XtraEdge for IIT-JEE 1 JANUARY 2011 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.
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"Faliure is Success if we learn from it"
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January, 2011 (Monthly Magazine)
NEXT MONTHS ATTRACTIONS
Much more IIT-JEE News.
Know IIT-JEE With 15 Best Questions of IIT-JEE Challenging Problems in Physics,, Chemistry & Maths Key Concepts & Problem Solving strategy for IIT-JEE.
Xtra Edge Test Series for JEE- 2011 & 2012 Mock Test Paper CBSE Class XII
S
Success Tips for the Months
• The greatest adventure is what lies ahead. • Fixers believe they can fix. Complainersbelieve 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
4The young innovator from IIT Kharagpur bagged first prize at IIT-Delhi fair
Engineering better managers
IITian ON THE PATH OF SUCCESS
6 Dr. Raghuram G. Rajan & Prof. Devang KhakharKNOW IIT-JEE
7Previous IIT-JEE Question
XTRAEDGE TEST SERIES
60
Class XII – IIT-JEE 2011 PaperClass XI – IIT-JEE 2012 Paper
Mock Test-2 (CBSE Board Pattern) [Class # XII] 78 Solution of Mock Test-1 (CBSE Pattern)
Regulars ...
DYNAMIC PHYSICS
16
8-Challenging Problems [Set# 9] Students’ Forum
Physics Fundamentals
Prism & Wave nature of Light Waves & Dopspler Effect
CATALYSE CHEMISTRY
33
Key ConceptCarbohydrates Salt Analysis
Understanding : Organic Chemistry
DICEY MATHS
43
Mathematical Challenges Students’ Forum Key Concept Differential Equations Trigonomatrical RatiosStudy Time...
Test Time ...
XtraEdge for IIT-JEE 4 JANUARY 2011 The young innovator from IIT
Kharagpur bagged first prize At IIT-Delhi fair
Indian Institutes of Technology (IITs) have always encouraged innovation and entrepreneurship. Shwetank Jain, studying BTech at IIT Kharagpur proved it by winning the first prize at the India Innovation Initiative (i3) National Fair by presenting his P2 Power Solutions – an Intelligent Power Conditioner with Hybrid System Integrator that economizes use of energy and enhances power quality. The fair was held at IIT Delhi and was organized by the Department of Science and Technology, Government of India, Agilent Technologies, and the Confederation of Indian Industry (CII).
Engineering better managers
“The faculty and the curriculum at DMS were excellent. Some of our courses were introduced only much later in other B-schools in the country. It was a fantastic experience,” says Saurav Sharma, alumni of IIT Delhi’s Department of Management Studies (DMS).
Sharma, now a director at United Bank of Switzerland and based in Hyderabad, is from the Class of 1997. DMS began functioning as a separate department from that year, though post-graduate management education was introduced in IIT Delhi in 1976.
Housed within the IIT Delhi campus, the institute offers, MBA (full time) with focus on management systems, MBA (full time) with focus on telecommunications system management and MBA (part-time) with focus on technology management. Apart from these, students can opt for specialisation in functional areas like finance, marketing, information technology and human resource. All IITs admit students to the two-year full time MBA programme through the Joint Management Entrance Test. “The faculty, infrastructure and courses offered here are unparalleled. Systems thinking and creative problem solving are two courses that are excellent and not taught in any other B-school,” says Tajinder Kohli, a first-year MBA student.
P.K. Jain, who teaches finance, with 35 years of teaching experience and 23 books and 120 research papers in national and international journals to his credit, is widely rated as one of the best teachers in India in his field. “Our objective is to turn a good engineer into a better engineer. We encourage students to do field work to increase their awareness of the real world work conditions,” says Vinayshil Gautam, who was the first head of the department of DMS and the founder director of IIM-Khozhikode.
Another reason why DMS is so sought after by students is the placements that they are offered. “The return on investment is really high and the brand name of IIT definitely helps. Including the hostel charges, one has to pay around Rs 2 lakh for the course. The minimum salary offered to graduates who pass out from here is at three times as much. In comparison, other comparable institutes charge 2-4 times these fees,” says Vipul Arora, a second year student and a member of the placement committee.“We find the
students of IIT-DMS very impressive. What we look for in our organisation is a combination of management skills, presentablity and communication skills,” said Vinay Tiwari, Campus Relations Manager, Citigroup. “The students here embody all these qualities.” The green campus and the canteens act as great value additions.“I love the greenery in the campus and it’s great to walk around. The ‘Uth Café’ is the most popular hangout here. From Indian to Italian, it offers all kinds of dishes and it remains abuzz till the wee hours of morning. Politics, studies, films… we discuss everything over cups of chai and coffee,” says Kavita Dara, a first year MBA student. Parivartan, the annual management festival and the magazine CHAOS (Creative and Holistic Approach to Organisational Systems) are among the various other platforms offered to students to nurture their creative spirits
IITs to assist, collaborate with other technical institutions
Bangalore: The Institutes of Technology Act, that is around forty years old, is being amended soon to allow the Indian Institutes of Technology (IITs) to take up the responsibility of supporting and collaborating with other technical education institutions and also to advise the state governments on technological problems within the zone where the IIT exists.
The Institutes of Technology (Amendments) Bill, 2010 will make it mandatory for all 15 IITs in the country to provide training, facilitate study visits, share laboratories as well as other resources with othertechnical institutions in their respective zones.
XtraEdge for IIT-JEE 5 JANUARY 2011 This will allow technical institutes,
that have been mushrooming all over the country but have often been criticized for poor standards, to collaborate with the IITs and better their educational system.
The bill has been given the go ahead by the Council of IITs. However, the parliamentary standing committee on human resource development has recommended that the clauses that make it compulsory for all IITs to collaborate with other technical institutions should not me made mandatory.
The parliamentary standing committee includes members such as Rahul Gandhi, Kanimozhi and Suresh Kalmadi. The committee had submitted a report last month regarding the proposed bill in which it had been stated that the issues of resources, capacity and faculty at the IITs must be addressed first. The report also said that the technical educational institutions must also have the vision to support and deal with the demands of the society as well as the industry.
The standing committee was headed by Oscar Fernandes and had relied heavily on the reservations of the finance ministry that had emphasized that the new clauses that were in the Institutes of Technology (Amendments) Bill were casting an obligation upon the IITs to meet the technological needs of the states where they were situated.
Innovators showcase designs at IIT Delhi fair
New Delhi: The India Innovation Initiative (i3) National Fair held at the Indian Institute of Technology, Delhi (IIT-D) showcased scientific innovations and inventions by several students from all over India.
The fair was organized in association with Agilent Technologies, Department of Science and Technology of the government of India and the Confederation of the Indian Industry. The first prize at the fair was won by Shwetank Jain, a 25 year old, whose project while pursuing B.Tech at IIT
Kharagpur helped him to gain entrepreneurial success. Jain's project was an Intelligent Power Conditioner with Hybrid System Integrator. Brainchild of Jain and his friends, the project called the P2 Power Solutions had the aim to provide innovative engineering solutions while focusing specifically on energy efficiency and power quality enhancement. The second prize at the fair was won by Nandan Kumar, Sudarshan Rajagopal and Sankamesh Ramaswamy for developing an automated machine that manufactures three-dimensional non-woven fibrous structures that have application in the medical field. Kumar, a textile engineer, informed that the cotton used in medical applications have short fibers that may be transferred to the surface of the wound and cause infection. "The fibers that we have created are hollow from inside and provide more space to absorb bodily fluids," he said. Over 50 innovators showcased their creations at the fair. They had been chosen from around 1000 entries from all over the country.
