Dear Students,
Find a mentor who can be your role model and your friend !
A mentor is someone you admire and under whom you can study. Throughout history, the mentor-protege relationship has proven quite fruituful. Socrates was one of the early mentors. Plato and Aristotle studied under him and later emerged as great philosophers in their own right.
Some basic rules to know mentors :
• The best mentors are successful people in their own field. Their behaviors are directly translatable to your life and will have more meaning to you. • Be suspicious of any mentors who seek to make you dependent on them.
It is better to have them teach you how to fish than to have them catch the fish for you. That way, you will remain in control.
• Turn your mentors into role models by examining their positive traits. Write down their virtues. without identifying to whom they belong. When you are with these mentors, look for even more behavior that reflect their success. Use these virtues as guidelines for achieving excellence in your field.
Be cautious while searching for a mentor :
• Select people to be your mentors who have the highest ethical standards and a genuine willingness to help others.
• Choose mentors who have and will share superb personal development habits with you and will encourage you to follow suit.
• Incorporate activities into your mentor relationship that will enable your mentor to introduce you to people of influence or helpfulness.
• Insist that your mentor be diligent about monitoring your progress with accountability functions.
• Encourage your mentor to make you an independent, competent, fully functioning, productive individual. (In other words, give them full permission to be brutally honest about what you need to change.)
Getting benefited from a role-mode :
Acquiring good habits from others will accelerate you towards achieving your goals. Ask yourself these questions to get the most out of your role model/mentors :
• What would they do in my situation?
• What do they do every day to encourage growth and to move closer to a goal ?
• How do they think in general ? in specific situations ?
• Do they have other facts of life in balance ? What effect does that have on their well-being ?
• How do their traits apply to me ?
• Which traits are worth working on first ? Later ?
A final word : Under the right circumstances mentors make excellent role models. The one-to-one setting is highly conducive to learning as well as to friendship. But the same cautions hold true here as for any role model. It is better to adapt their philosophies to your life than to adopt them.
Presenting forever positive ideas to your success. Yours truly
Pramod Maheshwari, B.Tech., IIT Delhi
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Editor : Pramod Maheshwari
No success is possible unless you believe that you can succeed.
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NEXT MONTHS ATTRACTIONS
Key Concepts & Problem Solving strategy for IIT-JEE. Know IIT-JEE With 15 Best Questions of IIT-JEE Challenging Problems in Physics, Chemistry & Maths Much more IIT-JEE News.
Xtra Edge Test Series for JEE – 2011 & 2012 AIEEE-2010 Examination Paper
S
Success Tips for the Months
• "All of us are born for a reason, but all of us don't discover why. Success in life has nothing to do with what you gain in life or accomplish for yourself. It's what you do for others."
• "Don't confuse fame with success. Madonna is one; Helen Keller is the other."
• "Success is not the result of spontaneous combustion. You must first set yourself on fire."
• "Success does not consist in never making mistakes but in never making the same one a second time."
• "A strong, positive self-image is the best possible preparation for success."
• "Failure is success if we learn from it." • "The first step toward success is taken
when you refuse to be a captive of the environment in which you first find yourself."
CONTENTS
INDEX PAGE
NEWS ARTICLE
4 IIT-Develops technology to produce stealth aircraftUrine-processing technologies yield rich cash flow potential
IITian ON THE PATH OF SUCCESS
8 Mr. Sujal PatelKNOW IIT-JEE
10 Previous IIT-JEE QuestionXTRAEDGE TEST SERIES
49 Class XII – IIT-JEE 2011 PaperClass XI – IIT-JEE 2012 Paper
IIT - 2010 Examination Paper with Solution
66
Regulars ...
DYNAMIC PHYSICS
17 8-Challenging Problems [Set# 1]Students’ Forum Physics Fundamentals Electrostatics-I
1- D Motion, Projectile Motion
CATALYSE CHEMISTRY
33 Key ConceptGaseous State & Real Gases General organic Chemistry
Understanding : Physical Chemistry
DICEY MATHS
41 Mathematical Challenges Students’ Forum Key Concept Complex Number Matrices & DeterminantsStudy Time...
IIT develops technology
to produce stealth
aircraft
Materials scientists at the Indian Institute of Technology in Roorkee (IIT-R) have developed microwave absorbing nanocom-posite coatings that could make aircraft almost invisible to radar. The technology for building invisible, or stealth aircraft, is a closely guarded secret of developed countries and a handful of laboratories in India are doing research in this area. Radars that emit pulses of microwave radiation identify flying aircraft by detecting the radiation reflected by the aircraft’s metallic body. The nanocomposite coatings developed by Rahul Sharma, R.C. Agarwala and Vijaya Agarwala at IIT-R absorb most of the incident radiation and reflect very little. Sharma, who revealed his team’s work at an international nanomaterials conference held recently at the Indian Institute of Science in Bangalore, believes their nano-product is a significant step in developing a technology to enable aircraft escape radar surveillance and protect its equipment from electronic “jamming”.Nanoparticles - so
called because of their very small size - are known to exhibit unique physical and chemical properties. The IIT team found that crystals of “barium hexaferrite” with particle size of 10-15 nanometres have the ability to absorb microwaves. (Human hair, for comparison, is 100,000 nanometres thick). They developed special processes for synthesizing the nanopowder and formulating it as a coating. Sharma said that the nanocomposite coating on the aluminium sheet absorbed 89 percent of incident microwaves at 15 giga hertz - the frequency normally used by radars — reflecting only 11 percent. A stealth aircraft should ideally absorb all the incident radiation and reflect nothing.
Urine-processing
technologies yield rich
cash flow potential
The stink is out of urine, literally and metaphorically, with a growing number of researchers spotting commercial and ecological value in a liquid most people consider waste.
The Indian Institute of Technology (IIT) Delhi, for instance, is working to harvest this human waste and
convert it into fertiliser. The Delhi government is willing to consider a revenue-share commercial venture selling the phosphates and nitrates in urine.
On the outskirts of Delhi, a little-known non-government organisa-tion, Fountain for Development Research and Action, is laying the ground for the first urine bank. It has diverted urine from two schools, where it has installed odour-free urinals, into a tank and transferred the run-off to a village nearby for use as fertiliser.
Director Madhab Nayak says the foundation is working towards making farmers aware of its potential as replacement for expensive urea.
"There is no such thing as waste," says Vijayaraghavan M Chariar, assistant professor at the Centre for Rural Development and Technology at IIT. "Urine consists of a lot of inorganic salts, which produce gases only when mixed with water. It is, in fact, pure fertiliser," he added.
IIT has come up with a cheap, odour-free, urinal which it has successfully tested on campus. The odour-free urinal combines technology with simple science to
translate into a significant water-saving initiative (urine smells only when mixed with water, which this technology eliminates).
Urine is collected through a tank placed underground, harvested and used as liquid fertiliser two to three metres below the ground on a five-acre field on campus, said Chariar, who can talk animatedly about this human waste and how its poor treatment alone has led to sanitation problems.
The public urinal at IIT uses a simple technology, called Zerodor, developed by Chariar, that fits into the waste coupler in the pan and diverts the urine through a drain where it is collected and harvested. The idea is not to allow it to mix with water at any stage.
