Dear Students,
All of us want to get organized. The first thing in getting organized is to find our obstacles and conquer them. Make a beginning by conquering obstacles for starting in right earnest.
Sometimes the quest for perfectionism holds us back. We occasionally feel that we should start when we have enough time to do a job thoroughly. One way to tackle this kind of mindset is to choose smaller projects or parts of projects that can be completed within 15 minutes to one hour or less. It is important to keep yourself motivated. Approach your projects as something which are going to give you pleasure and fun. Reward yourself for all that you accomplish no matter how small they may be. Never hesitate to ask for help from a friend. Make all efforts to keep your motivational level high. You might feel overwhelmed because you are focusing on every trivial thing that needs to be got done.
How do you act or react to your life? When you are merely reacting to events in your life, you are putting yourself in a weak position. You are only waiting for things to happen in order to take the next step in your life. On the other hand when you are enthusiastic about your happiness you facilitate great things to happen. It is always better to act from a position of power. Never be a passive victim of life. Be someone who steers his life in exactly the direction he wants it to go. It is all upto you now.
If you do what you have always done and in the way you have done it you shall get only such results which you have always got. Getting organized requires that not doing things that cause clutter, waste of time and hurt your chances adversely of realizing your goals. You should concentrate only on doing things that eliminate clutter, waste of times and hurt your chances adversely of realizing your goals. You should concentrate only on doing things that eliminate clutter, increase your productivity and provide the best chance for achieving your goals. The first step should be to stop leaving papers and other things on tables, desks, counter tops and in all kind of odd place. The more things you leave around in places other than rightful places the quicker the clutter will accumulate. Keep things in their assigned places after you have finished using them. It does not take along to put something away. If you leave things lying around they will build into a mountain of clutter. It could take hours if not weeks or months to trace them and declutter the atmosphere.
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.
• 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
If you can't make a mistake,
you can't make anything.
Volume - 5 Issue - 6 December, 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
Cover Design & Layout
Mohammed Rafiq 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-6 December, 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 Month • The greatest adventure is what lies ahead. • Fixers believe they can fix. Complainers
believe they can complain. They are both right.
• The tire model for motivation: People work best at the right pressure.
• Trust the force, Luke.
• Use your feelings or your feelings will use you.
• People who expect to fail are usually right. • The path to success is paved with
mistakes.
• You've got to cross that lonesome valley. You've got to cross it by yourself.
• Appreciate what your brain does. In case nobody else does.
• Learn to mock the woe-mongers.
• Be confident. Even if you are not, pretend to be. No one can tell the difference.
CONTENTS
INDEX PAGE
NEWS ARTICLE
4IIT-Jodhpur to begin functioning from February - March 2010
IIT-K students develop autonomous vehicle
IITian ON THE PATH OF SUCCESS
8Dr. Rajeewa Arya
KNOW IIT-JEE
10Previous IIT-JEE Question
XTRAEDGE TEST SERIES
58Class XII – IIT-JEE 2010 Paper Class XII – IIT-JEE 2011 Paper
Mock Test CBSE Pattern Paper -1 [Class # XII]
Regulars ...
DYNAMIC PHYSICS
188-Challenging Problems [Set# 8] Students’ Forum
Physics Fundamentals
Reflection at plane & curved surfaces Fluid Mechanics
CATALYST CHEMISTRY
35 Key Concept Carboxylic Acid Chemical KineticsUnderstanding : Inorganic Chemistry
DICEY MATHS
46Mathematical Challenges Students’ Forum Key Concept
Monotonicity, Maxima & Minima Function
Study Time...
IIT-Jodhpur to begin functioning from February - March 2010
Indian Institute of Technology in Rajasthan is all set to begin functioning from the MBM Engineering College, here from next academic session commencing in February-March 2010.
The decision to this effect was made after the visit of a central team headed by additional secretary, ministry of Human Resource Development, Ashok Thakur during recent visit to different sites, which included the sites proposed for the construction of the IIT-Jodhpur and the existing engineering college of the city.
So, if everything goes well, the next session of IIT-J, which is presently being run from the IIT-Kanpur campus, will start functioning from the campus of this college here.
Thakur, who himself approved the MBM college during a visit on Saturday, said, "The final decision is to be taken by the ministry, to whom we will submit the report in a 2-3 days."
Agarwal, who was quite ecstatic following the visit of the team, expressed hope that the next session of the IIT will start here from February–March 2010.
He added that the new building of the IIT will take atleast 2-3 years, but owing to repeated reminders by IIT-Kanpur citing its inability to continue running the session of IIT-J from its own campus, there is a growing pressure to shift it to Jodhpur.
IIT-K students develop autonomous vehicle
True to its reputation of being ahead in technological developments, the Indian Institute of Technology, Kanpur, in collaboration with the Boeing Company is all set to unveil a new autonomous vehicle `Abhyast' as part of its Golden Jubilee celebrations.
Prof Shantanu Bhattacharya, faculty in mechanical engineering department of IIT-K and co-ordinator of the project, while talking to TOI said, "The vehicle will set new landmarks in terms of applied robotics research. Besides being unique, it will probably be one of its kind in India. Such vehicles play important role in defence applications and disaster management plans." He added that quite a few modules of such vehicles have already been developed by the United States and some other countries. But, India has so far not produced vehicles of this nature.
Further, Prof Bhattacharya informed, `Abhyast' will serve as a
first running prototype for such vehicles in the country. "It has a very small footprint -- 30x30x15 cm -- and thus it can be easily carried by soldiers and relief workers for disaster management operations. This technology was widely used by the US in investigating the World Trade Centre attack as well as in the war zones of Afghanistan and Iraq," he added.
Earlier this year, as a part of its University relations programme, Boeing decided to actively collaborate with IIT-K to foster research and innovation among undergraduate students. As part of this endeavour, eight students of IIT-K -- Abhilash Jindal, Ankur Jain, Faiz Ahmed, Gaurav Dhama, Palash Soni, Shishir Pandya and Sriram Ganesan -- were selected by Prof Bhattacharya to work on the project. "My role was mainly confined to that of a mentor. `Abhyast' in fact the result of students' labour and skills and should serve as an inspiration for other students," said a beaming Prof Bhattacharya.
`Abhyast' is a fully autonomous vehicle capable of navigating in unstructured and unknown environments. The user needs to specify only the end coordinates where he wants the vehicle to reach, and the task of reaching there would be taken by the vehicle itself, requiring no intervention by the user. The
vehicle is equipped with high end sensors like GPS, IMU and SONARS to navigate and avoid obstacles in its vicinity. The vehicle has a tank like chassis design that allows it to move even on uneven and slippery terrain, thus making it a robust vehicle for warlike and other disastrous situations.
IIT forecast system for Oman, Maldives
Ocean State Forecasting System (OSFS), a technology developed by Indian Institute of Technology, Kharagpur, will now be adopted by Oman and Maldives. The World Meteorological Organisation (WMO) feels that OSFS should be immediately adopted by other nations that have sea coasts. The system, that will continuously measure height, direction and period of waves, will help shipping and navigation.
IIT-Kgp was given the prestigious project jointly by the union ministry of ocean development and the department of science and technology.
It was headed by the institute's former director, S K Dubey. The model was developed by the department of ocean engineering and naval architecture. It was completed about a year ago and handed over to the India Meteorological Department by the institute for adoption. Dubey is presently in IIT Delhi to coordinate the project.
"WMO was so impressed that it immediately referred the technology to all the coastal nations. But it cannot be transferred without proper training to end users, so we will
be training meteorologists from other countries in batches. This will start with Oman and Maldives. Their scientists will be on the campus for an intensive training," Dubey said.
