Dear Students,
One of the greatest virtues of human beings is their ability to think and act accordingly. The emergence of the techno savvy human from the tree swinging ape has really been a long journey. This transition has taken a span of countless centuries and lots of thinking caps have been involved. Inquisitiveness and aspiration to come out with the best have been the pillars for man's quest for development. Self-motivation is the sheer force, which pulled him apart and distinguished him from his primitive ancestors.
Many times, in our life, when we are reviving old memories we get into a phase of nostalgia. We feel that we could have done better than what we had achieved. But thinking back won’t rewind the tireless worker called time. All we can do is promise ourselves that we will give our very best in the future. But do we really keep up to our mental commitments? I can guess that 90% answers are in the negative. This is because of that creepy careless attitude which is slowly, but surely entering into our mind. We easily forget the pains of yesterday to relish the joys of today. This is the only time in our life, when we can control our fate, by controlling our mind. So it is time to pull up our socks and really motivate ourselves so that we can give our best shot in the future. Self-motivation is the need of the hour. Only we can control and restrict ourselves. It’s up to us, how we use our mental capabilities to the best of our abilities.
Here are some Fundas for self-motivation. Don't just read them digest each one of them and apply them and I bet it will make a better YOU. • The ultimate motivator is defeat. Once you are defeated, you
have nowhere to go except the top. • Then only thing stopping you is yourself.
• There is no guarantee that tomorrow will come. So do it today. • Intentions don't count, but action's do.
• Don't let who you are, stunt what you want to be. • Success is the greatest motivator.
• Your goals must be clear, but the guidelines must be flexible. Try to include these one liners in your scrapbook or on your favorite poster. You will be sub-consciously tuned to achieve what you want. Also do keep in mind that nothing can control your destiny but you! With Best Wishes for Your Future.
Yours truly
Pramod Maheshwari, B.Tech., IIT Delhi
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Editor : Pramod Maheshwari
Worry is a misuse of imagination
Volume - 5 Issue - 2
August, 2009 (Monthly Magazine)
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Volume-5 Issue-2 August, 2009 (Monthly Magazine)
NEXT MONTHS ATTRACTIONS Key Concepts & Problem Solving strategy for IIT-JEE. Know IIT-JEE With 15 Best Questions of IIT-JEE Challenging Problems in Physics, Chemistry & Maths Much more IIT-JEE News.
Xtra Edge Test Series for JEE-2010 & 2011
S
Success Tips for the Months
• "All of us are born for a reason, but all of us don't discover why. Success in life has nothing to do with what you gain in life or accomplish for yourself. It's what you do for others."
• "Don't confuse fame with success. Madonna is one; Helen Keller is the other."
• "Success is not the result of spontaneous combustion. You must first set yourself on fire."
• "Success does not consist in never making mistakes but in never making the same one a second time."
• "A strong, positive self-image is the best possible preparation for success."
• "Failure is success if we learn from it." • "The first step toward success is taken
when you refuse to be a captive of the environment in which you first find yourself."
CONTENTS
INDEX
PAGE
NEWS ARTICLE
3
69 of top 100 JEE rankers pick IIT-Bombay
IIT-JEE stars eye glory in International Physics Olympiad
IITian ON THE PATH OF SUCCESS
6
Abhay K. Bhushan
KNOW IIT-JEE
7
Previous IIT-JEE Question
XTRAEDGE TEST SERIES
45
Class XII – IIT-JEE 2010 PaperClass XI – IIT-JEE 2011 Paper
Regulars ...
DYNAMIC PHYSICS
13
8-Challenging Problems [Set# 4] Students’ Forum
Physics Fundamentals Capacitor - 2
Work, Energy, power & Conserv. law
CATALYST CHEMISTRY
26
Key ConceptReaction Mechanism Solid State
Understanding : Physical Chemistry
DICEY MATHS
35
Mathematical Challenges Students’ Forum Key Concept VectorPermutation & Combination
Study Time...
Cream of the crop: 69 of
top 100 JEE rankers
pick IIT-Bombay
MUMBAI: The composition of the elite technological club has changed. A decade ago admission to the IIT-Kanpur ensured demi-god treatment. Only the brightest and the best could get past the gates there. No longer. Mumbai is the new Kanpur, with Delhi and Chennai snapping at its heels. A look at the students’ choice of institute by the top 100 JEE rankers down the last half-a-decade reveals that preferences have changed dramatically. A number of factors have been responsible for the reordering, from geography to gastronomy and placement records to what coaching classes preach to students.
Of the top 100 JEE-2009 rankers, considered the elite group among engineering aspirants around the country, 69 students preferred to join IIT-Bombay over any other IIT. This was followed by Delhi — where 19 of the top-100 — have been admitted. While Bombay has been bettering its performance over the years, number of toppers going to Delhi has slipped.
"IIT-B's decision to introduce minors in all programmes has seen more students wanting to come to the Powai campus," reasoned the institute's JEE-2009 chairman A Pani. In 2008, the institute ushered in
academic reforms and permitted students to pick a minor course along with the core area of specialisation. This, explained Pani, has resulted most streams opening and closing admissions at higher ranks than previous years.
On each IIT campus, the top 100 students are considered as the rich creamy icing. Twenty years ago IIT-Kharagpur was the engineering mecca. The oldest IIT of the country, IIT-Kharagpur did not receive a single student from the top hundred this year; and before that, in 2004, only three of the top 100 went there.
A former JEE chairman explained, "While Bombay and Delhi were still building themselves, Kharagpur's students had already occupied top positions in big companies. Students looked at Kharagpur's illustrious alumni and rushed there. Now this has changed."
1,100 quota seats in IITs
not filled this year
MUMBAI: Every year, lakhs of students burn the midnight oil for months to get into the hallowed Indian Institutes of Technology. But as admissions closed on Wednesday, one startling fact emerged — there weren't enough qualified candidates to fill up the reserved seats on offer for the scheduled castes and scheduled tribes, or the physically challenged.
IIT heads told TOI that over 1,100 seats will now be transferred to the preparatory course. This course, which is like a feeder class, trains quota students for a year to equip them to qualify for the IITs. Students for the preparatory course are selected by reducing cut-offs even further. On the OBC (other backward classes) reservation front too, 53 seats were transferred to general category candidates, though the IITs are still only in the second year of the quotas (they are implementing 18% quota before moving to the total 27% reservation).
The IITs, in fact, had made various concessions to ensure they could fill the SC/ST seats. They lowered entry levels for these categories and even went as low as 50% below the last general category student's marks to do justice to the quota. Even this did not help them get the required number of backward category students.
Reservation for IIT
faculty to stay: Sibal
NEW DELHI: Reservations in faculty at the Indian Institutes of Technology will continue. HRD minister Kapil Sibal made it clear on Wednesday that efforts to exempt the elite institutions from quotas for SCs, STs and OBCs in the teaching staff had “proved infructuous’’.
He made the announcement at a meeting with IIT directors where he also told them to
explore the possibility of offering courses in medicine, law, social sciences and literature. As first reported by TOI on November 20, 2008, IITs too are keen to branch out to new subjects and multiple disciplines.
Sibal’s remark about quotas in the IIT faculty signals that the government may not make another push to bring in the Scheduled Caste and Scheduled Tribes (Reservation in Posts and Services) Bill, 2008. The bill had sought to exempt 47 elite institutions from faculty quota. It could not be passed in the Lok Sabha due to opposition from UPA allies like the RJD. Sibal’s remark came in response to a clarification sought by an IIT directors. The IITs are staunchly opposed to such a quota
Now, IIT-JEE stars eye
glory in International
Physics Olympiad
MUMBAI: After two years of poring over texts to ace the IIT-JEE, toppers now have to face another challenge. They are on their way to H1N1-hit Mexico where they will represent the country in the International Physics Olympiad in the first week of July.
The team that went in 2008 brought home four golds and a silver medal. This year's gang of boys would have a tough task cut out for them, professor Vijay Singh, national coordinator of the science Olympiads, said.
The team members-Nitin Jain (all-India Rank 1 in JEE), Shubham Tulsiani (AIR 2), Gopi Sivakanth (AIR 3), Priyank Parikh (AIR 6) and Vinit Atal (AIR 90)-are in the city, preparing for the big challenge.
