Dear Students,
It's the question you dreamed about when you were ten years old. It's the question our parents nagged you about during high school. It's the question that stresses most of us out more and more the older we get. "What do you want to be when you grow up?"
There are people who are studying political science but hate politics, nursing majors who hate biology, and accounting majors who hate math. Obviously, a lot of people are confused about what exactly it is that they want to spend their life doing. Think about it. if you work for 10 hours each day, you're going to end up spending over 50% of your awake life at work. Personally, I think it's important that we spend that 50% of your awake life at work. Personally, I think it's important that we spend that 50% wisely. But how can you make sure that you do? Here are some cool tips for how to decide that you really want to be when you grow up.
• Relax and Keep an Open Mind: Contrary to popular belief, you don't have to "choose a career" and stick with it for the rest of your life. You never have to sign a contract that says, "I agree to force myself to do this for the rest of my life" You're free to do whatever you want and the possibilities are endless. So relax, dream big, and keep an open mind.
• Notice Your Passions: Every one of us is born with an innate desire to do something purposeful with our lives. We long to do something that we're passionate about; something that will make a meaningful impact on the world.
• Figure Out How to Use Your Passions for a Larger Purpose: You notice that this is one of your passions, so you decide to become a personal trainer. Making a positive impact on the world will not only ensure that you are successful financially, it will also make you feel wonderful. It's proven principle: The more you give to the world, the more the world will give you in return.
• Figure Our How You Can Benefit: Once you've figured out what your passions are and how you can use those passions to add value to the world & to yourself, it's time to take the last step: figure out how you can make great success doing it. my most important piece of advice about this last step is to remember just that: It's the last part of the decision process. I feel sorry for people who choose an occupation based on the average income for that field. No amount of money can compensate for a life wasted at a job that makes you miserable. However, that's not to say that the money isn't important. Money is important, and I'm a firm believer in the concept that no matter what it is that you love doing, there's at least one way to make extraordinary money doing it. So be creative!
No matter how successful you become, how great your life is, or how beautiful you happen to be... there will still be times when you simply feel like you're an ugly mess. But when those times come, remember that all you need to get yourself back on track is a positive outlook, a dash of self confidence, and the willingness to make yourself feel better as soon as you know how.
Simply discover your passions, figure out how to use your passions to make an impact on the world & to yourself.
Presenting forever positive ideas to your success.
Yours truly
Pramod Maheshwari, B.Tech., IIT Delhi
Every effort has been made to avoid errors or omission in this publication. In spite of this, errors are possible. Any mistake, error or discrepancy noted may be brought to our notice which shall be taken care of in the forthcoming edition, hence any suggestion is welcome. It is notified that neither the publisher nor the author or seller will be responsible for any damage or loss of action to any one, of any kind, in any manner, there from.
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• All disputes are subject to the exclusive jurisdiction of the Kota Courts only.
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
If you can't make a mistake,
you can't make anything.
Volume - 5 Issue - 5 November, 2009 (Monthly Magazine) Editorial / Mailing Office :
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Volume-5 Issue-5 November, 2009 (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.
Xtra Edge Test Series for JEE-2010 & 2011
S
Success Tips for the Months
• If you can't make a mistake, you can't make anything.
• Sometimes a big step is safer; you can't cross a ditch in small jumps
• Self-confidence grows not from what you can do, but what you know you can do. • Children focus on what they can’t do.
Adults focus on what they can do.
• The secret of confidence is to know your resources.
• You never need to feel fear if you don't want to do anything.
• You got to know when to hold ‘em and know when to fold ‘em…
• An ounce of success is worth a pound of positive thinking.
• To understand motivation, know the power of the Hunter.
• Defeat is advance payment for victory.
CONTENTS
INDEX PAGE
NEWS ARTICLE
4IIT-K signs MoU with US university
White House names IIT-ian Arun Majumdar as America's Green Czar
IITian ON THE PATH OF SUCCESS
8 Mr. S. JanakiramanKNOW IIT-JEE
10Previous IIT-JEE Question
XTRAEDGE TEST SERIES
53Class XII – IIT-JEE 2010 Paper Class XII – IIT-JEE 2011 Paper
Regulars ...
DYNAMIC PHYSICS
188-Challenging Problems [Set# 7] Students’ Forum
Physics Fundamentals
Electromagnetic Induction & A.C. Simple Harmonic Motion
CATALYSE CHEMISTRY
32Key Concept Nitrogen Compounds Nitrogen Family
Understanding : Physical Chemistry
DICEY MATHS
44 Mathematical Challenges Students’ Forum Key Concept Differentiation Straight Line & CircleStudy Time...
IIT reviews expel rule The IITs are considering scrapping a weeding process they practice to expel weak students mid-course, stung by a string of legal challenges and allegations of caste discrimination.
Officials across the IITs have held two rounds of discussions on a proposal to replace expulsion of weak students with performance checks, administrators, including two directors
Around 10 students are dismissed from each IIT on an average every year for failing to earn a minimum number of credits required at mid-course stages, the administrators said.
An end to the system would mean that students, once admitted, would not be dismissed.
The move follows caste allegations against the IITs — a majority of students expelled during their courses for poor performance belong to Scheduled Castes or Scheduled Tribes. It also comes at a time when the Supreme Court has held that the IITs cannot “throw out” SC/ST students on the basis of poor performance. In cases when students are asked to leave, they are given the option to quit the BTech course, and instead opt for a less reputed diploma. Hardly any student opts for it, officials said.
The students argue that as they have cleared the IIT entrance test, their ability to pursue the BTech course cannot be challenged.
IIT-K signs MoU with US university
KANPUR: Indian Institute of Technology-Kanpur signed an MoU with the University of Texas, San Antonio, US, here on Saturday.
Director, IIT-K, SG Dhande represented the institute and CM Agarwal (an alumnus of IIT-K of the 1982 batch), dean, College of Engineering, represented the University of Texas. They hoped that the two educational institutes will be benefited from the agreement.
"The areas identified for collaborative research include fields like bio-material, where the focus will be on the research collaboration, faculty and student exchange programme. The co-operative research programme in other areas of mutual interest will be conceived later," Dhande said while talking to TOI.
Agarwal said that Indo-US center for bio-material has been established which will concentrate on the development of material which will be useful in treating patients suffering from knee problems like arthritis.
"The long-term goal is that the implants done in a human body should end as knee replacement in India is a costly affair and costs between Rs 30,000 and Rs 2 lakh. The best part would be that the human body itself reaches a stage where it can regenerate bone. This is called tissue engineering and doctors, scientists and biologists are working together for this cause," Agarwal said. Next month, a delegation will visit IIT-K to hold talks with Dhande, Agarwal added.
White House names IIT-ian Arun Majumdar as
America's Green Czar WASHINGTON: There was more than a hint of irony in the Obama White House on Friday naming Arun Majumdar, a product of the best engineering schools
in India and US, as the first Director of the US Department of Energy's Advanced Research Projects Agency-Energy (ARPA-E), an agency tasked with reducing America's reliance on foreign energy supplies, cutting greenhouse gas emissions, and improving energy efficiency.
That an (Indian) immigrant engineer-scientist should head the premier agency at a time Washington is hectoring the world, principally India and China, to cut emissions, amid a growing trade and job protectionism, says something about the United States – and Majumdar was quick to articulate it.
"It is a rare privilege and an honor when the President asks you to serve the nation in such a capacity," said Majumdar of his nomination, which, while needing to be confirmed by the Senate, sent ripples across the country's scientific-academic community. "I came to this country as an immigrant and am deeply appreciative and indebted to this nation for opening the doors and welcoming me with open arms. I have received so much. This is my way of stepping up and paying back."
Not that the IIT-Mumbai graduate has forsaken his roots – in fact, his roots may well have been responsible for his nomination.
