teachers open the door. you enter by yourself...civil engineer and river management specialist,...

XtraEdge for IIT-JEE 1 OCTOBER 2011

Dear Students,

Is examination a common cause of stress? In most Asian cultures, the great emphasis on academic achievement and high expectations of success make it especially stressful for students. The strong negative stigma attached to failure also adds to the pressure. Like it or not, we have to accept that examinations are necessary in any educational system. Even though it is debatable whether they are accurate measures of actual ability, no better alternatives have been proposed. Examinations remain necessary to motivate students’ learning, measure their progress and ultimately, serve as evidence of attainment of certain skills, standards or qualifications. Success at examinations provides opportunities to proceed with higher education and improves employment prospects, underlining their importance. No matter how well prepared, many factors may influence one’s performance at the time of the examination and there is seemingly, no definite guarantee of success. Essentially, it is this vital importance attached to success at examinations coupled with the element of uncertainty that makes them so stressful. As with other sources of stress, the stress of examinations is not all bad. It is a strong incentive for students to study and poses a challenge for individual achievement. However, when stress becomes excessive, performance begins to suffer. There is thus a need to control levels of stress before it becomes overwhelming and detrimental. Reliase of stress is necessary for optimum performance, the means of which is relasing. Learn to relax The stress responses produces muscle tension, which you would commonly experience as backache, neck ache or tension headache at the end of the day. Often this is unconscious. So to relax these muscles, you need to consciously practice relaxation exercises. These could involve muscle relaxation, deep breathing exercises, body massage or guided imagery. Like any particular skill, you need to practice them regularly in order to reap the benefits. Another way to relax is to maintain a quiet time as part of the daily routine. Quiet time refers to a time for you with no interruption from external sources or distractions. This is a time where you may choose to just think of nothing and relax. Finally, you can always take up a hobby to help you relax. Do something you enjoy, be it listening to music. Ideally, the drive to study should be internally driven by a desire to achieve one’s own personal goals. Instead, many are driven more by the fear of failure, which is more stress-provoking and leads easily to discouragement. Attending school should not merely revolve around preparation for examinations. Interacting with teachers, socializing with friends, participating in sports or other extra-curricular activities are all valuable aspects of a ‘well-rounded’ education. Instead of wishing things would get easier, start looking at how you can get better...

Yours truly

Pramod Maheshwari, B.Tech., IIT Delhi

Every effort has been made to avoid errors oromission in this publication. Inr spite of this, errors are possible. Any mistake, error ordiscrepancy noted may be brought to ournotice which shall be taken care of in theforthcoming edition, hence any suggestion iswelcome. It is notified that neither thepublisher nor the author or seller will beresponsible for any damage or loss of action toany one, of any kind, in any manner, there from.

• No Portion of the magazine can be published/ reproduced without the written permission of the publisher

• All disputes are subject to the exclusive jurisdiction of the Kota Courts only.

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

Teachers open the door. You enter by yourself Volume - 7 Issue - 4

October, 2011 (Monthly Magazine) Editorial / Mailing Office : 112-B, Shakti Nagar, Kota (Raj.) Tel. : 0744-2500492, 2500692, 3040000 e-mail : [email protected]

Editor :

Pramod Maheshwari [B.Tech. IIT-Delhi]

Cover Design

Satyanarayan Saini

Layout

Rajaram Gocher

Circulation & Advertisement

Praveen Chandna Ph 0744-3040000, 9672977502

Subscription

Himanshu Shukla Ph. 0744-3040000 © Strictly reserved with the publishers

Editorial

Unit Price ` 20/- Special Subscription Rates

6 issues : ` 100 /- [One issue free ] 12 issues : ` 200 /- [Two issues free] 24 issues : ` 400 /- [Four issues free]


Volume-7 Issue-4 October, 2011 (Monthly Magazine)

NEXT MONTHS ATTRACTIONS

Much more IIT-JEE News.

Know IIT-JEE With 15 Best Questions of IIT-JEE

Challenging Problems in Physics, Chemistry & Maths

Key Concepts & Problem Solving strategy for IIT-JEE.

Xtra Edge Test Series for JEE- 2012 & 2013

S

Success Tips for the Months

• If one asks for success and prepares for failure, he will get the situation he has prepared for.

• Loser's visualize the penalties of failure. Winner's visualize the rewards of success.

• Treat others as if they were what they ought to be and you help them to become what they are capable of being.

• You never achieve real success unless you like what you are doing

• The first step toward success is taken when you refuse to be a captive of the environment in which you first find yourself.

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

CONTENTS

INDEX PAGE

NEWS ARTICLE 3 • IITians sign up for online protest against govt • Bihar Professor gets distinguished IIT alumnus Award

IITian ON THE PATH OF SUCCESS 5 Mr. Ashwin Limaye & Ms. Amrita Sateesh Mahale

KNOW IIT-JEE 6 Previous IIT-JEE Question

XTRAEDGE TEST SERIES 47

Class XII – IIT-JEE 2012 Paper Class XI – IIT-JEE 2013 Paper

Regulars ..........

DYNAMIC PHYSICS 14

8-Challenging Problems [Set # 6] Students’ Forum Physics Fundamentals • Magnetic effects of current • Gravitation

CATALYSE CHEMISTRY 27

Key Concept • Halogen deravative • Halogen & Noble gases Family Understanding :Organic Chemistry

DICEY MATHS 36

Mathematical Challenges Students’ Forum Key Concept

• Limit, Continuity & Differentiability • Parabola, Ellipse & Hyperbola

Study Time........

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


IITians sign up for online protest against govt MUMBAI, INDIA: A group of IITians is protesting against the Indian government's proposal to change the IIT-JEE admission procedures. IITIANS.org, the online community has created a Facebook page under the name 'SAVE IITs' to call for global protest against the government's move to meddle with the admission process.

According to media reports, the HRD Ministry has proposed a new admission mechanism for education institutes, including IITs, for 2013-14 where students will be selected from a common merit list based on students' class 12 and entrance test marks.

The new proposal completely changes the existing admission system of Joint Entrance Exams (JEE), which was in place for decades.

The IITians feel that it will end the IITians' era since India has many states and respective education boards. They fear it will lead to corruption and politics will mask the significance of aptitude and knowledge in admissions to IITs.

The online protest has gathered support of 6,000 people, including IIT students and alumini.

“It's a peaceful protest and they are not trying to protest publicly. Through this, they are trying to spread the message to students and people. Based on that they will gather students' feedback and air their concerns to the government,” said a fourth year electrical engineering student - IIT Bombay.

The student said the government should provide logical reasons behind the proposed changes.

“Secondly, there are different state education boards which follow different syllabus. So this new proposed admission plan will not be fair to every students seeking admissions to IITs,” he stressed.

“Also, as the IITs are funded and run by government, do propose whatever they feel,” the student said.

Beside the IITIANS.org community, there are other community pages on Facebook such as 'We Hate Kapil Sibal', 'I Hate Kapil Sibal', 'Kapil Sibal Stinks'. These community pages have angry messages and remarks against the union minister and protest his proposed admission plan.

Bihar Professor gets distinguished IIT alumnus Award Prof. B B Pandey who hails from Bhojpur district of Bihar has received distinguished IIT Kharagpur alumnus award from PM Manmohan Singh for his contribution to innovation in transport infrastructure development. He is currently advisor (sponsored research and industrial consultancy) at IIT (Kharagpur).

Developed low cost technology for maintenance-free roads for villages and design guidelines for concrete and asphalt roads in India. He has also developed state-of-the-art equipment for testing of roads and highways.

Govt seeks IITians expertise to solve Kosi problem Bihar government has made a request to three experts, all IIT-Kharagpur alumni, to make their presentation about their cost-effective, innovative technology in breach-resistant earthen bund for river Kosi.

These experts have developed a technology which should ensure cent per cent success in preventing breaches in bunds. Since silting is a major

problem with most Indian rivers, application of this technology should prove extremely useful to tame floods, said one of the experts.

The three experts are T K Choudhary, civil engineer and river management specialist, Mihir Biswas, mechanical engineer, and Somnath Ghosh, former Hod, civil engineering, IIT-Kharagpur.

Mihir Biswas, an IIT-Kharagpur alumnus with considerable post-qualification experience in engineering and management The breach caused devastating flood and changed the course of the river.

Since the river bed of Kosi is at a higher level than the adjoining area, construction of a breach-resistant bund is the only solution to ensure safety from periodic floods, says an expert.

The bund comprises of a suitably designed bund section, concrete walls with precast concrete materials at toes of riverside earth slopes, and protective concrete tiles meant to avoid erosion of the bund.

The composite design will effectively prevent any breach in the bund. This innovative technology was developed by T K Choudhary based on his technical paper. The technology was since vetted by IIT-Kharagpur and received approval. Choudhary is an IIT alumnus and recognized river management expert.

“The technology is indigenous and labour-intensive, utilizing unskilled local labour, making execution very much cost-effective.

JNU makes it to top 100 in world university rankings The Jawaharlal Nehru University (JNU) has made it to the top 100 in the QS World University Rankings


this year. QS, for the first time, has released a subject-wise list in which the English department of JNU-Centre for English Studies – has been ranked among the top 100 English departments in the world. Apart from the English department, JNU's departments of geography and area studies, and politics and international studies too, have found a place in the top 100.

Cambridge University looking for partnerships in India

Aiming to strengthen its engagement with India, Cambridge University is looking to form high quality partnerships in the areas of drug discovery and disease management. One of the oldest academic institutions in the world, Cambridge University is the alma mater of many famed Indian personalities, including Prime Minister Manmohan Singh and Nobel laureate Amartya Sen. He pointed out the university is looking at some partnerships in the areas of drug discovery, apart from having interactions with the Council of Scientific and Industrial Research (CSIR) and the Department of Biotechnology.

Availing scholarships from Swiss govt The Swiss government awards a range of scholarships to Indian students through the Federal Commission for Scholarships for Foreign Students (FCS): university scholarships (Swiss universities and Federal Institutes of Technology). These scholarships provide graduates from all fields with the opportunity to pursue postgraduate studies or research in Switzerland at a public funded university or recognised institution. Scholarships are provided for a nine-month academic year. Scholarships for postgraduate or Masters programmes lasting up to two years (third or fourth semesters) may be extended depending on results from the first year.

University Special innovation council for biomedical engineering in Gujarat Technological University The Gujarat Technological University (GTU) Innovation Council has started the process of creating an innovation council specifically for biomedical engineering. The newly created council is aimed at working towards fostering innovations in the biomedical sector across the state. The process will involve industrialists, policymakers, students and faculties.

GTU has already conducted three meetings with students and faculties. The agenda and the other details of the initiative were discussed in the meeting.

"Most of the technologies for healthcare system in the country at present are imported ones. There is a great need to indigenize these technologies to meet our local needs and to make them affordable. Encouraging the students to apply there innovative minds in the area will lead to the emergence of such medical equipments," said Hiranmay Mahanta, managing director of Techpedia of Sristi which is partnering with GTU in encouraging the students to carry out innovative projects.

Benefits of solar energy discussed at IIT-K KANPUR: A two-day seminar on 'Micro solar energy generation and utilisation' at IIT-K . It is being held in association with the Institute of Electronics and Telecommunication Engineers (IETE), Kanpur.

The theme of the seminar is to make solar energy viable for domestic and commercial use. AK Chaturvedi said that the photo-voltaic cell is still the new research area as the best material from which the cells could be made. He said that a solar inverter capable of generating five kilowatt of power has been installed on the roof of a lab. "The inverter is generating power and it is being fed to the local grid situated within the campus," Chaturvedi added.

"The institute is working in the direction of increasing the efficiency and life of solar cells. IIT-K is working in several other areas of solar energy under the umbrella of Solar Energy Research Enclave.

Scientists and faculty members are working on various aspects of material, devices, system and policy related issues of solar energy. Anand told the gathering that the institute has been doing its bit by increasing awareness and educating the masses about solar technologies. IIT-K had organised a conference on 'Emerging trends in the photovoltaic generation and utilisation'.

In order to create awareness, an exhibition was held on the use of solar energy. In addition to solar lighting, heating and cooking systems, one solar-powered three-wheeler was also displayed. Tutorials were held to educate students about solar energy. Poster paper presentation and working model contest were also held. A panel discussion on 'Captive generation of solar power for domestic and commercial use' was also held.

IIT experts to help cut noise near Delhi Airport The central government Wednesday informed the Delhi High Court that it has sought help from Indian Institute of Technology (IIT) experts to reduce the noise level near the Indira Gandhi International (IGI) Airport.

The Directorate General of Civil Aviation (DGCA) have fixed the limits in airport zones at 105 decibels during the day and 95 decibels at night.

The maximum limit for noise level at night has been fixed at levels lower than most major international airports, including New York's JFK (99 decibels), Rio de Janeiro (118.4 decibels), Paris (104.5 decibels) and Osaka (107 decibels). Three airports in London have 94 decibels as the limit.


Success Story Success Story This article contains storys/interviews of persons who succeed after graduation from different IITs

He is simple, unassuming and, of course, intelligent! No doubt that Ashwin Limaye from Pune topped the M Tech dual degree. He has also topped in the last three years of the course. On the IIT experience The five years I spent at IIT were a great experience. The environment at IIT is totally different when compared to that of other colleges. We had a great hostel life too. The best thing about IIT is that it is not just about academics, it also involves a lot of extra-curricular activities and sports. The interaction with professors and students from diverse backgrounds is also an enriching experience. Although the class timings are from 8 a.m. to 5 p.m., we have all the time in the world to do what we like: this fine balance helps in overall personality development. My interests Badminton is my first love. I have headed the IIT team at various competitions in IIT and other inter-IIT competitions and won several medals. I'm very thankful to IIT for giving us such opportunities to play and practise. I also like reading both fiction and non-fiction. Next move I have joined McKinsey & Co as a consultant. It's a great place to work. I was interested in the people part of the businesses and I'm enjoying my work. My mantra for success The most important thing is to pay attention in the class, don't mix work and play, and stay focussed and organised. An engineering degree helps to have structured approach to problem-solving which is important. Advice to IIT aspirants Get your basics right, you should have problem-solving skills, you must learn to ask why and not how, and be focussed in what you do. I did not go for any formal coaching, I did a correspondence course along with 3 of my friends so it was mostly combined study. India or abroad? India, of course! I belong to this place and I would love to settle down in India. However, I would like to experience how the environment is abroad. India is a growing market

with plenty of opportunity, growth and innovation. It is the place to be now, this is where we can shine and excel. And I want to be a part of the great Indian growth story.

Ms. Amrita Sateesh Mahale, B Tech in Aerospace Engineering Lively, enthusiastic and cheerful, Amrita from Mumbai, is a topper all the way. She represents the growing number of women IIT toppers. She won the IIT silver medal for topping the Aerospace Engineering course and a gold medal for overall proficiency. On the IIT Experience It was a fantastic experience. The first year was very stressful as you are surround by toppers and child prodigies. It takes some time to get used to the environment. Studies are not as difficult as they are before joining IIT. You get the best facilities to study and work. Interests I like robotics and have done projects on robotics and won awards. I have also published research papers on aerospace and automobile fuel cells after an internship at Mercedes Benz. I like reading and writing. I was the editor of the department magazine. I also write poetry. I was also on the editorial board of the IIT newspaper. I learnt a lot of new things like drumming and playing squash at the IIT. I also like advertising and did two advertisements at O&M during an internship: Success mantra : Lots of hard work, two years of preparations, I attended coaching classes after my 10th standard. Advice to IIT aspirants One has to work really hard, put in lot of extra work for two years to get the branch of one's choice. But one should not be dejected if one doesn't get selected. One has a lot of opportunities and many interesting things to do now. Next move I have joined the Boston Consultancy Services. I would like to explore the corporate world. I enjoy my work. The job has lived up to my expectations. I used to think that the corporate world is a very cut-throat one, but I was surprised to find that it not so. Everyone is very helpful. And aerospace? Yes, it is on the radar, I would like to pursue research in aerospace in future.

Mr. Ashwin Limaye M Tech dual degree in Computer Science and Engineering


PHYSICS

1. A block X of mass 0.5 kg is held by a long massless string on a frictionless inclined plane of inclination 30º to the horizontal. The string is wound on a uniform solid cylindrical drum Y of mass 2 kg and of radius 0.2 m as shown in figure. The drum is given an initial angular velocity such that the block X starts moving up the plane.

X

Y

30º (i) Find the tension in the string during the motion. (ii) At a certain instant of time the magnitude of the

angular velocity of Y is 10 rad s–1 calculate the distance travelled by X from that instant of time until it comes to rest.

Sol. The drum is given an initial velocity such that the block X starts moving up the plane.

X

R

θ = 30º

m T

T

θ

mgmg cosθ mg sinθ

Y ω

As the time passes, the velocity of the block deceases.

The linear retardation a of the block X is given by mg sin θ – T = ma ...(i) The linear retardation of the block and the angular

acceleration of the drum (α) are related as a = Rα ...(ii) Where R is the Radius of the drum. The retarding torque of the drum is due to tension T

in the string. τ = T × R But τ = Iα where I = M.I. of drum about its axis of

rotation.

∴ T × R = 21 MR2α ...(iii)

= 2MR

21IQ

From (ii) TR = 21 MR2

ra ⇒ a =

MT2

Substituting this value in (i)

mg sin θ – T = m × MT2 ⇒ mg sin θ =

+

Mm21 T

∴ T = m2M

M)sinmg(+

×θ = 5.022

2º30sin8.95.0×+

×××

= 1.63 N (ii) The total kinetic energy of the drum and the block

at the instant when the drum is having angular velocity to 10 rads–1 gets converted into the potential energy of the block

θ

θ

mg

h = ssinθ S

[(K.E.)Rotational]drum + [(K.E.)Translational]block = mgh

21 Iω2 +

21 mv2 = mgS sin θ

21 Iω2 +

21 m(Rω)2 = mgS sin θ [Q v = Rω]

⇒ 21 MR2ω2 +

21 mR2ω2 = mgS sin θ

⇒ 21

θ+ω

sinmg)mM(R 22

= S

= 21 ×

º30sin8.95.0)5.02(10102.02.0

××+××× = 1.22 m

Alternatively

a = Rα = R × MR

T2 [From (iii)]

= 2.02

63.122.0×

××

V = 0S = S

a = a

Rω0 = u

KNOW IIT-JEE By Previous Exam Questions


Using v2 – u2 = 2as 02 – 22

0Rω = 2 (– a)s

⇒ S = 63.12

2.02.01010×

××× = 1.22 m

2. A right prism is to be made by selecting a proper

material and the angles A and B (B ≤ A), as shown in figure. It is desired that a ray of light incident on the face AB emerges parallel to the incident direction after two internal reflections. [IIT-1987]

A B

C (i) What should be the minimum refractive index

n for this to be possible ?

(ii) For n = 35 is it possible to achieve this with the

angle B equal to 30 degrees ? Sol. (i) Let x is the incident angle for reflection at AC.

Four total internal reflection x > ie (critical angle)

A B

C

x x

90–x M

y y 90–y

N

y

Let y be the incident angle of the ray on face CB. For

total internal reflection y > iC ∴ x + y > 2ic But x = ∠A and y = ∠B (from geometry) ∴ x + y = 90º ⇒ 90 > 2ic ⇒ ic < 45º The refractive index of the medium for this to

happen.

µ = cisin

1 = º45sin

1 = 2

(ii) For µ = 35

sin ic´ = µ1 =

3/51 =

53

⇒ ic´ = 37º y = 30º (Given)

∴ x = 60º x > ic´ but y < ic´ ⇒ Total internal reflection will take place on face

AC but not on CB. 3. (a) A charge of Q is uniformly distributed over a

spherical volume of radius R. Obtain an expression for the energy of the system.

(b) What will be the corresponding expression for the energy needed to completely disassemble the planet earth against the gravitational pull amongst its constituent particles ?

Assume the earth to be a sphere of uniform mass density. Calculate this energy, given the product of the mass and the radius of the earth to be 2.5 × 1031 kg-m.

(c) If the same charge of Q as in part (a) above is given to a spherical conductor of the same radius R, what will be the energy of the system ? [IIT-1992]

Sol. (a) In this case the electric field exists from centre of the sphere to infinity. Potential energy is stored in electric field with energy density

dr

u = 21

ε0E2 (Energy/Volume)

(i) Energy stored within the sphere (U1) Electric field at a distance r is

E = 04

1πε

. 3RQ . r

U = 21

∈0E2 = 20ε

2

30

rRQ.

41

πε

Volume of element dV = (4πr2)dr Energy stored in this volume dU = U(dr)

dU = (4πr2dr) 2

30

0 rRQ.

41

2

πεε

dU = 6

2

0 RQ.

401πε

.r4dr

∴ U1 = ∫R

0dU = ∫πε

R

0

46

0

2drr

R8Q

= R0

56

0

2]r[

R8Q

πε

U1 = R

Q.40

1 2

0πε ...(1)

(ii) Energy stored outside the sphere (U2) Electric field at a distance r is


E = 20 R

Q.4

1πε

∴ U = 21

ε0E2 = 2

20

0

RQ.

41

2

πεε

∴ dU = µ . dV = (4πr2dr)

πε

ε2

20

0

RQ.

41

2

dU = 0

2

8Qπε 2r

dr

∴ U2 = ∫α

RdU =

0

2

8Qπε

. ∫α

R 2rdr =

R8Q

0

2

πε ...(2)

Therefore, total energy of the system is

U = U1 + U2 = R40

Q

0

2

πε+

R8Q

0

2

πε

or U = 203

RQ

0

2

πε

(b) Comparing this with gravitational forces, the gravitational potential energy of earth will be

U = –53

RGM2

by replacing Q2 by M2 and 04

1πε

by G.

g = R

GM2

∴ G = M

gR 2

U = 53− MgR

Therefore, energy needed to completely disassemble the earth against gravitational pull amongst its constituent particle will be given by

E = |U| = 53 MgR

Substituting the values, we get

E = 53 (10m/s2) (2.5 × 1031 kg-m)

E = 1.5 × 1032 J (c) This is the case of a charged spherical conductor

of radius R, energy of which is given by = C

Q21 2

or U = R4

Q.21

0

2

πε =

R8Q

0

2

πε

4. A circular loop of radius R is bent along a diameter and given a shape as shown in figure. One of the semicircles (KNM) lies in the x-z plane and the other one (KLM) in the y-z plane with their centres at origin. Current I is flowing through each of the semicircles as shown in figure. [IIT-2000]

KI N

ML

I

z

x

y

(a) a particle of charge q is released at the origin with

a velocity →v = –v0 i . Find the instantaneous force

→F

on the particle. Assume that space is gravity free. (b) If an external uniform magnetic field B0 j is

applied determine the force 1F→

and 2F→

on the semicircles KLM and KNM due to the field and the

net force →F on the loop.

