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XtraEdge for IIT-JEE 1 JANUARY 2011

Dear Students,

Find a mentor who can be your role model and your friend ! A mentor is someone you admire and under whom you can study. Throughout history, the mentor-protege relationship has proven quite fruituful. Socrates was one of the early mentors. Plato and Aristotle studied under him and later emerged as great philosophers in their own right. Some basic rules to know mentors : • The best mentors are successful people in their own field. Their behaviors

are directly translatable to your life and will have more meaning to you. • Be suspicious of any mentors who seek to make you dependent on them. It

is better to have them teach you how to fish than to have them catch the fish for you. That way, you will remain in control.

• Turn your mentors into role models by examining their positive traits. Write down their virtues. without identifying to whom they belong. When you are with these mentors, look for even more behavior that reflect their success. Use these virtues as guidelines for achieving excellence in your field.

Be cautious while searching for a mentor : • Select people to be your mentors who have the highest ethical standards

and a genuine willingness to help others. • Choose mentors who have and will share superb personal development

habits with you and will encourage you to follow suit. • Incorporate activities into your mentor relationship that will enable your

mentor to introduce you to people of influence or helpfulness. • Insist that your mentor be diligent about monitoring your progress with

accountability functions. • Encourage your mentor to make you an independent, competent, fully

functioning, productive individual. (In other words, give them full permission to be brutally honest about what you need to change.)

Getting benefited from a role-mode : Acquiring good habits from others will accelerate you towards achieving your goals. Ask yourself these questions to get the most out of your role model/mentors : • What would they do in my situation? • What do they do every day to encourage growth and to move closer to a

goal ? • How do they think in general ? in specific situations ? • Do they have other facts of life in balance ? What effect does that have on

their well-being ? • How do their traits apply to me ? • Which traits are worth working on first ? Later ? A final word : Under the right circumstances mentors make excellent role models. The one-to-one setting is highly conducive to learning as well as to friendship. But the same cautions hold true here as for any role model. It is better to adapt their philosophies to your life than to adopt them . Presenting forever positive ideas to your success.

Yours truly

Pramod Maheshwari, B.Tech., IIT Delhi

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

Editor : Pramod Maheshwari

"Faliure is Success if we learn from it" Volume - 6 Issue - 7

January, 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-7 January, 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- 2011 & 2012

Mock Test Paper CBSE Class XII

S

Success Tips for the Months

• The greatest adventure is what lies ahead.

• Fixers believe they can fix. Complainers believe they can complain. They are both right.

• The tire model for motivation: People work best at the right pressure.

• Trust the force, Luke.

• Use your feelings or your feelings will use you.

• People who expect to fail are usually right.

• The path to success is paved with mistakes.

• You've got to cross that lonesome valley. You've got to cross it by yourself.

• Appreciate what your brain does. In case nobody else does.

• Learn to mock the woe-mongers.

• Be confident. Even if you are not, pretend to be. No one can tell the difference.

CONTENTS

INDEX PAGE

NEWS ARTICLE 4 The young innovator from IIT Kharagpur bagged first prize at IIT-Delhi fair Engineering better managers

IITian ON THE PATH OF SUCCESS 6 Dr. Raghuram G. Rajan & Prof. Devang Khakhar

KNOW IIT-JEE 7 Previous IIT-JEE Question

XTRAEDGE TEST SERIES 60

Class XII – IIT-JEE 2011 Paper Class XI – IIT-JEE 2012 Paper Mock Test-2 (CBSE Board Pattern) [Class # XII] 78 Solution of Mock Test-1 (CBSE Pattern)

Regulars ..........

DYNAMIC PHYSICS 16

8-Challenging Problems [Set# 9] Students’ Forum Physics Fundamentals Prism & Wave nature of Light Waves & Dopspler Effect CATALYSE CHEMISTRY 33

Key Concept Carbohydrates Salt Analysis Understanding : Organic Chemistry

DICEY MATHS 43

Mathematical Challenges Students’ Forum Key Concept

Differential Equations Trigonomatrical Ratios

Study Time........

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


The young innovator from IIT Kharagpur bagged first prize At IIT-Delhi fair Indian Institutes of Technology (IITs) have always encouraged innovation and entrepreneurship. Shwetank Jain, studying BTech at IIT Kharagpur proved it by winning the first prize at the India Innovation Initiative (i3) National Fair by presenting his P2 Power Solutions – an Intelligent Power Conditioner with Hybrid System Integrator that economizes use of energy and enhances power quality. The fair was held at IIT Delhi and was organized by the Department of Science and Technology, Government of India, Agilent Technologies, and the Confederation of Indian Industry (CII).

Engineering better managers

“The faculty and the curriculum at DMS were excellent. Some of our courses were introduced only much later in other B-schools in the country. It was a fantastic experience,” says Saurav Sharma, alumni of IIT Delhi’s Department of Management Studies (DMS).

Sharma, now a director at United Bank of Switzerland and based in Hyderabad, is from the Class of 1997. DMS began functioning as a separate department from that year, though post-graduate management education was introduced in IIT Delhi in 1976.

Housed within the IIT Delhi campus, the institute offers, MBA (full time) with focus on management systems, MBA (full time) with focus on telecommunications system management and MBA (part-time) with focus on technology management. Apart from these, students can opt for specialisation in functional areas like finance, marketing, information technology and human resource. All IITs admit students to the two-year full time MBA programme through the Joint Management Entrance Test. “The faculty, infrastructure and courses offered here are unparalleled. Systems thinking and creative problem solving are two courses that are excellent and not taught in any other B-school,” says Tajinder Kohli, a first-year MBA student.

P.K. Jain, who teaches finance, with 35 years of teaching experience and 23 books and 120 research papers in national and international journals to his credit, is widely rated as one of the best teachers in India in his field. “Our objective is to turn a good engineer into a better engineer. We encourage students to do field work to increase their awareness of the real world work conditions,” says Vinayshil Gautam, who was the first head of the department of DMS and the founder director of IIM-Khozhikode.

Another reason why DMS is so sought after by students is the placements that they are offered. “The return on investment is really high and the brand name of IIT definitely helps. Including the hostel charges, one has to pay around Rs 2 lakh for the course. The minimum salary offered to graduates who pass out from here is at three times as much. In comparison, other comparable institutes charge 2-4 times these fees,” says Vipul Arora, a second year student and a member of the placement committee.“We find the

students of IIT-DMS very impressive. What we look for in our organisation is a combination of management skills, presentablity and communication skills,” said Vinay Tiwari, Campus Relations Manager, Citigroup. “The students here embody all these qualities.” The green campus and the canteens act as great value additions.“I love the greenery in the campus and it’s great to walk around. The ‘Uth Café’ is the most popular hangout here. From Indian to Italian, it offers all kinds of dishes and it remains abuzz till the wee hours of morning. Politics, studies, films… we discuss everything over cups of chai and coffee,” says Kavita Dara, a first year MBA student. Parivartan, the annual management festival and the magazine CHAOS (Creative and Holistic Approach to Organisational Systems) are among the various other platforms offered to students to nurture their creative spirits

IITs to assist, collaborate with other technical institutions Bangalore: The Institutes of Technology Act, that is around forty years old, is being amended soon to allow the Indian Institutes of Technology (IITs) to take up the responsibility of supporting and collaborating with other technical education institutions and also to advise the state governments on technological problems within the zone where the IIT exists.

The Institutes of Technology (Amendments) Bill, 2010 will make it mandatory for all 15 IITs in the country to provide training, facilitate study visits, share laboratories as well as other resources with othertechnical institutions in their respective zones.


This will allow technical institutes, that have been mushrooming all over the country but have often been criticized for poor standards, to collaborate with the IITs and better their educational system.

The bill has been given the go ahead by the Council of IITs. However, the parliamentary standing committee on human resource development has recommended that the clauses that make it compulsory for all IITs to collaborate with other technical institutions should not me made mandatory.

The parliamentary standing committee includes members such as Rahul Gandhi, Kanimozhi and Suresh Kalmadi. The committee had submitted a report last month regarding the proposed bill in which it had been stated that the issues of resources, capacity and faculty at the IITs must be addressed first. The report also said that the technical educational institutions must also have the vision to support and deal with the demands of the society as well as the industry.

The standing committee was headed by Oscar Fernandes and had relied heavily on the reservations of the finance ministry that had emphasized that the new clauses that were in the Institutes of Technology (Amendments) Bill were casting an obligation upon the IITs to meet the technological needs of the states where they were situated.

Innovators showcase designs at IIT Delhi fair New Delhi: The India Innovation Initiative (i3) National Fair held at the Indian Institute of Technology, Delhi (IIT-D) showcased scientific innovations and inventions by several students from all over India.

The fair was organized in association with Agilent Technologies, Department of Science and Technology of the government of India and the Confederation of the Indian Industry.

The first prize at the fair was won by Shwetank Jain, a 25 year old, whose project while pursuing B.Tech at IIT

Kharagpur helped him to gain entrepreneurial success. Jain's project was an Intelligent Power Conditioner with Hybrid System Integrator. Brainchild of Jain and his friends, the project called the P2 Power Solutions had the aim to provide innovative engineering solutions while focusing specifically on energy efficiency and power quality enhancement. The second prize at the fair was won by Nandan Kumar, Sudarshan Rajagopal and Sankamesh Ramaswamy for developing an automated machine that manufactures three-dimensional non-woven fibrous structures that have application in the medical field. Kumar, a textile engineer, informed that the cotton used in medical applications have short fibers that may be transferred to the surface of the wound and cause infection. "The fibers that we have created are hollow from inside and provide more space to absorb bodily fluids," he said. Over 50 innovators showcased their creations at the fair. They had been chosen from around 1000 entries from all over the country.

Shourie: allow IITs to work autonomously The former Union Minister, Arun Shourie, said on Saturday that academic institutions of excellence such as the Indian Institutes of Technology should be left to work autonomously without government interference.

He was speaking at the Dr. Homi Bhabha Centenary Conclave, a three-day event that began at the Tata Institute of Fundamental Research (TIFR) here on Friday. The conclave brought together former recipients of the Homi Bhabha Fellowship. Mr. Shourie himself was a recipient.

Mr. Shourie said the TIFR, an institution of excellence, thrived because of its autonomy. “Mr. J.R.D. Tata entrusted Dr. Bhabha, and so did Pandit Nehru, with the job of setting up an institution of learning and research of the calibre of the TIFR.”

He said Dr. Bhabha’s approach was of “unobtrusive oversight.” Contrasting the TIFR with the institutions of today, Mr. Shourie said they suffered because

they were unduly “besieged by the government.” “Nine new IITs were announced despite a teacher shortage,” he said. “IIT-Rajasthan was announced two-and-a-half years ago. The head of the institute was appointed two years later. Still, the site for the institute was not decided. The Chief Minister wants it in his constituency, while his predecessor wanted it in hers. Therefore, it was decided that the IIT-Kanpur would house it, thereby burdening the parent institute.”

IIT Rajasthan PhD Programme Admission December 2010 Indian Institute of Technology (IIT), Rajasthan, has invited applications for admission into Doctorate of Philosophy (PhD) programme, to be offered in the academic session commencing from December 2010. Indian Institute of Technology (IIT) Rajasthan is one of the prestigious IITs of India. The initiative of establishing the institute was taken by Ministry of Human Resource Development (MHRD), Government of India, in 2008.

IIT students refraining from signing bonds with recruiters IIT-Madras academic affairs secretary said that most candidates these days were not willing to sign any bonds. "We have informed the recruiters that they would be required to incentivize students to make them work for them," he said.

At IIT-Bombay, companies that had insisted on students signing contracts with them had been politely declined from participating in the recruitment procedure.

IIT Joint Admission Test for M.Sc on May 8, 2011 New Delhi: The Joint Admission Test for M.Sc (JAM) for pursuing post-graduation at the Indian Institutes of Technology (IIT) will be conducted on May 8, 2011 at various test centres in the Bombay, Delhi, Guwahati, Kanpur, Kharagpur, Madras and Roorkee Zones.


Dr. Raghuram G. Rajan received his Bachelor’s degree in Electrical Engineering from IIT Delhi in 1985 along with the Director’s Gold Medal for all round performance. He graduated from I.I.M. Ahmedabad in 1987, where he received a gold medal for academic performance.

He joined the Graduate School of Business (GSB), University of Chicago in 1991 after obtaining a Ph.D. from MIT. In 1994, Dr. Rajan received tenure and appointment as Professor of Finance and is now the Joseph L. Gidwitz Distinguished Service Professor of Finance at the GSB. He has held a visiting chair at Northwestern (1996-97) and the Fischer Black visiting chair at M.I.T. (2000-2001).

Dr. Rajan is currently the Economic Counselor and Director of Research at the International Monetary Fund, USA. He has been on this position since 2003.

Dr. Rajan’s research interests focus primarily on economic institutions ranging from banks to property rights. He teaches courses on Corporate Finance and Banking at the GSB. His papers have been published in all the top economics and finance journals. He has also written a book with Luigi Zingales entitled Saving Capitalism from the Capitalists, which was published by Random House in February 2003.

Dr. Rajan is a Director of the American Finance Association and the Program Director for the Corporate Finance Program at the National Bureau of Economic Research. He has served as a consultant to a number of organizations including the World Bank, the Federal Reserve Board, the Reserve Bank of India, the Swedish Parliament, finance companies and banks.

In honouring Dr. Raghuram Rajan, IIT Delhi recognizes the outstanding contributions made by him as a Researcher, Educator and Financial Economist. Through his achievements, Dr. Raghuram G. Rajan has brought glory to the name of this Institute.

Prof. Devang Khakhar received his Bachelor’s Degree in Chemical Engineering from IIT Delhi in 1981. He subsequently received his Ph.D. in Chemical Engineering from the University of Massachusetts, Amherst in 1986.

Prof. Khakhar is presently Professor of Chemical Engineering and Dean of Faculty at IIT Bombay.

After spending one semester at IIT Kanpur as a Visiting Faculty Member, Prof. Khakhar joined IIT Bombay in January 1987 as a Lecturer. He became a Professor at IIT Bombay in 1996. He was also a Visiting Faculty Member at Northwestern University in 1996. Prof. Khakhar served as the Head of Chemical Engineering from 2002 to 2004 and became Dean of Faculty in 2005.

Prof. Khakhar has made pioneering research contributions in a number of areas of Chemical Engineering including polymers, granular mechanics and mixing. The work has been published in top international journals such as Nature and Science. Some of the work has had a direct impact on industrial practice. Prof. Khakhar teaches courses related to fluid mechanics, granular mechanics and polymers.

Prof. Khakhar has received several awards and recognitions for his work. He is a Fellow of the Indian Academic of Science, Bangalore, the Indian National Science Academy, New Delhi and the Indian National Academy of Engineering. He is a recipient of the Shanti Swarup Bhatnagar Prize, the Swarnajayanti Fellowship and the Excellence in Teaching Award of IIT Bombay.

Prof. Khakhar is currently a member of the governing council of the Indian National Academy of Engineering and a member of the Science and Engineering Research Council of the Department of Science and Technology, India.

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

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

Dr. Raghuram G. Rajan Electrical Engineering from IIT Delhi in 1985 Ph.D. from MIT. In 1994

Prof. Devang Khakhar BTech from IIT- Delhi in 1981, Ph.D. in Chemical Engineering from the University of Massachusetts, in 1986.


PHYSICS

1. A uniform wire having mass per unit length λ is placed over a liquid surface. The wire causes the liquid to depress by y(y << a) as shown in figure. Find surface tension of liquid. Neglect end effect. [IIT-2004]

y

a a

Sol. The free body diagram of wire is given below. If l is the length of wire, then for equilibrium 2F S sin θ = W.

F F

mg

θθ

F = S × l ⇒ 2S × l × Sin θ = λ × l × g

or S = θ

λsin2

g

also sin θ = y/a

∴ S = ay

g/2

λ = yga

2λ

⇒ Surface tension, S = yga

2+

2. A double-pane window used for insulating a room

thermally from outside consists of two glass sheets each of area 1 m2 and thickness 0.01 m separated by a 0.05 m thick stagnant air space. In the steady state, the room glass inter-face and the glass-outdoor interface are at constant temperatures of 27ºC and 0ºC respectively. Calculate the rate of heat flow through the window pane. Also find the temperatures of other interfaces. Given thermal conductivities of glass and air as 0.8 and 0.08 W m–1 K–1 respectively. [IIT-1997]

Sol.

IK1

IIK2

III K3

H H

θ1 θ2

d3d2d1

θB θA

θA > θB

At steady state

First material : H = 1

1

dK A (θA – θ1) ...(i)

Second material : H = 2

2

dK A (θ1 – θ2) ...(ii)

Third material : H = 3

3

dK

A (θ2 – θB) ...(iii)

From (i) θA – θ1 = AH

1

1

Kd ...(iv)

From (ii) θ1 – θ2 = AH

2

2

Kd ...(v)

From (iii) θ2 – θB = AH

3

3

Kd

...(vi)

⇒ Adding the above three equations we get

θA – θB =

++

3

3

2

2

1

1

Kd

Kd

Kd

AH

⇒ H =

3

3

2

2

1

1

)–(

Kd

Kd

Kd

ABA

++

θθ

Substituting the values

H =

8.001.0

08.005.0

8.001.0

1)0–27(

++ = 41.54 J/s

From (iv)

27 – θ1 = 154.41 ×

8.001.0 ⇒ θ1 = 26.48º

From (vi) θ2 – 0 = 154.41 ×

8.001.0 ⇒ θ2 = 0.52ºC

3. Two long straight parallel wires are 2 metres part,

perpendicular to the plane of the paper (see figure). The wire A carries a current of 9.6 amps, directed into the plane of the paper. The wire B carries a current such that the magnetic field of induction at the point P, at a

distance of 1110 metre from the wire B, is zero.

KNOW IIT-JEE By Previous Exam Questions


O

⊗

1.6 m

2 m

1.2m

10/11 m

P

S

A

B

Find : (i) The magnitude and direction of the current in B (ii) The magnitude of the magnetic field of induction

at the point S. (iii) The force per unit length on the wire B. [IIT-1987] Sol. (i) The magnetic field at P due to current in wire A

BA = π4

µ0

APrlA2 =

πµ4

0 ×

+

×

11102

6.92 ...(i)

(Direction P to M) The current in wire B should be in upward direction

so as to cancel the magnetic field due to A at P. (By right hand Thumb rule)

The magnetic field at P due to current in wire B ⊗

1.6 m

2 m

1.2m B

S

A

M BA BB

N 10/11m

BB = π

µ4

0

11102 Bl ...(ii)

From (i) and (ii)

π

µ4

0 ×

+

×

11102

6.92 = π

µ4

0 ×

11102 Bl

⇒ 32

116.9 × = 10

11×BI

⇒ IB = 3296 = 3A

(ii) The dimensions given shows that SA2 + SB2 = AB2 ⇒ ∠ASB = 90º Magnetic field due to A at S

BSA = π

µ4

0 . SA

A

rI2 =

πµ4

0 × 6.1

6.92×

(Directed S to B) Magnetic field due to B at S

BSA = π

µ4

0 . SB

B

rI2 =

πµ4

0

2.132×

(Directed S to A) The resultant magnetic field

B = 22SBSA BB + =

πµ4

022

6.03

8.06.9

+

= 10–7 × 13 = 1.3 × 10–6 T

(iii) Force per unit length on wire

B = π

µ4

0

AB

BA

rII2

= 2

36.9210 7– ××× = 28.8 × 10–7 N

This force will be repulsive nature. 4. In Figure S is a monochromatic point source emitting

light of wavelength λ = 500 nm. A thin lens of circular shape and focal length 0.10 m is cut into two identical haves L1 and L2 by a plane passing through a diameter. The two halves are placed symmetrically about the central axis SO with a gap of 0.5 m . The distance along the axis from S to L1 and L2 is 0.15 m while that from L1 and L2 to O is 1.30 m. The screen at O is normal to SO.

OA 0.5 mm

Screen

1.30 m0.15 m

L2

L1

S

(i) If the third intensity maximum occurs at the point A on the screen, find the distance OA.

(ii) If the gap between L1 and L2 is reduced from its original value of 0.5 mm, will the distance OA increase, decrease, or remain the same?

[IIT1993] Sol. (i) In this case, the two identical halves of convex

lens will create two seperate images S1 and S2 of the source S. These Images (S1 and S2) will behave as two coherant sources and the further dealing will be in accordance to Young's double slit experiment.

For lens L1 The object is S u = – 0.15 m


u

L2

L1

V

S1

S2

d

O

A

D

1.3m 0.15m

S O1

O2 0.3m

v = ? f = + 0.1 ,

v1 –

u1 =

f1

⇒ v1 =

f1 +

u1 =

1.01 +

15.0–1

⇒ v1 =

15.01.01.0–15.0

× =

15.01.005.0

×

⇒ v = 05.0

15.01.0 × = 0.3 m

∆SO1O2 and ∆SS1S2 are similar. Also the placement of O1 and O2 are symmetrical to S

∴ 21

21

OOSS =

uvu +

S1S2 =( )

u)OO(vu 21+ = ( )

15.03.015.0 + × 0.5 ×10–3

⇒ S1S2 = d = 1.5 × 10–3 m ∴ D = 1.3 – 0.3 = 1 m The fringe width

β = dDλ = 3–

9–

105.1110500

××× =

31 × 10–3

∴ Therefore,

OA = 3β = 3 × 31 × 10 m = 10–3 m

(ii) If the gap between L1 and L2 i.e., O1O2 is reduced. then d will be reduced. Then the fringe width with increase and hence OA will increase.

5. A hydrogen-like atom (described by the Bohr model) is

observed to emit six wavelengths, originating from all possible transitions between a group of levels. These levels have energies between – 0.85 eV and – 0.544 eV (including both these values).

(a) Find the atomic number of the atom. (b) Calculate the smallest wavelength emitted in

these transitions. (Take hc = 1240 eV-nm, ground state energy of

hydrogen atom = – 13.6 eV) [IIT-2002]

Sol. If x is the difference in quantum number of the states then x+1C2 = 6 ⇒ x = 3

Smallest λ n

n +3

Now, we have 2

2

n)eV6.13(Z– = – 0.85 eV ...(i)

and 2

2

)3n()eV6.13(Z–

+ = – 0.544 eV ...(ii)

Solving (i) and (ii) we get n = 12 and Z = 3 (b) Smallest wavelength λ is given by

λhc = (0.85 – 0.544) eV

Solving, we get λ ≈ 4052 nm.

CHEMISTRY

6. An ideal gas having initial pressure P, volume V and

temperature T is allowed to expand adiabatically until its volume becomes 5.66 V, while its temperature falls to T/2.

(a) How many degrees of freedom do the gas molecules have ?

(b) Obtain the work done by the gas during the expansion as a function of the initial pressure P and volume V. [IIT-1990]

Sol. (a) According to adiabatic gas equation, TVγ–1 = constant or T1V1

γ–1 = T2V2γ–1

Here, T1 = T ; T2 = T/2 V1 = V and V2 = 5.66 V

Hence, TVγ–1 = 2T × (5.66V)γ–1

= 2T × (5.66)γ–1 × Vγ–1

or (5.66)γ–1 = 2 ..(1) Taking log, (γ – 1)log 5.66 = log 2

or γ – 1 = 66.5log2log =

7528.03010.0 = 0.4

or γ = 1.4 If f, be the number of degrees of freedom, then

γ = 1 + f2 or 1.4 = 1 +

f2

or f2 = 1.4 – 1 = 0.4


or f = 4.0

2 = 5

(b) According to adiabatic gas equation, P1V1

γ = P2V2γ

Here, P1 = P V1 = V V2 = 5.66 V Hence, PVγ = P2 × (5.66V)γ = P2 ×(5.66)γ × Vγ

or P2 = γ)66.5(

P = 4.1)66.5(P =

32.11P [using eq.(1)]

Hence, work done by the gas during adiabatic expansion

= 1–

2211

γ− VPVP =

1–4.1

66.532.11

VPPV ×−

= 4.0

2PVPV −

= 4.02 ×

PV = 1.25 PV

7. (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

NH3H3N

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 pptBlack

CuS ↓ + H2SO4

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

– (yellow solution)


8. An organic compound A, C6H10O on reaction with CH3MgBr followed by acid treatment gives compound B. The compound B on ozonolysis gives compound C, which in presence of a base gives 1-acetyl cyclopentene D. The compound B on reaction with HBr gives compound E. Write the structures of A, B, C and E. Show how D is formed from C.

[IIT-2000] Sol. The given reaction can be summarised as below :

[A]C6H10O

(i) CH3MgBr

(ii) H+

O3[B] OH–[C] COCH3

HBr

[E][D]

Conclusions from the set of reactions (i) Carbon-hydrogen ratio of [A] indicates that it is

cyclic compound (ii) Reaction of [A] with CH3MgBr indicates that it

should have a ketonic group. (iii) As [B] undergoes ozonolysis to form [C], It must

have a double bond, and [C] must have two carbonyl groups.

(iv) Reaction of [C] a dicarbonyl compound) with a base gives a cyclic compound, it indicates that intramolecular condensation have occurred during this conversion. Thus [A] is cyclohexanone which explains all the given reactions.

(A)

(i) CH3MgBr

(ii) H+

O3 OH–

[C]

CHO

O CH3

(B)

O

–H2O HBr

BrCOCH3

OH

COCH3

; B

H3C

(D) (E)

9. A small quantity of solution containing 24Na radio-nuclide (half life 15 hours) of activity 1.0 micro-curie is injected into the blood of a person. A sample of the blood of volume 1 cm3 taken after 5 hours shows an activity of 296 disintegrations per minute. Determine the total volume of blood in the body of the person. Assume that the radioactive solution mixes uniformly in the blood of the person. (1 curie = 3.7 × 1010 disintegrations per second). [IIT-1994]

Sol. Given that, t1/2 = 15 hours = 15 × 3600 s

–dtdN = 1 micro-curie

= 1 × 10–6 × 3.7 × 1010 disintegrations s–1 = 3.7 × 104 disintegrations s–1

–dt

dN´ = 296 disintegrations min–1

= 60296 disintegrations s–1

t = 5 hours = 5 × 3600 s

Now, t1/2 = λ693.0

or λ = 2/1

693.0t

= 360015693.0

×s–1

= 1.283 × 10–5 s–1

and, –dtdN = λN0

or 3.7 × 104 = 1.283 × 10–5 × N0

or N0 = 5

4

10283.1107.3

−×× = 2.884 × 109

After 5 hours, N = N0e–λt

= 2.884 × 109 × 3600510283.1 5e ×××− −

= 2.884 × 109 × e–0.23 = 2.289 × 109

Again,–dt

dN´ = λN´0

or 60296 = 1.283 × 10–5 × N´0

or N´0 = 510238.160296

−×× = 3.845 × 105

If total volume of blood in the body be V, then

N

N 0 = V1

or V = 0N

N = 5

9

10845.310289.2

×× cm3

= 5953 cm3 = 5.953 litres 10. An organic compound (A), C6H6O, gives a specific

colour with FeCl3 solution and on heating with phthalic anhydride and conc. H2SO4 gives a white solid (B) which gives red colour with dilute NaOH (C). Compound (A) on heating with CCl4 and NaOH gives (D), which on acidification gives (E). Compound (E) gives violet colour with FeCl3. (E) on heating with (A) in presence of POCl3 gives (F), which is used as intestinal antiseptic. What are (A) to (F) ? Explain giving balanced equations.

Sol. (i) As it gives colour with FeCl3, hence it is phenol. (ii) On heating with phthalic anhydride it gives

phenolphthalein (B) which gives red colour of sodium salt of (B), i.e(C).

(iii) (A) undergoes Reimer-Tieman reaction with CCl4 and NaOH giving disodium salt (D), which on acidification gives (E) salicylic acid because violet colour is given by salicylic acid.

(iv) (E) on heating with phenol (A) gives salol (F) which is used as an intestinal antiseptic.

Reactions :


C — O

HO OH

H H O

C

O

H2SO4

∆; –H2O

C — O

HO OH

C

O

2NaOH

C

NaO O

COO–Na+

OH

+ CCl4 + 6NaOH ∆

(A)

ONa

(D)

COONa 2HCl–2NaCl

OH

(E)

COOH (A) C6H5OHPOCl3/∆; – H2O

OH

(F)

COOC6H5

MATHEMATICS

11. Sixteen players S1, S2, ... , S16 play in a tournament. They are divided into eight pairs at random from each pair a winner is decided on the basis of a game played between the two players of the pair. Assume that all the players are of equal strength.

(a) Find the probability that the players S1 is among the eight winners.

(b) Find the probability that exactly one of the two players S1 and S2 is among the eight winners.

[IIT-1997] Sol. (a) Probability of S1 to be among the eight winners

= (Probability of S1 being a pair) × (Probability of S1 winning in the group).

= 1 × 21 =

21 Q S1 is definitely in a

group (b) If S1 and S2 are in the same pair then exactly one

wins.

If S1 and S2 are in two pairs separately then exactly one of S1and S2 will be among the eight winners. If S1 wins and S2 loses or S1 loses and S2 wins. Now the probability of S1, S2 being in the same pair and one wins. = (Probability of S1, S2 being the same pair)

× (Probability of anyone winning in the pair). And the probability of S1, S2 being the same pair

= )()(

SnEn

Where n(E) = the number of ways in which 16 persons can be divided in 8 pairs.

∴ n(E) = !7.)!2(

)!14(7 and n(S) =

!8.)!2()!16(

8

∴ Probability of S1 and S2 being in the same pair

= )!16!.(7.)!2(!8.)!2)!.(14(

7

8 =

151

The probability of any one wining in the pairs of S1, S2 = P (certain event) = 1 ∴ The pairs of S1, S2 being in two pairs sparately

and S1 wins, S2 loses + The probability of S1, S2 being in two pairs separately and S1 loses, S2 wins.

=

!8.)!2()!16(

!7.)!2()!14(

–1

8

7×

21 ×

21 +

!8.)!2()!16(

!7.)!2()!14(

–1

8

7×

21 ×

21

= 21 ×

)!14(15)!14(14

×× =

157

∴ Required Probability = 151 +

157 =

158

12. Find the range of values of t for which

2sin t = 1–2–3

52–12

2

xxxx + , t ∈

ππ

2,

2– . [IIT-2005]

Sol. Here, 2 sin t = 1–2–3

52–12

2

xxxx + , t ∈

ππ

2,

2–

Put, 2sin t = y ⇒ – 2 ≤ y ≤ 2

∴ y = 1–2–3

52–12

2

xxxx +

⇒ (3y – 5)x2 – 2x(y – 1) – ( y + 1) = 0 Since x ∈ R – 1, – 1/3 as, 3x2 –2x – 1 ≠ 0 ∴ D ≥ 0 ⇒ 4(y – 1)2 + 4(3y – 5) (y +1) ≥ 0 ⇒ y2 – y – 1 ≥ 0

⇒ 2

21–

y –

45 ≥ 0


⇒

25–

21–y

+

25

21–y ≥ 0

⇒ y ≤ 2

5–1 or y ≥ 2

51+

or 2sin t ≤ 2

5–1 or 2sin t ≥ 2

51+

⇒ sin t ≤

π

10–sin or sin t ≥

π

103sin

⇒ t ≤ – 10π or t ≥

103π

Thus, Range for t ∈

ππ

10,–

2– ∪

ππ

2,

103

13. If f (x) is twice differentiable function such that f (a) = 0,

f (b) = 2, f (c) = 1, f (d) = 2, f (e) = 0, where a < b < c < d < e, then the minimum number of zero's of g(x) = f ' (x)2 + f '' (x). f (x) in the interval [a, e] is? [IIT-2006]

Sol. Let, g(x) = dxd [f (x) . f '(x)]

to get the zero of g(x) we take function h(x) = f (x). f ' (x)

between any two roots of h(x) there lies at least one root of h' (x) = 0

⇒ g(x) = 0 ⇒ h(x) = 0 ⇒ f (x) = 0 or f ' (x) = 0 If f (x) = 0 has 4 minimum solutions. f ' (x) = 0 has 3 minimum solutions. h(x) = 0 has 7 minimum solutions. ⇒ h' (x) = g(x) = 0 has 6 minimum solutions

14. The value of

∫

∫1

0

10150

1

0

10050

)–1(

)–1()5050(

dxx

dxx

[IIT-2006]

Sol. Let I1 = dxx∫1

0

10050 )–1( and I2 = ∫1

0

10150 )–1( dxx ,

using Integration by parts

I2 = 10

10150 ].)–1[( xx – (101) ∫1

0

4910050 )..50(–)–1( xxx dx

= 0 – ∫1

0

5010050 )(–)–1)(101)(50( dxxx

– I2 = 5050 ∫1

0

10050 )–1( x (– x50) dx

5050 I1 – I2 = 5050 ∫1

0

10050 )–1( x dx

+ 5050 ∫1

0

10050 )–1( x (– x50) dx

= 5050 ∫1

0

10150 )–1( x dx = 5050 I2

∴ 5050 I1 = I2 + 5050 I2

∴ 2

1)5050(I

I = 5051.