Ever since he joined the University of California (UC), Berkeley faculty in 1997, where he holds the Almy and Agnes Maynard Chair Professorship in the College of Engineering and heads the
Environmental Energy Technologies Division, Majumdar
has cemented the Lab's role as a world-renowned leader in energy efficiency research in close collaboration with India and China – a feat the White House has been quick to recognize and reward. Among the lab's partnerships is the Berkeley-India Joint Leadership on Energy and the Environment announced last year, which brings together researchers from Berkeley Lab and UC Berkeley, and other US and Indian universities and institutions, with a goal to reduce greenhouse gas emissions while maintaining sustained economic growth in both nations.
Another partnership between the Lab and China's Tsinghua University is to promote the shared development and implementation of building energy efficiency, a move intended to reduce energy consumption and greenhouse gas emissions in the US and China.
In fact, Majumdar's mentor in academia was Professor Chang-Lin Tien, a legendary Chinese Don who went to become the Chancellor of UC Berkeley in 1990, the first Asian to head a major university in the United States.
Majumdar's India-China connections is what appears to
have driven the Obama White House to choose him for the new job, considering the two countries are thought to be pivotal in the upcoming energy debate. "He has had a highly distinguished research career in the science and engineering of energy conversion, transport, and storage ranging from molecular and nanoscale level to large energy systems," the
White House said in its announcement. "At Berkeley Labs and UC Berkeley, he helped shape several strategic initiatives in the areas of energy efficiency, renewable energy as well as energy storage." For more than a decade, Majumdar, who is also the founding chair of the American Society
of Mechanical Engineers' Nanotechnology Institute, has been the country's leading materials scientist, making spectacular advances in energy conservation. He was recently credited with developing a way to use silicon nanowires to capture and use the energy lost as heat during the production of electricity. The futuristic technology could someday be used to convert the large amounts of waste heat into useful electricity. A graduate (mechanical engineering) of IIT-Mumbai, Majumdar came to the US in 1985 and received a Ph.D. in mechanical engineering from the University of California, Berkeley in 1989. His nomination continues the steady march of Indian geeks and academics in the higher echelons of administration. The Obama administration picked Indian-Americans for the post of White House Chief Information Officer (Vivek Kundra) and Chief Technology Officer (Aneesh Chopra). Majumdar will effectively be the Chief Energy Officer. IIT-B looks to solar power To help facilitate cost-effective solar thermal power generation, IIT Bombay plans to develop a megawatt-scale solar thermal power facility, which is being sponsored by the ministry of new and renewable energy.
"The idea is to help create cost-effective solar power. There is a huge gap between the demand and supply of electricity and one option worth developing is solar power," said a faculty member of
IIT Bombay. The plan to build the plant was proposed by IIT-B last year and it will come up at the Solar Energy Centre in Gurgaon. It will be connected to a grid and supply around a megawatt to the national grid. The test and simulation facility will be set up by a consortium involving different Indian industries and IIT-B.
"While the US and Europe have already built such consortia, it will be a first in India. This facility is expected to help in developing inexpensive solar power plants in the future," he said.
Even as the test facility will enable assessment of new technologies, components, and systems for solar thermal power, the simulation can be used to scale up designs and optimise use of solar power. "Besides developing indigenous capability, it is expected to provide the experience in concentrated solar power, which has the potential to provide a sustainable energy solution for India's power system," said the faculty member. The project, which will last for five years, is expected to start in another two years.
Signature drive launched for IIT status to ISM
RANCHI: A signature campaign seeking IIT status for the Indian School of Mines (ISM), Dhanbad, was launched , On the first day of the three-day campaign, students managed to collect 12,500 signatures in favour of their demand.
According to the campaign activists, a total of 1,25,000 signatures have been obtained so far in favour of the demand from across the state. Out of this, about 75,000 signatures were obtained from Dhanbad alone. The campaign was simultaneously launched at Dhanbad, Jamshedpur and Bokaro on Monday last to
garner the support of the people of Jharkhand towards the cause. "A delegation of ISM students met Congress MP Sachin Pilot on Friday and he too extended support to the cause," said a campaign activist.
Earlier, a delegation had called on Union minister of state for human resources development D Purandeshwari in this regard. The minister had promised to look into the matter.
Cong tussle holds up home for IIT
New Delhi: A tussle between Congress leaders is depriving IIT Rajasthan of a permanent home two years after its conception, replacing an earlier battle the party waged over the institute with the BJP when it ruled the state.
Coaching hub Kota, proposed by the former BJP state government as the venue for the IIT but dismissed by the UPA at the Centre, has now found powerful supporters within the Congress. The reason behind their demand: they had promised Kota an IIT in the Lok Sabha elections.
The problem: others in the Congress had campaigned on bringing the same IIT to Jodhpur, home of current chief minister Ashok Gehlot. The Congress won in both Kota and Jodhpur seats in the Lok Sabha polls and is well ensconced in Rajasthan, where it is also in power.
IIT Rajasthan, launched last year, however, continues to reside as a tenant on the IIT Kanpur campus. “It is a strange situation where we don’t know how to proceed. So we are likely to just sit on any decision for the time being,” a top government official said.
Senior Congress leader Digvijay Singh has, in his capacity as party
general secretary, written to human resource development minister Kapil Sibal, requesting that the new IIT be set up in Kota. Digvijay was a Congress observer for Rajasthan during the Lok Sabha polls and his letter represents concerns of several party MPs and MLAs from the southern parts of the state, sources said.
Kota MP Ijyaraj Singh, too, has met junior HRD minister D. Purandeswari, requesting that the IIT be set up in his constituency. But accepting Kota as the venue for the new IIT will not prove easy for the HRD ministry, sources said, because of a clear position it had earlier taken against the town’s eligibility.
The Rajasthan IIT is one of eight promised by the UPA under the Eleventh Five Year Plan and was announced in 2007. Vasundhara Raje Scindia, who was then chief minister, proposed Kota, the nearest big town to her family fief in Jhalawar, as the venue.
But a central team sent to examine prospective sites advised against Kota, arguing that it was poorly connected and would not attract top teachers, students and industry. Scindia accused then HRD minister Arjun Singh of “playing politics” over the IIT’s location. Last December, after the Congress wrested Rajasthan from the BJP in the Assembly polls, new chief minister Gehlot appointed a team to propose afresh a venue for the new institute.
The Gehlot-appointed panel recommended Jodhpur, the chief minister’s hometown.
In the Lok Sabha polls that followed a few months later, the Congress campaign for party candidate Chandresh Kumari in Jodhpur promised the IIT to the city.
IIT-K students learn the benefits of MEMS technology
KANPUR: "The MEMS (Micro-Electro Mechanical System) technology is widely used in the cellphones, which are being used by a large percentage of population nowadays. The major use of the MEMS technology is in the field of IT followed by the consumer electronics sector," said Dr V K Aatre, former scientific adviser to the defence minister, who was on his visit to the IIT-K, here on Thursday. Dr Aatre further informed that All India Institute of Medical Sciences (AIIMS) and IIT-Bombay are together working on a device called cardiac monitor (it's a temperature measuring device) which will provide the temperature of the arteries. This cardiac monitor would based on the MEMS technology.
"AIIMS and IIT-Bombay will now be going for the animal trials of the cardiac monitor before going for a human trial which is a genuine procedure," added Dr Aatre.
He also added that MEMS based technology is seeing its use in the fields of automobile, electronics, bio-medical, defence etc.
Dr Vikram Kumar, former director of NPL (it is one of the labs of CSIR) and professor, department of Physics at IIT-Delhi said, "We carry MEMS-based gadgets with us on a daily basis but we are hardly aware of it. The mobile phones and I-Pods are the best examples of the same. The MEMS technology is also used in Plasma TVs and other household consumer electronics."
Dr Kumar further said that the pressure sensors based on MEMS technology is widely used in the automobile industry. "At present the pressure sensors are being imported from the other companies of the world. The
pressure sensors have varied applications, especially in the fields like automobile, aerospace, acceleration etc", he added. "The pressure sensors are of various kinds and even used in a rocket," he said.
Dr N S Vyas, professor and HOD of the department of Mechanical
Engineering, IIT-Kanpur emphasised on the fact that MEMS
technology has several potential applications in the railways and automotive sector.
