → CH3COOH (D) (F) aq KOH CH3CHCH2OH OH heating CH3CH=CHCHO [H]
Alternatively, the compound A may be
CH3COOCH2CH2CH2CH3
(butylacetate) [O]
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,
Number of atoms per unit cell =
8
8 + 1 = 2
Area of the base = b × ON [From fig.(ii)] = b × a sin 60º = 2 3 a2 (Q b = a) a γ b α c β Figure (i)
Volume of the hexagonal cell = Area of the base × height =
2 3 a2.c But c = 3 2 2 a
∴ Volume of the hexagonal cell = 2 3 a2 . 3 2 2 a = a3 2
and radius of the atom, r = 2 a
Hence, fraction of total volume or atomic packing factor
= cell hexagonal the of Volume atoms 2 of Volume 60º N b O a figure (ii) = 2 a r 3 4 2 3 3 π × = 2 a 2 a 3 4 2 3 3 π × = 2 3 π = 0.74 = 74%
∴ The percentage of void space = 100 – 74 = 26% 8. 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. 0 Combustion H ∆ (ethane) = – 372.0 0 Combustion H ∆ (propane) = –530.0 ∆Hº for C (graphite) → C(g) = 172.0 Bond energy of H – H = 104.0 0 f H ∆ of H2O (l) = – 68.0 0 f H ∆ of CO2(g) = – 94.0 [IIT-1990] Sol. Given that,
C2H6(g) +
2
7O2(g) → 2CO2(g) + 3H2O(l)
∆H = – 372.0 kcal mol–1 ...(i) C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
∆H = – 530.0 kcal mol–1 ...(ii) ∆H0 for C(gr) → C(g) = 172.0 kcal mol–1
Bond energy of H – H = 104.0 kcal mol–1
0 f
H
∆ of H2O(l) = –68.0 kcal mol–1
0 f
H
∆ 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,
0 reaction
H
∆ = ∆H0product – ∆H0reactant
From eq. (i),
0 reactant ∆H = 2 × ∆H0CO2+ 3 ×∆HH02O – ∆H0C2H6 or – 372.0 = 2 × (–94.0) + 3 × (–68.0) – 0 H C2 6 ∆H
or ∆H0C2H6 = – 20 kcal mol–1 [for eq.(iii)] From eq. (ii)
0 reactant ∆H = 3 × ∆H0CO2+ 4 ×∆HH02O – ∆H0C3H8 or – 530.0 = 3 × (–94.0) + 4 × (–68.0) – 0 H C3 8 ∆H
or ∆H0C3H8 = – 24 kcal mol–1 [for eq.(iv)]
0 reactant
∆H = Sum of bond energies of reactants – Sum of bond energies of products From eq.(iii), 0 reactant ∆H = 2 × [C(s) → C(g)] + 6 × H → H 2 1 2 – [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 ...(v) From eq. (iv),
0 reactant ∆H = 3 × [C(s) → C(g)] + 8 × H → H 2 1 2 – [2 × C – C + 8 × C – H] or – 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
9. The electrode reactions for charging of a lead storage battery are
PbSO4 + 2e– → Pb + SO 42–
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]
Sol. Volume of 100 g of 15.7% H2SO4 =
11 . 1
100 = 90.9 ml
90.9 ml of 15.7% H2SO4 contains = 15.7 g H2SO4 ∴ 2.0 L of 15.7% H2SO4 contains = 9 . 90 0 . 2 1000 7 . 15 × × = 345.43 g H2SO4 Volume of 100 g of 36.9% H2SO4 = 28 . 1 100 = 78.125 ml