Chariar has already transferred this technology to Good Yield Environmental Technologies, a Kolkata firm, and filed for a patent. Chariar claims that Zerodor is a low cost product and would need replacement in only about two years.
Meanwhile, the Delhi government, which has already installed 200 such odourless urinals in different parts of the city, uses a different and perhaps more expensive technology. Amiya Chandra, deputy commissioner of the city’s municipal corporation, says, "Other than problems of
vandalism, these urinals are working perfectly."
In preparation for the Commonwealth Games, the Delhi government is planning to install 1,000 such urinals at a nominal cost of Rs 3 lakh.
Chariar is already working on the second phase of his project, which was initiated by Unicef and Stockholm Environment Institute, for setting up a small reactor to extract nitrates and phosphates from urine. "This could become a micro-enterprise from the urinal," says Chariar.
The Delhi government is also looking at installing Chariar’s technology at a few parks in the city, while harvesting urine in those places.
Chariar has even designed similar urinals for women. "We have filed for trademark registration and we are in discussion with companies for marketing it," he says. With a little more investment, he says, a hydrophobic coating on pans could make it water resistant and completely drain the urine, leaving no room for any oxidisation, which can also cause odour.
In the developed world, communities have been quick to realise the huge economic potential of urine. "Communities in Germany are exporting urine to neighbouring countriesthat are using it on their farms, says
Chariar, explaining how it could be diverted for use as a nutrient by a simple plumbing.
The urine tank could deliver the liquid nutrient directly to plants about two to three metres below the soil, he says.
The Centre for Banana Research in Trichy is already using it for banana plantations and the University of agriculture Sciences, Bangalore, too is looking at its varied uses.
IIT student produces
electricity from waste
water
Kolkata: Waste water
management is a big issue world wide and specially in India where there is acute shortage of the precious resource in many places but a 23-year-old student of IIT Kharagpur claims he has found a solution.
Apart from finding solutions management of waste water he has also demonstrated producing electricity from it, which could go a long way in protecting the earth's resources.
A look at reservoirs used
for water supply
Manoj Mandelia, who is pursuing integrated MTech at IIT Kharagpur, there was no policy in the country which examined
waste as part of a cycle of production-consumption-ecovery.
"Waste management still constituted a linear system of collection and disposal which creates health and environmental hazards," he said.
"I developed a product which uses the concept of microbial fuel cell (MFC is a bio-electrochemical system that drives a current by mimicking bacterial interactions found in nature), which could not only treat waste water but also produce electricity in the process," explains Mandelia who heads a team of five people in the project.
The Water Diviner
The project, named LOCUS which stands for Localised Operation of Bio-cells Using Sewage, can achieve chemical oxygen demand (COD) reduction levels in waste water to about 60-80 per cent.
IIT Kharagpur Calls for
Nominations for the
Nina Saxena Excellence
in Technology Award
IIT Kharagpur announces the fourth edition of Nina Saxena Excellence in Technology Award to invite entries of this year in areas of technical innovations. Nominations for entries to the award are open until April 30, 2010. Entries can be submitted either by post to the Director, IIT
Kharagpur, West Bengal,PIN -721302, India or by email at [email protected]. The nomination form is also available on the official award website The objective behind the award is to commemorate the spirit and drive of Dr. Nina Saxena, who personified technical excellence. The fourth edition will continue the tradition of rewarding pioneering innovations for betterment of society. A distinguished committee is being formed by IIT Kharagpur to adjudge the nominations for this award. The award committee is chaired by Director of IIT Kharagpur and is comprised of Deans and selected faculty members of IIT Kharagpur and well known alumnus, based in India and in the US.
IIT’s New On-Campus
Wind Turbine to
Support Green Jobs,
Research, and Education
The consortium members will research the wind energy challenges identified in the U.S. Department of Energy's "20% Wind Energy by 2030" report, including wind technology, grid system integration, and workforce
challenges. The consortium's plan relies on IIT experts in electrical and computer engineering; mechanical, materials, and aerospace engineering; architecture; business; and members of the Wanger Institute for Sustainable Energy Research to tackle these challenges.
Many of the university's departments and research centers will also work together to offer wind energy courses addressing the technical, operational, social, and environmental aspects of wind energy in consultation with industry. To ensure student involvement in the project, fellowships will be offered annually to undergraduate and graduate students in wind energy engineering fields of study. Faculty and students from international university consortium members will also be invited to IIT to attend workshops and to share ideas with their American counterparts. The wind energy consortium will work with small wind turbine manufacturer Viryd Technologies to procure and install an 8KW Viryd wind turbine on IIT's Main Campus, and to deliver a second turbine to one of IIT's engineering laboratories to perform turbine reliability studies. The consortium will also work with wind energy developer Invenergy to install a 1.5MW GE wind turbine adjacent to a wind farm in Marseilles, Ill. The close proximity of IIT's
Marseilles turbine to an existing wind farm provides an ideal opportunity to study turbine-to-turbine wake interaction, wind farm interaction, and wind energy efficiencies in addition to turbine reliability studies.
Hyderabad boy tops in
GATE 2010
HYDERABAD : Malladi Harikrishna, a final year computer science engineering student from the city, has topped the national level Graduate Aptitude Test in Engineering (GATE-2010). He achieved the feat in his first attempt, scoring 99.99 percentile by scoring 83.55 per cent in GATE.
The results were announced on March 15 by IIT-Guwahati, which
conducted GATE this year. Mallad, who topped the national-level
Graduate Aptitude Test in Engineering (GATE-2010), said, “My aim is to join ME at the Indian Institute of Science, Bengaluru. I want to become a scientist,” Harikrishna, 21, said.
Students who clear GATE are eligible for admission to masters’ degree courses in engineering,
technology, architecture,
pharmacy, science in premier institutes like IITs and NITs. Many students from the state made it to the top-100. Srujana (JNTU, Kukatpally) ranked 22, Karthik Nagarjuna 44, Mufaquam Ali 51 and Pavan Kishore got the 102nd rank in ECE stream.
Srinivas Reddy got the 68th rank in EEE, and V. Suryanarayana 31 in Mechanical.
GATE 2010 score is valid for two years from the date of announcement of the results, according to the details published on GATE website
Terminated IIT students
seek intervention of
Prez,PM
Terminated for "bad performance", 38 students of IIT Kanpur have taken the issue to President Pratibha Patil and Prime Minister Manmohan Singh with a plea that the institute reconsider the decision.
The students have written to Prime Minister and the PMO has forwarded the matter to the HRD Ministry for "appropriate action". The Ministry has again forwarded it to the institute for action, a ministry official said.
The 38 students, including 24 from under-graduate and 14
post-graduate levels, were denied admission into fresh semester in January this year for "bad academic record"
.Prof V N Pal, who is an alumni of the institute, has taken the students' issue to President Patil, who is the Visitor of the institute. Prof Pal met the President on April one at Rashtrapati Bhawan and discussed the issue of termination of students of IIT Kanpur at length.
He explained the socio-economic condition of these students.
