Dear Students,
It's the question you dreamed about when you were ten years old. It's the question our parents nagged you about during high school. It's the question that stresses most of us out more and more the older we get. "What do you want to be when you grow up?"
There are people who are studying political science but hate politics, nursing majors who hate biology, and accounting majors who hate math. Obviously, a lot of people are confused about what exactly it is that they want to spend their life doing. Think about it. if you work for 10 hours each day, you're going to end up spending over 50% of your awake life at work. Personally, I think it's important that we spend that 50% of your awake life at work. Personally, I think it's important that we spend that 50% wisely. But how can you make sure that you do? Here are some cool tips for how to decide that you really want to be when you grow up.
• Relax and Keep an Open Mind: Contrary to popular belief, you don't have to "choose a career" and stick with it for the rest of your life. You never have to sign a contract that says, "I agree to force myself to do this for the rest of my life" You're free to do whatever you want and the possibilities are endless. So relax, dream big, and keep an open mind.
• Notice Your Passions: Every one of us is born with an innate desire to do something purposeful with our lives. We long to do something that we're passionate about; something that will make a meaningful impact on the world.
• Figure Out How to Use Your Passions for a Larger Purpose: You notice that this is one of your passions, so you decide to become a personal trainer. Making a positive impact on the world will not only ensure that you are successful financially, it will also make you feel wonderful. It's proven principle: The more you give to the world, the more the world will give you in return.
• Figure Our How You Can Benefit: Once you've figured out what your passions are and how you can use those passions to add value to the world & to yourself, it's time to take the last step: figure out how you can make great success doing it. my most important piece of advice about this last step is to remember just that: It's the last part of the decision process. I feel sorry for people who choose an occupation based on the average income for that field. No amount of money can compensate for a life wasted at a job that makes you miserable. However, that's not to say that the money isn't important. Money is important, and I'm a firm believer in the concept that no matter what it is that you love doing, there's at least one way to make extraordinary money doing it. So be creative!
No matter how successful you become, how great your life is, or how beautiful you happen to be... there will still be times when you simply feel like you're an ugly mess. But when those times come, remember that all you need to get yourself back on track is a positive outlook, a dash of self confidence, and the willingness to make yourself feel better as soon as you know how.
Simply discover your passions, figure out how to use your passions to make an impact on the world & to yourself.
Presenting forever positive ideas to your success. Yours truly
Pramod Maheshwari, B.Tech., IIT Delhi
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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
• "The way to succeed is to double your errorrate."
• "Success is the ability to go from failure to failure without losing your enthusiasm." • "Success is the maximum utilization of the
ability that you have."
• We are all motivated by a keen desire for praise, and the better a man is, the more he is inspired to glory.
• Along with success comes a reputation for wisdom.
• They can, because they think they can. • Nothing can stop the man with the right
mental attitude from achieving his goal; nothing on earth can help the man with the wrong mental attitude.
• Keep steadily before you the fact that all true success depends at last upon yourself.
CONTENTS
INDEX PAGE
NEWS ARTICLE
4Dr. Abdul Kalam's Message to Every Indian Two Mumbai CAT toppers are from IIT-Bombay
IITian ON THE PATH OF SUCCESS
6Mr. Vineet Buch
KNOW IIT-JEE
7Previous IIT-JEE Question
XTRAEDGE TEST SERIES
50
Class XII – IIT-JEE 2011 PaperClass XI – IIT-JEE 2012 Paper
Mock Test-3 (CBSE Board Pattern) [Class # XII]
72
Solution of Mock Test-2 (CBSE Pattern)Solution of Mock Test-3 (CBSE Pattern)
Regulars ...
DYNAMIC PHYSICS
15
8-Challenging Problems [Set# 9] Students’ Forum
Physics Fundamentals
Matter Waves, Photo-electric Effect Thermal Expansion, Thermodynamics
CATALYSE CHEMISTRY
31
Key ConceptCarbonyl Compounds Co-ordination Compound & Metallurgy
Understanding : Physical Chemistry
DICEY MATHS
38
Mathematical Challenges Students’ Forum Key Concept Integration Trigonometrical EquationStudy Time...
