Dear Students,
The difference between success and failure is your attitude towards success and the strategies that you employ to achieve it. The difference between success and failure is only a few minutes or a few hours everyday. You have to keep on striving for success at every conceivable opportunity. Never postpone your happiness and zest for life and work. You should make it a habit to enjoy your profession and your job all the time. Never be a quitter because a quitter can never be a winner.
You should always remember that People live not by the reason of any care they have for themselves but by the love for them that is in other people. Have only those people for friends and companions who do their best to bring out the best in you. They will be of unlimited worth to you. Such persons understand what life means to you and your goal. They feel for you as you feel for yourselves. They are the ones who are bound to you in triumph and disaster. They provide a purpose to live and break the spell of loneliness. A true friend is worth befriending as he will always stand by you. But before you expect others to be the right person to be your friend you must also become one.
Be always committed to your cause. Be so engrossed in your work that you have hardly any time to think of anything else. The great secret of success is to do whatever you are to do and do it wholeheartedly. Make yourself the star of your workplace. For this you must have clear and precise objectives to be achieved within a definite time-frame. Always respect and value time. Be result-oriented and keep track of the hours. Respect the time of others as well as your own. Be always organized and write down everything you want to accomplish.
Always make an assessment of yesterday's "To Do" list to crosscheck how realistic it has turned out to be today. This will help you to avoid or rectify mistakes, if any, in your planning. Keep on visualizing your goals and lists of the task to be done.
Forever presenting 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.
• No Portion of the magazine can be published/ reproduced without the written permission of the publisher
• 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
Do not allow the quest for perfection to ruin your life because whatever you do you will always feel that you could have done better
Volume - 5 Issue - 3
September, 2009 (Monthly Magazine)
Editorial / Mailing Office :
112-B, Shakti Nagar, Kota (Raj.) 324009 Tel. : 0744-2500492, 2500692, 3040000 e-mail : [email protected]
Editor :
Pramod Maheshwari [B.Tech. IIT-Delhi]
Analyst & Correspondent
Mr. Ajay Jain [B.E]
Cover Design & Layout
Niranjan Jain
Om Gocher, Govind Saini
Circulation & Advertisement
Ankesh Jain, Praveen Chandna Ph (0744)- 3040007, 9001799502
Subscription
Sudha Jaisingh Ph. 0744-2500492, 2500692 © Strictly reserved with the publishers
Editorial
Unit Price Rs. 20/-
Special Subscription Rates 6 issues : Rs. 100 /- [One issue free ] 12 issues : Rs. 200 /- [Two issues free] 24 issues : Rs. 400 /- [Four issues free]Volume-5 Issue-3 September, 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
• "True success is overcoming the fear of being unsuccessful."
• "Get up one time more than you're knocked down."
• "Most people who succeed in the face of seemingly impossible conditions are people who simply don't know how to quit." • "The truth is that all of us attain the
greatest success and happiness possible in this life whenever we use our native capacities to their greatest extent."
• "When your physical environment is in alignment with your aspiration, success becomes the norm."
• "The most important single ingredient in the formula of success is knowing how to get along with people."
• "Dictionary is the only place that success comes before work. Hard work is the price we must pay for success. I think you can accomplish anything if you're willing to pay the price."
CONTENTS
INDEX
PAGE
NEWS ARTICLE
3
IIT-Kanpur students made Nano-satellite
IIT-K all set to start FM radio station
IITian ON THE PATH OF SUCCESS
7
Mr. Kannan M. Modgalya
KNOW IIT-JEE
8
Previous IIT-JEE Question
XTRAEDGE TEST SERIES
53
Class XII – IIT-JEE 2010 PaperClass XII – IIT-JEE 2011 Paper
Regulars ...
DYNAMIC PHYSICS
14
8-Challenging Problems [Set# 5] Students’ Forum
Physics Fundamentals Current Electricity
Circular Motion, Rotational Motion
CATALYST CHEMISTRY
33
Key ConceptAliphatic hydrocarbon
Oxygen Family & Hydrogen Family Understanding : Inorganic Chemistry
DICEY MATHS
42
Mathematical Challenges Students’ Forum Key Concept Probability Binomial TheoremStudy Time...
Test Time ...
IIT-Kanpur students made
Nano-satellite
Kanpur: The Indian Space Research Organization (ISRO) will launch Nano-satellite "Jugnu" made by students of the Indian Institute of Technology, Kanpur (IIT-K) in December 2009.
"A team of 20 students from our institute has made the nano-satellite," said Prof. S.G. Dhande, Director of IIT-Kanpur.
The weight of the satellite is less than 10 kilograms. The satellite is made worth of 2 ½ Crores. The satellite will give information related with drought, flood, agriculture and forestry.
In addition to this IIT-K will also celebrate its golden jubilee from August 2009 to December 2010. The inauguration is from August 8th-9th 2009. Inaugural address will be delivered on August 8. Mr. N. R. Narayana Murthy, the non-executive chairman and chief mentor of Infosys, will join the function as Chief Guest. He is a distinguished alumnus of IIT-K.
IIT-K all set to start FM
radio station
Kanpur: The Indian Institute of Technology, Kanpur (IIT-K) will soon start an FM radio station in the campus. This is for the first time an FM radio station is going to be operational from IIT campus. The FM station will broadcast scientific programmes and classical music.
IIT-K has already received green signal for the project from the Ministry of Information and Broadcasting. The FM station in IIT-K will operational following the model of America's National Public Radio (NPR).
Cultural programme will also air by the upcoming FM radio station at IIT-K.
"We have decided the land for the FM station and instruments worth Rs.22 lakh have been purchased. All formalities have also been completed to start the FM station," said Prof. Dhande, Director IIT-K.
A four-member committee headed by Prof. Sanjay Kasalkar, Registrar of the institute has been constituted to put the whole project into operation. The most important feature of the FM station will be the phone-in programme in which students from across the city can put their queries relating to science to the professors of the college.
IIT Delhi to hone marketing
skills at tourism ministry
NEW DELHI: The tourism ministry has, for the first time, organised a three-day training programme for its overseas officials by experts at the
Indian Institute of Technology (IIT) Delhi in an effort to sharpen their marketing skills.
Talking about the programme, Tourism Minister Kumari Selja said that the ministry's officials will be trained to promote India better.
'Our overseas officials who promote India as an ideal travelling destination will get this training on better and more effective marketing and promotion
technique from IIT Delhi's management wing. It will be a three-day training programme,' Selja said at the inaugural function of the ministry's two-day overseas marketing meet here. 'After this, we are planning a similar training programme for our domestic officials as well,' Selja added.
Twenty tourism officials will attend the first programme. 'This meet is very important, especially in the light of the global meltdown and its drastic effects on tourism. From January to June this year our foreign tourist arrival fell by 9.3 percent as compared to last year,' Selja said.
'However, our efforts to bring India back on the tourists' map, the several roadshows in different countries and promotion of niche tourism products has ensured that in June this year there was a positive growth of 0.2 percent in the foreign tourist arrival to India,' she added.
Selja further said that the meet, which will see overseas tourism officials and the private sector here interacting, is bound to throw up innovative ideas to attract more travellers to the country.
More than 500 candidates
refused to get into IITs
Mumbai: To get into Indian Institutes of Technology (IITs) is considered as "one of the hardest nuts to crack", but this year more than 500 students have refused to get into IITs after qualifying for the Joint Entrance Test, conducted by IITs.
All students have their own reason to refuse to study in IITs. Some of them have no confidence
in the new IITs, while some of them did not get their own choice of streams. An IIT official said, "This experience might force HRD minister Kapil Sibal to do a rethink on his expansion plans for the IITs."