Vidya Balan to address
IIT
She has been invited by one of the Indian Institutes of Technology to be a keynote speaker at an upcoming seminar. In an interview to a leading daily, she confirmed her invitation from one of the IITs. She said, “I have been approached by one of the IITs for being a keynote speaker at a seminar they are holding. The topic is ‘The changing face of the Indian heroine’. I am excited about it, but am figuring out dates and prior commitments at the moment. But, I hope this comes through.”
A MA in sociology, Vidya Balan believes her education has groomed her and given her the confidence to address seminars at these prestigious universities.
To his competitors, Sujal Patel is now a name to reckon with. His company Isilon Systems, in the clustered storage space, has not only earned its position as the fastest growing technology company in North America, but in the five years of selling its products, Patel has transformed the company from zero sales, into a company with a $100 million run rate, $80 million in cash, and no debt.
“Slightly better is not a good term,” says Patel, the CEO of Isilon Systems and a pioneer in the clustered storage space. “For a technology to be adopted in an existing market, you really need to have a technology that is substantially better — 10 (times) better than what is in the marketplace today,” adds Patel. “It has to be so much better that it is overwhelming for people who buy that product and service.” When Patel says so, he is not being merely theoretical. The unsatiated desire to bring out the best led this 35 year old entrepreneur to steer his company successfully, in a market space, which was crowded by biggies like EMC and NetApp; not to mention the 250 odd startups in the storage space.
Patel founded Isilon at the age of 26. Prior to this, Patel spent his initial career days at Real Networks. As an engineer, Patel used to solve some of Real Networks’
most complex back-end operational challenges. That experience gave him the insight for a new type of solution, a type of virtualized storage optimized for media. His experience gave him the insight to a real customer need, and his deep technical knowledge gave him the ability to spot a solution.
I did not want to wait 10-15 years, treading cautiously at every step before taking the plunge. So when I got the chance to found Isilon, I jumped at the opportunity,” beams Patel.
That’s also a reason which led venture capitalists to take this 26 year old lad seriously. A few months after Patel founded Isilon in 2001, the NASDAQ came down crashing, bringing the ‘Dot Com Burst’. The venture capital market was in disarray. With the existing companies dropping their revenues, there was not much hope for new companies to find potential investment. To make matters worse, there were about 250 other startups in the storage space. Patel was undeterred. Sure about his ideas, he approached close to 40 venture capitalists, and with perseverance, eventually he managed to gain their confidence. Five months after the company’s beginning, Patel had managed to raise $8.4 million venture
M
R
.
S
UJAL
P
ATEL
Success Story
capital in what was one of the toughest markets to raise money.
But all was not rosy yet. The road ahead proved to be more challenging than raising the funds. The next three years were spent in building the products. Developing the company’s IP not only took a tremendous amount of money, but also ate in to the time to get into the market. By the time the product was ready to debut in the markets in 2003, the company’s debt was near $20 million.
Cruising over Obstacles
Not only confident and unshakable, Patel is also a man of clear vision, with no illusions about his capabilities. At 26, as the Founder of the company, he served as the CEO for the initial three years. But he knew that, if he wanted to make Isilon the next big thing in storage space, he needed someone who knew the dynamics of running a large organization. He promptly stepped down and appointed as a new CEO, who knew what it takes to grow into the larger league. “I floated the company but knew my limitations in the business front. We had three products ready to debut in the market and if we wanted then to succeed, we needed an expert who knew the right strings to pull,” says Patel. His focus and timely decisions were fruitful in the subsequent years when Isilon was on a dream run, literally growing at 200 percent year-on-year. By the time the company went public, it was a $60 million company.
“My goal was to see Isilon become a $100 million company by 2007 and become a player to be reckoned in the $4 billion global storage market for its technology. And we were still $40 million short. I knew that despite the economic instability, the company I had founded had great potential if one could maximize it,” says Patel. To start with, he re-structured the entire organization, replacing every executive in the management, including CTO, CFO, Head of engineering, Head of operations and others. Such an action is quite unheard of, especially immediately after a company had gone public.
Next, Patel began revamping the business strategy by reaching out to the broader enterprise segments. It was not easy. The segment he wanted to target comprised of Fortune 50 companies who did not have much expectation from a startup like Isilon. With the organizational re-structuring, he completely overhauled the company’s services and products to meet the expectations of the large enterprise customers. Eventually, more and more Fortune 50 companies began to adopt Isilon’s products.
Finally he decided to increase the company’s focus on R&D. Innovation has always been an essential part of his life. A very innovative and inquisitive person, Patel is known to get at least six ideas every day. Even as a kid he was known for asking around 500 questions about everything under the Sun. Thus, Patel diverted about 25 percent of the company’s revenue towards R&D and fostered innovation within the organization.
So what drives this confident and zealous man? It is the self-belief, passion and the creative way of looking at problems and coming up with solutions, says Patel. Even during his college days, while working on the Internet, he looked for opportunities to think about new ways, solve problems, or to bring innovative techniques/technologies to the market place.
To run a business successfully, it is also important that one should be honest with oneself, the team and the stakeholders. The foremost thing that Patel did after taking over the company was to communicate with customers and partners and update them about the happenings within the organization, reinstating their faith in him and the company. “I went and talked to each of our investors and customers, telling what we planned and how much earnings we expected through the quarters. I believe that apart from your technology offerings, one reason companies want to do business with you is the goodwill you develop in tough times,” says Patel.
PHYSICS
1. In Searle's experiment, which is used to find Young's Modulus of elasticity, the diameter of experimental wire is D = 0.05 cm (measured by a scale of least count 0.001 cm) and length is L = 100 cm (measured by a scale of least count 0.1 cm). A weight of 50 N causes an extension of X = 0.125 cm (measured by a micrometer of least count 0.001 cm). Find maximum possible error in the values of Young's modulus. Screw gauge and meter scale are free from error.
[IIT-2004] Sol. Maximum percentage error in Y is given by
Y = 4 D W 2 π ×X L . max Y Y ∆ = 2 ∆ D D + x x ∆ + L L ∆ = 2 + + 110 1 . 0 125 . 0 001 . 0 05 . 0 001 . 0 = 0.0489
So maximum percentage error = 4.89%.
2. Particles P and Q of mass 20 gm and 40 gm respectively are simultaneously projected from points A and B on the ground. The initial velocities of P and Q make 45º and 135º angles respectively with the horizontal AB as shown in the figure. Each particle has an initial speed of 49 m/s. The separation AB is
245 m. [IIT-1982] A P Q 135º B 45º
Both particle travel in the same vertical plane and undergo a collision After the collision, P retraces its path, Determine the position of Q when it hits the ground. How much time after the collision does the particle Q take to reach the ground? Take g = 9.8 m/s2.
Sol. mp = 20 g mQ = 40 g
The horizontal velocities of both the particles are same and since both are projected simultaneously, these particle will meet exactly in the middle of AB (horizontally).
