Dear Students,
Find a mentor who can be your role model and your friend !
A mentor is someone you admire and under whom you can study. Throughout history, the mentor-protege relationship has proven quite fruituful. Socrates was one of the early mentors. Plato and Aristotle studied under him and later emerged as great philosophers in their own right.
Some basic rules to know mentors :
• The best mentors are successful people in their own field. Their behaviors are directly translatable to your life and will have more meaning to you. • Be suspicious of any mentors who seek to make you dependent on them.
It is better to have them teach you how to fish than to have them catch the fish for you. That way, you will remain in control.
• Turn your mentors into role models by examining their positive traits. Write down their virtues. without identifying to whom they belong. When you are with these mentors, look for even more behavior that reflect their success. Use these virtues as guidelines for achieving excellence in your field.
Be cautious while searching for a mentor :
• Select people to be your mentors who have the highest ethical standards and a genuine willingness to help others.
• Choose mentors who have and will share superb personal development habits with you and will encourage you to follow suit.
• Incorporate activities into your mentor relationship that will enable your mentor to introduce you to people of influence or helpfulness.
• Insist that your mentor be diligent about monitoring your progress with accountability functions.
• Encourage your mentor to make you an independent, competent, fully functioning, productive individual. (In other words, give them full permission to be brutally honest about what you need to change.)
Getting benefited from a role-mode :
Acquiring good habits from others will accelerate you towards achieving your goals. Ask yourself these questions to get the most out of your role model/mentors :
• What would they do in my situation?
• What do they do every day to encourage growth and to move closer to a goal ?
• How do they think in general ? in specific situations ?
• Do they have other facts of life in balance ? What effect does that have on their well-being ?
• How do their traits apply to me ?
• Which traits are worth working on first ? Later ?
A final word : Under the right circumstances mentors make excellent role
models. The one-to-one setting is highly conducive to learning as well as to friendship. But the same cautions hold true here as for any role model. It is better to adapt their philosophies to your life than to adopt them .
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
"Faliure is Success if we learn from it"
Volume - 5 Issue - 4 October, 2009 (Monthly Magazine)
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Volume-5 Issue-4 October, 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
• "The way to succeed is to double your error rate."
• "Success is the ability to go from failure to failure without losing your enthusiasm." • "Success is the maximum utilization of the
ability that you have."
• We are all motivated by a keen desire for praise, and the better a man is, the more he is inspired to glory.
• Along with success comes a reputation for wisdom.
• They can because they think they can. • Nothing can stop the man with the right
Nothing can stop the man with the right mental attitude from achieving his goal; nothing on earth can help the man with the wrong mental attitude.
• Keep steadily before you the fact that all true success depends at last upon yourself.
CONTENTS
INDEX
PAGE
NEWS ARTICLE
3
Class 12 marks may become critical for IIT admission IIT-K postpones launch of its dream satellite
IITian ON THE PATH OF SUCCESS
7
Dr. Rai Mahesh Kumar Sinha
KNOW IIT-JEE
8
Previous IIT-JEE Question
XTRAEDGE TEST SERIES
45
Class XII –IIT-JEE 2010 Paper Class XII –IIT-JEE 2011 Paper
Regulars ...
DYNAMIC PHYSICS
14
8-Challenging Problems [Set# 6] Students’ Forum Physics Fundamentals Magnetic Field Gravitation
CATALYST CHEMISTRY
27
Key Concept Halogen DerivativesHalogen & Noble Gases Family Understanding: Organic Chemistry
DICEY MATHS
36
Mathematical Challenges Students’ Forum Key Concept
Limit, Continuity & Differentiability Parabola, ElLipse & Hyperbola
Study Time...
Class 12 marks may
become critical for IIT
admission
CHENNAI: Marks scored in the Class XII board examinations are likely to become a key determining factor in addition to the performance in the
nerve-wracking Joint Entrance Examination (JEE) for admission into the prestigious Indian Institutes of Technology (IITs) by 2011.
In a couple of months, a pan-IIT committee constituted by the union human resources development (HRD) ministry to suggest reforms to the JEE is expected to submit its report recommending ways and means to factor in the marks scored by students in higher secondary examinations while preparing the IIT merit list. A meeting of all IIT directors and JEE representatives held in Chennai over the weekend discussed the proposed changes. The proposal comes amidst widespread concern among top academicians over the current IIT admission system which is entirely dependant on JEE scores and ignores academic performances in board exams. The inherent weakness of such a system is that the IITs have been able to largely attract only students who have been "conditioned for the JEE" by high profile coaching centres in Kota and Hyderabad. Such students who lack "raw intelligence", as described by IIT Madras director M S Ananth in the past, are sometimes at sea after entering the campus.
"We hope to devise a methodology to compute a normalised class XII cut-off eligibility score for each educational board (CBSE, ICSE, and State Boards). If it is approved, only students who have scored this cut-off mark would become eligible to appear for the JEE and consecutively for admission," IIT Madras deputy director V G Idichandy, who is heading the committee, said on Monday. The present eligibility norm of an aggregate score of 60% in Class XII determined by the IIT standing council, as opposed to 85% recommended by a JEE review committee four years ago, is considered too low a benchmark.
"We are collecting data on Class XII results of the past four to five years from different boards in all states to base our recommendation on. Much will depend on how we compute an acceptable method to normalise the marks scored in different boards. You have nearly 40 boards of education in India," Professor Idichandy said.
However, the more difficult part will be to convince authorities of all the boards to declare Class XII results within a specified timeframe every academic year. "This will be crucial for us as we have to base the JEE on the Class XII results. I personally think that this is where a common school board, at least at the level of higher secondary education, which has been proposed by the HRD minister Kapil Sibal, will be of help in determining any all-India merit list," he contended.
Idichandy acknowledged that the JEE cannot be abolished "but we want to give as much importance as possible, for the performance of students at the school level" in the IIT admissions.
Single-digit cutoffs continue
to dog IIT
NEW DELHI: In an unforeseen effect of RTI, globally respected IITs have been stuck in a spiral of low cut-offs in their joint entrance examinations
(JEE) for the last three years even for general candidates.
Despite all their efforts to pull out of the single-digit cut-offs they had fallen into in 2007 and 2008 (1,4 & 3 and 5,0 & 3 in Maths, Physics and Chemistry, respectively), IITs could improve only marginally this year, as evident from the marks announced earlier this month. Out of the maximum possible marks of 160 in each subject in 2009, the cut-offs in Maths and Chemistry barely broke into double digits (11 marks each) while it remained a single-digit score in Physics (8 marks).
This is even after IITs abandoned the cut-off formula they had adopted in 2007 and 2008 (20 percentile or the best of the bottom 20 per cent of the candidates) and tried a new one in 2009 (average or mean of the marks of all the candidates). Such ridiculously low cut-offs have been dogging IITs ever since they found themselves at a loss to explain to the Central Information Commission the basis on which they had arrived at the respectably high cut-offs of 37, 48 and 55 in the 2006 JEE, which was the first
to be held after RTI came into force in November 2005.
In their third and latest attempt to explain the 2006 cut-offs, they set up a committee last month consisting of directors of IIT Guwahati and IIT Bombay, Gautam Barua and Dewang Khakhar, to submit a report to the Calcutta high court showing the exact calculations.
The calculations contained in the 11-page report reveal that, in a major departure from the norms of fair selection, IITs had in the 2006 JEE excluded hundreds of high-aggregate scoring candidates even before arriving at the subject cut-offs, which was meant to be the first level of screening. It is because of this serious flaw in the implementation of the 2006 formula that IITs, in their two earlier attempts before the CIC and high court, could not account for the major mismatch between the stated cut-offs (37, 48 and 55) and those yielded by the two different formulas claimed by them (while the first formula produced cut-offs of -8, -3 & -6, the second resulted in 7,4 and 6).
