worry is a misuse of imagination. - career point · ravindra bharathi iit olympiad school conducted...

XtraEdge for IIT-JEE 1 MAY 2011

Dear Students,

All of us live and work within fixed patterns. These patterns and habits determine the quality of our life and the choices we make in life. There are a few vital things to know about ourselves. We should become aware of how much we influence others, how productive we are and what can help us to achieve our goals. It is important to create an environment which will promote our success. We should consciously create a system that would enable us to achieve our goals. Most of us live in systems which have come our way by an accident, circumstances or people we have met over a period of time. We are surrounded by our colleagues or subordinates who happened to be there by the fact of sheer recruitment earlier or later by the management. Our daily routines and schedules have been formed on the basis of convenience, coincidence, and the expectations of society and sometimes due to superstitions. The trick for success is to have an environment that helps in attaining our goals. Control your life. Make an effort to launch your day with a great start. A law of physics says that an object set in motion tends to remain in motion. It is the same thing with daily routine. To have a good start each morning will keep you upbeat during the day. If you begin the day stressed, you will tend to remain so that way. The best is to create a course of action or conditions where you are not hassled for being late for a meeting, worried about household affairs or distracted by happenings in the world. Aim to be highly successful. Control the direction of your life. Not only should you start the day on a cheerful note but also continue to do so during the day. Keep yourself stimulated and invigorated during the entire day. Start your day with a purpose. Have a daily direction and trajectory of action. It will keep you on your course all day long. Throughout the day reinforce your positive values and your choices. Anything that helps you in maintaining your highest values and your most important priorities should be welcome. Be in control of your life and work. Create and sustain a wonderful environment filled with beauty, peace, inspiration and hope. Plan your day in such a way that suits your plans objectives and makes you feel just right with the right amount of encouragement during the entire day. You should give a direction to your day and timing. Presenting forever positive ideas to your success.

Yours truly

Pramod Maheshwari, B.Tech., IIT Delhi

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Worry is a misuse of imagination. Volume - 6 Issue - 11

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

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Volume-6 Issue-11 May, 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

AIEEE 2011 Examination Paper & Solution

S

Success Tips for the Months

• If you can't make a mistake, you can't make anything.

• Sometimes a big step is safer; you can't cross a ditch in small jumps

• Self-confidence grows not from what you can do, but what you know you can do.

• Children focus on what they can’t do. Adults focus on what they can do.

• The secret of confidence is to know your resources.

• You never need to feel fear if you don't want to do anything.

• You got to know when to hold ‘em and know when to fold ‘em…

• An ounce of success is worth a pound of positive thinking.

• To understand motivation, know the power of the Hunter.

• Defeat is advance payment for victory.

CONTENTS

INDEX PAGE

NEWS ARTICLE 4 • Pollen, Nanoparticles, and Asthma : The Aerosol Science of Allergic Disease • Seminar on IIT olympiad Courses

IITian ON THE PATH OF SUCCESS 6 Mr. Narayanaswamy

KNOW IIT-JEE 7 Previous IIT-JEE Question

XTRAEDGE TEST SERIES 46

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

IIT - 2011 Examination Paper with Solution 62

Regulars ..........

DYNAMIC PHYSICS 14

8-Challenging Problems [Set # 1] Students’ Forum Physics Fundamentals • Electrostatics-I • 1-D Motion, Projectile Motion

CATALYSE CHEMISTRY 30

Key Concept • Gaseous State • General Organic Chemistry Understanding : Physical Chemistry

DICEY MATHS 38

Mathematical Challenges Students’ Forum Key Concept

• Complex Number • Matrices & Determinants

Study Time........

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


"Pollen, Nanoparticles, and Asthma: The Aerosol Science of Allergic Disease"

Pollen particles are so large that inhaled pollen deposits primarily in the nose and throat, yet allergists warn asthmatics that pollen may trigger airway inflammation deep in the lungs, leading to bronchoconstriction. At the same time, epidemiological studies yield conflicting results on the link between pollen and asthma, but do show links between exposures to fine particles and asthma. What began as an effort to unravel the physics behind this seemingly contradictory advice stimulated studies of wind-pollinating plants, pollen grains, and respirable allergens. Botanists had observed that live pollen grains will rupture when immersed in water, and speculated that the cytoplasmic material was the source of respirable allergen that could trigger asthma. Initial experiments confirmed our speculation that adhesion forces at the micrometer scale are too strong for the mechanism to explain the link, at least as it was originally stated. This seminar will examine how the structures of the rarely noticed flowers enable pollen antigen to be released in respirable particles. Along the way, we will examine some of the remarkable methods that wind-pollinating plants use to release their pollen. The new understanding of the ways that pollen antigen enters the air, and the wide range of sizes of particles in which it is found poses new challenges for the measurement of these airborne allergens. Present measurements are limited to manual microscopic analysis of samples collected from the air, a tedious and labor intensive undertaking. As a result, the number of locations where historical pollen counts have been recorded is very limited, challenging efforts both to quantify the association

between pollen and asthma and to understand how climate change may affect allergic disease in the coming decades. Measurement of the respirable allergen have only been reported for a few localized studies. A computer vision system has been developed to address the former problem. An ultrasensitive, label-free antibody assay based upon high-Q whispering gallery mode optical resonators is under development to address the latter.

Seminar on IIT Olympiad Courses Ravindra Bharathi IIT Olympiad School conducted an IIT Olympiad awareness seminar here on Saturday at Ravindra Bharathi School in Kakinada. The motto of the seminar is to create awareness about IIT Olympiad courses on students and parents. Speaking on this, Chief guest Ravindra Bharathi IIT Olympiad School Principal Mr. M.Ram Naresh explained about the importance of IIT Olympiad courses for the students and parents presided in the seminar.

IIT-Roorkee new campus inaugurated IIT-Roorkee today inaugurated its Greater Noida Campus at the Knowledge Park-II area here. It was inaugurated by Ashok Bhatnagar, Chairman, Board of Governors, IIT Roorkee in the presence of Director S C Saxena.

IITs in world’s top 100 universities on technical subjects The Indian Institutes of Technology (IITs) India’s premier technical institutes have made it to the world’s top 100 universities as per the recently-released subject-wise rankings by Quacquarelli Symonds (QS). In the engineering and technology rankings, IIT Bombay has bagged the 47th spot

while IIT Delhi is at 52nd position. IIT Kanpur follows at 63rd and IIT Madras at 68th.

HCL F O U ND E R SH I V NA D AR T O H E A D IIT-KH AR AG P U R’S B O A R D

Shiv Nadar, founder of software major HCL, was appointed the chairman of the board of governors of the Indian Institute of Technology (IIT), Kharagpur, a statement from the company said Thursday.

IIT-Kharagpur is the only engineering institution from India listed among top 500 universities worldwide in the Shanghai Jiao Tong University’s ‘Academic Ranking of World Universities’.

‘It is an honour to be entrusted the responsibility to further advance the global stature this great institution has built already,’ said Nadar.

Former chairpersons of IIT-Kharagpur include former West Bengal chief minister B.C. Roy, former chairman and managing director of Tata Steel Russi Mody, chairman emeritus of Ballarpur Industries Ltd. L.M. Thapar, and chairman emeritus of RPG Group R.P. Goenka.

IIT Delhi's golden jubilee lecture by Professor Susan Greenfield

IIT Delhi's Golden Jubilee Lecture by Professor Susan Greenfield, titled ‘The 21st century mind: How will technology change it?’ Susan Greenfield is a scientist,


writer, broadcaster and member of the House of Lords. She is also Director of the Institute for the Future of the Mind. She has been awarded 30 Honorary Degrees from British and Foreign universities and heads a multi-disciplinary research group exploring novel brain mechanisms linked to neurodegenerative diseases such as Alzheimer’s and Parkinson’s. In addition she has developed an interest in the impact of 21st Century technologies on how young people think and feel, as discussed in her book ID.

IIT student wins award for revolutionary research Mumbai: Prashanthi Kovur, a PhD student from the Centre for Excellence in Nanoelectronics, EE Department, IIT Bombay has been awarded the outstanding student researcher in the field of Physics, Chemistry of Material for nano-scale devices by the world largest semiconductor company TSMC.

Taiwan Semiconductor Manufacturing Company (TSMC) is one of the world's largest semi conductor foundry, which provides leading technologies for the semiconductor manufacturing industry. The purpose of the TSMC Outstanding Student Research Award is to recognise exceptional semiconductor related research carried out by graduate students. The competition for this award attracted hundreds of applications from students from various universities all over the world.

Prashanthi's research is based on the development of a novel multiferroic system for MEMS applications. Magnetism and ferroelectricity are essential to many forms of current technology, and the quest for multiferroic materials, where these two phenomena are intimately coupled, is of great technological and fundamental importance.

It has been shown for the first time that room temperature multiferrocity with significant coupling could be achieved in Dy modified BiFeO3 system. The current work focuses on

the fabrication of RF MEMS switch using this novel multiferroic based actuator.

This work was done under the guidance of Prof Vaijayanti R Palkar from EE Department, IIT Bombay. Prof Palkar who was invited as an honourable guest for the award ceremony delivered a lecture on 'Semiconductors in India and Functional Nano-oxides'.

IIT Bombay, Applied Materials launch State-of-the-Art Laboratory IIT Bombay, Applied Materials Collaboration and the inauguration of the "Chemistry Laboratory for Energy and Nanoelectronics" (CLEAN)

Applied Materials, Inc. the world's leading supplier of manufacturing solutions for the semiconductor, display and solar industries, and IIT Bombay, India's highest-rated niversity and a leader in education and research, today announced the state-of-the-art "Applied Materials Chemistry Laboratory for Energy and Nanoelectronics" (CLEAN) at the IIT Bombay (IITB) campus. This new laboratory expands the scope of collaboration between IIT Bombay and Applied Materials to include the development of new materials that can potentially be used in a variety of electronic and renewable energy-focused applications, including the fabrication of next-generation solar cells.

The lab was launched at a ceremony attended by Mike Splinter, chairman of the board, president and CEO of Applied Materials, Omkaram Nalamasu, chief technology officer of Applied Materials, and Prof. Devang Khakhar, director, IIT Bombay.

"This is a great example of the kind of university and corporate collaboration that is helping to advance technology by enabling world-class research, innovation and workforce development," said Mike Splinter. "Our goal is to serve as a catalyst for developing the critical technology needed to solve the many challenges of next-generation electronic and solar device manufacturing."

"Applied Materials has grown to become IIT Bombay's most important industry collaborator in terms of the scale of research collaboration," said Professor Devang Khakhar. "We welcome the establishment of the Applied Materials CLEAN laboratory through Applied's generous support. This will begin a new phase of the collaboration in areas related to renewable energy, which are a focus of IIT Bombay's research."

The event also celebrates a successful five-year relationship between IIT Bombay and Applied Materials for nanoelectronics and solar photovoltaic technology research. During the course of this special alliance, Applied Materials has endowed IIT Bombay with over $12 million for the following projects:

• The establishment of the Applied Materials Nanomanufacturing Laboratory, India's first 200mm semiconductor fabrication facility and one of the few university-based 200mm facilities worldwide.

• Collaborative research on nanoelectronics and solar PV technology.

• Applied's donation of three state-of-the-art physical vapour deposition and chemical vapour deposition process chambers to the National Centre for Photovoltaic Research and Education (NCPRE) for depositing thin films for solar cell applications.

• A solar PV and LED lighting system that lights up the main avenue at IITB's campus

• The establishment of The Applied Materials Chemistry Laboratory for Energy and Nanoelectronics


On passing out from IIT Chennai Mr. Narayanaswamy was offered scholarship by the prestigious Massachusetts Institute of Technology, USA. He who came from a middle class family believed that he had a moral obligation to give something in return for the lakhs of rupees the government spent on him as an IIT student. He had the intelligence and conviction to realize that this money came also from the poorest of the poor - who pay up the excise duty on textiles when they buy cloth, who pay up customs, excise and sales tax on diesel when they travel in a bus, and in numerous other ways indirectly pay the government. So he decided to join IAS hoping he could do something for the people of this country. How many young men have the will power to resist such an offer from USA? Narayanaswamy did never look at IAS as a black money spinner as his later life bears testimony to this fact.

After a decade of meritorious service in IAS, today, Narayanaswamy is being forced out of the IAS profession. Do you know why?

A real estate agent wanted to fill up a paddy field which is banned under law. An application came up before Narayanaswamy who was sub collector the, for an exemption from this rule for this plot of land. Upon visiting the site he found that the complaint from 60 poor families that they will face water logging due to the waste water from a nearby Government Medical College if this paddy field was filled up was correct. Narayanswamy came under intense political pressure but he did what was right - refused permission for filling up the paddy field. That was his first confrontation with politicians.

Soon after his marriage his father-in-law closed down a public road to build compound wall for his plot of land. People approached Narayanaswamy with complaint. When talking with his own father-in-law did not help, he removed the obstructing wall with police help. The result, his marriage broke up.

As district Collector he raided the house of a liquor baron who had defaulted Rupees 11 crores payment to

government and carried out revenue recovery. A Minister directly telephoned him and ordered to return the forfeited articles to the house of the liquor baron. Narayanswamy politely replied that it is difficult. The minister replied that Narayanaswamy will suffer.

In his district it was a practice to collect crores of rupees for earthen bunds meant for poor farmers, but which were never constructed. A bill for rupees 8 crores came up before Narayanaswamy. He inspected the bund. He found it very weak and said that he will pass the bill after the rainy season to ensure that the bund served the purpose. As expected the earthen bund was too weak to stand the rain and it disappeared in the rain. But he created a lot of enemies for saving 8 crores public money.

The net result of all such unholy activities was that he was asked to go on leave by the government. Later such an illustrious officer was posted as "State Co-Ordinator, Quality Improvement Programme for Schools". This is what the politician will do to a honest officer with backbone - post him in the most powerless position to teach him a lesson.

Since he found that nothing can be achieved for the people if he continued with the State Service he opted for central service. But that too was denied on some technical ground. What will you do when you have a brilliant computer career anywhere in the world you choose with the backing of several advanced technical papers too published in international journals to your credit? When you are powerless to do anything for the people, why should you waste your life as the Co-Ordinator for a Schools Programme?

Mr. Narayanaswamy is on the verge of leaving IAS to go to Paris to take up a well paid United Nations assignment. The politicians can laugh thinking another obstacle has been removed. But it is the helpless people of this country who will lose - not Narayanaswamy. But you have the power to support capable and honest bureaucrats like Narayaswamy, G.R.Khairnar and Alphons Kannamthanam who have suffered a lot under self seeking politicians who rule us. You have even the power to replace such politicians with these kind of people dedicated to the country. The question is will you do the little you can do NOW? At least a vote or word in support of such personalities?

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

Mr. Narayanaswamy IIT Chennai, IAS officer. (Secretary General Education, kerala)


PHYSICS

1. A non-conducting disc of radius a and uniform positive surface charge density σ is placed on the ground, with its axis vertical. A particle of mass m and positive charge q is dropped, along the axis of the disc, from a height H with zero initial velocity, the particle has q/m = ε0g/σ. [IIT-1990]

(a) Find the value of H if the particle just reaches the disc

(b) Sketch the potential energy of the particle as a function of its height and find its equilibrium position.

Sol. (A) Given that : a = radius of disc, σ = surface charge density, q/m = 4ε0g/s

The K.E. of the particle, which it react reaches the disc can be taken as zero.

Potential due to a charge disc at any axial point situated at a distance x from 0.

V(x) = ]x–xa[2

22

0+

εσ

Hence, V(H) = ]H–Ha[2

22

0+

εσ

and V(O) = 02a

εσ

According to law of conservation energy, Loss of gravitation potential energy = gain in electric potential energy.

H (m,q)

O a

H

mgH = q∆H = q[V(0) – V(H)]

mgH = q[a – H–)Ha( 22 + ] 02ε

σ ...(i)

From the given relation : 02q

εσ = 2 mg (given)

Putting this is equation (i), we get,

mgH = 2mg [a – H–)Ha( 22 + ]

or H = 2[a + H – )Ha( 22 + ]

or H = 2a + 2H – )Ha(2 22 +

or )Ha(2 22 + = H + 2a or 4a2 + 4H2 = H2 + 4a2 + 4aH

or 3H2 = 4aH or H = 3a4

[Q H = O is not valid] (B) Total potential energy of the particle at height h

U(x) = mgx + qV(x) = mgx + )x–xa(2q 22

0+

εσ ]

= mgx + 2mg ]x–)xa([ 22 +

= mg ]x–)xa(2[ 22 + ...(ii)

From equilibrium : dxdU = 0

This gives : x = 3

a

From equation (ii) graph between U(x) and x is and shown above

O a/ 3 H = 4a/3 X

U

2 mga

3 mga

2. Light is incident at an angle α on one planar end of a

transparent cylindrical rod of refractive index. Determine the least value of n so that the light entering the rod does not emerge from the curved surface of rod irrespective of the value α [IIT- 1992]

n

90 – ββ

α

Sol. The light entering the rod does not emerge from the curved surface of the rod when the angle 90 – r is greater than the critical angle.

KNOW IIT-JEE By Previous Exam Questions


i.e., µ ≤ Csin

1 where C is the critical angle.

Here C = 90 – r

⇒ µ ≤ )r–90sin(

1 ⇒ µ ≤ rcos

1

As a limiting case µ = rcos

1 ...(i)

Applying Snell's law at A

µ = rsin

sin α ⇒ sin r = µ

αsin ...(ii)

The smallest angle of incident on the curved surface

is when α = 2π . This can be taken as a limiting case

for angle of incidence on plane surface

From (ii) sin r = µπ 2/sin ⇒ µ =

rsin1 ...(iii)

From (i) and (ii) sin r = cos r ⇒ r = 45º

⇒ µ = º45cos

1 = 2/1

1

⇒ µ = 2 This is the least value of the refractive index of rod

for light entering the rod and not leaving it from the curved surface.

3. Two block A and B of equal masses are placed on

rough inclined plane as shown in figure. When and where will the two blocks come on the same line on the inclined plane if they are released simultaneously? Initially the block A is 2 m behind the block B. Co-efficient of kinetic friction for the blocks A and B are 0.2 and 0.3 respectively (g = 10 m/s2).

[IIT-2005] 2 m

A

B

B

A

45º

Sol. a = m

cosmg–sinmg k θµθ

∴ aA = g sin θ – µkA g cos θ ...(i) and aB = g sin θ – mkB g cos θ ...(ii) Putting values we get

aA = 289.0 and αB =

279.0

aAB is relative acceleration of A' w.r.t. B = aA – aB

L = 2 m ⇒ L = 21 aA/B t2

[Where L is the relative distance between A and B]

or t2 = B/Aa

L2 = BA a–a

L2

Putting values we get, t2 = 4 or t = 2s.] Distance moved by B during that time is given by

S = 2Bta

21

= 21 ×

279.0 × 4 =

27.02× × 10 = 27 m

Similarly for A = 28 m.

4. Two light springs of force constants k1 and k2 and a block of mass m are in one line AB on a smooth horizontal table such that one end of each spring is fixed on rigid on rigid supports and the other end is free shown in the figure. The distance CD between the free ends of the springs is 60 cms. If the block moves along AB with a velocity 120 cm/sec in between the springs, calculate the period of oscillation of the block.

(k1 = 1.8 N/m, k2 = 3.2 N/m. m = 200 gm) [IIT-1985]

k1 k2

BD m CA

60 cm V

Sol. The mass will strike the right spring, compress it.

The K.E. of the mass will convert into P.E. of the spring. Again the spring will return to its natural size thereby converting its P.E. to K.E. of the block. The

time taken for this process will be 2T . where

T = km2π .

∴ t1 = 2T = π

2Km = π

8.12.0 = 0.785

The block will move from A to B without any acceleration. The time taken will be

t2 = 12060 = 0.5

Now the block will compress the left spring and then the spring again attains ite natural length. The time taken will be

t3 = π 1K

m = π 8.12.0 = 1.05

Again the block moves from B to A, completing one oscillation. the time taken for doing so

t4 = 12060 = 0.5

∴ The complete time of oscillation will be = t1 + t2 + t3 + t4 = 0.785 + 0.5 + 1.05 + 0.5 = 2.83 (app.)


K1 K2

A m B

60 cm V

5. A solid sphere of copper of radius R and a hollow

sphere of the same material of inner radius r and outer radius A are heated to the same temperature and allowed to cool in the same environment. Which of them starts cooling faster ? [IIT-1982]

Sol. Since the temperature and surface area is same, therefore the Energy emitted per second by both spheres is same.

We know that Q = mc∆T Since Q is same and c is same (both copper)

∴ m ∝ T

1∆

Mass of hollow sphere is less ∴ Temperature change will be more. ∴ Hollow sphere will cool faster.

CHEMISTRY

6. The molar volume of liquid benzene (density = 0.877 g ml–1) increases by a factor of 2750

as it vaporizes at 20ºC and that of liquid toluene (density = 0.867 g ml–1) increases by a factor of 7720 at 20ºC. A solution of benzene and toluene at 20ºC has a vapour pressure of 46.0 torr. Find the mole fraction of benzene in vapour above the solution. [IIT-1996]

Sol. Given that, Density of benzene = 0.877 g ml–1 Molecular mass of benzene (C6H6) = 6 × 12 + 6 × 1 = 78

∴ Molar volume of benzene in liquid form =877.078 ml

= 877.078 ×

10001 L = 244.58 L

And molar volume of benzene in vapour phse

= 877.078 ×

10002750 L = 244.58 L

Density of toluene = 0.867 g ml–1 Molecular mass of toluene (C6H5CH3) = 6 × 12 + 5 × 1 + 1 × 12 + 3 × 1 = 92 ∴ Molar volume of toluene in liquid form

= 867.092 ml =

867.092 ×

10001 L

And molar volume of toluene in vapour phase

= 867.092 ×

10007720 L = 819.19 L

Using the ideal gas equation, PV = nRT

At T = 20ºC = 293 K

For benzene, P = 0BP =

VnRT

= 58.244

293082.01 ×× = 0.098 atm

= 74.48 torr (Q 1 atm = 760 torr) Similarly, for toluene,

P = 0TP =

VnRT

= 19.819

293082.01 ×× = 0.029 atm

= 22.04 torr (Q 1 atm = 760 torr) According to Raoult's law, PB = 0

BP xB = 74.48 xB

PT = 0TP xT = 22.04 (1 – xB)

And PM = 0BP xB + 0

TP xT or 46.0 = 74.48 xB + 22.04 (1 – xB) Solving, xB = 0.457 According to Dalton's law, PB = PM '

Bx (in vapour phase) or mole fraction of benzene in vapour form,

'Bx =

M

B

PP =

0.46457.048.74 × = 0.74

7. 0.15 mol of CO taken in a 2.5 L flask is maintained at

705 K along with a catalyst so that the following reaction takes place

CO(g) + 2H2(g) CH3OH(g) Hydrogen is introduced until the total pressure of the

system is 8.5 atm at equilibrium and 0.08 mol of methanol is formed. Calculate (a) Kp and Kc and (b) the final pressure if the same amount of CO and H2 as before are used, but with no catalyst so that the reaction does not take place. [IIT-1993]

Sol. We have CO(g) + 2H2(g) CH3OH(g) t = 0 0.15 mol teq 0.15 mol – x (

2Hn )0 – 2x x It is given that 0.08 mol of CH3OH is formed at

equilibrium. Hence OHCH3

n = x = 0.08 mol and nCO = 0.15 mol – x = 0.07 mol From the total pressure of 8.5 atm equilibrium, we

calculate the total amount of gases, i.e. CO, H2 and CH3OH at equilibrium.

ntotal = RTpV =

)K705)(molKLatm082.0()L5.2/mol08.0(

11– −

= 0.3676 mol Now, the amount of H2 at equilibrium is given as

2Hn = ntotal – nCO – OHCH3n

= (0.367 – 0.07 – 0.08) mol = 0.2176 mol


Hence, KC = 22

3

]H][CO[]OHCH[

= 2)L5.2/mol2176.0)(L5.2/mol07.0()L5.2)(atm5.8(

= 150.85 (mol L–1)–2 Now Kp = Kc(RT)∆vg = (150.85 mol–2L2)(0.082 L atm K–1 mol–1)(705 K)–2 = 0.04513 atm–2 Since

2Hn = (2Hn )0 – 2x, we have

(2Hn )0 =

2Hn + 2x = (0.2176 + 2 × 0.08)mol = 0.3776 mol Total amount of CO and H2 in the reacting system

before the reaction sets in is given as n0 = (nCO)0 + (

2Hn )0 = (0.15 + 0.3776)mol = 0.5276 mol

Hence, p0 = VRTn0

= )L5.2(

)K705)(molKatmL082.0)(mol5276.0( 11 −−

= 12.20 atm 8. An ester A(C4H8O2), on treatment with excess methyl

magnesium chloride followed by acidification, gives an alcohol B as the sole organic product. Alcohol B, on oxidation with NaOCl followed by acidification, gives acetic acid. Deduce the structures of A and B. Show the reactions involved. [IIT-1998]

Sol. The reactions of an ester with methyl magnesium chloride are as follows.

R–C–OR´ (A)

O CH3MgCl R–C–OR´

OMgCl

CH3

H+

–HOMgClR–C–CH3 + R´OH

O

CH3MgCl

R–C–CH3

OH H+

R–C–CH3

OMgCl

CH3

–HOMgClCH3

(B) Since the given ester (C4H8O2) produces only one alcohol B, it follows that RC(CH3)2OH and R´OH must be identical. Thus, the alkyl group R´ must be RC(CH3)2 – and the given ester A is

R – C – O – C – CH3

O CH3

R

(molecular formula R2C4H6O2 )

From the molecular formula of A, we conclude that R

must be H atom. Hence, the given ester is

H – C – O – CH – CH3

O

CH3

Isopropyl formate

The alcohol B is a secondary alcohol.

CH3 – CH – CH3

OH

Isopropyl alcohol

The oxidation of alcohol B with NaOCl will give a

ketone which further undergoes a haloform reaction.

CH3 – CH – CH3 + NaOCl

O

CH3 – C – CH3 + NaCl + H2OOH

CH3 – C – CH3 + 3NaOCl

O

CH3 – C – CCl3 + 3NaOHO

CH3 – C – CCl3 + NaOH

O

CH3 – C – O–Na+ + CHCl3O

The acidification of sodium acetate will produce

acetic acid. 9. (a) Write the chemical reaction associated with the

"brown ring test". (b) Draw the structures of [Co(NH3)6]3+, [Ni(CN)4]2–

and [Ni(CO)4]. Write the hybridization of atomic orbital of the transition metal in each case.

(c) An aqueous blue coloured solution of a transition metal sulphate reacts with H2S in acidic medium to give a black precipitate A, which is insoluble in warm aqueous solution of KOH. The blue solution on treatment with KI in weakly acidic medium, turns yellow and produces a white precipitate B. Identify the transition metal ion. Write the chemical reaction involved in the formation of A and B. [IIT-2000]

Sol. (a) NaNO3 + H2SO4 → NaHSO4 + HNO3 2HNO3 + 6FeSO4 + 3H2SO4 → 3Fe2(SO4)3 + 2NO + 4H2O [Fe(H2O)6]SO4.H2O + NO Ferrous Sulphate → [Fe(H2O)5NO] SO4 + 2H2O (Brown ring) (b) In [Co(NH3)6]3+ cobalt is present as Co3+ and its

coordination number is six. Co27 = 1s1, 2s22p6, 3s23p63d7, 4s2 Co3+ion = 1s2, 2s22p6, 3s23p63d6

3d 4s 4p

3d 4s 4p

d2sp3 hybridization

Hence

Co3+ion in Complex ion


3+

Co

NH3

NH3

NH3 H3N

H3N NH3

or

H3N

H3N

NH3 NH3

NH3NH3

Co3+

In [Ni(CN)4

2– nickel is present as Ni2+ ion and its coordination numbers is four

Ni28 =1s2, 2s22p6, 3s23p63d8, 4s2 Ni2+ ion = 1s2, 2s22p6, 3s23p63d8

3d 4s 4p

3d 4s 4p

dsp2 hybridization

Ni2+ ion =

Ni2+ion in Complex ion

Hence structure of [Ni(CN)4]2– is

Ni2+

N ≡ C

N ≡ C

C ≡ N

C ≡ N

In [Ni(CO)4, nickel is present as Ni atom i.e. its

oxidation number is zero and coordination number is four.

3d 4s 4p

sp3 hybridization

Ni in Complex

Its structure is as follows :

Ni

CO

CO

CO

OC

(c) The transition metal is Cu2+. The compound is

CuSO4.5H2O CuSO4 + H2S → mediumAcidic

pptBlackCuS ↓ + H2SO4

2CuSO4 + 4KI → Cu2I2 + I2 + 2K2SO4 (B) white I2 + I– → I3

– (yellow solution) 10. A basic volatile, nitrogen compound gave a foul

smelling gas when treated with CHCl3 and alcoholic KOH. A 0.295 g sample of the substance dissolved in aqueous HCl and treated with NaNO2 solution at 0ºC liberated a colourless; odourless gas whose volume

corresponds to 112 ml at STP. After the evolution of the gas was complete, the aqueous solution was distilled to give an organic liquid which did not contain nitrogen and which on warming with alkali and iodine gave yellow precipitate. Identify the original substance. Assume it contains one N-atom per molecule. [IIT-1993]

Sol. Clue 1. Nitrogen compound gave foul smelling gas when treated with CHCl3 and alc. KOH (carbylamine reaction), thus it is a primary amine.

Clue 2. This compound when treated with HCl + NaNO2 solution (nitrous acid test) at 0ºC liberates colourless and odourless gas.

CnH2n+1NH2 → + 2NaNOHCl AlcoholROH +

Nitrogen2N ↑

At STP, 112 ml of N2 is evolved from = 0.295 g CnH2n+1NH2 ∴ 22400 ml of N2 is evolved from

= 112

22400295.0 × = 59 g CnH2n+1NH2

∴ CnH2n+1NH2 = 59 or n × C + (2n + 1) × H + N + 2 × H = 59 or 12n + 2n + 1 + 14 + 2 × 1 = 59

or n = 1442 = 3

Thus the molecular formula of nitrogen compound is C3H7NH2.

Clue 3. Alcohol obtained gives iodoform test positive, thus it is a secondary alcohol and its structure should be

CH3CHCH3

OH2-propanol

and hence the structure of (A) should be

CH3CHCH3

NH2

Propan-2-amine

MATHEMATICS

11. With ususal notation, if in a triangle ABC

11

cb + = 12

ac + = 13

ba + , then prove that

7

Acos = 19

Bcos = 25

Ccos . [IIT-1984]

Sol. Let 11

cb + = 12

ac + = 13

ba + = λ

⇒ (b + c) = 11λ, c + a = 12λ, a + b = 13λ ⇒ 2(a + b + c) = 36λ or a + b + c = 18λ Now, b + c = 11λ and a + b + c = 18λ ⇒ a = 7λ c + a = 12λ and a + b + c = 18λ ⇒ b = 6λ a + b = 13λ and a + b + c = 18λ ⇒ c = 5λ


∴ cos A = bc

acb2

222 −+

= 2

222

)30(2492536

λλ−λ+λ =

51

cos B = ac

bca2

222 −+

= 2

222

70364925

λλ−λ+λ =

3519

cos C = ab

cba2

222 −+

= 2

222

84253649

λλ−λ+λ =

75

∴ cos A : cos B : cos C = 51 :

3519 :

75 = 7 : 19 : 25

12. Complex numbers z1, z2, z3 are the vertices A, B, C respectively of an isosceles right angled triangle with right angle at C. show that

(z1 – z2)2 = 2(z1 – z3) (z3 – z2) [IIT-1986] Sol. Since ∆ is right angled isosceles ∆. ∴ Rotating z2 about z3 in anticlock wise direction

through an angle of π/2, we get A(z1)

C(z2) B(z3)

31

32

zzzz

−− =

||||

31

32

zzzz

−− eiπ/2

where, |z2 – z3| = |z1 – z3| ⇒ (z2 – z3) = i(z1 – z3) squarring both sides we get, (z2 – z3)2 = – (z1 – z2)2 ⇒ 2

2z + 23z – 2z2z3 = –z1

2 – z32 + 2z1z3

⇒ 21z + 2

2z – 2z1z2 = 2z1z3 + 2z2z3 – 2 23z – 2z1z2

⇒ (z1 – z2)2 = 2(z1z3 – 23z ) + (z2z3 – z1z2)

⇒ (z1 – z2)2 = 2(z1 – z3)(z3 – z2)

13. Let n be a positive integer and (1 + x + x2)n = a0 + a1x + .... + a2nx2n Show that a0

2 – a12 + .... + a2n

2 = an [IIT-1994] Sol. (1 + x + x2)n = a0 + a1x + ... + a2nx2n ...(1)

Replacing x by –x1 , we obtain

n

xx

+− 2

111 = a0 – xa1 + 2

2

xa – 3

3

xa

+ ... + nn

xa

22

...(2)

Now, a02 – a1

2 – a32 + ... + a2n

2 = coefficient of the term independent of x in

[a0 + a1x + a2x2 + ... + a2nx2n]

+−+− nn

xa

xa

xaa 2

2221

0 ...