IITs come up with their RTI 'shield'
Stung by the exposure of admission anomalies in recent years, the IIT system has come up with an innovative method of blocking transparency even as it agreed to give data under RTI on the marks obtained by the four lakh candidates in this year’s joint entrance examination (JEE). It insisted on giving the data only in the hard copy running into hundreds of thousands of pages rather than in the more convenient form of a CD.
The information seeker, Rajeev Kumar, a computer science professor in IIT Kharagpur, is crying foul. For, the hard copy would not only result in a steep increase in the cost of information (running into six figures) but also make it almost impossible for him to detect irregularities in the latest JEE as he did in the three previous ones by analyzing the electronic data that had then by given to him under RTI.
As a result of this change in the strategy of the IIT system, central information commissioner Shailesh Gandhi fixed a hearing for November 6 specially to resolve this soft vs hard debate. The hearing follows the unusual reasons given by Gautam Barua, director of IIT Guwahati and overall in-charge of JEE 2009, for his failure to comply with the CIC’s disclosure direction passed on July 30.
In his first mail to CIC on October 2, Barua said that as there were a number of RTI applications seeking the CD, “we are apprehensive that this request for electronic data is to profit from it by using it for IIT JEE coaching purposes (planning, targeting particular cities, population segments, etc).”
The reference to the coaching institutes is reminiscent of the recent controversy over the move to raise the bar on 12th class marks to be eligible for IIT selection.
Asserting that IITs had “nothing to hide regarding the results”, Barua said, “We are ready to show the running of the software with the original data to the CIC, if it so desires.”
As a corollary, Barua made an issue of the fact that Kumar “has not asked to see the data, but he wants an electronic version delivered to him. Why is this so?” Kumar responded to that by pointing out that the irregularities he had uncovered in the JEE of the previous three years was on the basis of “compute intensive scientific calculations and analysis, which could not have been done just by looking at the data.”
Barua’s explanation in his subsequent mail on October 3 is: “By seeing, I meant that the appellant could come to IIT Guwahati and view the data, see the software being run, etc.” He added that if this option was unacceptable to CIC, “we will wish to provide the data in hard copy form, the costs of printing having to be borne by the appellant.”
If Kumar is pressing that the data be given to him “in the form in which it is originally available”, it is because the access to the electronic data of the previous three years helped him unearth, for instance, the shocking fact that general category candidates got into IITs after scoring in JEE as little as little as 5% in Physics and 6% in Mathematics.
Kerala seeks second IIT report on record
Kerala has filed an application in the apex court urging it to take on record the second part of the report titled Seismic Stability of Mullaiperiyar Composite Dam, submitted by D K Paul, Professor of earthquake engineering department, Indian Institute of Technology, Roorkee.
The first part of the report Structural Stability of Mullaiperiyar Dam Considering Seismic Effects — Part I — Seismic Hazard Assessment was submitted by IIT-Roorkee in May 2008. The said report is already before the apex court, it averred.
Kerala would like to file the second part of the report in support of its argument that Mullaiperiyar dam is not safe for storage, the application stated. The report stated that the earthquake safety of old concrete or masonry gravity dams under moderate to strong ground motions is of great concern. Although there is no evidence of catastrophic failure of gravity dams, yet the possibility of tensile cracking is never ruled out. The finite element analysis of dam
loading shows tensile stresses at the heel of dam-foundation interface, the report indicated. “The Mullaiperiyar dam is a composite gravity dam built during 1887-1895. The front and rear faces of the dam were built with un-coursed rubble masonry in lime surkhi mortar. The hearting is constructed of lime surkhi concrete. It lies in seismic zone III as per seismic zoning map of India. The 176 feet-high composite gravity dam is now over 114 years old,” the report went on to say. IITs want to be accredited by statutory body
As the government wants to make accreditation mandatory for all institutions, the IITs have said they would like to be accredited by a statutory body and not by the National Board of Accreditation (NBA).
The IIT directors have told the government that they have no objection to accreditation of the institutes, but insisted that the accreditation agency should be a statutory and autonomous organisation.
They expressed these views at the meeting of the IIT Council, the highest decision making body for the IITs, held here last month. HRD Minister Kapil Sibal, who is the chairman of the council, told them that the government would set up an accreditation agency by introducing a bill in the Parliament soon.
"The directors said the accreditation should be conducted by a statutory body
IITs told to reveal candidate details
The Central Information Commission has ordered the IITs to disclose most details of candidates who sat the 2009 entrance examination, rejecting the institutes’ argument that revealing candidates’ names would be a breach of their privacy. India’s apex watchdog for the Right to Information Act has ordered the IITs to reveal the names, addresses, pin codes and marks of all students who appeared in the Joint Entrance Examination this year.
In its November 6 order, the commission asked IIT Guwahati, the chief organiser among the IITs of the 2009 examination, to disclose by November 25 the information sought by the appellant.
The order follows efforts by the IIT to withhold information on candidates who appeared in the 2009 JEE despite earlier orders mandating the release of similar data on IIT candidates over the past three years.
The order is significant because a similar disclosure in 2006 revealed discrepancies between cutoff marks used by the IITs that year and the cutoffs arrived at by using the formula the institutes claimed to have used.
At least 994 students, who cleared the cutoffs arrived at by using the formula the IITs claimed to have used, were denied admission because the institutes used different cutoffs.
The IITs have till now been unable to explain how they arrived at the cutoffs — by using the formula they claim to have used.
The appellant in the 2009 case is the parent of a student who appeared in the controversial 2006 examination and is trying to use the RTI Act to ensure that the IITs do not repeat their errors. The IITs, in the 2009 case, argued that the release of candidate details sought by the appellant could compromise the privacy of these candidates.
The appellant had sought the registration numbers, names, gender, parents’ names, pin codes and marks in physics, chemistry and mathematics of all students who appeared in the 2009 examination.
The appellant has expressed concern that the IITs may be admitting the children of institute administrators or certain faculty members despite poor marks, and has argued that details he has sought would help clarify his doubts.
The commission, in its order, has argued that while providing email addresses and mobile phone numbers of candidates would constitute a violation of privacy, merely disclosing their names and addresses would not.
Open source software needs marketing
PUNE: There is a need for greater promotion of the use of open source software for information and communication technology (ICT)-based teaching and learning. Professor Kannan M Moudgalya of the Indian Institute of Technology,
Bombay (IIT-B), highlighted this on Monday. Moudgalya, who heads the Centre for Distance Engineering Education Programme (CDEEP) at the IIT, was delivering the keynote address at the launch of kPoint, a software solution for interactive learning and training. kPoint, developed by city-based Great Software Laboratory (GSL), was launched by noted computer expert Vijay Bhatkar, creator of India's Param series of supercomputers. Heads and professionals from leading IT companies as well as principals of engineering institutions were present at the occasion.
Open source software refers to computer software provided under a license that is in the public domain. "Open source software has a distinct cost advantage over the expensive commercial software packages. However, a considerable marketing effort is required to secure a greater and wider audience of students for courses transmitted live using ICT tools based on open source software," Moudgalya said.
"Open source software is often sufficient in most distance education programmes, except for some niche academic segments. However, academic institutions don't train students in using good open source software," he further stated.
Moudgalya gave an overview of the CDEEP's involvement in the Talk to teacher' programme, which is funded by the Union human resource development ministry and aims to train students as well as teachers in higher
education. IIT-B started disseminating its courses live on the internet nearly a decade ago. For the last two years, he stated, the CDEEP has been using the education satellite Edusat, provided by the Indian Space Research Organisation, and has raised a network of 75 centres for transmission of live courses. In his brief address, Bhatkar made out a strong case for Indian ICT professionals acknowledging and adopting technologies and innovations developed indigenously.
Sunil Gaitonde, chief executive of GSL, said, "Technology must bridge the gap between the growing number of learners and lesser number of teachers. The prevailing knowledge economy needs highly skilled workers and the existing faculty crunch can be tackled only through apt use of technology."
