Dear Students,
Is examination a common cause of stress?
In most Asian cultures, the great emphasis on academic achievement and high expectations of success make it especially stressful for students. The strong negative stigma attached to failure also adds to the pressure.
Like it or not, we have to accept that examinations are necessary in any educational system. Even though it is debatable whether they are accurate measures of actual ability, no better alternatives have been proposed. Examinations remain necessary to motivate students’ learning, measure their progress and ultimately, serve as evidence of attainment of certain skills, standards or qualifications.
Success at examinations provides opportunities to proceed with higher education and improves employment prospects, underlining their importance. No matter how well prepared, many factors may influence one’s performance at the time of the examination and there is seemingly, no definite guarantee of success. Essentially, it is this vital importance attached to success at examinations coupled with the element of uncertainty that makes them so stressful.
As with other sources of stress, the stress of examinations is not all bad. It is a strong incentive for students to study and poses a challenge for individual achievement. However, when stress becomes excessive, performance begins to suffer. There is thus a need to control levels of stress before it becomes overwhelming and detrimental. Reliase of stress is necessary for optimum performance, the means of which is relasing.
Learn to relax
The stress responses produces muscle tension, which you would commonly experience as backache, neck ache or tension headache at the end of the day. Often this is unconscious. So to relax these muscles, you need to consciously practice relaxation exercises. These could involve muscle relaxation, deep breathing exercises, body massage or guided imagery. Like any particular skill, you need to practice them regularly in order to reap the benefits.
Another way to relax is to maintain a quiet time as part of the daily routine. Quiet time refers to a time for you with no interruption from external sources or distractions. This is a time where you may choose to just think of nothing and relax. Finally, you can always take up a hobby to help you relax. Do something you enjoy, be it listening to music.
Ideally, the drive to study should be internally driven by a desire to achieve one’s own personal goals. Instead, many are driven more by the fear of failure, which is more stress-provoking and leads easily to discouragement.
Attending school should not merely revolve around preparation for examinations. Interacting with teachers, socializing with friends, participating in sports or other extra-curricular activities are all valuable aspects of a ‘well-rounded’ education.
Instead of wishing things would get easier, start looking at how you can get better...
Yours truly
Pramod Maheshwari, B.Tech., IIT Delhi
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CONTENTS
INDEX PAGE
NEWS ARTICLE
4Are Nanoparticles health hazard Nano is the new black
IITian ON THE PATH OF SUCCESS
6Ms. Padmasree Warrior & Dr. Krishan K. Sabnani
KNOW IIT-JEE
7Previous IIT-JEE Question
XTRAEDGE TEST SERIES
52Class XII – IIT-JEE 2011 Paper Class XI – IIT-JEE 2012 Paper
Regulars ...
DYNAMIC PHYSICS
148-Challenging Problems [Set# 3] Students’ Forum Physics Fundamentals Capacitor - 1 Friction
CATALYSE CHEMISTRY
29 Key Concept Reaction Mechanism EnergeticsUnderstanding : Organic Chemistry
DICEY MATHS
42 Mathematical Challenges Students’ Forum Key Concept 3-Dimensional GeometryProgression & Mathematical Induction
Study Time...
Are Nanoparticles a
Health Hazard?
Nanoparticles are a mega business opportunity for multinationals but they may pose a health hazard to users
There is a new industrial revolution taking place all around us. The only problem is we can’t see it. The building blocks, being developed at the cost of billions of dollars by scientists, governments and multinational corporations, are just a few atoms or molecules thick — nanoparticles. Many are less than 100 nanometres (nm) — one-billionth of a metre — thick. A single human red blood cell in comparison is around 500 nm in diametre. It’s a pity though that our eyesight isn’t good enough at nanometre level, for if it were, we would see that nanoparticles of precious metals like gold, silver and titanium have already made the jump from research labs to our homes. Manufactured nanoparticles are today present in thousands of consumer products around the world — silver in washing machines and water purifiers to kill bacteria, zinc in cosmetics to protect against ultraviolet rays, carbon nano-tubes in tennis rackets to make them stronger and lighter, titanium in household paints to decompose dust and grime without human intervention.
Nano is the New Black
“There’s Plenty of Room at the Bottom” So went the classic lecture by the Nobel prize-winning physicist Richard Feynman in 1959 that many nano-ficionados now consider the conceptual sun of the nanotechnology universe. “Whycannot we write the entire 24 volumes of the Encyclopedia Britannica on the head of a pin?” he asked.
“Because there isn’t much of a point, or money, in doing so!” is the answer he would have got today from nanotechnology researchers. Instead their time is mostly spent figuring out newer properties for nanoparticles which can then be embedded into commercial applications.
Nanoparticles are highly reactive and prone to unusual properties. Describing gold, a metal that is normally inert to all other chemicals, Prof. C.N.R Rao, Honorary President and Linus Pauling Professor at the Jawaharlal Nehru Centre for Advanced Scientific Research (JNCASR) and the head of its Nanoscience centre, says “At 200-300 nm thickness, gold is not metallic, it does not shine — in fact it is not gold. And at 1.5-2 nm, it reacts like mad!” Gold that is not gold when shrunk to nanometer size might sound like an absurdity to many, but it’s exactly this change in physical properties that make nanoparticles popular.
For example zinc oxide (ZnO) and titanium dioxide (TiO2) have been used as active ingredients in sunscreens for decades because of
their ability to absorb ultraviolet rays and reflect back much of the other remaining sunlight. But they are both white — the reason many sunscreens leave a white residue on the face. When shrunk to nanometre size however, they become transparent without losing their light reflecting or absorbing abilities.
Silver, an ornamental metal and a powerful bactericide, can be reduced to nanoparticle form to destroy disease-causing bacteria from all kinds of places — kitchen counters, contaminated water, dirty clothes and stinky underarms. Samsung claims its Silver Nano range of washing machines release hundreds of billions of silver nano-ions with each wash to kill over 99 percent of the bacteria found in dirty clothes, while the same technology when lined on the doors of their refrigerators kill bacteria that could spoil stored food. Eureka Forbes’ water purifiers use nanosilver-coated filters, developed by Prof. Pradeep, head of IIT-Madras’ Nanoscience department, to destroy harmful bacteria from drinking water. Swach, the mass-market water filter introduced by the Tata Group, also uses nanosilver (coated on rice husk particles) to purify drinking water. During the last flu pandemic threat authorities in Hong-Kong sprayed subways with nanosilver to disinfect them. L’Oreal, the world’s largest cosmetics company, reportedly spends over $600 million each year researching and patenting nanoparticles. The head of its nanotechnology unit also sits on the management board.
Therefore a metal that is a poor second cousin to gold in a world where we value yellow over white, is the undisputed metal of choice in the nanoparticle world. In fact, silver is more popular than any other material, according to the database of consumer products using nanoparticles maintained by the Woodrow Wilson International Center for Scholars.
Obama admn
nominates IIT alumnus
for post of NSF Director
IIT Madras alumnus Subra Suresh, popularly known as 'Bakthi Suresh' during his student days, has been nominated for the post of director of the National Science Foundation (NSF) by the Barrack Obama administration. An official relese from IIT Madras Alumni Association here said ''when confirmed by the Senate, Mr Suresh will become one of the highest ranking Indian-Americans ever to serve in an administration.'' An Indian-American technocrat, 53-year-old Subra Suresh completed his B.Tech Mechanical Engineering in 1977. Currently the dean of the MIT engineering school, he received the distinguished alumnus award in 1997. In a statement, President Obama said ''I am proud that such experienced and committed individuals have agreed to take on these important roles in my administration. I look forward to working with them in the coming months and years.'' The National Science Foundation is the funding source for nearly 20 per cent of all federally supported basic research and was an independent federal agency created by US Congress in 1950. Subra Suresh has been elected to the US National Academy of Engineering, the Indian National Academy of Engineering,the American Academy of Arts and Sciences, the Indian Academy of Sciences in Bangalore, the German National Academy of Sciences, the Royal Spanish Academy of Sciences and the Academy of Sciences of the Developing World based in Trieste, Italy.