78.125 ml of 36.9% H2SO4 contains = 36.9 g H2SO4 ∴ 2.0 L of 36.9% H2SO4 contains = 125 . 78 0 . 2 1000 9 . 36 × × = 944.64 g H2SO4
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 g H2SO4 requires = 98 21 . 599 F electricity = 98 21 . 599 × 96500 C electricity We know, Q = It Given that, t = 100 hrs = 100 × 3600 s ∴ I = t Q = 3600 100 98 96500 21 . 599 × × × A = 1.638 A ∴ Current used for charging battery = 1.638 ampere 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,
0 f
G
∆ (CO) = – 32.807 kcal mol–1
0 f
G
∆ (CO2) = –94.260 kcal mol–1
0 f
G
∆ [H2O (g)] = – 54.635 kcal mol–1
0 f
G
∆ [H2O(l) = – 56.69 kcal mol–1 (a) For the reaction,
CO(g) + H2O(g) CO2(g) + H2(g)
0 reaction
G
∆ = ∆G0f(CO2) –∆Gf0[H2O(g)] – ∆G0f(CO) = –94.260 – (– 54.635) – (–32.807) = – 6.818 kcal = – 6818 cal Also, ∆G0 = – 2.303 RT log 1 p K ∴ –6818 = –2.303 × 1.987 × 298 × log 1 p K or logKp1 = 5.00 or Kp1= Antilog 5.00 = 1.00 × 105 atm (b) For the reaction,
H2O(l) H2O(g) 0 reaction G ∆ = 0 f G ∆ [H2O(g) ] – 0 f G ∆ [H2O(l)] = – 54.635 – (– 56.69) = 2.055 kcal = 2055 cal Also, ∆G0 = –2.303 RT log 2 p K ∴ 2055 = –2.303 × 1.987 × 298 log Kp2 or logKp2= – 1.507 = 2.493 or Kp2= Antilog 2.493= 3.122 × 10–2 PH2O = Kp2= 3.122 × 10–2 atm
(c) CO(g) + H2O(l) CO2(g) + H2(g) Initial Pressure 1 1 1 Pressure at Equlibrium 1 – x 1 + x 1 + x Hence, Kp3 = Kp1. Kp2 = CO H CO P P . P 2 2 or 1.00 × 105 × 3.122 × 10–2 = x 1 ) x 1 )( x 1 ( − + + or x2 + 3124x – 3121 = 0 or x = a 2 ac 4 b b± 2− − = 1 2 ) 3121 ( 1 4 [ ) 3124 ( 3124 2 × − × × − ± − = 0.9987 ∴ PCO = 1 – x = 1 – 0.9987 = 1.3 × 10–3 atm 2 H P = PCO2 = 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
C(z2) B(z3) A(z1) 3 1 3 2 z z z z − − = | z z | | z z | 3 1 3 2 − − eiπ/2 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)
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]
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) = I2 + I2 = I2
= I (where I is any integer) Therefore, P = (I)2 = Integer 13. Evaluate :
∫
x sin ) x 2 (cos 1/2 dx [IIT-1987] Sol. I =∫
x sin ) x 2 (cos 1/2 dx =∫
− x sin x sin x cos2 2 dx =∫
cot2x−1dxputting, cot x = sec θ
⇒ –cosec2x dx = sec θ tan θ dθ.
We get, I =
∫
sec2x−1. ) sec 1 ( tan . sec 2θ + − θ θ dθ = –∫
θ + θ θ 2 2 sec 1 tan . sec dθ = –∫
θ + θ θ 3 2 cos cos sin dθ = –∫
θ + θ θ − ) cos 1 ( cos cos 1 2 2 dθ = –∫
θ + θ θ − θ + ) cos 1 ( cos cos 2 ) cos 1 ( 2 2 2 = –∫
secθ dθ + 2∫
θ + θ 2 cos 1 cos dθ= –log |sec θ + tan θ| + 2
∫
θ − θ 2 sin 2 cos dθ
= – log |sec θ + tan θ| +
∫
−t2
2
dt , where sin θ = t
= – log |sec θ + tan θ| + 2.