15. The position vectors of the vertices A, B and C of a

tetrahedron ABCD are î + ^j + ^k , î and 3 î ,

respectively. The altitude from vertex D to the opposite face ABC meets the median line through A of the triangle ABC at a point E. If the length of the side AD is 4 and the volume of the tetrahedron is

322 , find the position vector of the point E for all

its possible positions. [IIT-1996]

Sol. F is mid-point of BC i.e., F ≡ 23^^ ii+ = 2 î

and AE ⊥ DE (given)

A(i + j + k)D

C(3i)F(2i) B(i)

E1

λ

Let E divides AF in λ : 1. The position vector of E is given by

1

)(12 ^^^^

+λ+++λ kjii = ^

112 i

+λ+λ + ^

11 j+λ

+ ^

11 k+λ

Now, volume of the tetrahedron

= 31 (area of the base) (height)

⇒ 3

22 = 31 (area of the ∆ABC) (DE)

But area of the ∆ABC = |)(|21 →→

× BABC


= |)(2|21 ^^^

kji +× = |)|^^^^kiji ×+×

= |)–|^^jk = 2

Therefore, 3

22 = 31 )2( (DE)

⇒ DE = 2 Since ∆ADE is a right angle triangle, AD2 = AE2 + DE2 ⇒ (4)2 = AE2 + (2)2 ⇒ AE2 = 12

But →

AE =^

112 i

+λ+λ +

^

11 j+λ

+^

11 k+λ

– )(^^^kji ++

= ^

1i

+λλ –

^

1j

+λλ –

^

1k

+λλ

⇒ 2||→AE = 2)1(

1+λ

[λ2 + λ2 + λ2] = 2

2

)1(3+λλ

Therefore, 12 = 2

2

)1(3+λλ

⇒ 4(λ + 1)2 = λ2

⇒ 4λ2 + 4 + 8λ = λ2

⇒ 3λ2 + 8λ + 4 = 0

⇒ 3λ2 + 6λ + 2λ + 4 = 0

⇒ 3λ(λ + 2) + 2(λ + 2) = 0

⇒ (3λ + 2) (λ + 2) = 0

⇒ λ = – 2/3, λ = – 2

Therefore, when λ = – 2/3, position vector of E is

given by

^

112 i

+λ+λ +

^

11 j+λ

+ ^

11 k+λ

= ^

13/2–1)3/2.(–2 i

++ +

^

13/2–1 j

+ +

^

13/2–1 k

+

= ^

332–

13/4– i+

+ + ^

332–

1 j+

+ ^

332–

1 k+

= ^

3/13

34–

i

+

+ ^

3/11 j +

^

3/11 k

= –^i +

^3 j +

^3k

and when λ = – 2

Position vector of E is given by,

^

12–1)2(–2 i

++× +

^

12–1 j+

=^

12–1 k+

=^

1–14– i+ –

^j –

^k

= ^

3 i – ^j –

^k

Therefore, –^i +

^3 j +

^3k and +

^3 i –

^j –

^k are the

position vector of E.

MOTIVATION

• Pull the string, and it will follow wherever you wish. Push it, and it will go nowhere at all.

• Be the change that you want to see in the world.

• Efficiency is doing things right; effectiveness is doing the right things.

• Formula for success: under promise and over deliver.

• A life spent making mistakes is not only more honorable, but more useful than a life spent doing nothing.

• Discovery consists of seeing what everybody has seen and thinking what nobody else has thought.

• The best way to teach people is by telling a story.

• If you'll not settle for anything less than your best, you will be amazed at what you can accomplish in your lives.

• I had to pick myself up and get on with it, do it all over again, only even better this time.

• Improvement begins with I.

• Success depends upon previous preparation, and without such preparation there is sure to be failure.

• The man of virtue makes the difficulty to be overcome his first business, and success only a subsequent consideration.

• As a general rule the most successful man in life is the man who has the best information.

• The secret of success is constancy to purpose.

• One secret of success in life is for a man to be ready for his opportunity when it comes.


1. Four infinite thin current carrying sheets are placed in

YZ plane. The 2D view of the arrangement is as shown in fig. Direction of current has also been shown in the figure. The linear current density. i.e. current per unit width in the four sheets are I, 2I, 3I and 4I, respectively.

x

x

x

x

x

x

x

x

Y I II III IV

X

aa a a The magnetic field as a function of x is best

represented by

2. Match the column Column – I Column – II

(A) a charge particle is (P) Velocity of the moving in uniform particle may be electric and magnetic constant fields in gravity free space

(B) a charge particle is (Q) Path of the particle moving in uniform may be straight line electric, magnetic and gravitational fields

(C) a charge particle is (R) Path of the particle moving in uniform may be circular magnetic and gravitational fields (where electric field is zero)

(D) A charge particle is (S) Path of the particle moving in only may be helical uniform electric field (T) None 3. Magnetic flux in a circular coil of resistance Ω10

changes with time as shown in fig. Cross indicates a direction perpendicular to paper inwards.

Match the following

Column – I Column – II (A) At 1s, induced current is (P) Clockwise (B) At 5s, induced current is (Q) Anticlockwise (C) At 9s, induced current is (R) Zero (D) At 15s, induced current is (S) 2A

(T) None

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

By : Dev Sharma Director Academics, Jodhpur Branch

Physics Challenging Problems

Solut ions wil l be published in next issue

Set # 9


4. A conducting rod of length l is moved at constant velocity v0 on two parallel, concudting, smooth, fixed rails, which are placed in a uniform constant magnetic field B perpendicular to the plane of the rails as shown in figure. A resistance R is connected between the two ends of the rail. Then which of the following is/are correct ?

(A) The thermal power dissipated in the resistor is

equal to the rate of work done by an external person pulling the rod

(B) If applied external force is doubled, then a part of the external power increases the velocity of the rod

(C) Lenz’s law is not satisfied if the rod is accelerated by an external force

(D) If resistance R is doubled, then power required to maintain the constant velocity v0 becomes half

5. The x-z plane separates two media A & B of refractive indices .2&5.1 21 =µ=µ A ray of light travels from A to B. Its directions in the two media

are given by unit vectors ,jbia1

∧∧→+=µ .jbic2

∧∧→+=µ

Then

(A) 34

ca

= (B) 43

ca

=

(C) 34

db

= (D) 43

db

=

6. Two converging lenses of the same focal length f are

separated by distance 2f. The axis of the second lens is inclined at angle º60=θ with respect to the axis of the first lens. A parallel paraxial beam of light is incident from left side of the lens. Then

(A) Final image after all possible refraction will

formed at optical centre of first lens (B) Final image after all possible refraction will

formed at optical centre of second lens (C) Final image after all possible refraction will

formed at distance f from second lens (C) Final image after all possible refraction will

formed at distance f from first lens

7. If Cv for an ideal gas is given by Cv = (3 + 2T)R, where T is absolute temperature of gas, then the equation of adiabatic process for this gas is

(A) VT2 = constant (B) VT3e-2T = C (C) VT2e2T = constant (D) VT3e2T = constant

8. The pressure of one mole of ideal gas varies according to the law 2

0 VPP α−= where P0 & α are positive constant constants. The highest temperature that gas may attain:

(A) 2/1

00

3P

R3P2

α

(B) 2/1

00

3P

R2P3

α

(C) 2/1

00

3P

RP

α

(D) 2/1

00 PRP

α

Physics Facts

1. Due to gravity, the maximum speed a raindrop during a rain with falling speed can hit you is about 18 miles per hour (29 kilometers per hour).

2. The speed of light in meters is 299,792,458 meters per second. And how on Earth are you going to remember that? The number can be remembered from the number of letters in each word of the following phrase: "We guarantee certainty, clearly referring to this light mnemonic."

(The speed of light in miles per second is 186,282.397051221, or in miles per hour, 670,616,629.384395).

3. In air, at a temperature of 32 degrees Fahrenheit/0 degrees Celsius (freezing point of water) the speed of sound travels 1,087 feet (331 meters) per second. (It travels faster at higher temperatures).

(In 64 degrees Fahrenheit [18 degrees Celsius] the speed of sound travels 1,123 feet [342 meters] per second).

4. If an object floats on water, it displaces the water equal to its mass, but if the object sinks, it displaces water equal to its volume.

5. A calorie is defined as the amount of energy needed to raise one gram of water one degree Celsius (or from 14.5 degrees Celsius to 15.5 degrees Celsius).

f


1. As shown in graph, the relation of U v/s PV is linear So, aPVU +φ= ).(tan as b=φ)(tan

So, U = b, PV + a Using ideal gas equation PV = nRT U = b (nRT) + a Differentitate it, dU = nbRdT As dU = n CV dT

So bfRfbRCV =⇒==22

Degrees of freedom of the gas, as b = 3 So f = 6 Degrees of freedom are 6 so it is triatomic non-linear

gas.

2. bf 22121 +=γ⇒+=γ as f = 2b

So, 11 −+=γ b

3. As CV = bR it is not dependent on ‘a’ so if a varies

there is no change in the value of CV

a

CV

4. As CV = b R and ,2

210 tCCbf

+==

210 22 tCCf += and tC

dtdf

14=

Df/dt v/s t graph is a straight line with slope 4C1.

5. As →v and

→B are mutually perpendicular so path will

be circular but due to presence of resistive medium speed decreases and radius of circular path decreases.

So, path is spiral of decreasing radius Option (D) is correct

6. If vab= 0, then irrespective to the value of capacitance

C energy stored will be zero.

vab =21

2211

/1/1)/(/

RRRR

+ε−+ε

= 0

So, 1

1

Rε –

1

1

Rε = 0 ⇒

1

1

Rε =

2

2

Rε

⇒ 2

1

RR =

2

1

εε

ε2

R

R1

R2

ε1

C

Option (A) is correct 7. To calculate time constant Replace voltage source by short circuit mean by zero

resistance and then find Req with C and time

constant τ = Req.C

τ =

+

+R

RRRR

21

21 C

R

R1

R2

C⇒

R2

R1R2/R1+R2

C

Option (D) is correct

8. Maximum current through the resistance Imax

=R

vab =R

RRRR 212211 /1/1/)/()/( +ε−+ε =Reqε

Option (C) is correct.

Solution Physics Challenging Problems

Set # 8

8 Questions were Published in December Issue


1. A solid uniform sphere of radius r rolls without sliding along the inner surface of a fixed spherical shell of radius R and performs small oscillations. Calculate period of these oscillations.

Sol. Let the solid sphere be rotated through a very small angle θ, then displacement of its centre is equal to AB = r.θ. During this rotation point of contact changes from C to C' and inclination of radius OC' of the shell with vertical becomes equal to

φ = )–( rRradius

ABarc as shown in Figure.

Bφ

C'A

C

O

∴ φ = )–( rR

rθ ...(i)

When the solid sphere is released from this position, it starts to roll down towards equilibrium position C. Let its angular acceleration be α. Then acceleration of its centre will be equal to rα.

Consider free body diagram of the sphere Figure. mg cos φ

Iα

mg sin φ

Friction C'

N

since, C ' is instantaneous axis of rotation, therefore,

taking moments about it, (mg sin φ) r = Iα

or mgr . sinφ =

2

57 mr α

But φ is very small, therefore, sin φ ≈ φ

∴ mgr . φ = 57 mr2α

or α = 57

rg φ

Substituting value of φ from equation (1),

Angular acceleration, α = )–(7

5rR

g θ

Since, angular acceleration is restoring and is directly proportional to angular displacement θ, therefore, the sphere performs SHM. Hence, its period of oscillation is given by

T = onacceleratiangularntdisplacemeangular2π

or T = g

rR5

)–(72π Ans.

2. Three identically charged, small spheres each of mass

m are suspended from a common point by insulated light strings each of length l. The spheres are always on vertices of an equilateral triangle of length of the sides x (<< l). Calculate the rate dq/dt with which charge on each sphere increases if length of the sides of the equilateral triangle increases slowly according

to law dtdx =

xa .

Sol. Since dx/dt is given as x

a , therefore, to calculate

dtdq , first q is to be calculated in terms of x.

Since, side of equilateral triangle is x, therefore, electrostatic force between two particles is

F ' = 04

1πε 2

2

xq

Positions of particles A, B & C with respect to each other are shown in Figure (A).

F

F '

F 'A

x

x

G

C

60º

60º

B

Fig.(A)

Resultant electrostatic force on particle A is F = 2F' cos 30º.

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

Students' ForumPHYSICS


or F = 20

2

43x

qπε

In Figure (A) G is centroid of the triangle. Distance of centroid G from every particle is

r = 32 x cos 30º =

3x

Since, G is centroid, therefore, it always lies vertically below the point of suspension. Figure (B) shows a vertical plane passing through particle A and centroid G. All of the forces acting on particle A are shown in this figure.

G r

T

x

mg

F

Fig.(B)

θ

Considering horizontal forces on the particle A, T sin θ = F ...(i) For vertical forces, T cos θ = mg ...(ii) Dividing equation (i) by (ii)

tan θ = mgF ...(iii)

Since, x < < l, therefore, inclination θ of each thread with the vertical is very small.

Hence, tan θ ≈ sin θ = lr =

3lx

∴ From equation (iii) 3l

x = 20

2

43

mgxq

πε

or q = lmg

34 0πε

x3/2

∴ dtdq =

xtdxx

lmg

23

34 0πε

or dtqd =

lmga2

03πε Ans.

3. A coil of radius R carries current I. Another

concentric coil of radius r(r < < R) carries current i. Planes of two coils are mutually perpendicular and both the coils are free to rotate about common diameter. Find maximum kinetic energy of smaller coil when both the coils are released, masses of coils are M and m respectively.

Sol. If a magnetic dipole having moment M be rotated through angle 'θ' from equilibrium position in a uniform magnetic field B, work done on it is W = MB (1 – cos θ). This work is stored in the system in the form of energy. When system is released, dipole starts to rotate to occupy equilibrium position and the energy converts into kinetic energy and kinetic energy of the system is maximum when stored energy is completely released.

Magnetic induction, at centres due to current in larger

coil is B = RI

20µ

Magnetic dipole moment of smaller coil is iπr2. Initially planes of two coils are mutually

perpendicular, therefore θ is 90º or energy of the system is U = (iπr2) B (1 – cos 90º)

or U = R

rIi2

20 πµ

When coils are released, both the coils start to rotate about their common diameter and their kinetic energies are maximum when they become coplaner.

Moment of Inertia of larger coil about axis of rotation

is I1 = 21 MR2 and that of smaller coil is I2 =

21 mr2

Since, two coils rotate due to their mutual interaction only, therefore, if one coil rotates clockwise then the other rotates anticlockwise.

Let angular velocities of larger and smaller coils be numerically equal to ω1 and ω2 respectively when they become coplaner,

According to law of conservation of angular momentum,

I1ω1 = I2ω2 ...(i) and according to law of conservation of energy,

2112

1ωI + 2

2221

ωI = U ...(ii)

From above equations, maximum kinetic energy of smaller coil,

2222

1ωI =

21

1

IIUI+

= )(2 22

20

mrMRMRrIi+

πµ Ans.

4. RMS velocity of molecules of a di-atomic gas is to be

increased to 1.5 times. Calculate ratio of initial volume to final volume, if it is done

(i) Adiabatically ; (ii) Isobarically ; (iii) Calculate, also ratio of work done by gas during

these processes.


Sol. R.M.S. velocity of gas molecules ∝ T . where T is absolute temperature of the gas.

Therefore, to increase RMS velocity to 1.5 times, the temperature is to be increased to (1.5)2 = 2.25 times.

Let initial volume and temperature by V0 and T0 respectively and let final volume and temperature be V and T respectively, then T = 2.25. T0

(i) For adiabatic process, T.V(γ – 1) = constant

∴ )1–(

0γ

VV

= 0T

T = 2.25

But for diatomic gas γ = 1.4

VV0 = (2.25)2.5 = (1.5)5 = 7.594 Ans.(i)

Work done by gas during the process,

W1 = )1–(

–00

γPVVP

= )1–(

)T–T(nR 0

γ

(ii) For isobaric process, TV = constant

∴ VV0 =

TT0 =

25.21 =

94 Ans.(ii)

Work done by gas during the process, W2 = nR (T–T0)

(iii) Ratio of work done by gas during these processes is

2

1

WW =

)–1(1

γ = – 2.5 Ans.(iii)

5. Suppose potential energy between electron and

proton at separation r is given by U = K log r, where K is a constant. For such a hypothetical hydrogen atom, calculate radius of nth Bohr's orbit and energy levels.

Sol. If potential energy of a system of two particles, separated by a distance r is U, interaction force

between the particles is F = – drdU .

Since, potential energy of system of electron and proton in a hypothetical hydrogen atom is given by U = K log r, therefore, force between them,

F = – drdU = –

rK

Negative sign indicates force of attraction. This force of attraction between electron and proton

provides centripetal force required for circular motion of electron.

∴ r

mv2 =

rK where m is mass of electron and v is

its speed. or mv2 = K ...(i) According to Bohr's postulates, angular momentum,

mur = nπ2h ...(ii)

From equation (i) and (ii),

r = mK

nhπ2

Ans.

total energy of the atom is E = potential energy U +

kinetic energy

2

21 mv of electron

∴ E = K log r + 21 K =

2K (2 log r + 1)

= 2K (1 + log r2)

Substituting value of r,

E =

π+

mKhnK

2

22

2log1

2 Ans.

Birth of New Red Spot is the Thunderstorm on Jupiter

During the past few months, the astronomers have tracked an emerging second red spot on Jupiter, a growing rival about one-half the diameter of the planet's Great Red Spot. The Hubble Space Telescope has snapped the first detailed pictures of what some observers now call Red Spot Jr. Astronomers at the Space Telescope Science Institute in Boltimore said this was the first time scientists have witnessed the birth of these huge oval spots, presumably a convective phenomenon like a powerful thunderstorm. The Great spot was already present when the observers first looked with telescope at the planet some 400 years ago. Red Spot Jr. appeared in near-infrared images to be as bright in Jupiter's cloudy atmosphere as its big companion. The size of Red Spot Jr. is half the size of its big companion. The scientists say the new storm might rise higher above the main cloud deck than the older spot.

In a New Light: Jupiter, with its second red spot, in a picture released by NASA. Current observations, including Hubble pictures taken on May 12 and 18, show that the smaller red spot is drifting eastward in the Jovian Southern hemisphere and the Great Red Spot is moving westward. They should pass one another in early July. The pictures of the Red Spots are contrast-enhanced images taken in visible light and at near-infrared wave lengths. But the red spots, new and old, are really red.


Prism :

(i) Deviation 'δ' produced by the prism,

B C

A

Normal Normal

i i' Q P r r'

δ

δ = i + i' – A and A = r + r' (ii) For minimum deviation 'δm'

i = i' and r = r' and also PQ||BC and the refractive index for the material of prism is given by

µ =

δ+

2sin

2sin

A

A m

(iii) δ – i graph for prism

δm

i

δ

(iv) For not transmitting the ray from prism,

µ > cosec

2A

(v) For grazing incidence i = 90º and for grazing emergence i' = 90º. For maximum deviation i = 90º or i' = 90º

(vi) The limiting angle of prism = 2C when i = i' = 90º If the angle of prism A > 2C, then the rays are totally

reflected.

(vii) Right-angled prism : These prisms are used to turn a light beam to 90º or 180º. These are usually made of

crown glass for which µg = 1.5 and C =tan–1

gµ1 = 42º.

Such prisms are used in binoculars and submarine periscopes.

(viii) Deviation produced by a thin prism δ = (µ – 1)A (ix) Angular dispersion D = δv – δR = (µV – µR)A Where V and R stand for violet and red colours

respectively. Mean deviation δY = (µY – 1)A where µY is the refractive index of mean yellow

colour.

(x) Dispersive Power, ω =deviationMeandispersionAngular

=Y

RV

δδ−δ

ω =1−µ

µ−µ

Y

RV where µY =2

RV µ+µ

(xi) Pair of prisms (or crossed prism) : Two thin prisms of different material when placed crossed, i.e., with their refracting edges parallel and pointing in opposite directions as shown in figure, produce a total deviation δ given by

δ = δ1~ δ2 where δ1 and δ2 are the mean deviations produced by

the first and second prism respectively. Total angular dispersion D = D1 ~ D2 where D1 and D2 are the angular dispersions

produced by respective prisms. (xii) Dispersion without deviation : If the angle of two

prisms A and A' are so adjusted that the deviation produced by the mean ray by the first prism is equal and opposite to that produced by the second prism, then the total final beam will be parallel to the incident beam and there will be dispersion without deviation.

Prism & Wave Nature of Light

PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


A

Crown glass

Flint glass

A'

Here, δ = δ1 – δ2 = 0 or δ1 = δ2 i.e., (µ1 – 1)A = (µ2 – 1)A' This combination produces total angular dispersion. D = D1 – D2 = (µ1V – µ1R) A – (µ2V – µ2R)A' (xiii) Deviation without dispersion : If the combination is such that D = D1 ~ D2 = 0 or D1 = D2 or (µ1V – µ1R) A = (µ2V – µ2R)A' The combination is said to be achromatic and the

total mean deviation will be δ = δ1 ~ δ2 = (µ1 – 1)A ~ (µ2 – 1)A' Wave Nature of light Wave front : A point source produces a spherical wave front

∝

r1A or

∝ 2r

11

Where A = Amplitude, I = intensity and r = distance of point of observation from source. A line source produces a cylindrical form

∝

rA 1 or

∝

rI 1 .

Wave front is locus of points in the same phase. A distance source produce a plane wave front. Wave front for a parallel beam of light is plane. The angle between ray and wave front is 90º Huygen's principle: Huygen's principle is a geometrical method to find

secondary wave front produced by a primary wave front.

Thin lines shows the rays of light. Dotted line shows the wavefronts.

Interference of light : (a) Redistribution of light energy i.e. alternate

maximum and minima). Conditions for two light waves producing

interference is that (i) Wave should be of same wavelength/frequency. (ii) Waves should be travelling in the same direction. (iii)Wave should have a constant phase difference For the above conditions the two source must be

coherent and that is possible when we make two sources out of a single source of light.

For monochromatic light we get alternate maxima and minima of same colour. For white light we get white central fringe flanked by coloured fringes because fringe width of different colour is different due to different wavelengths.

(b) Resultant intensity at a point is

I = I1 + I2 + 2 21 II cos φ

When I1 = I2 = I0 then I = 4I0 cos2 φ/2 For constructive interference φ = ± 2nπ and ∆x = ±nλ

Imax = ( 1I + 2I )2 ∝ (A1 + A2)2 [Q I ∝ A2]

For destructive interference

φ = (2n + 1)π and ∆x =

−

21n λ

Imin = ( 1I – 2I )2 ∝ (A1 – A2)2

⇒ min

max

II =

)(

)(

21

21

II

II

−

+ = 2

21

221

)()(

AAAA

−+

The energy remains conserved during the process of interference.

P

α S1

S2

Intensity of light at any point P as shown the

figure I = I0cos2

λαπ tand

(c) The fringe width β = dDλ

Angular width θ = Dβ =

dλ

⇒ θ does not depend on D


(d) When the source of light is placed asymmetrical with respect to the slits then the central maxima also shifts.

α

S2

x y

Dx Dy

θ

S1

S

yD

y = xD

x and θ = – α

(e) If young's double slit experiment is done in a liquid of refractive index medium µ then the fringe width β´ = β/µ

(f)

PS1

S2

t

O´ O

If a transparent sheet of thickness t is placed in

front of upper slit then the central maxima shift upside. The new optical path becomes µt instead of t and the increase in optical path is (µ – 1)t.

The shift = dD (µ – 1) t =

λβ (µ – 1)t

(g) Interference in thin films :

r r r r

21i

t

Transmitted rays For reflected rays interference

Maxima 2 µt cos r = (2n – 1)2λ

Minima 2 µt cos r = nλ

Diffraction : Bending of light through an aperture / corner when

the dimension of aperture is comparable to the wavelength of light is called diffraction.

Fraunhoffer diffraction at a single slit Condition for minima : a sin θn = n λ Condition of secondary maxima :

a sin θn = 22

1 λ

+n Where n = n = 1, 2 ...

Width of central maxima = 2λD/a P

Oθ a

Angular width of central maxima = 2λ/a Angular width of secondary maxima = λ/a

Intensity at any point P = I0

2sin

αα

where α = λπ (a sin θ)

The ratio of intensities of secondary maxima are

221 ,

611 ,

1211 , ...

For a path difference of λ, the phase difference is 2π radian.

Polarisation :

I0

Unpolarisedlight

Polarised light

I cos2θ I = I0/2

rip Medium 1

Medium 2 ½ µ

ip = Angle of polarization, ip + r = 90º, µ = tan ip


Solved Examples

1. (i) A ray of light incident normally on one of the faces of a right-angled isosceles prism is found to be totally reflected. What is the minimum value of the refractive index of the material of prism ?

(ii) When the prism is immersed in water, trace the path of the emergent ray for the same incident ray indicating the values of all the angles (µω = 4/3)

Sol. (i) According to the problem, ∠A = 90º, ∠B = ∠C = 45º. At face BC, incident ray PQ is totally reflected

therefore i ≥ C fig. P

BA

R r=i N

i 45º

Q

N´

C Here i = 45º, ∴ Cmax = 45º; ∴ µ = 1/sin C

or µmin = (1/sin Cmax) = (1/sin 45º) = 2 = 1.414 (ii) When the prism is immersed in water, then for

normal incident ray, the ray passes undeviated up to PQ and becomes incident at face BC at angle of incidence 45º(fig.) The ray travels from glass to water, therefore from Snell's law,

rsinisin =

1

2

µµ , we have

rsinº45sin =

g

w

µµ

= gµw

∴ sin r = wgµ

º45sin = wµ

º45sin × µg

= 2

1 ×3/4

414.1 = 43 = 0.75

∴ r = sin–1(0.75) = 48º36´ P

BA

R

90º45º

45º

Q r

C

45ºR

The path of light ray is shown in fig.

2. The cross-section of a glass prism has the form of an isosceles triangle. One of the equal faces is coated with silver. A ray is normally incident on another unsilvered face and being reflected twice emerges through the base of the prism perpendicular to it. Find the angles of the prism.

Sol. Suppose refracting angle of prism be α and other two base angles of the isosceles prism be β. The light ray PQ, incident normally on the face AB, is refracted undeviated along QR. The refracted ray QR strikes the silvered face AC and gets reflected from it. The reflected ray RS now strikes the face AB from where it is again reflected along ST and emerges perpendicular to base BC.

A

B C

P

R S

Q

90º

N´2

N2

N´1i = α

ββ

θ=β

N1

T

i

α

It follows from fig. that angle of incidence on face

AC = i = α and also angle of incidence of face AB = θ = β As N2 N2´ is parallel to PR, hence θ = 2i

i.e., β = 2α Also α + 2β = 180º or α + 2(2α) = 180º or α = 36º so β = 2α = 72º

3. Two coherent light sources A and B with separation 2λ are placed on the x-axis symmetrically about the origin. They emit light of wavelength λ. Obtain the positions of maximum on a circle of large radius, lying in the xy-plane and with centre at the origin.

Sol. Distance between two coherent light sources = 2λ. Consider the interference of waves at some point C of

the circumference of circle.

BC = )cosr2r( 22 θλ−λ+

AC = )cosr2r( 22 θλ+λ+

∴ Path difference = AC – BC = λ (For Maxima) It is clear from figure, that Path difference = AC – BC = AP + OM = 2λ cos θ

Y

X

C

M r P

θ θ OA B

λ λ

∴ 2λ cos θ = λ or cos θ = 1/2 ∴ Possible values of angle θ = 60º, 120º, 180º, 240º,

300º, 360º.


4. Two point coherent sources are on a straight line d = nλ apart. The distance of a screen perpendicular to the line of the sources is D >> d from the nearest source. Calculate the distance of the point on the screen where the first bright fringe is formed.

Sol. Consider any point P on the screen at a distance x from O. Then

dO

D S2 S1

21D = D2 + x2 or D1 = D

2/1

2

21

+

Dx = D

+ 2

2

21

Dx ;

∴ D1 = D + D

x2

2

Similarly, D2 = (D + d) +)(2

2

dDx

+

∴ D2 – D1 (D + d) +)(2

2

dDx

+– D –

Dx2

2

= d +2

2x

−

+ DdD11 = d – d

)(2

2

dDDx

+

O D

D2

S1 S2

D1 x

P

d For the point O, D2 – D1 = d = nλ (given). Thus there is brightness at O of nth order. Since the

path difference decreases, the other fringes will be of lower order. The next bright fringe will be of (n – 1)th order. Hence for the next bright fringe

D2 – D2 = (n – 1)λ

d – d)(2

2

dDDx

+= (n – 1)λ

nλ – nλ)(2

2

λ+ nDDx = (n – 1)λ

∴ x =n

nDD )(2 λ+

5. One slit of a Young's experiment is covered by a glass plate (n = 1.4) and the other by another glass plate (n' = 1.7) of the same thickness t. The point of central maximum on the screen, before the plates were introduced, is now occupied by the previous fifth bright fringe. Find the thickness of the plates (λ = 4800 Å)

Sol. Path of the wave from slit S1 = D1 + n't – t Path of the wave from slit S2 = D2 + nt – t ∴ Path difference = D2 + nt – t – D1 – n't + t = (D2 – D1) + (n – n')t

O′

D

D2

S1

S2

D1

O d

x

But D2 – D1 = Dxd

∴ Path difference =Dxd + (n – n')t

Let O' be the point where paths difference is zero.

∴ Dxd = (n' – n)t

or, x =dD (n' – n) t =

λβ− tnn )'(

λ

=βdD

Q

Given that x = 5β ∴ 5β = λ

β− tnn )'(

or, t =n'n

5−λ or, t =

4.17.11048005 10

××× −

= 8 µm.

WATER CAN CRACK METAL

Did you know that water can expand in volume by around 1/9th of its total composition. This means that when frozen it occupies more space than when in liquid form.

The above phenomena is very common in houses where water pipes burst over the winter due to ice forming in the pipes. The pipes which were once wide enough for water to go through are now too small when the ice forms.

