XtraedgeNovember_2011_

XtraEdge for IIT-JEE 1 NOVEMBER 2011

Dear Students, Everyone wants success. Some people spend their every waking moment pursuing it, to the detriment of all else. For others, attaining success seems impossible. They conclude that it is destined for a select few. The rest of us are to remain "content with such things as we have". Having it all is not "in our stars". When you strive for success with the wrong assumptions, you will never reach it. It's like traveling somewhere with the wrong map. Zig Ziglar says that, "Success is a process, not an event," "a journey, not a destination." Jim Rohn describes it as " .... a condition that must be attracted not pursued." Success is something you must work hard and long to earn, for yourself. It has a price, sometimes a very high one. And most people aren't really and truly ready to pay that price, to do what success demands. If success has eluded you so far, perhaps you should try changing your assumptions. You need to accept that : • You must go through a growing process, which will require time and

patience, in order to achieve success. There are no short cuts. Anything else is a temporary illusion. Success that will remain with you, and bring you joy rather than sorrow, requires a learning process, a time to grow out of old habits and into new ones, a time to learn what works and what doesn't. And you must pay your dues, in full, in advance! so don't be in a hurry.

• You will need to acquire traits and skills that attract it. What does success mean to you ? Identify, in specific terms, what you regard as success. What traits or skills will you need to achieve this goal? Find 2 or 3 people who have what you want. Write down the habits that have made them successuf and resolve to copy them. This is called mentoring learning from others who have arrived where you want to go. Once you learn to do what it takes, you qualify. And when you qualify, success comes looking for you. You just can't be denied!

Remember, when parents try to teach their children to crawl, what they do? They put their favorite toy in front of them and teased them forward, inch by inch. They were after the toy, which kept them motivated. When they became good at reaching the toy, they had learned to crawl. After that, they could reach any destination they wanted. The DESTINATION was less important. They became champion crawlers in the PROCESS! When you are ready for success you attract it, with little effort. When you are not, it runs from you, no matter how hard you chase. In other words, you repel it! Most likely, this is the reason that success eludes people. Now that you know how to attract success, why not get started on the journey that will take you where you want to go. Any one can succeed, but unfortunately not every one will. Fate does not foist it upon you. You can have anything you want in life, if you're ready to pay the price. But if you consider the process too hard, too slow, or too long and lonely, you have qualified your self as a looser; painful but true. So don't short change yourself with short-cuts. Go out there today and start attracting success. It's literally yours for the taking! Presenting forever positive ideas to your success.

Yours truly

Pramod Maheshwari, B.Tech., IIT Delhi

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Editor : Pramod Maheshwari

Impatience never commanded success. Volume - 7 Issue - 5

November, 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-7 Issue-5 November, 2011 (Monthly Magazine)

NEXT MONTHS ATTRACTIONS

Much more IIT-JEE News.

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

Challenging Problems in Physics, Chemistry & Maths

Key Concepts & Problem Solving strategy for IIT-JEE.

Xtra Edge Test Series for JEE- 2012 & 2013

S

Success Tips for the Months

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

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

• "The ladder of success is best climbed by stepping on the rungs of opportunity."

• "Success is getting what you want. Happiness is wanting what you get."

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

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

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

CONTENTS

INDEX PAGE

NEWS ARTICLE 3 • Government plans to recover costs from IIT students • HRD to hand over 10,000 low-cost laptops to IIT Rajasthan

IITian ON THE PATH OF SUCCESS 5 Mr. Subrah S. Iyer

KNOW IIT-JEE 6 Previous IIT-JEE Question

XTRAEDGE TEST SERIES 47

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

Regulars ..........

DYNAMIC PHYSICS 14

8-Challenging Problems [Set # 7] Students’ Forum Physics Fundamentals • E.M.I. & A.C. • S.H.M.

CATALYSE CHEMISTRY 27

Key Concept • Nitrogen Compound • Nitrogen Family Understanding :Physical Chemistry

DICEY MATHS 38

Mathematical Challenges Students’ Forum Key Concept

• Differentiation • Straight Line & Circle

Study Time........

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


Government plans to recover costs from IIT students Students of the Indian Institutes of Technology (IITs) are set to be under a debt burden from their first day in class.

According to a new proposal, general category IIT graduates may soon have to pay back the money that the government incurs on their education as soon as they find a job after passing out. The 15 premier engineering institutes and the human resource development ministry gave their "in principle" nod to the proposal, which has been touted as a workable alternative to hiking tuition fees of the IITs.

The reimbursed amount will go to the IIT from where a student graduates. The step is believed to be in line with the government's efforts to give more administrative and financial autonomy to the IITs.The landmark decision, taken at Wednesday's meeting of the IIT Council, the highest decision-making body of these institutes, will not apply to students from SC, ST and OBC (noncreamy layer) categories. Nor will it apply to those going in for higher studies at the IITs.

HRD minister Kapil Sibal said the idea of giving back to the institute was agreed upon during a discussion on the Kakodkar Committee recommendation which had proposed that the IITs increase their annual tuition fee four times from Rs 50,000 each year to anything between Rs 2 lakh to Rs 2.5 lakh per annum.

"We have shot down that proposition as we do not want to burden the students' families. Students will continue to pay Rs 50,000 as their annual fee, but they will be expected to pay the difference between the tuition fee and the actual expenditure incurred by the institute once they start working," he said.

Sources said the payback proposal is "reasonable". IIT graduates are in great demand and command high salaries. Roughly half the students of an IIT are from the general category because 49.5 per cent of the seats are reserved. A rough estimate pegs the total expenditure on an IIT BTech graduate over four years at Rs 6 lakh to Rs 8 lakh. The student will not be liable to pay the difference between the tuition fee and the actual expenditure in case he studies further like pursue MTech, PhD and so on.

But the moment a student gets a job, irrespective of whether it is in the government or the private sector, the loan meter will start ticking. It was decided at the meeting that students will only need to return the amount in installments.

"A student who eventually becomes a researcher or joins an IIT as a faculty member will also be exempted as we want to encourage research and students to become teachers. In case the student remains unemployed, we won't expect him to pay," Sibal added. It is, anyway, rare for an IIT graduate to remain unemployed. Even though the ministry and the IITs have agreed upon this proposal "in principle", it can only be implemented if Sibal can get the finance ministry on board. Obviously then, as of now, there is no deadline for implementation of this decision. Another hurdle would be to ensure that students do not shirk their responsibility of paying back. Sibal said the shift to "demat" degrees and certificates will take care of this problem.

Last year, the minister had announced the start of a process for the establishment of a national database of academic qualifications (degrees or certificates from school to graduate and postgraduate levels, including professional degrees), which will be

created and maintained in a digital format by an identified, registered depository. For this purpose, the HRD ministry had constituted a task force under the supervision of IIT-Kanpur director Sanjay Dhande.

"The degree eventually goes to the employer. Once the demat system is in place, an IIT graduate's degree will reflect the obligation to pay the institute back and the money will come via the employer," said Sibal, adding that the details will be worked out in consultation with the IITs

HRD to hand over 10,000 low-cost laptops to IIT Rajasthan Jaipur: The much awaited low-cost laptops in India will be introduced through the IIT-Rajasthan. The Union HRD ministry announced that the first batch of low-cost laptops, designed for use by students, will be handed over to IIT Rajasthan.

"We would deliver the first lot of 10,000 laptops to IIT-Rajasthan in June," a ministry official said. The laptops would come for Rs 2,200 per unit. The original price band for these laptops was kept between Rs1,000-1,500 per unit.

"One lakh laptops have been ordered for students across the country. The remaining 90,000 units would be distributed in remaining states over the next four months," the official said.

The HRD ministry announced its plans of rolling out low-cost computing devices during a conference of education ministers from different states in New Delhi two days ago. The project had been in the pipeline for six years. The conference, presided over by Union HRD minister Kapil Sibbal, was attended by school education minister Bhanwar Lal Meghwal and


higher education minister Jitendra Singh.

Officials at the IIT Rajasthan in Jodhpur said the project of low-cost laptops was underway. But, they said they were not aware of the price band at which it would be available or the mode of distribution that would be adopted. "It's a Union government decision, an official of the institute said.

Officials said the government would subsidise 50 per cent of the cost and a student would pay only around Rs 1,000 for the device. On the basis of the feedback of the field trial, the computers will be made available for distribution among students under the National Mission On Education through Information and Communication Technology.

The computers will be equipped with WiFi connectivity, PDF Reader, Office applications, a web browser with remote device management capability and video streaming. They will come with 2 USB ports, built-in keyboard, a 7-inch touch-screen and 2 GB RAM.

IIT-Bombay announces Geomat12 IIT-B going to conduct Geomatrix’12, an International Conference on Geospatial Technologies and Applications to be held from 26th to 29th February 2012 at Indian Institute of Technology Bombay (IITB).

Geomatrix’12 is the third in a series of conferences on geoinformatics tools, techniques and applications being organized by the Centre of Studies in Resources Engineering (CSRE), IITB since 2009. The first two conferences were national level conferences; Geomatrix’12 is envisaged to be an international event with strong participation from national and international researchers and institutes of repute at both organization and delegate level.

We are expecting participation of delegates from various academic institutes, research organizations and industries to share their research findings and professional experiences

with fellow researchers, professionals and students from India and abroad. We are looking forward to your enthusiastic participation in Geomatrix’12.

The focus of the conference would be on:

(a) tools and techniques of GIS, remote Sensing, satellite image processing and GPS, and

(b) applications of GIS and remote sensing to exploration, assessment and management of natural resources (including minerals, water resources, forests, snow and glaciers, etc.), agriculture, natural hazard assessment and disaster management, environmental impact assessment including climate change studies, atmospheric studies, terrain studies, land-use planning, and related fields of earth sciences.

The specific themes include:

1. Advances in Tools and techniques of GIS

2. Recent advances in microwave remote sensing

3. State-of-the-art in satellite image processing

4. Global positioning systems and wireless sensor networks

5. Geospatial technology in mineral system studies and mineral exploration

6. Agroinformatics – tools and applications

7. Geoinformatics for coastal, marine, and urban environments

8. Geospatial technology for prediction and management of natural hazards and disasters

9. Monitoring and managing our glaciers and water resources using remote sensing

10. Education and educational technology in geoinformatics

Get a reality check: IITs well behind Chinese peers, says PM If you think that our Indian Institutes of Technology are producing the best brains of the world, think again. In a recent speech at IIT-Kharagpur, Prime Minister Manmohan Singh said the IITs

were well behind technology counterparts in China when it came to research and PhDs.

"The Kakodkar committee report noted the number of PhDs is very small in comparison to similar technology institutions in the USA and China," he said. It was important as it emphasised the challenge in creating an advanced research-based innovation ecosystem, with the involvement of industry and national technology programmes.

Talking about how Kakodkar panel was set up last year to come up with a report card of the progress of IITs in the country, Singh said that the recommendations of the committee will soon be considered by the Council of the IITs and then by the Government of India. Of many suggestions given by Kakodkar panel, the most significant one is about giving more autonomy to these institutions.

The PM also stressed on the fact that IITs need to take on a leadership role on innovations to stimulate long-term growth and development. He emphasises the need for a second Green Revolution. He said, "We have to usher in a soft revolution in our academic business and administrative cultureâ€¦our scientific and entrepreneurial energies should be channeled to spark the second Green Revolution, find new pathways for sustainable growth and living and make green growth a profitable business proposition."

PM Manmohan Singh was addressing the 60th anniversary and the 57th convocation of IIT-Kgp. At the event, 1,966 degrees were awarded, of which 235 were PhDs, 29 MS, 692 M Tech, 84 MBAs, 380 B. Tech and 216 MSc, among others. Several personalities, including Bharti Airtel chairman Sunil Mittal were awarded honourary doctorates.


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

Subrah S. Iyer (b.1957) is a leading technocrat, entrepreneur and Web conferencing pioneer of Indian origin. He is the founder and CEO of WebEx which has recently merged with Cisco Systems.

Early life Subrah S. Iyer was born and brought up in Mumbai. He had descended from Tamil immigrants who had migrated to Mumbai. He did his schooling in Mumbai and graduated from the Indian Institute of Technology. On completion of his graduation he moved to the US in the year 1982. He worked with Intel, Apple Inc., Quarterdeck, and Teleos Research prior to the establishment of WebEx.

Founding of WebEx In his childhood days, his father had sternly warned him against dreaming of becoming an entrepreneur. However, he overruled him when in 1996; he founded WebEx in partnership with Min Zhu.

The founding of the company by Subrah Iyar was fuelled by a new-found interest in Web Conferencing. Min Zhu, a Stanford-trained System Engineer had been struggling to develop a web-conferencing tool. Coincidentally, during this time, he befriended Subrah Iyar who was running Quarterdeck's research lab and the two formed a partnership.

Growth of WebEx WebEx struggled to make a profit in its early days, low bandwidth being one of the main reasons. Slowly, with the advancement of technology and the shift to broadband technology, WebEx began to emerge as a potent competitor with clients such as Hoover's Online, Oracle and Tibco Software. However, despite the below performance of Webex in its early days, it was generally a boom time for digital conferencing technology with the emergence of standards such as ISDN and Switched Digital Service. WebEx received its first funding of $25 million in December 1999.

Faced with a win or lose situation, the management of WebEx accepted the challenge with a brave heart. As a result of the new ideas propounded by Subrah Iyer, 2000 became a honeymoon year for WebEx. The revenues crossed the million mark and Subrah Iyar's own net worth rocketed from a paltry $450,000 in January 2000 to $129 million in November 2000. When enquired about it in an interview at a later stage, Subrah Iyar remarked, "It didn't get too scary, because I knew we had done everything based on fundamentals. You always have a feeling of uncertainty. But it was never a feeling of terror."

In 2003, when Microsoft purchased conferencing company Place ware it was thought to be the end of the road for Subrah Iyar and WebEx. However, WebEx survived and completed a $45 million acquisition of Intranets.com in 2005. As per the company website, more than 3.5 million people use Cisco’s WebEx products every month to communicate and collaborate online.

How WebEx went the Cisco route In Silicon Valley, being at the top of your game in a hot market means you can pretty much name your price. At least that's what seems to have happened to Web conferencing company WebEx.

In March 2007, Cisco Systems said it would pay $3.2 billion for the company. Cisco plans to integrate WebEx's online collaboration and meeting services into its unified communications business. Subrah Iyar, chairman and chief executive officer, has been with WebEx since the beginning as a co-founder. And through the years, he has established the company as a leader in the Web collaboration market, fending off tough competitors such as Microsoft.

WebEx had already been partnering with Cisco to integrate voice over IP capabilities into its Web conferencing services. So when potential suitors came knocking on WebEx's door, it made perfect sense for the company to talk to Cisco about a deal.

SUBRAH S. IYER Born : 1957, Mumbai, Maharashtra, India

Occupation : CEO, WebEx

Net worth : US$129 million (2000)


PHYSICS

1. Two square metal plates of side 1 m are kept 0.01 m apart like a parallel plate capacitor in air in such a way that one of their edges is perpendicular to an oil surface in a tank filled with an insulating oil. The plates are connected to a battery of emf 500 V. The plates are then lowered vertically into the oil at a speed of 0.001 ms–1. Calculate the current drawn from the battery during the process. (Dielectric constant of oil = 11, ε0 = 8.85 × 10–12C2N–1m–1)

[IIT-1994] Sol. The adjacent figure is a case of parallel plate

capacitor, The combined capacitance will be

1m d

1–x

x

+ –

V

C = C1 + C2

= d

)1x(k 0 ×ε +

d]1)x–1[(0 ×ε

C = d0ε

[kx + 1 – x] ...(i)

After time dt, the dielectric rises by dx. The new equivalent capacitance will be

C + dC = C1' + C2'

= d0ε

[(x + dx) × 1] + d

]1)dx–x–1[(0 ×ε

= d0ε

[kx + kdx + 1 – x – dx]

Change of capacitance in time dt

dC = d0ε

[kx + kdx + 1 – x – dx – kx – 1 + x]

= d0ε

(k – 1)dx ...(ii)

dtdC =

d0ε

(k – 1)dtdx =

d0ε

(k – 1)v ...(iii)

where v = dtdx

We know that q = CV

dtdq = V

dtdV ...(iv)

⇒ I = V d0ε

(k – 1)v

From (i) and (ii)

I = 01.0

1085.8500 12–×× (11 – 1) × 0.001

= 4.425 × 10–9 Amp. Ans. Alternatively We can differentiate eq. (i) w.r.t. 't', we get

dtdC =

d0ε

(K –1)dtdx and then proceed further.

2. An electrons gun G emits electrons of energy 2keV travelling in the positive x-direction. The electrons are required to hit the spot S where GS = 0.1m, and the line GS make an angle of 60º with the x-axis as shown in the figure. A uniform magnetic field →B parallel to GS in the region outside the electron

gun. Find →B parallel to GS exists in the region

outside the electron gun. Find the minimum value of B needed to make the electrons hit S. [IIT-1993]

S

xG

90ºv

→B

Sol. Let us resolve the velocity two rectangular components V1(= V cos θ) and V2(Vsin 60º), V1 component of velocity is responsible to move the charge particle in the direction of the magnetic field whereas V2 component is responsible for rotating the charged particle in circular motion. The overall path is helical. The condition for he charged particle to strike S with minimum value of B is

Pitch of Helix = GS

T × V1 = GS ⇒ qB

m2π × v cos 60º = 0.1

21 mv2 = E ⇒ V =

mE2

KNOW IIT-JEE By Previous Exam Questions


G V2 = v sin 60º

V1 = v cos 60ºv

r = 0.1 m

S →B

⇒ B = 1.0q

º60cosmV2×

π

⇒ B = 1.0q

m2×π ×

mE2 × cos 60º

= 1.0q

2×

π × mE2 × cos 60º

= 1.0106.1

14.3219– ××

×

= 19–331– 106.1102101.92 ×××××× × 21

= 19–108.149 × 0.316 × 10–23 = 47.37 × 10–4

= 4.737 × 10–3 T Ans.

3. A solenoid has an inductance of 10 henry and a resistance or 2 ohm. It is connected to a 10 volt battery. How long will it take for the magnetic energy to reach ¼ of its maximum value? [IIT-1996]

Sol. Let I0 be the current at steady state. The magnetic energy stored in the inductor at this state will be

10V

R = 2Ω L = 10H

E = 21 LI0

2 ...(i)

This is the maximum energy stored in the inductor. The current in the circuit for one fourth of this energy can be found as

21 × E =

21 LI0

2 ...(ii)

Dividing equation (i) and (ii)

4/E

E = 2

20

LI21

LI21

⇒ I = 2I0

Also, V = I0R ⇒ I0 = RV =

210 = 5 Amp.

∴ I = 2I0 =

25 = 2.5 Amp.

The equation for growth of current in L-R circuit is

I = I0 [1 – LRT–

e ] ⇒ 2.5 = 5 [1 – 10t2–

e ]

⇒ 21 = 1 – e–t/5 ⇒ e–t/5 =

21 ⇒ et/5 = 2

⇒ 5t = loge2

⇒ t = 5 loge2 = 2 × 2.303 × 0.3010 = 3.466 sec. Ans.

4. Two identical prisms of refractive index 3 are kept as shown in the figure. A light ray strikes the first prism at face AB. Find, [IIT-2005]

A

60º

60º

B

C

60º

60º

D

E

(a) the angle of incidence, so that the emergent ray from the first prism has minimum deviation.

(b) through what angle the prism DCE should be rotated about C so that the final emergent ray also has minimum deviation.

Sol. (a) For minimum deviation of emergent ray from the first prism MN is parallel to AC

∴ ∠ BMN = 90º ⇒ ∠ r = 30º Applying Snell's law at M

µ =rsinisin

sin i = µ sin r

sin i = 3 × sin 30º =23

⇒ i = 60º

A

60º

60º

B

C

i N QM

Pr

(b) When the prism DCE is rotated about C in anticlockwise direction, as shown in the figure, then the final emergent ray SR becomes parallel to the incident ray TM. Thus, the angle of deviation becomes zero.


5. A neutron of kinetic energy 65eV collides inelastically with a singly ionized helium atom at rest. It is scattered at an angle of 90º with respect of its original direction. [IIT-1993]

(i) Find the allowed values of the energy of the neutron and that of the atom after the collision.

(ii) If the atom get de-excited subsequently by emitting radiation, find the frequencies of the emitted radiation.

[Given: mass of He atom –4×(mass of neutron), Ionization energy of H atom = 13.6eV]

Sol.

m 4m

m K2

θ4m

K1

y

x

Applying conservation of linear momentum in horizontal direction

(Initial Momentum)x = (Final Momentum)x pix = pfx

⇒ Km2 = θcosK)m4(2 1 ...(i)

Now applying conservation of linear momentum in Y-direction

piy = pfy

0 = mK2 2 – θsinK)m4(2 1

⇒ mK2 2 = θsinK)m4(2 1 ...(ii)

Squaring and adding (i) and (ii) 2Km + 2K2m = 2(4m)K1 + 2(4m)K1 K1 + K2 = 4K1 ⇒ K = 4K1 – K2 ⇒ 4K1 – K2 = 65 ...(iii) When collision takes place, the electron gains

energy and jumps to higher orbit. Applying energy conservation K = K1 + K2 + ∆E ⇒ 65 = K1 + K2 + ∆E ...(iv) Possible value of ∆E For He+ Case (1) ∆E1 = – 13.6 – (54.4eV) = 40.8 eV ⇒ K1 + K2 = 24.2 eV from (4) Solving with (3), we get K2 = 6.36 eV; K1 = 17.84 eV Case (2) ∆E2 = – 6.04 – (–54.4 eV) = 48.36 eV ⇒ K1 + K2 = 16.64 eV from (4) Solving with (3), we get K2 = 0.312 eV; K1 = 16.328 eV Case (3)

∆E3 = – 3.4 – (–54.4eV) = 51.1 eV ⇒ K1 + K2 = 14 eV Solving with (3), we get K2 = 15.8 eV; K1 = – 1.8 eV But K.E. can never be negative therefore case (3)

is not possible. Therefore the allowed values of kinetic energies

are only that of case (1) and case (2) and electron can jump upto n = 3 only.

–54.4eVFor He+

n=4n=3

n=2

n=1

–13.6eV

–3.4eV–6.04eV

(ii) Thus when electron jumps back there are three possibilities

n3 → n1 or n3 → n2 and n2 → n1 The frequencies will be

ν1 = hEE 23 −

ν 2 = hEE 13 −

ν 3 = hEE 12 −

= 1.82×1015 Hz = 11.67×1015 Hz = 9.84×1015 Hz Ans.

