Dear Students,
All of us live and work within fixed patterns. These patterns and habits determine the quality of our life and the choices we make in life. There are a few vital things to know about ourselves. We should become aware of how much we influence others, how productive we are and what can help us to achieve our goals. It is important to create an environment which will promote our success. We should consciously create a system that would enable us to achieve our goals. Most of us live in systems which have come our way by an accident, circumstances or people we have met over a period of time. We are surrounded by our colleagues or subordinates who happened to be there by the fact of sheer recruitment earlier or later by the management. Our daily routines and schedules have been formed on the basis of convenience, coincidence, and the expectations of society and sometimes due to superstitions. The trick for success is to have an environment that helps in attaining our goals. Control your life. Make an effort to launch your day with a great start. A law of physics says that an object set in motion tends to remain in motion. It is the same thing with daily routine. To have a good start each morning will keep you upbeat during the day. If you begin the day stressed, you will tend to remain so that way. The best is to create a course of action or conditions where you are not hassled for being late for a meeting, worried about household affairs or distracted by happenings in the world.
Aim to be highly successful. Control the direction of your life. Not only should you start the day on a cheerful note but also continue to do so during the day. Keep yourself stimulated and invigorated during the entire day. Start your day with a purpose. Have a daily direction and trajectory of action. It will keep you on your course all day long. Throughout the day reinforce your positive values and your choices. Anything that helps you in maintaining your highest values and your most important priorities should be welcome. Be in control of your life and work. Create and sustain a wonderful environment filled with beauty, peace, inspiration and hope.
Plan your day in such a way that suits your plans objectives and makes you feel just right with the right amount of encouragement during the entire day. You should give a direction to your day and timing.
Presenting forever positive ideas to your success.
Yours truly
Pramod Maheshwari, B.Tech., IIT Delhi
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Volume-7 Issue-8 February, 2012 (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. IIT-JEE Mock Test Paper with Solution
AIEEE & BIT-SAT Mock Test Paper with Solution
S
Success Tips for the Months
• If one asks for success and prepares for failure, he will get the situation he has prepared for.
• Loser's visualize the penalties of failure. Winner's visualize the rewards of success. • Treat others as if they were what they
ought to be and you help them to become what they are capable of being.
• You never achieve real success unless you like what you are doing
• The first step toward success is taken when you refuse to be a captive of the environment in which you first find yourself.
• Believe in yourself ! Have faith in your abilities ! without a humble but reasonable confidence in your own powers you can not be successful or happy.
CONTENTS
INDEX PAGE
NEWS ARTICLE
3• 25% increase in number of girls from Bombay zone for IIT-JEE
• IIT placements : Companies Back on hiring track
IITian ON THE PATH OF SUCCESS
5 Dr. Amitabha GhoshKNOW IIT-JEE
6Previous IIT-JEE Question
XTRAEDGE TEST SERIES
47
Class XII – IIT-JEE 2012 PaperClass XI – IIT-JEE 2013 Paper
Mock Test-3 (CBSE Board Pattern) [Class # XII]
68
Solution of Mock Test-2 & 3 (CBSE Pattern)Regulars ...
DYNAMIC PHYSICS
14
8-Challenging Problems [Set # 10] Students’ Forum
Physics Fundamentals
Matter Waves, Photo-electric Effect Thermal Expansion, Thermodynamics
CATALYSE CHEMISTRY
29
Key ConceptCarbonyl Compound Co-ordination Compound & Metallurgy
Understanding : Physical Chemistry
DICEY MATHS
36
Mathematical Challenges Students’ Forum Key Concept Integration Trigonometrical EquationStudy Time...
Test Time ...
25% increase in number of girls from Bombay zone for IIT-JEE There are far more girls set to appear for the Indian Institute of Technology Joint Entrance Exam (IIT-JEE) in the Bombay zone in April 2012 than before. The number of applications by girls has increased by nearly 25% compared to last year. The Bombay zone includes Maharashtra, Goa, This year, girls account for nearly 27% of the total candidates applying, with around 20,300 girls from among 76,600 candidates. The IITs have been constantly trying to address the skewed male-female ratio at the IITs. “This is a significant increase,” said AV Mahajan, chairperson for JEE 2012, Bombay zone. “One reason for this could be that it was free for girls to apply online this year as well as cheaper to apply offline as compared to male applicants.”
Two IT-BHU students to start entrepreneurship club for school students
VARANASI: When they cracked IIT-JEE two years ago, getting attractive jobs in top IT companies was their dream. Now, they want to promote entrepreneurial skills in young school children (Class X onwards) by establishing entrepreneurship club for schools.
Bhanu Swami and Varun Agrawal, B Tech (III) students of electrical engineering of Institute of Technology, Banaras Hindu University (IT-BHU), are on their way to launch country's first entrepre-neurship club targeting students (Class X onwards). The initiative not only aims at tapping and promoting entrepreneurial skills in young students, but also promises to turn these students into entrepreneurs. This way they would become job providers rather than job seekers.
Akash 2 tablet to be launched by April 2012: Govt.
The government has expressed confidence it will be successful in bringing out the improved version of the Aakash tablet by April this year. The Aakash 2 is likely to have several improved specifications such as elongated battery life and faster processor. The government has been striving hard to make its Aakash project successful. The device, which is also touted as the world's cheapest tablet, faced harsh criticism from several quarters for its poor quality and dismal features.
IIT Rajasthan, which developed the prototype of the device along with Data Wind, found a series of faults in the device, prompting the government to reconsider extension of the LC to the Canadian company. Under pressure to provide better endowed low-cost device, the government has now its turned attention to the Aakash 2, which is likely to come at the same price tag of the original Aakash.
HRD Minister Kapil Sibal has said that the government will be looking for more manufacturers to manufacture the Aakash 2 tablets, attributing the massive demand for the move. “We want to make sure that the upgraded product caters to the need of the customers... We have involved ITI in order to upgrade it... We will be able to bring in Aakash-II by April," Kapil Sibal said.
Last week media reports had hinted that the government might shelve its Aakash project in view of the harsh criticism from most quarters. The reports were triggered by DataWind's opposition to certain test criteria suggested by the IIT. Read our previous coverage
IIT Techfest 2012 : All about Robotics
One of the highlights of the Techfest 2012, which held at IIT Bombay, was its international exhibitions arena. Evidently dominated by robots, the robotic section had covered several aspects - humanoid, surveillance, educational and industrial. The robotic platform at the Techfest 2012 showcased bots expressing emotions, playing intelligent games and serving as a model for military training, among others. With a great potential for applications in future in varying fields, lets look at these robots one by one.
Enjoy Moneycontrol.com on iPad and be prepared for a fantastic experience. Get real time stock quotes, interactive charts, market buzz, and watch CNBC-TV18, CNBC Awaaz live on your iPad. Check out the free moneycontrol app IIT-KGP celebrates the 9th annual Alumni Meet 2012 while its alumni commits to give back to their Alma meter
IIT Kharagpur celebrated its 9th Annual Alumni Meet. While it honoured its alumni with the Distinguished Service Award more alumni committed towards the cause of contributing to their Alma mater. A galaxy of alumni from world wide graced the 3-day dynamic event Auto Expo 2012 : Hydrogen-powered 3-wheeler at expo
NEW DELHI: The world's first hydrogen-powered three-wheeler, 'HyAlfa', was showcased at theAuto Expo on Monday. Part of a development project dubbed 'DelHy 3w', a fleet of 15 HyAlfa three-wheelers will run on an experimental basis at Pragati Maidan, where a
hydrogen refuelling station has also been set up.