Every year, the Homi Bhabha Centre for Science Education conducts a massive exercise to select the brightest brains from across the country who then represent India in the international Olympiads. Eighty countries will participate in the physics Olympiad. Last year's winning team was China. Mentor professor Singh said the team was putting in close to 12 hours a day at the camp. "Our students are champions in chemistry and maths as well. If there was a comprehensive Olympiad, the Indian team would win hands down,'' said Singh.
IIT-Patna to start PhD
programmes from July
PATNA: The newly set-up Indian Institute of Technology (IIT) in this Bihar city will start its doctoral programmes from next month, an official said on Friday.
"IIT-Patna will become the first among the eight new IITs set up last year to start PhD programmes," institute official Subhash Pandey said. The IIT will have PhD programmes in computer science, electrical engineering, mechanical engineering, chemistry, mathematics, physics, humanities and social sciences.
Pandey said that interviews of the applicants are underway and there are 30 vacancies. At present, the IIT is functioning from a polytechnic building here as a temporary campus. The process of land acquisition for a permanent campus is underway.
Plan panel favours IIT,
IIM offshore campuses
NEW DELHI: Doors may soon be open for Indian universities and government-run institutions like IIMs and IITs to set up campuses abroad to cross-subsidise higher education for vulnerable sections of society. The Planning Commission is in favour of formulating guidelines to allow Indian universities and government-run institutions to run business abroad to fund higher education for the poor back home and to expand the educational infrastructure in the country.
The move has come at the time when India is wooing foreign universities to set up campuses in the country.
Interestingly, as of now, there are no rules and regulations to
permit government-run institutions to set up offshore
campuses. So far, only private educational institutions were free to explore education opportunities abroad. Private institutions like Symbiosis and BITS, Pilani, have already opened campuses abroad. Only in May this year, Pune University became the first government-run institution to open its campus abroad, in UAE, after considerable legal and bureaucratic hurdles. The human resources development ministry had objected to the proposal of Pune University on the ground that there were no guidelines on opening campuses on foreign soil by government-run institutions. Pune University had to knock the doors of the PMO to get its proposal cleared.
Faculty divided over
location of IIT
JAIPUR: As the recommendation made by the state government-appointed Vyas committee on having the Indian Institute of Technology (IIT) in Jodhpur is a debate in itself, those who'll matter the most wherever the premier institute comes up - the faculty stand divided on whether the premier institute should come up in the capital city or somewhere else in the desert state.
Prem K Kalra, director, IIT-Rajasthan, reserves his opinion about the development. He says, "I am unaware of the grounds on which the Vyas committee has given nod to Jodhpur. I know what works for Jaipur, but will have to read the report to make a comment as this is a sensitive issue." While Kalra distances himself from making a comment, Nina Sabnani, who teaches animation and visual communication at IIT-Mumbai says, "An IIT is self sufficient to create its own brand. Its success doesn't depend on the place where it is located. If Mumbai is big and popular, IIT-Kharagpur too has made its mark."
Faculties across IIT's agree that the three basics behind the success of any IIT remain infrastructure, faculty and connectivity. "If these criteria are fulfilled, than the location, makes no difference," says Prof V K Vijay of IIT-Delhi.
But what might make a difference is that the IIT's reeling under deficit of trained faculty might find it tough to get the right kind of people to smaller city like Jodhpur. Not
willing to reveal her name a faculty at IIT-Mumbai says, "IIT anywhere will intellectually stimulate the place, but the place too needs to give back and stimulate those who will be there at the IIT campus. This is what that gives an edge to a bigger city which can provide better exposure to the faculty who are core to the success of any IIT."
Her thoughts are echoed by Kalra who feels that there is a complex matrix which has issues like the developmental prospects for the faculty, their family members, educational facility for their children and opportunities for their spouses which determines the success and feasibility of having an IIT anywhere. On these counts Pink City has an edge over any other center in the state. Other issues can be addressed, but managing faculty will be a challenge that will show its effect in the long run. As Vijay concludes, "There is an over all deficits of faculties across the board and to add to the woes the government in haste added seven more IIT's to the current ones. This will certainly dilute the brand in the long run."
Nachiket sets sights on IIT
AHMEDNAGAR: For Nachiket Kuntala, who emerged joint topper from Pune division in the SSC exams, securing first position comes as a matter of habit. Right from std I to IX, Nachiket secured the number one position and the SSC exam did not prove an exception. Nachiket, a student of the Shri Samarth Vidyamandir here, scored 627 marks (96.46%) to share the divisional top spot with Pune's Akshay Chate. Interested in an engineering
research career, Nachiket told TOI: "I wish to pursue my higher studies at the Indian Institute of Technology (IIT)." "Regular studies and focused approach were key to my success," Nachiket said. He did join a coaching class to hone his academic skills, but a routine of physical exercise, studies and extra-curricular activities kept him in good stead.
"I was particular about doing my home work and revisiting all those things taught at the coaching class," he said. Nachiket's father is a medical professional, while his mother teaches science in a school.
IIT Kanpur to open
extension centre in
Noida
The HRD Ministry has granted permission to IIT Kanpur to open an extension centre in Noida, work on which will start within a week.
IIT Kanpur Registrar Sanjiv Kashalkar told PTI that the work will be completed by 2012.
He said that a 'distance learning centre' will also be opened there.
Kashalkar said the centre will function on the lines of India International Centre with technocrats imparting technical
education through conferences.
It will also provide several short-term management courses and refresher courses meant for distance learning, he said.
The premier institute has been granted five acres of land in sector 62 of Noida.
Abhay K. Bhushan ( B.Tech. /Electrical Engg. / 1965 ) Chairman
A Square and serves on the boards of Point Cross and Mobile Web Surf
He obtained his B. Tech degree in Electrical Engineering from the Indian Institute of Technology Kanpur, in 1965. He obtained both his Masters in EE and Masters in Management degrees from the Massachusetts Institute of Technology. He has been the mentor of a host of start-up ventures in USA. He was a major contributor to the development of the Internet TCP/IP architecture, and was the author of FTP and the early versions of email protocols. He is co-holder of 12 US patents on semiconductor drying and cleaning technologies.
Mr. Bhushan initiated and managed the Environmental Leadership Program at Xerox and authored the widely acclaimed ‘Business Guide to Waste Reduction and Recycling'.
He was co-founder of YieldUP International, which went public on NASDAQ in 1995, and of Portola Communications, which was acquired by Netscape in 1997. In 1978-79 he worked on Rural Development in Allahabad, India, and was President of Indians for Collective Action, supporting grassroots development projects in India. He received the Community Service Award from the Indo-American Chamber of Commerce. He is presently the Chief Financial Officer of the IITK Foundation, USA, founding past president of PanIIT USA, and Coordinator for the PanIIT Global Committee. Mr. Abhay K. Bhushan has been conferred with the Distinguished Alumnus Award of IIT Kanpur, for excellence in entrepreneurship and his outstanding contributions to social activities.
Abhay K. Bhushan
B.Tech. /Electrical Engg. / 1965
Chief Financial Officer of the IITK Foundation, USA,
Adventure :
• Adventure is not outside man; it is within.
• There are two kinds of adventures : those who go truly hoping to find adventure and those who go secretly. hoping they won't.
• Life is either a daring adventure or nothing.
• Some people dream of worthy accomplishments while others stay awake and do them. • Life is an adventure. The greatest pleasure is doing what people say you cannot do.
Success Story
PHYSICS
1. One mole of an ideal monatomic gas is taken round
the cyclic process ABCA as shown in figure. Calculate. [IIT-1998] B C A 2V0 V0 P 3P0 P0
(a) the work done by the gas.
(b) the heat rejected by the gas in the path CA and the heat absorbed by the gas in the path AB;
(c) the net heat absorbed by the gas in the path BC; (d) the maximum temperature attained by the gas during the cycle.