Shourie: allow IITs to work autonomously
The former Union Minister, Arun Shourie, said on Saturday that academic institutions of excellence such as the Indian Institutes of Technology should be left to work autonomously without government interference.
He was speaking at the Dr. Homi Bhabha Centenary Conclave, a three-day event that began at the Tata Institute of Fundamental Research (TIFR) here on Friday. The conclave brought together former recipients of the Homi Bhabha Fellowship. Mr. Shourie himself was a recipient.
Mr. Shourie said the TIFR, an institution of excellence, thrived because of its autonomy. “Mr. J.R.D. Tata entrusted Dr. Bhabha, and so did Pandit Nehru, with the job of setting up an institution of learning and research of the calibre of the TIFR.”
He said Dr. Bhabha’s approach was of “unobtrusive oversight.” Contrasting the TIFR with the institutions of today, Mr. Shourie said they suffered because
they were unduly “besieged by the government.” “Nine new IITs were announced despite a teacher shortage,” he said. “IIT-Rajasthan was announced two-and-a-half years ago. The head of the institute was appointed two years later. Still, the site for the institute was not decided. The Chief Minister wants it in his constituency, while his predecessor wanted it in hers. Therefore, it was decided that the IIT-Kanpur would house it, thereby burdening the parent institute.”
IIT Rajasthan PhD Programme Admission December 2010 Indian Institute of Technology (IIT), Rajasthan, has invited applications for admission into Doctorate of Philosophy (PhD) programme, to be offered in the academic session commencing from December 2010. Indian Institute of Technology (IIT) Rajasthan is one of the prestigious IITs of India. The initiative of establishing the institute was taken by Ministry of Human Resource Development (MHRD), Government of India, in 2008.
IIT students refraining from signing bonds with recruiters IIT-Madras academic affairs secretary said that most candidates these days were not willing to sign any bonds. "We have informed the recruiters that they would be required to incentivize students to make them work for them," he said.
At IIT-Bombay, companies that had insisted on students signing contracts with them had been politely declined from participating in the recruitment procedure.
IIT Joint Admission Test for M.Sc on May 8, 2011
New Delhi: The Joint Admission Test for M.Sc (JAM) for pursuing post-graduation at the Indian Institutes of Technology (IIT) will be conducted on May 8, 2011 at various test centres in the Bombay, Delhi, Guwahati, Kanpur, Kharagpur, Madras and Roorkee Zones.
XtraEdge for IIT-JEE 6 JANUARY 2011
Dr. Raghuram G. Rajan received his Bachelor’s degree in Electrical Engineering from IIT Delhi in 1985 along with the Director’s Gold Medal for all round performance. He graduated from I.I.M. Ahmedabad in 1987, where he received a gold medal for academic performance.
He joined the Graduate School of Business (GSB), University of Chicago in 1991 after obtaining a Ph.D. from MIT. In 1994, Dr. Rajan received tenure and appointment as Professor of Finance and is now the Joseph L. Gidwitz Distinguished Service Professor of Finance at the GSB. He has held a visiting chair at Northwestern (1996-97) and the Fischer Black visiting chair at M.I.T. (2000-2001).
Dr. Rajan is currently the Economic Counselor and Director of Research at the International Monetary Fund, USA. He has been on this position since 2003.
Dr. Rajan’s research interests focus primarily on economic institutions ranging from banks to property rights. He teaches courses on Corporate Finance and Banking at the GSB. His papers have been published in all the top economics and finance journals. He has also written a book with Luigi Zingales entitled Saving Capitalism from the Capitalists, which was published by Random House in February 2003.
Dr. Rajan is a Director of the American Finance Association and the Program Director for the Corporate Finance Program at the National Bureau of Economic Research. He has served as a consultant to a number of organizations including the World Bank, the Federal Reserve Board, the Reserve Bank of India, the Swedish Parliament, finance companies and banks.
In honouring Dr. Raghuram Rajan, IIT Delhi recognizes the outstanding contributions made by him as a Researcher, Educator and Financial Economist. Through his achievements, Dr. Raghuram G. Rajan has brought glory to the name of this Institute.
Prof. Devang Khakhar received his Bachelor’s Degree in Chemical Engineering from IIT Delhi in 1981. He subsequently received his Ph.D. in Chemical Engineering from the University of Massachusetts, Amherst in 1986. Prof. Khakhar is presently Professor of Chemical Engineering and Dean of Faculty at IIT Bombay.
After spending one semester at IIT Kanpur as a Visiting Faculty Member, Prof. Khakhar joined IIT Bombay in January 1987 as a Lecturer. He became a Professor at IIT Bombay in 1996. He was also a Visiting Faculty Member at Northwestern University in 1996. Prof. Khakhar served as the Head of Chemical Engineering from 2002 to 2004 and became Dean of Faculty in 2005.
Prof. Khakhar has made pioneering research contributions in a number of areas of Chemical Engineering including polymers, granular mechanics and mixing. The work has been published in top international journals such as Nature and Science. Some of the work has had a direct impact on industrial practice. Prof. Khakhar teaches courses related to fluid mechanics, granular mechanics and polymers. Prof. Khakhar has received several awards and recognitions for his work. He is a Fellow of the Indian Academic of Science, Bangalore, the Indian National Science Academy, New Delhi and the Indian National Academy of Engineering. He is a recipient of the Shanti Swarup Bhatnagar Prize, the Swarnajayanti Fellowship and the Excellence in Teaching Award of IIT Bombay. Prof. Khakhar is currently a member of the governing council of the Indian National Academy of Engineering and a member of the Science and Engineering Research Council of the Department of Science and Technology, India.
In honouring Prof. Devang Khakhar, IIT Delhi recognizes the outstanding contributions made by him as a Researcher, Scientist and Educator. Through his achievements, Prof. Devang Khakhar has brought glory to the name of the Institute.
Success Story
Success Story
This article contains stories/interviews of persons who succeed after graduation from different IITs
Dr. Raghuram G. Rajan
Electrical Engineering from IIT Delhi in 1985
Ph.D. from MIT. In 1994
Prof. Devang Khakhar
BTech from IIT- Delhi in 1981, Ph.D. in Chemical Engineering from the University of Massachusetts, in 1986.