Gaitonde said: "Factors like grassroots videos, collaborations, mobile broadband, data mash-ups, collective intelligence and social operating systems are bound to make a sea change in the way education is delivered."
College of Engineering, Pune (CoEP) principal Anil Sahasrabudhe, Vishwakarma Institute of Technology principal Hemant Abhyankar, Persistent Systems chief Anand Deshpande and founder-CEO of music education web portal ShadjaMadhyam, Nandu Kulkarni, were among those present at the event.
Dr. Rajeewa (Rajiv) Arya, M. Tech., Ph. D is presently the Chief Executive Officer l at Moserbaer Photovoltaic (MBPV) in New Delhi, India. He was previously the COO & CTO for the Thin Film Vertical. He joined MBPV as a Senior Vice-President & CTO (Thin Film) in September, 2007. and Electrical Engineers (IEEE) and the Materials Research Society (MRS). Prior to that Dr. Arya was a founder, Vice-President and CTO at Optisolar (previously called Gen3Solar) in Hayward, California. Before founding Gen3solar, he was the Director of Oregon Renewable Energy Center (OREC), an academic/research center at the Oregon Institute of Technology (OIT).
Dr. Arya launched Arya International, Inc., a Solar Technology and Business Consulting firm, in 2003. Prior to that Dr. Arya worked at Solarex/ BPSolar for 19 years in various capacities, from Scientist to Executive Director, thin-film technology.
He has over 25 years experience in thin-film solar cells and modules. His R&D activities have centered on material and device aspects of three types of thin-film solar cells and modules – amorphous silicon, copper-indium-diselenide, and cadmium telluride. His work
includes product design, process scale-up, process transfer, piloting and start-up of a thin-film solar module plant. He has maintained a professional interest in many aspects of renewable energy components and systems. Dr. Arya holds a Masters of Science degree in Solid-State Physics from Jadavpur University, Calcutta, India and a Master of Technology degree in Material Science from the Indian Institute of Technology, Kanpur, India. He obtained his Ph.D. in Engineering from Brown University, Rhode Island, in 1983. He has extensive management training in Total Quality Management, Finance, Project management, and Technology Innovation Management.
Dr. Arya has authored and co-authored over 100 technical papers and holds 6 U.S. Patents. He is the recipient of the “Outstanding Paper Award” at the 7th PVSEC, the “Team of the Year” award from Solarex Quality Process, and his group received an R&D 100 award for the Power view Product. He chaired the Program Committee for the 29th IEEE Photovoltaic Specialists Conference in 2002. He is a member of the Institute of Electronics.
Dr. Rajeewa Arya
B.E., M.E.(IIT – Kanpur)Chief Executive Officer, Moser Bear (I), Ltd.
Success Story
This article contains story of a person who get succeed after graduation from different IIT'sPHYSICS
1. A circular disc with a groove along its diameter is placed horizontally on a rough surface. A block of mass 1 kg is placed as shown. The co-efficient of friction in contact is µ = 2/5. The disc has an acceleration of 25 m/s2 towards left. Find the acceleration of the block with respect to disc. Given cos θ = 4/5, sin θ = 3/5. [IIT-2006]
θ 25 m/s2
Sol. Applying pseudo force ma and resolving it. Applying Fnet = max for x-direction
ma cos θ – (f1 + f2+ = max ma cos θ – µN1 – µN2 = max ma cos θ – µma sin θ – µmg = max
⇒ ax = a cos θ – µa sin θ – µg = 25 × 5 4 – 5 2 × 25 × 5 3 – 5 2 × 10 = 10 m/s2
2. A wooden stick of length L, radius R and density ρ has a small metal piece of mass m (of negligible volume) attached to its one end. Find the minimum value for the mass m (in terms of given parameters) that would make the stick float vertically in equilibrium in liquid of density σ(>p). [IIT-1999] Sol. For the wooden stick-mass system to be in stable
equilibrium the centre of gravity of stick-mass system should be lower than the centre of buoyancy. Also in equilibrium the centre of gravity (G) and the centre of buoyancy (B) lie in the same vertical axis.
The above condition 1 will be satisfied if the mass is towards the lower side of the stick as shown in the figure.
The two forces will create a torque which will bring the stick-mass system in the vertical position of the stable equilibrium
Let l be the length of the stick immersed in the
FB=πR2hσg C (πR2Lρ)g mg θ θ h/2 L/2 Then OB = 2 l . Let OG = y
For vertical equilibrium FG = FB ⇒ (M + m)g = FB ⇒ πR2Lρg + mg = πR2 l σ g l = σ π + ρ π 2 2 R m L R ...(1)
Now using the concept of centre of mass to find y. Then y = n M my My1 2 + +
Since mass m is at O the origin, therefore y2 = 0 ∴ y = m M O m ) 2 / L ( M + × + = ) m M ( 2 ML + = ) m L R ( 2 L ) L R ( 2 2 + ρ π ρ π ...(2) Therefore for stable equilibrium
2 l > y ∴ ) R ( 2 m L R 2 2 σ π + ρ π > ) m L R ( 2 L ) L R ( 2 2 + ρ π ρ π ⇒ m ≥ π R2L ( ρσ – ρ) ∴ minimum value of m is πr2L ( ρσ – ρ) 3. A gaseous mixture enclosed in a vessel of volume V
consists of one mole of a gas A with λ (=Cp/Cv) = 5/3 and another gas B with λ = 7/5 at a certain temperature T. The relative molar masses of the gases A and B are 4 and 32, respectively. The gases A and B do not react with each other and are assumed to be ideal. The gaseous mixture follows the equation
KNOW IIT-JEE
(a) Find the number of moles of the gas B in the gaseous mixture.
(b) Compute the speed of sound in the gaseous mixture at T = 300 K.
(c) If T is raised by 1 K from 300 K, find the % change in the speed of sound in the gaseous mixture. (d) The mixture is compressed adiabatically to 1/5 of
its initial volume V. Find the change in its adiabatic compressibility in terms of the given quantities. Sol. (a) The ratio of specific heat of mixture of gases
γm = m v m p ) C ( ) C (
* m stands for mixture.
Also (Cp)m = B A pB B pA A n n C n C n + + and (Cv)m = B A vB B VA A n n C n C n + +
Also according to Mayer's relationship Cp – Cv = R ⇒ v p C C – 1 = v C R ⇒ γ – 1 = v C R ⇒ C v = 1 R − γ For Gas A (Cv)A =
1 3 / 5 R − = 2 R 3 ∴ (Cp)A = R + (Cv)A = R + 2 R 3 = 2 R 5 For Gas B (Cv)B = 1 5 / 7 R − = 2 R 5 ∴ (Cp)B = R + 2 R 5 = 2 R 7 γm = m v m p ) C ( ) C ( = vB B vA A pB B pA A C n C n C n C n + + = ) 2 / R 5 ( n ) 2 / R 3 ( 1 ) 2 / R 7 ( n ) 2 / R 5 ( 1 B B + × + × = B B n 5 3 n 7 5 + + According to the relationship
PV19/13 = constant we get γm = 19/13 ∴ B B n 5 3 n 7 5 + + = 13 19 ⇒ nB = 2 mol.