Sol. n = 1 = no. of moles, For monoatomic gas :
Cp = 2 R 5 , Cv = 2 R 3 Cyclic process A → B ⇒ Isochoric process C → A ⇒ Isobaric compression
(a) Work done = Area of closed curve ABCA during
cyclic process. i.e. ∆ABC ∆W = 2 1 × base × height = 2 1 V0 × 2P0 = P0V0
(b) Heat rejected by the gas in the path CA during
Isobaric compression process
∆QCA = nCp∆T = 1 × (5R/2)(TA – TC) TC = R I V P 2 0 0 × , TA = I R V P0 0 × , ∆QCA = − R V P 2 R V P 2 R 5 0 0 0 0 = – 2 5 P0V0
Heat absorbed by the gas on the path AB during Isochoric process ∆QAB = nCv∆T = 1 × (3R/2) (TB – TA) = × − × 1 R V P R 1 V P 3 2 R 3 0 0 0 0 = 3P0V0
(c) As ∆U = 0 in cyclic process, hence ∆Q = ∆W ∆QAB + ∆QCA + ∆QBC = ∆W, ∆QBC = P0V0 – 2 V P0 0 = 2 V P0 0
As net heat is absorbed by the gas during path BC, temp. will reach maximum between B and C.
(d) Equation for line BC P = – 0 0 V P 2 V + 5P0, As PV = RT hence, P = V RT
[For one mole] [as y = mx + c]
∴ RT = – 0 0 V P 2 V2 + 5P 0V ...(1) For maximum; dV dT = 0, – 0 0 V P 2 × 2V + 5P0 = 0; ∴ V = 4 V 5 0 ...(2) Hence from equation (1) and (2)
RTmax = – 0 0 V P 2 × 2 0 4 V 5 + 5P0 4 V 5 0 = –2P0V0 × 16 25 + 4 V P 25 0 0 = 8 25 P0V0 ∴ Tmax = 8 25 R V P0 0
2. A small body attached to one end of a vertically hanging spring is performing SHM about it's mean position with angular frequency ω and amplitude a. If at a height y* from the mean position, the body gets detached from the spring calculate the value of y* so that the height H attained by the mass is maximum. The body does not interact with the spring during it's subsequent motion after detachment. (aω2 > g)
[IIT-2005]
m y0
Sol. The total energy of the spring-mass system at any position of mass above the mean position is the sum of the follows.
(a) Gravitation potential energy of mass (b) Kinetic energy of mass
(c) Elastic potential of spring.
KNOW IIT-JEE
The mass will reach the highest point when its mechanical energy [Sum of (a) and (b)] is maximum. This is possible when elastic potential energy of system is zero.
⇒ The mass should detach when the spring is at its natural length.
Let L = Natural length of spring when mass m is hanging at equilibrium the
K mg l Kl L L Mean Position of oscillation mg = kl ; l = k mg ⇒ y = k mg ⇒ y = g2 ω [Q K = mω 2] where g2 ω < a (given)
3. A source of sound is moving along a circular orbit of radius 3 metres with an angular velocity of 10 rad/s. A sound detector located far away from the source is executing linear simple harmonic motion along the line BD with an amplitude BC = CD = 6 meters. The frequency of oscillation of the detector is 5/π per second. The source is at the point A when the detector is at the point B. If the source emits a continuous sound wave of frequency 340 Hz, find the maximum and the minimum frequencies recorded by the detector. [IIT-1990]
3m B C D A B C D A3 A2 A1 6m 6m
Sol. The angular frequency of the detector = 2πv Observer T Source T´ (Accoustic Image) Hill = 2π × π 5 = 10 rad/s
⇒ The angular frequency of the detector matches with that of the source.
A´
A´´
A B C D 6m 6m ω = 10 rad/sec
⇒ When the detector is at C moving towards D, the source is at A´ moving Left wards, It is in this situation that the frequency heard is minimum
v´ = v + − s 0 V V V V = 340 × ) 30 340 ( ) 60 340 ( + − = 257.3 Hz Again when the detector is at C moving towards B, the source is at A" moving rightwards. It is in this situation that the frequency heard is maximum
v´´ = v − + s 0 V V V V = 340 × ) 30 340 ( ) 60 340 ( − + = 438.7 Hz
4. A wire loop carrying a current I is placed in the x-y plane as shown in figure. [IIT-1991]
x y O v M +Q P a 120º N I
(a) If a particle with charge +Q and mass m is placed at the centre P and given a velocity →V along NP (see figure), find its instantaneous acceleration.
(b) If an external uniform magnetic induction field →
B = B iˆ is applied, find the force and the torque acting on the loop due to this field.
Sol. (a) Magnetic field at the centre P due to arc of circle, Subtending an angle of 120º at centre would be :
x y M +Q P a 60º N I 60º a r x y 60º v B1 = 3 1
(field dut to circle) = 3 1 a 2 I µ0 = a 6 I µ0 (outwards) = a I µ 16 . 0 0 (outwards) or B1 → = a I µ 16 . 0 0 kˆ
B2 = π 4 µ0 r I (sin 60º + sin 60º) Here, r = a cos 60º ∴ B2 = π 4 µ0 º 60 cos a I (2 sin 60º) or B2 = a I 2 µ0 π tan 60º = a I µ 27 . 0 0 (inwards) or B2 → = – a I µ 27 . 0 0 kˆ ∴B→net = → 1 B +B = –→2 a I µ 11 . 0 0 kˆ
Now, velocity of particle can be written as, →
v = v cos 60º iˆ + v sin 60º jˆ = 2 v iˆ + 2 v 3 jˆ Magnetic force → m F = Q(→v ×→B ) = a 2 IQv µ 11 . 0 0 jˆ – iˆ a 2 IQv µ 3 11 . 0 0 ∴ Instantaneous acceleration → a = m F→m = am 2 IQv µ 11 . 0 0 ) iˆ 3 jˆ ( −
(b) In uniform magnetic field, force on a current loop
is zero. Further, magnetic dipole moment of the loop will be, M =(IA) kˆ →
Here, A is the area of the loop. A = 3 1(πa2) – 2 1 [2 × a sin 60º] [a cos 60º] = 3 a2 π – 2 a2 sin 120º = 0.61 a2 ∴ M =(0.61 Ia→ 2) kˆ Given, →B = B iˆ ∴ →τ = M × → →B = (0.61 Ia2B) jˆ
5. In a series L-R circuit (L = 35 mH and R = 11 Ω), a variable emf source (V = V0 sin ωt) of Vrms = 220 V
and frequency 50 Hz is applied. Find the current amplitude in the circuit and phase of current with respect to voltage. Draw current time graph on given graph (π = 22/7) [IIT-2004] V = V0sinωt T/2 3T/2 2T T/4 O Sol. Given Vrms = 220 V v = 50 Hz, L = 35 mH, R = 11 Ω Impedance Z = (wL)2+R2 = 11 2Ω also I0 = Z V0 V0 = Vrms 2 ∴ I0 = Z 2 Vrms = 20A cos φ = Z R = 2 1 ∴ φ = 4 π ∴ graph is given by V = V0sin ωt I = I0sin(100πt–π/4) V = V0sin ωT
CHEMISTRY
6. One litre of a mixture of O2 and O3 at STP was
allowed to react with an excess of acidified solution if KI. The iodine liberated required 40 mL of M/10 sodium thiosulphate solution for titration. What is the mass percent of ozone in the mixture? Ultraviolet radiation of wavelength 300 nm can decompose ozone. Assume that one photon can decompose one ozone molecule, how many photons would have been required for the complete decomposition of ozone in the original mixture ? [IIT-1997] Sol. The reaction of O3 with I– in acidic medium is
O3 + 2I– + 2H+ → I2 + O2 + H2O
Hence, 1 mol O3 = 1 mol I2
The reaction of I2 with S2O32– is
2S2O32– + I2 → S4O62– + 2I–
Hence, 2 mol S2O32– ≡ 1 mol I2
Amount of S2O32– consumed = (40 × 10–3L) molL−1 10 1 = 40 × 10–4 mol Thus 40 × 10–4 mol S 2O32– ≡ 20 × 10–4 mol I2 ≡ 20 × 10–4 mol O3
Mass of O3 present in 1 L of mixture
= (20 × 10–4 mol) (48 g mol–1) = 9.6 × 10–2 g
Total amount of O2 an O3 present in 1 L of mixture at