India Trade Organisation Promotion ( ITPO) will use the vehicles on an experimental basis. The HyAlfa has been developed under a joint project by the United Nations Industrial Development Organisation ( UNIDO) International Centre for Hydrogen Energy Technologies ( ICHET), Mahindra & Mahindra and IIT-Delhi, with support from the Ministry of New and Renewable Energy. "The aim of this project is to convert vehicles so that they can carry and use hydrogen - a carbon-free fuel - and thus remove all pollutants," Mahindra & Mahindra president (automotive) Pawan Goenka said. He said the vehicle is not yet ready for commercial production and further fine-tuning will be required before moving in that direction. "Moreover, we also have to look at commercial viability of running a hydrogen-powered three-wheeler as the cost of hydrogen will be around Rs 250 per kg, which is not affordable at all," he said. This is a step in the right direction. Finally India Inc. with support from the likes of MNRE etc. is making headway in vehicles driven by alternative energy sources. Jury is still out on whether Hydrogen is the way of the future. Judging by the progress made in the USA and Germany on Hydrogen powered vehicles, we are not quite there yet. There are not only supply chain challenges, but Hydrogen continues to be a costly proposition. You can either produce Hydrogen from hydrocarbon cracking (which in turn has dependence on fossil fuels) or by splitting of water through electrolysis. This itself requires around 5-6 kWh of electricity. So unless the electricity source is a renewable source such as solar or wind, the electricity required to split water itself is most likely to come from thermal power plants, thereby not giving any benefit. However, as everyone knows, human ingenuity knows no bounds and technologies only develop incrementally. I firmly believe that we are one step away from a miracle where both Hydrogen consumption
(by means of fuel cells etc.) and Hydrogen generation (by means of hydrocarbon cracking or electrolyzing/splitting water) become viable for mass production and consumption. In the short term, we can and must focus immediately on "Hydrogen supplementation". By this I mean - Hydrogen CNG (or HCNG in short). New Delhi moved to CNG public transport a while back. In the short term, this did bring down the pollution levels and particulate matter in the atmosphere. After prolonged use, we are now becoming aware of other problems such as NOX emissions due to unclean or inefficient burning of CNG. NOX is highly carcinogenic and now the levels of NOX in New Delhi are far exceeding permissible limits of WHO. One way to solve this problem is to supplement CNG with Hydrogen. A blended product called Hythane (trade name for HCNG) is already under experimentation by Indian Oil. In conclusion, while Hydrogen powered vehicles are all a step in the right direction, the government should put impetus on technologies that are more feasible in solving current big-city problems that Indian cities face.
IITian tops CAT 2011
CHENNAI: Ajinkya Deshmukh, an IIT - Madras graduate, decided to put in 100% effort into preparing for the Common Admission Test this time, and got 100 percentile in return. He is one of nine MBA aspirants in the country to secure the top score this year. The IIMs published the CAT 2011 scores on their website on Wednesday.
On his success, Ajinkya, who wrote the test for the third time this year, said, "I always thought I haven't been making a full effort while preparing or writing the test. This time I was determined to give my best. I told myself that I had to do it this year."
Indian American sworn in as America's top science official
IIT Madras alumnus, Subra Suresh, has been sworn in as the director of America's National Science Foundation (NSF), the top US science body with a $7.4 billion budget to support scientific institutions.
"We are very grateful to have Subra taking this new task," said President Barack Obama at the White House Science after Suresh was sworn in as the 13th NSF director by John Holdren, Obama's science advisor. "He has been at MIT (Massachusetts Institute of Technology) and has been leading one of the top engineering programmes in the country, and for him now to be able to apply that to the National Science Foundation is just going to be outstanding," he said. "So we're very grateful for your service."
Suresh, 54, was confirmed by the US Senate for a six-year term.
He has served as dean of the engineering school and as Vannevar Bush Professor of Engineering at MIT.
A mechanical engineer, who later became interested in materials science and biology, Suresh has done pioneering work studying the biomechanics of blood cells under the influence of diseases such as malaria.
From 2000 to 2006, Suresh served as the head of the MIT Department of Materials Science and Engineering. He joined MIT in 1993 as the R.P. Simmons Professor of Materials Science and Engineering and held joint faculty appointments in the departments of mechanical engineering and biological engineering, as well as the division of health sciences and technology. Suresh holds a bachelor's degree from the Indian Institute of Technology in Madras and a master's degree from Iowa State University. Suresh was nominated by President Obama to become the new NSF director in place of Arden L. Bement Jr, who led the agency from 2004 until he resigned in May this yea
Dr Amitabha Ghosh was the only Asian on NASA's Mars Pathfinder mission. At present, he is a member of the Mars Odyssey Mission and the Mars Exploration Rover Mission.
During the Mars Pathfinder Mission, he conducted chemical analysis of rocks and soil on the landing site. The simple and unassuming 34-year-old planetary geologist has won several accolades, which include the NASA Mars Pathfinder Achievement Award in 1997 and the NASA Mars Exploration Rover Achievement Award in 2004. The journey from India to NASA.
It has been an intriguing experience. I was keen on geologic research data interpretation and solar system formation. During my geological research days in India, I had slept in railway stations while traveling to various places.
After my post graduation in applied geology from IIT Kharagpur, I wrote a letter to a professor at NASA expressing a desire to work at the space agency.
I made certain suggestions; in fact, it was a critical letter. In India, you can never imagine criticising your professor. My suggestions were approved, while I got an opportunity to work at NASA.
I think one requires luck and to put in sincere effort to achieve one's goals. Being in the right place at the right time is also important.
In Mumbai for the Pravasi Bharatiya Divas, he spoke about his work at NASA and his vision for India.
The Vision for India :
I feel there India has a great future. We have world-class companies. Today, companies like Infosys can be compared with world leaders like Oracle. Like the Information Technology revolution, we can have a science or space revolution. We have the potential to bring about revolutions in other sectors as well.
How India can we develop science and technology sector :
It should be treated as a business. There should be more private participation. We must have an external review to evaluate the system and make changes as science and technology can take the country forward.
We must check brain drain. About 80,000 students migrate to the US for further studies, and settle there. They find the facilities much better abroad. We need to reverse brain drain by enhancing and upgrading institutes in India.
The state of space research in India :
I don't want to make controversial statements. All I can say is India is not at the frontier of space research. We have made commendable progress but there is a long way to go. We can do much better. I would be glad to be of help in any way. Investment in research is investment in imagination. It is a matter of national pride and internal recognition. We need to allocate more funds to enhance research and development work.
We need good educational institutes like IITs and IIMs, but IITians don't rule the world. You must remember that Microsoft co-founder (Bill Gates does not have a college degree.
Youngsters must look around for role models and see what it is that they are doing right. Individuals must make use of their inherent strengths to succeed.
How can India become a leading global player : Globalisation will reap huge and long-term benefits and India must make the best use of the opportunities. At the PBD seminar, I found people presenting grandiose plans. Instead, we should look at the realities and immediate solutions.
The private sector has to be actively involved in the development of the country and the government has to respond to the needs of the people. Fifteen years ago, we didn't have an Infosys, today we have many global companies.
There should be drastic reduction in paper work. We need a scenario where one can start any business in a day, like in the US.
Can India have something like NASA:
The answer is: Yes, India can. All it requires is the right kind of investment, infrastructure, people and support from the government.
Success Story
Success Story
This article contains stories/interviews of persons who succeed after graduation from different IITs
Dr. Amitabha Ghosh
• Post graduation in applied geology from IIT Kharagpur, • Working at NASA
PHYSICS
1. A large open top container of negligible mass and uniform cross-sectional area A has a small holes of cross-sectional area A/100 in its side wall near the bottom. The container is kept on a smooth horizontal floor and contains a liquid of density ρ and mass m0.