CHEMISTRY

6. An organic compound CxH2yOy was burnt with twice

the amount of oxygen needed for complete combustion to CO2 and H2O. The hot gases when cooled to 0 ºC and 1 atm pressure, measure 2.24 L. The water collected during cooling weighed 0.9 g. The vapour pressure of pure water at 20ºC is 17.5 mm of Hg and is lowered by 0.104 mm when 50 g of the organic compound are dissolved in 1000 g of water. Give the molecular formula of the organic compound. [IIT-1983]

Sol. According to the question, an organic compound CxH2yOy was burnt with twice the amount of oxygen. Hence,

CxH2yOy + 2x O2 → xCO2 + yH2O + xO2 Volume of gases after combustion = 2.24 L (given) Volume of gases left after combustion = xCO2 + xO2

∴ x + x = 2.24 or x = 1.12 L 22.4 L CO2 = 1 mol CO2

∴ 1.12 L CO2 = 4.22

12.1 = 0.05 mol CO2

and 18 g H2O = 1 mol H2O


∴ 0.9 g H2O = 18

9.0 = 0.05 mol H2O

Thus, the empirical formula of the organic compound is CH2O.

Empirical formula mass = 12 + 2 + 16 = 30 Vapour pressure of the pure liquid,

0AP = 17.5 mm of Hg

Lowering in vapour pressure 0AP –PA =0.104mm of Hg

Mass of organic compound = 50 g Mass of water = 1000 g

∴ Mole fraction of organic compound =

181000

M50

M/50

+

where M is the molecular mass of the organic compound, the molecular mass of water being 18.

We know,

0A

A0A

PPP − = Mole fraction of organic compound

∴ 5.17

104.0 =

181000

M50

M/50

+

or 104.0

5.17 = 1 + 5018

M1000×

×

Solving, M = 150.5 ≈ 150

n = massformulaEmpirical

massMolecular = 30

150 = 5

∴ Molecular formula or organic compound

= 5(CH2O) = C5H10O5

7. At room temperature, the following reactions proceed nearly to completion :

2NO + O2 → 2NO2 → N2O4 The dimer, N2O4, solidified at 262 K. A 250 ml flask

and a 100 ml flask are separated by a stopcock. At 300 K, the nitric oxide in the larger flask exerts a pressure of 1.053 atm and the smaller one contains oxygen at 0.789 atm. The gases are mixed by opening the stopcock and after the end of the reaction the flasks are cooled to 200 K. Neglecting the vapour pressure of the dimer, find out the pressure and composition of the gas remaining at 220 K. (Assume the gases to behave ideally) [IIT-1992]

Sol. According to the gas equation, PV = nRT

or n = RTPV

At room temperature, For NO, P = 1.053 atm, V = 250 ml = 0.250 L

∴ Number of moles of NO = 3000821.0250.0053.1

××

= 0.01069 mol For O2, P = 0.789 atm, V = 100 ml = 0.1L

∴ Number of moles of O2 = 3000821.0

1.0789.0××

= 0.00320 mol According to the given reaction, 2NO + O2 → 2NO2 → N2O4 Composition of gas after completion of reaction, Number of moles of O2 = 0 1 mol of O2 react with = 2 mol of NO ∴ 0.00320 mol of O2 react with = 2 × 0.00320 = 0.0064 mol of NO Number of moles of NO left = 0.01069 – 0.0064 = 0.00429 mol Also, 1 mol of O2 yields = 1 mol of N2O4 ∴ Number of moles of N2O4 formed = 0.00320 mol N2O4 condenses on cooling, ∴ 0.350 L (0.1 + 0.250) contains only 0.00429 mol

of NO At T = 220 K, Pressure of the gas,

P = V

nRT = 350.0

2200821.000429.0 ×× = 0.221 atm

8. One gram of an alloy of aluminium and magnesium when treated with excess of dil. HCl forms magnesium chloride, aluminium chloride and hydrogen. The evolved hydrogen collected over mercury at 0ºC has a volume of 1.20 litres at 0.92 atm pressure. Calculate the composition of the alloy.

[At wt. Mg = 24, Al = 27] [IIT-1978] Sol. Let the alloy contains, Al = x g then Mg = (1 – x)g Step 1. 2Al + 6HCl → 2AlCl3 + 3H2 2 × 27 = 54 g 3 × 22.4 = 67.2 L At STP, 54 g of Al with HCl yields H2 = 67.2 L

∴ x g of Al with HCl yields H2 = 54

2.67 x L

Step 2. Mg + 2HCl → MgCl2 + H2 24 g 22.4 L At STP, 24 g of Mg with HCl yields H2 = 22.4 L ∴ (1 – x)g of Mg with HCl yields H2

= 24

4.22 × (1 – x)L


Step 3. Also given that, P1 = 0.92 atm P2 = 1 atm V1 = 1.20 L V2 = ? T1 = 0ºC = 273 K T2 = 273 K Total volume V2 of hydrogen collected over mercury

at STP is given by,

1

11

TVP =

2

22

TVP

or V2 = 1

11

TVP ×

2

2

PT =

27320.192.0 × ×

1273 = 1.104 L

Now, Volume of H2 evolved by Al + Volume of H2 evolved by Mg = Total volume of H2

or 54

2.67 x + 24

4.22 (1 – x) = 1.104

or 67.2 × 24x + 22.4 × 54 × 1 – 22.4 × 54 × x = 1.104 × 24 × 54 or 1612.8x + 1209.6 – 1209.6x = 1430.78 or 403.2 x = 21.18 or x = 0.5486 ∴ Mass of Al = 0.5486 g and Mass of Mg = 1 – 0.5486 = 0.4514 g Step 4. % composition of Al in 1 g alloy

= alloy of Mass

Al of Mass × 100

= 1

0.5486 × 100 = 54.86% Al

% composition of Mg in 1 g alloy

= alloy of MassMg of Mass × 100

= 1

0.4514 × 100 = 45.14 % Mg

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

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

(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

AgNO3Ag – C≡C – CH2CH2CH2OH + NH4NO3

White ppt. NH3

2CH3MgBrBr MgC≡C.CH2CH2CH2OMgBr + 2CH4

2H2/PtCH3CH2CH2CH2CH2OH

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


10. A white amorphous powder A when heated gives a colourless gas B, which turns lime water milky and the residue C which is yellow when hot but white when cold. The residue C dissolves in dilute HCl and the resulting solution gives a white precipitate on addition of potassium ferrocyanide solution. A dissolves in dilute HCl with the evolution of a gas which is identical in all respects with B. The solution of A as obtained above gives a white precipitate D on addition of excess of NH4OH and on passing H2S. Another portion of this solution gives initially a white precipitate E on addition of NaOH solution, which dissolves on further addition of the base. Identify the compound A to E. [IIT-1979]

Sol. The given information is as follows.

(a) powderwhiteA →heat

milkywaterlimeturnsgascolourless

B +

cold when hot white whenyellow

residueC

(b)C → HCldilute solution

→ 64 )CN(FeK white precipitate

(c) A dilute HCl Solution + B

(i) NH4OH

(ii) H2S

(i) NaOH

D white precipitate

Ewhite precipitate

dissolves

NaOH

From part (a), we conclude that B is CO2 as it turns lime water milky :

Ca(OH2) + CO2 →

thistoduemilky

3CaCO + H2O

and C is ZnO as it becomes yellow on heating and is white in cold. Hence, the salt A must be ZnCO3.

From part (b), it is confirmed that C is a salt of zinc (II) which dissolves in dilute HCl and white precipitate obtained after adding K4[Fe(CN)6 is due to Zn2[Fe(CN)6].

From part (c), it is again confirmed that A is ZnCO3 as on adding dilute HCl, we get CO2 and zinc (II) goes into solution. White precipitate is of ZnS which is precipitated in ammonical medium as its solubility product is not very low. White precipitate E is of Zn(OH)2 which dissolves as zincate, in excess of NaOH. Hence the given information is explained as follows.

(a) (A)

3ZnCO →heat (B)

2CO + (C)

ZnO

(b) (C)

ZnO → HCldil

solution2ZnCl → 64 )CN(FeK

eprecipitatWhite62 ])CN(Fe[Zn

(c) ZnCO3 → HCldil Solution

2ZnCl + CO2 + H2O

ZnCl2 + S2– → (D)

ZnS ↓ + 2Cl–

Zn2+ + 2OH– → (E)

2Zn(OH)

Zn(OH)2 + 2OH– → dissolves

-22ZnO + 2H2O

MATHEMATICS

11. Let f [(x + y)/2] = f (x) + f (y) / 2 for all real x and y,

If f ´(0) exists and equals –1 and f (0) = 1, find f (2). [IIT- 1995]

Sol. f

+

2yx =

2)()( yfxf + ∀ x, y ∈ R (given)

Putting y = 0, we get

f

2x =

2)0()( fxf + =

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

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

h

fhfh

)0()0(lim0

−+→

= – 1 ⇒ h

hfh

1)(lim0

−→

= –1

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

f ´(x)=h

xfhxfh

)()(lim0

−+→

=h

xfhxf

h

)(2

22

lim0

−

+

→

= h

xfhfxf

h

)(2

)2()2(

lim0

−+

→

= h

xfhfxf

h

)(12

2212

2221

lim0

−

−

+−

→

[Using equations (1)]

= h

xfhfxf

h

)(1)(21)(221

lim0

−−+−

→

= h

xfhfxfh

)(1)()(lim0

−−+→

= h

hfh

1)(lim0

−→

= –1

Therefore f ´(x) = – 1 ∀ x ∈ R

⇒ ∫ dxxf )´( = ∫ − dx1

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


12. Find all the solutions of : 4 cos2x sin x – 2 sin2x = 3 sin x [IIT-1983] Sol. 4 cos2x sin x – 2 sin2x = 3 sin x ⇒ 4(1 – sin2x) sin x – 2 sin2x – 3 sin x = 0 ⇒ 4 sin x – 4 sin3x – 2 sin2x – 3 sin x = 0 ⇒ – 4 sin3x – 2 sin2x + sin x = 0 ⇒ – sin x (4 sin2x + 2 sin x – 1) = 0 ⇒ sin x = 0 or 4 sin2x + 2 sin x – 1 = 0

⇒ sin x = sin 0 or sin x = )4(2

1642 +±−

⇒ x = nπ or sin x = 4

51±−

x = nπ or sin x = sin10π or sin x = sin

π−

103

⇒ x = nπ + (–1)n

10π , nπ + (–1)n

π

−103

∴ General solution set

⇒ (x : x = nπ ∪

π

−+π=10

)1(: nnxx

∪

π−

−+π=103)1(: nnxx

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

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

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

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

externally have a radius r and centre at O.

O2

C2 C1

O C

O1

r2

r

r1

Now, OO2 = r + r2 and OO1 = r1 – r ⇒ OO1 + OO2 = r1 + r2 which is greater than O1O2 as O1O2 < r1 + r2 (Q C2 lies inside C1) ⇒ locus of O is an ellipse with foci O1 and O2.

14. Find the smallest positive number p for which the equation cos (p sin x) = sin(p cos x) has a solution x ∈ [0, 2π]

[IIT-1995] Sol. cos (p sin x) = sin (p cos x) (given) ∀ x ∈[0, 2π]

⇒ cos (p sin x) = cos

−

π xp cos2

⇒ p sin x = 2nπ ±

−

π xp cos2

, n ∈ I

[Q cos θ = cos α ⇒ θ = 2nπ ± α, n ∈ I] ⇒ p sin x + p cos x = 2nπ + π/2 or p sin x – p cos x = 2nπ – π/2, n∈ I

⇒ p. 2

+ xx cos

21sin

21 = 2nπ + π/2

or p 2

− xx cos

21sin

21 = 2nπ – π/2, n ∈ I

⇒ p 2

π

+π xx cos

4sinsin

4cos = 2nπ +

2π

or p 2

π

−π xx cos

4sinsin

4cos = 2nπ –

2π , n ∈ I

⇒ p 2

π

+4

sin x = (4n + 1)2π , n ∈ I

or p 2

π

−4

sin x = (4n – 1)2π , n ∈ I

Now, –1 ≤ sin (x ± π/4) ≤ 1

⇒ – p 2 ≤ p 2 sin (x ± π/4) ≤ p 2

⇒ – p 2 ≤ 2

).14( π+n ≤ p 2 , n ∈ I

or – p 2 ≤ 2

)14( π−n ≤ p 2 , n ∈ I

Second inequality is always a subset of first, therefore, we have to consider only first.

It is sufficient to consider n ≥ 0, because for n > 0, the solution will be same for n ≥ 0.

If n ≥ 0, – 2 p ≤ (4n + 1) π/2

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

⇒ 2 p ≥ π/2 ⇒ p ≥ 22

π

Therefore least value of p = 22

π


15. Find the range of values of t for which

2 sin t = 123

5212

2

−−+−

xxxx , t ∈

ππ−

2,

2 [IIT-2005]

Sol. Here, 2 sin t = 123

5212

2

−−+−

xxxx , t ∈

ππ−

2,

2

Put, 2sin t = y ⇒ – 2 ≤ y ≤ 2 ∴ (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

21y

+−

25

21y ≥ 0

⇒ y ≤ 2

51− or y ≥ 2

51+

or 2 sin t ≤ 2

51− or sin t ≥ 2

51+

⇒ sin t ≤ sin

π−

10 or sin t ≥ sin

π

103

⇒ t ≤ – 10π or t ≥

103π

Thus, Range for t ∈

ππ−

2,

2 ∪

ππ

2,

103

Physics Facts

Wave Phenomena 1. Sound waves are longitudinal and mechanical.

2. Light slows down, bends toward the normal and has a shorter wavelength when it enters a higher (n) value medium.

3. All angles in wave theory problems are measured to the normal.

4. Blue light has more energy. A shorter wavelength and a higher frequency than red light (remember- ROYGBIV).

5. The electromagnetic spectrum (radio, infrared, visible. Ultraviolet x-ray and gamma) are listed lowest energy to highest.

6. A prism produces a rainbow from white light by dispersion (red bends the least because it slows the least).

7. Light wave are transverse (they can be polarized).

8. The speed of all types of electromagnetic waves is 3.0 x 108 m/sec in a vacuum.

9. The amplitude of a sound wave determines its energy.

10. Constructive interference occurs when two waves are zero (0) degrees out of phase or a whole number of wavelengths (360 degrees.) out of phase.

11. At the critical angle a wave will be refracted to 90 degrees.

12. According to the Doppler effect a wave source moving toward you will generate waves with a shorter wavelength and higher frequency.

13. Double slit diffraction works because of diffraction and interference.

14. Single slit diffraction produces a much wider central maximum than double slit.

15. Diffuse reflection occurs from dull surfaces while regular reflection occurs from mirror type surfaces.

16. As the frequency of a wave increases its energy increases and its wavelength decreases.

17. Transverse wave particles vibrate back and forth perpendicular to the wave direction.

18. Wave behavior is proven by diffraction, interference and the polarization of light.

19. Shorter waves with higher frequencies have shorter periods.

20. Radiowaves are electromagnetic and travel at the speed of light (c).

21. Monochromatic light has one frequency.

22. Coherent light waves are all in phase.

Geometric Optics

23. Real images are always inverted.

24. Virtual images are always upright.

25. Diverging lens (concave) produce only small virtual images.

26. Light rays bend away from the normal as they gain speed and a longer wavelength by entering a slower (n) medium frequency remains constant.

27. The focal length of a converging lens (convex) is shorter with a higher (n) value lens or if blue light replaces red.


1. A current I enters at A in a square loop of uniform resistance and leaves at B. The ratio of magnetic field at the centre of square loop due to segments AB and due to DC is

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

2. The potential difference across a 2H inductor is as a function of time is shown. At t = 0 current is zero. Then

(A) current at t = 2s is 5A (B) current at t = 2s is 10A (C) current versus time graph in inductor will be

(D) current versus time graph in inductor will be

3. A body is thrown from the surface of earth with

velocity 2

gR (when R is radius of earth) at some

angle from the vertical. If the maximum height reached by body is R/4 then angle of projection with vertical is

(A)

−

45sin 1 (B)

−

45cos 1

(C)

−

23sin 1 (D) None of these

4. There is an infinite straight chain of alternating charges

q and –q. The distance between the two neighbouring charges is equal to d. Then the interaction energy of any charge with all the other charges is

(A) d4

q2

0

2

∈π− (B)

d42logq2

0

e2

∈π

(C) d4

2logq2

0

e2

∈π−

(D) None of these

Passage # (Q. No. 5 to Q. No. 7) A extension of Young’s double slit interference

experiment its to increase the number of slits from two to a large number N. A three slit grating consists three slits of width a and separated by spacing d. Intensity due to single slit is I0.

5. The intensity pattern for a three-slit grating is given

by (assume a << λ and λ

θπ=φ

sind2 ) θ = angle of

diffraction (A) Iθ = Iθ(1 + 4 cosφ + 4cos2φ) (B) Iθ = I0(1 + 4 cosφ) (C) Iθ = 0 (D) Iθ = I0 (1 + 4 sin φ + 4 sin2 )

6. The maximum intensity for above three-slit grating is (A) I0 (B) 5I0 (C) 3I0 (D) 9I0

7. The value of φ where intensity is half of the maximum intensity is

(A) cos-1(0.56) (B) sin-1(0.56) (C) 30º (D) None of these

8. A satellite is revolving around earth in a circular orbit of m radius r0 with velocity v0. A particle of mass is projected from satellite in forward direction

with relative velocity 0v145v

−= . During

subsequent motion of particle match the following (assume M = mass of earth)

Column – I Column – II (A) Magnitude of total energy (P) 5GMm/8r0 of particle (B) Minimum distance of (Q) r0 particle from earth (C) Maximum distance of (R) 3/5r0 particle from earth (D) Minimum kinetic energy (S) 5GMm/8r0 of particle (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 # 7

D

A B

C

II


1. dQ = dW + dU as dQ = - dU CdT2dU2dW =−=∴ P.dV = 2CdT

R/C2VTCdT2dV.V

RT −⇒= = constant.

Now, dQ = -dU C.dT = -Cv. vCCdT −=⇒

Option [A] is correct 2. As =− R/C2VT constant

C2R

V.T−

⇒ = constant Option [B] is correct

3. x1TRW

−∆

= where PVx = constant

Option [B] is correct 4. SD;R,PC;S,QB;R,PA →→→→

Conceptual. 5. PD;TC;RB;QA →→→→

V596

733VV DA =×+

=−

V51

592VVV26

422VV BDBA =−=−∴=×+

=−

for no energy stored in capacitor VD = VB

BADA VVVV −=−∴

Ω=⇒×+

=3

14Y6Y2

259

6. )tT(at −=φ

)t2T(adtdE −=φ

=

∫=T

0

2dt.

REH

Option [D] is correct 7. 2

0 VTT α+=

2

0 PRTTT

α+= as PV = RT

2220

2 TRPTP.T α+=⇒

021

0 T2T0dTdP)TT(RT.P =⇒=⇒−α=

2/1000min )TT2(T2.R.P −−α=∴

[2]

8. ])t2x(34[

8.0)4t12xt12x3(

8.0y222 ++

=+++

=

[2]

Solution Physics Challenging Problems

Set # 6

8 Questions were Published in October Issue

Space Shuttle

OK here is the deal with the space shuttle. It has three rocket engines in the back, but there's absolutely no room inside for all the fuel it needs to launch itself up into space. All of that fuel is stored outside the shuttle, in the big brown cylinder, called the external tank.

The tank containing all the rocket fuel weighs seven times more than the space shuttle itself! That's a lot of really heavy fuel, and the space shuttle engines aren't quite strong enough to push the combined weight of the shuttle and the big bloated external tank up off the ground.

That's what the two long white solid rocket boosters strapped onto the sides of the external tank are for. They lift the tank! Fortunately, it was not necessary to strap an infinite series of smaller and smaller rockets to the sides of the solid rocket boosters.

It is not widely known that just behind the main flight deck of the space shuttle is a small Starbucks adapted for use in zero gravity.


1. One end of an ideal spring is fixed to a wall at origin O and axis of spring is parallel to x-axis. A block of mass m = 1 kg is attached to free end of the spring and its is performing SHM. Equation of position of the block in co-ordinate system shown in figure is x = 10 + 3. sin (10.t),t is in second and x in cm.

Another block of mass M = 3 kg, moving towards the origin with velocity 30 cm/sec collides with the block performing SHM at t = 0 and gets stuck to it Calculate

3 Kg 1 Kg

xO

(i) new amplitude of oscillations, (ii) new equation for position of the combined body

and (iii) loss of energy during collision. Neglect friction. Sol. Natural length of the spring is 10 cm and force

constant of the spring is K = 100 Nm–1 . Just before collision, velocities of 1 kg block and

3 kg block are (+ 0.30 ms–1) and (– 0.30 ms–1) respectively. Let velocity of combined body just after collision be v, then, according to law of conservation of momentum, (1 + 3) v = 1 (0.30) + 3 ( – 0.30)

or v = – 0.15 ms–1. Negative sign indicates that combined body starts to

move leftward. But at the instant of collision spring is in its natural length or combined body is in equilibrium position. Hence at t = 0, phase of combined body becomes equal to π.

Now angular frequency of oscillations of combined body is

ω' = Mm

K+

= 31

100+

= 5 rad sec–1 .

∴ New amplitude of oscillations is

a' = '|v|

ω =

515.0 = 0.3 m or 3 cm Ans. (i)

∴ Equation for position x of combined body is given by

x = l0 + a' sin(ω't + π) or x = 10 + 3 sin (5t + π) cm or x = 10 – 3 sin (5t) cm Ans. (ii) Kinetic energy of two blocks (Just before collision)

= 21 m(0.3)3 +

21 M(0.3)2 = 0.18 joule

Kinetic energy of combined body (just after collision)

= 21 (m + M) v2 = 0.045 Joule

∴ Loss of energy, during collision = 0.18 – 0.045 joule = 0.135 joule Ans.(iii)

2. Two identical blocks A and B of mass m = 3 kg are attached with ends of an ideal spring of force constant K = 2000 Nm–1 and rest over a smooth horizontal floor. Another identical block C moving with velocity v0 = 0.6 ms–1 as shown in fig. strikes the block A and gets stuck to it. Calculate for subsequent motion

(i) velocity of centre of mass of the system, (ii) frequency of oscillations of the system, (iii) oscillation energy of the system, and (iv) maximum compression of the spring.

m m mA BC v0

Sol. When block C collides with A and get stuck with it,

combined body moves to the right, due to which spring is compressed. Therefore, the combined body retards and block B accelerates. In fact, deformation of spring varies with time and the system continues to move rightwards. In other words, centre of mass of the system moves rightwards and combined body and block B oscillate about the centre of mass of the system.