In a bid to bridge this wide gap, the Barua-Khakhar committee took recourse to the "iterative process", which is used to increase the cut-offs "with every iteration" to get the desired number of candidates. But while determining the cut-off of one subject through the iterative process, the committee eliminated the candidates who had high marks in the other two subjects.
Thus, although they were supposed to be calculated separately through the iterative process, the cut-off of one subject affected the cut-offs of the other two subjects. The committee did not however admit this paradox anywhere in its report.
Had the IITs implemented their belatedly-disclosed iterative procedure in a fair and
transparent manner, the cut-offs would have actually been 42, 44 and 51, thereby reducing the deviation among the three subject cut-offs to 9 marks instead of 18 marks. This would have very significantly changed the composition of the merit list in 2006.
And had they applied the iterative process in the JEE of the past three years as well, the IITs would have been able to take their low cut-offs to a more respectable level and spared themselves the embarrassment of admitting general candidates who got, for instance, 5% in Physics in last year's JEE.
Now, IIT counselling
system goes online
MUMBAI: If you've made it to an Indian Institute of Technology, you no longer need to travel to the campus to book your seat. The tech schools have decided to take the counselling process online, thus allowing students to submit their preferences a mix of streams and IITs from home. Currently, students from across the country travel to the closest IIT after they make their mark in the Joint Entrance Exam. "Now, all general category students will be allowed to submit their preferences online. However, all other candidates will have to travel to the nearest IIT campus for the same as they have to submit their certificates to us,'' said IIT-Guwahati director Gautam Barua.
The decision to conduct the counselling online was taken when the directors recently met in Chennai to discuss plans for the upcoming JEE in April 2010. In another key decision, the IIT directors agreed to centrally conduct two or more rounds of seat allocation, to ensure that seats don't go abegging.
While this year, the IITs for the first time conducted a second
round of seat allotments, it was held at the institute level. Students who took admission were offered internal betterment before the second allotment had taken place. So, if a student with a ranking of 1,104 in JEE-2009 did not take the seat allotted to him in IIT-B, another candidate with a lower ranking got his place (if he had opted for that subject and IIT-B in his preference form). Also, if a candidate signed up at IIT-Delhi in the first round, s/he were not allowed to move to say IIT-Madras or IIT-Bombay even if a slot opened there and these institutes were listed in his/her choices. "Now, we want to remove that barrier. A student will be allowed to move out of one IIT and join another, if he prefers to do so in the later rounds of seat allotment,'' added Barua. In another relief to students, the IITs have decided to put out the answer key of the entrance exam, soon after the exam ends.
HRD allows IITs to take
non-PhDs as lecturers
NEW DELHI: Close to three decades ago, the Indian Institutes of Technology (IITs) upped the bar for selecting faculty: only PhDs were allowed totake classes. Diluting that lofty standard, the HRD ministry has now allowed non-PhDs to join as lecturers. What's more shocking is that at least 10% jobs have been reserved at the lecturer's level, an obsolete term that has been scrapped from academia around the world.
Making it tough for IITs to attract talent at the level of assistant professor is another clause that mandates the tech schools to take only those with three years' experience. IIT directors fear it might result in bright students preferring to take up posts at foreign universities where a fresher begins his career as an
assistant professor and not as a lecturer. Earlier, the IITs too were taking fresh, bright PhDs at assistant professor level.
While the directive on taking non-PhDs as lecturers is optional, the directors are clueless why it was inserted. "We don't need it. The four-tier recruitment concept is regressive and I don't understand why the government needs to disturb something that is in good equilibrium," asked an IIT director, who refused to be named.
Currently, none of the IITs has faculty members who are non-PhDs, barring a few of them who joined the tech schools in the 70's when the country did not have too many PhDs. But the ministry says the decision to take non-PhDs has not been thrust upon IITs. "There is no coercion involved. Faculty crunch is a fact," one official said.
"That clause was fine at the development stage. In the early years of the IITs, when we advertised for two posts, we used to get five applications. Now we get about 40 to 50, all of who are PhDs. But even now there are vacant posts for faculty merely because we are extremely choosy about who we pick," said a dean from IIT-Bombay. But some see no harm in this optional clause. "Allowing us to take non-PhDs is just an enabling clause. But what worries most of us is the provision that does not allow us to take bright PhDs fellows as assistant professors," said Gautam Barua, director of IIT-Guwahati. Several directors are seeing red over the fact that drawing up a rule to take 10% faculty as lecturers puts them in a "peculiar not-very-good position". Whether to take a candidate as a lecturer or as an assistant professor, said another director, "must be left to the good judgment of the selection panel".
The same rules apply to other central technical institutes like Indian Institutes of Management, National Institutes of Technology and the Indian Institutes of Science Education and Research.
IIT-G ranks 10th among
top tech institutes
The Indian Institute of Technology, Guwahati, an outcome of the Assam Accord, has earned the distinction of being ranked 10th in a list of 67 top engineering and technology institutes in India. The honour attests to the relatively young institution’s impeccable academic and research credentials.
The coveted top spot has been taken by Bangalore-based Indian Institute of Science, followed by IIT Kanpur in the second place and IIT Mumbai in the third slot. According to media reports the ranking has been made taking into account citations, publications and research record available between 1999 and 2008 in the Scopus International bibliographical database.
Published in the Current Science the list has been prepared by G Pratap and BM Gupta of the National Institute of Science Communication and Information Resources and National Institute of Science, Technology and Development Studies, reports stated.
Central team to study
IIT & IIM sites
A Central team will visit the desert state next week to study the sites proposed to set up IIT-Rajasthan and IIM. The final decision about the location of the institutes is likely only after the team's visit to the sites suggested by the state-constituted Vyas committee," Vipin Chandra Sharma, principal secretary, higher education, told TOI on Tuesday.
The state government had recently sent the committee report to the Union HRD ministry, which decided to see the proposed sites before taking a final call. "An HRD expert team would visit the state next week," Sharma said. He had gone to Delhi to discuss the setting up of the institutes in the state. "The Centre wants to expedite the site selection process as the project has already been delayed," Sharma pointed out.
The previous BJP government had proposed Kota as the location for the IIT. However, the HRD ministry rejected this on the ground that Kota is not connected by air and also cited the presence of tutorials as another deterrent. After assuming power, chief minister Ashok Gehlot constituted the Vyas committee, which recommended IIT at Jodhpur and IIM at Udaipur.
Interestingly, Rajasthan is the only one among seven states where the issue of IIT location is still dragging. This, despite the fact that the state encompasses 11% of the country's land and apparently it possesses the largest land bank. One of the tiniest, Himachal Pradesh has, however, identified the land near Mandi. The other states such as Andhra Pradesh, Bihar, Madhya Pradesh, Gujarat and Punjab too have sealed land for the new IITs. IIT-R at present is functional at IIT-Kanpur, which is overburdened with the presence of the two batches of IIT-R.
Two IIT-K profs part of
Chandrayan-II mission
Banking upon the rich expertise the IIT-Kanpur has in field of research and technology, the Indian Space Research Organisation (ISRO), which plans to send a lunar rover as a part of Chandrayan-II mission to the moon in the year 2012, has handed over the responsibility ofdevelopment and testing of computer vision based autonomous 3D map generation and development and validation of kinematic traction control models (a sub-controller which will correct the path of the rover due to slip and slide) to the two professors of this prestigious institute.
Dr Ashish Dutta, Associate Professor, Dept of Mechanical Engineering at IIT-K who is working on the development and validation of kinematic traction control models said, "In 2012 ISRO plans to send a lunar rover as a part of Chandrayan-II mission to the moon. The landing module would carry a mobile robot (rover) that would emerge out of the lander to explore the surface and also perform scientific experiments."