To find the vertical velocity at the time of collision let us consider the motion of P in vertical and horizontal directions. A P Q 135º B 45º 49m/s 49m/s s / m 2 49 s / m 2 / 49 49/ 2m/s 245m s / m 2 49 Horizontal direction Sx = 122.5 Tx = ? vx =49 2 ∴ velocity = time nt displaceme ⇒ 2 49 = x t 5 . 122 ∴ tx = 49 2 ) 5 . 122 ( Vertical Direction vy = ? uy = 2 49 ty = 49 2 ) 5 . 122 ( ay = – 9.8 m/s2 ∴ vy = uy + aty = 2 49 – 9.8 × 49 2 ) 5 . 122 ( = 0 mpVp mQVQ – + Before collision mpVp mQVQ′ After collision Also, v2 – u2 = 2as ∴ – 2 3 49 = 2 (– 9.8) × s ⇒ s = 61.25
KNOW IIT-JEE
⇒ The collision takes place at the maximum height where the velocities of both the particles will be in the horizontal direction.
Applying conservation of linear momentum in the horizontal direction with the information that P retraces its path therefore its momentum will be same in magnitude but different in direction.
Momentum of system after collision ∴ mpmp – mQvQ = – mPvP + mQv′Q ∴ v′Q = Q Q Q P P m v m v m 2 − = 040 . 0 ) 2 / 49 ( 040 . 0 ) 2 / 49 ( 02 . 0 2× × − = −1 040 . 0 04 . 0 2 49 = 2 49 × 0 = 0
New Path of Q after Collision
Considering vertical Motion of Q uy = 0 sy = – 61.25 ay = – 9.8 ty = ? S = ut + 2 1 at2 = 2 1 × (– 9.8) × t2 = (–61.25) ∴ t = 3.53 sec
Considering Horizontal motion of Q :
Since the VQ' = 0, therefore the particle Q falls down vertically so it falls down on the mid point of AB. 3. Three particles, each of mass m, are situated at the
vertices of an equilateral triangle of side length a. The only forces acting on the particles are their mutual gravitational forces. It is desired that each particle moves in a circle while maintaining the original mutual separation a. Find the initial velocity that should be given to each particle and also the time period of the circular motion. [IIT-1988] Sol. The radius of the circle
r = 4 a a 3 2 2− 2 = 3 a m m m a/ 3 v F FR F a
Let v be the velocity given. The centripetal force is provided by the resultant gravitational attraction of the two masses
FR = F2+F2+2F2cos60º = 3F = 3G 2 a m m× ∴ 3G 2 2 a m = r mv2 ⇒ v2 = 3 a ma G 3 2× ⇒ v = a Gm
Time period of circular motion
T = v r 2π = a Gm 3 a 2π = 2π Gm 3 a3
4. Two fixed charges –2Q and Q are located at the points with coordinates (–3a, 0) and (+3a, 0) respectively in
the x-y plane. [IIT-1991]
(a) Show that all points in the x-y plane where the electric potential due to the two charges is zero, lie on a circle. Find its radius and the location of its centre. (b) Give the expression V(x) at a general point on the x-axis
and sketch the function V(x) on the whole x-axis. (c) If a particle of charge +q starts form rest at the centre
of the circle, show by a short quantative argument that the particle eventually crosses the circle. Find its speed when it does so.
Sol. (a) Let P be a point in the X-Y plane with co-ordinates (x, y) at which the potential due to charges –2Q and +Q placed at A and B respectively be zero.
P(x,y) X O X′ A x (–3a,0) B y Y +Q C (5a,0) (3a-x) (–3a,0) Y′ (3a+x) ∴ 2 2 y ) x a 3 ( ) Q 2 ( K + + = (3a x)2 y2 ) Q ( K + − + ⇒ 2 (3a−x)2+y2 = (3a+x)2+y2 ⇒ 4[(3a – x)2 + y2] = [(3a + x)2 + y2]
⇒ 4[6a2 + x2 – 12ax + y2] = [6a2 + x2 + 12ax + y2]
⇒ 3x2 + 3y2 – 30ax + 27a2 = 0
⇒ x2 + y2 – 10ax + 9a2 = 0
This is the equation of a circle with centre at (5a, 0) and radius 4a. Thus c (5a, 0) is the centre of the circle.
(b) For x > 3a
To find V(x) at any point on X-axis let us consider a point (arbitrary) M at a distance x from the origin.
+Q M –2Q (3a,x) (–3a,0) O x
The potential at M will be
V(x) = a 3 x ) Q 2 ( K + − + ) a 3 x ( ) Q ( K − + where k = 0 4 1 πε ∴ V(x) = KQ + − − x 3a 2 a 3 x 1 For |x| > 3a Similarly, for 0 < |x| < 3a V(x) = KQ + − − 3a x 2 x a 3 1
Since circle of zero potential cuts the x-axis at (a, 0) and (9a, 0) hence V(x) = 0 at x = a at x = 9a
• From the above expressions
V(x) → ∞ at x → 3a and V(x) → – ∞ and x → – 3a. • V(x) → 0 as x → + ∞ • V(x) varies at x 1 in general. X V –3a a 3a
(c) Applying Energy Conservation
(K.E. + P.E.)centre = (KE. + P.E.)circumference
0 + K − a 8 Qq 2 a 2 Qq = 2 1mv2 + K − a 12 Qq 2 a 6 Qq ⇒ 2 1mv-2 = a 4 KQq ⇒ v = ma 2 KQq = πε 2ma Qq 4 1 0
5. A thin uniform wire AB of length 1m, an unknown resistance X and a resistance of 12Ω are connected by thick conducting strips, as shown in the figure. A battery and a galvanometer (with a sliding jockey connected to it) are also available. Connections are to be made to measure the unknown resistance X using the principle of Wheatstone bridge. Answer the following questions. [IIT-2002]
X B
A C D 12Ω
(a) Are there positive and negative terminals on the galvanometer ?
(b) Copy the figure in your answer book and show the battery and the galvanometer (with jockey) connected at appropriate points.
(c) After appropriate connections are made, it is found that no deflection takes place in the galvanometer when the sliding jockey touches the wire at a distance of 60 cm from A. Obtain the value of the resistance of X.
Sol. (a) No. There are no positive and negative terminals on the galvanometer.
(b) & (c) Q Bridge is balanced JB AJ R R = ρ ρ 4 . 0 6 . 0 = X 12Ω ⇒ X = 8 Ω
where ρ is the resistance per unit length.
J
A X C D G
12Ω
CHEMISTRY
6. The oxides of sodium and potassium contained in a 0.5 g sample of feldspar were converted to the respective chlorides. The weight of the chlorides thus obtained was 0.1180 g. Subsequent treatment of the chlorides with silver nitrate gave 0.2451 g of silver chloride. What is the percentage of Na2O and K2O in
the mixture ? [IIT-1979]
Sol. Mass of sample of feldspar containing Na2O and
K2O = 0.5 g.