Let just after the collision velocity of combined body formed by blocks C and A be v. Then, according to law of conservation of momentum,

(m + m)v = mv0

or v = 2

v0 = 0.3 ms–1

∴ Velocity of centre of mass of the system,

vc = mm2

0mvm2+

×+× = 0.2 ms–1

Now the system is as shown in fig. 2m m

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

Students' ForumPHYSICS


Its reduced mass, m0 = mm2

)m)(m2(+

= 3m2

∴ Frequency of oscillations,

f = 0m

K21π

= π105 Hz. Ans.

Since, just after the collision, combined body has velocity v, therefore, energy of the system at that

instant, E = 21 (2m)v2 = 0.27 joule

Due to velocity vC of centre of mass of the system, translational kinetic energy,

Et = 21 (3m) 2

cv = 0.18 joule

But total energy E of the system = its translational kinetic (Et) + oscillation energy (E0)

∴ E0 = E – Et = 0.09 joule At the instant of maximum compression, oscillation

energy is stored in the spring in the form of its strain energy. Let maximum compression of spring be x0.

then 20Kx

21 = E0

∴ x0 = 90 × 10–3 m or 103 mm Ans.

3. Calculate the inductance of a closely wound solenoid of length l whose winding is made of copper wire of mass m. The winding resistance is equal to R. The solenoid diametre is considerably less than its length.

Given, density of copper = d and resistivity = ρ. Sol. Let the radius of solenoid be a, total number of turns

N and let cross-sectional area of wire be S. Then length of the wire of which the solenoid is

made = 2π aN But its resistance is R

∴ R =

π

ρSaN2

or aN = πρ2

RS ...(i)

Mass of the wire is given as m ∴ (2πaN) S d = m

or aN = Sd2

mπ

...(ii)

From equation (i) and (ii)

(aN)2 = d4

mR2ρπ

...(iii)

Self-Inductance of a solenoid is L = l

20ANµ

where A = πa2

∴ L = l

220 Naπµ

= l

πµ0

d4mR

2ρπ

or L = d4

mR0

ρπµ

l Ans.

4. A wire frame of area A = 4 × 10–4 m2 and resistance R = 20 Ω is suspended freely by a thread of length l = 0.40 m. A uniform horizontal magnetic field of induction B = 0.8 T exists in the space such that plane of the frame is perpendicular to the magnetic field. At time t = 0, the frame is made to oscillate under gravity by displacing it through a = 2 × 10–2 m from its initial position along the direction of the magnetic field. The plane of frame is always along the thread and does not rotate about wire frame as a function of time. Calculate also, maximum current in the frame. (g = 10 ms–2)

Sol. Since, frame oscillates under gravity, therefore, it performs SHM with angular frequency

ω = l/g = 5 rad/sec Since the frame is released after displacing it

through a distance a at t = 0, therefore, initial phase of its harmonic oscillations is π/2. Hence, at time t, displacement x of frame from its mean position is given by x = a sin (ωt + π/2)

or x = 0.02 cos (5t) ...(i) Since, the frame always remains along the thread,

therefore, at time t, its inclination with the vertical is equal that of the thread as shown in figure.

x B

l

θ

This inclination θ is given by sin θ = l

x ...(ii)

Since, x << l, therefore, θ ≈ sin θ or θ = l

x ...(iii)

Since, magnetic field B is horizontal, therefore, its component normal to plane of the frame is equal to B cos θ.

Hence, at time t flux linked with the frame, φ = AB cos θ

since, inclination θ varies with time, therefore, flux φ also varies. Hence, an emf is induced in the frame.

Induced emf, e = – dtdφ = A B sin

dtdθ

or e = A B sin q

dtdx.1

l

∴ e = AB

l

x

l

1 ( – 02 × 5) sin (5t)

or e = – 2 × 10–6 sin (10 t) volt Ans.

Current, i = Re = – 10–7 sin (10 t) amp

∴ Maximum current (Imax) = 10–7 amp Ans.


5. A rod of mass m and having resistance R can rotate freely about a horizontal axis O, sliding along a metallic circular ring of radius a as shown in figure. The ring is fixed in a vertical plane with axis coinciding with the axis of rotation of the rod. A uniform magnetic field of induction B exists in the space parallel to the axis of rotation. A voltage source is connected across the ring and the axis. Neglecting self inductance of the circuit, calculate the law according to which emf of the source must vary so that rod rotates with constant angular velocity ω.

Sol. When the rod rotates, it cuts magnetic lines of flux. Hence, an emf is induced in the rod. Magnitude of this induced emf is equal to flux cut per second by the rod.

∴ Induced emf, e = B × area traced per second by

the rod = 21 Bωa2

According to Fleming's right hand rule, the induced emf tries to flow a radially outward current through the rod. It means that induced emf is in opposition to emf of the source.

Assuming tht the magnitude of induced emf is greater than that of the source, net emf of the circuit becomes equal to (e – E).

∴ Current through the rod is i = R

E–e .

(radially outward) Due to the current, force experienced by the rod is

equal to F = Bia. Direction of this force is such that it produces a retarding moment as shown in figure

E

i

F

O ×

– +

mg

ωt

This retarding moment, τ1 = F 2a

or τ1 = R2

Ba2(e – E)

Weight mg of the rod produces an accelerating moment

τ2 = mg 2a sin (ωt)

Since, angular velocity ω of the rod remains constant, therefore, the resultant torque on the rod must be equal to zero.

Hence, τ1 = τ2

or E = 21 Bωa2 –

BamgR sin (ωt) Ans.

Physics Facts

Modern Physics 1. The particle behavior of light is proven by the

photoelectric effect.

2. A photon is a particle of light wave packet.

3. Large objects have very short wavelengths when moving and thus can not be observed behaving as a wave. (DeBroglie Waves)

4. All electromagnetic waves originate from accelerating charged particles.

5. The frequency of a light wave determines its energy (E = hf).

6. The lowest energy state of a atom is called the ground state.

7. Increasing light frequency increases the kinetic energy of the emitted photo-electrons.

8. As the threshold frequency increase for a photo-cell (photo emissive material) the work function also increases.

9. Increasing light intensity increases the number of emitted photo-electrons but not their KE.

Internal Energy 10. Internal energy is the sum of temperature (ke)

and phase (pe) conditions.

11. Steam and liquid water molecules at 100 degrees have equal kinetic energies.

12. Degrees Kelvin (absolute temp.) Is equal to zero (0) degrees Celsius.

13. Temperature measures the average kinetic energy of the molecules.

14. Phase changes are due to potential energy changes.

15. Internal energy always flows from an object at higher temperature to one of lower temperature.

General 16. The most important formulas in the physics

regents are:

17. Physics is fun. (Honest!)

E

iO

×

–+

ωt


Electromagnetic Induction (E.M.I.) Faraday's law states that the induced emf in a closed

loop equals the negative of time rate of change of magnetic flux through the loop. This relation is valid whether the flux change is caused by a changing magnetic field, motion of the loop, or both.

ε = –dt

d BΦ

AB

ε

φ

Lenz's law states that an induced current or emf

always tends to oppose or cancel out the change that caused it. Lenz's law can be derived from Faraday's law, and is often easier to use.

ε

B (increasing)Change in B

Binduced

I

If a conductor moves in a magnetic field, a motional

emf is induced. ε = vBL (conductor with length L moves in uniform B

r field,

Lr

and vr

both perpendicular to Br

and to each other)

ε = ∫ × lrrr

d).Bv(

(all or part of a closed loop moves in a Br

field) × × × × × ×

× × × × × ×

× × × × × ×

× × × × × ×

× × × × × ×

× × × × × ×

× × × × × ×

a a

F=qvB

v

F = qE

q L

B+

–

When an emf is induced by a changing magnetic flux through a stationary conductor, there is an induced electric field E

rof non-electrostatic origin. This field

is non conservative and cannot be associated with a potential.

∫ lrr

d.E = – dt

d BΦ

G E

B

E

E

I

r

When a bulk piece of conducting material, such as a

metal, is in a changing magnetic field or moves through a field, currents called eddy currents are induced in the volume of the material.

I´I0

B0

B´

A time-varying electric field generates a displacement current iD, which acts as a source of magnetic field in exactly the same way as conduction current.

iD = εdt

d EΦ (displacement current)

Alternating Current (A.C.)

An alternator or ac source produces an emf varies sinusoidally with time. A sinusoidal voltage or current can be represented of the by a phasor, a vector that rotates counterclockwise with constant angular velocity ω equals to the angular frequency of the sinusoidal quantity. Its projection on the horizontal axis at any instant represents the instantaneous value of the quantity.

Electromagnetic Induction & A.C.

PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


i=I cos ωtO

ωI

ωt

For a sinusoidal current, the rectified average and rms

(root-mean-square) currents are proportional to the current amplitude I. Similarly, the rms value of a sinusoidal voltage is proportional to the voltage amplitude V.

Irav = π2 I = 0.637 I

Irms = 2I ; Vrms =

2V

In general, the instantaneous voltage between two points in an ac circuit is not in phase with the instantaneous current passing through points. The quantity φ is called the phase angle of the voltage relative to the current.

i = I cos ωt v = V cos(ωt + φ)

V cosφ

O

ωt

I

φ V

The voltage across a resistor R is in phase with the

current. The voltage across an inductor L leads the current by 90º (φ = + 90º), while the voltage across a capacitor C lags the current by 90º(φ = –90º). The voltage amplitude across each type of device is proportional to the current amplitude I. An inductor has inductive reactance XL = ωL and a capacitor has capacitive reactance XC = 1/ωC.

VR = IR; VL = IXL; VC = IXC

Resistor connected toac source

Inductor connected toac source

a R b a L b

i i

a C b

Capacitor connected toac source

i q –q i

In a general ac circuit, the voltage and current

amplitude are related by the circuit impedance Z. In an L-R-C series circuit, the values of L, R, C, and the

angular frequency ω determine the impedance and the phase angle φ of the voltage relative to the current.

V = IZ

Z = 2CL

2 )XX(R −+ = 22 )]C/1(L[R ω−ω+

tan φ = R

C/1L ω−ω

V = IZVL = IXL

VL – VC

VR = IR

VC = IXC

O

I

ωtφ

The average power input Pav to an ac circuit depend on the voltage and current amplitudes (or, equivalently, their rms values) and the phase angle φ of the voltage relative to the current. The quantity cos φ is called the power factor.

Pav = 21 VI cos φ = VrmsIrmscos φ

v, i, p

Pav = ½ VIcosφ

p

pi

tφ ω

In an L-R-C series circuit, the current becomes

maximum and the impedance becomes minimum at an angular frequency ω0 = 1/ LC called the resonance angular frequency. This phenomenon is called resonance. At resonance the voltage and current are in phase, and the impedance Z is equal to the resistance R.

I(A)200 Ω0.5

0.40.30.20.1

0 1000 2000 ω (rad/s)

500 Ω

2000 Ω

A transformer is used to transform the voltage and

current levels in an ac circuit. In an ideal transformer with no energy losses, if the primary winding has N1 turns and the secondary winding has N2 turns, the amplitudes (or rms values) of the two voltages are related by Eq. The amplitudes (or rms values) of the primary and secondary voltages and currents are related by Eq.


1

2

VV =

1

2

NN ; V1I1 = V2I2

Problem Solving Strategy (P.S.S.) : Faraday' Law Step 1: Identify the relevant concepts: Faraday's law

applies when there is a changing magnetic flux. To use the law, make sure you can identify an area through which there is a flux of magnetic field. This will usually be the area enclosed by a loop, usually made of a conducting material. As always, identify the target variable(s).

Step 2: Set up the problems using the following steps Faraday's law relates the induced emf to the rate

of change of magnetic flux. To calculate this rate of change, you first have to understand what is making the flux change. Is the conductor moving? Is it changing orientation? Is the magnetic field changing? Remember that it's not the flux itself that counts, but its rate of change.

Choose a direction for the area vector Ar

or Adr

. The direction must always be perpendicular to the plane of the area. Note that you always have two choice of direction. For instance, if the plane of the area is horizontal, A

r could point straight up

or straight down. It's like choosing which direction is the positive one in a problem involving motion in a straight line; it doesn't matter which direction you choose, just so you use it consistently throughout the problem.

Step 3: Execute the solution as follows : Calculate the magnetic flux using Eq.

φB = A.Brr

= BA cos φ if Br

is uniform over the

area of the loop or eq. φB = ∫ Ad.Brr

= ∫B dA cos φ

if it is not uniform, being mindful of the direction you chose for the area vector.

Calculate the induced emf using Eq. ε = –dt

d Bφ

(Faraday's law of induction) or ε = –Ndt

d Bφ .

If your conductor has N turns in a coil, do not forget multiply by N. Remember the sign rule for the positive direction of emf and use it consistently.

If the circuit resistance is known, you can calculate the magnitude of the induced current I using ε = IR.

Step 4: Evaluate your answer : Check your results for the proper units, and double-check that you have properly implemented the sign rules for calculating magnetic flux and induced emf.

P.S.S. :: Inductors in Circuits : Step 1: Identify the relevant concepts : An inductor is

just another circuit element, like a source of emf, a resistor, or a capacitor. One key difference is that when an inductor is included in a circuit, all the voltages, currents, and capacitor charges are in general functions of time, not constants as they have been in most of our previous circuit analysis. But Kirchhoff's rules, which we studied in section, are still valid. When the voltages and currents vary with time, Kirchhoff's rules hold at each instant of time.

Step 2: Set up the problem using the following steps Draw a large circuit diagram and label all

quantities known and unknown. Apply the junction rule immediately at any junction.

Determine which quantities are the target variables.

Step 3: Execute the solution as follows : Apply Kirchhoff's loop rule to each loop in the

circuit. As in all circuit analysis, getting the correct sign

for each potential difference is essential. To get the correct sign for the potential difference between the terminals of an inductor, remember Lenz's law and the sign rule described in section

in conjunction with eq. ε = –Ldtdi (self-induced

emf) and fig. In Kirchhoff's loop rule, when we go through an inductor in the same direction as the assumed current, we encounter a voltage drop equal to L di/dt, so the corresponding term in the loop equation is –L di/dt. When we go through an inductor in the opposite direction from the assumed current, the potential difference is reversed and the term to use in the loop equation is + L di/dt.

a

L

b

i

Vab = Ldtdi

Inductor with current i following from a to b: If di/dt > 0 : potential drops from a to b If di/dt < 0: potential increases from a to b If i is constant (di/dt = 0): no potential difference As always, solve for the target variables. Step 4: Evaluate your answer : Check whether your

answer is consistent with the way that inductors behave. If the current through an inductor is changing, your result should indicate that the potential difference across the inductor opposes the


change. If not, you probably used an incorrect sign somewhere in your calculation.

P.S.S. :: Alternating –Current Circuits : Step 1: Identify the relevant concepts: All of the

concepts that we used to analyze direct-current circuits also apply to alternating current circuits. However, we must be careful to distinguish between the amplitudes of alternating currents and voltages and their instantaneous values. We must also keep in mind the distinctions between resistance (for resistors), reactance (for inductors or capacitors), and impedance (for composite circuits).

Step 2: Set up the problem using the following steps Draw a diagram of the circuit and label all known

and unknown quantities. Determine the target variables. Step 3: Execute the solution as follows : In ac circuit problem it is nearly always easiest to

work with angular frequency ω. If you are given the ordinary frequency f, expressed in Hz, convert it using the relation ω = 2πf.

Keep in mind a few basic facts about phase relationships. For a resistor, voltage and current are always in phase, and the two corresponding phasor in a diagram always have the same direction. For an inductor, the voltage always leads the current by 90º (i.e., φ = + 90º), and the voltage phasor is always turned 90º counterclockwise from the current phasor. For a capacitor, the voltage always lags the current by 90º (i.e., φ = –90º), and the voltage phasor is always turned 90º clockwise from the current phasor.

Remember that with ac circuits, all voltages and currents are sinusoidal functions of time instead of being constant, but Kirchhoff's rules hold nonetheless at each instant. Thus, in a series circuit, the instantaneous current is the same in all circuit elements; in a parallel circuit, the instantaneous potential difference is the same across all circuit elements.

Inductive reactance, capacitive reactance, and impedance are analogous to resistance; each represents the ratio of voltage amplitude V to current amplitude I in a circuit element or combination of elements. Keep in mind, however, that phase relations play an essential role. The effect of resistance and reactance have to be combined by vector addition of the corresponding voltage phasors, as in fig(i) & (ii). When you have several circuit elements in series, for example, you can't just add all the numerical values of resistance and reactance to get the impedance; that would ignore the phase relations.

V = IZVL = IXL

VL – VC

VR = IR

VC = IXCO

I

ωt

Phasor diagram for thecase XL> XC

φ

V = IZ

VL = IXL

VR = IR

VL–VCO

I

ωt

Phasor diagram for thecase XL< XC

VC=IXC

φ

Fig. (i) Fig. (ii) Evaluate your answer : When working with a series

L-R-C circuit, you can check your results by comparing the values of the inductive reactance XL and the capacitive reactance XC. If XL > XC, then the voltage amplitude across the inductor is greater than that across the capacitor and the phase angle φ is positive (between 0 and 90º). If XL < XC , then the voltage amplitude across the inductor is less than that across the capacitor and the phase angle φ is negative between (0 and –90º).

1. A coil of 160 turns of cross-sectional area 250 cm2 rotates at an angular velocity of 300 rad/sec. about an axis parallel to the plane of the coil in a uniform magnetic field of 0.6 weber/metre2. What is the maximum e.m.f. induced in the coil. If the coil is connected to a resistance of 2 ohm, what is the maximum torque that has to be delivered to maintain its motion ?

Sol. We know that, emax = NABω = 160 × 0.6 × (250 × 10–4) × 300 = 720 volt.

Now imax = R

emax = 2

720 = 360 amp

τ(torque) = NiBA sin θ τmax = NiBA = 160 × 360 × 0.6 × (250 × 10–4) = 864 newton metre This torque opposes the rotation of the coil (Lenz's

Law). Hence to maintain the rotation of the coil, an equal torque must be applied in opposite direction. So the torque required is = 864 newton metre.

2. A closed coil having 50 turns, area 300 cm2, is

rotated from a position where it plane makes an angle of 45º with a magnetic field of flux density 2.0 weber/metre2 to a position perpendicular to the field in a time 0.1 sec. What is the average e.m.f. induced in the coil ?

Solved Examples


Sol. The flux linked initially with each turn of the coil is Φ = B.A = BA cos θ = BA cos 45º Substituting the values, we get

Φ = 2.0

2metreweber × (300 × 104 metre–2)×(0.7071)

= 4.24 × 10–2 weber The final flux linked with each turn of the coil Φ´ = BA cos 0º = BA = 2.0 × (300 × 10–4) = 6.0 × 10–2 weber Change in flux = Φ´ – Φ = (6.0 × 10–2) – 4.24 × 10–2 = 1.76 × 10–2 Weber This change is carried out in 0.1 sec. The magnitude

of the e.m.f. induced in the coil is given by

e = Ndt

)´(d Φ−Φ

= 50 × 1.01076.1 5−× = 8.8 volt.

3. A vertical copper disc of diameter 20 cm makes 10 revolution per second about a horizontal axis passing through it centre. A uniform magnetic field 10–2 weber/m2 acts perpendicular to the plane of the disc. Calculate the potential difference between its centre and rim in volt.

Sol. The magnetic flux Φ linked with the disc is given by

Φ = BA The induced e.m.f. (potential difference) between rim

and centre

∴ e = –dtdΦ = –

dtd (BA) = B

dtdA (numerically)

where dtdA is the area swept out by the disc in unit time.

∴ dtdA = πr2 × number of revolutions per second

= 3.14 × (0.1)2 × 10 = 0.314

∴ e = (10–2 weber/m2) × (0.314 m2/sec) = 3.14 × 10–3 volt.

4. A resistance R and inductance L and a capacitor C all are connected in series with an A.C. supply. The resistance of R is 16 ohm and for a given frequency, the inductive reactance of L is 24 ohm and capacitive reactance of C is 12 ohm, If the current in the circuit is 5 amp., find

(a) the potential difference across R, L and C (b) the impedance of the circuit (c) the voltage of A.C. supply (d) phase angle Sol. (a) Potential difference across resistance VR = iR = 5 × 16 = 80 volt Potential difference across inductance VL = i × (ωL) = 5 × 24 = 120 volt Potential difference across condenser VC = i × (1/ωC) = 5 × 12 = 60 volt

(b) Z =

ω−ω+

22

C1LR

= ])1224()16[( 22 −+ = 20 ohm

(c) The voltage of A.C. supply is given by E = iZ = 5 × 20 = 100 volt (d) Phase angle

φ = tan–1

ω−ω

R)C/1(L

= tan–1

−

161224

= tan–1(0.75) = 36º46´ 5. A 100 volt A.C. source of frequency 500 hertz is

connected to a L-C-R circuit with L = 8.1 millinery, C = 12.5 microfarad and R = 10 ohm, all connected in series. Find the potential difference across the resistance.

Sol. The impedance of L-C-R circuit is given by

Z = ])XX(R[ 2CL

2 −+

where XL = ωL = 2πfL = 2 × 3.14 × 500 × (8.1 × 10–3) = 25.4 ohm

and XC = C

1ω

= fC2

1π

= )105.12(50014.32

16−××××

= 25.4 ohm

∴ Z = ])4.254.25()10[( 22 −+ = 10 ohm

∴ ir.m.s. = ZE s.m.r =

ohm10volt100 = 10 amp.

Potential difference across resistance VR = ir.m.s. × R = 10 amp × 10 ohm = 100 volt.


Periodic motion is motion that repeats itself in a definite cycle. It occurs whenever a body has a stable equilibrium position and a restoring force that acts when it is displaced from equilibrium. Period T is the time for one cycle. Frequency f is the number of cycles per unit time. Angular frequency ω is 2π times the frequency.

f = T1 or T =

f1 ; ω = 2πf =

T2π

y

O xa

Fn

mg

y

Ox

nmg

y

Ox

a

Fnmg

If the net force is a restoring force F that is directly

proportional to the displacement x, the motion is called simple harmonic motion (SHM). In many cases this condition is satisfied if the displacement from equilibrium is small.

Fx = –kx; ax = mFx = –

mk x

Restoring force Fx

Fx = – kx

Displacement x

x > 0

Fx < 0

0

x < 0

Fx > 0

The circle of reference construction uses a rotating

vector called a phasor, having a length equal to the amplitude of the motion. Its projection on the horizontal axis represents the actual motion of a body in simple harmonic motion.

Q

P x A

O

y

x= A cos θ

Displacement

The angular frequency, frequency, and period in

SHM do not depend on the amplitude, but only on the mass m and force constant k.

ω = mk ; f =

πω2

= mk

21π

; T = f1 = 2π

km

In SHM, the displacement, velocity, and acceleration are sinusoidal functions of time. The angular frequency is ω = m/k ; the amplitude A and phase angle φ are determined by the initial position and velocity of the body.

x = A cos(ωt + φ)

Ax

O

–AT 2T

t

Energy is conserved in SHM. The total energy can be

expressed in terms of the force constant k and amplitude A.