In addition to existing seven IITs, eight IITs that are more new have been added during the eleventh five-year plan in the country. After not getting sufficient number of qualified student for the reserve categories, IITs have transferred 1100 reserved-category seats to the one-year preparatory course. The one-year 'prep course' trains quota students to bring them up to the mark.
But this year more than 500 candidates who have refused admission to IITs are from open category.
IT BHU soon to become IIT
New Delhi: The Institute of Technology, Banaras Hindu University (IT BHU) will soon join the league of Indian Institute of Technology (IIT).
The Ministry of Human Resource Development (MHRD) has decided to convert IT BHU into an IIT.
The decision was conveyed in the Lok Sabha on Monday by D. Purandeswari, Minister of State for HRD.
The decision was conveyed when she replied to a question on whether the government has taken any steps in developing technology institutions across the country on the lines of IITs. For a long time, IT BHU had been in a process of upgrading itself into an IIT.
The admission process for students is already being done through the IIT Joint Entrance Examination (IIT JEE).
As per S.N. Upadhyaya, Director, IT BHU, "The IIT status will not only improve the infrastructure and academics of the institute but
will also get better students and attract bright faculties".
"With more academic freedom, the institute will be able to introduce inter-disciplinary courses that were not possible earlier as we were tied up with the rules and regulations of the university, to go through various councils to get a new academic programme passed. But the status of IIT will enable us to introduce new courses without these limitations" he added. "Appointing faculties will also become much easier," he avers. Seven other institutes comprising of Bengal Engineering College; Howrah, Cochin University of Science and Technology; Kochi, Engineering and Technology Department of Jadavpur University and Zakir Hussain College of Engineering and Technology; Aligarh Muslim University apart from IT BHU are being considered to become a part of the IIT brigade.
IIT-D to conduct annual
convocation in Aug
New Delhi: The Indian Institute of Technology, Delhi (IIT-D) will award 180 PhD degrees to students,the highest ever. Scholars will be awarded their degree at the convocation taking place next month on August 8 and 9 in the campus.
Last year 147 students were awarded PhD.
IIT-D has taken several steps to make the institute as a reputed research centre in the country and attract students towards research programme.
Undergraduate and Postgraduate students will also be awarded their degrees during the convocation
IIT-Delhi to spend Rs.1
crore on tightening security
New Delhi: Trespassing is passé at the Indian Institute of Technology, Delhi (IIT-D) as the institute is all geared up to tighten their security levels.
IIT-D is planning to shell out Rs.1 crore on enhancing the level of security within next few months. This will include installation of close circuit television (CCTV) cameras, introducing electronic access system in hostels, academic areas such as library, reading room and canteen. "The raising number of population in the campus made us take this step. This being an educational campus it was practically not possible to keep a vigil on all those entering or leaving the campus. The new measures will help us in safeguarding the campus," said Surendra Prasad, Director, IIT-D. The surveillance at all the five entry gates of the campus by 17 CCTV cameras, will ensure that the 325 acres of the campus which is pretty permeable to the people from the neighbouring villages, cannot be easily breached.
With the new security measures, the institute hopes to overcome the menace of trespassers and petty thefts at the campus.
Nearly 5,500 students of IIT-D will be issued new identification cards before the end of this year. "The new cards will act as a single admit card to the library, hostel and reading room. Instead of the current cards that act as their identity cards as well as the library card. This will limit the entry and exit of the outsiders in the academic areas," said Captain B.N. Yadav, Security Officer, IIT-D.
Three new MTech
programmes at IIT this year
CHENNAI: Three new M.Tech programmes in Catalysis Technology, Nuclear Engineering and Petroleum Engineering will be offered at IIT-Madras in the 2009-10 academic year, director M S Ananth announced on Friday Delivering the director's report at the 46th convocation of the institute here on Friday, he said a five-year integrated dual degree programme would also be introduced in these disciplines.
"IIT-Madras is establishing a research park adjacent to its campus in Chennai, the first of its kind in the country. For this purpose, IIT-M has obtained 11.42 acres of land from the state government," he added. Pointing out that the facility was likely to be inaugurated in September, professor Ananth said they hoped to have the prime minister, the Tamil Nadu chief minister and Union Minister for Human Resource Development (MHRD) Kapil Sibal at the
inaugural event. A total of 1,439 students received
degrees 162 were awarded Ph.D, 121 M.S., 354 M.Tech, 83 M.Sc, 62 MBA, 313 B.Tech awards and 172
got dual degrees. Professor Jagdish N Bhagwati of the department of economics at Columbia University, who was conferred the degree of Doctor of Science (Honoris Causa) by the institute on the occasion, delivered the convocation address. "You should have learnt from your IIT education that great universities teach you two things: creation of knowledge and the practice of virtue. As scientists, you must imbibe knowledge and seek to extend its frontiers. As part of humanity, you must also learn to put it to good use," he said.
In addition to instructing students on the need for them to periodically adjust and revise their knowledge, professor Bhagwati reminded his audience that each of them owed his/her success equally to their parents. "Great opportunities are open now to put your knowledge, and your ingenuity, at the service of India and her poor," he said. Touching on the problem of limited availability of qualified faculty in the higher education system today, R Chidambaram, principal scientific advisor to Government of India, and chairman, board of governors, IIT-Madras, said online teaching could be one way of ensuring quality
technical education. "An alternative to the required growth in brick-and-mortar through expansion and addition is a massive online education system. A country like India therefore has no choice. The National Programme on Technology Enhanced Learning (NPTEL), liberally funded by the MHRD, is an opportunity to provide quality online engineering education," he said.
Bill Clinton to address IIT
summit in Chicago
Former American President Bill Clinton will be among many top leaders to address the seventh Pan-IIT Global Conference being held in Chicago in October. Kapil Sibal, Indian minister for human resource development; Sam Pitroda, Indian Knowledge Commission chairman; Aneesh Chopra, America's chief technology officer; and Meera Shankar, Indian ambassador in the US, will be among other keynote speakers at the Pan-IIT Global Conference to be held in Chicago from Oct 9-11.
According to conference chairman Ray Mehra, 'Entrepreneurship and Innovation in a Global Economy' is the theme of this year's techie summit. Over 3,000 IITians from around the world will attend the annual gathering to be opened by Sibal. Chopra will deliver the keynote address.
"We have invited President Clinton since the goals of his William J. Clinton Foundation and the Pan-IIT conference are the same. We have common areas like energy, climate change, health care and education to work on,'' Mehra told IANS.
"The president and other global leaders in their fields will discuss how we can transform ideas into action on both sides of the ocean (in the US and India),'' he said. Mehra said the Pan-IIT summit will take a holistic approach to
problems in areas like health and energy in India.
"We will discuss how the public health system (PHS) in India can be steered with inputs from the PHS in the US which is under massive changes now,'' he said. Mehra said the conference will also discuss its proposal called Panch Ratnas submitted to the Indian government to revolutionize higher education in the country.
"We presented a white paper titled PanIIT Panch Ratnas to President Pratibha Patil last month, proposing a five-point action plan to make India the global hub for knowledge creation and talent development by 2022. We will debate this in detail in Chicago,'' he said.
Pan-IIT's Panch Ratnas include implementation of wholesale policy reforms in education, quality control and increase capacity, and 'quantum improvement in faculty service conditions, deployment of technology for teaching and collaborative research, and the establishment of an industry-academia interface, according to Mehra.
Other prominent speakers at the summit include James Owens, chairman and CEO of Caterpillar Inc., Sharon Oster, dean of the Yale School of Management, Tulsi Tanti, chairman of Suzlon Energy, Carl Shramm, president and CEO of the Kauffman Foundation, and Prof Raghuram G Rajan of the University of Chicago and former chief economist of the IMF. There are said to be an estimated 35,000 IITians in the US.