Major decision regarding
Ganga at IIT-K
KANPUR: A collaboration between the consortium of seven IITs (IIT-Kanpur, Madras, Bombay, Delhi, Kharagpur, Guwahati and Roorkee) and Ministry of Environment and Forest, Government of India, is being worked out for the purpose of cleaning the national river Ganga. The two are expected to sign an important memorandum of understanding (MoU) in this regard during Prime Minister Manmohan Singh's visit to the Indian Institute of Technology-Kanpur. Singh is scheduled to visit IIT-K on July 3 to take part in its convocation ceremony.
It will be worth mentioning here that the Central government aims at meeting the formidable challenge of cleaning the Ganga. With this goal in mind, it had launched a new initiative and established the National Ganga River Basin Authority (NGRBA) last year. The Prime Minister, who is the Chairman of the NGRBA, asked the Ministry of Environment and Forest to involve IITs in the mega project. A joint meeting of all the seven IITs was convened on March 12, 2010 in which IIT-Kanpur was represented by Prof Vinod Tare, also the convener of the mission. It is for the very first time that the Central government has involved the seven IITs together in one single project of such a large magnitude
Prof Tare further informed TOI that the 'zero discharge' of both treated and untreated sewage waste into the river had been proposed to the government under which the waste would not be allowed into the Ganga.
"We will be doing this mega project in phases. The first phase will come to an end in 18 months wherein the concept of 'zero discharge' will be put into application. We also plan to apply the 'zero discharge' formula in four cities initially, viz. Hardwar, Rishikesh, Kanpur and Allahabad. If we are able to do so in these four cities, water of the river will become clean up to Allahabad and a major work will come to an end in the first phase."
Meanwhile, Prof SG Dhande, director, IIT-Kanpur, and Prof Vinod Tare from IIT-Kanpur, Prof Devang Khakhar, director, IIT-Bombay, took part in a meeting with the officials of the Ministry of Environment and Forest in New Delhi on May 19 and discussed all important aspects of the mega project. On the occasion, the team also handed over a set of proposals to the officials of the ministry. "A detailed project report will be given later as several social, legal aspects will have to be examined," said Prof Tare.
The consortium of the seven IITs have been sanctioned Rs 16 crore by the government for formulating a proper plan of action for the purpose of cleaning the Ganga. The authority formed by the Central government has both regulatory and developmental functions. The authority will take measures for effective abatement of pollution and conservation of the Ganga in keeping with sustainable develop-ment needs.
Science Research : Conventional solar cell efficiency could be increased from the current limit of 30 percent to more than 60 percent, suggests new research on semiconductor nanocrystals, or quantum dots, led by chemist Xiaoyang Zhu at The University of Texas at Austin.
Padmasree Warrior is senior vice president and chief technology officer for Motorola, with responsibility for Motorola Labs, the global software group and emerging early-stage businesses. Warrior's operational responsibilities include leading a global team of 4,600 technologists, prioritizing technology programs, creating value from intellectual property, guiding creative research from innovation through early-stage commercialization, and influencing standards and roadmaps. She also serves as a technology advisor to the office of the chairman and to the board's technology and design steering committee. Before assuming her current position in January 2003, Warrior was corporate vice president and general manager of Motorola's energy systems group, where she was responsible for profit and loss, sales, marketing, engineering and manufacturing. She also was general manager of Thought beam, Inc., a wholly owned subsidiary of Motorola, where she led the commercialization evaluation team related to compound semiconductor materials research.
Prior to these assignments, Warrior was corporate vice president and chief technology officer for Motorola's Semiconductor Products Sector (SPS).
A Motorola since 1984, she has been instrumental in driving innovative methods for technology commercialization realizing early "time to revenue" for the corporation. She has held many leadership positions within Motorola, was appointed vice president in 1999 and was elected a corporate officer in 2000.
Warrior received a M.S. degree in chemical engineering from Cornell University, and a B.S. degree in chemical engineering from the Indian Institute of Technology (IIT) in New Delhi, India.
Warrior served on the Texas Governor's Council for Digital Economy, and is a member of the Texas Higher Education Board review panel. She was one of six women nationwide selected to receive the "Women Elevating Science and Technology" award from Working Woman magazine in 2001. She also is a director of Ferro Corporation.
Krishan Sabnani is Senior Vice President of the Networking Research Laboratory at Bell Labs in New Jersey. For the past 23 years Krishan has been a member of Bell Labs Research. Krishan has conceived and launched several systems projects in the areas of Internetworking and wireless networking, led successful transfers of research ideas to products in Lucent and AT&T business units and conducted extensive personal research in data and wireless networking. He has built organizations known for technical excellence by recruiting and coaching the best people in the industry.
Krishan has received the 2005 IEEE Eric E. Sumner Award and the 2005 IEEE W. Wallace McDowell Award - the only person ever to receive both awards. Krishan is a Bell Labs Fellow. He is also a fellow of the Institute of Electrical and Electronic Engineers (IEEE) and the Association of Computing Machinery (ACM). He received the Leonard G. Abraham Prize Paper Award from the IEEE Communications Society in 1991. Krishan will receive the 2005 Distinguished Alumni Award from Indian Institute of Technology (IIT), New Delhi, India. He has also won the 2005 Thomas Alva Edison Patent Award from the R&D Council of New Jersey. He holds 37 patents and has published more than 70 papers.
In his personal research, Krishan has made major contributions to the communications protocols area. He has designed several protocols such as SNR, RMTP, and Airmail. He has also made significant contributions to conformance test generation, protocol validation, automated converter generation, and reverse engineering. Krishan received his Ph.D. in electrical engineering from Columbia University, New York, in 1981. He joined Bell Labs in 1981.
Key Awards and Honors
1. 2005 IEEE Eric E. Sumner Award, received for seminal contributions to networking protocols 2. 2005 IEEE Computer Society W. Wallace McDowell
Award
3. 2005 IIT Delhi Outstanding Alumni Award 4. 2005 Thomas Alva Edison Patent Award 5. 1991 Leonard G. Abraham Prize Paper Award 6. 1997 Bell Labs Fellow.
Success Story
Success Story
This articles contains stories of person who have succeed after graduation from different IIT's
Ms. Padmasree Warrior
B.Tech, IIT MadrasDr. Krishan K. Sabnani
B.Tech, IIT – KanpurPHYSICS
1. Masses M1, M2 and M3 are connected by strings of negligible mass which pass over massless and friction less pulleys P1 and P2 as shown in fig. The masses move such the portion of the string between P1 and P2 in parallel to the inclined plane and the portion of the string between P2 and M3 is horizontal. The masses M2 and M3 are 4.0 kg each and the coefficient of kinetic friction between the masses and the surfaces is 0.25. The inclined plane makes an angle of 37º with the horizontal. [IIT-1981]
P1
M1
M2
P2 M3
37º
If the mass M1 moves downwards with a uniform velocity, find
(a) the mass of M1
(b) The tension in the horizontal portion of the string (g = 9.8 m/sec2, sin 37º ≈ 3/5)
Sol. (a) Applying Fnet = ma on M1 we get
T – m1 . g = M1 × 0 = 0 ⇒ T = M1g ...(i) Applying Fnet = Ma on M2 we get
T – (T´ + M2g sin θ – f) = M2 × a
T = T´ + M2g sin θ + f = T´ + M2g sin θ + µN [Q f = µN = µM2 g cos θ] ∴ T = T´ + M2g sin θ + µM2g cos θ ...(ii)
P1 M1 M2 T´ P2 M2g M1g T T V θ M2gcosθ V f M2gsinθ θ T´ N M3g
Applying Fnet = Ma for M3 we get T´ – f ´ = M3 × 0
⇒ T´ = f ´ = µN´ = µM3g ...(iii) Putting the value of T and T´ from (i) and (iii) in (ii) we get
M1g = µM3g + M2g sin θ – µ M2g cos θ M1 = 0.25 × 4 + 4 × sin 37º + 0.25 × 4 × cos 37º = 4.2 kg
(b) The tension in the horizontal string will be T ´ = µM3g = 0.25 × 4 × 9.8 = 9.8 N
2. A 0.5 kg block slides from the point A (see fig.) on a horizontal track with an initial speed of 3 m/s towards a weightless horizontal spring of length 1 m and force constant 2 Newton/m. The part AB of the track is frictionless and the part BC has the coefficients of static and kinetic friction as 0.22 and 0.2 respectively. If the distances AB and BD are 2m and 2.14 m respectively, find the total distance through which the block moves before it comes to rest completely. (Take g = 10 m/s2) [IIT-1983]
A B D C
Sol. K.E. of block = work against friction + P.E. of spring
2 1 mv2 = µ k mg (2.14 + x) + 2 1 kx2 2 1 × 0.5 × 32 = 0.2 × 0.5 × 9.8(2.14 + x) + 2 1 2 × x2 2.14+ x + x2 = 2.25 ∴ x2 + x – 0.11 = 0 On solving we get x = – 10 11 or x = 10 1 = 0.1 (valid answer) Here the body stops momentarily.
