XtraedgeAugust_2010

"People with integrity do what they say they are going to do. Others have excuses."

Rudyard Kipling, the celebrated English author and poet, once said, "We have forty million reasons for failure, but not a single excuse." Yet today we are literally inundated with a tidal wave of excuses from every direction. In fact, it seems everyone has a reason, explanation or justification for not doing what they were supposed to do.

Why do so many of us crank out one excuse after another for virtually everything we fail to do? Well, for starters, excuses are easy. In fact, they're way too easy. After all, making excuses doesn't require any effort or commitment on our part. All we have to do is toss out excuse after excuse and we feel we're off the hook, since the best excuses always absolve us of any personal responsibility whatsoever.

While getting in the habit of making of excuses is easy, excuse making doesn't get any of us anywhere close to where we want to go in life. Sooner or later all of our years of excuses eventually catch up with us. Before we realize it, the best of life has slipped away in a lazy, hazy, crazy blur of excuses.

Ninety-nine percent of the failures come from people who have the habit of making excuses. Hold yourself responsible for a higher standard than anybody else expects of you, never excuse yourself.

The person who really wants to do something invariably finds a way to get it done. And for those who don't want to do something; well, one excuse is just as good as another I suppose. What it all boils down to is simply this: what kind of person do you really want to be? Do you want to make excuses - or make something happen instead?

It's time to turn all of your excuses loose once and for all. Each and every time you fall short, pick yourself up, learn from your mistakes and immediately get going again. No complaining, no explaining and absolutely no excuses allowed. You will find that the minute you stop making excuses and start finding a way to get the job done, you'll start making your life everything it could be and should be... and so much more.

Get rid of the excuses and you can get anywhere you've ever dreamed of going.

Presenting forever positive ideas to your success. Yours truly

Pramod Maheshwari, B.Tech., IIT Delhi

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Editor : Pramod Maheshwari

WORRY IS A MISUSE OF IMAGINATION.

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S

Success Tips for the Months

• " Always bear in mind that your own resolution to succeed is more important than any other thing."

• "God gave us two ends. One to sit on and one to think with. Success depends on which one you use; head you win -- tails, you lose."

• "The ladder of success is best climbed by stepping on the rungs of opportunity." • "Success is getting what you want.

Happiness is wanting what you get." • "The secret of success in life is for a man

to be ready for his opportunity when it comes."

• "I don't know the key to success, but the key to failure is trying to please everybody."

• "The secret of success is to be in harmony with existence, to be always calm… to let each wave of life wash us a little farther up the shore."

CONTENTS

INDEX PAGE

NEWS ARTICLE

Delhi power firm, IIT tie-up to reduce power loss

IITian ON THE PATH OF SUCCESS

KNOW IIT-JEE

XTRAEDGE TEST SERIES

Class XII – IIT-JEE 2011 Paper Class XI – IIT-JEE 2012 Paper

Regulars ...

DYNAMIC PHYSICS

Students’ Forum Physics Fundamentals Capacitor - 2

Work power energy & Conservation Law

CATALYSE CHEMISTRY

Key Concept Reaction Mechanism Solid State

Understanding : Physical Chemistry

DICEY MATHS

Permutation & Combination

Study Time...

22-year-old becomes

youngest IIT teacher

MUMBAI: IITians often liken the generation gap between themselves and their teachers to that between MS-DOS and Windows. This semester, however, the students on the Powai campus can look forward to someone much closer to their age: a physics teacher who has just entered his 20s.

At 22, Tathagat Avatar Tulsi, who has never studied in a classroom, plans to ask his students how they would want to be taught. "I have never taught in a class. But I believe I can come down to the level of a student and help them understand the subject," he said. Having completed high school when he was nine, his graduation in science at 10, an MSc in Physics at 12, and his PhD in Quantum Computing from the Indian Institute of Science (IISc), Bangalore, at 21, Tulsi says he is going to write to the Limca Book of Records to include him as the youngest faculty member in the country. Having achieved a lot pretty early in life, Tulsi may seem like a young man in hurry, but he has set a huge task for himself— to come up with an important scientific discovery, which will probably lead him to his ultimate dream: to own that shining piece of gold with Alfred Nobel on the obverse.

The "wonder boy", who suffered humiliation in August 2001 when a delegation of scientists taken by the department of science & technology to Lindau in Germany for an interaction with Nobel laureates, suggested that he was not a thinker, but a "fake prodigy" who had "mugged up" theories. Putting that behind, the Patna boy will stay on the Powai campus in the faculty quarters and work towards achieving that dream. That "not-so-distant" goal is probably why Tulsi chose teaching over a vocation. "I want to pursue my research and at IIT-B, I will have the leisure to continue my research and one day set up a lab focused on quantum computation in our country." Going to foreign shores is currently not on Tulsi’s plans. He chose the Powai college over Waterloo University, Canada, and the Indian Institute of Science Education & Research (IISER), Bhopal, both of which had also offered him teaching jobs.

Delhi power firm, IIT

tie-up to reduce power loss

New Delhi: The BSES Yamuna Power Limited (BYPL) and the Indian Institute of Technology (IIT)-Delhi have come together to improve the quality and reliability of power supply by reducing transmission and distribution losses.

A memorandum of understanding (MoU) was signed Monday between BYPL CEO Ramesh Narayanan and Anil Wali, managing director, Foundation for Innovation and Technology Transfer (FITT) - a society established by IIT-Delhi, to focus on how to bring the next-level or

"SMART" technology to the power distribution business and to keep pace with technological innovations taking place in the transmission business.