XtraEdge for IIT-JEE 7 JANUARY 2011
PHYSICS
1. A uniform wire having mass per unit length λ is placed over a liquid surface. The wire causes the liquid to depress by y(y << a) as shown in figure. Find surface tension of liquid. Neglect end effect. [IIT-2004]
y
a a
Sol. The free body diagram of wire is given below. If l is the length of wire, then for equilibrium 2F S sin θ = W. F F mg θ θ F = S × l ⇒ 2S × l × Sin θ = λ × l × g or S = θ λ sin 2 g
also sin θ = y/a ∴ S = a y g / 2 λ = y g a 2 λ ⇒ Surface tension, S = y g a 2 +
2. A double-pane window used for insulating a room thermally from outside consists of two glass sheets each of area 1 m2 and thickness 0.01 m separated by a 0.05 m thick stagnant air space. In the steady state, the room glass inter-face and the glass-outdoor interface are at constant temperatures of 27ºC and 0ºC respectively. Calculate the rate of heat flow through the window pane. Also find the temperatures of other interfaces. Given thermal conductivities of glass and air as 0.8 and 0.08 W m–1 K–1 respectively. [IIT-1997]
Sol. I K1 II K2 III K3 H H θ1 θ2 d3 d2 d1 θB θA θA>θB At steady state First material : H = 1 1 d K A (θA – θ1) ...(i) Second material : H = 2 2 d K A (θ1 – θ2) ...(ii) Third material : H = 3 3 d K A (θ2 – θB) ...(iii) From (i) θA – θ1 = A H 1 1 K d ...(iv) From (ii) θ1 – θ2 = A H 2 2 K d ...(v) From (iii) θ2 – θB = A H 3 3 K d ...(vi) ⇒ Adding the above three equations we get
θA – θB = + + 3 3 2 2 1 1 K d K d K d A H ⇒ H = 3 3 2 2 1 1 ) – ( K d K d K d A B A + + θ θ
Substituting the values
H = 8 . 0 01 . 0 08 . 0 05 . 0 8 . 0 01 . 0 1 ) 0 – 27 ( + + = 41.54 J/s From (iv) 27 – θ1 = 1 54 . 41 × 8 . 0 01 . 0 ⇒ θ 1 = 26.48º From (vi) θ2 – 0 = 1 54 . 41 × 8 . 0 01 . 0 ⇒ θ 2 = 0.52ºC 3. Two long straight parallel wires are 2 metres part,
perpendicular to the plane of the paper (see figure). The wire A carries a current of 9.6 amps, directed into the plane of the paper. The wire B carries a current such that the magnetic field of induction at the point P, at a distance of
11 10
metre from the wire B, is zero.
KNOW IIT-JEE
XtraEdge for IIT-JEE 8 JANUARY 2011 O ⊗ 1.6 m 2 m 1.2m 10/11 m P S A B Find :
(i) The magnitude and direction of the current in B (ii) The magnitude of the magnetic field of induction
at the point S.
(iii) The force per unit length on the wire B. [IIT-1987] Sol. (i) The magnetic field at P due to current in wire A
BA = π 4 µ0 AP r lA 2 = π µ 4 0 × + × 11 10 2 6 . 9 2 ...(i) (Direction P to M)
The current in wire B should be in upward direction so as to cancel the magnetic field due to A at P. (By right hand Thumb rule)
The magnetic field at P due to current in wire B
⊗ 1.6 m 2 m 1.2m B S A M BA BB N 10/11m BB = π µ 4 0 11 10 2lB ...(ii)
From (i) and (ii) π µ 4 0 × + × 11 10 2 6 . 9 2 = π µ 4 0 × 11 10 2lB ⇒ 32 11 6 . 9 × = 10 11 × B I ⇒ IB = 32 96 = 3A
(ii) The dimensions given shows that
SA2 + SB2 = AB2 ⇒ ∠ASB = 90º
Magnetic field due to A at S
BSA = π µ 4 0 . SA A r I 2 = π µ 4 0 × 6 . 1 6 . 9 2× (Directed S to B) Magnetic field due to B at S
BSA = π µ 4 0 . SB B r I 2 = π µ 4 0 2 . 1 3 2× (Directed S to A) The resultant magnetic field
B = BSA2 +BSB2 = π µ 4 0 2 2 6 . 0 3 8 . 0 6 . 9 + = 10–7 × 13 = 1.3 × 10–6 T (iii) Force per unit length on wire
B = π µ 4 0 AB B A r I I 2 = 2 3 6 . 9 2 10–7× × × = 28.8 × 10–7 N This force will be repulsive nature.
4. In Figure S is a monochromatic point source emitting light of wavelength λ = 500 nm. A thin lens of circular shape and focal length 0.10 m is cut into two identical haves L1 and L2 by a plane passing through a diameter. The two halves are placed symmetrically about the central axis SO with a gap of 0.5 m . The distance along the axis from S to L1 and L2 is 0.15 m while that from L1 and L2 to O is 1.30 m. The screen at O is normal to SO. O A 0.5 mm Screen 1.30 m 0.15 m L2 L1 S
(i) If the third intensity maximum occurs at the point A on the screen, find the distance OA.
(ii) If the gap between L1 and L2 is reduced from its original value of 0.5 mm, will the distance OA increase, decrease, or remain the same?
[IIT1993] Sol. (i) In this case, the two identical halves of convex
lens will create two seperate images S1 and S2 of the source S. These Images (S1 and S2) will behave as two coherant sources and the further dealing will be in accordance to Young's double slit experiment. For lens L1
The object is S u = – 0.15 m
XtraEdge for IIT-JEE 9 JANUARY 2011 u L2 L1 V S1 S2 d O A D 1.3m 0.15m S O1 O2 0.3m v = ? f = + 0.1 , v 1 – u 1 = f 1 ⇒ v 1 = f 1 + u 1 = 1 . 0 1 + 15 . 0 – 1 ⇒ v 1 = 15 . 0 1 . 0 1 . 0 – 15 . 0 × = 0.1 0.15 05 . 0 × ⇒ v = 05 . 0 15 . 0 1 . 0 × = 0.3 m
∆SO1O2 and ∆SS1S2 are similar. Also the placement of O1 and O2 are symmetrical to S
∴ 2 1 2 1 O O S S = u v u+ S1S2 =
(
)
u ) O O ( v u+ 1 2 =(
)
15 . 0 3 . 0 15 . 0 + × 0.5 ×10–3 ⇒ S1S2 = d = 1.5 × 10–3 m ∴ D = 1.3 – 0.3 = 1 m The fringe widthβ = d D λ = –3 9 – 10 5 . 1 1 10 500 × × × = 3 1 × 10–3 ∴ Therefore, OA = 3β = 3 × 3 1 × 10 m = 10–3 m
(ii) If the gap between L1 and L2 i.e., O1O2 is reduced. then d will be reduced. Then the fringe width with increase and hence OA will increase.
5. A hydrogen-like atom (described by the Bohr model) is observed to emit six wavelengths, originating from all possible transitions between a group of levels. These levels have energies between – 0.85 eV and – 0.544 eV (including both these values).
(a) Find the atomic number of the atom.
(b) Calculate the smallest wavelength emitted in these transitions.
(Take hc = 1240 eV-nm, ground state energy of hydrogen atom = – 13.6 eV) [IIT-2002]
Sol. If x is the difference in quantum number of the states
then x+1C 2 = 6 ⇒ x = 3 Smallest λ n n +3 Now, we have 2 2 n ) eV 6 . 13 ( Z – = – 0.85 eV ...(i) and 2 2 ) 3 n ( ) eV 6 . 13 ( Z – + = – 0.544 eV ...(ii) Solving (i) and (ii) we get n = 12 and Z = 3
(b) Smallest wavelength λ is given by λ
hc
= (0.85 – 0.544) eV Solving, we get λ ≈ 4052 nm.
CHEMISTRY
6. An ideal gas having initial pressure P, volume V and
temperature T is allowed to expand adiabatically until its volume becomes 5.66 V, while its temperature falls to T/2.
(a) How many degrees of freedom do the gas molecules have ?