Alternatively we may use the following formula 1 n m 1 − γ = 1 n 1 1 − γ + 1 n 2 2 − γ
where γm = Ratio of specific heats of mixture
(b) We know that velocity of sound in air is given by the relationship v = d P γ where d = density = v m Also, PV = (nA + nB)RT ⇒ PV = V ) n n ( A+ B RT ∴ v = V m V RT ) n n ( A B × + γ = m RT ) n n ( A+ B γ
Mass of the gas,
m = nAMA + nBMB = 1 × 4 + 2 × 32 = 68 g/mol mol = 0.068 kg/mol ∴ v = 068 . 0 13 300 314 . 8 ) 2 1 ( 19 × × × + = 400.03 ms–1 (c) We know that the velocity of sound
v = d P γ = M RT γ and v + ∆v = M ) T T ( R +∆ γ ⇒ v v v+∆ = T T T+∆ = 2 / 1 T T 1 +∆ ⇒ 1 + v v ∆ = 1 + 2 1 T T ∆ when ∆T << T then T T ∆ << 1 Percentage change v v ∆ × 100 = 2 1 × T T ∆ × 100 v v ∆ × 100 = 2 1 300 1 × 100 = 6 1 (d) PVγ = Constant
Differentiating the above equation Vγ(dP) – P(γ Vγ –1dV) = 0 ⇒ Vγ(dP) = γ PVγ–1 dV ⇒ dV dP = γ − γ γ V PV 1 = γPVγ–1 dV ⇒ V / dV dP − = –γ P ∴ Bulk Modulus B = γP ∴ Compressibility K = B 1 = P 1 γ ∴ K1 = 1 P 1 γ and K2 = P2 1 γ ∆K = K2 – K1 = 2 P 1 γ – P1 1 γ = − γ 2 P1 1 P 1 1
∴ Since the process is adiabatic, P2Vγ2 = P1Vγ1 ∴ P2 = P1 γ 2 1 V V = P1 γ 5 / V V 1 1 = P1Vγ1 ∴ ∆Κ = − γ γ 1 1 P 1 5 P 1 1 = − γ 5γ 1 1 P 1 1 P1 = V RT ) n n ( A+ B = V T 31 . 8 ) 2 1 ( + × × = V T 93 . 24 ⇒ ∆K = − × × 1 5 1 V T 93 . 24 13 19 1 13 / 19 = – T V 025 . 0 Pa–1
This unit Pa–1 is valid when V, T are taken in S.I. units.
4. A square loop of side 'a' with a capacitor of capa-citance C is located between two current carrying long parallel wires as shown. The value of I in the wires is given as I = I0 sin ωt. [IIT-2003]
a
a a
I I
(a) Calculate maximum current in the square loop. (b) Draw a graph between charges on the upper plate
of the capacitor vs time.
Sol. (a) Let us consider a small strip of thickness dx as shown in the figure.
The magnetic field at this strip
B = BA + BB (Perpendicular to the plane of paper directed upwards)
= π µ 2 0 x I + π µ 2 0 ) x a 3 ( I −
BA = Magnetic field due to current in wire A = π µ 2 I 0 − + x a 3 1 x 1
BB = Magnetic field due to current in wire B
x
I I
dx
Small amount of magnetic flux passing through the strip of thickness dx is
dφ = B × adx = ) x a 3 ( x 2 dx a 3 Ia 0 − π × µ
Total flux through the square loop φ =
∫
− π × µ a 2 a 2 0 ) x a 3 ( x dx 2 a 3 I = π µ Ia0 ln 2 = π µ0aln(2) (I0 sin ωt) The emf producede = dt dφ − = π ω µ0aI0 ln(2) cos ωt
Charged stored in the capacitor q = C × e = C ×
π ω
µ0aI0 ln(2) cos ωt ... (i) ∴ Current in the loop
i = dt dq = π ω µ × 2 0 0aI C ln(2) sin ωt ∴ imax = π ω µ0aI0 2Cln(2)
(b) From (1), the graph between charge and time is
Q0 Q π/2ω π/ω 2π/ω t 3π/2ω –Q0 Here q0 = π ω µ × aI ln(2) C 0 0
5. Highly energetic electrons are bombarded on a target of an element containing 30 neutrons. The ratio of the radii of nucleus to that of Helium nucleus is (14)1/3. Find (a) atomic number of the nucleus. (b) the frequency of Ka line of the X-ray produced. (R = 1.1 × 107 m– 1 and c = 3 × 108 m/s) [IIT-2005] Sol. (a) We know that radius of nucleus is given by the
formula
r = r0 A1/3 where r0 = constant and A = mass number.
For the Nucleus r1 = r0 41/3 ∴ 1 2 r r = 3 / 1 4 A ⇒ (14)1/3 = 1/3 4 A ⇒ A = 56
∴ No. of proton = A – no. of neutrons = 56 – 30 = 26 ∴ Atomic number = 26 (b) We know that ν = Rc (z – b)2 − 2 1 2 1 n 1 n 1 Here, R = 1.1 × 107, C = 3 × 108, Z = 26 b = 1 (for Ka), n1 = 1, n2 = 2 ∴ ν = 1.1 × 107 × 3 × 108 [26 – 1]2 − 4 1 4 1 = 3.3 × 1015 × 25 × 25 × 4 3 = 1.546 × 1018 Hz.
CHEMISTRY
6. The standard reduction potential of Ag+/Ag electrode at 298 K is 0.799V. Given that for AgI,
Ksp = 8.7 × 10–17, evaluate the potential of Ag+/Ag electrode in a saturated solution of AgI. Also calculate the standard reduction potential of I–electrode. [IIT-1994]
Sol. In the saturated solution of AgI, the half cell reactions are
At anode : Ag → Ag+ + e– At cathode : AgI + e– → Ag + I– Cell reaction AgI → Ag+ + I–
On applying Nernst equation Ecell = Eºcell – n 0591 . 0 log [Ag+] [I–] For electrode Ag+ + e– → Ag ∴ EAg+/Ag= EºAg+/Ag– n 0591 . 0 log ] Ag [ 1 + Ksp of AgI = [Ag+] [I–] Q [Ag+] = [I–] ∴ Ksp of AgI = [Ag+]2 ∴ [Ag+] of AgI = sp K of AgI [Ag+] = 8.7×10−17 = 9.3 × 10–9 M So EAg+/Ag = 0.799 – 1 0591 . 0 log 9 10 3 . 9 1 − × = + 0.799 + 0.0591 log 9.3 – 0.0591 × 9 log 10 = + 0.799 + 0.0591 × 0.9785 – 0.0591 × 9 = 0.325 V
For above cell reaction Ecell = Eºcell – n 0591 . 0 log [Ag+] [I–] = Eºcell – n 0591 . 0 log (Ksp of AgI) At equilibrium Ecell = 0 ∴ Eºcell = 1 0591 . 0 log(8.7 × 10–17) = –0.95 volt Eºcell = Eºcathode + Eºanode
–0.95 = –0.799 + EºAg/AgI/I–
(In form of cell reaction) EºAg/AgI/I– = – 0.95 + 0.799 = –0.151 V or EºI–/AgI/Ag = + 0.151 V
7. A sample of hard water contains 96 ppm of SO42– and 183 ppm of HCO3– with 60 ppm of Ca2+ as the only cation. How many moles of CaO will be required to remove HCO32– from 100 kg of this water ? If 1000 kg of this water is treated with the amount of CaO calculated above, what will be the concentration (in ppm) of residual Ca2+ ions ? (Assume CaCO3 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 106 g(= 1000 kg) of the given hard water, we have amount of SO42– ions = 96 g
amount of HCO3– ions = 183 g So amount of SO42– ions =
1 mol g 96 g 96 − = 1 mol
and 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 Ca2+ ions =
+ 2 3 1 = 2.5 mol The addition of CaO causes the following reactions:
CaO + Ca(HCO3)2 → 2CaCO3 + H2O
1.5 mol of CaO will be required for the removal of 1.5 mol of Ca(HCO3)2 in form of CaCO3.