Sol. (a) Magnetic field (→B ) at the origin = magnetic field

due to semicircle KLM + Magnetic field due to other semicircle KNM

∴ →B =

R4Iµ0 (– i ) +

R4Iµ0 ( j )

→B = –

R4Iµ0 i +

R4Iµ0 j

= R4Iµ0 (– i + j )

∴ Magnetic force acting on the particle

→F = q(

→v ×

→B )

= q(–v0 i ) × (– i + j )R4Iµ0

→F = – k

R4Iqvµ 00

(b) →F KLM =

→F KNM =

→F KM

And →F KM = BI(2R) i = 2BIR i

→1F =

→2F = 2BIR i

Total force on the loop,

→F =

→

1F + →

2F

or →F = 4BIR i

Note : If a current carrying wire ADC (of any shape)

is placed in a uniform magnetic field →B .

Then, →F ADC =

→F AC

or |→F ADC| = i (AC)B

From this we can conclude that net force on a current carrying loop in uniform magnetic field is zero. In the question, segments KLM and KNM also form a loop and they are also placed in a uniform magnetic field but in this case net force on the loop will not be zero. It would had been zero if the current in any of the segments was in opposite direction.


5. Assume that the de Broglie wave associated with an electron can form a standing wave between the atoms arranged in a one dimensional array with nodes at each of the atomic sites. It is found that one such standing wave is formed if the distance d between the atoms of the array is 2Å. A similar standing wave is again formed if d is increased to 2.5 Å but not for any intermediate value of d. Find the energy of the electrons in electron volts and the least value of d for which the standing wave of the type described above can form. [IIT-1997]

Sol. As nodes are formed at each of the atomic sites, hence

2Å = n

λ

2 ...(1)

[Q Distance between successive nodes = λ/2] Hence from the figure

2Å

N N N N N N n loops

2.5Å

N N N N N N (n+1) loops λ/2

N

and 2.5 Å = (n + 1)2λ ...(2)

∴ 25.2 =

n1n + ,

45 =

n1n + or n = 4

Hence, from equation (1),

2Å = 42λ i.e. λ = 1Å

d will be minimum, when

n = 1, dmin = 2λ =

2Å1 = 0.5 Å

Now, de broglie wavelength is given be

λ = mK2h or K =

m2.h

2

2

λ

∴ K = 1931210

234

106.1101.92)101()1063.6(

−−−

−

××××××× eV

= 6.11.98

)63.6( 2

×× × 102 eV = 151 eV

CHEMISTRY

6. A metallic element crystallizes into a lattice

containing a sequence of layers of ABABAB ......... . Any packing of spheres leaves out voids in the

lattice. What percentage by volume of this lattice is empty space ? [IIT-1996]

Sol. A unit cell of hcp structure is a hexagonal cell, which is shown in figure (i) & (ii). Three such cells form one hcp unit.

For hexagonal cell, a = b ≠ c; α = β = 90º and γ = 120º. It has 8 atoms at the corners and one inside,

Number of atoms per unit cell = 88 + 1 = 2

Area of the base = b × ON [From fig.(ii)] = b × a sin 60º

= 23 a2 (Q b = a)

a γ b α

cβ

Figure (i) Volume of the hexagonal cell

= Area of the base × height = 23 a2.c

But c = 322 a

∴ Volume of the hexagonal cell

= 23 a2 .

322 a = a3 2

and radius of the atom, r = 2a

Hence, fraction of total volume or atomic packing factor

= cell hexagonal theof Volume

atoms 2 of Volume

60º

N b

O

a

figure (ii)

= 2a

r342

3

3π× =

2a2a

342

3

3

π×

= 23

π = 0.74 = 74%

∴ The percentage of void space = 100 – 74 = 26%


7. An alkyl halide X, of formula C6H13Cl on treatment with potassium t-butoxide gives two isomeric alkenes Y and Z(C6H12). Both alkenes on hydrogenation give 2, 3-dimethyl butane. Predict the structures of X, Y and Z. [IIT-1996]

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

X136 ClHC

HCl–;

butoxidetK

∆

−− → 126HCZY +

Both, Y and Z have the same molecular formula C6H12(CnH2n). Since, both Y and Z absorb one mol of H2 to give same alkane 2, 3-dimethyl butane, hence they should have the skeleton of this alkane.

Y and Z (C6H12) Ni

H2→ CH3 – CH – CH – CH3

CH3 CH3

2,3-dimethyl butane

The above alkane can be prepared from two alkenes CH3 – C = C – CH3

CH3 CH3 2,3-dimethyl

butene-2 (Y)

and CH3 – CH – C = CH2

CH3 CH32,3-dimethyl butene-1

(Z)

The hydrogenation of Y and Z is shown below :

CH3 – C = C – CH3

CH3 CH3 (Y)

H2

Ni CH3 – CH – CH – CH3

CH3 CH3

CH3 – CH – C = CH2

CH3 CH3 (Z)

H2

Ni CH3 – CH – CH – CH3

CH3 CH3

Both, Y and Z can be obtained from following alkyl

halide :

CH3 – C – CH – CH3

CH3 CH3 2-chloro-2,3-dimethyl butane

(X)

K-t-butoxide

∆; –HCl

CH2 = C — CH – CH3

CH3 CH3

Cl

+ CH3 – C = C – CH3

CH3 CH3

(Z) 20% (Y) 80%

Hence, X,

CH3 – C – CH – CH3

CH3 CH3

Cl

Y, CH3 – C = C – CH3

CH3 CH3

Z, CH3 – CH – C = CH2

CH3 CH3

8. How is boron obtained from borax ? Give chemical equations with reaction conditions. Write the structure of B2H6 and its reaction with HCl.

[IIT-2002] Sol. When hot concentrated HCl is added to borax

(Na2B4O7.10H2O) the sparingly soluble H3BO3 is formed which on subsequent heating gives B2O3 which is reduced to boron on heating with Mg, Na or K

Na2B4O7(anhydrous) + 2HCl(hot, conc.) → 2NaCl + H2B4O7 H2B4O7 + 5H2O → 4H3BO3 ↓

2H3BO3 → heatingstrong B2O3 + 3H2O B2O3 + 6K → 2B + 3K2O or B2O3 + 6Na → 2B + 3Na2O or B2O3 + 3Mg → 2B + 3MgO

Structure of B2H6 B

H

H

B

H

H

H

H

1.19Å

97º

1.37Å

122º

1.77Å

Hydrogen bridge bonding (3C-2e bond)

B2H6 + HCl → B2H5Cl + H2 Normally this reaction takes place in the presence of

Lewis acid (AlCl3). 9. An organic compound A, C8H4O3, in dry benzene in

the presence of anhydrous AlCl3 gives compound B. The compound B on treatment with PCl5 followed by reaction with H2/Pd(BaSO4) gives compound C, which on reaction with hydrazine gives a cyclised compound D(C14H10N2). Identify A, B, C and D. Explain the formation of D from C. [IIT-2000]

Sol. The given reactions are as follows.

O

O +

O

AlCl3

O

O

OH

PCl5

H2/Pd (BaSO4)

C6H5

H

C

C

O O

H2NNH2

C6H5

N

N

The formation of D from C may be explained as

follows.


C6H5

C6H5

O O

NH2

NH2

O–

NH2

NH2 O–

+

+ C6H5O–

N – H

N – HOH

C6H5

N

N

10. The oxides of sodium and potassium contained in a 0.5 g sample of feldspar were converted to the respective chlorides. The weight of the chlorides thus obtained was 0.1180 g. Subsequent treatment of the chlorides with silver nitrate gave 0.2451 g of silver chloride. What is the percentage of Na2O and K2O in the mixture ? [IIT-1979]

Sol. Mass of sample of feldspar containing Na2O and K2O = 0.5 g.

According to the question, Na2O + 2HCl → 2NaCl + H2O ..(1) 2 × 23 + 16 = 62g 2(23 + 35.5) = 117 g K2O + 2HCl → 2KCl + H2O ..(2) 2 × 39 + 16 = 94g 2(39 + 35.5) = 149 g Mass of chlorides = 0.1180 g Let, Mass of NaCl = x g ∴ Mass of KCl = (0.1180 – x)g Again, on reaction with silver nitrate, NaCl + AgNO3 → AgCl + NaNO3 ..(3) 23 + 35.5 = 58.5g 108 + 35.5 = 143.5g KCl + AgNO3 → AgCl + KNO3 ...(4) 39 + 35.5 = 74.5g 108 + 35.5 = 143.5g Total mass of AgCl obtained = 0.2451 g Step 1. From eq. (3) 58.5 g of NaCl yields = 143.5 g AgCl

∴ x g of NaCl yields = 5.585.143 x g AgCl

And from eq. (4), 74.5 g of KCl yields = 143.5 g of AgCl ∴ (0.1180 – x)g of KCl yields

= 5.745.143 (0.1180 – x)g AgCl

Total mass of AgCl

5.585.143 x +

5.745.143 (0.1180 – x) = 0.2451

which gives, x = 0.0342 Hence, Mass of NaCl = x = 0.0342 g And Mass of KCl = 0.1180 – 0.0342 = 0.0838g Step 2. From eq.(1), 117 g of NaCl is obtained from = 62 g Na2O ∴ 0.0342 g NaCl is obtained from

= 11762 × 0.032 = 0.018 g Na2O

From eq. (2), 149 g of KCl is obtained from = 94 g K2O ∴ 0.0838 g of KCl is obtained from

= 14994 × 0.0838 = 0.053 g K2O

Step 3. % of Na2O in feldspar = 5.0

018.0 × 100 = 3.6%

% of K2O in feldspar =5.0

053.0 × 100 = 10.6 %

MATHEMATICS

11. Let α1, α2, β1, β2 be the roots of ax2 + bx + c = 0 and px2 + qx + r = 0 respectively. If the system of equations α1y + α2z = 0 and β1y + β2z = 0 has a non

trivial solution, then prove that 2

2

qb =

prac . [IIT-1987]

Sol. We are given α1, α2 are roots of ax2 + bx + c = 0 ⇒ α1 + α2 = – b/a and α1 α2 = c/a ...(i) and β1, β2 are roots of px2 + qx + r = 0

⇒ β1 + β2 = – pq and β1β2 =

pr ...(ii)

The system of equations, α1 y + α2 z = 0 β1 y + β2 z = 0, has non-trivial solution

∴ we have 22

11

βαβα

= 0

or 2

1

αα =

2

1

ββ applying componendo-dividendo

21

21

– ααα+α =

21

21

– βββ+β

⇒ (α1 + α2) (β1 – β2) = ( α1 – α2) (β1 + β2) ⇒ (α1 + α2)2 (β1 + β2)2 – 4β1β2 = (α1 + α2)2 – 4α1α2. (β1 + β2)2 using (i) and (ii), we get

2

2

ab

pr

pq 4–2

2 = 2

2

pq

ac

ab 4–2

2

⇒ 22

22

paqb –

parb

2

24 = 22

22

paqb – 2

24ap

cq

or a

rb2 =

pcq2

⇒ 2

2

qb =

prac

12. Let f (x) =

>+≤

0,–0,

32 xxaxxxxeax

where a is a positive constant. Find the interval in which f ' (x) is increasing. [IIT-1996]

Sol. At x = 0 L.H.L. = )(lim–0

xfx→

= 0lim–0

=→

ax

xxe

and R.H.L. = )(lim0

xfx +→

= +→0

limx

(x + ax2 – x3) = 0

Therefore, L.H.L. = R.H.L. = 0 = f (0) So f (x) is continuous at x = 0


Also, f ' (x) =

>+<+

0,3–210,.1

2 xifxaxxifaxee axax

and Lf '(0) = 0–

)0(–)(lim–0 x

fxfx→

= x

xeax

x

0–lim–0→

= –0

lim→x

eax = e0 = 1

and Rf '(0) = 0

)0(–)(lim0 ++→ x

fxfx

= x

xaxxx

0––lim32

0

++→

= +→0

limx

1 + ax – x2 = 1

Therefore, Lf ' (0) = Rf ' (0) = 1 ⇒ f ' (0) = 1

Hence, f ' (x) =

>+=<+

0,3–21010,)1(

2 xifxaxxifxifeax ax

Now, we can say without solving that f '(x) is continuous at x = 0 and hence on R. We have

f " (x) =

><++

0,6–20,)1(

xifxaxifeaxaae axax

and Lf " (0) = 0–

)0('–)('lim–0 x

fxfx→

= xeax ax

x

1–)1(lim–0

+→

=

+

→ xeae

axax

x

1–lim–0

= –0

lim→x

aeax + a . ax

eax

x

1–lim–0→

= ae0 + a(1) = 2a

and Rf " (0) = 0

)0('–)('lim0 ++→ x

fxfx

= x

xaxx

1–)3–21(lim2

0

++→

= x

xaxx

2

0

3–2lim+→

= +→0

limx

2a – 3x = 2a

Therefore Lf " (0) = Rf "(0) = 2a

Hence, f "(x) =

>=<+

06–20,20)2(

xifxaxifaxifeaxa ax

Now, for x < 0, f " (x) > 0 if ax + 2 > 0 ⇒ for x < 0, f "(x) > 0 if x > – 2/a

⇒ f ' (x) > 0 if – a2 < x < 0

and for x > 0, f " (x) > 0 if 2a – 6x > 0 ⇒ for x > 0, f " (0) if x < a/3 Thus, f (x) increases on [–2/a, 0] and on [0, a/3]

Hence, f (x) increases on

3,2– a

a.

13. Let C1 and C2 be two circles with C2 lying inside C1.

A circle C lying inside C1 touches C1 internally and C2 externally. Identify the locus of the centre of C.

[IIT-2001] Sol. Let the given circles C1 and C2 have centres O1 and

O2 and radii r1 and r2 respectively. Let the variable circle C touching C1 internally, C2

externally have a radius r and centre at O.

O2 r2

r

O C

O1

C2 C1

r1

Now, OO2 = r + r2 and OO1 = r1 – r ⇒ OO1 + OO2 = r1 + r2 which is greater then O1O2 and O1O2 < r1 + r2 (Q C2 lies inside C1) ⇒ locus of O is an ellipse with foci O1 and O2. Alternate solution : Let equations of C1 be x2 + y2 = r1

2 and of C2 be (x – a)2 + (y – b)2 = r2

2 Let cetnre C be (h, k) and radius r, then by the given

condition

22 )–()–( bkah + = r + r2 and 22 kh + = r1 – r

⇒ 22 )–()–( bkah + + 22 kh + = r1 + r2 Required locus is

22 )–()–( byax + + 22 yx + = r1 + r2 which represents an ellipse whose foci are at (a, b)

and (0, 0).

14. Evaluate

∫π

+

0

||cos cos21cos3cos

21sin2 xxe x sin x dx]

[IIT-2005]

Sol. I = ∫π

+

0

||cos cos21cos3cos

21sin2 xxe x sin x dx

⇒ I = ∫π

0

||cos cos21sin2.sin. dxxxe x


+ ∫π

0

||cos cos21cos3. xe x . sin x dx ...(i)

or I = I1 + I2

(Using ∫a

dxxf2

0

)( =

+=

=

∫ )()–2(,)(2

)(–)–2(,0

0

xfxafdxxf

xfxafa

∴ I1 = 6 ∫π

2/

0

cos cos21cos2.sin. xxe x . sin x = f (x)

and I2 = 0

xeas x cos

21cos3., ||cos . sin x = f (x)

⇒ )(–)–( xfxf =π ...(ii)

Now I1 = 6 dttet∫

1

02

sin2.

put, cos x = t

I2 =

+

1–

21sin

221cos

524 ee ...(iii)

from (i), (ii), (iii)

I =

+

1–

21sin

221cos

524 ee

15. A bag contains 12 red balls and 6 white balls. Six

balls are drawns one by one without replacement of which at least 4 balls are white. Find the probalility that in the next two drawns exactly one white ball is drawn. (Leave the answer in nCr). [IIT-2004]

Sol. Using Baye's theorem; P(B/A) =

∑

∑

=

=3

1

3

1

)(

)/().(

ii

iii

AP

ABPAP

where A be the event at least 4 white balls have been drawn.

A1 be the event exactly 4 white balls have been drawn. A2 be the event exactly 5 while balls have been drawn.

A3 be the event exactly 6 white balls have been drawn. B be the event exactly 1 white ball is drawn from two

draws.

∴P (B/A) =

618

66

012

618

56

112

618

46

212

212

11

111

618

56

112

212

12

110

618

46

212

...

......

CCC

CCC

CCC

CCC

CCC

CCC

CCC

++

+

= )...(

)...()...(

66

012

56

112

46

212

212

11

111

56

112

112

110

46

212

CCCCCCCCCCCCCCC

++

+

What is mercury poisoning?

CHEMICAL DANGER Too much mercury can make you sick, but sometimes the symptoms are hard to distinguish from other illnesses. What's mercury? There are three kinds of mercury. Depending on what the exposure is, you could have different symptoms and disease states. Elemental, or metal mercury, is found in thermometers. The problem with that is the inhalation of fumes that come off that mercury. Playing with it and ingesting it is not as toxic. That kind of mercury causes significant amounts of neurological damage. As the exposure gets longer, there may be additional changes in the bone marrow that affect the ability to produce blood cells, infertility and problems with heart rhythm. Mercury salts, which are basically industrial, if you breathe in or ingest them, gravitate more toward the kidney and not so much the nervous system. • The organic mercury is what gets into the food

chain. It's put into the water by chemical plants that are manufacturing things and they get into shellfish and fish, or elemental mercury that gets into the water is changed into organic mercury by sea life; we eat fish or shellfish and we get mercury exposure. That organic mercury acts very similarly to the elemental form. It affects a lot of nervous system damage. If a woman is pregnant, this can also cause birth defects and loss of the fetus if the levels get high enough.

Is mercury something we need in our diets, or is no amount nutritionally safe or necessary? No level is normal. Zero is normal. It doesn’t have a specific reason to be in our body. As long as we live on this Earth, because it's in Earth's crust and in the atmosphere, we're going to be exposed. But there is no specific function for that metal in our body. The issue is one of looking at the total body burden: How much mercury is in the body and what's known to be a normal background? Theoretically, there's going to be a baseline level, a general population average, but depending on where you live, that level may be higher or lower. If you live near a coast, you're more up to eating seafood. Or you may be in an industrial area where mercury is put into the water or the air.


Passage # (Q. No. 1 to Q. No. 3)

An ideal diatomic gas is expanded so that the amount of heat transferred to the gas is equal to the decrease in its internal energy.

1. The molar specific heat of the gas in this process is

given by C whose value is

(A) 2R5

− (B) 2R3

− (C) 2R (D) 2R5

2. The process can be represented by the equation TVn

= constant, where the value of n is

(A) 57n = (B)

51n = (C)

23n = (D)

53n =

3. If in the above process the initial temperature of the

gas be T0 and the final volume be 32 times the initial volume, the work done (in joules) by the gas during the process will be

(A) RT0 (B) 2

RT5 0 (C) 2RT0 (D) 2

RT0

4. For the photoelectric experiment, match the following column – I with column – II if frequency and intensity of incident photon are f and I while work function of metal is φ .

Column-I Column-II (A) If frequency is increased (P) Stopping potential keeping intensity and increases work function constant (B) If I is increased keeping (Q) Saturation f and φ are constant photocurrent increases (C) If φ is decreased keeping (R) Maximum kinetic f and I are constant energy of the photoelectrons increases (D) If the distance between (S) Stopping potential anode and cathode remains the same increases

(T) Maximum kinetic energy of the photoelectrons decreases

5. For the following circuit match the following columns at steady state.

X=2Ω3µF

R=3Ω

S=7Ω Y=4Ω

BD

A

C

6V

Column-I Column-II (A) Potential difference (in volts) (P) zero

across A and D

(B) Potential difference (in volts) (Q) 59

across capacitor

(C) Value of Y (in Ω ) for which (R) 51

no energy is stored in capacitor (D) Steady state current (in amp) (S) 14 in the branch containing capacitor

(T) 5

14 6. Magnetic flux linked with a stationary loop of

resistance R varies with time during time period T as follows: )tT(at −=φ

then the amount of heat generated in the loop during time T is (assume inductance of coil is negligible)

(A) R3

aT3 (B)

R3Ta 22

(C) RTa 32

(D) R3Ta 32

7. An ideal gas follows the process

,VTT 20 α+= where T0 and α are positive constant

and V is volume of one mole of gas. If this process the minimum pressure attained by gas is 0TR αβ then find the value of β .

8. A travelling wave is given by

)4t12xt12x3(

8.0y 22 +++= where x and y are m

and t is in sec then find the wave velocity in m/s.

This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety of possible twists and turns of problems in physics that would be very helpful in facing IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and we hope that this section would prove a rich resource for practicing challenging problems and enhancing the preparation level of IIT JEE aspirants.

By : Dev Sharma Director Academics, Jodhpur Branch

Physics Challenging Problems

Solut ions wil l be published in next issue

Set # 6


1. Path difference θ== cosdx

O

Screen

P

S1 S2 θ

Option [A,B,C,D] is correct

2. Conceptual

Option [A,C] is correct

3. )Rd(4

qVVV0

esargch.indqconductor +∈π=+=

Option [A,D] is correct 4. It temperature of gas at any time ‘t’ is T, then

l

)TT(kAq

dtdQ 0−

−=

Also dTR27ndT.C.ndQ P ××==

l

)TT(kAq

dtdT.

2nR7 0−

=∴

∫ ∫=−−

T

T

t

000

dtnR72

)TT(kAqdT

ll

Option [A] is correct

5. At maximum temperature Tmax.

)TT(kAq 0max −=l


6. max

max

0

0

TV

Tv

=


7. ∫ ∈=π×α==φ

→→

0

2 QR4R.Sd.E ,

Q = 3 × 10-10C So in terms of 10-10C Q = 3 8. )tT(at −=φ

)t2T(adtde −−=φ−

=

∫=T

0

2dt

ReH

Option [D] is correct

Solution Physics Challenging Problems

Set # 5

8 Questions were Published in September Issue

Physics Facts Magnetism

1. The direction of a magnetic field is defined by the direction a compass needle points. 2. Magnetic fields point from the north to the south outside the magnet and south to north inside the magnet. 3. Magnetic flux is measured in webers. 4. Left hands are for negative charges and right hands are for positive charges. 5. The first hand rule deals with the B-field around a current bearing wire, the third hand rule looks at the force on

charges moving in a B-field, and the second hand rule is redundant.

6. Solenoids are stronger with more current or more wire turns or adding a soft iron core.


1. In the circuit shown in fig. emf and internal resistance of battery are 6V and 0.5Ω respectively. Calculate charge on each capacitor in steady state.

2Ω

1µF

3Ω

2µF 3µF

3Ω 2Ω

Battery

+ –

Sol. In steady state no current flows through capacitors.

But due to circuit formed by resistors, a current is drawn from the battery. To calculate current through resistors in steady state, circuit may be analysed after removing capacitors. Then the circuit will be as shown in fig.