Both BYPL and IIT-Delhi will appoint one principal project investigator each. The cost of the projects will be borne by BYPL.

IANS.

Forests ministry teams

with IITs for Ganga

management plan

New Delhi: The Indian Institutes of Technology (IIT), the premier higher technical educational institutions of the Ministry of Human Resource Development (HRD) have committed themselves to the responsibility of development of a management plan of the Ganga river basin called the National Ganga River Basin Management Plan Project (NGBRM).

Seven IITs have come together for this purpose. The IITs have accepted this societal challenge as part of their response to the present-day challenges of the Indian society.

"It is required to ensure that the flow of the river Ganga must be continuous (Aviral Dhara), the river must have longitudinal and lateral connectivity, the river must have adequate space for its various

functions and the river must not be seen as a carrier of waste loads (Nirmal Dhara)," stated an official press release. The management plan will outline the strategy and the actions that need to be undertaken for the maintenance and restoration of the Ganga basin. The management plan should take into account the constraints of

population, urbanization, industrialization and agriculture

activities.

The IITs will form several thematic groups and each group will develop a detailed outline for the improvement of ecological health of the basin system. Besides the thematic groups, the IITs will also integrate in a holistic manner, all the issues into a comprehensive management plan.

In order to develop this plan, discussions will be held with local, state and other agencies who have to deal with the maintenance of the basin system. The management plan will also take into account the experience of earlier attempts of Ganga Action Plans. The HRD ministry and the Ministry of Environment & Forests are coming together to support the initiatives of the IITs. The work is estimated to be carried out in a period of 18 months. The funding for this project is estimated to be about Rs.15 crores. An agreement has been signed between the Directors of seven IITs and the Ministry of Environment & Forests in the presence of Minister of State for HRD, Kapil Sibal and Mister of State for Environments and Forest, Jairam Ramesh.

This initiative will involve not only faculty and students of seven IITs but will also take help from experts from other institutes and universities also.

IIT seeks robotic solution

in conflict

KOLKATA: A group of students from IIT-Kharagpur is working on a prototype of an Autonomous

Ground Vehicle (AGV) which, if it proves successful, may be developed by the Defence Research and Development Organisation (DRDO) for use by security agencies for Low-Intensity Conflict (LIC).

The six-member IIT team has bagged the first rank in Phase-I in the Student Robot Competition, 2010, organised by DRDO. As many as 240 colleges and institutes from across the country participated in the first phase of the contest, where third- and fourth-year students were asked to design an Autonomous Ground Vehicle for Low Intensity Conflict'. Only 14 teams were shortlisted for Phase-II. Among the entries from the east, the Indian School of Mines team from Dhanbad was ranked fourth. The National Institute of Technology, Rourkela, ranked 14th.

"Participants were told that the AGV will be a combat vehicle of the future and assigned tasks that a conventional manned vehicle cannot perform. These are basically autonomous robots to be used by security agencies engaged in LIC in urban and unstructured environments. These robots would be used in LIC and Explosive Ordnance Disposal (EOD) programmes in undesirable, hazardous and potentially life-threatening environments," said a senior DRDO official.

The AGV would have to be armed with sensors, software and other equipment to help it negotiate harsh terrain, identify and designate targets, engage and neutralise them. The vehicle would also have to detect minefields and neutralise them. In short, the AGV would be an autonomous off-road robotic platform that would navigate rough terrain and avoid natural and man-made obstacles in the shortest possible time.

According to the team of experts who judged the entries, the IIT-Kharagpur team, comprising Nalin

Gupta, Sarbartha Banerjee, Subhagato Dutta, Rahul Das, Anindita Bhattacharya and A Srinivas Reddy submitted an excellent design. Officials are waiting to find out how the prototype performs.

"The idea of the competition is to harness the innovative ideas of our student community to the National Robotics Program of India. The demands made from the participants are enormous. The robot will have to complete a closed loop obstacle course of 500 metres within an anticipated time of 20 minutes, using autonomous navigation. For the first 350 metres, the robot would have to navigate with the help of lane-following' by colour detection. While doing this, it would have to avoid static positive obstacles, cross over slopes, staircases and corrugations. It would have to navigate the remaining 150 metres with the help of GPS waypoints. Maximum width of the robot would have to be 1 metre, maximum speed of 10 km per hour and minimum turning radius of 5 metres. It would have to carry an additional payload of 10 kg. The robot would have to be self-powered in all respects," the official said. In Phase-I, teams submitted designs with system configuration details. In the next phase, the selected teams would have to build a prototype and make it perform before the judges. Initially, 10 teams were supposed to be shortlisted for Phase-II. Given the nature of the papers submitted, it was finally decided to shortlist 14. Each team received a cash award of Rs 1,00,000.

The teams are now busy building their prototypes. The final competition will be held at the Combat Vehicles Research and Development Establishment (CVRDE), Chennai, between September 27-29, where the prototypes would be tried out.

Prof. Mohit Randeria received his Bachelor’s Degree in Electrical Engineering from IIT Delhi in 1980. He obtained M.S. Degree from the California Institute of Technology, USA in 1982 and Ph.D. Degree from Cornell University, USA in 1987.