E = 21 mvx

2 + 21 kx2 =

21 kA2 = constant

E = K+U

Energy

U

KA

x O–A

In angular simple harmonic motion, the frequency and angular frequency are related to the moment of inertia I and the torsion constant k.

ω = Ik and f =

Ik

21π

A simple pendulum consists of a point mass m at the end of a massless string of length L. Its motion is approximately simple harmonic for sufficiently small amplitude; the angular frequency, frequency, and period depend only on g and L, not on the mass or amplitude.

T θ

L

mg sinθ

mx

θ mg cosθ

mg

Simple Harmonic Motion

PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


ω = Lg ; f =

πω2

= Lg

21π

T = ωπ2 =

f1 = 2π

gL

A physical pendulum is a body suspended from an axis of rotation a distance d from its center of gravity. If the moment of inertia about the axis of rotation is I, the angular frequency and period for small-amplitude oscillations are independent of amplitude.

ω = I

mgd ; T = 2πmgd

I

θ

mg sinθ

mg cosθ

mg

O z

d cg

d sinθ

Problem Solving Strategy : Simple Harmonic Motion I : Step 1: Identify the relevant concepts : An oscillating

system under goes simple harmonic motion (SHM) only if the restoring force is directly proportional to the displacement. Be certain that this is the case for the problem at hand before attempting to use any of the results of this section. As always, identify the target variables.

Step 2: Set up the problem using the following steps Identify the known and unknown quantities, and

determine which are the target variables. It's useful to distinguish between two kinds of

quantities. Basic properties of the system include the mass m and force constant k. (In some problems, m, k, or both can be determined from other information.) They also include quantities derived from m and k, such as the period T, frequency f, and angular frequency ω. Properties of the motion describe how the system behaves when it is set into motion in a particular way. They include the amplitude A, maximum velocity vmax, and phase angle φ, as well as the values of x,vx, and ax at the particular time.

If necessary, define an x-axis as. Step 3: Execute the solution as follows : Use the equations T = 1/f and ω = 2πf = 2π/T to

solve for the target variables. If you need to calculate the phase angle, be

certain to express it in radians. The quantity ωt in Eq. Fx = – kx is naturally in radians, so φ must be as well.

If you need to find the values of x, vx, and ax at various times, use Eqs.

f = π

ω2

= mk

21π

, vx = dtdx = – ωA sin (ωt + φ)

and ax = dt

dvx = 2

2

dtxd = –ω2A cos (ωt + φ).

If the initial position x0 and initial velocity v0x are both given, you can determine the phase angle

and amplitude from Eqs. φ = arctan

ω 0

x0

xv

(phase angle in SHM) and A = 2

2x02

0v

xω

+

(amplitude in SHM). If the body is given an initial positive displacement x0 but zero initial velocity (v0x = 0), then the amplitude is A = x0 and the phase angle is φ = 0. If it has an initial positive velocity v0x but no initial displacement (x0 = 0), the amplitude is A = v0x / ω and the phase angle is φ = –π/2.

Step 4: Evaluate your answer : Check your results to make sure that they're consistent. As an example, suppose you've used the initial position and velocity to find general expressions for x and vx at time t. If you substitute t = 0 into these expressions, you should get back the correct values of x0 and vvx.

Simple Harmonic Motion II The energy equation

E = 2xmv

21 +

21 kx2 =

21 kA2 = constant ...(i)

is a useful alternative relation between velocity and position, especially when energy quantities are also required. If the problem involves a relation among position, velocity, and acceleration without reference to time, it is usually easier to use Eq.

ax = 2

2

dtxd = –

mk x ...(ii)

(from Newton's second law) or eq. (i) (from energy conservation) than to use the general expressions for x, vx, and ax as functions of time [Eqs.

x = A cos (ωt + φ) (displacement in SHM),

vx = dtdx = –ωA sin (ωt + φ) (velocity in SHM) and

ax = dt

dvx = 2

2

dtxd = – ω2A cos (ωt + φ) (acceleration

in SHM), respectively ]. Because the energy equation involves x2 and vx

2, it cannot tell you the sign of x or vx, you have to infer the sign from the situation. For instance, if the body is moving from the equilibrium position towards the point of greatest positive displacement, then x is positive and vx is positive.


1. A body of mass 1 kg is executing simple harmonic motion which is given by x = 6.0 cos (100 t + π/4) cm. What is the (i) amplitude of displacement, (ii) frequency, (iii) initial phase, (iv) velocity, (v) acceleration, (vi) maximum kinetic energy ?

Sol. The given equation of S.H.M. is x = 6.0 cos (100 t + π/4) cm Comparing it with the standard equation of S.H.M.,

x = a cos (ωt + φ), we have (i) amplitude a = 6.0 cm (ii) frequency ω = 100 /sec (iii) initial phase φ = π/4

(iv) velocity v = ω )xa( 22 − = 100 )x36( 2− (v) acceleration = –ω2 x = – (100)2x = – 104 x

(vi) kinetic energy = 21 mv2 =

21 mω2(a2 – x2)

When x = 0, the kinetic energy is maximum, i.e.,

(K.E.)max = 21 mω2a2 =

21 × 1 × 104 ×

metre10036

= 18 joules

2. A particle of mass 0.8 kg is executing simple harmonic motion with an amplitude of 1.0 metre and periodic time 11/7 sec. Calculate the velocity and the kinetic energy of the particle at the moment when its displacement is 0.6 meter.

Sol. We know that, v = ω )ya( 22 − Further ω = 2π/T

∴ v = )ya(T2 22 −

π = )7/11(

14.32× ])6.0()0.1[( 22 −

= 3.2 m/sec Kinetic energy at this displacement is given by

K = 21 mv2 =

21 × 0.8 × (3.2)2 = 4.1 joule

3. A person normally weighing 60 kg stands on a platform which oscillates up and down harmonically at a frequency 2.0 sec–1 and an amplitude 5.0 cm. If a machine on the platform gives the person's weight against time, deduce the maximum and minimum reading it will show, take g = 10 m/sec2.

Sol. Acceleration of the platform a = ω2y Maximum acceleration amax = ω2A (A = Amplitude) ∴ amax = (2πν)2A (ν = frequency) = 4(3.14)2 (2)2 × 0.05 = 7.88 m/sec2

Maximum reading = g

)ag(m max+ = 10

)88.710(60 +

= 107.3 kg

Minimum reading = g

)ag(M max− =10

)88.710(60 −

= 12.7 kg.

4. A particle of mass m is located in a unidimensional potential field where the potential energy of the particle depends on the coordinate x as U(x) = U0(1 – cos C x); U0 and C are constants. Find the period of small oscillations that the particle performs about the equilibrium position.

Sol. Given that U(x) = U0(1 – cos C x)

We know that F = ma = – dx

)x(dV

∴ a = m1

−

dx)x(dU =

m1 [– U0 C sin C]

or a = – m

CU0 [Cx] = – mCU 2

0 x (Q sin Cx ≈ Cx)

Here acceleration is directly proportional to the negative of displacement. So, the motion is S.H.M. Time period T is given by

T = ωπ2 =

)m/CU(

22

0

π = 2π

2

0CUm

5. Find the period of small oscillations in a vertical plane performed by a ball of mass m = 40 g fixed at the middle of a horizontally stretched string l = 1.0 m in length. The tension of the string is assumed to be constant and equal to F = 10 N.

Sol. The situation is showing in fig. The components of T in upwards direction are T cos θ and T cos θ. Hence the force acting on the ball = 2 T cos θ

T Tθ θ

x

l/2 l/2

∴ma = – )x4/(

Fx222 +l

Q T = F and cos θ = )x4/(

x22 +l

As x is small, x2 can be neglected from the denominator.

∴ a = – )2/(m

Fx2l

= –

lmF4 x

or a = – ω2x where ω2 = (4 F/ml) Here acceleration is directly proportional to the

negative of displacement x. Hence the motion is S.H.M.

T = ωπ2 =

)m/F4(2

l

π = π

Fml

Substituting the given values, we get

T = 3.14 ×

× −

10)0.1)(104( 2

= 0.2 sec.

Solved Examples


Organic Chemistry

Fundamentals

Basicity of Amines : Amines are relatively weak bases. They are stronger

bases than water but are far weaker bases than hydroxide ions, alkoxide ions, and alkanide anions.

A convenient way to compare the base strengths of amines is to compare the acidity constants (or pKa values) of their conjugate acids, the corresponding alkylaminium ions. The expression for this acidity constant is as follows :

3HNR+

+ H2O RNH2 + H3O+

Ka = ]RNH[

]OH][RNH[

3

32+

+

pKa = – log Ka

The equilibrium for an amine that is relatively more basic will lie more toward the left in the above chemical equation than for an amine that is less basic. Consequently, the aminium ion from a more basic amine will have a smaller Ka (larger pKa) than the aminium ion of a less basic amine.

When we compare aminium ion acidities in terms of this equilibrium, we see that most primary alkylaminium ions (RNH3

+) are less acidic than ammonium ion (NH4

+). In other words, primary alkylamines (RNH2) are more basic than ammonia (NH3):

Conjugate acid pKa

NH3 CH3NH2 CH3CH2NH2 CH3CH2CH2NH2

9.26 10.64 10.75 10.67

We can account for this on the basis of the electron-releasing ability of an alkyl group. An alkyl group releases electrons, and it stabilizes the alkylaminium ion that results from the acid–base reaction by dispersing its positive charge. It stabilizes the alkylaminium ion to a greater extent than it stabilizes the amine :

R→– N – H + H – OH R →– N – H + OH

H

H

+

H By releasing electrons, R→– stabilizes

the alkylaminium ion through dispersal of charge

–

This explanation is supported by measurements

showing that in the gas phase the basicities of the

following amines increases with increasing methyl substitution :

(CH3)3N > (CH3)2NH > CH3NH2 > NH3 This is not the order of basicity of these amines in

aqueous solution, however. In aqueous solution the order is

(CH3)2NH > CH3NH2 > (CH3)3N > NH3 The reason for this apparent anomaly is now known.

In aqueous solution the aminium ions formed from secondary and primary amines are stabilized by salvation through hydrogen bonding much more effectively than are the aminium ions formed from tertiary amines. The aminium ion formed from a tertiary amine such as (CH3)3NH+ has only one hydrogen to use in hydrogen bonding to water molecules, whereas the aminium ions from secondary and primary amines have two and three hydrogens, respectively. Poorer solvation of the aminium ion formed from a tertiary amine more than counteracts the electron-releasing effect of the three methyl groups and makes the tertiary amine less basic than primary and secondary amines in aqueous solution. The electron-releasing effect does, however, make the tertiary amine more basic than ammonia.

Basicity of Arylamines Aromatic amines (e.g., aniline and p- toluidine) are

much weaker bases than the corresponding nonaromatic amine, cyclohexylamine :

Conjugate acid pKa

Cyclo-C6H11NH2 C6H5NH2 p-CH3C6H4NH2

10.64 4.58 5.08

We can account for this effect, in part, on the basis of

resonance contributions to the overall hybrid of an arylamine. For aniline, the following contributors are important :

NH2

1

NH2

2

NH2

3

NH2

4

NH2

5

–

+ +

–

+

–

Structures 1 and 2 are the Kekule structures that

contribute to any benzene derivative. Structures 3-5, however, delocalize the unshared electron pair of the nitrogen over the ortho and para positions of the ring. This delocalization of the electron pair makes it less

NITROGEN COMPOUND


available to a proton, and delocalization of the electron pair stabilizes aniline.

When aniline accepts a proton it becomes an anilinium ion :

C6H5 2HN⋅⋅

+ H2O C6H5 3HN+

+ HO–

Anilinium ion Once the electron pair of the nitrogen atom accepts

the proton, it is no longer available to participate in resonance, and hence we are only able to write two resonance structures for the anilinium ion – the two Kekule structures:

NH3 NH3 + +

Structures corresponding to 3 – 5 are not possible for

the anilinium ion, and, consequently, although resonance does stabilize the anilinium ion considerably, resonance does not stabilize the anilinium ion to as great an extent as it does aniline itself. This greater stabilization of the reactant (aniline) when compare to that of the product (anilinium ion) means that ∆Hº for the reaction

Aniline + H2O → anilinium ion + OH–

will be a larger positive quantity than that for the reaction

Cyclohexylamine + H2O → cyclohexylaminium ion + OH–

(See figure below) Aniline, as a result, is the weaker base.

Another important effect in explaining the lower basicity of aromatic amines is the electron-withdrawing effect of a phenyl group. Because the carbon atoms of a phenyl group are sp2 hybridized, they are more electronegative (and therefore more electron withdrawing) than the sp3-hybridized carbon atoms of alkyl groups.

Enth

alpy

NH3 + OH– +

Smaller resonance stabilization of anilinium ion

Larger resonance stabilization of aniline

:NH2 NH3 + OH–+

(1)

+ H2O

:NH2 (2)

+ H2O

∆Hº2 > ∆Hº1

∆Hº2

∆Hº1

Aminium Salts and Quaternary Ammonium Salts

When primary, secondary and tertiary amines act as bases and react with acids, they form compounds called aminium salts. In an aminium salt the positively charged nitrogen atom is attached to at least one hydrogen atom :

CH3CH2⋅⋅

NH2 + HCl → OH2

salt) aminium(an chloride umEthylamini

323 ClHNCHCH −+

(CH3CH2)⋅⋅

NH + HBr → OH2

bromide niumDiethylami2223 BrHN)CHCH( −

+

(CH3CH2)3⋅⋅

N + HI → OH2

bromide iniumTriethylam323 IHN)CHCH( −

+

When the central nitrogen atom of a compound is positively charged but is not attached to a hydrogen atom, the compound is called a quaternary ammonium salt. For example

CH2CH3

CH3CH2 – N+ – CH2CH3 Br–

CH2CH3

Tetraethylammonium bromide (a quaternary ammonium salt)

Quaternary ammonium halides – because they do not

have an unshared electron pair on the nitrogen atom – cannot act as bases :

acid) with reaction undergonot (doesbromide ammonium ethyl Tetra

−+

BrN)CHCH( 423

Quaternary ammonium hydroxides, however, are strong bases. As solids, or in solution, they consist entirely of quaternary ammonium cations (R4N+) and hydroxide ions (OH–); they are, therefore, strong bases – as strong as sodium or potassium hydroxide. Quaternary ammonium hydroxides react with acids to form quaternary ammonium salts :

(CH3)4

+NOH– + HCl → (CH3)4

+NCl– + H2O

Basicity of Aromatic Heterocyclic Amines In aqueous solution, aromatic heterocyclic amines

such as pyridine, pyrimidine, and pyrrole are much weaker bases than nonaromatic amines or ammonia. In the gas phase, however, pyridine and pyrrole are more basic than ammonia, indicating that solvation has a very important effect on their relative basicities;

NPyridine

pKa = 5.23

NPyrimidinepKa = 2.70

N

NQuinolinepKa = 4.5

H

(Conjugate acid pKa)

N

Pyrrole pKa = 0.40


Inorganic Chemistry

Fundamentals

Reaction of HNO3 on Metals. (a) Metals lying below hydrogen in the

electrochemical series : Metals such as Na, K, Ca, Mg, Al, Zn, etc., lying

below hydrogen in the electrochemical series normally displace hydrogen from dilute acids. Nitric acid also primarily behaves in the same manner. But, since it is a strong oxidising agent and hydrogen is a reducing agent, secondary reactions take place resulting in the reduction of nitric acid to give NO, N2O, N2 or NH3, depending upon the nature of the metal, the temperature and the concentration of the acid. Thus, dilute nitric acid reacts with zinc in the cold giving N2O or N2 according to the following eq.:

4Zn + 10HNO3 → 4Zn(NO3)2 + N2O + 5H2O 5Zn + 12HNO3 → 5Zn(NO3)2 + N2 + 6H2O Very dilute nitric acid gives NH3 which, of course, is

neutralised by nitric acid to form NH4NO3. 4Zn + 10HNO3 → 4Zn(NO3)2 + 3H2O + NH4NO3 Similarly, iron and tin also give NH4NO3 with dilute

nitric acid. Lead gives nitric oxide with dilute nitric acid in cold. Magnesium and manganese give hydrogen. Concentrated nitric acid essentially behaves as an oxidising agent and metals like aluminium, iron, chromium, etc., are rendered 'passive' probably due to surface oxidation.

(b) Metals lying above hydrogen in the electrochemical series. : Metals such as Cu, Bi, Hg, Ag, lying above hydrogen in the electrochemical series, do not liberate hydrogen from acids. In case of these metals, the action of nitric acid involves only the oxidation of the metals into the metallic oxides which dissolve in the acid to form nitrates accompained by evolution of NO or NO2 according as the acid is dilute or concentrated. For instance, concentrated acid attacks copper giving NO2.

Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 Dilute nitric acid gives NO. 3Cu + 8HNO3 → 3Cu(NO3)2 + 4H2O + 2NO (c) Noble metals : like Au, Pt, Rh and Ir are not

attacked by nitric acid. Gold and platinum, however, are atacked by aqua regia (3 parts conc. HCl and 1 part conc. HNO3) which contains free chlorine.

HNO3 + 3HCl → 2H2O + 2Cl + NOCl This chlorine attacks gold and platinum forming soluble

chlorides which form complexes with HCl, e.g., Au + 3Cl → AuCl3

AuCl3 + HCl → acidcAurochlori4HAuCl

Hydroxylamine, NH2OH : It may be regarded as derived from ammonia by the

replacement of one H atom by an OH group. It is prepared by the reduction of nitrites with sulphur

dioxide under carefully controlled conditions. A concentrated solution of sodium nitrite is mixed with a solution of sodium carbonate and sulphur dioxide at a temperature below 3ºC is passed till the solution becomes just acidic. The following reactions are supposed to take place :

Na2CO3 + SO2 + H2O → NaHSO3 + NaHCO3 NaNO2 + 3NaHSO3 →

sulphonate sodiummineHydroxyla

23 )NaSO(HON + Na2SO3 + H2O

The sulphonate can be easily hydrolysed to hydroxylamine.

HON(SO3Na)2 → OH2 NH2OH Alternatively, it is prepared by the electrolytic

reduction of nitric acid in 50% H2SO4 using amalgamated lead cathode.

NO2 – OH + 6H+ + 6e– → NH2OH + 2H2O It is a colourless solid melting at 33ºC. It is freely

soluble in water and lower alcohols. It is unstable and decomposes violently even at 20ºC.

3NH2OH → NH3 + N2 + 3H2O The aqueous solution of hydroxylamine is less basic

than ammonia (Kb = 1.8 × 10–5). Thus, NH2OH(aq) + H2O NH3OH+ + OH–; Kb = 6.6 × 10–9 Like H2O2, it acts as an oxidising as well as a

reducing agent depending upon circumstances. Nitrogen Trifluoride , NF3 : It is conveniently prepared by fluorinating ammonia.

4NH3 + 3F2 → catalystCu NF3 + 3NH4F It can also be prepared by the electrolysis of NH4F. It is a colourless gas (m.p. –207ºC; b.p. –129ºC)

which is quite stable thermodynamically. The gas acts as a fluorinating agent. It thus converts

Cu into CuF. 2NF3 + 2Cu → N2F4 + 2CuF As, Sb and Bi also get fluorinated by interaction with NF3.

NITROGEN FAMILY

KEY CONCEPT


NF3 has a pyramidal structure with FNF angle = ~ 102º and dipole moment = 0.24 D, compared with HNH angle = ~ 107º and dipole moment = 1.48 D in case of NH3. The difference in the dipole moments of NF3 and NH3 (both of which have pyramidal structure though) is due to the fact that while the dipole moments due to N – F bonds in NF3 are in opposite direction to the direction of the dipole moment of the lone pair on N atom, the dipole moments of N – H bonds in NH3 are in the same direction as the direction of the dipole moment of the lone pair on N atom, an illustrated below.

N

••

F F

F N

••

H H

H

Because of its lower dipole moment, NF3 is weaker ligand than NH3.

NF3 is known to form complexes such as [NF4]+ and F3N→O. Thus,

NF3 + 2F2 + SbF3 pressureHigh

º200 → [NF4+] [SbF6]–

2NF3 + O2 re temperatulowdischarge Electric → 2F3N→O

Dinitrogen Difluoride, N2F2 : Dinitrogen difluoride is best prepared by reacting

NHF2 with KF. KF + HNF2 etemperatur

low→ KF.HNF2

→ re temperatuRoom N2F2 + KHF2 The reaction yields both cis and trans isomers, viz.,

N N F F

N N F

F Both the isomeric forms are gases at room

temperature, cis form boiling at – 106ºC and trans form boiling at –112ºC. The cis form is thermodynamically more stable than the trans form. At above 70ºC, nearly 90% of N2F2 is present in the cis form : > 70ºCtrans N2F2 cis N2F2

(~ 90%)

If, however, the isomeric mixture is treated with AlCl3 or the chlorides of bivalent Mn, Co, Ni and Fe, at low temperature, the major product is trans N2F2.

The cis form reacts selectively with AsF5 to form [N2F]+[AsF6]– which when reacted with NaF – HF, regenerates the original compounds. The trans form does not react with AsF5.

isomerstransandcisofMixture

22FN + AsF6 →

onlyFN cisby Formed

62

22

]AsF[]F[N −+ + trans

22FN

[N2F]+[AsF6]– → −HFNaF Na+[AsF6]– + )cis(22FN

Dinitrogen Tetrafluoride, N2F4 : N2F4 is prepared by reacting HNF2 with NaClO.

2HNF2 →NaClO N2F4 + H2O HNF2, in turn, is obtained by first fluorinating urea

and then treating the fluorinated product with concentrated sulphuric acid.

N2F4 exists both in the staggered and the gauche conformations :

N

F F N

F F N

F

F

N

F

F

Staggered form (Side View) Gauche Form It is a colourless gas (b.p. – 73ºC; m.p. –164ºC). It is a strong fluorinating agent. Thus, SiH4 + N2F4 → SiF4 + N2 + 2H2 10 Li + N2F4 → 4LiF + 2Li3N AsF5 + N2F4 → [N2F3]+ [AsF6]– N2F4 reacts with sulphur to give a number of

fluorinated sulphur compounds such as SF4 and SF5.NF3.

N2F4 easily yields, at room temperature, the free radical. NF2 which reacts readily with a number of compounds. For example,

N2F4 2(.NF2)–2ClNF2

Cl2

2ON.NF22NO

Trisilylamine, N(SiH)3 : Trisilylamine is prepared by reacting

monochlorosilane with ammonia. 2SiH3Cl + 4NH3 → N(SiH3)3 + 3NH4Cl Trisilyamine is a trigonal planar compound with N

orbitals in sp2 hybrid state, unlike trimethyl or triethylamine which is pyramidal and has N orbitals in sp3 hybrid state. There is considerable π overlap between the p orbital (containing the lone pair) of N atom and the vacant dπ orbitals of Si atoms. The trigonal planar structure of N(SiH3)3 is, thus, strengthened due to pπ–dπ bonding.