Assuming that the liquid starts flowing out horizontally through the hole at t = 0, Calculate
[IIT-1997]
(i) the acceleration of the container, and
(ii) its velocity when 75% of the liquid has drained out.
Sol. (i) Let at any instant of time during the flow, the height of liquid in the container is x. The velocity of flow of liquid through small hole in the orifice by Toricelli's theorem is
v = 2gx ...(i)
The mass of liquid flowing per second through the orifice
= ρ × volume of liquid flowing per second
dt
dm = ρ × 2gx × 100
A ...(iii)
Therefore the rate of charge of momentum of the system in forward direction
= dt dm × v = 100 A gx 2 × ×ρ
(From (i) and (ii)) The rate of charge of momentum of the system in the backward direction
= Force on backward direction = m × a where m is mass of liquid in the container at the instant t m = Vol. × density
= A × x × ρ
x
ρ
v
∴ The rate of charge of momentum of the system in the backward direction
Axρ × a
By conservation of linear momentum Axp × a = 100 gxA 2 ρ ⇒ a = 50 g
(ii) By toricell's theorem v' = 2g×(0.25h) where h is the initial height of the liquid in the container m0 is the initial mass
∴ m0 = Ah × ρ ⇒ h = ρ A m0 ∴ v' = ρ × × A m 25 . 0 g 2 0 = ρ A 2 gm0
2. Two parallel plate capacitors A and B have the same separation d = 8.85 × 10–4 m between the plates. The
plate area of A and B are 0.04 m2 and 0.02 m2
respectively. A slab of dielectric constant (relative permittivity) K = 9 has dimensions such that it can exactly fill the space between the plates of capacitor
B. [IIT-1993] A 110 V B (c) A (b) (a) B
(i) The dielectric slab is placed inside A as shown in figure (a). A is then charged to a potential difference of 110 V. Calculate the capacitance of A and the energy stored in it.
(ii) The battery is disconnected and then the dielectric slab is moved from A. Find the work done by the external agency in removing the slab from A.
(iii) The same dielectric slab is now placed inside B, filling it completely. The two capacitors A and B are then connected as shown in figure (c). Calculate the energy stored in the system.
KNOW IIT-JEE
Sol. (i) The capacitor A with dielectric slab can be considered as two capacitors in parallel, one having dielectric slab and one not having dielectric slab each capacitor has an area of
2 A. The combined capacitance is A 110 V + + + + + – – – – + A/2 A/2 C = C1 + C2 = d ) 2 / A ( ε0 + d r ) 2 / A ( ε0ε = d 2 Aε0 [1 + ε r] = ––412 10 85 . 8 2 10 85 . 8 4 . 0 × × × × [1 + 9] = 2 × 10–9 F ∴ Energy stored = 2 1 CV2 = 2 1 × 2 × 10–9 × (110)2 = 1.21 × 10–5 J
(ii) Work done in removing the dielectric state = (Energy stored in capacitor without dielectric) – (Energy stored in capacitor with dielectric). It may be noted that while taking out the dielectric the charge on the capacitor plate remains the same. ∴ W = ' C 2 q2 – C 2 q2 Here, C = 2 × 10–9 F, C' = d Aε0 = –4–14 10 85 . 8 10 85 . 8 04 . 0 × × × = 0.4 × 10–9 F q = CV = 2 × 10–9 × 110 = 2.2 × 10–7C ∴ W = × × × 9 – 9 – 2 7 – 10 2 1 – 10 4 . 0 1 2 ) 10 2 . 2 ( = × × × × 4 . 0 2 4 . 0 – 2 10 2 10 2 . 2 2 . 2 9 – 14 – = 1.21 × 4 . 0 6 . 1 × 10–5 = 4.84 × 10–5 J
(iii) The capacitance of B = d AB r 0ε ε = –4 12 – 10 85 . 8 02 . 0 9 10 85 . 8 × × × × CB = 1.8 × 10–9 F
The charge on A, qA = 2.2 × 10–7 C gets distributed
into two parts.
∴ q1 + q2 = 2.2 × 10–7 C Also the potential
difference across A = p.d. across B
∴ A 1 C q = B 2 C q ⇒ q1 = B A C C . q 2 = –9 9 – 10 8 . 1 10 4 . 0 × × . q2 = 0.22 q2 ∴ 0.22 q2 + q2 = 2.2 × 10–7 ⇒ q2 = 22 . 1 2 . 2 × 10–7 = 1.8 × 10–7 C ⇒ q1 = 0.4 × 10–7 C
Total energy stored
= A 2 1 C 2 q + B 2 2 C 2 q = ––914 10 4 . 0 2 10 4 . 0 4 . 0 × × × × + ––814 10 8 . 1 2 10 8 . 1 8 . 1 × × × × = 0.2 × 10–5 + 0.9 × 10–5 = 1.1 × 10–5 J.
3. A particles of mass m = 1.6 × 10–27 kg and charge q =
1.6 × 10–19 C enters a region of uniform magnetic
field of strength 1 Tesla along the direction shown in figure. The speed of the particle is 107 m/s. (i) the
magnetic field is directed along the inward normal to the plane of the paper. The particle leaves the region of the field at the point F. Find the distance EF and the angle θ. (ii) If the direction of the field is along the outward normal to the plane of the paper, find the time spent by the particle in the region of the magnetic field after entering it at E. [IIT-1984]
F E 45º θ × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × Sol. (a) m = 1.6 × 10–27 kg, q = 1.6 × 10–19 C B = 1 T, v = 107 m/s F = q . v B sin α
(acting towards O by Fleming's left hand rule) F = qvB [Q α = 90º] But F = ma ∴ qvB = ma ∴ a = m qvB = –27 7 19 – 10 6 . 1 1 10 10 6 . 1 × × × × = 1015 m/s2
× × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × O F θ E 45º 45º 2 V V2= 2 V V2=
This is the centripetal acceleration ∴
r
v2 = 1015 ∠OEF = 45º
(Q OE, act as a radius) ⇒ r = 1514
10
10 = 0.1 m By symmetry ∠OFE = 45º ∴ ∠EOF = 90º (by Geometry)
If the magnetic field is in the outward direction and the particle enters in the same way at E, then according to Fleming's Left hand rule, the particle will turn towards clockwise direction and cover 3/4th of a circle as shown in the figure.
× × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × 45º 45º E O 45º F ∴ Time required = × π v r 2 4 3 = 4.71 × 10–8 sec.
4. A long solenoid of radius 'a' and number of turns per unit length n is enclosed by cylindrical shell of radius R, thickness d(d<<R) and length L. A variable current i = i0 sin ωt flows through the coil.
If the resitivity of the material of cylindrical shell is ρ, find the induced current in the shell. [IIT-2005] Sol. The magnetic field in the solenoid is given by
B = µ0ni
a
L
d R
⇒ B = µ0 n i0 sin ωt [Q i = i0 sin ωt given]
The magnetic flux linked with the solenoid φ = →B .→A = BA cos 90º
= (µ0 n i0 sin ωt) (πa2)
∴ The rate of change of magnetic flux through the solenoid
dt dφ
= π µ0 n a2 i0 ω cos ωt
The same rate of change of flux is linked with the cylindrical shell. By the principle of electromagetic induction, the induced emf produced in the cylinderical shell is I TOP VIEW e = – dt dφ = – πµ0 n a2 i0 ω cos ωt ...(i)
The resistance offered by the cylindrical shell to the flow of induced current I will be
R = ρ A l Here, l = 2πR, A = L × d ∴ R = ρ Ld R 2π ...(ii) The induced current I will be
I = R | e | = R 2 Ld ] t cos i na µ [ 0 2 0 π × ρ × ω ω π I = R 2 Ld t cos i na µ0 2 0 ρ × ω ω π Ans.