Prof. Mohit Randeria is presently Professor of Physics at the Ohio State University in Columbus, Ohio. Prof. Randeria started his academic career as an Assistant Professor of Physics at SUNY Stony Brook in 1989, after a few years of post-doctoral work at Cornell and Illinois. He then joined the Argonne National Lab in Illinois as a Staff Scientist in the Materials Science Division, and after four years there, resumed his academic career at the Tata Institute of Fundamental Research (TIFR), Mumbai in 1995. He spent nine years in the Theoretical Physics Department at TIFR as Reader, Associate Professor, and Professor. Since 2004, he has been a Professor of Physics at Ohio State University. Prof. Randeria is an acknowledged expert in the area of Theoretical Condensed Matter Physics. His research interests are focused, at present, on high temperature superconductivity, strongly correlated systems, and ultra-cold atomic gases.

Prof. Randeria is the author of over hundred research papers in Condensed Matter Physics. He is the recipient of a prize in Condensed Matter Physics, awarded in honour of Nobel Laureate Phillip Anderson by the International Center of Theoretical Physics, Trieste, Italy in 2002. He has also been awarded the Swarnajayanti Fellowship in 1998, the B.M. Birla Science prize in 1999, the S.S. Bhatnagar Award in 2002, and the George A. Miller Visiting Professorship at University of Illinois, Urbana-Champaign in 2002-2003.

In honouring Prof. Mohit Randeria, IIT Delhi recognizes the outstanding contributions made by him as a Researcher and Scientist. Through his achievements, Prof. Mohit Randeria has brought glory to the name of the

Dr. Rajiv Laroia received his Bachelor’s Degree in Electrical Engineering from IIT Delhi in 1985. He obtained his Ph.D. in Electrical Engineering from University of Maryland, College Park in 1992.

Dr. Laroia is presently Senior Vice President of Technology at Qualcomm, USA. Dr. Laroia joined Siemens Research Labs in Munich, Germany after graduating from IIT Delhi. In 1992, he joined AT&T Bell Labs after obtaining his Ph.D. In 2000, Dr. Laroia founded Flarion Technologies, a venture backed company to develop and commercialize a novel all-IP mobile wireless broadband technology. He served as the CTO of Flarion and in that role provided the vision and led the development of technology and products for the company. In 2006, Flarion was acquired by the wireless technology giant Qualcomm, where he currently serves as Senior Vice President (Technology).

Dr. Laroia is one of the world’s leading researchers and innovators in the field of communication. He holds more than 50 patents and has more than 100 others pending. His early research focused on wire-line communication. He has significant technology contributions to V.34 and V.90 International Standards for sending data over telephone lines. The technology he invented is used in virtually all dial-up modems in the world.

Since 1997, Dr. Laroia has been working in the field of mobile wireless communication. The technology he and his team developed at Flarion is now incorporated in all three major next generation international wireless standards UMTS LTE, UMB and Wimax. Dr. Laroia has won numerous industry awards. In 2006, he was inducted to the Innovation Hall of Fame at the University of Maryland. He is a fellow of the IEEE. In honouring Dr. Rajiv Laroia, IIT Delhi recognizes the outstanding contributions made by him as an Entrepreneur and Technologist. Through his achievements, Dr. Rajiv Laroia

Success Story

Success Story

This articles contains stories of persons who have succeed after graduation from different IIT's

Dr. Rajiv Laroia

PHYSICS

1. 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.

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)

2. 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 P_{0 0} × , ∆QCA =  −  R V P 2 R V P 2 R 5 0 0 0 0 _{= –} 2 5_P 0V0

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 P_{0 0} = 2 V P_{0 0}

As net heat is absorbed by the gas during path BC, temp. will reach maximum between B and C.

KNOW IIT-JEE

(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 _× 0 2 4 V 5 _      _{+ 5P} 0       4 V 5 0 = –2P0V0 × 16 25 + 4 V P 25 0 0 ₌ 8 25 P0V0 ∴ Tmax = 8 25 R V P0 0

3. Two isolated metallic solid spheres of radii R and 2R are charged such that both of these have same charge density σ. The spheres ares are located far away from each other, and connected by a thin conducting wire. Find the new charge density on the bigger sphere.

[IIT-1996] Sol. R 2R q2 σ σ q1 V q'2 σ q'1 V Connecting Wire

For sphere of radius R σ = 2 1 R 4 q π ∴ q1 = σ × 4πR2

For sphere of radius 2R σ = 2 ₂ ) R 2 ( 4 q π ⇒ q2 = σ × 16πR2

When the two spheres are connected then the potential on the two spheres will be same. There will be a rearrangement of charge for this to happen. Let q1' and q2' be the new charges on the two spheres.

Since the total charge remains the same

q'1 + q'2 = q1 + q2 = σ × 20 πR2 ...(1) Also Since V1 = V2 0 4 1 πε R ' q₁ = 0 2 1 πε 2R ' q₂ ⇒ q' = q2' _{. ..(2)}

Substituting the value of q1' from (2) in (1)

2 ' q₁ + q2' = σ × 20 πR2 ⇒ 2 ' q 3 2 _{= σ × 20 πR}2 ⇒ 2 ₂ ) R 2 ( 4 ' q π = 3 σ × 2 5