Science learning materials
on Net soon
By the end of this year, elementary school teachers and students can find an interactive, experiential science learning programme developed in association with IIT-Madras, freely available on the Internet. The
Kuruvila Jacob Initiative for promoting excellence in school education, which was set up in memory of the headmaster of the Madras Christian College High School, has been running a programme to develop multimedia-enhanced science learning materials for Classes VI to IX, in partnership with IIT-Madras since February 2007.
At the sixth annual function of the Initiative, launched on Kuruvila Jacob’s birth centenary, IIT-M Director M.S. Ananth launched the latest DVDs of Physics and Chemistry material, and announced that the whole project would be ready for a Web launch by the end of 2009.
“From the very beginning, it was decided to make it available in the public domain,” said Dr. Ananth, explaining how the modules were developed using the same equipment and team which are putting IIT’s engineering courses on YouTube.
About 80 teachers, from 22 schools in the city, were involved in developing the materials. “An experiment showing how the electron moves is worth three lectures in quantum mechanics,” said Dr. Ananth, discussing the experiential and demonstrative mode of teaching that the project has attempted to use in each module. He believes that distance education can help fill the gaps in the Indian school education system, just as it is starting to do in higher education as well. “The ratio of teachers to students is 1:100 or worse,” he pointed out. “We are moving from a gurukulam education at the home of the guru to what I call sishyakulam at the homes of the sishyas, no matter how scattered they may be,” he said, explaining that the Internet version of these science modules would allow students to study from the comfort of their own homes. Earlier, historian and writer Ramachandra Guha delivered the
keynote address at the function, on the topic ‘Why India is the most interesting country in the world.’
He explained that the country was
undergoing simultaneous revolutions in at least five areas –
industrial, urban, national, democratic and social – in a way that was unprecedented in international history.
IIT-K disappointed over
rejection of setting up of
n-reactor
In a disappointment to the officials of IIT Kanpur, the Department of Atomic Energy has rejected the college's proposal of setting up of a small nuclear reactor in its campus citing security reasons. However, the officials of IIT Kanpur said they were trying to clear the doubts and will try to convince the Department of Atomic Energy at the 'International Nuclear Meet' in November.
Director of IIT Kanpur Prof Sanjay Govind Dhande said that the college asked for a permission from the Department of Atomic Energy, Mumbai to set up a nuclear reactor for M.Tech students, who are doing Nuclear Engineering.
"We have asked for the permission from the Department of Atomic Energy, Mumbai, for a small reactor in the campus. But due to security reasons as cited by the Department it has been rejected," he said. He said the college was hopeful of
a positive reply after India signed the 123 agreement with the US. According to the college officials, since the last 35 years the college has been providing Nuclear engineering course to 15 students each year but they were able to give only theory classes. "The college has been giving only theory classes to the students. Due to the lack of a nuclear reactor facility in the college, the students are devoid of any practical training.
He added the Department has not only cited security problems but has also mentioned about the lack of nuclear fuel in the country.
IIT-D faculty gets National
Award for Atmospheric
Science and Technology
New Delhi: Prof. Shishir Kumar Dube, former Director of Indian Institute of Technology, Kharagpur and currently Professor at the Centre for Atmospheric Sciences, IIT, Delhi was awarded with National Award in Atmospheric Sciences for the year 2009.The award has been given to Prof. Dube by Ministry of Earth Sciences in recognition of his outstanding contributions in the field. He was born in Kalpi District of Uttar Pradesh on 4th October 1947. He worked at the India Meteorological Department from 1972-78 and then joined the faculty of IIT, Delhi. Research interests of Prof. Dube include Numerical Storm Surge Prediction, Ocean Wave Modeling, Coastal Marine Hazards and Regional Ocean State Forecasting Models. He is internationally recognized for his pioneering contributions in the field of storm surge prediction in the Bay of Bengal and the Arabian Sea. Prof. Dube is responsible for the development of real time operational surge prediction systems which under the auspices of World Meteorological Organisation have been transferred to the National Weather Services of Bangladesh, Maldives, Myanmar, Oman, Pakistan, Sri Lanka and Thailand for their operational use. Prof. Dube has received several honors and awards including the 12th MAUSAM Award for the year 1982-83.
Mr. Kannan M. Modgalaya completed his B.Tech in Chemical Engineering, IIT Madras, June 1980 and after that he was awarded by mastered degree as Master of Electrical Engineering from Rice University in May 1985 then achived Doctorate (Ph.D) in Chemical Engineering, Rice University, May 1985
Presently he is related to various reseach work and awarded with with various awards and fellowship that are described below:
Research Areas
Dynamical Systems, Discontinuity Detection, Discontinuity Sticking, Differential Algebraic Equations, Process Control, Digital Control, Reactor Control, Grade Transition.
Current Research Dynamics
He is studying discontinuous dynamical systems that exhibit the property of sliding. The concept of equivalent dynamics has been shown to speed up sliding DAE systems by 10,000 times.
Control
He is looking at the control of chemical processes, especially reactors. He also looked at control of systems, such as, an inverted pendulum through internet. Handling
of performance deterioration through network delays is being studied.
Simulation
He has worked on the topic of simulation methodologies and simulation environments through modern computing tools. This approach has been applied to simulation of neuro transmission in muscle cells. Automatic model derivation from first principles and data driven model generation are some focus areas.
Awards and Affiliations
1. Best Paper Presentation Award for "Control of a high index DAE system through a linear control law" by P. Vora,
2. K. Moudgalya and A. K. Pani, American Contro Conference, Anchorage, 8 May 2002.
3. Lovraj Kumar Industry-Academia Exchange Fellowship, April 1997
4. Best Poster Award for the paper ``An Integrated Simulation Environment'' by S. H. Rao, K. Moudgalya, K. V. Nori.
5. G. Sivakumar, International Conference on Advances in Chemical Engineering, 11-13 Dec. 1996, IIT Madras.
Mr. Kannan M. Modgalya
B.Tech, IIT Madras
Master of Electrical Engineering, Doctorate (Ph.D) in Chemical Engineering, Rice University
Dare to dream, dare to try, dare to fail, dare to succeed
Success Story
PHYSICS
1. A transverse harmonic disturbance is produced in a
string. The maximum transverse velocity is 3 m/s and maximum transverse acceleration is 90 m/s2. If the
wave velocity is 20 m/s then find the waveform. [IIT-2005]
Sol. The wave form of a transverse harmonic disturbance
y = a sin (ωt ± kx ± φ)
Given vmax = aω = 3 m/s ...(i)
Amax = aω2 = 90 m/s2 ....(ii)
Velocity of wave v = 20 m/s ...(iii) Dividing (ii) by (i)
ω ω a a 2 = 3 90 ⇒ ω = 30 rad/s ...(iv) Substituting the value of ω in (i) we get
a = 30 3 = 0.1 m ...(v) Now k = λ π 2 = v / v 2π = v v 2π = v ω = 20 30 = 2 3 ...(vi) From (iv), (v) and (vi) the wave form is y = 0.1 sin ± x±φ
2 3 t 30
2. A 5m long cylindrical steel wire with radius 2 × 10–3
m is suspended vertically from a rigid support and carries a bob of mass 100 kg at the other end. If the bob gets snapped, calculate the change in temperature of the wire ignoring radiation losses. (For the steel wire : Young's modulus = 2.1 × 1011 Pa; Density =
7860 kg/m3; Specific heat = 420 J/kg-K). [IIT-2001]
Sol. When the mass of 100 kg is attached, the string is
under tension and hence in the deformed state. Therefore it has potential energy (U) which is given by the formula.