IITs strategise for more PhD scholars
Efforts include joint MTech and PhD degrees and streamlining policies so that thesis papers are cleared within two months
With research becoming a clear focus area at all Indian Institutes of Technology (IITs) and with the 20-30 per cent growth in sponsored research, the premier technology institutes are now targeting a 10-30 per cent increase in PhD scholars.
Globally, China produces the maximum number of research scholars per year. It is widely recognised that there will be substantially more PhD engineers and scientists in China in 2010 than in the United States, as China produces three times the number of engineers per year. Smalley, a nobel prize-winning scientist from Rice University, recently concluded that by 2010, 90 per cent of all PhD physical scientists and engineers in the world will be Asians living in Asia, and among Asian PhD engineers and scientists, most will be produced by China.
India, therefore, is in a hurry to catch up. IIT-Kharagpur (IIT-KGP), for instance, awarded 212 PhDs this year, of which nearly 70 per cent had studied engineering. Last year, the institute had awarded 167 PhDs.
“We want at least 30 per cent of our students to be research scholars, double of what it is right now. We are making several enticements for that, like joint MTech and PhD degrees and streamlining policies so that thesis papers are cleared within two months instead of one year which is usually the norm,” said Damodar Acharya, director of IIT-KGP.
The institute has also introduced joint degree programmes with other reputed universities in India and abroad. A student admitted to such joint degree programmes has to spend upto two years in the partnering university and would have a joint guide. Through this programme, the institute aims at producing high quality faculty who will have exposure to at least two different environments.
The institute from its own fund supports written airfare up to two visits of the students to the partnering university. The local expenses of the student are taken care of by the partnering university. At IIT-Bombay, 140 PhDs graduated in 2007, 200 in 2008 and around 175 in 2009. "We are incubating our PhD students using their intellectual properties. This should encourage students and make them feel more secure about their research findings," said Rangan Banerjee, dean of research and development at IIT Bombay. At IIT Madras, from 2006 to 2009 there has been a 50 per cent increase in PhD intake. Currently the institute has around 1100 PhD scholars, informed Job Kurian, dean of sponsored research at IIT Madras. IIT Madras aims to have a 1:1 ratio between research scholars and undergraduates, from 1:5 ratio currently, said Kurian. IIT-Delhi has seen a 23 per cent increase in the number of PhD degrees given out this year. A total number of 181 PhD degrees was awarded as compared to 147 last year. "This is a phenomenal
achievement and is contrary to belief that we are very poor on research output," said M Balakrishnan, dean of post-graduate studies at IIT-Delhi. IIT Bhubaneswar, one of the newest IITs in India, is encouraging faculty to join the institute with their own research scholars. M Chakraborty, director of IIT Bhubaneswar, said that the institute is also making provision for upto Rs 5 lakh research grant to a faculty. This would help them to invest in necessary infrastructure they require to carry out their research, like softwares, hardwares, books and journals, etc.
Student researchers get a grant of Rs 15,000 per month. International exposure for faculty and student researchers and presenting their research papers at international conferences is another priority area for IIT Bhubaneswar.
IIT Gandhinagar (IIT-G), another new IIT, has also started focusing on establishing the institute as a preferred destination for research students by initiating quality research activities on the campus. The institute, which was established in 2008, has just received its second batch of undergraduate students, but is already working on lines of creating a centre for research. Human Resource Development Minister Kapil Sibal had recently said the country's premier Indian Institutes of Technology (IITs) must focus on quality research and act as a catalyst to boost technical education in India.
"This is not only necessary for the economic growth of the country but also for the IITs to make the transition as creator of knowledge. Without a large base of well educated undergraduates in the country it is difficult to imagine any significant growth in research output from these institutions," Sibal said.
Janakiraman (Jani) heads Product Engineering Services (PES) of MindTree as the President and Group CEO of the business unit. PES comprises of R&D Services (RDS) and Outsourced Product Development Services (OPD). PES under Jani covers whole spectrum of product and technology development services covering Semiconductor, Embedded System, Middleware and Application level Products for the Hi-Tech Product and Independent Software Vendor (ISV) organizations. In addition the Research Units under PES build ready to use Intellectual Properties (IPs) and re-usable Technology Building Blocks (TBBs) covering segments like short range wireless like Bluetooth and UWB, Video surveillance platforms including management and analytics, virtualization and cloud computing.
Starting from scratch, Jani has built the Product Engineering Services organization of MindTree through organic and inorganic means to over 3000 technology professionals executing projects in leading technology
areas for customers like Alcatel-Lucent, AOL, Apple, Cisco, Microsoft, Real Networks, Symantec, Texas Instruments, Toshiba, UTC, Vendavo and Volvo.
Jani has rich 28 years of experience in building R&D and Product Engineering Services organizations through setting up multiple dedicated development centers for the Engineering units for semiconductor, system and application product vendors. The services include IP licensing, Architecture-Design-Development, Independent Testing, Packaging, and Technical Support.
Jani holds Bachelor's degree in Electronics and Communications from the National Institute of Technology (NIT), Trichy, India, and Master's degree in Electronics from the Indian Institute of Technology (IIT), Chennai, India. He is the President of Indo Japan Chamber of Commerce & Industries (IJCCI), Karnataka and a member of India Semiconductor Association (ISA) Executive Council.
Mr. S. Janakiraman
B.E., M.E.(IIT – Chennai)
President & Group CEO – Product Engineering Services, Mindtree Ltd.
"Time is a circus, always packing up and moving away."
Success Story
This article contains story of a person who get succeed after graduation from different IIT'sPHYSICS
1. One gram mole of oxygen at 27º and one atmospheric pressure is enclosed in a vessel.
[IIT-1983] (i) Assuming the molecules to be moving with vrms,
Find the number of collisions per second which the molecules make with one square metre area of the vessel wall.
(ii) The vessel is next thermally insulated and moved with a constant speed v0. It is then suddenly stopped.
The process results in a rise of the temperature of the gas by 1ºC. Calculate the speed v0.
Sol. (i) We know that P = A F ∴ F = P × A = 105 × 1 = 105 N ... (i) But F = t p ∆ ∆ mv mv
∴ ∆p = F × ∆t = F × 1 = 105 [From (i)] ...(ii)
Let m be the mass of one molecule and v be the r.m.s. velocity.
Now Momentum change per second m2 (∆p) = n × 2mv ... (iii)
Where n is the number of collisions per second per square metre area
From (ii) and (iii)
n × 2mv = 105
∴ n =
mv 2
105
Root mean square velocity
v = M RT 3 = 1000 / 32 300 314 . 8 3× × = 483.4 m/s According to mole concept 6.023 × 1023
molecules will have mass 32 g
∴ 1 molecule will have mass 23 10 023 . 6 32 × g ∴ n = 4 . 483 32 2 10 023 . 6 105 23 × × × × = 1.97 × 1027
(ii) The kinetic energy of motion of molecules will be converted into heat energy.
K.E. of 1 gm mole of oxygen = 2
1 2
0
Mv ... (i) Where v0 is the velocity with which the vessel
was moving.
The heat gained by 1 gm mole of molecules at constant volume
= nCV ∆T= 1 × Cv × 1 = Cv ...(ii)
From (i) and (ii) 2 1 2 0 Mv = CV ... (iii) Now, Cp – CV = R ⇒ V p C C – V V C C = V C R ⇒ γ – 1 = V C R ∴ CV = 1 R − γ ... (iv)
From (iii) and (iv) 2 1 2 0 Mv = 1 R − γ ∴ v0 = ) 1 ( M R 2 − γ = (1.41 1) 100 32 314 . 8 2 − × ×
γ = 1.41 for O2 (diatomic gas)
⇒ v0 = 35.6 m/s
2. Hot oil is circulated through an insulated container with a wooden lid at the top whose conductivity K = 0.149 J/(m-ºC-sec), thickness t = 5 mm, emissivity = 0.6. Temperature of the top of the lid is maintained at Tl = 127ºC. If the ambient temperature Ta = 27ºC. [IIT-2003]
KNOW IIT-JEE
Tl = 127º
T0
Hot Oil Ta = 27ºC
Calculate :
(a) rate of heat loss per unit area due to radiation from the lid.