The IIT-K is involved in the following two aspects of the Chandrayan-II mission, first is the development and testing of computer vision based autonomous 3D terrain map generation and obstacle detection algorithms for path planning and second is development and validation of kinematic traction control models (a sub-controller which will correct the path of the rover due to slip and slide) for co-ordinating the six wheels of the rover based on wheels and surface interaction, said Dr Dutta "The lunar terrain consists of loose sand, dust, craters, ash etc. It is expected that due to slip, sinkage of the wheels the rover may not function as desired and drift from its desired path or may even overturn. Hence, terrain properties strongly influence rover mobility and eventually the success of the mission", he added. Dr KS Venkatesh, Associate Professor, Department of Electrical Engineering, IIT-Kanpur is working on the visual navigation of the lunar surface. Dr Dutta further elucidated that the vision
based map generation using a single stationary camera and structured light has already been completed. The system is capable of functioning in real time with reasonable computing resources. This method is now being extended to mobile platforms where the cameras would be mounted on a prototype rover moving on an uneven terrain. In order to identify the wheel and surface interaction parameters, a one-wheel test set up has been developed to study the variation of slip, friction etc for different types of lunar like terrain conditions. A kinematic control model of a six wheel rover with a rocker-bogey mechanism has also been developed. Finally, the vision based system would give us the 3D map of the terrain based on which the traction control algorithm would give the safest path for the rover, said he.
He concluded by saying that the projects are being funded through two MoUs signed between IIT-Kanpur and VSSC (Vikram Sarabhai Space Centre). VSSC is a centre of the Department of Space, Government of India.
IIT-K postpones launch
of its dream satellite
The launch of IIT-K's ISRO funded dream project, nano satellite 'Jugnu' has been postponed to next year, Director IIT-Kanpur said. "The project designed by the students and the scientists of the institution was scheduled to be launched by the end of this year but now it has been rescheduled for some time between Jan-March next year," Sanjay Govind Dhande, Director, IIT-Kanpur said today. Ruling out any link between the satellite's schedule with ISRO's Chandrayaan moon mission, Dhande said the institute will complete the project on time. "The students engaged in the project are a bit dejected by the jolt faced by Chandrayaan but we
hope that our project would not get delayed further," Dhande said. The satellite weighing around five kilograms is 34 cm long and 10 cm broad and has been designed to collect information regarding flood and drought situations in the country.
Robots play soccer at
IIT-KGP
On Sunday students, researchers and scientists of IIT Kharagpore (IIT-KGP) had gathered to watch a game of football. As they rooted for their teams in the five-a-side match, all the players have been built by them and by students from three other technical institutes.
The techies were witness to the prestigious RoboCup, being held in India for the first time. The RoboCup Challenge @ India 2009 was held from August 28 till 30 at the IIT campus, around 120 km southwest from Kolkata.
The host of this event, IIT Kharagpur, has become the first institute in the country to obtain an approval from the International RoboCup Federation.
“This is the first time that a RoboCup Challenge is being held in South-East Asia. It is adding yet another crown to IIT’s achievements in the field of robotics,” said its coordinator Mithilesh Gurujala.
Each team spent around Rs 80,000 to built their robots. Although IIT-KGP and Hyderabad-based institutes International Institute of Information Technology were finalists, the match could not be completed on Sunday due to some technical snag.
“There was also some problem with the batteries. We’re working on it and hopefully by 1 am on Monday we’ll be able to hold the final rounds,” said, Gurujala, also the referee for the matches.
Dr. Rai Mahesh Kumar Sinha completed his graduation Allahabad University, and after that he was awarded by master degree as Master of Technology in Industrial Electronics from I.I.T., Kharagpur in 1969 then achieved Ph.D. in Computer Science, Indian Institute of Technology, Kanpur in 1973. Presently, he is working as Professor in I.I.T., Kanpur and related to various research works
Areas of Interest: • Artificial Intelligence
• Natural Language Processing, Machine Translation, Speech to Speech Translation, Indian Language Technology
• Vision, Pattern Recognition, OCR, Document Processing
• Computer Architecture Research & Projects
R.M.K. Sinha works primarily in the area of Applied Artificial Intelligence. He applies AI techniques to document processing, text recognition, computer vision, speech processing, natural language processing and in design of knowledge based systems. Intercommunicating layers of knowledge and their integration is key to his design approach. R.M.K. Sinha also applies artificial neural networks and fuzzy computing techniques in pattern recognition. In natural language processing, one of the primary aims is to design machine aids for translation from English to Indian languages & vice-versa and among Indian languages. R.M.K. Sinha's approach is based on a new concept of using Pseudo-Interlingua, word expert model utilizing Karak theory, pattern directed rule base and hybrid example base. His investigations also include exploring design and development of special parallel architectures for computer vision and natural language processing.
R.M.K. Sinha has been working on R & D for Indian Language Technology for the last three decades and his research has touched and provided direction to almost all facets of providing technological solution to the problem of overcoming the language barrier in the country. The multi-lingual GIST technology and several other packages for Indian language processing have been developed under his supervision.
Some of the major projects that have been initiated and executed / currently being executed under his supervision are the following:
• Machine Translation
• Speech to Speech Translation • Optical Character Recognition • Vision Course Projects • Spell-checker design Honours & Recognition
• Associate of UNESCO Chair in Communication: ORBICOM, Quebec, Canada.
• Senior Member Institution of Electrical and Electronic Engineers (IEEE), USA.
• Member Technical Advisory Committee of Centre for Development of Advanced Computing (CDAC), India.
• Founder President, Society for Machine Aids for Translation and Communication (SMATAC) India. • Adjudged Best CS Teacher, Asian Institute of
Technology, Bangkok, 1983.
• Invited Expert on occasion of release of CD for Hindi fonts and Web-site by Smt. Sonia Gandhi, Vigyan Bhavan, June 20, 2005.
• Member Selection Committees for IITs, Universities and Ministries.
Dr. Rai Mahesh Kumar Sinha
Ph.D. (I.I.T. Kanpur) M.Tech (I.I.T. Kharagpur) M.Sc.Tech (Allabhad Univ.)
Success Story
PHYSICS
1. A wooden log of mass M and length L is hinged by a frictionless nail at O. A bullet of mass m strikes with velocity v and sticks to it. Find angular velocity of the system immediately after the collision about O.
[IIT-2005]
m v M
O
Sol. We know that →τ = dt
L d→ ⇒ →
τ × dt = d →L When angular impulse (→τ × d ) is zero, the angular →t momentum is constant. In this case for the wooden log-bullet system, the angular impulse about O is constant. Therefore,
[angular momentum of the system]initial
= angular momentum of the system]final
⇒ mv × L = I0 × ω ...(i)
where I0 is the moment of inertia of the wooden
log-bullet system after collision about O I0 = Iwooden log + Ibullet
=
3 1
ML2 + ML2 ...(ii)
From (i) and (ii) ω = + × 2 2 mL ML 3 1 L mv ⇒ ω = + mL 3 ML mv = L ) m 3 M ( mv 3 +
2. (a) A charge of Q is uniformly distributed over a spherical volume of radius R. Obtain an expression for the energy of the system.
(b) What will be the corresponding expression for the energy needed to completely disassemble the planet earth against the gravitational pull amongst its constituent particles ?
Assume the earth to be a sphere of uniform mass density. Calculate this energy, given the product of the mass and the radius of the earth to be 2.5 × 1031
kg-m.
