August 2009

Worry is a misuse of imagination Volume - 5 Issue - 2 August, 2009 (Monthly Magazine) Editorial / Mailing Office : 112-B, Shakti Nagar, Kota (Raj.) 324009

Editorial

Tel. : 0744-2500492, 2500692, 3040000 e-mail : [email protected] Editor : Pramod Maheshwari [B.Tech. IIT-Delhi] Analyst & Correspondent Mr. Ajay Jain [B.E] Mr. Narendra Garg [B.E] Cover Design & Layout Harendra Singh Solanki / Niranjan Jain Om Gocher / Govind Saini Circulation & Advertisement Ankesh Jain / Praveen Chandna Ph (0744)- 2500492, 2430505, 9001799503 Subscription Sudha Jaisingh Ph. 0744-2500492, 2500692 © Strictly reserved with the publishers • No Portion of the magazine can be published/ reproduced without the written permission of the publisher • All disputes are subject to the exclusive jurisdiction of the Kota Courts only. Every effort has been made to avoid errors or omission in this publication. In spite of this, errors are possible. Any mistake, error or discrepancy noted may be brought to our notice which shall be taken care of in the forthcoming edition, hence any suggestion is welcome. It is notified that neither the publisher nor the author or seller will be responsible for any damage or loss of action to any one, of any kind, in any manner, there from.

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.

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Yours truly Owned & Published by Pramod Maheshwari, 112, Shakti Nagar, Dadabari, Kota & Printed by Naval Maheshwari, Published & Printed at 112, Shakti Nagar, Dadabari, Kota.

Editor : Pramod Maheshwari XtraEdge for IIT-JEE

Pramod Maheshwari, B.Tech., IIT Delhi

1

AUGUST 2009

Volume-5 Issue-2 August, 2009 (Monthly Magazine) NEXT MONTHS ATTRACTIONS

CONTENTS INDEX

Regulars ..........

Key Concepts & Problem Solving strategy for IIT-JEE. Know IIT-JEE With 15 Best Questions of IIT-JEE Challenging Problems in Physics, Chemistry & Maths

NEWS ARTICLE

3

IITian ON THE PATH OF SUCCESS

6

KNOW IIT-JEE

7

69 of top 100 JEE rankers pick IIT-Bombay IIT-JEE stars eye glory in International Physics Olympiad

Much more IIT-JEE News. Xtra Edge Test Series for JEE-2010 & 2011

PAGE

Abhay K. Bhushan

Previous IIT-JEE Question

Study Time........ DYNAMIC PHYSICS 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."

8-Challenging Problems [Set# 4] Students’ Forum Physics Fundamentals Capacitor - 2 Work, Energy, power & Conserv. law

CATALYST CHEMISTRY

• "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."

DICEY MATHS

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35

Mathematical Challenges Students’ Forum Key Concept Vector Permutation & Combination

Test Time ..........

• "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."

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Key Concept Reaction Mechanism Solid State Understanding : Physical Chemistry

• "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."

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XTRAEDGE TEST SERIES

45

Class XII – IIT-JEE 2010 Paper Class XI – IIT-JEE 2011 Paper

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AUGUST 2009

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 JEE2009 chairman A Pani. In 2008, the institute ushered in

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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.

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 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, IITKharagpur 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.

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).

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."

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.

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. 3

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 AUGUST 2009

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 IITJEE, 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.

XtraEdge for IIT-JEE

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.

Plan panel favours IIT, IIM offshore campuses

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.

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.

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.

4

NEW DELHI: Doors may soon be open for Indian universities and government-run institutions like IIMs and IITs to set up campuses abroad to crosssubsidise higher education for vulnerable sections of society.

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.

AUGUST 2009

Faculty divided location of IIT

over

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, IITRajasthan, 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 IITMumbai is big and popular, IITKharagpur 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

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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 5

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. AUGUST 2009

Success Story This article contains story of a person who get succeed after graduation from different IIT's

Abhay K. Bhushan B.Tech. /Electrical Engg. / 1965 Chief Financial Officer of the IITK Foundation, USA,

Abhay K. Bhushan ( B.Tech. /Electrical Engg. / 1965 )

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.

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.

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.

Mr. Bhushan initiated and managed the Environmental Leadership Program at Xerox and authored the widely acclaimed ‘Business Guide to Waste Reduction and Recycling'.

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.

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AUGUST 2009

KNOW IIT-JEE By Previous Exam Questions

PHYSICS 1.

P0 V0 PV = 0 0 2 2 As net heat is absorbed by the gas during path BC, temp. will reach maximum between B and C. (d) Equation for line BC  2P  P = –  0  V + 5P0, As PV = RT hence,  V0  RT P= [For one mole] [as y = mx + c] V 2P ∴ RT = – 0 V2 + 5P0V ...(1) V0 ∆QBC

One mole of an ideal monatomic gas is taken round the cyclic process ABCA as shown in figure. Calculate. [IIT-1998] P B 3P0

P0

A

C

2V0 V0 (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 : 5R 3R , Cv = Cp = 2 2 Cyclic process A → B ⇒ Isochoric process C → A ⇒ Isobaric compression (a) Work done = Area of closed curve ABCA during cyclic process. i.e. ∆ABC 1 1 × base × height = V0 × 2P0 = P0V0 ∆W = 2 2 (b) Heat rejected by the gas in the path CA during Isobaric compression process ∆QCA = nCp∆T = 1 × (5R/2)(TA – TC) 2P0 V0 PV TC = , TA = 0 0 , I×R I× R 5 5R  P0 V0 2P0 V0  ∆QCA = = – P0V0 − 2 2  R R  Heat absorbed by the gas on the path AB during Isochoric process ∆QAB = nCv∆T = 1 × (3R/2) (TB – TA) 3R  3P0 V0 P0 V0  = = 3P0V0 − 2  1× R 1× R  (c) As ∆U = 0 in cyclic process, hence ∆Q = ∆W ∆QAB + ∆QCA + ∆QBC = ∆W,

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= P0V0 –

For maximum;

2P0 dT = 0, – × 2V + 5P0 = 0; V0 dV

5V0 4 Hence from equation (1) and (2) ∴ V=

...(2)

2

2P0  5V   5V  ×  0  + 5P0  0  V0 4  4    25P0 V0 25 25 = –2P0V0 × + = P0 V 0 4 16 8 25 P0 V0 ∴ Tmax = R 8 RTmax = –

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] y0

m 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.

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AUGUST 2009

⇒ The angular frequency of the detector matches with that of the source. A´ ω = 10 rad/sec

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 L

K

mg = kl ; ⇒y= ⇒ y= where 3.

ω2

⇒ 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 − V0  (340 − 60) v´ = v  = 257.3 Hz  = 340 × (340 + 30)  V + Vs  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 + V0  (340 + 60) v´´ = v  = 438.7 Hz  = 340 × (340 − 30) V V − s  

Kl

mg mg l= k

mg k g

ω2 g

A´´

L l

Mean Position of oscillation

[Q K = mω2]

4.

< a (given)

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] 6m 6m 3m

B

C

D

A

B

C

D

A wire loop carrying a current I is placed in the x-y plane as shown in figure. [IIT-1991] y v M

P

→

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 ˆi 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 : y M I

a 60º r 60º

a

N

x

+Q P

y

v 60º

x

1 1 µ I (field dut to circle) =  0  3 3  2a  0.16µ 0 I µ I (outwards) =  0  (outwards) = a  6a 

B1 =

T´ (Accoustic Image) Hill

5 = 10 rad/s π

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x

N (a) If a particle with charge +Q and mass m is placed

Observer Source

= 2π ×

120º

O

a

A3 Sol. The angular frequency of the detector = 2πv

T

+Q

I

A1 A2

6m 6m B C D

A

→

0.16µ 0 I ˆ k a Magnetic field due to straight wire NM at P : or B1 =

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AUGUST 2009

v = 50 Hz, L = 35 mH, R = 11 Ω Impedance

µ0 I (sin 60º + sin 60º) 4π r Here, r = a cos 60º µ I (2 sin 60º) ∴ B2 = 0 4π a cos 60º µ I 0.27µ 0 I or B2 = 0 tan 60º = (inwards) 2π a a → 0.27µ 0 I ˆ or B 2 = – k a → → → 0.11µ 0 I ˆ ∴ B net = B1 + B 2 = – k a Now, velocity of particle can be written as, B2 =

Z=

also

v ˆ 3v ˆ j i+ v = v cos 60º ˆi + v sin 60º ˆj = 2 2 Magnetic force →

V0 Z V0 = Vrms 2 I0 =

Vrms 2 = 20A Z R 1 = cos φ = Z 2 π ∴ φ= 4 ∴ graph is given by V = V0sin ωt ∴

→

→

( wL) 2 + R 2 = 11 2 Ω

→

I0 =

V = V0sin ωT

Fm = Q( v × B )

0.11µ 0 IQv ˆ 0.11 3µ 0 IQv ˆ j– i 2a 2a ∴ Instantaneous acceleration =

I = I0sin(100πt–π/4)

→

→

Fm 0.11µ 0 IQv ˆ = ( j − 3 ˆi ) m 2am (b) In uniform magnetic field, force on a current loop is zero. Further, magnetic dipole moment of the loop a=

CHEMISTRY

→

will be, M =(IA) kˆ Here, A is the area of the loop. 1 1 A = (πa2) – [2 × a sin 60º] [a cos 60º] 3 2 =

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 [IIT-1997] the original mixture ? 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  1  = (40 × 10–3L)  mol L−1   10  = 40 × 10–4 mol Thus 40 × 10–4 mol S2O32– ≡ 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 STP is

a2 πa 2 – sin 120º = 0.61 a2 2 3

→

∴ M =(0.61 Ia2) kˆ →

Given, B = B ˆi →

→

→

∴ τ = M × B = (0.61 Ia2B) ˆj

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 O

T/4

3T/2 2T

Sol. Given Vrms = 220 V

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9

AUGUST 2009

Since

pV (1 atm)(1L) = RT (0.082 atm LK −1mol −1 )(273K ) = 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 ntotal =

=

9.6 × 10 −2 −2

( n H 2 )0 = n H 2 + 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 + ( n H 2 )0 = (0.15 + 0.3776)mol = 0.5276 mol n RT Hence, p0 = 0 V

(0.5276mol)(0.082L atm K −1mol −1 )(705 K ) (2.5 L) = 12.20 atm

× 100 = 6.575

=

9.6 ×10 + 1.364 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

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. O OMgCl O 8.

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 CH3OH(g) CO(g) + 2H2(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 [IIT-1993] reaction does not take place. Sol. We have CH3OH(g) CO(g) + 2H2(g) t = 0 0.15 mol teq 0.15 mol – x ( n H 2 )0 – 2x x

R–C–OR´ CH3MgCl R–C–OR´ (A)

R–C–CH3 + R´OH CH3MgCl OMgCl

R–C–CH3

+

H –HOMgCl

CH3

R–C–CH3 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 CH3 O

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. (0.08 mol / 2.5L) pV = ntotal = RT (0.082 atm L K –1mol −1 )(705 K )

R – C – O – C – CH3 (molecular formula R2C4H6O2 ) R From the molecular formula of A, we conclude that R must be H atom. Hence, the given ester is O

= 0.3676 mol Now, the amount of H2 at equilibrium is given as n H 2 = ntotal – nCO – n CH 3OH

H – C – O – CH – CH3

Isopropyl formate

CH3 The alcohol B is a secondary alcohol. Isopropyl alcohol CH3 – CH – CH3

= (0.367 – 0.07 – 0.08) mol = 0.2176 mol [CH 3 OH ] Hence, KC = [CO][H 2 ]2

OH The oxidation of alcohol B with NaOCl will give a ketone which further undergoes a haloform reaction. CH3 – CH – CH3 + NaOCl

(8.5 atm)(2.5 L)

(0.07 mol / 2.5 L )(0.2176mol / 2.5L) 2 = 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

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CH3

H+ –HOMgCl

OH

It is given that 0.08 mol of CH3OH is formed at equilibrium. Hence n CH3OH = x = 0.08 mol

=

n H 2 = ( n H 2 )0 – 2x, we have

OH

CH3 – C – CH3 + NaCl + H2O O

10

AUGUST 2009

Sol. The given observations are as follows. (i) Hydrated metallic saltheat  → white anhydrous residue

CH3 – C – CH3 + 3NaOCl O

CH3 – C – CCl3 + 3NaOH

(ii) Aqueous solution of B →  dark brown compound ( C)

(iii) Salt B

CH3 – C – O–Na+ + CHCl3

O The acidification of sodium acetate will produce acetic acid.

H –H2O

(A)

COCH3 Base

COCH3 O

( E ) + ( F)

(D)

acidified KMnO4 Pink colour is discharged BaCl2 solution

dark brown

Hence, the salt A must be FeSO4.7H2O The observation (iii) is 2FeSO4 → Fe 2 O 3 + SO 2 + SO 3 14243 (D) brown

( E ) + ( F)

The gaseous mixture of SO2 and SO3 explains the observation (iv), namely, 2MnO −4 + 5SO2 + 2H2O → 2Mn 2+ + 5SO 24− + 4H+

HBr (B)

Brown residue + Two gases

White precipitate 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 → [Fe(H 2 O) 5 ( NO)]2 + + H2O

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 1acetyl 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. CH3 OMgBr O CH3 Br CH3 +

strong heating →

(iv) Gaseous mixture (E) + (F)

9.

CH3MgBr

( B)

NO

O

CH3 – C – CCl3 + NaOH O

(A)

(E)

no colour

pink colour

2H2O + SO2 + SO3 Ba2+ + SO32– → BaSO3

CH3

4H+ + SO32– + SO42–

white ppt

O O

Ba2+ + SO42– → BaSO 4 white ppt

(D)

(C)

Hence, the various compounds are (A) FeSO4.7H2O (B) FeSO4 (C) [Fe(H2O)5NO]SO4 (D) Fe2O3 (E) and (F) SO2 and SO3

The conversion of C into D may involve the following mechanism.