In the treated water, only CaSO4 is present now. 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, Molartiy of H+ in the treated water = 2 × 10–3 M Thus, pH = – log(2 × 10–3) = 2.7
8. A white precipitate was formed slowly when AgNO3 was added to compound (A) with molecular formula C6H13Cl. Compound (A) on treatment with hot alcoholic KOH gave a mixture of two isomeric alkenes (B) and (C), having formula C6H12. The mixture of (B) and (C) on ozonolysis, furnished four compounds (i) CH3CHO, (ii) C2H5CHO,
(iii) CH3COCH3 and
(iv) CH3 – CH(CH3)–CHO. What are the structures of (A) and (C) ? [IIT-1986] Sol. It is given that,
C6H13Cl
(A) Alkyl chloride
Alcoholic
KOH; –HCl Two alkenes (B) + (C) with Formula C
6H12 + CH3COCH3 + CH3 – CH – CHO CH3 C6H12 (B) and (C) (i) O3 ((ii) H2O/Zn CH3CHO + C2H5CHO
It is observed that during ozonolysis, no loss of carbon takes place, it may be concluded that CH3CHO and CH3 – CH(CH3) – CHO are the products of one alkene (B) and C2H5CHO and CH3COCH3 are the products of other alkene (say) (C). Thus, from the above we have :
CH3 – C = O + O = HC – CH – CH3 CH3 H –2[O] CH 3 – CH = CH – CH – CH3 CH3 (i) O3 CH 3CHO + OHC – CH – CH3 CH3 (ii) H2O/Zn (B) Similarly alkene (C) will be derived as :
C = O + O = CH.CH2CH3 –2[O] CH 3 – C = CH.CH2CH3 CH3 (i) O3 CH 3 – C = O + OHC.CH2CH3 CH3 (ii) H2O/Zn CH3 CH3 (C)
Since the compounds (B) and (C) are obtained when (A), C6H13Cl, is dehydrohalogenated by heating it with alcoholic KOH, as follows :
CH3CH2 – CH . CH – CH3 Alc.KOH CH3 – CH = CH – CH – CH3 + CH3CH2CH = C – CH3 CH3 (B) (20%) (minor) ∆; –HCl Cl CH3 CH3 (C) (80%) (major)
Since the Cl atom in (A) is an aliphatic chlorine, and it is attached to a secondary carbon atom which is adjacent to a tertiary cabon atom and one secondary carbon atom – CH
2 – CH . CH –
Cl CH3
, it will react
slowly with AgNO3 to give a white precipitate. Thus, CH3 – CH2 – CH – CH – CH3 A, Cl 3-chloro-2-methyl pentane CH3 CH3CH = CH – CH – CH3 B, CH3 4-methyl pentene -2 CH3CH2CH = C – CH3 C, CH3 2-methyl pentene-2
9. A hydrated metallic salt A, light green in colour, gives a white anhydrous residue B after being heated gradually. B is soluble in water and its aqueous solution reacts with NO to give a dark brown compound C. B on strong heating gives a brown residue and a mixture of two gases E and F. The gaseous mixture, when passed through acidified permanganate, discharges the pink colour and when passed through acidified BaCl2 solution, gives a white precipitate. Identify A, B, C, D, E and F.
[IIT-1988]
Sol. The given observations are as follows. (i) ) A ( salt metallic Hydrated heat → ) B ( residue anhydrous white
(ii) Aqueous solution of B NO→
) C ( compound brown dark (iii) Salt B heating Strong → ) D ( residue Brown + ) F ( ) E ( gases Two + (iv) Gaseous mixture (E) + (F) acidified KMnO4 BaCl2 solution Pink colour is discharged White precipitate
The observation (ii) shows that B must be ferrous sulphate since with NO, it gives dark brown compound according to the reaction
[Fe(H2O)6]2+ + NO → brown dark 2 5 2O) (NO)] H ( Fe [ + + H2O
Hence, the salt A must be FeSO4 . 7H2O The observation (iii) is
2FeSO4 → Fe2O3 + SO2 + SO3
(D)
brown (E) + (F)
The gaseous mixture of SO2 and SO3 explains the observation (iv), namely,
colour pink 4 MnO 2 − + 5SO2 + 2H2O → colour no 2 4 2 5SO Mn 2 ++ −+ 4H+ 2H2O + SO2 + SO3 4H+ + SO32– + SO42– Ba2+ + SO32– → . ppt white 3 BaSO ; Ba2+ + SO4– → . ppt white 4 BaSO Hence, the various compounds are
A. FeSO4 . 7H2O B. FeSO4 C. [Fe(H2O)5NO]SO4 D. Fe2O3
E and F SO2 and SO3
10. A white amorphous powder A when heated gives a colourless gas B, which turns lime water milky and the residue C which is yellow when hot but white when cold. The residue C dissolves in dilute HCl and the resulting solution gives a white precipitate on addition of potassium ferrocyanide solution. A dissolves in dilute HCl with the evolution of a gas
which is identical in all respects with B. The solution of A as obtained above gives a white precipitate D on addition of excess of NH4OH and on passing H2S. Another portion of this solution gives initially a white precipitate E on addition of NaOH solution, which dissolves on further addition of the base. Identify the compound A to E. [IIT-1979] Sol. The given information is as follows.
(a)
powder
whiteA →
heat
milkylimewater turnscolourlessgas
B +
cold when hot whiteyellow when residueC (b)C diluteHCl→ solution → K4Fe(CN)6 white precipitate (c) A dilute HCl Solution + B (i) NH4OH (ii) H2S (i) NaOH D white precipitate E white precipitate dissolves NaOH
From part (a), we conclude that B is CO2 as it turns lime water milky :
Ca(OH2) + CO2 →
this to due milky 3
CaCO + H2O
and C is ZnO as it becomes yellow on heating and is white in cold. Hence, the salt A must be ZnCO3. From part (b), it is confirmed that C is a salt of zinc (II) which dissolves in dilute HCl and white precipitate obtained after adding K4[Fe(CN)6 is due to Zn2[Fe(CN)6].
From part (c), it is again confirmed that A is ZnCO3 as on adding dilute HCl, we get CO2 and zinc (II) goes into solution. White precipitate is of ZnS which is precipitated in ammonical medium as its solubility product is not very low. White precipitate E is of Zn(OH)2 which dissolves as zincate, in excess of NaOH. Hence the given information is explained as follows. (a) (A) 3 ZnCO heat → (B)2 CO + (C) ZnO (b) (C) ZnO →dilHCl solution2 ZnClK4Fe(CN)6→ e precipitat White2 6 ] ) CN ( Fe [ Zn (c) ZnCO3 dil →HCl Solution2 ZnCl + CO2 + H2O ZnCl2 + S2– → (D) ZnS↓ + 2Cl– Zn2+ + 2OH– → (E) 2 Zn(OH) Zn(OH)2 + 2OH– → dissolves -2 2 ZnO + 2H2O
MATHEMATICS
11. Determine the name of the name of the curve described parametrically by the equations
x = t2 + t + 1, y = t2 – t + 1 [IIT-1998] Sol. We have, x = t2 + t + 1 and, y = t2 – t + 1 ⇒ x + y = 2(t2 + 1) and, x – y = 2t ⇒ x + y = 2 + − 1 2 y x 2 ⇒ 2(x + y) = (x – y)2 + 4 ⇒ x2 + y2 – 2xy – 2x – 2y + 4
Comparing this equation with the equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, we get
a = 1, b = 1, c = 4, h = –1, g = –1 and f = –1
∴ abc + 2fgh – af2 – bg2 – ch2 = 4 – 2 – 1 – 1 – 4 ≠ 0 and , h2 – ab = 1 – 1 = 0
Thus, we have
∆ ≠ 0 and h2 = ab
So, the given equations represent a parabola.