2Ω 3Ω

3Ω 2Ω

6V

+ –

0.5ΩI1

I2

(I1 – I2) (I1 – I2)I2

I1

Applying Kirchhoff's voltage law on lower mesh, 3I2 + 2I2 + 0.5I1 – 6 = 0 ...(1) Now applying KVL on upper mesh, 2(I1 – I2) + 3(I1 – I2) – 2I2 – 3I2 = 0 ...(2) From equations (1) and (2), I1 = 2A, I2 = 1A Let, in steady state, charges on capacitors of

capacitance 1 µF, 2 µF and 3 µF be q1, q2 and q3 respectively as shown in fig.

2Ω

1µF

3Ω

2µF 3µF

3Ω 2Ω

A

+ –

B C D E

F G H I J

(I1 – I2) (I1 – I2)

0.5Ω6V

I1

I2 I2 I2 I2

I1

+ – q1

+ – q2

+ – q3

Applying KVL on mesh ABIJA, + 61

101q

−× = 0

or q1 = 0 Ans. Now applying KVL on mesh BCHIB,

2(I1 – I2) + 62

102q

−× – 3I2 – 6

1

10q

− = 0,

q2 = 2 × 10–6 C or 2µC Ans. Now applying KVL on mesh DEFGD,

– 63

103q

−× = 0 or q3 = 0 Ans.

2. System shown in figure consists of two large parallel metallic plates carrying current in opposite directions. Current density in each plate is j per unit width. Calculate

× × ×

(i) Magnetic induction in space between the plates and

(ii) force acting per unit area of each plate. Sol. If a large plate carries a current which is uniformly

distributed over its width, then a uniform magnetic field is established around it.

If a section of plate, which is normal to the direction of flow of current, is considered then it will be as shown in figure (A)

× × × ×

P Q B

R SB

l

Fig. (A) Let magnetic induction of the field induced due to

current in one plate be B. Considering a length l in the section as shown in

figure(A) and applying Amperes's Circuital Law.

B.2l = m(lj) or B = 21

µ0j

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

Students' ForumPHYSICS


But there are two plates which carry equal current but in opposite directions. Therefore, magnetic fields due to these currents, in the space between the plates are unidirectional.

∴ Resultant magnetic field induction between the plates = 2B = µ0j Ans. (i)

Now consider an elemental width dx in the section of upper plate as shown in Figure(B). This elemental width is similar to a long straight conductor carrying current di = j dx

× × ×

dx

Fig. (B) Magnetic induction at this conductor due to current

in lower plate is B = 21

µ0j (leftward)

Hence, force on this conductor, dF = B di per unit length

or dF = 21

µ0j2 dx

per unit length But area of unit length of the conductor considered

= 1 . dx = dx

∴ Force per unit area of upper plate = dxdF

= 21

µ0j2 Ans. (ii)

3. A long straight wire is coplanar with a current carrying circular loop of radius R as shown in figure. Current flowing through wire and the loop is I0 and I respectively. If distance between centre of loop and wire is r = 2 R, calculate force of attraction between the wire and the loop.

I

r I0

Sol. Assuming that diameter of loop, which is perpendicular to the straight wire to be x-axis.

Considering two equal elemental are lengths Rdθ each of the loop at angles θ with negative x-axis as show in figure

Rdθ

2R

I0

Rdθ

θ θ x

dF'

dF'

Distance of each elemental length from straight wire is x = (r – R cos θ)

or x = R(2 – cos θ) ∴ Magnetic induction, due to current I0 through

straight wire, at positions of these elemental lengths is

B = x2I00

πµ =

)cos–2(R2I00

απµ

(along inward normal to this paper) Magnitude of force on each elemental length,

dF' = BI (R dθ) = )cos–2(2

IdI00

θπθµ

According to Fleming's left hand rule, directions of these two forces will be as shown in Figure.

Since, these forces are equally inclined with x-axis in opposite directions and have equal magnitude, therefore, their components normal to x-axis neutralise each other. Hence, their resultant is along negative x-axis or towards the straight wire.

This resultant force,

dF = dF'.2 cos θ =

θθ

πµ

cos–2cosII00 dx

∴ Net force on the loop,

F = π

µ II00 ∫π

θθ

0)cos–2(

cos dθ = µ0II0

33–2 Ans.

4. A plane wave propagates along positive x-direction in

a homogeneous medium of density ρ = 200 kg m–3. Due to propagation of the wave medium particles oscillate. Space density of their oscillation energy is E = 0.16 π2 Jm–3 and maximum shear strain produced in the medium is φ0 = 8π × 10–5. If at an instant phase difference between two particles located at point (1m, 1m, 1m) and (2m, 2m, 2m) is ∆θ = 144º, assuming at t = 0 phase of particles at x = 0 to be zero,

(i) calculate equation of the wave and (ii) state whether the wave is longitudinal or

transverse.


Sol. Since, the wave is a place travelling wave, therefore, intensity at every point will be the same.

Since, initial phase of particles at x = 0 is zero and the wave is travelling along positive x-direction, therefore, equation of the wave will be of the form

δ = α sin ω

vx–t ...(i)

Let intensity of the wave be I, then space density of oscillation energy of medium particles will be equal

to E = v1 .

But I = 2π2n2a2ρv, therefore, E = 2π2n2a2ρ = Jm–3 or a2n2 = 4 × 10–4 or an = 0.02 ...(ii)

Shear strain of the medium is φ = dxd .δ

Differentiating equation (i),

φ = – v

aω .cos ω

vx–t

Modulus of shear strain φ will be maximum when

coa ω

vx–t = ± 1

∴ Maximum shear strain φ0 = v

aω , but it is equal

to 8π × 10–5

∴ v

aω = 8π × 10–5 where ω = 2πn

∴ an = 4v × 10–5 ...(iii) Solving equation (ii) and (iii), v = 500 ms–1 Since, the wave is travelling along positive

x-direction, therefore, phase difference between particles at points (1m, 1m, 1m) and (2m, 2m, 2m) is due to difference between their x coordinates only.

The phase difference is given by ∆θ = 2π. λ

∆x

where ∆x = (x2 – x1) = (2 – 1) m = 1m

∴ λ = θ∆∆π x.2 = 2.5 m

But v = nλ, therefore, n = λv = 200 Hz

Substituting n = 200 Hz in equation (ii), a = 1 × 10–4 m Angular frequency, ω = 2πn = 400π rad/sec Substituting all these values in equation (i) δ = 10–4 sin π (400t – 0.8 x) m Ans. (i) Since, due to propagation of the wave, shear strain is

produced in the medium, therefore, the wave is a place transverse wave.

5. A hollow sphere of mass m = 510 gm is made of a material having specific heat capacity s = 450 J Kg–1 K–1. Inner and outer surfaces of the sphere have area a = 1 × 10–3 m2 and A = 6.931 × 10–3 m2 respectively. Sphere is suspended by a light non-conducting thread from ceiling of a room. Assuming sphere material to be highly conducting and its emissivity e = 0.80, calculate time taken by the sphere to cool from θ1 = 37º C to θ2 = 32ºC when room temperature, θ0 = 27ºC.

(Take, Stefan's constant σ = 3

17 × 10–8 Wm–2 K– 4).

Sol. Since, sphere is hollow, therefore, it has two surface (i) outer surface of area A, which is exposed to

atmosphere and (ii) inner surface of area a, which is not exposed to the atmosphere. Hence, no radiation takes place from inner surface or in other words, heat radiates from outer surface only.

Let at some instant t, difference between temperature of sphere and room be 'θ'.

Then net rate of radiation from surface of the sphere, E = eAσ[(T0 + θ)4 – T0

4] where T0 (273 + θ9)

or E = eAσT04

θ+ 1–

T1

4

0

But temperature difference θ varies from (θ1 – θ2) to (θ2 – θ0) which is always very small in comparison to

T0, therefore, higher powers of

θ

0T are negligible.

∴ E = eAσT04

θ+ 1–

T41

0 = 4 eAσT0

3. θ

Thermal capacity of the sphere, C = ms ∴ Rate of decrease of temperature difference is

θ

dtd– =

CE =

msTeA4 3

0σ . θ

or θθd = –

msTeA4 3

0σ . dt ...(i)

But at time t = 0, θ = (θ1 – θ0) and when θ = (θ2 – θ0), t = ?

Integrating equation (i) with these limits,

∫θθ

θθθθ

)–(

)–(

02

01

d = – msATe4 3

0σ∫t

0

dt

t = 30ATe4

msσ

.loge

θθθθ

02

01

–– = 4687.5 sec Ans.


Magnetic field : A magnetic field of strength B is said to exist at a

point if a current element or a moving charged particle passing through the point experiences a sideways force equal in magnitude to

∆F = I∆ lB sin θ or qvB sin θ

∆ I

θ B

v

θ B

+q where ∆l is the length of the current element, q is the

charge moving with velocity v, and θ is the angle between the direction of B and the current element, or between B and v, 0 < θ < π. The direction of the force ∆F is always perpendicular to the plane containing ∆l and B, or v and B. In the figures, this would mean the plane of the paper. The sense of ∆F is that in which a screw would move if rotated from ∆l or v to B through θ. In this case, this would mean a clockwise rotation, causing ∆F to be directed into the paper. In vector notation, this is summarized as

∆ Fr

= I(∆ lr

× Br

) or q( vr

× Br

) The unit of B is tesla (T) or newton per ampere metre

or weber per square metre. B is called the magnetic induction.

Biot-Savart Law : A moving charge or any current element give rise to a magnetic field. This is given by

(∆B)p = 20

rsinI

4µ θ∆

πl

P

rθ

I

∆l

where (∆B)p is the contribution of ∆l to the magnetic

field at P, while µ0 is a universal magnetic constant with the value µ0 = 4π × 10–7 weber/ampere metre or henry per metre. The direction of (∆B)P is perpendicular to the plane containing ∆l and r, with the same sense as the motion of a screw which is rotated from ∆ l

r towards r

r through the smaller

angle.

Magnetic field at the centre of a Circular Coil :

B = a2NIµ0 tesla (T)

where, a = radius of the coil, N = its number of turns, I = current. The direction of B is along the axis of the coil. Magnetic Field a Point on the Axis of a Coil :

B = 2/322

20

)xa(2NIaµ+

tesla (T)

where x is the distance of the point from its centre. Magnetic Field due to a Straight Conductor at a

Point :

B = dI

4µ0

π(sin θ1 + sin θ2)

where d is the perpendicular distance of the point from the conductor, θ1 and θ2 are the angles subtended by the upper and lower portions of the conductor at the point.

When the conductor is long B = dI2

4µ0

π

Magnetic Moment of a Loop : Magnetic moment of a current loop(m) = IS (current × area)

or m = IS ampere metre2 Torque on a current Loop τ = m B sin θ where θ is

the angle between normal to the loop and the magnetic field.

Energy of a Current Loop in a Magnetic Field U = Uθ = 0 + mB(1 – cos θ) Work Done in Turning a Current Loop W = mB(1 – cos θ)

Problem Solving Strategy : Magnetic Forces : Step 1 : Identify the relevant concepts : The right-

hand rule allows you to determine the magnetic force on a moving charged particle.

Step 2 : Set up the problem using the following steps :

Draw the velocity vector vr

and magnetic field Br

with their tails together so that you can visualize the plane in which these two vector lie.

Identify the angle φ between the two vectors.

Magnetic effects of current

PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


Identity the target variables. This may be the magnitude and direction of the force, or it may be the magnitude or direction of v

r or B

r.

Step 3 : Execute the solution as follows :

Express the magnetic force using Eq. BvqFrrr

×=

The magnitude of the force is given by Eq. F = qvB sin φ.

Remember that Fr

is perpendicular to the plane of the vectors v

r and B

r. The direction of Bv

rr× is

determined by the right-hand rule; keep referring to until you're sure you understand this rule. If q is negative, the force is opposite to Bv

rr× .

Step 4 : Evaluate your answer : Whenever you can, solve the problem in two ways. Do it directly from the geometric definition of the vector product. Then find the components of the vectors in some convenient axis system and calculate the vector product algebraically from the components. Verify that the results agree.

Problem Solving Strategy : Motion in Magnetic Fields : Step 1 : Identify the relevant concepts : In analyzing

the motion of a charged particle in electric and magnetic fields, you will apply Newton's second law of motion, Σ F

r= m a

r, with the net force given by

Fr

Σ = q )BvE(rrr

×+ . Often, other forces such as gravity can be neglected.

Step 2 : Setup the problem using the following steps : Determine the target variable(s). Often the use of components is the most efficient

approach. Choose a coordinate system and then express all vector quantities (including E

r, B

r, v

r,

Fr

and ar

) in terms of their components in this system.

Step 3 : Execute the solutions as follows : If the particle moves perpendicular to a uniform

magnetic field, the trajectory is circle with a radius and angular speed given by Eqs.

R = B|q|

mv and

ω = Rv = v

mvB|q| =

mB|q|

If your calculation involves a more complex trajectory, use F

rΣ = m a

r in component form:

ΣFx = max, and so forth. This approach is particularly useful when both electric and magnetic fields are present.

Step 4 : Evaluate your answer : Check whether your results are reasonable.

Problem Solving Strategy : Magnetic Field Calculations : Step 1 : Identify the relevant concepts : The law of

Biot and Savart allows you to calculate the magnetic field due to a current –carrying wire of any shape. The idea is to calculate the field due to a representative current element in the wire, then combine the contributions from all such elements to find the total field.

Step 2 : Setup the problem using the following steps : Make a diagram showing a representative current

element and the point P at which the field is to be determined (the field point).

Draw the current element lr

d , being careful to ensure that it points in the direction of the current.

Draw the unit vector r . Note that it is always directed from the current element (the source point) to the field point P.

Identify the target variables. Usually they will be the magnitude and direction of the magnetic field Br

. Step 3 : Execute the solution as follows :

Use eq. dB = 20

rsindI

4µ φ

πl or Bd

r= 2

0

rrdI

4µ ×

πlr

to express the magnetic field Bdr

at P from the representative current element.

Add up all the Bdr

's to find the total field at point P. In some situations the Bd

r's at point P have the

same direction for all the current elements; then the magnitude of the total B

r field is the sum of

the magnitudes of the Bdr

's. But often the Bdr

's have different direction for different current elements. Then you have to set up a coordinate system and represent each Bd

r in terms of its

components. The integrals for the total Br

is then expressed in terms of an integral for each component.

Sometimes you can use the symmetry of the situation to prove that one component of B

r must

vanish. Always be alert for ways to use symmetry to simplify the problem.

Look for ways to use the principle of superposition of magnetic fields. Later in this chapter we'll determine the fields produced by certain simple conductor shapes; if you encounter a conductor of a complex shape that can be represented as a combination of these simple shapes, you can use superposition to find the field of the complex shape. Examples include a rectangular loop and a semicircle with straight line segments on both sides.


Step 4 : Evaluate your answer : Often your answer will be a mathematical expression for B

r as a

function of the position of the field point. Check the answer by examining its behavior in as many limits as you can.

1. Two long wires a distance 2d apart carry equal antiparallel currents i, as shown in the figure. Calculate the magnetic induction at a point P equidistant from the wires at a distance D from a point midway between the wires.

A

D P 0

B

2d

Sol. The point is at a distance 22 dD + from each wire.

∴ magnitude of field due each = 22

0

dD

i24µ

+π

The direction of the field due to A is at right angles to AP and that due to B is at right angles to BP.

Resolving the field along OP and perpendicular to it, the normal components cancel out and the components along OP are added.

A

D P0

B

0

B sinθ B

θ

θ B cosθ B cosθ

B B sinθ

d

d

∴ B´ (field) at P = 2B cos θ along OP

= 2 × π4

µ022 dD

i2

+ 22 dD

d

+ =

)dD(idµ

220

+π

2. An alpha particle travels in a circular path of radius

0.45 m in a magnetic field with B = 1.2 Wb m–2. Calculate (a) its speed, (b) its kinetic energy, and (c) the potential difference through which if would have to be associated to achieve this energy. Mass of alpha particle = 6.64 × 10–27 kg.

Sol. (a) Bqv = mr2/r ⇒ v = Bqr/m

or v = m

Bqr = 27

19

1064.645.0102.32.1

−

−

××××

= 2.6 × 107 ms–1 (b) Bqv = mv2/r

⇒ Bq = mv/r = rmE2

= 2mv

21EQ

or E = m2

rqB 222= 27

22192

1064.6245.0)102.3(2.1

−

−

×××××

= 2.25 × 10–12 J (c) E(energy acquired) = Vq

or V = qE = 19

12

102.31025.2

−

−

×× ⇒ V = 7.0 × 106 V

3. Use Biot –Savart law to calculate the magnetic field

B at the common centre of the following circuits.

r2 r1

b

c

0

i

i

a

d

θ

r1

r2

i

i

d a 0 b c

Sol. The field due to the straight portions da and bc is zero

as the centre O is at end-on position relative to them. The field due to the curved parts are opposite as can seen by the screw rule. To find the magnitude due to either conductor, consider an element of width dl at angular distance α, from the radius Od.

b

c

0

i

a

d

α

dα

dl i

Then dB1 = 21

0

rº90sinid

4µ l

π perpendicular into the

plane of the paper

⇒ dB1 = 21

10

r)dr(i

4µ α

π (Q dl = r1dα)

⇒ B1 = ∫θ

απ 01

0 dri

4µ =

1

0

r4iµ

πθ

Similarly, B2, field at O due to cd = 2

0

r4iµ

πθ

out of the

plane of the paper.

∴ B, field due to the loop abcd =

−

πθ

21

0

r1

r1

4iµ

perpendicular into the plane. The second circuit is a special case of the above

when θ = π

∴ B =

−

210 r

1r1iµ

41

Solved Examples


4. A wire ring whose radius is 4 cm is at right angles to the general direction of a radically symmetrical diverging magnetic field as shown in the figure. The flux density in the region occupied by the wire itself is 0.1 Wb m–2 and the direction of the field everywhere is at an angle of 60º with the plane of the ring. Find the magnitude and direction of the force on the ring when the current in it is 15.9 A.

θ 0

Sol. Let us resolve the field along and perpendicular to the

axis of the ring. The resolved parts are B sin θ and B cos θ. The forces on the elements of the ring due to the 'Bsinθ' component are in the plane of the ring and are distributed symmetrically towards the centre all along the ring, so they sum up to zero. But the forces on the elements due to 'Bcosθ' component are along the normal to the ring, hence they sum up to a resultant along that direction.

∴ F = ΣI∆lB cos θ = BIl cos θ = BI2πR cos θ (∴ l = 2πR) or F = 2πBIR cos θ = 2π × 0.1 × 15.9 × (4 × 10–2) cos 60º = 0.2N 5. A long straight conductor carrying I1, is placed in the

plane of a ribbon carrying current I2 parallel to the previous one. The width of the ribbon is b and the straight conductor is at a distance 'a' from the near edge. Find the force of attraction between the two.

a I2

b I1

Sol. Consider a thin strip at a distance x and of thickness

dx. It is equivalent to a long straight conductor carrying (I2dx/b) current.

dF (force of attraction) = x2Iµ 10

π×

bdxI2

= b2IIµ 210

π×

xdx

∴ F = b2IIµ 210

π ∫+ba

x xdx

= b2IIµ 210

πln

bba +

1. Barium compounds are the source for the

different greens in fireworks. 2. There are 60,000 miles (97,000 km) in blood

vessels in every human. 3. The average person produces about 400 to

500 ml of cerebrospinal fluid every day. 4. Ernest Rutherford discovered that the atom

had a nucleus in 1911. 5. Impacts by comets or asteroids can also

generate giant tsunamis. 6. Basic surgery would cure 80% of the over

45 million blind people in the world. Sixty percent of whom live in sub-Saharan Africa, China and India.

7. Studies have confirmed that ginkgo

increases blood flow to the retina, and can slow retinal deterioration resulting in an increase of visual acuity. In clinical tests ginkgo has improved hearing loss in the elderly. It also improves circulation in the extremities relieving cold hands and feet, swelling in the limbs and chronic arterial blockage.

8. Venus may well once have had water like

Earth does, but because of the scorching surface temperature of 482 degrees C (900 degrees F). Any sign of it has long ago evaporated.

9. About 95 percent of every edible fat or oil

consists of fatty acids. Fatty acids all are based on carbon chains - carbon atoms linked together one after another in a single molecule. Different fatty acids are defined as saturated, monounsaturated, or polyunsaturated depending on how effectively hydrogen atoms have linked onto those carbon chains.

10. On average women cry 5.3 times a month.

Men only 1.4. 11. The Medal of Honor is the highest award for

valour in action against an enemy force which can be bestowed upon an individual serving in the Armed Services of the United States.


C Newton's Law of Gravitation : Two point masses m1 and m2, separated by' a distance

r, attract each other with a force

F = G 221

rmm

where G = 6.67 × 10–11 Nm2 kg–2 = universal constant of gravitation.

m1 F F m2

r This force between two masses acts equally on both

masses, acts though in opposite directions. It does not depend on the medium present between the two masses.

Gravitation Field : This is a region in space where any mass will

experience a force. The gravitational field strength (g) at a point is the force acting on a unit mass placed at that point. It is a vector.

m g

r r Any mass sets up a gravitational field around it. The

gravitational field strength at a distance r from a point mass m is

g = –G 2rm

The negative sign indicates that the gravitational field is always attractive.

Gravitational Potential : The gravitational potential (V) at a points is the work

that has to be done to bring a unit mass from infinity to that point. It is a scalar. The gravitational potential at a distance r from a point mass m is

V = – Grm

The negative sign arises because in bringing the unit mass from infinity, work is done by the system, so that its potential energy decreases.

The potential at a point does not depend on the actual path followed in bringing the unit mass from infinity. Thus, gravitational force is a conservative force.

Gravitational field (g) and potential (V) due to a spherical shell and a solid sphere :

R

M

rR

M

r

(i) Outside (i) Outside

g = – G 2rM g = – G 2r

M

gr

= – G 2rM r g

r = – G r

rM

2

V = – Gr

M V = – Gr

M

(ii) Inside (ii) Inside g = 0

g = – G 3RM r

where R is the radius of the sphere

V = – GRM

where R is the radius of the shell

V = – G 3R2M (3R2 – r2)

Escape Velocity : The minimum velocity to be imparted to a body on

the surface of a planet, so that it is carried beyond the gravitational field of that planet, is called the escape velocity of that planet. Obviously to carry the body beyond the gravitational field, the amount of energy needed is that which is required to bring it from infinity up to the surface of the planet. This is exactly the potential energy of the body. Potential energy per unit mass is equal to the potential of the field. So if m is the mass of the body and vc is the escape velocity, then

2emv

21 = Vm = G

RM m

where M is the mass of the planet and R is its radius.

or ve = RGM2

Gravitation

PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


If g is the gravitational field intensity, then

mg = 2RGM2 or GM = gR2

∴ ve = gR2

Satellites and Orbital speed (V0) : A satellite is a small body revolving around a larger

body under the gravitational attraction of the latter. The force of gravitational attraction provides the necessary centripetal force so that the satellite may be in rotational equilibrium. The speed at which rotational equilibrium is attained is called the orbital speed. Let it be v0. Then for rotational equilibrium

Fattraction = r

mv20

where r is the radius of the orbit, measured from the centre of the planet.