Test Time ...
Dr. Abdul Kalam’s Message to Every Indian
What does a system consist of? Very conveniently for us it consists of our neighbours, other households, other cities, other communities and the government. But definitely not me and YOU. When it comes to us actually making a positive contribution to the system we lock ourselves along with our families into a safe cocoon and look into the distance at countries far away and wait for a Mr.Clean to come along & work miracles for us with a majestic sweep of his hand or we leave the country and run away. ‘ASK WHAT WE CAN DO FOR INDIA AND DO WHAT HAS TO BE DONE TO MAKE INDIA WHAT AMERICA AND OTHER WESTERN COUNTRIES ARE TODAY’
Two Mumbai CAT toppers are from IIT-Bombay
Three people, who appeared for CAT from the city, scored 100 percentile. Two of them are from the computer science department of the IIT-B. The other is a faculty member of a city-based coaching institute.
Shashank Samant, 22, who scored 99.98 percentile on his last attempt in 2008, gave up the IIM seat to take up a job. “After a year and a half job in an investment firm, I was finally prepared to get in to an IIM. Though the 100 percentile was unexpected.” “After clearing the CAT, I spoke to my peers and seniors at IIT-B and decided that work experience would be important before getting in to the IIMs,” said Samant. He also added that IITs help in developing the aptitude to clear any competitive exams. Samant had graduated in computer science from IIT-B in 2009. About his choice of IIMs, Samant said, “I am ready to get in to any of the IIMs though I prefer
IIM-Ahmedabad and Bangalore over others.”
Gaurav Malpani, a fourth-year student of computer science at IIT-B has appeared for the entrance exam from Mumbai, he is originally from Kolkata. He managed to score100 percentile, without any coaching.
“I have never focused on developing my technical knowledge. I was only polishing my problem solving skills. I also focused on my vocabulary,” said 21-year-old Malpani. He insisted that he had never studied exclusively for CAT during the year. “I always knew that I had the aptitude to score well, but scoring 100 percentile was not expected,” he added.
“I would love to join IIM-Ahmedabad or Bangalore. Since I am from Kolkata, I will also consider seeking admission there. I am interested in pursuing an MBA in either finance or human resources,” he added. The faculty member of a city-based coaching institute Jose D’Abreu also got a perfect score.
IIT Techfest 2011: The Robots Raged
It was the perfect way to end a very well-planned event. On the first two days of Techfest, IIT Bombay was buzzing with exhibits of some cool robots and a few other inventions as well. On the final day, the robots became restless and just wanted to have a go at each other. What followed was a lengthy battle fought hard and long.
Mars Manoeuvre This tournament had
two robots moving around a grid collecting blocks. The team that collected the most blocks won. The last battle was between C2R and Black Beast, from Thailand and Australia respectively.
Black Beast: In all its glory
Black Beast was the creation of second year students from the Department of
Electrical & Computer Science at Swinburne University of Technology in Melbourne. Plagued with exams and other academic diseases, they still managed to build this in about two weeks.
Kids at Swinburne Uni, came second with a smile
Their robots communicated through radio frequency waves and everything, right from the wheels to the circuit boards was custom made. One robot would go and measure the dimensions of the grid. The second, after getting the information would start its mission of picking up blocks.
Can it say, "Thai, Robot"?
C2R was made by third year students of the computer engineering department of the Kasetsart University in Kapmphaeng Saen, Thailand. These kids spent two months and about Rs. 1,47,000 (100,000 Thai Bath) to make these robots.
Kasetsart University students with the C2R
Unlike the kids of Oz, they chose to use only one robot which would find its way and collect blocks. These robots also used radio frequency technology to communicate and had sensors, so that the robot never drifted away from the grid lines. The battle ensued and it was clear that spending more time with your robot makes them strong and obedient. In fact, they can also win you competitions! C2R won the Mars Manoeuvre competition and prize money of Rs. 1,50,000. The kids demonstrated how they won the battle. Check out the video below.