(c) If the same charge of Q as in part (a) above is given to a spherical conductor of the same radius R, what will be the energy of the system ? [IIT-1992] Sol. (a) In this case the electric field exists from centre of
the sphere to infinity. Potential energy is stored in electric field with energy density
dr
u =
2
1 ε0E2 (Energy/Volume)
(i) Energy stored within the sphere (U1)
Electric field at a distance r is
E = 0 4 1 πε . R3 Q . r U = 2 1 ∈0E2 = 2 0 ε 2 3 0 r R Q . 4 1 πε Volume of element dV = (4πr2)dr
Energy stored in this volume dU = U(dr) dU = (4πr2dr) 2 3 0 0 r R Q . 4 1 2 πε ε dU = 62 0 R Q . 40 1 πε .r 4dr ∴ U1 =
∫
R 0dU = πε∫
R 0 4 6 0 2 dr r R 8 Q = 6 5 R0 0 2 ] r [ R 8 Q πε U1 = R Q . 40 1 2 0 πε ...(1)(ii) Energy stored outside the sphere (U2)
Electric field at a distance r is
E = 2 0 R Q . 4 1 πε
KNOW IIT-JEE
∴ U = 2 1 ε0E2 = 2 2 0 0 R Q . 4 1 2 πε ε ∴ dU = µ . dV = (4πr2dr) πε ε 2 2 0 0 R Q . 4 1 2 dU = 0 2 8 Q πε r2 dr ∴ U2 =
∫
α RdU = 0 2 8 Q πε .∫
α R 2r dr = R 8 Q 0 2 πε ...(2) Therefore, total energy of the system isU = U1 + U2 = R 40 Q 0 2 πε +8 R Q 0 2 πε or U = 20 3 R Q 0 2 πε
(b) Comparing this with gravitational forces, the gravitational potential energy of earth will be
U = – 5 3 R GM2 by replacing Q2 by M2 and 0 4 1 πε by G. g = R GM2 ∴ G = M gR2 U = 5 3 − MgR
Therefore, energy needed to completely disassemble the earth against gravitational pull amongst its constituent particle will be given by
E = |U| = 5 3
MgR Substituting the values, we get
E =
5 3
(10m/s2) (2.5 × 1031 kg-m)
E = 1.5 × 1032 J
(c) This is the case of a charged spherical conductor of radius R, energy of which is given by =
C Q 2 1 2 or U = R 4 Q . 2 1 0 2 πε = 8 R Q 0 2 πε
3. A circular loop of radius R is bent along a diameter and given a shape as shown in figure. One of the semicircles (KNM) lies in the x-z plane and the other one (KLM) in the y-z plane with their centres at origin. Current I is flowing through each of the semicircles as shown in figure. [IIT-2000]
K I N M L I z x y
(a) a particle of charge q is released at the origin with a velocity →v = –v0iˆ . Find the instantaneous force
→
F on the particle. Assume that space is gravity free. (b) If an external uniform magnetic field B0jˆ is
applied determine the force F1 →
and F2 →
on the semicircles KLM and KNM due to the field and the net force →F on the loop.
Sol. (a) Magnetic field (→B ) at the origin = magnetic field due to semicircle KLM + Magnetic field due to other semicircle KNM ∴ →B = R 4 I µ0 (– iˆ ) + R 4 I µ0 ( jˆ ) →B = – R 4 I µ0 iˆ + R 4 I µ0 jˆ = R 4 I µ0 (– iˆ + jˆ )
∴ Magnetic force acting on the particle →F = q(→v ×→B ) = q{(–v0iˆ ) × (– iˆ + jˆ )} R 4 I µ0 →F = – kˆ R 4 I qv µ0 0 (b) →FKLM = → FKNM = → FKM
And →FKM = BI(2R) iˆ = 2BIR iˆ
→ 1
F = F = 2BIR→2 iˆ
Total force on the loop,
→
F = F + →1 F →2
or →F = 4BIR iˆ
Note : If a current carrying wire ADC (of any shape) is placed in a uniform magnetic field →B .
Then, →FADC =
→
FAC
or |→FADC| = iˆ (AC)B
From this we can conclude that net force on a current carrying loop in uniform magnetic field is zero. In the question, segments KLM and KNM also form a loop and they are also placed in a uniform magnetic field but in this case net force on the loop will not be zero. It would had been zero if the current in any of the segments was in opposite direction.
4. What will be the minimum angle of incidence such that the total internal reflection occurs on both the
surfaces? [IIT-2005]
µ1 = 2
µ2= 2
µ3 = 3
Sol. For total internal reflection on interface AB sin i = µ 1 1 2 = µ µ 2 1 = 2 2 = 2 1 ; i = 45º for total internal reflection on interface CD
sin i = µ 1 3 2 = µ µ 2 3 = 2 3 ⇒ i = 60º
⇒ The minimum angle for total internal reflection for both the interface is 60º.
5. Assume that the de Broglie wave associated with an electron can form a standing wave between the atoms arranged in a one dimensional array with nodes at each of the atomic sites. It is found that one such standing wave is formed if the distance d between the atoms of the array is 2Å. A similar standing wave is again formed if d is increased to 2.5 Å but not for any intermediate value of d. Find the energy of the electrons in electron volts and the least value of d for which the standing wave of the type described above
can form. [IIT-1997]
Sol. As nodes are formed at each of the atomic sites, hence 2Å = n λ 2 ...(1)
[Q Distance between successive nodes = λ/2] Hence from the figure
2Å N N N N N N n loops 2.5Å N N N N N N (n+1) loops λ/2 N and 2.5 Å = (n + 1) 2 λ ...(2) ∴ 2 5 . 2 = n 1 n+ , 4 5 = n 1 n+ or n = 4 Hence, from equation (1),
2Å = 4
2 λ
i.e. λ = 1Å
d will be minimum, when
n = 1, dmin = 2 λ = 2 Å 1 = 0.5 Å Now, de broglie wavelength is given be
λ = mK 2 h or K = m 2 . h 2 2 λ ∴ K = 10 2 34312 19 10 6 . 1 10 1 . 9 2 ) 10 1 ( ) 10 63 . 6 ( − − − − × × × × × × × eV = 6 . 1 1 . 9 8 ) 63 . 6 ( 2 × × × 10 2 eV = 151 eV
CHEMISTRY
6. A solution of 0.2 g of a compound containing Cu2+
and C2O42– ions on titration with 0.02 M KMnO4 in
presence of H2SO4 consumes 22.6 mL of the oxidant.
The resultant solution is neutralized with Na2CO3,
acidified with dilute acetic acid and treated with excess KI. The liberated iodine requires 11.3 mL of 0.05 M Na2S2O3 solution for complete reduction.
Find out the mole ratio of Cu2+ to C
2O42– in the
compound. Write down the balanced redox reactions involved in the above titration. [IIT-1991] Sol. The chemical equations involved in the titration of
C2O42– with MnO4– are :
MnO4– + 8H+ + 5e– → Mn2+ + 4H2O] × 2
C2O42– → 2CO2 + 2e–] × 5
2MnO4– + 5C2O42– + 16H+ →
2Mn2+ + 10CO
2 + 8H2O
From this equation, we conclude 2 mol MnO4–≡ 5
mol C2O4–. Hence,
Amount of C2O42– in the solution =
2 5 (22.6 mL) (0.02 M) = 2 5 L 1000 6 . 22 (0.02 mol L–1) = 0.00113 mol.