= coefficient of the term independent of x in

(1 + x + x2)nn

xx

+− 2

111

But R.H.S. = (1 + x + x2)nn

xx

+− 2

111

= n

nn

xxxxx

2

22 )1()1( +−++ = n

n

xxx

2

222 ])1[( −+

= n

n

xxxx

2

242 )21( −++ = n

n

xxx

2

42 )1( ++

Thus, a02 – a1

2 + a22 + ....a2n

2 = coefficient of the term independent of x in

nx21 (1 + x2 + x4)n

= coefficient of x2n in (1 + x2 + x4)n = coefficient of tn in (1 + t + t2)n = an

14. A rectangle PQRS has its side PQ parallel to the line y = mx and vertices P, Q and S lie on the lines y = a, x = b and x = –b, respectively. Find the locus of the vertex R. [IIT-1996]

Sol. Let the coordinates of R be (h, k). It is given that P lies on y = a. So, let the coordinates of P be (x1, a). Since PQ is parallel to the line y = mx. Therefore,

Slope of PQ = (Slope of y = mx) = m

And, Slope of PS = –mx) y of Slope(

1=

= –m1 [∴ PS ⊥ PQ]

Now, equation of PQ is y – a = m(x – x1) ...(i)

y y = 0

P

Q

x = b

(0, b)x

y´

R

O (0, – b)Sx´

x = –b

(0, a)


It is given that Q lies on x = b. So, Q is the point of intersection if (i) and x = b.

Putting x = b in (i), we get y = a + m(b – x1) So, coordinates of Q are (b, a + m(b – x1)).

Since PS passes through P(x1, a) and has slope – m1 .

So, Equation of PS is

y – a = –m1 (x – x1) ...(ii)

It is given that S lies on x = – b. So, S is the point of intersection of (ii) and x = –b.

Solving (ii) and x = – b, we get

y = a + m1 (b + x1)

So, coordinates of S are

++− )(1, 1xb

mab

Now, Slope of RS = bh

xbm

ak

+

+−− )(11

= m

But RS is parallel to PQ.

∴ bh

xbm

ak

+

+−− )(11

= m

⇒ b + x1 = m(k – a) – m2(h + b) ...(iii) Similarly,

Slope of RQ = bh

xbmak−

−−− )( 1

But, RQ is perpendicular to PQ whose slope is m.

∴ bh

xbmak−

−−− )( 1 = – m1

⇒ b – x1 = m1 (k – a) + 2

1m

(h – a) ...(iv)

We have only one variable x1. To eliminate x1, add (iii) and (iv) to obtain

2b = (k – a)

+

m1m – m2(h + b) + 2

1m

(h – b)

⇒ 2b = (k – a)

+m

m 12– h

+2

4 1m

m – b

+2

4 1m

m

⇒ (k – a)

+m

m 12– 2

22 )1)(1(m

mmh +−

– 2

22 )1(m

mb + = 0

⇒ (k – a) – m

mh )1( 2 − –m

mb )1( 2 + = 0

⇒ m(k – a) – h(m2 – 1) – b(m2 + 1) = 0 Hence, the locus of R(h, k) is m(y – a) – x(m2 – 1) – b(m2 + 1) = 0

15. Let f [(x + y)/2] = f (x) + f (y) / 2 for all real x and y, If f ´(0) exists and equals –1 and f (0) = 1, find f (2).

[IIT- 1995]

Sol. f

+

2yx =

2)()( yfxf + ∀ x, y ∈ R (given)

Putting y = 0, we get

f

2x =

2)0()( fxf + =

21 [1 + f (x)] [Q f (0) = 1]

⇒ 2f (x/2) = f (x) + 1 ⇒ f (x) = 2f (x/2) – 1 ∀x, y ∈ R ...(1) Since f´(0) = –1, we get

h

fhfh

)0()0(lim0

−+→

= – 1 ⇒ h

hfh

1)(lim0

−→

= –1

Now, let x ∈ R then applying formula of differentiability.

f ´(x) =h

xfhxfh

)()(lim0

−+→

= h

xfhxf

h

)(2

22

lim0

−

+

→

= h

xfhfxf

h

)(2

)2()2(

lim0

−+

→

= h

xfhfxf

h

)(12

2212

2221

lim0

−

−

+−

→

[Using equations (1)

= h

xfhfxf

h

)(1)(21)(221

lim0

−−+−

→

= h

xfhfxfh

)(1)()(lim0

−−+→

= h

hfh

1)(lim0

−→

= –1

Therefore f ´(x) = – 1 ∀ x ∈ R

⇒ ∫ dxxf )´( = ∫− dx1

⇒ f (x) = –x + k where k is a constant. But f (0) = 1, therefore f (0) = – 0 + k ⇒ f (x) = 1 – x ∀ x ∈ R ⇒ f (2) = – 1


1. A thin rod of length 2L is placed horizontally in a down ward uniform electric field E. Half of length is charged positively and other half with linear charge density λ . The rod of is hinged at its middle point and free to rotate in vertical plane, A weight w is hanged at a distance x from hinge to keep it horizontal. The value of x is

2L x

O

w

E

– – – – – – + + + + + + + λ λ

(A) E

wλ

(B) w2LE 2λ

(C) wLE 2λ (D)

wLE2 2λ

2. S is a cross section of a solenoid where magnetic

fields is increasing at a constant positive rate. V1 and V2 are two ideal voltmeters, then V1/V2 is

× × × × × × × × ×

× × V1 V22Ω1Ω

S

(A) 1 (B) 0.5 (C) 2 (D) 0.25

Passage # 1 (Q. No. 3 to Q. No. 4) In the given circuits three inductors L1, L2 and L3

and a resistance R are connected to a V volt battery neglecting mutual inductance. At t = 0 switch is closed and L3 > L2 > L1.

L2

L3 L1

R

V

3. Find current through L1 will be

(A) 321

2

LLLL

RV

++

(B) 321

1

LLLL

RV

++

(C) 321

3

LLLL

RV

++

(D)Can’t be determined

4. Time constant of the circuit is :

(A) R

LLL 321 ++ (B)

+

+213 LL

1L1

R1

(C)

+

+312 LL

1L1R

1 (D) None of these

Passage # 2 (Q. No. 5 to Q. No. 6) Hydrogen gas in the atomic state is excited to a

energy level such that the electrostatic potential energy of H-atom becomes -1.7eV. Now a photoelectric plate having work function W = 2.3eV is exposed to the emission spectra of this gas. Assuming all transitions to be possible. Answer the following questions.

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 avai lable in next issue

Set # 1


5. How many electric transition may result in an ejected photoelectron

(A) 3 (B) 4 (C) 5 (D) 2

6. The minimum de-broglie wavelength of the ejected

photo electrons is (A) 13.43Å (B) 3.43 Å (C) 2Å (D) None of these Passage # 3(Q. No. 7 to Q. No. 8) A charged dielectric sphere of negligible mass is

connected to a conducting square shaped loop of total resistance R and side ''l , by means of a light non conducting thread, which passes over two fixed pulleys as shown in the figure. The sphere carries a charge q and kept in uniform downward electric field E. The loop has mass ‘m’ (m < qE) and kept in a horizontal magnetic field B as shown. Answer the following.

Take 22 s/m20mqE,s/m10g == and

222

m/sN2mR

B−=

l

× × × × × ×× × × × × ×× × × × × ×× × × × × ×

BE

q

7. Choose the incorrect (wrong) statements

(A) The initial acceleration of loop is 10m/s2 (B) Loop’s one edge leaves magnetic field after 0.5

sec, its acceleration drops to zero (C) The magnetic force on loop is acting upwards (D) When loop is completely off the magnetic field its acceleration is 10m/s2

8. If after 2 sec, one sides of the loop emerges from the

magnetic field, the speed of the loop as a function of time is

(A) ]e1[5 t2−+ (B) ]e1[5 )4t2( −−+

(C) ]e1[5 t2−− (D) None of these

Black Holes-The Most Efficient Engines in the Universe

The scientists have just found the most energy-efficient engines in the universe. Black holes, whirling super dense centres of galaxies that suck in nearly everything. Jets of energy spurting out of older ultra-efficient black holes also seem to be playing a crucial role as zoning police in large galaxies preventing to many stars from sprouting. This explains why there are fewer burgeoning galaxies chock full of stars than previously expected. For the first time, the scientists have measured both the mas of hot gas that is being sucked into nine older black holes and the unseen super speedy jets of high energy particles split out, which essentially form a cosmic engine. Then they determined a rate of how efficient these older black hole engines are and were awe-struck. These black holes are 25 times more efficient than anything man has built, with nuclear power being the most efficient of man-made efforts, said the research's lead author, Professor Steve Allen of Stanford University. The galaxies in which these black holes live are bigger than the Milky way, which is the Earth's galaxy and are 50 million to 400 million light-years away.

Black holes are the most fuel-efficient engines in the universe. The results were surprising because the types of black holes studied were older, less powerful and generally considered boring, scientists said. But they ended up being more efficient than originally thought, possibly as efficient as their younger, brighter and more potent black hole siblings called quasars. One way the scientists measured the efficiency of black holes was by looking at the jets of high energy spewed out.


1. Two circular rings A and B, each of radius a = 30 cm are placed coaxially with their axes horizontal in a uniform electric field E = 105 NC–1 directed vertically upwards as shown in figure. Distance between centres of these rings A and B is h = 40 cm. Ring A has a positive charge q1 = 10 µC while ring B has a negative charge of magnitude q2 = 20 µC. A particle of mass m = 100 gm and carrying a positive charge q = 10 µC is released from rest at the centre of the ring A. Calculate its velocity when it has moved a distance of 40 cm.

B E A

a a

h Sol. Weight of the particle is W = mg = 1 newton Force on particle, due to electric field E is F = qE = 1

newton (upwards) It means weight of the particle is balanced by the

force F. Hence, net force on the particle is due to charge on rings.

Since particle is at centre of ring A therefore initially no force is exerted due to charge of this ring. But ring B is negatively charged. Therefore particle experiences a resultant force towards centre of ring B.

Separation between centres of rings is h = 40 cm and distance moved the particle is also 40 cm. It means velocity of particle is to be calculated when it reaches the centre of ring B.

According to law of conservation of energy, kinetic

energy

2mv

21 of particle at centre of B = Loss of

its electrical potential energy. Potential energy of particle at centre of A is

U1 = U0 + a

qq4

1 1

0πε +

222

0 ha

)q(–q4

1

+πε

Where U0 is potential energy due to electric field E. Potential energy at centre of B is

U2 = U0 + 22

1

0 ha

qq4

1

+πε +

a)q(–q.

41 2

0πε

But 2mv21 = U1 – U2

∴ v = 26 ms–1 2. Two thin similar convex glass pieces are joined

together, front to front, with rear portion silvered and the combination of glass pieces is placed at a distance a = 60 cm from a screen. A point object is placed on optical axis of the combination such that its m = 2 times the magnified image formed on the screen. If air between the glass pieces is replaced by water (µ = 4/3), calculate the distance through which the object must be displaced so that a sharp image is again formed on the screen.

Sol. A thin convex glass piece has both the surfaces equally curved, therefore, it works as a glass plate. When two thin similar convex glass pieces are joined front to front, though they form shape of an equi-convex lens as shown in figure(1), but its optical power is zero because medium inside this lens is air.

Air

Fig. (1) When its rear surface is silvered, it works as a

concave mirror. This concave mirror has focal length fm = – r/2 where r is the radius of curvature of each glass piece. Since, this concave mirror forms an image on screen, therefore the image is real. Hence screen and object both are on same side of the mirror as shown figure (2).

Air

Screen a

Object

Fig. (2) Since, image is m = 2 times magnified, therefore,

distance of object from the mirror is equal to a/m = 30 cm

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

Students' ForumPHYSICS


for this mirror, v = –a = – 60 cm, u = – 30 cm, Using mirror formula,

v1 +

u1 =

f1 , fm = – 20 cm or r = 40

when water is filled in space between glass pieces, an equi-convex lens of water is formed whose one surface is silvered. This silvered convex glass may be assumed as a combination of a convex lens and a concave mirror. Effective focal length of the

combination is given by F1 =

mf1 –

lf2 where fl is

focal length of the equiconvex lens of water. For the equiconvex lens, R1+ r , R2 = – r, µ = 4/3

Using, f1 = (µ–1)

21 R1–

R1 , f = f1 = 60 cm

Hence, effective focal length F is given by

F1 =

)20(–1 –

602

or F = – 12 cm since, a sharp image is again formed on the screen

therefore, fore this effective mirror, F = – 12 cm, v = – a = – 60 cm, u = ?

Using, mirror formula, F1 =

v1 +

u1 , u = – 15 cm

Initially the object was at a distance (–a/2) = – 30 cm from the mirror but now object must be at a distance (–15 cm). Hence, the object must be moved through [– 15 – (– 30)] cm = 15 cm towards the mirror (combination).

3. A uniform rod of length l = 75 cm is hinged at one of its ends and is free to rotate in vertical plane. It is a released from rest when the rod is horizontal. When rod becomes vertical, it is broken at mid point and lower part now moves freely. Calculate distance of the centre of lower part from hinge, when it again becomes vertical for the first time. (g = 10 ms–2 )

Sol. When rod is released, it rotates about upper end its gravitational potential energy converts into rotational kinetic energy. Initially centre of mass of the rod is in level of O while its depth from O is l/2 when rod becomes vertical as shown in figure. (1). If mass of rod is m, then loss of gravitational potential

energy = mg2l .

O

Fig. (1)

Moment of inertia of rod about O is I = 3

m 2l

Let angular velocity or rod be ω when it becomes vertical. According to law of conservation energy,

2I21

ω = mg 2l or ω =

l

g3 = 40 rad/sec

Now the rod gets broken at its mid point as shown in figure (2). At this instant velocity of centre of lower

part is v0 = 43l

ω (horizontally leftwards) and its

angular velocity is ω (clockwise). ω

ω v0

Fig. (2) Now this part moves freely under gravity. Therefore,

its angular velocity ω remains constant. This part again becomes vertical for the first time after completing half rotation. Therefore, time taken by the

lower part during this rotation is t = ωπ =

40π sec.

During this time. horizontal displacement centre of

lower part is x = v0t = 4

3 lπ = 1.77 m

and vertical displacement = 21 gt2 = 1.23 m

But initial depth of centre of lower part from hinge

was 43l , therefore vertical depth of its centre from O

at this instant is y =

+ 2gt

21

43l = 1.80 m.

Hence, distance of its centre from O is

r = 22 yx + = 2.52 m

4. Length of a horizontal arm of a U-tube is l = 21 cm and ends of both of the vertical arms are open to surroundings of pressure 10500 Nm–2. A liquid of density ρ = 103 kg m–3 is poured into the tube such that liquid just fills horizontal part of the tube. Now, one of the open ends is sealed and the tube is then rotated about a vertical axis passing through the other vertical arm with angular velocity ω0 = 10 radians/sec. If length of each vertical arm be a = 6 cm. calculate the length of air column in the sealed arm.


l = 21 cm

a =

6 cm

Sol. When tube is rotated, liquid starts to flow radially

outward and air in sealed arm is compressed. Let the shift of liquid be x as shown in figure.

(l – x)

a–x

x

x

A

B

Let cross-sectional area of tube be S. Initial volume of air, V0 = Sa and initial pressure

P0 = 10500 Nm–2 Final volume, V = S(a – x)

∴ Final pressure P = VVP 00 =

)x–a(.a.P0

or Pressure at B, P2 = P + xρg = )x–a(

aP0 + xρg

Centripetal force required for circular motion of vertical column of height x of liquid is provided by reaction of the tube while that to horizontal length (l – x) is provided by excess pressure at B.

Force exerted by pressure difference is

F1 = (PB – PA) S = (P2 – P0) S = Sgx)x–a(

xP0

ρ+

Mass of horizontal arm AB of liquid is, m = S (l – x)ρ Radius of circular path traced by its centre of mass is

r = x + 2

x–l =

+

2xl

∴ Centripetal force, F2 = mω02r

But F2 = F1 or S ρ (l – x) 202

xω

+l

= Sgx)x–a(

xP0

ρ+

or x = .01 m = 1cm ∴ Length of air column in sealed arm = (a – x)= 5cm 5. A reversible Carnot engine is coupled with three heat

reservoirs A,B and C as shown in Figure. Temperature of these reservoirs is

A

T1 = 400 K T2 = 300 K T3 = 200 K

Q1 = 1200 JEngine

Q2 Q3CB

T1 = 400 K, T2 = 300 K and T3 = 200K, respectively. During an integral number of complete cycles, the engine absorbs Q1 = 1200 joule of heat energy from reservoir A and performs W = 200 joule of mechanical work. Calculate quantities Q1 and Q2 of heat energy exchanged with the other two reservoirs B and C respectively. State whether the reservoirs absorb or lose heat energy.

Sol. Given engine can be assumed to work in two ways ; (i) engine working between a source of temperature

T1 and sink of temperature T2 and (ii) an engine working between source of temperature

T1 and sink of temperature T3. Hence, efficiency of these two is

η1 =1

21

TT–T =

41 and η2 =

1

31

TT–T

= 41 respectively

If an engine having efficiency η absorbs heat Q from source then heat energy converted into mechanical work by the engine will be equal to ηQ and remaining part (Q – ηQ) = (1 – η)Q should be rejected to the sink.

Since, heat rejected to sink by first engine, having efficiency η1, is Q2, therefore, heat absorbed by it

from source is equal to )–1(

Q

1

2

η =

3Q4 2

and work done by this engine is

W1 = )–1(

Q

1

12

ηη =

3Q2

Similarly, theat absorbed by the other engine from

source is equal to )–1(

Q

2

3

η = 2Q3

and work done by this engine is W2 = )–1(

.Q

2

23

ηη

= Q3

But W1 + W2 = 200 J

∴ 31 Q2 + Q3 = 200 J ...(1)

Total heat extracted from source A is

Q1 =

+ 32 Q2Q

34 or

+ 32 Q2Q

34 = 1200 J

...(2) Solving equation (1) and (2), Q2 = 1200 J and Q3 = – 200 J It means reservoir B absorbs 1200 J of heat and

reservoir C loses 200 J of heat. In fact engine works as an engine between reservoirs

A and B and as heat pump between reservoirs B and C.


• Coulomb's Law :

F0 = 221

0 rqq

41πε

(in vacuum)

Vectorially →F = 2

21

0 rqq

41πε

r

In any material medium F = 221

r0 rqq

41

επε

where εr is a constant of the material medium called its relative permittivity, and ε0 is a universal constant, called the permittivity of free space.

ε0 = 8.85 × 10–12 or 04

1πε

= 9 × 109

The unit of ε0 is C2N m–2 or farad per metre.

Also F = 221

r0 rqq

41

επε

Where ε is called the absolute permittivity of the medium. Obviously, εr = F0/F. Remember εr = ∞ for conductors. Conductors and insulators Each body contains enormous amounts of equal and opposite charges. A 'charged' body contains an excess of either positive or negative charge. In a conductor, some of the negative charges are free to move around. In an insulator (also called a dielectric), the charges cannot move. They can only undergto small localized displacements, causing polarization. Induction When a charged body A is brought near another body B, unlike charges are induced on the near surface of B (called bound charges) and like charges appear on the far surface of B (called free charges) If B is a conductor, the free charges can be removed by earthing B, e.g., by touching it. If B is an insulator, separation of like and unlike charges will still occur due to induction. However, the like charges cannot then be removed by earthing B.

• Electric Field And Potential Electric Field An electric field of strength E is said to exist at a point if a test charge ∆q at that point experiences a force given by

→→

∆=∆ FqF or qFE

∆∆

=

→→

The unit of electric field is Newton per coulomb or volt per metre. The electric field strength at a distance r from a point charge q in a medium of permittivity ε is given by

E =πε41

2rq

Vectorially →E =

πε41

2rq r

With reference to any origin

→E =

πε4q

3

rR

rR→→

→→

−

−

Where →R is the position vector of the field point and

→r , the position vector of q.

Due to a number of discrete charges

→E = 3

i

iNi

1i

1

rR

rR4q

→→

→→=

= −

−πε∑

Electric Potential The electric potential at a point is the work done by an external agent in bringing a unit positive charge from infinity up to that point along any arbitrary path.

VP =q

)agentexternalanby(W P

∆∆ →∞ volt(V) or JC–1

The potential difference between two points P and Q is given by

VP – VQ =q

)agentby(W PQ

∆

∆ → volt (V)

The potential at a distance r from a point charge q in a medium of permittivity ε is

ϕ or V =πε41

rq =

πε41

→→− rR

q

with reference to any arbitrary origin. Due to a number of charges

ϕ or V =→→

=

= −πε∑

i

1Ni

1i rR

q41

Electrostatics-I

PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


In a conductor, all points have the same potential. If charge q (coulomb) is placed at a point where the potential is V (volt), the potential energy of the system is qV (joule). It follows that if charges q1, q2 are separated by distance r, the mutual potential

energy of the system is r4

qq 21

πε.

• Relation Between Field (E) And Potential (V) The negative of the rate of change of potential along a given direction is equal to the component of the field that direction.

Er = –rV

∂∂ along r

and E⊥ =θ∂

∂r

V perpendicular to r

When two points have different potentials, an electric field will exist between them, directed from the higher to the lower potential.

• Lines of Force A line of force in an electric field is such a curve that the tangent to it at any point gives the direction of the field at that point. Lines of force cannot intersect each other because it is physically impossible for an electric field to have two directions simultaneously.

• Equipotential Surfaces The locus of points of equal potential is called an equipotential surface. Equipotential surfaces lie at right angles to the electric field. Like lines of force, they can never intersect.

Note: For solving problems involving electrostatic units, remember the following conversion factors:

3 × 109 esu of charge = 1 C 1 esu of potential = 300 V • Electric Flux

The electric flux over a surface is the product of its surface area and the normal component of the electric field strength on that surface. Thus,

dϕ = (E cos θ) ds = En ds =→E .

→ds

ds

O

E

N

The total electric flux over a surface is obtained by summing :

ϕE =∑→→

∆ s.E or ∫→→sd.E

Gauss's Theorem The total electric flux across a

closed surface is equal to 0

1ε

times the total charge

inside the surface.

Mathematically ∑→→

∆ s.E = q/ε0

where q is the total charge enclosed by the surface.

Problems in electrostatics can be greatly simplified by the use of Gaussian surfaces. These are imaginary surfaces in which the electric intensity is either parallel to or perpendicular to the surface every-where. There are no restrictions in constructing a Gaussian surface. The following results follow from Gauss's law

1. In a charged conductor, the entire charge resides only on the outer surface. (It must always be remembered that the electric field is zero inside a conductor.)

2. Near a large plane conductor with a charge density σ (i.e., charge per unit area), the electric intensity is

E = σ/ε0 along the normal to the plane 3. Near an infinite plane sheet of charge with a

charge density σ, the electric intensity is E = σ/2ε0 along the normal to the plane 4. The electric intensity at a distance r from the axis

of a long cylinder with λ charge per unit length (called the linear density of charge), is

E =02

1πε r

λ along→r

Problem solving strategy: Coulomb's Law : Step 1 : The relevant concepts : Coulomb's law

comes into play whenever you need to know the electric force acting between charged particles.

Step 2 : The problem using the following steps : Make a drawing showing the locations of the

charged particles and label each particle with its charge. This step is particularly important if more than two charged particles are present.

If three or more charges are present and they do not all lie on the same line, set up an xy-coordinate system.

Often you will need to find the electric force on just one particle. If so, identify that particle.

Step 3 : The solution as follows : For each particle that exerts a force on the particle

of interest, calculate the magnitude of that force

using equation F = 221

0 r|qq|

41πε

Sketch the electric force vectors acting on the particle(s) of interest due to each of the other particles (that is, make a free-body diagram). Remember that the force exerted by particle 1 on particle 2 points from particle 2 toward particle 1 if the two charges have opposite signs, but points from particle 2 directly away from particle 1 if the charges have the same sign.

Calculate the total electric force on the particle(s) of interest. Remember that the electric force, like any force, is a vector. When the forces acting on a charge are caused by two or more other charges, the total force on the charge is the vector sum of


the indivual forces. It's often helpful to use components in an xy-coordinate system. Be sure to use correct vector notation; if a symbol represents a vector quantity, put an arrow over it. If you get sloppy with your notation, you will also get sloppy with your thinking.

As always, using consistent units is essential. With the value of k = 1/4πε0 given above, distances must be in meters, charge in coulombs, and force in newtons. If you are given distance in centimeters, inches, or furlongs, donot forget to convert ! When a charge is given in microcoulombs (µC) or nanocoulombs (nC), remember that 1µC = 10–6C and 1nC = 10–9C.

Some example and problems in this and later chapters involve a continuous distribution of charge along a line or over a surface. In these cases the vector sum described in Step 3 becomes a vector integral, usually carried out by use of components. We divide the total charge distribution into infinitesimal pieces, use Coulomb's law for each piece, and then integrate to find the vector sum. Sometimes this process can be done without explicit use of integration.

In many situations the charge distribution will be symmetrical. For example, you might be asked to find the force on a charge Q in the presence of two other identical charges q, one above and to the left of Q and the other below and to the left of Q. If the distance from Q to each of the other charges are the same, the force on Q from each charge has the same magnitude; if each force vector makes the same angle with the horizontal axis, adding these vectors to find the net force is particularly easy. Whenever possible, exploit any symmetries to simplify the problem-solving process.

Step 4 : your answer : Check whether your numerical results are reasonable, and confirm that the like charges repel opposite charges attract.

Problem solving strategy : Electric-field calculations Step 1: the relevant concepts : Use the principle of

superposition whenever you need to calculate the electric field due to a charge distribution (two or more point charges, a distribution over a line, surface, or volume or a combination of these).

Step 2: The problem using the following steps : Make a drawing that clearly shows the locations

of the charges and your choice of coordinate axes. On your drawing, indicate the position of the field

point (the point at which you want to calculate the electric field E

r). Sometimes the field point will

be at some arbitrary position along a line. For example, you may be asked to find E

r at point on

the x-axis. Step 3 : The solution as follows :

Be sure to use a consistent set of units. Distances must be in meters and charge must be in coulombs. If you are given centimeters or nanocoulombs, do not forget to convert.

When adding up the electric fields caused by different parts of the charge distribution, remember that electric field is a vector, so you must use vector addition. Don't simply add together the magnitude of the individual fields: the directions are important, too.

Take advantage of any symmetries in the charge distribution. For example, if a positive charge and a negative charge of equal magnitude are placed symmetrically with respect to the field point, they produce electric fields of the same magnitude but with mirror-image directions. Exploiting these symmetries will simplify your calculations.

Must often you will use components to compute vector sums. Use proper vector notation; distinguish carefully between scalars, vectors, and components of vectors. Be certain the components are consistent with your choice of coordinate axes.

In working out the directions of Er

vectors, be careful to distinguish between the source point and the field point. The field produced by a point charge always points from source point to field point if the charge is positive; it points in the opposite direction if the charge is negative.

In some situations you will have a continuous distribution of charge along a line, over a surface, or through a volume. Then you must define a small element of charge that can be considered as a point, finds of all charge elements. Usually it is easiest to do this for each component of E

r

separately, and often you will need to evaluate one or more integrals. Make certain the limits on your integrals are correct; especially when the situation has symmetry, make sure you don't count the charge twice.

Step 4 : your answer : Check that the direction of Er

is reason able. If your result for the electric-field magnitude E is a function of position (say, the coordinate x), check your result in any limits for which you know what the magnitude should be. When possible, check your answer by calculating it in a different way.

Problem solving strategy : Gauss's Law Step 1 : Identify the relevant concepts : Gauss's law

is most useful in situations where the charge distribution has spherical or cylindrical symmetry or is distributed uniform over a plane. In these situations we determine the direction of E

r from the symmetry

of the charge distribution. If we are given the charge distribution. we can use Gauss's law to find the the magnitude of E

r. Alternatively, if we are given the

field, we can use Gauss's law to determine the details


of the charge distribution. In either case, begin your analysis by asking the question, "What is the symmetry ?"

Step 2 : Set up the problem using the following steps Select the surface that you will use with Gauss's

law. We often call it a Gaussian surface. If you are trying to find the field at a particular point, then that point must lie on your Gaussian surface.

The Gaussian surface does not have to be a real physical surface, such as a surface of a solid body. Often the appropriate surface is an imaginary geometric surface; it may be in empty space, embedded in a solid body, or both.

Usually you can evaluate the integral in Gauss's law (without using a computer) only if the Gaussian surface and the charge distribution have some symmetry property. If the charge distribution has cylindrical or spherical symmetry, choose the Gaussian surface to be a coaxial cylinder or a concentric sphere, respectively.

Step 3 : Execute the solution as follows : Carry out the integral in Eq.

ΦE = ∫ φ dAcosE = ∫ dAE = ∫ Ad.Err

= 0

enclQε

(various forms of Gauss's law) This may look like a daunting task, but the

symmetry of the charge distribution and your careful choice of a Gaussian surface makes it straightforward.

Often you can think of the closed surface as being made up of several separate surfaces, such as the side and ends of a cylinder. The integral ∫ dAE

over the entire closed surface is always equal to the sum of the integrals over all the separate surfaces. Some of these integrals may be zero, as in points 4 and 5 below.

If Er

is perpendicular (normal) at every point to a surface with area A, if points outward from the interior of the surface, and if it equal to EA. If instead E

r is perpendicular and inward, then E⊥ =

– E and ∫ ⊥dAE = – EA.

If Er

is tangent to a surface at every point, then E⊥ = 0 and the integral over that surface is zero.

If Er

= 0 at every point on a surface, the integral is zero.

In the integral ∫ ⊥dAE , E⊥ is always the

perpendicular component of the total electric field at each point on the closed Gaussian surface. In general, this field may be caused partly by charges within the surface and partly by charges outside it. Even when there is no charge within the surface, the field at points on the Gaussian

surface is not necessarily zero. In that case, however, the integral over the Gaussian surface – is always zero.

Once you have evaluated the integral, use eq. to solve for your target variable.

Step 4 : Evaluate your answer : Often your result will be a function that describes how the magnitude of the electric field varies with position. Examine this function with a critical eye to see whether it make sense.

1. Supposing that the earth has a charge surface density

of 1 electron/metre2, calculate (i) earth's potential, (ii) electric field just outside earths surface. The electronic charge is – 1.6 × 10–19 coulomb and earth's radius is 6.4×106 metre (ε0 = 8.9 × 10–12 coul2/nt–m2).

Sol. Let R and σ be the radius and charge surface density of earth respectively. The total charge, q on the earth surface is given by

q = 4 p R2 σ (i) The potential V at a point on earth's surface is same

as if the entire charge q were concentrated at its centre. Thus,

V =Rq.

41

0πε

=04

1πε

.RR4 2σπ =

0

.Rε

σ

Substituting the given values

V =)mnt/coul109.8(

)metre/coul106.1()metre104.6(2212

2196

−××−××

−

−−

= – 0.115coul

mnt − = – 0.115couljoule = – 0.115 volt.

(ii) E =04

1πε 2R

q =04

1πε

. 2

2

RR4 σπ =

0εσ

= 2212

219

mnt/coul109.8metre/coul106.1

−××−

−

−

= – 1.8 × 10–8 nt/coul.

The negative sign shows that E is radially inward. 2. Determine the electric field strength vector if the

potential of this field depends on x, y co-ordinates as (a) V = a(x2 – y2) and (b) V = axy.