G 2rMm =

rmv2

0

or v0 = r

GM

If a satellite is projected with velocity v < v0 the path is a small ellipse with point of projection as apogee and if v0 < v < 2 v0 the path is a bigger ellipse with point of projection as perigee. If v = 2 v0 the path is parabolic and if v > 2 v0 the path is hyperbolic.

1. Two masses M1 and M2 at an infinite distance from

each other and initially at rest, start interacting gravitationally. Find their velocity of approach when they are a distance s apart.

Sol. Since they move under mutual attraction and no external force acts on them, their momentum and energy are conserved.

∴ 0 = 21 M1v1

2 + 21 M2v2

2 – sMGM 21

(It is zero because in the beginning, both kinetic energy and potential energy are zero.)

0 = M1v1 + M2v2

Solving the equations v12 =

)MM(sGM2

21

22

+

and v22 =

)MM(sGM2

21

22

+

V(velocity of approach) = v1 – (–v2) = v1 + v2

= s

)MM(G2 21 +

2. A planet of mass m moves along a circle around the sun of mass ms with velocity v = 34.9 kms–1 with respect to the heliocentric frame of reference, that is, with the sun at the centre of the frame. Find the period of revolution of this planet around the sun and show that Kepler's third law, that is, the cube of the orbital radius is proportional to the square of the time revolution of planets, Given that ms = 1.97 × 1030 kg, G = 6.67 × 10–11 units.

Sol. From the dynamics of circular motion (assuming circular orbit)

mdv2

= Gm 2dm ⇒ v2 =

dGms

⇒ v = ωd

∴ v2 = v

Gmsω or ω = s

3

Gmv

or T = 3s

vGm2π

= 33

3011

)109.34(1097.11067.62

×××××π −

= 225 days

Taking the equation in terms of d and ω

2s

dGmm

= mω2d ⇒ T2 = s

2

Gm4π d3

∴ T2 ∝ d3 This is Kepler's third law. 3. An artificial satellite (of mass m) of the earth (radius

R and mass M) moves in an orbit whose radius is n times the radius of the earth. Assuming resistance to the motion to be proportional to the square of velocity, that is F = av2, find how long the satellite will take to fall on to the earth.

Sol. E (energy of the satellite) = – r

GMm + 21 mv2

By the dynamics of circular motion

2rGMm =

rmv2

⇒ v = r

GM

∴ E = –r

GMm + 21

rGMm = –

21

rGMm

⇒ dE = 2rGMm

21 dr

Also – dE = power × dt = Fv dt = av3dt

∴ 2rGMm

21 dr = –a

2/3

rGM

dt

⇒ dt = –GMa2

m r–1/2 dr

∴ t = – GMa2

m∫ −R

nR

2/1 drr = GMa

Rm ( n – 1)

Q GM = gR2, t = gRa

m ( n – 1)

Solved Examples


4. A spaceship approaches the moon (mass = M and radius = R) along a parabolic path which is almost tangential to its surface. At the moment of maximum approach, the brake rocket is fired to convert the spaceship into a satellite of the moon. Find the change in speed.

Sol. If v is the velocity at the vertex of the parabola, then v is also the escape velocity because if it is thrown with this velocity it will follow the parabolic path never to return to the moon.

M

v v orbit

Now vescape = RGM2

∆v = vfinal – vinitial = vorbit – vescape

⇒ ∆v = R

GM – RGM2 = –

RGM ( 2 – 1)

The negative sign means the speed has to be decreased.

∴ required change in speed = R

GM ( 2 – 1)

5. A satellite is revolving in a circular equatorial orbit of radius R = 2 × 104 km from east to west. Calculate the interval after which it will appear at the same equatorial town. Given that the radius of the earth = 6400 km and g (acceleration due to gravity) = 10 ms–2.

Sol. Let ω be the actual angular velocity of the satellite from east to west and ωc be the angular speed of the earth (west to east).

Then ωrelative = ω –(–ωe) = ω + ωe ⇒ ω = ωrel – ωe By the dynamics of circular motion

2RGMm = mω2r or ω2 = 3

2

RgR (Q GM= gRe

2)

⇒ ω = 3

2e

RgR ∴ ωrel = 3

2e

RgR + ωe

⇒ ωrel = 213

122

102104.610

××× + 7.27 × 10–5

(Q ωe = 86400

2π = 7.27 × 10–5)

⇒ ωrel = 22.6 × 10–5 + 7.27 × 10–5 = 30 × 10–5 rad s–1

∴ τ = rel

2ω

π = 510302

−×π

= 2.09 × 104s = 5 hr 48 min

Brief description: cobalt is a brittle, hard, transition metal with magnetic properties similar to those of iron. Cobalt is present in meteorites. Ore deposits are found in Zaire, Morocco and Canada. Cobalt-60 (60Co) is an artificially produced isotope used as a source of γ rays (high energy radiation). Cobalt salts colour glass a beautiful deep blue colour.

Basic information about and classifications of cobalt :

Name : Cobalt Symbol : Co Atomic number : 27 Atomic weight : 58.933195 (5) Standard state : solid at 298 K Group in periodic table : 9 Group name : (none) Period in periodic table : 4 Block in periodic table: d-block Colour : lustrous, metallic, greyish tinge Classification : Metallic Marmite, which we all eat here in England and which is what makes us English, is a source of vitamin B12, actually a compound containing cobalt. The equivalent, but altogether blander, in Australia is Vegemite. Marmite is available in the USA. Try mixing it with peanut butter.

This sample is from The Elements Collection, an attractive and safely packaged collection of the 92 naturally occurring elements that is available for sale.

ISOLATION :

Isolation: it is not normally necessary to make cobalt in the laboratory as it is available readily commercially. Many ores contain cobalt but not many are of economic importance. These include the sulphides and arsenides linnaeite, Co3S4, cobaltite, CoAsS, and smaltite, CoAs2. Industrially, however, it is normally produced as a byproduct from the produstion of copper, nickel, and lead.


Organic Chemistry

Fundamentals

Structure of aryl and vinyl halides : The low reactivity of aryl and vinyl halides towards

displacement has, like the stabilities of alkenes and dienes, been attributed to two different factors : (a) delocalization of electrons by resonance; and (b) differences in (σ) bond energies due to differences in hybridization of carbon.

Let us look first at the resonance interpretation. Chlorobenzene is considered to be a hybrid of not

only the two Kekule structures I and II, but also of three structures, III, IV, and V, in which chlorine is joined to carbon by a double bond ; in III, IV, and V chlorine bears a positive charge and the ortho and para positions of the ring bear a negative charge.

:Cl:

I

. . :Cl: . .

Cl: . .

⊕H

:Θ

Cl: . .

⊕

H :Θ

Cl:. .

⊕

H Θ:

II III IV V In a similar way, vinyl chloride is considered to be a

hybrid of structure VI and structure VII, in which chlorine is joined to carbon by a double bond ; in VII chlorine bears a positive charge and C–2 bears a negative charge. Other aryl and vinyl halides are considered to have structures exactly analogous to these.

H : C : : C : Cl : H .. H .. 2 1

VI

H : C : C : : Cl :H .. H .. 2 1

VII

.. ..

Θ ..

⊕..

Contribution from III, IV, and V, and from VII

stabilizes the chlorobenzene and vinyl chloride molecules, and gives double-bond character to the carbon-chlorine bond. Carbon and chlorine are thus held together by something more than a single pair of electrons, and the carbon-chlorine bond is stronger than if it were a pure single bond. The low reactivity of these halides toward nucleophilic substitution is due (partly, at least) to resonance stabilization of the halides (by a factor that in this case does not stabilize the transition state to the same extent); this stabilization increase the Eact for displacement, and thus slows down reaction. For aryl halides, another factor – which may will be the most important one–

is stabilization of the molecule by resonance involving the Kekule structures.

The alternative interpretation is simple. In alkyl halides the carbon holding halogen is sp3-hybridized. In aryl and vinyl halides, carbon is sp2-hybridized; the bond to halogen is shorter and stronger, and the molecule is more stable.

The carbon-halogen bonds of aryl and vinyl halides are unusually short. In chlorobenzene and vinyl chloride the C–C bond length is only 1.69 Å, as compared with a length of 1.77–1.80 Å in a large number of alkyl chlorides In bromobenzene and vinyl bromide the C–Br bond length is only 1.86 Å, as compared with a length of 1.91 – 1.92 Å in alkyl bromides.

Now, a double bond is shorter than a single bond joining the same pair of atoms; if the carbon–halogen bond in aryl and vinyl halides has double-bond character, it should be shorter than the carbon–halogen bond in alkyl halides. Alternatively, a bond formed by overlap of an sp2 orbital should be shorter than the corresponding bond involving an sp3 orbital.

Dipole moments of aryl and vinyl halides are unusually small. Organic halogen compounds are polar molecules; displacement of electrons towards the more electronegative element makes halogen relatively negative and carbon relatively positive. The dipole moments of a number of alkyl chlorides and bromides range from 2.02 D to 2.15 D. The mobile π electrons of the benzene ring and of the carbon –carbon double bond should be particularly easy to displace; hence we might have expected aryl and vinyl halides to have even larger dipole moments then alkyl halides.

Chlorobenzene and bromobenzene have dipole moments of only 1.7 D, and vinyl chloride and vinyl bromide have dipole moments of only 1.4 D. This is consistent with the resonance picture of these molecules. In the structures that contain doubly bonded halogen (III, IV, V, and VII) there is a positive charge on halogen and a negative charge on carbon; to the extent that these structures contribute to the hybrids, they tend to oppose the usual displacement of electrons toward halogen. Although there is still a net displacement of electrons towards halogen in aryl halides and vinyl halides, it is less than in other organic halides.

HALOGEN

DERIVATIVES


Elimination-addition mechanism for nucleophilic aromatic substitution. Benzyne : When an aryl halide like chlorobenzene is treated

with the very strong basic amide ion, NH2–, in liquid

ammonia, it is converted into aniline. This is not the simple displacement that, on the surface, it appears to be. Instead, the reaction involves two stages : elimination and then addition. The intermediate is the molecule called benzyne (or dehydrobenzene).

X

Aryl halide

NH2–

NH3

Benzyne

NH2–

NH3

Aniline

NH2

Benzyne has the structure shown in fig. in which an additional bond is formed between two carbons (the one originally holding the halogen and the one originally holding the hydrogen) by sideways overlap of sp2 orbitals. This new bond orbital lies along the side of the ring, and has little interaction with the π cloud lying above and below the ring. The sideways overlap is not very good, the new bond is a weak one, and benzyne is a highly reactive molecule.

H

H

H

H

Benzyne molecule. The sideways overlap of sp2 orbitals form a π bond out of the plane of the aromatic π cloud.

The elimination stage, in which benzyne is formed, involves two steps : abstraction of a hydrogen ion (step 1) by the amide ion to form ammonia and carbanion I, which then loses halide ion (step 2) to form benzyne.

(1)

X + NH2

–

I

X+ NH3

:–

(2)

X

I

+ X–

:– Benzyne

Elimination

The addition stage, in which benzyne is consumed, may also involve two steps : attachment of the amide ion (step 3) to form carbanion II, which then reacts with an acid, ammmonia, to abstract a hydrogen ion (step 4). It may be that step (3) and step (4) are

concerted, and addition involves a single step; if this is so, the transition state is probably one in which attachment of nitrogen has proceeded to a greater extent attachment of hydrogen, so that it has considerable carbanion character.

(3)

Benzyne

+ NH2–

II

NH2

:–

(4)

NH2

II

+ NH2–

:–Aniline

+ NH3

NH2

Addition

Some facts on which the above mechanism is based. (a) Fact. Labeled chlorobenzene in which 14C held the

chlorine atom was allowed to react with amide ion. In half the aniline obtained the amino group was held by 14C and in half it was held by an adjacent carbon.

NH2–*

(47%)NH3

Cl*

NH2 * NH2

+

(53%) Interpretation : In benzyne the labeled carbon and

the ones next to it become equivalent, and NH2– adds randomly (except for a small isotope effect) to one or the other.

NH2–*

NH3

Cl

*

*NH2

*

NH2–

NH3

NH2–

NH3

* NH2

*H2N

(b) Fact. Compounds containing two groups ortho to

halogen like 2-bromo-3-methyl anisole, do not react at all.

CH3No reaction

NH2–

BrCH3O

NH3

Interpretation : With no ortho hydrogen to be lost, benzyne cannot form.


Inorganic Chemistry

Fundamentals

Ionization Energy : The ionization energies of the halogens show the

usual trend to smaller values as the atoms increase in size. The values are very high, and there is little tendency for the atoms to lose electrons and form positive ions.

Ionization and hydration energies, electron affinity First ionization

energy (kJ mol–1)

Electron affinity

(kJ mol–1)

Hydration energy X–

(kJ mol–1)

F 1681 – 333 – 513

Cl 1256 – 349 – 370

Br 1143 – 325 – 339

I 1009 – 296 – 274

At – – 270 –

The ionization energy for F is appreciably higher than for the others because of its small size. F always has an oxidation state of (–1) except in F2. It forms compounds either by gaining an electron to form F–, or by sharing an electron to form a covalent bond.

Hydrogen has an ionization energy of 1311 kJ mol–1, and it forms H+ ions. It is at first surprising that the halogens Cl, Br and I have lower ionization energies than H, yet they do not form simple X+ ions. The ionization energy is the energy required to produce an ion from a single isolated gaseous atom. Usually we have a crystalline solid, or a solution, so the lattice energy or hydration energy must also be considered. Because H+ is very small , crystals containing H+ have a high lattice energy, and in solution the hydration energy is also very high (1091 kJ mol–1). The negative ions also have a hydration energy. Thus H+ ions are formed because the lattice energy, or the hydration energy, exceeds the ionization energy. In contrast the halide ions X+ would be large and thus have low hydration and lattice energies. Since the ionization energy would be larger than the lattice energy or hydration energy, these ions are not normally formed. However, a few compounds are know where I+ is stabilized by forming a complex with a Lewis base. for example [I(pyridine)2]+ NO3

–. The electron affinities for the halogens are all

negative. This shows that energy is evolved when a

halogen atom gains an electron, and X → X–. Thus, the halogen all form halide ions.

Bond energy in X2 Molecule : The elements all form diatomic molecules. It would

be expected that the bond energy in the X2 molecules would decrease as the atoms become larger, since increased size results in less effective overlap of orbitals. Cl2, Br2 and I2 show the expected trend (table) but the bond energy for F2 does not fit the expected trend.

Bond energy and bond lengths of X2 Bond energy (free

energy of dissociation) (kJ mol–1)

Bond length X2 (Å)

F 126 1.43

Cl 210 1.99

Br 158 2.28

I 118 2.66

The bond energy in F2 is abnormally low (126 kJmol–1), and this is largely responsible for its very high reactivity. (Other elements in the first row of the periodic table also have weaker bonds than the elements which follow in their respective groups. For example in Group 15 the N – N bond in hydrazine is weaker than P – P, and in Group 16 the O – O bond in peroxides is weaker than S – S.) Two different explanation have been suggested for the low bond energy :

(1) Mulliken postulated that in Cl2, Br2 and I2 some pd hybridization occurred, allowing some multiple bonding. This would make the bonds stronger than in F2 in which there are no d orbitals available.

(2) Coulson suggested that since fluorine atoms are small, the F – F distance is also small (1.48 Å), and hence internuclear repulsion is appreciable. The larger electron –electron repulsions between the lone pairs of electrons on the two fluorine atoms weaken the bond.

Pseudohalogens and Pseudohalides : A few ions are known, consisting of two or more atoms of which at least one is N, that have properties similar to those of the halide ions. They are therefore called pseudohalide ions. Pseudohalide ions are

HALOGEN & NOBLE GASES FAMILY

KEY CONCEPT


univalent, and these form salts resembling the halide salts.

For example, the sodium salts are soluble in water, but the silver salts are insoluble. The hydrogen compounds are acids like the halogen acids HX. Some of the pseudohalide ions combine to form dimers comparable with the halogen molecules X2. These include cyanogen (CN)2, thiocyanogen (SCN)2 and selenocyanogen (SeCN)2.

The important pseudohalogens Anion Acid Dimer

CN– : cyanide ion HCN hydrogen cyanide (CN)2 : cyanogen SCN– : thiocynate ion HSCN : thiocyanic acid (SCN)2 :

selenocyanogen SeCN– : selenocyanate ion (SeCN)2 :

selenocyanogen OCN– :cyanate ion HOCN : cyanic acid NCN2– : cyanamide ion H2NCN : cyanamide ONC– : fulminate ion HONC : fulminic acid N3

– : azide ion HN3 : hydrogen azide

The best known pseudohalide is CN–. This resembles Cl–, Br– and I– in the following respects :

1. It forms an acid HCN. 2. It can be oxidized to form a molecule cyanogen

(CN)2. 3. It forms insoluble salts with Ag+, Pb2+ and Hg+. 4. Interpseudohalogen compounds ClCN, BrCN and

ICN can be formed. 5. AgCN is insoluble in water but soluble in ammonia,

as is AgCl. 6. It forms a large number of complexes similar to

halide complexes. e.g. [Cu(CN)4]2– and [CuCl4]2–, and [Co(CN)6]3– and [CoCl6]3–.

Clathrate Compounds : Clathrate compounds of the noble gases are well

known. Normal chemical compounds have ionic or covalent bonds. However, in the clathrates atoms or molecules of the appropriate size are trapped in cavities in the crystal lattice of other compounds. Though the gases are trapped, they do not form bonds.

If an aqueous solution of quinol (1, 4-dihydroxybenzene) is crystallized under a pressure of 10 – 40 atmospheres of Ar, Kr or Xe, the gas becomes trapped in cavities of about 4Å diameter in the β-quinol structure. When the clathrate is dissolved, the hydrogen bonded arrangement of β-quinol breaks down and the noble gas escapes. Other small molecules such as O2, SO2, H2S, MeCN and CH3OH form clathrates as well as Ar, Kr and Xe. The smaller noble gases He and Ne do not form clathrate compounds because the gas atoms are small enough to escape from the cavities. The composition of these

clathrate compounds corresponds to 3 quinol : 1 trapped molecule, through normally all the cavities are not filled.

The gases Ar, Kr and Xe may be trapped in cavities in a similar way when water is frozen under a high pressure of the gas. These are clathrate compounds, but are more commonly called 'the noble gas hydrates'. They have formulae approximating to 6H2O : 1 gas atom. He and Ne are not trapped because they are too small. The heavier noble gases can also be trapped in cavities in synthetic zeolites, and samples have been obtained containing up to 20% of Ar by weight. Clathrates provide a convenient means of storing radioactive isotopes of Kr and Xe produced in nuclear reactors.

Structure and bonding in Xenon compounds : (i) Structure and bonding in XeF4 : The structure of

XeF4 is square planar, with Xe–F distances of 1.95 Å. The valence bond theory explains this by promoting two electrons as shown :

5s 5p 5d

(Electronic Structure of Xe-excited state) (four unpaired electrons form bonds to four fluorine

atoms six electron pairs form octahedral structure with two positions occupied by lone pairs)

The Xe atom bonds to four F atoms. The xenon 5px orbital forms a three-centre MO with 2p orbitals from two F atoms just as in XeF2. The 5py orbital forms another three-centre MO involving two more F atoms. The two three-centre obitals are at right angles to each other, thus giving a square planar molecule.

Xe

F

F F

F

(ii) Structure and bonding in XeF6 : The structure

of XeF6 is a distorted octahedron. The bonding in XeF6 has caused considerable controversy which is not completely resolved. The structure may be explained in valence bond terms by promoting three electrons in Xe :

5s 5p 5d

(Electronic structure of Xenon-exicted state) The six unpaired electrons form bonds with fluorine

atoms. The distribution of seven orbitals gives either a capped octahedron or a pentagonal bipyramid (as in


IF7). (A capped octahedron has a lone pair pointing through one of the faces of the octahedron) Since there are six bonds and one lone pair, a capped octahedron would give a distorted octahedral molecule. The molecular orbital approach fails with XeF6, since three three-centre molecular orbitals systems mutually at right angles would give a regular octahedral shape.

F

F

F

FXe F

F

The vibrational spectrum of gaseous XeF6 indicates

C3v, symmetry, i.e. an octahedron distorted by the lone pair at the centre of one triangular face. The structure of the molecule rapidly fluctuates between structures where the lone pair occupies each of the eight triangular faces. In various non-aqueous solvents, xenon hexafluoride forms a tetramer Xe4F24. Solid xenon hexafluoride is polymorphic. Except at very low temperatures it contains tetramers, where four square pyramidal XeF5

+ ions are joined to two similar ions by means of two bridging F–ions. The XeF distances are 1.84 Å on the square pyramidal units and 2.23 Å and 2.60 Å in the bridging groups.

Xenon Oxyfluorides : Structure of XeOF2 : Total number of electrons in valence shell of

Xe:12 (8 from Xe + 2 from O and 2 from F) Total number of electrons pairs

= 6(3σbp + 2lp + 1πbp)

F Xe F

O Hybridisation = sp3d (to accommodate 3bp and 2lp) Geometry = T-shaped Structure of XeOF4 : Total number of electron in valence shell of

Xe : 14 (8 from Xe + 2 from O + 4 from F)

Xe

F

F F

F

O

Total number of electron pairs = 7(5σbp + 1lp + 1dπ-pπbp) Hybridization : sp3d2 (to accommodate 5σbp and

1lp) Geometry : Square pyramidal Structure of XeO2F2 : Total number of electron in valence shell of

Xe = 14 (8 from Xe + 2 from F + 4 from O) Total number of electron pairs = 7(4σbp + 1lp + 2πbp)

F

Xe

F

O

O

Hybridization: sp3d (to accommodate 4bp + 1lp) Geometry : Trigonal bipyramidal or Sea-saw. Similarly : Structure of XeO3F2 and XeO2F4

• 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.

• 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.

• 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.


1. Compound (A), C3H6Cl2, on reduction with LiAlH4

gives propane. Treatment of (A) with aqueous alkali followed by oxidation gives (B) C3H4O4 which gives effervescence with NaHCO3. Esterification of (B) with ethanol gives (C), C7H12O4, which is well known synthetic reagent. When (B) is heated alone, the product is ethanoic acid, but while heating with soda-lime it gives methane. Compound (B) on reduction with LiAlH4 gives a diol which on reaction with SOCl2 gives back compound (A). Identify all the compounds and give balanced equation of the reactions.

Sol. Compound (B) gives effervescence with NaHCO3 solution. Hence it is a dicarboxylic acid, since it on heating alone gives acetic acid and with soda-lime CH4, it means two –COOH in it are at different carbon atoms.