5. A particle of charge equal to that of an electron, –e, and mass 208 times the mass of the electron (called a nu-meson) moves in a circular orbit around a nucleus of charge + 3e. (Take the mass of the nucleus to be infinite). Assuming that the bohr model of the atom is applicable to this system.
(i) Derive an expression for the radius of the nth Bohr orbit.
(ii) Find the value of n for which the radius of the orbit is approximately the same as that of the first Bohr orbit for the hydrogen atom.
(iii) Find the wavelength of the radiation emitted when the mu-meson jumps from the third orbit of the first orbit. [IIT-1988]
Sol. (i) Let m be the mass of electron. Then the mass of mu-meson is 208 m. According to Bohr's postualte, the angular momentum of mu-meson should be an integral multiple of h/2π.
+3e e r ∴ (208 m) vr = π 2 nh ∴ v = mr nh 208 2π× = mr nh π 416 …(1)
Since mu-meson is moving in a circular path therefore it needs centripetal force which is provided by the electrostatic force between the nucleus and mu-meson.
∴ r v ) m 208 ( 2 = 0 4 1 πε × 2 3 r e e× ∴ r = 2 0 2 mv 208 4 e 3 × πε
Substituting the value of v from (1) we get
r = 2 2 0 2 208 4 416 416 3 h n mr mr e × πε π × π × ⇒ r = 2 2 02 624 me h n π ε …(2)
(ii) The radius of the first orbit of the hydrogen atom = 2 2 0 me h π ε …(3)
To find the value of n for which the radius of
the orbit is approximately the same as that of the first Bohr orbit for hydrogen atom, we equate equation (2) and (3) 2 0 2 2 624 me h n π ε = 2 2 0 me h π ε ⇒ n = 624 ≈ 25 (iii) λ 1 = 208 R × z2 2 2 2 1 1 – 1 n n ⇒ λ 1 = 208 × 1.097 × 107 × 32 2 2 3 1 – 1 1 ⇒ λ = 5.478 × 10–11 m
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, measured 2.24 L. The water collected during cooling weighed 0.9 g. The vapour pressure of pure water at 20 ºC is 17.5 mm Hg and is lowered by 0.104 mm Hg when 50 g of the organic compound is dissolved in 1000 g of water. Give the molecular formula of the organic compound. [IIT-1983] Sol. The combustion reaction is
CxH2yOy + x O2 → x CO2 + y H2O
To start with, the amount of O2 taken is 2x. Hence,
after the combustion reaction, we will be left with the following amounts.
Amount of oxygen left unreacted = x Amount of carbon dioxide = x Amount of water = y
When this mixture is cooled to 0 ºC and 1 atm, we will be left with oxygen and carbon dioxide. Hence, the amount 2x occupies the given volume of 2.24 L at STP. Hence, Amount x = 1 Lmol 4 . 22 L ) 2 / 24 . 2 ( − = 0.05 mol Now, Mass of water collected = 0.9 g
Amount of water collected,
y = 1 mol g 18 g 9 . 0 − = 0.05 mol
Thus, the empirical formula of the compound is C0.05H2 × 0.05 O0.05, i.e. CH2O.
Now, according to Raoult's law – * p p ∆ = x2 i.e. mmHg 5 . 17 mmHg 104 . 0 = ) mol g 18 / g 1000 ( ) M / g 50 ( ) M / g 50 ( 1 − +
Solving for M, we get M = 150.5 g mol–1
Number of repeating units of CH2O in the
molecular formula = 16 2 12 5 . 150 + + ≈ 5
Hence, Molecular formula of the compound is C5H10O5.
7. The oxides of sodium and potassium contained in a 0.5 g sample of feldspar were converted to the respective chlorides. The weight of the chlorides thus obtained was 0.1180 g. Subsequent treatment of the chlorides with silver nitrate gave 0.2451 g of silver chloride. What is the percentage of Na2O and
Sol. Mass of sample of feldspar containing Na2O and
K2O = 0.5 g.
According to the question,
Na2O + 2HCl → 2NaCl + H2O ..(1)
2 × 23 + 16 = 62g 2(23 + 35.5) = 117 g K2O + 2HCl → 2KCl + H2O ...(2)
2 × 39 + 16 = 94g 2(39 + 35.5) = 149 g Mass of chlorides = 0.1180 g
Let, Mass of NaCl = x g
∴ Mass of KCl = (0.1180 – x)g Again, on reaction with silver nitrate,
NaCl + AgNO3 → AgCl + NaNO3 ...(3)
23 + 35.5 = 58.5g 108 + 35.5 = 143.5g
KCl + AgNO3 → AgCl + KNO3 ...(4)
39 + 35.5 = 74.5g 108 + 35.5 = 143.5g Total mass of AgCl obtained = 0.2451 g Step 1. From eq. (3)
58.5 g of NaCl yields = 143.5 g AgCl ∴ x g of NaCl yields = 5 . 58 5 . 143 x g AgCl
And from eq. (4),
74.5 g of KCl yields = 143.5 g of AgCl ∴ (0.1180 – x)g of KCl yields = 5 . 74 5 . 143 (0.1180 – x)g AgCl
Total mass of AgCl
5 . 58 5 . 143 x + 5 . 74 5 . 143 (0.1180 – x) = 0.2451 which gives, x = 0.0342
Hence, Mass of NaCl = x = 0.0342 g
And Mass of KCl = 0.1180 – 0.0342 = 0.0838g Step 2. From eq.(1),
117 g of NaCl is obtained from = 62 g Na2O
∴ 0.0342 g NaCl is obtained from
= 117 62 × 0.032 = 0.018 g Na2O From eq. (2), 149 g of KCl is obtained from = 94 g K2O ∴ 0.0838 g of KCl is obtained from = 149 94 × 0.0838 = 0.053 g K2O Step 3. % of Na2O in feldspar = 5 . 0 018 . 0 × 100 = 3.6% % of K2O in feldspar = 5 . 0 053 . 0 × 100 = 10.6 % 8. 5 ml of 8 N of nitric acid, 4.8 ml of 5 N hydrochloric acid and a certain volume of 17 M sulphuric acid are mixed together and made upto 2 litre. 30 ml of this mixture exactly neutralizes 42.9 ml of sodium carbonate solution containing one gram of Na2CO3.10H2O in 100 ml of water.