Restoring force at y = kx = 2 × 0.1 = 0.2 N Frictional force at
y = µs mg × x = 0.22 × 0.5 × 9.8 = 1.078 N Since friction force > Restoring force the body will stop here.
∴ The total distance travelled
= AB + BD + DY = 2 + 2.14 + 0.1 = 4.24 m. A B D Rough L C 2m 2.14m x Y
3. A small sphere rolls down without slipping from the top of a track in a vertical plane. The track in a vertical plane. The track has an elevated section and a horizontal part, The horizontal part is 1.0 meter above the ground level and the top of the track is 2.4 metres above the ground. Find the distance on the ground with respect to the point B(which is vertically
KNOW IIT-JEE
below the end of the track as shown in fig.) where the sphere lands. During its flight as a projectile, does the sphere continue to rotate about its centers of mass ?
Explain. [IIT-1987] A B 1.0m 2.4 m
Sol. Applying law of conservation of energy at point D and point A
P.E. at D = P.E. at A + (K.E.)T + (K.E.)R (K.E.)T = Translational K.E.
mg (2.4) = mg (1) + 2 1 mv2 + 2 1Iω2
(K.E.)R = Rotational K.E. Since the case is of rolling without slipping
D 2.4m 1m A B C ∴ v = rω ∴ ω = r
v where r is the radius of the sphere Also
I = 5 2 mr2 ∴ mg(2.4) = mg(1) + 2 1 mv2 + 2 1 × 5 2 mr2 × 2 2 r v ⇒ v = 4.43 m/s
After point A, the body takes a parabolic path. The vertical motion parameters of parabolic motion will be uy = 0 S = ut + 2 1 at2 Sy = 1m 1 = 4.9 ty2 ay = 9.8 m/s2 ∴ ty= ? ty = 9 . 4 1 = 0.45 sec Applying this time in horizontal motion of parabolic path, BC = 4.43 × 0.45 = 2m
During his flight as projectile, the sphere continues to rotate because of conservation of angular momentum. 4. Two square metal plates of side 1 m are kept 0.01 m apart like a parallel plate capacitor in air in such a way that one of their edges is perpendicular to an oil surface in a tank filled with an insulating oil. The plates are connected to a battery of emf 500 V. The plates are then lowered vertically into the oil at a
speed of 0.001 ms–1. Calculate the current drawn from the battery during the process. (Dielectric constant of oil = 11, ε0 = 8.85 × 10–12C2N–1m–1)
[IIT-1994] Sol. The adjacent figure is a case of parallel plate
capacitor. The combined capacitance will be
v + 1–x x 1m d C = C1 + C2 = d ) 1 x ( kε0 × + d ] 1 ) x 1 [( 0 − × ε C = d 0 ε [kx + 1 – x]
After time dt, the dielectric rises by dx. The new equivalent capacitance will be
C + dC = C1´ + C2´ = d kε0 [(x + dx) × 1] + d ] 1 ) dx x 1 [ 0 − − × ε dC = Change of capacitance in time dt
= d 0 ε [kx + kdx + 1 – x – dx – kx – 1 + x] = d 0 ε (k – 1)dx dt dC = d 0 ε (k – 1) dt dx = d 0 ε (k – 1)v ...(i) where v = dt dx We know that q = CV dt dq = V dt dC ...(ii) ⇒ I = V d 0 ε (k – 1)v From (i) and (ii)
I = 01 . 0 10 85 . 8 500× × −12 (11 – 1) × 0.001 = 4.425 × 10–9 Amp.
5. Two resistors, 400 ohms, and 800 ohms are connected in series with a 6-volt battery. It is desired to measure the current in the circuit. An ammeter of a 10 ohms resistance is used for this purpose. What will be the reading in the ammeter? Similarly, If a voltmeter of 10,000 ohms resistance is used to measure the potential difference across the 400-ohms resistor, What will be the reading in the voltmeter.
Sol. Applying Kirchoff's law moving in clockwise direction starting from battery we get
A 10Ω 400Ω 800Ω 6 volt + 6 – 10I – 400 I – 800 I = 0 ∴ 6 = 1210 I ∴ I = 1210 6 = 4.96 × 10–3 A
The voltmeter and 400 Ω resistor are in parallel and hence p.d. will be same
∴ 10,000 I1 = 400 I2 ...(i)
Applying Kircoff's law in loop ABCDEA starting from A in clockwise direction.
– 400 I2 – 800 I + 6 = 0 ∴ 6 = 400 I2 + 800 (I1 + I2) ∴ 6 = 400 I2 + 800(0.04 I2 + I2) From (i) putting the value of I1 ∴ 6 = 1232 I2 V I 400Ω 800Ω 6 volt 10,000Ω B F G C I D A E ∴ I2 = 4.87 × 10–3 Amp.
Potential drop across 400 Ω resistor = I2 × 400
= 4.87 × 10–3 × 400 = 1.948 volt ≈ 1.95 volt
∴ The reading measured by voltmeter = 1.95 volt
CHEMISTRY
6. At constant temperature and volume, X decomposes as 2X(g) → 3Y(g) + 2Z(g); Px is the partial pressure of X.
Observation
No. Time (in minute) Rx (in mm of Hg)
1 0 800 2 100 400 3 200 200 (i) What is the order of reaction with respect to X ? (ii) Find the rate constant.
(iii) Find the time for 75% completion of the reaction. (iv) Find the total pressure when pressure of X is 700 mm of Hg. [IIT-2005]
Sol.
(i) From the given data, it is evident that the t1/2 (half-life period)for the decomposition of X (g) is constant (100 minutes) therefore the order of reaction is one. (ii) Rate constant, K =
2 / 1 t 693 . 0 = 100 693 . 0 = 6.93 × 10–3 min–1
(iii) Time taken for 75% completion of reaction = 2t1/2 = 2 × 100 = 200 minutes
(iv) 2x → 2Y + 2Z Initial pressure 800 0 0 Ater time t (800 – 2p) 3P 2p
when the pressure of X is 700 mm of Hg the, 800 – 2P = 700
2P = 100; P = 50 mm of Hg
Total pressure = 800 – 2P + 3P + 2P = 800 + 150 = 950 mm of Hg.
7. A acid solution of Cu2+ salt containing 0.4 g of Cu2+ is electrolysed until all the copper is deposited. The electrolysis is continued for seven more minutes with the volume of solution kept at 100 ml. and the current at 1.2 amp. Calculate the volume of gases evolved at NTP during the entire electrolysis. [IIT-89] Sol. The chemical reactions taking place at the two
electrodes are
At cathode : Cu2+ + 2e– → Cu H2O H+ + OH–
However, note that only Cu2+ ions will be discharged so as these are present in solution and H+ ions will be discharged only when all the cu2+ ions have been deposited.