CH2

COOH COOH

2NaHCO3 CH2 COONa COONa + 2CO2 +2H2O

∆ CH3COOH + CO2 Soda-lime

∆CH4 + 2CO2

(B)

Acid (B) can be prepared from (A), C3H6Cl2, which

should be 1,3-dichloro propane.

CH2

CH2Cl CH2Cl

2NaOH(aq.) CH2 CH2OH CH2OH (–2NaCl)

Propane 1,3-diol

3[O] CH2 COOH COOH

(B)

+ H2O

(A)

Esterification of (B) with ethanol gives malonic ester

which is a synthetic reagent of high importance. CH2

COOH COOH

∆ CH2 COOC2H5

COOC2H5– 2H2OMalonic ester(B)

+ C2H5OH

LiAlH4 CH2 CH2OH CH2OH

CH2 COOH COOH

(B)–2H2O

2SOCl2 CH2 CH2ClCH2Cl–2SO2; –2HCl

(A)

Hence, (A) CH2

CH2Cl CH2Cl

(B)

2. A hydrocarbon (A) of the formula C8H10, on ozonolysis gives compound (B), C4H6O2, only. The compound (B) can also be obtained from the alkyl bromide, (C) (C3H4Br) upon treatment with Mg in dry ether, followed by treatment with CO2 and acidification. Identify (A), (B) and (C) and also give equations for the reactions.

Sol. A(C8H10) OH)ii(

O)i(

2

3 → )B(

264 OHC

Since compound (A) adds one mol of O3, hence it should have either a C = C or a – C ≡ C – bond. If it was alkene its formula should be C8H16 (CnH2n) and if it was alkyne it should have the formula C8H14; it is definite that the compound has an unsaturated group, it appears that it is a cyclosubstituted ethyne.

H – C ≡ C – H 106HC

H2

+

− → C3H5 – C ≡ C – C3H5

the C3H5 – correspond to cyclopropyl (∆) radical hence compound (A) is

CH – C ≡ C – CH CH2

CH2

CH2

CH2 1,2-dicyclopropyl ethane

The ozonolysis of above compound would give two moles of cyclopropane carboxylic acid (C4H6O2).

CH – C ≡ C – CH CH2

CH2

CH2

CH2 (A)

(i) O3

CH – C – C – CH CH2

CH2

CH2

CH2

H2O O

O – O

(B)

CH – C – C – CH CH2

CH2

CH2

CH2 +H2O2

O O

CH – COOH CH2

CH2 2

warm

Compound (B) is prepared from cyclopropyl bromide as follows :

CH – BrCH2

CH2 (C)

Mgeither

CH.MgBr CH2

CH2

C = O

∆

O

Cyclopropyl magnesium bromide

CH – COOH CH2

CH2

CH.COOMgBr CH2

CH2

Addition compound

HOH

dil. HCl; –MgBrOH

UNDERSTANDINGOrganic Chemistry


Hence, A, –C≡C–

B, –COOH 3. An organic compound (A) contains 69.42% C, 5.78%

H and 11.57% N. Its vapour density is 60.5. It evolves NH3 when boiled with KOH. On heating with P2O5, it gives a compound (B) having C = 81.55%, H = 4.85% and N= 13.59%. On reduction with Na + C2H5OH (B) gives a base, which reacts with HNO2 giving off N2 and yielding an alcohol (C). The alcohol can be oxidised to benzoic acid. Explain the above reactions and assign structural formulae to (A), (B) and (C)

Sol. (i) Calculation of empirical formula of (A) : Element At.

wt. % Relative no. of

atoms Simplest

ratio C 12 69.42

1242.69 = 5.785

826.0785.5 = 7

H 1 5.78 178.5 = 5.78

826.078.5 = 7

N 14 11.57 14

57.11 = 0.826 826.0826.0 = 1

O 16 13.23 16

23.13 = 1.827 826.0827.0 = 1

Hence, empirical formula of (A) = C7H7NO Empirical formula wt. = 84 + 7 + 14 + 16 = 121 (ii) Calculation of molecular weight of (A) : Molecular weight = 2 × V.D. = 2 × 60.5 = 121 (iii) Determination of molecular formula of (A):

n = .wtEmpirical.wtMolecular =

121121 = 1

Hence, molecular formula = empirical formula i.e., C7H7NO (iv) Calculate of empirical formula of (B) : Element At.

wt. % Relative no. of

atoms Simplest

ratio C 12 81.55

1255.81 = 6.80

97.080.6 = 7

H 1 4.85 185.4 = 4.85

97.085.4 = 5

N 14 13.59 14

59.13 = 0.97 97.097.0 = 1

Hence, empirical formula of (B) = C7H5N (v) Determination of structural formulae : (a) Since compound (A) on heating with KOH gives

NH3, a characteristic test of amide, hence the compound (A) is an amide (–CONH2).

(b) Since compound (B) is obtained by heating (A) with P2O5, a dehydrating agent.

OHNHCONHC 2)B(57

OP

)A(77

52 + →

The above reaction confirms that (A) is an amide, and the remaining reaction are :

C7H5N

COOH

Alcohol

[H] HNO2 N2 + (C) [O]

(B) The formula of benzoic acid indicates that the

compound (A) is an aromatic amide. Hence, the reactions are :

CONH2

Benzamide

KOH + NH3 ↑ Boil

(A)

COOK

CONH2

Benzonitrile

P2O5 + H2O ∆

(A)

C≡N

(B)

CH2NH2

Benzyl amine

Na +C2H5OH+ 4[H]

(Base)

C≡N

(B)

CH2NH2

Benzyl alcohol

HNO2 + N2 + H2O

CH2OH

(C)

CH2OH

Benzoic acid

2[O]– H2O

(C)

COOH

4.

)ClHC(CB)HC(A

136

HCl126 + →

B → KOH.alc D isomer of A D →Ozonolysis E (it gives negative test with Fehling solution but responds to iodoform test) A →Ozonolysis F + G Both gives positive Tollen's test but do not give iodoform test. F + G → NaOH.Conc HCOONa + primary alcohol Identify to A to G Sol.

)ClHC(CB)HC(A

136

HCl126 + →

B →Ozonolysis E (it gives negative test with Fehling solution but responds to iodoform test) A →Ozonolysis F + G Both gives positive Tollen's test but do not give iodoform test.


F + G → NaOH.Conc HCOONa + primary alcohol Both F and G are aldehydes because they give

positive Tollen's test and do not give iodoform test. These aldehydes give Cross Cannizzaro's reaction. So they do not have α-hydrogen atoms. In cross Cannizzaro's reaction HCOONa is formed along with p-alcohols. So in these an aldehyde is HCHO and another is (CH3)3C.CHO. F and G are obtained by ozonolysis of A. Therefore compound 'A' is

CH2 = CH – C(CH3)3. Structure of compound 'A' is

CH3 CH3 – C – CH = CH2 CH3

Compound 'A' on reaction with HCl gives comp. B and C which have molecular formula C6H13Cl. Thus,

CH3 – C – CH = CH2 → CH3 – C — CH – CH3

CH3

CH3

HCl

CH3

CH3 Cl

Comp. 'B'

CH3 – C – CH2 – CH2Cl

CH3

CH3

Comp. 'C'

+

Compound 'B' gives 'D' on dehydrohalogenation with

alc. KOH.

CH3 – C — CH – CH3

CH3

CH3

alc. KOH

CH3

CH3 Cl

Sec. carbonium ion

CH3 – C – CH

CH3

Compound 'D'

CH3 – C — CH – CH3 Boil (–Cl–)

+

+

CH3

CH3

H+

C = C CH3

CH3

CH3

CH3

Compound 'D' on ozonolysis to give compound 'E'

Compound 'E'

C = C CH3

CH3

CH3

CH3

Ozonolysis 2CH3 – C – CH3

O

Compound 'E' has methyl ketonic groups (–COCH3)

so it gives positive iodoform test and does not give the test with Fehling solution due to absence of –CHO group.

Compound 'A' on ozonolysis to give compounds F and G as follows :

(CH3)3CCH = CH2 → Ozonolysis

Comp. 'G'

(CH3)3C – CHO + CH2O'F'

Compound G and F gives crossed Cannizzaro's

reaction with conc. NaOH solution.

HCOONa + CH3 – C – CH2OH

Ozonolysis

CH3

CH3

Comp. 'G'

CH3 – C — CHO + CH2O + conc. NaOH →'F'

CH3

CH3 Hence,

Compound 'A' =

CH3 – C – CH = CH2 (C6H12)

CH3

CH3

Compound 'B' =

CH3 – C – CH – CH3

CH3

CH3 Cl

Compound 'C' =

CH3 – C – CH2 – CH2Cl

CH3

CH3

Compound 'D' =

C = C CH3

CH3

CH3

CH3

Compound 'E' =

CH3 – C – CH3

O

Compound 'F' =

H – C – H

O

Compound 'G' =

CH3 – C – CH2OH

CH3

CH3

5. An unsaturated hydrocarbon (A), C6H10 readily gives

(B) on treatment with NaNH2 in liquid NH3. When (B) is allowed to react with 1-chloro propane, a compound (C) is obtained. On partial hydrogenation in the presence of Lindlar catalyst (C) gives (D), C9H18. On ozonolysis (D) gives 2, 2-dimethyl propanal and butanal. Give structures of (A), (B), (C) and (D) with proper reasoning.

Sol. The structure of compound (D) can be obtained by joining the products of ozonolysis.

CH3 – C – CH = O + O = CH.CH2CH2CH3

CH3

CH3

2,2-dimethyl propanal

–2[O]

Butanal

CH3 – C – CH = CH.CH2CH2CH3

CH3

CH3 2,2-dimethyl heptene-3 (D)

Ozonolysis equation of (D) is :


CH3 – C – CH = CHCH2CH2CH3

CH3

CH3

(D)

(I) O3

CH3 – C – CHO + CH3CH2CH2CHO

CH3

CH3

(II) H2O/Zn

Alkene (D) is obtained by the partial hydrogenation

of (C), thus (C) contains a – C≡C – triple bond at C3.

CH3 – C – C ≡ C – CH2CH2CH3

CH3

CH3

(C)

H2


CH3

CH3

Lindlar catalyst

(D)

The starting compound (A) reacts with NaNH2 in presence of liquid NH3. It means it contains one –C≡CH at the terminal carbon, and, therefore gives a mono sodium derivative.

)A(106HC

3

2NH

NaNH → C4H9 – )B(

C ≡ C.Na

Compound (B) reacts with 1-chloro propane to give compound (C) as follows :

C4H9 – C ≡ C – Na + Cl – CH2CH2CH3 ∆

C4H9 – C≡C – CH2CH2CH3

–NaCl

(C)

(B) 1-chloro propane

But, (C) is CH3 – C – C ≡ C – CH2CH2CH3

CH3

CH3

Now, putting the value of C4H9 as a t-butyl radical, we have :

CH3 – C – C≡C – H

CH3

CH3 (A)

NaNH2 CH3 – C – C≡CNa + NH3

CH3

CH3 (B)

Hence,

CH3 – C – C≡CH

CH3

CH3 (3,3-dimethyl butyne-1)

(A)

CH3 – C – C≡CNa

CH3

CH3

(B)

CH3 – C – C≡C – CH2CH2CH3

CH3

CH3

(C)


CH3

CH3

(D)

CH – Br CH2

CH2

CH2

CH2

Cyclopropyl magnesium bromide

CH . MgBr Mg

ether

C=O∆

O

Pi Day On Pi Day (March 14 or 3-14) in 1879, a baby was born in Ulm, Germany to a German couple whose name meant "one stone". That baby was Albert Einstein!

occurs naturally in tables of death, in what is known as a Gaussian distribution of deaths in a population; that is, when a person dies, the event 'feels' Pi.

The symbol for Pi was introduced by the English mathematician William Jones in 1706.

Mathematician John Conway pointed out that if you break down the digits of Pi into blocks of ten, the probability that one of those blocks will contain ten distinct digits is about one in 40,000. Curiously, this first happens in the 7th block of ten digits.

There is the little rhyme to help the memorisation of twenty-one digits of :

Now, I wish I could recollect pi. "Eureka," cried the great inventor. Christmas Pudding; Christmas PieIs the problem's very center.


1. Find the greatest value of the expression (a – x) (b – y) (c – z) (ax + by + cz), where a, b, c are

known positive quantities and a – x, b – y, c – z are also positive?

2. Let f (x) satisfies the differential equation

xf '(x) + f (x) = g(x), where f (x) and g(x) are continuous functions. If f (x) is decreasing function for all x ∈ R+, then prove that

x.g (x) < ∫x

xg0

)( dx; for ∀ x > 0.

3. If a chord of the circle x2 + y2 = 32 makes equal

intercepts of length p on the coordinate axes, then find the range of p.

4. The arc AC of a circle subtends a right angle at the

centre O. B divides the arc AC in the ratio 2 : 1. If

OA = →a and OB = →b , find OC .

5. Out of 20 consecutive numbers 4 are chosen at

random. Prove that the chance of their sum being even is greater than that of their sum being odd.

6. Find a point P on the line 3x + 2y + 10 = 0 such that

|PA – PB| is maximum when A is (4, 2) and B is (2, 4).

7. Secants are drawn from a given point A to cut a given

circle at the pairs of points P1, Q1; P2, Q2; ...., Pn, Qn. Show that AP1 . AQ1 = AP2 . AQ2 = .... = APn . AQn 8. Let A & B be the matrices such that AAT = I and

AB = BA. Prove that ABT = ATB. 9. If a2 + b2 + c2 = 1, b + ic = (1 + a) z, prove that

ciba

++

1 =

iziz

−+

11 , where a, b, c are real numbers and z

is a complex number.

10. Let n is an odd positive integer, show that (without using mathematical induction) (n2 – 1)n is divisible by 24. Here n > 1.

`tàxÅtà|vtÄ VtÄÄxÇzxá This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety of possible twists and turns of problems in mathematics 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 : Shailendra Maheshwari Joint Director Academics, Career Point, KotaSolut ions wil l be published in next issue

6Set

INTERESTING SCIENCE FACTS

• The speed of light is 186,000 miles per second. • It takes 8 minutes 17 seconds for light to travel

from the Sun’s surface to the Earth. • In October 1999 the 6 billionth person was born. • 10 percent of all human beings ever born are

alive at this very moment. • The Earth spins at 1,000 mph but it travels

through space at an incredible 67,000 mph. • Every year over one million earthquakes shake

the Earth. • The largest ever hailstone weighed over 1 kg and

fell in Bangladesh in 1986. • Every second around 100 lightning bolts strike

the Earth. • Every year lightning kills 1000 people. • In October 1999 an Iceberg the size of London

broke free from the Antarctic ice shelf. • If you could drive your car straight up you would

arrive in space in just over an hour. • All the hydrogen atoms in our bodies were

created 12 billion years ago in the Big Bang. • The Earth is 4.56 billion years old…the same age

as the Moon and the Sun. • Alfred Nobel invented dynamite in 1866.


1. Let S1 ≡ a1x2 + 2h1xy + b1y2 + 2g1x + 2f 1y + c1 = 0 & S2 ≡ a2x2 + 2h2xy + b2y2 + 2g2x + 2f 2 y + c2 = 0 be the rectangular hyperbolas. So a1 + b1 = 0 & a2 + b2 = 0 Now S1 + λS2 = 0 represents the conics through their

points of intersection i.e. A, B, C and D. The sum of coefficient of x2 & y2 in it is

(a1 + λa2) + (b1 + λb2) = (a1 + b1) + λ (a2 + b2) = 0 Hence, it will also be rectangular hyperbola. Now for

λ when it represents pair of straight lines then also sum of coeff. of x2 & y2 will be zero. Hence those lines will be perpendicular. So AD & BC will be perpendicular. Similarly BD & AC and CD & AB will also pairs of perpendicular lines. Hence D will be orthocentre of triangle ABC. In fact orthocentre of triangle forms by any of 3 of these points will be the fourth point.

2. c = 2R; ∆4

abc = R; s∆ = r

A

BC

a2 + b2 = c2 a + b + c = 2s a + b = 2(s – R) as c = 2R a2 + b2 + 2ab = 4(s – R)2 = 4R2 + 2ab ab = 2(s – R)2 – 2R2 = 2s2 – 4sR 2∆ = 2s2 – 4sR

∆ = 2

2

s∆ – 2.

s∆ .R; 2r

∆ = 1 + rR2

∆ = r2 + 2Rr

3. m1(OQ) = hk & m2(PR) = –

1

1m

= h

k+β−

so m1m2 = –1

⇒ hk .

hk+β−

= – 1

k2 = +βh – h2; so βh = h2 + k2

β = h

kh 22 + so point R is

+ 0,22

hkh

O

(h,k) rR

(b,0)A(a,0)

Q(a,a)

Now (PR)2 = (RA)2

k2 + 222

+−

hkhh =

222

+−

hkha

⇒ k2 + h2 + 222

+h

kh – 2(h2 + k2)

= a2 + 222

+h

kh – 2ah

kh 22 +

⇒ k2 + h2 – a2 = 2(h2 + k2) – 2ah

kh 22 +

–a2 = (h – 2a) h

kh 22 +

(x2 + y2) (x – 2a) + a2x = 0 4. Let z = x be the purely real root then f (x) = x4 + 2x3 + 3x2 + 4x + 5 = 0 f ´(x) = 4x3 + 6x2 + 6x + 4 = 0 ⇒ 4(x + 1) (4x2 + 4 – 4x + 6x) = 0 ⇒ (x + 1) (4x2 + 2x + 4) = 0 ⇒ x = –1 is only real root & f (–1) = 1 – 2 + 3 – 4 + 5 = 3 > 0 no real root of f(x). Now let z = iy be the purely imaginary roots then y4 – 2iy3 – 3y2 + 4iy + 5 = 0 so y4 – 3y2 + 5 = 0 and 2y3 + 3y2 = 0 must have

simultaneous solution which is not possible. as y = 0, y = – 3/2 are the roots of 2nd but they do not

satisfy.

5. L.H.S. xa =

ay ⇒ x =

ya2

∫∞

+

0

ya

ayf

yayan

/)/(

2

2l

−2

2

ydya

I = ∫∞

+

0ya

ayf

y1 (ln a2 – lny) dy

MATHEMATICAL CHALLENGES SOLUTION FOR SEPTEMBER ISSUE (SET # 5)


I = ∫∞

+

0xa

axf (2ln a – lny)

ydy

2I = ∫∞

+

0xa

axf (lna)

xdx ⇒ I= ∫

∞

+

0xa

axf

xnal dx

6. Let the fixed points be P(α, 0) & Q (– α, 0) and

variable line be

y = mx + c as given 21

||

m

cm

+

+α .21

||

m

cm

+

+α− = a; where

'a' is a constant. so |c2 – m2α2| = a (1 + m2) ...(1) Now let foot of the perpendicular from (α, 0) be (h, k)

then c = k – mh & – m1 =

α−hk ⇒ m = –

kh α−

so c = k + k

hh )( α− = k

hhk α−+ 22

use these in (1)

22

2

2

222 )()(α

α−−

α−+k

hk

hhk = a 2

2)(1k

h α−+

|k2 + h2 – hα + hα – α2| |k2 + h2 – hα – hα + α2| = a(k2 + (h – α)2) so x2 + y2 = (α2 + a)

7. y = f (x) = ∫ −x

zzxe0

2dz = ∫ −

xzzx ee

0

2. dz

y´ = ∫ −x

zzx eze0

2. dz + 1 = – ∫ −−

xzzx zee

0

)2(21 2

dz + 1

= –

− ∫ −−

xzxzxzxz dzexeee

00 .).(

21 22

+ 1

= 21 xy + 1 ⇒

dxdy –

21 xy = 1

I.F. = ∫− dxx

e 2 = 4/2xe− solution is

y . 4/2xe− = ∫ − 4/2xe dx = ∫ −x

ze0

4/2dz

y = ∫ −x

zx ee0

4/4/ 22dz proved.

8. Since 1−α

α + 1−β

β = 47

2

2

−−

aa

)(1

)(2β+α−+αβ

β+α−αβ = 47

2

2

−−

aa

as given αβ = 4, so α + β = a2 + 1 Hence the equation is x2 – (a2 + 1) x + 4 = 0

Let f (x) = x2 – (a2 + 1)x + 4 Since both roots of f (x) = 0 lie in (1, 4), hence D = (a2 + 1)2 – 16 ≥ 0 ⇒ a ∈(–∞, – 3 ) ∪ ( 3 , ∞) ...(1) and f (1) > 0 ⇒ 1 – (a2 + 1) 4 > 0 ⇒ a ∈ (–2, 2) ...(2) and f (4) > 0 ⇒ 16 – (a2 + 1) 4 + 4 > 0 ⇒ a ∈ (–2, 2) ...(3)

and –a2

β ∈ (1, 4) ⇒ 1 < 2

12 +a < 4

⇒ a ∈ (– 7 , –1) ∪ (1, 7 ) ...(4)

Hence a ∈ (–2, – 3 ] ∪ [ 3 , 2)

9. f´(x) = 35 x2/3 –

310 x –1/3 = 3/1

)2(35

xx −

sign. dia of f´(x)

(2, 0)(5, 0)

(0,–3(4)1/3)

2 0+ +

x = 2 is local min. x = 0 is local max. f (x) is non diff. at x = 0 f (0) = 0 f (2) = 25/3 – 5.22/3.(2 – 5) = –3.22/3 = – 3(4)1/3 f (x) = x2/3 (x – 5) f (x) passes through (0, 0), (5, 0) If x5/3 – 5x2/3 = k has exactly one positive root then

from sketch. k > 0

10. There will be 99C44 subsets in which 1 will be least element

similarly there will be 98C49 subsets in which 2 will be least element

so ∑∈xp

pmin = 1.99C49 + 2.98C49 + 3.97C49 + .... + 51.49C49

= Coeff. of x49 in [(1 + x)99 + 2(1 + x)98 + ..... + 51(1+x)49]

= Coeff. of (1 + x)99

x

x

x

x

+−

+−

+−

+−

111

)1(51

111

)1(11 4851

= Coeff. of xx

xx

xx−+

−

+

+−+ 49

2

2

4899 )1(51

)1(

)1()1(

= coeff. of x51 in [(1 + x)101 – (1 + x)50] + coeff. of x50 in 51(1 + x)49 = 101C51 – 0 + 0 = 101C51


1. In a particular programming language, a valid variable name can consist of a sequence of one to six alphanumeric characters A, B, C, ... , Z, 0, 1, 2, ...., 9 beginning with a letter. Find the total number of valid variable names.

Sol. The number of valid variable names of one character = 26 (Q there are 26 letters and the variable name

must begin with a letter). The number of valid variable names of two

characters = 26 × 36 (Q the first place should have a letter and

the second place can have any one of the 26 + 10 alphanumeric characters, repetition being allowed).