U =
2 1
× stress × stain × volume
= 2 1 × Y ) Stress ( 2 × πr2 l = 2 1 Y ) r / Mg ( π 2 2 × πr2 l = 2 1 Y r g M 2 2 2 π l ...(i)
This energy is released in the form of heat, thereby raising the temperature of the wire
Q = mc ∆T ...(iii)
From (i) and (iii) Since U = Q Therefore ∴ mc∆T = Y r g M 2 1 2 2 2 π l ∴ ∆T = Ycm r g M 2 1 2 2 2 π l Here
m = mass of string = density × volume of string = ρ × πr2 l ∴ ∆T = ρ πr ) Yc ( g M 2 1 2 2 2 2 = 2 1 × 7860 420 10 1 . 2 ) 10 2 14 . 3 ( ) 10 100 ( 11 2 3 2 × × × × × × × − = 0.00457ºC
3. An unknown resistance X is to be determined using resistance R1, R2 or R3. Their corresponding null
points are A, B and C. Find which of the above will give the most accurate reading and why ? [IIT-2005] R = R1 or R2 or R3
Sol. All null point, the wheat stone bridge will be
balanced ∴ 1 r X = 2 r R ⇒ X = R 2 1 r r A X R G B C M N r1 r2
where R is a constant r1 and r2 are variable. The
maximum fraction error is X X ∆ = 1 1 r r ∆ + 2 2 r r ∆
Here ∆r1 = ∆r2 = y (say) then
For X X ∆ to be minimum r 1 × r2 should be max [Q r1 + r2 = c (Constt.] Let E = r1 × r2 ⇒ E = r1 × (r1 – c) ∴ 1 dr dE = (r1 – c) + r1 = 0
KNOW IIT-JEE
⇒ r1 = 2 c ⇒ r2 = 2 c ⇒ r1 = r2
⇒ R2 gives the most accurate value.
4. An electron in the ground state of hydrogen atom is revolving in anticlock-wise direction in a circular orbit of radius R.
30º nr
Br
(i) Obtain an expression for the orbital magnetic dipole moment of the electron.
(ii) The atom is placed in a uniform magnetic induction Brsuch that the plane-normal of the electron-orbit makes an angle of 30º with the magnetic induction. Find the torque experienced by the orbiting electron.
Sol. (i) Orbital magnetic dipole moment M = IA
where I is the current due to orbital motion of ⇒ M =
T e
× πR2
electron and A is the area of the loop made by electron. ⇒ M = π ω 2 e × πR2 ⇒ M = 2 1 eωR2 R ω e–
But according to Bohr's postulate mRω2 = π 2 nh ⇒ Rω2 = m 2 nh π ⇒ M = 2 e × m 2 nh π = 4 m nhe π
The direction of magnetic momentum is same as the direction of area vector, i.e. perpendicular to the plane of orbital motion.
→
τ
e–
30º nr Br
(ii) We know that torque →
τ = M × → →B
⇒ τ = MB sin θ
where θ is the angle between M and B ⇒ τ = m 4 nhe π × B sin 30º = 8 m B nhe π
The direction of torque can be found by right hand thumb rule.
The direction of torque is perpendicular to the plane containing nˆ and →B as shown.
5. A metal rod OA of mass 'm' and length 'r' is kept rotating with a constant angular speed ω in a vertical plane about a horizontal axis at the end O. The free end A is arranged to slide without friction along a fixed conducting circular ring in the same plane as that of rotation. A uniform and constant magnetic induction →B is applied perpendicular and into the plane of rotation as shown in the figure below. An inductor L and an external resistance R are connected through a switch S between the point O and a point C on the ring to form an electrical circuit. Neglect the resistance of the ring and the rod. Initially, the switch
is open. [IIT-1995] Y
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
A X C ω θ O L R S×
×
×
×
(a) What is the induced emf across the terminals of the switch ?
(b) The switch S is closed at time t = 0.
(i) Obtain an expression for the current as a function of time.
(ii) In the steady state, obtain the time dependence of the torque required to maintain the constant angular speed, given that the rod OA was along the positive X-axis at t = 0.
Sol. (a) Let us consider a small length of metal rod dx at a distance x from the origin. Small amount of emf (de) induced in this small length (due to metallic rod cutting magnetic lines of force) is
de = B(dx)v ...(i)
where v is the velocity of small length dx
v = xω ...(ii)
From (i) and (ii) de = B(dx)xω
∴ The total emf across the whole metallic rod OA is e =
∫
r ω 0Bx dx = Bω r 0 2 2 x = 2 Br2ω(b) i. The above diagram can be reconstructed as the adjacent figure e is a constant. O will accumulate positive charge and A negative when the switch S is
closed, transient current at any time t, when current I is flowing in the circuit,
S R L O A i I = I0(1 – et/τb) Here I0 = R e = R 2 r Bω 2 and TL = R L Therefore, I = R 2 r Bω 2 [1 – e–(R/L)t ] (ii) In steady state
In steady state I = R 2
r Bω 2
[Q t has a large value and L t
R e − → 0]
When current flows in the circuit in steady state, there is a power loss through the resistor. Also since the rod is rotating in a vertical plane, work needs to be done to keep it at constant angular speed.
Power loss due to current I will be P = I2R = 2 2 R 2 Br ω R ⇒ P = R 4 r B2 4ω2 t = t mg r/2 r/2 cosθθ
The torque required for this power ρ = τ1ω
⇒ τ1 =
R 4
r B2 4ω
Torque required to move the rod in circular motion against gravitation field
τ2 = mg ×
2 r
cos θ The total torque
τ = τ1 + τ2 (clock wise) τ = R 4 r B2 4ω + 2 mgr cos ωt
The required torque will be of same magnitude and in anticlockwise direction. The second term will change signs as the value of cos θ can be positive as well as negative.
CHEMISTRY
6. A sample of hard water contains 96 ppm of SO42– and
183 ppm of HCO3–, with Ca2+ as the only cation.
How many moles of CaO will be required to remove HCO3– from 1000 kg of this water ? If 1000 kg of this
water is treated with the amount of CaO calculated above, what will be concentration (in ppm) of residual Ca2+ ions? (Assume CaCO
3 to be completely
insoluble in water.) If the Ca2+ ions in one litre of the treated water are completely exchanged with hydrogen ions, what will be its pH ? (One ppm means one part of the substance in one million part of water, mass/mass. ) [IIT-1997] Sol. In 106g (= 1000 kg) of the given hard water, we will
have
Mass of SO42– ions = 96 g
Mass of HCO3– ions = 183 g
Thus Amount of SO42– ions = 1
mol g 96 g 96 − = 1 mol
Amount of HCO3– ions = 1
mol g 61 g 183 − = 3 mol
These ions are present as CaSO4 and Ca(HCO3)2.
Hence,
Amount of Ca+ ions = (1 + 1.5) mol = 2.5 mol.
The addition of CaO causes the following reaction : CaO + Ca(HCO3)2 → 2CaCO3 + H2O
To remove 1.5 mol of Ca(HCO3)2, 1.5 mol of CaO
will be required in the treated water. After this, the solution contains only CaSO4. Thus, 1 mol of Ca2+
ions will be present in 106 g of water. Hence its
concentration will be 40 ppm.