⇒ New charge density on bigger sphere

= 2 ₂ ) R 2 ( 4 ' q π = 6 5σ

4. In the given circuit

E1 = 3E2 = 2E3 = 6 volts ; R1 = 2R4 = 6 ohms

R3 = 2R2 = 4 ohms ; C = 5µF

Find the current in R3 and the energy stored in the

capacitor. [IIT-1988] R1 R2 R3 E1 E3 R4 E2 C Sol. R1 = 6Ω R2 = 2Ω R3 = 4Ω E1 = 6V E3 = 3V R4 = 4Ω G _A 5µF = C E _F 2V = E2 i2 i2i2+i1 i1 i2 B D C

Applying Kirchoff's law in ABFGA

6 – (i1 + i2) 4 = 0 …(1)

Applying Kichoff's law in BCDEFB

I2 × 3 – 3 – 2 + 2i2 + (i2 + i1) 4 = 0 …(2)

Putting the value of 4 (i1 + i2) = 6 in (2)

3i2 – 5 + 2i2 + 6 = 0

∴ i2 = –

5 1

A

Sybstituting this value in (i) we get i1 = 1.5 –       5 1 – = 1.7 A Therefore current in R3 = i1 + i2 = 1.7 – 0.2 = 1.5 A

To find the p.d. across the capacitor VE – 2 – 0.2 × 2 = VG

∴ VE – VG = 2.4 V

= 2

1 _{× 5 × 10}–6 _{× (2.4)}2

= 1.44 × 10–5 _J

5. 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

Magnetic field due to straight wire NM at P : 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 = B +→1 → 2 B = – a I µ 11 . 0 0 _kˆ

Now, velocity of particle can be written as, →

v= v cos 60ºiˆ + v sin 60ºjˆ =

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 Fm → = 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 Ia2_{B) jˆ}

CHEMISTRY

6. A sample of hard water contains 96 ppm of SO42– and

183 ppm of HCO3– with 60 ppm of Ca2+ as the only

cation. How many moles of CaO will be required to remove HCO32– from 100 kg of this water ? If 1000

kg of this water is treated with the amount of CaO calculated above, what will be the concentration (in ppm) of residual Ca2+ _{ions ? (Assume CaCO}

3 to be

completely insoluble in water). If the Ca2+ _{ions in one}

litre of the treated water are completely exchanged with hydrogen ions, what will be its pH ? (one ppm means one part of the substance in one million part of water, mass/mass) [IIT-1997] Sol. In 106 _{g(= 1000 kg) of the given hard water, we have}

amount of SO42– ions = 96 g

amount of HCO3– ions = 183 g

So amount of SO42– ions = ₁ mol g 96 g 96 − = 1 mol and amount of HCO3– ions = ₁

mol g 61 g 183 − = 3 mol These ions are present as CaSO4 and Ca(HCO3)2.

Hence, amount of Ca2+ _{ions = }      + 2 3 1 = 2.5 mol The addition of CaO causes the following reactions:

CaO + Ca(HCO3)2 → 2CaCO3 + H2O

1.5 mol of CaO will be required for the removal of 1.5 mol of Ca(HCO3)2 in form of CaCO3.

In the treated water, only CaSO4 is present now.

Thus, 1 mol of Ca2+ _{ions will be present in 10}6 _{g of}

Molarity of Ca2+ _{ions in the treated water will be 10}–3

mol l–1_.

If the Ca2+ _{ions are exchanged by H}+ _{ions then,}

Molartiy of H+ _{in the treated water = 2 × 10}–3 _M

Thus, pH = – log(2 × 10–3_{) = 2.7}

7. The vapour pressure of ethanol and methanol are 44.5 mm and 88.7 mm Hg respectively. An ideal solution is formed at the same temperature by mixing 60 g of ethanol and 40 g of methanol. Calculate the total vapour pressure of the solution and the mole fraction of methanol in the vapour. [IIT-1986] Sol. Given that,

For ethanol (C2H5OH), Pe0 = 44.5 mm Hg M(C2H5OH) = 2 × 12 + 5 × 1 + 1 × 16 + 1 × 1 = 46 m(C2H5OH) = 60 g ∴ Moles of ethanol, ne = M m ₌ 46 60 _{= 1.3} For methanol (CH3OH), Pm0= 88.7 mm Hg M(CH3OH) = 1 × 12 + 3 × 1 + 1 × 16 + 1 × 1 = 32 m(CH3OH) = 40 g ∴ Moles of methanol, nm = M m ₌ 32 40 _{= 1.25} ∴ xe = m e e n n n + = 1.3 1.25 3 . 1 + = 2.55 3 . 1 xm = m e m n n n + = 1.3 1.25 25 . 1 + = 2.55 25 . 1

According to Raoult's law, Pe = Pe0xe = 55 . 2 3 . 1 5 . 44 × = 22.69 mm Hg and Pm = Pm0 xm = 55 . 2 25 . 1 7 . 88 × = 43.48 mm Hg Hence, total vapour pressure of the solution,

PT = Pe + Pm

= 22.69 + 43.48 = 66.17 mm Hg According to Dalton's law,

Pm = PTx´m (in vapour form)

Hence, mole fraction of methanol in vapour form, x´m = T m P P = 17 . 66 48 . 43 = 0.66

8. An alkyl halide X, of formula C6H13Cl on treatment

with potassium t-butoxide gives two isomeric alkenes Y and Z(C6H12). Both alkenes on hydrogenation give

2, 3-dimethyl butane. Predict the structures of X, Y

and Z. [IIT-1996]

Sol. The alkyl halide X, on dehydrohalogenation gives two isomeric alkenes.