2 2 1 log θ − θ + sin 2 sin 2 +c
= – log |cot x + cot2x−1| + x tan 1 2 x tan 1 2 log 2 1 2 2 − − − + + c 14. Show that
∫
0π 2/ f(sin2x)sinxdx=∫
π 4/ 0 f(cos2x)cosxdx 2 [IIT-1990] Sol. Let, I =∫
π 2/ 0 f(sin2x)sinxdx ...(1) Then, I =∫
π 2/ 0 −π x 2 2 sin f .sin −π x 2 dx =∫
π 2/ 0 f{sin 2x} . cos x dx ...(2)adding (1) and (2), we get 2I =
∫
π 2/0 f(sin 2x) . (sin x + cos x) dx
= 2
∫
π 4/0 f(sin 2x) . (sin x + cos x) dx
= 22
∫
π 4/ 0 f(sin 2x) sin +π 4 x dx = 22∫
π 4/ 0 f −π x 4 2 sin . sin π− +π 4 x 4 dx = 2 2∫
π 4/ 0 f(cos 2x) . cos x dx ∴ I = 2∫
π 4/ 0 f(cos 2x) . cos x dx Hence,∫
π 2/ 0 f(sin(2x)) . sin x dx = 2∫
π 4/ 0 f(cos 2x) . cos x dx15. 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 5 4 2 O –2 –4 –5 x x2 + 16 ) x 4 ( − 2 = 25 ⇒ (x2 + 24) (x2 – 16) = 0 ⇒ x = ± 4 ∴ Required area = 2 − − − − −
∫
∫
∫
dx 4 4 x dx 4 x 4 dx x 25 2 0 4 2 2 2 4 0 2 = 2 + − − 4 0 1 2 5 x sin 2 25 x 25 2 x – − − − 4 2 3 2 0 3 x 4 3 x 4 1 3 x x 4 4 1 = 8+ π 4 25Passage # 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 V I φ V I φ
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.
V/f
R = 10Ω
V
~
LC
When source frequency get varied then the power factor becomes
2 1
(lagging) and freq. f2 and
2 1
(leading) at frequency f1 then.
1. 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 - (A) L R . 2 1 π (B) 2L R . 2 1 π (C) L R 2 . 2 1 π (D) R R 4 . 2 1 π
3. Value of watt less current when frequency is f1 (A) 5 2Amp. (B) 10 2Amp.
(C) 10 Amp. (D) Zero Amp.
4. 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 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.
5. 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
distance (in meter) y
x 1 2 3
(A) Acceleration of particle at x = 1 m is in +ve y-direction
(B) Velocity of particle at x = 1m is in-ve y-direction (C) Velocity of particle at x =1 m is zero
(D) None of these 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
Physics Challenging Problems
Solutions will be published in next issue
1. [D] As the magnetic field and area vector of the coil are in same direction so the magnetic flux passing through the coil is
0 cos Bs . S . B = = φ → → ) 1 )( r ( Bπ 2 = φ Induced emf. at d N
e− φ n = 1 single turn coil
) r B ( dt d ) ( dt d dt d e=− φ=− φ =− π 2 dt dB r e=−π 2 As B = B0 + B1 t2
[ ]
2Bt. . r e=−π 2 So 0 2B t dt dB 1 + = e = – πr2 2(–2)t = 4π2.t e = 4πr2t ... (i) Induced current R t. r 4 R e i= = π 2 as = π R r i = 4t ... (ii) Induced electric fieldr 2 t. r 4 r 2 e . dist e E 2 π π = π = = E = 2rt ... (iii)
Option (D) is correct, as induced electric field varies linearly with time for given value of radius.
2. [B] Induced current i = 4t So at t = 0 i = 0 3. [C] RMS value of current ) i ...( dt i 2 1 dt i T 1 I 2 0 2 T 0 2 RMS=
∫
=∫
As 0 16(8/3) 3 8 . 16 3 t 16 dt ) t 4 ( dt i 2 0 3 2 0 2 2 0 2 = − = = =∫
∫
. Amp 3 8 3 8 . 16 . 2 1 I RMS= =4. [A] Induced charge
2 0 2 2 0 2 0 2 t 4 tdt 4 idt q = = =
∫
∫
coulomb 8 0 2 4 . 4 − = = As 96500 C = 1 Faraday So 8Faraday 96500 1 C 8 = × As x 96500 1 =So induced charge = 8x faraday
5. [A] The forces working on metallic rod Fm = i L.B.