(b) temperature of the oil. (Given σ = 3
17× 10–8)
Sol. (a) The rate of heat loss per unit area per second due to radiation is given by Stefan's-Boltzmann law
E = εσ(T4 – 4 0 T ) = 0.6× 3 17 × 10–8 [(400)4 – (300)4] = 595 Watt/m2
(b) Let Toil be the temperature of the oil. Then rate of
heat flow through conduction = Rate of heat flow through radiation. l ) T T (
KA oil− = 595 × A where A is the area of the top of lid
⇒ Toil = k 595 l× + T = 149 . 0 10 5 595× × −3 + 400 = 419.83 K
3. A sphere of radius R is half submerged in liquid of density ρ. if the sphere is slightly pushed down and released, find the frequency of oscillation.
[IIT-2004] Sol. At equilibrium net force is zero
∴ Fmg = Fbuouncy or ρm × 3 4 π r3 × g = ρ × 3 2 π r3 g ⇒ ρm = 2 ρ
Let the sphere is slightly displaced downward by x.
∴ Fres = – π R2 xρg
[Q Volume of submerged portion of sphere increases by πR2 x, hence buouncy increases by π
R2 xρg] ∴ a = s 3 2 R 3 4 g R ρ π ρ π − = – R 2 g 3 x [Q acceleration a = m F and m = 3 4 π R3 ρ s] ∴ ω = R 2 g 3 [Comparing with a = – ω2 x] or v = π 2 1 R 2 g 3
4. Two identical prisms of refractive index 3 are kept as shown in the figure. A light ray strikes the first prism at face AB. Find, [IIT-2005]
A 60º 60º B C 60º 60º D E
(a) the angle of incidence, so that the emergent ray from the first prism has minimum deviation. (b) through what angle the prism DCE should be
rotated about C so that the final emergent ray also has minimum deviation.
Sol. (a) For minimum deviation of emergent ray from the first prism MN is parallel to AC
∴ ∠ BMN = 90º ⇒ ∠ r = 30º
Applying Snell's law at M µ = r sin i sin sin i = µ sin r sin i = 3 × sin 30º = 2 3 ⇒ i = 60º A 60º 60º B C i N Q M P r
(b) When the prism DCE is rotated about C in anticlockwise direction, as shown in the figure, then the final emergent ray SR becomes parallel to the incident ray TM. Thus, the angle of deviation becomes zero.
5. A neutron of kinetic energy 65eV collides inelastically with a singly ionized helium atom at rest. It is scattered at an angle of 90º with respect of its original direction. [IIT-1993] (i) Find the allowed values of the energy of the
neutron and that of the atom after the collision. (ii) If the atom get de-excited subsequently by
emitting radiation, find the frequencies of the emitted radiation.
[Given: mass of He atom –4×(mass of neutron), Ionization energy of H atom = 13.6eV]
Sol. m 4m m K2 θ 4m K1 y x
Applying conservation of linear momentum in horizontal direction
(Initial Momentum)x = (Final Momentum)x
pix = pfx
⇒ 2Km= 2(4m)K1cosθ ...(i) Now applying conservation of linear momentum in Y-direction
piy = pfy
0 = 2K2m– 2(4m)K1sinθ ⇒ 2K2m= 2(4m)K1sinθ ...(ii) Squaring and adding (i) and (ii)
2Km + 2K2m = 2(4m)K1 + 2(4m)K1
K1 + K2 = 4K1 ⇒ K = 4K1 – K2
⇒ 4K1 – K2 = 65 ...(iii)
When collision takes place, the electron gains energy and jumps to higher orbit.
Applying energy conservation K = K1 + K2 + ∆E
⇒ 65 = K1 + K2 + ∆E ...(iv)
Possible value of ∆E For He+
Case (1)
∆E1 = – 13.6 – (54.4eV) = 40.8 eV
⇒ K1 + K2 = 24.2 eV from (4)
Solving with (3), we get
K2 = 6.36 eV; K1 = 17.84 eV
Case (2)
∆E2 = – 6.04 – (–54.4 eV) = 48.36 eV
⇒ K1 + K2 = 16.64 eV from (4)
Solving with (3), we get
K2 = 0.312 eV; K1 = 16.328 eV
Case (3)
∆E3 = – 3.4 – (–54.4eV) = 51.1 eV
⇒ K1 + K2 = 14 eV
Solving with (3), we get
K2 = 15.8 eV; K1 = – 1.8 eV
But K.E. can never be negative therefore case (3) is not possible.
Therefore the allowed values of kinetic energies are only that of case (1) and case (2) and electron can jump upto n = 3 only.
–54.4eV For He+ n=4 n=3 n=2 n=1 –13.6eV –3.4eV –6.04eV
(ii) Thus when electron jumps back there are three possibilities
n3 → n1 or n3 → n2 and n2 → n1
The frequencies will be ν1 = h E E3− 2 ν 2 = h E E3− 1 ν 3 = h E E2− 1 = 1.82×1015 H z = 11.67×1015 Hz = 9.84×1015 Hz
CHEMISTRY
6. Estimate the difference in energy between 1st and 2nd
Bohr orbit for a H atom. At what minimum atomic no., a transition from n=2 to n = 1 energy level would result in the emission of X-rays with λ = 3.0×10–8 m.
Which hydrogen atom like species does this atomic no. corresponds to ? [IIT-1993] Sol. (a) For H atom,
Z = 1 ni = 2 nf = 1 En = – 2 19 n 10 76 . 21 × − J
Hence, difference in energy between first and second Bohr orbit for a H-atom is given by,
∆E = i n E – E = Enf 2 – E1 = – 2 19 2 10 76 . 21 × − + 2 19 1 10 76 . 21 × − = – 21.76 × 10–19 − 2 2 1 1 2 1 = 16.32 × 10–19J
(b) For λ = 3.0 × 10–8 m ∆E = λ hc = 34 8 8 10 0 . 3 10 3 10 626 . 6 − − × × × × = 6.626 × 10–18 J ....(i)
We know that, for H-like atoms,
En for H-like atom = En for H-atom × Z2
∴ ∆E for H-like atom = Z2 × ∆E for H-atom
= –Z2 × 21.76 × 10–19 − 2 2 1 1 2 1 = 16.32 × 10–19 Z2 ...(ii)
From eq. (i) and (ii),
16.32 × 10–19 Z2 = 6.626 × 10–18
or Z = 2
Thus, hydrogen atom like species for Z = 2 is He+.
7. At room temperature, the following reactions proceed nearly to completion :
2NO + O2 → 2NO2 → N2O4
The dimer, N2O4, solidified at 262 K. A 250 ml flask
and a 100 ml flask are separated by a stopcock. At 300 K, the nitric oxide in the larger flask exerts a pressure of 1.053 atm and the smaller one contains oxygen at 0.789 atm. The gases are mixed by opening the stopcock and after the end of the reaction the flasks are cooled to 200 K. Neglecting the vapour pressure of the dimer, find out the pressure and composition of the gas remaining at 220 K. (Assume the gases to behave ideally) [IIT-1992] Sol. According to the gas equation,
PV = nRT or n = RT PV At room temperature,
For NO, P = 1.053 atm, V = 250 ml = 0.250 L ∴ Number of moles of NO = 300 0821 . 0 250 . 0 053 . 1 × × = 0.01069 mol For O2, P = 0.789 atm, V = 100 ml = 0.1L ∴ Number of moles of O2 = 300 0821 . 0 1 . 0 789 . 0 × × = 0.00320 mol
According to the given reaction, 2NO + O2 → 2NO2 → N2O4
Composition of gas after completion of reaction, Number of moles of O2 = 0
1 mol of O2 react with = 2 mol of NO
∴ 0.00320 mol of O2 react with = 2 × 0.00320
= 0.0064 mol of NO Number of moles of NO left = 0.01069 – 0.0064
= 0.00429 mol
Also, 1 mol of O2 yields = 1 mol of N2O4
∴ Number of moles of N2O4 formed = 0.00320 mol
N2O4 condenses on cooling,
∴ 0.350 L (0.1 + 0.250) contains only 0.00429 mol of NO
At T = 220 K, Pressure of the gas,
P = V nRT = 350 . 0 220 0821 . 0 00429 . 0 × × = 0.221 atm
8. An organic compound CxH2yOy was burnt with twice
the amount of oxygen needed for complete combustion to CO2 and H2O. The hot gases when
cooled to 0 ºC and 1 atm pressure, measure 2.24 L. The water collected during cooling weighed 0.9 g. The vapour pressure of pure water at 20ºC is 17.5 mm of Hg and is lowered by 0.104 mm when 50 g of the organic compound are dissolved in 1000 g of water. Give the molecular formula of the organic
compound. [IIT-1983]
Sol. According to the question, an organic compound CxH2yOy was burnt with twice the amount of oxygen.