X13 6H Cl C HCl – ; butoxide t K ∆ − −_  →  12 6H C Z Y+

Both, Y and Z have the same molecular formula C6H12(CnH2n). Since, both Y and Z absorb one mol of

H2 to give same alkane 2, 3-dimethyl butane, hence

Y and Z (C6H12) Ni H2 →  CH3 – CH – CH – CH3 CH3 CH3 2,3-dimethyl butane

The above alkane can be prepared from two alkenes CH3 – C = C – CH3 CH3CH3 2,3-dimethyl butene-2 (Y) and CH3 – CH – C = CH2 CH3 CH3 2,3-dimethyl _butene-1 (Z)

The hydrogenation of Y and Z is shown below :

CH3 – C = C – CH3 CH3CH3 (Y) H2 Ni CH3 – CH – CH – CH3 CH3 CH3 CH3 – CH – C = CH2 CH3 CH3 (Z) H2 Ni CH3 – CH – CH – CH3 CH3 CH3

Both, Y and Z can be obtained from following alkyl halide : CH3 – C – CH – CH3 CH3CH3 2-chloro-2,3-dimethyl butane (X) K-t-butoxide ∆; –HCl CH2 = C — CH – CH3 CH3 CH3 Cl + CH3 – C = C – CH3 CH3CH3 (Z) 20% (Y) 80% Hence, X, CH3 – C – CH – CH3 CH3CH3 Cl Y, CH3 – C = C – CH3 CH3CH3 Z, CH3 – CH – C = CH2 CH3 CH3

9. An organic compound (X), C5H8O, does not react

appreciably with Lucas reagent at room temperatures but gives a precipitate with ammonical AgNO3

solution. With excess CH3MgBr; 0.42 g of (X) gives

224 ml of CH4 at STP. Treatment of (X) with H2 in

the presence of Pt catalyst followed by boiling with excess HI gives n-pentane. Suggest structure of (X) and write the equations involved. [IIT-1992] Sol. Lucas test sensitive test for the distinction of p, s, and

t-alcohol. A t-alcohol gives cloudiness immediately, while s-alcohol within 5 minutes. A p-alcohol does not react with the reagent at room temperature. Thus, the present compound (X) does not react with this

(X) = C4H6.CH2OH(p-alcohol)

Since the compound gives a ppt. with ammonical AgNO3, hence it is an alkyne containing one

–C≡ CH, thus (X) may be written as : HC≡C –C2H4 – CH2OH (X)

It is given that 0.42 g of the compound (which is 0.005 mol) produces 22.4 ml of CH4 at STP (which is

0.01 mol) with excess of CH3MgBr. This shows that

the compound (X) contains two active H atoms (H atom attached to O, S, N and –C≡CH is called active). Of these, one is due to the p-alcoholic group (–CH2OH) and the other is due to the –C≡CH bond,

since both these groups are present in (X), hence it evolves two moles of CH4 on reaction with

CH3MgBr. H – C≡C. ) X (2 4 H C – CH2OH + 2CH3MgBr → BrMgC≡C–C2H4 – CH2OMgBr + 2CH4

Moreover, the treatment of (X) with H2/Pt followed

by boiling with excess of HI gives n-pentane (remember that 2HI are required to convert one –CH2OH into CH3). This shows that the compound

(X) contains a straight chain of five carbon atoms. H – C≡C–C2H4 – CH2OH 2H2/Pt→ CH3CH2.C2H4 – CH2OH ∆   → 2HI _CH 3CH2CH2CH2CH3 + H2O + I2 n-pentane

On the basis of abvoe analytical facts (X) has the structure :

HC≡C.CH2 CH2 – CH2OH (X)

5 4 3 2 1

4-pentyne-1-ol The different equations of (X) are : ) X ( 2 2 2 OH CH CH CH C C H− ≡ − . temp Room HCl ZnCl_2_→  + No reaction AgNO3 Ag – C≡C – CH2CH2CH2OH + NH4NO3 White ppt. NH3 2CH3MgBr Br MgC≡C.CH2CH2CH2OMgBr + 2CH4 2H2/Pt CH3CH2CH2CH2CH2OH Pentanol-1 CH3CH2CH2CH2CH3 n-pentane 2 HI ∆, –H2O; –I2

The production of 2 moles of CH4 is confirmed as the

reactions give 224 ml of CH4. Q 84 g(X) gives = 2 × 22.4 litre CH4 ∴ 0.42 g (X) gives = 84 42 . 0 4 . 22 2× × = 224 ml of CH4

10. Compound (X) on reduction with LiAlH4 gives a

hydride (Y) containing 21.72% hydrogen along with other products. The compound (Y) reacts with air explosively resulting in boron trioxide. Identify (X) and (Y). Give balanced reactions involved in the formation of (Y) and its reaction with air. Draw the structure of (Y).