12. The circle x2 + y2 – 4x – 4y + 4 = 0 is inscribed in a triangle which has two of its sides along the coordinate axes. If the locus of the circumcentre of the triangle is
x + y – xy + k x2+y2 = 0,
find the value of k. [IIT-1987] Sol. Let OAB be the triangle in which the circle
x2 + y2 – 4x – 4y + 4 = 0 is inscribed. Let the equation of AB be b y a x + = 1 y B(0,b) y´ x´ O 2 C (a, 0)A x a+ 1 b y a x+ =
Since AB touches the circle x2 + y2 – 4x – 4y + 4 = 0. There fore, 2 2 b 1 a 1 1 b 2 a 2 + − + = 2 ⇒ – 2 2 b 1 a 1 1 b 2 a 2 + + − = 2
[Q O(0, 0) and C(2, 2) lie on the same side of AB Therefore, a 2 + b 2 – 1 < 0] ⇒ – 2 2 b a ) ab – a 2 b 2 ( + + = 2 ⇒ 2a + 2b – ab + 2 a2+b2 = 0 ...(i)
Let P(h, k) be the circumcentre of ∆OAB. Since ∆ OAB is a right angled triangle. So its circumcentre is the mid-point of AB.
∴ h = 2 a and k = 2 b ⇒ a = 2h and b = 2k ...(ii) From (i) and (ii), we get
4h + 4k – 4hk + 2 4h2+4k2 = 0 ⇒ h + k – hk + h2+k2 = 0
So, the locus of P(h, k) is
x + y – xy + x2+y2 = 0
But, the locus of the circumcentre is given to be x + y – xy + k x2+y2 = 0
Thus, the value of k is 1
13. Let C be any circle with centre (0, 2 ). Prove that at most two rational points can be there on C.
(A rational points is a point both of whose coordinates are rational numbers) [IIT-1997] Sol. The equation of any circle C with centre (0, 2 ) is
given by
(x – 0)2 + (y – 2 )2 = r2, where r is any positive real number.
or, x2 + y2 –2 2y = r2 – 2
If possible, let P(x1, y1), Q(x2, y2) and R(x3, y3) be three distinct rational points on circle C. Then,
2 2 y x21+ 12− y1 = r2 – 2 ...(ii) 2 2 y x22+ 22− y2 = r2 – 2 ...(iii) 2 2 y x23+ 23− y3 = r2 – 2 ...(iv)
We claim that at least two y1, y2, and y3 are distinct. For if y1 = y2 = y3, then P, Q and R lie on a line parallel to x-axis and a line parallel to x-axis does not cross the circle in more than two points. Thus, we have either y1 ≠ y2 or, y1 ≠ y3 or, y2 ≠ y3.
Subtracting (ii) from (iii) and (iv), we get )
y x
( 22+ 22 – (x12+y12) – 2 2(y2 – y1) = 0
and, (x23+y32) – (x12+y12) – 2 2(y3 – y1) = 0 ⇒ a1 – 2 b1 = 0 and a2 – 2 b2 = 0 ...(v)
where,
a1 = (x22+y22) – (x12+y12), b1 = 2(y2 – y1) a2 = (x23+y32) – (x12+y12), b2 = 2(y3 – y1)
Clearly, a1, a2, b1, b2 are rational numbers as x1, x2, x3, y1, y2, y3 are rational numbers.
Since either y1 ≠ y2 or, y1 ≠ y3 ∴ Either b1 ≠ 0 or, b2 ≠ 0 If b1 ≠ 0, then a1 – 2 b1 = 0 [From (v)] ⇒ 1 1 b a = 2 ,
which is not possible because
1 1
b a
is a rational number and 2 is an irrational number.
If b2 ≠ 0, then a2 – 2 b2 = 0 ⇒ 2 2 b a = 2 , which is not possible because
2 2
b a
is a rational number and 2 is an irrational number.
Thus, in both the cases we arrive at a contradiction. This means that our supposition is wrong. Hence, there can be at most two rational points on circle C. 14. A rectangle PQRS has its side PQ parallel to the line
y = mx and vertices P, Q and S lie on the lines y = a, x = b and x = –b, respectively. Find the locus of the
vertex R. [IIT-1996]
Sol. Let the coordinates of R be (h, k). It is given that P lies on y = a. So, let the coordinates of P be (x1, a). Since PQ is parallel to the line y = mx. Therefore, Slope of PQ = (Slope of y = mx) = m And, Slope of PS = – mx) y of Slope ( 1 = = – m 1 [∴ PS ⊥ PQ] Now, equation of PQ is y – a = m(x – x1) ...(i) y y = 0 P Q x = b (0, b) x y´ R O (0, – b) S x´ x = –b (0, a)
Putting x = b in (i), we get y = a + m(b – x1)
So, coordinates of Q are (b, a + m(b – x1)). Since PS passes through P(x1, a) and has slope –
m 1 . So, Equation of PS is y – a = – m 1 (x – x1) ...(ii) It is given that S lies on x = – b. So, S is the point of intersection of (ii) and x = –b.
Solving (ii) and x = – b, we get y = a +
m
1 (b + x1)
So, coordinates of S are − + (b+x ) m 1 a , b 1 Now, Slope of RS = b h ) x b ( m 1 a k 1 + + − − = m But RS is parallel to PQ. ∴ b h ) x b ( m 1 a k 1 + + − − = m ⇒ b + x1 = m(k – a) – m2(h + b) ...(iii) Similarly, Slope of RQ = b h ) x b ( m a k 1 − − − −
But, RQ is perpendicular to PQ whose slope is m. ∴ b h ) x b ( m a k 1 − − − − = – m 1 ⇒ b – x1 = m 1 (k – a) + 2 m 1 (h – a) ...(iv)
We have only one variable x1. To eliminate x1, add (iii) and (iv) to obtain
2b = (k – a) + m 1 m – m2(h + b) + 2 m 1 (h – b) ⇒ 2b = (k – a) + m 1 m2 – h + 2 4 m 1 m – b + 2 4 m 1 m ⇒ (k – a) + m 1 m2 – 2 2 2 m ) 1 m )( 1 m ( h − + – 2 2 2 m ) 1 m ( b + = 0 ⇒ (k – a) – m ) 1 m ( h 2− – m ) 1 m ( b 2+ = 0 ⇒ m(k – a) – h(m2 – 1) – b(m2 + 1) = 0 Hence, the locus of R(h, k) is
m(y – a) – x(m2 – 1) – b(m2 + 1) = 0
15. If A, B, C are the angles of a triangle ABC and the system of linear equations
x sin A + y sin B + z sin C = 0 x sin B + y sin C + z sin A = 0 x sin C + y sin A + z sin B = 0 has a non trivial solution, prove that
sin2A + sin2B + sin2C – (cos A + cos B + cos C + cos A cos B + cos B cos C + cos C cos A) = 0
[IIT-2002] Sol. The given system of linear equations has a non-trivial
solution. Therefore, B sin A sin C sin A sin C sin B sin C sin B sin A sin = 0 ⇒ B sin A sin B sin A sin C sin A sin C sin A sin C sin B sin C sin B sin C sin B sin A sin + + + + + + = 0 Applying C1 → C1 + C2 + C3 ⇒ (sin A + sin B + sin C)
B sin A sin 1 A sin C sin 1 C sin B sin 1 = 0 ⇒ B sin A sin 1 A sin C sin 1 C sin B sin 1 = 0 ≠ = + + 0 2 C cos 2 B cos 2 A cos 4 C sin B sin A sin Q ⇒ C sin B sin B sin A sin 0 C sin A sin B sin C sin 0 C sin B sin 1 − − − − = 0 Applying R2 → R2 – R1, R3 → R3 – R1 ⇒ –(sin B – sin C)2 – (sin A – sin C)
(sin A – sin B) = 0 ⇒ sin2B + sin2C – 2 sin B sin C + sin2A
– sin A sin B – sin C sin A + sin B sin C = 0 ⇒ sin2A + sin2B + sin2C – sin A sin B – sin B sin C
– sin C sin A = 0 ⇒ sin2A + sin2B + sin2C – cos A cos B
– cos B cos C – cos C cos A + cos (A + B) + cos (B + C) + cos (C + A) = 0 ⇒ sin2A + sin2B + sin2C – cos A cos B
– cos B cos C – cos C cos A – cos A – cos B – cos C = 0
Passage # 1 (Ques. 1 to 3)
Young's double slit experiment is conducted with the following conditions
1. Slit S1 and slit S2 are of same width
2. Slits are illuminated by monochromatic light source of wave length 'λb' which is of blue color.
3. Distance between slits and screen D. Distance between slits is 2d and D > > 2d
It is observed that 1st bright fringe is observed in front of one of the slit.