ntotal = RT pV = ) K 273 )( mol LK atm 082 . 0 ( ) L 1 )( atm 1 ( 1 1 − − = 4.462 × 10–2 mol Hence,
Amount of O2 present in 1 L of mixture
= (4.462 × 10–2 – 20 × 10–4) mol
= 4.262 × 10–2 mol
Mass of O2 present in 1 L of mixture
= (4.262 × 10–2 mol) (32 g mol–1) = 1.364 g
Mass percent of O3 in the mixture
= 364 . 1 10 6 . 9 10 6 . 9 2 2 + × × − − × 100 = 6.575 Amount of photons required to decompose O3
= Amount of O3 = 20 × 10–4 mol
Number of photons required
= (20 × 10–4 mol) (6.023 × 1023 mol–1) = 1.205 × 1021
7. 0.15 mol of CO taken in a 2.5 L flask is maintained at 705 K along with a catalyst so that the following reaction takes place
CO(g) + 2H2(g) CH3OH(g)
Hydrogen is introduced until the total pressure of the system is 8.5 atm at equilibrium and 0.08 mol of methanol is formed. Calculate (a) Kp and Kc and (b)
the final pressure if the same amount of CO and H2 as
before are used, but with no catalyst so that the reaction does not take place. [IIT-1993] Sol. We have
CO(g) + 2H2(g) CH3OH(g)
t = 0 0.15 mol
teq 0.15 mol – x (nH2)0 – 2x x
It is given that 0.08 mol of CH3OH is formed at
equilibrium. Hence nCH3OH = x = 0.08 mol
and nCO = 0.15 mol – x = 0.07 mol
From the total pressure of 8.5 atm equilibrium, we calculate the total amount of gases, i.e. CO, H2 and
CH3OH at equilibrium. ntotal = RT pV = ) K 705 )( mol K L atm 082 . 0 ( ) L 5 . 2 / mol 08 . 0 ( 1 1 – − = 0.3676 mol
Now, the amount of H2 at equilibrium is given as 2 H n = ntotal – nCO – nCH3OH = (0.367 – 0.07 – 0.08) mol = 0.2176 mol Hence, KC = 2 2 3 ] H ][ CO [ ] OH CH [ = 2 ) L 5 . 2 / mol 2176 . 0 )( L 5 . 2 / mol 07 . 0 ( ) L 5 . 2 )( atm 5 . 8 ( = 150.85 (mol L–1)–2 Now Kp = Kc(RT)∆vg
= (150.85 mol–2L2){(0.082 L atm K–1 mol–1)(705 K)}–2 = 0.04513 atm–2 Since nH2= (nH2)0 – 2x, we have ( 2 H n )0 = nH2+ 2x = (0.2176 + 2 × 0.08)mol = 0.3776 mol
Total amount of CO and H2 in the reacting system
before the reaction sets in is given as n0 = (nCO)0 + (nH2)0 = (0.15 + 0.3776)mol = 0.5276 mol Hence, p0 = V RT n0 = ) L 5 . 2 ( ) K 705 )( mol K atm L 082 . 0 )( mol 5276 . 0 ( −1 −1 = 12.20 atm
8. An ester A(C4H8O2), on treatment with excess methyl
magnesium chloride followed by acidification, gives an alcohol B as the sole organic product. Alcohol B, on oxidation with NaOCl followed by acidification, gives acetic acid. Deduce the structures of A and B. Show the reactions involved. [IIT-1998] Sol. The reactions of an ester with methyl magnesium
chloride are as follows.
R–C–OR´ (A) O CH3MgCl R–C–OR´ OMgCl CH3 H+ –HOMgCl R–C–CH3 + R´OH O CH3MgCl R–C–CH3 OH H+ R–C–CH3 OMgCl CH3 –HOMgCl CH3 (B)
Since the given ester (C4H8O2) produces only one
alcohol B, it follows that RC(CH3)2OH and R´OH
must be identical. Thus, the alkyl group R´ must be RC(CH3)2 – and the given ester A is
R – C – O – C – CH3
O CH3
R
(molecular formula R2C4H6O2 )
From the molecular formula of A, we conclude that R must be H atom. Hence, the given ester is
H – C – O – CH – CH3
O
CH3
Isopropyl formate
The alcohol B is a secondary alcohol. CH3 – CH – CH3
OH
Isopropyl alcohol
The oxidation of alcohol B with NaOCl will give a ketone which further undergoes a haloform reaction. CH3 – CH – CH3 + NaOCl
O
CH3 – C – CH3 + NaCl + H2O
CH3 – C – CH3 + 3NaOCl O CH3 – C – CCl3 + 3NaOH O CH3 – C – CCl3 + NaOH O CH3 – C – O–Na+ + CHCl3 O
The acidification of sodium acetate will produce acetic acid.
9. 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 reactions are as follows.
O CH3MgBr OMgBr CH3 H+ –H2O CH3 HBr CH3 Br (A) (B) (E) CH3 O O COCH3 O Base COCH3 (D) (C)
The conversion of C into D may involve the following mechanism. COCH3 (C) CH2 O –BH+ B+ COCH3 HC O COCH3 HC O– –B BH+ COCH3 OH –BH+ +B COCH3 OH – –OH– COCH3 (D)
10. A hydrated metallic salt A, light green in colour, gives a white anhydrous residue B after being heated gradually. B is soluble in water and its aqueous solution reacts with NO to give a dark brown compound C. B on strong heating gives a brown residue and a mixture of two gases E and F. The gaseous mixture, when passed through acidified permanganate, discharge the pink colour and when passed through acidified BaCl2 solution, gives a
white precipitate. Identify A, B, C, D, E and F.
[IIT-1988]
Sol. The given observations are as follows. (i) ) A ( salt metallic Hydrated heat → ) B ( residue anhydrous white
(ii) Aqueous solution of B →NO
) C ( compound brown dark
(iii) Salt B heatingstrong →
) D ( residue Brown + ) F ( ) E ( gases Two + Gaseous mixture (E) + (F) acidified KMnO4 BaCl2 solution Pink colour is discharged White precipitate (iv)
The observation (ii) shows that B must be ferrous sulphate since with NO, it gives dark brown compound according to the reaction
[Fe(H2O)6]2+ + NO → brown dark 2 5 2O) (NO)] H ( Fe [ + + H2O
Hence, the salt A must be FeSO4.7H2O
The observation (iii) is 2FeSO4 → brown(D) 3 2O Fe + 43 42 1 ) F ( ) E ( 3 2 SO SO + +
The gaseous mixture of SO2 and SO3 explains the
observation (iv), namely,
colour pink 4 MnO 2 −+ 5SO2 + 2H2O → colour no 2 Mn 2 + + 2− 4 SO 5 + 4H+ 2H2O + SO2 + SO3 4H+ + SO32– + SO42– Ba2+ + SO 32– → ppt white 3 BaSO Ba2+ + SO 42– → ppt white 4 BaSO
Hence, the various compounds are (A) FeSO4.7H2O (B) FeSO4
(C) [Fe(H2O)5NO]SO4 (D) Fe2O3
(E) and (F) SO2 and SO3
MATHEMATICS
11. Prove that
tan α + 2 tan 2α + 4 tan 4α + 8 cot 8α = cot α
[IIT-1988] Sol. We know that
cot θ – tan θ = θ θ − tan tan 1 2 = 2 θ θ − tan 2 tan 1 2 = 2 cot 2θ ∴ L.H.S. = tan α + 2 tan 2α + 4 tan 4α + 8 cot 8α = –{cot α – tan α – 2tan 2α – 4 tan 4α}
+ 8 cot 8α + cot α = –{2 cot 2α – 2 tan 2α – 4 tan 4α}
+ 8 cot 8α + cot α = –{2(2 cot 4α) – 4 tan 4α} + 8 cot 8α + cot α = – 4 {cot 4α – tan 4α} + 8 cot 8α + cot α = – 8 cot 8α + 8 cot 8α + cot α
12. Find the smallest positive number p for which the equation
cos (p sin x) = sin(p cos x) has a solution x ∈ [0, 2π]
[IIT-1995] Sol. cos (p sin x) = sin (p cos x) (given) ∀ x ∈[0, 2π]
⇒ cos (p sin x) = cos −π pcosx 2 ⇒ p sin x = 2nπ ± −π x cos p 2 , n ∈ I [Q cos θ = cos α ⇒ θ = 2nπ ± α, n ∈ I] ⇒ p sin x + p cos x = 2nπ + π/2 or p sin x – p cos x = 2nπ – π/2, n∈ I ⇒ p. 2 + cosx 2 1 x sin 2 1 = 2nπ + π/2 or p 2 − cosx 2 1 x sin 2 1 = 2nπ – π/2, n ∈ I ⇒ p 2 π + π x cos 4 sin x sin 4 cos = 2nπ + 2 π or p 2 π − π x cos 4 sin x sin 4 cos = 2nπ – 2 π, n ∈ I ⇒ p 2 +π 4 x sin = (4n + 1) 2 π, n ∈ I or p 2 −π 4 x sin = (4n – 1) 2 π, n ∈ I Now, –1 ≤ sin (x ± π/4) ≤ 1 ⇒ –p 2 ≤ p 2 sin (x ± π/4) ≤ p 2 ⇒ –p 2 ≤ 2 ). 1 n 4 ( + π ≤ p 2 , n ∈ I or –p 2 ≤ 2 ) 1 n 4 ( − π ≤ p 2 , n ∈ I
Second inequality is always a subset of first, therefore, we have to consider only first.