Since the lone pairs of electrons of N atom are engaged in pπ-dπ bonding between N and Si, they are no longer available for donation to Lewis acids. Trisilylamine, therefore, behaves as much weaker base compared to trimethylamine or triethylamine. Hence, trisilylamine does not form adducts with BH3 even at low temperature whereas trimethylamine or triethylamine does so readily. Due to the same reason, N(SiH3)3 acts as a much weaker ligand compared to N(CH3)3 and N(C2H5)3.


1. A container whose volume is V contains an

equilibrium mixture that consists of 2 mol each of PCl5, PCl3 and Cl2 (all as gases). The pressure is 3 bar and temperature is T. A certain amount of Cl2(g) is now introduced, keeping the pressure and temperature constant, until the equilibrium volume is 2V. Calculate the amount of Cl2 that was added and the value of 0

pK .

Sol. At equilibrium, we have

mol25PCl

mol23PCl +

mol22Cl

Total amount = 6 mol

Thus ºKp = )ºp/pPCl(

)ºp/pCl)(ºp/pPCl(

5

23 =

ºp/p62

ºp/p62 2

Substituting p = 3 bar, we get 0

pK = 1

Let x be the amount of PCl3 that combines when the amount y of chlorine is added keeping p and T constant. Thus, the amounts of PCl3, Cl2 and PCl5 become

n(PCl3) = 2 mol – x n(Cl2) = y + 2 mol – x n(PCl5 = 2 mol + x Since the final volume after the addition of Cl2 is

twice the initial volume, it follows that the total amount of gases in 2V is 2 × 6 mol = 12 mol. Since n(PCl3) + n(PCl5) is 4 mol, the total amount of chlorine is 8 mol.

Total amount = y + 6 mol – x = 12 mol Their partial pressures are

3PClp =

mol12xmol2 − p =

mol12xmol2 − × 3 bar

2Clp =

mol12mol8 p =

mol12mol8 × 3 bar = 2 bar

5PClp =

mol12xmol2 + p =

mol12xmol2 + × 3 bar

Substituting these in the expression

ºKp = )ºp/pPCl(

)ºp/pCl)(ºp/pPCl(

5

23 (where pº = 1 bar)

we get ºKp =

+

−

mol4xmol2

)2(mol4

xmol2

= )xmol2(

)2)(xmol2(+

− = 1

(as ºK p = 1)

or 4 – 2 (x/ mol) = 2 + (x/mol) or 3(x/mol) = 2 ⇒ x/mol = 2/3 = 0.67 Therefore, the amount of Cl2 added y = 6 mol + x = 6.67 mol

2. A solution comprising 0.1 mol of naphthalene and 0.9 mol of benzene is cooled until some solid benzene freezes out. The solution is then decanted off from the solid, and warmed to 353 K, where its vapour pressure is found to be 670 Torr. The freezing and normal boiling points of benzene are 278.5 K and 353 K, respectively, and its enthalpy of fusion is 10.67 kJ mol–1. Calculate the temperature to which the solution was cooled originally and the amount of benzene that must have frozen out. Assume conditions of ideal solution.

Sol. The given data are : n1 = 0.9 mol, n2 = 0.1 mol,

*fT = 278.5 K, *

bT = 353 K,

p* = 760 Torr, p = 670 Torr, ∆fusH1,.m = 10.67 kJ mol–1 From the relative lowering of vapour pressure,

we obtain the amount fraction of the solute (i.e. naphthalene).

x2 = *p

p*p − = Torr 760

Torr 670– Torr 760 = 0.1185

Since x2 = n2/(n1 + n2), we get

n1 + n2 = 2

2

xn =

0.1185mol 1.0 = 0.844 mol

Since n2 = 0.1 mol, we get n1 = 0.844 mol – n2 = 0.844 mol – 0.1 mol = 0.744 mol Hence, the amount of benzene frozen out

0.9 mol – 0.744 mol = 0.156 mol The freezing point depression constant of benzene is

UNDERSTANDINGPhysical Chemistry


Kf = m,1fus

2*f1

HRTM

∆

= )molJ10670(

)K5.278)(molKJ314.8)(molkg078.0(1

2111

−

−−−

= 4.714 K kg mol–1 Molality of the solution is

m = 1

2

mn =

11

2

Mnn

= )molkg078.0()mol744.0(

)mol1.0(1−

= 1.723 mol kg–1

Finally – ∆Tf = Kfm

= (4.714 K kg mol–1)(1.723 mol kg–1) = 8.12 K 3. What is the solubility of AgCl in 0.20 M NH3 ? Given : Ksp(AgCl) = 1.7 × 10–10 M2 K1 = [Ag(NH3)+] / [Ag+] [NH3] = 2.33 × 103 M–1 and K2 = [Ag(NH3)2

+]/[Ag(NH3)+][NH3] = 7.14 × 103 M–1

Sol. If x be the concentration of AgCl in the solution, then [Cl–] = x From the Ksp for AgCl, we derive

[Ag+] = ]Cl[

Ksp−

= x

M107.1 210−×

If we assume that the majority of the dissolved Ag+ goes into solution as Ag(NH3)2

+ then [Ag(NH3)2+] = x

Since two molecules of NH3 are required for every Ag(NH3)2

+ ion formed, we have [NH3] = 0.20 M – 2x Therefore,

Kinst =])NH(Ag[

]NH][Ag[

23

23+

+

= x

)x2M20.0(x

M107.1 2210

−

× −

= 6.0 × 10–8 M2 From which we derive

2

2

x)x2M20.0( − = 210

28

M107.1M100.6

−

−

×× = 3.5 × 102

which gives x = [Ag(NH3)2+] = 9.6 × 10–3 M, which

is the solubility of AgCl in 0.20 M NH3

4. The critical temperature and pressure for NO are 177 K and 6.485 MPa, respectively, and for CCl4 these are 550 K and 4.56 MPa, respectively. Which gas (i) has smaller value for the van der Walls constant b; (ii) has smaller value of constant a; (iii) has larger critical volume; and (iv) is most nearly ideal in behaviour at 300 K and 1.013 MPa.

Sol. We have Tc(NO) = 177 K Tc(CCl4) = 550 K pc(NO) = 6.485 MPa pc(CCl4) = 4.56 MPa

(i) Since c

c

Tp

= Rb27/a8b27/a 2

= b8

R therefore, b = c

c

p8RT

Thus,

b(NO) = )MPa485.6)(8(

)molKcmMPa314.8)(K177( 11–3 −

= 28.36 cm3 mol–1 and

b(CCl4) = )MPa56.4)(8(

)molKcmMPa314.8)(K550 113 −−

= 125.35 cm3 mol–1 Hence b(NO) < b(CCl4) (ii) Since a = 27pcb2 therefore a(NO) = (27) (6.485 MPa) (28.36 cm3 mol–1)2 = 140827 MPa cm6mol–2 ≡ 140.827 kPa dm6 mol–2 a(CCl4) = (27) (4.56 MPa) (125.35 cm3 mol–1)2 = 1934538 MPa cm6 mol–2 ≡ 1934.538 KPa dm6mol–2 Hence a(NO) < a(CCl4) (iii) Since Vc = 3b therefore, Vc(NO) = 3 × (28.36 cm3 mol–1) = 85.08 cm3 mol–1 Vc(CCl4) = 3 × (125.35 cm3 mol–1) = 376.05 cm3 mol–1 Hence Vc(NO) < Vc(CCl4) (iv) NO is more ideal in behaviour at 300 K and

1.013 MPa, because its critical temperature is less than 300 K, whereas for CCl4 the corresponding critical temperature is greater than 300 K.

5. Calculate ∆rU, ∆rH and ∆rS for the process 1 mole H2O (1,293 K, 101.325 kPa) → 1 mol H2O (g, 523 K, 101.325 kPa) Given the following data : Cp,m (1) = 75.312 J K–1 mol–1 ; Cp,m(g) = 35.982 J K–1 mol–1

∆vapH at 373 K, 101.325 kPa = 40.668 kJ mol–1 Sol. The changes in ∆rU, ∆rH and ∆rS can be calculated

following the reversible paths given below. Step I: 1 mole H2O(1,293 K, 101.325 kPa) → 1 mole H2O(1,373 K, 101.325 kPa) qp = ∆rH = Cp,m(1) ∆T = (75.312 J K–1 mol–1 ) (80 K) = 6024.96 J mol–1



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∆rS = Cp,m ln 1

2

TT

= (75.312 J K–1 mol–1) × 2.303 × log

K293K373

= 18.184 J K–1 mol–1

∆rU = ∆rH – p∆rV ~– ∆rH Step II: 1 mol H2O(1,373 K, 101.325 kPa) → 1 mol H2O (g, 373K, 101.325 kPa) qp = ∆vapH = 40.668 kJ mol–1

∆rS = K373molJ40668 1–

= 109.03 J K–1 mol–1

∆rU = ∆rH – p∆rV = 40668 J mol–1 – (101.325 kPa)

×

K273K373)moldm414.22( 1–3

= 40668 J mol–1 – 3 103 J mol–1 = 37565 J mol–1 Step III: 1 mol H2O(g, 373 K, 101.325 kPa) → 1 mol H2O(g, 523 K, 101.325 kPa) ∆rH = Cp,m (g) ∆T = (35.982 J K–1 mol–1) (150 K) = 5397.3 J mol–1

∆rS = Cp,m (g) ln 1

2

TT

= (35.982 J K–1 mol–1) × 2.303 × log

K373K523

= (35982 J K–1mol–1) × 2.303 × 0.1468 = 12.164 J K–1 mol–1 ∆rU = ∆rH – R(∆T) = 5397.3 J mol–1 – (8.314 J K–1 mol–1) (150 K) = 5397.3 J mol–1 – 1247.1 J mol–1 = 4 150.2 J mol–1 Thus ∆Utotal

= (6024.96 + 37565 + 4150.2) J mol–1

= 47740.16 J mol–1 ∆Htotal = (6024.96 + 40668 + 5397.3) J mol–1

= 52090.26 J mol–1 ∆Stotal = (18.184 + 109.03 + 12.164) J K–1 mol–1

= 139.378 J K–1 mol–1

Modern depiction of benzene

As is standard for resonance diagrams, a double-headed arrow is used to indicate that the two structures are not distinct entities, but merely hypothetical possibilities. Neither is an accurate representation of the actual compound, which is best represented by a hybrid (average) of these structures, which can be seen at right. A C=C bond is shorter than a C−C bond, but benzene is perfectly hexagonal—all six carbon-carbon bonds have the same length, intermediate between that of a single and that of a double bond.

A better representation is that of the circular π bond (Armstrong's inner cycle), in which the electron density is evenly distributed through a π-bond above and below the ring. This model more correctly represents the location of electron density within the aromatic ring.

The single bonds are formed with electrons in line between the carbon nuclei—these are called σ-bonds. Double bonds consist of a σ-bond and a π-bond. The π-bonds are formed from overlap of atomic p-orbitals above and below the plane of the ring. The following diagram shows the positions of these p-orbitals:

Benzene electron orbitals :

Since they are out of the plane of the atoms, these orbitals can interact with each other freely, and become delocalised. This means that instead of being tied to one atom of carbon, each electron is shared by all six in the ring. Thus, there are not enough electrons to form double bonds on all the carbon atoms, but the "extra" electrons strengthen all of the bonds on the ring equally. The resulting molecular orbital has π symmetry.

Benzene orbital delocalisation :


1. Show that the lines 4x + y – 9 = 0, x – 2y + 3 = 0, 5x – y – 6 = 0 make equal intercepts on any line of gradient 2.

2. ABC is a triangle with ∠A = 90°, AD is altitude. a acts along AB such that | a | =1/AB, b acts along

AC such that | b | = AC1 . Prove that a + b is a

vector along AD and | a + b | = AD1 .

3. A circle passes through the origin O and cuts two lines x + y = 0 and x – y = 0 in P and Q respectively. If the straight line PQ always passes through a fixed point, find the locus of the centre of the circle.

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

3x +....+ an tan

nx , where a1, a2, a3,...an ∈ R and n ∈ N. If |f (x)| ≤

|tan x| for ∀ x ∈

ππ

−2

,2

, prove that ∑=

n

i

i

ia

1

≤ 1.

5. Three digit numbers are formed. What is the

probability that the middle digit is largest. 6. Prove that area of the region bounded by the curve

y = log2 (2 – x) and containing the points satisfying the inequality (x – |x|)2 + (y – |y|)2 ≤ 4 is

−

π+

27log

42

2

2ee sq. units.

7. r1, r2, r3 be the radii of the circles drawn on the

altitudes respectively MD, ME and MF of the triangles respectively ∆MBC, ∆MCA, ∆MAB, as their diameters, where M be the circumcentre of the acute angled triangle ∆ABC. Prove that

21

2

ra + 2

2

2

rb + 2

3

2

rc ≥ 144.

8. Equilateral triangles are described externally on the sides BC, CA and AB of a given triangle ABC. Prove using complex numbers that their centroids form an equilateral triangle

9. Let a be a fixed real number satisfying 0 < a < π,

such that Tr = ∫−

+−−

a

a rurur

2cos21cos1 du

Prove that +→1

limr

Tr ,T1, −→1r

lim Tr form an A.P.

10. Let a, b, c be real numbers such that the roots of the

cubic equation x3 + ax2 + bx + c = 0 are all real. Prove that no one of these roots is greater than (2 ba 32 − – a)/3.

`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

7Set

Maths Funda

1.62

is the Golden Number, also called Phi. Golden Number property: ( + 1)/ = /1 The fraction 1/998999 contains Fibonacci numbers, i.e.: 1/998999=0.000001001002003005008013021034055089... Radii at 0° and approximately 222.49° divide a circle in the golden ratio: B/A = /1 = ( 5 + 1)/2 = 1 + (1 / (1 + (1 / (1 + (1 / (1 + ... ))))))= ( 4 + (4! - 4))/4 = -2sin(666) ≈ Fn+1 / Fn (F = Fibonacci numbers) ≈ 1.61803 39887 49894 84820 45868 34365 63811... The 3184th Fibonacci number is an apocalypse number (Apocalpyse numbers are numbers having exactly 666 digits).


1. Utilize the formula : If a1 + a2 + ....... + an = k (constant), then a1a2 ..... an has the greatest value

when a1 = a2 = ...... = an = nk , where a1, a2, ......, an

are all positive. (Using the concept of A.M. ≥ G.M.) Let E = (a – x) (b – y) (c – z) (ax + by + cz) Then abc E = (a2 – ax)(b2 – by)(c2 – cz)(ax + by + cz) Now we have (a2 – ax) + (b2 – by) + (c2 – cz) + (ax + by + cz) = a2 + b2 + c2 (constant) a2 – ax = b2 – by = c2 – cz = ax + by + cz

= 4

222 cba ++

∴ the greatest value of abc E = abc

cba256

)( 4222 ++

2. xf ´(x) + f (x) = g(x) ...(1) xf ´(x) = g(x) – f (x) < 0; because f ´(x) < 0 & x > 0 So g(x) < f (x) x g(x) < x f (x); as x > 0 ...(2)

Now from (1) dxd (xf (x)) = g(x)

so xf (x) = dxxgx

∫0)( use it is (2)

xg(x) < dxxgx

∫0)( ; for x > 0

3. Let the eqn of chord be x + y = p ...(1)

O (p, 0)

135º32

In the limiting condition the line (1) will touch the

circle , Therefore p = 8, so as required |p| < 8

4. Let OC = →c |a|2 = |b|2 = |c|2

since ACAB =

12 So,

θ−

πθ

2||

||

a

a = 12

O

C A2:1 B

ba

where ∠AOB = θ θ = π – 2θ ⇒ θ = π/3

Hence ba. = |a|2 cos 3π =

21 |a|2

cb . = |b|2 cos 6π =

23 |b|2 =

23 |a|2 & ca. = 0

Let →→→

+= byaxc

So ca. = x|a|2 + y b.a ⇒

+

2yx |a|2 = 0

⇒ x = – 2y

and cb . = x ba. + y|b|2

⇒ 23 |a|2 =

21 |a|2x + y|b|2

So, x + 2y = 3

& x = – 2y

So, x = – 2y + 2y = 3

⇒ 2

3y = 3 ⇒ y = 3

2

Hence x = – 2y = –

31

So →c =

→→

+− ba

32

3

MATHEMATICAL CHALLENGES SOLUTION FOR OCTOBER ISSUE (SET # 6)


5. Sum will be odd if 1 out of 4 chosen numbers is odd and others are even or 1 is even & others are odd.

P(O) = 4

203

101

10

CC.C2

= 323160

P(E) = 1 – 323160 =

323163

Hence P(E) > P(O)

6. Let the point be P (x, y) so, 3x + 2y + 10 = 0 ...(1) since |PA – PB| is maximum hence P, A, B must be colinear

1241421yx

= 0

⇒ – x – y + 6 = 0 ....(2) from (1) & (2) x = –22 & y = 28 So, point P is (–22, 28)

7. Let the point A be (x1, y1) and the circle be x2 + y2 = a2

P1

P2

Q1

Q2 A

Line AP1 is θ

−cos

1xx = θ

−sin

1yy = r

Solve it with circle. (x1 + r cos θ)2 + (y1 + r sin θ)2 = a2 x1

2 + y12 + 2rx1 cos θ + 2ry1 sin θ + r2 = a2

so r2 + (2x1 cos θ + 2y1 sin θ)r + x12 + y1

2 – a2 = 0

r1 . r2 = x12 + y1

2 – a2 = ( )21S

so, AP1 . AQ1 = ( )21S since

P1A . Q1A is independent on n, hence AP1 . AQ1 = AP2 . AQ2 = ........ = APn . AQn

8. (AB)T = (BA)T

BTAT = ATBT so BTATA = ATBTA BT = ATBTA (as AAT = I) ABT = AATBTA ABT = BTA (again as AAT = I) Hence proved.

9. iz = acbi

+−

1

11

−+

iziz =

acbiacbi

−−−++−

11 =

)()1()1(ibac

cabi−−+−

−++ ...(i)

Now as given (a + ib) (a – ib) = 1 – c2 = (1 – c) (1 + c)

1

1+−

+iz

iz =

ibacc

cbaabi

+−

++

++

++

2

22

1)1(

1)(

= ]1[)1(

])1[()(2

2

cibacibaciba

−+++

−+++

= 2

1

++

ciba .

11

+−

iziz (using (1))

iz

iz−+

11 =

1++

ciba (Hence proved)

10. Let n(n2 – 1) = (n – 1) n (n + 1) Since n is odd so (n – 1) (n + 1) is the product of two

consecutive even numbers, so it is divisible by 8. Since (n – 1) n (n + 1) is the product of 3 consecutive integers so it is divisible by 3 also Hence n(n2 – 1) is divisible by 24.

Interesting Science Facts

• The dinosaurs became extinct before the Rockies or the Alps were formed.

• Female black widow spiders eat their males after mating.

• When a flea jumps, the rate of acceleration is 20 times that of the space shuttle during launch.

• The earliest wine makers lived in Egypt around 2300 BC.

• If our Sun were just inch in diameter, the nearest star would be 445 miles away.

• The Australian billy goat plum contains 100 times more vitamin C than an orange.

• Astronauts cannot belch - there is no gravity to separate liquid from gas in their stomachs.

• The air at the summit of Mount Everest, 29,029 feet is only a third as thick as the air at sea level.

• One million, million, million, million, millionth of a second after the Big Bang the Universe was the size of a …pea.

• DNA was first discovered in 1869 by Swiss Friedrich Mieschler.


1. Let S(λ) be the area included between the parabola y = x2 + 2x – 3 and the line y = λx + 1. Find the least value of S(λ).

Sol.

β α

Solve y = x2 + 2x – 3 & y = λx + 1. To get x2 + (2 – λ) x – 4 = 0 ⇒ α + β = λ – 2, αβ = – 4

S(λ) = ∫β

α[(λx + 1) – (x2 + 2x – 3)] dx

S´(λ) = ∫β

α λdd [(λx + 1) – (x2 + 2x – 3)]dx

= ∫β

αdxx =

2

22 α−β

S´(λ) = 2

))(( α+βα−β

= 16)2(2

)2( 2 +−λ−λ = 0

⇒ λ = 2 Area is minimum for λ = 2

minm Area = 3

32 (find your self)

2. Find the area enclosed by the curve, max |x + y|, |x – y| = 1 Sol. Case I : If |x + y| ≥ |x – y|

A

C

D

B

⇒ |x + y| = 1 ⇒ locus of point (x, y) is two line segments AB and

CD. Case II If |x + y| ≤ |x – y| ⇒ |x – y| = 1 ⇒ locus of point (x, y) is two line segments BD and

AC. Then area bounded by the locus of (x, y) point is

2 (unit)2, (because locus is a square of side one unit).

3. If normals at the points P and Q of the parabola y2 = 4ax meet at the point R of the parabola. Show that the locus of centroid of the ∆ PQR is a ray. Find the equation of the ray.

Sol. Let P = (at2, 2at). Then Q is

ta

ta 22,2 2

=

ta

ta 4,42

and R is (aT2, 2aT). where T = – t – t2

centroid of the ∆PQR

=

++++3

2/42,3/4 222 aTtaataTtaat

y co-ordinates of the centroid

= 2at + ta4 + 2a

−−

tt 2 = 0

Thus centrocid of the ∆PQR for any choice of P on the parabola lies on the axis of the parabola.

x-coordinates of the centroid

=

+++

2

22 24

3 tt

tta

=

++ 24

32

22

tta ≥

32a (4 + 2) = 4a

Hence equation of the ray is given by y = 0, x > 4a

4. In a class of 20 students, the probability that exactly x students pass the examination is directly proportional to x2 (0 ≤ x ≤ 20). Find out the probability that a student selected at random has passed the examination. If a selected students has been found to pass the examination find out the probability that he/she is only student to have passed the examination.

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

Students' ForumMATHS


Sol. Let Ex : event that exactly x out of 20 students pass the examination and A : event that a particular student passes the examination ⇒ P(Ex) = kx2 (k is the proportionality constant) Now, E0, E2, ....., E20 are mutually exclusive and

exhaustive events. ⇒ P(E0) + P(E1) + P(E2) + ... + P(E20) = 1 ⇒ 0 + k(1)2 + k(2)2 + .... + k(20)2 = 1

⇒ k

++

6)140)(120)(20( = 1

⇒ k = 2870

1

Now, P(A) = ∑=

20

0

)/().(x

xx EAPEP

= ∑=

20

0

2

x

kx . 20x = ∑

=

20

0

3

20 x

xk

= 2

2)120(20

2870201

+

× =

8263

and P(E1/A) = )A(P

)E/A(P).E(P 11

=

8263

201.)1(

28701 2

= 44100

1

5. Let ABCD be any arbitrary plane quadrilateral in the space having E as the point of intersection of its diagonals. If ∆1 and ∆2 be the areas of triangles DEC and AEB, using vector method prove that

∆ ≥ 1∆ + 2∆ , where ∆ is the area of the quadrilateral ABCD. Also discuss the case when the equality holds.