The number of valid variable names of three characters

= 26 × 36 × 36 and so on. ∴ the total number of valid variable names of one to

six characters = 26 + 26 × 36 + 26 × 362 + ... + 26 × 365 = 26(1 + 36 + 362 + ... + 365)

= 26. 1–361–366

= 3526 (366 – 1).

2. Sum to n terms 2–1 xx + 4

2

–1 xx + 8

4

–1 xx + ...

Also find the sum to infinite terms when |x| < 1.

Sol. Here tn = n

n

x

x2

2

–1

1–

= )1)(–1(

1–)1(1–1–

1–

22

2

nn

n

xx

x

+

+

= 1–2–1

1n

x – n

x2–1

1

putting n = 1, 2, 3, .... we get

t1 = x–1

1 – 2–11x

t2 = 2–11x

– 4–11x

t3 = 4–11x

– 8–11x

..........................................

tn = 1–2–1

1n

x – n

x2–1

1

____________________________ Adding,

sum = x–1

1 – nx2–1

1 = )–1)(–1(

–2

2

n

n

xx

xx

= )–1)(–1(

)–1(2

1–2

n

n

xx

xx

∴ sum to infinite terms

= )–1)(–1(

)–1(lim2

1–2

n

n

xx

xxn ∞→

= x

x–1

for n

x2 → 0 as n → ∞ when |x| < 1. 3. Let O be a point in the ∆ABC such that

∠OAC = ∠OCB = ∠OBA = α. Prove that cot α = cot A + cot B + cot C.

Sol.

A

B C Oα

α

α

From ∆AOC, AOCsin

AC = αsin

OC .

But ∠AOC = π – (∠OCA + ∠OAC) = π – (C – α + α) = π – C

∴ )C–sin(

ACπ

= αsin

OC

or Csin

b = αsin

OC ...(i)

Similarly, from ∆OBC we get

Bsin

a = )–Bsin(

OCα

...(ii)

(i) ÷ (ii) ⇒ CsinBsin

ab =

αα

sin)–Bsin(

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

Students' ForumMATHS


or CsinAsin

Bsin 2 =

ααα

sinsin.Bcos–cos.Bsin

=

AsinBsin

ab

Q

= cot α. sin B – cos B

or CsinAsin

Bsin = cot α – cot B

or CsinAsin)CAsin( + = cot α – cot B Q π – (A + C) = B

or Csin.Asin

Csin.AcosCcos.Asin + = cot α – cot B

or cot C + cot A = cot α – cot B ∴ cot α = cot A + cot B + cot C. 4. Let f (x) be a real function not identically zero such

that f (x + y2n+1) = f (x) + f (y)2n+1, n ∈ N and x, y are any real numbers and f ' (0) ≥ 0. Find the values of f (5) and f ' (10).

Sol. Here, f (x + y2n+1) = f (x) + f (y)2n+1 ...(i) Putting x = 0, y = 0 we get f (0) = f (0) + f (0)2n+1 ∴ f (0) = 0.

f ' (0) ≥ 0 ⇒ x

fxfx

)0(–)(lim0→

≥ 0

⇒ xxf

x

)(lim0→

≥ 0

∴ if x > 0, f (x) ≥ 0 ...(ii) Putting x = 0, y = 1 in (i), f (1) = f (0) + f (1)2n+1 or f (1) [1 – f (1)2n] = 0 ∴ f (1) = 0 or 1, from (2). Putting y = 1 in (1), for all real x, f (x + 1) = f (x) + f (1)2n+1 ...(iii) ∴ f(1) = 0 ⇒ f (x + 1) = f (x) ⇒ f (1) = f (2) = f (3) = ... = 0, i.e., f (x) is identically zero. ∴ f (1) ≠ 0. Hence f (1) = 1. So from (3), f (x + 1) = f (x) + 1 ...(iv) ∴ f (5) = f (4) + 1 = f (3) + 1 + 1 = f (2) + 1 + 2 = f (1) + 1) + 3 = f (1) + 4 = 1 + 4 = 5 Also (iv) ⇒ f (x) is a function whose value

increases by 1 when variable x is increases by 1. ∴ f (x) = x; ∴ f ' (x) = 1 ∴ f ' (10) = 1.

5. A man is standing on a straight bridge over a river and another man on a boat is on the river just below the man on the bridge. If the first man starts walking at the uniform speed of 4 m/min and the boat moves perpendicularly to the bridge at the speed of 5 m/min then at what rate are they separating after 4 minutes if the height of the bridge above the boat is 3 m?

Sol. In the beginning, the man is at A on the bridge and the boat is at B on the river. After t minutes, the man is at C and the boat is at D.

From the question, AB ⊥ AC and BD ⊥ AC. ∴ AC is perpendicular to the plane of AB and BD. ∴ AC is perpendicular to each line in the plane

ABD. ∴ AC ⊥ AD.

A C

bridge

B

D Also, AC = 4t m and BD = 5t m and AB = 3 m ∴ from the right-angled ∆ABD,

AD = 22 BDAB + = 22 )5(3 t+ m

and from the right-angled ∆DAC,

DC = 22 ACAD +

If DC = z m, z = 222 )4()5(3 tt ++ = 2419 t+

∴ dtdz =

24192

1

t+ × 41 × 2t

∴ at t = 4 min.,

dtdz =

244192

4241

×+

×× m/min = 665

164 m/min.

6. A cubic function f (x) vanishes at x = –2 and has

relative maximum/minimum at x = –1 and x = 31 .

If ∫1

1–

)( dxxf = 3

14 , find f (x).

Sol. Let f (x) = ax3 + bx2 + cx + λ; then f ' (x) = 3ax2 + 2bx + c. From the question we have f (– 2) = 0 ∴ – 8a + 4b – 2c + λ = 0 ...(i)

Also, f '(x) = 0 when x = – 1, 31 (using condition for

relative maximum/minimum). ∴ f '(–1) = 0 ⇒ 3a – 2b + c = 0 ...(ii)


and f '

31 = 0 ⇒

3a +

32b + c = 0,i.e.,

a + 2b + 3c = 0 ...(iii) (ii) + (iii) ⇒ 4a + 4c = 0, i.e., c = – a ∴ (iii) ⇒ a + 2b – 3a = 0, i.e., b = a ∴ from (i), – 8a + 4a + 2a + λ = 0, i.e., λ = 2a. ∴ f (x) = ax3 + ax2 – ax + 2a = a(x3 + x2 – x +2) ...(iv)

∴ ∫1

1–

)( dxxf = 3

14 gives

∫ ++1

1–

23 )2–( xxxa dx = 3

14

∴ 1

1–

2242

2–

34

++ xxxxa =

314

or

++ 2–

21–

31–

41–2

21–

31

41a =

314

∴ a . 3

14 = 3

14 , i.e., a = 1.

∴ from (iv), f (x) = x3 + x2 – x + 2.

Know about Pie

3.14

= Perimeter / Diagonal, of any circle. Pi expanded to 45 decimal places: 3.14159 26535 89793 23846 26433 83279 50288 41971 69399 Pi expanded to 52 binary places: 11.0010010000111 1110110101010 0010001000010 1101000111001 You cannot square a disc using just a ruler and compasses because is a transcendental number. = 4(1/1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + ... ) = 2(2/1 x 2/3 x 4/3 x 4/5 x 6/5 x 6/7 x 8/7 x 8/9 x ... ) ≈ 355/113 (a real good rational approximation of ) ≈ (6 2)/5 In the late 18th century, James Stirling, a Scottish mathematician, developed an approximation for factorials using the transcendental numbers 'Pi' and 'e': n! ≈ (2 n)1/2 (n/e)n The most famous formula for calculating Pi is Machin's formula:

/4 = 4 arctan(1/5) – arctan(1/239) This formula, and similar ones, were used to push the accuracy of approximations to Pi to over 500 decimal places by the early 18th century (this was all hand calculation!). Interestingly, there are no occurrences of the sequence 123456 in the first million digits of Pi. - posted by George Pantazis Bamboozlement with Pi Does Pi equal 3? No? Then have a look on the algebraic equation below: x = ( + 3)/2 2x = + 3 2x( - 3) = ( + 3)( - 3) 2 x - 6x = 2 - 9 9 - 6x = 2 - 2 x 9 - 6x + x2 = 2 - 2 x + x2 (3 - x)2 = ( - x)2 3 - x = - x 3 = We use Pi to: Describe the DNA double helix, Determining the distribution of primes - the probability that two randomly selected integers are relatively prime (i.e. have no common factors) is 6 / p2, Analyzing the ripples on water Checking for accuracy - as there are now millions upon millions of known decimal places of Pi, by asking a super computer to compute this many figures its accuracy can be tested.

THE COLOURS OF COMPLEX METAL IONS

This page is going to take a simple look at the origin of colour in complex ions - in particular, why so many transition metal ions are coloured. Be aware that this is only an introduction to what can grow into an extremely complicated topic. Why do we see some compounds as being coloured? White light You will know, of course, that if you pass white light through a prism it splits into all the colours of the rainbow. Visible light is simply a small part of an electromagnetic spectrum most of which we can't see - gamma rays, X-rays, infra-red, radio waves and so on. Each of these has a particular wavelength, ranging from 10-16 metres for gamma rays to several hundred metres for radio waves. Visible light has wavelengths from about 400 to 750 nm. (1 nanometre = 10-9 metres.) The diagram shows an approximation to the spectrum of visible light.


Limits : Theorems of Limits : If f (x) and g(x) are two functions, then (i) )]()([lim xgxf

ax±

→ = )(lim xf

ax→ ± )(lim xg

ax→

(ii) )]().([lim xgxfax→

= )(lim xfax→

. )(lim xgax→

(iii)

→ )(

)(limxgxf

ax =

)(lim

)(lim

xg

xf

ax

ax

→

→ if )(lim xgax→

≠ 0

(iv) )]([lim xkfax→

= k )(lim xfax→

, where k is constant.

(v) )(lim xfax→

= )(lim xfax→

(vi) qp

axxf /)(lim

→ =

qp

axxf

/)(lim

→ , where p and q are

integers. Some important expansions :

(i) sin x =

+−+− ....!7!5!3

753 xxxx

(ii) cos x =

+−+− ....!6!4!2

1642 xxx

(iii) sin h x =

∞+++ ....!5!3

53 xxx

(iv) cos h x =

∞+++ ....!4!2

142 xx

(v) tan x =

+++ ....152

3

53 xxx

(vi) log(1 + x) =

+−+− ....4

x3

x2

xx432

(vii) ex =

++++ ....!3!2

132 xxx

(viii) ax =

+++ ....)(log!2

log1 22

axax

(ix) (1 – x)–1 = 1 + x + x2 + x3 + ......

(x) sin–1x =

++++ ......7

.65.

43.

21

5.

43.

21

3.

21 753 xxxx

(xi) tan–1x =

−+− .....

51

31 53 xxx

Some important Limits : (i) x

xsinlim

0→ = 0

(ii) xx

coslim0→

= 1

(iii) x

xx

sinlim0→

= 1 = x

xx sinlim

0→

(iv) x

xx

tanlim0→

= 1 = x

xx tanlim

0→

(v) x

xx

)1log(lim0

+→

= 1

(vi) xx

e0

lim→

= 1

(vii) x

ex

x

1lim0

−→

= 1

(viii) x

a x

x

1lim0

−→

= logea

(ix) axax nn

ax −−

→lim = nan–1

(x) x

x x

+

→∞

11lim = e = x

x x

+

−∞→

11lim

(xi) xx

x /1

0)1(lim +

→ = e

(xii) x

x xa

+

→∞1lim = ea

(xiii) nx

a∞→

lim =

<>∞

1,01,

aifaif

i.e. a∞ = ∞, if a > 1 and a∞ = 0, if a < 1

(xiv) x

x n

x

1)1(lim0

−+→

= n

(xv) x

xx

1

0

sinlim−

→ = 1 =

xx

x

1

0

tanlim−

→

LIMIT,CONTINUITY & DIFFERENTIABILITY

Mathematics Fundamentals MA

TH

S


(xvi) xax

1sinlim −

→ = sin–1a, |a| ≤ 1

(xvii) xax

1coslim −

→= cos–1a, |a| ≤ 1

(xviii) xax

1tanlim −

→ = tan–1a, – ∞ < a < ∞

(xix) xeexloglim

→ = 1

(xx) 20

cos1limx

xx

−→

= 21

Let )(lim xfax→

= l and )(lim xgax→

= m, then

(xxi) )())((lim xgax

xf→

= lm

(xxii) If f (x) ≤ g(x) for every x in the deleted neighbourhood (nbd) of a, then )(lim xf

ax→ ≤ )(lim xg

ax→.

(xxiii) If f (x) ≤ g(x) ≤ h(x) for every x in the deleted nbd of a and )(lim xf

ax→ = l = )(lim xh

ax→, then )(lim xg

ax→ = l.

(xxiv) )(lim xfogax→

= f

→)(lim xg

ax = f (m)

In particular (a) )(loglim xfax→

= log

→)(lim xf

ax= logl

(b) )x(f

axelim

→ =

)x(flimaxe → = el

(xxv) If )(lim xfax→

= + ∞ or – ∞, then )(

1limxfax→

= 0.

Evaluation of Limits (Working Rules) :

By factorisation : To evaluate )()(lim

xx

ax ψφ

→, factorise

both φ(x) and ψ(x), if possible, then cancel the common factor involving a from the numerator and the denominator. In the last obtain the limit by substituting a for x.

Evaluation by substitution : To evaluate )x(flimax→

,

put x = a + h and simplify the numerator and denominator, then cancel the common factor involving h in the numerator and denominator. In the last obtain the limit by substituting h = 0.

By L – Hospital's rule : Apply L-Hospital's rule to

the form 00 or

∞∞ .

)()(lim

xgxf

ax→=

)´()´(lim

xgxf

ax→ =

)()(lim

xgxf

n

n

ax→

By using expansion formulae : The expansion formulae can also be used with advantage in simplification and evaluation of limits.

By rationalisation : In case if numerator or denominator (or both) are irrational functions,

rationalisation of numerator or denominator (or both) helps to obtain the limit of the function.

Continuity : f (x) is continuous at x = a if )(lim xf

ax→exists and is

equal to f (a) i.e. if )(lim–

xfax→

= f (a) = )(lim xfax +→

.

Discontinuous functions : A function f is said to be discontinuous at a point a of its domain D if is not continuous there at. The point a is then called a point of discontinuity of the function. The discontinuity may arise due to any of the following situations:

(a) )(lim xfax +→

or )(lim xfax −→

of both may not exist.

(b) )(lim xfax +→

as well as )(lim xfax −→

may exist but are

unequal. (c) )(lim xf

ax +→ as well as )(lim xf

ax −→both may exist

but either of the two or both may not be equal to f(a). We classify the point of discontinuity according to

various situations discussed above. Removable discontinuity : A function f is said to

have removable discontinuity at x = a if )(lim xf

ax −→= )(lim xf

ax +→but their common value is not

equal to f (a). Such a discontinuity can be removed by assigning a suitable value to the function f at x = a.

Discontinuity of the first kind : A function f is said to have a discontinuity of the first kind at x = a if

)(lim xfax −→

and )(lim xfax +→

both exist but are not equal.

f is said to have a discontinuity of the first kind from the left at x = a if )(lim xf

ax −→exists but not equal to

f (a). Discontinuity of the first kind from the right is similarly defined.

Discontinuity of second kind : A function f is said to have a discontinuity of the second kind at x = a if neither )(lim xf

ax −→nor )(lim xf

ax +→exists.

f if said to have discontinuity of the second kind from the left at x = a if )x(flim

ax −→does not exist.

Similarly, if )x(flimax +→

does not exist, then f is said to

have discontinuity of the second kind from the right at x = a.

Differentiability : f (x) is said to be differentiable at x = a if R´ = L´

i.e. h

afhafLth

)()(0

−+→

= h

afhafLth −

−−→

)()(0

Note : We discuss R, L or R´, L´ at x = a when the function is defined differently for x > a or x < a and at x = a.


Parabola : The locus of a point which moves such that its

distance from a fixed point is equal to its distance from a fixed straight line, i.e. e = 1 is called a parabola.

X´ Z O S(a, 0)

M Y

Y

L

L´ P´

P

N X

Its equation in standard form is y2 = 4ax (i) Focus S (a, 0) (ii) Equation of directrix ZM is x + a = 0 (iii) Vertex is O (0, 0) (iv) Axis of parabola is X´OX Some definitions : Focal distance : The distance of a point on parabola

from focus is called focal distance. If P(x1, y1) is on the parabola, then focal distance is x1 + a.

Focal chord : The chord of parabola which passes through focus is called focal chord of parabola.

Latus rectum : The chord of parabola which passes through focus and perpendicular to axis of parabola is called latus rectum of parabola. Its length is 4a and end points are L(a, 2a) and L´(a, – 2a).

Double ordinate : Any chord which is perpendicular to the axis of the parabola is called its double ordinate.

Equation of tangent at P(x1, y1) is yy1 = 2a(x + x1) and equation of tangent in slope form is

y = mx + ma

Here point of contact is

ma

ma 2,2

Equation of normal at P (x1, y1) is

y – y1 = ay

21− (x – x1)

and equation of normal in slope form is y = mx – 2am – am3

Here foot of normal is (am2, –2am) The line y = mx + c may be tangent to the parabola if

c = a/m and may be normal to the parabola if c = –2am – am3.

Chord of contact at point (x1, y1) is yy1 = 2a (x + x1) Ellipse : If a point moves in a plane in such a way that ratio of

its distances from a fixed point (focus) and a fixed straight line (directrix) is always less than 1, i.e. e < 1 called an ellipse

Standard equation of an ellipse is 2

2

ax + 2

2

by = 1

where b2 = a2 (1 – e2) Now, When a > b

S´ (–ae,0)A´

(–a,0)A´

(–a,0)

O (0,0)

(ae,0) S XZ

M

L1 B(0,b)

Y

L

Z´X´

M´

L´

Y´

B´(0,– b)

In this position, (i) Major axis 2a and minor axis 2b (ii) Foci, S´(– ae, 0) and S(ae, 0) and centre O(0, 0) (iii) Vertices A´ (– a, 0) and A(a, 0) (iv) Equation of directries ZM and Z´M´ are

x ± ea = 0, Z

0,

ea and Z´

− 0,

ea

(v) Length of latus rectum is ab22 = LL´ = L1L1´

The coordinates of points of intersection of line y = mx + c and the ellipse are given by

++

−222

2

222

2,

mab

b

mab

ma

PARABOLA, ELLIPSE & HYPERBOLA

Mathematics Fundamentals MA

TH

S


Equation of tangents of ellipse in term of m is

y = mx ± 222 mab + and the line y = mx + c is a tangent of the ellipse, if

c = ± 222 mab + The length of chord cuts off by the ellipse from the

line y = mx + c is

222

22222 .12mab

cbmamab+

−++

The equation of tangent at any point (x1, y1) on the ellipse is

21

axx + 2

1

byy = 1

and at the point (a cos φ, b sin φ) on the ellipse, the tangents is

a

x φcos + b

y φsin = 1

Parametric equations of the ellipse are x = a cos θ and y = b sin θ.

The equation of normal at any point (x1, y1) on the ellipse is

1

21)(

xaxx − =

1

21)(

ybyy −

also at the point (a cos φ, b sin φ) on the ellipse, the equation of normal is

ax sec φ – by cosec φ = a2 – b2 Focal distance of a point P(x1, y1) are a ± ex1 Chord of contact at point (x1, y1) is

21

axx + 2

1

byy = 1

Chord whose mid-point is (h, k) is

2ahx + 2b

ky = 2

2

ah + 2

2

bk i.e. T = S1

The locus of point of intersection of two perpendicular tangents drawn on the ellipse is x2 + y2 = a2 + b2. This locus is a circle whose centre is the centre of the ellipse and radius is length of line joining the vertices of major and minor axis. This circle is called "director circle".

The eccentric angle of point P on the ellipse is made by the major axis with the line PO, where O is centre of the ellipse.

(a) The sum of the focal distance of any point on an ellipse is equal to the major axis of the ellipse.

(b) The point (x1, y1) lies outside, on or inside the ellipse f (x, y) = 0 according as f (x1, y1) > = or < 0.

The locus of mid-point of parallel chords of an ellipse

is called its diameter and its equation is y = maxb

2

2−

which is passes through centre of the ellipse.

The two diameter of an ellipse each of which bisect the parallel chords of others are called conjugate diameters. Therefore, the two diameters y = m1x and

y = m2x will be conjugate diameter if m1m2 = – 2

2

ab .

Hyperbola : When the ratio (defined in parabola and ellipse) is

greater than 1, i.e. e > 1, then the conic is said to be hyperbola.

Since the equation of the hyperbola 2

2

ax – 2

2

by = 1

differs from that of the ellipse 2

2

ax + 2

2

by = 1 in

having –b2, most of the results proved for the ellipse are true for the hyperbola, if we replace b2 by – b2 in their proofs. We therefore, give below the list of corresponding results applicable in case of hyperbola.

Standard equation of hyperbola is 2

2

ax – 2

2

by = 1

where b2 = a2 (e2 – 1)

M´ Y M L

L´

XX´

L1

L1´ B´

Y´

A´Z´ O Z (–ae,0)S´

(–a,0)

(0,b

) (0

,0) A (a,0)

S(ae,0)

(0,b

)

B

In this case,

Foci are S (ae, 0) and S´(– ae, 0). Equation of directrices ZM and Z´M´ are

x m ea = 0, Z

0,

ea and Z´

− 0,

ea

Transverse axis AA´ = 2a, conjugate axis BB´ = 2b. Centre O (0, 0).

Length of latus rectum LL´ = L1L1´ = ab22

The difference of focal distance from any point P(x1, y1) on hyperbola remains constant and is equal to the length of transverse axis. i.e.

S´P ~ SP = (ex1 + a) – (ex1 – a) = 2a The equation of rectangular hyperbola

x2 – y2 = a2 = b2 i.e. in standard form of hyperbola put a = b. Hence e = 2 for rectangular hyperbola.


PHYSICS

Questions 1 to 4 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer.

1. In order to measure the spring constant of a spring, mass A, B and C observe the observations in an experiment using the same spring, but blocks of different masses as shown. Assume that Hook's law

is valid and time period is given by T = kM

21π

.

The observations are shown in the table. Least count for mass = 1 gm. Least count for time = 0.1 sec.