Hence,
CxH2yOy + 2x O2 → xCO2 + yH2O + xO2
Volume of gases after combustion = 2.24 L (given) Volume of gases left after combustion = xCO2 + xO2
∴ x + x = 2.24 or x = 1.12 L 22.4 L CO2 = 1 mol CO2 ∴ 1.12 L CO2 = 4 . 22 12 . 1 = 0.05 mol CO2 and 18 g H2O = 1 mol H2O ∴ 0.9 g H2O = 18 9 . 0 = 0.05 mol H2O
Thus, the empirical formula of the organic compound is CH2O.
Empirical formula mass = 12 + 2 + 16 = 30 Vapour pressure of the pure liquid,
0 A
P = 17.5 mm of Hg
Lowering in vapour pressureP –PA0 A =0.104mm of Hg
Mass of organic compound = 50 g Mass of water = 1000 g
∴ Mole fraction of organic compound =
18 1000 M 50 M / 50 + where M is the molecular mass of the organic compound, the molecular mass of water being 18.
We know, 0 A A 0 A P P P −
= Mole fraction of organic compound
∴ 5 . 17 104 . 0 = 18 1000 M 50 M / 50 + or 104 . 0 5 . 17 = 1 + 50 18 M 1000 × × Solving, M = 150.5 ≈ 150 n = mass formula Empirical mass Molecular = 30 150 = 5 ∴ Molecular formula or organic compound
= 5(CH2O) = C5H10O5
9. Compound (A) (C6H12O2) on reduction with LiAlH4
yielded two compounds (B) and (C). The compound (B) on oxidation gave (D) which on treatment with alkali (aqueous) and subsequent heating furnished (E). The later on catalytic hydrogenation gave (C). The compound (D) was oxidised further to give (F) which was found to be monobasic acid (m. wt. 60.0). Deduce structures of (A) to (E). [IIT – 1990] Sol. Clue 1. Compound (F) is a monobasic acid molecular
mass 60. ∴ CnH2n+1COOH = 60 or n × C + (2n + 1) × H + C + 2 × O + H = 60 or 12n + 2n + 1 + 12 + 2 × 16 + 1 = 60 or n = 14 46 60− = 1 ∴ (F) is CH3COOH.
Clue 2. (D) on oxidation gives (F), therefore (D) is CH3CHO.
Clue 3. (B) on oxidation gives (D), therefore (B) is CH3CH2OH.
Clue 4. (D) undergoes aldol condensation and on heating gives (E), therefore (E) is CH3CH = CHCHO.
Clue 5. (E) on reduction gives (C), therefore (C) is CH3CH2CH2CH2OH.
Clue 6. (A) having formula C6H12O2 on reduction
yields (B) and (C).
∴ (A) is CH3CH2CH2COOCH2CH3.
The reactions are :
CH3CH2CH2COOCH2CH3 LiAlH4 CH3CH2CH2CH2OH + CH3CH2OH Ethyl butanoate (A) Reduction Butanol (C) CH3CH = CHCHO Crotonaldehyde CH3CH(OH)CH2CHO Aldol [O] CH3CHO Ethanal (D) Ethanol (B) [O] CH3COOH Ethanoic acid (F) aq.NaOH
10. A white substance (A) reacts with dil. H2SO4 to
produce a colourless gas (B) and a colourless solution (C). The reaction between (B) and acidified K2Cr2O7
solution produces a green solution and a slightly coloured precipitate (D). The substance (D) burns in air to produce a gas (E) which reacts with (B) to yield (D) and a colourless liquid. Anhydrous copper sulphate is turned blue on addition of this colourless liquid. Addition of aqueous NH3 or NaOH to (C)
produces first a precipitate, which dissolves in the excess of the respective reagent to produce a clear solution in each case. Identify (A), (B), (C), (D) and (E). Write the equations of the reactions involved.
[IIT-2001] Sol. (A) is ZnS, ) A ( ZnS + H2SO4 → ) C ( 4 ZnSO + ) B (2 S H ) B ( 2 S H 3 + K2Cr2O7 + 4H2SO4 → K2SO4 + Cr2(SO4)3 + 7H2O + ) D ( grey White3S ) D (S + O → Air2 SO (E)2 ) E ( 2 SO + ) B ( 2 S H 2 → ) C ( liquid Colourless2 O H 2 + ) (DS 3 5H2O + White4 CuSO → Blue4 2 O H 5 . CuSO
ZnSO4 + 2NaOH → Zn(OH)2 + Na2SO4
Zn(OH)2 + 2NaOH → Na2ZnO2 (soluble) + 2H2O
Also in excess of NH4OH it forms soluble complex
MATHEMATICS
11. Find the centre and radius of the circle formed by all the points represented by z = x + iy satisfying the relation β − α − z
z =k(k ≠ 1), where α and β are constant
complex numbers given by α = α1 + iα2, β = β1 + iβ2.
[IIT-2004] Sol. As we know; |z|2 = z. z ⇒ 22 | z | | z | β − α − = k2 ⇒ (z – α)( z – α ) = k2(z – β)( z – β ) |z|2 – α z – α z + |α|2 = k2(|z|2 – β z – β z+ |β|2) or |z|2 (1 – k2) – (α – k2β) z – ( α – β k2) z + (|α|2 – k2|β|2) = 0 ⇒ |z|2 – ) k 1 ( ) k ( 2 2 − β − α z – ) k 1 ( ) k ( 2 2 − β − α z + ) k 1 ( | | k | | 2 2 2 2 − β − α = 0 ...(i) On comparing with equation of circle,
|z|2 + a z + α z + b = 0
whose centre is (–a) and radius = |a|2− b ∴ centre for (i)
= 2 2 k 1 k − β − α and radius = 22 22 22 k 1 k k 1 k k 1 k − β β − α α − − β − α − β − α radius = 2 k 1 ) ( k − β − α
12. A is targeting to B, B and C are targeting to A. Probability of hitting the target by A, B and C are
3 2 , 2 1 and 3 1
respectively. If A is hit, then find the probability that B hits the target and C does not.
[IIT-2003] Sol. Here,
P(A) = probability that A will hit B = 3 2
P(B) = probability that B will hit A = 2 1
P(C) = probability that C will hit A = 3 1 P(E) = probability that A will be hit ⇒ P(E) = 1 – P( B ). P( C ) = 1 – 2 1 . 3 2 = 3 2
Probability if A is hit by B and not by C.
⇒ P(B ∩ C / E) ⇒ ) E ( P ) C ( P ). B ( P = 3 23 2 . 2 1 = 2 1
13. Find the equation of the normal to the curve y = (1 + x)y + sin–1 (sin2 x) at x = 0 [IIT-1993]
Sol. y = (1 + x)y + sin–1 (sin2 x) (given)
Let y = u + v, where u = (1 + x)y, v = sin–1 (sin2 x).