(b) Obtain the work done by the gas during the expansion as a function of the initial pressure P and volume V. [IIT-1990] Sol. (a) According to adiabatic gas equation,
TVγ–1 = constant or T1V1γ–1 = T2V2γ–1 Here, T1 = T ; T2 = T/2 V1 = V and V2 = 5.66 V Hence, TVγ–1 = 2 T × (5.66V)γ–1 = 2 T × (5.66)γ–1 × Vγ–1 or (5.66)γ–1= 2 ..(1) Taking log, (γ – 1)log 5.66 = log 2 or γ – 1 = 66 . 5 log 2 log = 7528 . 0 3010 . 0 = 0.4 or γ = 1.4
If f, be the number of degrees of freedom, then γ = 1 + f 2 or 1.4 = 1 + f 2 or f 2 = 1.4 – 1 = 0.4
XtraEdge for IIT-JEE 10 JANUARY 2011 or f = 4 . 0 2 = 5
(b) According to adiabatic gas equation,
P1V1γ = P2V2γ Here, P1 = P V1 = V V2 = 5.66 V Hence, PVγ = P2 × (5.66V)γ = P2 ×(5.66)γ × Vγ or P2 = γ ) 66 . 5 ( P = 4 . 1 ) 66 . 5 ( P = 32 . 11 P [using eq.(1)]
Hence, work done by the gas during adiabatic expansion = 1 – 2 2 1 1 γ − VP V P = 1 – 4 . 1 66 . 5 32 . 11 V P PV− × = 4 . 0 2 PV PV − = 4 . 0 2× PV = 1.25 PV
7. (a) Write the chemical reaction associated with the "brown ring test".
(b) Draw the structures of [Co(NH3)6]3+, [Ni(CN)4]2– and [Ni(CO)4]. Write the hybridization of atomic orbital of the transition metal in each case.
(c) An aqueous blue coloured solution of a transition metal sulphate reacts with H2S in acidic medium to give a black precipitate A, which is insoluble in warm
aqueous solution of KOH. The blue solution on treatment with KI in weakly acidic medium, turns yellow and produces a white precipitate B. Identify the transition metal ion. Write the chemical reaction involved in the formation of A and B. [IIT-2000] Sol. (a) NaNO3 + H2SO4 → NaHSO4 + HNO3
2HNO3 + 6FeSO4 + 3H2SO4 →
3Fe2(SO4)3 + 2NO + 4H2O [Fe(H2O)6]SO4.H2O + NO
Ferrous Sulphate
→ [Fe(H2O)5NO] SO4 + 2H2O
(Brown ring)
(b) In [Co(NH3)6]3+ cobalt is present as Co3+ and its coordination number is six.
Co27 = 1s1, 2s22p6, 3s23p63d7, 4s2 Co3+ion = 1s2, 2s22p6, 3s23p63d6 3d 4s 4p 3d 4s 4p d2sp3 hybridization Hence Co3+ion in Complex ion 3+ Co NH3 NH3 NH3 H3N H3N NH3 or H3N H3N NH3 NH 3 NH3 NH3 Co3+
In [Ni(CN)42– nickel is present as Ni2+ ion and its coordination numbers is four
Ni28 =1s2, 2s22p6, 3s23p63d8, 4s2 Ni2+ ion = 1s2, 2s22p6, 3s23p63d8 3d 4s 4p 3d 4s 4p dsp2 hybridization Ni2+ ion = Ni2+ion in Complex ion
Hence structure of [Ni(CN)4]2– is
Ni2+
N ≡ C
N ≡ C
C ≡ N
C ≡ N
In [Ni(CO)4, nickel is present as Ni atom i.e. its oxidation number is zero and coordination number is four.
3d 4s 4p
sp3 hybridization
Ni in Complex
Its structure is as follows :
Ni CO
CO
CO OC
(c) The transition metal is Cu2+. The compound is CuSO4.5H2O
CuSO4 + H2S Acidicmedium→ ppt Black CuS↓ + H2SO4 2CuSO4 + 4KI → Cu2I2 + I2 + 2K2SO4 (B) white I2 + I– → I3– (yellow solution)
XtraEdge for IIT-JEE 11 JANUARY 2011 8. An organic compound A, C6H10O on reaction with
CH3MgBr followed by acid treatment gives compound B. The compound B on ozonolysis gives compound C, which in presence of a base gives 1-acetyl cyclopentene D. The compound B on reaction with HBr gives compound E. Write the structures of A, B, C and E. Show how D is formed from C.
[IIT-2000] Sol. The given reaction can be summarised as below :
[A] C6H10O (i) CH3MgBr (ii) H+ O3 [B] [C] OH– COCH3 HBr [E] [D]
Conclusions from the set of reactions
(i) Carbon-hydrogen ratio of [A] indicates that it is cyclic compound
(ii) Reaction of [A] with CH3MgBr indicates that it should have a ketonic group.
(iii) As [B] undergoes ozonolysis to form [C], It must have a double bond, and [C] must have two carbonyl groups.
(iv) Reaction of [C] a dicarbonyl compound) with a base gives a cyclic compound, it indicates that intramolecular condensation have occurred during this conversion. Thus [A] is cyclohexanone which explains all the given reactions.
(A) (i) CH3MgBr (ii) H+ O3 OH– [C] CHO O CH3 (B) O –H2O HBr Br COCH3 OH COCH3 ; B H3C (D) (E)
9. A small quantity of solution containing 24Na radio-nuclide (half life 15 hours) of activity 1.0 micro-curie is injected into the blood of a person. A sample of the blood of volume 1 cm3 taken after 5 hours shows an activity of 296 disintegrations per minute. Determine the total volume of blood in the body of the person. Assume that the radioactive solution mixes uniformly in the blood of the person. (1 curie = 3.7 × 1010 disintegrations per second). [IIT-1994] Sol. Given that,
t1/2 = 15 hours = 15 × 3600 s – dt dN = 1 micro-curie = 1 × 10–6 × 3.7 × 1010 disintegrations s–1 = 3.7 × 104 disintegrations s–1 – dt dN´= 296 disintegrations min–1 = 60 296 disintegrations s–1 t = 5 hours = 5 × 3600 s Now, t1/2 = λ 693 . 0 or λ = 2 / 1 693 . 0 t = 15 3600 693 . 0 × s –1 = 1.283 × 10–5 s–1 and, – dt dN = λN 0 or 3.7 × 104 = 1.283 × 10–5 × N 0 or N0 = 5 4 10 283 . 1 10 7 . 3 − × × = 2.884 × 109 After 5 hours, N = N0e–λt = 2.884 × 109 × e−1.283×10−5×5×3600 = 2.884 × 109 × e–0.23 = 2.289 × 109 Again,– dt dN´= λN´ 0 or 60 296 = 1.283 × 10–5 × N´ 0 or N´0 = 5 10 238 . 1 60 296 − × × = 3.845 × 10 5 If total volume of blood in the body be V, then
N N0´ = V 1 or V = 0 ´ N N = 5 9 10 845 . 3 10 289 . 2 × × cm3 = 5953 cm3 = 5.953 litres
10. An organic compound (A), C6H6O, gives a specific colour with FeCl3 solution and on heating with phthalic anhydride and conc. H2SO4 gives a white solid (B) which gives red colour with dilute NaOH (C). Compound (A) on heating with CCl4 and NaOH gives (D), which on acidification gives (E). Compound (E) gives violet colour with FeCl3. (E) on heating with (A) in presence of POCl3 gives (F), which is used as intestinal antiseptic. What are (A) to (F) ? Explain giving balanced equations.