M k

Smooth surface

ManMass of block

M (kg)Number of

oscillations nTotal time for (n) oscillations (sec)

Time period (sec)

A 1 20 10 0.5B 1 10 5 0.5C 2 20 14 0.7

If eA, eB, eC, are the percentage errors in calculation

of k, i.e.,

×

∆ 100kk for man A, B, C respectively,

then : (A) eA = eB = eC (B) eB is minimum (C) eA is maximum (D) eC is minimum 2. Which one of the following statements is wrong

about X-rays (A) Frequency of Kα (characteristics) X-ray of Zn

is greater in comparison to Kα X-ray to Ni (B) Cut off wavelength of continuous X-ray

depends on the kinetic energy of the slowest electron in the X-ray tube

(C) For same value of applied accelerating voltage in the X-ray tube, cut off wavelength is same for Ni and Cu target

(D) For constant applied voltage if number of electrons striking per second on the target increases, then intensity of X-ray increases

IIT-JEE 2012

XtraEdge Test Series # 6

Based on New Pattern

Time : 3 Hours

Syllabus : Physics : Calorimetry, K.T.G., Thermodynamics, Heat Transfer, Thermal expansion, Transverse wave, Sound wave, Doppler's effect, Atomic Structure, Radioactivity, X-ray, Nuclear Physics, Matter Waves, Photoelectric Effect, Practical Physics. Chemistry : Chemical Equilibrium, Acid Base, Ionic Equilibrium, Classification & Nomenclature, Isomerism , Hydrogen Family, Boron Family & Carbon Family, S-block elements, Nitrogen Family, Oxygen Family, Halogen Family & Noble Gas, Salt Analysis, Metallurgy, Co-ordination Compounds, Transitional Elements. Mathematics : Point, Straight line, Circle, Parabola, Ellipse, Hyperbola, Vector, 3-D, Probability, Determinants, Matrices.

Instructions : [Each subject contain] Section – I : Question 1 to 4 are multiple choice questions with only one correct answer. +3 marks will be

awarded for correct answer and -1 mark for wrong answer. Section – II : Question 5 to 9 are multiple choice question with multiple correct answer. +4 marks will be awarded for

correct answer and -1 mark for wrong answer.

Section – III : Question 10 to 11 are Column Matching type questions. +8 marks will be awarded for the complete correctly matched answer (i.e. +2 marks for each correct row) and No Negative marks for wrong answer.

Section – IV : Question 12 to 19 are Numerical Response Question (single digit Ans. type) +4 marks will be awarded for correct answer and No Negative marks for wrong answer.


3. A transverse sinusoidal wave is traveling along a string in the negative x-axis. At particular time t, for the shown snap shot of wave, magnitude of displacements of point A, B and C are same. Take velocity of point in positive y direction as positive and in negative direction as negative. At given instant of time velocities of points A, B, C are vA,

vB, vC respectively. Then respective values of B

A

vv

and C

A

vv are -

A B

C x

y

(A) + 1, – 1 (B) + 1, + 2 (C) –1, 1 (D) –1, –1 4. A radioactive sample S1 has activity A and another

sample S2 has activity 3A. The half lives of S1 and S2 are 2T and 5T respectively. The ratio of number of nuclei present in S1 to S2 is -

(A) 5/2 (B) 5/6 (C) 2/15 (D) 3/1

Questions 5 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and –1 mark for each wrong answer. 5. Energy liberated in the de-excitation of hydrogen

atom from 3rd level to 1st level falls on a photo-cathode. Later when the same photo-cathode is exposed to a spectrum of some unknown hydrogen like gas, excited to 2nd energy level, it is found that the de-Broglie wavelength of the fastest photoelectrons, now ejected has decreased by a factor of 3. For this new gas, difference of energies of 2nd Lyman line and 1st Balmer line if found to be 3 times the ionization potential of the hydrogen atom. Select the correct statement(s)

(A) The gas is lithium (B) The gas is helium (C) The work function of photo-cathode is 8.5 eV (D) The work function of photo-cathode is 5.5 eV 6. In a kitchen experiment, you empty a tray of ice

cubes into a bowl of water. After an hour or so, when the mixture has come to thermal equilibrium, you notice a little more water in the bowl than you started with and fewer ice cubes in the bowl than you started with. One can say that –

(A) the temperature of the water is slightly higher than the remaining ice cubes

(B) the temperature of the water is slightly lower than the remaining ice cubes

(C) the temperature of the water is the same as the temperature of the remaining ice cubes

(D) the temperature of the water or the ice cubes depends on the exact mass of water and ice cubes in the bowl.

7. AB : Isothermal (TA = 300 K) BC : Adiabatic (Work = 5J) CD : Constant pressure (5 atm) DE : Isothermal EA : Adiabatic (Change in internal energy 8J) –

P

VC

BA

ED

(A) Work done in the cycle is 13 J (B) Change in internal energy in path CD is – 13J (C) Heat transferred in path DE is –11J (D) Work done in EA is – 8J 8. A standing wave of time period T is set up in a

string clamped between two rigid supports. At t = 0 antinode is at its maximum displacement 2A.

(A) The energy of a node is equal to energy of an antinode for the first time at t = T/8

(B) The energy of node and antinode becomes equal after every T/2 second

(C) The displacement of the particle of antinode at

8Tt = is A2 .

(D) The displacement of the particle of node is zero 9. Density (ρ) veruses temperature (T) graph of a

thermodynamic cycle of an ideal gas is as shown. If BC and AD are the part of rectangular hyperbola then which of the following graphs will represent the same thermodynamic cycle?

Temperature (T)

Density (ρ)

C A

D

B


(A)

Volume (V)

Pres

sure

P C

A D

B

(B)

Volume (V)

Pres

sure

C

A D

B P

(C)

Temperature (T)

Pres

sure

C

A D

B P

(D)

Temperature (T)Pr

essu

re C

A D

B P

This section contains 2 questions (Questions 10, 11). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

A B C D

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T

Mark your response in OMR sheet against the question number of that question in section-II. + 8 marks will be given for complete correct answer (i.e. +2 marks for each correct row) and NO NEGATIVE MARKING for wrong answer.

10. Match the Column : Column-I Column-II (A) ) Ice formation in a (P) Molecular transfer lake leads to uniform heat content (B) The mode not (Q) heat transfer by associated with solid convection (C) Water heated in a (R) Time taken is round Bottomed flask inversely proportional to ambient temperature (D) Land breeze (S) process of heat transfer by conduction (T) free electrons play the role for heat transfer

11. Match the matrix for the nuclear reaction given below :

Column-I Column-II (A) 92U235 + 0n1 → 54Xe

139 + 38Sr94 + 3 0n1 + Energy (B) 4 1H1 = 2He4 + 2β + 2υ + Energy (C) 5B10 + 2He4 → 6C13 + 1H1 + Energy (D) 0n1 → 1H1 + β– + ν + Energy Column-II (P) Converts some matter into energy (Q) Fission (R) Fusion (S) Involves weak nuclear forces (T) is nuclear reaction This section contains 8 questions (Q.12 to 19). +4 marks will be given for each correct answer and –1 mark for each wrong answer. The answer to each of the questions is a SINGLE-DIGIT INTEGER, ranging from 0 to 9. The appropriate bubbles below the respective question numbers in the OMR have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following :

012

3

4

56

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

X Y Z W

12. Figure shows, displacement of a particle on a string transmitting wave along x-axis as a function of time. At t = 0 particle is at half its maximum displacement. Amplitude of wave (in cm) is.

displacement(in meter)

60° 12.5 25.0 37.5 50.0

x

y

t (in π × 10–3 sec)

13. Two boats are floating on a pond in same direction and with the same speed v. Each boat sends through the water, a signal to the other. The frequencies ν0 of the generated signals are the same. Find the ratio of frequencies received by the boats.


14. Two identical stationary sound sources, emit sound waves of frequency 10 Hz, and speed 300 m/sec as shown. An observer is moving between the sources with a velocity 30 m/sec. Find the beat frequency as recorded by the observer (Hz).

S1 O V0 S2

15. The photoelectric current in a vacuum photocell is reduced to zero when its cesium (φ = 1.89eV) electrode is irradiated by radiation of wavelength λ = 2700 Å and a decelerating voltage V = 3V is applied. Then the magnitude of outer contact potential difference is N × 10–1 V, then the value of N is –

16. The De-Broglie wavelength of electron in the third Bohr orbit of hydrogen in 10–9 m is (given radius of first Bohr orbit is 5.3 × 10–11 m) –

17. The binding energy of an electron in the ground state of He atom is equal to 24.6 eV. The energy required to remove both the electrons (if the ionisation energy of hydrogen is 13.6 eV) is N × 101 eV then N is equal to –

18. The nucleus 92U238 is unstable against α-decay with a half-life of about 4.5 × 109 years. Then the kinetic energy of the emitted α-particle in MeV is [m (92U238) = 238.05081u ; m(2He4) = 4.00260 u; m(-90Th234) = 234.04363u]

19. A polonium (84P0209) nucleus transforms into one of

lead (82Pb207) by emitting an α-particle, then the kinetic energy of the α-particle in MeV is -

[m (P0) = 209.98297u ; m (Pb) = 205.97446 m (α-particle) = 4.00260 u]

CHEMISTRY

Questions 1 to 4 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer. 1. Give the IUPAC name of

CH3–CH2–CH2–CH–CH– CH2–CH2–CH3

H3C–CH

CH3

H3C–C–CH3

CH3

(A) 4-isopropyl-5-ter. butyl octane (B) 4-ter. butyl-5-isopropyl octane (C) 2-methyl-3-propyl-4-ter. butyl heptane (D) 2, 2-dimethyl-3-propyl-4-isopropyl heptane

2. Correct order of occlusion property is - (A) Pd > Pt > Au > Colloidal Pd > Pt (B) Colloidal Pd > Pd > Pt > Au > Ni (C) Ni < Au > Pt > Pd > Colloidal Pd (D) Au > Pt > Pd > Ni > Colloidal Pd

3. Metallic sulphates can be obtained by reacting the metals (above hydrogen in ECS), or its oxide, hydroxide or carbonate with dil.H2SO4. Gp IA metals also form hydrogen sulphates which can be isolated in solid. In general metal sulphates are soluble in water and crystallizes with water of crystallization. Sulphate are thermally more stable than nitrates. Select the stable hydrogen sulphate which can be obtained in solid state-

(A) KHSO4 (B) CaHSO4 (C) FeHSO4 (D) All of these

4. The geometrical shapes of XeF5+ , XeF6 and

XeF82– respectively are

(A) trigonal bipyramidal , octahedral and square planar

(B) square based pyramidal , distorted octahedral and octahedral

(C) planar pentagonal , octahedral and square anti prismatic

(D) square based pyramidal, distorted octahedral and square anti prismatic

Questions 5 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and –1 mark for each wrong answer. 5. Which of the following is correct for the given

structures ?

H

HHO

OH

CH3

CH3

(I)

HOH

CH3

H OH

(II)

CH3

H

OH

CH3

CH3

H

OH

(III)


(A) I and II diastereomers (B) I and III enantiomers (C) II and III diastereomers (D) II and III are identical

6. Which of the following is/are correct ?

(A) In COCl2, C Cl bond length is shorter than single bond

(B) Si C bond strength is greater than C C bond strength

(C) Due to smaller size of carbon CO44– does not

exist

(D) CO2 can act as ligand

7. Which of the following is/are correct regarding the active nitrogen ?

(A) It is produced by the passage of electric discharge through the molecular nitrogen

(B) It reacts with hydrocarbon to produce HCN (C) It reacts with H2S to give blue solid (D) When electric discharge is stopped it produces

a yellow afterglow due to recombination of atomic nitrogen

8. K2CrO4 is used identify– (A) Pb2+ (B) Ba2+ (C) Ag+ (D) Ca2+

9. Reduction of a metal oxide by excess carbon at high temperature is a method for the commercial preparation of some metals. This method can be successfully applied in the case of–

(A) BeO and Al2O3 (B) ZnO and Fe2O3 (C) CaO and Cr2O3 (D) BaO and U3O8


A B C D

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T


10. Match the column : Column-I Column-II (A) Complex exhibit no (P) Cis-[Co(en)2

Geometrical isomerism (H2O)2]3+ (B) complex exhibit no (Q) Cis-]Pt(NH3)2 Cl2] optical activity (C) Complex exhibit optical (R) Trans-[Co(en)2

activity (D) Complex exhibit no (S) Ni(CO)4 paramagnetism (T) CaCO3

11. Match the column: Column-I (A) A(g) + B(g) AB(g) (B) A2(g) + B2(g) 2AB(g) (C) A(s) B(g) + C(g)

(D) 21 A2(g) +

23 B2(g) AB3(g)

Column-II (P) Forward shift by addition of inert gas at

constant (Q) Backward shift by addition of inert gas at

constant pressure (R) Not affected by addition of inert gas at constant

volume and also unaffected by increase in pressure

(S) Not affected by addition of small amount of reactant pressure

(T) None of these This section contains 8 questions (Q.12 to 19). +4 marks will be given for each correct answer and –1 mark for each wrong answer. The answer to each of the questions is a SINGLE-DIGIT INTEGER, ranging from 0 to 9. The appropriate bubbles below the respective question numbers in the OMR have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following :

012

3

4

56

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

X Y Z W


12. The total number of geometrical isomers exist for the complex M (AA) B2C2 is …………

13. 1 mole oxygen and 0.375 moles of solid A4 are heated to react completely in a sealed vessel to produce only one gaseous compound B. After the formation of compound B, the vessel is brought to initial temperature, the pressure is found to be half of the initial pressure. Find out number of oxygen atoms per molecules of B ?

14. KSP of Mg(OH)2 is 4.0 × 10–12. The no. of moles of Mg2+ ions in one litre of it's saturated solution in 0.1 M NaOH is report answer in term of ans × 10–10

15. A mixture of KOH and Ca(OH)2 weighing 6.13 gram is completely neutralised by an acid. If weight percentage of KOH in mixture is 45.68 and normality of acid is 20N then find the volume (in ml) of acid used in neutralisation.

16. In borax (Na2 B4O7, 10H2O) the number of B-O-B bonds is ………

17. On heating CaC2O4 , ………type of gases are produced

18. How many cyclic structural isomers are possible with molecular formula C4H7Cl ?

19. In molecule of nitro glycerin the number of N atoms present are …………

MATHEMATICS

Questions 1 to 4 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer. 1. If acute angle between the lines represented by

2x2 + 5xy + 3y2 + 6x + 7y + 4 = 0 is tan–1 m and a2 + b2 – ab – a – b + 1 ≤ 0 then 3a + 2b equals -

(A) m21 (B)

m1 (C) m (D) 2m

2. A rhombus is inscribed in the region common to the two circles x2 + y2 – 4x – 12 = 0 and

x2 + y2 + 4x – 12 = 0 with two of its vertices on the line joining the centres of the circles. The area of the rhombus is -

(A) 38 sq. units (B) 34 sq. units

(C) 36 sq. units (D) None of these

3. A parabola y = ax2 + bx + c crosses the x-axis at (α, 0) and (β, 0) both to the right of the origin. A circle also passes through these two points. The length of a tangent from the origin to the circle is-

(A) abc (B) ac2 (C) b/a (D)

ac

4. Let →a ,

→b &

→c be non-coplanar unit vectors

equally inclined to one another at an acute angle θ.

Then |[→a

→b

→c ]| in terms of θ is equal to -

(A) (1 + cos θ) θ2cos

(B) (1 + cos θ) θ− 2cos21

(C) (1 – cos θ) θ+ cos21 (D) None of these

Questions 5 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and –1 mark for each wrong answer. 5. Let PN be perpendicular from point P(1, 2, 3) to xy

plane if OP makes an angle α with positive direction of z-axis and ON makes an angle β with positive direction of x-axis, where O is origin (α, β are acute angles), then-

(A) sin α sin β = 142 (B) cos α cos β =

141

(C) tan α = 35 (D) None of these

6. A(1, 2) and B(7, 10) are two points. If P(x, y) is a point such that the angle APB is 60º and the area of the triangle APB is maximum then which of the following is (are) true ?

(A) P lies on any line perpendicular to AB (B) P lies on the right bisector of AB (C) P lies on the straight line 3x + 4y = 36 (D) P lies on the circle passing through the points

(1, 2) and (7, 10) and having a radius of 10 units

7. A circle centred at O has radius 1 and contains the point A. Segment AB is tangent to the circle at A and ∠AOB = θ. If point C lies on OA and BC bisects the angle ABO then OC equals

O C A

B

θ


(A) sec θ (sec θ – tan θ) (B) θ+

θsin1

cos2

(C) θ+ sin1

1 (D) θθ−

2cossin1

8. Let the pair of tangents drawn from (0, 2) to the

parabola y2 – 2y + 4x + 5 = 0 and the normal at the point of contact of tangents form a quadrilateral

(A) quadrilateral will be square (B) tangents drawn are intersecting at 90º (C) quadrilateral will be a rectangle (D) the point (0, 2) lies on the directrix of the parabola

9. If P is a point of the ellipse 2

2

2

2

by

ax

+ = 1, whose

focii are S and S ′. Let ∠PSS′ = α and ∠PS′S = β, then

(A) PS + PS′ = 2a, If a > b (B) PS + PS ′ = 2b, if a < b

(C) tan 2α tan

2β =

ee

+−

11

(D) tan2α tan

2β = 2

22

bba − [a – 22 ba − ]

when a > b


A B C D

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T


10. Match the Column : Column-I Column-II (A) If (α, β) is a point on circle whose (P) 22 centre is on x-axis which touches line y = – x at (2, –2) then greatest value of α is (B) The sum of squares of length (Q) 0 of chords intercepted by line x + y = n, n ∈ N on circle x2 + y2 = 4 (C) If point (a, a) lies between lines (R) 1 | x + y | = 6 then [ | a | ] is ([ . ] denote greatest integer) (D) No. of points from where (S) 4+ 2 2 perpendicular tangents can be

drawn on 154

22=−

yx (T) 2

11. Match the column : Column-I Column-II

(A) ) If two numbers are chosen (P) 25693

from odd natural numbers less than hundred and multiplied together in all possible ways, then the probability of the product to be divisible by 5 is

(B) A bag 'A' contains 2 white (Q) 66324

and 3 red balls another bag 'B' contains 4 white and 5 red balls. If one ball is drawn at random from one bag and it is found to be red, then the probability that it was drawn from the bag B is

(C) Eight coins are tossed at a time, (R) 24589

the probability of getting heads up in majority is

(D) From a well shuffled pack of (S) 5225

52 playing cards, if cards are drawn one by one without replacement till black ace comes, then probability that the black ace comes in 4th draw is

(T) 66393


This section contains 8 questions (Q.12 to 19). +4 marks will be given for each correct answer and –1 mark for each wrong answer. The answer to each of the questions is a SINGLE-DIGIT INTEGER, ranging from 0 to 9. The appropriate bubbles below the respective question numbers in the OMR have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following :

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

X Y Z W

12. Locus of the poles of the tangents to the hyperbola

x2 – y2 = (39)2 with respect to the parabola y2 = 156x is the ellipse 4x2 + y2 = k. 'S' is sum of digits in k. Find number of divisors of S.

13. cbarrr

,, are three unit vectors and every two are inclined to each other at an angle cos–1(3/5). If

crbqapbarrrrr

++=× , where p, q, r are scalars, then 55q2 is equal to

14. Let ai , i = 1, 2, 3 ..., n denotes the integers in the

domain of function f (x) =

−−

21254log 2/1 x

x

where ai < ai+1 ∀ i ∈ N. If the line

L : 4

2 1ax − = 2

1

aay + =

5

3

aaz − meets the xy, yz

and zx planes at A, B and C respectively, and if volume of the tetrahedron OABD is V, where 'O', is the origin and D is the image of C in the x-axis,

then find the value of 7V9 .

15. Two rays with common end point 'O' form a 30° angle. Point A lies on one ray. Point B on the other ray and AB = 1. Find the maximum possible length of OB.

16. A line 'L' is drawn from (4, 3) to meet the lines L1 : 3x + 4y + 5 = 0 and L2 : 3x + 4y + 15 = 0 at points A and B respectively. From 'A' a line, perpendicular to L is drawn meeting the line L2 at A1. Similarly, from point 'B' a line, perpendicular to L is drawn meeting the line L1 at B1. Thus parallelogram AA1BB1 is formed. Least value of area of parallelogram AA1BB1 is.....

17. If the tangent at P to the parabola y2 = 7x is parallel

to the line 6y – x + 11 = 0, then square of the distance of P from the vertex of the parabola is D.

find

1000

D where [ . ] = GIF

18. An ellipse of eccentricity 32 is inscribed in an

ellipse of equal eccentricity and area equals to 9 square units in such a way that both the ellipse touch each other at end of their common major axis. If length of major axis of smaller ellipse is equal to length of minor axis of bigger ellipse, find the area of the bigger ellipse outside the smaller ellipse.

19. Nine tiles are numbered 1, 2, 3, 4, 5, 6, 7, 8, 9

respectively. Each of the three players A, B and C randomly selects 3 tiles and they sum up those three values as marked on the tiles. The probability

that all three players obtain an odd sum is nm ,

where m and n are relatively prime positive

integers. Compute the value of 17

)( nm + .

Puzzle : Crates of Fruit

• You are on an island and there are three crates

of fruit that have washed up in front of you. One crate contains only apples. One crate contains only oranges. The other crate contains both apples and oranges.

• Each crate is labeled. One reads "apples", one reads "oranges", and one reads "apples and oranges". You know that NONE of the crates have been labeled correctly - they are all wrong.

• If you can only take out and look at just one of the pieces of fruit from just one of the crates, how can you label ALL of the crates correctly?


PHYSICS

Questions 1 to 4 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer. 1. There are two parts of a vessel. The pressure in one

part is P and its volume is V. The volume of another part is 4V and there is vacuum in it. If the intervening wall is ruptured, then work done by the gas and change in its internal energy will be –

P

V

Vacuum

4V (A) δW = 2PV, dU = – ve

(B) δW = 3PV, dU = 0 (C) δW = 0, dU = + ve

(D) δW = 0, dU = 0

2. A mixture of 8gm of helium and 14gm of nitrogen is enclosed in a vessel of constant volume at 300K. The quantity of heat absorbed by the mixture to double the root mean velocity of its molecules is –

(R = universal gas constant ) (A) 2725 R (B) 3630 R (C) 3825 R (D) 5625 R

3. Two particles of medium disturbed by the wave

propagation are at x1 = 0 and x2 = 1 cm. The respective displacement (in cm) of the particles can be given by equations

t3sin2y1 π=

π

−π=8

t3sin2y2

wave velocity can be - (A) 16 cm/s (B) 24 cm/s (C) 12 cm /s (D) 8 cm/s 4. An open organ pipe of length ‘L’ vibrates in second

harmonic mode. The pressure variation is maximum: (neglect end corrections)

(A) at the two ends

(B) at a distance 4L from either end inside the tube

(C) at the mid-point of the tube (D) none of these Questions 5 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and –1 mark for each wrong answer.