Calculate the amount in grams of the sulphate ions in solution. [IIT-1985] Sol. Given that,
3 HNO N = 8 N 3 HNO V = 5 ml NHCl = 5 N VHCl = 4.8 ml 4 2SO H M = 17 M
Step 1. Meq. of HNO3 in 2L solution
= NHNO3× VHNO3 = 8 × 5 = 40 ∴ Meq. of HNO3 in 30 ml solution
=
2000
40 × 30 = 0.6 Step 2. Meq. of HCl in 2L solution
= NHCl × VHCl = 5 × 4.8 = 24 ∴ Meq. of HCl in 30 ml solution = 2000 24 × 30 = 0.36 Step 3. Meq. of H2SO4 in 2L solution
= Valency factor × MH2SO4 × VH2SO4 = 2 × 17 × VH2SO4 ∴ Meq. of H2SO4 in 30 ml solution = 2000 30 V 34× H2SO4× = 0.51VH2SO4 Step 4. Also given that,
Volume of Na2CO3.10H2O = 100 ml Mass of Na2CO3.10H2O = 1 g Equivalent mass of Na2CO3.10H2O = 2 mass Molecular = 2 286 = 143 g equiv–1 We know, Normality = (ml) Volume solute of mass Equivalent 1000 solute of Mass × × = 100 143 1000 1 × × = 0.070 N Meq. of Na2CO3.10H2O = NNa2CO3.10H2O × VNa2CO3.10H2O = 0.070 × 42.9 = 3.003
Step 5. 1 gram equivalent acid neutralizes 1 gram equivalent of base. ∴ 0.6 + 0.36 + 0.51VH2SO4 = 3.003 Solving , VH2SO4 = 4 ml Step 6. 1000 ml of 1 M H2SO4 contains = 96 g SO42–ions ∴ 4ml of 17 M H2SO4 contains = 1000 4 17 96× × = 6.528g SO42– ions
9. The equilibrium constant Kp of the reaction
2SO2(g) + O2(g) 2SO3(g) is 900 atm–1 at
800 K. A mixture containing SO3 and O2 having
initial partial pressures of 1 atm and 2 atm, respectively, is heated at constant volume to equilibrate. Calculate the pressure of each gas at 800 K. [IIT- 1989]
Sol. Since to start with SO2 is not present, it is expected
that some of SO3 will decompose to give SO2 and
O2 at equilibrium. If 2x is the partial pressure of
SO3 that is decreased at equilibrium, we would
have 2SO2(g) + O2(g) 2SO3(g) t = 0 0 2 atm 1 atm teq 2x 2 atm + x 1 atm – 2x Hence, Kp = ) p ( ) p ( ) p ( 2 2 3 O 2 SO 2 SO = ) x atm 2 ( ) x 2 ( ) x 2 atm 1 ( 2 2 + − = 900 atm–1
Assuming x << 2 atm, we get
) atm 2 ( ) x 2 ( ) x 2 atm 1 ( 2 2 − = 900 atm–1 or 2 2 ) x 2 ( ) x 2 atm 1 ( − = 1800 or x 2 atm 1 – 1 = 42.43 or x = 43 . 43 2 1 × atm = 0.0115 atm Hence, p(SO2) = 2x = 0.023 atm;
p(O2) = 2atm + x = 2.0115 atm and
p(SO3) = 1 atm – 2x = 0.977 atm
10. A white substance (A) reacts with dil. H2SO4 to
produce a colourless gas (B) and a colourless solution (C). The reaction between (B) and acidified K2Cr2O7 solution produces a green
solution and a slightly coloured precipitate (D). The substance (D) burns in air to produce a gas (E) which reacts with (B) to yield (D) and a colourless liquid. Anhydrous copper sulphate is turned blue on addition of this colourless liquid. Addition of aqueous NH3 or NaOH to (C) produces first a
precipitate, which dissolves in the excess of the respective reagent to produce a clear solution in each case. Identify (A), (B), (C), (D) and (E). Write the equations of the reactions involved. [IIT-2001] Sol. (A) is ZnS, ) A ( ZnS+ H2SO4 → ) C ( 4 ZnSO + ) B (2 S H ) B ( 2 S H 3 + K2Cr2O7 + 4H2SO4 → K2SO4 + Cr2(SO4)3 + 7H2O + ) D ( grey White3S ) D (S + OAir2 → SO(E)2 ) E ( 2 SO + ) B ( 2 S H 2 → ) C ( liquid Colourless2 O H 2 + ) (DS 3 5H2O + White4 CuSO → Blue4 2 O H 5 . CuSO
ZnSO4 + 2NaOH → Zn(OH)2 + Na2SO4
Zn(OH)2 + 2NaOH → Na2ZnO2 (soluble) + 2H2O
Also in excess of NH4OH it forms soluble
complex [Zn(NH3)4](OH)2.
MATHEMATICS
11. If f (x) is twice differentiable function such that f (a) = 0, f (b) = 2, f (c) = 1, f (d) = 2, f (e) = 0,
where a < b < c < d < e, then the minimum number
of zero's of g(x) = {f ' (x)}2 + f '' (x). f (x) in the
interval [a, e] is? [IIT-2006] Sol. Let, g(x) =
dx d
[f (x) . f '(x)]
to get the zero of g(x) we take function h(x) = f (x). f ' (x)
between any two roots of h(x) there lies at least
one root of h' (x) = 0
⇒ g(x) = 0 ⇒ h(x) = 0 ⇒ f (x) = 0 or f ' (x) = 0 If f (x) = 0 has 4 minimum solutions. f ' (x) = 0 has 3 minimum solutions. h(x) = 0 has 7 minimum solutions.
⇒ h' (x) = g(x) = 0 has 6 minimum solutions 12. The position vectors of the vertices A, B and C of a
tetrahedron ABCD are ^i + j^ + ^k, i^ and 3^i, respectively. The altitude from vertex D to the
opposite face ABC meets the median line through A of the triangle ABC at a point E. If the length of
the side AD is 4 and the volume of the tetrahedron
is 3
2
2 , find the position vector of the point E for all its possible positions. [IIT-1996]
Sol. F is mid-point of BC i.e., F ≡ 2 3^ ^ i i+ = 2^i and AE ⊥ DE (given) A(i + j + k) D C(3i) F(2i) B(i) E 1 λ
Let E divides AF in λ : 1. The position vector of E is given by 1 ) ( 1 2 ^ ^ ^ ^ + λ + + + λi i j k = ^ 1 1 2 i + λ + λ + ^ 1 1 j + λ + ^ 1 1 k + λ Now, volume of the tetrahedron
= 3 1
(area of the base) (height)
⇒ 3 2 2 = 3
1 (area of the ∆ABC) (DE)
But area of the ∆ABC = |( )| 2 1 → → × BA BC = |2 ( )| 2 1 ^ ^ ^ k j i× + = |i^×j^+^i×^k)| = |^k–j^)| = 2 Therefore, 3 2 2 = 3 1 ) 2 ( (DE) ⇒ DE = 2
Since ∆ADE is a right angle triangle,
AD2 = AE2 + DE2 ⇒ (4)2 = AE2 + (2)2 ⇒ AE2 = 12 But AE→ = ^ 1 1 2 i + λ + λ + ^ 1 1 j + λ + ^ 1 1 k + λ –( ) ^ ^ ^ k j i+ + = ^ 1i + λ λ – ^ 1j + λ λ – ^ 1k + λ λ ⇒ |AE = → |2 2 ) 1 ( 1 + λ [λ 2 + λ2 + λ2] = 2 2 ) 1 ( 3 + λ λ Therefore, 12 = 2 2 ) 1 ( 3 + λ λ ⇒ 4(λ + 1)2 = λ2 ⇒ 4λ2 + 4 + 8λ = λ2 ⇒ 3λ2 + 8λ + 4 = 0 ⇒ 3λ2 + 6λ + 2λ + 4 = 0 ⇒ 3λ(λ + 2) + 2(λ + 2) = 0 ⇒ (3λ + 2) (λ + 2) = 0 ⇒ λ = – 2/3, λ = – 2
Therefore, when λ = – 2/3, position vector of E is given by ^ 1 1 2 i + λ + λ + ^ 1 1 j + λ + ^ 1 1 k + λ = ^ 1 3 / 2 – 1 ) 3 / 2 .(– 2 i + + + ^ 1 3 / 2 – 1 j + + ^ 1 3 / 2 – 1 k + = ^ 3 3 2 – 1 3 / 4 – i + + + ^ 3 3 2 – 1 j + + ^ 3 3 2 – 1 k + = ^ 3 / 1 3 3 4 – i + + ^ 3 / 1 1 j + ^ 3 / 1 1 k = –i^ + 3j^+ 3k^ and when λ = – 2
Position vector of E is given by,
^ 1 2 – 1 ) 2 (– 2 i + + × + ^ 1 2 – 1 j + + ^ 1 2 – 1 k + = ^ 1 – 1 4 – i + – ^j –k ^ = 3 i^ – j – ^ ^k Therefore, –^i + 3 j + ^ 3k^ and +3 i^ – j – ^ ^k are
the position vector of E.