Molarity of Ca2+ ions in the treated water will be 10–3
mol L–1. If the Ca2+ ions are exchanged by H+ ions,
then
Molarity of H+ in the treated water = 2 × 10–3 M. Thus, pH = – log (2 × 10–3) = 2.7
7. The equilibrium constant Kp of the reaction
2SO2(g) + O2(g) 2SO3(g) is 900 atm–1 at 800
K. A mixture containing SO3 and O2 having initial
partial pressures of 1 atm and 2 atm, respectively, is heated at constant volume to equilibrate. Calculate the pressure of each gas at 800 K. [IIT- 1989] Sol. Since to start with SO2 is not present, it is expected
that some of SO3 will decompose to give SO2 and O2
at equilibrium. If 2x is the partial pressure of SO3 that
is decreased at equilibrium, we would have 2SO2(g) + O2(g) 2SO3(g) t = 0 0 2 atm 1 atm teq 2x 2 atm + x 1 atm – 2x Hence, Kp = ) p ( ) p ( ) p ( 2 2 3 O 2 SO 2 SO = ) x atm 2 ( ) x 2 ( ) x 2 atm 1 ( 2 2 + − = 900 atm–1
Assuming x << 2 atm, we get ) atm 2 ( ) x 2 ( ) x 2 atm 1 ( 2 2 − = 900 atm–1 or 2 2 ) x 2 ( ) x 2 atm 1 ( − = 1800 or x 2 atm 1 – 1 = 42.43 or x = 43 . 43 2 1 × atm = 0.0115 atm Hence, p(SO2) = 2x = 0.023 atm;
p(O2) = 2atm + x = 2.0115 atm and
p(SO3) = 1 atm – 2x = 0.977 atm
8. The sodium salt of a carboxylic acid, A, was produced by passing a gas, B, into an aqueous solution of caustic alkali at an elevated temperature and pressure. A, on heating in presence of sodium hydroxide followed by treatment with sulphuric acid gave a dibasic acid, C.A sample of 0.4 g of acid C, on combustion gave 0.08 g of water and 0.39g of carbon dioxide. The silver salt of the acid weighing 1.0 g on ignition yielded 0.71 g of silver as residue. Identify
A, B and C. [IIT- 1990]
Sol. From the given data, we can determine the molar mass of dicarboxylic acid. Since on ignition, 2 mol of Ag (≡ 215.8 g Ag) will be left per mole of dicarboxylic acid, we have
Ag g 0.71 acid ic dicarboxyl of silver of g 1.0 × 215.8 g Ag = 303.94 g of silver salt of dicarboxylic acid. Hence, Macid = 303.94 g mol–1 – 2MAg + 2MH
= 303.94 g mol–1 – 215.8 g mol–1 + 2 g mol–1
= 90.14 g mol–1
Now, since the molar mass of two carboxylic groups (2COOH) is 90 g mol–1, the dicarboxylic acid (C) is
oxalic acid (HCOOH – COOH). The nature of dicarboxylic acid may be confirmed from the combustion data.
COOH COOH + 2
1
O2 2CO2 + H2O
Molar mass 90 g mol–1 44 g mol–1 18g mol–1
Given data 0.4 g 0.39 g 0.08 g The calculated values of masses of CO2 and H2O
produced during the combustion of 0.4 g of oxalic acid are acid oxalic g 90 CO g 44 2× 2 × 0.4 oxalic acid = 0.39 g CO 2 acid oxalic g 90 O H g 8 1 2 × 0.4 g oxalic acid = 0.08 g H 2O
These values tally with the given data. Hence, the compound C is confirmed to be oxalic acid. The production of C from A is as follows.
A heat NaOH → Intermediate compound →H2SO4 COOH |COOH The compound A must be formic acid as the above reactions are used in the manufacture of oxalic acid. The reactions are
2HCOOH → 2HCOONa →−H2 (COONa)
2 → H+ ) C ( 2 ) COOH (
The sodium salt of A is produced by passing a gas B into an aqueous solution of caustic alkali at an elevated temperature and pressure. The reaction involved here is NaOH + CO e r pressu and rature High tempe→ HCOONa
Hence, the gas B is carbon monoxide. Thus, the structures of A, B and C are
A is HCOOH; B is CO, C is (COOH)2
9. An organic compound A, C8H4O3, in dry benzene in
the presence of anhydrous AlCl3 gives compound B.
The compound B on treatment with PCl5 followed by
reaction with H2/Pd(BaSO4) gives compound C,
which on reaction with hydrazine gives a cyclised compound D(C14H10N2). Identify A, B, C and D.
Explain the formation of D from C. [IIT-2000] Sol. The given reactions are as follows.
O O + O AlCl3 O O OH PCl5 H2/Pd (BaSO4) C6H5 H C C O O H2NNH2 C6H5 N N The formation of D from C may be explained as follows. C6H5 C6H5 O O NH2 NH2 O– NH2 NH2 O– + + C6H5 O – N – H N – H OH C6H5 N N
10. An aqueous solution of salt A gives a white crystalline precipitate B with NaCl solution. The filtrate gives a black precipitate C when H2S is
passed through it. Compound B dissolves in hot water and the solution gives yellow precipitate D on treatment with potassium iodide and cooling. The compound A does not give any gas with dilute HCl
but liberates a raddish brown gas on heating. Identify the compounds A to D giving the involved equation.
[IIT-1976] Sol. The given information's are as follows.
(a) Aqueousof Asolution →NaCl
B ofprecipitate White
Filter Residue Soluble Yellow precipitate (D)
hot water KI
Filtrate H2S Black precipitate (C)
(b) No gas ← HCl Salt →
heat Brown coloured gas
From the information given in part (a), it may be concluded that the compound A is a lead salt. B is lead chloride as it is soluble in hot water. Yellow precipitate D is due to PbI2. Black precipitate C is
due to PbS. Lead chloride being spraingly soluble in water, a little of it remains in the filtrate which is subsequently precipitated as PbS.
From the information given in part (b), it may be concluded that anion associated with lead (II) is nitrate because lead nitrate dissociates on heating as 2Pb(NO3)2 → 2PbO + 4NO2 + O2
Brown coloured gas Hence, A is Pb(NO3)2, B is PbCl2, C is PbS and D is
PbI2.