Sol. (i) As it gives colour with FeCl3, hence it is phenol. (ii) On heating with phthalic anhydride it gives phenolphthalein (B) which gives red colour of sodium salt of (B), i.e(C).
(iii) (A) undergoes Reimer-Tieman reaction with CCl4 and NaOH giving disodium salt (D), which on acidification gives (E) salicylic acid because violet colour is given by salicylic acid.
(iv) (E) on heating with phenol (A) gives salol (F) which is used as an intestinal antiseptic.
XtraEdge for IIT-JEE 12 JANUARY 2011 C — O HO OH H H O C O H2SO4 ∆; –H2O C — O HO OH C O 2NaOH C NaO O COO–Na+ OH + CCl4 + 6NaOH ∆ (A) ONa (D) COONa 2HCl –2NaCl OH (E) COOH (A) C6H5OH POCl3/∆; – H2O OH (F) COOC6H5
MATHEMATICS
11. Sixteen players S1, S2, ... , S16 play in a tournament. They are divided into eight pairs at random from each pair a winner is decided on the basis of a game played between the two players of the pair. Assume that all the players are of equal strength.
(a) Find the probability that the players S1 is among the eight winners.
(b) Find the probability that exactly one of the two players S1 and S2 is among the eight winners.
[IIT-1997]
Sol. (a) Probability of S1 to be among the eight winners = (Probability of S1 being a pair) × (Probability of S1 winning in the group).
= 1 × 2 1 = 2 1 {Q S1 is definitely in a group}
(b) If S1 and S2 are in the same pair then exactly one wins.
If S1 and S2 are in two pairs separately then exactly one of S1and S2 will be among the eight winners. If
S1 wins and S2 loses or S1 loses and S2 wins.
Now the probability of S1, S2 being in the same pair and one wins.
= (Probability of S1, S2 being the same pair)
× (Probability of anyone winning in the pair). And the probability of S1, S2 being the same pair
= ) ( ) ( S n E n
Where n(E) = the number of ways in which 16
persons can be divided in 8 pairs. ∴ n(E) = ! 7 . ) ! 2 ( )! 14 ( 7 and n(S) = (2!) .8! )! 16 ( 8
∴ Probability of S1 and S2 being in the same pair = )! 16 !.( 7 . ) ! 2 ( ! 8 . ) ! 2 )!.( 14 ( 7 8 = 15 1
The probability of any one wining in the pairs of S1,
S2 = P (certain event) = 1
∴ The pairs of S1, S2 being in two pairs sparately and S1 wins, S2 loses + The probability of S1, S2 being in two pairs separately and S1 loses, S2 wins. = ! 8 . ) ! 2 ( )! 16 ( ! 7 . ) ! 2 ( )! 14 ( – 1 8 7 × 2 1 × 2 1 + ! 8 . ) ! 2 ( )! 16 ( ! 7 . ) ! 2 ( )! 14 ( – 1 8 7 × 2 1 × 2 1 = 2 1 × )! 14 ( 15 )! 14 ( 14 × × = 15 7 ∴ Required Probability = 15 1 + 15 7 = 15 8
12. Find the range of values of t for which
2sin t = 1 – 2 – 3 5 2 – 1 2 2 x x x x+ , t ∈ π π 2 , 2 – . [IIT-2005]
Sol. Here, 2 sin t =
1 – 2 – 3 5 2 – 1 2 2 x x x x+ , t ∈ π π 2 , 2 – Put, 2sin t = y ⇒ – 2 ≤ y ≤ 2 ∴ y = 1 – 2 – 3 5 2 – 1 2 2 x x x x+ ⇒ (3y – 5)x2 – 2x(y – 1) – ( y + 1) = 0 Since x ∈ R – {1, – 1/3} {as, 3x2 –2x – 1 ≠ 0} ∴ D ≥ 0 ⇒ 4(y – 1)2 + 4(3y – 5) (y +1) ≥ 0 ⇒ y2 – y – 1 ≥ 0 ⇒ 2 2 1 – y – 4 5 ≥ 0
XtraEdge for IIT-JEE 13 JANUARY 2011 ⇒ 2 5 – 2 1 – y + 2 5 2 1 – y ≥ 0 ⇒ y ≤ 2 5 – 1 or y ≥ 2 5 1+ or 2sin t ≤ 2 5 – 1 or 2sin t ≥ 2 5 1+ ⇒ sin t ≤ π 10 – sin or sin t ≥ π 10 3 sin ⇒ t ≤ – 10 π or t ≥ 10 3π
Thus, Range for t ∈
π π 10 ,– 2 – ∪ π π 2 , 10 3
13. If f (x) is twice differentiable function such that f (a) = 0, f (b) = 2, f (c) = 1, f (d) = 2, f (e) = 0, where a < b < c < d < e, then the minimum number of
zero's of g(x) = {f ' (x)}2 + f '' (x). f (x) in the interval [a, e] is? [IIT-2006] Sol. Let, g(x) =
dx d
[f (x) . f '(x)]
to get the zero of g(x) we take function h(x) = f (x). f ' (x)
between any two roots of h(x) there lies at least one
root of h' (x) = 0
⇒ g(x) = 0 ⇒ h(x) = 0
⇒ f (x) = 0 or f ' (x) = 0
If f (x) = 0 has 4 minimum solutions. f ' (x) = 0 has 3 minimum solutions. h(x) = 0 has 7 minimum solutions.