Atcathode : 2OH–→ H
2O + O + 2e– O + O → O2
Thus in first case, Cu2+ ion will be discharged at the cathode and O2 gas at the anode. Let us calculate the volume of gas (O2) discharged during electrolysis. According to Faraday's cecond law
31.75 g Cu ≡ 8 g of oxygen ≡ 5.6 litres of O2 at NTP 0.4 g Cu = 75 . 31 6 . 5 × 0.4 litres of O 2 at NTP = 0.07055 litres = 70.55 ml
As explained earlier, when all the Cu2+ ion will be deposited at cathode, H+ ions will start going to cathode liberating hydrogen (H2) gas i.e.
H+ + e–
H H + H → H2
However, the anode reaction remains same as previous. Thus in the second (latter) case, amount of H2 collected at cathode should be calculated.
8 g of O2 = 1 g of H2
5.6 litres of O2 at NTP = 11.2 litres of hydrogen Quantity of electricity passed after 1st electrolysis, i.e. Q = i × t = 1.2 × 7 × 60 = 504 coulombs
504 coulombs will liberate = 96500 504 6 . 5 × = 29.24 ml of O2. Similarly, H2 liberated by 504 coulombs = 11.2 ×
96500 504
= 58.45 ml
Total volume of O2 liberated = 70.55 + 29.24 = 99.79 ml vol. of H2 liberated = 58.48 ml.
8. Cyclobutyl bromide on treatment with magnesium in dry ether forms an organometallic (A). The organometallic reacts with ethanal to give an alcohol (B) after mild acidification Prolonged treatment of alcohol (B) with an equivalent amount of HBr gives 1-bromo 1-methylcyclopentane (C). Write the structures of (A), (B) and explain how (C) is obtained
from (B). [IIT-2001] Sol. Br ether dry→ MgBr Cyclobutymagnesium Bromide (A) → CH3CHO,H3O+ CH–OH 1-Cyclobutylethanol (B) CH3 CH–CH3 (B) OH → H+ CH–CH3 Oxonium ion OH2 ⊕ –H2O+→ CH–CH3 (2º carbocation 4-membered ring) + alkylshift 2 , 1 through ansion exp ring − → H (ring expansion) (2º carbocation in 5 membered ring ⊕ H CH3 Hydrideshift – 2 , 1 → CH3 (3º carbocation) ⊕ H H → Br– Br 1-bromo-1-methyl Cyclopentane (C) ⊕ H H CH3
9. A basic, volatile nitrogen compound gave a foul smelling gas when treated with chloroform and alcoholic potash. A 0.295 g sample of the substance. Dissolved in aq. HCl and treated with NaNO2 solution at 0ºC, liberated a colorless, 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-93]
Sol. Let us summarise the given facts.
112 ml of colourless, odourless gas at S.T.P + Residue (i) aq. HCl (ii) NaNO20ºC Basic Nitrogen Compound (0.295 g) CHCl3 KOH Foul smelling gas
Distilled aq. Sol.
Organic liquid (no N)
OH–/I 2
Yellow ppt.
Reaction of the original compound with alcoholic potash and chloroform to give foul smelling gas indicates that it contains a primary –NH2 group. R–NH2 + CHCl3 + KOH → R–NC↑ (Basic compound) Carbylamine
(foul smelling) Determination of mol. Weight of the amine.
112 ml. of gas is evolved at S.T.P. by 0.295 g of amine 22400 ml. of gas is evolved by = 112 295 . 0 × 22400 = 59 Hence the mol. wt. of the amine = 59
∴ Mol. wt. of the alkyl group = 59 – 16 = 43 Nature of alkyl gp. of mol. wt.= 43 = C3H7– Thus the amine may be either
CH3CH2CH2NH2 or CHNH2
CH3
CH3
The reaction of amine with NaNO2 at 0ºC and all other reactions may thus be written as below.
CH3CH2CH2NH2 C º 0 / NaNO ) ii ( HCl ) i ( 2 → CH3CH2CH2OH + N2↑ n-Propylamine → aq.sol.distill CH 3CH2CH2OH → OH–,I2 No yellow ppt. (CH3)CHNH2 → (CH3)2CHOH + N2↑ Isopropylamine → (CH3)2CHOH ) reaction Haloform ( I , OH–2→ CHI3↓ (yellow)
Since the given reactions correspond to isopropylamine, the original compound is isopropylamine, (CH3)2CHNH2
10. Interpret the non-linear shape of H2S molecule and non-planar shape of PCl3 using valence shell electron pair repulsion (VSEPR) theory. (Atomic numbers : H = 1, P = 15, S = 16, Cl = 17.) [IIT-98] Sol. In H2S, no. of hybrid orbitals =
2
1(6 + 2 – 0 + 0) = 4 Hence here sulphur is sp3 hybridised, so
16S = 1s2, 2s22p6, 4 4 3 4 4 2 1 ion hybridisat sp 1 z 1 y 2 x 2 3 p 3 p 3 p 3 s 3
or S H H H H S
Due to repulsion between lp - lp; the geometry of H2S is distorted from tetrahedral to V-shape.
In PCl3, no. of hybrid orbitals =
2
1[5 + 3 – 0 + 0] = 4
Hence, here P shows sp3-hybridisation 15P = 1s2, 2s22p6, 4 4 3 4 4 2 1 ion hybridisat sp 1 z 1 y 1 x 2 3 p 3 p 3 p 3 s 3 P or P Cl Cl Cl Cl Cl Cl
Thus due to repulsion between lp – bp, geometry is distorted from tetrahedral to pyramidal.
MATHEMATICS
11. The circle x2 + y2 – 4x – 4y + 4 = 0 is inscribed in a triangle which has two of its sides along the coordinate axes. If the locus of the circumcentre of the triangle is
x + y – xy + k x2+y2 = 0,
find the value of k. [IIT-1987] Sol. Let OAB be the triangle in which the circle
x2 + y2 – 4x – 4y + 4 = 0 is inscribed. Let the equation of AB be b y a x + = 1 y B(0,b) y´ x´ O 2 C (a, 0)A x 1byax=+ 1 b y a x+ =
Since AB touches the circle x2 + y2 – 4x – 4y + 4 = 0. There fore, 2 2 b 1 a 1 1 b 2 a 2 + − + = 2 ⇒ – 2 2 b 1 a 1 1 b 2 a 2 + + − = 2
[Q O(0, 0) and C(2, 2) lie on the same side of AB Therefore, a 2 + b 2 – 1 < 0] ⇒ – 2 2 b a ) ab – a 2 b 2 ( + + = 2 ⇒ 2a + 2b – ab + 2 a2+b2 = 0 ...(i)
Let P(h, k) be the circumcentre of ∆OAB. Since ∆ OAB is a right angled triangle. So its circumcentre is the mid-point of AB.