Sol. (a) V = a(x2 – y2)

Hence, Ex = –xV

∂∂ = – 2ax, Ey = –

yV

∂∂ = + 2ay

∴ E = – 2axi + 2ayj or E = – 2a(xi – yj) (b) V = a x y

Solved Examples


Hence, Ex = –xV

∂∂ = –ay, Ey = –

yV

∂∂ = – ax

∴ E = – ayi – axj = – a[yi + xj]

3. A charge Q is distributed over two concentric hollow spheres of radii r and R (> r) such the surface densities are equal. Find the potential at the common centre.

q

O

R

r

q′

Sol. Let q and q′ be the charges on inner and outer sphere.

Then q + q′ = Q …(1) As the surface densities are equal, hence

2r4qπ

= 2R4'q

π

(∴ Surface density = charge/area) ∴ q R2 = q′ r2 …(2) From eq. (1) q′ = (Q – q), hence q R2 = (Q – q)r2 q(R2 + r2) = Q r2

∴ q = 22

2

rRrQ+

and q′ = Q – q = 22

2

rRRQ+

Now potential at O is given by

V =04

1πε r

q +04

1πε r

'q

=04

1πε r)rR(

rQ22

2

++

041πε r)rR(

rQ22

2

+

=04

Qπε )rR(

)Rr(22 +

+

4. S1 and S2 are two parallel concentric spheres enclosing charges q and 2q respectively as shown in fig.

(a) What is the ratio of electric flux through S1 and S2 ? (b) How will the electric flux through the sphere S1

change, if a medium of dielectric constant 5 is introduced in the space inside S1 in place of air ?

q

2q

S1 S2

Sol. (a) Let Φ1 and Φ2 be the electric flux through spheres S1 and S2 respectively.

Φ1 =0

qε

and Φ2 = 00

q3q2qε

=ε+

∴ 2

1

ΦΦ =

0

0

/q3/q

εε

=31

(b) Let E be the electric field intensity on the surface of sphere S1 due to charge q placed inside the sphere. When dielectric medium of dielectric constant K is introduced inside sphere S1, then electric field intensity E′ is given by

E′ = E/K Now the flux Φ′ through S1 becomes

Φ′ = ∫∫ ε==

0

'

KqdS.E

K1dS.E

∴ Φ′ = 05

qε

5. A charge of 4 × 10–8 C is distributed uniformly on the surface of a sphere of radius 1 cm. It is covered by a concentric, hollow conducting sphere of a radius 5 cm. (a) Find the electric field at a point 2 cm away from the centre. (b) A charge of 6 × 10–8 C is placed on the hollow sphere. Find the surface charge density on the outer surface of the hollow sphere.

Sol. (a) See fig. (a) Let P be a point where we have to calculate the electric field. We draw a Gaussian surface (shown dotted) through point P. The flux through this surface is

2cmP5cm

q = 6 × 10–8 C

Fig. (a) Fig. (b) Φ = ∫∫ −×π== E)102(4dSEdS.E 22

According to Gauss's law, Φ = q/ε0 ∴ 4π × (2 × 10–2)2 E = q/ε0

or E = 220 )102(4

q−××πε

= 4

89

104)104()109(

−

−

××××

= 9 × 105 N/C (b) See fig. (b) We draw a Gaussian surface (shown

dotted) through the material of hollow sphere. We know that the electric field in a conducting material is zero, therefore the flux through this Gaussian surface is zero. Using Gauss's law, the total charge enclosed must be zero. So, the charge on the inner surface of hollow sphere is 6 × 10–8 C. So, the charge on the outer surface will be 10 × 10–8 C.


Kinematics :

Velocity (in a particular direction)

= takenTime

)directionthatin(ntDisplaceme

xAB)V(r

= AxVr

– BxVr

and xAB)V(r

= t

dx

Where dx is the displacement in the x direction in time t.

Swimmer crossing a river

d vscosθ vs θ

vssinθ vr

Time taken to cross the river = θcosV

ds

For minimum time, θ should be zero.

vR vs

vr

x

in this case resultant velocity

VR = 2r

2s vV + and t =

svd .

Also x = vr × t For reaching a point just opposite the horizontal

component of velocity should be zero. v sinθ = Vr

|Average Velocity| = time

|ntDisplaceme|

ar

= t

u–vrr

= t

)u(–vrr

+

⇒ |a| = t

cosuv2–uv 22 θ+

Where θ is the angle between v and u. The direction of acceleration is along the resultant of

vr

and (– ur

).

Graphs During analysis of a graph, the first thing is see the

physical quantities drawn along x-axis and y-axis. If y = mx, the graph is a straight line passing through

the origin with slope = m. [see fig. (a)]

θ

Y

X(i)

m = tanθ θ is acute and m is positive

Y

Xθ

(ii)

m = tanθ θ is abtuse and m is positive

fig.(a) if y = mx + c, the graph is a straight line not passing

through the origin and having an intercept c which may be positive or negative [see fig. (b,) (c)]]

c

Y

X(i)

m is '+' ve

c

Y

X(ii)

c is negativem is '+' ve

Y

X (ii)

c is positive m is negative

fig(b)

fig(c) For y = kx2, where k is a constant, we get parabola

[see fig (d)]

Y

XParabola Fig.(d)

x2 + y2 = r2 is equation of a circle with centre at origin and radius r.

1-D Motion, Projectile Motion PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


For (x – a)2 + (y – b)2 = r2, the motion is in a circular path with centre at (a, b) and radius r

2

2

2

2

by

ax

+ = 1 is equation of an ellipse

x × y = constant give a rectangular hyperbola.

Note : To decide the path of motion of a body, a

relationship between x and y is required. Area under-t graph represents change in velocity.

Calculus method is used for all types of motion (a = 0 or a = constt or a = variable)

a = f(t)

Differentiate w.r.t time

v = f(t) s = f(t)

Differentiate w.r.t time

integrate w.r.t. time

integrate w.r.t. time

S stand for displacement

a = v dsdv =

dtdv = 2

2

dtsd

Also vx = dtdx ⇒ vy =

dtdy

ax = dt

dvx = 2

2

dtxd and ay =

dtdvy = 2

2

dtyd

The same concept can be applied for z-co-ordintae. Projectile motion :

ucosθ

ucosθ

ucosθ θ

usinθ

usinθ

θ u

Q

P

u

ucosθ

ucosθ

F g F

g F g

F g

Projectile motion is a uniformly accelerated motion. For a projectile motion, the horizontal component of

velocity does not change during the path because there is no force in the horizontal direction. The vertical component of velocity goes on decreasing with time from O to P. At he highest point it becomes zero. From P to Q again. the vertical component of velocity increases but in downwards direction. Therefore the minimum velocity is at the topmost

point and it is u cos θ directed in the horizontal direction.

The mechanical energy of a projectile remain constant throughout the path.

the following approach should be adopted for solving problems in two-dimensional motion :

Resolve the 2-D motion in two 1-D motions in two mutually perpendicular directions (x and y direction) Resolve the vector quantitative along these directions. Now use equations of motion separately for x-direction and y-directions.

If you do not resolve a 2-D motions in two 1-D motions in two 1-D motion then use equations of motion in vector form

vr

= ur

+ at ; sr

= ut + 21 a

rt2 ; v

r. vr

– ur

. u = 2 ar

sr

s = 21 ( u

r + v

r)t

When y = f(x) and we are interested to find (a) The values of x for which y is maximum for

minimum (b) The maximum/minimum values of y then we may

use the concept of maxima and minima. Problem solving strategy : Motion with constant Acceleration : Step 1: Identify the relevant concepts : In most

straight-line motion problems, you can use the constant-acceleration equations. Occasionally, however, you will encounter a situation in which the acceleration isn't constant. In such a case, you'll need a different approach

ax = dt

d xυ = dtd

dtdx = 2

2

dtxd

Step 2: Set up the problem using the following steps: You must decide at the beginning of a problem

where the origin of coordinates are usually a matter of convenience. If is often easiest to place the particle at the origin at time t = 0; then x0 = 0. It is always helpful to make a motion diagram showing these choices and some later positions of the particle.

Remember that your choice of the positive axis direction automatically determines the positive directions for velocity and acceleration. If x is positive to the right of the origin, the vx and ax are also positive toward the right.

Restate the problem in words first, and then translate this description into symbols and equations. When does the particle arrive at a certain point (that is, what is the value of t)? where is the particle when its velocity has a specified value (that is, what is the value of x when vx has the specified value)? "where is the motorcyclist when his velocity is 25m/s?"


Translated into symbols, this becomes "What is the value of x when vx = 25 m/s?"

Make a list of quantities such as x, x0,vx,v0x,ax and t. In general, some of the them will be known quantities, and decide which of the unknowns are the target variables. Be on the lookout for implicit information. For example. "A are sits at a stoplight" Usually means v0x = 0.

Step 3 : Execute the solution : Choose an equation from Equation vx = v0x + axt

x = x0 + v0xt + 21 axt2 (constant acceleration only)

2xv = 2

x0v + 2ax(x – x0) (constant accelerations only)

x – x0 =

+

2vv xx0 t (constant acceleration only)

that contains only one of the target variables. Solve this equation for the equation for the target variable, using symbols only. then substitute the known values and compute the value of the target variable. sometimes you will have to solve two simultaneous equations for two unknown quantities.

Step 4 : Evaluate your answer : Take a herd look at your results to see whether they make sense. Are they within the general range of values you expected?

Problem solving strategy :

Projectile Motion : Step 1 : Identify the relevant concepts : The key

concept to remember is the throughout projectile motion, the acceleration is downward and has a constant magnitude g. Be on the lookout for aspects of the problem that do not involve projectile motion. For example, the projectile-motion equations don't apply to throwing a ball, because during the throw the ball is acted on by both the thrower's hand and gravity. These equations come into play only after the ball leaves the thrower's hand.

Step 2 : Set up the problem using the following steps Define your coordinate system and make a sketch

showing axes. Usually it's easiest to place the origin to place the origin at the initial (t = 0) position of the projectile. (If the projectile is a thrown ball or a dart shot from a gun, the thrower's hand or exits the muzzle of the gun.) Also, it's usually best to take the x-axis as being horizontal and the y-axis as being upward. Then the initial position is x0 = 0 and y0 = 0, and the components of the (constant) acceleration are ax = 0, ay = – g.

List the unknown and known quantities, and decide which unknowns are your target variables. In some problems you'll be given the initial

velocity (either in terms of components or in terms of magnitude and direction) and asked to find the coordinates and velocity components as some later time. In other problems you might be given two points on the trajectory and asked to find the initial velocity. In any case, you'll be using equations

x = (v0 cosα0)t (projectile motion) through ...(1)

vy = v0 sin α0 – gt (projectile motion) ...(2) make sure that you have as many equations as

there are target variables to be found. It often helps to state the problem in words and

then translate those words into symbols. For example, when does the particle arrive at a certain point ? (That is at what value of t?) Where is the particle when its velocity has a certain value? (That is, what are the values of x and y when vx or vy has the specified value ?) At the highest point in a trajectory, vy = 0. so the question "When does the particle reach its highest points ?" translates into "When does the projectile return to its initial elevation?" translates into "What is the value of t when y = y0 ?"

Step 3 : Execute the solution use equation (1) & (2) to find the target variables. As you do so, resist the

temptation to break the trajectory into segments and analyze each segment separately. You don't have to start all over, with a new axis and a new time scale, when the projectile reaches its highest point ! It's almost always easier to set up equation (1) & (2)

at the starts and continue to use the same axes and time scale throughout the problem.

Step 4 : Evaluate your answer : As always, look at your results to see whether they make sense and whether the numerical values seem reasonable.

Relative Velocity : Step 1 : Identify the relevant concepts : Whenever

you see the phrase "velocity relative to" or "velocity with respect to", it's likely that the concepts of relative will be helpful.

Step 2 : Set up the problem : Label each frame of reference in the problem. Each moving object has its own frame of reference; in addition, you'll almost always have to include the frame of reference of the earth's surface. (Statements such as "The car is traveling north at 90 km/h" implicitly refer to the car's velocity relative to the surface of the earth.) Use the labels to help identify the target variable. For example, if you want to find the velocity of a car (C) with respect to a bus (B), your target variable is vC/B.

Step 3 : Execute the solution : Solve for the target variable using equation

vP/A = vP/B + vB/A (relative velocity along a line) ...(1)


(If the velocities are not along the same direction, you'll need to use the vector from of this equations, derived later in this section.) It's important to note the order of the double subscripts in equation (1) vA/B always means "velocity of A relative to B." These subscripts obey an interesting kind of algebra, as equation (1) shown. If regard each one as a fraction, then the fraction on the left side is the product of the fractions on the right sides : P/A = (P/B) (B/A). This is a handy rule you can use when applying Equation (1) to any number of frames of reference. For example, if there are three different frames of reference A, B, and C, we can write immediately.

vP/A = vP/C + vC/B + VB/A Step 4 : Evaluate your answer : Be on the lookout for

stray minus signs in your answer. If the target variable is the velocity of a car relative to a bus (vV/B), make sure that you haven't accidentally calculated the velocity of the bus relative of the car (vB/C). If you have made this mistake, you can recover using equation.

vA/B = – vB/A

1. A small glass ball is pushed with a speed V from A.

It moves on a smooth surface and collides with the wall at B. If it loses half of its speed during the collision, find the distance, average speed and velocity of the ball till it reaches at its initial position.

V 0.5V BA

d Sol. The ball moves from A to B with a constant speed V.

Since it loses half of its speed on collision, it returns from B to A with a constant speed V/2.

∴ V1 = V and V2 = V/2

Using the formula, VaV =)V/d()V/d(

dd

2211

21

++

Putting d1 = d2 = d; V1 = V and V2 = V/2

We obtain, VaV =)V5.0/d()dV(

dd 21

++ =

3V2

From the formula,

average velocity =net

net

2

2

1

1

21aV

t|s|

Vs

Vs

|ss|V→→→

→=

+

+=

Since s1 = s2 = d and snet = |ss| 21→→

+ = 0

(As the ball returns to its initial position, the change in position, the change in position vector of the ball, that is the net displacement will be zero).

∴ |V| aV→

= 0.

2. A long belt is moving horizontally with a speed of 4 Km/hour. A child runs on this belt to and fro with a speed of 9 Km/hour (with respect to the belt) between his father and mother located 50 m apart on the moving belt. For an observer on a stationary platform outside, what is the

(a) speed of the child running in the direction of motion of the belt,

(b) speed of the child running opposite to the direction of motion of the belt and

(c) time taken by the child in case (a) and (b) ? Which of the answers change, if motion is viewed by

one of the parents ? Sol. Let us consider positive direction of x-axis from left

to right (a) Here, vB = + 4 Km/hour Speed of child w.r.t. belt, vC = = 9 Km/hour ∴ Speed of child w.r.t. stationary observer, vC′ = vC + vB or vC′ = 9 + 4 = 13 Km/hour (b) Here, vB = + 4 Km/hour, vC = – 9 Km/hour ∴ Speed of child w.r.t. stationary observer, vC′ = vC + vB or vC′ = – 9 + 4 = –5 Km/hour The negative sign shows that the child appears to run

in a direction opposite to the direction of motion of the belt.

(c) Distance between the parents, s = 50 m = 0.05 Km Since parents and child are located on the same belt,

the speed of the child as observe by stationary observer in either direction (either father to mother or from mother to father) will be 9 Km/hour.

Time taken by the child in case (a) and (b),

t =hour/km9km50.0 = 20 sec.

If the motion is observed by one of parents, answer to case (a) case (b) gets altred. It is because the speed of the child w.r.t. either of mother or father is 9 Km/hour.

3. A particle is projected with velocity v0 = 100 m/s at

an angle θ = 30º with the horizontal. Find : (a) velocity of the particle after 2 sec. (b) angle between initial velocity and the velocity after 2 sec. (c) the maximum height reached by the projectile (d) horizontal range of the projectile

Solved Examples


Sol. (a) jvivv ytxtt

→→→+=

where i and j are the unit vectors along +ve x and +ve y-axis respectively

→

tv =(ux + axt) i + (uy + ayt) j

Here, ux = v0 cos θ = 50 3 m/s, ax = 0 uy = v0 sin θ = 50 m/s, ay = – g (Q g acts downwards)

→

tv = 50 i3 + (50 – 10 × 2) j

=[50 i3 + 30 j ] m/s

∴ |→

2v | = )vv( 2y

2x + = 22 )30()350( +

(b) →

0v = 50 i3 + 50 j

→

2v = 50 i3 + 30 j

∴ →

0v .→

2v = 7500 + 1500 = 9000

If α is the angle between →

0v and→

2v

Then, cos α =|v||v|

v.v

20

20→→

→→

×=

65.911009000×

α = cos–1 (0.98) = 10.8º (c) vy

2 – u2y = 2ayy

At y = ymax, vy = 0 ∴ 0 – v0

2 sin2 θ = 2 (–g)ymax

∴ ymax = g2sinv 22

0 θ = 125 m

(d) R =g

2sinu 2 θ = 1732 m

4. A ball starts falling with zero initial velocity on a smooth inclined plane forming an angle α with the horizontal. Having fallen the distance 'h', the ball rebounds elastically off the inclined plane. At what distance from the impact point will the ball rebound for the second time ?

α

α α

Sol. Just before impact magnitude of velocity of the ball,

v = )gh2(

As the ball collides elastically and the inclined plane is fixed, the ball follows the law of reflection.

Now along the incline, velocity component after impact is v sin α and acceleration is g sin α. Perpendicular to the incline, velocity component is vcos α and acceleration (– g cos α). Hence, if we measure x and y-coordinate along the incline and perpendicular to the incline, then

x = (v sin α) t + ½ (g sin α)t2 and y = (v cos α) t – ½ (g cos α)t2 When the ball hits the plane for a second time, y = 0, (v cos α)t – ½(g cos α)t2 or t = (2v/g) Putting this value of t in x,

x =gsinv4 2 α = 8h sin α

5. A batsman hits a ball at a height of 1.22m above the ground so that ball leaves the bat at an angle 45º with the horizontal. A 7.31 m high wall is situated at a distance of 97.53 m from the position of the batsman. Will the ball clear the wall if its range is 106.68 m. Take g = 10 m/s2

Sol. R(range) =g

2sinv20 θ

or, 20v =

θ2sinRg = Rg as θ = 45º

1.22mA

v0

45º

B106.68m

or, v0 = )Rg( …(1)

Equation of trajectory

y = x tan 45º –º45cosv2

gx22

0

2

or, y = x –.½Rg2

gx2= x –

Rggx2

Putting x = 97.53, we get

y = 97.53 –1068.106

)53.97(10 2

×× = 8.35 cm

Hence, height of the ball from the ground level is h = 8.35 + 1.22 = 9.577 m As height of the wall is 7.31 m so the ball will clear

the wall.


Physical Chemistry

Fundamentals

Real Gases : Deviation from Ideal Behaviour : Real gases do not obey the ideal gas laws exactly

under all conditions of temperature and pressure. Experiments show that at low pressures and moderately high temperatures, gases obey the laws of Boyle, Charles and Avogadro approximately, but as the pressure is increased or the temperature is decreased, a marked departure from ideal behaviour is observed.

V

p

H2

Ideal gas

Plot of p versus V of hydrogen, as compared to that of an ideal gas The curve for the real gas has a tendency to coincide

with that of an ideal gas at low pressures when the volume is large. At higher pressures, however, deviations are observed.

Compressibility Factor : The deviations can be displayed more clearly, by

plotting the ratio of the observed molar volume Vm to the ideal molar volume Vm,ideal (= RT/p) as a function of pressure at constant temperature. This ratio is called the compressibility factor Z and can be expressed as

Z = ideal,m

m

VV

= RTp Vm

Plots of Compressibility Factor versus Pressure : For an ideal gas Z = 1 and is independent of pressure

and temperature. For a real gas, Z = f(T, p), a function of both temperature and pressure.

A graph between Z and p for some gases at 273.15 K, the pressure range in this graph is very large. It can be noted that:

(1) Z is always greater than 1 for H2. (2) For N2, Z < 1 in the lower pressure range and is

greater than 1 at higher pressures. It decreases with increase of pressure in the lower pressure region, passes through a minimum at some pressure and then

increases continuously with pressure in the higher pressure region.

(3) For CO2, there is a large dip in the beginning. In fact, for gases which are easily liquefied, Z dips sharply below the ideal line in the low pressure region.

100Z

ideal gas CH4

N2 H2 t = 0ºC

p/101.325 bar200 300

1.0

0

CO2

Plots of Z versus p of a few gases This graph gives an impression that the nature of the

deviations depend upon the nature of the gas. In fact, it is not so. The determining factor is the temperature relative to the critical temperature of the particular gas; near the critical temperature, the pV curves are like those for CO2, but when far away, the curves are like those for H2 (below fig.)

200Z

ideal gas T1

T1>T2>T3>T4

p/101.325 kPa400 600

1.0

0

T2

T3T4

Plots of Z versus p of a single gas

at various temperatures Provided the pressure is of the order of 1 bar or less,

and the temperature is not too near the point of liquefaction, the observed deviations from the ideal gas laws are not more than a few percent. Under these conditions, therefore, the equation pV = nRT and related expressions may be used.

Van der Waals Equation of state for a Real gas Causes of Deviations from Ideal Behaviour : The ideal gas laws can be derived from the kinetic

theory of gases which is based on the following two important assumptions:

GASEOUS STATE

KEY CONCEPT


(i) The volume occupied by the molecules is negligible in comparison to the total volume of the gas.

(ii) The molecules exert no forces of attraction upon one another.

Derivation of van der Waals Equation : Van der Waals was the first to introduce

systematically the correction terms due to the above two invalid assumptions in the ideal gas equation piVi = nRT. His corrections are given below.

Correction for volume : Vi in the ideal gas equation represents an ideal

volume where the molecules can move freely. In real gases, a part of the total volume is, however, occupied by the molecules of the gas. Hence, the free volume Vi is the total volume V minus the volume occupied by the molecules. If b represents the effective volume occupied by the molecules of 1 mole of a gas, then for the amount n of the gas Vi is given by

Vi = V – nb ...(1) Where b is called the excluded volume or co-volume.

The numerical value of b is four times the actual volume occupied by the gas molecules. This can be shown as follows.

If we consider only bimolecular collisions, then the volume occupied by the sphere of radius 2r represents the excluded volume per pair of molecules as shown in below Fig.

excluded volume

2r

Excluded volume per pair of molecules Thus, excluded volume per pair of molecules

= 34

π(2r)3 = 8

π 3r

34

Excluded volume per molecule

=

π 3r

348

21 = 4

π 3r

34

= 4 (volume occupied by a molecule) Since b represents excluded volume per mole of the

gas, it is obvious that

b =

π 3

A r344N

Correction for Forces of Attraction : Consider a molecule A in the bulk of a vessel as

shown in Fig. This molecule is surrounded by other molecules in a symmetrical manner, with the result that this molecule on the whole experiences no net force of attraction.

AB

Arrangement of molecules within and near the surface of a vessel Now, consider a molecule B near the side of the

vessel, which is about to strike one of its sides, thus contributing towards the total pressure of the gas. There are molecules only on one side of the vessel, i.e. towards its centre, with the result that this molecule experiences a net force of attraction towards the centre of the vessel. This results in decreasing the velocity of the molecule, and hence its momentum. Thus, the molecule does not contribute as much force as it would have, had there been no force of attraction. Thus, the pressure of a real gas would be smaller than the corresponding pressure of an ideal gas, i.e.

pi = p + correction term ...(2) This correction term depends upon two factors: (i) The number of molecules per unit volume of the

vessel Large this number, larger will be the net force of attraction with which the molecule B is dragged behind. This results in a greater decrease in the velocity of the molecule B and hence a greater decrease in the rate of change of momentum. Consequently, the correction term also has a large value. If n is the amount of the gas present in the volume V of the container, the number of molecules per unit volume of the container is given as

N' = V

nNA or N' ∝ Vn

Thus, the correction term is given as : Correction term ∝ n/V ...( 2a) (ii) The number of molecules striking the side of the

vessel per unit time Larger this number, larger will be the decrease in the rate of change of momentum. Consequently, the correction term also has a larger value,. Now, the number of molecules striking the side of vessel in a unit time also depends upon the number of molecules present in unit volume of the container, and hence in the present case:

Correction term ∝ n / V ...(2b) Taking both these factors together, we have


Correction term ∝

Vn

Vn

or Correction term = a 2

2

Vn ...( 3)

Where a is the proportionality constant and is a measure of the forces of attraction between the molecules. Thus

pi = p + a 2

2

Vn ...(4)

The unit of the term an2/V2 will be the same as that of the pressure. Thus, the SI unit of a will be Pa m6 mol–2. It may be conveniently expressed in kPa dm6 mol–2.

When the expressions as given by Eqs (1) and (4) are substituted in the ideal gas equation piVi = nRT, we get

+ 2

2

Vanp (V – nb) = nRT ...(5)

This equation is applicable to real gases and is known as the van der Waals equation.

Values of van der Waals Constants : The constants a and b in van der Waals equation are

called van der Waals constants and their values depend upon the nature of the gas. They

Van Der Waals Constants

Gas 2–6 moldmkPaa 1–3 moldm

b

H2 He N2

O2 Cl2 NO NO2 H2O CH4 C2H6 C3H8 C4H10(n) C4H10(iso) C5H12(n) CO CO2

21.764 3.457

140.842 137.802 657.903 135.776 535.401 553.639 228.285 556.173 877.880 1466.173 1304.053 1926.188 150.468 363.959

0.026 61 0.023 70 0.039 13 0.031 83 0.056 22 0.027 89 0.044 24 0.030 49 0.042 78 0.063 80 0.084 45 0.122 6 0.114 2 0.146 0 0.039 85 0.042 67

are characteristics of the gas. The values of these constants are determined by the critical constants of the gas. Actually, the so-called constant vary to some extent with temperature and this shows that the van

der Waals equation is not a complete solution of the behaviour of real gases.

Applicability of the Van Der Waals Equation : Since the van der Waals equation is applicable to real

gases, it is worth considering how far this equation can explain the experimental behaviours of real gases. The van der Waals equation for 1 mole of a gas is

+ 2

mVap (Vm – b) = RT ..(i)

At low pressure When pressure is low, the volume is sufficiently large and b can be ignored in comparison to Vm in Eq. (i). Thus, we have

+ 2

mVap Vm = RT or pVm +

mVa =RT

or Z = 1 – RTV

a

m ...(ii)

From the above equation it is clear that in the low pressure region, Z is less than 1. On increasing the pressure in this region, the value of the term (a/VmRT) increase as V is inversely proportional to p. Consequently, Z decreases with increase of p.

At high pressure When p is large , Vm will be small and one cannot ignore b in comparison to Vm. However, the term 2

mV/a may be considered negligible in comparison to p in Eq. (i) Thus,

p(Vm – b) = RT or Z = 1 + RTpb ...(iiii)

Here Z is greater than 1 and increases linearly with pressure. This explains the nature of the graph in the high pressure region.

A high temperature and low pressure If temperature is high, Vm will also be sufficiently large and thus the term 2

mV/a will be negligibly small. At this stage, b may also be negligible in comparison to Vm. Under these conditions, Eq. (i) reduces to an ideal gas equation of state:

pVm = RT Hydrogen and helium The value of a is extremely

small for these gases as they are difficult to liquefy. Thus, we have the equation of state as p(Vm – b) = RT, obtained from the van der Waals equation by ignoring the term 2

mV/a . Hence, Z is always greater than 1 and it increases with increase of p.

The van dar Waals equation is a distinct improvement over the ideal gas law in that it gives qualitative reasons for the deviations from ideal behaviour. However, the generality of the equation is lost as it contains two constants, the values of which depend upon the nature of the gas.


Organic Chemistry

Fundamentals

Stability of different types of carbocations in decreasing order :

C ⊕

> CH ⊕

>

⊕

>

(Ph)3

⊕C > (Ph)2

⊕CH > Ph –

⊕C H2 ≥

CH2 = CH – CH2 ≥ R – C – R > R – CH – R

⊕ ⊕ ⊕

R

> R – CH2 > CH2 = CH ⊕ ⊕

A special stability is associated with cycloproyl

methyl cations and this stability increases with every additional cyclopropyl group.

This is undoubtedly because of conjugation between the bent orbitals of the cyclopropyl ring and the vacant p-orbital of the cation carbon.

C H

H

H H

H

H Cyclopropyl methyl cation orbital representation conjugation

with the p-like orbital of the ring Nucleophilicity versus basicity : If the nucleophilic atoms are from the same period of

the periodic table, strength as a nucleophile parallels strength as a base. For example :

H2O < NH3 CH3OH ≈ H2O < CH3CO2

Θ < CH3OΘ ≈ OHΘ

Increasing base strength Increasing nucleophile strength ⇒ Nucleophile strength increases down a column of the

periodic table (in solvents that can have hydrogen bond, such as water and alcohols). For example :

ΘOR <

ΘSR

R3N < R3P

ΘF <

ΘCl <

ΘBr <

ΘI

increasing nucleophilic strength decreasing base strength ⇒ Steric bulk decreases nucleophilicity. For

example :

H3C – C – O

CH3

CH3

Θ < HOΘ

weaker nucleophileStronger base

stronger nucleophileweaker base

Leaving Groups : A good leaving groups is the one which becomes a

stable ion after its departure. As most leaving groups leave as a negative ion, the good leaving groups are those ions which stabilize this negative charge most effectively. The weak bases do this best, thus the best groups are weak bases. If a group is a weak base i.e., the conjugate base of a strong acid, it will generally be a good leaving group. In an SN2 reaction the leaving group begins to gain negative charge as the transition state is reached. The more the negative charge is stabilized, the lower is the energy of the transition state; this lowers the energy of activation and thereby increases the rate of reaction.

The acids HCl, HBr, HI and H2SO4 are all strong acids since the anions Cl–, Br–, I– and HSO4

– are stable anions these anions (weak bases) are also good leaving groups in SN2 reactions. Of the halogens, an iodide ion is the best leaving group and the fluoride ion is the poorest :

I– > Br– > Cl– > F– The order of basicity is opposite : F– > Cl– > Br– > I–,

the reason that alkyl fluorides are ineffective substrates in SN2 reactions is related, to the relatively low acidity of HF (pKa = 3). Sulfonic acids, R SO2OH are similar to sulfuric acid in acidity and the sulfonate ion RSO3

– is a very good leaving group. Alky benzenesulfonates, alkyl p-toluenesulfonates are therefore, very good substrates in SN2 reactions.

The triflate ion (CF3SO3–) is one of the best leaving

groups known, it is the anion of CF3SO3H which is a strong acid much stronger than sulfuric acid.

GENERAL ORGANIC CHEMISTRY

KEY CONCEPT


Kinetic Isotope Effects : The kinetic isotope effect is a change of rate that

occurs upon isotopic substitution and is generally expressed as a ratio of the rate constants, k light/k heavy. A normal isotope effect is one where the ratio of k light to k heavy is greater than 1. In an inverse isotope effect, the ratio is less than 1. A primary isotope effect is one which results from the making or breaking of a bond to an isotopically substituted atom and this must occur in the rate determining step. A secondary isotope effect is attributable to isotopic substitution of an atom not involved in bond making or breaking in the rate determining step. Thus when a hydrogen in a substrate is replaced by deuterium, there is often a change in the rate. Such changes are known as deuterium, isotope effects and are expressed by the ratio kH/kD, the typical value for this ratio is 7. The ground state vibrational energy (the zero-point vibrational energy) of a bond depends on the mass of the atoms and is lower when the reduced mass is higher. Consequently, D – C, D – O, D – N bonds, etc., have lower energies in the ground state than the corresponding H – C, H – O, H – N bonds, etc. Thus, complete dissociation of deuterium bond would require more energy than that for a corresponding hydrogen bond in the same environment. In case a H – C, H –O, or H – N bond is not broken at all in a reaction or is broken in a non-rate-determining step, substitution of deuterium for hydrogen generally does not lead to a change in the rate, however, if the bond is broken in the rate-determining step, the rate must be lowered by the substitution. This helps in determination of mechanism. In the bromination of acetone, the rate determining step is the tautomerization of acetone which involves cleavage of a C–H bond. In case this mechanistic assignment is correct, one should observe a substantial isotop effect on the bromination of deuterated acetone. Indeed kH/kD was found to be around 7.