1. If monochromatic light source of blue color is replaced by the white colored light source then maximum wavelength which is missing in front of one of the slit is -
(A) Never of indigo and violet colors (B) It is always of less than blue color
(C) Missing wave lengths can have wave length more or less than blue color
(D) b ) g sin mis ( max. . 3 2 λ = λ
2. Relation of the maximum wavelength missing with wave length of Blue light is -
(A) 2λb (B) 3λb
(C) 4λb (D) 5λb
3. With blue light if the slit widths are made unequal then -
(A) Position of 1st bright fringe will not be in front of one of the slit
(B) Dark fringe which was of black colour earlier with same slit widths now is of blue colour
(C) YDSE can not be conducted with unequal slit widths
(D) Dark fringe is always black either the slit widths are equal or not
Passage # 2 (Ques. 4 to 5) For the given circuit
R a 2R 3R R R b 4R 12R 8R R R
4. What should be the value of R so that equivalent resistance between terminals a and b is 1Ω -
(A) 5 2 Ω (B) 2 5 Ω (C) 2Ω (D) 15Ω
5. If current passing through the circuit is 1 amp then- (A) Potential difference across 4R and 8R is in the
ratio of 1 : 2
(B) Potential difference across ab is 1 volt when measured by an ideal voltmeter if R =
5 2 Ω (C) Maximum potential difference will appear
across 12R resistance
(D) Maximum potential difference will appear across 3R resistance
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
Passage # 3 (Ques. 6 to 8)
Behaviour of capacitor in electric circuits is very typical because of it's energy storing nature. Capacitor behaves in just opposite manner to inductor, Inductor 'L' which is measured in Henary in SI system stores the energy in magnetic field instead of capacitor which stores in electric field Inductor opposes the change in current and capacitor opposes change in voltage.
Behaviour of inductor: Steady State t = 0 open switch t → ∞ Closed switch
L/R is known as time constant of R-L series circuit which is measured in ohm
For the electric circuit shown
C R1 ε1
R
ε2 R2
6. If capacitance C varies even after that energy stored in capacitor is zero at steady state then -
(A) 2 1 2 1 R R ε ε = (B) 1 2 2 1 R R ε ε = (C) ε1 + ε2 = 0 (D) ε1R1 + ε2R2 = 0
7. Time constant for the circuit -
(A) RC (B) R1C if ε1 > ε2 (C) R2C if ε1 < ε2 (D) + + 2 1 2 1 R R R R R C where εeq = 2 1 2 1 1 1 R / 1 R / 1 R / R / + ε − ε Req = 2 1 2 1 R R R R +
8. Maximum current passing through resistance R - (A) eq eq R ε (B) eq eq R R+ ε (C) R eq ε (D) R | |ε1−ε2
SCIENCE TIPS
• What is the expression for growing current, in LR
circuit ? ® I = I0 − − tL R e 1
• What is the range of infrared spectrum ?
® This covers wavelengths
from 10–3 m down to 7.8 × 10–7 m • What is the nature of graph between electric field
and potential energy (U) ?
® The nature of the graph
will be parabola having symmetry about U-axis • Why no beats can be heard if the frequencies of
the two interfering waves differ by more than ten ? ® this is due to persistence of hearing • Why heating systems based on steam are more
efficient than those based on circulation of hot water ? ® This is because steam has more heat than water
a the same temperature • Can the specific heat of a gas be infinity ? ® Yes • What is the liquid ascent formula for a capillary ?
® h = pg cos T 2 γ θ – 3 r where h is the height through
which a liquid of density ρ and surface tension T rises in a
capillary tube of radius r • What is the expression for total time of flight (T)
for oblique projection ? ® T = g sin u
2 θ
• The space charge limited current iP in the diode value is given by ® iP = k Vp3/2 • What is an ideal gas ? ® An ideal gas is one in
which intermolecular forces are absent • Can a rough sea be calmed by pouring oil on its
surface ? ® Yes
• What is the expression for fringe width (β) in Young's double slit experiment? ® β=Dλ/d where
D is the distance between the source and screen and d is
1. As the resistances of voltmeters in upper branch are R, R/2, R/4 ...
the equivalent circuit is as shown below
a ... b V Lower Branch R R/2 R/4 upper Branch
the resistance of upper branch is = R + R/2 + R/4 + ... up to infinite = R + + +... 4 1 2 1 1 = R − 21/ 1 1 = 2R
further the equivalent circuit is
a b
V
Lower Branch R upper branch
the resistance of voltmeter V should be 2R so that current in upper and lower branch is same.
2. Entire upper branch is having the resistance of 2R and voltmeter V1 is having the resistance of R so we can conclude that equivalent resistance of all the voltmeters in upper branch except V1 is R and the upper branch is as follows:
i R a V1 V2 V3 ...up to infinite b a b C R V1=X V2=Y
As reading of voltmeter V1 is X = i.R sum of the readings of voltmeters is Y = i.R Except V1 in upper branch
So, X = Y
3. From current division formula we can conclude that current in upper and lower branch are in the ratio of 1 : 2. a b R R i C 2i R′=R voltmeter V
Reading of voltmeter V1 is i.R Reading of voltmeter V is (2i.)R So V = 2V1 4. x. A B a. b. dx. l. l = length of rod = b – a
charge on element of length dx is dq dq = λdx as λ = 3x dq = 3xdx
Equivalent current due to element of length dx di = ω.dq =
π ω 2 (3xdx)
Total equivalent current i = (3xd ) 2 d x b a i
∫ ∫
π ω = = π ω 2 3 b a 2 2 x = π ω 2 3 − 2 a b2 2 = 2 3 . π ω 2 (b 2 – a2) = π ω 4 3 (b2 – a2) Option A is correct (B) Equivalent currentSolution
Physics Challenging Problems
Set # 7
= π ω 4 3 (b2 – a2) = π ω 4 3 (b – a)(b + a) = π ω 4 3 (b + a)(b – a) = π 43 ω. (b + a).l As ω = 4π/3 So, Equivalent current = π 4 3 . 3 4π . (b + a).l = (b + a).l = const.l i ∝ l Option B is correct. (D) Charge on rod q =
∫
∫
= = b a b a 2 x q 3xd 3. x2 d = 2 3 (b2 – a2) Option D is correct Ans. A, B, D 5. For part B q > qclosed cone open cone for part A
q = q
closed cone open cone Equivalent current i = π ω 2 .q i = π ω 2 .q , i = π ω 2 .q
cone - C1 (closed cone) cone-C3 (closed cone) = π ω 2 .q i = π ω 2 .(σ) (closed cone) (Surface area of
closed cone)
If σ varies then charge on cone C1 differs from C3 So their currents will be different.