∴ h = 2 a and k = 2 b ⇒ a = 2h and b = 2k ...(ii) From (i) and (ii), we get
4h + 4k – 4hk + 2 4h2+4k2 = 0 ⇒ h + k – hk + h2+k2 = 0 So, the locus of P(h, k) is
x + y – xy + x2+y2 = 0
But, the locus of the circumcentre is given to be x + y – xy + k x2+y2 = 0
Thus, the value of k is 1
12. If A, B, C are the angles of a triangle ABC and the system of linear equations
x sin A + y sin B + z sin C = 0 x sin B + y sin C + z sin A = 0 x sin C + y sin A + z sin B = 0 has a non trivial solution, prove that
sin2A + sin2B + sin2C – (cos A + cos B + cos C + cos A cos B + cos B cos C + cos C cos A) = 0
[IIT-2002] Sol. The given system of linear equations has a non-trivial
solution. Therefore, B sin A sin C sin A sin C sin B sin C sin B sin A sin = 0 ⇒ B sin A sin B sin A sin C sin A sin C sin A sin C sin B sin C sin B sin C sin B sin A sin + + + + + + = 0 Applying C1 → C1 + C2 + C3
⇒ (sin A + sin B + sin C)
B sin A sin 1 A sin C sin 1 C sin B sin 1 = 0 ⇒ B sin A sin 1 A sin C sin 1 C sin B sin 1 = 0 ≠ = + + 0 2 C cos 2 B cos 2 A cos 4 C sin B sin A sin Q
⇒ C sin B sin B sin A sin 0 C sin A sin B sin C sin 0 C sin B sin 1 − − − − = 0 Applying R2 → R2 – R1, R3 → R3 – R1 ⇒ –(sin B – sin C)2 – (sin A – sin C)
(sin A – sin B) = 0 ⇒ sin2B + sin2C – 2 sin B sin C + sin2A
– sin A sin B – sin C sin A + sin B sin C = 0 ⇒ sin2A + sin2B + sin2C – sin A sin B – sin B sin C
– sin C sin A = 0 ⇒ sin2A + sin2B + sin2C – cos A cos B
– cos B cos C – cos C cos A + cos (A + B) + cos (B + C) + cos (C + A) = 0 ⇒ sin2A + sin2B + sin2C – cos A cos B
– cos B cos C – cos C cos A – cos A – cos B – cos C = 0 13. Determine the name of the name of the curve
described parametrically by the equations
x = t2 + t + 1, y = t2 – t + 1 [IIT-1998] Sol. We have, x = t2 + t + 1 and, y = t2 – t + 1 ⇒ x + y = 2(t2 + 1) and, x – y = 2t ⇒ x + y = 2 + − 1 2 y x 2 ⇒ 2(x + y) = (x – y)2 + 4 ⇒ x2 + y2 – 2xy – 2x – 2y + 4
Comparing this equation with the equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, we get
a = 1, b = 1, c = 4, h = –1, g = –1 and f = –1
∴ abc + 2fgh – af2 – bg2 – ch2 = 4 – 2 – 1 – 1 – 4 ≠ 0 and , h2 – ab = 1 – 1 = 0
Thus, we have
∆ ≠ 0 and h2 = ab
So, the given equations represent a parabola.
14. Let C be any circle with centre (0, 2 ). Prove that at most two rational points can be there on C.
(A rational points is a point both of whose coordinates are rational numbers) [IIT-1997] Sol. The equation of any circle C with centre (0, 2 ) is
given by
(x – 0)2 + (y – 2 )2 = r2, where r is any positive real number.
or, x2 + y2 –2 2y = r2 – 2
If possible, let P(x1, y1), Q(x2, y2) and R(x3, y3) be three distinct rational points on circle C. Then,
2 2 y x 2 1 2 1 + − y1 = r2 – 2 ...(ii) 2 2 y x 2 2 2 2+ − y2 = r2 – 2 ...(iii) 2 2 y x 2 3 2 3+ − y3 = r2 – 2 ...(iv)
We claim that at least two y1, y2, and y3 are distinct. For if y1 = y2 = y3, then P, Q and R lie on a line parallel to x-axis and a line parallel to x-axis does not cross the circle in more than two points. Thus, we have either y1 ≠ y2 or, y1 ≠ y3 or, y2 ≠ y3.
Subtracting (ii) from (iii) and (iv), we get
) y x ( 22+ 22 – (x12+y12) – 2 2(y2 – y1) = 0 and, (x y2) 3 2 3+ – (x12+y12) – 2 2(y3 – y1) = 0 ⇒ a1 – 2 b1 = 0 and a2 – 2 b2 = 0 ...(v) where, a1 = (x22+y22) – (x12+y12), b1 = 2(y2 – y1) a2 = (x32+y32) – (x12+y12), b2 = 2(y3 – y1) Clearly, a1, a2, b1, b2 are rational numbers as x1, x2,
x3, y1, y2, y3 are rational numbers. Since either y1 ≠ y2 or, y1 ≠ y3 ∴ Either b1 ≠ 0 or, b2 ≠ 0 If b1 ≠ 0, then a1 – 2 b1 = 0 [From (v)] ⇒ 1 1 b a = 2 ,
which is not possible because 1 1
b
a is a rational
number and 2is an irrational number. If b2 ≠ 0, then a2 – 2 b2 = 0 ⇒ 2 2 b a = 2 ,
which is not possible because 2 2
b a
is a rational number and 2 is an irrational number.
Thus, in both the cases we arrive at a contradiction. This means that our supposition is wrong. Hence, there can be at most two rational points on circle C. 15. A rectangle PQRS has its side PQ parallel to the line
y = mx and vertices P, Q and S lie on the lines y = a, x = b and x = –b, respectively. Find the locus of the
vertex R. [IIT-1996]
Sol. Let the coordinates of R be (h, k). It is given that P lies on y = a. So, let the coordinates of P be (x1, a). Since PQ is parallel to the line y = mx. Therefore, Slope of PQ = (Slope of y = mx) = m And, Slope of PS = – mx) y of Slope ( 1 = = – m 1 [∴ PS ⊥ PQ] Now, equation of PQ is y – a = m(x – x1) ...(i)
y y = 0 P Q x = b (0, b) x y´ R O (0, – b) S x´ x = –b (0, a)
It is given that Q lies on x = b. So, Q is the point of intersection if (i) and x = b.
Putting x = b in (i), we get y = a + m(b – x1)
So, coordinates of Q are (b, a + m(b – x1)). Since PS passes through P(x1, a) and has slope –
m 1 . So, Equation of PS is y – a = – m 1 (x – x1) ...(ii) It is given that S lies on x = – b. So, S is the point of intersection of (ii) and x = –b.
Solving (ii) and x = – b, we get y = a +
m 1
(b + x1) So, coordinates of S are
− + (b+x ) m 1 a , b 1 Now, Slope of RS = b h ) x b ( m 1 a k 1 + + − − = m But RS is parallel to PQ. ∴ b h ) x b ( m 1 a k 1 + + − − = m ⇒ b + x1 = m(k – a) – m2(h + b) ...(iii) Similarly, Slope of RQ = b h ) x b ( m a k 1 − − − −
But, RQ is perpendicular to PQ whose slope is m. ∴ b h ) x b ( m a k 1 − − − − = – m 1 ⇒ b – x1 = m 1 (k – a) + 2 m 1 (h – a) ...(iv)
We have only one variable x1. To eliminate x1, add (iii) and (iv) to obtain
2b = (k – a) + m 1 m – m2(h + b) + 2 m 1 (h – b) ⇒ 2b = (k – a) + m 1 m2 – h + 2 4 m 1 m – b + 2 4 m 1 m ⇒ (k–a) + m 1 m2 – 2 2 2 m ) 1 m )( 1 m ( h − + – 22 2 m ) 1 m ( b + =0 ⇒ (k – a) – m ) 1 m ( h 2− – m ) 1 m ( b 2+ = 0 ⇒ m(k – a) – h(m2 – 1) – b(m2 + 1) = 0
Hence, the locus of R(h, k) is m(y – a) – x(m2 – 1) – b(m2 + 1) = 0
BEWARE OF THE BATTERIES
THAT YOU USE!
Have you ever noticed that the batteries are becoming smaller and smaller day after day? Many scientists and researchers have been finding the effective way to shrink the batteries into the smallest size as possible!
In this case, Jae Kwon, an assistant professor of Electrical and computer engineering has recently developed a nuclear energy source, which is smaller, lighter and more efficient than the common batteries.
Kwon’s described that the new discovered radioisotope battery can provide power density as much as six orders of magnitude higher than chemical batteries.
Kwon and his research team members have been cooperated and working on building a small nuclear battery. According to the information, the radioisotope batteries are having the size and thickness of a penny, but it’s powerful enough to power various micro or nanoelectromechanical systems.
Even though the nuclear power sources have always been a safety concern, they’ve claimed to be safe, as the nuclear power sources have been used for powering many types of devices, including the pace-makers, space satellites and underwater systems. Kwon’s battery is in a liquid semiconductor rather than a solid semiconductor, as he believed that the liquid semiconductor can overcome the problem, where the lattice structure of the semiconductor being damaged, if it’s in the solid semiconductor form!