CH2

COCH3

COCH3

COCH3 O

B+ –BH+

HC

O

HC

O–

MATHEMATICS

BH+ –B

11. Prove that tan α + 2 tan 2α + 4 tan 4α + 8 cot 8α = cot α [IIT-1988] Sol. We know that  1 − tan 2 θ  1 − tan 2 θ  = 2 cot 2θ = 2 cot θ – tan θ =  2 tan θ  tan θ   ∴ 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 = cot α

(C)

COCH3 OH +B –BH+

COCH3 –

COCH3 OH –OH– (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]

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11

AUGUST 2009

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] π  ⇒ cos (p sin x) = cos  − p cos x  2 

  1 = cos tan–1 sin  sin −1    1+ x2

 1 = cos  tan −1  1+ x2 

  =  

  1 1 or p 2  sin x − cos x  = 2nπ – π/2, n ∈ I 2   2 π π π   ⇒ p 2  cos sin x + sin cos x  = 2nπ + 2 4 4   π π π   or p 2  cos sin x − sin cos x  = 2nπ – , n ∈ I 2 4 4  

  π π  ⇒ p 2 sin  x +  = (4n + 1) , n ∈ I 4  2     π  π or p 2 sin  x −  = (4n – 1) , n ∈ I 4  2   Now, –1 ≤ sin (x ± π/4) ≤ 1

= R.H.S.

Sol. Let us take a point P( 6 cos θ, 2

3 sin θ) on

2

x y = 1. Now to minimise the distance from P + 6 3 to given straight line x + y = 7, shortest distance exists along the common normal. Y x+y=7

If n ≥ 0, – 2 p ≤ (4n + 1) π/2

P

⇒ (4n + 1) π/2 ≤ 2 p For p to be least, n should be least ⇒ n=0 π ⇒ 2 p ≥ π/2 ⇒ p ≥ 2 2 π Therefore least value of p = 2 2

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x2 + 2

15. Find a point on the curve x2 + 2y2 = 6 whose distance from the line x + y = 7, is minimum. [IIT-2003]

⇒ –p 2 ≤ p 2 sin (x ± π/4) ≤ p 2 (4n + 1).π ⇒ –p 2 ≤ ≤p 2 ,n∈I 2 (4n − 1)π or –p 2 ≤ ≤p 2 ,n∈I 2 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.

Sol. L.H.S. = cos tan–1{sin(cot–1x)}

x2 +1

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

π  ⇒ p sin x = 2nπ ±  − p cos x  , n ∈ I 2  [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  1  1 ⇒ p. 2  sin x + cos x  = 2nπ + π/2 2  2 

13. Prove that cos tan–1{(sin cot–1x)} =

   

X

O

Slope of normal at P =

x2 +1

6 sec θ 6 cos ecθ

=

2 tan θ = 1

1 2 and sin θ = 3 2 Hence, P(2, 1) So, cos θ =

x2 + 2 [IIT-2002]

12

AUGUST 2009

Physics Challenging Problems

Set # 4

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

So l ut i ons wi l l b e pub l i s he d i n ne x t i s s ue 6.

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

rr

r r

rr

r r

r r

r

r r

r r

r r

r

A

D rough smooth

b

r

r r

rr

r r

rr

7. 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

8.

10Ω 5Ω

a 5Ω

5Ω

key-k

C

B

r r

r r

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

b

(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 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 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 –σ

10Ω

D

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

(A) Magnitude of electric field at a distance h above the negatively charged sheet is ρD − σ 2ε 0

(B) Magnitude of electric field inside the slab at a distance h below the negatively charged sheet (h < D) is σ + ρ(D − 2h )

y

2ε 0

m

g

(C) Magnitude of electric field at a distance h below the bottom of the slab is ρD − σ

x2 = 4ay

4ε 0

x

(A)

ga

(B)

XtraEdge for IIT-JEE

+ρ

2ga

(C)

g / a (D)

(D) Magnitude of electric field at a distance h below the bottom of the slab is ρD − σ

g / 2a

2ε 0

13

AUGUST 2009

8 R/2

1.

Q1 =

∫

ρ(r )4πr 2 dr =

0 R

∫ ρ(r) × 4πr

Q2 =

R/2 R

2

Physics Challenging Problems Qu e s t i ons we r e P ub l i s h ed i n J ul y I ss ue R /2

∫

α 4πr 2 dr

5.

At maximum temperature KA (Tmax – T0) q= l Option [A] is correct

6.

As the process is isobaric V V V T ∴ 0 = max ⇒ max = max T0 Tmax V0 T0 Option [A] is correct

0

dr

r 



Solution

∫ 2α1 − R 4πr dr

=

2

R /2

Q1 Q2 Option [A] is correct Fraction =

2.

7.

ρL R= A

−eαr 3mε 0 a∝r

Heat produced =

eα 3ε 0 m

V V = d L V E´ = 2L E E´ = 2 ∴ option [D] is correct E=

2π ω Option [B] is correct

T=

∫

Q1 + ε0

r

2

r  4πr dr  2α1 −  R  ε0  R/2

∫

8.

Option [A] is correct 4.

Let at any instant t temperature is T. The net rate at which heat is absorbed by the gas is KA(T − T0 ) dQ =q– .......(1) dt l 7 .......(2) Now, dQ = nCpdT = n × R × dT 2 KA(T − T0 ) 7 dT ∴n× R =q– 2 dt l T

of

∫

T0

2L ρ × 2L R R´ = = 4A 2

V2 as R become half R ∴ heat produced is doubled

a=

E.ds =

4A

L

q × ρr −eρr = = 3ε 0 3ε 0

3.

⇒

A

F = qE

∴ω=

Set # 3

Cartoon 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 directedpressure explosions and of reckless cowards who are so eager to escape that they exit directly through the wall of a house, leaving a cookiecutout-perfect hole. The threat of skunks or matrimony often catalyzes this reaction.

t

2 dT = dt 7 nRl ql − KA(T − T0 )

∫ 0

Option [A] is correct

XtraEdge for IIT-JEE

Option [A,C, D] is correct

14

AUGUST 2009

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

PHYSICS 1.

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. 1 Using s = ut + at2, t = 1 second 2 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

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. m1

m3

m2 37º

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, m2g m1 g T T N2

∴ Velocity of m2 is v2 =

= 13 ms–1 Ans. Velocity of wedge at this instant = bt = 2 ms–1 Ans. 2.

C

N2

m2(0.8a – 0.2b) N1 N1 ...(1) For horizontal forces on m1, T = m1a For vertical forces on wedge, T – T cos 37º + N2 sin 37º = m3b ...(2) For horizontal forces on m2 ...(3) N2sin 37º – T cos 37º = m2(0.8a – 0.2b) For vertical forces on m2,

XtraEdge for IIT-JEE

m v0

A

m B

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.

37

m3b

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

m2(0.6a + 0.6b)

m1a

v 22 x + v 22 y

15

AUGUST 2009

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 v or v = 0 = 0.3 ms–1 2 ∴ Velocity of centre of mass of the system, 2m × v + m × 0 = 0.2 ms–1 vc = 2m + m Now the system is as shown in fig. 2m m

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, Mg T1 T2 1α T1

(2m)(m) 2m = 2m + m 3 ∴ Frequency of oscillations,

Its reduced mass,

m0 =

1 K 5 10 = Hz. Ans. 2π m 0 π Since, just after the collision, combined body has velocity v, therefore, energy of the system at that 1 instant, E = (2m)v2 = 0.27 joule 2 Due to velocity vC of centre of mass of the system, translational kinetic energy, 1 Et = (3m) v c2 = 0.18 joule 2 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. 1 Kx 02 = E0 then 2 f=

∴ x0 = 3.

90 × 10–3 m or 3 10 mm

A

O m1g

B

m2 (b – a) α

Friction

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. +10–10C +2 × 10–10C

b

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

B A

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T2

m2g N For forces on block A, ...(1) T1 – m1g = m1(2bα) For forces on block B, ...(2) m2g – T2 = m2(b – a)α Taking moments of forces acting on the spool, about O, ...(3) T2(b – a) – T1(2b) = Iα 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.

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

m1 (2bα)

16

AUGUST 2009

+q

–q

p

5.

(2 × 10 −10 − q ) =0 2ε 0 S

or q = 5 × 10–11 coulomb or 50 pC Hence, the charges are as shown in fig. +

–

150 pC +

–

+

–

q2 – +

+

+

+ 150 pC

50 pC +

1

–

+

+

–

+

+ 150 pC

–

+

+

–12

150 pC + +

(50×10 –q´)

+

q1

–

–

+

6

–

(q2 – q3)

4

–

– + q3 + – (q1 – q2 + 2q3)

+ q3

+

– +

q2

5 (q1 + q2)

–

Applying Kirchhoff´s voltage 1 – 2 – 6 –1, q q2 q + 3 – 1 =0 C C C or q1 = (q2 + q3)

law

on

mesh

...(i)

For mesh 2 – 3 – 6 – 2, q 2 − q3 q − q 2 + 2q 3 q – 1 – 3 =0 C C C 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, q − q3 q1 q + 2 + 2 –V=0 C C C 1 CV. Substituting q1 = 6q3 and q2 = 5q3, q3 = 15 But charge drawn by the arrangement from battery is 11 q = (q1 + q2) = 11q3 = CV 15 q 11C = = 11µF Ans. ∴ Equivalent capacitance = V 15

+ – E q Capacitance of the capacitor is ε S C = 0 = 5 × 10–12 F d Applying Kirchhoff's voltage law,

–

XtraEdge for IIT-JEE

–

+

q´

(50 × 10 −12 − q´) –E=0 C ∴ q´ = 1 × 10–10 coulomb ∴ Energy supplied by battery = q´E = 10–9 joule

+

q1

3

(q1 + q2)

+

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. +

q2 – q3

2

+

5

6

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.

(10 −10 + q ) q q – – 2ε 0S 2ε 0 S 2ε 0 S –

4

3

1

Considering a point P inside the plate B, Electric field on it is E=

Nine identical capacitors, each of capacitance C = 15 µF are connected as shown in fig. Calculate equivalent capacitance between terminals 1 and 4. 2

+(2 × 10–10 – q)

+(10–10 + q)

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.

Ans.

17

AUGUST 2009

P HYSICS F UNDAMENTAL F OR IIT-J EE

Capacitor-2 KEY CONCEPTS & PROBLEM SOLVING STRATEGY Capacitors in Series : +Q –Q

+Q –Q

+Q –Q

A C1

C2

V1

If CpV is the net capacitance for the parallel combination of capacitors then

C3

V2

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

B

V3

V

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. ...(2) V = V1 + V2 + V3 If Cs is the net capacitance of the series combination, then

  V1 =    

U=

Capacitors in Parallel :

A

–Q1 C1

+Q2

–Q2 C2

+Q3

–Q3 C3

⇒ U=

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. ...(1) i.e. V = V1 = V2 = V3

1 C3 1 Cs

  V   

1 1 1 1 1 = + + + .... + Cs C1 C 2 C 3 Cn

 Q  

 ε0A σ and V = Ed where E =  d ε 0 

1 ε0A 2 2 Ed 2 d

1 ε0E2τ 2 where τ is volume of the capacitor

⇒ U=

⇒

...(2)

Electrostatic Energy U = Ue = τ Volume = Electrostatic Pressure

(b) The total charge Q is shared by each capacitor in the direct ratio of the capacitances.

=

⇒ Q = Q1 + Q2 + Q3

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  V3 =    

 1 ⇒ U =  ε 0 E 2  (Ad)  2

V

⇒

   V;   

1 CV2 2

where C =

B

Q Q Q V= 1 = 2 = 3 C1 C2 C3

1 C2 1 Cs

C  C  C Q1 =  1  Q; Q2 =  2  Q; Q3 =  3  Cp   Cp   Cp      and so on, where Cp = C1 + C2 + C3 + ... + Cn Energy Density : For a parallel plate capacitor

Q Q and V = C1 Cs

+Q1

  V2 =    

(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

1 1 1 1 = + + Cs C1 C 2 C 3

Further V1 =

   V;   

and so on, where

Q Q Q Q = + + Cs C1 C 2 C 3 ⇒

1 C1 1 Cs

18

σ2 1 ε0E2 = 2 2ε 0

 σ Q E =  ε0   AUGUST 2009

Energy for series and parallel combinations : Series Combination : For a series combination of capacitor Q = constant and

⇒

F=

 σ2 Q2 =   2ε 2ε 0 A  0

 σ Q Q = σA, E =  ε 0 

1 1 1 1 = + + + ... C1 C 2 C 3 Cs ⇒

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

Q2 Q2 Q2 Q2 = + + + .... 2C s 2C1 2C 2 2C 3

⇒ Us = U1 + U2 + U3 + ...... Parallel Combination : For a parallel combination of capacitors V = constant and Cp = C1 + C2 + C3 + ....

∑q = 0 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

1 1 1 1 CPV2 = C1V2 + C2V2 + C3V2 + ... 2 2 2 2 ⇒ 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 + – + – + – + – + – – + ⇒

⇒ x = q1 – q2 +

⇒

ε0A x

Q2 Q2 = (x – dx) 2C´ 2ε 0 A

If dU is the change in potential energy, then dU = Uf – Ui Q2 Q2 (x – dx) – x 2ε 0 A 2ε 0 A

dU =

⇒

dU = –

Q2 dx 2ε 0 A

Further since F=–

dU dx

XtraEdge for IIT-JEE

–

+

C

∑V = 0

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 q capacitor i.e. ∆V = + . C (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 q capacitor i.e. ∆V = – . C

 Q2  Q x U= ⇒U=   2ε A  2C  0 

⇒

–

B

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.