Option A incorrect q = q (cone C1) (cone - C2) ) coneC ( i 1 = π ω
2 . (ConeCq 1) and (coneC2)
i = π ω 2 .(ConeCq 2) i = i (cone C1) (cone C2) Option B is correct
As charge on cone C3 ≠ charge on cone C4 Option C correct
Part-A and part-B will have different charges so option D incorrect
Ans. B, C
6. The circuit is as follows
CT R1 a R2 b R3 c 10Ω
Full scale deflection current for galvanometer is ig = Ω 10 m 50 = 5mA
For terminals CT and a range is 5V so using R = g i V– G ⇒ R 1 = 3 10 5 5 − × – 10 = 990Ω R1 = 990Ω
7. Range between CT and b is 10 volt so, Using R = g i V – G ⇒ R1 + R2 = 3 10 5 10 − × – 10 990 + R2 = 2000 – 10 R2 = 2000 – 1000 = 1000Ω R2 = 1000Ω
8. Range between CT and c is V so Using R = g i V – G R1 + R2 + R3 = 3 10 5 V − × – 10 ⇒ 990 + 1000 + 3000 = 3 10 5 V − × – 10 ⇒ 5000 = 3 10 5 V − × ⇒ V = 25 volt
1. A homogeneous sphere of radius r rolls without slipping with constant angular speed ω' over bigger sphere, of radius R, which is in pure rotation with constant angular speed ω about its centre O. (Fig.) Find the time taken.
r R C O ω ω′
(i) for the centre C of the rolling sphere to return to its initial position (with respect to O), and
(ii) for the point of contact of the rolling sphere to make one full revolution over the bigger rotating sphere. (b)(i) Determine the acceleration of the contact point of the
rolling sphere, and
(ii) the point of greatest acceleration of the rolling sphere (both w.r.t. the centre O)
Sol. Suppose that at t = 0, the contact points of lie along the fixed line (reference line) OX. Let at time t the line OC makes the angle θ with OX (Fig.)
C θ P O ω ω′ P′ X t^
It is better to express the velocity and acceleration of any point including contact point P (say) at an arbitrary instant of time t as,
v po = v co and apo = aco = apc + apc + aco (1) We know that in the case of pure rolling the velocity of contact point of the rolling body has zero velocity and zero tangential acceleration relative to the contact point of the surface on which it rolls. So if P′ be the contact point of rotating sphere at time t, then
vp′o = vpo
So, vp'o = vpc + vco (2)
(a) If t is the direction of common tangent, then from eqn. (2) vp'o (t) = vpc (t) + vco (t) ωR = – ω′R + θ dt d (R + r) or, dt dθ = r R r ' R + ω + ω (3) so, dt = ) r ' R ( r R ω + ω + dθ (4)
(ii) It is simple to observer if rotating sphere were at rest, the CM of rolling sphere C will turn by the angle 2 π, with angular speed
θ dt
d – ω to satisfy the condition
of the problem.
dθ
ω dt
Hence the sought time t′ (say) is given by t′ = ω − θ π dt d 2 = ) ' ( r ) r R ( 2 ω − ω + π [using Eq. (4)]
(b) From apo = apc + aco = apc (tangential) + apc (normal) + aco In our case apc (tangential) = 0, because ω′ = constant Hence, apo = apc (normal) + aco (5)
apo = + θ + ω − (R r) dt d r ' 2 2 (using dt dθ = r R r ' R + ω + ω
from equation (3) or part (a)) (ii) From eqn. (5) it is obvious that the maximum value |apo |max = |apc (normal) | + | aco |
= – ω′2r + 2 dt d θ (R + r) where dt
dθwill be substituted from Eqn. (3).
Expert’s Solution for Question asked by IIT-JEE Aspirants
Students' Forum
2. Show that the temperature of a planet varies inversely as the square root of its distance from the Sun. Sol. Let Rs be the radius of the Sun. Consider the Sun as a
black body.
Energy emitted per sec by it equals 4 π 2 s
R σ 4 s
T This energy falls uniformly on the inner surface of spheres centred on the Sun. If d is the distance of the planet from the Sun, then energy falling on unit area of the sphere of radius d is :
2 4 s 2 s d 4 T R 4 π σ π = 2 4 s 2 s d T R σ
This energy received by the planet is given by Q = πr2 = 2 4 s 2 s d T R σ
where r is the radius of planet. If T is the temperature of the planet, then energy lost by it per sec is 4 π r2 σ T4
In the steady state the rate of reception of energy is equal to the loss of energy
Hence, 4 π r2 σ T4 = 2 4 s 2 s 2 d T R r σ π Thus, T ∝ d 1
3. A sphere of specific gravity s just fits into a vertical cylinder with lower end closed. The sphere is allowed to drop slowly until it is held in equilibrium by the thrust of the compressed air. There is no leakage of air. If the diameter of the sphere is d, the length of the cylinder is L and the height of the water barometer is h, then what will be the position of sphere ?
Sol. Initially, the cylinder contained air at atmospheric pressure. When the sphere comes down into the cylinder by the action of its own weight, it presses the air downwards. Suppose the sphere comes to position C which is at height x above the closed end. Let the sphere remain in equilibrium in this position.
Volume of sphere = (4/3)πr3 Weight of the sphere = (4/3)πr3sg Volume of cylinder = πr2L
Volume of air inside the cylinder when the sphere is in position A = πr2L – (2/3)πr3
Volume of compressed air when the sphere is in position C = [πr2x – (2/3)πr3]
Atmospheric pressure = h cm of water
Let the pressure of compressed air be p cm of water.
A
C
d
x L
According to Boyle's Law (assuming no change in the temperature of compressed air)
p[πr2x – (2/3)πr3] = h[πr2L – (2/3)πr3] or p[x – (2r/3) = h[L – (2r/3)] ...(1)
In the equilibrium position C, the weight of the sphere is balanced by the difference of vertical thrust on either side due to atmospheric and compressed air. Hence, πr2(p – h)g = mg = (4/3)πr3sg
or p – h = (4/3)rs ...(2) From equation (1) and (2), we get
x = p 3 ) r 2 L 3 ( h − + 3 r 2 or x = rs 4 h 3 ) r 2 L 3 ( h + − + 3 r 2 = ) rs 4 h 3 ( 3 s r 8 hL 9 2 + + or x = ds 6 h 9 s d 2 hL 9 2 + +
4. Given the position of the object O and the image I as shown in the figure. Find (a) the position of the convex lens (b) its focal length and the magnification of the image. Verify graphically.
1 2 3 4 5 6 7 8 9 10 11 12 13 0 1 2 –1 –2 Y O X I 14
Sol. A ray of light from the object passes undeviated through the optic centre of the lens (C) and also the image I. So join OI.
So it cuts the principal axis XY and C. So AB is the position of the lens. A ray parallel to the principal axis from the object after refraction meets the principal axis at F. F is the focus.
A B I 14 C O 4 X 8 F X 10.4 F is at a distance 2.4 cm from C.
∴ Focal length of lens = 2.4 × 10 = 24 cm By calculation f 1 = v 1 – u 1 = 60 1 – 40 1 − = 2400 100 = 24 1 ∴ f = 24 cm Magnification = u v = 40 60 = 2 3 = object of length image of length = mm 2 mm 3 = 2 3
5. Calculate the magnetic field B at the point P shown in the figure. Assume that i = 10 A and a = 8.0 cm.
C i D A i i i P B a/4 a 4
Sol. First of all, we determine the expression of B at a distance R from a straight conductor of length l.
R r dx x x1 x2 φ
Consider a typical element dx. The magnitude of the contribution dB of this element to the magnetic field at P as found from Biot-Savart's law is
dB = 0 2 r 4 sin dx i µ π θ ...(i) Since, the direction of the contribution dB at the point P for all elements are identical viz, at right angle into the plane of the figure, the resultant field is obtained by simply integration equation (i), which gives B =
∫
2 1 x x dB =∫
θ π 2 1 x x 2 0 r dx sin 4 i µ where, sin θ = r R and r = (x2 + R2)1/2 ∴ B =(
)
∫
+ π 2 1 x x 2 / 3 2 2 0 R x dx 4 iR µ ...(ii)Let x = R tan φ, such that dx = R sec2 φ dφ At x = x1 φ1 = tan–1 R x1 and x = x2, φ2 = tan–1 R x2
Also, x2 + R2 = R2 tan2φ + R2 = R2 sec2 φ Hence, equation (ii) becomes
B =