Time : 3 Hours Syllabus : Physics : : Calorimetry, K.T.G., Thermodynamics, Heat Transfer, Thermal expansion, Transverse wave, Sound wave, Doppler's effect. Chemistry : Chemical Equilibrium, Acid Base, Ionic Equilibrium, Classification & Nomenclature, Isomerism, Hydrogen Family, Boron Family & Carbon Family, S-block elements. Mathematics : Point, Straight line, Circle, Parabola, Ellipse, Hyperbola, Vector, 3-D Instructions : [Each subject contain] Section – I : Question 1 to 4 are multiple choice questions with only one correct answer. +3 marks will be awarded

for correct answer and -1 mark for wrong answer. Section – II : Question 5 to 9 are multiple choice question with multiple correct answer. +4 marks will be awarded for correct

answer and -1 mark for wrong answer.

Section – III : Question 10 to 11 are Column Matching type questions. +8 marks will be awarded for the complete correctly matched answer (i.e. +2 marks for each correct row) and No Negative marks for wrong answer.

Section – IV : Question 12 to 19 are Numerical Response Question (single digit Ans. type) +4 marks will be awarded for correct answer and No Negative marks for wrong answer.

IIT-JEE 2013

XtraEdge Test Series # 6

Based on New Pattern


5. One gram molecule of nitrogen occupies 2 × 104 cm3 at a pressure of 106 dyne cm–2. Given : NA = 6 × 1023. Which of the following is correct ?

(A) The value of kT is 1/3 × 10–13 erg (B) The value of kT is 1/4 × 10–13 erg (C) Mean kinetic energy per molecule is 5 × 10–14 erg (D) Mean kinetic energy per molecule is 9.8 erg

6. Three identical rods of same material are joined to form a triangular shape ABC as shown. Angles at edge A and C are respectively θ1 and θ2 as shown. When this triangular shape is heated then –

A

B C

θ1

θ2

(A) θ1 decreases and θ2 increases (B) θ1 increases and θ2 decreases (C) θ1 increases (D) θ2 increases

7.

y(mm)

x(m)

105

–5–10

figure 1

t = 0

y(mm)

x(m)

105

–5–10

figure II

1

t = 124

s

The figures represent two snaps of a traveling wave on a string of mass per unit length µ = 0.25 kg/m. The two snaps are taken at time t =

0 and at s241t = . Then -

(A) speed of wave is 4 m/s (B) the tension in the string is 4 N (C) the equation of the wave is

π

+π−π=6

t4xsin10y

(D) the maximum velocity of the particle = s/m25π

8. Choose the correct statement(s) from the following- (A) any function of the form y (x, t) = f (vt – x)

represents a travelling wave (B) The velocity, wavelength and frequency of a

wave do not undergo any change when it is reflected from a surface

(C) When an ultrasonic wave travels from air into water, it bends towards the normal to the air-water interface

(D) the velocity of sound is generally greater in solids than in gases at NTP

9. A source of sound of constant frequency f is moving in a circular track of constant radius R with constant angular velocity ω. A stationary observer is observing the motion as shown. θ is the angle made by source corresponding to sound which is heard by observer later on -

R

θ

ω

Observer

(A) Observer detects minimum frequency at θ = π/2 (B) Observer detects minimum frequency at θ = ( 3π/2) (C) Observer detects minimum frequency at θ = π/2 (D) Observer detects minimum frequency at θ = (3π/2)


ABCD

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T


10. Match the column: Column-I

(A) Longitudinal waves can (B) Transverse wave can (C) The speed of sound in a medium depends on (D) Light wave can Column-II (P) Have a unique wavelength (Q) Have a unique wave velocity (R) Be polarized (S) Elastic property as well as inertia property (T) None of these

11. Match the column : Column-I (A) Ice formation in a lake (B) The mode not associated with solids (C) Water heated in a round Bottomed flask (D) Land breeze


Column-II (P)Molecular transfer leads to uniform heat content

(Q) heat transfer by convection (R) Time taken is inversely proportional to ambient

temperature (S) process of heat transfer by conduction (T) None of these


0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

X Y Z W

12. Aluminium container of mass 100 g contains 200 g

of ice at –20ºC. Heat is added to the system at the rate of 100 calories per second. Temperature of the system after four minutes is T then what is the value of T/5?

Given : Specific heat of ice = 0.5 cal g–1(ºC)–1 Specific heat of aluminium = 0.2 cal g–1(ºC)–1 Latent heat of fusion of ice = 80 cal g–1

13. A mixture of 250 g of water and 200 g of ice at 0ºC

is kept in a calorimeter of water equivalent 50 g. If 200 g of steam at 100ºC is passed through this mixture. Mass of contents of the calorimeter is m then find the value of m/5. Latent heat of fusion of ice = 80 cal g–1 and latent heat of vaporisation of water = 540 cal g–1.

14. The temperature of equal masses of three different

liquids A,B and C are 12ºC, 19ºC and 28ºC respectively. The temperature when A and B are mixed is 16ºC and when B and C are mixed it is 23ºC. The temperature when A and C are mixed is TºC then find the value of T/5?

15. About 5 g of water at 30ºC and 5 g of ice at –20ºC are mixed together in a calorimeter. Find the final temperature of the mixture. Water equivalent of the calorimeter is negligible.

Specific heat of ice = 0.5 cal g–1(ºC)–1 Latent heat of fusion of ice = 80 cal g–1

16. A lead bullet just melts when stopped by an obstacle. Assuming that 25% of heat is absorbed by the obstacle. Velocity of the bullet is v then find the value of v/100. (If its initial temperature is 27ºC.) Given,

Melting point of lead = 327ºC Specific heat of lead = 0.03 cal g–1(ºC)–1 Latent heat of fusion of lead = 6 cal g–1

and J = 4.2 joule per calorie.

17. An ice cube of mass 0.1 kg of 0ºC is placed in an isolated container which is at 227ºC. The specific heat S of the container varies with temperature T according to the empirical relation S = A + BT, where A = 100 cal/kg-K and B = 2 0 × 10–2cal/kg-K2. If the final temperature of the container is 27ºC. The mass of the container is x then find the value of 2x. (Latent heat of fusion of water = 8 × 104 cal/kg, sp. heat of water = 103cal/kg-K).

18. In a quink tube experiment a tuning fork of frequency 300 Hz is vibrated at one end. It is observed that intensity decreases from maximum to 50 % of its maximum value as tube is moved by 6.25 cm. Velocity of sound (in m/s) is v then find the value of v/100.

19. A long spring such as slinky is often used to demonstrate longitudinal waves. If mass of spring is m, length L and force constant K, then find the speed of longitudinal waves on the spring where m = 0.250 kg, L = 2.00 m K = 1.50 N/m.

CHEMISTRY

Questions 1 to 4 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer. 1. The incorrect IUPAC name is - (A) CH3–CH–CH–CH3

Cl Br2-Bromo-3-Chlorobutane

(B) CH3–CH–CH–CH3

CH3 CH2–CH3

2,3-Dimethylpentane


(C) CH3–C≡C–CH(CH3)2 4-Methylpent-2-yne

(D) CH3–C–CH–CH3

O

CH3

2-Methyl-3-butanone

2. Maleic acid and fumaric acid are - (A) position isomers (B) Functional isomers (C) Geometrical isomers (D) optical isomers

3. Which of the following is not correct - (A) [H+] = [OH–] = wK for a neutral solution at

all temperature (B) [H+] = [OH–] = 10–7 for a neutral solution at all

temperature (C) [H+] > wK and [OH–] < wK for an acidic

solution (D) H+ < wK and [OH–] > wK for an alkaline

solution

4. Which of the following constitutes a set amphoteric species -

(A) H3O+ , −42POH , −

3HCO

(B) H2O , −24HPO , −

22POH

(C) H2O , −32POH , −2

4HPO

(D) −42OHC , −

42POH , −24SO

Questions 5 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and –1 mark for each wrong answer. 5. Which one is correct sets only for heterocyclic

aromatic compounds ? (A) Oxirane, Furan, Oxitane (B) Pyridine, Aziridine, Benzene (C) 2-Bromothiophene, Quinoline, Furan (D) Isoquinoline, Pyrrole, Quinoline

6. For the reaction : A2(g) + 3B2(g) 2 AB3(g) ∆H = –329 KJ, Dissociation of AB3(g) will be favoured by : (A) increasing the temperature (B) increasing the volume of the container (C) adding of B2(g) gas (D) adding of inert gas at constant pressure

7. Atomic numbers of Cr and Fe are 24 and 26 respectively, which of the following is diamagnetic?

(A) Cr(CO)6 (B) Fe(CO)5 (C) Fe(CN)6

4– (D) Cr(NH3)63+

8. Which of the following hydrides are covalent and polymeric ?

(A) Ca (B) Ba (C) Be (D) Mg

9. Select correct statements - (A) Ca3(PO4)2 is part of bones and 3Ca3(PO4)2.

CaF2 is part of enamel in teeth (B) Ca2+ ions are important in blood clotting (C) BeH2 and MgH2 are covalent and polymeric

while CaH2 , SrH2 and BaH2 are ionic (D) BeH2 contain three-centre two-electron bonds


ABCD

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T


10. Match the Column : Column-I

(A)

NH2

H CH2OH

CH3

and

OH

H3C CH2NH2

H

(B)

Et

H Cl

CH3

and

Et

H CH3

Cl

(C)

Et

H OH

CH3

and

OH

H3C EtH

(D)

H

H3C2

OH

and

OH

H3C2

H

Column-II (P) Structural (Q) Identical (R) Enantiomers (S) Diastereomers (T) Chain ring isomer


11. Match the column: Column-I Column-II (A) Two electron (P) (BN)x

three centre bond (B) Four electron (Q) B2H6

three centre bond

(C) sp3 hybrid orbitals (R) AlCl3 (D) Inorganic graphite (S) B4H10 (T) P4O10


0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

X Y Z W

12. The total number of Acyclic isomers possible for

C3H6O is/are …………

13. Consider the reaction AB2(g) ABg + B(g). It the initial pressure of AB2 is 100 torr and equilibrium pressure is 120 torr. The equilibrium constant Kp in terms of torr is.

14. Fe(CO)5 + n NO → The value of n is ………

15. 1 mole of B2H6 (g) on hydrolysis yields ……… moles of H2(g)

16. In the equilibrium MgCO3(s) MgO(s) + CO2(g), 2COP = 1 atm, hence, Kp = ......

17. Find total number of stereocentre. H

H

OH NH2

18. The total number of geometrical isomers of 2, 4-hexadienoic acid is.

19. What is OH– ion concentration in 0.01M solution of aniline (Kb for aniline = 4 × 10–10). Report your answer in terms of ans × 10–11 (if ans is x × 10–11 then report x)

MATHEMATICS

Questions 1 to 4 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer. 1. If G is the centroid and I the incentre of the triangle

with vertices A(–36, 7), B(20, 7) and C(0, – 8), then GI is equal to -

(A) 3/173 (B) 3/397

(C) 3/205 (D) None of these

2. The distance of the point B with position vector i + 2j + 3k from the line passing through the point A whose position vector is 4i + 2j + 2k and parallel to the vector 2i + 3j + 6k is

(A) 10 (B) 5 (C) 6 (D) 8

3. The locus of the points of intersection of the tangents at the extremities of the chords of the ellipse x2 + 2y2 = 6 which touch the ellipse x2 + 4y2 = 4 is -

(A) x2 + y2 = 4 (B) x2 + y2 = 6 (C) x2 + y2 = 9 (D) none of these

4. PN is an ordinate of the parabola y2 = 9x. A straight line is drawn through the mid-point M of PN parallel to the axis of the parabola meeting the parabola at Q. NQ meets the tangent at the vertex A, at a point T, then AT/NP =

(A) 3/2 (B) 4/3 (C) 2/3 (D) ¾

Questions 5 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and –1 mark for each wrong answer. 5. For all values of θ, the lines represented by the

equation (2 cos θ + 3 sin θ) x + (3 cos θ – 5sin θ) y

– (5 cos θ – 2sin θ) = 0 (A) pass through a fixed point (B) pass throught the point (1, 1) (C) reflection of the fixed point in the line x + y = 2 is ( 2 – 1, 2 – 1) (D) pass through the origin if tan θ = 5/2


6. If the circle x2 + y2 + 2gx + 2fy + c = 0 cuts each of the circles x2 + y2 – 4 = 0, x2 + y2 – 6x – 8y + 10 = 0 and x2 + y2 + 2x – 4y – 2 = 0 at the extremities of a diameter, then -

(A) c = – 4 (B) g + f = c – 1 (C) g2 + f 2 – c = 17 (D) g f = 6

7. If the two lines represented by x2 (tan2θ + cos2θ) – 2x y tan θ + y2 sin2 θ = 0 make angle α, β with the x-axis, then

(A) tan α + tan β = 4 cosec 2θ (B) tan α tan β = sec2 θ + tan2θ (C) tan α – tan β = 2

(D) βα

tantan =

θθ+

2sin–22sin2

8. If P, Q, R are three points on a parabola y2 = 4ax whose ordinates are in geometrical progression, then the tangents at P and R meet on

(A) the line through Q parallel to x-axis (B) the line through Q parallel to y-axis (C) the line joining Q to the vertex (D) the line joining Q to the focus.

9. The coordinates of the end point of the latus rectum of the parabola (y – 1)2 = 2(x + 2), which does not lie on the line 2x + y + 3 = 0 are

(A) (– 2, 1) (B) (–3/2, 1) (C) (– 3/2, 2) (D) (– 3/2, 0)


A B C D

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T


10. Match the column: Column-I Column-II (A) Length of the tangent (P) a/2 from (– a, 0) to y2 = 4ax (B) Length of a latus rectum (Q) 17/5a of x2 + 4y2 = a2

(C) Length of the chord of (R) 2a the ellipse x2 + 4y2 = a2 joining an end point of the major axis with an end point of the minor axis

(D) Length of the perpendicular (S) 2/5a from the centre of the ellipse x2 + 4y2 = a2 on the tangent at the point where it intersects the line y = x (T) 5/2

11. Match the column: Column-I Column-II (A) A point on the plane (P) (1, 2, 3) 2x + 4y – 5z + 15 = 0 is

(B) A point on the line (Q) (2, 4, 7)

2

1+x = 4

2+y = 5

2+z is

(C) A point on both (R) (5, 10, 13) 2x + 4y – 5z + 15 = 0

and 2

1+x = 4

2+y = 5

2+z is

(D) Mid-point of the points (S) (3/2, 3, 5) joining (A) and (B) (T) None This section contains 8 questions (Q.12 to 19). +4 marks will be given for each correct answer and –1 mark for each wrong answer. The answer to each of the questions is a SINGLE-DIGIT INTEGER, ranging from 0 to 9. The appropriate bubbles below the respective question numbers in the OMR have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following :

012

3

4

56

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

X Y Z W


12. If a = i + 2j – 3k, b = 2i + j – k and u is a vector satisfying a × u = a × b and a . u = 0 then 2 |u|2 is equal to ____ .

13. If the tangent at the point P on the circle x2 + y2 + 6x + 6y = 2 meets the straight line 5x – 2y + 6 = 0 at a point Q on the y-axis then the length of PQ is

14. The distance between the parallel lines given by (x + 7y)2 + 24 (x + 7y) – 42 = 0 is

15. The radius of the circle passing through the foci of

the ellipse 916

22 yx+ = 1 and having its centre at

(0, 3) is

16. Vertices of a triangle are (0, 0) (41α, 37) and (– 37, 41β) where α and β are the roots of the equation 3x2 – 16x + 15 = 0. The area of the

triangle is k, then 543k is equal to

17. The volume of the tetrahedron whose vertices are the points with position vectors i – 6j + 10k, – i – 3j + 7k, 5i – j + λk and 7i – 4j + 7k is 11 cubic units if the value of λ is (λ > 0)

18. If the planes x = cy + bz, y = az + cx and z = bx + ay pass through a line, then a2 + b2 + c2 + 2abc is equal to

19. If the sum of the intercepts made by the lines y = x + 2, y = 2x + 3, y = 3x + 4, ... y = 50x + 51, on

y-axis is 2

)50(51 k+× – 1, then k is.

Brief description: nickel is found as a constituent in most meteorites and often serves as one of the criteria for distinguishing a meteorite from other minerals. Iron meteorites, or siderites, may contain iron alloyed with from 5 to nearly 20% nickel. The USA 5-cent coin (whose nickname is "nickel") contains just 25% nickel. Nickel is a silvery white metal that takes on a high polish. It is hard, malleable, ductile, somewhat ferromagnetic, and a fair conductor of heat and electricity. Nickel carbonyl, [Ni(CO)4], is an extremely toxic gas and exposure should not exceed 0.007 mg M-3.

Basic information Name: Nickel Symbol: Ni Atomic number: 28 Atomic weight: 58.6934 (2) Standard state: solid at 298 K Group in periodic table: 10 Group name: (none) Period in periodic table: 4 Block in periodic table: d-block Colour: lustrous, metallic, silvery tinge Classification: Metallic Small and large samples of nickel foil like this, as well as sheet, wire, mesh and rod (and nickel alloys in foil, sheet, wire, insulated wire and rod form) can be purchased from Advent Research Materials via their web catalogue. ISOLATION : Isolation: it is not normally necessary to make nickel in the laboratory as it is available readily commercially. Small amounts of pure nickel can be islated in the laborotory through the purification of crude nickel with carbon monoxide. The intermediate in this process is the highly toxic nickel tetracarbonyl, Ni(CO)4. The carbonyl decomposes on heating to about 250°C to form pure nickel powder.

Apples and Friends

You have a basket containing ten apples. You have ten friends, who each desire an apple. You give each of your friends one apple.After a few minutes each of your friends has one apple each, yet there is an apple remaining in the basket. How ?


XtraEdge Test Series ANSWER KEY

PHYSICS Ques 1 2 3 4 5 6 7 8 9 Ans D B C C B,C C B,D C,D A,C

Ques 10 11 Column Match Ans (A) → R,S,T (B) → Q (C) → Q (D) → Q (A) → P,Q (B) →P,R (C) →P,R(D) → P,S

Ques 12 13 14 15 16 17 18 19 Numerical Response Ans 5 2 2 3 1 8 4 5

CHEMISTRY

Ques 1 2 3 4 5 6 7 8 9 Ans B B A D A,C A,B,C,D A,B,C,D A,B A,C

Ques 10 11 Column Match

Ans (A) → S (B) → Q,R,S (C) → P (D) → P,Q,RS

(A) → Q (B) → R (C) →P,S (D) → Q


MATHEMATICS

Ques 1 2 3 4 5 6 7 8 9 Ans B A D C A,C B,C A,C,D B,C,D A,B,C

Ques 10 11 Column Match Ans (A) → S (B) → P (C) → Q,R,T (D) → Q (A) → R (B) → S (C) → P (D) → Q


PHYSICS Ques 1 2 3 4 5 6 7 8 9 Ans D C B B A,C C,D A,B,C,D A,B,D C,D

Ques 10 11 Column Match

Ans (A) → P,Q (B) → P,Q,R (C) → S (D) → P,Q,R

(A) → R,S (B) → Q (C) → P,Q (D) → Q


CHEMISTRY

Ques 1 2 3 4 5 6 7 8 9 Ans A C B C C,D A,B,D A,B,C C,D A,B,C,D

Ques 10 11 Column Match Ans (A) → P (B) → R (C) → Q (D) → S (A) → Q,R (B) → S (C) → Q,R (D) → P


MATHEMATICS Ques 1 2 3 4 5 6 7 8 9 Ans C A C C A,B,C,D A,B,C,D A,C,D B C

Ques 10 11 Column Match Ans (A) → R (B) → P (C) → S (D) → Q (A) → Q (B) → P (C) → R (D) → S


IIT- JEE 2012 (October issue)

IIT- JEE 2013 (October issue)


'XtraEdge for IIT-JEE

IIT JEE becoming more competitive examination day by day. Regular change in pattern making it more challenging.

"XtraEdge for IIT JEE" magazine makes sure you're updated & at the forefront. Every month get the XtraEdge Advantage at your door step.

Magazine content is prepared by highly experienced faculty members on the latest trend of IIT JEE. Predict future paper trends with XtraEdge Test Series every month to give students practice, practice & more practice. Take advantage of experts' articles on concepts development and problem solving skills Stay informed about latest exam dates, syllabus, new study techniques, time management skills and much more XtraFunda. Confidence building exercises with Self Tests and success stories of IITians Elevate you to the international arena with international Olympiad/Contests problems and Challenging Questions.

SUBSCRIPTION FORM FOR “EXTRAEDGE FOR IIT-JEE

The Manager-Subscription, “XtraEdge for IIT-JEE” Career Point Infosystems Ltd, 4th Floor, CP-Tower, IPIA, Kota (Raj)-324005 I wish to subscribe for the monthly Magazine “XtraEdge for IIT-JEE”

Half Yearly Subscription (Rs. 100/-) One Year subscription (Rs. 200/-) Two year Subscription (Rs. 400/-) I am paying R. …………………….through

Money Order (M.O) Bank Demand Draft of No………………………..Bank………………………………………………………..Dated

(Note: Demand Draft should be in favour of "Career Point Infosystems Ltd" payable at Kota.) Name: _______________________________________________________________________________________ Father's Name: _______________________________________________________________________________________

Address: _______________________________________________________________________________________

________________________City_____________________________State__________________________

PIN_________________________________________Ph with STD Code __________________________

Class Studying in ________________E-Mail: ________________________________________________

From months: ____________________to ________________________________________________

Subscription Offer for Students

Special Offer

C


XtraEdge for IIT-JEE

IIT JEE becoming more competitive examination day by day. Regular change in pattern making it more challenging.

"XtraEdge for IIT JEE" magazine makes sure you're updated & at the forefront. Every month get the XtraEdge Advantage at your door step.

Magazine content is prepared by highly experienced faculty members on the latest trend of the IIT JEE. Predict future paper trends with XtraEdge Test Series every month to give students practice, practice & more practice. Take advantage of experts' articles on concepts development and problem solving skills Stay informed about latest exam dates, syllabus, new study techniques, time management skills and much more XtraFunda. Confidence building exercises with Self Tests and success stories of IITians Elevate you to the international arena with international Olympiad/ Contests problems and Challenging Questions.

FREE SUBSCRIPTION FORM FOR “EXTRAEDGE FOR IIT-JEE

The Manager-Subscription, “XtraEdge for IIT-JEE” Career Point Infosystems Ltd, 4th Floor, CP-Tower, IPIA, Kota (Raj)-324005 We wish to subscribe for the monthly Magazine “XtraEdge for IIT-JEE”

Half Yearly Subscription One Year subscription Two year Subscription

Institution Detail:

Graduate Collage Senior Secondary School Higher Secondary School Name of the Institute: _____________________________________________________________________________ Name of the Principal: _____________________________________________________________________________

Mailing Address: _____________________________________________________________________________

__________________City_________________________State__________________________

PIN_____________________Ph with STD Code_____________________________________

Fax_______________________________ E-Mail_____________________________________

Board/ University: _____________________________________________________________________________________

School Seal with Signature

Subscription Offer for Schools

C

C

teachers open the door. you enter by yourself...civil engineer and river management specialist,...

Documents