CH3COCH3 + Br2 → CH3COCH2Br

rate-determining step Bromoacetone

CH3COCH3 CH3C = CH2

OH

Several mechanisms get support from kinetic isotope

effect. Some of these are, oxidation of alcohols with chromic acid and electrophilic aromatic substitution. An example of a secondary isotope effect, where it is sure that the C – H bond does not break at all in the reaction. Secondary isotope effects for kH/kD are generally between 0.6 and 2.0.

(CZ3)2CHBr + H2O → (CZ3)2CHOH + HBr the solvolysis of isopropyl bromide where Z = H or D, kH/kD is

1.34 Secondary isotope effect.

The substitution of tritium for hydrogen gives isotope effects which are numerically larger (kH/kT = 16).

E2 elimination like SN2 process takes place in one step (without the formation of any intermediates). As the attacking base begins to abstract a proton from a carbon next to the leaving group, the C – H bond begins to break, a new carbon-carbon double bond begins to form and leaving group begins to depart. In confirmation with this mechanism, the base induced elimination of HBr from (I) proceeds 7.11 times faster than the elimination of DBr from (II). Thus C–H or C – D bond is broken in the rate determining step. If it was not so there would not have been any rate difference.

– C – CH2Br – CH = CH2Base

Faster reaction

H

H1-Bromo-2-phenylethane (I)

– C – CH2Br – CD = CH2Base

Slower reaction

D

D1-Bromo-2,2-dideuterio-2-phenylethane (II)

No deuterium isotope effect is found in E1 reactions since the rupture of C – H (or C – D) bond occurs after the rate determing step, rather than during it. Thus no rate difference can be measured between a deuterated and a non deuterated substrate.

Mechanism Review : Substitution versus Elimination

SN2 Primary substrate Back-side attack of Nu : with respect to LG Strong/polarizable unhindered nucleophile

Bimolecular in rate-determining step Concerted bond forming/bond breaking Inverse of stereochemistry Favored by polar aprotic solvent.

SN1 and E1

Tertiary substrate Carbocation intermediate Weak nucleophile/base (e.g., solvent)

Unimolecular in rate-determining step Racemization if SN1 Removal of β-hydrogen if E1 Protic solvent assists ionization of LG Low temperature (SN1)/high temperature (E2)

SN2 and E2 Secondary or primary substrate Strong unhindered base/nucleophile leads to SN2 Strong hindered base/nucleophile leads to E2 Low temperatrue (SN2)/high temperature (E2)

E2 Tertiary or secondary substrate Concerted anti-coplanar TS Bimolecular in rate-determining step Strong hindered base High temperature


1. The freezing point of an aqueous solution of KCN

containing 0.189 mol kg–1 was – 0.704 ºC. On adding 0.095 mol of Hg(CN)2, the freezing point of the solution became –0.530ºC. Assuming that the complex is formed according to the equation

Hg(CN)2 + x CN– → –x2x)CN(Hg +

Find the formula of the complex. Sol. Molality of the solution containing only KCN is

m = f

f

K)T(–∆

= )molkgK86.1(

)K704.0(1– = 0.379 mol kg–1

This is just double of the given molality ( = 0.189 mol kg–1) of KCN, indicating complete dissociation of KCN. Molality of the solution after the formation of the complex

m = f

f

K)T(–∆

= )molkgK86.1(

)K530.0(1– = 0.285 mol kg–1

If it be assumed that the whole of Hg(CN)2 is converted into complex, the amounts of various species in 1 kg of solvent after the formation of the complex will be

n(K+) = 0.189 mol, n(CN–) = (0.189 – x) mol

))CN(Hg(n –x2x+ = 0.095 mol

Total amount of species in 1 kg solvent becomes ntotal = [0.189 + (0.189 – x) + 0.095] mol = (0.473 – x) mol Equating this to 0.285 mol, we get (0.473 – x) mol = 0.285 mol i.e. x = (0.473 – 0.285) = 0.188

Number of CN– units combined = mol095.0mol188.0 = 2

Thus, the formula of the complex is –24)CN(Hg .

2. The equilibrium constant Kp of the reaction 2SO2(g) + O2(g) 2SO3(g) is 900 atm–1 at 800

K. A mixture containing SO3 and O2 having initial partial pressures of 1 atm and 2 atm, respectively, is heated at constant volume to equilibrate. Calculate the pressure of each gas at 800 K. [IIT- 1989]

Sol. Since to start with SO2 is not present, it is expected that some of SO3 will decompose to give SO2 and O2

at equilibrium. If 2x is the partial pressure of SO3 that is decreased at equilibrium, we would have

2SO2(g) + O2(g) 2SO3(g) t = 0 0 2 atm 1 atm teq 2x 2 atm + x 1 atm – 2x

Hence, Kp = )p()p(

)p(

22

3

O2

SO

2SO =

)xatm2()x2()x2atm1(

2

2

+−

= 900 atm–1 Assuming x << 2 atm, we get

)atm2()x2(

)x2atm1(2

2− = 900 atm–1

or 2

2

)x2()x2atm1( − = 1800

or x2

atm1 – 1 = 42.43

or x = 43.432

1×

atm = 0.0115 atm

Hence, p(SO2) = 2x = 0.023 atm; p(O2) = 2atm + x = 2.0115 atm and p(SO3) = 1 atm – 2x = 0.977 atm 3. Equal weights of CH4 and O2 are mixed in an empty

container of 1 L at 27ºC, Calculate the (a) fraction of total pressure exerted by O2, (b) total pressure if the weights of gases are 32 g each [IIT-1980] Sol. Given that, V = 1 L T = 27ºC = 300 K R = 0.0821 L atm K–1 mol–1 (constant) Mass of CH4 = Mass of O2 = mg (let)

(a) Moles of O2 = 32m

Moles of CH4 = 16m

Total no. of moles = 32m +

16m

UNDERSTANDINGPhysical Chemistry


Moles fraction of O2 = )16/m()32/m(

)32/m(+

= 31

According to Dalton's law of partial pressure,

2O´P = P × mole fraction of O2

∴ Fraction of total pressure exerted by O2,

= P´P

2O = Mole fraction of O2 = 31

(b) Given that, Mass of O2 = Mass of CH4 = 32 g

No. of moles of O2 = 3232 = 1

No. of moles of CH4 = 1632 = 2

According to Dalton's law of partial pressure,

2O´P = P × mole fraction of O2

= V

nRT × Mole fraction of O2

= 1

3000821.03 ×× × 31 = 24.63 atm

Similarly,

4CH´P = P × mole fraction of CH4

= V

nRT × Mole fraction of CH4

= 1

3000821.03 ×× × 32 = 49.26 atm

Total pressure, P = 24.63 + 49.26 = 73.89 atm 4. A solution contains Na2CO3 and NaHCO3.10 ml of

this requires 2.0 ml of 0.1 M H2SO4 for neutralization using phenolphthalein as indicator. Methyl orange is then added when a further 2.5 ml of 0.2 M H2SO4 was required. Calculate the strength of Na2CO3 and NaHCO3 in solution. [IIT-1978]

Sol. Step 1.

Equivalent mass of Na2CO3 = 2

massMolecular

= 2

106 = 53

Meq. of Na2CO3 in solution = 53m1 × 1000

Step 2.

Equivalent mass of NaHCO3 = 1

massMolecular

= 84

Meq. of NaHCO3 in solution = 84m2 × 1000

Step 3. Meq. of H2SO4 used with phenolphthalein = Valency factor × Molarity × Volume (ml) = 2 × 0.1 × 2.0 = 0.4 2Na2CO3 + H2SO4 → 2NaHCO3 + Na2SO4 Meq. of H2SO4 used with phenolphthalein

= 21 Meq. of Na2CO3 ∴

21 Meq. of Na2CO3 = 0.4

Step 4. Meq. of H2SO4 used with methyl orange = Valency factor × molarity × volume(ml) = 2 × 0.2 × 2.5 = 1 Meq. of H2SO4 used with methyl orange

= Meq. of NaHCO3 + 21 Meq. of Na2CO3

∴ Meq. of NaHCO3 + 21 Meq. of Na2CO3 = 1

∴ Meq. of NaHCO3 = 1 – 0.4 = 0.6 and Meq. of Na2CO3 = 2 × 0.4 = 0.8

Step 5.

53m1 × 1000 = 0.8 or m1 =

1000538.0 × = 0.0424

∴ Strength of Na2CO3 solution = 10

10000424.0 ×

= 4.24 g L–1

Step 6.

84m2 × 1000 = 0.6 or m2 =

1000846.0 × = 0.0504

∴ Strength of NaHCO3 solution = 10

10000504.0 ×

= 5.04 g L–1

5. Using the data given below, calculate the bond enthalpy of C–C and C–H bonds.

∆CHº(ethane) = –1556.5 kJ mol–1 ∆CHº (propane) = –2117.5 kJ mol–1 C(graphite) → C(g); ∆H = 719.7 kJ mol–1 Bond enthalpy of H–H = 435.1 kJ mol–1 ∆fHº(H2O, 1) = –284.5 kJ mol–1 ∆fHº(CO2, g) = –393.3 kJ mol–1 [IIT-1990]

Sol. From the enthalpy of combustion of ethane and propane, we write


(1) C2H6(g) + 27 O2(g) → 2CO2(g) + 3H2O(1) :

∆CH = 3∆fH(H2O, 1) + 2∆fH(CO2, g) – ∆fH(C2H6, g) Thus, ∆fH(C2H6,g) = – ∆CH + 3∆fH(H2O, 1)+ 2∆fH(CO2, g) = (1556.5 – 3 × 284.5 – 2 × 393.3) kJ mol–1 = – 83.6 kJ mol–1 (2) C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(1) ∆CH = 3∆f H(CO2, g)+ 4∆fH(H2O), 1) – ∆fH(C3H8, g) Thus ∆fH(C3H8, g) = –∆CH + 3∆fH(CO2, g) + 4∆fH(H2O, 1) = (2217.5 – 3 × 393.5 – 4 × 284.5) kJ mol–1 = –101.0 kJ mol–1 To calculate the εC–H and εC–C, we carry out the

following manipulations. (i) 2C(graphite) + 3H2(g) → C2H6(g) ∆H = – 83.6 kJ mol–1 2C(g) → 2C (graphite) ∆H = –2 × 719.7 kJ mol–1 6H(g) → 3H2(g) ∆H = –3 × 435.1 kJ mol–1 Add 2C(g) + 6H(g) → C2H6(g) ∆H(i) = (–83.6 – 2 × 719.7 – 3 × 435.1) kJ mol–1 = – 2828.3 kJ mol–1 (ii) 3C(graphite) + 4H2(g) → C3H8(g) ∆H = –101.0 kJ mol–1 3C(g) → 3C (graphite) ∆H = –3 × 719.7 kJ mol–1 8H(g) → 4H2(g) ∆H = – 4 × 435.1 kJ mol–1 Add 3C(g) + 8H(g) → C3H8(g) ∆H(ii) = (– 101 – 3 × 719.7 – 4 × 435.1) kJ mol–1 = – 4000.5 kJ mol–1 Now, ∆H(i) = εC–C – 6εC–H = –2828.3 kJ mol–1 ∆H(ii) = –2εC–C – 8εC–H = –4000.5 kJ mol–1 Solving for εC–C and εC–H, we get εC–H = 414.0 kJ mol–1 and εC–H = 344.3 kJ mol–1

Galena

Galena is the natural mineral form of lead sulfide. It is the most important lead ore mineral.

Galena is one of the most abundant and widely distributed sulfide minerals. It crystallizes in the cubic crystal system often showing octahedral forms. It is often associated with the minerals sphalerite, calcite and fluorite. Galena deposits often contain significant amounts of silver as included silver sulfide mineral phases or as limited solid solution within the galena structure. These argentiferous galenas have long been the most important ore of silver in mining. In addition zinc, cadmium, antimony, arsenic and bismuth also occur in variable amounts in lead ores. Selenium substitutes for sulfur in the structure constituting a solid solution series. The lead telluride mineral altaite has the same crystal structure as galena. Within the weathering or oxidation zone galena alters to anglesite (lead sulfate) or cerussite (lead carbonate). Galena exposed to acid mine drainage can be oxidized to anglesite by naturally occurring bacteria and archaea, in a process similar to bioleaching [3] Galena uses : One of the earliest uses of galena was as kohl, which in Ancient Egypt, was applied around the eyes to reduce the glare of the desert sun and to repel flies, which were a potential source of disease.[4] Galena is a semiconductor with a small bandgap of about 0.4 eV which found use in early wireless communication systems. For example, it was used as the crystal in crystal radio sets, in which it was used as a point-contact diode to detect the radio signals. The galena crystal was used with a safety pin or similar sharp wire, which was known as a "cat's whisker". Making such wireless sets was a popular home hobby in the North of England during the 1930s. Derbyshire was one of the main areas where Galena was mined. Scientists that were linked to this application are Karl Ferdinand Braun and Sir Jagdish Bose. In modern wireless communication systems, galena detectors have been replaced by more reliable semiconductor devices, though silicon point-contact microwave detectors still exist in the market.


Passage : A bag contains ‘n’ cards marked 1, 2, 3, ......, n. ‘X’

draws a card from the bag and the card is put back into the bag. Then ‘Y’ draws a card. The probability that ‘X’ draws.

1. The same card as ‘Y’ is –

(A) n1 (B)

n21

(C) 21

n (D)

n2

2. a higher card than ‘Y’ is –

(A) n

n 1− (B) n

n2

1−

(C) 21

nn − (D) 22

1n

n −

3. a lower card than ‘Y’ is –

(A) n

n 1− (B) n

n2

1−

(C) 21

nn − (d) 22

1n

n −

4. Evaluate : 228

∫∫

−

−

1

0

4924

1

0

5025

)1(

)1(

dxxx

dxxx = ?

5. Find the minimum value of

(x1 – x2)2 + 2

22

21 )13)(17(

20

−−− xx

x

where x1 ∈ R+ and x2 ∈ (13, 17).

6. Let f (x) = a1 tan x + a2 tan 2x + a3 tan

3x + ...........

........ + an tannx , where a1, a2, a3, ... an ∈ R and

n ∈ N. If | f (x) | ≤ | tan x | for ∀ x ∈

ππ−

2,

2, Prove

that ∑=

n

i

i

ia

1

≤ 1

7. Let az2 + bz + c be a polynomial with complex coefficients such that a and b are non zero. Prove that the zeros of this polynomial lie in the region.

| z | ≤ + ab +

bc

8. Find the fifth degree polynomial which leaves remainder 1 when divided by (x – 1)3 and remainder –1 when divided by (x + 1)3.

9. A quadrilateral ABCD is inscribed in a circle of radius R such that AB2 + CD2 = 4R2. Using vector method prove that its diagonals are at right angle.

10. Through a focus of an ellipse two chords are drawn and a conic is described to pass through their extremities, and also through the centre of the ellipse. Prove that it cuts the major axis in another fixed point.

`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

1Set

Honesty

• To be persuasive, You must be believable. To be believable, You must be credible. To be credible, You must be truthful.

• An honest man is the noblest work of God. • If I am honesty in all my dealings, I can never

experience fear. • Prefer a loss to a dishonest gain; one brings pain

for the moment, the other for all time.


1. Let S be the coefficients of x49 in given expression

f (x) and if P be product of roots of the equation

f (x) = 0, then find the value of PS , given that :

f (x) = (x – 1)2

− 2

2x

−

21x

− 3

3x

−

31x ,

.........

− 25

25x

−

251x

Sol. Here we can write f(x) as :

f (x) =

−

−

−− 25

25...3

32

2)1( xxxx

×

−

−

−−

251...

31

21)1( xxxx

Now roots of f (x) = 0 are;

12, 22, 32, ..... , 252 and 1, 21 ,

31 , .....,

251

Now f (x) is the polynomial of degree 50, So coefficient of x49 will be : S = – (sum of roots)

= – (12 + 22 + ... + 252) –

++++

251....

31

211

= –

+

×× K6

512625 where, K = ∑=

25

1n n1

⇒ S = –(K + 5525). Product of roots :

12 . 22 . 32 .... 252 . 1 . 21 .

31 ....

251 = 1 . 2 . 3 ...25

∴ P = 25 !

Hence PS =

!25)5525K( +− , where K = ∑

=

25

1n n1

2. From an external point P(α, 2) a variable line is

drawn to meet the ellipse 4

y9

x 22+ = 1 at the points

A and D. Same line meets the x-axis and y-axis at the points B and C respectively. Find the range of values of 'α' such that PA. PD = PB.PC.

Sol. We have been given,

xB

D

CO

A

yP(α,2)

PA.PD = PB.PC Equation of any line through point 'P' is :

θα−

cosx =

θ−

sin2y = r

or x = α + r cos θ, y = 2 + r sin θ Putting this point in the equation of given ellipse, we get 4(r cos θ + α)2 + 9(2 + r sin θ)2 = 36 ⇒ r2 (4 cos2 θ + 9 sin2θ) + 4r (9 sin θ + 2 α cos θ) + 4 α2 = 0 Since PA and PD are the roots of this quadratic in r,

we get

PA.PD = )sin9cos4(

422

2

θ+θα ...(i)

Similarly, putting x = r cos θ + α, y = r sin θ + 2 in the equation of coordinate axis i.e. xy = 0

(r cos θ + α). (r sin θ + 2) = 0 ⇒ r2 sin θ cos θ + r (2 cos θ + α sin θ) + 2α = 0 Since PB and PC and the roots of this quadratic in 'r',

we get, PB.PC = θθ

αcossin

2 = α

α2sin

4 ...(ii)

Thus, we get

θ

α2sin

4 = θ+θ

α22

2

sin9cos44 from (i) and (ii)

⇒ θ

α2sin

4 = )2cos1(9)2cos1(4

8 2

θ−+θ+α

⇒ θ2sin

1 = θ−

α2cos513

2

⇒ 13 = 5 cos 2θ + 2α sin 2θ

where; 5 cos 2θ + 2α sin 2θ ≤ 2425 α+

∴ 254 2 +α ≥ 13 ⇒ α2 ≥ 4

25169 − = 36

⇒ α ∈ (–∞, – 6] ∪ [6, ∞)

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

Students' ForumMATHS


3. Find the sum of the terms of G.P. a + ar + ar2 + ..... + ∞ where a is the value of x for which the function

7 + 2x loge25 – 5x – 1 – 52–x has the greatest value and

r is the ∫ +π→

x

x xxdttLt

0 2

2

0 )tan(

Sol. S = r1

a−

,

To get the greatest value f´(x) = 2log e25 – 5x – 1 log 5

+ 52–x log 5

f ´(x) = 4 loge5 – 5x–1 loge5 + 5.51 – xloge5

⇒ f ´(x) = 0 put 5x – 1 t(> 0)

t2 – 4t – 5 = 0 ⇒ t = 5 ⇒ 5x –1 = 5 ⇒ x = 2

to evaluate r :

r = ∫ +π→

x

x xxdttLt

0 2

2

0 )tan( =

π1

since a = 2, r = π1 ⇒ sum of G.P. =

12−ππ

4. Let [x] stands for the greatest integer function find

the derivative of f (x) = xxxx sin3 2])1[( +++ , where it

exists in (1, 1.5). Indicate the point(s) where it does not exist. Give reason(s) for your conclusion.

Sol. The greatest integer [x3 + 1] takes jump from 2 to 3 at 3 2 and again from 3 to 4 at 3 3 in [1, 1.5] and therefore it is discontinuous at these two points. As a result the given function is discontinuous at 3 2 and hence not differentiable.

To find the derivative at other points we write :

in (1, 3 2 ), f (x) = xxx sin2)2( ++

⇒ f ´(x) = 1sin2)2( −++ xxx

x2 + sin x + (x + 2) (2x + cos x) log (x + 2)

in ( 3 2 , 3 3 ), f (x) = xxx sin2)3( ++ ,

f ´(x) = 1sin2)3( −++ xxx x2 + sin x

+ (2x + cos x) (x + 3) × loge (x + 3)

in ( 3 5 , 1.5), f (x) = xxx sin2)4( ++ ,

f ´(x) = 1sin2)4( −++ xxx , x2 + sin x + (2x + cos x)

(x + 4) × loge(x + 4)

5. For three unit vectors a , b and c not all collinear given that ca ˆˆ× = bc ˆˆ× and ab ˆˆ× = ca ˆˆ× . Show that cosα + cos β + cos γ = –3/2, where α, β and γ are the angles between a and b , b and c and c and a respectively.

Sol. a × c = bc ˆˆ× ⇒ ( a + b ) × c = →0

⇒ c is collinear with a + b ⇒ a + b = λ c for same λ ∈ R

Similarly b + c = µ a for some scalar u

Now a + b = λ c ⇒ a + b + c = (λ + 1) →c

Similarly ⇒ a + b + c = (µ + 1) a Hence (λ + 1) c = (µ + 1) a , either λ + 1 = µ + 1 = 0 or c is collinear with a . But c can not be collinear to a other wise ac ˆˆ × = 0

⇒ bc ˆˆ × = 0

⇒ b is collinear to with c

⇒ a b and c are collinear. Hence c is not collinear to a ⇒ λ + 1 = µ + 1 = 0 ⇒ λ ± µ = –1 Hence b + c = µ a

⇒ a + b + c = →0

⇒ ( a + b + c ) . ( a + b + c ) = 0

⇒ 1 + 1 + 1 + 2 ( a . b + b . c + c . a ) = 0

⇒ a . b + b . c + c . a = –23

⇒ cos α + cos β + cos γ = – 23

6. If a, b, c and n are positive integers such that

a + b + c = n, show that (aabbcc)1/n + (abbcca)1/n + (acbacb)1/n ≤ n. Sol. Since a, b, c are integers, from A.M. – G.M.

inequality we can write

cbatimesccctimesbbbtimesaaa

++++++++++ )...()...()....(

≥ [(a.a....a times)(b.b....b times)(c.c...c times)]1/a +b+c

⇒

cbaccbbaa

++++ ... ≥ ( ) cbacba cba ++

1

Similarly, bac

cbbaac++++ ... ≥ ( ) cbacac cba ++

1

and acb

cabcab++++ ...

≥ ( ) acbacb cba ++1

Adding these three inequalities, we get

cba

cabcabcba++

+++++ 222222 ≥ (aabbcc)1/n

+ (acbacb)1/n + (abbcca)1/n

where LHS = cba

cba++

++ 2)(

= a + b + c Hence proved


1− is denoted by ‘i’ and is pronounced as ‘iota’.

i = 1− ⇒ i2 = –1, i3 = –i, i4 = 1.

If a, b ∈ R and i = 1− then a + ib is called a complex number. The complex number a + ib is also denoted by the ordered pair (a, b)

If z = a + ib is a complex number, then : (i) a is called the real part of z and we write Re (z) = a. (ii) b is called the imaginary part of z and we write Im (z) = b

Two complex numbers z1 and z2 are said to be equal complex numbers if Re (z1) = Re (z2) and Im (z1) = Im (z2).

If z = x + iy is a non zero complex number, then 1/z is called the multiplicative inverse of z.

If x + iy is a complex number, then the complex number x – iy is called the conjugate of the complex number x + iy and we write iyx + = x – iy.

Algebra of Complex Numbers (i) Addition : (a + ib) + (c + id) = (a + c) + i(b + d) (ii) Subtraction : (a + ib) – (c + id) = (a – c) + i(b – d) (iii) Multiplication : (a + ib) + (c + id) = (ac – bd) + i(ab + bc) (iv) Division by a non-zero complex number :

idciba

++ = 22 dc

bdac++ + i 22 dc

adbc+− , (c + id) ≠ 0

Properties : If z1, z2 are complex numbers, then

(i) )( 1z = z1

(ii) z + z = 2 Re (z) (iii) z – z = 2i Im (z) (iv) z = z iff z is purely real (v) z = z iff z is purely imaginary

(vi) 2121 zzzz +=+

(vii) 2121 zzzz +=+

(viii) 2121 –– zzzz =

(ix) 2

1

2

1

zz

zz

=

provided z2 ≠ 0

If x + iy is a complex number, then the non-negative

ral number 22 yx + is called the modulus of the complex number x + iy and write

| x + iy| = 22 yx +

Properties : If z1, z2 are complex numbers, then (i) | z1 | = 0 iff z1 = 0 (ii) | z1 | = | z 1 | = | – z1 | (iii) – | z1 | ≤ Re (z1) ≤ | z1 | (iv) – | z1 | ≤ Im (z1) ≤ | z1 | (v) | z1 z 1 | = | z1 |2 (vi) | z1 + z2 | ≤ | z1 | + | z2 | (vii) | z1 – z2 | ≥ | z1 | – | z2 | (viii) | z1 z2 | = | z1 | | z2 |

(ix) 2

1

zz =

||||

2

1

zz , provided z2 ≠ 0

(x) | z1 + z2 |2 = | z1 |2 + | z2 |2 + 2 Re (z1 2z )

(xi) | z1 – z2 |2 = | z1 |2 + | z2 |2 – 2 Re (z1 2z )

(xi) | z1 + z2 |2 + | z1 – z2 |2 = 2 [| z1 |2 + | z2 |2 ]. De Moivre’s Theorem

(i) If n is any integer (positive or negative), then (cos θ + i sin θ)n = cos nθ + i sin nθ (ii) If n is a rational number, then the value or one of

the values of (cos θ + i sin θ)n is cos nθ + i sin nθ Euler’s Formula

eiθ = cos θ + i sin θ and e–iθ = cos θ – i sin θ

Square root of complex number Square root of z = a + ib are given by

±

−

+

+

2||

2|| aziaz for b > 0 and

−

+

±2||–

2|| aziaz for b < 0.

COMPLEX NUMBER

Mathematics Fundamentals MA

TH

S


If ω = 2

31 i+− , then the cube roots of unity are 1,

ω and ω2. We have: (i) 1 + ω + ω2 = 0 (ii) ω3 = 1

Let z = x + iy be any complex number. Let z = r (cos θ + i sin θ) where r > 0. ∴ x = r cos θ and y = r sin θ ∴ x2 + y2 = r2

⇒ r = 22 yx + (Q r > 0)

∴ cos θ = 22 yx

x

+ and sin θ =

22 yx

y

+

The value of θ is found by solving these equations. θ is called the argument (or amplitude) of z.

If – p < θ ≤ π, then θ is called the principal argument of z.

Identification of θ –

x y arg(z) Interval of θ

+ + θ

π

<θ<2

0

+ – –θ

<θ<

π 02

–

– + (π – θ)

π<θ<

π2

– – –(π – θ)

π

<θ<π2

––

If z1 and z2 are two complex numbers then

(i) | z1 – z2 | is the distance between the points with affixes z1 and z2.

(ii) nmnzmz

++ 12 is the affix of the point dividing the

line joining the points with affixes z1 and z2 in the ratio m : n internally.

(iii) nmnzmz

–– 12 is the affix of the point dividing the

line joining the points with affixes z1 and z2 in the ratio m : n externally where m ≠ n.

(iv) If z1, z2, z3 are the affixes of the vertices of a

triangle then the affix of its centroid is 3

321 zzz ++ .

(v) z = tz1 + (1 – t)z2 is the equation of the line joining points with affixes z1 and z2. Here ‘t’ is a parameter.

(vi) 12

1

12

1

zzzz

zzzz

−−

=−− is the equation of the line

joining points with affixes z1 and z2. Three points with affixes z1, z2, z3 are collinear if

111

33

22

11

zzzzzz

= 0.

The general equation of a straight line is 0=++ bzaza , where b is any real number.

(i) | z – z1 | < r represents the circle with centre z1 and radius r.

(ii) | z – z1 | < r represents the interior of the circle with centre z1 and radius r.

1

1

zzzz

−− = k represents a circle line which is the

perpendicular bisector of the line segment joining points with affixes z1 and z2.

(z – z1) )( 2zz − + )( 1zz − + (z – z2) = 0 represents the circle with line joining points with affixes z1 and z2 as a diameter.

| z – z1 | + | z – z2 | = 2k, k ∈ R+ represents the ellipse with foci at points with affixes z1 and z2.

If z1, z2, z3 be the affixes of the points A, B, C respectively, then the angle between AB and AC is

given by arg

−−

12

13

zzzz .

If z1, z2, z3, z4 are the affixes of the points A, B, C, D respectively, then the angle between AB and CD is

given by arg

−−

34

12

zzzz .

nth roots of a complex number Let z = r (cos θ + i sin θ), r > 0 be any complex

number. nth root o z = z1/n

= r1/n

θ+π

+θ+π

nki

nk 2sin2cos ,

where k = 0, 1, 2, ………, n – 1. There are n distinct values and sum of all these

values is 0. Logarithm of a complex number

Let z = reiθ be any complex number. Then log z = log reiθ = log r + log eiθ = log r + iθ log e = log r + iθ. ∴ log z = log | z | + i amp (z).


Matrices :

An m × n matrix is a rectangular array of mn numbers (real or complex) arranged in an ordered set of m horizontal lines called rows and n vertical lines called columns enclosed in parentheses. An m × n matrix A is usually written as :

A =

mnmjmm

inijii

nj

nj

aaaa

aaaa

aaaaaaa

......

......

......

......a

21

21

222221

111211

MM

MM

Where 1 ≤ i ≤ m and 1 ≤ j ≤ n and is written in compact form as A = [aij]m× n

A matrix A = [aij]m × n is called (i) a rectangular matrix if m ≠ n (ii) a square matrix if m = n (iii) a row matrix or row vector if m = 1 (iv) a column matrix or column vector if n = 1 (v) a null matrix if aij = 0 for all i, j and is denoted by

Om× n (vi) a diagonal matrix if aij = 0 for i ≠ j (vii) a scalar matrix if aij = 0 for i ≠ j and all diagonal

elements aii are equal Two matrices can be added only when thye are of same

order. If A = [aij]m × n and B = [bij]m × n, then sum of A and B is denoted by A + B and is a matrix [aij + bij]m × n

The product of two matrices A and B, written as AB, is defined in this very order of matrices if number of columns of A (pre factor) is equal to the number of rows of B (post factor). If AB is defined , we say that A and B are conformable for multiplication in the order AB.

If A = [aij]m × n and B = [bij]n × p, then their product AB is a matrix C = [cij]m × p where

Cij = sum of the products of elements of ith row of A with the corresponding elements of jth column of B.

Types of matrices : (i) Idempotent if A2 = A (ii) Periodic if Ak+1 = A for some positive integer k.

The least value of k is called the period of A.

(iii) Nilpotent if Ak = O when k is a positive integer. Least value of k is called the index of the nilpotent matrix.

(iv) Involutary if A2 = I. The matrix obtained from a matrix A = [aij]m × n by

changing its rows into columns and columns of A into rows is called the transpose of A and is denoted by A′.

A square matrix a = [aij]n × n is said to be (i) Symmetric if aij = aji for all i and j i.e. if A′ = A. (ii) Skew-symmetric if aij = – aji for all i and j i.e., if A′ = –A.

Every square matrix A can be uniquely written as sum of a symmetric and a skew-symmetric matrix.

A = 21 (A + A′) +

21 (A – A′) where

21 (A + A′) is

symmetric and 21 (A – A′) is skew-symmetric.

Let A = [aij]m × n be a given matrix. Then the matrix obtained from A by replacing all the elements by their conjugate complex is called the conjugate of the matrix A and is denoted by ][ ijaA = .

Properties :

(i) ( )A = A

(ii) )BA( + = A + B

(iii) )A(λ = λ A , where λ is a scalar

(iv) BA)BA( = . Determinant : Consider the set of linear equations a1x + b1y = 0 and

a2x + b2y = 0, where on eliminating x and y we get the eliminant a1b2 – a2b1 = 0; or symbolically, we write in the determinant notation

22

11

baba

≡ a1b2 – a2b1 = 0

Here the scalar a1b2 – a2b1 is said to be the expansion

of the 2 × 2 order determinant 22

11

baba

having 2

rows and 2 columns. Similarly, a determinant of 3 × 3 order can be

expanded as :

MATRICES & DETERMINANTS

Mathematics Fundamentals MA

TH

S


333

222

111

cbacbacba

= a133

22

cbcb

– b133

22

caca

+ c133

22

baba

= a1(b2c3 – b3c2) – b1(a2c3 – a3c2) + c1(a2b3 – a3b2) = a1(b2c3 – b3c2) – a2(b1c3 – b3c1) + a3(b1c2 – b2c1) = Σ(± aibjck)

To every square matrix A = [aij]m × n is associated a number of function called the determinant of A and is denoted by | A | or det A.