It is sufficient to consider n ≥ 0, because for n > 0, the solution will be same for n ≥ 0.
If n ≥ 0, – 2 p ≤ (4n + 1) π/2 ⇒ (4n + 1) π/2 ≤ 2 p
For p to be least, n should be least ⇒ n = 0
⇒ 2 p ≥ π/2 ⇒ p ≥ 2 2
π
Therefore least value of p = 2 2
π
13. Prove that cos tan–1{(sin cot–1x)} = 2 x 1 x 2 2 + + [IIT-2002] Sol. L.H.S. = cos tan–1{sin(cot–1x)}
= cos tan–1 + − 2 1 x 1 1 sin sin = cos + − 2 1 x 1 1 tan = 2 x 1 x 2 2 + + = R.H.S.
14. Let f be a one-one function with domain {x, y, z} and range {1, 2, 3}. It is given that exactly one of the following statement is true and the remaining two are false f(x) = 1, f(y) ≠ 1, f(z) ≠ 2 determine f–1(1)
[IIT-1982] Sol. It gives three cases :
Case I. When f(x) = 1 is true In this case remaining two are false ∴ f(y) = 1 and f(z) = 2
This means x and y have the same image so f(x) is not an injective, which is a contradiction
Case II. When f(y) ≠ 1 is true. If f(y) ≠ 1 and f(z) = 2
i.e. both x and y are not mapped to 1. So either both associate to 2 or 3, Thus, it is not injective
Case III. When f(z) ≠ 2 is true
If f(z) ≠ 2 is true then remaining statements are false ∴ If f(x) ≠ 1 and f(y) = 1
But f is injective
Thus we have f(x) = 2, f(y) = 1 and f(z) = 3 Hence, f–1(1) = y
15. Find a point on the curve x2 + 2y2 = 6 whose distance
from the line x + y = 7, is minimum. [IIT-2003]
Sol. Let us take a point P( 6 cos θ, 3 sin θ) on
3 y 6 x2 2
+ = 1. Now to minimise the distance from P to given straight line x + y = 7, shortest distance exists along the common normal.
X O P Y x + y = 7 Slope of normal at P = θ θ ec cos 6 sec 6 = 2 tan θ = 1 So, cos θ = 3 2 and sin θ = 2 1 Hence, P(2, 1)
Passage # 1 (Q. 1 & Q. 2)
A battery of 10V/1Ω is connected between the terminals of 'a' and 'b' of an infinite planner ladder network of resistances then find the followings. Take r = 10Ω
r r r r r r r r r r r r b r r r r r r r r r r r r r r r r r r r r r r r r r a
1. What will be the value of terminal voltage of the battery.
2. Find the heat developed inside the battery in 1sec.
Passage # 2 (Q. 3 & Q. 4)
If a non ideal battery of 50V/0.5Ω is connected between terminals 'a' and 'b' then find the ratio voltmeter reading to the emf of the battery
10Ω 5Ω 5Ω 5Ω 10Ω a key-k b
3. When key K is open
4. When key K is closed
5. A particle of mass m is allowed to oscillate near the minimum point of a vertical parabolic path having the equation x2 = 4ay, then the angular frequency of
small oscillation of particle is
y
m g x
2 = 4ay
x
(A) ga (B) ga2 (C) g/a (D) g/2a
6. ABC is a fixed incline plane with D mid point of AC. Part AD of incline plane is rough such that when a sphere released from A starts rolling, while the part DC is smooth. The sphere reaches the bottom point C, then
A
B C
rough smooth
D
(A) It is in pure rolling in the part DC
(B) Work done by friction on the sphere is negative when it moves from A to D
(C) Mechanical energy of sphere remains constant for its motion from A to C
(D) All of the above
7. A parallel plate capacitor of plate area A and separation d is provided with thin insulating spacers to keep its plates aligned in an environment of fluctuating temperature. If the coefficient of thermal expansion of material of plate is α then the coefficient of thermal expansion (αS) of the spacers
in order that the capacitance does not vary with temperature (ignore effect of spacers on capacitance) (A) αS = α/2 (B) αS = 3α (C) αS = 2α (D) αS = α
8. We have an infinite non-conducting sheet of negligible thickness carrying a uniform surface charge density –σ and next to it an infinite parallel slab of thickness D with uniform volume charge density +ρ. All charges are fixed
D
–σ +ρ
(A) Magnitude of electric field at a distance h above the negatively charged sheet is
0 2 D ε σ − ρ
(B) Magnitude of electric field inside the slab at a distance h below the negatively charged sheet (h < D) is 0 2 ) h 2 D ( ε − ρ + σ
(C) Magnitude of electric field at a distance h below the bottom of the slab is
0 4 D ε σ − ρ
(D) Magnitude of electric field at a distance h below the bottom of the slab is
0 2 D ε σ − ρ
This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety of possible twists and turns of problems in physics that would be very helpful in facing IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and we hope that this section would prove a rich resource for practicing challenging problems and enhancing the preparation level of IIT JEE aspirants.
By : Dev Sharma Director Academics, Jodhpur Branch
Physics Challenging Problems
Soluti ons wil l be published in next issue
1. Q1 =
∫
ρ π 2 / R 0 2dr r 4 ) r ( =∫
α π 2 / R 0 2dr r 4 Q2 =∫
ρ × π R 2 / R 2dr r 4 ) r ( =∫
π − α R 2 / R 2dr r 4 R r 1 2 Fraction = 2 1 Q QOption [A] is correct
2. F = qE = 0 3 r q ε ρ × = 0 3 r e ε ρ − a = 0 m 3 r e ε α − a ∝ r ∴ ω = m 3 e 0 ε α T = ω π 2 Option [B] is correct 3.
∫
E.ds=∫
ε π − α + ε r 2 / R 0 2 0 1 4 r dr R r 1 2 QOption [A] is correct
4. Let at any instant t temperature is T. The net rate at which heat is absorbed by the gas is
dt dQ = q – l ) T T ( KA − 0 ...(1) Now, dQ = nCpdT = n × 2 7 R × dT ...(2) ∴ n × 2 7 R dt dT = q – l ) T T ( KA − 0 of
∫
− − T T0q KA(T T0) dT l =∫
t 0 dt nR 7 2 l Option [A] is correct5. At maximum temperature
q =
l KA
(Tmax – T0)
Option [A] is correct
6. As the process is isobaric ∴ 0 0 T V = max max T V ⇒ 0 max V V = 0 max T T
Option [A] is correct
7. L 2L 4A A ⇒ R = A L ρ R´ = A 4 L 2 × ρ = 2 R Heat produced = R V2 as R become half ∴ heat produced is doubled
E = d V = L V E´ = L 2 V E´ = 2 E ∴ option [D] is correct
8. Option [A,C, D] is correct
Solution
Physics Challenging Problems
Set # 38
Qu estions were Publish ed in J ul y IssueCartoon Law of Physics
Any body passing through solid matter will leave a perforation conforming to its perimeter.
Also called the silhouette of passage, this phenomenon is the specialty of victims of directed-pressure explosions and of reckless cowards who are so eager to escape that they exit directly through the wall of a house, leaving a cookie-cutout-perfect hole. The threat of skunks or matrimony often catalyzes this reaction.