The chemical equations involved during the treatment of KI and the titration with Na2S2O3 are
2Cu2+ + 4I– → Cu
2I2 + I2
and I2 + 2S2O32– → 2I– + S4O62–
From these equations, we conclude 2 mol Cu2+ ≡ 4 mol I– ≡ 1 mol I
2
and 1 mol I2 ≡ 2 mol S2O32–
Now, Amount of S2O32– consumed = (11.3 mL)
(0.05 M) = L 1000 3 . 11 (0.05 mol L–1) = 1000 3 . 11 (0.05) mol = 0.000565 mol Amount of Cu2+ equivalent to the above amount of
S2O32– = 0.000565 mol Hence, 2+− 4 2 2 O C of Amount Cu of Amount = 00113 . 0 000565 . 0 = 2 1
7. Using the data given below, calculate the bond enthalpy of C–C and C–H bonds.
∆CHº(ethane) = –1556.5 kJ mol–1
∆CHº (propane) = –2117.5 kJ mol–1
C(graphite) → C(g); ∆H = 719.7 kJ mol–1
Bond enthalpy of H–H = 435.1 kJ mol–1
∆fHº(H2O, 1) = –284.5 kJ mol–1
∆fHº(CO2, g) = –393.3 kJ mol–1 [IIT-1990]
Sol. From the enthalpy of combustion of ethane and propane, we write (1) C2H6(g) + 2 7 O2(g) → 2CO2(g) + 3H2O(1) : ∆CH = 3∆fH(H2O, 1) + 2∆fH(CO2, g) – ∆fH(C2H6, g) Thus, ∆fH(C2H6,g) = – ∆CH + 3∆fH(H2O, 1)+ 2∆fH(CO2, g) = (1556.5 – 3 × 284.5 – 2 × 393.3) kJ mol–1 = – 83.6 kJ mol–1 (2) C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(1) ∆CH = 3∆f H(CO2, g)+ 4∆fH(H2O), 1) – ∆fH(C3H8, g) Thus ∆fH(C3H8, g) = –∆CH + 3∆fH(CO2, g) + 4∆fH(H2O, 1) = (2217.5 – 3 × 393.5 – 4 × 284.5) kJ mol–1 = –101.0 kJ mol–1
To calculate the εC–H and εC–C, we carry out the
following manipulations. (i) 2C(graphite) + 3H2(g) → C2H6(g) ∆H = – 83.6 kJ mol–1 2C(g) → 2C (graphite) ∆H = –2 × 719.7 kJ mol–1 6H(g) → 3H2(g) ∆H = –3 × 435.1 kJ mol–1 Add 2C(g) + 6H(g) → C2H6(g) ∆H(i) = (–83.6 – 2 × 719.7 – 3 × 435.1) kJ mol–1 = – 2828.3 kJ mol–1 (ii) 3C(graphite) + 4H2(g) → C3H8(g) ∆H = –101.0 kJ mol–1 3C(g) → 3C (graphite) ∆H = –3 × 719.7 kJ mol–1 8H(g) → 4H2(g) ∆H = – 4 × 435.1 kJ mol–1 Add 3C(g) + 8H(g) → C3H8(g) ∆H(ii) = (– 101 – 3 × 719.7 – 4 × 435.1) kJ mol–1 = – 4000.5 kJ mol–1 Now, ∆H(i) = εC–C – 6εC–H = –2828.3 kJ mol–1 ∆H(ii) = –2εC–C – 8εC–H = –4000.5 kJ mol–1 Solving for εC–C and εC–H, we get
εC–H = 414.0 kJ mol–1
and εC–H = 344.3 kJ mol–1
8. When 3.06 g of solid NH4HS is introduced into a
two-litre evacuated flask at 27ºC, 30% of the solid decomposes into gaseous ammonia and hydrogen sulphide. (i) Calculate Kc and Kp for the reaction at
27ºC. (ii) What would happen to equilibrium when more solid NH4HS is introduced into the flask ?
Sol. The reaction along with the given data is NH4HS(s) NH3(g) + H2S(g)
t = 0 3.06g (= 0.06mol) 0 0
teq 0.7 × 0.06 mol 0.3 × 0.06 mol 0.3 × 0.06 mol
= 0.018 mol = 0.018 mol
(i) The equilibrium constant KC is
KC = [NH3][H2S] = L 2 mol 018 . 0 L 2 mol 018 . 0 = 8.1 × 10–5 (mol/L)2
The equilibrium constant Kp is
Kp = Kc(RT)∆vg
= (8.1 × 10–5 mol2/L2)
(0.082 atm L mol–1 K–1) (300 K)2
= 4.90 × 10–2 atm2
(ii) There will not be any effect on the equilibrium by introducing more of solid NH4HS as the equilibrium
constant is independent of the quantity of solid. 9. A basic volatile nitrogen compound gave a foul
smelling gas when treated with choroform and alcoholic potash. A 0.295 g sample of the substance, dissolved in aqueous HCl, and treated with NaNO2
solution at 0 ºC liberated a colourless, odourless gas whose volume corresponded to 112 mL at STP. After the evolution of the gas was complete, the aqueous solution was distilled to give an organic liquid which did not contain nitrogen and which on warming with alkali and iodine gave a yellow precipitate. Identify the original substance. Assume that it contains one N
atom per molecule. [IIT-1993]
Sol. Since the compound gives a foul smellings gas on
treating with CHCl3 and alcoholic KOH, the
compound must be a primary amine. RNH2 + CHCl3 + 3KOH → gas) smelling (foulisocyanide alkylRNC + 3KCl + 3H2O ...(1)
Since the compound on treating with NaNO2/HCl at
0 ºC produce a colourless gas, the compound must be an aliphatic primary amine.
RNH2 + HNO2 → ROH + N2 + H2O
Thus, the gas produced is nitrogen.
Amount of gas liberated = 1
mol mL 22400 mL 112 − =200 1 mol From the above equation, it is obvious that
Amount of compound RNH2 =
200 1
mol If M is the molar mass of RNH2, then
M g 295 . 0 = 200 1 mol or M = 0.295 × 200 g mol–1 = 59 g mol–1.
Thus, the molar mass of alkyl group R is (59 – 16)g, i.e. 43 g mol–1. Hence, R must be C3H7.
From Eq. (2), it is obvious that the liquid obtained after distillation is ROH. Since this gives yellow precipitates with alkali and iodine (iodoform test), it must contain CH3 – C
OH group.
Hence, it is concluded that ROH is CH3 – CH – CH3
OH .
Thus, the original compound is CH3 – CH – CH3
NH2
Isopropylamine 10. A certain hydrocarbon A was found to contain 85.7
per cent carbon and 14.3 per cent hydrogen. This compound consumes 1 molar equivalent of hydrogen to give a saturated hydrocarbon B. 1.0 g of hydrocarbon A just decolourized 38.05 g of a 5 percent solution (by mass) of Br2 in CCl4. Compound
A, on oxidation with concentrated KMnO4, gave
compound C (molecular formula C4H8O) and
compound C could easily be prepared by the action of acidic aqueous mercuric sulphate on 2-butyne. Determine the molecular formula of A and deduce
the structures A, B and C. [IIT-1984]
Sol. The ratio of atoms in the compound A is C : H : : 12 7 . 85 : 1 3 . 14 : : 7.14 : 14.3 : : 1 : 2 Thus, Empirical formula of A is CH2.
Since the compound A consumes 1 mol of hydrogen, the molecule of A contains only one carbon-carbon double bond. From the data on the absorption of bromine, we can calculate the molar mass of A as shown in the following.
Mass of bromine absorbed by 1.0 g of hydrocarbon =
100 5
× 38.05 g
Mass of hydrocarbon absorbing 160 g (= 1 mol) of Br2 = ) 100 / 05 . 38 5 ( 0 . 1 × × 160 g = 84.1 g.
Hence, Molar mass of A is 84.1 g mol–1
The number of repeating CH2 group in one molecule
of A will be 6(= 84.1/14). Hence, Molecular formula of A is C6H12. Now, it is given that
) A ( 12 6H C conc.KMnO4→ ) C ( 8 4H O C + CH3COOH
The compound C is obtained by the hydration of 2-butyne. Hence, its structure obtained from the reaction is CH3C ≡ CCH3 +H2O HgSO4/H2SO4 CH3C = CHCH3 OH → CH3CCH2CH3 O 2-butanone (C) 2-butyne
Finally, the structure of A would be
C = O + HOOCCH3 CH3 CH3CH2 [O] C = CHCH3 CH3 CH3CH2 (C) 3-methylpent-2-ene (A) The structure of B is CH3CH2C = CHCH3 CH3 H2 CH3CH2CHCH2CH3 CH3 3-methyl pentane (B) Hence, Molecular formula of A is C6H12
Structure of A is CH3CH2C = CHCH3 CH3 3-methyl pent-2-ene Structure of B is CH3CH2CHCH2CH3 CH3 3-methyl pentane Structure of C is CH3CCH2CH3 O 2-butanone
MATHEMATICS
11. An unbiased die, with faces numbered 1, 2, 3, 4, 5, 6, is thrown n times and the list on n numbers showing up is noted. What is the probability that among the numbers 1, 2, 3, 4, 5, 6 only three numbers appear in
this list ? [IIT-2001]
Sol. Let us define at onto function F from A : [r1, r2 ... rn]
to B : [1, 2, 3] where r1r2 .... rn are the readings of n
throws and 1, 2, 3 are the numbers that appear in the n throws.