According to the question,
Na2O + 2HCl → 2NaCl + H2O ..(1)
2 × 23 + 16 = 62g 2(23 + 35.5) = 117 g
K2O + 2HCl → 2KCl + H2O ...(2)
2 × 39 + 16 = 94g 2(39 + 35.5) = 149 g Mass of chlorides = 0.1180 g
Let, Mass of NaCl = x g
∴ Mass of KCl = (0.1180 – x)g Again, on reaction with silver nitrate,
NaCl + AgNO3 → AgCl + NaNO3 ...(3)
23 + 35.5 = 58.5g 108 + 35.5 = 143.5g
KCl + AgNO3 → AgCl + KNO3 ...(4)
39 + 35.5 = 74.5g 108 + 35.5 = 143.5g Total mass of AgCl obtained = 0.2451 g Step 1. From eq. (3)
58.5 g of NaCl yields = 143.5 g AgCl ∴ x g of NaCl yields = 5 . 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 Na 2O 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 K 2O Step 3. % of Na2O in feldspar = 5 . 0 018 . 0 × 100 = 3.6% % of K2O in feldspar = 5 . 0 053 . 0 × 100 = 10.6 % 7. A metallic element crystallizes into a lattice
containing a sequence of layers of ABABAB ... Any packing of spheres leaves out voids in the lattice. What percentage by volume of this lattice is empty
space ? [IIT-1996]
Sol. A unit cell of hcp structure is a hexagonal cell, which is shown in fig. Three such cells form one hcp unit. For hexagonal cell, a = b ≠ c; α = β = 90º and γ = 120º. It has 8 atoms at the corners and one inside, hence
Number of atoms per unit cell = 8 8 + 1 = 2 N b a O 60º
Area of the base = b × ON = b × a sin 60º =
2 3
a2
( Q b = a) Volume of the hexagonal cell
= Area of the base × height =
2 3 a2. c But c = 3 2 2 a c b α β a γ
∴ Volume of the hexagonal cell = 2 3 a2 . 3 2 2 a = a3 2
and radius of the atom, r = a/2
Hence, fraction of total volume of atomic packing factor = cell hexagonal the of Volume atoms 2 of Volume = 2 a r 3 4 2 3 3 π × = 2 a 2 a 3 4 2 3 3 π × = 2 3 π = 0.74 = 74%
∴ The percentage of void space = 100 – 74 = 26% 8. A basic nitrogen compound gave a foul smelling gas
when treated with CHCl3 and alc. KOH. A 0.295 g
sample of the substance, dissolved in aq. HCl and treated with NaNO2 solution at 0ºC liberated a
colourless, odourless gas whose volume corresponds to 112 ml at STP. After the evolution of gas was completed, the aq. solution was distilled to give an organic liquid which did not contain nitrogen and which on warming with alkali and I2 gave a yellow
precipitate. Identify the original sustance. Assume that it contains one N atom per molecule. [IIT-1993] Sol. As the compound on heating with CHCl3 and alc.
KOH gives foul smelling gas, it should be any primary amine.
RNH2 + CHCl3 + 3KOH →∆ RN C
(Alkyl isocyanide (foul smelling gas)
+ 3KCl + 3H2O
Since the compound on treating with NaNO2 and HCl
at 0ºC produces a colourless gas, the compound must be a p-aliphatic amine, because if it was aromatic diazonium salt might have been produced
NaNO2 + HCl → NaCl + HNO2
RNH2 + HNO2 → ROH + N2 + H2O
Thus, the gas liberated is N2.
Amount of gas produced =
ml 400 , 22 ml 112 = 200 1 ml
From the above equation, it is obvious that amount of compound RNH2 =
200 1 mol. If M is the molar mass of RNH2, then
) M ( mass Molar g 295 . 0 = 200 1 mol ∴ M = 0.295 × 200 g mol–1 = 59 g mol–1
Thus, the molar mass of alkyl group (R—) will be, 59 – 16 = 43 g mol–1.
Hence, R = C3H7 —, i.e., CH3CH2CH2 –
or CH3
CH3
CH–
The original compound may be either CH3CH2CH2NH2 or CH3 – CH – CH3
NH2
From equation, it is clear that the liquid obtained after distillation is ROH. Since it gives yellow ppt. with NaOH and I2, it must have CH3 — C —
OH
group.
Hence, it is concluded that ROH is CH3 – CH – CH3
OH
.
Thus, the original compound is CH3 – CH – CH3
NH2
.
The different equations are :
CH3 Isopropyl amine CH– NH2 + CHCl3 + 3KOH ∆ CH3 CH3 CH3 CH – N C + 3KCl + Isopropyl isocyanide CH3 CH3 CH – NH2 + HNO2 → Isopropyl alcohol CH3 CH3 CH – OH + N2 + H2O CH3 – CH – CH3 + I2 + 2NaOH → OH O Acetone CH3 – C – CH3 + 2NaI + 2H2O CH3 – C – CH3 + 3I2 + 3NaOH → O O CI3 – C – CH3 + 3NaI + 3H2O
CI3 – C – CH3 + NaOH → CH3COONa + CHI3
O Yellow ppt.
∆
9. An organic compound (A) C8H6, on treatment with
dil. H2SO4 containing HgSO4 gives a compound (B),
which can also be obtained from a reaction of benzene with an acid chloride in the presence of anhydrous AlCl3. The compound (B) when treated
with iodine in aq. KOH, yields (C) and a yellow compound (D). Identify (A), (B), (C) and (D) with justification. Show, how (B) is formed from (A).
[IIT-1994] Sol. The given reactions may be formulated as follows :
C8H6 (A) Dil H2SO4 ∆ (B) I2 + KOH (C) + (D) AlCl3 ∆ C6H6 + Acid chloride HgSO4
The reaction of compound (B) with I2 in KOH is
iodoform reaction. The compound (B) must have a –COCH3 group so as to exhibit iodoform reaction.
Since (B) is obtained from benzene by Friedal-Crafts reaction, it is an aromatic ketone (C6H5COCH3). The
compound (C) must be potassium salt of an acid. The compound (A) may be represented as C6H5C2H.
Since it gives C6H5COCH3 on treating with dil.
H2SO4 and HgSO4, it must contain a triple bond
(–C ≡ CH) in the side chain. Here, the given reactions may be formulated as follows :
C≡CH (A) dil H2SO4 HgSO4; H2O C = CH2 OH COCH3 Acetophenone (B) Benzene CH3COCl AlCl3; – HCl
– C – CH3 O + 3I2 + 4KOH –3KI;–3H2O ∆ (C) Potassium benzoate COOK + CHI3 (D) (B) Hence. (A) C≡CH Phenyl acetylene (B) COCH3 Acetophenone (C) COOK Potassium benzoate (D) Idoform3 CHI
10. When 20.02 g of a white solid (X) is heated, 4.4 g of an acid gas (A) and 1.8g of a neutral gas (B) are evolved leaving behind a solid residue (Y) of weight 13.8 g. Gas (A) turns lime water milky and (B) condenses into a liquid which changes anhydrous CuSO4 blue. The aqueous solution of (Y) is alkaline
to litmus and gives 19.7 g of a white precipitate (Z) with BaCl2 solution. (Z) gives CO2 with an acid.
Identify (A), (B), (X), (Y) and (Z). [IIT-1989] Sol. (i) Since acidic gas (A) turns lime water milky hence
it is CO2 or SO2, both of which form white insoluble
compound with Ca(OH)2
(ii) Neutral gas (B) condenses into a liquid which turns anhydrous CuSO4 (white) into blue
(CuSO4.5H2O), hence (B) is H2O.
(iii) (Y) gives alkaline solution and its solution forms white precipitate (Z) with BaCl2 solution. (Z) on
heating gives the acid gas CO2, hence (Z) is BaCO3
and therefore (Y) is a metal carbonate.