13. Evaluate
∫
π π − π + − + π 3 / 3 / 3 3 | | cos 2 4 x x dx [IIT-2004] Sol. Let, I =∫
π π − +π − π 3 / 3 / 3 | | cos 2 x dx + 4∫
−ππ +π − 3 / 3 / 3 3 | | cos 2 x dx x Using∫
− a a f(x)dx= = − − = −∫
( ) , ( ) ( ) 2 ) ( ) ( , 0 0 f xdx f x f x x f x f a ∴ I = 2∫
π +π − π 3 / 0 3 | | cos 2 x dx + 0 +π −∫
−ππ 3 / 3 / 3 3 | | cos 2 odd is x dx x as I = 2π∫
π π + − 3 / 0 2 cos(x /3) dx = 2π∫
π π − 3 / 2 3 / 2 cost dt , where x + 3 π = t= 2π
∫
π π + 3 / 2 3 / 2 2 2 tan 3 1 2 sec t dt t = 2π∫
+ 3 3 / 1 1 3 2 2 u du = 3 4π .{
3tan−1 3u}
1/33 = 3 4π (tan–1 3 – tan–11) = 3 4π tan–1 2 1 ∴∫
π π − +π − + π 3 / 3 / 3 3 | | cos 2 4 x x dx = 3 4π tan–1 2 1 .14. An unbiased die, with faces numbered 1, 2, 3, 4, 5, 6, is thrown n times and the list on n numbers showing up is noted. What is the probability that among the numbers 1, 2, 3, 4, 5, 6 only three numbers appear in this list ? [IIT-2001] Sol. Let us define at onto function F from A : [r1, r2 ...
rn] to B : [1, 2, 3] where r1r2 .... rn are the readings
of n throws and 1, 2, 3 are the numbers that appear in the n throws.
Number of such functions,
M = N – [n(1) – n(2) + n(3)] where N = total number of functions and
n(t) = number of function having exactly t elements
in the range.
Now, N = 3n, n(1) = 3.2n, n(2) = 3, n(3) = 0
⇒ M = 3n – 3.2n + 3
Hence the total number of favourable cases = (3n – 3.2n + 3). 6C 3 ⇒ Required probability = n n n C 6 ) 3 2 . 3 3 ( − + ×6 3
15. Show that the value of tanx/tan3x, wherever defined never lies between 1/3 and 3. [IIT-1992] Sol. y = x x 3 tan tan = x x x x 2 3 tan 3 – 1 tan – tan 3 tan = x x x x 3 2 tan – tan 3 ) tan 3 – 1 ( tan = x x 2 2 tan – 3 tan 3 – 1 [Q tan 3x ≠ 0 ⇒ 3x ≠ 0] ⇒ x ≠ 0 ⇒ tan x ≠ 0 0 ∞ + – + 3 1/3 Let tan x = t ⇒ y = 22 – 3 3 – 1 t t ⇒ 3y – t2y = 1 – 3t2 ⇒ 3y – 1 = t2y – 3t2 ⇒ 3y – 1 = t2 (y – 3) ⇒ 3 – 1 – 3 y y = t2 ⇒ 3 – 1 – 3 y y ≥ 0, t2 ≥ 0 ∀ t ∈ R ⇒ y ∈ (– ∞, 1/3) ∪ (3, ∞)
Therefore, y is not defined in between (1/3, 3).
Physics Facts
Nuclear Physics
1. Alpha particles are the same as helium nuclei and have the symbol . 1. The atomic number is equal to the number of protons (2 for alpha) 2. Deuterium ( ) is an isotope of hydrogen ( )
3. The number of nucleons is equal to protons + neutrons (4 for alpha)
4. Only charged particles can be accelerated in a particle accelerator such as a cyclotron or Van Der Graaf generator.
5. Natural radiation is alpha ( ), beta ( ) and gamma (high energy x-rays) 6. A loss of a beta particle results in an increase in atomic number.
7. All nuclei weigh less than their parts. This mass defect is converted into binding energy. (E=mc2)
8. Isotopes have different neutron numbers and atomic masses but the same number of protons (atomic numbers).
1. A particle of charge Q and of negligible initial speed is accelerated through a potential difference of U. The particle reaches a region of uniform magnetic field of induction B, where it undergoes circular motion. If potential difference is doubled & B is also doubled then magnetic moment of the circular current due to circular motion of charge Q will become
(A) double (B) half (C) four times (D) remain same
2. Three long rod AA, AB, CC are moving with a speed v in a uniform magnetic field B perpendicular to the plane of paper as shown in figure. The triangle formed between the three wires is always an equilateral triangle. If resistance per unit length of wire is λ, then the induced current in the triangle is
A B C A C B v v B0 × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × B0 v (A) λ 3 v B0 (B) λ 3 v B 2 0 (C) λ 3 v B0 (D) λ v B0
3. In the diagram shown, the wires P1Q1 and P2Q2 are
made to slide on the rails with same speed of 5m/s. In this region a magnetic field of 1T exists. The electric current in 9Ω resistor is
P1 P2 Q2 Q1 2Ω 2Ω 9Ω 4cm × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × ×
(A) zero if both wires slide towards left
(B) zero if both wires slide in opposite direction (C) 0.2mA if both wires move towards left
(D) 0.2mA if both wires move in opposite direction 4. A circular current loop is shown in the adjacent
figure. The magnetic field in the region is along x-axis and its magnitude in the space is increasing with increasing y-coordinate. The net magnetic force on the loop is
y
B
x
(A) along + z-axis (B) along –z-axis (C) along +y-axis (D) None of these
5. A point charge q moves from A to B along a parabolic path (AB is latus rectum). Electric field is along x-axis. The work done by the field is
y2 = 4ax
x E
A B
y
(A) – qEa (B) –2qEa (C) +qeA (D) 2qEa
6. In a cylindrical zone of radius R, magnetic field →B is present perpendicular to the plane of the paper into it. Magnitude of →B is varying with time as B0(p + qt)
where p & q are positive constant. Consider a static triangle loop OAM. Emf induced in the triangular loop 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
Solutions will be published in next issue
A M O B R 60º 60º × × × × × × × × × × × × × × × × × × × × × × (A) 3 qR B 2 0π 2 (B) 3 qR B0 2 (C) 3 PR B0 2 (D) 2 R PB0 2
7. Consider cylindrical region of the magnetic field shown in the figure. Region I and II have fields directed perpendicularly outward and inward respectively. Fields are varying with time as
Region I : B = 3B0t
Region II : B = B0t
There is no net induced electric field in the region outside the magnetic field then the ratio of =
2 1 2 r r Region II × × × × × × × × × × × × × × × × Region I r1 r2 8. Match the column:
Column – I Column – II
(A) In a series RLC (P) Z may increase circuit if C decreases
(B) In a series RLC (Q) Z may decrease circuit if L increases
(C) In a series RLC (R) resonance
circuit if R decreases frequency increases (D) In a series RLC (S) power factor circuit if C increases decrease
• After travelling 2.4 billion miles in just over 6 years to reach Jupiter, Galileo missed its target at the Jovian moon Io by only 67 miles. That's like shooting an arrow from Los Angeles at a bull's-eye in New York and missing by only 6 inches!
• Utopia ia a large, smooth lying area of Mars. • The biggest star has a diameter of 1800 million
miles, making it 2000 times bigger than the Sun. • 15% of the world's fresh water flows doen the
Amazon.