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 =

+q1 –q1 +q2 –q2

1 2

2

Also

–

A

At any instant let the plate separation be x, then C=

  A =  1 ε 0 E 2  A  2  

19

AUGUST 2009

2.

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.

C1

C2

A

C C B C

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. 1 C 3

A

2C

2C

C C

C 4

C

P

P

C

B

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 3C Resultant of 3C and (3) is 4 3C 7C and (4) is Resultant of 4 4 So total capacitance across AB is C 7C CAB = AP ⇒ CAB = 2 8

XtraEdge for IIT-JEE

C2 C1 Q1 –Q1 Q–Q1 –(Q–Q1) a b D (2Q1–Q) i –(2Q1–Q) C3 j

B

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) Q Q − Q1 or VA – VB = 1 + ...(1) C1 C2 Also, VA – VB = (VA – VD) + (VD – VE) + (VE – VB) Q 2Q1 − Q Q + 1 ...(2) or VA – VB = 1 + C1 C3 C1 We have to eliminate Q1 from these equation to get Q . the equivalent capacitance (VA − VB ) The first equation may be written as  1 Q 1   + VA – VB = Q1  − C2  C1 C 2  C1C 2 C1 (VA – VB) = Q1 + Q ...(3) or C 2 − C1 C 2 − C1 The second equation may be written as  1 Q 1   – VA – VB = 2Q1  +   C1 C 3  C 3

C

C

C1

e f E g h Q–Q1 –(Q–Q1) Q1 –Q1 C2 C1

C

A

B

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.

Find the net capacitance of the circuit shown between the points A and B.

C

C2

C3

A

Solved Examples 1.

Find the equivalent capacitance between the point A and B in figure.

20

AUGUST 2009

Further C34 is again in parallel. Hence the effective capacity C × 2C 0 5 5 A = C0 + 0 = C0 = Kε0 . C 0 + 2C 0 3 3 d (ii) Charge on the plate 5 = charge on the uper half of parallel combination 2 Kε 0 AV0 2  ∴Q5 = V0  C 0  = 3 d 3   Charge on plate 3 on the surface facing 4 kε 0 AV0 ∴ V0C0 = d Charge on plate 3 on the surface facing 2 = [potential difference across (3 – 2)]C0 C0 AV0 C0 = Kε0 = V0 C 0 + 2C 0 3d

C1C 2 C1 (VA – VB) = Q1 – Q ...(4) 2(C1 + C 3 ) 2(C1 + C 3 ) Subtracting (4) from (3)  CC C1C 3  (VA – VB)  1 2 −   C 2 − C1 2(C1 + C 3 )  or

 C1  C1 =  + Q  C 2 − C1 2(C1 + C 3 )  or (VA – VB)[2C1C2(C1 + C3) – C1C3(C2 – C1)] = C1[2(C1 + C3) + (C2 – C1)]Q 2C1C 2 + C 2 C 3 + C 3C1 Q or C = = VA − VB C1 + C 2 + 2C 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 3.

Kε 0 AV0 AV0 + Kε0 d 3d Kε 0 AV0  1  A 4 = 1 + 3  = 3 Kε0 d V0 d  

∴ Q3 =

4.

3µF

3µF

20Ω

C

A

Sol. The circuit is redrawn in fig (a, b, c)

2

3µF

1µF B

3

2

3

4

5

4

3µF

Q2 (–)

(+)

A

(b)

10Ω C

100 V Fig.(a) B 2µF

6µF

i.e., C12, C32, C34 and C54. The capacity of each  Kε A  capacitor is  0  = C0. The effective capacity  d  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 C 0 × 2C 0 C 0 + 2C 0

1µF 1µF

20Ω

Q1

XtraEdge for IIT-JEE

10Ω

100 V

(Q2/2)

Q

1µF

1µF

(+)

(a)

(Q2/2)

1µF

B

(–)

4 3 2 1 1

In diagram find the potential difference between the points A and B and between the points B and C in the steady state.

3/2 µF

R

1µF

1µF

P

Q 20Ω

A

P

10Ω C 100 V

Fig.(b)

S

Q 20Ω

A

10Ω C 100 V

Fig.(c)

From fig. (c). potential difference between P and Q = Potential difference between R and S = 100 volt 21

AUGUST 2009

2

3 × 10–6 × 100 2 = 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.

1 CV 2 V × (3C)   = 2 6 3 ∴ Total final energy

∴ Q = capacity × volt =

=

Ef = ∴

Q 150 × 10 −6 = = 25 volt. capacity 6 × 10 −6 Again potential difference between C and D = potential difference across the two ends of condenser of capacity 2µF ∴ V1 =

V2 = 5.

150 × 10 −6 2 × 10 −6

3 CV 2 5 CV2 + = CV2 2 6 3

Ei CV 2 3 = = 2 Ef 5 (5 / 3)CV

Space Shuttle

= 75 volt

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. S

V

A

C

B

C

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.

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 K C1 = 1 C = 3C. K 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 V ∴ V1 = volt 3 1 Initial energy of capacitor A = CV2 2 1 2 energy of capacitor B = CV 2 1 1 ∴ Total energy Ei = CV2 + CV2 = CV2 2 2 Final energy of capacitor A 1 3 × (3C)V2 = CV2 = 2 2 Final energy of capacitor B

XtraEdge for IIT-JEE

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.

22

AUGUST 2009

P HYSICS F UNDAMENTAL F OR IIT-J EE

Work, Power, Energy & Conservation Law KEY CONCEPTS & PROBLEM SOLVING STRATEGY

The potential energy of particle in the gravitational field is given by

Work, Energy and Power :

Work is done when a force (F) is displaced.

U = U0 + mgh where U0 = potential energy of the body at the ground level.

dr θ

This is true only for objects near the surface of the earth because g is uniform only near the surface of the earth.

F

The work done is

The strain potential energy of a spring is given by U 1 = kx2, where k is the force constant of the spring 2 and x is the charge in length of the spring. This change in length may be either a compression or on extension.

dW = F dr cos θ Using vector notation rr dW = F.dr When the force and the displacement are in the same direction, θ = 0, cos θ = +1, work done is positive.

Potential Energy and force

When the force and the displacement are in opposite directions, θ = 180º, cos θ = –1, work done is negative.

Fx = –

When the displacement is perpendicular to the direction of the force, θ = 90º, cos θ = 0, no work is done. r r For a system of particles the quantity F.dx cm is

Principle of Conservation of energy : Conservative and Non-conservative Force : If the work done by a force in moving a body from one point to another depends only on the positions of the body and not on the process or the path taken, the force is said to be conservative. Gravitational force, spring force, elastic forces, electric and magnetic forces are examples of conservative forces. If the work done depends on the paths taken, the force is said to be non-conservative. Frictional force is a nonconservative force. Work-energy Theorem : The work by external forces on a body is equal to the change of kinetic energy of the body. This is true for both constant forces and variable forces (variable in both magnitude and direction).

∫

called pseudo work. At times actual work may be zero but not pseudo work. Work is a scalar quantity. Its unit is joule. Power is the rate of doing work. Thus Power =

work done time taken

The unit of power is the watt (= joules/second). rr The power of an agent is given by P = F.v where F is the force applied by the agent and v is the velocity of the body on which the agent applies the force.

For a particle W = ∆K. For a system of particles Wnet = Wreal + Wpseudo = ∆Kcm

The energy of a system is its capacity of doing work. Mechanical energy may be of two types :

Principle of Conservation of Energy : Energy can neither be created nor destroyed by any process.

(i) kinetic energy and (ii) potential energy. The kinetic energy of a particle is T =

1 mv2. 2

The kinetic energy of a system is T =

1 2 Mv cm 2

XtraEdge for IIT-JEE

∂U ∂x

For a particle K + U = a constant. For a system of particles Kcm + Uext + Eint = a constant. However, energy can be transformed from one form into another.

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AUGUST 2009

Collision of Bodies : Elastic Collision : When two bodies meet a with certain relative speed they are said to collide with each other. In a collision, kinetic energy is transferred, from one body to another. When the transfer of kinetic energy takes place in such a way that the total kinetic energy is conserved, the collision is said to be perfectly elastic, or simply elastic. When kinetic energy is not conserved the collision is said to be inelastic. Further, in a collision, if one body gets embedded in the other and kinetic energy is not conserved, it is a completely inelastic collision. In inelastic and completely inelastic collisions there is always a loss of kinetic energy and this energy is converted into other forms of energy, mostly heat. A collision is said to be direct or head-on if the relative motion before and after the collision is in the same direction; if not it an oblique collision. Remember the following points while solving problems on the collision of bodies. (i) Apply the principle of conservation of momentum. In one-dimensional direct collisions, one equation is obtained by equating momenta before and after collision in the direction of motion. In two-dimensional collisions, select the line of impact as the X-axis and the line perpendicular to it as the Y-axis and obtain two equations by equation by equating momenta before and after the collision along the X- and Yaxes. Remember that momentum is a vector quantity. It may be positive or negative depending on the direction. Choose any one direction as positive; the opposite will be negative. (ii) If it is an elastic collision, apply the principle of conservation of kinetic energy. For inelastic collisions, apply the principle of conservation of energy to obtain an additional equation. (iii) Remember there is no change in momentum along the common tangent to the colliding bodies. Coefficient of restitution : According to Newton, the relative velocity of a body after collision is proportional to its relative velocity in the same direction before collision, with a reversal of sign. Here, relative velocity means the velocity of any one of the colliding bodies (say A) with respect to the other colliding body (say B). The constant of proportionality is called the coefficient of restitution (e). That is VAB (after collision) = –e × V´AB (before collision) This is Newton's law of collision. For elastic collisions, e = 1. For inelastic collisions, e 4 case II : If D ≥ 0 ⇒ k ≤ 5/4 (and both roots α, β should be –ve) 1 α+β 0 2 ⇒k>1 or k < –1 common values of k are such that 5  x ∈ (–∞, –1) ∪  , ∞  4  (ii) for exactly two real solutions one root of the equation f(t) = 0 should be positive and other negative, as for one positive t we have two real x. For this D > 0 and f(0) < 0 (or c < αβ < 0) 5 and k2 – 1 < 0 ⇒ k< 4 ⇒ – 1 < k < 1, Thus k ∈ (–1, 1).

f ( x + h ) − f (x ) h f ( x ) + f (h ) − f ( x ) = lim h →0 h f (h ) − f (0) = lim = f ´(0) h →0 h ⇒ f(x) = x . f ´(0) + c for x = 0 ⇒ c = 0 h →0

∫

0

n 3f ´(0) 2

x . f ´(0) dx =

Hence, I1 + I2 + I3 + I4 + I5 f ´(0) 3 = (1 + 23 + 33 + 43 + 53) 2 f ´(0)  900  450 =   2  4  f ´(0) = 4 ⇒ f(x) = 4x 2.

Let [x] stands for the greatest integer function find 2

the derivative of f(x) = ( x + [ x 3 + 1]) x +sin x , where it exists in (1, 1.5). Indicate the point(s) where it does not exist. Give reason(s) for your conclusion. Sol. The greatest integer [x3 + 1] takes jump from 2 to 3 at 3

2 and again from 3 to 4 at 3 3 in [1, 1.5] and therefore it is discontinuous at these two points. As a

4.

3

result the given function is discontinuous at 2 and hence not differentiable. To find the derivative at other points we write : in (1,

3

2 ), f(x) = ( x + 2)

1 f(n) – 2

2

⇒ f ´(x) = ( x + 2) x + sin x −1 {x2 + sin x + (x + 2) (2x + cos x) log (x + 2)} in ( 3 2 , 3 3 ), f(x) = ( x + 3) x x 2 + sin x −1

+ sin x

Sol. L.H.S. =

,

∫

n

∑ f (r) , [.] denotes greatest integer. r =0

∫

n

0

f ´(x )[ x ] dx –

∫

n 0

x.f ´(x ) dx +

2

f´(x) = ( x + 3) {x + sin x + (2x + cos x) (x + 3) × loge (x + 3)}

XtraEdge for IIT-JEE

Prove that for a differentiable function f(x) n n 1 1  f ´(x )[ x ] − x +  dx = f ´(x ) dx – f(0) + 0 0 2 2  

∫

x 2 + sin x

2

,

2

3.