Sol. Since B2O3 is formed by reaction of (Y) with air, (Y)

therefore should be B2H6 in which % of hydrogen is

21.72. The compound (X) on reduction with LiAlH4

gives B2H6. Thus it is boron trihalide. The reactions

are shown as: ) X ( 3 BX 4 + 3LiAlH4 → ) Y (2 6 H B 2 + 3LiX + 3AlX3 (X = Cl or Br) ) Y (2 6 H B + 3O2 → B2O3 + 3H2O + heat Structure of B2H6 is as follows: B Ht B Ht Ht Ht Hb Hb or B Ht Ht Hb Hb B Ht Ht 121.5º 1.33Å 97º 1.19Å 1.77Å

Thus, the diborane molecule has four two-centre-two electron bonds (2c-2e– _{bonds) also called usual bonds}

and two three-centre-two-electron bonds (3c-2e–

bonds) also called banana bonds. Hydrogen attached to usual and banana bonds are called Ht (terminal H)

and Hb (bridged H) respectively

MATHEMATICS

11. Find the values of a and b so that the function

f(x) =      π ≤ < π − π ≤ ≤ π + π ≤ ≤ + x 2 / , x sin b x 2 cos a 2 / x 4 / , b x cot x 2 4 / x 0 , x sin 2 a x

is continuous for 0 ≤ x ≤ π [IIT-1989] Sol. As, f(x) is continuous for 0 ≤ x ≤ π

∴ R.H.L.       π = 4 x at = L.H.L.       π = 4 x at ⇒       π π₊ b 4 cot 4 . 2 =      π₊ π 4 sin . 2 a 4 ⇒ 2 π + b = 4 π + a

⇒ a – b = 4 π ....(i) also, R.H.L       π = 2 x at = L.H.L       π = 2 x at ⇒       π₋ π 2 sin b 2 2 cos a =       π π₊ b 2 cot . 2 . 2 ⇒ – a – b = b ⇒ a + 2b = 0 ...(ii)

From (i) and (ii), a =

2 3π and b = 4 3π − 12. Find dx dy at x = –1, when x 2 sin ) y (sin π + 2 3 sec–1_{(2x) + 2}x _{tan ln (x + 2) = 0} [IIT-1991] Sol. Here, x 2 sin ) y (sin π + 2 3

sec–1_{(2x) + 2}x _{tan (log (x + 2)) = 0}

Differentiating both sides, we get x 2 sin ) y (sin π . log(sin y) . cos 2 π x . 2 π +       π x 2

sin _(siny) sin2x−1     π . cos y . dx dy + 2 x 4 |) x | 2 ( 2 . 2 3 2₋ + (x 2) )) 2 x (log( sec . 2x 2 + +

+ 2x _{log 2 . tan (log(x + 2)) = 0}

putting, _       π − = − = 1,y 3 x , we get         π − −       3 , 1 dx dy = 2 2 3 1 3         π −         π − = 3 3 2₋ π π

13. ABC is a triangle such that

sin(2A + B) = sin(C – A) = –sin(B + 2C) = 2 1 If A, B and C are in Arithmetic Progression, determine the values of A, B and C. [IIT-1990] Sol. Given that in ∆ABC, A, B and C are in A.P.

A + C = 2B also A + B + C = 180º ⇒ B = 60º Also given that,

sin (2A + B) = sin (C – A) = – sin (B + 2C) = ½ ...(1) ⇒ sin (2A + 60º) = sin (C – A) = – sin (60º + 2C) =

2 1 ⇒ 2A + 60º = 30º, 150º

{neglecting 30º, as not possible} ⇒ 2A + 60º = 150º

⇒ A = 45º

again from (1), sin (60º + 2c) = –1/2 ⇒ 60º + 2C = 210º, 330º ⇒ C = 75º or 135º Also from (1) sin (C – A) = ½

C – A = 30º, 150º, 195º

for A = 45º, C = 75º and C = 195º (not possible) ∴ C = 75º

Hence, A = 45º, B = 60º, C = 75º

14. If exp {(sin2_{x + sin}4_{x + sin}6_{x + ... ∞). ln 2} satisfies}

the equation x2 _{– 9x + 8 = 0, find the value of}

x sin x cos x cos + , 0 < x < 2 π . [IIT-1991] Sol. exp {(sin2_{x + sin}4_{x + sin}6_{x + ... ∞) log}

e2 ⇒ 1 sin x.log 2 x sin e 2 2 e− _{⇒ e}loge2tan2x ⇒ tan2x 2 satisfy x2 _{– 9x + 8 = 0 ⇒ x = 1, 8} ∴ tan2x 2 = 1 and 2tan2x = 8 ⇒ tan2_{x = 0 and tan}2_{x = 3}

⇒ x = nπ and tan2_{x =} 2 3 tan       π and x = nπ ± 3 π Neglecting x = nπ as 0 < x < 2 π ⇒ x = 3 π _∈       π 2 , 0 ∴ x sin x cos x cos + = 2 3 2 1 2 1 + = 3 1 1 + × 3 1 1 3 − − = 2 1 3− ∴ x sin x cos x cos + = 2 1 3−

15. Find the value of : cos (2 cos–1 _{x + sin}–1_{x) at x =} 5 1

, where 0 ≤ cos–1_{x ≤ π and –π/2 ≤ sin}–1_{x ≤ π/2}

[IIT-1981] Sol. cos{2cos–1_{x + sin}–1_x}

= cos       π + − 2 x

cos 1 _{, as cos}–1_{x + sin}–1_{x =} 2 π = – sin(cos–1_{x )} = – sin(sin–1 _1−x2 ₎ = – sin _       − − 2 1 5 1 1 sin = – sin _       ₋ 5 6 2 sin 1 = 5 6 2

1. In the circuit shown in figure C₁ =C₂ =2µF. Then charge stored in (steady state)

C1 C2

2Ω 1Ω 3Ω

120 V 1Ω 2Ω 3Ω

(A) capacitor C1 is zero

(B) capacitor C2 is zero

(C) both capacitors is zero (D) capacitor C1 is 40µC Passage # (Q. No. 2 to Q. No. 4)

A charged metal sheet is placed into uniform electric field E, perpendicularly to the electric field

lines. After placing the sheet into the field, the electric field on the left side of the sheet will be E1 = 5.6 × 105 V/m and

on the right it will be E2 = 3.1 × 105 V/m.