MATHEMATICS
11. Let n be a positive integer and (1 + x + x2)n = a0 + a1x + .... + a2nx2n
Show that a02 – a12 + .... + a2n2 = an [IIT-1994]
Sol. (1 + x + x2)n = a 0 + a1x + ... + a2nx2n ...(1) Replacing x by – x 1 , we obtain n 2 x 1 x 1 1 − + = a 0 – x a1 + 22 x a – 33 x a + ... + 22nn x a ...(2) Now, a02 – a12 – a32 + ... + a2n2 = coefficient of the
term independent of x in [a0 + a1x + a2x2 + ... + a2nx2n] 0− 1+ 22− + 22nn x a ... x a x a a
= coefficient of the term independent of x in (1 + x + x2)n n 2 x 1 x 1 1 − + But R.H.S. = (1 + x + x2)n n 2 x 1 x 1 1 − + = 2n n 2 n 2 x ) 1 x x ( ) x x 1 ( + + − + = 2n n 2 2 2 x ] x ) 1 x [( + − = 2n n 2 4 2 x ) x x x 2 1 ( + + − = 2n n 4 2 x ) x x 1 ( + + Thus, a02 – a12 + a22 + ....a2n2
= coefficient of the term independent of x in
n 2 x 1 (1 + x2 + x4)n = coefficient of x2n in (1 + x2 + x4)n = coefficient of tn in (1 + t + t2)n = a n
12. Find the range of values of t for which 2 sin t = 1 x 2 x 3 x 5 x 2 1 2 2 − − + − , t ∈−π π 2 , 2 [IIT-2005]
Sol. Here, 2 sin t =
1 x 2 x 3 x 5 x 2 1 2 2 − − + − , t ∈ −π π 2 , 2 Put, 2sin t = y ⇒ – 2 ≤ y ≤ 2 ∴ (3y – 5)x2 – 2x(y – 1) – (y + 1) = 0 Since x ∈ R – {1, –1/3} {as, 3x2 – 2x – 1 ≠ 0} ∴ D ≥ 0 ⇒ 4(y – 1)2 + 4(3y – 5) (y + 1) ≥ 0 ⇒ y2 – y – 1 ≥ 0 ⇒ 2 2 1 y − – 4 5 ≥ 0 ⇒ − − 2 5 2 1 y + − 2 5 2 1 y ≥ 0 ⇒ y ≤ 2 5 1− or y ≥ 2 5 1+ or 2 sin t ≤ 2 5 1− or sin t ≥ 2 5 1+ ⇒ sin t ≤ sin π− 10 or sin t ≥ sin π 10 3 ⇒ t ≤ – 10 π or t ≥ 10 3π Thus, Range for t ∈
−π π 2 , 2 ∪ π π 2 , 10 3 13. Evaluate
∫
+ + 2 x) xe 1 ( x ) 1 x ( dx [IIT-1996] Sol. I =∫
+ + 2 x) xe 1 ( x ) 1 x ( dx =∫
+ + 2 x x x ) xe 1 ( xe ) 1 x ( e dx Put 1 + xex = t ⇒ (ex + xex)dx = dt Therefore I =∫
−1)t2 t ( dt Let 2 t ) 1 t ( 1 − ≡ (t 1) A − + t B + 2 t C ⇒ 1 ≡ At2 + Bt(t – 1) + C(t – 1) Put t = 0 1 = – C Put t = 1 1 = A0 = A + B ⇒ 0 = 1 + B Therefore, I =
∫
− − − t2 1 t 1 1 t 1 dt = ln|t – 1| – ln|t| + t 1 + c = ln t 1 t− + t 1 + c = ln x x xe 1 xe + + 1 xex 1 + + c 14. Evaluate∫
π π − +π − + π 3 / 3 / 3 3 | x | cos 2 x 4 dx [IIT-2004] Sol. Let, I =∫
π π − +π − π 3 / 3 / 3 | x | cos 2 dx + 4∫
π π − +π − 3 / 3 / 3 3 | x | cos 2 dx x Using∫
− a af(x)dx = = − − = −∫
f(x)dx, f( x) f(x) 2 ) x ( f ) x ( f , 0 a 0 ∴ I = 2∫
π +π − π 3 / 0 3 | x | cos 2 dx + 0 +π −∫
−ππ 3 / 3 / 3 odd is 3 | x | cos 2 dx x as I = 2π∫
π π + − 3 / 0 2 cos(x /3) dx = 2π∫
π π − 3 / 2 3 / 2 cost dt , where x + 3 π= t = 2π∫
π π + 3 / 2 3 / 2 2 2 t tan 3 1 dt 2 t sec = 2π∫
+ 3 3 / 1 1 3u2 du 2 = 3 4π .{
3tan−1 3u}
1/33 = 3 4π (tan–1 3 – tan–11) = 3 4π tan–1 2 1 ∴∫
π π − +π − + π 3 / 3 / 3 3 | x | cos 2 x 4 dx = 3 4π tan–1 2 1 .15. Sketch the curves and identify the region bounded by x = 1/2, x = 2, y = ln x and y = 2x. Find the area of
this region. [IIT-1991]
Sol. The required area is the shaded portion in following figure. y = 2x y = logex 2 1 1/2 O y In the region 2 1
≤ x ≤ 2 the curve y = 2x lies above as
compared to y = logex
Hence, the required area =
∫
2 − 2 / 1 x logx)dx 2 ( = 2 2 / 1 x ) x x log x ( 2 log 2 − − = 2 log 2 4− – 2 5 log 2 + 2 3GLOBAL WARMING IS REAL
The arctic ice is receding and global warming is no longer a theory but a reality. Scientists predict that by the year 2100, the average surface temperature will jump up by 6 degrees Fahrenheit. Nighttime temperatures will be higher and there will be hotter days.
Since air temperature is a powerful component of climate, there will be unavoidable climate changes in the future. Some climate changes involve extreme weather disturbances such as more severe hurricanes and longer droughts. There will be an increased precipitation of snow and rain during winter. The faster melting of snow during the spring will result in flooding. All these climate changes are predicted based on the assumption that changes will be relatively gradual.
Passage # 1
Emission of electrons from the surface of the metal, when the metallic surface is illuminated by light of proper frequency / wavelength / energy is known as photoelectric effect.
The current flow due to photo electrons is known as photo current and it is dependent on the intensity of incident light photons.
The proper polarity applied potential difference which can stop the motion of most energetic photo electron and ultimately the photo current is known as stopping potential.
– Variation of electric field for incident light E = E1 sin ω1t + E2 sin ω2t + E3 sin ω3t
– Energy related with freq. ω1, ω2, ω3 are 2eV, 2.8eV,
3eV and intensity are Int1, Int2 and Int3 then
M E T A L W = 2.5 eV Photo Electrons Incident Light
1. Write the value of stopping potential for the given metallic surface.
2. If the intensity of light of frequency ω1 decreases and
then ceases then what will be the effect on stopping potential and photo current.
3. If the intensity of light of frequency ω2 decreases and
then ceases then what will be the effect on stopping potential and photo current.
4. If the intensity of light of frequency ω3 decreases and
then ceases then what will be the effect on stopping potential and photo current.
Passage # 2
If the Angular momentum of the revolving electron is quantized then it is exempted from the law of electro dynamics and it will not emit electromagnetic radiations. Its energy remains constant and it is the ultimate cause of stability of the atom.
If hydrogen atom electron get excited, it will return to the ground state and emits the energy spectra which is known as hydrogen spectra which consists of various series like Lymen, Balmer ...
Hydrogen atom HA-1 and Hydrogen atom HA-2 are in ground state and in first excited state. Accepts energies 12.1 eV and 1.9 eV respectively.
H YDR OGE N SPEC T . TU B E ABS O RBT ION L IQUI D COL U M N Emission spectra M E T A L W = 2eV Photo electrons
5. Find the number of possible lines in emission spectra of Hydrogen spectrum tube
6. Name the spectral lines in Hydrogen spectra.
7. If the "Absorbtion Column" Absorbs 2nd line of
Lymen series then what will be the stopping potential.
8. If "Absorbtion Column" get removed then how photo current & stopping potential get changed.
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
Soluti ons wil l be published in next issue
1. As req(ab) = 0
req (ab)
a b
10V 1Ω
so terminal voltage of the battery = 0
2. current through the battery i = 10/1 = 10 amp H = i2Rt = (10)2 (1) (1) = 100 Watt 3. Key –k open Req(ab) = 7.5 Ω 10Ω 5Ω 5Ω a 5Ω 5Ω b 50 V 0.5 Ω i = ) b , a ( eq R r+ ε = 5 . 7 5 . 0 50 + = 8 50 = 4 25 Amp V = ε – i r = 50 – 4 25 × 0.5 = 50 – 8 25 = 8 175 ⇒ so ε : V = 50 : 8 175 = 400 : 175 ⇒ 16 : 9 4. ) b , a ( eq R = ) 5 ( 2 5 10 ) 5 )( 10 ( 2 ) 5 ( 5 ) 5 ( 10 + + + + = 25 175 = 9 i = ) b , a ( eq R r+ ε = 9 5 . 0 50 + = 9.5 50 = 95 500 = 19 100 ⇒ V = ε – i r = 50 – 19 100 × 0.5 = 50 – 38 100 ⇒ 50 – 19 50 = 50 19 18 ε : V = 50 : 50 19 18 = 19 : 18 5. y N x mg cos θ mg sin θ θ ma = – mg sin θ a = – g sin θ or a = – g tan θ ...(1) (as θ is small) Now, x2 = 4ay; ∴ dx dy = a 2 x ∴ a = – g a 2 x –ω2x = – a 2 gx ⇒ ω = a 2 g Option [D] is correct.