1. A circuit consisting of a constant e.m.f. ‘E’, a self induction ‘L’ and a resistance ‘R’ is closed at t = 0. The relation between the current I in the circuit and time t is as shown by curve ‘a’ in the figure. When one or more of parameters E, R and L are changed, the curve ‘b’ is obtained. The steady state current is same in both the cases. Then it is possible that
(a) (b)
t I
(A) E and R are kept constant and L is increased (B) E and R are kept constant and L is decreased (C) E and R are both halved and L is kept constant (D) E and L are kept constant and R is decreased 2. Consider a resistor of uniform cross section area
connected to a battery of internal resistance zero. If the length of the resistor is doubled by stretching it then
(A) current will become four times
(B) the electric field in the wire will become half (C) the thermal power produced by the resistor
will become one fourth
(D) the product of the current density and conductance will become half
3. In front of an earthed conductor a point charge +q is placed as shown in figure
+q
(A) On the surface of conductor the net charge is always negative
(B) On the surface of conductor at the same points charges are negative and at some points charges may be positive distributed non uniformly (C) Inside the conductor electric field due to point
charge is non-zero (D) None of these
Passage # (Q. No. 4 to Q. No. 6)
Resistance value of an unknown resistor is calculated using the formula R= V/I where V and I be the readings of the voltmeter and the ammeter respectively. Consider the circuits below. The internal resistances of the voltmeter and the ammeter (RV and RG respectively) are finite and non zero.
A R V E r A R V E r
Fig. (A) Fig. (B)
Let RA and RB the calculated values in the two cases A and B respectively.
4. The relation between RA and the actual value R is (A) R > RA (B) R < RA
(C) R = RA (D) dependent upon E and r 5. The relation between RB and the actual value R is
(A) R< RB (B) R > RB
(C) R = RB (D) dependent upon E and R 6. If the resistance of voltmeter is RV = 1 KΩ and
that of ammeter is RG = 1Ω, the magnitude of the percentage error in the measurement of R (the value of R is nearly 10 Ω) is
(A) zero in both cases
(B) non-zero but equal in both cases (C) more in circuit A
(D) more in circuit B Passage # (Q. No. 7 to Q. No. 8)
The figure shows the interference pattern obtained in a double-slit experiment using light of wavelength 600 nm. 1,2,3,4 and 5 are marked on five fringes.
7. The third order bright fringe is (A) 2 (B) 3 (C) 4 (D) 5
8. Which fringe results from a phase difference of 4π
between the light waves incidenting from two slits (A) 2 (B) 3 (C) 4 (D) 5
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
1.[C] Initially the potential at centre of sphere is x Q 4 1 x Q 2 4 1 x Q 4 1 V 0 0 0 C 3 πε = πε + πε =
After the sphere grounded, potential at centre becomes zero. Let the net charge on sphere finally be q. x Qr 3 q or 0 x Q 3 4 1 r q 4 1 0 0 = = πε + πε ∴
∴The charge flowing out of sphere is
x Qr 3
2. [D] The velocity is maximum at mean position. Hence the magnetic force on block is maximum, at its mean position.
The magnetic force on the block while it crosses the mean position towards right and left is as shown N1 vmax Mg + qvmaxB Case-1 N2 vmax mg Case-2 qvmaxB
Hence normal reaction is maximum in case-1 and minimum in case-2. Hence correct option is D. 3. (A) → Q,R, (B) → P,S, (C) → P,R, (D) → Q,S 4. [A,B,C]
Total charge =
∫
Idt=areaunder curve=10CAverage current = 5A dt Idt =
∫
∫
Total heat produced =
∫
I2Rdt∫
− + = =2 0 2 J 3 200 dt . 1 . ) 10 t 5 (Maximum power = I2R, when I is maximum current = 100 × 1 =100W. 5. [B,D] Equivalent circuit x x x r/2 ε r/2 ε r Induced emf 2 a B 2 r B e= ω 2 = ω 2 (∴ Radius = a) By nodal equation, nodal
0 r 0 x r e x 4 = − + − 5x = 4e x = 4e/5 and 5 a B 2 r x i= = ω 2 = + = r 5 e 4 4 / r r e i 6. (A) → R , (B) → Q,S, (C) → P, (D) → Q,R (A) →F constant and = →u×→F=0
Therefore initial velocity is either in direction of constant force or opposite to it. Hence the particle will move in straight line and speed may increase or decrease.
(B) →u.→F=0 and →F constant =
Initial velocity is perpendicular to constant force, hence the path will be parabolic with speed of particle increasing.
(C) →v.→F=0 means instantaneous velocity is always perpendicular to force. Hence the speed will remain constant. And also |→F|=constant. Since the particle moves in one plane, the resulting motion has to be circular.
(D) →u=2∧i−3∧j and →a=6∧i−9∧j. Hence initial velocity is in same direction of constant acceleration, therefore particle moves in straight line with increasing speed.
7. [B] r O C θ v0 q r R qB mv r= 0 ;R= sin2r θ θ = sin m 2 qBR v0 8. [B,C] × ×××× × h dr r r 2 NIhdr hdr B d 0 π µ = × = φ I sin t R b R log 2 h N max 0 total × × ω + π × µ = φ dt d e= φtotal
Solution
Physics Challenging Problems
Set # 2
1. A particle of mass m moves along a horizontal circle of radius R such that normal acceleration of particle varies with time as an = Kt2 , where K is a constant. Calculate
(i) tangential force on particle at time t, (ii) total force on particle at time t,
(iii) Power developed by total force at time t, and (iv) average power developed by total force over
first t second.
Sol. Since, the particle is moving along a circle, therefore, its normal acceleration is centripetal acceleration i.e. v2/R, where v is velocity of particle at time t.
∴ R
v2 = Kt2 or v = t. KR
…(1)
Due to centripetal acceleration, particle follows a circular path but due to it velocity its magnitude does not change. Velocity magnitude increases due to tangential acceleration alone.
∴ Tangential acceleration, at = dt d . v = KR ∴ Tangential force, Ft = mat = m KR Ans. (i) Normal force, Fn = man = mKt2 ∴ Resultant force on particle,
F = Ft2+Fn2
= m K(R K4) t + Ans. (ii) Since, power developed by force →F is given by P = →F→v, therefore, power developed by normal force Fn is always zero because its direction is always perpendicular to the instantaneous direction of motion of the particle. Hence, power is developed by tangential force alone. Figure
Ft
Fn
v
i.e. P = F →t →v = mkRt Ans.(iii) Since, resultant force equals (mass × acceleration), therefore, resultant force is used to accelerate the
body. It means that velocity of the body increases due to resultant force. Hence power developed by resultant force is used to increase kinetic energy of the body.
∴ Average power developed by resultant force = Average rate of increase of KE
Initial kinetic energy (at t = 0), E0 = 0 Kinetic energy at time t, E =
2 1 mv2 = 2 1 mKRt2 …(2) ∴ Average power = t E – E 0 = 2 1 mKRt` Ans. (iv) 2. Figure Shows a particle of mass m = 100 gm,
attached with four identical springs, each of length l = 10 cm. Initial tension in each spring is F0 = 25 newton. Neglecting gravity, calculate period of small oscillations of the particle along a line perpendicular to the plane of the figure
B C P A D m
Sol. Let the particle be displaced slightly through x along a line normal to plane of the figure. Then each spring is further elongated. Since, springs are identical, therefore, increase in tension of each spring will be the same. Let this increase be dE0.
P (F0 + dF0) (F0 + dF0) C A C l l θ θ
First considering forces exerted by spring AP and CP only as shown in Figure.