Q. 2 Find the total charge of the sheet if a electric force of 0.08N is exerted on it

(A) 0.28µC (B) 0.32µC

(C) 0.24µC (D) 0.38µC

Q. 3 Find the area of sheet of one side

(A) 0.02m2 _{(B) 0.03m}2 _{(C) 0.04 m}2 _{(D) 0.05 m}2 Q. 4 Find the value of E

(A) 2.5 × 104 _{V/m (B) 12.5 × 10}4 _V/m

(C) 3.5 × 104 _{V/m (D) 8.7 × 10}4 _V/m Q. 5 A capacitor consists of two parallel metal plate of

area A separated by a distance d. A dielectric slab of area A, thickness b & dielectric constant K is placed inside the capacitor. If CK is the

capacitance of capacitor with dielectric. How much K and b are restricted so that CK = 2C,

where C is capacitance without dielectric

(A) b d 3 d & d b 2 b 4 K < ≤ − = (B) b d 2 d & d b 2 b 2 K < ≤ − = (C) b 2d 2 d & d b 2 b 2 K ≤ ≤ − = (D) b d 4 d & d b 2 b 2 K ≤ ≤ − =

Q. 6 In the circuit shown in figure S1 S1 1Ω 2Ω 10V3Ω

20V i

(A) i = 2.5A when S1 is closed and S2 is open

(B) A

3 20

i= when S1 is open and S2 is closed

(C) A

3 5

i= when S1 and S2 both are open

(D) i = 20A when both S1 and S2 are closed Q. 7 A charged particle of unit mass and unit charge

moves with velocity →v=(8∧i+6∧j)m/s in a magnetic field of B=2k∧T. Choose the correct alternative(s)

(A) The path of the particle may be x2 _{+ y}2 _{– 4x – 24 = 0}

(B) The path of the particle may be x2 _{+ y}2 _{= 25}

(D) The time period of the particle will be 3.14s Q. 8 In the diagram

shown, the wires P1Q1 and

P2Q2 are made

to slide on the rails with same

speed of 5m/s. In this region a magnetic field of 1T exists. The electric current in 9Ω resistor is (A) zero if both wires slide towards left

(B) zero if both wires slide in opposite direction (C) 20mA if both wires move towards left 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

Set # 4 E1 E2 × P1 P2 9Ω Q1 Q2 2Ω 2Ω × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × 4 cm

1. [A,C] (1 e ) R E I= − −t/τ E and R is constant. , R L = τ if L increases then τ will increase hence the curve will shift towards right if E and R are halved then τ will increase hence the curve will shift towards right.

2. [B,C] (Moderate) As the length is double, the cross section area of the wire becomes half, thus the resistance of the wire

A L

R=ρ becomes four times the previous value. Hence after the wire is elongated the current becomes one fourth. Electric field is potential difference per unit length and hence becomes half the initial value. The power delivered to resistance is

R V p

2 = and hence becomes one fourth.

3. [A,B,C]

Charge is distributed over the surface of conductor in such a way that net field due to the charge and outside charge q is zero inside. Field due to only q is non-zero.

4. [A] R R R R . R R V V A = ₊ < 5. [A] RB =R+RG >R

6. [A] (Tough)% error in case A

100 1 R R R 100 R R R V V A _×       − + = × − % 1 100 R R R V − = × + − = %error in case B % 10 100 R R 100 R R R_{B G} = × = × −

Hence percentage error in circuit B is more than that in A.

7. [D] Order of the fringe can be counted on either side of the central maximum for example fringe no. 3 is first order bright fringe.

8. [C]

Solution

Physics Challenging Problems

Set # 3

8

• In late 2001, Associated Press reported, "NASA might allow McDonald's to put its logo on the international space station galley in exchange for McDonald's promoting space exploration to kids". Err...Mine's a Big Mac Please.

• A 10 pound sack of flour on the moon would bake six times as much bread as a sack weighing 10 pounds on earth.

• The Comets that pass close to the Sun originally came from one of two places; either the Oort Cloud or the Kuiper Belt. Approximately a dozen 'new' Comets are discovered every year. Because they are so far from the Sun, the Comets in the Oort Cloud take over 1 million years to make a single revolution around the Sun.

• There are stars as much as 400,000 brighter than the sun and others as much as 400,000 time fainter if they could all be seen at the same distance.

• A pulsar is a small star made up of neutrons so densely packed together that if one the size of a silver dollar landed on Earth, it would weigh approximately 100 million tons.

• An exploding supernova can outshine an entire galaxy of stars.

• There are 17 bodies in the solar system whose radius is greater than 1000 km.

• Over 90 per cent of the Universe consists of invisible 'dark matter'.

• In 1719 Mars was closer to Earth than it would be until the year 2003.

• By 2100, in the absence of emissions control policies, carbon dioxide concentrations are projected to be 30-150% higher than today's levels.

1. A block B of mass m = 0.5 kg is attached with upper end of a vertical spring of force constant K = 1000 Nm–1 _{as shown}

in Figure. Another identical block A falls from a height h = 49.5 cm on the block B and gets stuck with it. The combined body starts to perform vertical oscillations. Calculate amplitude of these oscillations. (g = 10 ms–2₎

Sol. Since block A falls freely under gravity through a height h before colliding with block B, therefore, its velocity just before collision is v0 = 2gh.