1. In the arrangement shown in fig. a wedge of mass m3
3.45 kg is placed on a smooth horizontal surface. A small and light pulley is connected on its top edge, as shown. A light, flexible thread passes over the pulley. Two blocks having mass m1 = 1.3 kg and m2 = 1.5 kg
are connected at the ends of the thread. m1 is on
smooth horizontal surface and m2 rests on inclined
surface of the wedge. Base length of wedge is 2m and inclination is 37º. m2 is initially near the top edge of
the wedge.
m2
37º m3
m1
If the whole system is released from rest, calculate (i) velocity of wedge when m2 reaches its bottom,
(ii) velocity of m2 at that instant and tension in the
thread during motion of m2.
All the surface are smooth. (g = 10 ms–2)
Sol. Let acceleration of m1 be a (rightwards) and that of
wedge be b (leftwards). Acceleration of m2 (relative
to wedge) becomes (a + b), down the plane. Therefore, resultant acceleration of m2 is vector sum
of the two accelerations
(i) (a + b) down the plane and (ii) b leftwards. Hence, components of this resultant acceleration are (i) {(a + b) cos 37º – b} = (0.8a – 0.2b) horizontally rightward and
(ii) (a + b) sin 37º = (0.6a + 0.6b) vertically downward.
Considering free body diagrams, T m1g m1a N1 m3b N1 N2 N2 37 m2g m2(0.6a + 0.6b) m2(0.8a – 0.2b) T
For horizontal forces on m1, T = m1a ...(1)
For vertical forces on wedge,
T – T cos 37º + N2 sin 37º = m3b ...(2)
For horizontal forces on m2
N2sin 37º – T cos 37º = m2(0.8a – 0.2b) ...(3)
For vertical forces on m2,
m2g – N2cos 37º – T.sin 37º = m2(0.6a + 0.6b) ...(4)
From above equations, a = 3 ms–2, b = 2ms–2
and T = 3.9 newton
Since, base angle and base length of wedge are 37º and 2m respectively, therefore, height of its vertical face is 2. tan 37º = 1.5 m.
Now considering vertical motion of m2 from top to
bottom of the wedge,
u = 0, acceleration = (0.6a + 0.66b) = 3ms–2 and
displacement = 1.50 m. Using s = ut +
2 1
at2, t = 1 second
At this instant, horizontal component of velocity of m2 is
v2x = (0.8a – 0.2b) t = 2 ms–1
and vertical component,
v2y = (0.6a + 0.6b) t = 3 ms–1
∴ Velocity of m2 is v2 = v22x+v22y = 13 ms–1 Ans.
Velocity of wedge at this instant = bt = 2 ms–1 Ans.
2. Two identical blocks A and B of mass m = 3 kg are attached with ends of an ideal spring of force constant K = 2000 Nm–1 and rest over a smooth
horizontal floor. Another identical block C moving with velocity v0 = 0.6 ms–1 as shown in fig. strikes
the block A and gets stuck to it. Calculate for subsequent motion
(i) velocity of centre of mass of the system, (ii) frequency of oscillations of the system, (iii) oscillation energy of the system, and (iv) maximum compression of the spring.
m
m m
A B
C v0
Sol. When block C collides with A and get stuck with it, combined body moves to the right, due to which spring is compressed. Therefore, the combined body retards and block B accelerates. In fact, deformation of spring varies with time and the system continues to move rightwards. In other words, centre of mass of the system moves rightwards and combined body and block B oscillate about the centre of mass of the system.
Expert’s Solution for Question asked by IIT-JEE Aspirants
Students' Forum
PHYSICS
Let just after the collision velocity of combined body formed by blocks C and A be v. Then, according to law of conservation of momentum,
(m + m)v = mv0
or v = 2 v0
= 0.3 ms–1
∴ Velocity of centre of mass of the system, vc = m m 2 0 m v m 2 + × + × = 0.2 ms–1
Now the system is as shown in fig.
2m m
Its reduced mass, m0 =
m m 2 ) m )( m 2 ( + = 3 m 2 ∴ Frequency of oscillations, f = 0 m K 2 1 π = π 10 5 Hz. Ans.
Since, just after the collision, combined body has velocity v, therefore, energy of the system at that instant, E =
2 1
(2m)v2 = 0.27 joule
Due to velocity vC of centre of mass of the system,
translational kinetic energy, Et =
2 1
(3m)v = 0.18 joule c2
But total energy E of the system = its translational kinetic (Et) + oscillation energy (E0)
∴ E0 = E – Et = 0.09 joule
At the instant of maximum compression, oscillation energy is stored in the spring in the form of its strain energy. Let maximum compression of spring be x0.
then Kx20 2 1
= E0
∴ x0 = 90 × 10–3 m or 3 10mm Ans.
3. In the arrangement shown in fig. mass of blocks A and B is m1 = 0.5 kg and m2 = 10 kg, respectively and
mass of spool is M = 8 kg. Inner and outer radii of the spool are a = 10 cm and b = 15 cm respectively. Its moment of inertia about its own axis is I0 = 0.10 kg m2. If friction be sufficient to prevent
sliding, calculate acceleration of blocks A and B.
a b
B A
Sol. Since, friction is sufficient to prevent sliding, therefore, the spool has tendency to roll about the instantaneous axis of rotation which is line of contact of spool surface with the horizontal plane. About this line, tension in left thread produces anticlockwise moment and that in right thread produces clockwise moment. Since, moment produced by weight of block B is greater than that produced by weight of block A, therefore, the spool rotates clockwise.
Let angular acceleration of spool be α clockwise, then accelerations of blocks A and B will be 2bα(upwards) and (b – a)α downwards respectively. Moment of inertia of spool, about instantaneous axis of rotation, O,
I = I0 + Mb2 = 0.28 kg m2
Consider free body diagrams,
A T1 m1g m1(2bα) B T2 m2g m2(b – a) α N Friction T1 Mg 1α T2 O
For forces on block A,
T1 – m1g = m1(2bα) ...(1)
For forces on block B,
m2g – T2 = m2(b – a)α ...(2)
Taking moments of forces acting on the spool, about O,
T2(b – a) – T1(2b) = Iα ...(3)
From equations (1), (2) & (3), T1 = 6.5 N, T2 = 95 N
and α = 10 rad/sec2
∴ Acceleration of block A = 2bα = 3 ms–2 (upward)
and acceleration of block B = (b – a)α = 0.5 ms–2
(downward) Ans.
4. Each plate of a parallel plate air capacitor has are area S = 5 × 10–3 m2 and are d = 8.85 mm apart as shown
in fig. Plate A has a positive charge q1 = 10– 10
coulomb and plate B has charge q2 = +2 × 10–10
coulomb. Calculate energy supplied by a battery of emf E = 10 volt when its positive terminal is connected with plate A and negative terminal with plate B.
+10–10
C +2 × 10–10C
A d B
Sol. Charges q1 and q2 get distributed such that charges
appearing on inner surfaces of two plates become numerically equal but opposite in nature. Since charge q1 on plate A is less than charge q2 on plate B,
negatively charged and that of B become positively charged.
Let magnitude of this charge be q. Then distribution of charge on various surfaces will be as shown in fig. But the plates are metallic, therefore electric field inside the plates will be zero.
+(10 –10 + q) +( 2 × 10 –10 – q) –q +q p
Considering a point P inside the plate B, Electric field on it is E = S 2 ) q 10 ( 0 10 ε + − – S 2 q 0 ε – 2 S q 0 ε – S 2 ) q 10 2 ( 0 10 ε − × − = 0 or q = 5 × 10–11 coulomb or 50 pC
Hence, the charges are as shown in fig.
150 pC – – – + + + + + + + + + 50 pC 150 pC
When battery is connected with the plates, a charge flows through the circuit. Due to flow of this charge, charges on inner surfaces are changed while charges on outer surfaces remain unchanged.
Let charge flowing through the battery be q´. Then charges on various surfaces become as shown in fig.
150 pC – – – + + + + + + + + + (50×10–12–q´) 150 pC + – q E q´
Capacitance of the capacitor is
C = d S 0 ε = 5 × 10–12 F
Applying Kirchhoff's voltage law,
– C ´) q 10 50 ( × −12− – E = 0 ∴ q´ = 1 × 10–10 coulomb
∴ Energy supplied by battery
= q´E = 10–9 joule Ans.