Number of such functions,
M = N – [n(1) – n(2) + n(3)] where N = total number of functions and
n(t) = number of function having exactly t elements in the range.
Now, N = 3n, n(1) = 3.2n, n(2) = 3, n(3) = 0
⇒ M = 3n – 3.2n + 3
Hence the total number of favourable cases = (3n – 3.2n + 3). 6C 3 ⇒ required probability = n nn 6 3 6 C ) 3 2 . 3 3 ( − + ×
12. A straight line L through the origin meets the line x + y = 1 and x + y = 3 at P and Q respectively. Through P and Q two straight lines L1 and L2 are
drawn, parallel to 2x – y = 5 and 3x + y = 5 respectively. Lines L1 and L2 intersect at R, shown
that the locus of R as L varies, is a straight line. [IIT-2002]
Sol. Let the equation of straight line L be y = mx P ≡ + + m 1 m , 1 m 1 ; Q ≡ + + m 1 m 3 , 1 m 3 Now equation of L1 : y – 2x = 1 m 2 m + − ...(1) equation of L2 : y + 3x = 1 m 9 m 3 + + ...(2) By eliminating 'm' from equation (1) and (2), we get locus of R as x – 3y + 5 = 0, which represents a straight line.
13. From a point A common tangents are drawn to the circle x2 + y2 = a2/2 and parabola y2 = 4ax. Find the area of the quadrilateral formed by the common tangents, the chord of contact of the circle and the chord of contact of the parabola. [IIT-1996] Sol. Equation of any tangent to the parabola, y2 = 4ax is
y = mx + a/m.
This line will touch the circle x2 + y2 = a2/2
A(–a, 0) C O B y D L E πx = – a /2 x = a x If 2 m a = 2 a2 (m2 + 1) ⇒ 2 m 1 = 2 1 (m2 + 1) ⇒ 2 = m4 + m2 ⇒ m4 + m2 – 2 = 0 ⇒ (m2 – 1)(m2 + 2) = 0
⇒ m2 – 1 = 0, m2 = – 2 (which is not possible).
⇒ m = ± 1
Therefore, two common tangents are y = x + a and y = –x – a
These two intersect at A(–a, 0)
The chord of contact of A(–a, 0) for the circle x2 + y2 = a2/2 is
(–a)x + 0.y = a2/2 or x = – a/2
and chord of contact of A(–a, 0) for the parabola y2 = 4ax is 0.y = 2a(x – a) or x = a Again length of BC = 2BK = 2 OB2−OK2 = 2 4 a 2 a2 2 − = 2 4 a2 = a
and we know that DE is the latus rectum of the parabola so its length is 4a.
Thus area of the trapezium
BCDE = 2 1 (BC + DE) (KL) = 2 1 (a + 4a) 2 a 3 = 4 a 15 2
14. Let V be the volume of the parallelopiped formed by the vectors → a = a1iˆ + a2jˆ + a3kˆ ; → b = b1iˆ + b2jˆ + b3kˆ → c = c1iˆ + c2jˆ + c3kˆ
If ar, br, cr, where r = 1, 2, 3 are non-negative real
numbers and
∑
= + + 3 1 r r r r b c ) a ( = 3L. Show that V ≤ L3. [IIT-2002] Sol. V = |→a.(→b×→c)| ≤ 2 3 2 2 2 1 a a a + + 2 3 2 2 2 1 b b b + + 2 3 2 2 2 1 c c c + + ...(1) Now, L = 3 ) c c c ( ) b b b ( ) a a a ( 1+ 2+ 3 + 1+ 2+ 3 + 1+ 2+ 3 [(a1 + a2 + a3) (b1 + b2 + b3) (c1 + c2 + c3)]1/3 ∴ L3 ≥ [(a 1 + a2 + a3)(b1 + b2 + b3)(c1 + c2 + c3)]..(2) Now, (a1 + a2 + a3)2 = a +12 a +22 a + 2a23 1a2 + 2a1a3 + 2a2a3 ≥ a +12 a +22 a 23 ⇒ (a1 + a2 + a3) ≥ a12+a22+a32 Similarly, (b1 + b2 + b3) ≥ b12+b22+b23 and (c1 + c2 + c3) ≥ c12+c22+c32 ∴ from (1) and (2) L3 ≥ [( 2 1 a + 2 2 a + 2 3 a )( 2 3 2 2 2 1 b b b + + )( 2 3 2 2 2 1 c c c + + )]1/3 ≥ V15. T is a prallelopiped in which A, B, C and D are vertices of one face and the just above it has corresponding vertices A´, B´, C´, D´, T is now compressed to S with face ABCD remaining same and A´, B´, C´, D´ shifted to A´´, B´´, C´´, D´´ in S. The volume of parallelopiped S is reduced to 90% of T. Prove that locus of A´´ is a plane. [IIT-2004] Sol. Let the equation of the plane ABCD be
ax + by + cz + d = 0, the point A´´ be (α, β, γ) and the height of the parallelopiped ABCD be h.
⇒ 2 2 2 b c a | d c b a | + + + γ + β + α = 90%. h ⇒ aα + bβ + cγ + d = ± 0.9h a2+b2+c2 ⇒ locus is, ax + by + cz + d = ±0.9h a2+b2+c2
Passage # 1 (Ques. 1 to 4)
The internal energy 'U' v/s PV graph where P is the pressure and V is the volume of an ideal gas filled up in a piston cylinder system is shown below
If tan φ = b then
PV φ
U
(0, a)
1. What is the atomocity and the shape of the Gaseous molecule if b = 3 and a = 2.
2. Write the relation of adiabatic index of the gas in terms of a or b or in terms of both a and b.
3. If 'a' start varying with respect to time as
a(t) = 2(3 + t) and b remains constant then draw the graph of CV v/s a where CV is the molar specific heat
at constant volume.
4. If b start varying with respect to time as
b(t) = c0 + c1t2 where c0 and c1 are positive constants
then find the slope of dt df
v/s t graph where f is the degrees of freedom for the gas.
5. A particle enters in the given magnetic field
→
B = B0kˆ where B0 is a constant with the velocity of
jˆ b iˆ a v= + →
where a and b are the positive constants. The place where the magnetic field exists and the particle moves is filled with the resistive medium then path followed by the particle is-
(Charge on particle q and mass m)
1. Circular path with radius of the circular path is r = 0 2 2 B . q b a m +
2. Helix and the pitch of the Helix is
0 B . q m 2π .a 3. Helix and the pitch of the Helix is
0 B . q m 2π .b 4. Same path as followed by circulating electrons
which is responsible for the unstable Rutherford atomic model, means spiral path of decreasing radius.
Passage # 2 (Ques. 6 to 8)
Two conducting wires are sliding in two separate portions, the details of motion are given along with the figure. If terminals a and d are grounded then
a b Y R x d v c l l Part-2 Part-1 C/R-1 C/R-2 2v C/R-3 B ⊗ B C/R – Conducting Rail R = 10 Ω vBl = 30 volts
6. Current passing through resistance R and it's direction.
7. P/d across terminals a and c.
8. Energy of deutron accelerated by potential difference across b and c.
This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety of possible twists and turns of problems in physics that would be very helpful in facing IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and we hope that this section would prove a rich resource for practicing challenging problems and enhancing the preparation level of IIT JEE aspirants.