The brittle nature of the metal when it feezes combined with the ice expanding causes them to crack. In fact the enormous power of frozen water is also used in quarries to loosen up huge blocks of rocks.


Key Concepts : 1. Equation of a harmonic wave is y = a sin(kx ± ωt ± φ). Here y is measure of disturbance from zero level.

y may represent as electric field, magnetic field, pressure etc. Also K = 2π/λ = wave number.

Note : The positive sign between kx and ωt shows that the wave propagates is the + x direction. If the wave travels in the –X direction then negative sign is used between kx and ωt.

2. Particle Velocity :

v = dtdy = aω cos (kx ± ωt ± φ)

∴ Maximum particle velocity = aω = velocity amplitude Particle velocity is different from wave velocity.

The wave velocity v = vλ. 3. Particle acceleration :

A = dtdv = 2

2

dtyd = – aω2 cos(kx + ωt ± φ) = – ω2y

Max acceleration = acceleration amplitude = –ω2a

4. Velocity of transverse wave on a string = mT

Where m = mass per unit length = ρ × 4

2Dπ

Where ρ = density of the wire material and D = diameter of wire More the tension, more is the velocity 5. A wave, after reflection from a free end, suffers no

change of π. A wave, after reflection from a free end, suffers no

change in the phase.

6. Velocity of sound in a fluid = ρB

For air B = γP ∴ v = ργP =

MRTγ

Velocity of sound in general follows the order Vsolid > Vliquid > Vgas

⇒ Velocity of sound ∝ T

Also velocity of sound ∝ M/γ

and velocity of sound ∝ ρ/1 .

But velocity of sound does not depend on pressure because P/ρ becomes constant.

Velocity of sound depend on the frame of reference. 7. (a) According to principle of superposition

→y =

→1y +

→2y + ….

(b) Interference of waves y1 = A1 sin(kx – ωt) y2 = A2 sin(kx – ωt + φ) For constructive interference φ = 2nπ n = 0, 1, 2, ……

(i) Imax = ( 1I + 2I )2

(waves should be in same phase) (ii) Amax = A1 + A2 For destructive interference φ = (2n + 1)π n = 0, 1, 2, ……

(i) Imin = ( 1I – 2I )2

(waves should be in opposite phase) (ii) Amin = A1 – A2

(c) I = I1 + I2 + 2 21II cos φ

Where φ is the phase difference between the two waves.

8. Beats : When two waves of same amplitude with slight difference in frequency (<10), traveling in the same direction superpose, beats are produced.

The equation for beats is

y =

ω−ω

2)cos(2 21 tA sin

ω−ω

221 t

Where amplitude at a given location

= 2Acos

ω−ω

221 t

The above expression shows that amplitude change with time.

Beat frequency = no of maxima / minima per second = v1 – v2

Waves & Doppler Effect

PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


9. Standing waves (stationary) When two waves of same amplitude and frequency

moving in opposite direction superimpose, standing waves are produced.

Nodes are the point where the displacement is always zero.

The amplitudes of different particles different and is maximum at antinodes.

The equation of standing waves is The above expression shows that the amplitude is

different for different values of x and varies sinusoidally.

For a node to occur at position x, y = 0 ⇒ kx = 0 For an antinode two occur at position x, y should

be max ⇒ kx = π/2 , …. In terms of pressure ∆P = ∆P0 cos kx cos ωt. 10. For standing waves on strings (and both end open

organ pipe)

Fundamental frequency v0 = 0λ

v = l2

v

First mode of vibration v1 = 1λ

v = 2

l2v

= 2v0 = 2nd harmonic

nth mode of vibration vn = n

l2v = nv0

= nth harmonic where v = mT for string.

Also more the tension in the same string, higher is the value of v0

11. For closed organ pipe :

Fundamental frequency v0 = 0λ

v = l4

v

First mode of vibration v1 = 1λ

v = 3

l4v = 3v0

= Third harmonic

nth mode of vibration vn = (2n + 1)l4

v

where n = 1, 2, ….. In case where end correction is taken replace l by

(l + e) 12. (a) Intensity of sound at a distance r from a point

source is I = 24 rPπ

where P = power of source.

(b) For a line source I = lr

Pπ

where l is the length of source

(c) I = 21

ρv(4π2v2)A2 = ρν2

)amplitudeessure(Pr 2

13. Doppler's effect :

v = v0

±±

s

L

vvvv vL = velocity of listener

The above formula is valid when vs < v Replace v by (v ± vm) if it is given that the

medium also moving. When listener and source are not moving along

the line joining the two, then the component of velocity along the line joining the two are taken as velocity of listener or source.

14. If the source and listener are on the same vehicle and the sound is reflected from a stationary object towards which the vehicle is approaching then the frequency of sound as heard by the observer is

v´ = v0

++

s

L

vvvv

15. For a path difference of λ, the phase difference is 2π for harmonic waves.

16. For a transverse wave the energy per unit length possessed by a string is given as

ld

dE = m(4π2f2)A2cos2 (kx – ωt)

17. Equation for a wave pulse is y = f(x + vt) 18. When a wave on reaching on interface is partly

reflected and partly transmitted then for no power loss.

Pi = Pt + Pr where Pi = Power of incident wave Pt = Power of transmitted wave Pr = Power of reflected wave.

Also in this case Ar =

+−

12

12

vvvv Ai; At =

+ 21

22vv

v Ai

Where Ai, Ar and At are amplitudes of incident reflected and transmitted waves v1 is the velocity in the medium of incidence and v2 is the velocity in the medium where transmitted wave is present.

Problem Solving Strategy : Mechanical Waves Identify the relevant concepts : Wave problems fall

into two broad categories. Kinematics problems are concerned with describing wave motion; they involve wave speed v, wave length λ(or wave number k), frequency f (or angular frequency ω), and amplitude A. They may also involve the position, velocity, and acceleration of individual particles in the medium. Dynamics problems also use concepts from Newton's laws such as force and mass. In this chapter we'll encounter problems that involve the relation of wave speed to the mechanical properties of the wave medium. We'll get into these relations.

As always, make sure that you identify the target variable(s) for the problem. In some cases it will be the wavelength, frequency, or wave speed; in other


cases you'll be asked to find an expression for the wave function.

Set up the problem using the following steps : Make a list of the quantities whose value are

given. To help you visualize the situation, you'll find it useful to sketch graphs of y versus x (fig. a) and of y versus (fig. b). Label your graphs with the values of the known quantities.

y

Wave displacementversus coordinate x

at time t = 0

Wavelength λ

A

A

(a)

y Wave displacement

versus time t at coordinate x = 0

Period T

A

A

(b)

t t

Decide which equations you'll need to use. If any

two of v, f, and λ are given, you'll need to use eq. v = λf (periodic wave) to find the third quantity. If the problem involves the angular frequency ω and / or the wave number k, you'll need to use the definitions of those quantities and eq. (ω = vk). You may also need the various forms of the wave function given in Eqs.

y(x, t) = A cos

−ω t

vx = A cos 2πf

− t

vx ,

y(x, t) = A cos 2π

−

λ Ttx

and y(x, t) = A cos (kx – ωt). If the wave speed is not given, and you don't have

enough information to determine it using v = λf, you may be able to find v using the relationship between v and the mechanical properties of the system.

Execute the solution as follows : Solve for the unknown quantities using the equations you've selected. In some problems all you need to do is find the value of one of the wave variables.

If you're asked to determine the wave function, you need to know A and any two of v, λ and f(or v, k and ω). Once you have this information, you can use it in eq. (ω = vk). You may also need the various forms of the wave function given in Eqs.

y(x, t) = A cos

−ω t

vx = A cos 2πf

− t

vx ,

y(x, t) = A cos 2π

−

λ Ttx and y(x, t) = A cos (kx – ωt)

to get the specific wave function for the problem at hand. Once you have that, you can find the value of y at any point (value of x) and at any time by substituting into the wave function.

Evaluate your answer : Look at your results with a critical eye. Check to see whether the values of v, f, and λ (or v, ω, and k) agree with the relationships given in eq. . v = λf or w = vk. If you've calculated the wave function, check one or more special cases for which you can guess what the results ought to be.

Problem Solving Strategy : Standing waves Identify the relevant concepts : As with traveling

waves, it's useful to distinguish between the purely kinematic quantities, such as wave speed v, wavelength λ, and frequency f, and the dynamic quantities involving the properties of the medium, such as F and µ for transverse waves on a string. Once you decide what the target variable is, try to determine whether the problem is only kinematic in nature or whether the properties of the medium are also involved.

Set up the problem using the following steps : In visualizing nodes and antinodes in standing

waves, it is always helpful to draw diagrams. For a string you can draw the shape at one instant and label the nodes N and antinodes A. The distance between two adjacent nodes or two adjacent antinodes is always λ/2, and the distance between a node and the adjacent antinode is always λ/4.

Decide which equation you'll need to use. The wave function for the standing wave is almost always useful ex. y(x, t) = (ASW sin kx) sin ωt.

You can compute the wave speed if you know either λ and f (or, equivalently, k = 2π/λ and ω = 2πf) or the properties of the medium (for a string. F and µ.)

Execute the solution as follows: Solve for the unknown quantities using the equations you've selected. Once you have the wave function, you can find the value of the displacement y at any point in the wave medium (value of x) and at any time. You can find the velocity of a particle in the wave medium by taking the partial derivative of y with respect to time. To find the acceleration of such a particle, take the second partial derivative of y with respect to time.

Evaluate your answer : Compare your numerical answers with your diagram. Check that the wave function is compatible with the boundary conditions (for example, the displacement should be zero at a fixed end).

Problem Solving Strategy : Sound Intensity Identify the relevant concepts : The relationships

between intensity and amplitude of a sound wave are rather straightforward. Quite a few other quantities are involved in these relationships, however, so it's particularly important to deciede which is your target variable.

Set up the problem using the following steps :


Sort the various physical quantities into categories. The amplitude is described by A or pmax, and the frequency f can be determined from ω, k, or λ. These quantites are related through the wave speed v, which in turn is determined by the properties of the medium: B and ρ for a liquid; γ, T, and M for a gas.

Determine which quantities are given and which are the unknown target variables. Then start looking for relationships that take you where you want to go.

Execute the solution as follows: Use the equations you've selected to solve for the target variables. Be certain that all of the quantities are expressed in the correct units. In particular, if temperature is used to calculate the speed of sound in a gas, make sure that it is expressed in Kelvins (Celsius temperature plus 273.15).

Evaluate your answer: There are multiple relationships among the quantities that describe a wave. Try using an alternative one to check your results.

Problem Solving Strategy : Doppler Effect Identify the relevant concepts : The Doppler effect is

relevant whenever the source of waves, the wave detector (listener), or both are in motion.

Set up the problem using the following steps : Establish a coordinate system. Define the positive

direction to be the direction from the listener to the source, and make sure you know the signs of all relevant velocities. A velocity in the direction from the listener toward the source is positive; a velocity in the opposite direction is negative. Also, the velocities must all be measured relative to the air in which the sound is traveling.

Use consistent notation to identify the various quantities: subscript S for source, L for listener.

Determine which unknown quantities are your target variables.

Execute the solutions :

Use eq. fL = S

L

vvvv

++ fS to relate the frequencies at

the source and the listener, the sound speed, and the velocities of the source and the listener. If the source is moving, you can find the wavelength measured by the listener using Eq.

λ = Sf

v – S

S

fv =

S

S

fvv − or λ =

S

S

fvv + .

When a wave is reflected from a surface, either stationary or moving, the analysis can be carried out in two steps. In the first, the surface plays the role of listener; the frequency with which the wave crests arrive at the surface is fL. Then think of the surface as a new source, emitting waves

with this same frequency fL. Finally, determine what frequency is heard by a listener detecting this new wave.

Evaluate your answer: Ask whether your final result makes sense. If the source and the listener are moving towards each other, fL > FS; if they are moving apart, fL < fS. If the source and the listener have no relative motion, fL = fS.

1. A stationary wave is given by

y = 5 sin 3xπ cos 40 πt

where x and y are in cm and t is in seconds. (a) What are the amplitude and velocity of the

component waves whose superposition can give rise to this vibration ?

(b) What is the distance between the nodes ? (c) What is the velocity of a particle of the string at

the position x = 1.5 cm when t = 9/8 s ? Sol. Using the relation 2 sin C cos D = sin (C + D) +

sin(C – D)

y = 5 sin 3xπ cos 40 πt =

25 × 2 sin

3xπ cos 40πt

⇒ y =

π−

π+

π+

π txtx 403

sin403

sin25

= 25 .sin

π

+π3

40 xt – 25 sin

π

−π3

40 xt

= 25 .sin

π

+π3

40 xt +25 sin

π+

π−π

340 xt

Thus, the given stationary wave is formed by the superposition of the progressive waves

y1 = 25 sin

π

+π3

40 xt and y2 = 25 sin

π+

π−π

3xt40

Comparing each wave with the standard form of the progressive wave

y = a sin

α+

λπ

−ω2t ; a =

25 = 2.5 cm

ω = 40π or n = 20

and λπ2 =

32π or λ = 6 cm = 0.06 m

∴ c = nλ = 20 × 0.06 = 1.2 ms–1

Distance between the nodes = 2λ =

206.0 = 0.03 m

Q y = 5 sin3xπ cos 40 πt

v = dtdy = – 5 × 40π sin

3xπ sin 40 πt

⇒ v = – 200 π sin3xπ sin 40πt

Solved Examples


∴ At x = 1.5 cm and t = 89 s

v = – 200π sin 45π = 0 2. An engine blowing a whistle of frequency 133 Hz

moves with a velocity of 60 m s–1 towards a hill from which an echo is heard. Calculate the frequency of the echo heard by the driver. (Velocity of sound in air = 340 ms–1.)

Sol. The 'image' of the source approaches the driver at the same speed. Here, the image or echo is the source.

∴ vs = + 60 ms–1, v0 = – 60 ms–1

n´ = svc

vc−− 0 × n

∴ n´ = 60340

)60(340−−− × 133 = 190 Hz

3. A source of sound of frequency 1000 Hz moves to

the right with a speed of 32 ms–1 relative to the ground. To its right is reflecting surface moving to the left with a speed of 64 ms–1 relative to the ground. Take the speed of sound in air to be 332 m s–1 and find

(a) the wavelength of the sound emitted in air by the source

(b) the number of waves per second arriving at the reflecting surface

(c) the speed of the reflected waves, and (d) the wavelength of the reflected waves Sol. (a) Due to the motion of the source, the wavelength

(and hence, the frequency) is actually changed from λ to λ´ such that if n = actual frequency

λ´ = nvc S− =

100032332 − = 0.3 m

(b) The number of waves arriving at the reflecting surface is the same as the number of waves received by an observer moving towards the source. This is given by the apparent frequency.

n´ = Svc

vc−− 0 × n =

32332)64(332

−−− × 1000 = 1320 Hz

(c) Same as that of the incident wave because the speed of a wave depends only on the characteristics of the medium.

∴ speed of the reflected wave = 332 ms–1 (d) To calculate the wavelength of the reflected wave,

we may consider the source to be stationary and emitting waves of wavelength 0.3 m. If the reflector were stationary, waves in a tube of length c would reach the reflector and the same number of reflected waves would be contained in a tube of the same length, so the wavelength of the reflected wave would also be the same as that of the incident wave. But when the reflector moves towards the source with speed vref´ it would reflect additional waves contained in vref and the total number of waves reflected would be

contained in a tube of length c – vref. If λ´is the changed wavelength of the wave due to the motion of the source

λ ´´ =

λ

+λ

−´´

)( refref

vcvc

or λ´´ = ref

ref

vcvc

+

− × λ´ =

6433264332

+− × 0.3 = 0.2 m

4. Find the ratio of the fundamental frequencies of two

identical strings after one of them is stretched by 2% and the other by 4%.

Sol. n = mT

l21 . If l0 be the initial length and f be

fractional increase in length, l = l0 + fl0. Since tension is proportional to the increase in length,

T = k × fl0 where k is a constant.

m = 00 ll f

M+

where M is the mass of the string

∴ n = )1(2

1

0 f+l )1(/

0

0

fMkf

+l

l = )1(2

1 0

0 fMfk

+l

l

Since l0, k and M are constants n ∝ f1

f+

∴ 2

1

nn =

)1()1(

12

21

ffff

++ =

)02.01(04.0)04.01(02.0

++ = 0.71

5. An open organ pipe has a fundamental frequency of

300 Hz. The first overtone of a closed organ pipe has the same frequency as the first overtone of the open pipe. How long is each pipe ? The velocity of sound in air = 350 ms–1.

Sol. For a closed pipe n = l4

c and 3n, 5n, 7n, ... are the

overtones. For an open pipe n = l2

c and 2n, 3n, 4n,

... are the overtones.

⇒ l = nc

2 =

3002350×

= 0.58 m

The frequency of the first overtone = 2n = 2 × 300 = 600 Hz ∴ the frequency of the first overtone of the closed

pipe = 600 = 3n ∴ n = 200 Hz

∴ 200 = l4

350 or l = 2004

350×

= 0.44 m


Definition and Classification : Carbohydrates are polyhydroxy aldehydes,

polyhydroxy ketones, or compounds that can be hydrolyzed to them. A carbohydrate that cannot be hydrolyzed to simpler compounds is called a monosaccharide. A carbohydrate that can be hydrolyzed to two monosaccharide molecules is called a disaccharide. A carbohydrate that can be hydrolyzed to many monosaccharide molecules is called a polysaccharide.

A monosaccharide may be further classified. If it contains an aldehyde group, it is known as an aldose; if it contains a keto group, it is known as a ketose. Depending upon the number of carbon atoms. It contains, a monosaccharide is known as a triose, tetrose, pentose, hexose, and so on. An aldohexose, for example, is a six-carbon monosaccharide containing an aldehyde group; a ketopentose is a five-carbon monosaccharide containing a keto group. Most naturally occurring monosaccharides are pentoses or hexoses.

Carbohydrates that reduce Fehling’s (or Benedict’s) or Tollens’ reagent are known as reducing sugars. All monosaccharides, whether aldose or ketose, are known as reducing sugars. Most disaccharides are reducing sugars; sucrose (common table sugar) is a notable exception, for it is a non-reducing sugar.

(+)-Glucose : an aldohexose : Because it is the unit of which starch, cellulose, and

glycogen are made up, and because of its special role in biological processes, (+)-glucose is by far the most abundant monosaccharide- there are probably more (+)-glucose units in nature than any other organic group–and by far the most important monosaccharide.

Cyclic structure of D-(+)-glucose. Formation of glucosides : D-(+)-glucose is a pentahydroxy aldehyde. D-(+)-

glucose had been definitely proved to have structure. CHO

OH H OH OH

H HO

H H

CH2OH D-(+)-Glucose

By 1895 it had become clear that the picture of D-(+)-glucose as a pentahydroxy aldehyde had to be modified.

Among the facts that had still to be accounted for were the following:

(a) D-(+)-Glucose fails to undergo certain reactions typical of aldehydes. Although it is readily oxidized, it gives a negative Schiff test and does not form a bisulfite addition product.

(b) D-(+)-Glucose exists in two isomeric forms which undergo mutarotation. When crystals of ordinary D-(+)-glucose of m.p. 146ºC are dissolved in water, the specific rotation gradually drops from an initial + 112º to + 52.7º. On the other hand, when crystals of D-(+)-glucose of m.p. 150ºC (obtained by crystallization at temperatures above 98ºC) are dissolved in water, the specific rotation gradually rises from an initial + 19º to + 52.7º. The form with the higher positive rotation is called α-D-(+)-glucose and that with lower rotation β-D-(+)-glucose. The change in rotation of each of these to the equilibrium value is called mutarotation.

(c) D-(+)-Glucose forms two isomeric methyl D-glucosides. Aldehydes react with alcohols in the presence of anhydrous HCl to form acetals. If the alcohol is, say methanol, the acetal contains two methyl groups :

–C=O

H

–C–OCH3

H

OH

–C–OCH3

H

OCH3

CH3OH,H+ CH3OH,H+

Aldehyde Hemiacetal Acetal When D-(+)-glucose is treated with methanol and

HCl, the product, methyl D-glucoside, contains only one –CH3 group; yet it has properties resembling those of a full acetal. It does not spontaneously revert to aldehyde and alcohol on contact with water, but requires hydrolysis by aqueous acids.

Furthermore, not just one but two of these monomethyl derivatives of D-(+)-glucose are known, one with m.p. 165ºC and specific rotation + 158º, and the other with m.p. 107 ºC and specific rotation –33º. The isomer of higher positive rotation is called methyl α-D-glucoside, and the other is called methyl β-D-glucoside. These glucosides do not undergo mutarotation, and do not reduce Tollens’ or Fehling’s reagent.

Organic Chemistry

Fundamentals

CARBOHYDRATES KEY CONCEPT


D-(+)-Glucose has the cyclic structure represented crudely by IIa and IIIa, more accurately by IIb and IIIb.

H H

HO H H

OH OH H OH

1 23 4 5

6 CH2OH H OH

OHOH H

HHH

HO

CH2OH O

123

4 5

6

IIa IIb

α-D-(+)-Glucose (m.p. 146 ºC, [α] = +112º)

O

HO H

HO H H

H OH H OH

1 23 4 5

6 CH2OH H OH

HOH H

HOHH

HO

CH2OH O

123

4 5

6

IIIa IIIb

β-D-(+)-Glucose (m.p. 150 ºC, [α] = +19º)

O

D-(+)-Glucose is the hemiacetal corresponding to

reaction between the aldehyde group and the C-5 hydroxyl group of the open-chain structure. It has a cyclic structure simply because aldehyde and alcohol are part of the same molecule.

There are two isomeric forms of D-(+)-glucose because this cyclic structure has one more chiral centre than Fisher’s original open-chain structure. α-D-(+)-Glucose and β-D-(+)-glucose are diastereomers, differing in configuration about C-1. Such a pair of distereomers are called anomers.

As hemiacetals, α-and β-D-(+)- glucose are readily hydrolyzed by water. In aqueous solution either anomer is converted –via the open-chain form–into an equilibrium mixture containing both cyclic isomers. This mutarotation results from the ready opening and closing of the hemiacetal ring.

The typical aldehyde reactions of D-(+)-glucose –osazone formation, and perhaps reduction of Tollens’ and Fehling’s reagents– are presumably due to a small amount of open-chain compound, which is replenished as fast as it is consumed. The concentration of this open-chain structure, however, is too low (less than 0.5%) for certain easily reversible aldehyde reactions like bisulfite addition and the Schiff test.

Disaccharides : Disaccharides are carbohydrates that are made up of

two monosaccharide units. On hydrolysis a molecule of disaccharide yields two molecules of monosaccharide.

We shall study four disaccharides : (+)-maltose (malt sugar), (+)-cellobiose, (+)-lactose (milk sugar), and (+)-sucrose (cane or beet sugar).

(+)-Maltose : (+)-Maltose can be obtained, among other products,

by partial hydrolysis of starch in aqueous acid. (+)-Maltose is also formed in one stage of the fermentation of starch to ethyl alcohol; here hydrolysis is catalyzed by the enzyme diastase, which is present in malt (sprouted barley).

Let us look at some of the facts from which the structure of (+)-maltose has been deduced.

(+)-Maltose has the molecular formula C12H22O11. It reduces Tollens’ and Fehling’s reagents and hence is a reducing sugar. It reacts with phenylhydrazine to yield an osazone, C12H20O9(=NNHC6H5)2. It is oxidized by bromide water to a monocarboxylic acid, (C11H21O10)COOH, maltobionic acid. (+)-Maltose exists in alpha ([α] = + 168º) and beta ([α] = + 112º) forms which undergo mutarotation in solution (equilibrium [α] = + 136º).

(+)-Cellobiose : When cellulose (cotton fibers) is treated for several

days with sulfuric acid and acetic anhydride, a combination of acetylation and hydrolysis takes place; there is obtained the octaacetate of (+)-cellobiose. Alkaline hydrolysis of the octaacetate yields (+)-cellobiose itself.

Like (+)-maltose, (+)-cellobiose has the molecular formula C12H22O11, is a reducing sugar, forms an osazone, exists in alpha and beta forms that undergo mutarotation, and can be hydrolyzed to two molecules of D-(+)-glucose. The sequence of oxidation, methylation, and hydrolysis (as described for (+)-maltose) shows that (+)-cellobiose contains two pyranose rings and glucoside linkage to an –OH group on C–4.

(+)-Cellobiose differs from (+)-maltose in one respect : it is hydrolyzed by the enzyme emulsin (from bitter almonds), not by maltase. Since emulsin is known to hydrolyze only β-glucoside linkages.

(+)-Lactose : (+)-Lactose makes up about 5% of human milk and

of cow’s milk. It is obtained commercially as a by-product of cheese manufacture, being found in the whey, the aqueous solution that remains after the milk proteins have been coagulated. Milk sours when lactose is converted into lactic acid (sour, like all acids) by bacterial action (e.g., by Lactobacillus bulgaricus).


(+)-Lactose has the molecular formula C12H22O11, is a reducing sugar, forms an osazone, and exists in alpha and beta forms which undergo mutarotation. Acidic hydrolysis or treatment with emulsin (which splits β linkages only) converts (+)-lactose into equal amounts of D-(+)-glucose and D-(+)-galactose. (+)-Lactose is evidently a β-glycoside formed by the union of a molecule of D-(+)glucose and a molecule of D-(+)-galactose.

(+)-Sucrose : (+)-Sucrose is our common table sugar, obtained

from sugar cane and sugar beets. Of organic chemicals, it is the one produced in the largest amount in pure form.

(+)-Sucrose has the molecular formula C12H22O11. It does not reduce Tollen’s or Fehling’s reagent. It is a non-reducing sugar, and in this respect it differs from the other disaccharides we have studied. Moreover, (+)-sucrose does not form an osazone, does not exist in anomeric forms, and does not show mutarotation in solution. All these facts indicate that (+)-sucrose does not contain a “free”aldehyde or ketone group.

(+)-Sucrose is made up of a D-glucose unit and a D-fructose unit; since there is no “free” carbonyl group, if must be both a D-glucoside and a D-fructoside.

Polysaccharides : Polysaccharides are compounds made up of many-

hundreds or even thousands-monosaccharide units per molecule.

Polysaccharides are naturally occurring polymers, which can be considered as derived from aldoses or ketoses by polymerization with loss of water. A polysaccharide derived from hexoses, for example, has the general formula (C6H10O5)n.

The most important polysaccharides are cellulose and starch. Both are produced in plants from carbon dioxide and water by the process of photosynthesis.

Starch : Starch occurs as granules whose size and shape are

characteristic of the plant from which the starch is obtained. When intact, starch granules are insoluble in cold water; if the outer membrane has been broken by grinding, the granules swell in cold water and form a gel.

In general, starch contains about 20% of a water-soluble fraction called amylose, and 80% of a water-insoluble fraction called amylopectin. These two fractions appear to correspond to different carbohydrates of high molecular weight and formula (C6H10O5)n. Upon treatment with acid or under the

influence of enzymes, the components of starch are hydrolyzed progressively to dextrin (a mixture of low-molecular-weight polysaccharides), (+)-maltose, and finally D-(+)-glucose. (A mixture of all these is found in corn sirup, for example.) Both amylose and amylopectin are made up of D-(+)-glucose units, but differ in molecular size and shape.

Cellulose : Cellulose is the chief component of wood and plant

fibers; cotton, for instance, is nearly pure cellulose. It is insoluble in water and tasteless; it is a non-reducing carbohydrate. These properties, in part at least, are due to its extremely high molecular weight.

Cellulose has the formula (C6H10O5)n. Complete hydrolysis by acid yields D-(+)-glucose as the only monosaccharide. Hydrolysis of completely methylated cellulose gives a high yield of 2, 3, 6-tri-O-methyl-D-glucose. Like starch, therefore, cellulose is made up of chains of D-glucose units, each unit joined by a glycoside linkage to C–4 of the next.

Cellulose differs from starch, however, in the configuration of the glycoside linkage. Upon treatment with acetic anhydride and sulfuric acid, cellulose yields octa-O-acetylcellobiose.

Reactions of cellulose : Like any alcohol, cellulose form esters. Treatment

with a mixture of nitric and sulfuric acid converts cellulose into cellulose nitrate. The properties and uses of the product depend upon the extent of nitration.

In the presence of acetic anhydride, acetic acid, and a little sulfuric acid, cellulose is converted into the triacetate. Partial hydrolysis removes some of the acetate groups, degrades the chains to smaller fragments (of 200–300 units each), and yields the vastly important commercial cellulose acetate (roughly a diacetate). Cellulose acetate is less flammable than cellulose nitrate and has replaced the nitrate in many of its applications, in safety-type photographic film, for example. When a solution of cellulose acetate in acetone is forced through the fine holes of a spinnerette, the solvent evaporates and leaves solid filaments. Threads from these filaments make up the material known as acetate rayon. Industrially, cellulose is alkylated to ethers by action of alkyl chlorides (cheaper than sulfates) in the presence of alkali. Considerable degradation of the long chains is unavoidable in these reactions. Methyl, ethyl, and benzyl ethers of cellulose are important in the production of textiles, films, and various plastic objects.


Inorganic Chemistry

Fundamentals

Identification of acidic radicals For the identification of the acidic radicals, the

following scheme is followed. Group I : The radicals which are analysed by dilute

H2SO4 or dilute HCl. These are (i) carbonate (ii) sulphite, (iii) sulphide, (iv) nitrite, and (v) acetate

Group II : The radicals which are analysed by concentrated H2SO4 . These are (i) chloride, (ii) bromide, (iii) iodide (iv) nitrate, and (v) oxalate

Group III : The radicals which are not analysed by dilute and concentrated H2SO4. These are (i) sulphate, (ii) Phosphate, (iii) borate, and (iv) fluoride.

Group I : Add dilute HCl or H2SO4 to a small amount of

substance and warm gently, observe. 1. Carbonate or CO3

2– : The carbonates are decomposed with the

effervescence of carbon dioxide gas. Na2CO3 + H2SO4 → Na2SO4 + H2O + CO2 ↑ When this gas is passed through lime water, it

turns milky with the formation of calcium carbonate.

Ca(OH)2 + CO2 → CaCO3 + H2O Lime water White ppt.

If the CO2, gas is passed in excess, the milky solution becomes colourless due to the formation of soluble calcium bicarbonate.

CaCO3 + H2O + CO2 → Ca(HCO3)2 White ppt. Soluble Note : Carbonates of bismuth and barium are not easily

decomposed by dilute H2SO4. Dilute HCl should be used.

Sulphur dioxide evolved from sulphites also turns lime water milky.

Ca(OH)2 + SO2 → CaSO3 + H2O White ppt. However SO2 can be identified by its pungent

odour of burning sulphur. PbCO3 reacts with HCl or H2SO4 to give in the

initial stage some effervescence but the reaction slows down due to formation of a protective insoluble layer of PbCl2 or PbSO4 on the surface of remaining salt or mixture.

2. Sulphite : The sulphites give out sulphur dioxide gas having

suffocating smell of burning sulphur. CaSO3 + H2SO4 → CaSO4 + H2O + SO2 ↑ When acidified potassium dichromate paper is

exposed to the gas, it attains green colour due to the formation of chromic sulphate.