Let velocity of combined body just after collision be v.

Applying law of conservation of momentum for collision, 2mv = mv0 or v = 2 v₀ = 2 1 _2gh

Since, before collision, block B was in static equilibrium, therefore, compressive force in spring was exactly equal to its weight.

Suppose, initially spring was compressed through y0, then

Ky0 = mg or y0 = K mg

Hence, initial strain energy in spring, U0 = 2 1 Ky02 = K 2 g m2 2

After collision, combined body starts to move vertically downward due to velocity v. Therefore, spring is further compressed. Let maximum contraction of spring be y.

Then according to law of conservation of energy, Maximum strain energy in spring = initial strain energy U0

+ kinetic energy of combined body just after collision + further loss of potential energy of combined body. ∴ 2 1 _Ky2 ₌ K 2 g m2 2 ₊ 2 1 _{(2m) v}2 _{+ 2mg (y – y} 0) or 2 1 Ky2 _{– 2mgy +}         2 2 ₂ 0 –mv K 2 g m – mgy 2 = 0 or 500 y2 _{– 10y – 1.2 = 0}

From above equation y = + .6 or – .04

It means during vertical oscillations of combined body, its lowest position corresponds to a contraction of 0.06 m or 6 cm of spring and uppermost position corresponds to an elongation of 0.04 m or 4 cm of the spring.

Distance between these two extreme positions is 2a = (6 + 4) cm

Or amplitude of oscillations, a = 5 cm 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.

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

Expert’s Solution for Question asked by IIT-JEE Aspirants

Students' Forum

P H Y S I C S

h A B

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)v2_c = 0.18 joule

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. An artificial satellite of mass m of a planet of mass M, revolves in a circular orbit whose radius is n times the radius R of the planet. In the process of motion the satellite experiences a slight resistance due to cosmic dust. Assuming resistance force on satellite to depend on velocity as F = a.v2 _{where a is}

a constant, calculate how long the satellite will stay in orbit before it falls onto the planet's surface. Sol. Due to slight resistance offered by cosmic dust,

energy of satellite decreases slowly but continuously and setellite follows a spiral path of decreasing radius and ultimately falls onto the planet's surface as shown in Figure

Planet

Since, energy of satellite decreases slowly, therefore, its radial velocity is negligible in comparison to tangential velocity.

Let at an instant, distance of satellite from centre of planet be x,

Velocity of satellite is v =

x GM

…(1) and energy of satellite is E = –

2 1 x m . GM _...(2) Resisting force, F = av2 ₌ x aGM

During a very small time interval dt, Distance travelled by the satellite is vdt =

x GM

dt ∴ ` Work done by satellite against resistance of cosmic

dust is dW = F . (v.dt) = a 2 / 3 x GM       dt

But due to the work done by stellite, its energy decreases by the same amount and radius x also decreases simultanceously.

Or Increase in energy of satellite is dE = – dW. But from equation (2)

dE = 2 1 2 x GMm_dx. ∴ 2 1 2 x GMm . dx = – a 2 / 3 x GM       dt or dt = – GM a 2 m x dx …(3) At initial moment (t = 0), x = nR and we have to calculate time t when x becomes equal to R. Integrating equation (3) and substituting above limits. Gm a 2 m – dt? t 0 t =

∫

∫

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 = 10–10coulomb 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}–10_C

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,

therefore inner surface of plate. A becomes 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 × 1 0 –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 q_{2 +} C q3 _– C q_{1 = 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 E2_d2 ⇒ U =       ε 2 0E 2 1 (Ad) ⇒ U = 2 1_ε 0E2τ

where τ is volume of the capacitor ⇒ τ U _{= U} e = Volume Energy tic Electrosta = Electrostatic Pressure = 2 1 _ε 0E2 = 0 2 2ε σ       ε σ = 0 E Q

Capacitor-2

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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

∑

Important 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.

⇒

∑

Conventions 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 C_{AP ⇒ 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 + C₂ 1C₁ 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 − = 1C2₁ C₂2 32C₃3 1 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_Kε 0 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 ε_{0 0}

Charge on plate 3 on the surface facing 4 ∴ V0C0 =

d AV kε_{0 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 Kε_{0 0} + Kε0 d 3 AV₀ = d AV Kε_{0 0}     + 3 1 1 = 3 4_Kε 0 d A_V 0

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 = ₆ 6 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 _{= CV}2

Final energy of capacitor A

=

2

1 _{× (3C)V}2 ₌ 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

Believe

• Four things for success: work and pray, think and believe.

• The future belongs to those who believe in the beauty of their dreams.

• I believe in being an innovator. • You can do it if you believe you can!

• Believe deep down in your heart that you're destined to do great things.

• He who believes is strong; he who doubts is weak. Strong convictions precede great actions.

• I can't believe that God put us on this earth to be ordinary.

• Believe in yourself! Have faith in your abilities! Without a humble but reasonable confidence in your own powers you cannot be successful or happy.

• When you believe in a thing, believe in it all the way, implicitly and unquestionable.

• To love means loving the unlovable. To forgive means pardoning the unpardonable. Faith means believing the unbelievable. Hope means hoping when everything seems hopeless. • If you want to be confident, but don’t normally

act that way, today, just this once, act in the physical world the way you believe a confident person would.