5. Nine identical capacitors, each of capacitance C = 15 µF are connected as shown in fig. Calculate equivalent capacitance between terminals 1 and 4.
1 6 5
4 3
2
Sol. Given arrangement of capacitors is symmetric about mid-point of arm 3–6. If the arrangement is rotated through 180º about this point, given arrangement is obtained again. Let a battery of emf V be connected across terminals 1 and 4 of the arrangement. Then, in steady state, charges on various capacitors will be as shown in fig. 1 6 5 4 3 2 q2 – + + + + – – + + – + – + – + – – – q1 (q2 – q3) (q1 + q2) (q1 + q2) + – (q1 – q2 + 2q3) q2 – q3 q1 q3 q3 q2
Applying Kirchhoff´s voltage law on mesh 1 – 2 – 6 –1, C q2 + C q3 – C q1 = 0 or q1 = (q2 + q3) ...(i) For mesh 2 – 3 – 6 – 2, C q q2− 3 – C q 2 q q1− 2+ 3 – C q3 = 0 or q1 = (2q2 – 4q3) ...(2)
From equation (1) and (2), q2 = 5q3 and q1 = 6q3
Now applying Kirchhoff's voltage law on mesh 1 – 6 – 5 – 4 – V – 1, C q1 + C q q2− 3 + C q2 – V = 0 Substituting q1 = 6q3 and q2 = 5q3, q3 = 15 1 CV. But charge drawn by the arrangement from battery is
q = (q1 + q2) = 11q3 = 15 11 CV ∴ Equivalent capacitance = V q = 15 C 11 = 11µF Ans.
Capacitors in Series : V1 V2 V3 V A +Q–Q +Q–Q +Q–Q C1 C2 C3 B
In this arrangement of capacitor the charge has no alternative path(s) to flow.
(a) The charges on each capacitor are equal i.e. Q = C1V1 = C2V2 = C3V3 ...(1)
(b) The total potential difference across AB is shared by the capacitors in the inverse ratio of the capacitances.
V = V1 + V2 + V3 ...(2)
If Cs is the net capacitance of the series combination,
then s C Q = 3 2 1 C Q C Q C Q + + ⇒ s C 1 = 3 2 1 C 1 C 1 C 1 + + Further V1 = 1 C Q and V = s C Q Capacitors in Parallel : V A +Q1 –Q1 +Q2 –Q2 C1 C2 +Q3 –Q3 C3 B
In such an arrangement of capacitors the charge has an alternative path(s) to flow
(a) The potential difference across each capacitor is same and equals the total potential applied.
i.e. V = V1 = V2 = V3 ...(1) ⇒ V = 1 1 C Q = 2 2 C Q = 3 3 C Q ...(2) (b) The total charge Q is shared by each capacitor in the direct ratio of the capacitances.
⇒ Q = Q1 + Q2 + Q3
If CpV is the net capacitance for the parallel
combination of capacitors then
CpV = C1V+ C2V + C3V ⇒ Cp = C1 + C2 + C3
Important terms :
(a) If C1, C2, C3 .... are capacitors connected in series
and if total potential across all is V, then potential across each capacitor is
V1 = s 1 C 1 C 1 V; V2 = s 2 C 1 C 1 V; V3 = s 3 C 1 C 1 V
and so on, where s C 1 = n 3 2 1 C 1 .... C 1 C 1 C 1 + + + +
(b) If C1, C2, C3 ... are capacitors connected in
parallel and if Q is total charge on the combination, then charge on each capacitor is
Q1 = p 1 C C Q; Q2 = p 2 C C Q; Q3 = p 3 C C Q
and so on, where Cp = C1 + C2 + C3 + ... + Cn
Energy Density :
For a parallel plate capacitor
U = 2 1 CV2 where C = d A 0 ε and V = Ed ε σ = 0 E where ⇒ U = d A 2 1ε0 E2d2 ⇒ U = ε 2 0E 2 1 (Ad) ⇒ U = 2 1 ε0E2τ
where τ is volume of the capacitor ⇒ τ U = Ue = Volume Energy tic Electrosta = Electrostatic Pressure = 2 1 ε0E2 = 0 2 2ε σ ε σ = 0 E Q
Capacitor-2
P
HYSICS
F
UNDAMENTAL
F
OR
IIT-J
EE
Energy for series and parallel combinations :
Series Combination : For a series combination of capacitor Q = constant and
s C 1 = ... C 1 C 1 C 1 3 2 1 + + + ⇒ s 2 C 2 Q = 1 2 C 2 Q + 2 2 C 2 Q + 3 2 C 2 Q + .... ⇒ Us = U1 + U2 + U3 + ...
Parallel Combination : For a parallel combination of capacitors V = constant and
Cp = C1 + C2 + C3 + .... ⇒ 2 1 CPV2 = 2 1 C1V2 + 2 1 C2V2 + 2 1 C3V2 + ... ⇒ Up = U1 + U2 + U3 + ....
Electrostatic force between the plates of a parallel plate capacitor :
The plates of the capacitor each carry equal and opposite charges, hence they must attract each other with a force, say F.
+ + + + + + – – – – – – +Q –Q
At any instant let the plate separation be x, then
C = x A 0 ε Also U = C 2 Q2 ⇒ U = ε A 2 Q 0 2 x
Let the plates be moved towards each other through dx, such that the new separation between the plates is (x – dx). If Uf is the final potential energy, then
Uf = ´ C 2 Q2 = A 2 Q 0 2 ε (x – dx)
If dU is the change in potential energy, then dU = Uf – Ui ⇒ dU = A 2 Q 0 2 ε (x – dx) – 2 A Q 0 2 ε x ⇒ dU = – A 2 Q 0 2 ε dx Further since F = – dx dU ⇒ F = A 2 Q 0 2 ε = ε σ 0 2 2 A = ε 2 0E 2 1 A ε σ = σ = 0 E , A Q Q
Kirochhoff's laws for capacitor circuits :
Kirchhoff's first law or junction law : Charge can never accumulate at a junction i.e. at the junction
∑
q = 0Important terms : This law is helpful in determining the nature of charge on an unknown capacitor plate. Charge on capacitor C can be determined by using this rule. As no charge must accumulate at the junction O, so if x is charge on plate 1 of C, then
–q1 + q2 + x = 0 ⇒ x = q1 – q2 + – +q1 – + – –q1 +q2 –q2 B 1 2 A C
i.e. plate 1 has a charge (q1 – q2) and plate 2 has a
charge –(q1 – q2).
Kirchhoffs second law or loop law :
In a closed loop (a closed loop is the one which starts and ends at the same point), the algebraic sum of potential differences across each element of a closed circuit is zero.
⇒
∑
V = 0Conventions followed to apply loop law :
(a) In a loop, across a battery, if we travel from negative terminal of battery to the positive terminal then there is a potential rise and a +ve sign is applied with voltage of the battery.
(b) In a loop, across a battery, if we travel from positive terminal of the battery to the negative terminal then there is a potential fall and a –ve sign is applied with voltage of the battery.
(c) In a loop, across a capacitor, if we go from negative plate to the positive plate of the capacitor then there is a potential rise and a +ve sign is to be taken with potential difference across the capacitor i.e. ∆V = +
C q .
(d) In a loop, across a capacitor, if we go from positive plate to the negative plate of the capacitor then there is a potential fall and a –ve sign is to be taken with the potential difference across the capacitor i.e. ∆V = –
C q .
Finding net capacitance of circuits : A. Simple Circuits :
Analyse the circuit carefully to conclude which pair of capacitors are in series and which are in parallel (This all should be done keeping in mind the points across which net capacitance has to be calculated). Find their net capacitance and again draw an equivalent diagram to apply the above specified technique repeatedly so as to get the total capacitance between the specified points.
B. Concept of line of symmetry :
Line of symmetry (L.O.S.) is an imagination of our mind to divide a highly symmetric circuit into two equal halves such that the points of the circuit through which LOS passes are at equal potential.