Now f ´(x) = lim

h

+ sin x

, {x + sin x + (2x + cos x) f ´(x) = ( x + 4) (x + 4) × loge(x + 4)}

n

find f(x) Sol. Putting x = y = 0 ⇒ f(0) = 0

so In = n

2

38

1 2

∫

n

0

f ´(x ) dx

AUGUST 2009

n

=

 f ´(x )[ x ] dx –  xf ( x ) | 0n − r −1 

∑∫ r =1

r

n

=

∑

r

∫

(r − 1) f ´(x)dx – n f(n)+

r =1

r −1



n

0

→ Sol. aˆ × cˆ = cˆ × bˆ ⇒ ( aˆ + bˆ ) × cˆ = 0 ⇒ cˆ is collinear with aˆ + bˆ ⇒ aˆ + bˆ = λ cˆ for same λ ∈ R Similarly bˆ + cˆ = µ aˆ for some scalar u

n 0

1

n

∫ f (x)dx + 2 (f(n) – f(0)) 0

n

=

1

∫ f (x)  + 2 f(x) |

→ Now aˆ + bˆ = λ cˆ ⇒ aˆ + bˆ + cˆ = (λ + 1) c Similarly ⇒ aˆ + bˆ + cˆ = (µ + 1) aˆ Hence (λ + 1) cˆ = (µ + 1) aˆ , either λ + 1 = µ + 1 = 0 or cˆ is collinear with aˆ . But cˆ can not be collinear to aˆ other wise cˆ × aˆ = 0 ⇒ cˆ × bˆ = 0

n

∑ (r − 1) {f(r) – f(r – 1)} – n f(n) + ∫ f (x) dx 0

r =1

1 (f(n) – f(0)) 2 = (f(2) – f(1)) + 2{f(3) – f(2)} + .... + (n – 1) (f(n) n 1 1  – f(n –1)) –  n −  f(n) – f(0) + f ( x ) dx 0 2 2  = –f(1) – f(2) – f(3) .... f(n – 1) – f(n) n 1 1  +  n − n +  f(n) – f(0) + f ( x ) dx 0 2 2  +

∫

⇒ bˆ is collinear to with cˆ ⇒ aˆ bˆ and cˆ are collinear. Hence cˆ is not collinear to aˆ ⇒λ+1=µ+1=0 ⇒ λ ± µ = –1 Hence bˆ + cˆ = µ aˆ

∫

n

=–

1

1

n

∑ f (r) + 2 f(n) – 2 f(0) + ∫ f (x) dx r =0

0

→ ⇒ aˆ + bˆ + cˆ = 0 ⇒ ( aˆ + bˆ + cˆ ) . ( aˆ + bˆ + cˆ ) = 0 ⇒ 1 + 1 + 1 + 2 ( aˆ . bˆ + bˆ . cˆ + cˆ . aˆ ) = 0

If normals at the points P and Q of the parabola y2 = 4ax meet at the point R of the parabola. Show that the locus of centroid of the ∆ PQR is a ray. Find the equation of the ray. Sol. Let P = (at2, 2at). Then Q is   2 2   a   , 2a  2  =  4a , 4a   t  t   t 2 t   2 and R is (aT2, 2aT). where T = –t – t centroid of the ∆PQR  at 2 + 4a / t 2 + aT 2 2at + 4a / t + 2aT  , =   3 3   y co-ordinates of the centroid 2 4a  = 2at + + 2a  − t −  = 0 t t  Thus centrocid of the ∆PQR for any choice of P on the parabola lies on the axis of the parabola. x –coordinates the centroid 2 a 4  2  = t 2 + 2 +  t +   3  t   t  2a  2 4 2a  (4 + 2) = 4a = t + 2 + 2 ≥ 3  3 t  Hence equation of the ray is given by y = 0, x > 4a 5.

6.

3 ⇒ aˆ . bˆ + bˆ . cˆ + cˆ . aˆ = – 2 ⇒ cos α + cos β + cos γ = –

Know About 1.73 is also known as Theodorus' constant (it is named after Theodorus of Cyrene, who proved that the square roots of the numbers from 3 to 17, excluding 4, 9, and 16, are irrational). is the diagonal of a cube having 1-unit sides. is the height of an equilateral triangle having 2-unit sides. The shape 'Vesica piscis' (fish bladder) has a major axis/minor axis ratio equal to the square root of 3, this can be shown by constructing two equilateral triangles within it. ≈ 1.7320508075 6887729352 7446341505 8723669428 0525381038 0628055806 9794519330 1690880003 7081146186 7572485756 7562614141 5406703029 9699450949 9895247881 1655512094...

For three unit vectors aˆ , bˆ and cˆ not all collinear given that aˆ × cˆ = cˆ × bˆ and bˆ × aˆ = aˆ × cˆ . Show that cosα + cos β + cos γ = –3/2, where α, β and γ are the angles between aˆ and bˆ , bˆ and cˆ and cˆ and aˆ respectively.

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3 2

= 2sinus(60°) = 2sinus(30°) = 1 + (1 / (1 + (1 / (2 + (1 / (1 + (1 / 2 + ... ))))))) ≈ 97/56

39

AUGUST 2009

MATH

VECTOR Mathematics Fundamentals Properties of vectors : (i) Addition of vectors

Representation of vectors : Geometrically a vector is represent by a line segment.

Triangle law of addition : If in a ∆ABC,

For example, a = AB . Here A is called the initial point and B, the terminal point or tip. Magnitude or modulus of a is expressed as

AB = a, BC = b and AC = c, then AB + BC = AC i.e., a + b = c

|a | = | AB | = AB.

C

B c=a+b

a

B a A Parallelogram law of addition : If in a

A

Types of Vector: Zero or null vector : A vector whose magnitude is zero is called zero or null vector and it is

parallelogram OACB, OA = a, OB = b and

OC = c B

represented by O . Unit vector : A vector whose modulus is unity, is called a unit vector. The unit vector in the direction of a vector a is denoted by aˆ , read as “a cap”. Thus, | aˆ | = 1.

aˆ =

C c=a+b

b

a Vector a = a | | Magnitude of a

O

a

A

Then OA + OB = OC i.e., a + b = c, where OC is a diagonal of the parallelogram OABC. Addition in component form : If the vectors are defined in terms of i, j, and k, i.e., if a = a1i + a2j + a3k and b = b1i + b2j + b3k, then their sum is defined as a + b = (a1 + b1)i + (a2 + b2)j + (a3 + b3)k. Properties of vector addition : Vector addition has the following properties. Binary operation : The sum of two vectors is always a vector. Commutativity : For any two vectors a and b, a + b = b + a. Associativity : For any three vectors a, b and c, a + ( b + c ) = ( a + b) + c Identity : Zero vector is the identity for addition. For any vector a, 0 + a = a = a + 0 Additive inverse : For every vector a its negative vector –a exits such that a + (–a) = (–a) + a = 0 i.e., (–a) is the additive inverse of the vector a.

Like and unlike vectors : Vectors are said to be like when they have the same sense of direction and unlike when they have opposite directions. Collinear or parallel vectors : Vectors having the same or parallel supports are called collinear or parallel vectors. Co-initial vectors : Vectors having the same initial point are called co-initial vectors. Coplanar vectors : A system of vectors is said to be coplanar, if their supports are parallel to the same plane. Two vectors having the same initial point are always coplanar but such three or more vectors may or may not be coplanar. Negative of a vector : The vector which has the same magnitude as the vector a but opposite direction, is called the negative of a and is denoted by –a. Thus, if PQ = a, then QP = –a.

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b

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AUGUST 2009

Subtraction of vectors : If a and b are two vectors, then their subtraction a – b is defined as a – b = a + (–b) where –b is the negative of b having magnitude equal to that of b and direction opposite to b. If a = a1i + a2j + a3k, b = b1i + b2j + b3k Then a – b = (a1 – b1)i + (a2 – b2)j + (a3 – b3)k. B a+b O

Scalar or Dot product Scalar or Dot product of two vectors : If a and b are two non-zero vectors and θ be the angle between them, then their scalar product (or dot product) is denoted by a . b and is defined as the scalar |a | |b| cosθ, where |a| and |b| are modulii of a and b respectively and 0 ≤ θ ≤ π. Dot product of two vectors is a scalar quantity. B

b a

b

A

θ

–b

A O a Angle between two vectors : If a, b be two vectors inclined at an angle θ, then a . b = |a | |b| cos θ

a + (–b) = a – b

B´ Properties of vector subtraction :

(i) a – b ≠ b – a

⇒ cos θ =

(ii) (a – b) – c ≠ a – (b – c) (iii) Since any one side of a triangle is less than the sum and greater than the difference of the other two sides, so for any two vectors a and b, we have

 a.b   θ = cos–1   | a || b |  If a = a1i + a2j + a3k and b = b1i + b2j + b3k; then

⇒

(a) |a + b| ≤ |a | + |b| (b) |a + b| ≥ |a | – |b|

 a 1 b1 + a 2 b 2 + a 3 b 3 θ = cos–1   2 2 2 2 2 2  a 1 + a 2 + a 3 b1 + b 2 + b 3

(c) |a – b| ≤ |a | + |b| (d) |a – b| ≥ |a | – |b| Multiplication of a vector by a scalar : If a is a vector and m is a scalar (i.e., a real number) then ma is a vector whose magnitude is m times that of a and whose direction is the same as that of a, if m is positive and opposite to that of a, if m is negative. Properties of Multiplication of vector by a scalar : The following are properties of multiplication of vectors by scalars, for vector a, b and scalars m, n. (i) m(–a) = (–m)a = –(ma) (ii) (–m) (–a) = ma (iii) m(na) = (mn)a = n(ma) (iv) (m + n)a = ma + na (v) m(a + b) = ma + mb Position vector :

   

Properties of scalar product Commutativity : The scalar product of two vector is commutative i.e., a . b = b . a Distributivity of scalar product over vector addition : The scalar product of vectors is distributive over vector addition i.e., (a) a . (b + c) = a . b + a . c, (Left distributivity) (b) (b + c) . a = b . a + c . a, (Right distributivity) Let a and b be two non-zero vectors a . b = 0 ⇔ a ⊥ b. As i, j, k are mutually perpendicular unit vectors along the coordinate axes, therefore, i . j = j . i = 0; j . k = k . j = 0; k . i = i . k = 0. For any vector a . a . a = |a|2. As i. j. k are unit vectors along the co-ordinate axes, therefore i . i = |i|2 = 1, j . j = |j|2 = 1 and k . k = |k|2 = 1 If m, n are scalars and a . b be two vectors, then ma . nb = mn(a . b) = (mna) . b = a .(mnb) For any vectors a and b, we have (a) a . (–b) = –(a . b) = (–a). b (b) (–a) . (–b) = a . b For any two vectors a and b, we have (a) |a + b|2 = |a|2 + |b|2 + 2a.b (b) |a – b|2 = |a|2 + |b|2 – 2a.b

AB in terms of the position vectors of points A and B : If a and b are position vectors of points A and B respectively. Then, OA = a, OB = b ∴ AB = (Position vector of B)– (Position vector of A) = OB – OA = b – a Position vector of a dividing point : The position vectors of the points dividing the line AB in the ratio m : n internally or externally are mb + na mb − na or . m+n m−n

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a.b | a || b |

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AUGUST 2009

(c) (a + b) . (a – b) = |a|2 – |b|2

Right handed system of vectors : Three mutually perpendicular vectors a, b, c form a right handed system of vector iff a × b = c, b × c = a, c × a = b Left handed system of vectors : The vectors a, b, c mutually perpendicular to one another form a left handed system of vector iff c × b = a, a × c = b, b × a = c. Area of parallelogram and triangle : The area of a parallelogram with adjacent sides a and b is |a × b|. The area of a plane quadrilateral ABCD is 1 | AC × BD | , where AC and BC are its 2 diagonals.

(d) |a + b| = |a| + |b| ⇒ a | | b (e) |a + b|2 = |a|2 + |b|2 ⇒ a ⊥ b (f) |a + b| = |a – b| ⇒ a ⊥ b Vector or Cross product Vector product of two vectors : Let a, b be two non-zero, non-parallel vectors.

b θ O

a

The area of a triangle ABC is

Then a × b = |a| |b| sin θ ηˆ , and |a × b| = |a | |b| sin θ, where θ is the angle between a and b, ηˆ is a unit vector perpendicular to the plane of a and b such that a, b, ηˆ form a right-handed system.

1 1 | BC × BA | or | CB × CA | 2 2 Scalar triple product Scalar triple product of three vectors : If a, b, c are three vectors, then scalar triple product is defined as the dot product of two vectors a and b × c. It is generally denoted by a . (b × c) or [a b c]. Properties of scalar triple product : If a, b, c are cyclically permuted, the value of scalar triple product remains the same. i.e., (a × b) . c = (b × c) . a = (c × a). b or [a b c] = [b a c] = [c a b] Vector triple product Let a, b, c be any three vectors, then the vectors a × (b × c) and (a × b) × c are called vector triple product of a, b, c. Thus, a × (b × c) = (a . c) b – (a . b)c Properties of vector triple product : The vector triple product a × (b × c) is a linear combination of those two vectors which are within brackets. The vector r = a × (b × c) is perpendicular to a and lies in the plane of b and c. The formula a × (b × c) = (a . c)b – (a . b)c is true only when the vector outside the bracket is on the left most side. If it is not, we first shift on left by using the properties of cross product and then apply the same formula. Thus, (b × c) × a = –{a × (b × c)} = {(a . c)b – (a . b)c} = (a . b)c – (a . c)b Vector triple product is a vector quantity.

Properties of vector product : Vector product is not commutative i.e., if a and b are any two vectors, then a × b ≠ b × a, however, a × b = –(b × a) If a, b are two vectors and m, n are scalars, then ma × nb = mn(a × b) = m(a × nb) = n(ma × b). Distributivity of vector product over vector addition. Let a, b, c be any three vectors. Then (a) a × (b + c) = a × b + a × c (left distributivity) (b) (b + c) × a = b × a + c × a (Right disributivity) For any three vectors a, b, c we have a × ( b – c ) = a × b – a × c. The vector product of two non-zero vectors is zero vector iff they are parallel (Collinear) i.e., a × b = 0 ⇔ a| | b, a, b are non-zero vectors. It follows from the above property that a × a = 0 for every non-zero vector a, which in turn implies that i × i = j × j = k × k = 0. Vector product of orthonormal triad of unit vectors i, j, k using the definition of the vector product, we obtain i × j = k, j × k = i, k × i = j, j × i = –k, k × j = –i, i × k = –j. Vector product in terms of components : If a = a1i + a2j + a3k and b = b1i + b2j + b3k.

i Then, a × b = a 1 b1

j a2 b2

k a3 b3

Angle between two vectors : If θ is the angle between a and b then sin θ =

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1 | AB × AC | or 2

| a×b | | a || b |

a × (b × c) ≠ (a × b) × c

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MATH

PERMUTATION & COMBINATION Mathematics Fundamentals

Permutation : Definition : The ways of arranging or selecting a smaller or an equal number of persons or objects at a time from a given group of person or objects with due regard being paid to the order of arrangement or selection are called the (different) permutations. Number of permutations without repetition : Arranging n objects, taken r at a time equivalent to filling r places from n things. r-places :

1

2

3

4

Condition permutations : Number of permutations of n dissimilar things taken r at a time when p particular things always occur = n – pCr – p r !. Number of permutations of n dissimilar things taken r at a time when p particular things never occur = n – pCr r !. The total number of permutations of n different things taken not more than r at a time, when each thing may be repeated any number of times, is n (n r − 1) . n −1 Number of permutations of n different things, taken all at a time, when m specified things always come together is m ! × (n – m + 1) !. Number of permutation of n different things, taken all at a time, when m specified things never come together is n ! – m! × (n – m + 1) !. Let there be n objects, of which m objects are alike of one kind, and the remaining (n – m) objects are alike of another kind. Then, the total number of mutually distinguishable permutations that can be formed from these objects is n! . ( m !) × ( n − m ) !

r

Number of choice n (n–1)(n–2) (n–3)

n–(r – 1)

The number of ways of arranging = The number of ways of filling r places. = n(n – 1) (n – 2) ....... (n – r + 1) =

n! n ( n − 1)(n − 2)....(n − r + 1)((n − r )!) = (n − r )! (n − r ) !