By : Dev Sharma Director Academics, Jodhpur Branch
Physics Challenging Problems
Soluti ons wil l be published in next issue
1. As energies of light photons falling on metallic surface because of trichromatic light are 2eV, 2.8eV and 3eV
The work function of the metal is 2.5eV so light waves/photons corresponding to frequency ω2 and ω3
are able to have photo electric effect but not of frequency ω1.
so KEmax. due to light of frequency ω2 is
(KEmax.)2 = E2 – W = 2.8 – 2.5 = 0.3eV
KEmax. due to light of frequency ω3 is
(KEmax.)3 = E3 – W = 3 – 2.5 = 0.5eV
As (KEmax.)3 > (KEmax.)2 so fastest photo electron is
related with (KEmax.)3 and the stopping potential will
be 0.5 volt.
2. As light waves/photons corresponding to frequency ω1 are not able eject the photo electrons so there is no
effect on stopping potential and photo current. 3. As light waves/photons corresponding to frequency
ω2 are able to have the photo electric effect but not
ejecting the fastest moving photoelectron so, Photo current decreases but there is no effect on stopping potential.
4. As light corresponding to frequency ω3 is able to
have photo electric effect and responsible to eject fastest photoelectron so,
- Photo current decreases
- As no photo electric effect takes place due to this light intensity = 0
So stopping potential is not 0.5 volt instead of that it is 0.3 volt because now the fastest electron is due to light of frequency ω2. 5. As hydrogen atom HA1 Ground state –13.6 eV n = 1 Nucleus
Accepts energy photon of 12.1eV so the final energy = –13.6 + 12.1
= –1.5 eV corresponds to n = 3 As hydrogen atom HA2
First excited state
–3.4 eV
n = 1 Nucleus
n = 2
Accepts energy photon of 1.9eV so the final energy = –3.4 + 1.9 = –1.5 eV corresponds to n = 3 electron n = 3 n = 2 n = 1 –1.5 eV –3.4 eV –13.6 eV Position of electron in hydrogen atom HA1 and HA2 & hydrogen spectra
No. of spectral lines emitted = 2 ) 1 n ( n − = 2 ) 1 3 ( 3 − = 3
6. First and second spectral lines of Lymean series and first spectral line of Balmer series.
7. Energy of photons from hydrogen spectrum tube are- 1.9 eV Ist line of Balmer series 10.2 eV Ist line of Lymen series 12.1 eV IInd line of Lymen series If 2nd line of Lymen series get aabsorbed by absorption column then
KEmax. of fastest elect. = E – W = 10.2 – 2
= 8.2 eV
(No photo electric emission due to 1st line of Balmer series)
So stopping potential is 8.2 volt.
8. If absorption column get removed then photon of energies 1.9eV, 10.2eV and 12.1eV falls on to metal. Now, KEmax of fastest elect.
KEmax. = E – W = 12.1-2
= 10.1eV and stopping potential = 10.1volt so stopping potential increases from 8.2 volt to 10.1volt and as no. of photoelectrons ejected will be more because of simultaneous presence of 10.2eV and 12.1eV photons so photocurrent also increases.
Solution
Physics Challenging Problems
Set # 5
1. Two particles of masses m1 and m2 separated a
distance L from each other are released from their initial rest state. What will their velocity be when the distance between them is l ?
Sol. Notice that when the masses were released, the velocity of the center of mass was
cm vr = 2 1 2 1 m m m . 0 m . 0 + + = 0 ...(1)
Because both of the initial velocities are zero. Thus, the total momentum of the system is zero. We denote the velocities of the masses m1 and m2 as v1 and v2,
respectively. Using the law of conservation of linear momentum, derived from the absence of external forces, we obtain :
m1v1 = m2v2 ...(2)
The gravitational force between the masses is conservative. Calculating the potential energy between the two masses at the moment of release, we arrive at : Ep = –
∫∞
L dr F = –∫∞
L 12 2 r m Gm dr = – L m Gm1 2 ...(3) When the masses arrive at distance l we have : – L m Gm1 2 = – l 2 1m Gm + 2 1 m1v12 + 2 1 m2v22 ..(4)Using this result along with Eq. (2), we obtain
− + = − + = L l 1 m m m G 2 v L l 1 m m m G 2 v 2 1 2 1 2 2 2 2 2 1 2 1 l l ..(5)
We are also interested in the relative velocity, which can be expressed :
12
vr = vr – 1 vr = (v2 1 + v2) rˆ
We can solve this equation simply by plugging the calculated v1 and v2 into it, or as following :
|vr |12 2 = |vr – 1 vr |22 = (vr – 1 vr ). (2 vr – 1 vr ) 2 = 2 1 vr + 2 2 vr – 2vr .1 vr 2 = v + 12 v + 2v22 1v2 ...(6)
Notice that the directions of the velocities are opposite. Using Eq. (2) and Eq. (5), we have :
|vr |12 2 = 2 1 v + 2 2 v + 2 2 1 m m 2 1 v = 2 1 m m G 2 + − L 1 1 l ( 2 1 m +m + 2m22 1m2) = 2G − L 1 1 l (m1 + m2) ...(7) and therefore, |vr | = 12 (m m ) L 1 1 G 2 1+ 2 − l ..(8)
Another method of finding v12 is to use one of the
masses as the frame of reference; a0
m1 m2
for example, m1 (fig. 1). The frame of m1 is not
inertial. Its acceleration, a0, is :
a0 = 1 m F = 12 2 1 r m Gm m 1 = 22 r Gm ...(9) The force exerted on m2 in this frame, F´, is the sum
of the gravity and D'alembert's force, F´ = – 12 2 r m Gm – m2 2 2 r Gm = – Gm2(m1 + m2) 2 r 1
The difference between the initial and the final kinetic energies, ∆Ek, must equal the work done by
the force F´, ∆Ek = W,
where ∆Ek = 2 1 m2v . 22 Therefore, W = r2F.d 1 r r r r r r
∫
=∫
l− + L 2 1 2 r2 1 ) m m ( Gm dr Hence, v12 = (m m ) L 1 1 G 2 1+ 2 − l as expected2. A mass m1 is hung on an ideal (massless) spring.
Another mass m2 is connected to the other end of the
spring (see figure). The whole system is held at rest. At t = 0, m2 is released and the system falls freely due
to gravity. Assume that the natural length of the spring is L0, its initial stretched length (before t = 0)
is L and the acceleration due to gravity is g.
(i) Find the position of the centre of mass as a function of time.
Expert’s Solution for Question asked by IIT-JEE Aspirants
Students' Forum
(ii) Write the equations of motion for the two masses in the frame of the laboratory.
(iii) Find the distance between m1 and m2 as a
function of time.
m1
k m2
g
Sol. (i) Denoting the position of the center of mass at t = 0 by x0, we can write: x(t) = x0 +
2 1
gt2
where the downwards direction is defined as positive. (ii) Let us first find the force constant of the spring from the force equation of the initial state :
k = 0 1 L L g m − ...(1)
The equations of motion are :
m1 L0 m2 x1 x m1•x = m•1 1g – k(x1 – x2 – L0) ...(2) m2•x = m•2 2g + k(x1 – x2 – L0) ...(3)
Note that (x1 – x2 – L0) > 0 implies positive
acceleration (downwards) for m2, and negative
acceleration (upwards) for m1.