Sol. Let the position vector of the points A, B, C and and

D with respect to E be →a , →b , – λ1.

→a and –λ2→b ;

where λ, λ2 ∈ R+

Now, ∆1 = 21 |DECE|

→→× = ||

221 →→

×λλ

ba

)0(E

)a,(C 1λ−

)b(B)a(A

)b,(D 2λ−

⇒ 1∆ = |2|||21

1λλ×→→ba =

2||

→→× ba

21λλ ...(i)

and ∆2 = |EABE|21 →→

× = 21 ||

→→× ba

⇒ 2∆ = 2

||→→

× ba ...(ii)

also ∆ = |DBCA|21 →→

× = 2

)1)(1( 21 λ+λ+ ||→→

× ba

∴ ∆ = 2

||→→

× ba )1)(1( 21 λ+λ+

∆ = 2

||→→

× ba21211 λλ+λ+λ+

where, 2

21 λ+λ ≥ 21λλ

∴ ∆ = 2

||→→

× ba212121 λλ+λλ+

≥ ||21 →→

× ba + 221 )1( λλ+

= 2

||→→

× ba + 2

|||| 21λλ×→→ba

∴ ∆ ≥ 2∆ + 1∆ using (i) and (ii)

It is clear that equality holds if λ1 = λ2 and in this case side AB and DC will become parallel.

6. Let a1, a2, ......, an be real constant, x be a real variable

and f (x) = cos(a1 + x) +21 cos(a2 + x) +

41 cos(a3 + x) +

...... + 121

−n cos(an + x). Given that f (x1) = f (x2) = 0,

prove that (x2 – x1) = mπ for integer m. Sol. f (x) may be written as,

f (x) = ∑=

−

n

kk

112

1 cos(ak + x)

= ∑=

−

n

kk

112

1 cosak. cos x – sin ak . sin x

= cos x .

∑

=−

n

kk

ka

112

cos – sin x

∑

=−

n

kk

ka

112

sin

= A cos x – B sin x, where A = ∑=

−

n

kk

ka

112

cos and

B = ∑=

−

n

kk

ka

112

sin since f (x1) = f (x2) = 0

⇒ A cos x1 – B sin x1 = 0 and A cos x2 – B sin x2 = 0

⇒ tan x1 = BA ⇒ tan x2 =

BA

⇒ tan x1 = tan x2 ⇒ (x2 – x1) = mπ


Differentiation and Applications of Derivatives :

If y = f (x), then

1. dxdy =

hxfhxf

h

)()(lim0

−+→

2. axdx

dy

=

=

axafxf

ax −−

→

)()(lim

3. axdx

dy

=

=

hafhaf

hx

)()(lim −+→

If u = f (x), v = φ(x), then

1. dxd (k) = 0

2. dxd (ku) = k

dxdu

3. dxd (u ± v) =

dxdu ±

dxdv

4. dxd (uv) = u

dxdv + v

dxdu

5.

vu

dxdu = 2v

dxdvu

dxduv −

6. If x = f (t), y = φ (t), then dxdy =

dtdx

dtdy

7. If y = f[φ(x)], then dxdy = f´[φ(x)].

dxd [φ(x)]

8. If w = f (y), then dxdw = f ´(y)

dxdy

9. If y = f (x), z = φ(x), then dzdy =

dxdy .

dzdx

10. dxdy .

dydx = 1 or

dxdy =

dydx /1

1. dxd (k) = 0

2. nxdxd = nxn–1

3. dxd

nx1 = – 1+nx

n

4. dxd x =

x21

5. dxd ex = ex

6. dxd ax = ax log a

7. dxd log x =

x1

8. dxd logax =

x1 logae

9. dxd sin x = cos x

10. dxd cos x = – sin x

11. dxd tan x = sec2x

12. dxd cot x = – cosec2x

13. dxd sec x = sec x tan x

14. dxd cosec x = – cosec x cot x

15. dxd sin–1x =

21

1

x−

16. dxd cos–1x = –

21

1

x−

17. dxd tan–1x = 21

1x+

18. dxd cot–1x = – 21

1x+

19. dxd sec–1x =

1

12 −xx

20. dxd cosec–1x = –

1

12 −xx

DIFFERENTIATION Mathematics Fundamentals M

AT

HS


Suitable substitutions : The functions any also be reduced to simplar forms by the substitutions as follows.

1. If the function involve the term )( 22 xa − , then

put x = a sin θ or x = a cos θ.

2. If the function involve the term )( 22 xa + , then

put x = a tan θ or x = a cot θ.

3. If the function involve the term )( 22 ax − , then

put x = a sec θ or x = a cosec θ.

4. If the function involve the term xaxa

+− , then put

x = a cos θ or x = a cos 2θ

All the above substitutions are also true, if a = 1

Differentiation by taking logarithm :

Differentiation of the functions of the following types are obtained by taking logarithm.

1. When the functions consists of the product and quotient of a number of functions.

2. When a function of x is raised to a power which is itself a function of x.

For example, let y = [f (x)]φ(x)

Taking logarithm of both sides, log y = φ(x) log f (x)

Differentiating both sides w.r.t 'x',

y1

dxdy = φ´(x) log f (x) + φ(x).

)()´(

xfxf

= [f (x)]φ(x) log f(x).φ´(x) + φ(x) . [f (x)φ(x) – 1.f´(x)

dxdy = Differential of y treading f (x) as constant

+ Differential of y treating φ(x) as constant.

It is an important formula.

Differentiation of implicit functions :

1. If f(x, y) = 0 is a implicit function, then

dxdy = –

y/fx/f

∂∂∂∂

= – constant x keepingy w.r.t.f of Diff.constanty keeping x w.r.t.f of .Diff

For example, consider f (x, y) = x2 + 3xy + y2 = 0, then

dxdy = –

yfxf

∂∂∂∂

// = –

yxyx

2332

++

1. If y = f (x), then

dxdy = y1 = f´(x), 2

2

dxyd = y2 = f´´(x), .....

ndxyd 2

= yn = f n(x)

2. n

n

dxd (ax + b)n = n ! an

3. n

n

dxd (ax + b)m = m(m – 1)

.... (m – n + 1) an(ax + b)m–n

4. n

n

dxd emx = mnemx

5. n

n

dxd amx = mnamx (log a)n

6. n

n

dxd log(ax + b) = n

nn

baxna)(

!)1()1( 1

+−− −

7. n

n

dxd sin (ax + b) = an sin

π

++2

nbax

8. n

n

dxd cos (ax + b) = an cos

π

++2

nbax

Leibnitz's theorem : If u and v are any two functions of x such that their desired differential coefficients exist, then the nth differential coefficient of uv is given by

Dn(uv) = (Dnu)v + nC1(Dn–1u)(Dv)

+ nC2(Dn–2u)(D2v) +...... + u(Dnv)

Ability

• We can accomplish almost anything win tin our ability if we but think we can.

• He is the best sailor who can steer within fewest points of the wind, and exact a motive power out of the greatest obstacles.

• Our work is the presentation of our capabilities.

• The wind and the waves are always on the side of the ablest navigator.


Different standard form of the equation of a straight line :

General form : Ax + By + C = 0 where A, B, C are any real numbers not all zero.

Gradient (Tangent) form : y = mx + c It is the equation of a straight line which cuts off an

intercept c on y-axis and makes an angle with the positive direction (anticlockwise) of x-axis such that tan θ = m. The number m is called slope or the gradient of this line.

Intercept form :

ax +

by = 1

It is the equation of straight line which cuts off intercepts a and b on the axis of x and y respectively.

Normal form (Perpendicular form) : x cos α + y sin α = p It is the equation of a straight line on which the

length of the perpendicular from the origin is p and α is the angle which , this perpendicular makes with the positive direction of x-axis.

One point form : y – y1 = m(x – x1) It is the equation of a straight line passing through a

given point (x1, y1) and having slope m. Parametric equation :

θ

−cos

1xx = θ

−sin

1yy = r

It is the equation of a straight line passes through a given point A(x1, y1) and makes an angle θ with x-axis.

Two points form :

y – y1 = 12

12

xxyy

−− (x – x1)

It is the equation of a straight line passing through

two given points (x1, y1) and (x2, y2), where 12

12

xxyy

−−

is its slope. Point of intersection of two lines a1x + b1y + c1 = 0

and a2x + b2y + c2 = 0 is given by

−−

−−

1221

2112

1221

1221 ,babacaca

babacbcb

Angle between two lines : The angle θ between two lines whose slopes are m1

and m2 is given by

tan θ = 21

21

1 mmmm

+−

If θ is angle between two lines then π – θ is also the angle between them.

The equation of any straight line parallel to a given line ax + by + c = 0 is ax + by + k = 0.

The equation of any straight line perpendicular to a given line, ax + by + c = 0 is bx – ay + k = 0.

The equation of any straight line passing through the point of intersection of two given lines l1 ≡ a1x + b1y + c1 = 0 and l2 ≡ a2x + b2y + c2 = 0 is l1 + λl 2 = 0

where λ is any real number, which can be determined by given additional condition in the question.

The length of perpendicular from a given point (x1, y1) to a given line ax + by + c = 0 is

)( 22

11

ba

cbyax

+

++ = p (say)

In particular, the length of perpendicular from origin

(0, 0) to the line ax + by + c = 0 is 22 ba

c

+

Equation of Bisectors : The equations of the bisectors of the angles between

the lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are

21

21

111

ba

cybxa

+

++ = ±

22

22

222

ba

cybxa

+

++

Distance between parallel lines : Choose a convenient point on any of the lines

(put x = 0 and find the value of y or put y = 0 and find the value of x). Now the perpendicular distance from this point on the other line will give the required distance between the given parallel lines.

Pair of straight lines : The equation ax2 + 2hxy + by2 = 0 represents a pair of

straight lines passing through the origin.

STRAIGHT LINE & CIRCLE

Mathematics Fundamentals MA

TH

S


Let the lines represented by ax2 + 2hxy + by2 = 0 be y – m1x = 0 and y – m2x = 0, then

m1 + m2 = – bh2 and m1m2 =

ba

General equation of second degree in x, y is ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 ...(i) This equation represents two straight lines, if ∆ = abc + 2fgh – af2 – bg2 – ch2 = 0

or cfgfbhgha

= 0

and point of intersection of these lines is given by

−−

−−

22 ,habafhg

habbghf

The angle between the two straight lines represented by (i) is given by

tan θ = ± ba

abh+

−22

If ax2 + 2hxy + by2 + 2gx + 2f y + c = 0 represents a pair of parallel straight lines, then the distance between them is given by

2)(

2

baaacg

+− or 2

)(

2

babbcf

+−

Circle: Different forms of the equations of a circle :

Centre radius form : the equation of a circle whose centre is the point (h, k) and radius 'a' is

(x – h)2 + (y – k)2 = a2 General equation of a circle : It is given by

x2 + y2 + 2gx + 2f y + c = 0 ...(i) Equation (i) can also be written as

|x – (– g)|2 + |y – (–f )|2 = | cfg −+ 22 |2

which is in centre-radius form, so by comparing, we get the coordinates of centre (– g, – f ) and radius is

cfg −+ 22 .

Parametric Equations of a Circle : The parametric equations of a circle (x – h)2 + (y – k)2 = a2 are x = h + a cos θ and y = k + a sin θ, where θ is a parameter.

Lengths of intercepts on the coordinate axes made by

the circle (i) are 2 cg −2 and 2 cf −2

Equation of the circle on the line joining the points A(x1, y1) and B(x2, y2) as diameter is given by

−−

1

1

xxyy

−−

2

2

xxyy

= 1

If C1, C2 are the centres and a1, a2 are the radii of two circles, then

(i) The circles touch each other externally, if C1C2 = a1 + a2 (ii) The circles touch each other internally, if C1C2 = |a1 – a2| (iii) The circles intersects at two points, if |a1 – a2| < C1C2 < a1 + a2 (iv) The circles neither intersect nor touch each other, if C1C2 > a1 + a2 or C1C2 < |a1 – a2|

Equation of any circle through the point of intersection of two given circles S1 = 0 and S2 = 0 is given by S1 + λS2 = 0 (λ ≠ –1) and λ can be determined by an additional condition.

Equation of the tangent to the given circle x2 + y2 + 2gx + 2fy + c = 0 at any point (x1, y1) on it,

is xx1 + yy1 + g(x + x1) + f (y + y1) + c = 0 The straight line y = mx + c touches the circle x2 + y2

= a2, if c2 = a2(1 + m2) and the point of contact of the

tangent y = mx ± a 21 m+ , is

+

±

+ 22 1,

1 m

a

m

mam

Length of tangent drawn from the point (x1, y1) to the circle S = 0 is 1S , where

S1 = x12 + y1

2 + 2gx1 + 2f y1 + c The equation of pair of tangents drawn from point

(x1, y1) to the circle S = 0 i.e. x2 + y2 + 2gx + 2f y + c = 0, is SS1 = T2,

where T ≡ xx1 + yy1 + g(x + x1) + f (y + y1) + c and S1 as mentioned above.

Chord with a given Middle point : the equation of the chord of the circle S = 0 whose

mid-point is (x1, y1) is given by T = S1, where T and S1 as defined a above.

If θ be the angle at which two circles of radii r1 and r2 intersect, then

cos θ = 21

222

21

2 rrdrr −+

where d is distance between their centres. Note: Two circles are said to be intersect

orthogonally if the angle between their tangents at their point of intersection is a right angle i.e.

r12 + r2

2 = d2 or 2g1g2 + 2f1 f2 = c1 + c2

Radical axis : The equation of the radical axis of the two circle is S1 – S2 = 0 i.e.

2x(g1 – g2) + 2y(f1 – f2) + c1 – c2 = 0


PHYSICS

Questions 1 to 10 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. Given are four arrangements of three fixed electric

charges. In each arrangement, a point labeled P is also identified a test charge, +q, is placed at point P. All of the charges are the same magnitude Q. But they can be either +ve or –ve as shown. The charges and point P all lie on a straight line. The distance between adjacent items, either between two charges or between a charge and point P are all the same

(I) P (II) P (III) P (IV) P Correct order of choices in a decreasing order of

magnitude force on P is - (A) II > I > III > IV (B) I > II > III > IV (C) II > I > IV > III (D) III > IV > I > II

2. A segment of angle θ is cut from a half disc, symmetrically as shown. If the centre of mass of the remaining part is at a distance 'a' from O and the centre of mass of the original half disc was at distance d then it can be definitely said that :

O

θ

(A) a = d (B) a > d (C) a < d (D) all the above are possible depending on θ

3. The moment of inertia of a hollow thick spherical shell of mass M and its inner radius R1 and outer radius R2 about diameter is :

(A) )(5

)(231

32

51

52

RRRRM

−

− (B) )(3

)(231

32

51

52

RRRRM

−

−

(C) )(5

)(431

32

51

52

RRRRM

−

− (D) )(3

)(431

32

51

52

RRRRM

−

−

4. A cyclic process ABCA is shown in a V-T diagram. The corresponding PV diagram will be

V

T

B C

A

(A)

P

V

A C

B (B)

P

V

A B

C

(C)

P

V

B C

A (D)

P

VC

B

A

IIT-JEE 2012

XtraEdge Test Series # 7

Based on New Pattern

Time : 3 Hours

Syllabus : Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus

Instructions : Section – I : Question 1 to 10 are multiple choice questions with only one correct answer. +3 marks will be

awarded for correct answer and -1 mark for wrong answer.

Section – II : Question 11 to 15 are multiple choice question with multiple correct answer. +4 marks will be awarded for correct answer and -1 mark for wrong answer.

Section – III : Question 16 to 21 are passage based single correct type questions. +3 marks will be awarded for correct answer and -1 mark for wrong answer

Section – IV : Question 22 to 23 are Column Matching type questions. +8 marks will be awarded for the complete correctly matched answer (i.e. +2 marks for each correct row) and No Negative marks for wrong answer.


5. A rectangular metal plate of dimension (a × b) having two holes of radii r1 and r2 (r1 > r2) and their positions are shown in the figure, now the plate is heated such that its temperature rises by ∆T. Then separation between the holes :

d

a (A) decreases (B) increases (C) remain constant (D) can not say

6. Assuming all surface to be smooth minimum value of horizontal acceleration 'a' so that sphere losses contact at P is –

P Q α α

a

(A) g sinα (B) g tanα (C) g cotα (D) g cosecα

7. A uniform rod of mass M and length L lies flat on a frictionless horizontal surface. Two forces F and 2F are applied along the length of the rod as shown. The tension in the rod at the point P is

L

P

L/4

2FF

(A) 4

3F (B) 3F (C) 4

5F (D) 4

7F

8. The Kα wavelength of an element with atomic number z is λz. The kα wavelength of an element with atomic number 3z is λ3z. Then

(A) λz > 9λ3z (B) λz < 9λ3z

(C) λz = 9λ3z (D) Depending on z, λz can be greater or smaller

then (9λ3z).

9. The BE per nucleon of deutron (1H2) and helium nucleus (2He4) is 1.1 MeV and 7 MeV respectively. If two deutron react to form a single helium nucleus, then energy released is -

(A) 23.6 MeV (B) 4.8 MeV (C) 25.8 MeV (D) None of these

10. In a radioactive decay, let N represent the number of residual active nuclei, D the number of daughter nuclei, and R the rate of decay at any time t. Three curves are shown in Fig. The correct ones are –

t(1)

N NR

t(2)

t(3)

D

(A) 1 and 3 (B) 2 and 3 (C) 1 and 2 (D) all three Questions 11 to 15 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and – 1 mark for each wrong answer.

11. Consider a hypothetical binding energy per nucleon curve. Which of the following fission or fusion may occur ?

50 100 150 200

2468

10

(BE)n (MeV/n)

(A) X250 → Y160 + Z90 (B) X220 → Y180 + Z40

(C) X40 + Y70 → Z110 (D) X40 + Y120 → Z160

12. In series R-C circuit : R C

V = V0 sin ωt (A) current leads the applied voltage by

φ = tan–1

RCω1

(B) current lags the applied voltage by φ < 90º (C) 00

VVc < , where 0cV and V0 are the maximum

values of the voltage across the capacitor and applied voltage respectively

(D) applied voltage, voltage across the resistor and current are in phase.

13. A sound wave of frequency 'f ' travels horizontally to the right. It is reflected from a large vertical plane surface moving to the left with speed v0. The speed of sound in the medium is v. Choose the correct statement.

(A) The number of waves striking the surface per

second is

+vvv 0 f

(B) The wavelength of the reflected wave is

−+

0

0

vvvv

f


(C) The frequency of the reflected wave is

−+

0

0

vvvv

f

(D) The number of beats heard by a stationary listener to the left of the reflecting surface is

− 0

0

vvv

f

14. Illustrated below is a uniform cubical block of mass M and side a. Mark the correct statement(s)

p A

C B

M

D

a

(A) The moment of inertia about axis A, passing

through the centre of mass is IA = 61 Ma2

(B) The moment of inertia about axis B, which

bisect one of the cube faces is IB = 125 Ma2

(C) The moment of inertia about axis C, along one

of the cube edges is IC = 32 Ma2

(D) The moment of inertia about axis D, which bisects one of the horizontal cube faces is

127 Ma2

15. In the figure the block on the smooth table is set into motion in a circular orbit of radius r round the centre hole. The hanging mass is identical to the mass on the table and remains in equilibrium, neglect friction. The string connecting the two blocks is massless and intextensible. Select the correct answer.

m r

m

(A) the angular speed ω of the block in its circular motion is rg /

(B) kinetic energy as function of r is given by mgr/2

(C) angular momentum about the hole is conserved even if hanging block is pulled down

(D) The block on table is in equillibrium

This section contains 3 paragraphs; each has 2 multiple choice questions. (Questions 16 to 21) 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.

Passage # 1 (Ques. 16 - 17)

Consider an electron moving in a circular orbit of radius 'r' in an external uniform and steady magnetic field B. Assume that Bohr's quantization principle regarding the angular momentum is true in this case. Now answer the following question :

16. If radius of nth orbit is rn and speed in this orbit is vn then correct relationship between them

(A) rn ∝ vn (B) rn ∝ vn2

(C) rn2 ∝ vn (D) rn ∝ 1/vn

17. If total energy of e– in moving these orbit is sum of KE and potential energy of interaction between the magnetic moment of orbital current and magnetic field B. Then total energy in nth orbit is

(e = charge of electron, m = mass of e–, h = plank's constant)

(A) En = m

nheBπ2

(B) En = m

nheBπ4

(C) En = zero (D) En = m

nheBπ

2

Passage # 2 (Ques. 18 - 19) When the strain is small (say < 0.01), the stress is

proportional to the strain. This is the region where Hook's law is valid and where young's modulus is defined. The point a on the curve represents the proportional limit up to which stress and strain are proportional. If the strain is increased a little bit, the stress is not proportional to the strain. However, the wire still remains elastic. This means, if the stretching force is removed, the wire acquires its natural length. This behaviour is shown up to a point b on the curve known as the elastic limit on the yield point. If the wire is stretched beyond the elastic limit, the strain increases much more rapidly. If the stretching force is removed, the wire does not come back to its natural length. Some permanent increase in length takes places. In figure, we have shown this behaviour by the dashed line from C. The behaviour of the wire is now plastic. If the deformation is increased further, the wire breaks at a point d known as fracture point. If large deformation takes place between the elastic limit and the fracture point, the material is called ductile. If it breaks soon after the elastic limit is crossed, it is called brittle.


d

plastic behaviourIII

II

c b a

IElastic behaviour

o 0.3

stress

strain Permanent set

a = proportional limitb = Elastic limit d = fracture point

18. A metal wire will retain its original shape when

load is removed in region. (A) oc (B) cd (C) ac (D) od

19. Yield point and elastic limit point coincide for (A) ductile material (B) malleable material (C) brittle material (D) elastic material

Passage # 3 (Ques. 20 - 21) A uniform dense solid cylinder of mass m and

radius R is released from rest on an inclined plane. After releasing from rest it starts performing pure rolling (i.e. rolling without slipping). As there is no slipping the friction force acts is static in nature. Therefore the relative velocity between the points in contact is zero.