Differentiating dx dy= dx du + dx dv ...(1) Now, u = (1 + x)
take logarithm of both sides loge u = loge (1 + x)y ⇒ loge u = y loge (1 + x) ⇒ u 1 dx du = x 1 y + +dx dy . {loge (1 + x)} ⇒ dx du =(1+x)y + + + dxlog (1 x) dy x 1 y e ...(2) Again, v = sin–1 sin2 x
⇒ sin v = sin2 x ⇒ cos v dx dv= 2. sin x cos x ⇒ dx dv= v cos 1 [2 sin x cos x] ⇒ = v sin 1 x cos x sin 2 2 − = 1 sin x x cos x sin 2 4 − ...(3) Put these values in equation (1)
dx dy = (1 + x)y + + + dxlog (1 x) dy x 1 y e + x sin 1 x cos x sin 2 4 − ⇒ dx dy = ) x 1 ln( ) x 1 ( 1 x sin 1 / x cos x sin 2 ) x 1 ( y y 4 1 y + + − − + + − At x = 0 y = (1 + 0)y + sin–1 sin (0) = 1
⇒ dx dy = ) 0 1 ln( ) 0 1 ( 1 ) 0 sin 1 ( / 0 cos . 0 sin 2 ) 0 1 ( 1 1 4 1 1 + + − − + + − ⇒ dx dy= 1
Again the slope of the normal is m = – dx / dy 1 = – 1
Thus, the required equation of the normal is y – 1 = (– 1) (x – 0)
i.e., y + x – 1 = 0.
14. Determine the equation of the curve passing through the origin in the from y = f(x), which satisfies the differential equation
dx dy
= sin (10x + 6y) [IIT-1996] Sol. dx dy = sin (10x + 6y) Let 10x + 6y = t (given) ...(1) ⇒ 10 + 6 dx dy= dx dt ⇒ dx dy = 6 1 −10 dx dt
Now the given differential equation becomes sin t = 6 1 −10 dx dt ⇒ 6sin t = dx dt – 10 ⇒ dx dt = 6 sin t + 10 ⇒ 10 t sin 6 dt
+ = dx apply variable separable Integrating both the sides, we get
∫
=∫
+10 dx t sin 6 dt ⇒ 2 1∫
3sint+ 5 dt = x + c ...(2) Let I1 =∫
+ 5 t sin 3 dt Put tan t/2 = u ⇒ 2 1 sec2 t/2 dt = du ⇒ dt = 2 / t sec du 2 2 ⇒ dt = 2 / t tan 1 du 2 2 + ⇒ dt = 2 u 1 du 2 + Also, I1 =∫
+ 5 t sin 3 dt =∫
+ +tan t/2 5 1 2 / t tan 2 3 dt 2 =∫
+ + + ) 2 t tan 5 5 2 t tan 6 ( dt ) 2 / t tan 1 ( 2 2 =∫
+ + + + ) 5 u 6 u 5 )( u 1 ( du ) u 1 ( 2 2 2 2 = 5 2∫
u2+(6du/5)u+1 = 5 2∫
+ + − +1 25 9 25 9 u 5 6 u du 2 = 5 2∫
+ + 25 16 5 3 u du 2 = 5 2∫
+ + 2 2 5 4 5 3 u du = 5 2 . 4 5 tan–1 + 5 / 4 5 / 3 u = 2 1 tan–1 + 4 3 u 5 = 2 1 tan–1 + 4 3 2 / t tan 5 Putting this in (2) Now 2 1I 1 = x + c ⇒ 4 1 tan–1 + 4 3 2 t tan 5 = x + c ⇒ tan–1 + 4 3 2 t tan 5 = 4x + 4c ⇒ 41[5 tan (5x + 3y) + 3] = tan (4x + 4c) ⇒ 5tan (5x + 3y) + 3 = 4 tan (4x + 4c) When x = 0, y = 0 we get 5 tan 0 + 3 = 4 tan (4c) ⇒ 4 3= tan 4c ⇒ 4c = tan–1 4 3
Then, 5 tan (5x + 3y) + 3 = 4 tan (4x + tan–1 3/4) ⇒ tan (5x + 3y) = 5 4 tan (4x + tan–1 3/4) – 5 3 ⇒ 5x + 3y = tan–1 + − − 5 3 } 4 / 3 tan x 4 {tan( 5 4 1 ⇒ 3y = tan–1 + − − 5 3 } 4 / 3 tan x 4 {tan( 5 4 1 –5x ⇒ y = 3 1 tan–1 + − − 5 3 } 4 / 3 tan x 4 {tan( 5 4 1 – 3 x 5
15. A tangent to the ellipse x2 + 4y2 = 4 meets the ellipse
x2 + 2y2 = 6 at P and Q. Prove that the tangents at
P and Q of the ellipse x2 + 2y2 = 6 arc at right angles.
[IIT-1997] Sol. x2 + 4y2 = 4 (given) ⇒ 4 x2 + 1 y2 = 1 ...(1)
Equation of any tangent to the ellipse on (1) can be written as
2
xcos θ + ysinθ = 1 ...(2) Equation of second ellipse
1 O –2 2 –1 3 – 3 – 6 6 x Q A P 90º y x2 + 2y2= 6 (given) ⇒ 6 x2 + 3 y2 = 1 ...(3)
Suppose the tangents at P and Q meet in A(h, k). 6
x h +
3
ky = 1 ...(4)
But (4) and (2) represent the same straight line, so comparing (4) and (2) 2 / cos 6 / h θ =sinθ 3 / k = 1 1
⇒ h = 3cos θ and k = 3sinθ
Equation of the chord of contact of the tangents through A(h, k) is
Therefore, coordinates of A are (3cosθ.3sinθ) Now, the joint equation of the tangents at A is given by T2 = SS 1 i.e., 2 1 3 ky 6 hx + − = − + 1 3 y 6 x2 2 − + 1 3 k 6 h2 2 ...(5) In equation (5) coefficient of x2 = 36 h2 – 6 1 − + 1 3 k 6 h2 2 = 36 h2 – 36 h2 – 18 k2 + 6 1 = 6 1– 18 k2 coefficient of y2 = 9 k2 – 3 1 − + 1 3 k 6 h2 2 = 9 k2 – 18 h2 – 9 k2 + 3 1 = – 18 h2 + 3 1 Again coefficient of x2 + coefficient of y2
= – 18 1 (h2 + k2) + 6 1 + 3 1 = – 18 1 (9cos2 θ + 9sin2 θ) + 2 1 = – 18 9 + 2 1 = – 2 1 + 2 1 = 0
which shows that two lines represent by (5) are at right angles to each other.
SCIENCE TIPS
• An electron is moving along X-axis in a magnetic field acting along Y-axis. What is the direction of magnetic force acting on it. ® Along Z-axis • What is the equation of a plane progressive simple
harmonic wave traverlling in + x direction? ® y = a sin λ 2π (vt – x) = a sin 2π − λ x T t • What type of magnetic material is used in making
permanent magnets? ® Ferromagnetic • A wire kept along north-south is allowed to fall
freely. Will an induced emf be set up? ® No • Which of A.C. or D.C. is blocked by a capacitor? ® D.C.
Passage # 1 (Ques. 1 to 3) a Upper Branch ...up to infinite. V1 V1 V1 b V Lower Branch
Resistance of volt meters V1, V2, V3 ... are
R, 2 e ln R , 4 R, 3 ) 2 ( e ln R , 16 R , 5 ) 2 ( e ln R ... respectively then
1. Find the resistance of voltmeter V such that the current in upper branch is same as in lower branch. 2. If the reading of voltmeter is V1 is X and the sum of
reading of all the voltmeters in upper branch except voltmeter X is Y then. Is X = Y or not. Write the reason to support your answer.