Net force working on rod Fnet = Mg – i. LB = .LB R vBL Mg− = R L vB Mg 2 2 − a Fm = i L.B. b X i Fg = Mg Fnet = R L vB Mg Ma= − 2 2
a = acceleration of metallic rod ab or acceleration = MR L vB g a= − 2 2 or acceleration g kV MR L B v g dt dv 2 2 − = − − = Here MR L B k 2 2 = dt kv g dv kv g dt dv = − ⇒ − =
Integrating on both sides
∫
=∫
− v 0 t 0 dt kv g dv t g ln k 1 ) kv g ln( k 1 t ) kv g ln( k 1 v 0 = − − − − ⇒ = − − ⇒ e kt g kv g t g kv g ln k 1 − = ⇒ − = − −Solution
Physics Challenging Problems
Set # 10
⇒ kt kt 1 e g v k e g v k 1− = − ⇒ = − − and v = k g (1 – e–kt) acceleration
[
kt]
e kt k g ) k ( e k g dt dv a= = − − − = − acceleration a =g
e–kt ) e 1 ( k g v= − −kt where MR L B k 2 2 = 6. [A] Acceleration a = ge–kt When t = 1/k a g.e k g.e1 .37g 1 . k = = = − −Acceleration a = 37 (maximum acceleration)
So time after which acceleration is 37% of maximum acceleration is t = 1/k = MR/B2L2 7. [A] Velocity (1 e ) k g v= − −kt for t = 1/k ) velocity imum (max 63 . ) e 1 ( k g v= − −1 =
Time after which velocity is 63% of maximum velocity is t = 1/k = MR / B2L2
8. [A] Area traversed by rod = x.L x distance travelled down and in time t So, Areal velocity =
dt dA = L.v dt dx . L ) xL ( dt d ) A ( dt d = = = (1 e ) k g . L − −kt = Areal velo.= k gL (1 – e–kt)
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.
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. (i) Calculate initial acceleration of the particle. (ii) Calculate velocity of particle when it reaches at
the centre of upper ring B. (g = 10 ms–2) a A B a h
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) (ii) force F exerted by the charge on ring B.
This force, F = q ) h a ( h q 4 1 2 / 3 2 2 2 0 + πε
where q2 = – 20 µC (charge on ring B) ∴ F = 5.76 Newton (upwards) Let acceleration of the particle be a.
F
mg ma
Considering its free body diagram (Fig.), ma = F – mg
or a = 47.6 ms–2 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 –
(i) to increase its gravitational potential energy by mgh and
(ii) to provide kinetic energy
2
1 mv2 to the particle.
But work done by electric field is W = q (V1 – V2) where V1 is potential at centre of ring A
and V2 is potential at centre of ring B. V1 = a q 4 1 1 0 πε + 2 2 2 0 a h q 4 1 + πε = – 6 × 10 4 volt V2 = . 4 1 0 πε 2 2 1 h a q + +4 . 1 0 πε a q2 = – 42 × 104 volt ∴ W = q (V1 – V2) = 3.6 joule But W = mgh + 2 1 mv2 ∴ v = 8 ms–1. Ans. (ii) 2. 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. A + – C3 C4 C1 C2 C5 B
Expert’s Solution for Question asked by IIT-JEE Aspirants
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Sol. In the given circuit capacitors C4 and C5 are in series with each other while capacitor C2 is in parallel with this series combination. Then capacitors C3, C1 and above said combination are in series with each other. When steady state is reached, no current flows through the circuit. To analyse the given circuit, it may be assumed that a charge q1 leaves the battery from its positive terminal and flows from A to B. This charge first reaches lower plate of capacitor C3. Hence, this plate becomes positively charged and upper plate negatively charged. Now charge q1 reaches the capacitor C1. Its left plate becomes positively charged and right plate negatively charged. Now charge q1 reaches the junction B. From where it gets divided into two parts. Let a charge q2 flow through series combination of capacitors C4 and C5. Then a charge (q1 – q2) flows through capacitor C2. Hence, in steady state charges on different capacitors will be as shown in Fig.