Solved Examples
1. Find the net capacitance of the circuit shown between the points A and B.
C C C C C C C A B
Sol. This circuit is highly symmetric and so we can consider the line of symmetry to pass through the circuit to divide it into two equal (identical) halves. If line of symmetry passes through a branch possessing a capacitor, then on each side of line of Symmetry the capacitance will become 2C (2C and 2C in series will gives C), as shown. 2C 1 C C C 3 4 A P P C B 2C C C LOS
Now, the concept of line of Symmetry makes our job easy to calculate capacitance across AP. (1) and (2) are in parallel further in series with (3), whose resultant capacitance is in parallel with (4).
Resultant of (1) and (2) is 3C Resultant of 3C and (3) is 4 C 3 Resultant of 4 C 3 and (4) is 4 C 7 So total capacitance across AB is
CAB = 2 CAP ⇒ C AB = 8 C 7
2. Find the equivalent capacitance between the point A and B in figure. A C2 C1 B C3 C2 C1
Sol. Let us connect a battery between the points A and B. The charge distribution is shown in figure. Suppose the positive terminal of the battery supplies a charge +Q and the negative terminal a charge –Q. The charge Q is divided between plates a and e.
A C2 C1 B C3 C2 C1 Q1 –Q1 Q–Q1 –(Q–Q1) Q1 –Q1 Q–Q1 –(Q–Q1) (2Q1–Q) –(2Q1–Q) i j e f E g h a b D
Let a charge Q1 goes to the plate a and the rest Q – Q1
goes to the plate e. The charge –Q supplied by the negative terminal is divided between plates d and h. Using the symmetry of the figure, charge –Q1 goes to
the plate h (as it has a capacitance C1) and –(Q – Q1)
to the plate d (as it has a capacitance C2). This is
because if we look into the circuit from A or from B, the circuit looks identical. The division of charge at A and at B should, therefore, be similar. The charges on the other plates may be written easily. The charge on the plate i is 2Q1 – Q which ensures that the total
charge on plates b, c and i remains zero as these three plates form an isolated system.
We have VA – VB = (VA – VD) + (VD – VB) or VA – VB = 1 1 C Q + 2 1 C Q Q− ...(1) Also, VA – VB = (VA – VD) + (VD – VE) + (VE – VB) or VA – VB = 1 1 C Q + 3 1 C Q Q 2 − + 1 1 C Q ...(2) We have to eliminate Q1 from these equation to get
the equivalent capacitance
) V V ( Q B A− . The first equation may be written as
VA – VB = Q1 − 2 1 C 1 C 1 + 2 C Q or 1 2 2 1 C C C C − (VA – VB) = Q1 + 2 1 1 C C C − Q ...(3)
The second equation may be written as VA – VB = 2Q1 + 3 1 C 1 C 1 – 3 C Q
or ) C C ( 2 C C 3 1 2 1 + (VA – VB) = Q1 – 2(C C ) C 3 1 1 + Q ...(4) Subtracting (4) from (3) (VA – VB) + − − 2(C C ) C C C C C C 3 1 3 1 1 2 2 1 = + + − 2(C C ) C C C C 3 1 1 1 2 1 Q or (VA – VB)[2C1C2(C1 + C3) – C1C3(C2 – C1)] = C1[2(C1 + C3) + (C2 – C1)]Q or C = B A V V Q − = 1 2 3 1 3 3 2 2 1 C 2 C C C C C C C C 2 + + + +
3. Five identical conducting plates 1, 2, 3, 4 and 5 are fixed parallel to and equidistant from each other as shown in fig. Plates 2 and 5 are connected by a conductor while 1 and 3 are joined by another conductor. The junction of 1 and 3 the plate 4 are connected to a source of constant e.m.f. V0. Find
(i) The effective capacity of the system between the terminals of the source
(ii) the charge on plates 3 and 5.
Given d = distance between any two successive plates and A = are of either face of each plate.
Sol. (i) The equivalent circuits is shown in fig. The system consists of four capacitors.
5 4 3 2 1 (–) (+) (a) (b) 1 2 3 2 3 4 5 4 (Q2/2) (Q2/2) Q2 Q1 Q (–) (+)
i.e., C12, C32, C34 and C54. The capacity of each
capacitor is ε d A K 0
= C0. The effective capacity
across the source can be calculated as follows : The capacitors C12 and C32 are in parallel and hence
their capacity is C0 + C0 = 2C0. The capacitor C54 is
in series with effective capacitor of capacity 2C0.
Hence the resultant capacity will be
0 0 0 0 C 2 C C 2 C + ×
Further C34 is again in parallel. Hence the effective
capacity = C0 + 0 0 0 0 C 2 C C 2 C + × = 3 5 C0 = 3 5 Kε0 d A .
(ii) Charge on the plate 5 = charge on the uper half of parallel combination ∴Q5 = V0 0 C 3 2 = d AV K 3 2 ε0 0
Charge on plate 3 on the surface facing 4 ∴ V0C0 =
d AV kε0 0
Charge on plate 3 on the surface facing 2 = [potential difference across (3 – 2)]C0
= V0 0 0 0 C 2 C C + C0 = Kε0 3d AV0 ∴ Q3 = d AV Kε0 0 + Kε0 d 3 AV0 = d AV Kε0 0 + 3 1 1 = 3 4 Kε0 d A V0
4. In diagram find the potential difference between the points A and B and between the points B and C in the steady state. B 3µF 3µF 1µF 1µF 1µF 20Ω 100 V 10Ω A C
Sol. The circuit is redrawn in fig (a, b, c)
1µF 20Ω 10Ω A C 3µF 1µF 3µF 1µF B 100 V Fig.(a) 1µF 20Ω 10Ω A C 100 V Fig.(b) 6µF 2µF P Q B 1µF 20Ω 10Ω A C 100 V Fig.(c) R P Q 3/2 µF S From fig. (c).
potential difference between P and Q = Potential difference between R and S = 100 volt
∴ Q = capacity × volt = 2 3
× 10–6 × 100 = 150 × 10–6 coulomb
Now according to fig.(b), the charge flowing through capacitors of capacity 6 µF and 2 µF is 150 × 10–6
coulomb because they are connected in series. Potential difference between A and B = Potential difference across the two ends of condenser of capacity 6 µF. ∴ V1 = capacity Q = 66 10 6 10 150 − − × × = 25 volt.
Again potential difference between C and D = potential difference across the two ends of condenser of capacity 2µF V2 = 6 6 10 2 10 150 − − × × = 75 volt
5. Fig. shows two identical parallel plate capacitors connected to a battery with switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant (or relative permittivity) 3. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.
V A C B C
S
Sol. Initially the charge on either capacitor, i.e. qA or qB is
CV coulomb.
When dielectric is introduced, the new capacitance of either capacitor
C1 =
K K1
C = 3C.
After the opening of switch S, the potential across capacitor A is volt.
Let the potential across capacitor B is V1
∴ qB = CV = C1V1 or CV = 3CV1
∴ V1 =
3 V
volt
Initial energy of capacitor A = 2 1 CV2 energy of capacitor B = 2 1 CV2 ∴ Total energy Ei = 2 1 CV2 + 2 1 CV2 = CV2
Final energy of capacitor A
= 2 1 × (3C)V2 = 2 3 CV2
Final energy of capacitor B
= 2 1 × (3C) 2 3 V = 6 CV2
∴ Total final energy Ef = 2 3 CV2 + 6 CV2 = 3 5 CV2 ∴ f i E E = 2 2 CV ) 3 / 5 ( CV = 5 3
Space Shuttle
OK here is the deal with the space shuttle. It has three rocket engines in the back, but there's absolutely no room inside for all the fuel it needs to launch itself up into space. All of that fuel is stored outside the shuttle, in the big brown cylinder, called the external tank.
The tank containing all the rocket fuel weighs seven times more than the space shuttle itself! That's a lot of really heavy fuel, and the space shuttle engines aren't quite strong enough to push the combined weight of the shuttle and the big bloated external tank up off the ground.
That's what the two long white solid rocket boosters strapped onto the sides of the external tank are for. They lift the tank! Fortunately, it was not necessary to strap an infinite series of smaller and smaller rockets to the sides of the solid rocket boosters.
It is not widely known that just behind the main flight deck of the space shuttle is a small Starbucks adapted for use in zero gravity.