⇒ h' (x) = g(x) = 0 has 6 minimum solutions
14. The value of
∫
∫
1 0 101 50 1 0 100 50 ) – 1 ( ) – 1 ( ) 5050 ( dx x dx x [IIT-2006] Sol. Let I1 =∫
x dx 1 0 100 50) – 1 ( and I2 =∫
1 0 101 50) – 1 ( x dx, using Integration by partsI2 =[(1–x50)101.x]10 – (101)
∫
1 0 49 100 50) (–50. ). – 1 ( x x x dx = 0 –∫
1 0 50 100 50) (– ) – 1 )( 101 )( 50 ( x x dx – I2 = 5050∫
1 0 100 50) – 1 ( x (– x50) dx 5050 I1 – I2 = 5050∫
1 0 100 50) – 1 ( x dx + 5050∫
1 0 100 50) – 1 ( x (– x50) dx = 5050∫
1 0 101 50) – 1 ( x dx = 5050 I2 ∴ 5050 I1 =I2 + 5050 I2 ∴ 2 1 ) 5050 ( I I = 5051.15. The position vectors of the vertices A, B and C of a tetrahedron ABCD are i + ^ ^j + ^k , i and 3^ ^i , respectively. The altitude from vertex D to the opposite face ABC meets the median line through A of the triangle ABC at a point E. If the length of the side AD is 4 and the volume of the tetrahedron is
3 2 2
, find the position vector of the point E for all its possible positions. [IIT-1996] Sol. F is mid-point of BC i.e., F ≡
2 3^ ^ i i+ = 2^i and AE ⊥ DE (given) A(i + j + k) D C(3i) F(2i) B(i) E 1 λ
Let E divides AF in λ : 1. The position vector of E is given by 1 ) ( 1 2 ^ ^ ^ ^ + λ + + + λi i j k = ^ 1 1 2 i + λ + λ + ^ 1 1 j + λ + ^ 1 1 k + λ
Now, volume of the tetrahedron =
3
1 (area of the base) (height)
⇒ 3
2 2 =
3
1 (area of the ∆ABC) (DE)
But area of the ∆ABC = |( )| 2
1 → →
× BA
XtraEdge for IIT-JEE 14 JANUARY 2011 = |2 ( )| 2 1 ^ ^ ^ k j i× + = |i^×j^+^i×^k)| = |^k–j^)| = 2 Therefore, 3 2 2 = 3 1 ( 2) (DE) ⇒ DE = 2
Since ∆ADE is a right angle triangle,
AD2 = AE2 + DE2 ⇒ (4)2 = AE2 + (2)2 ⇒ AE2 = 12 But AE→ = ^ 1 1 2 i + λ + λ + ^ 1 1 j + λ + ^ 1 1 k + λ –( ) ^ ^ ^ k j i+ + = ^ 1i + λ λ – ^ 1j + λ λ – ^ 1k + λ λ ⇒ |AE→ |2 = 2 ) 1 ( 1 + λ [λ 2 + λ2 + λ2] = 2 2 ) 1 ( 3 + λ λ Therefore, 12 = 2 2 ) 1 ( 3 + λ λ ⇒ 4(λ + 1)2 = λ2 ⇒ 4λ2 + 4 + 8λ = λ2 ⇒ 3λ2 + 8λ + 4 = 0 ⇒ 3λ2 + 6λ + 2λ + 4 = 0 ⇒ 3λ(λ + 2) + 2(λ + 2) = 0 ⇒ (3λ + 2) (λ + 2) = 0 ⇒ λ = – 2/3, λ = – 2
Therefore, when λ = – 2/3, position vector of E is given by ^ 1 1 2 i + λ + λ + ^ 1 1 j + λ + ^ 1 1 k + λ = ^ 1 3 / 2 – 1 ) 3 / 2 .(– 2 i + + + ^ 1 3 / 2 – 1 j + + ^ 1 3 / 2 – 1 k + = ^ 3 3 2 – 1 3 / 4 – i + + + ^ 3 3 2 – 1 j + + ^ 3 3 2 – 1 k + = ^ 3 / 1 3 3 4 – i + + ^ 3 / 1 1 j + ^ 3 / 1 1 k = – ^i + 3 j^+ 3k^ and when λ = – 2
Position vector of E is given by, ^ 1 2 – 1 ) 2 (– 2 i + + × + ^ 1 2 – 1 j + = ^ 1 2 – 1 k + = ^ 1 – 1 4 – + i – ^j–^k = 3 i^ – ^j – ^k Therefore, –^i + 3 j^+ 3k^ and +3 i^ – j^– k^ are the
position vector of E.
MOTIVATION
• Pull the string, and it will follow wherever you wish. Push it, and it will go nowhere at all. • Be the change that you want to see in the
world.
• Efficiency is doing things right; effectiveness is doing the right things.
• Formula for success: under promise and over deliver.
• A life spent making mistakes is not only more honorable, but more useful than a life spent doing nothing.
• Discovery consists of seeing what everybody has seen and thinking what nobody else has thought.
• The best way to teach people is by telling a story.
• If you'll not settle for anything less than your best, you will be amazed at what you can accomplish in your lives.
• I had to pick myself up and get on with it, do it all over again, only even better this time. • Improvement begins with I.
• Success depends upon previous preparation, and without such preparation there is sure to be failure.
• The man of virtue makes the difficulty to be overcome his first business, and success only a subsequent consideration.
• As a general rule the most successful man in life is the man who has the best information. • The secret of success is constancy to purpose. • One secret of success in life is for a man to be
XtraEdge for IIT-JEE 15 JANUARY 2011
XtraEdge for IIT-JEE 16 JANUARY 2011 1. Four infinite thin current carrying sheets are placed in
YZ plane. The 2D view of the arrangement is as shown in fig. Direction of current has also been shown in the figure. The linear current density. i.e. current per unit width in the four sheets are I, 2I, 3I and 4I, respectively.
x x x x x x x x Y I II III IV X a a a a
The magnetic field as a function of x is best represented by
2. Match the column
Column – I Column – II
(A) a charge particle is (P) Velocity of the moving in uniform particle may be electric and magnetic constant fields in gravity free
space
(B) a charge particle is (Q) Path of the particle moving in uniform may be straight line electric, magnetic
and gravitational fields
(C) a charge particle is (R) Path of the particle moving in uniform may be circular magnetic and
gravitational fields (where electric field is zero)
(D) A charge particle is (S) Path of the particle moving in only may be helical uniform electric field
(T) None
3. Magnetic flux in a circular coil of resistance 10 Ω changes with time as shown in fig. Cross indicates a direction perpendicular to paper inwards.
Match the following
Column – I Column – II (A) At 1s, induced current is (P) Clockwise (B) At 5s, induced current is (Q) Anticlockwise (C) At 9s, induced current is (R) Zero
(D) At 15s, induced current is (S) 2A (T) None This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety
of possible twists and turns of problems in physics that would be very helpful in facing IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and we hope that this section would prove a rich resource for practicing challenging problems and enhancing the preparation level of IIT JEE aspirants.
By : Dev Sharma Director Academics, Jodhpur Branch
Physics Challenging Problems
Solutions will be published in next issue
XtraEdge for IIT-JEE 17 JANUARY 2011 4. A conducting rod of length l is moved at constant
velocity v0 on two parallel, concudting, smooth, fixed rails, which are placed in a uniform constant magnetic field B perpendicular to the plane of the rails as shown in figure. A resistance R is connected between the two ends of the rail. Then which of the following is/are correct ?
(A) The thermal power dissipated in the resistor is equal to the rate of work done by an external person pulling the rod
(B) If applied external force is doubled, then a part of the external power increases the velocity of the rod
(C) Lenz’s law is not satisfied if the rod is accelerated by an external force
(D) If resistance R is doubled, then power required to maintain the constant velocity v0 becomes half 5. The x-z plane separates two media A & B of
refractive indices µ1=1.5 &µ2 =2. A ray of light travels from A to B. Its directions in the two media are given by unit vectors µ→1=ai∧+b∧,jµ→2 =ci∧+b∧j.
Then (A) 3 4 c a = (B) 4 3 c a = (C) 3 4 d b = (D) 4 3 d b =
6. Two converging lenses of the same focal length f are separated by distance 2f. The axis of the second lens is inclined at angle θ=60º with respect to the axis of the first lens. A parallel paraxial beam of light is incident from left side of the lens. Then
(A) Final image after all possible refraction will formed at optical centre of first lens
(B) Final image after all possible refraction will formed at optical centre of second lens
(C) Final image after all possible refraction will formed at distance f from second lens
(C) Final image after all possible refraction will formed at distance f from first lens
7. If Cv for an ideal gas is given by Cv = (3 + 2T)R, where T is absolute temperature of gas, then the equation of adiabatic process for this gas is
(A) VT2 = constant (B) VT3e-2T = C (C) VT2e2T = constant (D) VT3e2T = constant
8. The pressure of one mole of ideal gas varies according to the law 2
0 V
P
P= −α where P0 & α are positive constant constants. The highest temperature that gas may attain:
(A) 2 / 1 0 0 3 P R 3 P 2 α (B) 2 / 1 0 0 3 P R 2 P 3 α (C) 2 / 1 0 0 3 P R P α (D) 2 / 1 0 0 P R P α
Physics Facts
1. Due to gravity, the maximum speed a raindrop during a rain with falling speed can hit you is about 18 miles per hour (29 kilometers per hour).