6. As surface DC is smooth it will slip but as in path AD it is pure rolling therefore work done by friction is zero. Hence mechanical energy is conserved. Option [C] is correct. 7. C = x A 0 ε
, where x is separation between plates dT dC C I = dT dA A 1 – dT dx x 1 For dT dC = 0, x 1 dT dx = dT dA A 1 ⇒ αS = 2α Option [C] is correct.
8. At a distance h above the sheet E = Esheet + Eslab = 0 2 – ε σ + 0 2 D ε ρ = 0 2 D ε σ − ρ
At a distance h below the top surface of slab Eslab = 0 2 ) h 2 D ( ε − ρ E = Esheet + Eslab = 0 2ε σ + 0 2 ) h 2 D ( ε − ρ = 0 2 ) h 2 D ( ε − ρ + σ
At a distance h below the bottom surface of the slab
= 0 2 – ε σ + 0 2 D ε ρ = 0 2 D ε σ − ρ Option [A,B,D] is correct
Solution
Physics Challenging Problems
Set # 4
1. A ball with mass M/2 filled with gas (whose mass is M/2) is standing on a frictionless table. A bullet of mass m = M/4 and velocity v0xˆ penetrates the ball,
and is rests inside at t = 0 (see figure). Assume that the amount of gas emitted during the collision can be neglected. The compressed gas is emitted at a constant velocity v0/2 relative to the ball and at an
even rate gas dt dM = K(K is a positive constant). m,v0 R
1. What is the velocity of the ball after the collision
with the bullet ?
2. Find the velocity of the ball v(t) as a function of time. As same that the emission of gas starts at t = 0. What is the final velocity of the ball ?
Note : The penetration of the bullet into the ball is the horizontal plane.
Sol. 1. From the law of conservation of linear
momentum, we can write : (m + M)v1 = mv0
we define v1 as the velocity of M + m (the ball, bullet
and all the gas). Hence, vr = 1 M m m + v0xˆ = 5 xˆ v0
2. The force that the emitted gas applied on the ball
is : Fgas = dt ) v m ( d r = dt ) v m ( d r vr = k 2 v0 (right) where dt v dr
= 0. Notes that vr is the relative velocity between the gas and the ball. The mass of the ball and the bullet is time dependent. It equals m + M – kt. Therefore, the equation of motion is :
(m + M – kt) dt dv = –k 2 v0
Rearranging and integrating this expression we obtain:
vr (t) – vr = – 1 2 v0
ln(m + M – kt) xˆ|t0
Substituting the boundaries and v1 from Eq- we
derive : vr (t) = v0 − + + + kt M m M m ln 2 1 5 1 xˆ
The final velocity of the ball is determined according to the time that passes until the force applied by the emitted gas stops, or when the mass of the gas in the ball is zero, 0 = 2 M – kT. Hence, T = k 2 M
Substituting Eq. (7)in Eq. (6) yields vfinal = v(T) = v0 + + + 2 / M m M m ln 2 1 5 1 = 0.455 v0
2. Two masses, m1 and m2, are tied to the ends of a
spring whose force constant is k, and whose natural length is a. This system placed horizontally on a
perfectly smooth table, as shown in fig. At t = 0, m1
is bumped and receives a linear momentum of
1
pr = p0xˆ , where p0 is a constant.
m2 k m1
a
(i) Write the equations of motion for m1 and m2.
What is the velocity of the center of mass?
(ii) Prove that the harmonic oscillation equation of
the system is : µ(x&& – 2 x&& ) = – k(x1 2 – x1)
where µ = 2 1 2 1 m m m m +
(iii) What is the oscillation amplitude of (x2 – x1) ?
Sol. The motion of the system of the two bodies can be
conveniently described by using the center of mass frame of reference. The centre of mass moves in a straight line with constant velocity, due to the conservation of linear momentum. In the centre of mass frame the two bodies perform simple harmonic oscillations. Denoting the position of the masses m1
and m2 by x1 and x2, respectively, we can express the
distance between the masses as x2 – x1. The change in
the length of the spring is then x = x2 – x1 – a.
(i) The forces applied by the spring on the two systems are :
Expert’s Solution for Question asked by IIT-JEE Aspirants
Students' Forum
m1x&& 1= kx ...(1)
m2x&& = 2 – kx ...(2)
The signs are used according to the position of the mass relative to the spring. Multiplying Eq.(1) by m2,
Eq. (2) by m1, and subtracting Eq. (1) from Eq. (2)
we have
m1m2(x&& – 2 x&& ) = – k(m1 1 + m2)x ...(3)
Since x = x2 – x1 – a, we have x& =x& – 2 x& . And 1
x&& = x&& – 2 x&& . Therefore, 1
m1m2x&& = –k(m1 + m2)x ...(4)
and the solution to this equation is :
x(t) = A cos(ω0t + φ) ...(5) where ω0 = k m m m m 2 1 2
1+ . Notice that x does not
denote the position of any of the masses. It denotes the difference between the distance between the masses and the initial state, so that x = (x2 – x1) – a.
The velocity of the centre of mass is given by :
cm vr = 2 1 0 m m xˆ p + ...(6)
(ii) The equation specified in the problem is easily
derived from Eq. (3), which we found in the first section. The constant µ is called the "reduced mass" of the system, and is defined as
µ 1 ≡ 1 m 1 + 2 m 1 ...(7) or in a different form, µ = 2 1 2 1 m m m m + ...(8)
(iii) Taking energy into consideration, we have
Ek = 2 1 m1 2 1 0 m p = 2 1 1 2 0 m p ...(9)
where Ek is the initial kinetic energy. The kinetic
energy of the center of mass is : Ek(cm) = 2 1 2 0 m m p 2 1 + ...(10) and therefore, the total kinetic energy in the centre of mass frame becomes :
E´k(cm) = Ek – Ek(cm) = 20 2 1 1 2 p ) m m ( m 2 m +
The kinetic energy is proportional to the square of ampitude, E´k = 2 1 kA2 Therefore, A = ) m m ( km p m 2 1 1 2 0 2 +
3. A thin rod of length L and mass m moves on a smooth horizontal surface at a velocity vr in the 0
direction of its length. The end of the rod hits the end of a second rod of the same length and mass, which is
initially at rest, perpendicular to the first rod (see figure.) At the moment they touch, the two poles stick together, while the angle between them remain π/2. V0 L X L A
(i) What is the center of mass velocity of the system ? Where is the center of mass ? Express Rcm
r
at the moment of collision.
(ii) What is the angular velocity of the system about the centre of mass ?
(iii) What is the maximal velocity of the pole's end, denoted by A, after the collision ? When does the point A reach this velocity for the first time ?