K2Cr2O7 + H2SO4 + 3SO2 → K2SO4 + Cr2 (SO4)3 + H2O The sulphite also gives white precipitate with

BaCl2, Soluble in dil. HCl Na2SO3 + BaCl2 → 2 NaCl + BaSO3 ↓ 3. Sulphide, S–2: The sulphide salts form H2S which smells like

rotten eggs. Na2S + H2SO4 → Na2SO4 + H2S ↑ On exposure to this gas, the lead acetate paper

turns black due to the formation of lead sulphide. Pb(CH3COO)2 + H2S → PbS ↓ + 2CH3COOH black ppt.

The sulphides also turn sodium nitroprusside solution violet (use sodium carbonate extract for this test).

Na2S + Na2[FeNO(CN)5] → Na4 [Fe(NOS) (CN)5] Sulphides of lead, calcium, nickel, cobalt,

antimony and stannic are not decomposed with dilute H2SO4. Conc. HCl should be used for their test.However brisk evolution of H2S takes place even by use of dilute H2SO4 if a pinch of zinc dust is added.

Zn + H2SO4 → ZnSO4 + 2H HgS + 2H →Hg + H2S ↑

4. Nitrite, NO2– :

The nitrites yield a colourless nitric oxide gas which in contact with oxygen of the air becomes brown due to the formation of nitrogen dioxide.

2KNO2 + H2SO4 →`K2SO4 + 2HNO2 Nitrous acid 3HNO2 → H2O + 2NO + HNO3 2 NO + O2 → 2NO2 ↑ brown coloured gas

SALT ANALYSIS

KEY CONCEPT


On passing the gas through dilute FeSO4 solution, brown coloured complex salt is formed.

FeSO4.7H2O + NO → [Fe(H2O)5NO].SO4 + 2H2O Brown coloured (panta aquo nitroso ferrous sulphate) When a mixture of iodide and nitrite is treated

with dilute H2SO4, the iodides are decomposed giving violet vapours of iodine, which turns starch iodide paper blue.

2NaNO2 + H2SO4 → Na2SO4 + 2HNO2 2KI + H2SO4 → K2SO4 + 2HI 2HNO2+ 2HI → 2H2O + I2 + 2NO Violet vapours I2 + Starch → Blue colour

5. Acetate : Acetates decompose to give acetic acid vapours

having characteristic smell of vinegar. 2CH3COONa + H2SO4 → 2CH3COOH + Na2SO4 All acetates are soluble in water and their aqueous

solution on addition to neutral FeCl3 solution develops a blood red colour due to the formation of ferric acetate.

FeCl3 + 3CH3COONa → (CH3COO)3Fe + 3NaCl

Blood Red colour Acetates are also decomposed with oxalic acid

and give off acetic acid. 2CH3COONa + H2C2O4 → Na2C2O4 + 2CH3COOH Note : The ferric chloride solution supplied in the

laboratory is always acidic due to hydrolysis. It is made neutral by the addition of dilute solution of NH4OH drop by drop with constant stirring till the precipitate formed does not dissolve. The filtrate is called neutral ferric chloride solution.

Before testing acetate in the aqueous solution by FeCl3, it must be made sure that the solution does not contain

(i) CO3–2, (ii) SO3

–2 (iii) PO4

–3, (iv) I–

Since these also combine with Fe+3. Therefore , the test of acetate should be performed by neutral ferric chloride solution only after the removal of these ions by AgNO3 solution.

Group II: Add concentrated H2SO4 to a small amount of the salt

or mixture and warm gently, observe.

1. Chloride Cl–: Colourless pungent fumes of hydrogen chloride are

evolved. NaCl + H2SO4 → NaHSO4 + HCl ↑ The gas evolved forms white fumes of ammonium

chloride with NH4OH. NH4OH + HCl → NH4Cl + H2O White fumes The gas evolved or solution of chloride salt forms

a curdy precipitate of silver chloride with silver nitrate solution.

AgNO3 + HCl → AgCl ↓ + HNO3 Yellowish : green chlorine gas with suffocating

odour is evolved on addition of MnO2 to the above reaction mixture.

NaCl + H2SO4 –→ NaHSO4 + HCl MnO2 + 4HCl –→ MnCl2 + 2H2O + Cl2 Note : The curdy precipitate of AgCl dissolves in

ammonium hydroxide forming a complex salt. AgCl + 2NH4OH → Ag(NH3)2Cl + 2H2O The solution having the silver complex on

acidifying with dilute nitric acid gives again a white precipitate of silver chloride.

Ag(NH3)2Cl + 2HNO3 → AgCl + 2NH4NO3 Chromyl chloride Test : When solid chloride is

heated with conc. H2SO4 in presence of K2Cr2O7, deep red vapours of chromyl chloride are evolved.

NaCl + H2SO4 → NaHSO4 + HCl K2Cr2O7 + 2H2SO4 → 2KHSO4 + 2CrO3 + H2O CrO3 + 2HCl → CrO2Cl2 + H2O Chromyl chloride These vapours on passing through NaOH

solution, form the yellow solution due to the formation of sodium chromate.

CrO2Cl2 + 4NaOH → Na2CrO4 +2NaCl+ 2H2O Yellow colour The yellow solution neutralised with acetic acid

gives a yellow precipitate of lead chromate with lead acetate.

Na2CrO4 + Pb(CH3COO)2 –→ PbCrO4 + 2CH3COONa

Yellow ppt. Note : This test is not given by the chloride of mercuric,

tin, silver, lead and antimony. The chromyl chloride test is always to be

performed in a dry test tube otherwise the


chromyl chloride vapours will be hydrolysed in the test tube.

CrO2Cl2 + 2H2O →H2CrO4 + 2HCl Bromides and iodides do not give this test. 2. Bromide, Br– : Reddish- brown fumes of bromine are formed. NaBr + H2SO4 → NaHSO4 + HBr 2HBr + H2SO4 → Br2 + 2H2O + SO2 More reddish brown fumes of bromine are

evolved when MnO2 is added. 2NaBr + MnO2 + 3H2SO4 → 2NaHSO4 + MnSO4 + 2H2O + Br2 The aqueous solution of bromide or sodium

carbonate extract gives pale yellow precipitate of silver bromide which partly dissolves in excess of NH4OH forming a soluble complex.

NaBr + AgNO3 → AgBr ↓ + NaNO3 Pale yellow ppt. AgBr +2NH4OH → Ag(NH3)2Br + 2H2O 3. Iodide, I– : Violet vapours of iodine are evolved. 2KI + H2SO4 → 2KHSO4 + 2HI 2 HI + H2SO4 → I2 + SO2 + 2H2O Violet vapours with starch produce blue colour. I2 + Starch → Blue colour More violet vapours are evolved when MnO2 is

added. 2KI + MnO2 + 3H2SO4 –→ 2KHSO4 + MnSO4 + 2H2O + I2

Aqueous solution of the iodide or sodium carbonate extract gives yellow precipitate of AgI with silver nitrate solution which does not dissolve in NH4OH.

NaI + AgNO3 → AgI + NaNO3 Yellow ppt.

Note : Sodium carbonate extract of bromide and iodide

on addition of CHCl3 and chlorine water gives brown or violet layer to CHCl3 respectively.

2NaBr + Cl2 → 2NaCl + Br2 ; Br2 + CHCl3 → Brown 2NaI + Cl2 → 2NaCl + I2 ; I2 + CHCl3 → Violet Excess of chlorine water should be avoided as the

layer may become colour less due to conversion of Br2 into HBrO and I2 into HIO3.

Br2 + 2H2O + Cl2 → 2HBrO + 2HCl

I2 + 5Cl2 + 6H2O → 2HIO3 + 10 HCl

4. Nitrate, NO3– :

Light brown fumes of nitrogen dioxide are evolved.

NaNO3 + H2SO4 → NaHSO4 + HNO3 4 HNO3 → 2H2O + 4 NO2 + O2 These fumes intensify when copper turnings are

added. Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2O Ring Test : An aqueous solution of salt is mixed

with freshly prepared FeSO4 solution and conc. H2SO4 is poured in test tube from sides, a brown ring is formed on account of the formation of a complex at the junction of two liquids.

NaNO3 + H2SO4 → NaHSO4 + HNO3 6 FeSO4 + 2HNO3 + 3H2SO4 →

3Fe2 (SO4)3 + 4H2O + 2NO [Fe(H2O)6]SO4. H2O + NO → Ferrous sulphate [Fe(H2O)5 NO]SO4 + 2H2O Brown ring

The nitratess can also be tested by boiling nitrate with Zn or Al in presence of concentrated NaOH solution when ammonia is evolved which can be detected by the characteristics odour.

Zn + 2NaOH → Na2ZnO2 + 2H Al + NaOH + H2O → NaAlO2 + 3H NaNO3 + 8H → NaOH + 2H2O + NH3 Note : Ring test is not reliable in presence of

nitrite, bromide and iodide. 5. Oxalate, C2O4

–2 : A mixture of CO and CO2 is given off. The CO

burns with blue flame. Na2C2O4 + H2SO4 → Na2SO4 + H2C2O4 H2C2O4 + [H2SO4] → CO + CO2 + H2O + [H2SO4] 2CO + O2 → 2CO2

A solution of oxalates give the white precipitate with CaCl2 solution. This precipitate get dissolved in dil. H2SO4 and decolourises KMnO4 (acidified) solution.

Na2C2O4 + CaCl2 → CaC2O4 ↓ + 2NaCl CaC2O4 + H2SO4 → CaSO4 + H2C2O4 5H2C2O4 + 2KMnO4 + 3H2SO4 → 2 MnSO4 + K2SO4+ 8 H2O + 10CO2

Group III : 1. Sulphate ,SO4

–2 : Add conc. HNO3 to a small amount of substance or

take sodium carbonate extract and then add BaCl2 solution. A white precipitate of BaSO4 insoluble in conc. acid is obtained.


Na2SO4 + BaCl2 → 2NaCl + BaSO4 White ppt. Note : Silver and lead if present, may be precipitated

as silver chloride and lead chloride by the addition of barium chloride. To avoid it, barium nitrate may be used in place of barium chloride.

2. Borate : To a small quantity of the substance (salt or mixture),

add few ml. of ethyl alcohol and conc. H2SO4. Stir the contents with a glass rod. Heat the test tube and bring the mouth of the test tube near the flame. The formation of green edged flame indicates the presence of borate.

2Na3BO3 + 3H2SO4 → 3Na2SO4 + 2H3BO3 H3BO3 + 3C2H5OH → (C2H5)3BO3 + 3H2O Ethyl borate

3. Phosphate : Add conc. HNO3 to a small amount of substance or

take sodium carbonate extract, heat and then add ammonium molybdate. A canary yellow precipitate of ammonium phospho molybdate is formed.

Ca3(PO4)2 + 6HNO3 → 3Ca (NO3)2 + 2H3PO4 H3PO4 + 12 (NH4)2 MoO4 + 21 HNO3 → (NH4)3 PO4. 12 MoO3 + 21 NH4NO3 + 12 H2O Canary yellow ppt. Note : Arsenic also yields a yellow precipitate of

(NH4)3. AsO4.12 MoO3 (Ammonium arseno molybdate).Thus in presence of As, phosphate is tested in the filtrate of second group.

The precipitate of ammonium phosphomolybdate dissolves in excess of phosphate. Thus, the reagent (ammonium molybdate) should always be added in excess.

HCl interferes in this test. Hence, before the test of phosphate is to be performed, the solution should be boiled to remove HCl.

Reducing agents such as sulphites, sulphides, etc., interfere as they reduce Mo+6 to molybdenum blue (Mo3O8.xH2O). The solution, therefore, turns blue. In such cases, the solutions should be boiled with HNO3 so as to oxidise them before the addition of ammonium molybdate.

4. Fluoride : Take small amount of the substance in dry test tube

and add an equal amount of sand and conc. H2SO4.Heat the contents and place a glass rod moistened with water over the mouth of the test tube. A gelatinous waxy white deposit on the rod is formed.

2NaF + H2SO4 → Na2SO4 + H2F2 SiO2 + 2H2F2 → SiF4 + 2H2O

3SiF4 + 4H2O → H4SiO4 + 2H2SiF6 Silicic acid (white)

Note : The test should be performed in perfectly dry test

tube otherwise waxy white deposit will not be formed on the rod.

HgCl2 and NH4Cl also give white deposits under these conditions, but these are crystalline in nature.

Sodium carbonate extract : One part of the given substance is mixed with

about 3 parts of sodium carbonate and nearly 10 to 15 ml. of distilled water. The contents are then heated for 10-15 minutes and filtered. The filtrate is known as sodium carbonate extract or soda extract and this contains soluble sodium salts due to exchange of partners in between sodium carbonate and salts.

CaCl2 + Na2CO3 → CaCO3 + 2NaCl Insoluble Sodium chloride (soluble) PbSO4 + Na2CO3 → PbCO3 + Na2SO4 Insoluble Sodium sulphate (Soluble) BaCl2 + Na2CO3 → BaCO3 + 2 NaCl Insoluble Sodium chloride (Soluble)

The carbonates of the cations of the mixtures are mostly insoluble in water and are obtained in the residue. On the other hand, sodium salts of the anions (acidic radicals) of the mixture being soluble in water are obtained in the filtrate.

The sodium carbonate extract is basic in nature and before it is used for the analysis of a particular acidic radical, it is first neutralised by the addition of small quantity of an appropriate acid. The acid is added to the extract till the effervescence cease to evolve.

Advantages of preparing sodium carbonate extract-

The preparation of sodium carbonate extract affords a convenient method for bringing the anions of the mixture into solution which were otherwise insoluble with cation of salt.

It removes the basic radicals (usually coloured) which interferes in the usual tests of some of the acidic radicals.

The residue can be used for the tests of basic radicals of I to VI groups. Such a solution does not involve the problem of removing interfering radicals like oxalate, fluoride, borate and phosphate.


1. 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

OH 2-propanol

and hence the structure of (A) should be

CH3CHCH3

NH2

Propan-2-amine

2. An organic compound A, C8H4O3, in dry benzene in the presence of anhydrous AlCl3 gives compound B. The compound B on treatment with PCl5 followed by reaction with H2/Pd(BaSO4) gives compound C, which on reaction with hydrazine gives a cyclised compound D(C14H10N2). Identify A, B, C and D. Explain the formation of D from C. [IIT-2000]

Sol. The given reactions are as follows.

O

O +

O

AlCl3

O

O

OH

PCl5

H2/Pd (BaSO4)

C6H5

H

C

C

O O

H2NNH2

C6H5

N

N

The formation of D from C may be explained as

follows.

C6H5 C6H5

OO

NH2

NH2

O–

NH2

NH2 O–

+

+ C6H5O–

N – H

N – HOH

C6H5

N

N

3. An organic compound (X), C5H8O, does not react appreciably with Lucas reagent at room temperatures but gives a precipitate with ammonical AgNO3 solution. With excess CH3MgBr; 0.42 g of (X) gives 224 ml of CH4 at STP. Treatment of (X) with H2 in the presence of Pt catalyst followed by boiling with excess HI gives n-pentane. Suggest structure of (X) and write the equations involved. [IIT-1992]

Sol. Lucas test sensitive test for the distinction of p, s, and t-alcohol. A t-alcohol gives cloudiness immediately, while s-alcohol within 5 minutes. A p-alcohol does not react with the reagent at room temperature. Thus, the present compound (X) does not react with this reagent, hence it is a p-alcohol.

UNDERSTANDINGOrganic Chemistry


(X) = C4H6.CH2OH(p-alcohol) Since the compound gives a ppt. with ammonical

AgNO3, hence it is an alkyne containing one –C≡ CH, thus (X) may be written as :

HC≡C –C2H4 – CH2OH (X) It is given that 0.42 g of the compound (which is

0.005 mol) produces 22.4 ml of CH4 at STP (which is 0.01 mol) with excess of CH3MgBr. This shows that the compound (X) contains two active H atoms (H atom attached to O, S, N and –C≡CH is called active). Of these, one is due to the p-alcoholic group (–CH2OH) and the other is due to the –C≡CH bond, since both these groups are present in (X), hence it evolves two moles of CH4 on reaction with CH3MgBr.

H – C≡C.)X(

42HC – CH2OH + 2CH3MgBr →

BrMgC≡C–C2H4 – CH2OMgBr + 2CH4 Moreover, the treatment of (X) with H2/Pt followed

by boiling with excess of HI gives n-pentane (remember that 2HI are required to convert one –CH2OH into CH3). This shows that the compound (X) contains a straight chain of five carbon atoms.

H – C≡C–C2H4 – CH2OH → Pt/H2 2 CH3CH2.C2H4 – CH2OH

∆ → HI2 CH3CH2CH2CH2CH3 + H2O + I2

n-pentane On the basis of abvoe analytical facts (X) has the

structure :

HC≡C.CH2 CH2 – CH2OH (X) 5 4 3 2 1

4-pentyne-1-ol The different equations of (X) are :

)X(

222 OHCHCHCHCCH −≡− .tempRoomHClZnCl2 → + No reaction

AgNO3 Ag – C≡C – CH2CH2CH2OH + NH4NO3

White ppt. NH3

2CH3MgBr Br MgC≡C.CH2CH2CH2OMgBr + 2CH4

2H2/Pt CH3CH2CH2CH2CH2OH

Pentanol-1

CH3CH2CH2CH2CH3

n-pentane

2 HI ∆, –H2O; –I2

The production of 2 moles of CH4 is confirmed as the reactions give 224 ml of CH4.

Q 84 g(X) gives = 2 × 22.4 litre CH4

∴ 0.42 g (X) gives = 84

42.04.222 ××

= 224 ml of CH4 4. Outline the accepted mechanism of the following

reaction. Show the various steps including the charged intermediates.

+ CH3–C–Cl|| O

→ 3AlCl C–CH3+HCl

|| O

Sol. Ozonolysis of (A) to acetone and an aldehyde

indicates the following part structure of alkene (A) :

C=CHRCH3

CH3Alkene (A)

→ 3O

C=O + OCH.RCH3

CH3 aldehyde acetone

As per problem : RCHO → ]O[ RCOOH [B] → 2Br/P Bromo compound [C] → OH2 Hydroxy acid [D]

Structure of (D) is determined by the reaction :

C = OCH3

CH3

→HCN C CH3

CH3

OH

CN →

+H/OH2

CCH3

CH3

OH

COOH (D) The compound (D) is obtained by hydrolysis of (C)

with aqueous alkali since (C) is a bromo compound, therefore it has a bromo group where the compound (D) has a hydroxyl group. Therefore, structure of C is

C CH3

CH3

Br

COOH

The compound (C) is formed by bromination of compound (B), therefore, the compound (B) is

C CH3

CH3

H

COOH

The compound (B) is formed by oxidation of an aldehyde therefore the structure of the aldehyde is

C CH3

CH3

H

CHO

The aldehyde and acetone are formed by ozonolysis of alkene. Therefore, the double bond in alkene should be between the carbon atoms of the two carbonyl compounds (the aldehyde and acetone). Therefore, the compounds and the reactions are identified as

C–C=CCH3

CH3

CH3

CHO| |HH

(A)

→ozonolysis C

CH3

CH3

H

CHO


+ O = C

CH3

CH3

C

CH3

CH3

H

CHO → ]O[ C

CH3

CH3

H

COOH

→ P/Br2 C CH3

CH3

Br

COOH →hydrolysis

C

CH3

CH3

OH

COOH(D)

5. A hydrocarbon (A) of the formula C8H10, on

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

Sol. A(C8H10) OH)ii(

O)i(

2

3 →)B(

264 OHC

Since compound (A) adds one mol of O3, hence it should have either C = C or a – C ≡ C – bond. If it was alkene its formula should be C8H16 (CnH2n), and if it was alkyne it should have the formula C8H14; it means it is neither a simple alkenen or simple alkyne. However it is definite that the compound has an unsaturated group, it appears that it is a cyclosubstituted ethyne.

H – C ≡ C – H 106HC

H2

+

− → C3H5 – C ≡ C – C3H5

the C3H5 – corresponds to cyclopropyl (∆) radical, hence compund (A) is

CH – C≡C – CH

CH2

CH2

CH2

CH2 1,2-dicyclopropyl ethyne

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

CH – C≡C – CH

CH2

CH2

CH2

CH2 (A)

(i) O3

CH – C — C – CH CH2

CH2

CH2

CH2 (A)

H2O

O

O O

Warm

CH – C – C – CH

CH2

CH2

CH2

CH2 O O

(B)

+ H2O2

CH2

CH2

CH – COOH 2

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

CH – Br CH2

CH2

CH2

CH2

Cyclopropylmagnesium bromide

CH . MgBr Mg

ether

C=O∆

O

CH .COOMgBr

CH2

CH2

CH2

CH2

Addition compound

CH–COOH HOH

dil. HCl; –MgBrOH

Ozone-layer Healing Itself The ozone-layer is showing signs of recoring, but it is unlikely to stabilise at pre-1980 levels. The depletion of the earth's protective ozone-layer is caused by the chemical action of chlorine and bromine released by man-made chlorofluorocarbons (CFCs), which are used in aerosol sprays and cooling equipments.

Ozone-depleting chemicals were banned by the 1987 Montreal Protocol which has now been ratified by 180 nations. Ozone-layer is getting better due to Montreal Protocol.

Dr. Weatherhead and Dr. Signe Bech Anderson of the Danish Meteorological Institute in Copenhagen analysed data from satellites and ground stations and information from 14 modelling studies. They found that ozone levels have stabilised or increased slightly in the last 10 years. The full recovery is still decades away.

Process of Ozone-layer Healing

The research team leader Dr. S. B. Anderson said, "The depletion has been most severe at the poles and to a lesser extent at mid-latitudes covering bands of North America, South America and Europe." Shifting temperatrues, green house gases, nitrous oxide and atmospheric dynamics, which can influence ozone-levels, are going to change inthe future. Volcanic activities on the Earth can also have an impact. The 1993 Mount Pinatubo eruption in the Phillippines caLI sed ozone-levels to backslide for several years.


1. Given that φ (x) = )()()()(

cabxcxbx

−−−− f(a) +

)()()()(

abcbaxcx

−−−− f (b)+

)()()()(

bcacbxax

−−−− f (c) – f (x)

where a < c < b and f '' (x) exists at all points in (a, b). Prove that there exists a number µ, a < µ < b , such that

)()(

)(caba

af−−

+ )()(

)(abcb

bf−−

+ )()(

)(bcac

cf−−

= 21 f '' (µ).

2. An unbiased die is tossed until it lands the same way

up twice running. Find the probability that it requires r tosses.

3. Given the base of a triangle and the sum of its sides

prove that the locus of the centre of its incircle is an ellipse.

4. Let f (x) = ax2 + bx + c & g (x) = cx2 + bx + a, such

that | f (0) | ≤ 1, | f (1) | ≤ 1 and |f (–1) | ≤ 1 , prove that | f (x) | ≤ 5/4 and | g (x) | ≤ 2.

5. In order to find the dip of an oil bed below the

surface of the ground, vertical borings are made from the angular points, A, B, C of a triangle ABC which is in horizontal plane. The depth of the bed at these points are found to be x, x + y and x + z respectively. Show that the dip θ (angle with horizontal) of the oil bed which is assumed to be a plane is given by tan θ .

sin A = Abcyz

bz

cy cos2

2

2

2

2−+ where b and c are the

lengths of the sides CA and AB respectively and A is the angle between CA and AB.

6. Evaluate : ∫ +−

xxx

5cos217cos8cos

7. Let f (x) be an even function such that f′ (x) is

continuous, find y for which 2

2

dxyd = ∫

−

x

x

dttf )(

8. Prove the inequality (aα + bα)1/α < (aβ + bβ)1/β, for a > 0, b > 0 & α > β > 0. 9. A circle of radius 1 rolls (without sliding) along

the x-axis so that its centre is of the form (t, 1) with t increasing. A certain point P touches the x-axis at the origin as the circle rolls. As the circle rolls further, the point P passes through the point (x, 1/2). Find x, when it passes through (x, 1/2) first time.

10. Find all positive integers n for which 1−n + 1+n is rational.

`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

9Set

SCIENCE TIPS

• An electron is moving along X-axis in a magnetic field acting along Y-axis. What is the direction of magnetic force acting on it. Along Z-axis

• What is the equation of a plane progressive simple harmonic wave traverlling in + x direction?

y = a sinλ

2π (vt – x) = a sin 2π

−

λx

Tt

• What type of magnetic material is used in making permanent magnets? Ferromagnetic

• A wire kept along north-south is allowed to fall freely. Will an induced emf be set up? No

• Which of A.C. or D.C. is blocked by a capacitor? D.C.


1.

A

B

C

d

a

d b

c Plane through mid pt of AB, ⊥ to CD is

( rr

– 1/2 ( ar

+ br

)).( cr

– dr

) = 0; Let centroid

4

dcbarrrr

+++ = 0 at origin

( rr

+ 1/2 ( cr

+ dr

)). ( cr

– dr

) = 0

| rr

+ cr

|2 = | rr

+ dr

|2

it is the locus of pt. equidistance from – cr

& – dr

similarly.

| rr

+ cr

|2 = | rr

+ dr

|2 = | rr

+ ar

|2 = | rr

+ br

|2

so the pt. is equidistant from – ar

, – br

, – cr

, – dr

(i.e. circumcentre of tetrahedron – ar

, – br

, – cr

, – dr

)

2. As the function is symmetrical about x = a & x = b

lines

so f (a + x) = f (a − x) ................(1) &

f (b + x ) = f (b − x) ................(2)

As it is defined for x ∈ R.

Let x = b − a − t in (1)

f (b − t) = f (2a − b + t)

use (2) in it

f (b + t) = f (2a − 2b + b + t)

so the function is periodic & its possible period

may be |2a − 2b| = 2b – 2a (as b > a).

3. If A is the area of the triangle with sides a, b and c, then A2 = s (s − a) (s − b) (s − c) ;

where 2s = a + b + c.

using AM - GM inequality for s − a, s − b, s − c, we have

A2 ≤ s 3

3)()()(

−+−+− csbsas

A2 ≤ s3

323

− ss = 3

4

3s ⇒ A ≤

33

2s

Let 2s = p , then A ≤ 312

2p

Amax = 312

2p , As condition of equality holds iff

s − a = s − b = s − c which happen if a = b = c.

so Amax = 312

2p ; for a = b = c

Now again p ≥ A312

pmin = A312 , and

again equality holds if a = b = c.

4. acb 42 − ≤ | b | 241bac

− ≤ | b | 241bac

+

≤ | b |

+

221bac

so acb 42 − ≤ | b | + bac2

so that a

acba

b2

42

2 −±− ≤

ab2

+ a

b2

+ bc

= ab +

bc

MATHEMATICAL CHALLENGES SOLUTION FOR DECEMBER ISSUE (SET # 8)


Hence the solutions of az2 + bz + c = 0 satisfy

condition | z | ≤ + ab +

bc .

5. P (a cos θ, b sin θ)

Equation of AC ⇒ ax cos θ +

by sin θ = 1

A

B C D

P(θ)

Point A : (0 , b cosec θ)

Equation of BC ⇒ y = −b

Point C = ax cos θ − sin θ = 1

x = θθ+

cos)sin1( a

Point C

−θ

θ+ ba ,cos

)sin1(

Area A = 21 AD . BC = AD . DC

=

θ+

sinbb .

θθ+

cos)sin1(a

= θθθ+

cossin)sin1( 2ab

A = ab θθ

θ+cossin

)sin1( 2

θd

dA =

ab . θθ

θ−θθ+−θθθ+22

2222

cossin)sin(cos)sin1(sincos)sin1(2

= θθ

θ+22 cossin

)sin1(ab[2cos2θ sinθ − (1 + sinθ) cos2θ

+ (1 + sinθ) sin3θ)]

= θθ

θ+22 cossin

)sin1(ab

[– cos2θ + sin2θ + sin3θ + cos2θ sinθ]

for max./min . θd

dA = 0

sinθ (cos2θ + sin2θ) + sin2θ − cos2θ = 0

sinθ + sin2θ − (1 − sin2θ) = 0

⇒ 2sin2θ + sinθ – 1 = 0

(2 sinθ − 1) (sin θ + 1) = 0

as sin θ ≠ −1

sin θ = 1/2 ; θ = π/6

when θ > π/6 ; θd

dA > 0

when θ < π/6 ; θd

dA < 0

so θ = 6π ; is the pt of min.

min. area.

Amin = θθθ+

cossin)sin1( 2ab

= 2/3.2/1)2/11(. 2+ab

=

23.

21.4

9×ab

= 3 3 ab sq. units. 6. ax2 + 2hxy + by2 = 0; (y – M1x) (y – M2x) = 0

where M1 + M2 = – bh2 & M1M2 =

ba

Now as given the second pair must be given by

(y – M1x)(M2y + x) = 0

M2y2 + (1 – M1M2)xy – M1x2 = 0

Compare it with a´x2 + 2h´xy + b´y2 = 0

´2

bM

= ´2

1 21

hMM−

= ´

1

aM−

so ´2

bM

= –´

1

aM−

= ´´

21

abMM

−+

= ´)´(

2abbh

−−

= ´2

1 21

hMM−

= ´2/1

hba−


M2 = –´)´(

´2abb

hb−

& M1 = – bh

aab´2

´)( −

Since M1M2 = ba so

´)´(´2abb

hb−

.bh

aab´2

´)( − = ba

Thus ´´´´

abbha

− =

ababh−´

7. LHS = coeff. of xn in [nC0(1 + x)m + nC1(1 + x)m+1 +

.... + nCn(1 + x)m + n]

= coeff. of xn in (1 + x)m[nC0 + nC1(1 + x) + ........

..... + nC0 (1 + x)n]

= coeff. of xn in (1 + x)m(2 + x)n

= coeff of xn in (1 + x)m ∑=

−n

r

rrnr

n xC0

2.

= nC0 . mC0 + nC1 .mC1 . 2 + nC2 . mC2 . 22 + .... + nCnmCn . 2n

8. In = ∫−

−1

1

2 )1( nx cos mx dx

= 1

1

2 sin)1(−

−

mmxx n + ∫

−

−−1

1

12 )1(2 nxxmn sin mx dx

= 0 + mn2

( )

−+−−+

−− ∫

−

−−

−

−1

1

122221

1

12 cos)1()1()1(2(1cos1( dxmxxxxnmn

mxxx nnn

= 22m

n∫−

−−1

1

22 )1( nx [ ]22 1)22( xxn −++− cosmx dx

= 22m

n∫−

−−1

1

22 )1( nx [ ]1)12( 2 ++− xn cos mx dx

= 22m

n

−−−−− ∫∫

−

−

−

−1

1

221

1

12 cos)1()22(cos)1()12( dxmxxndxmxxn nn

m2In = (2n (2n − 1) In–1 − 4n (n − 1) In–2.

Hence proved.

9. tn = 2+

+)1(.

122 nn

n

= 21

n − 2)1(

1+n

Sn = 1 − 221 + 22

1 – 231 + 23

1− 24

1 .................

Sn = 1 − 2)1(1+n

Required sum = →∞n

Lim Sn = 1.

10. Let the given circle be x2 + y2 = r2 & parametric

angles of A, B, C are respectively θ1 , θ2 & θ3. Let the slopes of the given two lines are m1 & m2. Sides AB & BC are parallel to these lines.

A(θ1)

C(θ3) B(θ2)

Equation of AB;

x cos 2

21 θ+θ + y sin 2

21 θ+θ = r cos 2

21 θ−θ

so m1 = – cot 2

21 θ+θ = θ1 + θ2 = α

similarly : m2 = − cot 2

32 θ+θ = θ2 + θ3 = β

Here α, β are constants as m1 & m2 are constants.