• In 1995, each American used an annual average of 731 pounds of paper, more than double the amount used in the 1980's. Contrary to predictions that computers would displace paper, consumption is growing.
• The term 'black hole' was coined in 1968 when John Wheeler described how an in-falling object 'becomes dimmer millisecond by millisecond...light and particles incident from outside...go down the black hole only to add to its mass and increase its gravitational attraction.' • The 'Red Planet' isn't really red at all, Nasa
photographs indicate that it is more of a tan or butterscotch colour.
• The International Space Station orbits at 248 miles above the Earth.
• The axis of orbit of the planet Uranus is tilted at a 90 degree angle.
• Astronomers have discovered over 10,000 asteroids - but put them together and they would be smaller than the Moon.
• Have you ever seen a ring around the moon? Folklore has it that this means bad weather is coming.
1. For the process A→B V M 1 RT 2 3 1 U ⇒ ∝ ρ ∝ or T ∝ V
So AB is isobaric process (pressure in constant) = = 0 0 A A 2RT 3 U 4 As R 3 U 8 T 0 A 2 M V ρ = so M 3 U 16 PA = 0ρ0 and 0 B 4 M V ρ = so WAB = P[VB - VA] 0 0 0 0 0 0 0 4 M M 3 U 16 2 M 4 M M 3 U 16 ρ × ρ − = ρ − ρ ρ ⇒ Option [D] is correct 2. The slope of the graph is
0 0 V 2 P m=− The equation of this graph is given by
2 P 3 V 2 V P P 0 0 0 + − = 2 P 3 V 2 V P V nRT 0 0 0 + − = ⇒ V nR 2 P 3 V nRV 2 V P T 2 0 0 0 + − − = ⇒ 0 nR 2 P 3 nRV 2 V P 2 dV dT 0 0 0 0 + = − = 2 V 3 V nR 2 P 2 nRV V P 0 0 0 0 = ⇒ = ⇒ At , 2 V 3
V= 0 temp will be maximum [n = 1]
2 V 3 R 2 P 3 4 V 9 RV 2 P T 02 0 0 0 0 max =− × + × R V P 4 9 R V P 8 9 0 0 + 0 0 − ⇒ R V P 8 9 0 0 ⇒ Option [C] is correct 3. TV-3 = k V k nR PV 3 = ⇒ −
for this polytrophic process, PV-2 = constant
so x = -2 0 1 2 1 2 nRT 3 2 3 ) T T ( nR x 1 ) T T ( nR w = − = − − = ∆U = nCV∆T = n 3/2 R(3T0 – T0) = 3nRT0 0 0 113 nRT nRT 3 3 2 W U Q = + = = ∆ = ∆ ∆Q = nC∆T R 6 11 ) T 2 ( n nRT 3 11 T n Q C 0 0 = = ∆ ∆ = ⇒
Option [A,C,D] is correct
4. y R F Fx Fy ωt x F = mω2R Fy = F sin ωt Fy = mω2R sin (ωt) * In figure as cos ωt = R x ⇒ x=Rcosωt * For one complete circle, θ=ωt
* y R vx ωt x vx = – ωR sin ωt v = ωR Option [A,C] is correct 5. From 1st law, ∆Q = ∆U + W nCαT + nCvαT + pαV ⇒ nCαT = nCVαT + nRT/V αV V e T V nRT V e T C V e T n V 0 V 0 V V 0 α = α α + α α ⇒ α α α ⇒ αC = αCV + R/V ⇒ C = CV + R/αV Option [B] is correct 6. Option [C] is correct 7. Option [B] is correct 8. Option [D] is correct
Solution
Physics Challenging Problems
Set # 9
1. AB and CD are two ideal springs having force constant K1 and K2 respectively. Lower ends of these springs are
attached to the ground so that the springs remain vertical. A light rod or length 3a is attached with upper ends B and C of springs. A particle of mass m is fixed with the rod at a distance a from end B and in equilibrium, the rod is horizontal. Calculate period of small vertical oscillations of the system.
B a m 2a C
K2
K1
A D
Sol. Let, in equilibrium, compressive forces in left and right springs be F1 and F2, respectively.
Considering free body diagram of rod BC, (figure)
B C F1 a mg 2a F2 F1 + F2 = mg ...(i)
Taking moments about B, mg × a = F2 × 3a
or F2 =
3 1mg
Substituting this value in equation (i), F1 =
3 2mg
Let the particle be pressed from its equilibrium position by applying a force 'F'. Let left and right springs be further compressed through y1 and y2,
respectively. Increase in compressive forces in the spring will be K1y1 and K2y2 respectively. In other
words, total compressive forces in two springs will be (F1 + K1y1) and (F2 + K2y2) respectively.
Now considering new free body diagram (figure) of the rod BC,
(F1 + K1y1) + (F2 + K2y2) = F + mg ...(ii)
Taking moments about B, (figure)
B C (F1 + K1y1) a mg 2a (F2 + K2y2) F (F + mg) × a = (F2 + K2y2) × 3a ...(iii)
Solving above equations, K1y1 = 3 F 2 and K2y2 = 3 F . or y1 = 1 K 3 F 2 and y 2 = 2 K 3 F
Since, distance of the particle from left spring is 'a' and that from right spring is '2a', therefore, downward displacement of the particle from equilibrium position will be
y = 3 y 2 1 + 3 y2 = l K 9 F 4 + 2 K 9 F = 2 1 2 1 K K 9 F ) K 4 K ( + or F = ) K 4 K ( K K 9 2 1 2 1 + ...(iv) Now, if the particle be released, it starts accelerating upwards due to excess compressive force in springs. Hence, the restoring force is (K1y1 + K2y2), which is
numerically equal to F. or Restoring force = F = ) K 4 K ( K K 9 2 1 2 1 + . y or Restoring acceleration, a = m F = ) K 4 K ( m K K 9 2 1 2 1 + .y
Since, acceleration is restoring and is directly proportional to displacement y, therefore, the particle performs SHM. Its period of oscillations is
T = a y 2π or T = 2 1 2 1 K K ) K 4 K ( m 3 2π + Ans.
Expert’s Solution for Question asked by IIT-JEE Aspirants
Students Forum
2. A plane spiral coil is made on a thin insulated wire and has N = 100 turns. Radii of inside and outside turns are
a = 10 cm and b = 20 cm respectively. A magnetic field
normal to the plane of spiral exists in the space. The magnetic field increases at a constant rate α = 0.3 tesla/second. Calculate potential difference between the ends of the spiral.
Sol. Since, magnetic field is increasing, therefore flux linked with the coil is also increasing. Due to increase in flux an emf is induced in it. Potential difference between ends of the spiral coil is equal to magnitude of emf induced in it.
Since there are N turns in a radial width (b – a), therefore number of turns in the spiral coil per unit radial width is n = ) a – b ( N
Consider a concentric circular ring of radius x and radial thickness dx
Number of turns in this ring = n dx.
Let at some instant magnetic field strength be B, Flux linked with the ring, φ = πx2B
Emf induced in this ring = (n . dx) dt dφ = (n . dx) π dt dB x2 = πnα x2dx
∴ Total emf induced in the spiral coil,
e =
∫
= = α π b x a x 2dx) x n ( = 3 1 πn α (b3 – a3) Substituting value of n, e = 31 πN α(a2 + b2 + ab) = 2.2 volt Ans.
3. A spherical shell of radius R is filled with water. Temperature of atmosphere is (– θ) ºC. The shell is exposed to atmosphere and all water comes down to 0ºC and then it starts freezing from outer surface towards the centre of the shell. Assuming shell to be highly conducting, calculate time for whole mass of water to freeze. Thermal conductivity of ice is K and latent heat of its fusion is L. Density of water is ρ. Neglect expansion during freezing.