= n Pr The number of arrangements of n different objects taken all at a time = nPn = n ! (i)

n

P0 =

n! = 1; nPr = n. n–1Pr – 1 n!

(ii) 0 ! = 1;

1 = 0 or (–r) ! = ∞ (r ∈ N) (– r ) !

The above theorem can be extended further i.e., if there are n objects, of which p1 are alike of one kind; p2 are alike of another kind; p3 are alike of 3rd kind; .....; pr are alike of rth kind such that p1 + p2 + .... + pr = n; then the number of permutations of these n n! . objects is (p1 !) × (p 2 !) × .... × (p r !)

Number of permutations with repetition : The number of permutations (arrangements) of n different objects, taken r at a time, when each object may occur once, twice, thrice, ....... upto r times in any arrangement = The number of ways of filling r places where each place can be filled by any one of n objects. r-places :

1

Number of choices : n

2

3

4

(n) (n) (n)

Circular permutations : Difference between clockwise and anti-clockwise arrangement : If anti-clockwise and clockwise order of arrangement are not distinct e.g., arrangement of beads in a necklace, arrangement of flowers in garland etc. then the number of circular permutations (n − 1) ! . of n distinct items is 2 Number of circular permutations of n different things, taken r at a time, when clockwise and n pr anticlockwise orders are taken as different is . r

r n

The number of permutations = The number of ways of filling r places = (n)r. The number of arrangements that can be formed using n objects out of which p are identical (and of one kind) q are identical (and of another kind), r are identical (and of another kind) and the rest n! . are distinct is p !q ! r !

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43

AUGUST 2009

Number or circular permutations of n different things, taken r at a time, when clockwise and n pr . anticlockwise orders are not different is 2r Theorems on circular permutations : Theorem (i) : The number of circular permutations on n different objects is (n – 1) !. Theorem (ii) : Then number of ways in which n persons can be seated round a table is (n – 1) !. Theorem (iii) : The number of ways in which n different beads can be arranged to form a 1 necklace, is (n − 1) ! . 2 Combinations : Definition : Each of the different groups or selection which can be formed by taking some or all of a number of objects, irrespective of their arrangements, called a combination. Notation : The number of all combinations of n things, taken r at a time is denoted by C(n, r) or nCr or  n  . nC is always a natural number. r  r   Difference between a permutation and combination : In a combination only selection is made whereas in a permutation not only a selection is made but also an arrangement in a definite order is considered. Each combination corresponds to many permutations. For example, the six permutations ABC, ACB, BCA, CBA and CAB correspond to the same combination ABC. Number of combinations without repetition The number of combinations (selections or groups) that can be formed from n different objects taken n! . Also r (0 ≤ r ≤ n) at a time is nCr = r !( n − r ) ! n Cr = nCn –r. Let the total number of selections (or groups) = x. Each group contains r objects, which can be arranged in r ! ways. Hence the number of arrangements of r objects = x × (r!). But the number of arrangements = npr. n! ⇒ x(r !) = npr ⇒ x = = n Cr r !( n − r ) !

The total number of ways in which it is possible to make groups by taking some or all out of n = (n1 + n2 + .....) things, when n1 are alike of one kind, n2 are alike of second kind, and so on is {(n1 + 1) (n2 + 1) .....} – 1. The number of selections taking at least one out of a1 + a2 + a3 + .... + an + k objects, where a1 are alike (of one kind), a2 are alike (of second kind) and so on ............... an are alike (of nth kind) and k are distinct = [(a1 + 1) (a2 + 1) (a3 + 1) ......... (an + 1)]2k – 1 Conditional combinations : The number of ways in which r objects can be selected from n different objects if k particular objects are (i) Always included = n – kCr–k (ii) Never included = n – kCr The number of combinations of n objects, of which p are identical, taken r at a time is

Number of combinations with repetition and all possible selections : The number of combinations of n distinct objects taken r at a time when any object may be repeated any number of times. = Coefficient of xr in (1 + x + x2 + ...... + xr)n = Coefficient of xr in (1 – x)–n = n + r – 1Cr The total number of ways in which it is possible to form groups by taking some or all of n things at a time is nC1 + nC2 + .... + nCn = 2n – 1.

Derangement : Any change in the given order of the things is called a derangement. If n things form an arrangement in a row, the number of ways in which they can be deranged so that no one of them occupies its original place is  1 1 1 1 n ! 1 − + − + ... + (−1) n .  n!  1! 2! 3!

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n–p

Cr + n – pCr – 1 +...........+ n – pC0, if r ≤ p and Cr + n – pCr – 1 + ........... + n – pCr – p, if r > p. Division into groups The number of ways in which n different things can be arranged into r different groups is n + r – 1Pn or n ! n –1Cr – 1 according as blank group are or are not admissible. Number of ways in which m × n different objects can be distributed equally among n persons (or numbered groups) = (number of ways of dividing into groups) × (number of groups)! n–p

=

(mn ) ! n ! (mn ) ! = ( m !) n n ! ( m !) n

If order of group is not important: The number of ways in which mn different things can be divided (mn ) ! equally into m groups is . ( m !) m m ! If order of groups is important : The number of ways in which mn different things can be divided equally into m distinct groups is

(mn )! m

(n !) m!

44

×m!=

(mn )! (n !) m

AUGUST 2009

Based on New Pattern

a

IIT-JEE 2010 XtraEdge Test Series # 4

Time : 3 Hours Syllabus : Physics : Essential Mathematics, Vector, Units & Dimension, Motion in One dimension, Projectile motion, Circular motion, Electrostatics & Gauss's Law, Capacitance, Current electricity, Alternating Current, Magnetic Field, E.M.I. Chemistry : Mole Concept, Chemical Bonding, Atomic Structure, Periodic Table, Chemical Kinetics, Electro Chemistry, Solid state, Solutions, Surface Chemistry, Nuclear Chemistry. Mathematics: Trigonometric Ratios, Trigonmetrical Equation, Inverse Trigonmetrical Functions, Properties of Triangle, Radii of Circle, Function, Limit, Continiuty, Differentiation, Apllication of Differentiation (Tangent & Normal, Monotonicity, Maxima & Minima).

Instructions : Section - I • Question 1 to 6 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct answer and -1 mark for wrong answer. • Question 7 to 10 are Multiple choice questions with multiple correct option. 4 marks will be awarded for each correct answer and –1 mark for each wrong answer. • Question 11 to 16 are Passage based multiple correct type questions. 5 marks will be awarded for each correct answer and –1 mark for each wrong answer. Section - II • Question 17 to 19 are numerical response questions. Each question carries +6 marks. There is no negative marking in this section.

PHYSICS

(A) ∆tL – ∆tU =

Questions 1 to 6 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1.

2.

(C)

A helicopter takes off along the vertical with an acceleration a = 3 m/s2 and zero initial velocity. In a certain time t1, the pilot switches off the engine. At the point of take off, the sound dies away in a time t2 = 30 s. If the speed of sound is 320 m/s, then the velocity of the helicopter at the moment when its engine is switched off is – (A) 120 m/s (B) 60 m/s (C) 80 m/s (D) 720 m/s

3.

∆t L = ∆t U

(D)

8H g

8gH

A ball of mass m having charge q and attached to a string of length L can rotate in a circle in a vertical plane about the centre O, where another charge q is kept stationary. Then the minimum velocity of the ball at its lowest position, so that the string may not slacken in any position, should be –

m, q

45

q2 4πε 0 mL

(A)

5gL −

(C)

 q2 (5gL) +   4πε mL2 0 

(D)

5gL

H

∆tL

XtraEdge for IIT-JEE

8H g

(B) ∆t 2L – ∆t 2U =

q L

A glass ball is thrown as a projectile in an evacuated tube. Let ∆tL be the time interval between two successive passages across the lower level and ∆tu the time interval between the two passages across the upper level, and H the distance between the two levels. Then – ∆tU

∆t L = ∆t U

8H g

2

5gL +

(B)

   

q2 4πε 0 mL

2

AUGUST 2009

4.

An isolated, conducting spherical shell of radius R carries a negative charge Q. A small charged spherical metal ball, carrying positive charge q is placed inside the shell and connected by a wire to the inner wall of the shell. Then the common potential of the two is – (A) positive (B) negative (C) zero (D) any of the above

5.

Two large parallel copper plates are L metre apart and have a uniform electric field between them. An electron is released from the negative plate at the same time that a proton is released from the positive plate. The gravity and the force of the particles on each other are neglected. The two particles cross each other at a distance x from the positive plate. Then x is– P

7.

h θ B R (A) the path of the ball is a parabola (B) the path of the ball is a straight line (C) the angle θ = tan–1(g/a) (D) the range R = h a/g

8.

An uncharged capacitor C is fully charged by a constant emf E in series with a resistor R. Then the correct statement (s) is (are) – (A) Final energy stored in the capacitor is half of the energy supplied by the emf. (B) Internal energy dissipated by the resistor is equal to the final energy stored in the capacitor (C) Final energy stored in the capacitor is equal to the energy supplied by the emf. (D) All of the above.

9.

A rectangular N-turn coil in the-yz plane (see figure) carrying a current I is placed in a uniform magnetic field B = B0( xˆ + yˆ ). Then the maximum torque will act when the coil is rotated by an angle θ – Z b

–

+ e L

(A) x = (C) x =

6.

mp me

L

m pme mp + me

L

(B) x =

L 2

(D) x =

me L me + mp

A ball is dropped from a height h. Wind is blowing horizontally in such a manner that it imparts the ball a horizontal constant acceleration of a m/s2 to the ball. The ball strikes the ground at B (see figure). Then – a(m/s2) A

The conducting bar of length L shown in figure, oscillates across the rails with the velocity v = v0 cos ωt in a magnetic field B = B0 cos ωt (in a direction perpendicular to the rails and into the paper). At the mean position the rails are connected by a conductor. If at t = 0, the rod is at the mean position, then the instantaneous emf induced in the loop is –

a

I

conducting rails

conducting bar

Y X L

(A) θ = –π/4 (C) θ = + π/4

v B

mean position

10. Mark the correct statement (s) : (A) The potential difference between two points in an electric field must be zero (B) The potential difference two points in an electric filed may be zero (C) If a positively charged metal sphere is brought near a conducting sphere, it is found that a force of attraction occurs between the two, this means that one conducting sphere must be charged (D) In (C) the conducting sphere may be charged

rod oscillates between this region

(A) B0V0L sin ωt (C) B0V0L cos ωt

(B) B0V0L cos ωt (D) B0V0L sin (2ωt)

Questions 7 to 10 are multiple choice questions. Each questions has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct.

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(B) θ = + 3π/4 (D) θ = –3π/4

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AUGUST 2009

This section contains 2 paragraphs; each has 3 multiple choice questions. (Question 11 to 16) Each question has 4 choices (A), (B), (C) and (D) out of which ONE OR MORE THAN ONE is correct.

(C) ˆi = – rˆ sin θ + ˆrt cos θ (D) ˆi = – rˆr cos θ + rˆt sin θ

Passage : II (Q. No. 14 to 16)

Passage : I (Q. No. 11 to 13)

→

As a charged particle 'q' moving with a velocity v →

Consider a point object of mass 'm' moving in a circle of radius a = 1 m. For any instantaneous position of the object, θ is the angle that the radial line joining the object and the centre makes with the positive Xaxis of a cartesian coordinate system with the centre of circle O as the origin. ˆi and ˆj are unit vectors

enters a uniform magnetic field B , it experiences a →

→

→

→

v = (8 ˆi – 6 ˆj + 4 kˆ ) × 106 m/s and force acting on it has a magnitude 1.6 N.

P O

θ

→

containing a uniform magnetic field B = – 0.4 kˆ T. At any instant, velocity of the particle is

rˆr

rˆt

→

angle between v and B , force experienced is zero and the particle passes undeflected. For θ = 90º, the particle moves along a circular arc and the magnetic force (qvB) provides the necessary centripetal  mv 2   . For other values of θ(θ ≠ 0º, 180º, force   r    90º), the charged particle moves along a helical path which is the resultant motion of simultaneous circular and translational motions. Suppose a particle, that carries a charge of magnitude q and has a mass 4 × 10–15 kg, is moving in a region

along X-axis and Y-axis, respectively. Suppose that the motion is a 'Uniform Circular Motion' with a π constant angular speed rad/sec and that the sense 36 of rotation is counter clock with θ = 0 at t = 0. For an object which moves in a circle, it is usually convenient to introduce two mutually perpendicular unit vectors, rˆr and rˆt , as shown in fig. Here rˆr is the radial unit vector and rˆt , the tangential unit vector.