(iii) An easy way to solve this problem is by transforming to frame which accelerates with our system at g. In this frame, D' alernbert's force exists, balancing gravity so that we are left only with the force applied by the spring. Also, in this frame, the center of mass x´0 is at rest, so we choose x´0 is at
rest, so we choose x´0 = 0. Therefore,
2 1 2 2 1 1 m m ) t ( ´ x m ) t ( ´ x m + + = x0(t) = x0(0) = 0 ...(4)
This leads us to:
x´t(t) = – 1 2 m m x2(t) ...(5)
The distance between the two masses is given by : R(t) = R´(t) = x´1(t) – x´2(t) = x´1(t) + 2 1 m m x´1(t) = 2 2 1 m m m + x´1(t) ...(6)
Now in the current frame of reference, the equations of motion are defined as:
1 1x´ m && = – k(x´1 – x´2 – L0) ...(7) 2 2x´ m && = + k(x´1 – x´2 – L0) ...(8)
We define r(t) ≡ R(t) – L0 as the displacement from
equilibrium. From Eq. (6), we have :
+ = + = + = ) t ( ´ x m m m ) t ( r ) t ( ´ x m m m ) t ( r L – ) t ( ´ x m m m ) t ( r 1 2 2 1 1 2 2 1 0 1 2 2 1 && && & & ...(9)
From Eq. (5) we have
= = − = ) t ( x m m – ´ x ) t ( x m m – ´ x ) t ( x m m ´ x 2 1 2 1 2 1 2 1 2 1 2 1 && && & & ...(10)
Using the above six equations we can rewrite Eqs. (7-8) using only terms of r(t). The two equations will be identical, since using eq. (5) leaves us with only one degree of freedom. Therefore,
r m m m m 2 1 2 1 && + = µ r&& ...(11) where µ = 2 1 2 1 m m m m
+ is the reduced mass of the
system.
Another way to write the equation of motion is by dividing Eq. (2) by m1, dividing Eq. (3) by m2 and
subtracting Eq. (3) from Eq. (2). The resulting equation is : 1 x&& –x&& = – k2 + 2 1 m 1 m 1 (x1 – x2 – L0) = – µ k (x1 – x2 – L0) ...(12)
The solution to these equation is:
r(t) = A cos φ + t µ k ...(13)
where A and φ are determined from the initial conditions, R(0) = L0, and R(0) = 0. Hence,
R(t) = r(t) + L0 = L0 + (L – L0) cos t µ k ..(14)
Note : the transformation to the frame of the centre of mass used here is quite common in two-body problems. In general, we replaced:
m1, m2 → µ = 2 1 2 1 m m m m + x1, x2 → r = x2 – r1
This reduces the number of equations and variables. 3. A cylindrical rigid body of mass M, radius R and
moment of inertia I about its center of mass is thrown horizontaly onto a plane at a velocity v0. Initially, the
body slides and does not roll. Gradually, it starts to roll as a result of the friction of the plane, until it finally rolls at a velocity v without sliding.
(i) Compute the final velocity v. (ii) Find v for a spherical body.
(iii) Later, the sphere reaches a perfectly smooth area of the plane. Find the angular velocity and the velocity of the center of mass on the smooth area.
v0 t = 0 v t = t0 ω v´ µ = 0 ω´
Sol. (i) We use the principal of conservation of angular momentum about the point of contact between the body and the plane A to solve the problem. The torque relative to this point vanishes ( Nr = 0); so,
dt J dr = 0. + ω = + + = = R v M I p R J J R Mv J cm , in final 0 initial & r r ...(1)
We arrive at Eq. (1) by relying upon the fact that the final angular momentum about the point A equals the angular momentum in the center of mass frame, Iω, plus the angular momentum of the center of mass point about the point A. At t = 0 the only motion of the mass is the rolling; therefore, v = ωR. Hence,
Jf = I
R v
+ MvR ...(2)
Applying the principle of conservation of momentum, Ji = Jf, we obtain: v = 2 0 MR I 1 v + ...(3) (ii) For a spherical rigid body we know that I = 5 2 MR2. Therefore, v = 5 / 7 v0 = 7 5 v0 ...(4)
Note that this problem cannot be solved easily by using the principle of conservation of energy. The reason is the existence of the force of friction. When the mass stops sliding, the friction does not vanish, but it does not do any work, because the point of contact between the body and the plane, A, is temporarily at rest. Therefore,
W =
∫
f rr.dr = 0 for dr = 0. r(iii) On a frictionless surface, the linear and angular momentum are conserved. Therefore, ω and v are unchanged, or : ω = ω = ´ v ´ v
4. A hollow steel sphere, weighing 200 kg is floating on water. A weight of 10 kg is to be placed on it in order to submerge when the temperature is 20ºC. How much less weight is to be placed when temperature increases to 25ºC ?
Given γwater = 1.5 × 10–4 / ºC, αsteel = 1 × 10–5/ ºC
Sol. At the instant of submergence,
Total mass of sphere and weight placed on it
= mass of water displaced
∴ mass of water displaced at 20ºC
= (200 + 10)kg = 210 kg and
volume of the sphere = volume of water displaced by it.
∴ volume of sphere at 20ºC is V0 =
0
210 ρ where ρ0 is density of water at 20ºC.
volume of sphere at 25º C becomes equal to V = V0(1 + 3αs∆θ) = 0 210 ρ [1 + 3 × 10 –5 (25 – 20)] = 0 0315 . 210 ρ Density of water at 25ºC becomes equal to
ρ = ρ0(1 – γw∆θ) = 0.99925 ρ0
Mass of water to be displaced at 25ºC in order to submerge the sphere = V.ρ = 209.847 kg
∴ Required difference of weight to be placed on it = 210 – 209.87 = 0.126 kg Ans. 5. Suppose a nucleus X, initially at rest, undergoes
α-decay according to the equation, X
225
92 → Y + α
The emitted α-particle is found to move along a helical path in a uniform magnetic field of induction B = 5T. Radius and pitch of the helix traced by the
α-particle are R = 5 cm and p = 7.5 π cm, respectively. Calculate binding energy per nucleon of nucleus X.
Given that, m(Y) = 221.03 u m(α) = 4.003 u m(n) = 1.009 u m(p) = 1.008 u Mass of α-particle = 3 2 × 10–26 kg 1 u = 931 MeV/c2
Sol. Let velocity of emitted α-particle be v at angle θ with the direction of magnetic field.
Then radius of helical path traced by the α-particle, R = qB sin mv θ or v sin θ = m RqB = 1.2 × 107 ms–1
where q (charge of α-particle) = 3.2 × 10–19 coulomb.
and pitch, p = qB cos mv 2π θ or v cos θ = m 2 pqB π = 9 × 10 6 ms–1.
∴ velocity of emitted α-particle, v = (vcosθ)2+(vsinθ)2
= 1.5 × 107 ms–1.
When an α-particle is emitted with velocity v from a stationary nucleus X, decay product (nucleus Y) recoils. According to law of conservation of momentum, that recoil velocity V of Y is given by
myV = mav ...(1)
where mass of nucleus Y, my = 003 . 4 03 . 221 × 3 2 × 10–26 kg ∴ V = 2.715 × 105 ms–1
∴ Total energy released during α-decay of nucleus X is E = kinetic energy of nucleus Y
+ kinetic energy of α-particle
or E = 2 1 myV2 + 2 1 mαv2 = 4.77 MeV
Hence, mass lost during α-decay, ∆m = 931 E u = 0.005 u ∴ mass of nucleus, X, mx = my + mα + ∆m = 225.038 u mass defect in nucleus X,
md = [92mp + (225 – 92) mn] – mx
∴ md = 1.895 u
∴ Binding energy per nucleon in nucleus X
= 225 931 md× MeV = 7.84 MeV Ans.
Regents Physics
You Should Know
Nuclear Physics :
•
Alpha particles are the same as helium nuclei and have the symbol .•
The atomic number is equal to the number of protons (2 for alpha)•
Deuterium ( ) is an isotope of hydrogen( )