Now equation of AC ;

x cos

θ+θ2

31 + y sin 2

31 θ+θ = r cos

231 θ−θ

x cos

θ+θ2

31 + y sin 2

31 θ+θ= rk

where k = cos 2

β−α (i . e. constant)

so foot of the perpendicular from centre of given

circle on AC

θ+θθ+θ2

,2

cos 3131 krkr is

which lies on x2 + y2 = (rk)2.


1. A traveller starts from a certain place on a certain day

and travels 1 km on the first day and on subsequent days, he travels 2 km more than the previous day. After 3 days, a second traveller sets out from the same place and on his first day he travels 12 km and on subsequent days he travels 1 km more than the previous day. On how many days will the second traveller be ahead of the first?

Sol. The first traveller travels on different days as follow (in km) 1, 1 + 2, 1 + 2 + 2, ... .

After 3 days the first traveller is already ahead by (1 + 3 + 5) km, i.e., 9 km.

1 3 5

The second traveller travels on different days as

follows : 0, 0, 0, 12, 13, 14, ... After n days from the day the second traveller starts,

the distance covered by the first = 1 + 3 + 5 + (7 + 9 + ... to n terms) = 1 + 3 + 5 + ... to (n + 3) terms = (n + 3)2 and the distance covered by the second = 12 + 13 + 14 + ... to n terms

= 2n 24 + (n – 1).1 =

2)23( +nn

The second traveller is ahead of the first on the nth day (after the second sets off) if

2

)23( +nn > (n + 3)2

or n2 + 23n > 2(n2 + 6n + 9) or n2 – 11n + 18 < 0 or (n – 2) (n – 9) < 0. So n – 2 > 0 and n – 9 < 0 ...(i) or n – 2 < 0 and n – 9 > 0 ...(ii) (i) ⇒ n > 2 and n < 9 (ii) ⇒ n < 2 and n > 0 (absurd) Thus, from the begining of the 3rd day to the end of

the 9th day the second traveller is ahead of the first. So, the second is ahead of the first on the 3rd, 4th, 5th,

..., 9th days (after the second sets off). Hence, the required number of days = 7.

2. If two lines cut the circle |z| = r at the points representing the complex numbers a, b and c, d, and they meet at the point z, prove that

z(a–1b–1 – c–1d–1) = a–1 + b–1 – c–1 – d–1.

Sol. Clearly |a| = |b| = |c| = |d| = r. ∴ aa = bb = cc = dd = r2 (Q |z|2 = zz ).

Now a, b and z are collinear. Let they lie on the line zα + zα + β = 0, β ∈ R.

∴ zα + zα + β = 0 aα + aα + β = 0 bα + bα + β = 0.

Eliminating α , α, β we get 111

bbaazz

= 0.

Similarly, 111

ddcczz

= 0. On simplification,

)–()–(–)–( abbabazbaz + = 0

and )–()–(–)–( cbdcdczdcz + = 0 By cross-multiplication,

)–)(–()–)(–(– abbadccddcba

z+

= )–)(–()–)(–(–

1dcbadcba +

∴

)–(–––)–(

2222dc

br

ar

dr

crbaz

= (c – d)

arb

bra

22–. – (a – b)

crd

drc

22–.

Q aa = r2, etc.

or

)–.(–––).–( dc

abab

cdcdbaz

= (c – d) ab

ba 22 – – (a – b) cd

dc 22 –

or z(a – b) (c – d)

+

abcd11–

= (a – b)(c – d)

++

cddc

abba –

∴ z(a–1b–1 – c–1d–1) = a–1 + b–1 – c–1 – d–1 Q a ≠ b, c ≠ d.

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

Students' ForumMATHS


3. If a

α4sin + b

α4cos = ba +

1 , show that

3

8sina

α + 3

8cosb

α = 3)(1ba +

.

Sol. Here a

ba + sin4α + b

ba + cos4α = 1

or sin4α + cos4α + ab sin4α +

ba cos4α = 1

or (sin2α + cos2α)2 – 2 sin2α . cos2α

+ ab sin4α +

ba cos4α = 1

or 2

2sin

α

ab – 2 .

ab sin2α .

ba cos2α

+ 2

2cos

α

ba = 0

or 2

22 cos–sin

αα

ba

ab = 0

∴ ab sin2α =

ba cos2α

or sin2α = ba cos2α

∴ a

α2sin = b

α2cos = ba +

α+α 22 cossin

∴ sin2α = ba

a+

, cos2α = ba

b+

∴ 3

8sina

α + 3

8cosb

α = 31a

. 4

4

)( baa+

+ 31

b. 4

4

)( bab+

= 4)( baa

+ + 4)( ba

b+

= 4)( baba

++ = 3)(

1ba +

4. In the ∆ABC, a similar ∆A'B'C' is inscribed so that

B'C' = λ. BC. If B'C' is inclined at an angle θ with

BC, prove that λ cos θ = 21 .

Sol. ∆ABC and ∆A'B'C' are similar where ∠B'A'C' = ∠BAC = A, ∠A'B'C' = ∠ABC = B, ∠B'C'A' = ∠BCA = C.

A

B'

CA' BO

C' C BA

θ

In ∆AC'B', ∠AB'C' = ∠B'OC + ∠ACB = θ + C

∴ )sin(

'C

AC+θ

= A

CBsin

''

or )sin(

'C

AC+θ

= A

BCsinλ = λ .

Aa

sin= 2λR.

In ∆BA'C', ∠BA'C' = ∠A'C'B' – ∠A'OC' = C – θ

∴ )–sin(

'θC

BC = B

CAsin

''

∴ )–sin(

'θC

BC = B

ACsin.λ =

Bb

sinλ = 2λR

λ== trianglessimilarfrom

ACCA

BCCB ''''

Q

Thus, we get AC' = 2λR sin (θ + C) and BC' = 2λR sin(C – θ) ∴ c = AB = AC' + BC' = 2λR sin(C + θ)

+ sin (C – θ) = 2λR . 2sin C . cos θ

∴ cos θ = CR

csin4λ

= Rλ4

1 . C

csin

= Rλ4

1 . 2R = λ21

∴ λ cos θ = 21 .

5. A ray of light is sent along the line x – 2y – 3 = 0. On

reaching the line 3x – 2y – 5 = 0 the ray is reflected from it. Find the equation of the line containing the reflected ray.

Sol. Let A be the point of incidence. ∴ A is the point of intersection of x – 2y – 3 = 0 an d 3x – 2y – 5 = 0. Solving these, we get x = 1, y = –1. So A = (1, – 1).

x – 2y–3=0

P

Q

A 3x – 2y–5=0

Let P be any point on the line of incidence

x – 2y – 3 = 0. So, we take P = (3, 0). Let Q be the image of P by the

line 3x – 2y – 5 = 0. Let Q = (α, β). Clearly PQ ⊥ the line 3x – 2y – 5 = 0

and the middle point of PQ is on 3x – 2y – 5 = 0.

∴ 3–0–

αβ .

23 = – 1 ...(i)

and 3 . 2

3+α – 2 . 2

0+β – 5 = 0 ...(ii)

(i) ⇒ 3β + 2(α – 3) = 0 or 2α + 3β – 6 = 0 ...(iii) (ii) ⇒ 3(α + 3) – 2β – 10 = 0 or 3α – 2β – 1 = 0. ...(iv)


Solving (iii) and (iv), 12–3–

α = 218– +

β = 9–4–

1

∴ 15–α =

16–β =

13–1 ; ∴ α =

1315 , β =

1316 .

∴ Q =

1316,

1315 .

The line containing the reflected ray is the joining the

points A(1, –1) and Q

1316,

1315 .

∴ The required equation is y + 1 = 1–

1315

11316

+(x – 1)

or y + 1 = 2

29 (x – 1)

or 2y + 2 = 29x – 29 ; ∴ 29x – 2y – 31 = 0.

6. Prove that the angle subtended by any chord of a rectangular hyperbola at the centre is the supplement of the angle between the tangents at the ends of the chord.

Sol. Let P(x1, y1) and Q(x2, y2) be two ends of a chord of the rectangular hyperbola

x2 – y2 = 1 ...(i) R

Q P

O

Now, 'm' of OP = 1

1

xy

'm' of OQ = 2

2

xy

∴ tan θ =

2

2

1

1

2

2

1

1

.1

–

xy

xy

xy

xy

+=

2121

2112 –yyxxyxyx

+,

where ∠POQ = θ. The equation of tangents at P and Q are xx1 – yy1 = 1 and xx2 – yy2 = 1.

Their slopes are 1

1

yx and

2

2

yx .

∴ tan φ =

2

2

1

1

2

2

1

1

.1

–

yx

yx

yx

yx

+ =

2121

1221 –xxyyyxyx

+

∴ tan θ and tan φ are equal in magnitude but opposite in sign

∴ tan θ = – tan φ = tan (π – φ) ∴ θ + φ = π.

Puzzle : Marble Mix Up

• Years ago, to puzzle his friends, a scientist gave one of four containers containing blue and/or yellow marbles to each of the friends; Tom, Dick, Harry, and Sally.

• There were 3 marbles in each container, and the

number of blue marbles was different in each one. There was a piece of paper in each container telling which color marbles were in that container, but the papers had been mixed up and were ALL in the wrong containers.

• He then told all of his friends to take 2 marbles

out of their container, read the label, and then tell him the color of the third marble.

• So Tom took two blue marbles out of his

container and looked at the label. He was able to tell the color of the third marble immediately.

• Dick took 1 blue marble and 1 yellow marble from his container. After looking at his label he was able to tell the color of his remaining marble.

• Harry took 2 yellow marbles from his

container. He looked at the label in his container, but could not tell what color the remaining marble was.

• Sally, without even looking at her marbles or her label, was able to tell the scientist what color her marbles were. Can you tell what color marbles Sally had? Can you also tell what color marbles the others had, and what label was in each of their containers?


Differential Equation :

An equation involving independent variable x, dependent variable y and the differential coefficients

dxdy , 2

2

dxyd , .... is called differential equation.

Examples :

(1) dxdy = 1 + x + y

(2) dxdy + xy = cot x

(3) 3

4

4

dxyd – 4

dxdy + 4y = 5 cos 3x

(4) x22

2

dxyd +

2

1

+

dxdy = 0

Order of a Differential Equation :

The order of a differential equation is the order of the highest derivative occurring in the differential equation. For example, the order of above differential equations are 1, 1, 4 and 2 respectively.

Degree of a Differential Equation :

The degree of the differential equation is the degree of the highest derivative when differential coefficients are free from radical and fraction. For example, the degree of above differential equations are 1, 1, 3 and 2 respectively.

Linear and Non-linear Differential Equation :

A differential equation in which the dependent variable and its differential coefficients occurs only in the first degree and are not multiplied together is called a linear differential equation. The general and nth order differential equation is given below :

a0(x) n

n

dxyd + a1(x) 1

1

−

−

n

n

dxyd + .... + an – 1 dx

dy

+ an(x)y + φ(x) = 0

Those equations which are not linear are called non-linear differential equations.

Formation of Differential Equation :

(1) Write down the given equation.

(2) Differentiate it successively with respect to x that number of times equal to the arbitrary constants.

(3) And hence on eliminating arbitrary constants results a differential equation which involves x, y,

dxdy , 2

2

dxyd .....

Solution of Differential Equation :

A solution of a differential equation is any function which when put into the equation changes it into an identity.

General and particular solution :

The solution which contains a number of arbitrary constant equal to the order of the equation is called general solution by giving particular values to the constants are called particular solutions.

Several Types of Differential Equations and their Solution :

(1) Solution of differential equation

dxdy = f(x) is y = ∫ + cdxxf )(

(2) Solution of differential equation

dxdy = f(x) g(y) is ∫ )(yg

dy = ∫ + cdxxf )(

(3) Solution of diff. equation dxdy = f(ax + by + c) by

putting ax + by + c = v and dxdy =

− a

dxdv

b1

)(vbfa

dv+

= dx

Thus solution is by integrating

∫ + )(vbfadv = ∫dx

DIFFERENTIAL EQUATIONS

Mathematics Fundamentals MATHS


(4) To solve the homogeneous differential equation

dxdy =

),(),(

yxgyxf , substitute y = vx and so

dxdy = v + x

dxdv .

Thus v + xdxdv = f(v)

⇒ x

dx = vvf

dv−)(

Therefore solution is ∫ xdx = ∫ − vvf

dv)(

+ c

Equation reducible to homogeneous form :

A differential equation of the form

dxdy =

222

111

cybxacybxa

++++ ,

where 2

1

aa ≠

2

1

bb , can be reduced to homogeneous

form by adopting the following procedure :

Put x = X + h, y = Y + k,

so that dXdY =

dxdy

The equation then transformed to

dXdY =

)()(

22222

11111

ckbhaYbXackbhaYbXa

++++++++

Now choose h and k such that a1h + b1k + c1 = 0 and a2h + b2k + c2 = 0. Then for these values of h and k, the equation becomes

dXdY =

YbXaYbXa

22

11

++

This is a homogeneous equation which can be solved by putting Y = vX and then Y and X should be replaced by y – k and x – h.

Special case :

If dxdy =

´´´ cybxacbyax

++++ and

´aa =

´bb = m (say), i.e.

when coefficient of x and y in numerator and denominator are proportional, then the above equation cannot be solved by the discussed before because the values of h and k given by the equations will be indeterminate.

In order to solve such equations, we proceed as explained in the following example.

Solve dxdy =

43762

+−+−

yxyx =

437)3(2

+−+−

yxyx

== 2

´´ bb

aaobviously

Put x – 3y = v

⇒ 1 – 3dxdy =

dxdv (Now proceed yourself)

Solution of the linear differential equation :

dxdy + Py = Q, where P and Q are either constants or

functions of x, is

∫ dxPye = ∫

∫ dxPQe dx + c

Where ∫ dxPe is called the integrating factor.

Equations reducible to linear form :

Bernoulli's equation : A differential equation of

the form dxdy + Py = Qyn, where P and Q are

functions of x alone is called Bernoulli's equation.

Dividing by yn, we get y–n

dxdy + y–(n – 1). P = Q

Putting y–(n – 1) = Y, so that nyn)1( −

dxdy =

dxdY ,

we get dxdY + (1 – n)P. Y = (1 – n)Q

which is a linear differential equation.

If the given equations is of the form

dxdy + P. f(y) = Q . g(y), where P and Q are

functions of x alone, we divide the equation by g(y) and get

dxdy

yg )(1 + P.

)()(

ygyf = Q

Now substitute )()(

ygyf = v and solve.

Solution of the differential equation :

2

2

dxyd = f(x) is obtained by integrating it with respect

to x twice.


Some Important Definitions and Formulae : Measurement of angles : The angles are measured

in degrees, grades or in radius which are defined as follows:

Degree : A right angle is divided into 90 equal parts and each part is called a degree. Thus a right angle is equal to 90 degrees. One degree is denoted by 1º.

A degree is divided into sixty equal parts is called a minute. One minute is denoted by 1´.

A minute is divided into sixty equal parts and each parts is called a second. One second is denoted by 1´´.

Thus, 1 right angle = 90º (Read as 90 degrees) 1º = 60´ (Read as 60 minutes) 1´ = 60´´ (Read as 60 seconds). Grades : A right angle is divided into 100 equal parts

and each part is called a grade. Thus a right angle is equal to 100 grades. One grade is denoted by 1g.

A grade is divided into 100 equal parts and each part is called a minute and is denoted by 1´.

A minute is divided into 100 equal parts and each part is called a second and is denoted by 1"

Thus, 1 right angled = 100g (Read as 100 grades) 1g = 100´ (Read as 100 minutes) 1´ = 100´´ (Read as 100 seconds) Radians : A radian is the angle subtended at the

centre of a circle by an arc equal in length to the radius of the circle.

Domain and Range of a Trigono. Function : If f : X → Y is a function, defined on the set X, then

the domain of the function f, written as Domf is the set of all independent variables x, for which the image f(x) is well defined element of Y, called the co-domain of f.

Range of f : X → Y is the set of all images f(x) which belongs to Y, i.e.,

Range f = f(x) ∈ Y : x ∈ X ⊆ Y The domain and range of trigonmetrical functions are

tabulated as follows :

Trigo. Function

Domain Range

sin x R, the set of all the real number

–1 ≤ sin x ≤ 1

cos x R – 1 ≤ cos x ≤ 1

tan x

R –

∈

π+ Inn ,

2)12(

R

cosec x R – n π, n ∈ I R – x : –1 < x < 1

sec x

R –

∈

π+ Inn ,

2)12(

R – x : –1 < x < 1

cot x R – n π, n ∈ I R

Relation between Trigonometrically Ratios and identities:

tan θ = θθ

cossin ; cot θ =

θθ

sincos

sin A cosec A = tan A cot A = cos A sec A = 1 sin2θ + cos2θ = 1 or sin2θ = 1 – cos2θ or cos2θ = 1 – sin2θ 1 + tan2θ = sec2θ or sec2θ – tan2θ = 1 or sec2θ – 1 = tan2θ 1 + cot2θ = cosec2θ or cosec2θ – cot2θ = 1 or cosec2θ – 1 = cot2θ Since sin2A + cos2A = 1, hence each of sin A and

cos A is numerically less than or equal to unity. i.e.

| sin A| ≤ 1 and | cos A | ≤ 1 or –1 ≤ sin A ≤ 1 and – 1 ≤ cos A ≤ 1 Note : The modulus of real number x is defined as

|x| = x if x ≥ 0 and |x| = – x if x < 0. Since sec A and cosec A are respectively

reciprocals of cos A and sin A, therefore the values of sec A and cosec A are always numerically greater than or equal to unity i.e.

sec A ≥ 1 or sec A ≤ – 1 and cosec A ≥ 1 or cosec A ≤ – 1 In other words, we never have

TRIGONOMETRICAL RATIOS

Mathematics Fundamentals MATHS


–1 < cosec A < 1 and –1 < sec A < 1.

Trigonometrical Ratios for Various Angles :

θ 0 6π

4π

3π

2π

π 2

3π 2π

sin θ

0 21

2

1

23

1 0 –1 0

cos θ

1 23

2

1

21

0 –1 0 1

tan θ

0 3

1 1 3 ∞ 0 ∞ 0

Trigonometrical Ratios for Related Angles :

θ – θ

2π

±

θ

π ± θ

23π

± θ

2π ± θ

sin – sin θ

cos θ m sin θ

– cos θ

± sin θ

cos cos θ m sin θ

– cos θ

± sin θ

cos θ

tan – tan θ

m cot θ

± tan θ

m cot θ

± tan θ

cot – cot θ

m tan θ

± cot θ

m tan θ

± cot θ

Addition and Subtraction Formulae : sin (A ± B) = sin A cos B ± cos A sin B cos (A ± B) = cos A cos B m sin A sin B

tan (A ± B) = BABA

tantan1tantan

m

±

cot (A ± B) = AB

BAcotcot

1cotcot±

m

sin (A + B) sin (A – B) = sin2A – sin2B = cos2B – cos2A cos (A + B) cos (A – B) = cos2A – sin2B = cos2B – sin2A Formulae for Changing the Sum or Difference into Product :

sin C + sin D = 2 sin 2

DC + cos 2

DC −

sin C – sin D = 2 cos 2

DC + sin 2

DC −

cos C + cos D = 2 cos 2

DC + cos 2

DC −

cos C – cos D = 2 sin 2

DC + sin 2

CD −

Formulae for Changing the Product into Sum or Difference : 2 sin A cos B = sin (A + B) + sin (A – B)

2 cos A sin B = sin (A + B) – sin (A – B) 2 cos A cos B = cos (A + B) + cos (A – B) 2 sin A sin B = cos (A – B) – cos(A + B)

Formulae Involving Double, Triple and Half Angles :

sin 2θ = 2 sin θ cos θ = θ+

θ2tan1

tan2

cos 2θ = cos2 θ – sin2 θ = 2 cos2θ – 1

= 1 – 2 sin2θ = θ+θ−

2

2

tan1tan1

sin 2θ = ±

2cos1 θ− ; cos

2θ = ±

2cos1 θ+

tan 2θ = ±

θ+θ−

cos1cos1

tan 2θ = θ−

θ2tan1

tan2

sin 3θ = 3 sin θ – 4 sin3θ

or sin3θ = 41 (3 sin θ – sin 3θ)

cos 3θ = 4 cos3θ – 3 cos θ

or cos3θ = 41 (3 cos θ + cos 3θ)

tan 3θ = θ−

θ−θ2

3

tan31tantan3

π

+π≠θ6

n

Trigonometrical Ratios for Some Special Angles :

θ 21º7 15º 22

21º

sin θ 22

624 −−2213 −

222 −

cos θ 22

624 ++2213 +

222 +

tan θ

( 3 – 2 )

( 2 –1) 2 – 3 2 – 1

θ 18º 36º

sin θ 4

15 −

45210 −

cos θ 4

5210 + 4

15 +

tan θ 5

51025 − 525 −

Important Points to Remember :


Maximum and minimum values of

a sin x + b cos x are + 22 ba + , – 22 ba + respectively.

sin2x + cosec2x ≥ 2 for every real x. cos2x + sec2x ≥ 2 for every real x. tan2x + cot2x ≥ 2 for every real x

If x = sec θ + tan θ, then x1 = sec θ – tan θ

If x = cosec θ + cot θ, then x1 = cosec θ – cot θ

cos θ . cos 2θ . cos 4θ . cos 8θ

.... cos 2n–1θ = θθ

sin22sin

n

n

sin θ sin (60º – θ) sin (60º + θ) = 41 sin 3θ

cos θ cos (60º – θ) cos (60º + θ) = 41 cos 3θ

tan θ tan (60º – θ) tan (60º + θ) = tan 3θ

Conditional Identities : 1. If A + B + C = 180º, then sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C sin 2A + sin 2B – sin 2C = 4 cos A cos B sin C sin (B + C – A) + sin (C + A – B) + sin (A + B – C) = 4 sin A sin B sin C cos 2A + cos 2B + cos 2C = –1 – 4 cos A cos B cos C cos 2A + cos 2B – cos 2C = 1 – 4 sin A sin B cos C 2. If A + B + C = 180º, then

sin A + sin B + sin C = 4 cos2A cos

2B cos

2C

sin A + sin B – sin C = 4 sin2A sin

2B sin

2C

cos A + cos B + cos C = 1 + 4 sin2A sin

2B sin

2C

cos A + cos B – cos C = –1 + 4cos2A cos

2B sin

2C

CB

Asinsin

cos + AC

Bsinsin

cos + BA

Csinsin

cos = 2

3. If A + B + C = π, then sin2A + sin2B – sin2C = 2 sin A sin B cos C cos2A + cos2B + cos2C = 1 – 2 cos A cos B cos C sin2A + sin2B + sin2C = 2 + 2 cos A cos B cos C cos2A + cos2B – cos2C = 1 – 2 sin A sin B cos C 4. If A + B + C = π, then

sin2

2A + sin2

2B + sin2

2C = 1 – 2 sin

2A sin

2B sin

2C

cos2

2A +cos2

2B + cos2

2C = 2 + 2 sin

2A sin

2B sin

2C

sin2

2A + sin2

2B – sin2

2C = 1 – 2cos

2A cos

2B cos

2C

cos2

2A + cos2

2B – cos2

2C = 2cos

2A cos

2B sin

2C

5. If x + y + z = π/2, then sin2x + sin2y + sin2z = 1 – 2 sin x sin y sin z cos2x + cos2y + cos2z = 2 + 2 sin x sin y sin z sin 2x + sin 2y + sin 2z = 4 cos x cos y cos z 6. If A + B + C = π, then tan A + tan B + tan C = tan A tan B tan C cot B cot C + cot C cot A + cot A cot B = 1

tan2B tan

2C + tan

2C tan

2A + tan

2A tan

2B = 1

cot 2A + cot

2B + cot

2C = cot

2A cot

2B cot

2C

7. (a) For any angles A, B, C we have sin (A + B + C) = sin A cos B cos C + cos A sin B cos C

+ cos A cos B sin C – sin A sin B sin C cos (A + B + C) = cos A cos B cos C – cos A sin B sin C – sin A cos B sin C – sin A sin B cos C tan(A + B + C)

= ACCBBA

CBACBAtantantantantantan1tantantan–tantantan

−−−++

(b) If A,B, C are the angles of a triangle, then sin(A + B + C) = sin π = 0 and cos (A + B + C) = cos π = –1 then (a) gives sin A sin B sin C = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C and (a) gives 1 + cos A cos B cos C = cos A sin B sin C + sin A cos B sin C + sin A sin B cos C Method of Componendo and Dividendo :

If qp =

ba , then by componendo and dividendo we

can write

qpqp

+− =

baba

+− or

pqpq

+− =

abab

+−


or qpqp

−+ =

baba

−+ or

pqpq

−+ =

abab

−+


a

PHYSICS

Questions 1 to 9 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer.

1. A 6 kg block rests as shown on the upper surface of a 15 kg wedge neglecting friction, the system is released from rest. The acceleration of B relative to A is (g = 10 m/sec2)

B

A

6 kg

15 kg

30º

(A) 5.5 m/sec2 (B) 2.75 m/sec2 (C) 11 m/sec2 (D) 8.25 m/sec2 2. A ideal gas having molar specific heat at constant

volume CV. If is undergoing a process where temperature is varying as T = T0eαV where α is constant and ''V'' is the volume occupied by the gas. The molar specific heat of the gas for the given process as a function of volume is given by -

(A) CV + VRα (B) CV +

VR

α

(C) CV + V

Rα2 (D) CV + V

Rα2

3. An earthen pitcher loses 1 gm of water per minute due to evaporation. If the water equivalent of pitcher is 0.5 kg and pitcher contains 9.6 kg of water, calculate the time required for the water in pitcher to cool to 25º C from original temperature 30ºC. Neglect radiation effects. Latent heat of vaporization in this range of temperature is 580 cal/gm and specific heat of water is 1 cal/gm ºC

(A) 30.5 min (B) 41.2 min (C) 38.6 min (D) 34.5 min

4. A point object and a plane mirror are moving as shown figure at t = 0. The x-comp. of velocity of image at t = 1 sec will be -

(A) 6 m/s toward left (B) 6 m/s towards right (C) 33 m/s towards left (D) None of these 5. A convex lens of focal length 10 m starts falling with

its plane horizontal. There is a stationary point object 20 m above it in its axis. Find the velocity of its image at t = 1 sec

20 m/s

g = 10 m/s2

at t = 0

IIT-JEE 2011

XtraEdge Test Series # 9

Based on New Pattern

Time : 3 Hours Syllabus : Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus Instructions : Section - I • Question 1 to 9 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct

answer and -1 mark for wrong answer. • Question 10 to 13 are Reason and Assertion type question with one is correct answer. +3 marks and –1 mark for

wrong answer. • Question 14 to 19 are passage based single type questions. +4 marks will be awarded for correct answer and

-1 mark for wrong answer.. Section - II • Question 20 to 22 are Column Matching type questions. +6 marks will be awarded for the complete correctly

matched answer and No Negative marks for wrong answer. However, +1 mark will be given for a correctly marked answer in any row.

2m/s

1m/s

1m/s2

2m/s2

30º


(A) 7.5 m/s↑ (B) 7.5 m/s↓ (C) 12.5 m/s↑ (D) 5 m/s↑

6. 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

2F

4V 2Ω

k3

k2

k1

0.5Ω

2H

(A) Cbe

1–14 (B) Cb

e

1–124

(C) Cbe

1–18 (D) zero

7. 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) –R

KQ2

8. A wire loop ABCDE carrying a current I is placed in x-y plane as shown in figure. A particle of mass m and charge q is projected from origin with velocity

Vr

= 20V )ji( ^^+ m/s. The instantaneous acceleration

acts along the (r-radius of circular arc ABC) y

45º xO

D

E

90º

r/2

B

C

A

(A) AO (B) OA (C) x-axis (D) OP 9. A source is moving across a circle given by equation

x2 + y2 = R2 with constant speed 36

330π m/sec in

anticlockwise sense. A detector is at rest at point (2R, 0) w.r.t. centre of circle. If the frequency emitted by the source is f0 and speed of sound is 330 m/sec. Then

(A) The position of source when detector records the

maximum frequency is

+

2R,–

2R3

(B) The coordinate of source when detector records minimum frequency is [R, R]

(C) Maximum frequency recorded by detector is

3636

+π f0

(D) Minimum frequency recorded by detector is

π–3636 f0

This section contains 4 questions numbered 10 to 13, (Reason and Assertion type question). Each question contains Assertion (A) and Reason (R). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. +3 marks will be given for each correct answer and – 1 mark for each wrong answer.

The following questions given below consist of an "Assertion" (A) and "Reason" (R) Type questions. Use the following Key to choose the appropriate answer.

(A) If both (A) and (R) are true, and (R) is the correct explanation of (A).

(B) If both (A) and (R) are true but (R) is not the correct explanation of (A).

(C) If (A) is true but (R) is false. (D) If (A) is false but (R) is true.

10. Assertion (A) : In case of an electron and a photon having same momentum, wavelength associated with electron is smaller.

Reason (R) : Electron cannot move with a speed of photon.

11. Assertion (A) : In the process of photoelectric emission by monochromatic light, all the emitted photo-electrons possess the same kinetic energy.

Reason (R) : In photoelectric effect a single photon interacts with a single electron and electron is emitted only if energy of each of incident photon is greater than the work function.

12. Assertion (A) : During phase transformation internal energy of material doesn't changes.

Reason (R) : Temperature of material remains same during phase-transformation.

13. Assertion (A) : At the temperature T1, vrms of hydrogen may be equal to vrms of oxygen at temperature T2.

Reason (R) : vrms speed of a gas is proportional to the square root of its absolute temperature.


This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 14 to 19) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. +4 marks will be given for each correct answer and – 1 mark for each wrong answer.

Passage # 1 (Ques. 14 to 16) In a YDSE setup (see figure) the light source

executes SHM between P and Q according to equation n = A sin ωt. S being the mean position. Assume d <<< D, A <<< L and ω small enough to neglect Doppler effect. If source were stationary at S. intensity at O would be I0

x

S1

S2

d O

D L

Q S P

O : point of central 14. The fractional change in intensity of central maxima

as a function of time is -

(A) L

tA ωsin (B) L

tA ωsin2

(C) L

tA ωsin3 (D) L

tA2sin ω

15. When source comes towards Q - (A) The bright fringes will be less bright (B) The bright fringes will be more bright (C) The fringe width will increase (D) The fringe width will decrease 16. The fringe width β can be expressed as - (A) β = β0 sin ωt (B) β = β0 cos ωt (C) β = β0 cos 2 ωt (D) None of these Passage # 2 (Ques. 17 to 19) Circuit shown in figure is a part of circuit. The

potential at different points are as indicated in the diagram. Capacitance of all capacitors are in µF.

7V

10V

5V 20V

5V1µF

8µF

4µF 10µF

2µF

5µF 6µF

G H

10V

17. Potential at G is approximately - (A) 9.82 V (B) 8.32 V (C) 6.25 V (D) 2.5 V 18. Potential at H is approximately - (A) 9.75 V (B) 13.18 V (C) 9 V (D) 11 V

19. Charge (in µC) across 10 µF capacitor is nearly - (A) 18.2 (B) 33.6 (C) 16.4 (D) 42.8

This section contains 3 questions (Questions 20 to 22). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S, then the correctly bubbled 4 × 4 matrix should be as follows :

ABCD

P Q R S

S

P

P P Q R

R R

Q Q

S S

P Q R S

Mark your response in OMR sheet against the question number of that question in section-II. + 6 marks will be given for complete correct answer and No Negative marks for wrong answer. However, 1 mark will be given for a correctly marked answer in any row.