We know that rolling is combined rotation and translation. During its downward journey along the incline the cylinder moves distance l along the incline. The angle of inclination from horizontal is α and the coefficient of friction is given as µ. The acceleration due to gravity is g downwards. Air resistance is not present.

20. The acceleration of centre of mass of cylinder is- (A) g sin α – µg cos α (B) g sin α

(C) 3

2g sin α (D) none of these

21. (The final angular speed of cylinder is

(A) 2sin

34

Rg αl (B) 2

sin32

Rg αl

(C) 2sin

31

Rg αl (D) none of these

This section contains 2 questions (Questions 22, 23). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

ABCD

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T

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

22. Assume that 2 bodies collide head on. The graph of their velocities with time are shown in column-I match them with appropriate situation in column II

Column -I Column-II

(A)

v

t

(1) (2) (P)

m1 m2

m1 < m2, 0 < e <1

(B)

v

t

(1)

(2)

(Q) m1 wall

2nd body is massive

2

(C) v

t

(1) (2) (R) putty ball

(D) v

t

(1) (2) (S) v1 v2

m1 = m2 e = 1 v1 > v2

(T)

m1 m2

m1 > m2 e = 1

23. Match the Column : Column -I Column-II (A)The charge q is projected

perpendicular to the electric field. Then it moves through the magnetic field

E B

q

(P)

(B) The charge is released from rest in a crossed B

r and E

r

B

E

q

(Q)

α


(C) The charge is projected perpendicular to E in a crossed E

r and E

r

BE

q

(R)

(D) The charge is projected at a nonzero angle θ(< 90º) with the magnetic induction

Bv

q

(S)

(T)

CHEMISTRY

Questions 1 to 10 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer.

1. The IUPAC name of compound HO – C = O

NH2 – C ==== C ––– C – H

NH2

CH3

Cl

is –

(A) 2, 3 diamino-4-chloro-2-pentenoic acid (B) 4-chloro-3, 3-diamino-2-pentenoic acid (C) 3, 3–diamino-3-chloro-pentenoic acid (D) All of the above

2. Identify the correct statements -

(A) The compound

H3C CH3

COOH O

fails to undergo

decarboxylation (B) A Grignard reagent can be successfully made

from the following dibromide Br

Br (C) Cyclopentan –1, 3– dione exists almost 100%

in the enol form whereas diacetyl (CH3COCOCH3) can exist in the keto form as well as the enol form

(D) Among the following resonance structure given below, (ii) will be the major contributor to the resonance hybrid.

C–CH3 ↔ C–CH3

O : O–: ..

Θ ..

(i) (ii)

3. Which of the following graphs are properly represented ?

(A) For normal reaction

Rat

e

Temperature

(B)

For explosive reaction

Rat

e

Temperature

(C)

For all normal reactionRat

e

Temperature

(D) For explosive reaction

Rat

e

Temperature45ºC

4. A hydrogen electrode is placed in a buffer solution of sodium acetate and acetic acid in the ratio a : b and other in the ratio b : a was taken. If electrode potential values are found to be E1 and E2 then which of the following is/are correct for the pKa value of the acid.

(A) 118.0

EE 21 − (B) –118.0

EE 21 +

(C) 2

1

EE × 0.118 (D)

118.0EE 12 −

5. Identify the incorrect statement – (A) In solid state N2O5 exist as NO2

+ and NO3–ions

(B) Solid PCl5 contains PCl4+ and PCl6

– ions (C) Solid PBr5 has PBr4

+ and Br– (D) In N4S4 all the bond angles are equal

6. A bulb of constant volume is attached to a manometer tube open at other end as shown in figure. The manometer is filled with a liquid of density (1/3rd) that of mercury. Initially h was 228 cm.

Gash

Through a small hole in the bulb gas leaked

assuming pressure decreases as dtdp = – kP.


If value of h is 114 cm after 14 minutes. What is the value of k (in hour–1) ?

[Use : ln(4/3) = 0.28 and density of Hg = 13.6 g/mL]

(A) 0.6 (B) 1.2 (C) 2.4 (D) None of these

7. The dipole moment of HCl is 1.03D, if H–Cl bond distance is 1.26Å, what is the percentage of ionic character in the H–Cl bond ?

(A) 60% (B) 29% (C) 17% (D) 39% 8. Arrange NH4

+, H2O, H3O+, HF & OH– in increasing order of acidic nature -

(A) OH– < H2O < NH4+ < HF < H3O+

(B) H3O+ > HF > H2O > NH4+ > OH–

(C) NH4+ < HF < H3O+ < H2O < OH–

(D) H3O+ < NH4+ < HF < OH– < H2O

9. A compression of an ideal gas is represented by curve AB, which of the following is wrong

B(vB)

A(vA)

log V

log P

(A) number of collision increases B

A

VV times

(B) number of moles in this process is constant (C) it is isothermal process (D) it is possible for ideal gas

10. A compound containing only sodium, nitrogen and oxygen has 33.33% by weight of sodium. What is the possible oxidation number of nitrogen in the compound?

(A) –3 (B) + 3 (C) –2 (D) + 5 Questions 11 to 15 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and – 1 mark for each wrong answer. 11. Which of the following is/are correct regarding the

periodic classification of elements ? (A) The properties of elements are the periodic

function of their atomic number (B) Non metals are lesser in number than metals (C) The first ionization energies of elements in a

period do not increase with the increase in atomic numbers

(D) For transition elements the d-subshells are filled with electrons monotonically with the increase in atomic number

12. In the purification Zr and B, which of the following is/are true ?

(A) Zr + 2I2 → ZrI4(g) overpassedWhotwhitethe →

the pure Zr is deposited on W

(B) 2B + 3I2 → 2BI3(g) overpassedWhotwhitethe →

the pure B is deposited on W

(C) Zr + 2I2 → ZrI4 (s) Wwithmixedheatedthen& →

ZrI4 is reduced to ZrI2 (D) none of these

13. Which of the following statements is correct ? (A) At 273ºC, the volume of a given mass of a gas

at 0ºC and 1 atm. pressure will be twice its volume

(B) At –136.5ºC, the volume of a given mass of a gas at 0ºC and 1 atm. pressure will be half of its volume

(C) The mass ratio of equal volumes of NH3 and H2S under similar conditions of temperature and pressure is 1 : 2

(D) The molar ratio of equal masses of CH4 and SO2 is 4 : 1

14. Dopamine of a drug used in the treatment of

parkinson's disease. CH2–CH

Dopamin

NH2 COOH HO

HO

Which of the following statements about this compound are correct ?

(A) It can exist in optically active forms. (B) One mole will react with three moles of sodium

hydroxide to form a salt (C) It can exist as a Zwitter ion in the aqueous solution (D) It gives nitroso compound on treatment with

HNO2.

15. In the given table types of H bonds and some H bond energies are given and other H bond energies are not given. You are to perdit the unknown H-bond energies.

Types of H-bonds H-bond energies in KJ/mol

(I) F – H …….. O – F – H …….. F 30 (II) O – H …….. O – O – H …….. F 15 (III) F – H …….. F– – (IV) N – H …….. N – Correct prediction are – (A) H-bond energy for (I) may be 20 kJ/mol (B) H-bond energy for (II) may be 25 kJ/mol (C) H-bond energy for (III) may be 113 kJ/mol (D) H-bond energy for (IV) may be 12 kJ/mol



Passage # 1 (Ques. 16 - 17) Freezing point of a liquid is defined as that

temperature at which it is in equilibrium with its solid phase.

solvent P = 1 atm

Solid

solution ∆Tf

Vapour Pressure

T T0

T0 > T

Temperature

Phase diagram for a pure solvent and solution for depression in freezing point.

16. The freezing point of the solvent is -

(A) ST

G–H∆

∆∆ (B) SH

∆∆

(C) SG

∆∆ (D)

HS

∆∆

17. Freezing point of the solution is smaller than the freezing point of the solvent. Because -

(A) ∆H of solution and ∆H of solvent are almost same due to identical intermolecular forces

(B) ∆S of the solution is larger than ∆S of solvent (C) ∆S of the solution is smaller than the ∆S of

solvent (D) ∆H of the solution is much higher than the ∆H

of solvent but ∆S of solution is smaller than that of the solvent

Passage # 2 (Ques. 18 - 19) Effect of temperature on the equilibrium process is

analysed by using the thermodynamics. From the thermodynamics relation ∆G° = – 2.3 RT logK........(1) ∆G° = Standard free

energy change ∆G° =∆H° – T∆S°….(2) ∆H° = Standard heat of

the reaction From (1) & (2) – 2.3 RT log K = ∆H° – T∆S° ; ∆S° : Standard

Entropy change,

log K = RT3.2H°∆ +

R3.2S°∆ ........(3)

Clearly if a plot of log K vs 1/T is made then it is a straight line having slope

= R3.2H– °∆ & y–intercept =

R3.2S°∆ .

If at a temperature T1 equilibrium constant be K1 and at temperature T2 equilibrium constant be K2 then, the above equation reduces to :

⇒ log K1 = 1RT3.2

H– °∆ + R3.2

S°∆ ........(4)

⇒ log K2 = 2RT3.2

H– °∆ + R3.2

S°∆ ........ (5)

Subtracting (4) from (5) we get.

⇒ log 1

2

KK =

R3.2H°∆

21 T1–

T1

18. If standard heat of dissociation of PCl5 is 230 Cal.

then the slope of the graph of log K vs T1 is -

(A) + 50 (B) – 50

(C) 10 (D) None of these

19. For exothermic reaction of ∆So < 0 then the sketch

of log K vs T1 may be -

(A)

1/T

log K (B)

1/T

log K

(C)

1/T

log K (D)

1/T

log K

Passage # 3 (Ques. 20 - 21) A pleasant smelling optically active compound,

monoester 'F' has molecular weight 186. It doesn't react with Br2 in CCl4. Hydrolysis of 'F' gives two optically active compounds 'G', which is soluble in NaOH and 'H'. H gives a positive iodoform test, but on warming with conc. H2SO4 gives I with no disastereomers. When the Ag+ salt of 'G' is reacted with Br2, racemic 'J' is formed. Optically active J is formed when 'H' is treated with tosyl chloride (TsCl), and then with NaBr.

20. The pleasant smelling optically active compound, F is -

(A) (CH3)2CH–

3CH|

–CHC||O

O–

3CH|

–CH CH(CH3)2

(B) (CH3)3C–CH2 C||O

–O–

3CH|

–CH CH(CH3)2


(C) CH3CH2CH2

3CH|

–CH C||O

–O–

3CH|

–CH CH2CH2CH3

(D) CH3CH2

3CH|

–CH CH2– 2COCH||O

–

3

2

CH|

CHCH –CH3

21. How would be the structure of F if I exists as diastereomers ?

(A) (CH3)2CH

3CH|

CHCO||O

3

23

CH|

)CH(CHCH

(B) (CH3)3CCH2 C||O

O

3

23

CH|

)CH(CHCH

(C) CH3CH2CH2

3CH|CHCO

||O

3

322

CH|

CHCHCHCH

(D) CH3CH2

3

2

HC|

CHCH C||O

OCH2

3

32

CH|

CHCHCH

This section contains 2 questions (Questions 22, 23). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows:

A B C D

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T


22. Match the Column : Column-I Column-II (A) 5.4 g of Al (P) 0.5 NA electrons (B) 1.2 g of Mg2+ (Q) 15.9994 amu (C) Exact atomic weight (R) 0.2 mole atoms of mixture of oxygen isotopes (D) 0.9 mL of H2O (S) 0.05 moles (T) 3.1 × 1023 electrons

23. Match the Column : Column -I Column-II

(A) Two electron three centre bond

(P) (BN)x

(B) Four electron three centre bond

(Q) B2H6

(C) sp3 hybrid orbitals (R) AlCl3 (D) Inorganic graphite (S)

(T) B4H10 HF

MATHEMATICS


1. The differential equation of all ellipse centred at the origin is

(A) y2 + xy12 – yy1 = 0

(B) xyy2 + xy12 – yy1 = 0

(C) yy2 + xy12 – xy1 = 0

(D) none of these 2. The value of x for which the matrix

A =

− 121010702

is inverse of

B =

−−

−

xxx

xxx

24010

714 is

(A) 21 (B)

31 (C)

41 (D)

51

3. If x = (7 + 34 )2n = [x] + f , then x(1– f ) if equal to (A) 1 (B) 2 (C) 3 (D) 4 4. The number of values of x ∈[0, nπ], n ∈ I that

satisfy log|sinx|(1 + cos x) = 2 is (A) 0 (B) n (C) 2n (D) none 5. Reflection of the line za + za = 0 in the real axis is

(A) za + az = 0 (B) aa =

zz

(C) (a + a ) (z + z ) = 0 (D) None of these

6. The tangent to the curve x = a θ2cos cos θ,

y = a θ2cos sin θ at the point corresponding to θ = π/6 is -

(A) parallel to the x-axis (B) parallel to the y-axis (C) parallel to line y = x (D) none of these


7. The exponent of 7 in 100C50 is - (A) 0 (B) 2 (C) 4 (D) none of these

8. Let f (x) =

=≤<

02||0

1||

xforxforx

, then at x = 0, f has -

(A) a local maximum (B) no local maximum (C) a local minimum (D) no extremum 9. If f (x) is a polynomial satisfying f (x).f (1/x) = f (x) + f (1/x), and f (3) = 28, then f (4)

is given by - (A) 63 (B) 65 (C) 67 (D) 68

10. The integer n for which n

x

x xexx ))(cos1(coslim

0

−−→

is a finite nonzero number is - (A) 1 (B) 2 (C) 3 (D) 4 Questions 11 to 15 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and – 1 mark for each wrong answer.

11. If f (x) = 211 x−− , then (A) f is continuous on [–1, 1] (B) f is continuous at x = 0 (C) f is not differentiable at x = 0 (D) f is differentiable everywhere 12. The line y = mx + c intersects the circle x2 + y2 = r2

at two real distinct points if

(A) – r 21 m+ < c ≤ 0 (B) 0 ≤ c < r 21 m+

(C) – c 21 m− < r (D) r < c 21 m+

13. Let α, β be the roots of x2 – 4x + A = 0 and γ, δ be the roots of x2 – 36x + B = 0. If α, β, γ, δ forms an increasing G.P., then

(A) B = 81 A (B) A = 3 (C) B = 243 (D) A + B = 251

14. Given an isosceles triangle with equal sides of length b, base angle α < π/4, R, r the radii and O, I the centres of the circumcircle and incircle, respectively. Then

(A) R = 21 b cosec α (B) ∆ = 2b2 sin 2α

(C) r = )cos1(2

2sinbα+

α (D) OI = )2/cos(sin2

)2/3cos(αα

αb

15. If ∫α

α−0 coscos1 xdx =

αsinA + B (a ≠ 0). Then

possible values of A and B are

(A) A = 2π , B = 0 (B) A =

4π , B =

απ

sin4

(C) A = 6π , B =

απ

sin (D) A = π, B =

απ

sin


Passage # 1 (Ques. 16 - 17) We can derive reduction formulas for the integral of

the form ∫ ,sin dxxn ∫ dxxncos , ∫ ,tan dxxn

∫ dxxncot and other integrals of these form using

integration by parts. In turn these reduction formulas can be used to compute x etc.

16. If ∫ dxx5sin = –51 sin4 x cos x + A sin2 x cos x

–158 cos x + C then A is equal to -

(A) – 2/15 (B) – 3/5 (C) – 4/15 (D) – 1/15

17. If ∫ dxx6tan =51 tan5 x + A tan3 x + tan x – x + C

then A is equal to - (A) 1/3 (B) 2/3 (C) – 2/3 (D) – 1/3

Passage # 2 (Ques. 18 - 19) Using differentiability and continuity of a function

f which satisfies certain functional equation, we can determine in some cases the function explicity. E.g. If f satisfies f (x + y) = f (x) f (y) for all x, y ∈ R and f (x) ≠ 0 for any x ∈ R and f ′(0) = 1 then f (x) = ex.

18. If a function f satisfy f

+

3yx =

3)()(2 yfxf ++

for real x and y and f ′ (2) = 3, then f (x) is equal to -

(A) –121 x3 + x2 (B) 24 log (3x + 2)

(C) 3x + 2 (D) 43 x2 + 2


19. If f is a differentiable function on R and f ′(0) = 2

satisfying f (x + y) =)()(1)()(

yfxfyfxf

−+ then f (π/8) is

equal to - (A) 1/2 (B) 1 (C) 3/2 (D) tan π/8 Passage # 3 (Ques. 20 - 21) Let k be the length of any edge of a regular

tetrahedron. (A tetrahedron whose edges are all equal in length is called a regular tetrahedron.) The angle between a line and a plane is equal to the complement of the angle between the line and the normal to the plane whereas the angle between two planes is equal to the angle between the normals. Let O be the origin of reference and A, B and C vertices with position vectors a, b and c respectively of the regular tetrahedron.

20. The angle between any edge and a face not containing the edge is

(A) cos–1(1/2) (B) cos–1 (1/4) (C) cos–1 (1/ 3 ) (D) π/3

21. The value of [a b c]2 is (A) k2 (B) (1/2)k2 (C) (1/3)k2 (D) k3

This section contains 2 questions (Questions 22, 23). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows:

A B C D

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T


22. Match the following- Column -I

Column -II

(A) If the lines 1

2–x = 1

3–y = λ

4–z

and λ

1–x = 2

4–y = 1

5z −1

5–z

intersect at (α, β, γ) then λ =

(P) 0

(B) If

++π

∞→ 21tan–

44lim 1–

xxx

x =

y2 + 4y + 5 then y =

(Q) –1

(C) If chord x + y + 1= 0 of parabola y2 = ax subtends 90º at (0, 0) then a =

(R) –3

(D) If →a =

^i +

^j +

^k ,

→a .

→b = 1

and →a ×

→b =

^j –

^k , then |

→b | is

equal to

(S) 1 (T) 2

23. A is a set containing n elements. A subset P of A is

chosen at random. The set A is reconstructed by replacing the elements of the subset P. A subset Q of A is again chosen at random. The probability that (where |x| = number of elements in X)

Column-I Column-II (A) P ∩ Q = φ (P) n(3n–1)/4n

(B) P ∩ Q is a singleton (Q) (3/4)n

(C) P ∩ Q contains 2 (R) 2nCn/4n

elements (D) |P| = |Q| (S) 9n(n–1)/2(4n) (T) None

COMPLEMENTARY COLOURS

If you arrange some colours in a circle, you get a "colour wheel". The diagram shows one possible version of this. An internet search will throw up many different versions!

Colours directly opposite each other on the colour wheel are said to be complementary colours. Blue and yellow are complementary colours; red and cyan are complementary; and so are green and magenta. Mixing together two complementary colours of light will give you white light. What this all means is that if a particular colour is absorbed from white light, what your eye detects by mixing up all the other wavelengths of light is its complementary colour. Copper(II) sulphate solution is pale blue (cyan) because it absorbs light in the red region of the spectrum. Cyan is the complementary colour of red. The origin of colour in complex ions Transition metal v other metal complex ions


PHYSICS


1. When sound wave is refracted from air to water, which of the following will remain unchanged ?

(A) wave number (B) wavelength (C) wave velocity (D) frequency

2. In stationary wave – (A) All the particles of the medium vibrate in

phase (B) All the nodes vibrate in phase (C) All the antinodes vibrate in phase (D) All the particles between consecutive nodes

vibrate in phase

3. A 20 N metal block is suspended by a spring balance. A beaker containing some water is placed on a weighing machine which reads 40 N. The spring balance is now lowered so that the block gets immersed in water. The spring balance now reads 16N. The reading of the weighing machine will be -

(A) 36 N (B) 44 N (C) 60 N (D) None

4. The springs shown in figure is unstretched when James bond starts pulling on the cord. The mass of the block is m. If he exerts a constant force F. The amplitude of the motion of the block will be-

m

k

k F(const.)

Smooth

(A) k

F2

(B) kF (C)

kF2 (D) None

5. Ideal fluid flows along a flat tube of constant cross-section, located in a horizontal plane and bent as shown in figure (top view). The flow is steady. The velocities of fluid at point 1 and at point 2 are v1 and v2 respectively then correct relation is –

12

(A) v1 > v2 (B) v2 > v1

(C) v1 = v2 (D) None of these

6. A body is moving down a long inclined plane of inclination θ. The coefficient of friction between the body and the plane varies as µ = 0.5 x, where x is the distance moved down the plane. The body will have maximum velocity, when it has travelled a distance x given by -

(A) x = 2 tan θ (B) x = θtan

2

(C) x = 2 cot θ (D) x = θcot

2

Time : 3 Hours Syllabus : Syllabus : Physics : Full Syllabus, Chemistry : Full Syllabus, Mathematics : Full syllabus

Instructions : Section – I : Question 1 to 10 are multiple choice questions with only one correct answer. +3 marks will be

awarded for correct answer and -1 mark for wrong answer.

Section – II : Question 11 to 15 are multiple choice question with multiple correct answer. +4 marks will be awarded for correct answer and -1 mark for wrong answer.

Section – III : Question 16 to 21 are passage based single correct type questions. +3 marks will be awarded for correct answer and -1 mark for wrong answer

Section – IV : Question 22 to 23 are Column Matching type questions. +8 marks will be awarded for the complete correctly matched answer (i.e. +2 marks for each correct row) and No Negative marks for wrong answer.

XtraEdge Test Series # 7

IIT-JEE 2013

Based on New Pattern


7. A system consists of three particles, each of mass m and located at (1, 1), (2, 2) and (3, 3). The co-ordinates of the centre of mass are -

(A) (1, 1) (B) (2, 2) (C) (3, 3) (D) (6, 6) 8. A thin wire of length l and mass m is bent in the

form of a semicircle. Its moment of inertia about an axis joining its free ends will be –

O m

P (A) ml2 (B) zero

(C) 2

2mlπ

(D) 2

2

2ml

π

9. What is the velocity v of a metallic ball of radius r falling in a tank of liquid at the instant when its acceleration is one half that of a freely falling body? (The densities of metal and of liquid are ρ and σ respectively and the viscosity coefficient of the liquid is η) -

(A) η9

2 gr (ρ – 2σ) (B) η9

2 gr (2ρ – σ)

(C) η9

2 gr (ρ – σ) (D) η9

2 2 gr (ρ – σ)

10. An anisotropic material has coefficient of linear thermal expansion α1 and α2 along x and y respectively. Coefficient of superficial expansion of its material will be equal to –

(A) α1 + α2 (B) α1 + 2α2

(C) 3α1 + 2α2 (D) 2

21 α+α

Questions 11 to 15 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and – 1 mark for each wrong answer. 11. A man is standing on a road and observes that the

rain is falling at angle 45º with the vertical. The man starts running on the road with constant acceleration 0.5 m/s2. After a certain time from the start of the motion, it appears to him that the rain is still falling at angle 45º with the vertical, with speed

22 m/s. Motion of the man is in the same vertical plane in which the rain is falling. Then which of the following statement(s) are true -

(A) It is not possible. (B) Speed of the rain relative to the ground is

2 m/s (C) Speed of the man when he finds rain to be

falling at angle 45º with the vertical, is 4 m/s (D) the man has travelled a distance 16 m on the

road by the time he again finds rain to be falling at angle 45º

12. All the blocks shown in the figure are at rest. The pulley is smooth and string is light. Coefficient of friction at all the contacts is 0.2. A frictional force of 10N acts between A and B. The block A is about to slide on block B. The normal reaction and frictional force exerted by the ground on the block B is -

B 5kg A

C

(A) The normal reaction exerted by the ground on the block B is 110 N.