Sol. (i) We know that vr = v0yˆ . According to the
definition : cm vr = i i i i i m v m Σ Σ r = v0yˆ 2 1 ...(1) By definition : cm Rr = i i 1 i i m r m Σ Σ r = m 2 yˆ L 2 1 m xˆ L 2 1 m − + = 4 1 L( xˆ – yˆ ...(2) where we choose the origin to be the point of collision (see figure)
1 1/4 ω θ 1/4 1 ) yˆ xˆ )( 4 / x ( ) 0 t ( Rcm = = − r
(ii) In the center of mass frame, the angular momentum about the centre of mass before the collision is : cm J r = (rr – 1 Rrcm) × m1(vr – 1 vrcm) + rr – 1 Rcm r ) × m2(vr – 2 vr ) cm = – 4 L mv0zˆ ...(3) where we substitute rr = –1 2 1 L yˆ and rr = L xˆ . We 2
IC = 2 mL2 3 1 – 2m + 2 2 4 L 4 L = 12 5 mL2 ...(4) Hence, ωr = cm C cm I J r = – zˆ mL 12 5 L mv 4 1 2 0 = – zˆ L v 5 3 0 ...(5)
(iii) The maximal velocity is reached when the velocity of the point A in the centre of mass frame is in the direction of vr ; i.e., + yˆ (see figure). cm Therefore, 1 1 c.m. RAC A VA vA,max = vcm + ωRAC ...(6)
By using simple geometry, we obtain : RAC = 2 2 4 L 4 L 3 + = 4 10 L ...(7) Notice that the distance RAC is constant in the center
of mass frame. The whole system rotates at an angular velocity of ωr about a fixed axis which cm passes through the center of mass. Therefore,
vA,max = 2 1 v0 + 4 10 L v 5 3 0 L = 2 1 v0 + 10 3 1
The time required for RA,cm to rotate to an angle θ
(see figure) is given by :
t = | |ω θ r = L v 5 3 ) 3 ( tan 0 1 − = 0 v L 3 5 tan–1(3)
4. Two long parallel conducting rails are fixed at a distance l apart in a horizontal plane. A rod of mass m and resistance R can slide along the rails. A uniform magnetic field of induction B exists along vertically downward direction and coefficient of friction between rails and rod is µ.
If a battery of emf E and negligible internal resistance is connected between the rails at left end, as shown in fig. calculate minimum value , Fmijn of force required
to be applied horizontally rightwards on the rod to move it to the right.
F
×
B E + – l If F = 2Fmin, calculate(i) steady state velocity of the rod,
(ii) power required to keep the rod moving with that steady velocity and
(iii) thermal power generated in the rod.
Sol. Initially, due to battery, a current I = E/R flows through the rod. According to Fleming's left rule, the rod experiences a force Bil towards left, due to the current.
Considering free body diagram of the rod as shown in fig. Fmin mg F´ N µN
For vertical forces,
N = mg
For horizontal forces, Fmin = Bil + µN or Fmin = +µmg R BEl Ans.
Where a force F = 2 Fmin is applied, the rod begins to
accelerate rightwards and due to its motion, an emf is induced in it. According to Fleming's right hand rule this induced emf tries to flow an anti-clockwise current in the circuit. It means induced emf reinforces emf of the battery. Hence, leftward force on rod increases or retarding force increases. Therefore, acceleration of the rod decreases and it becomes zero or velocity becomes constant when resultant force on the rod become zero.
Let the steady state velocity of the rod be v0.
Induced emf, e = Blv0.
∴ Total emf of the ciruit = E + e = (E + Blv0)
∴ Current through the circuit, i = R ) v B E ( + l 0 Due to current, leftward force on the rod,
F´ = Bil = R ) v B E ( Bl + l 0
Considering free body diagram of the rod shown in fig. F = Fmin mg Bil N µN
For horizontal forces, F´ + µN = 2Fmin ∴ R Bl (E + Blv0) + µmg = 2 +µmg R BEl or v0 = + 2 2 B BE µmgR l l Ans. (i) Power required to keep the rod moving with this steady velocity v0 is P = Fv0 = 2 2 2 RB ) BE µmgR ( 2 l l + Ans. (ii) Thermal power generated in the rod = i2R
= 2 2 2 RB ) BE 2 µmgR ( l l + Ans. (iii) 5. Due to a point isotropic sonic source, loudness at a
point is L = 40 dB. If density of air is ρ = 15/11 kg m–3 and velocity of sound in air, v = 330 ms–1,
calculate
(i) pressure oscillation amplitude at the point of observation and
(ii) ratio of oscillation amplitude of particles of medium to wavelength of sonic waves.
Sol. Due to propagation of longitudinal waves, pressure of the medium varies with time. Maximum change in pressure is given by (∆P)max = v Paω γ ...(1) Speed of sound waves in a gas is v =
ρ γP
∴ γP = ρv2 ...(2)
Loudness at a point is given by L = 10 log10
10
I I
dB where I0 = 10–12 Wm–2
∴ Intensity of sound waves at the point is I = I0. antilog10 10 L or I = 1 × 10–8 Wm–2 But intensity I = 2π2n2a2ρv = 2 1 (4π2n2a2)ρv ∴ 4π2n2a2 = v I 2 ρ
But 4π2n2a2 = (aω)2, hence, aω =
v I 2
ρ ...(3) Substituting γρ = ρv2 and aω =
v I 2
ρ in equation (1), (∆P)max = 2Iρ = 3 × 10v –3 Nm–2 Ans. (i)
Substituting ω = 2πn in equation (3), 2πna = v I 2 ρ But n = λ v
∴ Ratio of oscillation amplitude to wavelength is λ a = v I 2 v 2 1 ρ π = π − 99 10 6 Ans. (ii)
Regents Physics
You Should Know
Modern Physics :
• The particle behavior of light is proven by the photoelectric effect.
• A photon is a particle of light {wave packet}. • Large objects have very short wavelengths
when moving and thus can not be observed behaving as a wave. (DeBroglie Waves)
• All electromagnetic waves originate from accelerating charged particles.
• The frequency of a light wave determines its energy (E = hf).
• The lowest energy state of a atom is called the ground state.
• Increasing light frequency increases the kinetic energy of the emitted photo-electrons.
• As the threshold frequency increase for a photo-cell (photo emissive material) the work function also increases.
• Increasing light intensity increases the number of emitted photo-electrons but not their KE.
Mechanics :
• Centripetal force and centripetal acceleration vectors are toward the center of the circle- while the velocity vector is tangent to the circle.
• An unbalanced force (object not in equilibrium) must produce acceleration.
• The slope of the distance-tine graph is velocity. • The equilibrant force is equal in magnitude but
opposite in direction to the resultant vector. • Momentum is conserved in all collision
systems.
• Magnitude is a term use to state how large a vector quantity is.
Review of Concepts :
Electric current is the rate of transfer of charge through a certain surface.
The direction of electric current is as that of flow of positive charge.
If a charge ∆q cross an area in time ∆t, then the average current = ∆q/∆t
Its unit is C/s or ampere.
Electric current has direction as well as magnitude but it is a scalar quantity.
Electric current obeys simple law of algebra. i.e., I = I1 + I2 α I1 I2 I1 Types of Current :
Steady state current or constant current : This type of current is not function of time.
Transient or variable current : This type of current passing through a surface depends upon time. i.e., I = f(t) or I = t q lim 0 t ∆ ∆ → ∆ ⇒ dt dq
Electric charge passing a surface in time t = q =
∫
t 0Idt Average current I =∫
∫
t 0 t 0 dt dt IConvection Current : The electric due to
mechanical transfer of charged particle is called convection current. Convection current in different situation.
Case I : If a point charge is rotating with constant angular velocity ω. I = T q ; T = ω π 2 ⇒ I = π ω 2 q
Case II : If a non-conducting ring having λ charge per unit length is rotating with constant angular velocity ω about an axis passing through centre of ring and perpendicular to the plane of ring.
I = R λω θ Jˆ ∆S or J = θ ∆ ∆ cos S I Its unit A/m2
Electric current can be defined as flux of current density vector.
i.e., i =
∫
→j.dS→Relation between drift velocity and current density v = – d
en j
→
Here, negative sign indicates that drifting of electron takes place in the opposite direction of current density.
The average thermal velocity of electron is zero. Electric resistance : Electric resistance (R) is
defined as the opposition to the flow of electric charge through the material.
It is a microscopic quantity. Its symbol is
Its unit is ohm. (a) (b) R = A l ρ where, R = resistance, ρ = resistivity of the material, l = length of the conductor, A = area of cross section Continuity Equation :
∫
c→j.dS→ = – dtdq
The continuity equation is based on conservation principle of charge.
Drift Velocity (vd) : When a potential difference
is applied between ends of metallic conductor, an