20. A block placed on rough inclined plane. Angle of inclination θ of the plane as shown in varied starting from zero. The coefficient of static friction and kinetic friction between the block and the plane is µs and µk respectively (µs > µk). Column-II shows the graphs which necessarily contains θ taken on x-axis. Column-I represents the quantities taken on y-axis of column-I

m

θ

Column-I Column-II (A) Friction force between (P) the block and plane (B) Normal force between (Q) the block and plane (C) Total contact force (R) between the block and plane (D) Acceleration of block (S)

21. The figure shows two mirrors placed as shown near

monochromatic light source S. (wavelength λ and intensity at point P = I0). If Inet is the resultant intensity at point P due to super position of direct and reflected waves. Then match the following columns.


P S d/2

d/2

D

Column-I Column-II

(A) If Dd 2

= λ (P) Inet = 9I0

(B) If Dd 2

= 2λ (Q) Inet = 5I0

(C) If Dd 2

<<< λ (R) Inet = I0

(D) If one mirror is (S) Inet = 4I0 removed and

Dd 2

= λ

22. Twelve identical metal rods are joined to form a cube as shown in figure. A1, A2 ... A4 are corner of cube P, Q and R are mid-point of A1A5, A4A8 and A5A8 respectively. When A1 and A3 are maintained at 120º C and 0ºC respectively, heat flown across A1 and A3 is 80 Joule/sec.

Column-I contains different location in cube Column-II contains temperature at and current through that locations.

A1

P

A5 R

A8 A7

A6

A3

A2

A4

Q

Column-I Column-II

(A) P (P) 60ºC (B) Q (Q) 100ºC (C) R (R) 70ºC (D) A4 (S) 10 J/sec

CHEMISTRY


1. Vander Waal's equation for a real gas is

+ 2

2

VanP (V – nb) = nRT

Plot of quantity Q = ba

ab+

with temperature is

(A)

T

Q (B)

Q

T

(C)

T

Q (D)

T

Q

2. A solution containing NaOH and Na2CO3 was titrated

against HCl using phenolphthalein as an indicator. The tire value of HCl solution was found to be x ml. At the end point, methyl orange was added and the titration continued. A further y ml of HCl solution was required to get the end point with methyl orange. The volume of HCl solution used with Na2CO3 during the whole process is

(A) 2x (B) 2y (C) x (D) y – x 3. Equal volumes of 1 M KMnO4 and 1 M K2Cr2O7 are

used to oxidise ferrous ion (Fe2+) solution in acidic medium. The amount of ferrous ions used with KMnO4 is

(A) less than that used with K2Cr2O7 (B) more than that used with K2Cr2O7 (C) equal to that used with K2Cr2O7 (D) cannot be compared 4. For crystallisation of a solid from the aqueous

solution, if the values of ∆H and ∆S are –x J mol–1 and – y J K–1 mol–1 respectively, which of the following relationships is correct

(A) x = T × y (B) x > T × y (C) x < T × y (D) None of these 5. Hexamethylenediamine, C6H16N2, is one of the

starting materials for the production of nylon. It can be prepared from adipic acid C6H10O4, by the following reaction

C6H10O4(l) + 2NH3(g) + 4H2(g) → C6H16N2(l) + 4H2O(l) If 385 g of hexamethylenediamine is made from

5.00 × 102 g of adipic acid, the percent yield is (A) 24.2 % (B) 75.0%

(C) 96.9 % (D) 99.9%


6. One volume of an organic gaseous compound containing carbon, hydrogen and nitrogen yielded 3 volumes of CO2, 4.5 volumes of water vapours and half volume of N2 on complete oxidation. The molecular formula of the organic compound is

(A) C3H9N (B) C3H9N2 (C) C3H3N3 (D) CH2N2 7. Reduction of but-2-yne with Na and liquid NH3 gives

an alkene which upon catalytic hydrogenation with D2/Pt gives an alkane. The alkene and alkane formed respectively are

(A) cis but-2-ene and racemic-2, 3-dideuterobutane (B) trans but-2-ene and meso 2, 3-dideuterobutane (C) trans but-2-ene and racemic 2, 3-dideuterobutane (D) cis but-2-ene and meso 2, 3-dideuterobutane 8. The order of Keq values for the following keto-enol

equilibrium constants is

CH3–CHO K1

CH2 = CH–OH

K2

CH3–C–CH2–C–CH3

O O

CH3–C=CH–C–CH3

OH O

K3 CH3–C–CH3

O

CH2=C–CH3

OH

(A) K1 > K2 > K3 (B) K2 > K3 > K1 (C) K2 > K1 > K3 (D) K1 > K3 > K2

9. +

O 1. AlCl3

2. H+/H2O A

The product 'A' is

(A) Me

PhOH

(B)

O

Ph

OH

(C)

OH

H

MePh (D)

PhH

This section contains 4 questions numbered 10 to 13, (Reason and Assertion type question). Each question contains Assertion (A) and Reason (R). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. +3 marks will be given for each correct answer and – 1 mark for each wrong answer.




(C) If (A) is true but (R) is false. (D) If (A) is false but (R) is true. 10. Assertion (A) : The heat absorbed during the

isothermal expansion of an ideal gas against vacuum is zero.

Reason (R) : The volume occupied by the molecules of an ideal gas is zero.

11. Assertion (A) : The value of van der Waals constant

'a' is larger for ammonia than for nitrogen. Reason (R): Hydrogen bonding is present in

ammonia. 12. Assertion (A): 3-hydroxy - butan-2-one on treatment

with [Ag(NH3)2]⊕ cause precipitation of silver. Reason (R): [Ag(NH3)2] ⊕ oxidises 3-hydroxy butan-

2-one to butan-2-3-dione 13. Assertion (A): HBr adds to 1,4-pentadiene at a faster

rate than to 1,3-pentadiene Reason (R): 1,4-pentadiene is less stable than

1,3-pentadiene. This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 14 to 19) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. +4 marks will be given for each correct answer and – 1 mark for each wrong answer.

Passage # 1 (Ques. 14 to 16)

The temperature dependence of the dissociation constant for the formic acid in aqueous is given by

log Ka = – T

1400 + 5 – 0.01 T

The enthalpy of neutralisation between strong acid and strong base at 27°C is equal to –56 kJ/equal.

14. What is the enthalpy of neutralisation of HCOOH

against NaOH at 27°C ? (A) –56 kJ/mol (B) + 56 kJ/mol (C) –46.427 kJ/mol (D) –9.573 kJ/mol

15. What is the standard entropy change for the dissociation of HCOOH at 300 K ?

(A) –19.147 J K–1 mol–1

(B) +19.147 kJ K–1 mol–1 (C) –73 J K–1 mol–1 (D) – 40 JK–1 mol–1


16. At what approximate temperature the dissociation for formic acid is maximum ?

(A) 300 K (B) 374 K (C) 474 K (D) There is no maximum temperature for dissociation Passage # 2 (Ques. 17 to 19)

OH

CH3

CHCl3 + NaOH

P HCN/NH4Cl H3O+/∆ (R) α-amino acid

(zwitter ion)O

CH3 CHCl2

H⊕

Isomerisation

OHΘ Q

17. Product P is -

(A)

OH

CHO

(B) OH

CHO

(C)

OH

CH3

OHC CHO (D)

OH CHO

CH3

18. Product R is -

(A)

OH

CH3

CH–CH2–COOΘ

NH3 ⊕ (B)

OH

CH3

CH–COOΘ

NH3

⊕

(C)

OΘ

CH3

CH–COOH

NH3 ⊕

(D) OH

CH–COOΘ

NH3 ⊕

19. Product Q is -

(A)

OH

CHO CH3

(B)

OH

CHCl2

CH3

(C)

OH

CHO

CH3 (D)

O

CHOCH3


ABCD

P Q R S

S

P

P P Q R

R R

Q Q

S S

P Q R S

Mark your response in OMR sheet against the question number of that question in section-II. + 6 marks will be given for complete correct answer and No Negative marks for wrong answer. However, 1 mark will be given for a correctly marked answer in any row. 20. Column –I Column II (A) NH3 → NO3

– (P) M/20 (B) Fe2S3 → 2FeSO4 + SO2 (Q) M/5 (C) KMnO4 in acidic medium (R) M/8 (D) CuS → CuSO4 (S) M 21. Column –I Column II (A) Cyclopropenyl carbocation (P) Hyperconjugation (B) Cyclopentadienyl anion (Q) All carbon atoms are sp2 hybridized (C) Benzyne (R) Aromatic nature (D) t-Butyl carbocation (S) Diamagnetic 22. Column –I Column II (A) Pyrrole (P) Explosive (B) Nitrobenzene (Q) Aromatic (C) Trinitrotoluene (R) sp2 hybrid oxygen (D) Benzaldehyde (S) less reactive benzene in electrophilic substitution reaction

MATHEMATICS



1. If α, β, γ, δ are four complex numbers such that δγ is

real and αδ – βγ ≠ 0, then z = tt

δ+γβ+α , t ∈ R

represents a (A) circle (B) parabola (C) ellipse (D) straight line

2. Let S denote the set of all real values of a for which the roots of the equation (1 + a) x2 – 3ax + 4a = 0 exceed 1, then S equals

(A)

0,

716– (B)

1,–

716–

(C) (– 1 , 2) (D) (–1, 3) 3. Sum of the coefficients of the terms of degree m in

the expansion of (1 + x)n (1 + y)n (1 + z)n is (A) (nCm)3 (B) 3(nCm) (C) nC3m (D) 3nCm

4. The system of linear equations x + y + z = 6, x + 2y + 3z = 14 and 2x + 5y + λz = µ(λ, µ ∈ R) has a unique solution

(A) λ ≠ 8 (B) λ = 8, µ ≠ 36 (C) λ = 8, µ = 36 (D) none of these

5. The plane 2x – y + 3z + 5 = 0 is rotated through 90º about its line of intersection with the plane 5x – 4y – 2z + 1 = 0. The equation of the plane in the new position is

(A) 6x – 9y – 29z – 31 = 0 (B) 27x – 24y – 26z – 13 = 0 (C) 43x – 32y – 2z + 27 = 0 (D) 26x – 43y – 151z – 165 = 0

6. If f(x) – f(x) =

xyyxf

–1– and f has domain (–1, 1)

then a function satisfying the above functional equation is

(A) 2 + log

+

xx

–11 (B) log

+ xx

1–1

(C) log 2–12

xx (D) tan–1

+

xx

–11

7. Tangent is drawn to ellipse 27

2x + y2 = 1 at

(3 3 cos θ, sin θ) (where θ ∈ (0, π/2)). Then the value of θ such that sum of intercepts on axes made by this tangent is least is

(A) π/3 (B) π/6 (C) π/8 (D) π/4

8. If I = ( )∫

+2/322 xax

dx , then I is equal to

(A) – 22 xax

ax

+

+ + C (B) – 22

1

xax

axa +

+ + C

(C) – 22

2

1

xax

axa +

+ + C (D) – 33

2

1

xax

axa +

+ + C

9. If for k ∈ N,

xkx

sin2sin = 2[cos x + cos 3x + ... + cos (2k – 1)x],

then value of I = ∫π 2/

0

2sin kx cot x dx is

(A) – π/2 (B) 0 (C) π/2 (D) π This section contains 4 questions numbered 10 to 13, (Reason and Assertion type question). Each question contains Assertion (A) and Reason (R). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. +3 marks will be given for each correct answer and – 1 mark for each wrong answer.




(C) If (A) is true but (R) is false. (D) If (A) is false but (R) is true. 10. Assertion (A) : cosec–1 (3/2) + cos–1 (2/3) – 2 cot–1(1/7)

– cot–1 7 is equal to cot–1 7. Reason (R) : sin–1 x + cos–1 x = π/2, tan–1x + cot–1 x

= π/2, cosec–1 x = sin–1 (1/x), cot–1 (x) = tan–1 (1/x) 11. Assertion (A) : Equation of a circle through the origin

and belonging to the co-axial system, of which the limiting points are (1, 1) and (3, 3) is

2x2 + 2y2 – 3x – 3y = 0 Reason (R) : Equation of a circle passing through

the points (1, 1) and (3, 3) is x2 + y2 – 2x – 6y + 6 = 0 12. Assertion (A) : The absolute minimum value of

|x – 1| + | x – 2| + |x – 3| is 2. Reason (R) : The function |x –1| + |x – 2| + |x – 3| is

differentiable on R ~ 1, 2.


13. Assertion (A) : Curve satisfying the differential equation y' = y/2x passing through (2, 1) is a parabola with focus (1/4, 0).

Reason (R) : The differentiable equation y' = y/2x is of variable separable.

This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 14 to 19) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. +4 marks will be given for each correct answer and – 1 mark for each wrong answer.

Passage # 1 (Ques. 14 to 16) There are n urns each containing (n +1) balls such

that the ith urn contains 'i' white balls and (n + 1 – i) red balls. Let ui be the event of selecting ith urn, i = 1, 2, 3, ..., n and W denotes the event of getting a white balls.

14. If P(ui) ∝ i, where i = 1, 2, 3, ..., n, then ∞→n

lim P(W) is

equal to :

(A) 1 (B) 32

(C) 41 (D)

43

15. If P(ui) = c, where c is a constant, then P(ui/W) is equal to :

(A) 1

2+n

(B) 1

1+n

(C) 1+n

n (D) 21

16. If n is even and E denotes the event of choosing even

numbered urn

=

nuP i

1)( , then the value of

P(W/E) is :

(A) 122

++

nn (B)

)1(22

++

nn

(C) 1+n

n (D) 1

1+n

Passage # 2 (Ques. 17 to 19)

For every function f(x) which is twice differentiable, these will be good approximation of

∫

≅

b

a

cbdxxf2–)( f(a) + f(b). Now if we take

c = 2

ba + , then using above again, we get

∫b

a

dxxf )( = ∫c

a

dxxf )( + ∫b

c

dxxf )(

≅ 4– ab f(a) + f(b) + 2f(c)and so on.

We get approximation for value of ∫b

a

dxxf )( .

17. Good approximation of ∫π 2/

0

sin x dx, is :

(A) 4π (B) )12(

4+

π

(C) )12(8

+π (D)

8π

18. If f " (x) < 0, ∀ x ∈ (a, b), and (c, f(c)) is point of

maxima where c ∈ (a, b), then f ' (c) is :

(A) ab

afbf–

)(–)( (B)

abafbf

–)(–)(3

(C)

abafbf

–)(–)(2 (D) 0

19. If 3)–(

)()(2

)–(–)(

limat

aftfatdxxft

aat

∫ +

→ = 0,

then degree of polynomial function f(x) at-most is: (A) 0 (B) 1 (C) 3 (D) 2 This section contains 3 questions (Questions 20 to 22). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S, then the correctly bubbled 4 × 4 matrix should be as follows :

ABCD

P Q R S

S

P

P P Q R

R R

Q Q

S S

P Q R S



20. Column-I Column-II (A) Two rays in I quadrant (P) 2 x + y = |a| and ax – y = 1 intersect each other in the interval a ∈ (a0, ∞), then value of (2a0/3) is (B) Point (α, β, γ) lie on the (Q) 3/2 plane x + y + z = 2, let

→a =

^iα +

^jβ +

^kγ

^k × )(

^akr

× = 0, then γ is

(C) ∫1

0

2 )–1( dyy + ∫0

1

2 )1–( dyy (R) ∫1

0

–1 dxx

+ ∫ +0

1–

1 dxx

(D) If sin A sin B sin C (S) 1 + cos A cos B = 1, then the value of sin C = 21. Column-I Column-II (A) Let f(x) = [x – 1] + [1 – x], (P) continuous at a [x] is the G.I.F., a is an integer (B) Let f be as in (A) a is not (Q)

0lim→x

f(x) does not

an integer exist

(C) f(x) = xx

a xx

+

+

][1–][

, x ≠ 0 (R) f(a) = 0

(D) f(x) = xx

aa xx

cos–cot– coscot

(S)2/

limπ→x

f(x) = log a

22. Let ak = nCk for 0 ≤ k ≤ n and Ak =

k

k

aa

001– and

B = 1

1–

1

. +=

∑ k

n

kk AA =

b

a0

0

Column-I Column-II

(A) a (P) 1

2+nn (2nCn)

(B) a – b (Q) 0 (C) a + b (R) 2nCn+1

(D) ba (S) 1

SCIENCE TIPS

• What is the expression for growing current, in LR

circuit ? I = I0

−

− tLR

e1

• What is the range of infrared spectrum ? This covers wavelengths from 10–3 m down to 7.8 × 10–7 m

• What is the nature of graph between electric field and potential energy (U) ?

The nature of the graph will be parabola having symmetry about U-axis

• Why no beats can be heard if the frequencies of the two interfering waves differ by more than ten ? this is due to persistence

of hearing

• Why heating systems based on steam are more efficient than those based on circulation of hot water ? This is because steam

has more heat than water a the same temperature

• Can the specific heat of a gas be infinity ? Yes

• What is the liquid ascent formula for a capillary ?

h = pgcosT2

γθ –

3r

where h is the height through which a liquid of density ρ and

surface tension T rises in a capillary tube of radius r

• What is the expression for total time of flight (T)

for oblique projection ? T = gsinu2 θ

• The space charge limited current iP in the diode value is given by iP = k Vp

3/2

• What is an ideal gas ? An ideal gas is one in which intermolecular

forces are absent

• Can a rough sea be calmed by pouring oil on its surface ? Yes

• What is the expression for fringe width (β) in Young's double slit experiment? β=Dλ/d where

D is the distance between the source and screen and d is distance between two slits


PHYSICS


1. A man is throwing bricks of mass 2 kg each onto a floor of height 2 m. Bricks reach to floor with speed

102 m/s. If man throws 10 bricks in a minute, power of man is –

(A) 3

40 Watt (B) 445 Watt

(C) 3

20 Watt (D) 3

35 Watt

2. Figure shows a sphere of radius R from which a sphere of radius R/2 is cut. There is a point 'P' on x-axis. Let E1 and E2 respectively be gravitation field at point P when it is at distance r1 and r2 from centre. Then–

y

x

(A) r1 < r2 < R then E1 = E2 (B) r1 < r2 < R then E1 < E2 (C) r1 < R < r2 then E1 > E2 (D) R < r1 < r2 then E1 < E2

3. A ball is thrown in air making some angle with horizontal. Considering buoyancy due to air which is

equal to 501 th of weight of ball, percentage change

in range of ball is – (A) Zero (B) – 2% (C) 2% (D) 5% 4. Two blocks of mass m1 and m2 are kept on

frictionless inclined plane as shown in figure. Friction coefficient between blocks is 'µ'. Friction on block 1 is –

21

θ (A) µm1g cos θ up the plane (B) µm1g cos θ down the plane (C) m1g sin θ up the plane (D) Zero

5. A block is projected up an incline plane having angle of inclination 60º with speed gh6 . Maximum height attained by block is –

uh

(A) 2.5 h (B) 2 h (C) 1.5 h (D) 0.5 h

IIT-JEE 2012

XtraEdge Test Series # 9

Based on New Pattern

Time : 3 Hours Syllabus : Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus Instructions : Section - I • Question 1 to 9 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct

answer and -1 mark for wrong answer. • Question 10 to 13 are Reason and Assertion type question with one is correct answer. +3 marks and –1 mark for

wrong answer. • Question 14 to 19 are passage based single type questions. +4 marks will be awarded for correct answer and

-1 mark for wrong answer.. Section - II • Question 20 to 22 are Column Matching type questions. +6 marks will be awarded for the complete correctly

matched answer and No Negative marks for wrong answer. However, +1 mark will be given for a correctly marked answer in any row.


6. AA′ and BB′ are two parallel axis through a body of mass 'M' shown in figure. Point 'C' is center of mass of the body. If IAA′ = I0, then IBB′ is equal to –

B′ A′

B A

d

d

C

(A) I0 + Md2 (B) I0 + 2Md2 (C) I0 – Md2 (D) I0 + 3Md2 7. An ice-cube is put into a glass of water. Water in

glass will get cooled primarily due to – (A) Convection (B) Conduction (C) Radiation (D) Conduction & Radiation

8. Minimum height through which an ice cube must fall so that all ice melt (assume that any loss of mechanical energy get converted into thermal energy and Lf : Latent heat of fusion) is –

(A) g

L f

2

(B) g

L f

(C) gL f2

(D) will depend on mass of ice-ball 9. A liquid X is kept in calibrated container 'C'.

Coefficient of cubical expansion of liquid and container is γl and γc respectively. If level of liquid is at 200 cc mark, then –

(A) volume of container over liquid surface will not change if γl = γc

(B) level of liquid will remain at 200 cc mark, if γl = γc

(C) level of liquid will may go above 200 cc mark, if γl < γc

(D) none of these This section contains 4 questions numbered 10 to 13, (Reason and Assertion type question). Each question contains Assertion (A) and Reason (R). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. +3 marks will be given for each correct answer and – 1 mark for each wrong answer.




(C) If (A) is true but (R) is false. (D) If (A) is false but (R) is true. 10. Assertion (A) : A block moving with constant speed

on a horizontal surface along a straight line cannot be in translational equilibrium.

Reason (R) : Translational equilibrium means net force acting on it is zero

11. Assertion (A) : At the highest position of a parabolic

trajectory of a projectile, time rate of change of speed is zero.

Reason (R) : Angle between velocity vector and

acceleration vector is 2π at this position.

12. Assertion (A) : large normal force is required to

move apart two glass plates enclosing a thin water film.

Reason (R) : Due to surface tension, water pulls each plate towards each other.

13. Assertion (A) : A perfectly black coloured body

must acts as a black body. Reason (R) : A perfectly black coloured body absorb

all light falling on it. This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 14 to 19) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. +4 marks will be given for each correct answer and – 1 mark for each wrong answer.

Passage # 1 (Ques. 14 to 16) A uniform disc of radius R and mass M has a round

cut as shown in figure. The mass of remaining portion has mass 'm'.

B

C′

B′

AA′C


14. Moment of inertia of body about BB′ is –

(A) 8

11 2MR (B) 8

15 MR2

(C) 4815 MR2 (D)

4811 MR2

15. Moment of inertia of body about CC′ is –

(A) 8

MR15 2 (B)

811 MR2

(C) 4815 MR2 (D)

7127 MR2

16. If body is made free to rotate about horizontal axis AA′, angular acceleration of body when centre of body and axis AA′ lie on same horizontal plane –

(A) Rg

207 (B)

R15g8

(C) Rg

4528 (D) None of these

Passage # 2 (Ques. 17 to 19) A cylindrical vessel of radius 'R' filled with water is

rotated about its vertical axis with a constant angular velocity 'ω'.

17. Shape of the meniscus will be – (A) circular away from centre (B) parabolic away from centre (C) convex (D) none of these

18. Difference of height of the liquid level at centre and surface of cylinder is –

(A) g2

2ω . R (B) g

2ω R2

(C) g2

2ω . R2 (D) 23

g

2ω R2

19. If pressure at centre is P0 and density of water is 'ρ' then pressure at distance x from centre is –

(A) P0 + 2

22 ρω r (B) P0 – 2

22 ρω r

(C) P0 + 2

2 ρω r (D) P0 + 2

3 22 ρω r


ABCD

P Q R S

S

P

P P Q R

R R

Q Q

S S

P Q R S


20. A ball tied with a string of length 10 cm is free to rotate in vertical circle. It is given velocity

27 meter/sec in horizontal direction when it is at

its lowest position. Then match the following : Column-I Column-II

(A) Height at which the (P) 16027

ball loses circular path (in meter)

(B) Maximum height (Q) 2

1

reached by ball (in meter)

(C) Velocity of ball at the (R) 203

time of leaving circular path (in meter/sec)

(D) Minimum velocity of (S) 22

1

ball (in meter/sec)

21. Column I contain different process and column II contains molar heat capacity. Match the following:

Column-I Column-II

(A) T = T0 eαV (P) CV +V1

Rα+

(B) P = P0 eαV (Q) 1

R−γ

γ + VPR

0

α

(C) P = P0 + α/V (R) CV +V

Rα

(D) T = T0 + α/V (S) CV + R(1 +V

T0

α)

22. Column-I contains some phenomena of heat transfer and column-II contains the mechanism of heat transfer.

Column-I Column-II (A) Heating up water in bucket (P) Convection with immersion rod slightly immersed in it (B) Freezing of lakes in colder (Q) Conduction region (C) Cooling water in container (R) Radiation with an ice cube fixed at the bottom (D) Heating up atmosphere (S) None of these


CHEMISTRY


1. The shape of TeCl4 is - (A) Linear (B) Square planar (C) Tetrahedral (D) See-Saw

2. Oxidation states of carbon and nitrogen in KCN are, respectively -

(A) – 3, + 2 (B) + 2, – 3 (C) + 1, – 2 (D) zero each 3. How many moles of nitrogen is produced by the

oxidation of one mole of hydrazine by 2/3 mole bromate ion

(A) 31 (B) 1 (C) 1.5 (D)

4. The IUPAC name of the compound is :

C–CH CH3

CH3

H

NH2 (A) 1-amino-1-phenyl-2-methyl propane (B) 2-methyl-1-phenyl propanamine (C) 2-methyl-1-aminol-1-phenyl propane (D) 1-isopropyl-1-phenyl methyl amine

5. If C2 in above compound is rotated by 120º angle in anticlockwise direction along C2–C3, Which of the following form will be produced

H H

H H

CH3

(1) CH3

(2) (3)

(4)

(A) Partial eclipsed (B) Perfectly eclipsed (C) Staggered (D)Gauche conformation 6. Which of the following compounds will exhibit

geometrical isomerism ? (A) 1-phenyl-2-butene (B) 3-phenyl-1-butene (C) 2-phenyl-1-butene (D)1,1-dephenyl-1-propene

7. Which of the following is/are correct regarding π acceptors and π donors ?

(A) π acceptor have empty π orbitals with correct symmetry whereas π donors have filled π orbitals with correct symmetry

(B) π acceptor involve the metal ions with lower oxidation state whereas π donor favour the central metal ion at higher oxidation state

(C) π acceptor increases CFSE whereas π donor decreases CFSE

(D) All of the above are correct 8. An isotone of Ge76

32 is :

(A) Ge7732 (B) As77

33

(C) Se7734 (D) Se79

34 9. The energy of an electron in the first Bohr orbit of H

atom is –13.6 eV. The possible energy value(s) of the excited state(s) for electrons in Bohr orbits of hydrogen is (are)

(A) – 3.4 eV (B) – 4.2 eV (C) – 6.8 eV (D) + 6.8 eV This section contains 4 questions numbered 10 to 13, (Reason and Assertion type question). Each question contains Assertion (A) and Reason (R). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. +3 marks will be given for each correct answer and – 1 mark for each wrong answer.




(C) If (A) is true but (R) is false. (D) If (A) is false but (R) is true. 10. Assertion (A) : For each ten degree rise of temperature

the specific rate constant is nearly doubled. Reason (R) : Energy-wise distribution of molecules

in a gas is an experimental function of temperature. 11. Assertion (A) : Dimethyl sulphide is commonly used

for the reduction of an ozonide of an alkene to get the carbonyl compounds.

Reason (R) : It reduces the ozonide giveing water soluble dimethyl sulphoxide and excess of it evaporates.


12. Assertion (A) : In the titration of Na2CO3 with HCl using methyl orange indicator, the volume required at the equivalence point is twice that of the acid required using phenolphthalein indicator.

Reason (R) : Two moles of HCl are required for the complete neutralization of one mole of Na2CO3.

13. Assertion (A) : Band gap in germanium is small. Reason (R) : The energy spread of each germanium

atomic energy level is infinitesimally small. This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 14 to 19) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. +4 marks will be given for each correct answer and – 1 mark for each wrong answer.

Passage # 1 (Ques. 14 to 16) Several short-lived radioactive species have been

used to determine the age of wood or animal fossils. One of the most interesting substances is 6C14 half-life 5760 years) which is used in determining the age of carbon-bearing materials (e.g. wood, animal fossils, etc.) Carbon-14 is produced by the bombardment of nitrogen atoms present in the upper atmosphere with neutrons (from cosmic rays).

7N14 + 0n1 → 6C14 + 1H1 Thus carbon-14 is oxidised to CO2 and eventually

ingested by plants and animals. The death of plants or animals put an end to the intake of C14 from the atmosphere. After this the amount of C14 in the dead tissues starts decreasing due to its disintegration as per the following reaction : (

6C14 → 7N14 + –1β0 The C14 isotope enters the biosphere when carbon

dioxide is taken up in plant photosynthesis. Plant are eaten by animals, which exhale C14 as CO2. Eventually, C14 participates in many aspects of the carbon cycle. The C14 lost by radioactive decay is constantly replenished by the production of new isotopes in the atmosphere. In this decay-replenishment process, a dynamic equilibrium is established whereby the ratio of C14 to C12 remains constant in living matter. But when an individual plant or an animal dies, the C14 isotope in it is no longer replenished, so the ratio decreases as C14

decays. So, the number of C14 nuclei after time t (after the death of living matter) would be less than in a living matter. The decay constant can be calculated using the following formula,

t1/2 = λ693.0

The intensity of the cosmic rays have remain the same for 30,000 years. But since some years the changes in this are observed due to excessive burning of fossil fuel and nuclear tests.

14. Why do we use the carbon dating to calculate the age of the fossil?

(A) Rate of exchange of carbon between atmosphere and living is slower than decay of C14

(B) It is not appropriate to use C14 dating to determine age

(C) Rate of exchange of C14 between atmosphere and living organism is so fast that an equilibrium is set up between the intake of C14 by organism and its exponential decay

(D) none of the above 15. What should be the age of the fossil for meaningful

determination of its age? (A) 6 years (B) 6000 years (C) 60,000 years (D) can be used to calculate any age 16. A nuclear explosion has taken place leading to

increase in concentration of C14 in nearby areas. C14

concentration is C1 in nearby areas and C2 in areas far away. If the age of the fossil is determined to be T1 and T1 at the respective places then

(A) The age of the fossil will increase at the place where explosion has taken place and

T1 – T2 = λ1 ln

2

1

CC

(B) The age of the fossil will decrease at the place where explosion has taken place and

T1 – T2 = λ1 ln

2

1

CC

(C) The age of fossil will be determined to be same

(D) 2

1

TT =

2

1

CC

Passage # 2 (Ques. 17 to 19) The noble gases have closed-shell electronic

configuration and are monoatomic gases under normal conditions. The low boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions.

The direct reaction of xenon with fluorine leads to a series of compounds with oxidation numbers +2, +4 and +6. XeF4 reacts violently with water to given XeO3. The compound of xenon exhibit rich stereochemistry and their geometries can be deduced considering the total number of electron pairs in the valence shell.