3. If the resistance of the voltmeter V is R then write the relation between the reading of voltmeter V1 and V.
4. A rod is rotating about axis YY′ as shown below the linear charge density at distance x is λ(x) = 3x and it is rotating with angular speed ω about axis YY′ then
Y x. A B a. b. Y'
(A) Equivalent current i = π ω 4 3
(b2 – a2)
(B) If length of the rod varies keeping (a + b) as constant and angular speed ω =
3 4π
then equivalent current i is directly proportional to length of rod
(C) Charge on rod is 3(b2 – a2)
(D) Charge on rod is 3/2(b2 – a2)
5. Part-A and Part-B, a pair of closed and open cone is shown. In Part - A the charge on both the cones is same and in part - B surface charge density on both the cones is same
All the cones are rotating with angular speed ω as shown in figure then
ω C-1 Q ω C-2 Q Part-A ω C-3 σ ω C-4 σ Part-B
(A) Equivalent current of C-1 and C-3 is same irrespective to value of σ
(B) Equivalent current of C-1 and C-2 is same (C) Equivalent current of C-3 and C-4 is different (D) Equivalent current of Part - A and Part-B
can be same dependent on value of σ Passage # 2 (Ques. 6 to 8)
A multirange voltmeter is shown below. The galvanometer is having the resistance of it's coil as 10Ω and the maximum potential difference that can be applied across the galvanometer is 50mv then
CT R1 G a 5V R2 b R3 c CT is the common terminal
6. If the range between CT and a is 5volt then the value of resistance R1.
7. If the range between CT and b is just double as the range between CT and a then the value of resistance R2.
8. If the value of R3 is 3000Ω then what will be the
range between CT and c. 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
Sol ut i ons w il l be publ i s hed i n next is s ue
1. As shown in graph, the relation of U v/s PV is linear So, PV φ U (0, a) U = (tan φ).PV + a as (tan φ) = b So, U = b. PV + a
Using ideal gas equation PV = nRT U = b (nRT) + a Differentiate it, dU = nbRdT As dU = n CV dT So CV = bR = 2 f R ⇒ 2 f = b Degrees of freedom of the gas, f = 2b as b = 3
So f = 6
Degrees of freedom are 6 so it is tri-atomic non- linear gas.
2. γ = 1 + f 2 ⇒ γ = 1 + b 2 2 as f = 2b So, γ = 1 + b–1
3. As CV = bR it is not dependent on 'a' so if a varies
there is no change in the value of CV.
CV a 4. As CV = b R and 2 f = b = C0 + C1t2, f = 2C0 + 2C1t2 and dt df = 4C 1t
df / dt v/s t graph is a straight line with slope 4C1.
5. As →v and are →B mutually perpendicular so path will be circular but due to presence of resistive medium speed decreases and radius of circular path decreases.
So, path is spiral of decreasing radius Option (D) is correct
[For Ans. 6, 7, 8]
The equivalent circuit diagram is As 'a' is grounded a As 'a' grounded e1 = 2v(B)1 = 2v BI = 60V b x Y e2=vB(2e)=2vBI = 60 volt c d As 'd' is grounded R=10Ω
6. Current through 'R' is from Y to x
i =
10 120
= 12 Amp.
7. Potential difference across a and c is 60 volt. 8. Charge on deutron is e so energy of deutron is
120eV.
Solution
Physics Challenging Problems
Set # 6
1. A stone is projected with velocity v0 at an angle θ0
from the horizontal. Find the angular velocity of the stone relative to the point of projection, while it is at its maximum height.
Sol. Method 1 :
The position vector of the particle at any arbitrary instant when it is at point P is given by :
r(P) = xi + yj, r = y tanθi + yi Differentiating w.r.t. time we get,
dt dr
= dt
dy tanθ i + y sec2θ θ&i +
dt dy
j At the highest point of trajectory.
dt dy = 0 and y = H = g 2 sin v20 2θ0 So, at the sought point,
x O x R y Huθ v ur P θ y
v0 cos θ0 i = H sec2θ . θ&i
So, θ& = H cos v0 θ0 cos2θ or, θ&= θ θ 0 2 0 0 sin v cos g 2 cos2θ Method 2 :
At an arbitrary instant, when the particle is at point P tan θ = x/y
Differentiating Eqn. (1) w.r.t. time we get
sec2θ θ&= 2 y dt dx y dt dy x − θ&= − 2 y dt dx y dt dy x cos2θ At position P, dt dx = v 0 cos θ0, dt dy = 0, x = R/2 and y = H where R = g 2 sin v20 θ0 and H = g 2 sin v20 2θ0 Substituting the above values in Eqn.(2)
θ& = θ θ 0 2 0 0 sin v cos g 2 cos2θ
2. A bead of mass m is fitted onto a rough rod of length of 2l and can move along it only. At the initial moment the bead is in the middle of the rod. The rod moves translationally in space with the constant acceleration a in a direction forming an angle α with the rod (fig.)
m a
l α
(a) Find the time when the bead will leave the rod if the co-efficient of friction between bead and the rod is µk. (Neglect the weight of the bead)
(b) Do the part (a) of this problem without neglecting the weight of the bead.
Sol. Let us work in the frame of the rod.
(a) In this case in the absence of friction, under the action of the inertial force – ma the bead will move up along the rod, hence the kinetic friction will act along the rod down (fig.)
From Fx = max
ma cos α – µN = marel (1)
From, Fy = may
N = ma sin α (2)
Using (2) in (1), we get arel = a(cos α – µ sin α)
m a N y x fr α
From kinematic equation in the frame of the rod. ∆x = v0xt + 2 1 axt2 or, l = 0 + 2 1 arel t2 t = rel a 2l = ) sin (cos a 2 α µ − α l
Expert’s Solution for Question asked by IIT-JEE Aspirants
Students' Forum
(b) In the frame of rod, the bead is under the action of three forces N, –ma and mg, except kinetic friction. Case (i): If a cos α > g sin α, then the kinetic friction having the value of µkN will act to resist the upward
motion of the bead along the rod. (Fig.) From, Fy = may (for the bead)
N – mg cos α – ma sin α = 0
N = m a (sin α + g cos α) (1)
and from
Fx = cosα – µN – mg sinα = marel (2)
a N y x fr α ma mg Using (1) in (2), we get
arel = a cosα – µ(a sinα + g cosα) – g sinα
or, arel = a(cosα + µsinα) – g(µ cosα + sinα) (3)
From kinematic equation, (in the frame of rod) ∆x = v0xt + 2 1 axt2 or, l = 0 + 2 1 arel t2 t = rel a 2l = ) sin cos ( g ) sin (cos a 2 α + α µ − α µ − α l
Case (ii): When g sinα > a cosα , the bead will move down along the rod, so the kinetic friction will act upward along the rod. (Fig.)
From,
Fy' = may'
N = m (g cosα + a sinα) (4) and from Fx' = max'
mg sinα – µN – mg cosα = m arel (5)
Using (4) in (5), we get
a'rel = g sinα – µ(g cosα + a sinα) – a cosα
x'
N fr
α
ma mg
or, a'rel =g(sinα – µ(g cosα) – a(cosα + sinα)) (6)
From the kinematic equation, ∆x' = v0x' t + 2 1 ax' t2 or, l = 0 + 2 1 a'rel t2 So, t' = rel ' a 2l = ) sin cos ( a ) cos (sin g 2 α µ + α µ − α µ − α l
[In this part of problem a cosα = g sinα is not relevant] 3. Determine the period of oscillations of mercury of
mass m poured into a bent tube whose arms form the angles θ1 and θ2 with the vertical respectively (Fig.
a). The cross-sectional area of the tube is s. neglect the viscosity of the mercury.
θ1 θ2 θ1 θ2 x x Sol. Method 1: (x = dx/dt, • •x = d• 2x/dt2)
If at an arbitrary moment during the oscillations, the mercury rises in the left arm by x it must fall by the same length in the other arm. As the viscosity of the mercury is negligible, the mechanical energy of oscillations of the given system is conserved, i.e. U(x) + T(x, x) = Constant.
Taking the P.E. of oscillations zero at the equilibrium position and using the mathematical trick of negative mass, the conservation of M.E. of oscillations gives: (s x ρ) g 2 cos x θ1 – θ ρ − 2 cos x g ) ( sx 2 + 2 1 m(x )• 2 = Constant or, 2 g sρ x2 (cosθ 2 +cosθ2) + 2 1 m(x )• 2= Constant
Differentiating w.r.t. time, we get 2 ) cos (cos g sρ θ1+ θ2 2xx +• 2 1 m 2x = 0 ••x• or •x = –• m ) cos (cos g sρ θ1+ θ2 x Thus, the sought time period,
T = 2π ) cos (cos g s m 2 1+ θ θ ρ Method 2:
If the mercury rises in the left arm by x, obviously it must fall by the same length in the right arm. At this position the total pressure difference in the two arms will be
ρg x cos θ1 + ρg x cos θ2 = ρg x (cos θ1 + cos θ2)
This will give rise to a restoring force – ρg x (cos θ1 + cos θ2) s
This must equal mass times acceleration which can be obtained from work-energy principle.