(iv) Since (Y) and (A) are produced from (X), thus (X) is a metal bicarbonate. g 02 . 20 ) X ( → g 4 . 4 ) A ( + g 80 . 1 ) B ( + g 8 . 13 ) Y (
From the above values we may write a general equation for a bicarbonate.
) X ( 3 MHCO 2 →∆ ) A ( 2 CO + H2 ) B (O + M2(YCO) 3 Q 4.4g CO2 is obtained from 20.02 g of MHCO3
∴ 4g CO2 is obtained from 200.2 g of MHCO3
Molecular weight of MHCO3 =
2 2 . 200 = 100.1 Atomic weight of M = 39
Thus, the metal M is potassium and then (X) is KHCO3. The equations are :
) X ( 3 KHCO 2 →∆ ) Y ( 3 2CO K + ) A ( 2 CO + ) B (2 O H ) Y ( 3 2CO K + BaCl2 → 2KCl + ) Z ( 3 BaCO ) Z ( 3 BaCO →∆ BaO + ) A ( 2 CO ↑
Hence, (A) is CO2 (B) is H2O (X) is KHCO3 (Y) is
K2CO3 and (Z) is BaCO3.
MATHEMATICS
11. From a point A common tangents are drawn to the circle x2 + y2 = a2/2 and parabola y2 = 4ax. Find the
area of the quadrilateral formed by the common tangents, the chord of contact of the circle and the chord of contact of the parabola. [IIT-1996] Sol. Equation of any tangent to the parabola, y2 = 4ax is
y = mx + a/m.
This line will touch the circle x2 + y2 = a2/2
A(–a, 0) C O B y D L E πx = – a/2 x = a x If 2 m a = 2 a2 (m2 + 1) ⇒ 2 m 1 = 2 1(m2 + 1) ⇒ 2 = m4 + m2 ⇒ m4 + m2 – 2 = 0 ⇒ (m2 – 1)(m2 + 2) = 0
⇒ m2 – 1 = 0, m2 = – 2 (which is not possible).
⇒ m = ± 1
Therefore, two common tangents are y = x + a and y = –x – a
These two intersect at A(–a, 0)
The chord of contact of A(–a, 0) for the circle x2 + y2 = a2/2 is
(–a)x + 0.y = a2/2 or x = – a/2
and chord of contact of A(–a, 0) for the parabola y2 = 4ax is
0.y = 2a(x – a) or x = a Again length of BC = 2BK
= 2 OB2−OK2 = 2 4 a 2 a2 2 − = 2 4 a2 = a
and we know that DE is the latus rectum of the parabola so its length is 4a.
Thus area of the trapezium
BCDE = 2 1 (BC + DE) (KL) = 2 1(a + 4a) 2 a 3 = 4 a 15 2
12. Let V be the volume of the parallelopiped formed by the vectors → a= a1iˆ + a2jˆ + a3kˆ; → b= b1iˆ + b2jˆ + b3kˆ → c= c1iˆ + c2jˆ + c3kˆ
If ar, br, cr, where r = 1, 2, 3 are non-negative real
numbers and
∑
= + + 3 1 r r r r b c ) a ( = 3L. Show that V ≤ L3. [IIT-2002] Sol. V = |→a.(→b×→c)| ≤ 2 3 2 2 2 1 a a a + + 2 3 2 2 2 1 b b b + + 2 3 2 2 2 1 c c c + + ...(1) Now, L = 3 ) c c c ( ) b b b ( ) a a a ( 1+ 2+ 3 + 1+ 2+ 3 + 1+ 2+ 3 [(a1 + a2 + a3) (b1 + b2 + b3) (c1 + c2 + c3)]1/3 ∴ L3 ≥ [(a 1 + a2 + a3)(b1 + b2 + b3)(c1 + c2 + c3)]..(2) Now, (a1 + a2 + a3)2 = 2 1 a + 2 2 a + 2 3 a + 2a1a2 + 2a1a3 + 2a2a3 ≥ a12+a22+a32 ⇒ (a1 + a2 + a3) ≥ a12+a22+a32 Similarly, (b1 + b2 + b3) ≥ b12+b22+b23 and (c1 + c2 + c3) ≥ c12+c22+c32 ∴ from (1) and (2) L3 ≥ [( 2 1 a + 2 2 a + 2 3 a )( 2 3 2 2 2 1 b b b + + )( 2 3 2 2 2 1 c c c + + )]1/3 ≥ V13. T is a prallelopiped in which A, B, C and D are vertices of one face and the just above it has corresponding vertices A´, B´, C´, D´, T is now compressed to S with face ABCD remaining same and A´, B´, C´, D´ shifted to A´´, B´´, C´´, D´´ in S. The volume of parallelopiped S is reduced to 90% of T. Prove that locus of A´´ is a plane. [IIT-2004] Sol. Let the equation of the plane ABCD be
ax + by + cz + d = 0, the point A´´ be (α, β, γ) and the height of the parallelopiped ABCD be h.
⇒ 2 2 2 b c a | d c b a | + + + γ + β + α = 90%. h ⇒ aα + bβ + cγ + d = ± 0.9h a2+b2+c2 ⇒ locus is, ax + by + cz + d = ±0.9h a2+b2+c2 ⇒ locus of A´ is a plane parallel to the plane ABCD 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 nn 6 3 6 C ) 3 2 . 3 3 ( − + ×
15. A straight line L through the origin meets the line x + y = 1 and x + y = 3 at P and Q respectively. Through P and Q two straight lines L1 and L2 are
drawn, parallel to 2x – y = 5 and 3x + y = 5 respectively. Lines L1 and L2 intersect at R, shown
that the locus of R as L varies, is a straight line. [IIT-2002] Sol. Let the equation of straight line L be y = mx
P ≡ + + m 1 m , 1 m 1 ; Q ≡ + + m 1 m 3 , 1 m 3 Now equation of L1 : y – 2x = 1 m 2 m + − ...(1) equation of L2 : y + 3x = 1 m 9 m 3 + + ...(2) By eliminating 'm' from equation (1) and (2), we get locus of R as x – 3y + 5 = 0, which represents a straight line.
1. Two capacitors C1 and C2, can be charged to a
potential V/2 each by having
C1 C2 O R R S2 S1 V
(A) S1 closed and S2 open
(B) S1 open and S2 closed
(C) S1 and S2 both closed
(D) cannot be charged at V/2
2. Energy liberated in the de-excitation of hydrogen atom from 3rd level to 1st level falls on a
photo-cathode. Later when the same photo-cathode is exposed to a spectrum of some unknown hydrogen like gas, excited to 2nd energy level, it is
found that the de-Broglie wavelength of the fastest photoelectrons, now ejected has decreased by a factor of 3. For this new gas, difference of energies of 2nd Lyman line and 1st Balmer line if
found to be 3 times the ionization potential of the hydrogen atom. Select the correct statement(s) (A) The gas is lithium
(B) The gas is helium
(C) The work function of photo-cathode is 8.5eV (D) The work function of photo-cathode is 5.5eV
3. In the figure shown there exists a uniform time varying magnetic field B = [(4T/s) t + 0.3T] in a cylindrical region of radius 4m. An equilateral triangular conducting loop is placed in the magnetic field with its centroide on the axis of the field and its plane perpendicular to the field.