Thus, | A | =

nnnn

n

n

aaa

aaaaaa

...

...

...

21

22221

11211

MMM

If A = [aij]n × n, then the matrix obtained from A after deleting ith row and jth column is called a submatrix of A. The determinant of this submatrix is called a minor or aij.

Sum of products of elements of a row (or column) in a det with their corresponding cofactors is equal to the value of the determinant.

i.e., ∑=

n

iija

1

Cij = | A | and ∑=

n

jija

1

Cij = | A |.

(i) If all the elements of any two rows or two columns of a determinant ate either identical or proportional, then the determinant is zero.

(ii) If A is a square matrix of order n, then | kA | = kn | A |. (iii) If ∆ is determinant of order n and ∆′ is the

determinant obtained from ∆ by replacing the elements by the corresponding cofactors, then

∆′ = ∆n–1 (iv) Determinant of a skew-symmetric matrix of odd

order is always zero. The determinant of a square matrix can be evaluated

by expanding from any row or column. If A = [aij]n × n is a square matrix and Cij is the

cofactor of aij in A, then the transpose of the matrix obtained from A after replacing each element by the corresponding cofactor is called the adjoint of A and is denoted by adj. A.

Thus, adj. A = [Cij]′. Properties of adjoint of a square matrix (i) If A is a square matrix of order n, then A . (adj. A) = (adj . A) A = | A | In. (ii) If | A | = 0, then A (adj. A) = (adj. A) A = O. (iii) | adj . A | = | A |n –1 if | A | ≠ 0 (iv) adj. (AB) = (adj. B) (adj. A). (v) adj. (adj. A) = | A |n – 2 A.

Let A be a square matrix of order n. Then the inverse of

A is given by A–1 = |A|

1 adj. A.

Reversal law : If A, B, C are invertible matrices of same order, then

(i) (AB)–1 = B–1 A–1 (ii) (ABC)–1 = C–1 B–1 A–1

Criterion of consistency of a system of linear equations (i) The non-homogeneous system AX = B, B ≠ 0 has

unique solution if | A | ≠ 0 and the unique solution is given by X = A–1B.

(ii) Cramer’s Rule : If | A | ≠ 0 and X = (x1, x2,..., xn)′

then for each i =1, 2, 3, …, n ; xi = |A||A| i where

Ai is the matrix obtained from A by replacing the ith column with B.

(iii) If | A | = 0 and (adj. A) B = O, then the system AX = B is consistent and has infinitely many solutions.

(iv) If | A | = 0 and (adj. A) B ≠ O, then the system AX = B is inconsistent.

(v) If | A | ≠ 0 then the homogeneous system AX = O has only null solution or trivial solution

(i.e., x1 = 0, x2 = 0, …. xn = 0) (vi) If | A | = 0, then the system AX = O has non-null

solution. (i) Area of a triangle having vertices at (x1, y1), (x2, y2)

and (x3, y3) is given by 111

21

33

22

11

yxyxyx

(ii) Three points A(x1, y1), B(x2, y2) and C(x3, y3) are collinear iff area of ∆ABC = 0.

A square matrix A is called an orthogonal matrix if AA′ = AA′ = I.

A square matrix A is called unitary if AAθ = AθA = I (i) The determinant of a unitary matrix is of modulus

unity.

(ii) If A is a unitary matrix then A′, A , Aθ, A–1 are unitary.

(iii) Product of two unitary matrices is unitary. Differentiation of Determinants :

Let A = | C1 C2 C3 | is a determinant then

dxdA = | C′1 C2 C3 | + | C1 C′2 C3 | + | C1 C2 C′3 |

Same process we have for row. Thus, to differentiate a determinant, we differentiate one

column (or row) at a time, keeping others unchanged.


a

PHYSICS

Question 1 to 7 are multiple choice questions. Each questions has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. A rectangular piece of aluminum is 5.10 ± 0.01 cm

long and 1.90 ± 0.01 cm wide. The area of the rectangle is written as -

(A) (9.69 ± 0.03) cm2 (B) (9.69 ± 0.01) cm2 (C) (9.69 ± 0.02) cm2 (D) (9.69 ± 0.07) cm2

2. One body fall freely from a point P at a height H + h as shown in figure whilst another body is projected upwards with an initial velocity v0. from point R at the same time as the first body begins to fall and second body meet at a point Q at a height h. The maximum height attained by of the second body for the given initial velocity is -

P

Q

Rh

H

(A) H + h (B) h4

)hH( 2+

(C) H4

)hH( 2+ (D) h

)hH( 2+

3. The velocity of a car is given by v = (10 m/s2)t – (5 m/s2) t2

Initially car at x = 0 at t = 0. Time taken by the car to reach its maximum positive x-coordinates is -

(A) 0 s (B) 1 s (C) 2 s (D) 1.5 s

4. For the circuit shown here keys k1 and k3 are closed

for 1 second. Key k2 is closed at the instant k1 and k3 are opened. Maximum charge on the capacitor after key k2 is closed is –

2V k1

0.5 Ω2F

k2

k3 2 Ω4 V

2 H

(A) Cbe1–12

(B) Cb

e1–124

(C) Cbe1–18

(D) zero

IIT-JEE 2012

XtraEdge Test Series # 1

Based on New Pattern

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

Instructions : [Each subject contain] Section – I : Question 1 to 7 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 8 to 11 are multiple choice questions with multiple correct answer. +4 marks will be

awarded for correct answer and No Negative marks for wrong answer. Section – III : Question 12 to 16 are passage based single correct type questions. +3 marks will be awarded for

correct answer and -1 mark for wrong answer. Section – IV : Question 17 to 23 are Numerical Response Question (single digit Ans. type) +4 marks will be

awarded for correct answer and No Negative marks for wrong answer.


5. The circuit shown in the diagram extends to the right into infinity. Each battery has emf ε(unknown) and internal resistance r. Each resistor has the resistance 4r. The reading of ideal ammeter shown in the diagram is I. Find the value of ε is terms of I and r

4 r

ε,r

4 r 4 r

ε,r ε,rA

(A) 4.82 I r (B) 6.82 I r (C) 1.82 I r (D) 10.82 I r

6. The metal (hollow) sphere of radius R, 2R and 3R are placed into each other such that their centres are at the same point. The inner sphere is given a charge of Q. the middle one is charged to 2Q and the outer one is charged to 3Q. Find the potentials, measured from the common centre of the circles, at a distance 4R, if the potential at the centre is taken to be zero -

(A) R

KQ23 (B) –

RKQ

23

(C) R

KQ (D) – R2

KQ

7. Which of these is not correct regarding eddy current?

(A) Eddy currents result due to motion of a metallic plate in magnetic field

(B) Eddy currents are minimised in transformer by using laminated core with metal laminations

(C) In induction furnace, eddy currents raise temperature of metal upto melting point

(D) Eddy currents are named so, as they propagate similar to swirling eddies in water

Questions 8 to 11 are multiple choice questions. Each questions has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. 8. Acceleration-time graph of particle moving along

x-axis is given by

t(s)6 4 2

2

4 a (m/s2)

Initially particle is at rest and at origin -

(A) Velocity of the particle at t = 6 s is 14 m/s (B) Position of the particle at t = 2 s is x = 4 m (C) Velocity of the particle at t = 4 s is 4 m/s (D) Position of the particle at t = 2 s is x = 8 m

9. A particle is projected with a speed v and an angle θ with horizontal. Choose the correct statements -

(A) Speed of the particle is zero at highest point (B) Speed of the particle is minimum at highest

point (C) Acceleration of particle through out he motion

is constant (D) Acceleration of particle varies with time

10. A

C

150 Ω

30 Ω20 Ω

100 Ω

B D

(A) VB – VC is greatest (B) VA = VC (C) VB – VD is greatest (D) VA = VD

11. Two charged particles M and N enters a space of uniform magnetic field with velocities perpendiculars to the magnetic field. The paths are as shown in the figure. The possible reasons for different paths may be -

×××××××××

×××××××××

×××××××××

× × × × ××× × ×

× × ×× × × × ××

× × × × × ×× × ×

× × ×× × × × ××

× × × × × ×× × ×

× ××× × × × × ×

× × × × × ×× × ×

× ××× × × × × ×

M N

(A) The charge of M is greater than that of N. (B) The momentum of M is greater than that of N (C) Specific charge of M is greater than that of N (D) The speed of M is less than that of N

This section contains 2 paragraphs, one has 3 multiple choice questions and other has 2 multiple choice questions (Question 12 to 16). Each questions has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Passage : I (Q. No. 12 to 14)

Velocity of the river with respect to ground is given v0 with of the river is d. A swimmer swims (with respect to water) perpendicular to the current with acceleration a = t (where t is time) starting from rest from the origin O at t = 0


d

O

y

v0

x

12. The time of crossing the river is - (A) (d)1/3 (B) (6d)1/3 (C) (2d)1/3 (D) None of these

13. The drift of the swimmer is - (A) v0(d)1/3 (B) v0(2d)1/3 (C) v0(6d)1/3 (D) None of these

14. The equation of trajectory of the path followed by the swimmer -

(A) y = 0v

x (B) y = 20

2

v2x

(C) y = 30

3

v6x (D) None of these

Passage : II (Q. No. 15 to 16) A metal ring having three metallic spokes of length

r = 0.2 m is in vertical plane and can spin around a fixed horizontal axis in a homogeneous magnetic field of a magnetic induction B = 0.5 T. The lines of magnetic field are perpendicular to the plane of metal ring. Between the axis of the metal ring and its perimeter we connect a consumer of resistance of 0.15 Ω with the help of two sliding contact. We fix a thread of negligible mass to the rim of the ring and wind it several times around the ring and to its end we fix a body of a mass of 20 g. At a given moment we release the body of mass m. the friction is negligible everywhere, the resistance of the ring, the spokes and the connected wiring is also negligible.

15. What is the torque exerted on the ring with spokes by the magnetic force when the body of mass m is moving with a constant velocity -

B

m (A) 0.02 N-m (B) 0.04 N-m (C) 0.06 N-m (D) 0.08 N-m

16. What current is flowing through the consumer when the velocity of the body of mass m is 3m/s -

(A) 1 Amp (B) 3 Amp (C) 2 Amp (D) 4 Amp

This section contains 7 questions (Q.17 to 23). +4 marks will be given for each correct answer and no negative marking. 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

17. The speed of a motor launch with respect to the flow of water in stream was v = 7 m/s, the speed of the stream was u = 3 m/s. When the launch began traveling upstream, a float was dropped from it. The launch traveled l = 4.2 km upstream, turned about and caught up with the float. How long was it before the launch reached the float (Answer in ................× 101 minutes)

18. A man is running with a speed 8 m/s constant in magnitude and direction passes under a lantern hanging at a height 10 m above the ground. Find the velocity which the edge of the shadow of the man's head moves over the ground with if has height is 2 m. (Answer in ............ × 101 m/sec)

19. Two motor vehicles run at constant speeds 5 m/s each along highways intersecting at an angle 60º. In what time after they meet at the intersection will the distance between the vehicles be 310 m.

20. A boy is standing on an open truck. Truck is moving with an acceleration 2m/s2 on horizontal road. When speed of truck is 10 m/s, boy projected a ball with velocity 10 m/s in vertical upward direction relative to himself (take g = 10 m/s2). After how many seconds of projection of ball, ball is moving backward horizontally as seen by boy.


21. A current carrying loop is placed in a uniform magnetic field pointing in negative z direction. Branch PQRS is a three quarter circle, while branch PS is straight. If force on branch PS is

2 F. Force on branch PQR is given a × F. Then value of a is.

P

y

x S

R

Q

22. One of the sides of the frame shown in the figure can move. The frame is in uniform magnetic field of B0. Which is perpendicular to the plane of the frame. At the time t= 0. the magnetic field begins

to decrease as B(t) = kt1

B0

+. Side of the frame is

moving with velocity v in order not to induce electric current in it. If the value of k is 3 sec–1, then v = ........... × a.

t = 0

B

a

b B

v

23. Two circular rings of identical radii and resistance of 36 Ω each are placed in such a way that they cross each other's centre C1 and C2 a shown in figure. Conducting joints are made at intersection points A and B of the rings. An ideal cell of emf 20 V is connected across AB. The power delivered by cell is ........ × 25 watt.

A

B

C2 C1

CHEMISTRY

Question 1 to 7 are multiple choice questions. Each questions has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

1. For a general nth order process A → P with initial concentration of reactant "a" and rate constant k, the expression for time for 75% completion of reaction is -

(A) k1

a2–2

1–n1

1–n

1–n

(B)

k1

a2–2

1–n1

1–n

2–n2

(C) k1

a1–2

1–n1

1–n

2–n2

(D)

k1

a1–2

2–n1

1–n2

2–n2

2. An important application of radioactive decay is in the dating of rocks, fossils, and ancient objects. Naturally occurring radioactive uranium -238 decays in a series of U92

238 steps that finally lead to the stable isotope 206Pb.

A sample of rock is found to contain 23.2 mg of uranium -238 and 1.42 mg of lead -206. If the half-life of 238U92 is 4.51 × 109 years, what is the age of rock ?

(A) 1.8 × 1010 years (B) 4.4 × 108 years (C) 1.7 × 109 years (D) 2.8 × 107 years

3. For a reaction A → Product, half-life measured for two different values of initial concentrations 5 × 10–3 M and 25 × 10–4 M are 1.0 and 8.0 hrs respectively. If initial concentration is adjusted to 1.25 × 10–3 M, the new half-life would be :

(A) 16 hrs (B) 32 hrs (C) 64 hrs (D) 256 hrs

4. Consider the following standard reduction potentials :

Half reaction (V) Ni2+(aq) + 2e– Ni (s) Eº = – 0.23 V Fe2+(aq) + 2e– Fe (s) Eº = – 0.41 V Mn2+(aq) + 2e– Mn (s) Eº = – 1.03 V Co2+(aq) + 2e– Co (s) Eº = – 0.28 V Cr2+(aq) + 2e– Cr(s) Eº = – 0.74 V

Which of the following metals could be used successfully to galvanize steel?

(A) Ni only (B) Ni and Co (C) Fe only (D) Mn and Cr

5. A beaker contains a small amount of gold dust (Au(s)). Which of the following aqueous solution, when added to the beaker, would dissolve the gold dust (ie, convert Au (s) to + Au3+ (aq))?

Half reaction Eº (at 25ºC) Zn2+ + 2e– → Zn – 0.76 Al3+ + 3e– → Al –1.66 O2 + 2H+ + 2e– → H2O2 0.70 Cr2O7

2– + 6e– + H– → 2Cr3+ 1.23 Au3+ + 3e– → Au 1.50 O2 + 4H+ + 4e– → 2Η2Ο 1.30

(A) Cr2O72– (acidic solution)

(B) H2O2 (acidic solution) (C) Al3+ (D) Zn2+


6. Choose the INCORRECT statement from the following

(A) Hexagonal closest packed (HCP) and cubic closest packed (CCP) structures have coordination numbers of 12 and 7, respectively

(B) HCP and CCP structures involve ABABAB and ABCABCABC stacking, respectively

(C) HCP and CCP structures involve the same net packing density

(D) Metals can crystallize in either HCP or CCP structures, depending on the element

7. C60 (buckyball) is cubic closest packed (face-centred cubic) in its crystalline form. If you insert potassium atoms into all the tetrahedral and octahedral holes of the C60 structure, the formula would become KxC60. What is the value of x ?

(A) 1 (B) 2 (C) 3 (D) None

Questions 8 to 11 are multiple choice questions. Each questions has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. 8. Which one of the following statements is (are) in a

concentration cell made of Fe in 0.010 M Fe2+ (aq) and Fe in 0.10 M Fe2+ (aq)?

(A)The Eºcell = 0 (B) ∆G = 0 (C) The cell reaction is Fe2+ (aq 0.010 M) → Fe2+ (aq, 0.10 M) (D) At equilibrium, the [Fe2+]anode = [Fe2+]cathode

9. The emf of a galvanic cell depends upon (A) surface area of electrodes (B) concentration of electrolyte (C) volume of electrolyte (D) temperature

10. To 20 mL of 0.1 M barium chloride solution 0.2 mL of 3 M chromium sulphate is added BaSO4 formed is completely insoluble and all the salts are completely dissociated (Change in volume is ignored). Then,

(A) boiling point is raised (B) boiling point is lowered (C) freezing point does not change (D) freezing point is increased

11. An atomic solid has hexagonal arrangement of unit cell with height of hexagonal (in close packed arrangement) as "h" The radius of atom in terms of height is

(A) h324 (B)

32

4h

(C) 23

2h (D)

23

4h

This section contains 2 paragraphs, one has 3 multiple choice questions and other has 2 multiple choice questions (Question 12 to 16). Each questions has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Passage : I (Q. No. 12 to 14)

The structure of unit cell of perovskite – a salt of lanthanum(La), manganese (Mn) and oxygen, has Mn2+ at the each corner, oxide on every edge centre and a lanthanum ion at the body centre.

12. What is the empirical formula of the salt? (A) LaMnO2 (B) LaMn2O3 (C) LaMnO3 (D) Both (B) and (C)

13. What are the coordination ion numbers of Mn; La and O, respectively?

(A) 6, 2, 8 (B) 6, 4, 12 (C) 8, 4, 8 (D) 6, 12, 2

14. Considering the closed packed arrangement of ions, which of the following statement regarding the ionic radii is most the appropriate?

(A) +4Lar > –2O

r

(B) +4Lar > +2Mn

r

(C) Edgelength of unit cell = +2Mnr + –2O

r

(D) None of the above

Passage : II (Q. No. 15 to 16) Properties such as vapour pressure, boiling point,

freezing point of pure solvent changes when solute molecules are added in order to prepare homogeneous solution. These are called colligative properties. Application of colligative properties are very useful in day to day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing agent in the radiator of automobiles. A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.90. Also given are the following information :

Kf of water = 1.86 K kg mol–1; Kf of ethanol = 2.00 K kg mol–1; Kb of water = 0.52 K kg mol–1, kb of ethanol = 1.20 K kg mol–1; Standard freezing point of water = 273 K; Standard freezing point of ethanol = 155.7 K

Standard boiling point of water = 373 K; Standard boiling point of ethanol = 351.5 K

Vapour pressure of pure water = 32.8 mm of Hg; vapour pressure of pure ethanol = 40 mm of Hg. In answering the following questions, consider the solutions to be ideal dilute solutions and solute to be non-volatile, non-electrolytic.

15. The freezing point of solution M is (A) 268.7 K (B) 268.5 K (C) 234.2 K (D) 150.9 K


16. The vapour pressure of solution M is (A) 39.3 mm of Hg (B) 36.0 mm of Hg (C) 29.5 mm of Hg (D) 28.8 mm of Hg


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

17. A body-centred cubic lattice is made-up of two types of atoms, A and B in which A occupy the body-centre, while B is the corners. Due to some imperfections in the solid, one corner is left unoccupied per unit cell and formula of the solid is AxBy. Here x is.

18. In HCP arrangement of atoms, coordination number of atoms in the same layer is,

19. If an aqueous NaCl solution is electrolysed using a current of 5 A for 200 min, Volume of Cl2(g) in litres, produced under STP condition is (F = 96000 C).

20. Equivalent conductance of 0.2 aqueous solution of a weak monobasic acid (HA) is 10 S cm2 equiv–1 and that at infinite dilution is 200 S cm2 equiv–1. Hence, pH of this solution is.

21. How many Faradays of electricity is required for electrolysis of 72 g of acidified water?

22. A 0.4 molal aqueous solution of NaxA has freezing point of –3.27º C. If Kf of water is 1.86 K kg mol–1, the value of x is : (Assume complete of ionization of salt NaxA).

23. Solubility product of PbCl2 at 25ºC is 4 × 10–6 . Kf of water is 2K kg mol–1. When PbCl2 is dissolved in water its saturated solution, the freezing point

decreases by º100m

. The value of "m" is -

MATHEMATICS


1. If tan2

π

9x f (x) = xπ+ 2cos1 + | f (x)|, then

f (3) is equal to -

(A) 2

1 (B) – 22

1

(C) 21– (D)

221

2. If y = x3 – c and y = x2 + ax + b touch each other at

the point (1, 2), then |–|

||||||cba

cba+

++ is equal to -

(A) 1 (B) 0 (C) 2 (D) None of these

3. If the domain of the function

f (x) = )3–92(

12 aaxxn ++l

is R then number of

integral values of a is - (A) 8 (B) 10 (C) 5 (D) 6

4. If the vertical distance between the graphs

y = 3

53

3–25–3532xx

xxx+

++ & y = k approaches to zero

as x approaches to infinity, then the value of 'k' is -

(A) 3

2 (B) 23 (C) 1 (D) 0

5. The length of the sub tangent to the curve

y = 35

23

–5)1(

xxx + at x = 1 is -

(A) 2081 (B)

8120 (C)

2027 (D)

2720

6. If (f (x))(n) represents nth order derivative of the function y = f (x) & g(x) = (ln (x))(5) , then g(x) is always -

(A) increasing and concave up (B) increasing & concave down (C) decreasing and concave up (D) decreasing & concave down


7. If the equation 3x4 – 16x3 + 30x2 – 24x + 12 = 12a has exactly 2 solution then the complete set of value of 'a' is -

(A)

∞,

31 (B) (3, ∞)

(C) (– 1, ∞) (D) (1, ∞)

Questions 8 to 11 are multiple choice questions. Each questions has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct.

8. If f (x) = sgn || xa xa ; g(x) = ]sgn[ || xa x

a for a > 1 and x ∈ R0, where . & [.] denotes the fractional part and integral part functions respectively, then which of the following statements holds good for the function h(x), where (lna) h(x) = (ln f (x) + lng(x))

(A) h is even (B) h is odd (C) h is decreasing (D) h is increasing

9. If f (x) = max (sin–1x, cos–1x) and

g(x) = f

+

21x –

4π , then identify correct

statements

(A) range of g(x) is

π

43,0

(B) g(x) is not differentiable at x = 0

(C) g(x) = k have only one solution is k ∈

ππ

43,

4

(D) f (cos 5) + f (sin 5) = 7π – 15

10. Identify surjective functions -

(A) f :

π

2,0 → [12, ∞), f (x) = 4 tan2x + 9 cot2x

(B) f : [0, 2π] → [0, 50], f (x) = 24cosx – 7sinx +25 (C) f : R → R, f (x) = (2x + 5) (3x + 7) (4x + 9)

(D) f : R → [0, 1), f (x) = 1||

||+x

x

11. If f is a function defined from R to R such that f(f (f (x))) – (f (f (x)))3 = f (f (x)), then -

(A) f (x) has exactly two real roots (B) f (x) is continuous everywhere (C) f (x) is surjective function (D) f (x) = f –1(x) at x = 0


Passage : I (Q. No. 12 to 14) Consider a function f (x) satisfying a functional rule f (x) + 2f (1 – x) = x2 + 1, ∀ x ∈ R. Let g(x) be another function such that g'(x) = 3 f (x) ∀ x ∈ R and g(0) = 0. x-axis touches the graph of y = g(x) at (a, 0).

12. The range of the function f (sin x) is - (A) [0, 8] (B) [1, 8]

(C)

38,

31 (D)

38,0

13. If the number of solutions of the equation g(|x|) = k be 2, then the range of values of k, are -

(A)

34,0 (B)

34,0

(C)

∞,

34 (D)

∞,

34

14. If the area of triangle formed by the points (k, 0), (k, g(k)) & (a, 0) is maximum, where k ∈ (0, 3) then the value of k is

(A) 43 (B) 1 (C)

34 (D)

25

Passage : II (Q. No. 15 to 16) Let the function f and g be defined by

f (x) =

+∈∈

xxfxxxx

allfor)2()2,1[for–2)1,0[for

and g(x) = 4f (3x) + 1 for all x.

15. Sum of all the solutions of the equation f (x) = 0.6 for 0 ≤ x ≤ 7 is

(A) 23.64 (B) 13.44 (C) 31.04 (D) None of these

16.

35tan2tan 1– f is equal to -

(A) 43 (B)

31 (C)

32 (D)

34



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17. Let f (x) =

≥+<

0If,3][0If,–][2

xxxxxx

, where x and

[x] are the fractional part and greatest integer of x respectively. The number of solutions of the equation f (x) – x = x ∀ x ∈ [– 8, 8] are.

18. If f (x) = ∞→h

lim h

+

xhxn

sin)/1sin(

l ; 0 < x < 2π ,

then number of points where f (x) is discontinuous is.

19. If set of value of 'p' for which the equation

21–

12sin

xx

+ + px = 0 has exactly three solutions is

(l, m) then 8m – l is.

20. If the equation 2|x – 2| – |x + 1| + x = k, has infinite solutions then the value of k is.

21. The value of

+

∑=

∞→

n

rn

nrn 1 2cos1

12lim , where [.]

represents greatest integer function, is.

22. The greatest value of the function

)104(log + (cos2θ – 6 sinθ cosθ + 3 sin2θ + 2) is.

23. From a point on the curve

y = sin–1 x + cos–1 x– a tangent is drawn to the

curve g(x) = x + cos x + 2π – 1. Then sum of its

intercepts on coordinate axes is.

WHAT ARE EARTHQUAKES?

Earthquakes like hurricanes are not only super destructive forces but continue to remain a mystery in terms of how to predict and anticipate them. To understand the level of destruction associated with earthquakes you really need to look at some examples of the past.

If we go back to the 27th July 1976 in Tangshan, China, a huge earthquake racked up an official death toll of 255,000 people. In addition to this an estimated 690,000 were also injured, whole families, industries and areas were wiped out in the blink of a second. The scale of destruction is hard to imagine but earthquakes of all scales continue to happen all the time.

So what exactly are they ? Well the earths outer layer is made up of a thin crust divided into a number of plates. The edges of these plates are referred to as boundaries and it’s at these boundaries that the plates collide, slide and rub against each other. Over time when the pressure at the plate edges gets too much, something has to give which results in the sudden and often violent tremblings we know as earthquakes.

The strength of an earthquake is measured using a machine called a seismograph. It records the trembling of the ground and scientists are able to measure the exact power of the quake via a scale known as the richter scale. The numbers range from 1-10 with 1 being a minor earthquake (happen multiple times per day and in most case we don’t even feel them) and 7-10 being the stronger quakes (happen around once every 10-20 years). There’s a lot to learn about earthquakes so hopefully we’ll release some more cool facts in the coming months.


PHYSICS


1. xtandxd

(A) 2 sec2x (tan x)–1/2 (B) 21 sec2x (tan x)–1/2

(C) 21 (tan x)–1/2 (D) 2 (tan x )–1/2

2. 1x

)1x(dxd 2

++

(A) 2

2

)1x(1–x2x

++ (B) 2

2

)1x(1x2–x

++

(C) 1x

1–x2x 2

++ (D) 2

2

)1x(1x2x

+++

3. If |→A ×

→B | = )B.A(3

→→, then the value of )BA(

→→+

is -

(A) (A2 + B2 + AB)1/2 (B) 2/1

22

3ABBA

++

(C) A + B (D) )A2 + B2 + 3 AB)1/2

4. Correct graph of 3x + 4y + 1 = 0 is -

(A)

1

X

Y

(B)

1/4

X

Y

(C)

–1/4X

Y

(D)

–1/4X

Y

5. Correct graph of y – 1 = x2 is -

(A)

X

Y

(B)

X

Y

(C)

X

Y

(D)

X

Y

6. If →r = bt2 ^

i + ct3 ^j , where b and c are positive

constant, the time at which velocity vector makes an angle θ = 60º with positive y-axis is -

(A) bc (B)

c33b2

(C) b3c2 (D)

c3b2

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

Instructions : [Each subject contain] Section – I : Question 1 to 7 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 8 to 11 are multiple choice questions with multiple correct answer. +4 marks will be awarded

for correct answer and No Negative marks for wrong answer. Section – III : Question 12 to 16 are passage based single correct type questions. +3 marks will be awarded for correct

answer and -1 mark for wrong answer. Section – IV : Question 17 to 23 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 # 1

Based on New Pattern


7. A man throws a stone in vertical upward direction from ground and reach maximum height H. During ascent, he travels a distance of H/5 in the last second. Then H is equal to : (Take g = 10 m/s2)

(A) 25 m (B) 20 m (C) 10 m (D) data is insufficient


8. Which of the following graph(s) is/are not possible?

(A)

Dis

tanc

e

O Time

(B)

D

ista

nce

O Time

(C) Dis

tanc

e

O t1 Time

(D)

Dis

tanc

e

O Time

9. If the velocity of a body is constant - (A) |Velocity | = speed (B) |Average velocity| = speed (C) Velocity = average velocity (D) Speed = average speed

10. The two vectors →A and

→B are drawn from a

common point and →C =

→A +

→B , then angle

between →A and

→B is -

(A) 90º if C2 = A2 + B2 (B) greater than 90º if C2 < A2 + B2 (C) greater than 90º if C2 > A2 + B2 (D) less than 90º if C2 > A2 + B2

11. Which of the following quantities are independent of the choice of orientation of the co-ordinate axes?

(A) →a +

→b (B) ax + by

(C) |→a +

→b –

→c | (D) Angle between

→a and

→b

This section contains 2 paragraphs, one has 3 multiple choice questions and other has 2 multiple choice questions (Question 12 to 16). Each questions has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Passage : I (No. 12 to 14)

An object is rotating on a horizontal circular track with constant speed 'u0' in clock-wise direction. Initially object was moving in east direction.

12. Change in velocity as object rotates by an angle 90º is -

(A) zero (B) 2u0

(C) 2 u0 (D) 3 u0

13. Direction of change in velocity in above question is-

(A) south-east (B) south-west (C) south (D) west

14. Charge in velocity as objects rotates by an angle 180º is -

(A) zero (B) 2 u0 towards west (C) 2 u0 towards west (D) 2 u0 towards south

Passage : II (No. 15 to 16) A particle is projected from ground with a velocity

u and an angle 53º with the horizontal. During rising after t= 1 s angle made by the velocity vector with the horizontal is 37º. (Take g = 10 m/s2)

15. Value of u is

(A) 7

100 m/s (B) 7

200 m/s

(C) 7

300 m/s (D) date is insufficient

16. Speed of particle at the moment when angle is 37º -

(A) 100 m/s (B) 7

150 m/s

(C) 7

100 m/s (D) date is insufficient


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17. A stone is projected from level ground. At a height of 0.4 m above ground its velocity is found to be →v = (

^i6 +

^j2 ) m/s. x-axis is along horizontal

and y-axis vertically upwards). The angle of projection with the vertical is found to be (10 m) degree. Find the m.

18. A particle starts from the origin goes along the x-axis to the point (30 m, 0 m) and then returns along the same line to the point (–30 m, 0 m). The distance travelled by the particle is (10 p) metre. Find the value of 'p'.

19. A body initially at rest moving along x-axis in such a way so that its acceleration Vs displacement plot is as shown in figure. what will be the maximum velocity of particle in m/sec.

S 1m 0.5

1 m/s2

a

20. If a cone of radius R and height R is taken out of a solid hemi sphere of radius R as shown in figure. Find the volume of remaining shaded part.