(B) The normal reaction exerted by the ground on the block B is 50 N.

(C) The frictional force exerted by the ground on the block B is 20 N

(D) The frictional force exerted by the ground on the block B is zero.

13. The value of mass m for which the 100 kg block remain is static equilibrium is –

37º

µ = 0.3 100

(A) 35 kg (B) 37 kg (C) 83 kg (D) 85 kg 14. A particle is describing circular motion in a

horizontal plane in contact with the smooth inside surface of a fixed right circular cone with its axis vertical and vertex down. The height of the plane of motion above the vertex is h and the semi vertical angle of the cone is α. The period of revolution of the particle –

α h

(A) increases as h increases (B) decreases as h increases (C) increases as α increases (D) decreases as α increases


15. A wire of density 9 × 103 kg / m3 is stretched between two clamps 1 m apart and is stretched to an extension of 4.9 × 10–4 metre. Young's modulus of material is 9 × 1010 N/m2 then -

(A) The lowest frequency of standing wave is 35 Hz (B) The frequency of 1st overtone is 70 Hz (C) The frequency of 1st overtone is 105 Hz (D) The stress in the wire is 4.41 × 107 N/m2


Passage # 1 (Ques. 16 - 17) In the figure shown a uniform solid sphere is

released on the top of a fixed inclined plane of inclination 37º and height ‘h’. It rolls without sliding (take sin 37º = 3/5)

37º 16. The acceleration of the centre of the sphere is -

(A) 5

3g (B) 5

4g (C) 7

4g (D) 7

3g

17. The speed of the point of contact of the sphere with the inclined plane when the sphere reaches the bottom of the incline is -

(A) gh2 (B) 7

10gh

(C) Zero (D) gh22

Passage # 2 (Ques. 18 - 19) In a standing wave experiment, a 1.2 kg horizontal

rope is fixed in place at its two ends (x = 0 and x = 2.0 m) and made to oscillate up and down in the fundamental mode at frequency 5.0 Hz. At t = 0, the point at x = 1.0 m has zero displacement and is moving upward in the positive direction of y axis with a transverse velocity 3.14 m/s

18. Speed of the participating travelling wave on the rope is -

(A) 6 m/s (B) 15 m/s (C) 20 m/s (D) 24 m/s

19. What is the correct expression of the standing wave equation -

(A) (0.1) sin

π

2 × sin ( )π10 t

(B) (0.1) sin (π) × sin (10π) t

(C) (0.05) sin

π

2× cos ( )π10 t

(D) (0.04) sin (π) × sin (10π) t

Passage # 3 (Ques. 20 - 21) Transverse and longitudinal standing waves are

easily represented by sine waves. The distance between an adjacent node and antinode (N – A) is a

quarter of the wavelength, 4λ .

A

A N N N

λ/4λ/4

λ Fundamental and first overtone for a pipe closed at

one end L

λ/4A N

λ/4

L

A NN A

L/3

4λ = L

34L

=λ

Therefore λ0 = 4L Therefore, λ1= 4

3L

and ν0 = L

V4

& v1=

34LV = 3

Lv

4

overtones are for the first overtone multiplies of ν0 : n = 3, the third

nµ0 = n

Lv

4 harmonic.

This implies that the ratio of natural frequency is n = 1 : 3 : 5.........

20. When an organ pipe is open at both ends, it resonates with a fundamental frequency of 240 Hz. What is the fundamental frequency of the same pipe if it is closed at one end-

(A) 64 Hz (B) 120 Hz (C) 360 Hz (D) 480Hz

21. A pipe resonates at 60 Hz, 100 Hz and 140 Hz consecutive frequencies. How long is the pipe ?

(A) 1.4 m (B) 2.8 m (C) 4.3 m (D) 8.5 m



A B C D

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T


22. Let V and E denote the gravitational potential and gravitational field respectively at a point due to certain uniform mass distribution described in four different situation of column – I, then

Column -I Column -II (A) At centre of thin

spherical shell (P) E = 0

(B) At centre of solid

sphere (Q) E ≠ 0

(C) At the centre of a

thick hemi spherical shell

(R) V ≠ 0

(D) At centre of line joining two point masses of equal magnitude

(S) (T)

V = 0 None

23. Two blocks A and B of mass m and 2m respectively

are connected by a massless spring of spring constant K. This system lies over a smooth horizontal surface. At t = 0 the block A has velocity u towards right as shown while the speed of block B is zero, and the length of spring is equal to its natural length at that instant. In each situation of column-I, certain statements are given and corresponding results are given in column-II, Match the statements in column-I to the corresponding results in column-II :

m 2m

K A B u

smooth horizontal surface

Column I Column II (A) The velocity of (P) Can never be zero block A (B) The velocity of (Q) may be zero at block B certain instants of time (C) The kinetic energy (R) is minimum at of system of maximum two blocks compression of spring (D) The potential (S) is maximum at energy of spring maximum extension of spring (T) None

CHEMISTRY


1. At constant pressure P, A dissociates on heating according to the equation

A(g) B(g) + C(g) The equilibrium partial pressure of A at T K is 1/9

P, the equilibrium Kp at TK is

(A) 98 P (B)

964 P (C)

916 P (D) 9 P

2. Calculate the pH of 6.66 × 10–3 M solution of Al(OH)3. Its first dissociation is 100% where as second dissociation is 50% and third dissociation is negligible.

(A) 2 (B) 12 (C) 11 (D) 13

3. pH of the blood in the body is maintained by buffer solution of

(A) glucose and salt concentration (B) protein and salt concentration (C) CO3

3– and HCO3–

(D) Salt and carbonate ion

4. IUPAC name of the following compound is : OH

CH3

(A) 2-methyl-3-cyclohexenol (B) 3-methyl-1-cyclohexen-4-ol (C) 4-hydroxy-3-methyl-1-cyclohexene (D) 2-hydroxy-1-methylcyclohexene


5. Which will form geometrical isomers ?

(A)

Cl

Cl

(B) CH3CH = NOH

(C)

(D) All

6. The dissolution of Al(OH)3 by a solution of NaOH results in the formation of

(A) [Al(H2O)4(OH)2]+ (B) [Al(H2O)3(OH)3] (C) [Al(H2O)2(OH)4]– (D) [Al(H2O)6](OH)3

7. Helium-oxygen mixture is used by deep sea divers in preference to nitrogen-oxygen mixture because

(A) helium is much less soluble in blood than nitrogen

(B) nitrogen is highly soluble in water (C) helium is insoluble in water (D) nitrogen is less soluble in blood than helium

8. SF4 + BF3 → (A). The compound 'A' is (A) [SF5]–[BF2]+ (B)[SF3]+[BF4]– (C) SF6 (D) S2F4

9. Which of the following reactions is a redox reaction?

(A) Cr2O3 + 6HCl → 2CrCl3 + 3H2O (B) CrO3 + 2NaOH → Na2CrO4 + H2O (C) 2CrO4

2– + H+ Cr2O72– + OH–

(D) Cr2O72– + 6I– + 14H+ 2Cr3+ + 3I2

+ 7H2O

10. The combustion reaction occurring in an automobile is 2C8H18(s) + 5O2(g) → 16CO2(g) + 18H2O(1). This reaction is accompanied with

(A) ∆H = –ve, ∆S = + ve, ∆G = + ve (B) ∆H = + ve, ∆S = –ve, ∆G = + ve (C) ∆H = – ve, ∆S = +ve, ∆G = – ve (D) ∆Η = +ve, ∆S = +ve, ∆G = – ve

Questions 11 to 15 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and – 1 mark for each wrong answer. 11. Which of the following species are correctly

matched with their geometries according to the VSEPR theory -

(A) ClF2– → linear

(B) IF4+ → see – saw

(C) SnCl5– → trigonal bipyramidal

(D) 33

••)SiH(N → pyramidal

12. KCl has a dipole moment of 10 D. The inter ionic distance in KCl is 2.6 Å. Which of the following statement are true for this compound ?

(A) The theoretical value of dipole moment, if the compound were completely ionic is 12.5 D.

(B) The % ionic character of the compound is 85 % (C) It is a poor conductor of electricity (D) The forces operating in this molecule are

coulombic type 13. The major product of reaction

→ 2Br is –

(A)

Br

Br

(B)

Br

Br

(C)

BrBr

(D) None of these

14.

Br

Br

+ KOH (alc) —→

Which of the following can be formed.

(A) (B)

Br

(C)

Br

Br

(D)

Br

15. 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 recemic-2, 3-dideuterobutane (B) trans-but-2-ene and meso-2, 3-dideuterobutane (C) trans-but-2-ene and recemic-2, 3-dideuterobutane (D) cis-but-2-ene and meso-2, 3-dideuterobutane



Passage # 1 (Ques. 16 - 17) In order to explain the existance of doublets in the spectra of alkali metals, Goudsmit and Uhlenbeck in 1925 proposed that electron has an intrinsic angular momentum due to spining about its own axis. The value of spining a angular momentum of electron can be described by 2 spin quantum number s and ms. The physical significance of s and ms is similar as of l and ml.

16. The possible value of s for electron is - (A) 1/2 (B) – 1/2 (C) 0 (D) 1

17. Relation between s and ms is :

(A) )1s(s +π2

h . cos θ = ms

(B) )1s(s + cos θ = ms

(C) π4h3 = ms

(D) None of these

Passage # 2 (Ques. 18 - 19) The expression for the reaction quotient, Q, is

similar to that for equilibrium constant K. The value of Q for the given composition of a reaction mixture helps us to know whether the reaction will move forward or backward or remain in equilibrium. It also helps to predict the effect of pressure on the direction of the gaseous reaction. In certain reactions, addition of inert gas also favours either the formation of reactants or products. The value of equilibrium constant of a reaction changes with change of temperature and the change is given by van't Hoff equation, d ln Kp/dT = ∆Hº/RT2 where enthalpy change, ∆Hº, is taken as constant in the small temperature range.

18. The reaction N2(g) + 3H2(g) 2NH3(g) is in equilibrium. Now the reaction mixture is compressed to half the volume

(A) More of ammonia will be formed (B) Ammonia will dissociate back into N2 and H2 (C) There will be no effect on equilibrium (D)Equilibrium constant of the reaction will

change

19. For the above reaction in equilibrium, helium gas was added but the mixture was allowed to expand to keep the pressure constant. Then

(A) More of ammonia will be formed (B) Ammonia will dissociate back into N2 and H2 (C) There will be no effect on equilibrium (D) Equilibrium constant of the reaction will change

Passage # 3 (Ques. 20 - 21) Lithium only forms monoxide when heated in

oxygen. Sodium forms monoxide and peroxide in excess of oxygen. Other alkali metals form super oxide with oxygen i.e., MO2. The abnormal behaviour of lithium is due to small size. The larger size of nearer alkali metals also decides the role in formation of superoxides. The three ions related to each other as follows :

ionOxide

2O − → 2O2/1 ionPeroxide

22O − → 2O

ionSuperoxide2O2 −

All the three ions abstract proton from water.

20. Consider the following reaction : M + O2 →

oxide)per su(2MO (M = alkali metal)

Select the correct statement : (A) M can not be Li and Na (B) M can not be Cs and Rb (C) M can not be Li and Rb (D) None of these

21. Lithium does not form stable peroxide because : (A) of its small size (B) d-orbital are absent in it (C) it is highly reactive and form superoxide in

place of peroxide (D) covalent nature of peroxide


ABCD

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T



22. Match the Column : Column –I Column II (A) pKb of X– (Ka of HX = 10–6) (P) 6.9 (B) pH of 10–8 M HCl (Q) 8 (C) pHof 10–2M acetic acid solution (R) 3.3 (Ka = 1.6 × 10–5) (D) pH of a solution obtained by (S) 3.4 mixing equal volumes of solution with pH 3 & 5. (t)

23. Match the Column : Column I Column II

(A) Br

O

(p) Nucleopilic substitution

(B) OH (q) Elimination

(C)

OH

CHO (r) Nucleophilic addition

(D)

NO2

Br

(s) Esterification with acetic anhydride

(t) Dehydrogenation

MATHEMATICS


1. In a certain test there are n questions. In this test 2k students gave wrong answers to at least (n – k) questions, where k = 0, 1, 2, ...... , n. If the total number of wrong answers is 4095, then value of n is

(A) 11 (B) 12 (C) 13 (D) 15

2. The values of x between 0 and 2π which satisfy the

equation sin x x2cos8 = 1 are in A.P. The common difference of the A.P. is

(A) π/8 (B) π/4 (C) 3π/8 (D) 5π/8

3. If p1, p2, p3 are respectively the perpendicular from the vertices of a triangle to the opposite sides, then

1

cosp

A + 2

cosp

B + 3

cosp

C is equal to

(A) 1/r (B) 1/R (C) 1/∆ (D) None

4. The positive integer n for which 2 × 22 + 3 × 23 + 4 × 24 + .... + n × 2n = 2n + 10 is (A) 510 (B) 511 (C) 512 (D) 513

5. A vector c, directed along the internal bisector of the angle between the vectors a = 7i – 4j – 4k and

b = –2i – j + 2k, with |c| = 5 6 , is -

(A) 35 (i – 7j + 2k) (B)

35 (5i + 5j + 2k)

(C) 35 (i + 7j + 2k) (D)

35 (–5i + 5j + 2k)

6. The locus of the mid-point of the line segment joining the focus to a moving point on the parabola y2 = 4ax is another parabola with directrix -

(A) x = – a (B) x = – a/2 (C) x = 0 (D) x = a/2

7. A flagstaff stands in the centre of a rectangular field whose diagonal is 1200 m, and subtends angles 15º and 45º at the mid points of the sides of the field. The height of the flagstaff is -

(A) 200 m (B) 300 32 + m

(C) 300 32 − m (D) 400 m

8. If two vertices of a triangle are (–2, 3) and (5, – 1), orthocentre lies at the origin and centroid on the line x + y = 7, then the third vertex lies at

(A) (7, 4) (B) (8, 14) (C) (12, 21) (D) None

9. If tan x + tan(x + π/3) + tan (x + 2π/3) = 3, then - (A) tan x = 1 (B) tan 2x = 1 (C) tan 3x = 1 (D) none of these

10. The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only, is

(A) 55 (B) 66 (C) 77 (D) 88

Questions 11 to 15 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and – 1 mark for each wrong answer. 11. The coordinates of a point on the line

2

1−x = 31

−+y

= z at a distance 144 from the

point (1, –1, 0) are (A) (9, – 13, 4) (B) ( 148 + 1, –12 14 – 1, 4 14


(C) (–7, 11, – 4) (D) (– 148 + 1, 12 14 – 1, –4 14 12. If x2 + 2hxy + y2 = 0 represents the equations of the

straight lines through the origin which make an angle α with the straight line y + x = 0, then

(A) sec 2α = h

(B) cos α = hh

21+

(C) 2 sin α = h

h+1

(D) cot α = 1

1−+

hh

13. If z1, z2, z3, z4 are the vertices of a square in that order, then

(A) z1 + z3 = z2 + z4 (B) |z1 – z2| = |z2 – z3| = |z3 – z4| = |z4 – z1| (C) |z1 – z3| = |z2 – z4| (D) (z1 – z3)/(z2 – z4) is purely imaginary

14. The equation of a tangent to the hyperbola 3x2 – y2 = 3, parallel to the line y = 2x + 4 is -

(A) y = 2x + 3 (B) y = 2x + 1 (C) y = 2x – 1 (D) y = 2x + 2

15. If m is a positive integer, then ])13[( 2m+ + 1, where [x] denotes greatest integer ≤ n, is divisible by-

(A) 2m (B) 2m+1 (C) 2m+2 (D) 22m


Passage # 1 (Ques. 16 - 17)

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)

16. The value of x for which f (x) = 0 is (A) –1/2 (B) 0 (C) 1/2 (D) 1

17. If a2 = 26/51, then b2 is equal to - (A) 1/25 (B) 24/25 (C) 25/26 (D) 50/51

Passage # 2 (Ques. 18 - 19) To solve equation or inequality involving

exponential expression f(x)g(x), we may use logarithm or the identity xy = xy aa log

where a > 0, a ≠ 1.

18. Solution set of the inequality 3x(0.333 ....)x–3 ≤ (1/27)x is (A) [3/2, 5] (B) (– ∞, 3/2] (C) (0, ∞) (D) None of these

19. Solution set of the inequality 2(25)x – 5(10x) + 2(4x) ≥ 0 is (A) (–1, ∞) (B) (0, ∞) (C) (2, ∞) (D) None of these Passage # 3 (Ques. 20 - 21)

C : x2 + y2 = 9, E : 9

2x + 4

2y = 1, L : y = 2x

20. P is a point on the circle C, the perpendicular PL to the major axis of the ellipse E meets the ellipse at

M, then PLML is equal to

(A) 1/3 (B) 2/3 (C) 1/2 (D) none of these

21. Equation of the diameter of the ellipse E conjugate to the diameter represented by L is

(A) 9x + 2y = 0 (B) 2x + 9y = 0 (C) 4x + 9y = 0 (D) 4x – 9y = 0


ABCD

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T



22. Match the following -

Column- I Column- II (A) If a, b, c are unequal

positive numbers and b is A.M. of a and c then the roots of

ax2 + 2bx + c = 0 are

(P) of opposite signs

(B) If a ∈ R, then the roots of the equation

x2 – (a +1) x – a2 – 4 = 0 are

(Q) rational numbers

(C) If a, b, c are unequal positive numbers and b is H.M. of a and c then the roots of ax2 + 2bx + c = 0 are

(R) real and unequal

(D) If |a ± b| < c and a = 0 then the roots of a2x2 + (b2 + a2 – c2)x

+ b2 = 0 are

(S) imaginary (T) None

23. Match the Column : Column-I Column-II (A) Equation of the polar (P) 8x + 2y – 23 = 0 of (–7, –9) with respect to the circle x2 + y2 – 12x – 8y – 48 = 0 (B) Equation of the (Q) 13x + 13y – 30 = 0 common chord of the circles x2 + y2 + 2x + 2y + 1 = 0 and x2 + y2 + 4x + 3y + 2 = 0 (C) Equation of the (R) 2x + y + 1 = 0 tangent at (–7, –9) to the circle x2 + y2 + 12x + 8y + 26 = 0 (D) Equation of the radical (S) x + 5y + 52 = 0 axis of the circles 2x2 + 2y2 + 4x + 4y + 9 = 0 and x2 + y2 + 6x+3y – 7 = 0 (T) x + y – 1 = 0

Regents Physics You Should Know

Nuclear Physics : • Alpha particles are the same as helium nuclei

and have the symbol .

• The atomic number is equal to the number of protons (2 for alpha)

• Deuterium ( ) is an isotope of hydrogen

( )

• The number of nucleons is equal to protons + neutrons (4 for alpha)

• Only charged particles can be accelerated in a particle accelerator such as a cyclotron or Van Der Graaf generator.

• Natural radiation is alpha ( ), beta ( ) and gamma (high energy x-rays)

• A loss of a beta particle results in an increase in atomic number.

• All nuclei weigh less than their parts. This mass defect is converted into binding energy. (E=mc2)

• Isotopes have different neutron numbers and atomic masses but the same number of protons (atomic numbers).

• Geiger counters, photographic plates, cloud and bubble chambers are all used to detect or observe radiation.

• Rutherford discovered the positive nucleus using his famous gold-foil experiment.

• Fusion requires that hydrogen be combined to make helium.

• Fission requires that a neutron causes uranium to be split into middle size atoms and produce extra neutrons.

• Radioactive half-lives can not be changed by heat or pressure.

• One AMU of mass is equal to 931 meV of energy (E = mc2).


XtraEdge Test Series ANSWER KEY

PHYSICS Ques 1 2 3 4 5 6 7 8 9 10 11 Ans C C A A B B D A A B A,C Ques 12 13 14 15 16 17 18 19 20 21 Ans A,C,D A A,B,C A,B,C A A D C C A

Ques 22 23 Column Match Ans A→ S; B→ R; C→ Q; D→ P A→ R; B→ P; C→ S; D→ Q

CHEMISTRY

Ques 1 2 3 4 5 6 7 8 9 10 11 Ans A A B B D B C A A B A,B,D Ques 12 13 14 15 16 17 18 19 20 21 Ans A,B A,B,C,D A B,C B B B B A C

Ques 22 23 Column Match Ans A→ R; B→ P,S,T; C→ Q; D→ P,T A→ Q,R; B→ S; C→ Q,R; D→ P

MATHEMATICS

Ques 1 2 3 4 5 6 7 8 9 10 11 Ans B D A A A A A D B C A,B,C Ques 12 13 14 15 16 17 18 19 20 21 Ans A,B A,B,C A,C,D A,B C D C B C B

Ques 22 23 Column Match Ans A→ P,Q; B→ Q,R; C→ Q; D→ S A→ R; B→ P; C→ S; D→ R

PHYSICS Ques 1 2 3 4 5 6 7 8 9 10 11 Ans D D B A B A B D A C C,D Ques 12 13 14 15 16 17 18 19 20 21 Ans A,D B,C A,C A,B D C C A B C

Ques 22 23 Column Match Ans A→ P,R; B→ P,R; C→ Q,R; D→ P,R A→ P; B→ Q; C→ P,R; D→ Q,S

CHEMISTRY

Ques 1 2 3 4 5 6 7 8 9 10 11 Ans C B C A D C A B D C A,B,C Ques 12 13 14 15 16 17 18 19 20 21 Ans A,C,D A A,B,C,D C,D A B A B A A

Ques 22 23 Column Match Ans A→ Q; B→ P; C→ S; D→ R A→ P,Q; B→ P,S,T; C→ R,S; D→ P

MATHEMATICS Ques 1 2 3 4 5 6 7 8 9 10 11 Ans B B B B A C C D A C A,C Ques 12 13 14 15 16 17 18 19 20 21 Ans A,B,D A,B,C,D B,C A,B A B B D B B

Ques 22 23 Column Match Ans A→ R; B→ P,R; C→ S; D→ R A→ Q; B→ R; C→ S; D→ P

IIT- JEE 2012 (November issue)

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