17. Argon is used in are welding because of its (A) low reactivity with metal (B) ability to lower the melting point of metal (C) Flammability (D) high calorific value


18. The structure of XeO3 is (A) linear (B) planar (C) pyramidal (D) T-shaped 19. XeF4 and XeF6 are expected to be (A) oxidizing (B) reducing (C) unreactive (D) strongly basic This section contains 3 questions (Questions 20 to 22). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, B-Q, B-R, C-P, C-Q and D-S, then the correctly bubbled 4 × 4 matrix should be as follows :

A B C D

P Q R S

S

P

P P Q R

R R

Q Q

S S

P Q R S

Mark your response in OMR sheet against the question number of that question in section-II. + 6 marks will be given for complete correct answer and No Negative marks for wrong answer. However, 1 mark will be given for a correctly marked answer in any row. 20. Match the extraction processes listed in Column-I

with metals listed in Column-II Column-I Column-II (A) Self-reduction (P) Lead (B) Carbon reduction (Q) Silver (C) Complex formation (R) Copper and displacement by metal (D) Decomposition of (S) Boron iodide 21. Match the column : Column-I Column-II (A) Efflorescent (P) NaOH (B) Deliquescent (Q) KOH (C) Fusion mixture (R) Na2CO3 and K2CO3 (D) washing soda (S) Na2CO3. 10H2O (T) Na2SO4 22. Match the half-reaction (in column I) with equivalent

mass (molar mass = M) (in column II) Column –I Column II (A) Cr2O7

2– → Cr3+ (P) M (B) C2O4

2– → CO2 (Q) M/2 (C) MnO4

– → MnO2 (R) M/6 (D) HC2O4

– → C2O42– (S) M/3

MATHEMATICS


1. Sn, the sum to n terms of the series (n2 – 12) + 2(n2 – 22) + 3(n2 – 32) + ... is

(A) 41 n2 (n2 – 1) (B)

41 n(n + 1)2

(C) 0 (D) 2n (n2 – 1) 2. The number of ways of dividing 15 men and 15

women into 15 couples, each consisting of a man and a woman, is

(A) 1240 (B) 1840 (C) 1820 (D) 2005 3. If cos (θ – α) = a and sin (θ – β) = b (0 < θ – α, θ – β

< π/2), then cos2 (α – β) + 2ab sin (α – β) is equal to (A) 4a2 b2 (B) a2 – b2 (C) a2 + b2 (D) – a2 b2. 4. If A and B are acute positive angles satisfying the eqn.

3 sin2 A + 2 sin2 B = 1 and 3 sin 2A – 2 sin 2B = 0, then A + 2B is equal to

(A) π/4 (B) π/2 (C) 3π/4 (D) 2π/3

5. In a triangle ABC, C

rrcos1

21

++ is equal to

(A) ∆abc2 (B)

∆+

cba

(C) ∆2

abc (D) 2∆abc

6. If P is a point (x, y) on the line y = – 3x such that P

and the point (3, 4) are on the opposite sides of the line 3x – 4y = 8, then

(A) x > 8/15, y < – 8/5 (B) x > 8/5, y < – 8/15 (C) x = 8/15, y = – 8/5 (D) none of these 7. C1 and C2 are circles of unit radius with centres at

(0, 0) and (1, 0) respectively. C3 is a circle of unit radius, passes through the centres of the circles C1 and C2 and have its centre above x-axis. Equation of the common tangent to C1 and C3 which does not pass through C2 is

(A) x – 3 y + 2 = 0 (B) 3 x – y + 2 = 0

(C) 3 x – y – 2 = 0 (D) x + 3 y + 2 = 0


8. If the tangent at a point (a cos θ, b sin θ) on the ellipse x2/a2 + y2/b2 = 1 meets the auxillary circle in two points, the chord joining them subtends a right angle at the centre; then the eccentricity of the ellipse is given by

(A) (1 + cos2 θ)–1/2 (B) 1 + sin2 θ (C) (1 + sin2 θ)–1/2 (D) 1 + cos2 θ 9. If pth, qth, rth term of a G.P. are the positive numbers

a, b, c, then the angle between the vectors log a2i + log b2j + log c2k and (q – r)i + (r – p)j + (p – q)k is

(A) π/3 (B) π/2

(C) sin–1 222/1 cba ++ (D) none of these This section contains 4 questions numbered 10 to 13, (Reason and Assertion type question). Each question contains Assertion (A) and Reason (R). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. +3 marks will be given for each correct answer and – 1 mark for each wrong answer.




(C) If (A) is true but (R) is false. (D) If (A) is false but (R) is true. 10. Assertion (A) : If a and b are integers and roots of

x2 + ax + b = 0 are rational then these roots must be integers.

Reason (R) : If a and b are integers then roots of x2 + ax + b = 0 are necessarily integers.

11. Assertion (A) : The coefficient of the term of

independent of x in the expansion of n

xx

++ 69 is

!!)!2(3

nnnn

.

Reason (R) : The coefficient of xr in the expansion

of (1 + x)n is

rn

.

12. Assertion (A) : If in a triangle ABC sin2 A + sin2 B + sin2 C = 2, then one of the angles must

be 90º. Reason (R) : In any triangle ABC cos 2A + cos 2B + cos 2C = – 1 – 4 cos A cos B cos C

13. Assertion (A) : P is a point (a, b, c). Let A, B, C be the images of P in yz, zx and xy planes respectively then equation of the plane passing through the points A, B

and C is ax +

by +

cz = 1.

Reason (R) : The image of a point P in a plane is the foot of the perpendicular draw from P on the plane.

This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 14 to 19) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. +4 marks will be given for each correct answer and – 1 mark for each wrong answer. Passage # 1 (Ques. 14 to 16) f(x) = sin cot–1(x + 1) – cos (tan–1 x) a = cos tan–1 sin cot–1 x b = cos (2 cos–1 x + sin–1 x) 14. The value of x for which f(x) = 0 is (A) – 1/2 (B) 0 (C) 1/2 (D) 1 15. If f(x) = 0 then a2 is equal to (A) 1/2 (B) 2/3 (C) 5/9 (D) 9/5 16. If a2 = 26/51, then b2 is equal to (A) 1/25 (B) 24/25 (C) 25/26 (D) 50/51 Passage # 2 (Ques. 17 to 19) Let a > 0, a ≠ 1 and x > 0. (i) loga x > 0, a > 1 ⇔ x > 0, a > 1 (ii) loga x > 0, 0 < a < 1 ⇔ 0 < x < 1, 0 < a < 1 (iii) loga x < 0, a > 1 ⇔ 0 < x < 1, a > 1 (iv) loga x < 0, 0 < a < 1 ⇔ x > 1, 0 < a < 1 17. The solution set of x log1/10 (x2

+ x + 1) > 0 is (A) (– ∞, – 1) (B) (0, 1)

(C) (1, ∞) (D)

∞,101

18. The solution set of

log3 8–1212–5

xx + log1/3 x ≤ 0 is

(A)

32,

125 (B)

21,

125

(C)

31,

125 (D)

21,

125

19. The solution set of

113

10–log1 29/1 ≤

+

πxx is


(A)

31,0 ∪

310,3 (B) (π, ∞)

(C)

31,0 ∪ (π, ∞) (D) (3, π) ∪ (π, ∞)


A B C D

P Q R S

S

P

P P Q R

R R

Q Q

S S

P Q R S


20. Normals at P, Q, R are drawn to y2 = 4x which intersect at (3, 0). Then :

Column-I Column-II (A) Area of ∆PQR (P) 2

(B) Radius of circumcircle (Q) 25

of ∆PQR

(C) Centroid of ∆PQR (R)

0,

25

(D) Circumcentre of ∆PQR (S)

0,

32

21. Value of Column-I Column-II (A) 4 α (β4 – α4) (P) 0 if α + iβ, β ≠ 0 is a root of z5 = 1

(B) 212

z+ + |2 – z|2 (Q) 1

if zz = 1

(C) wzz ++ )(21

21 (R) 4

+ wzz –)(21

21 +

if w = 21zz , |z1| = 3, |z2| = 1

(D) z1 – z2 + z3 – z4 if (S) 10 z1, z2, z3, z4 represent vertices of a parallelogram 22. Column-I Column-II (A) Centroid of the triangle (P) (1, 6, 5) with vertices A(2, 3, 7), B(6, 7, 5), C(1, 2, 3) (B) Mid-point of the line (Q) (3, 4, 5) joining the points A(7, 9, 11) and B(–5, 3, –1) (C) A point on the line (R) (3, 3, 2)

2x =

3y =

5z , at a

distance 2 from the origin (D) Coordinates of the point (S) ( 38/4 , 38/6 ,

dividing the join (5, 5, 0) 38/10 ) and (0, 0, 5) in the ratio 2 : 3.

• William Bottke at Cornell University in the US has calculated that at least 900 asteroids of a kilometre or more across regularly sweep across Earth's path.

• The Dutch astronomer Christiaan Huygens (1629 - 1695) drew Mars using an advanced telescope of his own design. He recorded a large, dark spot on Mars, probably Syrtis Major. He noticed that the spot returned to the same position at the same time the next day, and calculated that Mars has a 24 hour period. (It is actually 24 hours and 37 minutes)

• Space debris travels through space at over 18,000 mph.

• The nucleus of Comet Halley is approximately 16x8x8 kilometers. Contrary to prior expectations, Halley's nucleus is very dark: its albedo is only about 0.03 making it darker than coal and one of the darkest objects in the solar system.

• A car travelling at a constant speed of 60 miles per hour would take longer than 48 million years to reach the nearest star (other than our Sun), Proxima Centauri. This is about 685,000 average human lifetimes.


PHYSICS 1. Define the term electric dipole moment. Is it a scalar

or a vector quantity ?

2. The variation of potential difference V with length l in case of two potentiometers P and Q is as shown. Which one of these two will you prefer for comparing e.m.f's of two primary cells ?

PQ

l

V

3. Arrange the given electromagnetic radiations in the descending order of their frequencies : Infra-red, X-rays, Ultraviolet and Gamma rays.

4. The de Broglie wavelengths, associated with a

proton and a neutron, are found to be equal. Which of the two has a higher value for kinetic energy ?

5. Carbon and silicon are known to have similar lattice structures. However, the four bonding electrons of carbon are present in second orbit while those of silicon are present in its third orbit . How does this difference result in a difference in their electrical conductivities ?

6. An unknown input (A) and the input (B) shown here,

are used as the two inputs in a NAND gate. The output Y, has the form shown below. Identify the intervals over which the input ‘A’ must be ‘low’.

B

YO O

O O

0 L1 L2 L3 L4 L5

General Instructions : Physics & Chemistry • Time given for each subject paper is 3 hrs and Max. marks 70 for each. • All questions are compulsory. • Marks for each question are indicated against it. • Question numbers 1 to 8 are very short-answer questions and carrying 1 mark each. • Question numbers 9 to 18 are short-answer questions, and carry 2 marks each. • Question numbers 19 to 27 are also short-answer questions, and carry 3 marks each. • Question numbers 28 to 30 are long-answer questions and carry 5 marks each. • Use of calculators is not permitted.

General Instructions : Mathematics • Time given to solve this subject paper is 3 hrs and Max. marks 100. • All questions are compulsory. • The question paper consists of 29 questions divided into three sections A, B and C. Section A comprises of 10 questions of one mark each. Section B comprises of 12 questions of four marks each. Section C comprises of 7 questions of six marks each. • All question in Section A are to be answered in one word, one sentence or as per the exact requirement of the question. • There is no overall choice. However, internal choice has been provided in 4 questions of four marks each and

2 question of six marks each. You have to attempt only one of the alternatives in all such questions. • Use of calculators is not permitted.

MOCK TEST-2

CBSE BOARD PATTERN

CLASS # XII

SUBJECT : PHYSICS , CHEMISTRY & MATHEMATICS Solut ions wil l be published in next issue


7. Does a beam of white light give a spectrum on passing through a hollow prism ?

8. Why do we prefer a magnifying glass of smaller

focal length ?

9. Write the relation for the force →F acting on a charge

carrier q moving with a velocity →v through a

magnetic field →B in vector notation. Using this

relation, deduce the conditions under which this force will be (i) maximum (ii) minimum.

10. A galvanometer has a resistance of 30 Ω. It gives full

scale deflection with a current of 2 mA. Calculate the value of the resistance needed to convert it into an ammeter of range 0-0.3 A.

11. Two capacitors of capacitance 6 µF and 12 µF are connected in series with a battery. The voltage across the 6 µF capacitor is 2 V. Compute the total battery voltage.

OR A parallel plate capacitor with air between the plates

has a capacitance of 8 pF. The separation between the plates is now reduced by half and the space between them is filled with a medium of dielectric constant 5. Calculate the value of capacitance of the capacitor in the second case.

12. What do the terms ‘depletion region’ and ‘barrier potential’ mean for a p-n junction ?

13. We do not choose to transmit an audio signal by just directly converting it to an e.m. wave of the same frequency. Give two reasons for the same.

14. Define the term ‘Activity’ of a radioactive substance. State its SI unit.

Two different radioactive elements with half lives T1 and T2 have N1 and N2 (undecayed) atoms respectively present at a given instant. Determine the ratio of their activities at this instant.

15 Draw the block diagram of a communication system.

16. State Bohr's postulate for the ‘permitted orbits’ for the electron in a hydrogen atom.

Use this postulate to prove that the circumference of the nth permitted orbit for the electron can ‘contain’ exactly n wave lengths of the de Broglie wavelength associated with the electron in that orbit.

17. Four double convex lenses with the following specifications are available.

cm2cm2cm5cm10

cm5cm10cm100cm100

DCBA

AperturelengthFocalLens

Which two of the given four lenses should be selected as the objective and eyepiece to construct an astronomical telescope and why ? What will be the magnifying power and normal length of the telescope tube so constructed ?

18. In Young's experiment on interference, what shall

happen if monochromatic source is replaced by a source of white light ?

19. Using Biot-Savart law, derive an expression for the intensity of magnetic field at a point near a long current carrying straight conductor.

20. A copper circular coil of 20 turns and radius 10 cm is

placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. A current of 5.0 A is flowing through the coil. Calculate :

(a) total torque acting on the coil (b) average force acting on each electron in the coil. Given : Free electron density in copper = 1029 m–3

and cross-sectional area of copper wire = 10–5 m2. OR

A galvanometer with a coil of resistance 12 Ω shows a full scale deflection for a current of 2.5 mA.Calculate the value of the resistance required to convert it into (a) an ammeter of range 0 to 7.5 A and (b) a voltmeter of range 0 to 10 V. Draw the diagrams to show how you will connect this resistance to the galvanometer in each case.

21. A 10 m long wire of uniform cross-section and 20

Ω resistance is used in a potentiometer. The wire is connected in series with a battery of 5V along with an external resistance of 480 Ω. If an unknown e.m.f. E is balanced at 6.0 m length of the wire, calculate :

(i) the potential gradient of the potentiometer wire (ii) the value of unknown e.m.f. E.

22. Define the term electrical resistivity of a material.

Write its S.I. units. Derive an expression for the resistivity of a metal in terms of number density and mass of free electrons present in it.

23. Devices A and B are connected independently to a

variable frequency voltage source as shown. The current in A is ahead of the applied voltage whereas it lags behind the voltage in B.

A

~

B

~

(i) Identify the devices A and B (ii) How will the current in each of these devices

change on decreasing the frequency of the applied voltage ?

Give reasons to support your answer in each case.


24 Using a suitable combination from a NOR, an OR and a NOT gate, draw circuits to obtain the truth table given below :

0100

1010

1100

YBA

1011

1010

1100

YBA

(i) (ii)

25. Calculate the de-Broglie wavelength of (i) an electron (in the hydrogen atom) moving with a speed of 1/100 of the speed of light in vacuum and (ii) a ball of radius 5 mm and mass 3 × 10–2 kg moving with a speed of 100 ms–1. Hence show that the wave nature of matter is important at the atomic level but is not really relevant at the macroscopic level.

26. The spectrum of a star in the visible and the ultraviolet region was observed and the wavelength of some of the lines that could be identified were found to be:

824 Å, 970 Å, 1120 Å, 2504 Å, 5173 Å, 6100 Å Which of these lines cannot belong to

hydrogen atom spectrum ? Given Rydberg constant

R = 1.03 × 107m–1 and R1 = 970 Å. Support your

answer with suitable calculations 27. Show that central maximum is twice as wide as the

other maxima and the pattern becomes narrower as the width of the slit is increased.

28. The given circuit diagram shows a series LCR

circuit connected to a variable frequency 230 V source

40 Ω

~

80 Ω 5.0 H

230 V (a) Determine the source frequency which drives the

circuit in resonance. (b) Obtain the impedance of the circuit and the

amplitude of current at the resonating frequency. (c) Determine the rms potential drops across the

three elements of the circuit. (d) How do you explain the observation that the

algebraic sum of the voltages across the three elements obtained in (c) is greater than the supplied voltage ?

OR The primary coil of an ideal step-up transformer as 100 turns and the transformation ratio is also 100. The input voltage and power are 220 V and 1100 W respectively. Calculate :

(i) number of turns in the secondary (ii) the current in the primary (iii) voltage across the secondary (iv) the current in the secondary (v) power in the secondary. 29. What is a dielectric ? Why does the capacitance of a

parallel plate capacitor increase on introduction of a dielectric in between its two plates ? Derive an expression for the capacitance of such a capacitor having two identical plates each of area A and separated by distance x. The space between the plates has a medium of dielectric constant k.

OR Derive an expression for the energy stored in a

parallel plate capacitor with air as the medium between its plates.

Air is now replaced by a dielectric medium of dielectric constant k. How does it change the total energy of the capacitor if

(i) the capacitor remains connected to the same battery ?

(ii) the capacitor is disconnected from the battery ? 30. Discuss the phenomenon of refraction through a

prism and derive the relation

µ =

2sin

2sin

A

A m

δ+

.

Where ‘A’ is the angle of prism, ‘µ’ is the refractive index of the material of the prism and ‘δm’ is the angle of minimum deviation.

OR With the help of a labelled ray diagram, show the

image formation by a compound microscope. Derive an expression for its magnifying power.

CHEMISTRY

1. Why PCl5 is found but NCl5 do not ? 2. Give two ores of aluminium ? 3 Define Enzymes . 4. What is the coordination number of atoms in BCC,

HCP and CCP and simple cubic lattices ?

5. The rate constant of a zero order reaction is 0.25 Mh–1. What will be the initial concentration of the reactant if after 30 minutes its concentration is 0.075 M ?

6. How much electricity in terms of Faraday is required to produce 20 g of Ca from molten CaCl2 ?

7. Why is adsorption always exothermic ?


8. Which reagents are required for the one step

conversion of chlorobenzene into toluene ? Also name the reaction.

9. Why Carbon monoxide is poisonous ? 10. What are the different oxidation states exhibited by

lanthanides ? 11. Explain baeyer's process to remove impurities in

metallurgy of aluminium ? 12. Explain the following terms in brief (i) Colligative property; (ii) Reverse osmosis 13. What is known as 'activation energy' ? How is the

activation energy affected by (i) the use of a catalyst and (ii) a rise in temperature ? 14. The rate of a particular reaction doubles when

temperature changes from 27ºC to 37ºC. Calculate the activation energy of such reaction.

15. In the button cell, widely used in watches and other

devices, the following reaction takes place : Zn (s) + Ag2O(s) + H2O(l)

→ Zn2+(aq) + 2 Ag (s) + 2 OH¯ (aq) Determine Eº and ∆rGº for the reaction Given : 0

Ag/AgE + = + 0.80 V, 0Zn/Zn2E + =–0.76 V

16. What are fuel cells ? Write the electrode reactions of a fuel cell which uses the reaction of hydrogen with oxygen.

17. Identify X, Y and Z in the following reactions : C6H5Cl

OCu2

3NH → X → HClandNaNO2 Y

WarmKI→ Z.

18. How are the following conversions carried out ? (i) Ethyl bromide to butan -1-ol (ii) Propanone to 2-methylpropan-2-ol. 19. Describe the preparation of potassium dichromate

from chromite ore. What is effect of increasing pH on solution of potassium dichromate ?

20. Give IUPAC names of : (i) [CoCl2 (en)2] Cl (ii) K2[Zn(OH)4] (iii) [Co(NH3)4(H2O) Cl] Cl2 21. What are polymers. Give the structure of terylene

and Nylon 6, 6. 22. What are dissachharides. Give three examples and

their monosaccharide unit - 23. What are antibiotics. Give types with examples.

24. An element having atomic mass 52, occurs in body

centred cubic (bcc) structure with a cell edge of 288 pm. The density of the element is 7.2 g cm–3. Evaluate Avogadro's number.

25. (a) The vapour pressure of water is 12.3 kPa at 300

K. Calculate vapour pressure of 1 molal solution of a solute in it.

(b) Calculate the mass of a nonvolatile solute (molecular mass 40) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.

26. (a) What are micelles ? How do they differ from

ordinary colloidal particles ? Give two examples of micelles forming substances

(b) State Hardy – Schulze rule. (c) State the principle of electrodialysis. 27. Arrange the following compounds in increasing order

of their property as indicated : (a)Acetaldehyde, Acetone, Di-tert-butyl ketone,

Methyl tert-butyl ketone (reactivity towards HCN) (b)CH3CH2CH(Br)COOH, CH3CH(Br)CH2COOH,

(CH3)2CHCOOH, CH3CH2CH2COOH (acid strength)

(c)Benzoic acid, 4-Nitrobenzoic acid, 3, 4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)

28. If the starting material for the manufacture of

silicones is RSiCl3, write the structure of the product formed ?

29. An aromatic compound 'A' gives ammonia gas on

boiling with caustic potash. On heating with P2O5 A give B which on further reduction with sodium / alcohol C forms a base which reacts with HNO2 giving off nitrogen and yielding alcohol C. The alcohol C can be oxidised to benzoic acid. Give the structures of A, B, C.

30. Write chemical tests to distinguish between formic acid and acetic acid.

MATHEMATICS

Section A

1. Find trace of matrix A =

−−

2468575114

2. Find x, y so that

−++40210 2 yyx =

−

+yy

x50

3432


3. Evaluate °°°−°

80cos80sin10cos10sin

4. If ex + ey = ex + y then prove that

dxdy =

)1()1(

+−−

xy

yx

eeee .

5. If f(x) = 3223

−−

xx then find f(f(x)).

6. Evaluate ∫

−− dx

xxe x

cos1sin1

7. Solve : cos2 x dxdy + y = tan x

8. Show that ( ar

× br

)2 = bbbabaaarrrr

rrrr

....

9. Find the area of the parallelogram determined by the

vectors i + 2 j + 3 k and 3 i –2 j + k . 10. Find the angle between the line

rr

= ( i + 2 j – k ) + λ( i – j + k ) and the plane

rr

. (2 i – j + k ) = 4.

Section B

11. Using properties of determinant prove that

cbaabacbacbccba

++−−−++−−−++

= 2 (a + b) (b + c) (c + a) 12. An integer is chosen at random from first two

hundred natural number. What is the probability that the integer chosen is divisible by 6 or 8?

OR A candidate has to reach the examination centre in

time. Probability of him going by bus or scooter or

by other means of transport are 103 ,

101 ,

53

respectively. The probability that he will be late is 41

and 31 respectively, if he travels by bus or scooter.

But he reaches in time if the uses any mode of transport. He reached late at the centre. Find the probability that he travelled by bus.

13. Differentiate tan–1

−+

xx 11 2

w.r.t. tan–1 x, x ≠ 0.

14. If x = a

+

tt 1 , y = a

−

tt 1 then prove that

dxdy =

yx

OR

If xpyq = (x + y)p+q, prove that dxdy =

xy

15. Using Langrange's theorem, prove that

b

ab − < log (b/a) < a

ab − where 0 < a < b.

16. For what choice of a, b function

f(x) =

>+≤

cxbaxcxx

,,2

is differentiable at x = c.

17. Show that f : N → N defined by

f(n) =

→

→+

evennn

oddnn

,2

,2

1

is many-one onto function.

18. Evaluate ∫ dxx

x4sin

sin

19. Evaluate ∫ θθ+θ dcottan

OR

Evaluate ∫ −+++

)5)(3()4)(1(

22

22

xxxx dx

20. The normal lines to a given curve at each point pass

through (2, 0). The curve passes through (2, 3). Formulate the differential equation and hence find out the equation of the curve.

21. Find λ and µ if (2 i + 6 j + 27 k ) × ( i + λ j + µ k ) = 0

r

22. Find the equation of the plane passing through the

intersection of the planes 4x – y + z = 10 and x + y – z = 4 and parallel to the line with direction ratios proportional to 2, 1, 1. Find also the perpendicular distance of (1, 1, 1) from this plane.

OR

Find the vector equation of the line parallel to the line

51−x =

23 y− =

41+z and passing through (3, 0, –4).

Also find the distance between these two lines.


Section C

23. Evaluate ∫b

a

dxxsin as limit of sums.

(Using by first principal method of integration) 24. A bag contains 4 white balls and 2 black balls.

Another contains 9 white balls and 5 black balls. If one ball is drawn from each bag. Find the probability that

(A) both are white (B) one is white and one is black. 25. Sketch the region common to the circle

x2 + y2 = 16 and the parabola x2 = 6y. Also, find the area of the region using integration.

26. Find the shortest distance between the lines whose

vector equations are r

r= ( i +2 j + 3 k ) + λ(2 i +3 j + 4 k ) and,

rr

=(2 i +4 j + 5 k ) + µ (4 i +6 j + 8 k ) 27. An oil company requires 12,000; 20,000 and 15,000

barrels of high-grade, medium grade and low grade oil, respectively. Refinery A produces 100, 300 and 200 barrels per day of high grade, medium grade and low grade oil, respectively. While refinery B produces 200, 400 and 100 barrels per day of high grade, medium grade and low grade oil, respectively. If refinery A costs Rs.400 per day and refinery B costs Rs.300 per day to operate, how many days should each be run to minimize costs while satisfying requirements.

28. If A =

−−

321112111

, then find A–1.Using A–1, solve

the following system of linear equations x + y + z = 3 2x – y + z = 2 x – 2y + 3z = 2

OR

If A =

−−

−−

7321311154

, then find A–1. Using A–1,

solve the following system of linear equations 4x – 5y –11z = 12 x – 3y + z = 1 2x +3y – 7z = 2

29. Show that the volume of largest cone that can be

inscribed in a sphere of radius R is 278 of the volume

of sphere. OR Find the maximum and minimum values of

f(x) = sin x + 22cos x , x ∈

π

2,0

Chemistry Facts

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

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

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

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

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

6. Hafnium was named after the city of Copenhagen, Denmark.

7. The heaviest type of lepton is the tau.


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PHYSICS 1. A straight line passing through pole and centre of

curvature is called principal axis.

2. Frequency

3. (i) N/C (ii) Coul-meter

4. It states about conservation of energy. According to it direction of induced current would be such that it opposes the cause of change of magnetic flux.

5. E0 = CB0

B0 = CE0 = 8

6

103103

×× = 0.01 T

6. The charge of the ‘excess’ charge carriers gets balanced by an equal and opposite charge of the ionized cores in the lattice.

7. We have

=

∞=

=

=

→∞

−

41.const1–

21.constEand

43.const

21–

11.constE

222

2212

∴ Ratio = 3 : 1

8. Infrared radiations get readily absorbed by water molecules in most materials. This increases their thermal motion and heats them up.

(i) visible light (ii) Microwaves

9. Focal length do not depend upon the medium in which the mirror is held. So focal length will not change.

10. As width of central maxima, βc = a

fλ2

So, βc ∝ a1 , βc ∝ f & βc ∝ λ thus,

(a) width increases with decrease in a (b) width increases with increase in f (c) width decreases with decrease in λ

11. –3nC –2nC

1nC2nC 2 m

r

r = 22 = 1 m

Vcentre = rk [2 + 1 – 2 – 3] × 10–9

= 1109 9× [–2 × 10–9]

= –18 volt 12. Blue – 6 (A) Black – 0 (B) yellow – 4 (C) R = AB × 10C = 60 × 104

∴ i = RV = 41060

30×

= 0.5 × 10–4 = 0.05 mA

13. Microwave → In microwave oven Ultraviolet rays → To sterlizing sergical instruments Gamma rays → In medical science.

14. Instantaneous Activity = R = – dtdN = λN

∴ dtdR =

dtd (λN) =

dtdN

λ

= λ (–λ N) = – λ2 N

= – 2

2/1

2log

Te N

∴ dtdR ∝ 2

2/1 )T(1

MOCK TEST-1 (SOLUTION) MOCK TEST– 1 PUBLISHED IN DECEMBER ISSUE


15. Energy of a photon of the incident radiation

= λhc = 19–9–

834–

106.110300103104.6

×××××× eV = 4eV

This being less than the work function of Mo, there would be no photo-emission from Mo.

16. eVs = hν – w ∴ e (V2 – V1) = h (ν2 – ν1)

or V2 – V1 = eh (ν2 – ν1)

∴ V2 – V1 = 19–

34–

106.1104.6

×× × (8 – 4) × 1015 volt = 16 volt

17. We have

m

hcλ

= Eg

or λm = gE

hc

18. LOS → line of sight Waves used → space waves It is both-the height of transmitting antenna as well

as the height of receiving antenna that affects the range of the mode of communication.

19. Imax = ( 1I + 2I )2 = ( I + II δ+ )2

= I 2

11

δ++

II

= I

2

21

11

δ

++II = I

2

211

δ

++II

= I 2

22

δ

+II = I.4

2

41

δ

+II = 4I

δ

+II

21 ≈ 4I

Imin = ( 1I – 2I )2 = ( I – II δ+ )2

= I 2

1–1

δ+

II

= I

2

21

1–1

δ

+II = I

2

21–1

δ

+II

= I 2

2–

δ

II = I. 2

2

4)(

IIδ

= I

I4

)( 2δ

20.

q+

++++

+ ++r

(a) r > R

0

.∈

=∫→→

inqdsE

E. 4πr2 = 0∈

q

∴ 204 r

qE∈π

=

(b) Q In r < R, qin = 0 ∴ E = 0

21. i = eqR

V

5 Ω 10Ω

5 Ω

10Ω

10Ω 10Ω

10 V

22. (i) W = MB (cosθ1 – cosθ2) = MB (cos0° – cos 180°) = 2 MB (ii) W = MB (cosθ1 – cosθ2) = MB (cos0° – cos 90°) = MB

23. φ = BA

∴ ε = – dt

Ndφ

= –N td (BA)

= –NA dtd B0 cos ωt

tBNA ωω=ε sin0

24. 0.5 = 22 )(

12

LR ω+ ……. (i)

and tanφ = R

X L

tan π/3 = RLω ………… (ii)


OR

F = i B l =

RBv l Bl =

RBv 22l

25. Total B.E. of parent Nucleus = 7.8 × 235 MeV = 1833 MeV Total B.E. of daughter nucleus = 7.835 × 231 MeV = 1809.9 MeV Total B.E. of α-particle = 7.07 × 4 MeV = 28.28 MeV Increase in B.E. after the reaction = [(180.9 + 28.28) – (1833)] MeV = 5.18 MeV This is the energy released in the reaction, since it assumed to be taken up totally by the α-particle,

21 mv2 = 5.18 × 1.6 × 10–13 J

∴ v2 = 68.6

2.318.5 × × 1014 m2s–2

= 48.2 × 107 ms–1

1.58 × 107 ms–1

26. The modulation index (µ) for an AM wave equals the ratio of the peak value of the modulating signal (Am) to the peak value of the carrier wave

(Ac) µ = c

m

AA

Given that

=

+=

mc

mc

AAbAAa

and –

∴ =

+=

2–

2baAandbaA mc

∴ µ = baba

+–

27. Postulates of Bohr's Model –(i) In atom, electron moves round the nucleus in stable orbits. Attraction force b/w nucleus and electron given the required centripetal force for electron

mass of electron = m charge on electron = e atomic no. = z radii of orbit = r speed of electron = v Centripetal force = Attraction force

r

mv2 = 2

2

rkze

⇒ r

kzemv2

2 =

(ii) In stable orbits electron does not emit any radiation. But at jumping from one orbit to another. It emits or absorbs energy.

This energy equals difference of energy of two energy levels.