Y

→

force F = q( v × B ). For θ = 0º or 180º, θ being the

X

14. Pitch of the helical path is (A) 122.4 cm (B) 62.8 cm (C) 24.6 cm (D) 74.4 cm 11. For any instantaneous position of the object P, the radial unit vector rˆr can be expressed as

15. Which of the following is correct ? (A) Motion of the particle is non-periodic but y and z-position co-ordinates vary in a periodic manner (B) Motion of the particle is non-periodic but x and y-position co-ordinates vary in a periodic manner (C) Motion of the particle is non-periodic but x and z-position co-ordinates vary in a periodic manner (D) Motion of the particle is periodic and all the position co-ordinates vary in a periodic manner

(A) rˆr = ˆi sin θ + ˆj cos θ (B) rˆ = ˆi cos θ + ˆj sin θ (C) rˆ = ˆi sin θ – ˆj cos θ (D) rˆr = – ˆi cos θ – ˆj sin θ

12. For any position of the object P, the tangential unit vector can be expressed as (A) rˆ = ˆi cos θ – ˆj sin θ

16. If the co-ordinates of the particle at t = 0 are (2m, 1m, 0), co-ordinate at a time t = 3T, where T is a time period of circular component of motion, will be (A) (2m, 1m, 400 m) (B) (0.142 m, 130 m, 0) (C) (2m, 1m, 1.884 m) (D) (142 m, 130m, 628 m)

t

(B) rˆt = ˆi sin θ – ˆj cos θ (C) rˆt = – ˆi cos + ˆj sin θ (D) rˆt = – ˆi sin θ + ˆj cos θ

Numerical response questions (Q. 17 to 19). Answers to this Section are to be given in the form of nearest integer-in four digits. Please follow as per example : (i.e. for answer : 1492.2 write 1492; 491.8 write 0492; 92.5 write 0092; 2.1 write 0002)

13. In terms of rˆr , rˆt and θ, ˆi can be expressed as (A) ˆi = rˆr cos θ − rˆt sin θ (B) ˆi = – rˆr sin θ + rˆt cos θ

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AUGUST 2009

17. A current of 4 A flows in a coil when connected to a 12V dc source. If the same coil is connected to a 12V, 50 rad/s ac source a current of 2.4 A flows in the circuit. Also find the power developed in the circuit if a 2500 µF capacitor is connected in series with the coil.

4.

A reaction follows the given concentration (C) vs time graph. The rate for this reaction at 20 seconds will be – 0.5 0.4

0.3 0.2

18. A capacitor of capacity 2 µF is charged to a potential difference of 12V. It is then connected across an inductor of inductance 0.6 mH. What is the current in the circuit at a time when the potential difference across the capacitor is 6.0 V ?

0.1 0 20 40 60 80 100 Time/second (A) 4 × 10–3 Ms–1 (C) 2 × 10–2 Ms–1

19. A ball of mass 100 g is projected vertically upwards from the ground with a velocity of 49 m/s. At the same lime another identical ball is dropped from a height of 98 metre to fall freely along the same path as that followed by the first ball. After some time the two balls collide and stick together and finally fall to ground. Find the time of flight of the masses. (g = 9.8 m/s2)

5.

The potential of the Daniell cell, ZnSO 4 CuSO 4 Zn Cu was reported by Buckbee, (1M ) (1M ) Surdzial, and Metz as Eº = 1.1028 – 0.641 × 10–3 T + 0.72 × 10–5 T2, where T is the celcius temperature. Calculate ∆Sº for the cell reaction at 25º C (A) – 45.32 (B) – 34.52 (C) – 25.43 (D) – 54.23

6.

In a hypothetical solid C atoms form CCP lattice with A atoms occupying all the Tetrahedral voids and B atoms occupying all the octahedral voids. A and B atoms are of the appropriate size such that there is no distortion in the CCP lattice. Now if a plane is cut (as shown) then the cross section would like –

CHEMISTRY Questions 1 to 6 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1.

2.

A mixture of CO and CO2 having a volume of 20 ml is mixed with x ml of oxygen and electrically sparked. The volume after explosion is (16 + x) ml under the same conditions. What would be the residual volume if 30 ml of the original mixture is treated with aqueous NaOH ? (A) 12 ml (B) 10 ml (C) 9 ml (D) 8 ml Rutherford’s experiment, which estabilished nuclear model of the atom, used a beam of (A) β–particles, which impinged on a metal foil got absorbed (B) γ–rays, which impinged on a metal foil ejected electrons (C) helium atoms, which impinged on a metal and got scattered (D) helium nuclei, which impinged on a metal and got scattered

Plane

CCP unit cell

the C

and (A)

and

C

C

CC

B

(B)

B

B

B

B

B

C

C

C

C

C

CC

B

B

B

C

C

C

A

C

foil foil

C B

B

B

(D)

A

A

(B) CO2 >N2O5 MgO >Al2O3

C A

A

C

C

B

C

The correct order of acidic strength is – (A) Cl2O7 >SO2>P4O10

C

C

Questions 7 to 10 are multiple choice questions. Each questions has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct.

(D) K2O >CaO >MgO

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C

B A

(C)

3.

(B) 8 × 10–2 Ms–1 (D) 7 × 10–3 Ms–1

48

AUGUST 2009

7.

8.

9.

Select the correct statements from the following regarding sols – (A) Viscosity of lyophilic sols (emulsoid) is much higher than that of solvent (B) Surface tension of lyophobic sols (suspensoid) is usually low. (C) The particles of lyophilic sols always carry a characteristics charge either positive or negative (D) Hydrophobic sols can easily be coagulated by addition of electrolytes

Eº = 0.02 V

Dish A anode Dish B cathode Given NO3– + 3H3O + 2e– → HNO2 + 4H2O Eº = 0.94 V

Which of the following is/are correct ? (A) α-rays are more penetrating than β-rays (B) α-rays have greater ionizing power than β-rays (C) β-particles are not present in the nucleus, yet they are emitted from the nucleus (D) γ-rays are not emitted simultaneously with α and β-rays.

11. Which of the following statements must be true of the solutions in order for the cell to operate with the voltage indicated ? (A) The solution in Dish A must be acidic (B) The solution in Dish B must be acidic (C) The solutions in both Dish A and Dish B must be acidic (D) No acid may be in either Dish A or Dish B

Choose the correct statement(s) (A) At the anode, the species having minimum reduction potential is formed from the oxidation of corresponding oxidizable species (B) In highly alkaline medium, the anodic process during the electrolytic process is 4OH– → O2 + 2H2O + 4e– (C) The standard potential of Cl– | AgCl | Ag half– cell is related to that of Ag+ | Ag through the expression RT º + ln Ksp (AgCl) E ºAg + / Ag = E Cl – |AgCl|Ag F (D) Compounds of active metals (Zn, Na, Mg) are reducible by H2 whereas those of noble metals (Cu, Ag, Au) are not reducible.

12. At what pH will the cell potential be zero if the activity of other components are equal to one ? 0.02 0.02 (A) (B) – 2 × 0.059 0.059 0.04 0.02 2 (D) × (C) 0.059 0.059 3 13. How many moles of electrons pass through the circuit when 0.6 mole of Hg2+ and 0.30 mole of HNO2 are produced in the cell that contains 0.5 mole

10. The co-ordination number of FCC structure for metals is 12, since (A) each atom touches 4 others in same layer, 3 in layer above and 3 in layer below (B) each atom touches 4 others in same layer, 4 in layer above and 4 in layer below (C) each atom touches 6 others in same layer, 3 in layer above and 3 in layer below (D) each atom touches 3 others in same layer, 6 in layer above and 6 in layer below

of Hg22+ and 0.40 mole of NO3– at the begining of the reaction ? (A) 0.6 mole (B) 0.8 mole (C) 0.3 mole (D) 1 mole

Passage : II (Q. No. 14 to 16) According to Raoult’s law partial pressure of any volatile constituent of a solution at a constant temperature is equal to the vapour pressure of pure constituent multiplied by mole-fraction of that constituent in the solution. This law is the major deciding factor whether a solution will be ideal or non-ideal. Ideal solution always obey Raoult’s law at energy range of concentration. A liquid mixture of benzene & toluene is composed of 1 mol of benzene & 1 mol of toluene. Benzene & toluene mixture behave

This section contains 2 paragraphs; each has 3 multiple choice questions. (Question 11 to 16) Each question has 4 choices (A), (B), (C) and (D) out of which ONE OR MORE THAN ONE is correct. Passage : I (Q. No. 11 to 13) The cell potential for the unbalanced chemical reaction : Hg22+ + NO3– + H3O+ → Hg2+ +HNO2 + H2O is measured under standard state conditions in the electrochemical cell shown in the accompanying diagram. The cell voltage is positive : EºCell = 0.02 V

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ideally. Given p 0t = 4.274 KN/m2; p 0b = 13.734 KN/m2 where subscript t & b stands for toluene & benzene respectively. 49

AUGUST 2009

14. If pressure over mixture at 300 K is reduced at what pressure does the first vapour form ? (A) 18.008 KN/m2 (B) 9.004 KN/m2 2 (D) 3.003 KN/m2 (C) 13.05 KN/m 15. What is the composition of first trace of vapour formed in terms of mole-fraction of toluene – (A) 0.2373 (B) 0.7627 (C) 0.333 (D) 0.67 16. If the pressure is further reduced, at what pressure does last trace of liquid disappear – (A) 9.004 KN/m2 (B) 6.519 KN/m2 (D) None of these (C) 3.03 KN/m2

2.

If A and B are acute positive angles satisfying the equation 3 sin2A + 2 sin2B = 1 and 3 sin 2A – 2 sin 2B = 0, then (A + 2B) is (A) 0 (B) π/2 (C) π/4 (D) π/3

3.

If α, β are the roots of the equation; 6x2 + 11x + 3 =0 then : (A) both cos–-1 α and cos–1 β are real (B) both cosec–1 α and cosec–1 β are real (C) both cot–1α and cot–1β are real (D) None of these

4.

If f(x) = 64x3 +

Numerical response questions (Q. 17 to 19). Answers to this Section are to be given in the form of nearest integer-in four digits. Please follow as per example : (i.e. for answer : 1492.2 write 1492; 491.8 write 0492; 92.5 write 0092; 2.1 write 0002)

1 = 2 then x (A) f(α) = –64 (C) f(β) = – 16

and α, β are the roots of

(B) f(β) = – 8 (D) f(α) = –24

5.

A differentiable function f is defined for all x > 0 and satisfies f(x2) = x4 for all x > 0, then f´(16) is equal to (A) 64 (B) 16 (C) 8 (D) None of these

6.

Let f(x) = xex(1 – x), then f(x) is (A) increasing on [–1/2, 1] (B) decreasing on R (C) increasing on R (D) decreasing on [–1/2, 1]

18. Graph between log x/m and log p is straight line inclined at an angle of 45. When pressure is 0.5 atm and ln k = 0.693, what will be the amount of solute adsorbed per gm of adsorbent ?

Questions 7 to 10 are multiple choice questions. Each questions has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct.

19. A current of dry air was passed through a series of bulbs containing 1.25 g of a solute A2B in 50 g of water and then through pure water. The loss in weight of the former series of bulbs was 0.98 g and in the later series 0.01 g. If the molecular weight of A2B is 80 calculate degree of dissociation of A2B.

7.

MATHEMATICS Questions 1 to 6 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

8.

The value of 3π 5π 7π 9π 11π 13π π sin sin .sin .sin .sin .sin .sin 14 14 14 14 14 14 14 is equal to : (A) 1 (B) 1/16 (C) 1/64 (D) None of these

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x3

4x +

17. The specific rate constant of the decomposition of N2O5 is 0.008 min-1. The volume of O2 collected after 20 minutes is 16 ml. Find the volume (in ml) that would be collected at the end of reaction. NO2 formed is dissolved in CCl4.

1.

1

1  The lim x 8  3  (where [x] is greatest integer x →0 x  function) is (A) a nonzero real number (B) a rational number (C) an integer (D) zero Let f(x) = g´(x)

e1/ x − e −1/ x

where g´ is the e1 / x + e −1 / x derivative of g and is a continuous function then lim f ( x ) exist if x→0

(A) g(x) is polynomial (B) g(x) = x (C) g(x) = x2 (D) g(x) = x3 h(x) where h(x) is a polynomial 50

AUGUST 2009

9.

16. If f(x) is the function required to find largest term in Q. 14 then (A) f is increase for all x (B) f decreases for all x

For x > 1, y = log x – (x – 1) satisfies the inequality (A) x – 1 > y (B) x2 – 1 > y x −1 (C) y > x – 1 (D) 0 (B) a > 0, b < 0 (C) a > 0, b > 0 (D) a < 0, b < 0

17. Let Pn(x) be a function satisfying ∞

Passage : I (Q. No. 11 to 13) Let f(x) be a real valued function not identically zero, such that f(x + yn) = f(x) + (f(y))n ∀ x, y ∈ R where n ∈ Nn ≠ 1 and f´(0) ≥ 0. We may get an explicit form of the function f(x).

(B) n (D) 2

12. The value of f(5) is (A) 2 (C) 5n

(B) 3 (D) 5

13.

400

Numerical response questions (Q. 17 to 19). Answers to this Section are to be given in the form of nearest integer-in four digits. Please follow as per example : (i.e. for answer : 1492.2 write 1492; 491.8 write 0492; 92.5 write 0092; 2.1 write 0002)

This section contains 2 paragraphs; each has 3 multiple choice questions. (Question 11 to 16) Each question has 4 choices (A), (B), (C) and (D) out of which ONE OR MORE THAN ONE is correct.

11. The value of f´(0) is (A) 1 (C) 1 + n

3

∑h

n

Pn ( x ) = (1 – 2hx + h2)–1/2, |x| ≤ 1, |h| < 1/3.

n =0

Find P3(10).

18. The value

tan 6 x − 2 tan 5 x − 3 tan 4 x

lim

x → tan

−1

3

tan 2 x − 4 tan x + 3

is.