(given that R = π

153 )

R

21. If y = 4x2 – 4x + 7. Find the minimum value of y.

22. The area of a rectangle of size 1.25 cm × 1.55 cm is 1.9 y, where y is single digit numbers. Find y.

23. A 2m wide truck is moving with a speed of 55 m/s along a straight horizontal road. A man starts crossing the road with a uniform speed v when the truck is 4m away from him The minimum value of v (in m/s) to cross the truck safely is -

CHEMISTRY


1. Which of the following have identical geometry with same number of lone pair (s)?

(I) PCl5 (II) PF3Br2 (III) BrF5 (IV) SbF5

2– (A) I and II (B) II and III (C) III and IV (D) III, II and IV

2. Polyethylene can be produced from calcium carbide according to the following sequence of reactions

CaC2 + H2O → CaO + HC ≡ CH

nHC ≡ CH + nH2 → —CH2–CH2—

n

The mass of polyethylene which can be produced from 40.0 kg of CaC2 is -

(A) 6.75 kg (B) 17.5 kg (C) 8.75 kg (D) 9.75 kg

3. sp3 hybridisation is in - (A) AlH4

– (B) CH3–

(C) Be4(CH3COO)6O (D) All of these

4. What will be the maximum spin multiplicity of 3d-orbital?

(A) 4 (B) 6 (C) 5 (D) 10

5. Photons having energy equivalent to binding energy of 2nd state of Li+ ion are used at metal surface of work function 10.6 eV. If the ejected electrons are further accelerated through the potential difference of 5 V then the minimum value of de-Broglie wavelength associated with the electron is -

(A) 2.45 Å (B) 9.15 Å (C) 5 Å (D) 11 Å

6. An impure sample of Ba(OH)2 (mol. wt. = 171) of mass 1 gram allowed to react with 80 ml of 0.20 M HCl (aq). When excess acid was titrated with NaOH, 20 ml of NaOH (aq) was required. 10 ml of the same NaOH (aq) required 30 ml of the 0.1 M HCl (aq.) in a separate titration percentage purity of Ba(OH)2 sample is -

(A) 4.275 (B) 42.75 (C) 85.5 (D) None of these


7. What may be the correct orbital notation if the wave function is –

Ψ = 2/3

0a1

6811

π σ2e–σ/3 (3cos2θ – 1); Here σ =

r/a0 and a0 = 20

2

mehπ

∈

(A) 4s (B) 2xP2

(C) 3Py (D) 3dz


8. Which of the following statements is/are true for P4S3 molecule -

(A) It contains six P–S bonds and three P-P-bonds (B) It contains six P-S bonds and ten lone pairs (C) It has all atoms sp3-hybridised (D) It contains six P-P bonds and ten lone pairs

9. Which of the following species are correctly matched with their geometries according to the VSEPR theory -

(A) ClF2– → linear

(B) IF4+ → see – saw

(C) SnCl5– → trigonal bipyramidal

(D) 33

••)SiH(N → pyramidal

10. Sodium borohydride (NaBH4) reacts with iodine to form boron triodide, Sodium iodide and HI. In an experiment, 76 gram sodium borohydride is mixed with 300 gram of Iodine. In this reaction -

(A) NaBH4 is the excess reagent (B) Iodine is the excess reagent (C) If 39.2 gram of BI3 are formed, its percentage

yield is 40 (D) The weight of the reagent remaining in excess

is 64.78 gram

11. KCl has a dipole moment of 10 D. The inter ionic distance in KCl is 2.6 Å. Which of the following statement are true for this compound ?

(A) The theoretical value of dipole moment, if the compound were completely ionic is 12.5 D.

(B) The % ionic character of the compound is 85 % (C) It is a poor conductor of electricity (D) The forces operating in this molecule are

coulombic type


Passage : I (Q. No. 12 to 14) In order to explain the existance of doublets in the spectra of alkali metals, Goudsmit and Uhlenbeck in 1925 proposed that electron has an intrinsic angular momentum due to spining about its own axis. The value of spining a angular momentum of electron can be described by 2 spin quantum number s and ms. The physical significance of s and ms is similar as of l and ml.

12. The possible value of s for electron is - (A) 1/2 (B) – 1/2 (C) 0 (D) 1

13. Relation between s and ms is :

(A) )1s(s +π2

h . cos θ = ms

(B) )1s(s + cos θ = ms

(C) π4h3 = ms

(D) None of these

14. Spin angular momentum of electron has magnitude

equal to :

(A) π4h3 (B)

π2h3

(C) π2

nh (Here n is any positive integer)

(D) 21

Passage : II (Q. No. 15 to 16) Ozone in the upper atomsphere absorbs ultraviolet

radiation which induce the following chemical reaction in –

O3(g) → O2(g) + O(g) O2 produced in the above photochemical

dissociation undergoes further dissociation into one normal oxygen atom and one more energetic oxygen atom O*.

O2(g) → O + O* If O* has 3eV more energy than O and normal

dissociation energy of O2 is 480 KJ/mol. (Given 1 eV/photon = 96 KJ/mol)

15. What is maximum wavelength effective for the photochemical dissociation of O2 molecule -

(A) 2440 Å (B) 1547 Å (C) 1000 Å (D) 155 Å


16. Half a mole of photon is used to break the O3 molecule completely according to Ist reaction. O2 produced in this reaction dissociates according to 2nd reaction, then what is the total energy required in KJ to carry out the dissociation of O2.

(A) 384 KJ (B) 280 KJ (C) 480 KJ (D) 300 KJ


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17. A compound exists in the gaseous phase both as monomer (A) and dimer (A2). The molecular weight of A is 60. In an experiment 240 g of the compound was confined in a vessel of volume 32.84 litre and heated to 127º C. Calculate the pressure (in atm) developed if the compound exists as dimer to the extent of 50% by weight under these conditions.

18. Li2+ ion in its ground state absorbs a photon of energy 183 electron volt. Electron of Li2+ ion strikes the He+ ion. (in Å) after being struck by electron of Li2+ ion.

19. First ionization energy of Li is 13.6 eV. If it is

assumed that outermost electron of Li revolves under the influence of nucleus which is shielded by inner two electrons in first orbit of Li, then find out by what amount of charge inner two electrons shield the nucleus?

20. A photon of 50 eV energy strikes a metal surface

(having work function 1.64 eV). Photoelectron ejected from metal with maximum kinetic energy strikes He+ ion (in ground state).

Find out de-Broglie wavelength of electron of He+

ion finally in Å.

21. Find number of 120º bond angles in O = CF2. 22. Give number of H-atoms in H2C = SF4 which are

in plane of axial F-atoms. 23. For an element Zeff for its outermost electron is

3.55. If atomic radius of element is 0.75 Å, then find out its electro negativity.

MATHEMATICS


1. If f (θ) = θ

θ+θ2cos2

2cos2sin–1 , then value of

f (11º) f (34º) equals -

(A) 21 (B)

43

(C) 41 (D) None of these

2. If 3 ≤ a < 4 then value of sin–1(sin [a]) + tan–1(tan [a]) + sec–1(sec[a]) (where [.] denotes greatest integer function) is - (A) 2π (B) 2π – 3 (C) 3 (D) 2π + 3

3. If sum of all solutions of equation

3cot2θ + 10 cotθ + 3 = 0 in [0, 2π] is 2πk where

k ∈ I then k equals - (A) 3 (B) 6 (C) 10 (D) 15

4. Range of k for which k cos2x – k cos x + 1 ≥ 0 ∀ x ∈ R is -

(A) k < – 21 (B) k > 4

(C) – 21 ≤ k ≤ 4 (D) None of these

5. If sin (sin x + cos x) = cos (cos x – sin x ) then

largest possible value of sin x is -

(A) 2

1 (B) 1

(C) 4–16 2π (D)

4π


6. If ∆ is area of ∆ABC and length of two sides are 3 & 5 respectively, if third side is c, then

(A) ∆ ≤ 312

64162 ++ cc (B) ∆ = 8

54162 ++ cc

(C) ∆ > 34

74162 ++ cc (D) None of these

7. A point Q is selected at random inside the equilateral triangle. If sum of lengths of perpendicular dropped on sides from Q is P. Then altitude of triangle is -

(A) 2P (B)

3P

(C) P (D) None of these

Questions 8 to 11 are multiple choice questions. Each questions has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. 8. If cotθ + tanθ = x and secθ – cosθ = y, then - (A) x sinθ . cosθ = 1 (B) sin2θ = y cosθ (C) (x2y)1/3 + (xy2)1/3 = 1 (D) (x2y)2/3 – (xy2)2/3 = 1

9. If 0 ≤ θ ≤ π and sin 2θ = θ+ sin1 – θsin–1

then possible values of tanθ is/are -

(A) 34 (B) –

43

(C) – 34 (D) 0

10. If

+

xx 2

2

cos1cos (1 + tan22y) (3 + sin 3z) = 4

then - (A) x is a multiple of π (B) x is a multiple of 2π (C) y is a multiple of π/2 (D) None of these

11. In a ∆ABC, if r = 1, R = 3, s = 5 which of the following is/are correct (where a, b, c represent sides of triangle) -

(A) ar ∆ABC = 5 (B) Product of sides of ∆ABC is 60 (C) a2 + b2 + c2 = 24 (D) Sum of exradii of ∆ABC is 13


Passage : I (Q. No. 12 to 14) Let α ± β is not an odd multiple of π.

If cos α + cos β = b, sin α + sin β = a, θ = 2

β+α

and sin 2θ + cos 2θ = 1 + 22)–(

babana

+ where n ∈ I

then 12. Value of n is - (A) 0 (B) 1 (C) – 2 (D) None of these

13. If cosecnx = A then sin 3A. sin A is polynomial in x, whose degree is equal to -

(A) 5 (B) 4 (C) 2 (D) 3

14. If degree of polynomial obtained in above question is p then max. value of (p + 1) sin x + (p + 2) cos x is -

(A) p + 2 (B) p + 3 (C) p + 1 (D) None of these

Passage : II (Q. No. 15 to 16) Let sides of a triangle are a = n + 1, b = n + 2 &

c = n with sin C = 54 , then answer the following.

15. Area of ∆ABC is - (A) 84 (B) 72 (C) 60 (D) None o these

16. Largest exradius of circle escribing ∆ABC - (A) 12 (B) 14 (C) 16 (D) None of these


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17. If 772342cos11–3sin

θθ

= 0

Then no. of values of θ in [0, 2π].

18. Minimum value of y = (sin x + cosec x)2 + (cos x + sec x)2 ∀ x ∈ R is.

19. If sin–1(sin 5) > x2 – 4x then number of possible integral values of x will be.

20. Find no. of solution of x for ex cot x = 1 where x ∈ (0, 2π).

21. In ∆ABC if angles A, B, C are in A. P. & ∠A exceeds lowest angle by 30º & D divides BC

internally in 1 : 3 then CADBAD

∠∠ =

k1 . Find k

22. The number of solution of cos(2sin–1(cot(tan–1(sec(6cosec–1x))))) + 1 = 0 where x > 0 is

23. Circumcentre is at origin & a ≤ sin A. P(x, y) lie

inside the circumcircle & k = |xy|8

1 , then least

integer value of k can be.

Prefix Symbol Powers of 10

deci d 10–1

centi c 10–2

milli m 10–3

micro µ 10–6

nano n 10–9

pico p 10–12

femto f 10–15

atto a 10–18

zepto z 10–21

yocto y 10–24

deca da 101

hecto h 102

kilo k 103

mega (or million) M 106

giga (or billion) G 109

tera (or trillion) T 1012

peta P 1015

exa E 1018

zetta Z 1021

Yotta Y 1024

Sub multiplies

Multiples

PREFIXES

Facts on the Sun The Milky Way galaxy is home to Planet Earth and is energized by the enormous ball of energy, the Sun. This big star is much closer to us than the millions of stars that twinkle in the night sky and it supports life with the heat generated. The quest to know the space or void that surrounds the planet and the celestial bodies that intrigue man by their shimmer and movements have led scientists to delve deep into the study of the universe, with the support of harnessed technology. Research reveals that the Sun is average in size and features, compared to the million others that just appear small in the night sky due to their proximity to earth. The Sun is the center of our galaxy and is made up of hot gases that comprise elements like calcium, hydrogen, helium, sodium, iron and magnesium. The Sun is a big star that simply looks small in the sky because it is approximately 93 million miles or 150 million km away. While the Earth is about 13 thousand kilometers across, the sun is 1.4 million kilometers and the difference in size between the Earth and the Sun is better understood with the calculation that it would take more than a million Earths to fill the Sun, if possible! Considering that a beam of sunlight takes approximately 8 minutes to reach us, scientists have calculated that it would take man about a hundred and seventy six years to reach the sun; this is in conjunction to the speeds we allow ourselves on the planet. Although actually weighing the Sun with a scale is impossible, scientists compute its weight with the understanding that it contains the same material mass that is weighable in our solar system. Certain comparisons and the updated study that the Sun is over 300,000 times heavier than Earth, defines an approximate gravitational pull and the weight increase factor. The Sun has no solid surface! The Sun is estimated to be more than four billion years old. In comparison to the time frame within which earliest evidence of life forms on the planet have been recorded, the Big Bang seemed to have taken place much, much earlier. And, the Sun is not expected to change in form or size at least another 5 billion years. Once the hydrogen exhausts, or burns out, scientists expect this Center of the Galaxy to enter a new phase of existence, burning out the helium component next. The burning of helium will result in the expansion of the star to nearly 100 times its current size! The prediction and study reveals that the resultant Red Giant will then collapse into a smaller White Dwarf!


CHEMISTRY

SECTION – I Single Correct Choice Type

This section contains 7 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

1. Geometrical shapes of the complexes formed by the reaction of Ni2+ with Cl–, CN– and H2O, respectively, are

(A) octahedral, tetrahedral and square planar (B) tetrahedral, square planar and octahedral (C) square planar, tetrahedral and octahedral

(D) octahedral, square planar and octahedral Ans. [B] Sol. Complexes are : [NiCl4]–2, [Ni(CN)4]–2 &

[Ni(H2O)6]+2 Ni+2 = 3d84s0 [NiCl4]–2 : Now Since, Cl– is a weak legand so no

pairing of electron take place and geometry is tetrahedral

[Ni(CN)4]–2 : Since, CN– is a strong legand so pairing of electron will take place & geometry is square planar. [Ni(H2O)6]+2 : It will formed octahedral complex since C.N. = 6

2. AgNO3 (aq.) was added to an aqueous KCl solution gradually and the conductivity of the solution was measured. The plot of conductance ( Λ ) versus the volume of AgNO3 is -

Λ

volume(P)

Λ

volume(Q)

Λ

volume(R)

Λ

volume(S)

(A) (P) (B) (Q) (C) (R) (D) (S) Ans. [D] Sol. Because in the beginning of the reaction no of

ions remain constant so conductivity remains constant but after complete precipitation of Cl– the no. of ions increases in solution. So conductivity increases.

3. Bombardment of aluminum by α-particle leads to its artificial disintegration in two ways, (i) and (ii) as shown. Products X, Y and Z respectively are -

Al2713 YP30

15 +

ZSi3014 +XSi30

14 +

(ii)

(i)

(A) proton, neutron, positron

(B) neutron, positron, proton (C) proton, positron, neutron

(D) positron, proton, neutron Ans. [A]

IIT-JEE 2011 PAPER-I (PAPER & SOLUTION)

Time : 3 Hours Total Marks : 240

Instructions : [Each subject contain]

Section – I : Multiple choice questions with only one correct answer. +3 marks will be awarded for correct answer and -1 mark for wrong answer. [No. of Ques. : 7]

Section – II : Multiple choice questions with multiple correct answer. +4 marks will be awarded for correct answer and No Negative marking for wrong answer. [No. of Ques. : 4]

Section – III : Passage based single correct type questions. +3 marks will be awarded for correct answer and -1 mark for wrong answer. [2 passage, No. of Ques. : 5]

Section – IV : Numerical Response Question (single digit Ans. type) +4 marks will be awarded for correct answer and No Negative marking for wrong answer. [No. of Ques. : 7]


Sol. nPAl 10

3015

He2713

42 + →

He42

Si3014 + H1

1 Si3014 + e0

1+

4. Extra pure N2 can be obtained by heating - (A) NH3 with CuO (B) NH4NO3

(C) (NH4)2Cr2O7 (D) Ba(N3)2

Ans. [D]

Sol. Ba(N3)2 →∆ Ba + 3N2 ↑

5. Among the following compounds, the most acidic is -

(A) p-nitrophenol (B) p-hydroxybenzoic acid

(C) o-hydroxybenzoic acid (D) p-toluic acid

Ans. [C] Sol. o-hydroxy benzoic acid is stronger acid due to

ortho effect

6. The major product of the following reaction is -

C O

NHC O

(i) KOH

(ii) Br CH2Cl

(A)

C O

N–CH2C O

Br

(B)

C O

NC O

CH2Cl

(C)

C O

N

O–CH2 Br

(D)

C O

N

O CH2Cl

Ans. [A]

Sol.

CO

NHCO

KOH

CO

NKCO

Cl–CH2– – Br

CO

N – CH2– –BrCO

Θ ⊕

7. Dissolving 120 g of urea (mol. wt. 60) in 1000 g

of water gave a solution of density 1.15 g/mL. The molarity of the solution is -

(A) 1.78 M (B) 2.00 M (C) 2.05 M (D) 2.22 M

Ans. [C]

Sol. M = wtmol10dx ××

= 60

1015.17.10 ×× = 2.05 M

x = percentage by weight

x = 1001000120

120×

+

SECTION – II Multiple Correct Choice Type

This section contains 4 multiple choice questions. Each questions has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct.

8. Extraction of metal from the ore cassiterite involves -

(A) carbon reduction of an oxide ore (B) self-reduction of a sulphide ore

(C) removal of copper impurity (D) removal of iron impurity

Ans. [A,C,D] Sol. Cassiterite is SnO2. To reduce SnO2 into Sn, carbon reduction process

is used. Sn has iron impurity. SnO2 + C → Sn + CO2


9. Amongst the given options, the compound(s) in which all the atoms are in one plane in all the possible conformations (if any), is (are)

(A) C–CH

CH2H2C

H (B) H–C≡C–C

H

CH2 (C) H2C=C=O (D) H2C=C=CH2

Ans. [B,C] Sol. Factual.

10. The correct statement(s) pertaining to the adsorption of a gas on a solid surface is (are)

(A) Adsorption is always exothermic (B) Physiosorption may transform into

chemisorption at high temperature (C) Physiosorption increases with increasing

temperature but chemisorption decreases with increasing temperature

(D) Chemisorption is more exothermic than physiosorption, however it is very slow due to higher energy of activation

Ans. [A, B, D] Sol. Factual.

11. According to kinetic theory of gases - (A) collisions are always elastic (B) heavier molecules transfer more momentum

to the wall of the container (C) only a small number of molecules have very

high velocity (D) between collisions, the molecules move in

straight lines with constant velocities Ans. [A, D] Sol. Factual.

SECTION – III Paragraph Type

This section contains 2 paragraphs. Based upon the first paragraph 3 multiple choice questions and based upon the second paragraph 2 multiple choice questions have to be answered. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Paragraph for Questions No. 12 to 14 When a metal rod M is dipped into an aqueous

colourless concentrated solution of compound N, the solution turns light blue. Addition of aqueous NaCl to the blue solution gives a white precipitate O. Addition of aq. NH3 dissolves O and gives an intense blue solution.

12. The metal rod M is - (A) Fe (B) Cu

(C) Ni (D) Co

Ans. [B] Sol. Metal rod M is Cu

13. The compounds N is - (A) AgNO3 (B) Zn(NO3)2

(C) Al(NO3)3 (D) Pb(NO3)2

Ans. [A] Sol. Cu + AgNO3(conc.) —→ Cu(NO3)2 + Ag light blue

14. The final solution contains - (A) [Pb(NH3)4]2+ and [CoCl4]2–

(B) [Al(NH3)4]3+ and [Cu(NH3)4]2+

(C) [Ag(NH3)2]+ and [Cu(NH3)4]2+ (D) [Ag(NH3)2]+ and [Ni(NH3)6]2+

Ans. [C] Sol. AgCl + NH2(aq) → [Ag(NH3)2]+ Cu+2 + NH3(aq) → [Cu(NH3)4]+2 Intense blue

Paragraph for Question No. 15 to 16

An acyclic hydrocarbon P, having molecular formula C6H10, gave acetone as the only organic product through the following sequence of reactions, in which Q is an intermediate organic compound.

P

(C6H10)

(i) dil. H2SO4/HgSO4

(ii) NaBH4/ethanol (iii) dil. acid

Q

(i) conc. H2SO4 (catalytic amount) (–H2O) (ii) O3 (iii) Zn/H2O

2 CH3C CH3

O

15. The structure of compound P is - (A) CH3CH2CH2CH2–C≡C–H

(B) H3CH2C–C≡C–CH2CH3

(C) H–C–C≡C–CH3

H3C

H3C

(D) H3C–C–C ≡C–H

H3C

H3C

Ans. [D]


Sol.

CH3–C–C≡C–H (i) dil. H2SO4/HgSO4

(ii) NaBH4 (iii) dil. HCl

CH3

CH3

(P)

CH3–C–CH–CH3

CH3

H3C OH

(Q)

–H2O

(i) conc. H2SO4 CH3–C=C

CH3

CH3

(ii) O3 (iii) Zn/H2O

C=O CH3

CH3

H3C

2

16. The structure of compound Q is -

(A) H–C–C–CH2CH3

H3C

H3C

OH

H

(B) H3C–C–C–CH3

H3C

H3C

OH

H

(C) H–C–CH2CHCH3

H3C

H3C

OH

(D) CH3CH2CH2CHCH2CH3

OH

Ans. [B] Sol. Factual.

SECTION – IV

Numerical Response Type This section contains. 7 questions. The answer to each questions is a single digit integer ranging from 0 to 9. The bubble corresponding to the correct answer is to be darkened in the ORS.

17. Reaction of Br2 with Na2CO3 in aq. solution gives sodium bromide and sodium bromate with evolution of CO2 gas. The number of sodium bromide molecules involved in the balanced chemical equation is.

Ans. [5] Sol. 3 Na2CO3 + 3Br2 → 5NaBr + NaBrO3 + 3CO2

18. The difference in the oxidation numbers of the

two types of sulphur atoms in Na2S4O6 is. Ans. [5]

Sol.

NaO–S–S–S–S–ONa

O#

O

⊕ Θ Θ ⊕ ∗ ∗ O

O

#

O.N. S* = 0 O.N. S# = + 5 ∴ Difference = 5

19. The maximum number of electrons that can have principal quantum number, n = 3 and spin quantum number, ms = – 1/2, is.

Ans. [9] Sol. For n = 3, max e– = 2n2 = 18

Half of them can have ms = – 1/2 20. A decapeptide (mol. wt. 796) on complete

hydrolysis gives glycine (mol. wt. 75), alanine and phenylalanine. Glycine contributes 47.0% to the total weight of the hydrolysed products. The number of glycine units present in the decapeptide is.

Ans. [6]

Sol. Decapeptide water

molecule9 → (x) glycine + (y)

alanine + (z) phenylalanine Mass of hydrolysed product = 796 + 18 × 9

mass of glycine = 958 × 10047

= 450.26

No. of glycine unit = 675

26.450=

21. To an evacuated vessel with movable piston under external pressure of 1 atm., 0.1 mol of He and 1.0 mol of an unknown compound (vapour pressure 0.68 atm, at 0ºC) are introduced. Considering the ideal gas behaviour, the total volume (in litre) of the gases at 0ºC is close to.

Ans. [7] Sol.

He + compounds

Pext = 1 atm

Vapour pressure of compound = 0.68 ∴ PHe

= 1 – 0.68 = 0.32 ∴ By PV = nRT, for He

V = 32.0

2730821.01.0P

RTn

He

He ××=

V ~ 7L


22. The total number of alkenes possible by dehydrobromination of 3-bromo-3 cyclopentylhexane using alcoholic KOH is.

Ans. [5]

Sol. C–C–C–C–C–C

1 2 3 4 5 6

Br

alc.KOH

C–C–C–C= C– C

+ G.I

C–C–C=C– C– C

+ G.I

+

C–C–C–C– C– C 23. The work function (φ) of some metals is listed

below. The number of metals which will show photoelectric effect when light of 300 nm wavelength falls on the metal is.

Metal Li Na K Mg Cu Ag Fe Pt W φ (eV) 2.4 2.3 2.2 3.7 4.8 4.3 4.7 6.3 4.75

Ans. [4]

Sol. Efalling = 199

834

106.1103001031062.6hc

−−

−

××××××

=λ

= 4.137 eV

The metals having less work function will show photoelectric effect

Hence Li, Na, K, Mg

PHYSICS


This section contains 7 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. 24. 5.6 liter of helium gas at STP is adiabatically

compressed to 0.7 liter. Taking the initial temperature to be T1, the work done in the process is-

(A) 1RT89

(B) 1RT23

(C) 1RT8

15 (D) 1RT

29

Ans. [A] Sol. T1V1

γ–1 = T2V2γ–1

T2 = T1

1

2

1

VV

−γ

= T1

135

7.07.5 −

= T1(8)2/3 = 4T1

no. of mole = 41

W = 1

)TT(nR 21

−γ−

= 3/2

)TT4(R41

11 −× =

89

RT1

25. A ball of mass (m) 0.5 kg is attached to the end of a string having length (L) 0.5 m. The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of angular velocity of ball (in radian/s) is-

m

L

(A) 9 (B) 18

(C) 27 (D) 36 Ans. [D] Sol.

L

Tθ

θ

mg T cos θ = mg

T sin θ = mω2L sin θ T = mω2L

ω2max =

mLTmax

ωmax = mLTmax =

5.05.0324×

= 4324×

= 36 rad/s

26. Consider an electric field xEE 0=r

, where E0 is a constant. The flux through the shaded area (as shown in the figure) due to this field is-

(a,a,a)

(0,0,0) (0,a,0)

(a,0,a)

z

x

y


(A) 2E0a2 (B) 20aE2

(C) E0a2 (D) 2aE 2

0

Ans. [C] Sol.

(a,a,a)

(0,0,0) (0,a,0)

(a,0,a)

z

x

y

H(a,a,a)

(0,a,0)

E(a,0,a)

x

y

(a,a,0)D(a,0,0)

G (0,a,a)F(0,0,a)

A B

C

z

flux through EHBA

= flux through EHDC = E0a2

27. A police car with a siren of frequency 8 kHz is moving with uniform velocity 36 km/hr towards a tall building which reflects the sound waves. The speed of sound in air is 320 m/s. The frequency of the siren heard by the car driver is-

(A) 8.50 kHz (B) 8.25 kHz (C) 7.75 kHz (D) 7.50 kHz

Ans. [A]

Sol. Vs = 3600

1036 3× m/s = 10 m/s, ν = 8 KHz

V0 = 320 m/s V0 = Vs = 10 m/s

Vs

Observer

ν′ =

′−

+

s

0VVVV

ν

=

−+

1032010320

8 KHz

= 310330

× 8 = 8.51 KHz

28. A meter bridge is set-up as shown, to determine an unknown resistance 'X' using a standard 10 ohm resistor. The galvanometer show null point when tapping-key is at 52 cm mark. The end-corrections are 1 cm and 2 cm respectively for the ends A and B. The determine value of 'X' is-

X 10Ω

A B

(A) 10.2 ohm (B) 10.6 ohm (C) 10.8 ohm (D) 11.1 ohm

Ans. [B]

Sol. 10x

= 248152

++

= 50

1053×= 10.6

29. A 2 µF capacitor is charged as shown in figure. The percentage of its stored energy dissipated after the switch S is turned to position 2 is-

1 2

S

2µF 8µF V

(A) 0 % (B) 20 %

(C) 75 % (D) 80 % Ans. [D] Sol.

2µF 8µF


∆U = 21

× 8282

+×

[V – 0]2

= 21

× 1016

× V2 = 10V8 2

Ui′ =

21

× 2 × V2

= V2

% dissipated = iU

U∆ =

108

× 100

= 80% 30. The wavelength of the first spectral line in the

Balmer series of hydrogen atom is 6561 Å. The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is-

(A) 1215 Å (B) 1640 Å (C) 2430 Å (D) 4687 Å

Ans. [A]

Sol. 6561hc

= 13.6

− 22 3

121

λhc

= 13.6 × 4

− 22 4

121

6561hc

= 13.6 × 365

λhc

= 13.6 × 4 × 163

6561hc

= 365

× 34

λ = 275

λ = 27

56561× = 243 × 5

= 1215 Å


This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or More may be correct.

31. A spherical metal shall A of radius RA and a solid metal sphere B of radius RB(<RA) are kept far apart and each is given charge '+Q'. Now they are connected by a thin metal wire. Then-

(A) insideAE = 0 (B) QA > QB

(C) B

A

σσ

= A

B

RR

(D)

surfaceonB

surfaceonA EE <

Ans. [A, B, C, D]

Sol.

RB

Q

RA

Q S

Ans.(A)

A

A

RkQ

= B

B

RkQ

[Final potential will be same]

B

A

QQ

= B

A

RR

as RA > RB

∴ QA > QB Ans. [B]

2B

B

2A

A

R4QR4

Q

π

π =

2B

B

2A

A

RRRR

= B

A

σσ

= A

B

RR

Ans [C]

as RA > RB ∴ σB > σA

∴ EB > EA Ans. [D] 32. A metal rod of length 'L' and mass 'm' is pivoted

at one end. A thin disk of mass 'M' and radius 'R' (<L) is attached at its centre to the free end of the rod. Consider two ways the disc is attached :(case A) The disc is not free to rotate about its center and (case B) the disc is free to rotate about its center. The rod-disc system performs SHM in vertical plane after being released from the same displaced position. Which of the following statement(s) is (are) true ?

(A) Restoring torque in case A = Restoring torque in case B

(B) Restoring torque in case A < Restoring torque in case B

(C) Angular frequency for case A > Angular frequency for case B

(D) Angular frequency for case A < Angular frequency for case B

Ans. [A,D]

33. An electron and a proton are moving on straight parallel paths with same velocity. They enter a semi-infinite region of uniform magnetic field perpendicular to the velocity. Which of the following statement(s) is/are true ?


(A) They will never come out of the magnetic field region

(B) They will come out traveling along parallel paths

(C) They will come out at the same time (D) They will come out at different times Ans. [B, D] Sol.

e–

p

v

v

Rp

Re

× × × × × × × × × × × ×

× ×

×

×

×

×

× ×

×

× ×

×

eBvm

R

Rvm

evB

eBVmR

RvmevB

pp

p

2p

ee

e

2e

=

=

=

=

Re Rp >

eBm

vReT e

eπ

=π

= ⇒ Te Tp >

eBm

T pp

π=

34. A composite book is made of slabs A, B, C, D and E of different thermal conductivities (given in terms of a constant K) and sizes (given in terms of length, L) as shown in the figure. All slabs are of same width. Heat 'Q' flow only from left to right through the blocks. Then in steady state-

heat 0 1L 5L 6L

A B

C

D

4K

3K

5K

E

6K

3L

4L

1L

(A) heat flow through A and E slabs are same (B) heat flow through slab E is maximum (C) temperature difference across slab E is

smallest (D) heat flow through C = heat flow through B +

heat flow through D Ans. [A, C, D]

Sol.

A B

C

D

4K

3K

5K

E

6K

2A

A

4A

A

2K

4A

L 4L L

H

16

H3

2H

16H5

H

All the three system shown are in series hence rate of heat flow will be same through both A & E.

)KA(8

LR A = ; KA3L4R B = ;

KA8L4RC = ;

KA5L4R D = ;

KA24LR E =

Using parallel combination rate of heat flow across C = rate of heat flow through B+ rate of heat flow through D.

KA8HLHRAA ==θ∆

KA4HL

KA3L4

16H3R

16H3

BB =

==θ∆

( )KA4HL

KA8L4

2HR

2H

CC =

==θ∆

KA4HL

KA5L4

16H5

D =

=θ∆

KA24

HLKA24LHE =

=θ∆


This section contains 2 paragraphs.. Based upon the first paragraph 3 multiple choice question and based upon on the other paragraph 2 multiple choice questions have to be answered. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Paragraph for Questions 35 to 37 Phase space diagrams are useful tools in analyzing all kinds of dynamical problems. They are especially useful in studying the changes in motion as initial position and momentum are changed. Here we consider some simple dynamical systems in one-dimension. For such systems, phase space is a plane in which position is plotted along horizontal axis and momentum is plotted along vertical axis. The phase space diagram is x(t) vs p(t) curve in this plane. The arrow on the curve indicates the time flow. For example, the phase space diagram for a particle moving with constant velocity is a straight line as


shown in the figure. We use the sign convention in which position or momentum upwards (or to right) is positive and downwards (or to left) is negative.

Mom

entu

m

Position 35. The phase space diagram for a ball thrown

vertically up from ground is-

(A)

Position

Momentum

(B)

Position

Momentum

(C)

Position

Momentum

(D)

Position

Momentum

Ans. [D] Sol. From conservation of mechanical energy

22 mu21mgxmv

21

=+

gxm2umvm 22222 =− p2 – p0

2 = 2m2gx p2 = p0

2 + 2m2gx

36. The phase space diagram for simple harmonic motion is a circle centered at the origin. In the figure, the two circles represent the same oscillator but for different initial conditions, and E1 and E2 are the total mechanical energies respectively. Then-

Momentum

E1E2

a

2a

Position

(A) E1 = 2 E2 (B) E1 = 2E2

(C) E1 = 4E2 (D) E1 = 16 E2 Ans. [C] Sol.