2. The speed of light in meters is 299,792,458 meters per second. And how on Earth are you going to remember that? The number can be remembered from the number of letters in each word of the following phrase: "We guarantee certainty, clearly referring to this light mnemonic."
(The speed of light in miles per second is 186,282.397051221, or in miles per hour, 670,616,629.384395).
3. In air, at a temperature of 32 degrees Fahrenheit/0 degrees Celsius (freezing point of water) the speed of sound travels 1,087 feet (331 meters) per second. (It travels faster at higher temperatures).
(In 64 degrees Fahrenheit [18 degrees Celsius] the speed of sound travels 1,123 feet [342 meters] per second).
4. If an object floats on water, it displaces the water equal to its mass, but if the object sinks, it displaces water equal to its volume.
5. A calorie is defined as the amount of energy needed to raise one gram of water one degree Celsius (or from 14.5 degrees Celsius to 15.5 degrees Celsius).
XtraEdge for IIT-JEE 18 JANUARY 2011 1. As shown in graph, the relation of U v/s PV is linear So,
a PV
U =(tanφ). + as b(tanφ)=
So, U = b, PV + a
Using ideal gas equation PV = nRT
U = b (nRT) + a Differentitate it, dU = nbRdT As dU = n CV dT So CV =bR= f R⇒ f =b 2 2
Degrees of freedom of the gas, as b = 3 So f = 6
Degrees of freedom are 6 so it is triatomic non-linear gas. 2. b f 2 2 1 2 1+ ⇒γ= + = γ as f = 2b So, γ=1+b−1
3. As CV = bR it is not dependent on ‘a’ so if a varies
there is no change in the value of CV
a CV 4. As CV= b R and , 2 2 1 0 Ct C b f = = + 2 1 0 2 2C Ct f = + and Ct dt df 1 4 =
Df/dt v/s t graph is a straight line with slope 4C1.
5. As →v and →B are mutually perpendicular so path will be circular but due to presence of resistive medium speed decreases and radius of circular path decreases. So, path is spiral of decreasing radius
Option (D) is correct
6. If vab= 0, then irrespective to the value of capacitance
C energy stored will be zero.
vab = 2 1 2 2 1 1 / 1 / 1 ) / ( / R R R R + ε − + ε = 0 So, 1 1 R ε – 1 1 R ε = 0 ⇒ 1 1 R ε = 2 2 R ε ⇒ 2 1 R R = 2 1 ε ε ε2 R R1 R2 ε1 C
Option (A) is correct 7. To calculate time constant
Replace voltage source by short circuit mean by zero resistance and then find Req with C and time constant τ = Req.C τ = + +R R R R R 2 1 2 1 C R R1 R2 C
⇒
R2 R1R2/R1+R2 C Option (D) is correct8. Maximum current through the resistance Imax = R vab = R R R R R1 2 2 1 2 1/ ) ( / )/1/ 1/ (ε + −ε + = R eq ε Option (C) is correct.
Solution
Physics Challenging Problems
Set # 8
XtraEdge for IIT-JEE 19 JANUARY 2011 1. A solid uniform sphere of radius r rolls without
sliding along the inner surface of a fixed spherical shell of radius R and performs small oscillations. Calculate period of these oscillations.
Sol. Let the solid sphere be rotated through a very small angle θ, then displacement of its centre is equal to
AB = r.θ. During this rotation point of contact changes from C to C' and inclination of radius OC' of the shell with vertical becomes equal to
φ = ) – (R r radius AB arc as shown in Figure. B φ C' A C O ∴ φ = ) – (R r rθ ...(i) When the solid sphere is released from this position, it starts to roll down towards equilibrium position C. Let its angular acceleration be α. Then acceleration of its centre will be equal to rα.
Consider free body diagram of the sphere Figure.
mg cos φ Iα mg sin φ Friction C' N
since, C ' is instantaneous axis of rotation, therefore, taking moments about it,
(mg sin φ) r = Iα or mgr . sinφ = 2 5 7mr α
But φ is very small, therefore, sin φ ≈ φ ∴ mgr . φ = 5 7 mr2α or α = 5 7 r g φ
Substituting value of φ from equation (1),
Angular acceleration, α = ) – ( 7 5 r R g θ
Since, angular acceleration is restoring and is directly proportional to angular displacement θ, therefore, the sphere performs SHM. Hence, its period of oscillation is given by T = on accelerati angular nt displaceme angular 2π or T = g r R 5 ) – ( 7 2π Ans.
2. Three identically charged, small spheres each of mass
m are suspended from a common point by insulated light strings each of length l. The spheres are always on vertices of an equilateral triangle of length of the sides x (<< l). Calculate the rate dq/dt with which charge on each sphere increases if length of the sides of the equilateral triangle increases slowly according to law dt dx = x a .
Sol. Since dx/dt is given as
x
a , therefore, to calculate dt
dq
, first q is to be calculated in terms of x.
Since, side of equilateral triangle is x, therefore, electrostatic force between two particles is
F ' = 0 4 1 πε 2 2 x q
Positions of particles A, B & C with respect to each other are shown in Figure (A).
F F ' F ' A x x G C 60º 60º B Fig.(A)
Resultant electrostatic force on particle A is
F = 2F' cos 30º.
Expert’s Solution for Question asked by IIT-JEE Aspirants
Students' Forum
XtraEdge for IIT-JEE 20 JANUARY 2011 or F = 2 0 2 4 3 x q πε
In Figure (A) G is centroid of the triangle. Distance of centroid G from every particle is
r =
3
2x cos 30º = 3
x
Since, G is centroid, therefore, it always lies vertically below the point of suspension. Figure (B) shows a vertical plane passing through particle A and centroid G. All of the forces acting on particle A are shown in this figure.
G r T x mg F Fig.(B) θ
Considering horizontal forces on the particle A,
T sin θ = F ...(i)
For vertical forces, T cos θ = mg ...(ii) Dividing equation (i) by (ii)
tan θ =
mg
F ...(iii)
Since, x < < l, therefore, inclination θ of each thread with the vertical is very small.
Hence, tan θ ≈ sin θ =
l r = 3 l x
∴ From equation (iii)
3 l x = 2 0 2 4 3 mgx q πε or q = l mg 3 4πε0 x3/2 ∴ dt dq = xt dx x l mg 2 3 3 4πε0 or dt qd = l mga2 0 3πε Ans.
3. A coil of radius R carries current I. Another concentric coil of radius r(r < < R) carries current i. Planes of two coils are mutually perpendicular and both the coils are free to rotate about common diameter. Find maximum kinetic energy of smaller coil when both the coils are released, masses of coils are M and m respectively.