Restoring force produced by these two springs = (F0 + dF0) 2 sin θ
Since x is very small, therefore, sin l x ≈ θ Neglecting product of very small quantities, restoring force produced by these two springs
Expert’s Solution for Question asked by IIT-JEE Aspirants
Students' Forum
= 2F0 l x
Similarly, restoring force produced by two remaining springs BP and DP will also be equal to
l x F 2 0
∴ Resultant restoring force, F = 2 × l x F 2 0 = l 0 F 4 .x ∴ Restoring acceleration is directly proportional to
displacement x, therefore, the particle executes SHM, ∴ Its period T = on accelerati nt displaceme 2π or T = 0 F 4 m 2π l = 0 F ml π = 0.02π sec Ans. 3. Find the electric potential, at any point on the axis of
a uniformly charged circular disc, whose surface charge density is σ, radius, a.
Sol. Let us consider a small elemental thin ring of width dy.
Area of the ring = 2πy dy
Charge on this elemental ring = (2πy dy)σ
Again, we can consider that this ring is divided into a large number of small elements. Each such element e is at the same distance from P. Hence, potential produced by this ring of width dy at the point P is given by dV, where
dV = 2 2 0 r y dy y 2 4 1 + σ π πε e y O a r P 2 2 y r +
Again, each ring as we go from centre to rim, produces different contributions. Since the distance of each ring from P changes as y increases from 0 to a, hence, total potential produced by the whole ring, V =
∫
= a 0 ydV =∫
πε πσ a 04 0 2 . 2 2 y r ydy + =∫
+ ε σ a 0 2 2 0 r y ydy 2 Put (r2+y2) = P ∴ r2 + y2 = p2 or 2ydy = 2pdp ∴∫
+ 2 2 y r ydy =∫
p pdp =∫
dp= p = r2+y2 ∴∫
+ ε σ a 0 2 2 0 r y ydy 2 = a 0 2 2 0 y r + ε σ = + − ε σ r a r 2 2 2 0 ∴ V = + − ε σ r a r 2 2 2 0 As a special case, if r >> a r2+a2 = r 2 / 1 2 r a 1 + ≈ r + 22 r a 2 1 1 or, V = 0 2ε σ . r 2 a2 × π π = r 4 q 0 πε [Q πa2 = A and Aσ = q]i.e., the result is the same as if all the charge is concentrated at the centre of the ring.
4. Three concentric, conducting spherical shells A,B and C have radii a = 10 cm, b = 20 cm and c = 30 cm respectively. The innermost shell A is earthed and charges q2= 4µC and q3 = 3 µC are given to shells B and C respectively. Calculate charge q1 induced on shell A and energy stored in the system. Sol. System of three concentric shells is as shown in
Figure(A) Since, Shell A is earthed, therefore its potential is zero. But its potential is
a b A q1 q2 q3 C B c (A) a b A (–3µC) C B c (B) (+3µC) (–1µC) (+1µC) (+4µC) V = + + πε c q b q a q 4 1 1 2 3 0 ∴ + + c q b q a q1 2 3 = 0 or q1 = – 3µC Ans.
Now, charges on different surface will be as shown in Figure(B) to calculate energy stored in the system, it can be as considered in three parts : (i) a spherical capacitor having radii a and b and
Its capacitance, C1 = ) a – b ( ab 4πε0
(ii) a spherical capacitor having radii b and c and having charge (q2 + q1) = 1 µC. Its capacitance. C2 = ) b – c ( bc 4πε0
(iii) an isolated sphere of radius c and having charge (q3 + q2 + q1) = 4µC
∴ Energy stored in the system = 1 2 1 C 2 q + 2 2 2 1 C 2 ) q q ( + + 3 2 3 2 1 C 2 ) q q q ( + + = 0.45 joule Ans.
5. In the circuit shown in Figure emf of each battery is E = 20 volts and capacitance is C1 = 5 µF, C2 = 3 µF and C3 = 6 µF. Calculate charge on capacitor C3 when switch S is closed and steady state is reached. Calculate also, heat generated in the circuit.
E +– C1 C2 C3 C1 C2 S + –E
Sol. When Switch S is closed and steady state is reached, the circuit becomes symmetric about the dotted line shown in Figure(A).
E +– C1 C2 C3 C1 C2 +–E (A)
Right part of the circuit is exactly mirror image of the left part. Hence, charges on both plates of capacitor C3 should be identical. But charges on plates of a capacitor are always opposite to each other. It means one of the plates is always positively charged and the other is negatively charged. Both these conditions can be satisfied only if charge on capacitor C3 is zero.
To calculate heat generated in the circuit initial and final charges on all the capacitors must be known. First analyse the circuit when switch S was open Charges on capacitors will be as shown in Figure (B) E + – C1 C2 C3 C1 C2 S + – E – + – + – + + – + –(q1 –q2) q2 (B)
Applying Kirchhoff's voltage law on left mesh, of Figure (B) 1 1 C q + 2 2 1 C q – q – E = 0 …(1) For middle mesh,
3 2 C q + 2 2 C q – 2 2 1 C q – q = 0 …(2) From equation (1) and (2), q1 = 50 µC and q2 = 20 µC
Now consider the circuit when switch S is closed and steady state is reached. The circuit will be as shown in Figure (C) E + – C1 C2 C3 C1 C2 +–E – – + + + + – –q q q q (C)
Applying Kirchhoff's voltage law on left mesh of Figure(C) 1 C q + 2 C q – E = 0 or q = 37.5 µC. After shorting of switch S, increase in charge on capacitor C1 of
Left mesh, ∆q1 = (q – q1) = – 12.5 µC That for capacitor C2 of left mesh,
∆q2 = q – (q1 –q2) = 7.5 µC That for capacitor C3 , ∆q3 = 0 – q2 = – 20µC That for capacitor C2 of right mesh,
∆q4 = (q –q2) = 17.5 µC That for capacitor C1 or right mesh,
∆q5 = q – 0 = 37.5 µC Since, heat generated in the circuit is given by
H =
∑
∆ C 2 ) q ( 2 H = 1 2 1 C 2 ) q (∆ + 2 2 2 C 2 ) q (∆ + 3 2 3 C 2 ) q (∆ + 2 2 4 C 2 ) q (∆ + 1 2 5 C 2 ) q (∆ = 250 × 10–6 joule Ans.Capacitance :
Whenever charge is given to a conductor of any shape its potential increases. The more the charge (Q) given to the conductor the more is its potential (V) i.e. Q ∝ V
⇒ Q = CV
where C is constant of proportionality called capacitance of the conductor C = Q/V, C = Q
SI unit of capacitance is farad (F) and 1 F = 1 coulomb/volt (1CV–1)
Energy stored in a charged capacitor : W = CV02 2 1 = C 2 Q2 = 2 1QV 0 Capacitance of an isolated sphere :
Let a conducting sphere of radius a acquire a potential V when a charge Q is given to it. The potential acquired by the sphere is
V = a 4 Q 0 πε ⇒ C = V Q = 4πε 0a
Charge sharing Between two charged conductors :
C1 V1 C2 V2 q1 = C1V1 q2 = C2V2 (Initially) C1 V C2 V q´1 = C1V q´2 = C2V (Finally) V = 2 1 2 2 1 1 C C V C V C + +
There is always a loss in energy during the sharing process as some energy gets converted to heat. Loss = – ∆U = + 2 1 2 1 C C C C 2 1 (V 1 – V2)2 Capacitor or Condenser :
An arrangement which has capability of collecting (and storing) charge and whose capacitance can be varied is called a capacitor (or condenser)
The capacitance of a capacitor depends.
(a) directly on the size of the conductors of the capacitor.
(b) directly on the dielectric constant K of the medium between the conductors.
(c) inversely on the distance of separation between the conductor.
Principle of a condenser :
Consider a conducting plate A which is given a charge Q such that its potential rises to V. Then
C = Q/V
Let us place another identical conducting plate B parallel to it such that charge is induced on plate B (as shown in figure).
+ + + + + + + + + + + + + + + + A Q
If V– is the potential at A due to induced negative charge on B and V+ is the potential at A due to induced positive charge on B, then
+ + + + + + + + + + + + + + + + A + + + + + + + + – – – – – – – – B C´ = ´ V Q = − +− +V V V Q
Since V´ < V (as the induced negative charge lies closer to the plate A in comparison to induced positive charge).