∆E = E1 – E2 hν = E2 – E1 h = Plank const. ν = frequency of radiation (iii) Electron moves only in those orbits where

angular momentum of electrons

mvr = π2

nh

n = principle energy level = 1,2,3, …… ∞ Hydrogen spectrum –– According to Bohr's theory

radii of nth orbit = 22

22

4 kmehn

π

velocity of electron v = nhkme224π

total energy in nth orbit En = 2220

2

8–

nhme

∈

hence 21–

nEn ∝ ⇒ 2

'nREn =

while electrons jumps from one orbit to another E = E2 – E1

∆E = R'

21

22

1–1nn

⇒ λhc = R'

21

22

1–1nn

⇒ λ1 =

hcR'

21

22

1–1nn

⇒ λ1 = R'

21

22

1–1nn

Hydrogen Spectrum Different series – (1) Lymann - (Ultraviolet) (2) Balmer - (Visible) (3) Paschen - (Infra Red) (4) Brackett - (Infra Red) (5) Pfund - (Infra Red) (6) Hemfry - (Infra Red) Shortcomings of Bohr's Model –


(1) Applicable only for atoms having one electrons and not applicable for others.

(2) Nucleous is not stable. (3) Quantisation of angular momentum is not

explained (4) Orbits of electrons are elliptical, not circular. (5) Does not explain Stork's and Zeeman's effect.

28. (a) The sources of light, which emit light waves of

same frequency with constant phase difference. That is phase difference do not change with time.

Interference pattern is not observed as two physically different light sources can never be coherent.

(b) For bright fringe, ∆x = nλ ; n = 0, 1,2,3, ……

or Dyd = nλ

or y = dDnλ

So, bright fringes are obtained at

y = 0, dDλ ,

dDλ2 …..

For dark fringes , ∆x = (2n + 1) 2λ ; n = 0,1,2,3 …..

or Dyd = (2n + 1)

2λ

or y = d

Dn2

)12( λ+

So, dark fringes are obtained at

y = dD

2λ ,

dD

23λ ,

dD

25λ ….

So, dark fringes are obtained at

y = dDλ ,

dD

23λ ,

dD

25λ ……

And the distance between two consecutive bright or

dark fringe i.e, fringe width, β = dDλ .

OR (a) In the figure, AB is object placed beyond F

perpendicular to the principal axis of a concave lens. A' B' is the erect, virtual and diminished image.

B

A A' P

v f

u

c F

∆ABC ≡ ∆A' B' C

⇒ '' BA

AB = CB

BC'

…… (1)

∆PCF ≡ ∆A' B' F

'' BA

PC = FB

CF'

……… (2)

From Eq. (1) & (2)

CB

BC'

= FB

CF'

⇒ vu

–– =

)–(––

vff

or uf – uv = vf

or v1 –

f1 =

u1

or v1 –

u1 =

f1 proved

(a) P = P1 + P2 = + 10 – 5 = + 5D

(b) f = P1 =

51 m = 20 cm.

m = uv

–– = 2 ⇒ v = 2u ⇒

u21 –

u1 =

201

⇒ – u21 =

201

⇒ u = –10 cm. 29. Step transfer works on the principle of mutual

induction .

Secondary coilPrimary coil

Vp = – Np dtdφ

Vs = –Ns dtdφ

∴ p

s

p

s

NN

VV

=

For step-up transformer, p

s

NN

> 1

Flux does not change in the case of d.c. so no emf is induced.

OR Principle : A.C. Generator works on the principle of

E.M.I. Working : When coil is rotated in the magnetic field

(produced in the poles), magnetic flux linking with the coil change.


N Sω

∴ φ = BA cos ωt

∴ ε = –dt

Ndφ = –N dtd (BA cos ωt)

∴ ε = NB ω sin ωt.

30. U = a

kq2 × 12 +

2

2

akq × 12 +

3

2

akq × 4

or Total Electric flux from the surface

φ = ∫→→dsE .

= ∫ θcosEds

= ∫ θ∈π

cos4 2

0 rqds

= 04 ∈π

q ∫θ

2cosr

ds

= 04 ∈π

q × 4π = 0∈

q

CHEMISTRY

1. Holme signal is mixture of calcium carbide (CaC2) and calcium phosphide (Ca3P2)

2. (a) Copper Pyrite or Chalcopyrite ⇒ CuFeS2 (b) Copper glance ⇒ Cu2S (c) malachite ⇒ CuCo3. Cu (OH)2 (d) cuprite ⇒ Cu2O

3. Glycine (H2N – CH2 – COOH) is simplest amino acid.

4. t1/2

[A0]

5. Eºcell = 1.229 V

6. (a) The conc. above which associative colloid is formed is called CMC.

(b) Emulsifier stabalise the emulsion → soap or detergents

7. Fluoboric acid

8. CH3 – CH2 – CH – CH2OH + CH3CH2I

CH3

9. Structure of N2O5 is –

O

ON – O – N

O

O

10. d-block elements are used as catalyst due to two reasons

(A) They show variable oxidation states and hence they are converted to unstable intermediate species while they convert reactants to products this reaction occurs via a path of lower activation energy.

(B) They adsorb reactant molecules on their surface which provide a suitable surface area for the reaction to occur.

11. Small amount of very pure titanium or zirconium metal can be prepared by this method. Impure metal is heated in an evacuated vessel with I2. TiI4 or ZrI4 is formed which. vaporizes leaving behind impurities. The gaseous MI4 is decomposed on a white hot tungsten filament.

⇒ Zr + 2 I2 → K870 ZrI4 → K2075 Zr + 2I2

⇒ Ti + 2I2 → K525 TiI4 → K1675 Ti + 2I2

12. Rate constant can be defined as rate of chemical reaction when conc. of all the reactants become unity Factors on which it depends –

(a) Nature of reactants (b) Temperature (c) Catalyst 13. Ist Method : Given K = 2.303

∴ t1/2 = K

3.0303.2 ×

= 303.2

3.0303.2 × = 0.3 sec.

t90 = 3

10 × t1/2

= 3

10 × 0.3 = 1 sec

q

θ

→ds

→E


II method

K = t303.2 ln

− xa

a

0

0

2.303 = 90t303.2 ln

10100

t90 = 1 sec

14. (i) N – Bromo succinimide

(ii) sodium acetylide H – C ≡ +

NaC–

15. A = Cresol; B = Benzyl alcohol. 16. (i) 4-Methylhex-2-en-3-ol (ii) 4-chloro-4-phenyl-1-butanol 17. Factual

18. (i) A = B = ;

SO3H

19. The steady decrease in the size of lanthanide ions

(M3+) with the increase in atomic number is called Lanthanide contraction.

Cause of Lanthanide Contractions : As the atomic number increases due to addition of protons nuclear charge increases. And in case of lanthanides electrons are filled in 4f orbital which have poorest screening effect. Thus the increased nuclear charge occurs much on 5d and 6s shell and the electron cloud shrinks. This results in gradual decrease in size of lanthanides with increase in atomic number.

20. )s4d3( 28

Ni =

3d8 4s2 4p

)s4d3(

208

Ni + =

3d8 4s0 4p

[NiCl4] 2– = x x x x x x x x

Cl¯ Cl¯ Cl¯ Cl¯

sp3

since Cl¯ is weak ligand, thus, it unable to pair electrons and hence it is paramagnetic

[Ni(CN)4]2– = x x x x x x

CN¯ CN¯CN¯

dsp2

x x

CN¯

Since, CN¯ is strong ligand, thus it able to pair electrons and hence it is diamagnetic

21. (i) Buna S –CH2 – CH = CH – CH2 – CH – CH2 –

C6H5

( )n

(ii) Buna N –CH2 – CH = CH – CH2 – CH – CH2 –

CN

( )n

(iii) Neoprene

–CH2 – C = CH – CH2 –

Cl

( )n

22. Amino acids can be classified as - (a) Essential – Not produced in body

Ex – Lysine, leucine (b) Semi Essential – Partially produced in body Ex – Histidine & Arginine (c) Non essential – Produced in body Ex – Glycine, Alanine

23. Tranquilizers are drugs used to reduce stress. These are of following types :-

(i) Hypnotic Ex – Barbituric acid and it's derivatives (ii) Non-Hypnotic Ex – Chlordiazepoxide, Meprobamate

24. Eºcell = Eºcathode – Eºanode = 0.34 – (– 0.25) = +0.59 V

ln Kc = 059.0n Eºcell

= 059.02 × 0.59 = 20

Kc = 1 × 1020 25. (a) Brownian motion : Random movement of

colloidal particles in the medium is called brownian movement. Due to brownian motion particles colloide with each other and it imparts stability in the colloidal solution.

(b) Tyndall effect : Scattering of light when rays fall on the colloidal particles is called Tyndall effect. Due to this scattering of light the path of light in the colloidal solution get illuminated. Application : colour of sky is blue.

(c) Electrophoresis : Movement of colloidal particles in the presence of electric field is called electrophoresis as charge is present on the colloidal particles and ∴ movement of particles takes place toward opposite pole.


26. A is

CH3 – C = CH – CH – CH3

CH3

CH3

;

B is

CH3 – CH – COOH

CH3

;

C is CH3 – C – COOH

Br

CH3

;

D is

CH3 – C – COOH

OH

CH3

27. A is (CH3)3COOH; B is (CH3)C – NH2; C is (CH3)3C – NC; D and E are (CH3)3C – OH and (CH3)2C = CH2

28. The state of hybridisation of boron in diborane is sp3, In between boron atoms and bridging hydrogens, three centre two electron bond is present. The structure of diborane is –

x x

x x

x

x B B

Ht

Ht

Hb

Hb Ht

Ht

Hb = Bridging hydrogens Ht = Terminal hydrogens

29. (i) Rate of rxn gives us how fast reactants are converted into products. Per unit time change in the conc. of reactants or products. Rate constant is equals to rate of rxn when conc. of all the reactants become unity or one.

Rate Rate const. Depends upon Depend upon - Conc. of reactants - Nature of reactants - Nature of reactants - Temperature - Temperature - Catalyst - Pressure - Surface area - Catalyst - pH of medium

(ii) m = 62

6.222 × 200

1000 = 17.95 m

d = mM +

1000MMB

1.072 = M

+

100062

95.171

1.072 = M (0.117) M = 9.107 mol/lit (iii) Mixture which boils at const temperature is called

as azeotropic mixture. There are two types of azeotropes.

(a) maximum boiling point azeotrope – The azeotropic mixture in which boiling point of the mixture is greater than its constituents.

(b) Minimum boiling point azeotrope – The azeotrope in which B.P. of the mixture is less than its constituents

30. (a) (i) 3 a = 4r (ii) 2 a = 4r (iii) a = 2r

(b) The non-stoichiometric point defect responsible for colour in alkali metal is Metal excess defect in which alkali halide when heated in metal vapour results in removal of halogen atom from its position and in place of halogen e– get situated. This results in development of colour in the compound. ↑ ↑ ↑ ↑ (direction of dipoles) Ex [CrO2, Fe, Co, Ni etc]

↑ ↓ ↑ ↓ ( magnetic dipoles) Ex MnO, Mn2O3 etc.

MATHEMATICS

Section A

1. A2 × 2 =

2221

1211

aaaa

aij = 232 ji −

a11 = 2

32 −=

21 ; a12 =

262 −

= 24 = 2

a21 = 2

34 −=

21 ; a22 =

264 −

= 22 = 1

A =

12/122/1

2. 1,26132

==⇒

=+=−

yxyx

yxand

5,329432–

==⇒

=+=−

baba

ba

Hence x = 2, y = 1, a = 3, b = 5


3. logab . logba – 1

= ab

loglog ×

ba

loglog –1 = 1 – 1 = 0

4. Let a ∈ R ⇒ 1 + a. a = 1 + a2 > 0 ⇒ (a, a) ∈ R1 ⇒ R1 is reflexive.

5. dxdy = ax ln a + ex + axa–1 + xx(1 + ln x)

6. We have

I = dxxxx

222 222

∫

Putting x222 = t and

x222x22 x2 (log 2)3 dx = dt,

we get I = ∫ 3)2(log1 dt = 3)2(log

1 t + C

= 3)2(log1 x222 + C

7. We have, y = A cos (x + B), …(i) Since the given equation contains two arbitrary

constants, we shall differentiate it two times and we shall get a differential equation of second order.

Differentiating (i) w.r.t. x, we get

dxdy = –A sin (x + B) …(ii)

Differentiating (ii) w.r.t. x, we get

2

2

dxyd = – A cos (x + B)

⇒ 2

2

dxyd = – y

⇒ 2

2

dxyd + y = 0, [Using (i)]

which is the required differential equation of the given family of curves.

8. The vectors ar

and br

are perpendicular to each other ∴ a

r. b

r = 0

⇒ ( 2 i + λ j + k ). ( i –2 j +3 k ) = 0 ⇒ (2) (1) + λ (–2) + (1) (3) = 0 ⇒ –2λ + 5 = 0 ⇒ λ = 5/2

9. Let θ be the angle between vectors ar

and br

We have, [ a

r ] = | br

| = 2 and ar . b

r= – 1

∴ cos θ = ||||

.babarr

rr

⇒ cos θ = 22

1×

− = –21

⇒ cos θ = cos 2π/3 ⇒ θ = 2π/3 [Q 0 ≤ θ ≤ π] Hence, the angle between a

r and b

r is 2π/3.

10. Let the direction ratios of the required line be a, b, c. Since it is perpendicular to the two given lines.

Therefore, a + 2b + 3c = 0 … (i) and –3a + 2b + 5c = 0 … (ii) solving (i) and (ii) by cross-multiplication, we get

4a =

14−b =

8c or,

2a =

7−b =

4c = k (say)

Thus, the required line passes through (–1, 3, –2) and has direction ratios proportional to 2, –7, 4. So its

equation is 2

1+x = 73

−−y =

42+z

Section B

11. C1 → C1 – C2 – C3

∆ = yxxzxqpprpbaaca

++−++−++−

222

= – 2 yxxzxqppqpbaaca

++++++

C2 → C2 – C1, C3 → C3 – C1

= – 2 yzxqrpbca

C2 ↔ C3

= – (–2) zyxrqpcba

= 2 zyxrqpcba

12. A → Getting a heart card B → Getting a queen card Required probability ⇒ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= 5213 +

524 –

131 =

134

OR p = P (correct forcasting) = 1/3 q = P (two incorrect forecasting) = 2/3 n = 4 Let r be the number of correct forecast P (at least three correct results) = P ( r = 3) + P (r = 4)

= 4C3 444

44

334

31

32C

31

32

+

−−

[Q P (r) = nCr qn– r Pr]

= 4C1

40

04

31

31

32

31

32

+

C


= 4 · 32 ·

271 +1 · 1 ·

811

= 91

819

811

818

==+

13. Q f(0) = f(0+)

k = 0

lim→h

h

hhh

cos1214936

+−

+−− .

++

++

h

h

cos12

cos12

= 0

lim→h 2

2cos1

)19)(14(

hh

h

hh

−

−− . [ 2 + hcos1+ ]

= 0

lim→h

−

−

−

2cos1

1914

hhhh

hh

[ 2 + hcos1+ ]

= )22.()2/1(

)9()4( nn ll

k = 16 2 (ln2 ln3)

14. dxdy = n [x + 22 ax + ] n–1

++

222

21ax

x

dxdy = n [x + 22 xa + ] n –1

22

22 ][

xa

xxa

+

++

= 22

22 ][

xa

xaxn n

+

++

dxdy =

22 xa

ny

+

15. Put x = a cos θ

y = tan–1

θ+θ−

cos1cos1

y = tan–1

2/cos22/sin2

2

2

θθ

y = tan–1 |tan θ/2| Q – a < x < a ⇒ – a < a cos θ < a ⇒ – 1 < cos θ < 1 ⇒ θ ∈ (0, π) ∴ y = tan–1(tan θ/2)

y = 21 cos–1 (x/a) ⇒

dxdy =

222

1

xa −

−

OR Given : x = a (θ – sin θ); y = a (1 – cosθ)

⇒ θd

dx = a (1– cos θ); θd

dy = a sin θ

⇒ )cos1(

sin//

θ−θ

=θθ

=a

addxddy

dxdy = cot

2θ .

Now dxd

dd

dxyd θ

θ

θ= .

2cot2

2

= –21 cosec2

)cos1(1.

2 θ−θ

a

= 2

cos41 4 θ

− eca

.

At θ =2π ;

π

−=4

cos41 4

2

2ec

adxyd = –

a1 .

16. dxdy = 3x2 + a

)6,1( −

dxdy = 3 + a

slope of line = 1 ∴ a + 3 = 1 ⇒ a = –2 From curve y = x3 – 2x + b passes through (1, –6) – 6 = 1 –2 + b ⇒ b = – 5

17. one-one : Let x1, x2 ∈ R – –1 f(x1) = f(x2)

11

1

+xx =

12

2

+xx ⇒ x1x2 + x1 = x1x2 + x2

⇒ x1 = x2 ⇒ one-one onto :

Let y = 1+x

x⇒ x =

yy−1

Q Range = R – 1 = codomain ⇒ onto function For f–1

Let y =1+x

x

⇒ x =y

y−1

⇒ f–1(x) =x

x−1

18. = dxx∫

−

22cos1

= ∫ − x2cos21(41 + cos22x) dx

= ∫ − x2cos21(41 + )

24cos1 x+ dx

= ∫ − x2cos42(81 + 1 + cos 4x) dx

= ∫ − x2cos43(81 + cos 4x) dx

=

+−

44sin2sin23

81 xxx + C


19. = ∫ dxxx 2cos4cos221

= ∫ + dxxx )2cos6(cos21

=

+

22sin

66sin

21 xx + C

20. The equation of the family of circles of radius r is (x – a)2 + (y – b)2 = r2 where a and b are parameters. Since equation (i) contains two arbitrary constants,

we differentiate it two times w.r.t x and the differential equation will be of second order.

Differentiating (i) w.r.t. x, we get

2(x – a) + 2(y – b)dxdy = 0

⇒ (x – a) + (y – b)dxdy = 0

Differentiating (ii) w.r.t. x, we get

1 + (y – b) 2

2

dxyd +

2

dxdy = 0

From (iii), we have

y – b = – 22

2

/)/(1

dxyddxdy+

Putting the value of (y – b) in (ii), we obtain

x – a = 22

2

/

1

dxyd

dxdy

dxdy

+

Substituting the values of (x – a) and (y – b) in (i), we get

222

222

)/(

1

dxyd

dxdy

dxdy

+

+ 222

2

)/(

1

dxyd

dxdy

+

= r2

⇒ 32

1

+

dxdy = r2

2

2

2

dxyd

This is the required differential equation. OR

The given diff. eq. is

x .02 3 =−− xydxdy

⇒ 221 xyxdx

dy=− ...(1)

This is a linear diff. eq.

On comparing by, QPydxdy

=+

Here, P = – x1 , Q = 2x2

I.F. = xdxxPdx

eee log1

−−== ∫∫

= 1log −xe = 1−x

∴ The reqd. sol. of eq. (1) is

y . x–1 = ∫ +− cdxxx 12.2

= ∫ + cdxx2 = x2 + c

⇒ y = x3 + cx.

21. We have, a

r . br

= ar . c

r and ar ≠ 0

r

⇒ ar . b

r– a

r . cr = 0 and a

r ≠ 0r

⇒ a

r . ( br

– cr ) = 0 and a

r ≠ 0r

⇒ b

r– c

r = 0r

or, ar

⊥ ( br

– cr ) Q [ a

r≠ 0

r]

⇒ br

= cr or , a

r⊥ ( b

r– c

r ) …(i) Again, a

r × br

= ar × c

r and ar ≠ 0

r

⇒ ar × b

r– a

r × cr = 0

r and a

r ≠ 0r

⇒ a

r × ( br

– cr ) = 0

rand a

r ≠ 0r

⇒ b

r– c

r = 0r

or, ar || ( b

r– c

r ) [Q ar

≠ 0r

] ⇒ b

r= c

r or, ar || ( b

r– c

r ) …(ii) From (i) and (ii), it follows that b

r= c

r , because ar

cannot be both parallel and perpendicular to ( br

– cr

) OR

Q || →a = || →b = 1 (Given)

|→a +

→b |2 = |

→a |2 + ||

→b 2 +2

→→ba

= 1 + 1 + 2 ||||→→ba cos θ

= 2 + 2 (1) (1) cos θ = 2 (1 + cosθ)

= 2 . 2 cos2 2θ

2||→→

+ ba = 4 cos2 2θ

||→→

+ ba = 2 cos 2θ

cos2θ =

21 ||

→→+ ba Proved.

22. We have,

3

1−x = 2

1+y = 5

1−z = λ (say)

⇒ x = 3λ +1, y = 2λ –1, z = 5λ + 1 So, the coordinates of a general point on this line are

(3λ +1, 2λ –1, 5λ + 1) The equation of the second line is

4

2+x = 3

1−y = 21

−+z = µ (say)

⇒ x = 4µ –2, y = 3µ + 1, z = –2µ–1 So the coordinates of a general point on this line are (4µ –2; 3µ + 1, –2µ –1) If the line intersect, then they have a common point. So, for some values of λ and µ, we must have


3λ + 1 = 4µ –2, 2λ–1 = 3µ + 1 and 5λ + 1 = –2µ –1 ⇒ 3λ – 4µ = –3 … (i) 2λ – 3µ = 2 …(ii) and, 5λ + 2µ = –2 …(iii) Solving (i) and (ii), we obtain λ = –17 and µ = –12. These values of λ and µ do not satisfy the third equation. Hence, the given lines do not intersect.

Section C

23. Let I = ∫ +−−1

0

21 )1(cot dxxx

[Q cot–1 x = tan–1 1/x]

Then, I = ∫

+−−

1

02

1

11tan dx

xx

⇒ I = ∫

−−−

1

0

1

)1(11tan dx

xx

⇒ I = ∫

−−−+−

1

0

1

)1(1)1(tan dx

xxxx

⇒ I = ∫ −+ −−1

0

11 )1(tantan dxxx

⇒ I = ∫ ∫ −+ −−1

0

1

0

11 )1(tantan dxxdxx

−=∫ ∫

a a

dxxafdxxf0 0

)()(Q

⇒ I = ∫ ∫ −−+ −−1

0

1

0

11 )1(1tantan dxxdxx

⇒ I = ∫ ∫ −− +1

0

1

0

11 tantan dxxdxx

⇒ I = 2 ∫ −1

0

1tan dxx ⇒ I = 2III

dxx∫ −1

0

1 1.tan

⇒ I = 2 [ ]10

1tan xx − – 2 ∫ +

1

021 x

x dx

⇒ I = 2 [ ]10

1tan xx − – ∫ +

1

021

2xx dx

⇒ I = 2 [ ]10

1tan xx − – [ ]10

2 )1log( x+

⇒ I = 2

−

π 04

– (log2 – log1)

⇒ I = 2π – log 2

24. Let the two men are A and B A → A speaks truth B → B speaks truth

P(A) = 10080 , P )A( =

10020

P(B) = 10090 , P )B( =

10010

(A) required probability

= P(A). P )B( + P(B). P )A( = 5013

(B) required probability

= P(A). P(B) + P )A( . P )B( =5037

25. Area = ∫4

2

dxy = ∫

+−

4

22

8 dxx

= 21

∫ ∫+−4

2

4

2

4 dxdxx

B

y

(0, 4)

(8, 0) x

2y = –x + 8A x = 4 x = 2 O

|

= 21

4

2

2

2x

− + 4 4

2]x[

= 21

+−

24

216 + 4[4 –2]

= 21 (–6) + 4(2)

= 8 –3 = 5 sq. units. 26. Let Q be the image of the point P( i + 3 j + 4 k ) in

the plane rr

. (2 i – j + k ) + 3 = 0. Then, PQ is normal to the plane. Since PQ passes through P and is normal to the given plane, therefore equation of line PQ is

rr

= ( i + 3 j + 4 k ) + λ (2 i – j + k ) Since Q lies on line PQ, so let the position vector of

Q be ( i +3 j + 4 k ) + λ (2 i – j + k )

= (1 + 2λ) i + (3 – λ) j + (4+ λ) k Since R is the mid-point of PQ. Therefore, position vector of R is

2

]ˆ4ˆ3ˆ[]ˆ)4(ˆ)–3(ˆ)21[( kjikji +++λ++λ+λ+


= (λ +1) i +

λ

−2

3 j +

λ

+2

4 k

Since R lies on the plane rr

. (2 i – j + k )+ 3 = 0

∴

λ

++

λ

−++λ kji ˆ2

4ˆ2

3ˆ)1( . (2 i – j + k ) + 3

= 0

⇒ 2λ + 2 – 3 + 2λ + 4 +

2λ + 3 = 0

⇒ λ = –2

P

R

Q

( i + 3 j+ 4 k )

Thus, the position vector of Q is ( i +3 j +4 k ) – 2 (2 i – j + k ) = –3 i + 5 j +2 k 27. Suppose the dealer buys x fans and y sewing

machines. Since the dealer has space for at most 20 items. Therefore,

x + y ≤ 20 A fan costs Rs.360 and a sewing machine costs

Rs.240. Therefore, total cost of x fans and y sewing machines is Rs.(360x +240y). But the dealer has only Rs.5760 to invest . Therefore,

360x + 240y ≤ 5760 Since the dealer can sell all the items that he can buy

and the profit on a fan is of Rs.22 and on a sewing machine the profit is of Rs.18. Therefore, total profit on selling x fans.

Let Z denote the total profit. Then, Z = 22x + 18y. Clearly x, y ≥ 0 Thus, the mathematical formulation of the given

problem is Maximize Z = 22x + 18y S. t. x + y ≤ 20 360 x + 240y ≤ 5760 and x ≥ 0, y ≥ 0 To solve this LPP graphically, we first convert the

inequations into equations and draw the corresponding lines. The feasible region of the LPP is shaded in fig. The corner points of the feasible region OA2PB1 are O(0, 0), A2(16, 0), P(8,12) and B1 (0, 20).

These points have been obtained by solving the corresponding intersecting lines, simultaneously.

(0, 0)

A2 (16, 0)

360x + 240y = 5760

X

B1(0,20)

B2 (0,24)

y

P (8,12)

A1 (20, 0)

x + y = 20

The values of the objective function Z at corner

points of the feasible region are given in the following table.

Points (x, y) Value of the objective function

Z = 22x + 18y O (0, 0) Z= 22 × 0 + 18 × 0 = 0 A2(16, 0) Z = 22×16 + 18 × 0 = 352 P (8, 12) Z = 22 × 8 + 18 × 12 = 392 B1(0, 20) Z = 22 × 0 + 20 × 18 = 360

Clearly, Z is maximum at x = 8 and y = 12. The maximum value of Z is 392.

Hence, the dealer should purchase 8 fans and 12 sewing machines to obtain the maximum profit under given conditions.

28. A11 = 12

10−−

= 2; A12= – 10

12−

= 2;

A13 = 20

02−

= –4

A21 = – 12

11−−

= –1; A22 = 1012

−= –2;

A23 = – 20

12−

= 4

A31= 1011

= 1; A32 = – 1212

= 0;

A33 = 0212

= –2

∴ adj A = ′

−−−

−

201421422

=

−−−−

244022112

|A| = 2(2) + (1) (2) + 1(–4) = 4 + 2 –4 = 2 ≠ 0

∴ A–1 = |A|Aadj =

21

−−−−

244022112

The given system of equations can be written


AX = B where X =

zyx

, B =

153

∴ X = A–1B

zyx

=21

−−−−

244022112

153

= 21

−+−−

+−

22012106

156=

21

−64

2=

−32

1

∴ x = 1, y = –2, z = 3 OR

A11 = 43

23−−

= –12 + 6 = – 6;

A12 = – 43

22−

= 14

A13 = 33

32−

= – 6 – 9 = –15;

A21 = – 4332

−−−

= – (–8 –9) = 17

A22 = 4331

−−

= –4 + 9 = 5;

A23 = – 33

21−

= – (–3 –6) = 9

A31 = 2332 −

= 4 + 9 = 13;

A32 = –2231 −

= – (2 + 6) = –8

A33 = 3221

= 3 – 4 = –1

∴ adj A =′

−−

−−

1813951715146

=

−−−

−

19158514

13176

|A| = 1 (–6) + 2(14) + (–3) (–15) = – 6 + 28 + 45 = 67

∴ A–1 = |A|Aadj =

671

−−−

−

19158514

13176

The given system of equations can be written as

AX = B where X =

zyx

, B =

−

1124

∴ X = A–1 B

∴

zyx

=671

−−−

−

19158514

13176

−

1124

=671

−+−+−

++

111860881054

1433424=

671

−

67134

201=

−12

3

Hence x = 3, y = –2, z = 1 29. 1st Part: f ′(x) = cos x –sin x = 0 ⇒ tan x = 1 ⇒ x = π/4 f ′′(x) = –sin x – cos x

f ′′(x)|x = π/4 = –1/ 2 – 1/ 2 = – 2 < 0

Local maxima at x = π/4 OR 2nd Part: Let x > 0, y > 0 x + y = 14 … (1) Let S = x2 + y2 S = x2 + (14 –x)2

dxdS = 2x + 2(14 – x) (–1) = 0 ⇒ x = 7; y = 7

2

2

dxSd = 2 + 2 = 4 > 0 is minimum ∴ x = 7 = y

Cartoon Law of Physics

As speed increases, objects can be in several places at once.

This is particularly true of tooth-and-claw fights, in which a character's head may be glimpsed emerging from the cloud of altercation at several places simultaneously. This effect is common as well among bodies that are spinning or being throttled.

A `wacky' character has the option of self- replication only at manic high speeds and may ricochet off walls to achieve the velocity required.


XtraEdge Test Series ANSWER KEY

PHYSICS Ques 1 2 3 4 5 6 7 8 9 10 A n s A C B B B B B B A D Ques 11 12 13 14 15 16 17 18 19 A n s D D B B B D D A B

20 A → S B → R C → Q D → P 21 A → P,Q,R B → S C → R,S D → P,Q Column

Matching 22 A → Q B → P C → R,S D → P

CHEMISTRY

Ques 1 2 3 4 5 6 7 8 9 10 A n s A B A B C A C B A C Ques 11 12 13 14 15 16 17 18 19 A n s A A D C A B D B C

20 A → R B → P C → Q D → R 21 A → Q,R,S B → Q,R,S C → R,S D → P,S Column

Matching 22 A → Q,R,S B → Q,R,S C → P,Q,R D → Q,R,S

MATHEMATICS

Ques 1 2 3 4 5 6 7 8 9 10 A n s D B D A B B A C C D Ques 11 12 13 14 15 16 17 18 19 A n s B C D B A B C D B

20 A → Q B → P C → R D → S 21 A → R B → P C → Q D → S Column

Matching 22 A → R B → Q C → P D → S

PHYSICS

Ques 1 2 3 4 5 6 7 8 9 10 A n s A A C D A D A B B D Ques 11 12 13 14 15 16 17 18 19 A n s A A D D C C B C A

20 A → R B → P C → Q D → S 21 A → R B → P C → Q D → S Column

Matching 22 A → Q,R B → P,Q C → Q D → P

CHEMISTRY

Ques 1 2 3 4 5 6 7 8 9 10 A n s D B B B D A D B A A Ques 11 12 13 14 15 16 17 18 19 A n s A B B C B A A C A

20 A → P,R B → P,R C → Q D → S 21 A → S B → P,Q C → R D → S Column

Matching 22 A → R B → P C → S D → Q

MATHEMATICS

Ques 1 2 3 4 5 6 7 8 9 10 A n s A A C B C A B C B C Ques 11 12 13 14 15 16 17 18 19 A n s A A C A C B A B A

20 A → P B → Q C → S D → R 21 A → Q B → S C → R D → P Column

Matching 22 A → Q B → P C → S D → R

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