K' EMax 1

K' Emax 2

K.E1= 0

K.E2 = 0

maximal position

maximal position

4)a(k

21

)a2(k21

EE

2

2

2

1 ==

E1 = 4E2 37. Consider the spring-mass system, with the mass

submerged in water, as shown in figure. The phase space diagram for one cycle of this system is-


(A)

Position

Momentum

(B)

Position

Momentum

(C)

Position

Momentum

(D)

Position

Momentum

Ans. [B] Paragraph for Questions 38 to 39 A dense collection of equal number of electrons

and positive ions is called neutral plasma. Certain solids containing fixed positive ions surrounded by free electrons can be treated as neutral plasma. Let 'N' be the number density of free electrons, each of mass 'm'. When the electrons are subjected to an electric field, they are displaced relatively away from the heavy positive ions. If the electric field becomes zero, the electrons begin to oscillate about the positive ions with a natural angular frequency 'ωp', which is called the plasma frequency. To sustain the oscillations, a time varying electric field needs to be applied that

has an angular frequency ω, where a part of the energy is absorbed and a part of it is reflected. As ω approaches ωp, all the free electrons are set to resonance together and all the energy is reflected. This is the explanation of high reflectivity of metals.

38. Taking the electronic charge as 'e' and the permittivity as 'ε0', use dimensional analysis to determine the correct expression for ωp.

(A) 0mε

Ne (B)

Nemε0

(C) 0

2

mεNe

(D) 20

Nemε

Ans. [C]

Sol. F = mω2l ≡ 20

2

4e

lπε

ω2 ≡ 30

2

4e

lπε ≡

ε0

2

me

3

3Nl

l

ω = 0

2

mNe

ε

39. Estimate the wavelength at which plasma reflection will occur for metal having the density of electrons N ≈ 4 × 1027 m–3. Take ε0 = 10–11 and m ≈ 10–30, where these quantities are in proper SI unit-

(A) 800 nm (B) 600 nm (C) 300 nm (D) 200 nm

Ans. [B] c = λf

ωp = ω = λπc2

= 0

2

εmNe

20

Nemc2 ε

π=λ N

me

c2 0επ=

= 27

1130

19

8

104)10)(10(

106.110314.32

××××× −−

−

= 589 × 10–9 m ≈ 600 nm

SECTION – IV Numerical Response Type

This section contains 7 questions. The answer to each question is a single-digit integer, ranging from 0 to 9. The bubble corresponding to the correct answer is to be darkened in the ORS.


40. A block is moving on an inclined plane making an angle 45º with the horizontal and the coefficient of friction is µ. The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If we define N = 10 µ, then N is.

Ans. [5] Sol.

45ºN=mgcosθ

mgsinθ + µmgcosθ = 3(mgsinθ – µmgcosθ)

⇒ 2

µ2

1+ =

2µ3

23

−

⇒ 2µ4

= 2

2

µ = 21

N = 10µ =

2110 = 5

41. A boy is pushing a ring of mass 2 kg and radius 0.5 m with a stick as shown in the figure. The stick applies a force of 2 N on the ring and rolls it without slipping with an acceleration of 0.3 m/s2. The coefficient of friction between the ground and the ring is large enough that rolling always occurs and the coefficient of friction between the stick and the ring is (P/10). The value of P is.

Stick

Ground Ans. [4] Sol.

2N2N

f2 f1= µ×2 =

5P

102P

=×

mg

2 – f2 = Macm …….(1) f2 = 2 – 2 × 0.3 = 1 .4 N

(f2 – f1) R = Icm α

(f2 – f1) R = R

aMR cm2 ×

f2 – f1 = Macm f1 = f2 – macm = 1.4 – 2 × 0.3 = 0.8 N

5P8.0 = ⇒ P = 4

Note : It has been assumed that the stick applies horizontal force of 2N (only normal reaction)

42. Four point charge, each of +q, are rigidly fixed at the four corners of a square planar soap film of side 'a'. The surface tension of the soap film is γ. The system of charges and planar film are in equilibrium, and

a = k1/N2

γq

, where 'k' is a constant. Then N is.

Ans. [3] Sol.

FABFAC

FAD A D

B C

+q

+q +q

+q

FAC = 20

2

a8q

πε

FAD = FAB = 20

2

a4q

πε

FR =

+

πε 21º45cos2

a4q

20

2

= r (2) (BD) = 2 r( 2 a)

⇒ a3 = r28212q

0

2

πε

+

a = 3/12

rqk

⇒ N = 3

where k =

3/1

28212

π

+


43. Four solid spheres each of diameter 5 cm and mass 0.5 kg are placed with their centers at the corners of a square of side 4 cm. The momentum of inertia of the system about the diagonal of the square is N × 10–4kg-m2, then N is.

Ans. [9] Sol.

m

m m

m

l

r = 25

cm = 25

× 10–2 m

m = 21

kg

l = 4 × 10–2 m Using parallel axis theorem

Itotal =

×××× −410

45

21

524 +

××× −4108

212

⇒ 10–4 + 8 × 10–4 ⇒ 9 × 10–4 kg m2

44. The activity of a freshly prepared radioactive sample is 1010 disintegrations per second, whose mean life is 109 s. The mass of an atom of this radioisotope is 10–25 kg. The mass (in mg) of the radioactive sample is.

Ans. [1] Sol. A = λN

1010 = 9101

× N

N = 1019 mass of sample = 1019 × 10–25 × 1 × 106 = 1 mg 45. A long circular tube of length 10 m and radius 0.3

carries a current I along its curved surface as shown. A wire-loop of resistance 0.005 ohm and of radius 0.1 m is placed inside the tube with its axis coinciding with the axis of the tube. The current varies as I = I0cos(300 t) where I0 is constant. If the magnetic moment of the loop is N µ0 I0 sin(300 t), then 'N' is.

I

Ans. [6]

Sol. B = 10

t300cosIµ 00

φ = 10

Iµ 00 × 3.14 × 0.01 cos 300 t

φ = 3.14 × µ0I0 cos 300 t × 10–3

e = dtdφ

− = 3.14 × 300 µ0I0 sin 300t × 10–3

i = Re

= 005.0

10t300sinIµ30014.3 300

−××

i = 3.14 × 60 µ0I0 sin 300 t

Magnetic moment = 3.14 × 60 × 3.14 × 100

01.0 ×

µ0I0 sin 300 t = 5.9 µ0I0sin 300 t = 6 µ0 I0 sin 300 t 46. Steel wire of length 'L' at 40ºC is suspended from

the ceiling and then a mass 'm' is hung from its free end. The wire is cooled down from 40ºC to 30ºC to region its original length 'L'. The coefficient of linear thermal expansion of the steel is 10–5/ºC, Young's modulus of steel is 1011 N/m2 and radius of the wire is 1 mm. Assume that L >> diameter of the wire. Then the value of 'm' in kg is nearly.

Ans. [3] Sol.

m

∆L = AYmgL

= Lα(∆θ)

⇒ m = g

)(AY θ∆α =

1010101010 5116 ××××π −−

m = 3.14 kg ⇒ 3 kg


MATHEMATICS


This section contains 7 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

47. The value of ∫ +

3

222

2

)–6sin(sinsinn

nxnx

xxl

ll

dx is

(A) 41

ln 23

(B) 21

ln 23

(C) ln 23

(D) 61

ln 23

Ans. [A]

Sol. Let x2 = t xdx = 2dt

I = ∫ −+

3

2)6sin(sin

sin21 n

n

dttnt

tl

ll

... (1)

I = ∫ −+−

3

2)6sin(sin

)6(sin21 n

n

dttnt

tnl

ll

l … (2)

Add (1) & (2)

2I = ∫3

221 n

n

dtl

l

I = 41

(ln3 – ln2) = 23

41 nl

48. Let the straight line x = b divide the area enclosed by y = (1 – x)2, y = 0, and x = 0 into two parts R1(0 ≤ x

≤ b) and R2(b ≤ x ≤ 1) such that R1 – R2 = 41

.

Then b equals

(A) 43

(B) 21

(C) 31

(D) 41

Ans. [B]

Sol.

0 b 1

R2 R1

R1 – R2 = 41

∫ −b

dxx0

2)1( – ∫ −1

2)1(b

dxx = 41

b

x

0

3

3)1(

− –

13

3)1(

b

x

− =

41

3

)1( 3−b +

31

– 0 + 3

)1( 3−b =

41

3

)1(2 3−b =

41

– 31

= 12

1−

(b – 1)3 = – 81

b – 1 = 21

− ⇒ b = 21

49. Let →a = i + j + k ,

→b = i – j + k and

→c =

i – j – k be three vectors. A vector →v in the

plane of →a and

→b , whose projection on

→c is

31 , is given by

(A) i – 3 j + 3 k (B) – 3 i – 3 j – k

(C) 3 i – j + 3 k (D) i + 3 j – 3 k Ans. [C]

Sol. Let →ν = x i + y j + z k

Q [→a

→b

→ν ] = 0

zyx111111

− = 0

On solving x = z ….(1)

Q projection of →ν on

→c is

31

So, 3

1=

||

.→

→→ν

c

c ⇒

3zyx −−

= 3

1

⇒ x – y – z = 1 ….(2) So solving (1) & (2) y = –1 & x = z

50. Let (x0, y0) be the solution of the following equations

(2x)ln 2 = (3y)ln 3 3ln x = 2ln y Then x0 is


(A) 61

(B) 31

(C) 21

(D) 6

Ans. [C] Sol. (2x)ln 2 = ( 3y)ln 3 ln 2 (ln 2 + ln x) = ln 3 (ln 3 + ln y) ln 2 . ln x – ln 3 ln y = (ln 3)2 – (ln 2)2 .....(1) 3ln x = 2ln y ln x . ln 3 = lny . ln 2

ln y = ln x 23

nn

l

l .....(2)

Solving (1) & (2)

ln x = – ln 2 ⇒ x = 21

51. Let α and β be the roots of x2 – 6x – 2 = 0, with α > β. If an = αn – βn for n ≥ 1, then the value of

9

810

22–

aaa

is

(A) 1 (B) 2 (C) 3 (D) 4

Ans. [C] Sol. ∴ x2 – 6x – 2 = 0 has roots α, β So, α2 – 2 = 6α & β2 – 2 = 6β an = αn – βn

So, 9

810

22

aaa −

= )(2

)(2)(99

881010

β−α

β−α−β−α

= )(2

)2()2(99

2828

β−α−ββ−−αα

= )(2

)6()6(99

88

β−αββ−αα

= 3.

52. A straight line L through the point (3, –2) is inclined at an angle 60° to the line 3 x + y = 1. If L also intersects the x-axis, then the equation of L is

(A) y + 3 x + 2 – 3 3 = 0

(B) y – 3 x + 2 + 3 3 = 0

(C) 3 y – x + 3 + 2 3 = 0

(D) 3 y + x – 3 + 2 3 = 0 Ans. [B] Sol. Let the slope of the line is m

tan 60º = m

m31

3−

+

3 = m

m31

3−

+

So, m + 3 = ± 3 (1 – 3 m)

m + 3 = 3 – 3m m + 3 = – 3 + 3m

m = 0 m = 3 hence line hence line

y = – 2 y + 2 = 3 (x – 3)

y – 3 x + 2 + 33 = 0 as line intersects x-axis

So line will be, y – 3 x + 2 + 33 = 0

53. Let P = θ : sin θ – cos θ = 2 cos θ and Q = θ : sin θ + cos θ = 2 sin θ be two sets. Then

(A) P ⊂ Q and Q – P ≠ ∅ (B) Q ⊄ P

(C) P ⊄ Q (D) P = Q

Ans. [D]

Sol. P : sin θ – cos θ = 2 cos θ

sin θ = ( 2 + 1) cos θ

tan θ = 2 + 1

tan θ = tan 6721

°

θ = nπ + 8

3π, n ∈ I …(1)

Q : sin θ + cos θ = 2 sin θ

cos θ = ( 2 – 1) sin θ

tanθ = 12

1−

= 2 + 1

θ = nπ + 8

3π, n ∈ I …(2)

∴ P = Q


This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONE OR MORE may be correct.

54. The vector(s) which is/are coplanar with vectors i + j + 2 k and i + 2 j + k , and perpendicular

to the vector i + j + k is /are

(A) j – k (B) – i + j

(C) i – j (D) – j + k


Ans. [A,D]

Sol. kzjyixr ˆˆˆ ++= is coplanar with the given vector so

∴ 121211zyx

= 0

So, 3x = y + z ...(1)

∴ kjir ˆˆˆ ++⊥→

So, 0)ˆˆˆ(. =++→

kjir So, x + y + z = 0 ...(2) On solving (1) & (2) So, x = 0 ∴ y + z = 0 ∴ (A) & (D) Satisfy 55. Let f : R → R be a function such that f(x + y) = f(x) + f(y), ∀x, y ∈ R If f(x) is differentiable at x = 0, then (A) f(x) is differentiable only in a finite interval

containing zero (B) f(x) is continuous ∀ x ∈ R (C) f '(x) is constant ∀ x ∈ R (D) f(x) is differentiable except at finitely many

points Ans. [B,C] Sol. f(x + y) = f(x) + f(y) By Partial differentiation with respect to x f ' (x + y) = f ' (x) f ' (y) = f '(0) f(y) = (f '(0))y + c f(y) = ky +c ∴ f(y) = ky as f(0) = 0 ∴ f(x) = kx Alternate

f '(x) = 0

lim→h

h

xfhxf )()( −+

= 0

lim→h

h

xfhfxf )()()( −+ =

hhf

h

)(lim0→

= λ (let) f(x) = λx + c As f(0) = 0 ⇒ c = 0 f(x) = λx

56. Let M and N be two 3 × 3 non-singular skew-symmetric matrices such that MN = NM. If PT denotes the transpose of P, then M2N2 (MTN)–1(MN–1)T is equal to

(A) M2 (B) –N2 (C) –M2 (D) MN

Ans. [C]

Sol. Q MN = NM M2 N2 = MN MN Q (MT)–1 = (–M)–1 = –M–1 Given, M2N2(MTN)–1. (MN–1)T = – MN MN N–1 M–1 N–1

I

M

= –M NN–1 M = – M2 The most suitable answer is (C), although

given information is contradictory as Skew symmetric matrix of odd order cannot be non singular

57. Let the eccentricity of the hyperbola 2

2

ax

– 2

2

by

= 1

be reciprocal to that of the ellipse x2 + 4y2 = 4. If the hyperbola passes through a focus of the ellipse, then

(A) the equation of the hyperbola is 3

2x–

2

2y= 1

(B) a focus of the hyperbola is (2, 0)

(C) the eccentricity of the hyperbola is 35

(D) the equation of the hyperbola is x2 – 3y2 = 3 Ans. [B,D] Sol. Let e1 = eccentricity of hyperbola e2 = eccentricity of ellipse

∴ e1 = 2

1e

so eccentricity of ellipse = 23

= e2

eccentricity of ellipse = 3

2= e1

Now focus of ellipse is (± ae2, 0) ≡ (± 3 , 0) Hyperbola passes through it

So, 2

2)3(a

– 0 = 1 ⇒ a2 = 3

also b2 = a2 (e12 – 1)

b2 = 3

1–

34

= 1

and hyperbola

113

22=−

yx

also focus (± ae1, 0) ≡ (± 2, 0)



This section contains 2 paragraphs. Based upon one of the paragraphs 3 multiple choice questions and based on the other paragraph 2 multiple choice questions have to be answered. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Paragraph for Question No. 58 to 60 Let a, b and c be three real numbers satisfying

[a b c]

737728791

= [0 0 0] …….(E)

58. If the point P(a, b, c), with reference to (E), lies on the plane 2x + y + z = 1, then the value of 7a + b + c is

(A) 0 (B) 12 (C) 7 (D) 6 Ans. [D]

Sol. [a b c]

737728791

= [0 0 0]

a + 8b + 7c = 0 9a + 2b + 3c = 0 7a + 7b + 7c = 0 On solving above equation

(a, b, c) ≡

λ

λ−

λ− ,

76,

7

∴ (a, b, c) lies on the plane 2x + y + z = 1

So 7

2λ− –

76λ

+ λ = 1

on solving λ = – 7 So 7a + b + c = 6

59. Let ω be a solution of x3 – 1 = 0 with Im(ω) > 0. If a = 2 with b and c satisfying (E), then the value

of aω3

+ bω1

+ cω3

is equal to

(A) –2 (B) 2 (C) 3 (D) –3

Ans. [A]

Sol. ∴ (a, b, c) ≡

λ

λ−

λ− ,

76,

7

∴ a = 2 is given so λ = – 14 So (a, b, c) ≡ (2, 12, – 14)

So cba ω

+ω

+ω

313 = –2

60. Let b = 6, with a and c satisfying (E). If α and β are the roots of the quadratic equation

ax2 + bx + c = 0, then n

n∑

∞

=

+

0

11βα

is

(A) 6 (B) 7 (C) 76

(D) ∞

Ans. [B]

Sol. ∴ (a, b, c) ≡

λ

λ−

λ− ,

76,

7

∴ b = 6 so λ = – 7. So (a, b, c) ≡ (1, 6, –7) So the equation ax2 + bx + c = 0 x2 + 6x – 7 = 0 So α = 1, β = – 7

S = ∑∞

=

β

+α0

11

n

n

= n

∑

−

71

11

= n

∑

76

= 1 + 76

+ 2

76

+ .... ∞

=

761

1

− = 7

Paragraph for Question Nos. 61 and 62 Let U1 and U2 be two urns such that U1 contains 3

white and 2 red balls, and U2 contains only 1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from U1 and put into U2. However, if tail appears then 2 balls are drawn at random from U1 and put into U2. Now 1 ball is drawn at random from U2

61. The probability of the drawn ball from U2 being white is

(A) 3013

(B) 3023

(C) 3019

(D) 3011

Ans. [B]

Sol. RW

23 W1

U1 U2 Required probability = P(H)[P(W/H) × P(W2) +

P(R/H)P(W2)] + P(T) [P

TWboth

P(W2) +

P

TRboth P(W2) +

TWRP 11 &

P(W2)]


=

×

×+×+×+

×+×

32

311

21

21

521

53

21

25

12

13

25

22

25

23

CCC

CC

CC

=

+++

+

52

301

103

21

51

53

21

= 3023

3011

52

=+

62. Given that the drawn ball from U2 is white, the

probability that head appeared on the coin is

(A) 2317

(B) 2311

(C) 2315

(D) 2312

Ans. [D] Sol. Required probability =

+

+

+

+

+

)W(PT

W&RP)W(PT

RbothP

)W(PT

WbothP)T(P)W(PHRP)W(P

HWP)H(P

)W(PHRP)W(P

HWP)H(P

211

2

221

21

21

21

= 1312

3023

21

521

53

21

=

×+×

SECTION – IV

Numerical Response Type This section contains 7 questions. The answer to each of the questions is a single-digit integer, ranging from 0 to 9. The bubble corresponding to the correct answer is to be darkened in the ORS.

63. Let a1, a2, a3, .., a100 be an arithmetic progression

with a1 = 3 and Sp = ∑=

p

iia

1

, 1 ≤ p ≤ 100. For any

integer n with 1 ≤ n ≤ 20, let m = 5n. If n

m

SS

does

not depend on n, then a2 is Ans. [9] Sol. a1 = 3

n

m

SS

= n

n

SS5 =

])1(2[2

])15(2[2

5

1

1

dnan

dnan

−+

−+

= nddndd

+−+−

)6(]5)6[(5

Q n

n5

SS

is independent of n so d = 6

So a2 = a1 + d = 3 + 6 = 9

64. Consider the parabola y2 = 8x. Let ∆1 be the area of the triangle formed by the end points of its

latus rectum and the point P

2,

21

on the

parabola, and ∆2 be the area of the triangle formed by drawing tangents at P and at the end points of

the latus rectum. Then 2

1

∆∆

is

Ans. [2] Sol. It is a property that area of triangle formed by

joining three points lying on parabola is twice the area of triangle formed by tangents at these points

Alternate : y2 = 8x

P

2,

21

• •

• A P

B(2, –4)

3/2

(2, 4)

∆1 = 21

|Base × Height|

= 21

×23

× 8 = 6

Also

Equation of tangent at P

2,

21

• • P

•

•

y (2) = 4.

+

21x

y = 2x + 1 ...(1) Tangent at A : y = x + 2 Tangent at B : – y = + x + 2 ⇒ y = – x – 2 Point of intersection L(–2, 0), M (1, 3), N (–1, –1)

∆2 = 111131102

21

−−

−

= |21

[–2(4) + (–1 + 3)]|


= ]138[21

−+− = 3

So, 2

1

∆∆

= 36

= 2

65. The positive integer value of n > 3 satisfying the

equation

π

nsin

1 =

π

n2sin

1 +

π

n3sin

1 is

Ans. [7]

Sol. Let nπ

= θ

θsin

1 =

θ2sin1

+ θ3sin

1

θsin

1 –

θ3sin1

= θ2sin

1

[sin 3θ – sin θ] sin 2θ = sin θ sin 3θ 2 sin θ cos 2θ sin 2θ = sin θ sin 3θ Q sin θ ≠ 0 2 cos 2θ sin 2θ = sin 3θ sin 4θ = sin 3θ so either 4θ = 3θ or 4θ = π – 3θ

so θ = 0 or θ = 7π

so n = 7

66. Let f(θ) = sin

θθ2cos

sintan 1– , where

–4π

< θ < 4π

. Then the value of )(tan θd

d (f (θ))

is Ans. [1]

Sol. Q tan–1

θθ2cos

sin = sin–1 tan θ

so f (θ) = sin (sin–1 tan θ) = tan θ

Q )(tan

))((θ

θd

fd =

)(tan)(tan

θθ

dd

= 1

67. If z is any complex number satisfying | z – 3 – 2i| ≤ 2, then the minimum value of | 2z – 6 + 5i| is

Ans. [5] Sol.

A

P

25,–3

Min value

(3, 2) B

So, Min of |2z – 6 + 5i| = PA

= Min 2253– iz + = 2 ×

25

= 5

68. The minimum value of the sum of real numbers a–5, a–4, 3a–3, 1, a8 and a10 with a > 0 is

Ans. [8] Sol. A.M. ≥ G.M.

81 1083–3–3–4–5– aaaaaaa +++++++

≥ (a–5. a–4. a–3. a–3. a–3. 1. a8 .a10)1/8 a–5 + a–4 + a–3 + a–3 + a–3 + 1 + a8 + a10 ≥ 8 so minimum value is 8

69. Let f : [1, ∞) → [2, ∞) be a differentiable function

such that f(1) = 2. If 6 ∫x

dttf1

)( = 3x f(x) – x3 for

all x ≥ 1, then the value of f (2) is Ans. [6]

Sol. 6 ∫x

dttf1

)( = 3xf(x) – x3

6 f(x) = 3f(x) + 3xf '(x) –3x2 3f(x) = 3x f '(x) – 3x2

3y = 3xdxdy

– 3x2

xdxdy

– y = x2

dxdy

– xy

= x

I.F. = ∫ dx

xe1–

= e–ln x = x1

y .x1

= ∫ dxx

x 1. ⇒ xy

= x + c ⇒ y = x2 + cx

Q f(1) = 2 ⇒ c = 1 y = x2 + x f(2) = 4 + 2 = 6


CHEMISTRY


This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

1. Oxidation states of the metal in the minerals haematite and magnetite respectively, are

(A) II, III in haematile and III in magnetite (B) II, III in haematite and II in magnetite (C) II in haematite and II, III in magnetite (D) III in haematite and II, III in magnetite Ans. [D]

Sol. Haematite is 323

OeF+

Magnetite is Fe3O4 or 3232

OeF.eOF++

2. The following carbohydrate is

H OH H

O OH

OH H H

HO HO

H

(A) a ketohexose (B) an aldohexose (C) an α-furanose (D) an α-pyranose Ans. [B] Sol. Aldohexose

3. The major product of the following reaction is RCH2OH

H+ (anhydrous) O

(A) a hemiacetal (B) an acetal (C) an ether (D) an ester

Ans. [B]

Sol.

O H (anhydrous)

R- CH2 –OH ⊕

OCH2–RO Acetal

4. Amongst the compounds given, the one that would form a brilliant colored dye on treatment with NaNO2 in dill. HCl followed by addition to an alkaline solution of β -naphthol is

(A) N(CH3)2 (B)

NHCH3

(C) NH2

H3C

(D) CH2NH2

Ans. [C] Sol. NaNO2 + HCl (dil.) CH3 NH2

0–5°C CH3 N2 Cl

Alkaline solution OH

CH3 N = N

HO

Coloured dye

5. The freezing point (in °C) of a solution containing 0.1 g of K3[Fe(CN)6] (mol. wt. 329) in 100 g of water (kf = 1.86 K kg mol–1) is

(A) –2.3 × 10–2 (B) –5.7 × 10–2

(C) –5.7 × 10–3 (D) –1.2 × 10–2

IIT-JEE 2011 PAPER-II (PAPER & SOLUTION)

Time : 3 Hours Total Marks : 240

Instructions : [Each subject contain]

Section – I : Multiple choice questions with only one correct answer. +3 marks will be awarded for correct answer and -1 mark for wrong answer. [No. of Ques. : 8]

Section – II : Multiple choice questions with multiple correct answer. +4 marks will be awarded for correct answer and No Negative marking for wrong answer. [No. of Ques. : 4]

Section – III : Numerical Response Question (single digit Ans. type) +4 marks will be awarded for correct answer and No Negative marking for wrong answer. [No. of Ques. : 6]

Section – IV : Column Matching type question +2 marks for each correctly matched row and No Negative marking for wrong answer. [No. of Ques. : 2]


Ans. [A] Sol. ∆T = kf × m × i × 1000

= 100041003291.086.1 ××

××

= 2.26 × 10–2 ≈ 2.3 × 10–2

6. Consider the following cell reaction :

)(OH2Fe2

H4OFe2

2)aq(2

)aq()g(2)s(

l+→

+++

+

E0 = 1.67 V

At [Fe2+] = 10–3 M, P(O2) = P(O2) = 0.1 atm and pH = 3, the cell potential at 25 °C is

(A) 1.47 V (B) 1.77 V (C) 1.87 V (D) 1.57 V Ans. [D]

Sol. Ecell = E°cell 4O

2

]H][P[]Fe[log

40591.0

2

+

+−

= 43

23

]10][1.0[]10[log

4591.067.1

−

−

−

= 1.57 V 7. Passing H2S gas into a mixture of Mn2+, Ni2+,

Cu2+ and Hg2+ ions in an acidified aqueous solution precipitates.

(A) CuS and HgS (B) MnS and CuS (C) MnS and NiS (D) NiS and HgS Ans. [A] Sol. Cu+ 2, Hg+2 are group II basic radicals 8. Among the following complexes (K–P) K3[Fe(CN)6] (K), [CO(NH3)6]Cl3 (L), Na3

[Co(oxalate)3] (M), [Ni(H2O)6]Cl2 (N), K2[Pt(CN)4] (O) and [Zn(H2O)6] (NO3)2 (P) the

diamagnetic complexes are (A) K, L, M, N (B) K, M, O, P (C) L, M, O, P (D) L, M, N, O Ans. [C] (L) : [Co(NH3)6]Cl3 (M) : Na3[Co(Ox)3] (O) : K2[Pt(CN)4] (P) : [Zn(H2O)6] (NO3)2

SECTION – II Multiple correct Choice Type

This section contains 4 multiple choice questions. Each questions has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct.

9. Reduction of the metal centre in aqueous permanganate ion involves

(A) 3 electrons in neutral medium (B) 5 electrons in neutral medium (C) 3 electrons in alkaline medium (D) 5 electrons in acidic medium Ans. [A, C, D] Sol. → In alkaline solution, KMnO4 is first reduced to

mangnate and then to insoluble MnO2

→

]O[3KOH2MnO2

OHKMnO24

4

Neutral2

7

4

++

→++

+

→ I OH4Mne5H8MnO 22acidic

4 + →++ +−+−

10. For the first order reaction : 2N2O5(g) → 4NO2(g) + O2(g)

(A) the concentration of the reactant decreases exponentially with time.

(B) the half-life of the reaction decreases with increasing temperature

(C) the half-life of the reaction depends on the initial concentration of the reactant

(D) the reaction proceeds to 99.6 % completion in eight half-life duration.

Ans. [A, B, D]

Sol. ]A[eCC ktAA 0

−=

]B[eA693.0

eA693.0

K693.0t RT/E

0RT/E

021

a

a===

−

]D[8

21log

1004log

n100

421

1004.0 4

=

===

=

11. The equilibrium: 2Cu' Cu° + Cu'' in aqueous medium at 25 °C shifts towards the

left in the presence of

(A) −3NO (B) −Cl

(C) SCN– (D) CN– Ans. [B, C, D] Sol. Cu2 Cl2, Cu2(CN)2 and Cu2 (SCN)2 are stable

12. The correct functional group X and the reagent / reaction conditions Y in the following scheme are

X– (CH2)4 – XO

(i) Y

(ii) C–(CH2)4–C

HO

O

OHHeat


(A) X = COOCH3, Y = H2/Ni/Heat (B) X = CONH2, Y = H2 / Ni / heat (C) X = CONH2, Y = Br2/ NaOH (D) X = CN, Y = H2 / Ni / heat

Ans. [A, B, C, D] Sol. Factual

SECTION – III

Integer Answer Type

This section contains 6 Question. The answer to each of the question is a single-digit integer, ranging from 0 to 9. The total bubble corresponding answer it to be darkened in the ORS.

13. In 1 L saturated solution of AgCl [Ksp (AgCl) = 1.6 × 10–10], 0.1 mol of CuCl [Ksp (CuCl) = 1.0 × 10–6] is added. The resultant concentration of Ag+ in the solution is 1.6 × 10–x. The value of "x" is

Ans. [7]

Sol. 21

1

KKK

]Ag[+

=+ Q K1 < < K2

∴ 221 KKK ≅+

∴ 6

10

100.1

106.1]Ag[−

−+

×

×= = 1.6 × 10–7

x = 7

14. The maximum number of isomers (including stereoisomers) that are possible on monochlorination of the following compound, is

C

CH3

CH3CH2 CH2CH3 H

Ans. [8] Sol.

CH3

CH3–CH2 –CH –CH2–CH3 Cl2 / hν

CH3

Cl–CH2 –CH2 –CH–CH2– CH3

(2)

*

or

CH3

CH3 –CH –CH–CH2– CH3 (4) Cl

**

or

CH3

CH3 –CH2 –C–CH2– CH3 (1)

Cl

or

CH2Cl

CH3 –CH2 –CH–CH2– CH3 (1)

15. Among the following the number of compounds than can react with PCl5 to give POCl3 is

O2, CO2, SO2, H2O, H2SO4, P4O10 Ans. [4] Sol. PCl5 + H2O → POCl3 + 2HCl PCl5 + H2SO4 → POCl3 + H2O + SO2Cl2 6PCl5 + P4O10 → 10POCl3 PCl5 + SO2 → POCl3 + SO2Cl2

16. The number of hexagonal faces that are present in a truncated octahedron is

Ans. [8] Sol.