⇒ C´ > C
Further, if B is earthed from the outer side (see figure) then Vn = V – V
– as the entire positive charge flows to the earth. So
C" = n V Q = − −V V Q ⇒ Cn >> C
So, if an identical earthed conductor is placed in the viscinty of a charged conductor then the capacitance of the charged conductor increases appreciably. This is the principle of a parallel plate capacitor.
Capacitor-1
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Parallel Plate Capacitor : –σ A B + σ d + + + + + + + + – – – – – – – – + + + + + + + + A B A = Area of plate d = Separation between the plates E = ε σ= 0 kε σ
It consists of two metallic plates A and B each of area A at separation d. Plate A is positively charged and plate B is earthed. If K is the dielectric constant of the material medium and E is the field that exists between the two plates, then
E = ε σ = 0 Kε σ = σ= A q and d V E Q ⇒ d V = A K q 0 ε ⇒ C = V q = d A Kε0
If medium between the plates is air or vacuum, then K = 1 ⇒ C0 = d A 0 ε Special Case I :
When the space between the parallel plate capacitor is partly filled with a dielectric of thickness t(<d) If no slab is introduced between the plates of the capacitor, then a field E0 given by E0 =
0 ε σ , exists in a space d. E= K E0 +σ dt
On inserting the slab of thickness t, a field E = K E0 exists inside the slab of thickness t and a field E0 exists in remaining space (d – t). If V is total potential then V = E0(d – t) + Et ⇒ C = V q = − − ε K 1 1 t d A 0 Special Case II :
When the space between the parallel plate capacitor is partly filled by a conducting slab of thickness t(<d).
It no conducting slab is introduced between the plates, then a field E0 =
0
ε σ
exists in a space d. If C0 be the capacitance (without the introduction of conducting slab), then C0 =
d A 0 ε +σ d t E= 0 E0
On inserting the slab, field inside it is zero and so a field E0 =
0
ε σ
now exists in a space (d – t) ⇒ V = E0(d – t) ⇒ V = 0 ε σ (d – t) ⇒ V = 0 A q ε (d – t) ⇒ C = V q = t d A 0 − ε ⇒ C = − ε dt t 1 d A 0 ⇒ C = − d t 1 C0 Since d – t < d ⇒ C > C0
i.e. Capacitance increases on insertion of conducting slab between the plates of capacitor.
Charge induced on a dielectric :
+ + + + + + + + – – – – – – – – –qp + + + + + + + + – – – – – – – – +qp –q +q E 0 Ep E = E0 – Ep
Resultant dielectric field within the plates is E = E0 – Ep
⇒ E = 0 1 ε (σ – σp) ...(1) Also E = 0 Kε σ ...(2) Compare (1) and (2), we get
0 1 ε (σ – σp) = Kε0 σ ⇒ σp = σ − K 1 1 ⇒ A qp = − K 1 1 A q ⇒ qp = q − K 1 1 Spherical capacitor : B C2 A a C1 b
let C1 be the capacitance in between the two conductors and C2 be capacitance out side both. To find C1 :
Imagine the outer surface of B to be earthed. Then –q is the charge induced on the inner surface of B. If V is the potential difference between the two surfaces, then V = Ka 4 q 0 πε + 4 Kb q 0 πε − ⇒ V = K 4 q 0 πε − b 1 a 1 ⇒ C = V q = 4πε 0K − a b ab ...(1) To find C,
Imagine A to be made open circuited (i.e. made non conducting), then
C2 = 4πε0Kb ...(2)
Case I : When battery is connected to B and A is earthed. Then C1 and C2 are in parallel
⇒ C = C1 + C2 ⇒ C = 4πε0K − a b ab + 4πε 0Kb ⇒ C = 4πε0K − a b b2
Case II : When battery is connected to A, then C1 and C2 are in series.
⇒ C 1 = 1 C 1 + 2 C 1 ⇒ C 1 = ab a b− K 4 1 0 πε + 4 Kb 1 0 πε ⇒ C 1 = − + πε a 1 a b Kb 4 1 0 ⇒ C 1 = πε a b Kb 4 1 0 ⇒ C = 4πε0Ka
Case III : When battery connected to A and B is earthed. Then C2 can be omitted as it will not receive any charge. So, C = C1 ⇒ C = 4πε0K − a b ab
Case IV : When battery connected to B and A is open circuited (or made non conducted) then C1 can be omitted (as it is open circuited). So,
C = C2 ⇒ C = 4πε0Kb Cylindrical capacitor :
Let inner cylinder be given a charge per unit length of λ = l q
. A charge – q is induced on length l at inner surface of outer cylinder
b a q –q l E = r 2πε0 λ for a < r < b ⇒ – dr dV = Kr 2πε0 λ ⇒
∫
surface outer surface inner dV = –∫
= = πε λ r b a r 0 r dr K 2⇒ Vinner surface – Vouter surface = K 2πε0 λ loge a b
Since, inner surface is at higher potential and outer at lower potential, so
–q b
a +q
Gaussian surface
⇒ Vouter surface – Vinner surface = K 2πε0 λ loge a b
⇒ Vinner surface – Vouter surface =
K 2 q 0l πε loge a b ⇒ C = surface outer surface inner V V q − = πε a b log K 2 e 0l ⇒ C = πε a b log K 2 e 0l
(A) Energy stored in a capacitor E = 2 1CV2 = 2 1QV = C Q 2 1 2
and energy stored per unit volume = 2 1 ε0E2 Note: The energy is stored in a capacitor is in the form of electric field between the plates.
(B) A parallel plate capacitor is charged by a battery and then the capacitor disconnected from the battery (a) If the distance between plates of the capacitor is increased then the new parameter of the capacitors as compared to the previous parameters is
q' = q; C' = ' d A 0 ε , V' = ' C ' q , E' = ' d '
V (charge will not change) Energy = 2 1C'V'2 If a dielectric slab (dielectric constant k) is introduced between the plates then
q' = q, C' = kC, V' = k V , E' = k E U'(Energy) = k U (charge will not change)
(C) A parallel plate capacitor is charged by a battery. (a) If the distance between plates of the capacitor is Increased (with the battery connected) then the new parameters of the capacitors as compared to the previous parameters is V' = V, C' = ' d A 0 ε , q' = C'V', E' = d ' V , Energy U' = 2 1C'V'2
(b) If a dielectric slab (dielectric constant k) is introduced between two plates then
V' = V, C' = kC, q' = kq, E' = k
E; U' = KU (p.d.) will not change)
1. A capacitor of 20 µF and charged to 500 volt is connected in parallel with another capacitor of 10 µF charged to 200 volt. Find the common potential. Sol. Charge on one capacitor q1 = C1V1
∴ q1 = 20 × 10–6 × 500 = 0.01 coulomb Charge on second capacitor
q2 = 10 × 10–6 × 200 = 0.002 coulomb The charge on the two capacitors
q = q1 + q2 = 0.01 + 0.002 = 0.003 coulomb Total capacity C = C1 + C2 = 20 × 10–6 + 10 × 10–6 = 30 × 10–6 Farad. Common potential = q/C = 6 10 30 012 . 0 − × = 400 Volt.
2. A battery of 10V is connected to a capacitor of capacity of 0.1 F. The battery is now removed and this capacitor is connected to a second uncharged capacitor. If the charge distributes equally on these two capacitors, find the total energy stored in the two capacitors. Further, compare this energy with the initial energy stored in the first capacitor.
Sol. The initial energy stored in the first capacitor. U0 = 2 1 CV2 = 2 1× 0.1 × (10)2 = 5.0 J
When this capacitor is connected to the second uncharged capacitor, the charge distributes equally. This shows that the capacitance of the second capacitor is also C. The voltage across each capacitor will be V/2. If U be the energy stored in the two capacitors, then U = 2 1 C 2 2 V + 2 1 C 2 2 V = 4 1CV2 = 2.5 J 0 U U = 0 . 5 5 . 2 = 2 1