+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + C B A
(A) e.m.f. induced in any one rod is 16V
(B) e.m.f. induced in the complete ∆ABC is
V 3 48
(C) e.m.f. induced in the complete ∆ABC is 48V (D) e.m.f. induced in any one rod is 16 3V
4. 6 parallel plates are arranged as shown. Each plate has an area A and Distance between them is as shown. Plate 1-4 and plates 3-6 are connected equivalent capacitance across 2 and 5 can be writted as
d nA∈0
. Find min value of n. (n, d are natural numbers) d d d d 2d 1 2 3 4 5 6
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
5. Match the following
Column – I Column – II (A) A light conducting (P) Magnetic field B
circular flexible is doubled. loop of wire of
radius r carrying current I is placed in uniform magnetic field B, the tension in the loop is doubled if
(B) Magnetic field at a (Q) Inductance is point due to a long increased by four straight current times.
carrying wire at a point near the wire is doubled if
(C) The energy stored (R) Current I is in the inductor will doubled become four times
(D) The force acting on a (S) Radius r is moving charge, doubled moving in a constant
magnetic field will be doubled if
(T) Velocity v is Doubled Passage # (Q. No. 6 to Q. No. 8 )
A solid, insulating ball of radius ‘a’ is surrounded by a conducting spherical shell of inner radius ‘b’ and outer radius ‘c’ as shown in the figure. The inner ball has a charge Q which is uniformly distribute throughout is volume. The conducting spherical shell has a charge –Q.
Answer the following questions.
–Q Q
a c b
6. Assuming the potential at infinity to be zero, the potential at a point located at a distance a/2 from the centre of the sphere will be
(A) − πε b 1 a 2 4 Q 0 (B) − πε b 1 a 8 11 4 Q 0 (C) − πε b 1 a 1 4 Q 0 (D) None of these
7. Work done by external agent in taking a charge q slowly from inner surface of the shell to surface of the sphericalball will be
(A) − c 1 a 1 kQq (B) − a 1 b 1 kQq (C) − b 1 a 1 kQq (D) − a 1 c 1 kQq
8. Now the outer shell is grounded, i.e., the outer surface is fixed to be zero. Now the charge on the inner ball will be
(A) zero (B) Q (C) + − b 1 c 1 a 1 C Q (D) + − b 1 c 1 a 1 b Q
Cartoon Law of Physics
Any body passing through solid matter will leave a perforation conforming to its perimeter.Also called the silhouette of passage, this phenomenon is the specialty of victims of directed-pressure explosions and of reckless cowards who are so eager to escape that they exit directly through the wall of a house, leaving a cookie-cutout-perfect hole. The threat of skunks or matrimony often catalyzes this reaction.
1. A fighter plane flies at a velocity of 300
sec m . On
the fighter plane there is a gun which shoots at a rate of 40 rounds per second with a muzzle velocity of 1200
sec
m . The shots are aimed at another fighter
plane flying at a velocity of 200
sec m
. Find the rate at which the projectiles hit the target plane :
(a) When the two planes move in the same direction, and the target plane is in front of the shooting plane. (b) The same as (a), when the target plane is in the rear
of the shooting plane.
(c) When the two planes move towards one another. (d) When the two planes move away from one another. Sol. Denote by vs the velocity of the plane from which the
shots are fired, by vt the velocity of the target plane
and by L the distance between them at the certain moment of time when the shooting plane starts to shoot. Denote by r the rate of fire of the gun and by v the muzzle velocity.
(a) The time it takes for the first projectile to reach the target plane is:
t1 = t s v v v L − + …(1) After a time of r 1
the second projectile is shot, and the distance between the planes at this time is given by: L' = L – r v vs− t …(2)
Thus, the time it takes the second projectile to arrive at the target plane is:
t2 = t s t s v v v r v v L − + − − …(3) which is ∆t = t2 + r 1 – t1 = r 1 – ) v v v ( r v v t s t s − + − = ) v v v ( r v t s− + …(4)
after the first shot. Naturally, the time increment does not depend on the initial distance; thus the rate of hitting is: r′ = t 1 ∆ = r v ) v v v ( + s− t = 40 × 1200 1300= 43.33 ond sec hits … (5)
(b) Using the same reasoning for this case, we obtain: r′ = r v v v v− s+ t = 40 × 1200 1100 = 36.66 ond sec hits … (6) (c) In this case: r′ = r v v v v+ s+ t = 40 × 1200 1700 = 56.67 ond sec hits … (7) (d) Here, r′ = r v v v v− s− t = 40 × 1200 700 = 23.33 ond sec hits … (8) 2. Consider the system described in figure.
m1
m2
(a) Use the equations of energy conservation to find the velocities of masses m1 and m2 after they are
released from rest and pass a distance y (assume m2 > m1).
(b) Use the expression obtained in the first section to find the acceleration of the masses.
Sol. (a) We write the change in the potential energy of the masses :
m1 : = ∆Ep = m1gy ...(1)
m2 : ∆Ep = – m2gy ...(2)
The change in the kinetic energy is ∆Ek = 2 1 m 1v12 + 2 1m 2v22 ...(3)
Expert’s Solution for Question asked by IIT-JEE Aspirants
Students' Forum
Because |vr1| = |vr2| ≡ v, we obtain ∆Ek =
2 1
(m1 + m2)v2 ...(4)
Since there are no external forces except gravity, which is a conservative force, we know, using energy conservation that : ∆Ep + ∆Ek= 0 ...(5) By substituting (m1 – m2)gy + 2 1 (m1 + m2)v2 = 0 ...(6) Hence, v(y) = 2 / 1 2 1 1 2 m m ) m m ( g 2 + − y ...(7) (b) By definition, a = dt dv = 2 / 1 2 1 1 2 m m ) m m ( g 2 + − . dt y d = y 2 1 m m ) m m ( g 2 1/2 2 1 1 2 + − . dt dy ...(8) and because dt dy
= v we substitute Eq. (7) into Eq. (8) and obtain. a = 1 2 1 2 m m m m + − g
3. A smooth incline of lift angle α is accelerated at a rate α. A block of mass m is placed on the incline. At t = 0 the block is released, and begins moving (see figure.) N m mg a y´ x´ α
(a) Write the equations of motion for the block. (b) What is the maximal value of 'a' for which the
block will remain attached to the incline ?
(c) How much time is required for the block to slide a distance L along the incline ?
(d) If we accelerate the incline in the opposite direction, what is the minimal value of a necessary for the block to slide up the incline ?