19. Let (a – b cos y) (a + b cos x) = a2 – b2 and sin x f ( y) dy = . If a2 – b2 = 192 then f(π/2) = dx (a + b cos x ) 2

1

∫ f (x) dx is equal to

•

The Andromeda Galaxy is 2.3 million light years away.

•

Pluto lies at the outer edge of the planetary system of our sun, and at the inner edge of the Kuiper Belt, a belt of icy comets that are the remnants of the formation of the solar system.

•

14. The largest term of an = n2/(n3 + 200) is (A) 29/453 (B) 49/543 (C) 43/543 (D) 41/451

On June 8 2004, Venus passed directly between the Earth and the Sun, appearing as a large black dot travelling across the Sun's disk. This event is known as a "transit of Venus" and is very rare: the last one was in 1882, the next one is in 2012.

•

15. The largest term of the sequence an = n/(n2 + 10) is (A) 3/19 (B) 2/13 (C) 1 (D) 1/7

A sunbeam setting out through space at the rate of 186,000 miles a second would describe a gigantic circle and return to its origins after about 200 billion years.

•

Mercury can only be seen from the Earth at twilight.

0

(A) 1/2n (C) 1/2

(B) 2n (D) 2

Passage : II (Q. No. 14 to 16) Among several applications of maxima and minima is finding the largest term of a sequence. Let be a sequence. Consider f(x) obtained by replacing x by n x n e.g. let an = consider f(x) = on [1, ∞] n +1 x +1 x f´(x) = > 0 for all x. ( x + 1) 2 Hence max f(x) = lim f ( x ) = 1. x →∞

XtraEdge for IIT-JEE

51

AUGUST 2009

Based on New Pattern

IIT-JEE 2011 XtraEdge Test Series # 4

Time : 3 Hours Syllabus : Physics : Essential Mathematics, Vector, Units & Dimension, Motion in One dimension, Projectile motion, Circular motion. Chemistry : Mole Concept, Chemical Bonding, Atomic Structure, Periodic Table . Mathematics: Trigonometric Ratios, Trigonmetrical Equation, Inverse Trigonmetrical Functions, Properties of Triangle, Radii of Circle.

Instructions : Section - I • Question 1 to 6 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct answer and -1 mark for wrong answer. • Question 7 to 10 are Multiple choice questions with multiple correct option. 4 marks will be awarded for each correct answer and –1 mark for each wrong answer. • Question 11 to 16 are Passage based multiple correct type questions. 5 marks will be awarded for each correct answer and –1 mark for each wrong answer. Section - II • Question 17 to 19 are numerical response questions. Each question carries +6 marks. There is no negative marking in this section.

3.

PHYSICS

Fig shows six successive positions of a particle moving along the x axis. At t = 0, it is at position x = 1 meter. The arrow shows velocity. At t = 0, it has velocity v = 2 m/s in the forward direction. Then the average velocity in the intervals AD and AF are, respectively –

Questions 1 to 6 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1.

2.

t(s) 0 A

A balance can weight with a precision of 0.5 kg. Your weight on this balance is 55 kg. wt and when you hold your books bag, your weight is 64 kg. What is the percentage uncertainty in your bag's weight ? (A) 0.9% (B) 0.8% (C) 5.6% (D) 18%

B

2

C

2.5 D 3.5 E

It is given that Planck's time tp is t p ∝ ci G j h k where c = speed of light, G = gravitational constant and h = Planck's constant. Then the exponents i, j and k are – 5 1 1 (A) i = – , j = , k = 2 2 2 1 3 (B) i = , j = – , k = 1 2 2 3 2 (C) i = – , j = 1, k = – 2 5 1 1 (D) i = 0, j = – , k = 2 2

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1

2 m/s

4

F

–4 m/s 0

1

2

3

4

5

6

7

x (m) (A) 2 m/s and – 4 m/s (B) 1 m/s and – 1/4 m/s (C) 1/4 ms and 2ms (D) 2 m/s and 1/4 m/s

4.

52

A particle moves so that its position as a function of time in SI units is r r ( t ) = ˆi + t2 ˆj +t kˆ Then the velocity of the particle varies with time as

AUGUST 2009

(B)

velocity (v)

velocity (v)

(A)

5.

time (t)

(D)

45º time (t)

velocity (v)

velocity (v)

time (t)

(C)

Questions 7 to 10 are multiple choice questions. Each questions has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. r r r 7. For the three vectors d , d1 and d 2 , suppose r r r d = d1 + d 2 then

60º

(A) d must be greater than d1, d2 (B) d may be greater than d1, d2 (C) d must be smaller than d1, d2 (D) d may be smaller than d1, d2

time (t)

A particle A moves along the line y = 30 m with a constant velocity v = 3.0 m/s directed parallel to the x-axis. A second particle B starts at the origin with zero speed and constant acceleration a = 0.4 m/s2 at the same instant that the particle A passes the y-axis. If B makes an angle θ with the x-axis, then the value of θ for collision to occur between A and B should be– Y A

v a θ

B

(A) 30º

X (D) 72º

(C) 60º

An object moves in a straight line as described by the velocity time graph in fig. 20 Velocity (m/s)

6.

(B) 45º

A jumbo jet needs to reach a speed of 360 km/h on the runway for takeoff. If we assume constant acceleration and a runway of length 1.8 km, then – (A) the minimum acceleration required from rest is 2.8 m/s2 (B) the minimum time from rest to takeoff is 36 second (C) the minimum acceleration required is 9.8 m/s2 (D) the minimum time for takeoff is 72 second

9.

The engineer of a train moving at a speed v1 sights a freight train a distance d ahead of him on the same track moving in the same direction with a slower speed v2. The puts on the brakes and gives his train a constant deceleration a. Then – (A) if d > (v1 – v2)2/2a, there will be no collision (B) if d < (v1 – v2)2/2a, there will be a collision (C) The x versus t graph of the two trains if no collision occurs, may be as shown here in fig.

x

15 10

O (D) All of these

5 0

1

2

3

4

5

6

a

(A)

0 1 2 3 4 5 times (s)

XtraEdge for IIT-JEE

0 1 2 3 4 5 times (s)

a

a 0 1 2 3 4 5 times (s)

t (s) 0 1 2 3

a

(B)

(D)

time

10. A body is dropped from rest and falls freely. The position (y), the velocity (v) and the acceleration (a) of the body at t = 0, 1, 2, 3, (second) are given in the table. Then the correct statements are –

Times (s) Then the graph that represent the acceleration of the object as a function of time is –

(C)

8.

y (m) 0 y1 y2 y3

v (m/s) 0 v1 v2 v3

a (m/s2) g a1 a2 a3

(A) y1 : y2 : y3 = 1 : 3 : 5 (B) y1 : y2 : y3 = 1 : 4 : 9 (C) v1 : v2 : v3 = 1 : 2 : 3 (D) a1 : a2 : a3 = 1 : 3 : 5

0 1 2 3 4 5 times (s)

53

AUGUST 2009

This section contains 2 paragraphs; each has 3 multiple choice questions. (Question 11 to 16) Each question has 4 choices (A), (B), (C) and (D) out of which ONE OR MORE THAN ONE is correct.

Passage : II (Q. No. 14 to 16) Consider a simple pendulum–an object attached to one end of a thread is suspended vertically from the other end. If the object is released from a slight displacement from the mean position, it oscillates in a simple harmonic manner with a time period

Passage : I (Q. No. 11 to 13) Suppose that a point mass 'm' is moving under a

T = 2π

→

constant force F = 2 ˆi – ˆj + kˆ newton. At some instant, t = 0, point P (x, m, ym,) [m – metre] is the instantaneous position of the mass. We know that torque can be expressed as the cross-product of position vector and force vector, i.e., Y Q(x1, y1, z1)

Here l is the distance of centre of mass of the object from the point of suspension. Obviously, it is possible to determine acceleration due to gravity 'g' by measuring time period of oscillation of a given pendulum. Let us perform this experiment. Length of the thread is measured using a metre scale. Six measurements of length of thread give its values as 20 cm, 20.5 cm, 18.5 cm, 21 cm, 19.5 cm and 20.5 cm. We take a spherical object of uniform density, and its diameter, if required, is measured with the help of a vernier callipers. In the given vernier callipers, 20 VSD ≡ 19 MSD and a part of the main scale is as shown if fig. Using the least count (vernier constant) of vernier callipers as the mean absolute error, percentage error in the measurement of diameter is found to be 1%. Finally, time of 20 oscillations is measured and five measurements give its value as : 17 sec, 18 sec, 19 sec, 17.5 sec, 18.5 sec 3 (cm) 4 (cm)

P(x, y, –1) X

O Z →

→

→

τ= r × F At P, torque can be expressed as, →

τ = –4 ˆj – 4 kˆ newton-metre At some other instant, t = 3 sec, the point mass has another instantaneous position Q(x, y, z) such that the displacement vector between point P and Q and the given force are mutually perpendicular. Also, xcomponent of torque at Q is zero and y and zcomponents are equal in magnitude and directed along the negative directions of the respective axes. Using a definite scale, if we construct a parallelogram

14. Percentage error in the measurement of length is nearly (A) 10.6% (B) 5.8% (C) 3.3% (D) 1.8 %

→

with the position vector of Q and the given force F as its adjacent sides, area of this parallelogram is

15. Vernier constant of the given vernier callipers is (A) 2.5 × 10–3 cm (B) 2.5 × 10–2 cm (C) 0.25 cm (D) 2.5 × 10–4 cm

2 2 m2. Area of the given parallelogram, in fact, represents a physical quantity whose magnitude in SI system can be expressed as 5 times the given area.

16. Effective length of the pendulum, i.e., distance between the point of suspension and the centre of mass of the object, can be written as (A) (20.5 ± 0.6) cm (B) (20.125 ± 0.446) cm (C) (20.125 ± 0.669) cm (D) (20.25 ± 0.54) cm

11. Coordinates of point P (measured in metre) are (A) (1, 0, –1) (B) (2, 1, –1) (C) (1, 1, –1) (D) (2, 2, –1) 12. At Q, torque acting on the mass can be expressed as (A) –5 ˆj – 5 kˆ newton-metre

Numerical response questions (Q. 17 to 19). Answers to this Section are to be given in the form of nearest integer-in four digits. Please follow as per example : (i.e. for answer : 1492.2 write 1492; 491.8 write 0492; 92.5 write 0092; 2.1 write 0002)

(B) –8 ˆj – 8 kˆ newton-metre (C) – 10 ˆj – 10 kˆ newton-metre (D) –12 ˆj – 12 kˆ newton-metre

17. The width of a river is 25 m and in it water is flowing with a velocity of 4 m/min. A boatman is standing on the bank of the river. He wants to sail the boat to a point at the other bank which is directly opposite to him. In what time will he cross the river, if he can sail the boat at 8 m/min. relative to the water ?

13. Coordinates of point Q(measured in metre) are (A) (4, 3, –3) (B) (2, –1, 1) (C) (3, –2, 2) (D) (2, 4, –2)

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AUGUST 2009

18. At a picnic spot on a hill, a boy falls over the cliff. Suddenly Shaktiman arrives and dives off the edge 2.0 s after the start of the boy's fall. If the cliff is 100 m high, what must Shaktiman's initial velocity be if he is to catch the boy just before he reaches the ground ?

(D) The positive charge should be present on the electropositive element and the negative charge on the electronegative element

5.

Light of wavelength λ shines on a metal surface with intensity x and the metal emits y electrons per second of average energy, z. What will happen to y and z if x is doubled ? (A) y will be doubled and z will become half (B) y will remain same and z will be doubled (C) both y and z will be doubled (D) y will be doubled but z will remain same

6.

In two H atoms X and Y the electrons move around the nucleus in circular orbits of radius r and 4r respectively. The ratio of the times taken by them to complete one revolution is (A) 1 : 4 (B) 1 : 2 (C) 1 : 8 (D) 2 : 1

19. On a cricket field, the batsman is at the origin of coordinates and a fielder stands in position →

→

(46 i + 28 j )m. The batsman hits the ball so that it rolls along the ground with constant velocity →

→

(7.5 i + 10 j ) m/s. The fielder can run with a speed of 5 m/s. If he starts to run immediately the ball is hit, what is the shortest time in which he could intercept the ball?

CHEMISTRY Questions 1 to 6 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

Questions 7 to 10 are multiple choice questions. Each questions has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct.

1.

Iron forms two oxides, in first oxide 56 gram. Iron is found to be combined with 16 gram oxygen and in second oxide 112 gram . Iron is found to be combined with 48 gram oxygen. This data satisfy the law of (A) Conservation of mass (B) Reciprocal proportion (C) Multiple proportion (D) Combining volume

7.

Which of the following statements is/are correct ? (A) Group 12(IIB) elements do not show characteristic properties of transition metals (B) Among transition elements, tungsten has the highest melting point (C) Among transition elements, group 3 (IIIB) elements have lowest densities (D) Transition metals are more electropositive than alkaline earth metals.

2.

A carbon compound containing carbon and oxygen has molar mass equal to 288. On analysis it is found to contain 50% by mass of each element. Therefore molecular formula of the compound is(B) C4O3 (A) C12O9 (C) C3O4 (D) C9O12

8.

3.

Lattice energy of BeCO3 (I) , MgCO3 (II) and CaCO3 (III) are in the order (A) I > II > III (B) I < II < III (C) I < III < II (D) II < I < III

11.2 g of mixture of MCl (volatile) and NaCl gave 28.7 g of white ppt with excess of AgNO3 solution. 11.2 g of same mixture on heating gave a gas that on passing into AgNO3 solution gave 14.35 g of white ppt. Hence ? (A) Ionic mass of M+ is 18 (B) Mixture has equal mol fraction of MCl and NaCl (C) MCl and NaCl are in 1 : 2 molar ratio (D) Ionic mass of M+ is 10

9.

4.