+ – C3 C4 C1 C2 C5 B C D A E q1 – + E q1 + – (q1 – q2) + – q2 + – + –q2
Since terminal A is connected with positive terminal of the battery, therefore, potential of A is higher than that of B.
It is given that potential difference (VA – VB) is equal to 11 volt, ∴ + 3 1 C q + 1 1 C q = 11 or q1 = 3 1 3 1 C C C C 11 + = 30 µC
Now applying Kirchhoff's voltage law on mesh BCDEB, ∴ + 2 2 1 C ) q q ( − – 5 2 C q – 4 2 C q = 0 or q2 = 12.6 µC
Potential difference across C5 =
5 2
C
q = 1.8 volt Ans.
3. System shown in Fig. consists of two large parallel metallic plates carrying current in opposite directions.
× × ×
Current density in each plate is j per unit width. Calculate
(i) magnetic induction in space between the plates and
(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 × Q S R B B l Fig. (1)
Let magnetic induction of the field induced due to current in one plate be B.
Considering a length l in the section as shown in Fig.(1) and applying Amperes's Circuital Law, B. 2l = µ0 (lj) or B =
2 1
µ0j
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.
∴ Resultant magnetic field induction between the plates = 2B = µ0j Ans. (i) 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
× × ×
dx
Fig. (2)
Magnetic induction at this conductor due to current in lower plate is B =
2
Hence, force on this conductor, dF = B di per unit length
or dF =
2 1µ0j2 dx
per unit length
But area of unit length of the conductor considered = 1. dx = dx
∴ Force per unit area of upper plate = dx dF
=
2
1µ0j2 Ans. (ii)
4. 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.
Asume that area of contact of tyre with road remains unchanged. (g = 10 ms–2)
Sol. Atmospheric pressure, P0 = 105 Nm–2
Increase in pressure (due to weight of bicycle and its rider) = A mg= ) 10 24 ( 10 120 4 − × × Nm–2 = 5 × 105 Nm–2 ∴ Final pressure of air in the tube,
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
n = RT V P2 2 = RT ) 10 2 ( ) 10 6 ( × 5 × × −3
Volume of these moles at atmospheric pressure, V1 = 0 P nRT = 5 5 3 10 ) 10 2 ( ) 10 6 ( × × × − m3 = 12 × 10–3 m3
Initially, volume of air in tube (at atmospheric pressure) is V0 = 0.75 × 2000 cm3
= 1.5 × 10–3 m3
∴ Volume (at atmospheric pressure) of air to be pumped in is
∆V = V1 – V0 = 10.5 × 10–3 m3
Volume (at atmospheric pressure) of air pumped in each stroke is
v = 500 cm3 = 0.5 × 10–3 m3 ∴ Number of strokes required
= 33 10 5 . 0 10 5 . 10 × × = 21 Ans.
5. 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 co-ordinate system shown in Fig.
O Air
θ
x y
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.
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. i (µ1) (µ2) A θθ r Fig. (1) Applying Snell's law at point A,
θ sin i sin = µ1 or sin θ = 1 µ i sin … (i)
Refractive index of second slab with respect to first slab = 1 2 µ µ .
Now applying Snell's law at point B, r sin sinθ = 1 2 µ µ
Substituting value of sin θ from equation (1),
r sin
i sin
= µ2.
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. O Air θ x y P (90 –α α ) 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,
) 90 sin( sin α − θ = 1 µ
Substituting values of θ and µ,
α cos 2 1 = 2 / 1 2 y k 4 1 + or sec2 α = 4 + 2 y k 4 1 ∴ 1 + tan2 α = 1 + 2 y k 4
But k = 1, therefore, tan α = y 2 But tan α = dx dy ∴ y dy = 2 dx
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,
∫
0yydy =∫
x 0 dx 2 or y2 = 4x Ans.SCIENCE TIPS
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