Sol. (a) D'alembert's force exists in the acclerated system, on the plane of the incline. Therefore,
m ar = mg xˆ ´ sin α – mg yˆ ´ cos α + N yˆ ´ – m ar ar is expressed by the unit vectors of the accelerated system as :
ar = – a xˆ ´ cos α – a yˆ ´ sin α and in component form :
≥ α + α − = α + α = ) 0 N ( sin a cos g m N a cos a sin g a ´ y ´ x
(b) The maximal value is obtained when N = 0 and ay´ = 0. (The block is still upon the inclined surface,
as stipulated.) Therefore, a = g cot α
(c) The equation of motion with a constant acceleration 'a' is :
x(t) = x0 + v0t +
2 1at2
In our case, we have L = 12ax´t2. Therefore,
t = α + α acos sin g L 2
(d) Because 'a' is changed to (–a), the equation of motion in the ´xˆ direction becomes :
´ x
a = g sin α – a cos α
In order for the mass to slide up the incline, the condition ax´> 0 must be met. Thus,
a > g tan α
4. Two long wires are placed on a smooth horizontal table. Wires have equal but opposite charges. Magnitude of linear charge density on each wire is λ. Calculate (for unit length of wires) work required to increase the separation between the wires from a to 2a. Sol. Since, wires have opposite charge, therefore, they
attract each other. To increase separation between the wires, work is to be done against this force of attraction.
Let at some instant separation between the wires be x as shown in Fig.
To calculate force of attraction between the wires, first electric field due to charge on one wire at position of the second wire is to be calculated.
Therefore, considering a cylindrical surface of radius x and of unit length, co-axial with positively charged wire, x + + + + – – – – Its area = 2π x × 1 Charge enclosed within it = λ
∴ Flux passing through the cylindrical surface = 0
ε λ
Electric field, E = Flux passing per unit area. = x 2 ) / ( 0 π ε λ = x 2πε0 λ
Magnitude of charge on unit length of second wire = λ ∴ Force of attraction per unit length is F = λ E
or F = x 2 0 2 πε λ
To increase the separation, wires are to be pulled apart by applying an infinitesimally greater force (F + dF). ∴ Work done to increase separation from x to (x + dx),
dW = (F + dF) dx ≈ F. dx or Total work done =
∫
== a 2 x a x Fdx=
∫
= = πε λ a 2 x a x 0 2 dx x 2 = 0 2 2πε λ loge 25. Two short electric dipoles having dipole moment p1
and p2 are placed co-axially and uni-directionally, at
a distance r apart. Calculate nature and magnitude of force between them.
Sol. Let second dipole having dipole moment p2 consist of
charges (+ q2) and (– q2) which are separated by an
elemental distance 2dr as shown in Fig.
Then p2 = q2 2dr …(1) r p1 → + – q2 q2 2.dr
Since, dipoles are separated by a distance r, it means distance between their centres is r. Distance of charges (– q2) & (+ q2) from centre of first dipole is
(r – dr) & (r + dr), respectively. If electric field strength due to and at distance r from dipole having dipole moment p1 is E, then electric field strength at
position of two charges will be (E – dE) & (E + dE), respectively. Where E = 0 4 1 ε π r31 p 2 (rightward) ∴ dE = – 3 . 0 4 1 ε π . r41 p 2 dr
Force on charge (– q2) is F1 = (E – dE) q2 (leftward)
and that on charge (+ q2) is F2 = (E + dE) q2
(rightwards)
Hence, net force on second dipole is F = F2 – F1 (rightward) or F = dE 2q2 = – 3 0 4 1 ε π . 4 1 r p 2 dr 2q2 But 2q2 dr = p2 ∴ Net force = – 0 4 1 πε 4 2 1 r p p 6
(–ve) sign indicates that actual direction of force on second dipole is leftward or force between two dipoles is of attraction and its magnitude is F = 0 4 1 πε . 4 2 1 r p p 6
1. The typical size of a meteor is about one cubic centimeter, which is equivalent to the size of a sugar cube.
2. Each day, Earth accumulate 10 to 100 tons of material.
3. There are over 100 billion galaxies in the universe.
4. The largest galaxies contain nearly 400 billion stars.
5. The risk of a falling meteorite striking a human occurs once every 9,300 years. 6. A piece of a neutron star the size of a pin
point would way 1 million tons.
7. Europa, Jupiter’s moon, is completely covered in ice.
8. Light reflecting off the moon takes 1.2822 seconds to reach Earth.
9. There has only been one satellite destroyed by a meteor, it was the European Space Agency’s Olympus in 1993.
10. The International Space Station orbits at 248 miles above the Earth.
11. The Earth orbits the Sun at 66,700mph. 12. Venus spins in the opposite direction
compared to the Earth and most other planets. This means that the Sun rises in the West and sets in the East.
13. The Moon is moving away from the Earth at about 34cm per year.
14. The Sun, composed mostly of helium and hydrogen, has a surface temperature of 6000 degrees Celsius.
15. A manned rocket reaches the moon in less time than it took a stagecoach to travel the length of England.
16. The nearest known black hole is 1,600 light years (10 quadrillion miles/16 quadrillion kilometers) away.
• Coulomb's Law : F0 = 122 0 r q q 4 1 πε (in vacuum) Vectorially →F = 122 0 r q q 4 1 πε rˆ
In any material medium F = 122 r 0 r q q 4 1 ε πε
where εr is a constant of the material medium called
its relative permittivity, and ε0 is a universal constant,
called the permittivity of free space. ε0 = 8.85 × 10–12 or 0 4 1 πε = 9 × 10 9
The unit of ε0 is C2N m–2 or farad per metre.
Also F = 122 r 0 r q q 4 1 ε πε
Where ε is called the absolute permittivity of the medium.
Obviously, εr = F0/F. Remember εr = ∞ for
conductors.
Conductors and insulators Each body contains enormous amounts of equal and opposite charges. A 'charged' body contains an excess of either positive or negative charge.
In a conductor, some of the negative charges are free to move around. In an insulator (also called a dielectric), the charges cannot move. They can only undergto small localized displacements, causing polarization.
Induction When a charged body A is brought near another body B, unlike charges are induced on the near surface of B (called bound charges) and like charges appear on the far surface of B (called free charges) If B is a conductor, the free charges can be removed by earthing B, e.g., by touching it. If B is an insulator, separation of like and unlike charges will still occur due to induction. However, the like charges cannot then be removed by earthing B. • Electric Field And Potential
Electric Field An electric field of strength E is said to exist at a point if a test charge ∆q at that point experiences a force given by
→ → ∆ = ∆F qF or q F E ∆ ∆ = → →
The unit of electric field is Newton per coulomb or volt per metre. The electric field strength at a distance r from a point charge q in a medium of permittivity ε is given by E = πε 4 1 2 r q Vectorially →E= πε 4 1 2 r q rˆ With reference to any origin
→ E = πε 4 q 3 r R r R → → → → − −
Where →R is the position vector of the field point and →
r , the position vector of q.
Due to a number of discrete charges → E= 3 i i N i 1 i 1 r R r R 4 q → → → → = = − − πε
∑
Electric Potential The electric potential at a point is the work done by an external agent in bringing a unit positive charge from infinity up to that point along any arbitrary path.
VP = q ) agent external an by ( W P ∆ ∆ ∞→ volt(V) or JC–1
The potential difference between two points P and Q is given by VP – VQ = q ) agent by ( WQ P ∆ ∆ → volt (V)
The potential at a distance r from a point charge q in a medium of permittivity ε is ϕ or V = πε 4 1 r q= πε 4 1 → → − r R q
with reference to any arbitrary origin. Due to a number of charges
ϕ or V = → → = = πε −