Sol. Heat flows from water to atmosphere because atmospheric temperature is below 0ºC. Water is filled in spherical shell, therefore heat flows in radial direction.
Let at some instant t, thickness of ice layer be x. Then a concentric sphere of radius (R – x) is in liquid form as shown in figure. Heat flows from this sphere to atmosphere through ice layer.
To calculate rate of heat flow, first we have to calculate thermal resistance of this ice layer.
Considering a concentric spherical shell of radius r [(R – x) < r < R] and radial thickness dr as shown in figure.
r
R
(R–x)
Its thermal resistance = K r 4 dr 2 π ∴ Total thermal resistance of ice layer
∫
π R ) x – R ( 2 Kr 4 dr = ) x – R ( KR 4 x πTemperature just inside and outside the ice layer is 0ºC and (– θ)ºC, respectively.
∴ Temperature difference = 0 – (– q) = θº C Hence, rate of heat flow, H =
π θ ) x – R ( KR 4 x = x ) x – R ( KR 4π θ
rate of freezing of mass = L H
∴ Rate of freezing of volume = ρ L H But it is equal to 4π(R – x)2 . dt dx ∴ x L ) x – R ( KR 4 ρ θ π = 4p (R – x)2 . dt dx or θ ρ KR ) x – R ( x L . dx = dt
At instant t = 0, thickness of ice layer was equal to x = 0 and we have to calculate time t when whole mass has frozen or when x = R
Substituting these limits,
∫
θ∫
ρ = t 0 R 0 dx ) x – R ( x KR L dt or t = θ ρ K 6 LR2 Ans.4. In a Young's experiment, the upper slit is covered by a thin glass plate of refractive index µ1 = 1.4 while the
lower slit is covered by another glass plate having the same thickness as the first one but having refractive index µ2 = 1.7. Interference pattern is observed using
light of wavelength λ = 5400 Å. It is found that the point P on the screen where the central maxima fell before the glass plates ware inserted, now has 3/4 the original intensity. It is further observed that what used to be the fifth maxima earlier, lies below the point P while the sixth minima lies above the point P. Neglecting absorption of light by glass plates, calculate thickness of the glass plates.
Sol. If glass plates are not inserted in slits, central maximum occurs at perpendicular bisector of slits. Hence, point P lies on perpendicular bisector of S1S2
as shown in figure.
S1
S2
P Screen
When a plate is inserted in upper slit, the interference pattern shifts upwards and due to plate inserted in lower slit, it shifts downwards.
Since, distance of nth maxima (nth bright fringe) from central bright fringe is nω, where ω is fringe width and fifth maxima now lies below the point P, therefore, shift of fringe pattern is downwards and it is greater that 5 ω.
Since, sixth minima (sixth dark fringe) lies above point P and distance of mth minima from central
bright fringe is 2 1 –
m ω, therefore, shift of fringe pattern is less than 5.5 ω. Hence, the shift lies between 5 ω and 5.5 ω.
Let intensity of light due to each slit be I0. Then
l1 = l2 = l0.
Since, before insertion of glass plates, a bright fringe (central bright fringe) was formed at P, therefore, original intensity at P was equal to Imax.
But Imax = ( I1 + I2 )2 = 4I0
But now intensity at P is 3/4 of the original intensity, therefore now intensity at P is
I = 3/4 Imax = 3I0.
But I = I1 + I2 + 2 I1I2 cos φ where φ is phase difference between two rays reaching P.
Substituting I1 = I2 = I0 and I = 3I0 in above
equation, cos φ =
2
1 or φ = (2nπ + π/3) where n is an integer
But phase difference, φ = 2π ω s where s is shift of fringe pattern. Hence, s = + 6 1 n ω.
But s lies between 5ω and 5.5 ω, therefore,
s = + 6 1 5 ω = 6 31 ω where ω = d D λ ∴ s = d 6 D 31λ
where D is distance of screen from slits and d is distance between slits.
Let thickness of each glass plate be t.
∴ Upward shift due to upper glass plate, s1 = d tD ) 1 – (µ1
and downward shift due to lower glass plate, s2 = d tD ) 1 – (µ2
∴ Resultant downward shift s = s2 – s1.
Substituting values of s, s1 and s2,
t = ) – ( 6 31 1 2 µ µ λ = 9.3 µm Ans.
5. A point isotropic light source of power P = 12 watt is located on the axis of a circular mirror plate of radius R = 3 cm. If distance of source from the plate is a = 39 cm and reflection coefficient of mirror plate is α = 0.70, calculate force exerted by light rays on the plate. Sol. When light rays are incident on the mirror plate, a part of then is reflected and a part is absorbed by the plate. Therefore, momentum of light rays changes. Due to change in the momentum, a force is experienced by the plate. Magnitude of the force is equal to rate of change of momentum. To calculate rate of change of momentum, first we have to consider a circular ring coplanar and co-axial with the plate. Let the radius of that ring be x and radial thickness dx as shown figure
Source θ
a
Distance of every point of this ring from the source is
r = a2+x2 .
Hence, intensity of light incident on the ring is
I = 2
r 4
P π
Direction of incident rays is inclined at angle θ with normal to the plane of ring. Therefore, power incident on the ring, p = (2πxdx)I cos θ
or p = 2 r 2 dx x cos P θ
Since, incident power is in the form of light rays which are incident at angle θ with normal to the plate, therefore, net rate of incidence of momentum on the ring considered
= c cos P θ = c r 2 dx x cos P 2 2θ
Since, 70% of the rays are reflected and 30% are absorbed by the plate, therefore, rate of change of momentum from the ring considered
= c
1[(0.7 p cos θ) × 2 + (0.3 p cos θ)] But it is equal to force dF on the ring. ∴ dF = 0.85 2 2 2 2 ) x a ( c dx x Pa + or Total force on the plate,
F =
∫
= = + R x 0 x 2 2 2 2 ) x a ( dx x c Pa 85 . 0 = +R ) a ( 1 – a 1 c 2 Pa 85 . 0 2 2 2 2 = c ) R a ( 2 PR 85 . 0 2 2 2 + = 1 × 10 –10 N Ans.NUCLEAR REACTOR TYPES – ARE THEY REALLY SAFE?
The nuclear power industry along with the reactor technology has been constantly developing for more than five decades now. Nuclear reactors can be classified based on their nuclear reaction, the moderator material used, generation of the reactor, fuel phase, fuel type, coolant used, etc. The fission nuclear reactors are mostly dealt with because the fusion reactors are still in the developing stages and the fission reactors are already being used for the past six decades.
• Based on nuclear reaction
This type refers to the thermal (slow) reactors and the fast reactors based on the speed of neutrons. Thermal reactors are the most affordable and common as they use the natural and raw uranium; and the neutrons are decelerated from their natural speed when emitted from the broken atomic nuclei, and uses a moderating material in the process. The Fast reactors are very expensive that require more enriched fuel.
• Based on moderator material
Thermal reactors (because of the presence of the moderating material), and Graphite, Normal water and Heavy water are also used as moderators. The moderating materials in the Graphite and the Heavy water reactors thermalize the neutrons and keep the natural uranium intact without any enrichment.
• Based on generation
Generation I reactors were the first prototype reactors, Generation II used standard designs till 50s, Generation III were more modern, lightweighted, more efficient and were used till late 90s, the latest i.e. Generation IV reactors targeting on economical and minimal waste, are still in the research and development stage which may officially work until late 2020s.
• Based on fuel phase and fuel type
It is Solid, Liquid or Gas reactor where Solid is most typical. The fuel type reactors also come with fuel phase- uranium or thorium, which are available in abundant quantities on the land.