Sol. If a magnetic dipole having moment M be rotated through angle 'θ' from equilibrium position in a uniform magnetic field B, work done on it is W = MB (1 – cos θ). This work is stored in the system in the form of energy. When system is released, dipole starts to rotate to occupy equilibrium position and the energy converts into kinetic energy and kinetic energy of the system is maximum when stored energy is completely released.
Magnetic induction, at centres due to current in larger coil is B = R I 2 0 µ
Magnetic dipole moment of smaller coil is iπr2.
Initially planes of two coils are mutually perpendicular, therefore θ is 90º or energy of the system is U = (iπr2) B (1 – cos 90º)
or U = R r Ii 2 2 0 π µ
When coils are released, both the coils start to rotate about their common diameter and their kinetic energies are maximum when they become coplaner. Moment of Inertia of larger coil about axis of rotation is I1 =
2 1
MR2 and that of smaller coil is I 2 =
2 1
mr2 Since, two coils rotate due to their mutual interaction only, therefore, if one coil rotates clockwise then the other rotates anticlockwise.
Let angular velocities of larger and smaller coils be numerically equal to ω1 and ω2 respectively when they become coplaner,
According to law of conservation of angular momentum,
I1ω1 = I2ω2 ...(i) and according to law of conservation of energy,
2 1 1 2 1 ωI + 2 22 2 1 ωI = U ...(ii) From above equations, maximum kinetic energy of smaller coil, 2 2 2 2 1 ωI = 2 1 1 I I UI + = ) ( 2 2 2 2 0 mr MR MRr Ii + π µ Ans.
4. RMS velocity of molecules of a di-atomic gas is to be increased to 1.5 times. Calculate ratio of initial volume to final volume, if it is done
(i) Adiabatically ; (ii) Isobarically ;
(iii) Calculate, also ratio of work done by gas during these processes.
XtraEdge for IIT-JEE 21 JANUARY 2011 Sol. R.M.S. velocity of gas molecules ∝ T .
where T is absolute temperature of the gas. Therefore, to increase RMS velocity to 1.5 times, the temperature is to be increased to (1.5)2 = 2.25 times. Let initial volume and temperature by V0 and T0 respectively and let final volume and temperature be
V and T respectively, then T = 2.25. T0 (i) For adiabatic process, T.V(γ – 1) = constant ∴ ) 1 – ( 0 γ V V = 0 T T = 2.25 But for diatomic gas γ = 1.4
V V0
= (2.25)2.5 = (1.5)5 = 7.594 Ans.(i) Work done by gas during the process,
W1 = ) 1 – ( – 0 0 γ PV V P = ) 1 – ( ) T – T ( nR 0 γ
(ii) For isobaric process,
T V = constant ∴ V V0 = T T0 = 25 . 2 1 = 9 4 Ans.(ii) Work done by gas during the process,
W2 = nR (T–T0)
(iii) Ratio of work done by gas during these processes is 2 1 W W = ) – 1 ( 1 γ = – 2.5 Ans.(iii)
5. Suppose potential energy between electron and proton at separation r is given by U = K log r, where
K is a constant. For such a hypothetical hydrogen atom, calculate radius of nth Bohr's orbit and energy levels.
Sol. If potential energy of a system of two particles, separated by a distance r is U, interaction force between the particles is F = –
dr dU
.
Since, potential energy of system of electron and proton in a hypothetical hydrogen atom is given by
U = K log r, therefore, force between them,
F = – dr dU = – r K
Negative sign indicates force of attraction.
This force of attraction between electron and proton provides centripetal force required for circular motion of electron. ∴ r mv2 = r K
where m is mass of electron and v is its speed.
or mv2= K ...(i)
According to Bohr's postulates, angular momentum,
mur = n
π 2
h
...(ii) From equation (i) and (ii),
r =
mK nh
π
2 Ans.
total energy of the atom is E = potential energy U + kinetic energy 2 2 1mv of electron ∴ E = K log r + 2 1 K = 2 K (2 log r + 1) = 2 K (1 + log r2) Substituting value of r, E = π + mK h n K 2 2 2 2 log 1 2 Ans.
Birth of New Red Spot is the
Thunderstorm on Jupiter
During the past few months, the astronomers have tracked an emerging second red spot on Jupiter, a growing rival about one-half the diameter of the planet's Great Red Spot. The Hubble Space Telescope has snapped the first detailed pictures of what some observers now call Red Spot Jr.
Astronomers at the Space Telescope Science Institute in Boltimore said this was the first time scientists have witnessed the birth of these huge oval spots, presumably a convective phenomenon like a powerful thunderstorm. The Great spot was already present when the observers first looked with telescope at the planet some 400 years ago. Red Spot Jr. appeared in near-infrared images to be as bright in Jupiter's cloudy atmosphere as its big companion. The size of Red Spot Jr. is half the size of its big companion. The scientists say the new storm might rise higher above the main cloud deck than the older spot.
In a New Light: Jupiter, with its second red spot, in a picture released by NASA.
Current observations, including Hubble pictures taken on May 12 and 18, show that the smaller red spot is drifting eastward in the Jovian Southern hemisphere and the Great Red Spot is moving westward. They should pass one another in early July. The pictures of the Red Spots are contrast-enhanced images taken in visible light and at near-infrared wave lengths. But the red spots, new and old, are really red.
XtraEdge for IIT-JEE 22 JANUARY 2011 Prism :
(i) Deviation 'δ' produced by the prism,
B C A Normal Normal i i' Q P r r' δ δ = i + i' – A and A = r + r'
(ii) For minimum deviation 'δm'
i = i' and r = r' and also PQ||BC and the refractive index for the material of prism is given by
µ = +δ 2 sin 2 sin A A m
(iii) δ – i graph for prism
δm
i δ
(iv) For not transmitting the ray from prism, µ > cosec 2 A
(v) For grazing incidence i = 90º and for grazing emergence i' = 90º. For maximum deviation i = 90º or i' = 90º
(vi) The limiting angle of prism = 2C when i = i' = 90º
If the angle of prism A > 2C, then the rays are totally reflected.
(vii) Right-angled prism : These prisms are used to turn a light beam to 90º or 180º. These are usually made of crown glass for which µg = 1.5 and C =tan–1
g µ 1 = 42º. Such prisms are used in binoculars and submarine periscopes.
(viii) Deviation produced by a thin prism δ = (µ – 1)A (ix) Angular dispersion D = δv – δR = (µV – µR)A
Where V and R stand for violet and red colours respectively.
Mean deviation δY = (µY – 1)A
where µY is the refractive index of mean yellow colour. (x) Dispersive Power, ω = deviation Mean dispersion Angular = Y R V δ δ − δ ω = 1 − µ µ − µ Y R V where µ Y = 2 R V +µ µ
(xi) Pair of prisms (or crossed prism) : Two thin prisms of different material when placed crossed, i.e., with their refracting edges parallel and pointing in opposite directions as shown in figure, produce a total deviation δ given by
δ = δ1~ δ2
where δ1 and δ2 are the mean deviations produced by the first and second prism respectively.
Total angular dispersion
D = D1 ~ D2
where D1 and D2 are the angular dispersions produced by respective prisms.
(xii) Dispersion without deviation : If the angle of two prisms A and A' are so adjusted that the deviation produced by the mean ray by the first prism is equal and opposite to that produced by the second prism, then the total final beam will be parallel to the incident beam and there will be dispersion without deviation.