8 Hexagonal faces

17. The volume (in mL) of 0.1 M AgNO3 required for complete precipitation of chloride ions present in 30 mL of 0.01 M solution of [Cr(H2O)5Cl]Cl2, as silver chloride is close to

Ans. [6] Sol. 0.1 V = 30 × 0.01 × 2

1.0

23.0V ×= = 6 ml

18. The total number of contributing structures showing hyper conjugation (involving C-H bonds) for the following carbocation is

H3C CH2CH3 ⊕

Ans. [6] Sol. 6 (α – H → 6)


SECTION – IV

Matrix match Type This section contain 2 questions. Each question has four statements (A, B, C and D) given in column I and five statements (p, q, r, s and t) in column II. Any given statement in column I can have correct matching with ONE or MORE statement (s) given in Column II. For example, if for a given question, statement B matches with the statements given in q and r, then for the particular question, against statement B, darken the bubbles corresponding to q and r in the ORS

19. Match the transformation in column I with

appropriate options in column II Column-I (A) CO2(s) → CO2 (g) (B) CaCO3 (s) → CaO(s) + CO2(g) (C) )g(HH2 2→• (D) P(white solid) → P(red, solid)

Column-II (p) phase transition (q) allotropic change (r) ∆H is positive (s) ∆S is positive (t) ∆S is negative

Ans. [A → p, r, s; B → r, s; C → t ; D → p, q, t] Sol. [A] CO2(s) → CO2(g) p, r, s [B] CaCO3 (s) → CaO (s) + CO2(g) r, s [C] )g(HH2 2→• t [D] Pwhite → Pred p, q, t 20. Match the reactions in column I with appropriate

type of steps/reactive intermediate involved in these reactions as given in column II Column-I (A)

O

aq NaOH

O OH3C

(B)

O CH3MgI CH2CH2CH2Cl CH

O

(C) 18

OH2SO4 CH2CH2CH2OH

18 O

(D)

H2SO4 CH2CH2CH2C(CH3)2 OH

H3C CH3 Column-II (p) Nucleophilic substitution (q) Electrophilic substitution (r) Dehydration (s) Nucleophilic addition (t) Carbanion

Ans. [A → r, t, s; B → p, s, t; C → r, s ; D → r, q] Sol. Factual

PHYSICS



21. A light ray travelling in glass medium is incident on glass-air interface at an angle of incidence θ. The reflected (R) and transmitted (T) intensities, both as function of θ, are plotted. The correct sketch is :

(A)

100%

T

R

0

Inte

nsity

θ 90º

(B)

100%

T

R

0

Inte

nsity

θ 90º


(C)

100%

T

R

0 In

tens

ityθ 90º

(C)

100%

T

R

0

Inte

nsity

θ 90º Ans. [C]

e θ θ

When θ > θC, no ray will transmit ⇒ T = 0, T + R = 100 % and R > 0

T

θc

22. A wooden block performs SHM on a frictionless surface with frequency, v0. The block carries a charge + Q on its surface. If now a uniform electric field E

r is switched-on as shown, then the

SHM of the block will be

Er

+Q

(A) of the same frequency and with shifted mean

position. (B) of the same frequency and with the same

mean position (C) of changed frequency and with shifted mean

position (D) of changed frequency and with the same

mean position Ans. [A]

In order to have net force zero, the mean position will be shifted towards right but the time period will remain unaffected.

23. The density of solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale. The reading on the main scale is 2.5 mm and that on the circular scale is 20 division. If the measured mass of the ball has a relative error of 2%, the relative percentage error in the density is -

(A) 0.9% (B) 2.4% (C) 3.1% (D) 4.2% Ans. [C] Pitch = 0.5 mm

divisions on the = 50

circular scale

⇒ least count of screw gauge = 50

5.0 = 0.01

main scale, reading = 2.5 mm

circular scale reading = 20

⇒ reading = 2.5 mm + (20 × 0.01) mm

= 2.5 mm + 0.2 mm = 2.7 mm

ρ = 3

2D

34

m

π

ρρ∆ =

mm∆ + 3

DD∆

%error = ρρ∆ × 100

= 2 % + 3

7.201.0 × 100 = 3.1.

24. A ball of mass 0.2 kg rests on a vertical post of

height 5m. A bullet of mass 0.01 kg. travelling with a velocity V m/s in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 m and the bullet at a distance of 100 m from the foot of the post. The initial velocity V of the bullet is

V m/s

H = 5m

20 100 0


Ans.[D]

v

H = 5m

Ball Bullet

20 m

100 m

m1 = 0.01 kg m2 = 0.2 kg

T = gH2

= 1 sec

Let v1 & v2 be velocity of bullet & ball respectively just after collision.

v2 × 1 = 20 ⇒ v2 = 20 & v1 = 100 From conservation of momentum 0.01 × v = (0.01 × 100) + (0.2 × 20) 0.01 v = 1 + 4 = 5

v = 2105

− = 500 m/sec.

25. Which of the field patterns given below is valid for electric field as well as for magnetic field ?

(A)

(B)

(C)

(D)

Ans.[C] Electric lines of force for induced electric field is

closed loop. 26. A point mass is subjected to two simultaneous

sinusoidal displacements in x-direction, x1(t) = A

sin ωt and x2(t) = A sin

π

+ω3

2t .Adding a third

sinusoidal displacement x3(t) = B sin(ωt + φ) brings the mass to a complete rest. The values of B and φ are :

(A) 4

3,2 πA (B) A, 3

4π

(C) 6

5,3 πA (D) 3

, πA

Ans.[B]

θ

30º120º

A

A

B Here φ = π + θ

A cos 30º = B sin θ ⇒ B sin θ = 2A3 and A sin

30º + B cosθ = A ⇒ B cos θ = 2A

Solving above, B = A and θ = 60º = 3π .

Hence φ = 240º = 34π

27. A long insulated copper wire is closely wound as

spiral of ‘N’ turns. The spiral has inner radius ‘a’ and outer radius ‘b’. The spiral lies in the X-Y plane and a steady current ‘I’ flows through the wire. The Z-component of the magnetic field at the centre of the spiral is -

x

y

a b

I

(A)

−µ

ab

abNI ln

)(20 (B)

−+

−µ

abab

abNI ln

)(20

(C)

µ

ab

bNI ln

20 (D)

−+µ

abab

bNI ln

20


Ans.[A]

dr

a r

b

x

y

No. of turns per unit thickness = ab

N−

magnetic field at centre due to element = r2

i)dN(0µ

dB = r2i0µ

− abN dr

B = )ab(2

iN0

−µ

∫b

ardr =

)ab(2iN0

−µ

ln 3

28. A satellite is moving with a constant speed ‘V’ in

a circular orbit about the earth. An object of mass ‘m’ is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of its ejection, the kinetic energy of the object is -

(A) 221 mV (B) mV2

(C) 223 mV (D) 2mV2

Ans.[B] M V

•

r

⇒ r

mv2 = 2r

GmMe

⇒ r = 2VGMe ...(1)

If K.E. of mass m = was k then from

E = K – r

GmMe = 0

⇒ K = m

rGMe = mv2



29. Two solid spheres A and B of equal volumes but of different densities dA and dB are connected by a string. They are fully immersed in a fluid of density dF. They get arranged into an equilibrium state as shown in the figure with a tension in the string. The arrangement is possible only if –

A

B

(A) dA < dF (B) dB > dF (C) dA > dF (D) dA + dB = 2dF Ans.[A,B,D]

Vdfg

VdAg

VdBg

Vdfg

A

B

system will be in equilibrium with tension in string only if df > dA and dB > df. If both A & B are considered as a system then

2Vdfg = V (dA + db)g ⇒ dA + dB = 2df

30. Which of the following statement(s) is/are correct?

(A) If the electric field due to a point charge varies as r–2.5 instead of r–2, then the Gauss law will still be valid.

(B) The Gauss law can be used to calculate the field distribution around an electric dipole.

(C) If the electric field between two point charges is zero somewhere, then the sign of the two charges is the same.

(D) The work done by the external force in moving a unit positive charge from point A at potential VA to point B at potential VB is (VB – VA).

Ans.[C,D]


31. A series R-C circuit is connected to AC voltage source. Consider two cases ; (A) when C is without a dielectric medium and (B) when C is filled with dielectric of constant 4. The current IR through the resistor and voltage VC across the capacitor are compared in the two cases. Which of the following is/are true -

(A) RR

AR II > (B) R

RAR II <

(C) BC

AC VV > (D) B

CA

C VV <

Ans.[B,C]

~

R C

2

21 C

1RZ

ω+=

~

R 4C

22

2 C41RZ

ω+=

z1 > z2 ∴ BR

AR II <

C

IV

AKA

C ω= ;

C4I

VBKB

C ω= ; A

CBC VV <

32. A thin ring of mass 2 kg and radius 0.5 m is rolling without slipping on a horizontal plane with velocity 1 m/s. A small ball of mass 0.1 kg, moving with velocity 20 m/s in the opposite direction, hits the ring at a height of 0.75 m and goes vertically up with velocity 10 m/s. Immediately after the collision.

10 m/s

1 m/s

20 m/s

0.75 m

(A) the ring has pure rotation about its stationary C.M

(B) the ring comes to a complete stop (C) friction between the ring and the ground is to

the left (D) there is no friction between the ring and the

ground Ans. [A,C]

SECTION – III Integer Answer Type

This section contains 6 questions. The answer to each of the questions is a single-digit integer, ranging from 0 to 9. The bubble corresponding to the correct answer is to be darkened in the ORS

33. Two batteries of different emfs and different internal resistances are connected as shown. The voltage across AB in volts is -

3V 2Ω

6V 1Ω

A B

Ans.[5]

VA – VB =

21

11

23

16

+

+ =

5.15.16 +

= 5.15.7

= 5V

34. A series R-C combination is connected to an AC voltage of angular frequency ω = 500 radian/s. If the impendence of the R-C circuit is 25.1R , the time constant (in millisecond) of the circuit is.

Ans. [4]

Z = R 25.1 τ = RC

R2 + 2

5001

C= Z2

R2 + 2

5001

C= R2 × 1. 25

2

5001

C= 0.25 R2 ⇒

C5001

= 0.5R

2500

1= RC

2501

= RC

0.004 sec = RC RC = 4 mill second.

35. A train is moving along a straight line with a constant acceleration ‘a’. A boy standing in the train throws a ball forward with a speed of 10 m/s, at an angle of 60º to the horizontal. The boy has to move forward by 1.15 m inside the train to catch the ball back at the initial height. The acceleration of the train, in m/s2, is.


Ans. [5]

T = 102

3102×

××= 3 sec

x = 10 cos 60° (T) = 5 3 m In frame of train,

5 3 = 21

× a × ( 3 )2 + 1.15

(a : acceleration of train) a = 5 m/sec2

36. Water (with refractive index = 4/3) in a tank is 18 cm deep. Oil of refractive index 7/4 lies on water making a convex surface of radius of curvature ‘R = 6 cm’ as shown. Consider oil to act as a thin lens. An object ‘S’ is placed 24 cm above water surface. The location of its image is at ‘x’ cm above the bottom of the tank. Then ‘x’ is –

R = 6 cmµ = 7/4

µ = 1.0

S

µ = 4/3

Ans.[2]

uv13 µ

−µ

= 1

12R

µ−µ+

2

23R

µ−µ

V34

– 241

−=

6

147

−+

∞

−47

34

⇒ V34

+ 241

= 81

⇒ V34

= 121

⇒ V = 16 cm ∴ Ans. = (18 –16) cm = 2 cm

37. A block of mass 0.18 kg is attached to a spring of force-constant 2 N/m. The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is un-stretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block in m/s is V = N/10. Then N is

Ans.[4]

m = 0.18 kg

µ = 0.1

k = 2N/m

Using W – E theorem

21 × m(u)2 =

21 K (x)2 + µmg (x)

21 × (0.18) u2 =

21 × 2 × 36 × 10–4 + 0.1 × 0.18

× 10 × 0.06 ⇒ u = 0.4 m/sec.

⇒ 104 m/sec.

38. A silver sphere of radius 1 cm and work function 4.7 eV is suspended from an insulating thread in free-space. It is under continuous illumination of 200 nm wavelength light. As photoelectrons are emitted, the sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the sphere is A × 10z (where 1 < A< 10). The value of ‘Z’ is.

Ans. [7]

Energy of photon ≈200

1240 eV = 6.2 eV

Maximum KE of a electron = 6.2 eV – 4.7 eV

When potential on surface of sphere becomes equal to 1.5V

r

q

04 ∈π= 1.5 V ⇒ q = 1.5 × (4π ε0) × r

No. of photoelectron emitted n = 190

106.1r)4(5.1

−×

πε×

= 1. 04 × 107

SECTION – IV

Matrix match Type

This section contains 2 questions. Each question has four statements (A, B, C and D) given in Column I and five statements (p, q, r, s and t) in Column II. Any given statement Column I can have correct matching with ONE or MORE statement(s) given in Column II. For example, if for a given question, statement B matches with the statements given in q and r, then for the particular question, against statement B, darken the bubbles corresponds to q and r in the ORS.

39. One mole of a monoatomic ideal gas is taken through a cycle ABCDA as shown in the P-V diagram. Column-II gives the characteristics involved in the cycle. Match them with each of the processes given in Column-I.


B A 3P

P

C 1P

0 1V 3V 9V V

D

Column-I Column-II (A) Process A → B (p) Internal energy decreases (B) Process B → C (q) Internal energy increases (C) Process C → D (r) Heat is lost (D) Process D → A (s) Heat is gained (t) Work is done on the gas

Ans. (A) → p,r,t; (B) → p,r; (C) → q,s; (D) → r,t Process AB : (Pressure is constant)

If TA = T ⇒ TB = 3T

So ∆U = Negative [Q ∆U = nCv∆T] ∆W = nR∆T = Negative ∆Q = ∆U + ∆W = Negative Process BC : (Volume is constant)

If TB = 3T then TC =

9T

∆U = nCv∆T = Negative ∆W = Zero ∆Q = Negative Process C → D : (Pressure is constant)

If TC = 9T then TD = T

∆U = nCv∆T = positive ∆W = positive ∆Q = positive Process D → A : TD = T and TA = T Hence process is isothermal ∆U = 0 ∆W = negative ∆Q = negative

40. Column-I shows four systems, each of the same length L, for producing standing waves. The lowest possible natural frequency of a system is called its fundamental frequency, whose wavelength is denoted as λ1. Match each system with statements given in Column-II describing the nature and waves.

Column-I

(A) Pipe closed at one end 0 L

(B) Pipe open at both ends 0 L

(C) Stretched wire clamped at both ends

0 L (D) Stretched wire clamped at both ends and at

mid-point

0 L L/2 Column-II

(p) Longitudinal waves (q) Transverse waves (r) λ1 = L (s) λf = 2L (t) λf = 4L Ans. (A) → p,t; (B) → p,s; (C) → q,s; (D) → q,r

(A)

L4

f =λ

⇒ λf = 4L

(B) Longitudinal waves

L2

f =λ

(C) Stretched wire clamped at both ends

2fλ = L ⇒ λf = 2L

(D)

2fλ + L

2f =

λ

⇒ λf = L


MATHEMATICS



41. Let P(6, 3) be a point on the hyperbola

12

2

2

2=−

by

ax . If the normal at the point P

intersects the x-axis at (9, 0), then the eccentricity of the hyperbola is -

(A) 25 (B)

23

(C) 2 (D) 3 Ans. [B] Sol. Equation of the normal at (6, 3) is

6

2xa+

3

2 yb= a2 + b2

it passes through (9, 0)

so 6

9 2a= a2 + b2

⇒ b2 = 2

2a

Now b2 = a2 (e2 –1)

∴ e2 –1 = 21

e2 = 23

⇒ e = 23

42. Let (x, y) be any point on the parabola y2 = 4x. Let P be the point that divides the line segment from (0, 0) to (x, y) in the ratio 1 : 3. Then the locus of P is

(A) x2 = y (B) y2 = 2x (C) y2 = x (D) x2 = 2y Ans. [C]

Sol. h = 4

2t, k =

42t

(t2, 2t)

(0, 0)

• P(h, k) 3

1

t2 = 4h, t = 2k so 4k2 = 4h ∴ k2 = h hence required locus is y2 = x

43. Let f(x) = x2 and g(x) = sin x for all x∈R. Then the set of all x satisfying (f o g o g o f) (x) = (g o g o f) (x), where (f o g) (x) = f(g(x)), is

(A) ...2,1,0, ∈π± nn

(B) ...2,1, ∈π± nn

(C) ....,2,1,0,1,2....,,22

−−∈π+π nn

(D) 2nπ, n∈…., –2, –1, 0, 1, 2, …. Ans. [A] Sol. gof(x) = gf(x) = g(x2) = sin x2 go (gof(x)) = g(sin x2) = sin (sin x2) fo(gogof(x)) = f(sin (sin x2)) = (sin(sin x2))2 ∴ (sin (sin x2))2 = sin (sin x2) sin (sin x2) (sin (sin x2) –1) = 0 sin (sin x2) = 0 or sin (sin x2) = 1

sin x2 = nπ sin x2 = 2nπ + 2π

At n = 0 At n = 0

sin x2 = 0 sin x2 = 2π

x2 = nπ Not possible

x = ± πn ; n ∈ 0, 1, 2, ….

44. Let f : [–1, 2] → [0, ∞) be a continuous function such that f(x) = f(1 – x) for all x∈[–1, 2]. Let

∫−

=2

11 )( dxxfxR , and R2 be the area of the region

bounded by y = f(x), x = – 1, x = 2, and the x-axis. Then

(A) R1 = 2R2 (B) R1 = 3R2

(C) 2R1 = R2 (D) 3R1 = R2 Ans. [C]

Sol. R1 = ∫−

2

1

)( dxxfx … (i)

R1 = ∫−

−−2

1

)1()1( dxxfx = ∫−

−2

1

)()1( dxxfx ...(ii)

(i) + (ii)

2R1 = ∫−

2

1

)( dxxf = R2

∴ 2R1 = R2


45. If x/12

0x)]b1(nx1[lim ++

→l = 2b sin2 θ, b > 0 and

θ∈(– π, π], then the value of θ is -

(A) 4π

± (B) 3π

±

(C) 6π

± (D) 2π

±

Ans. [D] Sol.

0lim→x

(1 + x ln(1 + b2)]1/x = 2b sin2 θ b > 0;

θ ∈ (–π, π)

0

lim→x

)1(

)1(1

2

2

2

)]1(1[

bn

bnxbnx

+

+

++

l

ll = 2b sin2 θ

)1( 2bne +l = 2b sin2 θ 1 + b2 = 2b sin2 θ

2 sin2θ = b +b1

RHS = b + b1

≥ 2 as b > 0

But LHS = 2 sin2 θ ≤ 2 Only possibility 2 sin2 θ = 2 sin2θ = 1

θ = ±2π

46. The circle passing through the point (–1, 0) and touching the y-axis at (0, 2) also passes through the point -

(A)

− 0,

23 (B)

− 2,

25

(C)

−

25,

23 (D) (– 4, 0)

Ans. [D] Sol. ∴ (h – 0)2 + (2 –2)2 = (h + 1)2 + (2 – 0)2 h2 = h2 + 1 + 2h + 4

(–1, 0)

(h, 2) (0, 2)

h = 25

−

Equation of circle is

2

25

+x + (y –2)2 =

20

25–

−

x2 + 425

+ 5x + y2 + 4 – 4y = 425

x2 + y2 + 5x – 4y + 4 = 0 from given points only point (– 4, 0) satisfies this

equation. 47. Let ω ≠ 1 be a cube root of unity and S be the set

of all non-singular matrices of the form

ωωω

11

1

2cba

, where each of a, b and c is either

ω or ω2. Then the number of distinct matrices in the set S is :

(A) 2 (B) 6 (C) 4 (D) 8 Ans. [A]

Sol. 1

11

2 ωωω c

ba ≠ 0

(1 – ωc) – a (ω – ω2c) + b(ω2 – ω2) ≠ 0 1 – ωc – aω + acω2 ≠ 0 (1 – ωc) – aω (1 – ωc) ≠ 0 (1 – ωc) (1 – aω) ≠ 0 c ≠ ω2 & a ≠ ω2 & b = ω or ω2 (a, b, c) ≡ (ω, ω, ω) or (ω, ω2, ω) 48. A value of b for which the equations x2 + bx – 1 = 0, x2 + x + b = 0 have one root in common is -

(A) 2− (B) 3i−

(C) 5i (D) 2 Ans. [B] Sol. x2 + bx –1 = 0 … (i) x2 + x + b = 0 … (ii)

(i) – (ii) we get x = 1–1

bb +

Put this value in (i)

2

1–1

+

bb

+ b

+

1–1

bb

–1 = 0

⇒ b3 + 3b = 0 ⇒ b(b2 + 3) = 0

⇒ b = 0 or b = ± i 3




49. If

>≤<−

≤<π

−−

π−≤

π−−

=

1,ln101

02

,cos2

,2

)(

xxxx

xx

xx

xf , then

(A) f(x) is continuous at x = 2π

−

(B) f(x) is not differentiable at x = 0

(C) f(x) is differentiable at x = 1

(D) f(x) is differentiable at x = 23

−

Ans. [A, B, C, D]

Sol. At x = 2π

−

LHL = 0, RHL = 0, f

π−

2= 0, So f(x) is

continuous at x = 2π

−

At x = 0 LHD = 0; RHD = 1 So f(x) is not differentiable at x = 0 At x = 1 LHD = 1, RHD = 1 So f(x) is differentiable at x = 1

in

π− 0,

2; f(x) = – cos x

so f(x) is differentiable at x = 23

−

50. Let L be a normal to the parabola y2 = 4x. If L passes through the point (9, 6), then L is given by

(A) y – x + 3 = 0 (B) y + 3x – 33 = 0 (C) y + x – 15 = 0 (D) y – 2x + 12 = 0 Ans. [A, B, D] Sol. y = mx – 2m – m3 It passes through (9, 6) 6 = 9m –2m – m3 m3 – 7m + 6 = 0 (m –1) (m –2) (m + 3) = 0 ∴ m = –3, 1, 2 Hence equations will be y = x – 3, y = 2x –12 and y = –3x + 33

51. Let E and F be two independent events. The probability that exactly one of them occurs is 11/25 and the probability of none of them occurring is 2/25. If P(T) denotes the probability of occurrence of the event T, then -

(A) 53)(,

54)( == FPEP

(B) 52)(,

51)( == FPEP

(C) 51)(,

52)( == FPEP

(D) 54)(,

53)( == FPEP

Ans. [A, D]

Sol. P(E) (1 – P(F)) + (1 – P(E)) P(F) = 2511

P(E) + P(F) –2P (E) P(F) = 2511

… (1)

(1 – P(E)) (1 – P(F)) = 252

1 – P(E) – P(F) + P(E) P(F) = 252

P(E) + P(F) – P(E) P(F) = 2523

... (2)

From (1) & (2)

P(E) P(F) = 2512

and P(E) + P(F) = 57

so either

P(E) = 54

, P(F) = 53

and P(E) = 53

, P(F) =54

52. Let f : (0, 1) → R be defined by bxxbxf

−−

=1

)( ,

where b is a constant such that 0 < b < 1. Then (A) f is not invertible on (0, 1)

(B) f ≠ f –1 on (0, 1) & f '(b) =)0('

1f

(C) f = f –1 on (0, 1) and f '(b) =)0('

1f

(D) f –1 is differentiable on (0, 1) Ans. [A, B] Sol. f : (0, 1) → R

f(x) = bxxb

−−

1 ∀ b ∈ (0, 1)

f ′(x) = 2

2

)1(1

bxb−

− = (–) ve


So f(x) is monotonically decreasing for x ∈ (0, 1) so for x ∈ (0, 1) f(x) ∈ (f(1), f(0)) f(x) ∈ (–1, b) so f(x) is not onto. so f(x) is not invertible function.

SECTION – III

Integer Answer Type

This section contains 6 questions. The answer to each of the questions is a single-digit integer, ranging from 0 to 9. The bubble corresponding to the correct answer is to be darkened in the ORS

53. Let y'(x) + y(x)g'(x) = g(x)g'(x), y(0) = 0, x∈R,

where f '(x) denotes dx

xfd )(and g(x) is a given

non-constant differentiable function on R with g(0) = g(2) = 0. Then the value of y(2) is.

Ans. [0]

Sol. dgdy

+ y = g

I. F. = ∫ dg.1 = g

y.eg = ∫ dggeg . = geg – ∫ dgeg .

yeg = geg – eg + c y = g –1 + ce–g Q y(0) = 0 & g(0) = 0 at x = 0 0 = 0 –1 + Ce–0 C = 1 y = g –1 + e–g at x = 2 y(2) = 0 – 1 + e–0 = 0

54. Let kia ˆˆ −−=r

, jib ˆˆ +−=r

and kjic ˆ3ˆ2ˆ ++=r

be three given vectors. If r

ris a vector such that

bcbrrrrr

×=× and 0. =arrr

, then the value of brrr

. is.

Ans. [9]

Sol. ar

= – i – k , br

= – i + j , cr

= i + 2 j + 3 k

bcrbcrbcrrrrrrrrrr

λ+=⇒λ=−⇒=×− 0)(

Q 0=⋅ arrr

⇒ 0.. =λ+ abcarrrr

⇒ 4..

=−=λabcarrrr

⇒ 9||.. 2 =λ+= bbcbrrrrrr

55. Let ω = eiπ/3, and a, b, c, x, y, z be non-zero complex numbers such that a + b + c = x, a + bω + cω2 = y, a + bω2 + cω = z. Then the

value of 222

222

||||||||||||

cbazyx

++++

is.

Ans. [*]

Sol. wrong question if ω = 3/2πie then ans is 3. If ω = 3/πie then no integral solution is possible.

56. Let M be a 3× 3 matrix satisfying

−=

321

010

M ,

−=

−

111

01

1M and

=

1200

111

M . Then the sum

of the diagonal entries of M is. Ans. [9]

Sol. Let M =

ihgfedcba

Q M

010

=

−

321

⇒ b = –1, e = 2, h = 3

∴ M

−01

1=

−111

⇒ a = 0, d = 3, g = 2

M

111

=

1200

⇒ c = 1, f = –5, i = 7

So a + e + i = 0 + 2 + 7 = 9

57. The number of distinct real roots of x4 – 4x3 + 12x2 + x – 1 = 0 is

Ans. [2] Sol. Let f(x) = x4 – 4x3 + 12x2 + x – 1

Let α, β, γ, δ are the root of equation. ∴ αβγδ = –1 so the equation has at least two

real roots. ...(i) f '(x) = 4x3 – 12x2 + 24x + 1 f "(x) = 12x2 – 24x + 24 = 12((x + 1)2 + 1) so f "(x) > 0 so f '(x) = 0 has only one real roots

so f(x) = 0 has at most two real roots. ...(ii) from (i) & (ii) f(x) = 0 has exactly two real roots.


58. The straight line 2x – 3y = 1 divides the circular region x2 + y2 ≤ 6 into two parts. If

−

=

41,

81,

41,

41,

43,

25,

43,2S , then

the number of point(s) in S lying inside the smaller part is -

Ans. [2] Sol.

(2, 3)

0,

21

Pont (x1, y1) lies inside the region if

0621

21 ≤−+ yx & 2x1 – 3y1 – 1 ≤ 0.

≡

43,21P 06

1694 ≤−+ True

01494 >−− True

≡

43,

25

2P 06169

425

≤−+ False

−

≡41,

41

3P 06161

161

≤−+ True

0143

42

>−+ True

≡

41,

81

4P 06161

641

≤−+ True

0143

82

>−− False

So P1 & P3 lies in the interval

SECTION – IV Matrix match Type

This section contains 2 questions. Each question has four statements (A, B, C and D) given in Column I and five statements (p, q, r, s and t) in Column II. Any given statement Column I can have correct matching with ONE or MORE statement(s) given in Column II. For example, if for a given question, statement B matches with the statements given in q and r, then for the particular question, against statement B, darken the bubbles corresponds to q and r in the ORS.

59. Match the statements given in Column-I with the values given in Column-II.

Column-I

(A) If kja ˆ3ˆ +=r

, kjb ˆ3ˆ +−=r

and kc ˆ32=r

form a triangle, then the internal angle of the

triangle between ar

and br

is

(B) If ∫ −=−b

a

badxxxf 22)3)(( , then the value

of

π

6f is

(C) The value of ∫ ππ

6/5

6/7

2)sec(

3lndxx is

(D) The maximum value of

− zArg

11

for

|z| = 1, z ≠ 1 is given by

Column-I (p) π/6 (q) 2π/3 (r) (r) π/3 (s) π (t) π/2 Ans. [A → q ; B → p ; C → s ; D → s] Sol. (A)

cr b

r

θar

21

||||cos −=

−⋅−

=θbabarr

rr

⇒ 3

2π=θ

(B) 22))(3)(( badxxxfb

a

−=−∫

differentiating w.r.t (b). f(b) – 3b = – 2b

bbf =)(

So 66π

=

πf


(C) dxxn

I ∫ ππ

=6/5

6/7

2sec

3l

6/56/7

2|tansec|

3xxn

nI π+π

ππ

= ll

π=⋅π

= 33

nn

I ll

(D) ∴ |z| = 1 z = cos θ + i sin θ. ∀ θ∈ (– π. π] and θ ≠ 0.

)1(

1z

Arg−

θ

+=

θ−θ−=

22

cot

21

sincos11 i

Argi

Arg

=2

θ−π so maximum value is π.

60. Match the statements given in Column-I with the intervals/union of intervals given in Column-II.

Column-I (A) The set

±≠=

−1z,1|z|

number,complexaisz;z1

iz2Re 2 is

(B) The domain of the function

−=

−

−−

)1(2

21

31)3(8sin)( x

xxf is

(C) If1tan1

tan1tan1tan1

)(θ−−

θθ−θ

=θf , then the

set

π

<θ≤θ2

0:)(f is

(D) If 0),103()( 2/3 ≥−= xxxxf , then f(x) is increasing in

Column-II (p) (– ∞, –1) ∪ (1, ∞) (q) (– ∞, 0) ∪ (0, ∞) (r) [2, ∞) (s) ( – ∞, –1] ∪ [1, ∞) (t) (– ∞, 0] ∪ [2, ∞) Ans. [A → p, r, s ; B → r, t ; C → r ; D → r]

Sol. (A) Let z = cos θ + i sin θ

so θ−=θ−θ−

θ+=

−ec

iii

ziz cos

2sin2cos1)sin(cos2

12

2

∀ θ ≠ (2n + 1) 2π

so ),1[]1,(eccosz1

iz2Re 2 ∞∪−−∞∈θ−=

−

(B) x

x

x

x

222

2

3938

3138

−×

=−×

−

−

Let 3x = t

So f(x) = sin–1

−=

−× −

21

2 98sin

3938

tt

x

x

19

81 2 ≤−

≤−tt on solving

x ∈ (– ∞, 0] ∪ [2, ∞) ∪ 1 (C) f(θ) = 2 sec2θ so f(θ) ∈[2, ∞) (D) f(x) = 3x5/2 – 10x3/2

f'(x) = ( )22

15−xx

So f(x) is increasing for f '(x) ≥ 0 x ∈ [2, ∞)

Random Facts • As a gas' temperature is raised to over

10,000°, its molecules collide so violently that they are broken apart into individual atoms.

• When the tsunami reaches the coast and moves inland, the water level can rise many meters. In extreme cases, water level has risen to more than 15 m (50 ft) for tsunamis of distant origin and over 30 m (100 ft) for tsunami waves generated near the earthquake’s epicenter.

• Some minerals, notably quartz, are piezoelectric--that is, they produce electricity when subjected to pressure or stress. This same phenomenon is probably also responsible for "earthquake lights," the luminescence sometimes reported (and, on occasion, photographed) in the sky during earthquakes.


XtraEdge Test Series ANSWER KEY

PHYSICS

Ques 1 2 3 4 5 6 7 8 Ans D C C B A B A A,B,C Ques 9 10 11 12 13 14 15 16 Ans B,C B,C A,B,C,D B C C B C

Ques 17 18 19 20 21 22 23 Numerical Response Ans 3 1 2 1 2 3 4

CHEMISTRY

Ques 1 2 3 4 5 6 7 8 Ans C B D D B A C B,C Ques 9 10 11 12 13 14 15 16 Ans B,D B,D D C D B D B


MATHEMATICS

Ques 1 2 3 4 5 6 7 8 Ans A A D A C C A B,D Ques 9 10 11 12 13 14 15 16 Ans A,B,C A,B,C,D B,D D C A A A


PHYSICS Ques 1 2 3 4 5 6 7 8 Ans B A A D A B A A,B,C,D Ques 9 10 11 12 13 14 15 16 Ans A,B,C A,B,D A,C,D C B C B B


CHEMISTRY

Ques 1 2 3 4 5 6 7 8 Ans A B D B A C D A,B,C Ques 9 10 11 12 13 14 15 16 Ans A,B,C A,D A,C,D A B A B A


MATHEMATICS

Ques 1 2 3 4 5 6 7 8 Ans A C C C D A C A,B,D Ques 9 10 11 12 13 14 15 16 Ans C,D A,C A,B,C,D C C B A B


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