Which of the following conditions is not correct for resonating structures ? (A) The contributing structures must have the same number of unpaired electrons (B) The contributing structures should have similar energies (C) The contributing structures should be so written that unlike charges reside on atoms that are far apart

Specify the coordination geometry around the hybridization of N and B atoms in a 1 : 1 complex of [IIT- 2002] BF3 and NH3 -

XtraEdge for IIT-JEE

(A) N : tetrahedral, sp3 ;B : tetrahedral, sp3 (B) N : pyramidal, sp3 ; B : pyramidal, sp3 (C) N : pyramidal, sp3 ; B : planar, sp2 (D) N : pyramidal, sp3 ; B : tetrahedral, sp3

55

AUGUST 2009

10. The radii of two of the first four Bohr orbits of the hydrogen atom are in the ratio 1 : 4. The energy difference between them may be : (A) Either 12.09 eV or 3.4 eV (B) Either 2.55 eV or 10.2 eV (C) Either 13.6 eV or 3.4 eV (D) Either 3.4 eV or 0.85 eV

Passage : II (Q. No. 14 to 16) Electronegativity is defined as the power of an atom in a molecule to attract the sigma bonded electron pair itself. Pauling established a scale of electronegativity based on the excess bond energy in heteronuclear covalent bond AB. The energy of the bond DA–B can be considered to be the sum of the nonpolar contribution, Dnp and a polar contribution Dp which gives additional stability to the covalent bond due to coulombic attraction arising out of the partial ionic nature of the bond. DA–B = Dnp + Dp

This section contains 2 paragraphs; each has 3 multiple choice questions. (Question 11 to 16) Each question has 4 choices (A), (B), (C) and (D) out of which ONE OR MORE THAN ONE is correct.

Pauling assumed that the nonpolar bond energy Dnp is average energy of the bond energies DA–A & DB–B

Passage : I (Q. No. 11 to 13) The radius of the nucleus of an atom can be approximately determined as, rnu = (1.4×10–13)A1/3 where A is mass number of the atoms and M is the charge of the electron = 4.8 × 10–10 esu. The mass of α-particle = 4 × mass of H-atom 10 ×10–24 gm. mass of hydrogen atom = 6 Consider during collision kinetic energy of α-particle just equal to coulombic force of repulsion. The mass number of Au = 197 The mass number of He = 4 The atomic number of Au = 79 Given : (4) 1 /3 = 1.59 and (197) 1 /3 = 5.82

2 × 79 × (4.8) 10.374

so that ,

1 ( D A − A + D B− B ) 2 The polar contribution Dp, also called ionic resonance energy ∆, is a relative measure of the polarity of the A–B bond. As per pauling electronegativity difference , ∆x is proportional to the ∆, as = DA––B –

∆x = (xA –xB) = 0.088 ∆ where xA and xB are electronegativities of A & B respectively xA > xB . The bond energies are in KJ/mol. On further study it is seen electronegativity of the atom also linearly increases with the partial ionic charge (q). Thus electronegativity, x = a + bq. Large and soft atoms have low value of b, while small and hard atoms have a large value of b. The % ionic character related with the % ioinic character = 3.5 (xA–xB )2 + 16 |(xA–xB )|

2

= 351

3.51× 3 = 3.245 Plank’s constant, h = 6.625 × 10 – 3 4 JS 11. What is the distance between the α-particle and Au nucleus during the collision – (A) 10.374 × 10–13 cm (B) 10.374 Å (C) 10.374 × 10–10 cm (D) 10.374 nm

14. Enthalpy of formation AB(g) from A2(g) & B2(g) is represented by ∆fH(AB) in KJ/mol. Thus electronegativity difference (∆x) between A and B is related with ∆f H (AB) in KJ/mol as – (A) –∆fH = 129.13 × (∆x)

12. What should be the minimum velocity of the α-particle to strike the nucleus of 79Au197 ? (A) 3.245 × 108 m/s (C) 3.245 × 105 m/s

(B) 3.245 × 109 m/s (D) 3.245 × 107 m/s

(B) ∆fH = 129.13 × (∆x)

13. What is the de-broglie’s wave length associated with a α-particle while it is moving to colloide with the Au nucleus ? 6.625 × 6 6.625 × 6 (A) ×1025m (B) ×10–15m 4 × 3.245 3.245 6.625 × 6 6.625 × 6 (C) ×10–15m (D) ×10–15m 4 × 3.245 5 × 3.245

XtraEdge for IIT-JEE

1 ( D A − A + D B− B ) 2 Dp =DA–B–Dnp

Dnp =

(C) –∆fH = 129.13 × (∆x)2 (D) ∆fH = 129.13 × (∆x)2

15. What is the order of polarity of the following bond? (i) C – H (ii) F – H (iii) Br – H (iv) Na – I (v) K – F (vi) Li – Cl 56

AUGUST 2009

(A) Li– Cl >K– F>F–H > Br – H > C–H >Na – I (B) K–F>Li– Cl > F– H >Br –H > Na – I >C–H (C) K –F >F – H > Li– Cl >Br–H>Na –I >C–H (D) K–F>Li–Cl > F– H >Na – I > Br –H > C–H

16. As per the relation, x = a + bq which of the given plot describes the electronegativity variation of F & Cl accurately ? Given : xF = 4, xCl = 3 & xH = 2; 4 (A) x 3 2

2.

If tan x – tan2x = 1, then the value of tan4x – 2 tan3x – tan2x + 2 tan x + 1 is : (A) 1 (B) 2 (C) 3 (D) 4

3.

The number of roots of the equation |sinx| = |cos 3x|, belonging to [–2π, 2π] are : (A) 32 (B) 28 (C) 24 (D) 30

4.

The value 'a' for which the equation 4 cosec2(π (a + x)) + a2 – 4a = 0 has a real solution is: (A) a = 1 (B) a = 2 (C) a = 3 (D) None of these

5.

If p1, p2, p3 are respectively the perpendicular from the vertices of a triangle to the opposite sides, then cos A cos B cos C + + is equal to p1 p2 p3

Cl

F

4 (B) x 3 2

Cl q

q F

4 (C) x 3 2

F

Cl

F 4 (D) x 3 2

q

Cl

(A) 1/r

q

6.

Numerical response questions (Q. 17 to 19). Answers to this Section are to be given in the form of nearest integer-in four digits. Please follow as per example : (i.e. for answer : 1492.2 write 1492; 491.8 write 0492; 92.5 write 0092; 2.1 write 0002)

(B) 1/R

(C) 1/∆

(D) None

The number of solution of the equation; 1 + x2 + 2x sin (cos–1y) = 0 is (A) 1 (B) 2 (C) 3 (D) 4

Questions 7 to 10 are multiple choice questions. Each questions has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct.

17. The vapour density of a mixture containing NO2 and N2O4 is 38.3 at 27ºC. Calculate the millimoles of NO2 in 100 g mixture.

7.

18. A sample of a mixture of CaCl2 and NaCl weighing 4.22 g was treated with sodium carbonate to precipitate all the calcium ion as CaCO3 which was heated and quantitatively converted to 0.959 g of CaO. Calculate % of CaCl2 in the mixture.

If A and B are acute angles such that A + B and A – B satisfy the equation tan2θ – 4 tan θ + 1 = 0, then (A) A = π/4 (B) A = π/6 (C) B = π/4 (D) B = π/6

8.

19. The line at 434 nm in the Balmer series of the hydrogen spectrum corresponds to a transition of an electron from the nth to second Bohr orbit. What is the value of n ?

If the numerical value of tan (cos–1(4/5) + tan–1(2/3) is a/b then (A) a + b = 23 (B) a – b = 11 (C) 3b = a + 1 (D) 2a = 3b

9.

A solution of the equation 2

(1 – tan θ) (1 + tan θ) sec2θ + 2 tan θ = 0 where θ lies in the interval (–π/2, π/2) is given by (A) θ = 0 (B) θ = π/3 (C) θ = –π/3 (D) θ = π/6

MATHEMATICS

10. Given an isosceles triangle with equal sides of length b, base angle α < π/4, R, r the radii and O, I the centres of the circumcircle and incircle, respectively. Then 1 (A) R = b cosec α (B) ∆ = 2b2 sin 2α 2

Questions 1 to 6 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1.

The numerical value of sin

5π 7π π .sin .sin is equal 18 18 18

to (A) 1

(B)

1 8

XtraEdge for IIT-JEE

(C)

1 4

(D)

1 2

(C) r =

57

b sin 2α 2(1 + cos α )

(D) OI =

b cos(3α / 2) 2 sin α cos(α / 2) AUGUST 2009

18. If y = sin(cot–1x) and x = 99, then 1/y2 is equal to

This section contains 2 paragraphs; each has 3 multiple choice questions. (Question 11 to 16) Each question has 4 choices (A), (B), (C) and (D) out of which ONE OR MORE THAN ONE is correct.

19. If sin 3 α = 4 sin α sin(x + α) sin (x – α), then 864 sin2 x + 3620 cos2 x is equal to

Passage : I (Q. No. 11 to 13) Consider the equations 5 sin2x + 3 sin x cos x – 3 cos2x = 2 ...(1) sin2x – cos 2x = 2 – sin 2x ...(2) 11. If α is root of (1) and β is a root of (2) then tan α + tan β can be equal to (A) 1 +

69 6

− 3 + 69 (C) 6

Chemistry Facts

69 6

(B) 1 –

− 3 − 69 (D) 3

12. If tan α, tan β satisfy (1) and cos γ, cos δ satisfy (2) then tan α tan β + cos γ + cos δ can be equal to 5 2 (A) –1 (B) – + 3 13 (C)

5 2 − 3 13

1.

The element with the lowest boiling point is also helium at -452.07 degrees Fahrenheit (268.93 degrees Celsius.

2.

The word "atom" comes from the Greek word atomos, meaning "uncut."

3.

In 1964, scientists in Russia discovered element 104, and suggested the name Kurchatovium and symbol Ku in honor of Igor Vasilevich Kurchatov. Then in 1969, scientists in the U.S. also found element 104, and propsed the name Rutherfordium (symbol Rf), in the honor of the New Zealand physicist Ernest R. Rutherford. To get the names past the I.U.P.A.C., it won with rutherfordium.

4.

The first and relatively pure atom of tantalum was produced by von Bolton in 1907.

5.

Andres Manual del Rio discovered what we call today vanadium. He called it panchromium, and then changed it to erythronium (red), after noting that upon heating it turned red. In 1831, Nils Gabriel Sefström (a Swedish chemist) was working with some iron ores and this matter was lead to honor the Northern Germanic tribes' goddess Vanadis due to its inspiration in multi-colors. In the same year, Friedrich Wöhler came into posession of del Rio's erythronium, and confirmed it to be vanadium, after Vanadis. The name Vanadium is now being used instead of del Rio's erythronium.

6.

Hafnium was named Copenhagen, Denmark.

7.

The heaviest type of lepton is the tau.

5 2 (D) – − 3 13

13. The number of solutions common to (1) and (2) is (A) 0 (B) 1 (C) finite (D) infinite Passage : II (Q. No. 14 to 16) α = cos–1(4/5), β = tan–1(2/3), 0 < α, β < π/2 14. α + β is equal to (A) tan–1(17/6) –1

(C) sin (3/5)

(B) sin–1(17/ 5 13 ) –1

(D) cos (3/ 13 )

15. α – β is equal to (A) cos–1(18/ 5 13 )

(B) sin–12/ 13 )

(C) tan–1(1/18)

(D) cos–1(3/ 13 )

16. cos–1(44/125) is equal to (A) 2α (B) 3α (C) π – 3α (D) π – 2α Numerical response questions (Q. 17 to 19). Answers to this Section are to be given in the form of nearest integer-in four digits. Please follow as per example : (i.e. for answer : 1492.2 write 1492; 491.8 write 0492; 92.5 write 0092; 2.1 write 0002) 17. In a triangle ABC, if r1 = 2r2 = 3r3 then [100 – (4a/5b)]2 is equal to

XtraEdge for IIT-JEE

58

after

the

city

of

AUGUST 2009

XtraEdge Test Series ANSWER KEY IIT- JEE 2010 (August) issue) PHYSICS Ques Ans Ques Ans

1 C

2 B

3 B

4 D

5 D

6 C

7 B , C ,D

8 A ,B

9 A ,B

10 A ,D

11 B

12 D

13 A

14 B

15 B

16 C

17 0017

18 0001

19 0007

6

7

8

9

10 B,C

C H E MI S T R Y Ques

1

2

3

4

5

Ans

A

D

A

D

D

C

A ,D

B , C ,D

A ,B

Ques

11

12

13

14

15

16

17

18

19

Ans

C

D

A

B

A

B

0108

0001

0041

MATHEMATICS Ques

1

2

3

4

5

6

7

8

9

10

Ans

C

C

C

C

D

A

B , C ,D

C,D

A,B,D

A ,B

Ques

11

12

13

14

15

16

17

18

19

Ans

A

D

C

B

A

C

2485

0162

0192

IIT- JEE 2011 (August issue)

PHYSICS Ques

1

2

3

4

5

6

7

8

9

10

Ans

C

A

D

C

A

D

B,D

A ,B

D

B,C

Ques

11

12

13

14

15

16

17

18

19

Ans

B

C

A

C

A

C

0004

0027

0004

C H E MI S T R Y Ques Ans Ques Ans

1 C

2 A

3 A

4 C

5 D

6 C

7 A , B ,C

8 A ,B

9 A

11 A

12 D

13 C

14 C

15 D

16 D

17 0437

18 0045

19 0005

10 B

MATHEMATICS Ques Ans Ques Ans

1 B

2 D

3 C

4 B

5 B

6 A

7 A ,D

8 A , B ,C

9 B,C

11 A,B

12 B,D

13 A ,C

14 A ,B

15 A ,C

16 C

17 9801

18 9802

19 1553

XtraEdge for IIT-JEE

59

10 A .C . D

AUGUST 2009