The K.E. of the mercury in the tube is clearly :
2 1
So mass times acceleration must be : m•x • Hence,
m•x + ρ g x(cos θ• 1 + cos θ2) s = 0
which is the Eq. of S.H.M. with time period T = 2π ) cos (cos s g m 2 1+ θ θ ρ
4. A sphere and a cube of the same material and same total surface area are placed in the same space turn by turn, after heating them to the same temperature. Compare their initial rate of cooling in the enclosure. Sol. Rate of emission of energy
σ T4 S = m sphere c sphere dt dT − Also σ T4 S = m cube c cube dt dT − So, cube sphere ) dt / dT ( ) dt / dT ( − − = cube sphere m m = cube sphere V V But Vsphere= 3 r 4π 3 = 3 4π 3/2 4 S π , because, S=4πr 2 and Vcube = a3 = 3/2 2 / 3 6 S , because, S = 6a2
Hence the required ratio =
3 ) 6 4 ( π−1/2× 3/2 = 1.38 5. A positive point charge q of mass m, kept at a
distance x0 (in the same plane) from a fixed very long
straight current i is projected normally away from it with speed v. Find the maximum separation between the wire and the particle.
Sol. We know that a moving charge in magnetic field experiences a side way force given by the formula F = q (v × B) at a certain instant of time. As the magnetic field is not uniform, the particle does not follow the circular path but the speed (v) of the particle is constant. Here the magnetic field set up by the straight current is directed normally into the page (i.e., along the negative z-axis) and the initial velocity of the particle is along x-axis and further the force F is always in the x-y plane, so the motion of the particle is confined in the xy plane. The force at time t (Fig.) after starting from point P is.
x i O x. y x0. F = q (v × B) or, F = q
{
}
− π µ × + ( k) x i 2 j v i v 0 y x or, = π µ 2 0 x i q (–vy i + vxj) (A) so, Fx = – π µ 2 0 mx v i q y so, ax = – mx v i q 2 y 0 π µ or, dx dv vx x = π µ 2 0 mx v i q y (1) But v +2x v = v2y 2 so, 2vxdvx = 2vydvy = 0 or, vxdvx = – vydvy (2)From Eqns. (1) and (2)
qi m 2 dv 0 y µπ = x dx or,
∫
µ π v 0 0qi m 2 dvy =∫
x x0 x dx or, qi m 2 0 µ π v = ln 0 x x Hence, x = x0 e2 π mv/µ0 qiNote : Instead of Fx we may write Fy from Eqn. (A)
and then proceed in similar fashion.
MEMORABLE POINTS
• The vector relation between linear velocity and angular velocity is ® →v =→ω ×→r • In the case of uniform circular motion the angle between
→
ω and→r is always ® 90º(hence |→v | = ωr • The relation between Faraday constant F, Avogadro number N and the electronic charge e is ® F = Ne • Depolariser used in Lechlanche cell is
® Manganese dioxide • The absorption or evolution of heat at a junction of two dissimilar metals when a current is passed is
known as ® Peltier effect
• The part of the human ear where sound is transduced
is the ® Cochlea
• Similar trait resulting from similar selection pressure acting on similar gene pool is termed
® Parallel evolution
• Group of related species with the potential, directly or indirectly, of forming fertile hybrids with one another is called ® Coenospecies
Electromagnetic Induction (E.M.I.)
Faraday's law states that the induced emf in a closed loop equals the negative of time rate of change of magnetic flux through the loop. This relation is valid whether the flux change is caused by a changing magnetic field, motion of the loop, or both.
ε = – dt dΦB A B ε φ
Lenz's law states that an induced current or emf always tends to oppose or cancel out the change that caused it. Lenz's law can be derived from Faraday's law, and is often easier to use.
ε
B (increasing)
Change in B
Binduced
I
If a conductor moves in a magnetic field, a motional emf is induced.
ε = vBL
(conductor with length L moves in uniform Br field, Lr and vr both perpendicular to Br and to each other)
ε =
∫
(vr×Br).dlr(all or part of a closed loop moves in a Br field)
× × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × a a F=qvB v F = qE q L B + –
When an emf is induced by a changing magnetic flux through a stationary conductor, there is an induced electric field Erof non-electrostatic origin. This field is non conservative and cannot be associated with a potential.
∫
Er.dlr = – dt dΦB G E B E E I rWhen a bulk piece of conducting material, such as a metal, is in a changing magnetic field or moves through a field, currents called eddy currents are induced in the volume of the material.
I´ I0
B0
B´
A time-varying electric field generates a displacement current iD, which acts as a source of
magnetic field in exactly the same way as conduction current.
iD = ε
dt
dΦE (displacement current)
Alternating Current (A.C.)
An alternator or ac source produces an emf varies sinusoidally with time. A sinusoidal voltage or current can be represented of the by a phasor, a vector that rotates counterclockwise with constant angular velocity ω equals to the angular frequency of the sinusoidal quantity. Its projection on the horizontal axis at any instant represents the instantaneous value of the quantity.
Electromagnetic Induction & A.C.
PHYSICS F
UNDAMENTAL F
OR
IIT-JEE
i=I cos ωt O
ω I ωt
For a sinusoidal current, the rectified average and rms (root-mean-square) currents are proportional to the current amplitude I. Similarly, the rms value of a sinusoidal voltage is proportional to the voltage amplitude V. Irav = π 2 I = 0.637 I Irms = 2 I ; Vrms = 2 V
In general, the instantaneous voltage between two points in an ac circuit is not in phase with the instantaneous current passing through points. The quantity φ is called the phase angle of the voltage relative to the current.
i = I cos ωt v = V cos(ωt + φ) V cosφ O ωt I φ V
The voltage across a resistor R is in phase with the current. The voltage across an inductor L leads the current by 90º (φ = + 90º), while the voltage across a capacitor C lags the current by 90º(φ = –90º). The voltage amplitude across each type of device is proportional to the current amplitude I. An inductor has inductive reactance XL = ωL and a capacitor has
capacitive reactance XC = 1/ωC.
VR = IR; VL = IXL; VC = IXC
Resistor connected to
ac source Inductor connected toac source
a R b a L b i i a C b Capacitor connected to ac source i q –q i
In a general ac circuit, the voltage and current amplitude are related by the circuit impedance Z. In an L-R-C series circuit, the values of L, R, C, and the angular frequency ω determine the impedance and the phase angle φ of the voltage relative to the current.
V = IZ Z = R2+(XL−XC)2 = R2+[ωL−(1/ωC)]2 tan φ = R C / 1 L− ω ω V = IZ VL = IXL VL – VC VR = IR VC = IXC O I ωt φ
The average power input Pav to an ac circuit depend
on the voltage and current amplitudes (or, equivalently, their rms values) and the phase angle φ of the voltage relative to the current. The quantity cos φ is called the power factor.
Pav = 2 1VI cos φ = V rmsIrmscos φ v, i, p Pav = ½ VIcosφ p p i t φ ω ?
In an L-R-C series circuit, the current becomes maximum and the impedance becomes minimum at an angular frequency ω0 = 1/ LC called the
resonance angular frequency. This phenomenon is called resonance. At resonance the voltage and current are in phase, and the impedance Z is equal to the resistance R. I(A) 200 Ω 0.5 0.4 0.3 0.2 0.1 0 1000 2000 ω (rad/s) 500 Ω 2000 Ω
A transformer is used to transform the voltage and current levels in an ac circuit. In an ideal transformer with no energy losses, if the primary winding has N1
turns and the secondary winding has N2 turns, the
amplitudes (or rms values) of the two voltages are related by Eq. The amplitudes (or rms values) of the primary and secondary voltages and currents are related by Eq. 1 2 V V = 1 2 N N ; V1I1 = V2I2