IIT-Bombay gets $3 million gift
More than 40 years ago a quiet student named Victor Menezes graduated from the Indian Institute of Technology Bombay (IIT-B). He went on to become, among other things, the senior vice-chairperson of Citigroup Inc. His “small way to say thank you” to the institute has translated into a $3 million
(about Rs 13.5 crore) towards a state-of-the-art convention centre on the institute’s Powai campus
“I received priceless education from IIT Bombay and this is a small way to say thank you”, said Menezes. “I hope the centre will help support the exchange of ideas at IIT Bombay.”
IIT-JEE candidates to get performance cards now
Students appearing for the next Joint Entrance Examination (JEE) for admission to IITs will get performance cards specifying marks and the ranks secured by them in the test. However, as per the new provision, they cannot seek regrading or re-totalling.
For the first time, the JEE Board would issue performance cards which can be considered as certificates by many other institutions wanting to give admission to JEE candidates. The board will also put out the answers of the questions on its website to help students make assessment of their performance.
IIT Guwahati Director Prof Gautam Baruah said the board had urged for issuing such performance cards which would serve as certificates for the students. “Many other institutes, which want to take JEE candidates, can give admission to students on the basis of these performance cards,” Baruah said.
Indian institute of Science will start management course
Indian Institute of Science (IISc.) is planning to start a two-year Master programme in management from this academic session. The new courses will be very advance as it will concentrate more on technology management and business analytics.
IISc registrar R Mohan Das said the course would concentrate on synergies between managing science and technology. Das said, “India, in recent times, has emerged as one of the global hubs of technology and research and development (R&D) units. Such technology-based and R&D-intensive industries need executives with exposure and training in technology management and business analytics. The program has been specially designed to train students in technology management and business analytics.” The course will be conducted by the department of management studies, which was established in the year 1848, and is one of the oldest schools in the country. An official at the dept. said that application forms for the course will be available from the month of February. Candidates who have passed the Joint Entrance Test (JMET) with first class BE/B.Tech degree/equivalent is eligible for the course. The department will conduct group discussion and personal interview before selecting students for the course.
IIT Mandi to formulate plan for solving technical problems pertaining to agriculture in Himachal Pradesh
Shimla: Shri Ram Subhag Singh,
Secretary, Agriculture and Information and Public Relations said that H.P. Agriculture Department and IIT Mandi
would formulate a long term scheme for solving technical problems pertaining to agriculture and a Joint Working Group at State level would be formed for solving the problems relating to farm technology.
Star Donor of the Month - Mr. Rajesh Achanta [1987/BT/ME]
I have been donating off and on as a way of keeping the connection with IITM going & also to express gratitude for the many ways in which the institute shaped me in my formative years. I'll be transiting through Chennai in early January - I would like to stop by at IITM & relive old memories for a little while!
Orissa CM confers award to IIT-Kanpur Prof. Dr Devi Prasad Mishra
In recognition of his research work, Dr. Mishra received Sir Rajendranath Mookerjee Memorial and Aerospace Engineering Division Prize from The Institution of Engineers (India), Kolkata, India. Dr. Mishra has more than 15 years of teaching and research experience. He has served as Visiting Professor in 2002 at the Tokyo-Denki University, Japan. Presently, he is working as an Associate Professor in the Department of Aerospace Engineering at Indian Institute of Technology (IIT) Kanpur, Kanpur, India where he was instrumental in establishing a combustion laboratory. His areas of research interest include combustion, computational fluid dynamics, atomization, nanomaterial synthesis etc. He is an Associate Editor, Journal of Natural Gas Science and Engineering, Elsevier, USA and Assistant Editor, International Journal of Hydrogen Energy, Elsevier, USA. Currently he is serving as Editor, Asia Pacific Conference on Combustion, 2010. Dr. Mishra has four Indian patents and more than 154 publications in referred Journals and in conference proceedings to his credit.
Vineet Buch still remembers 10 June 1987. Bhopal. The Indian Institute of Technology All India Joint Entrance Exam (IIT-JEE) results were announced. Buch, then a 15-year-old dabbling with career choices, scanned through the rank-holders list. Then he scanned it again. Soon he made up his mind. He would try and finish No. 1 in the entrance exam. “It seemed like a cool thing to do.”
Every year thousands of Indian students aspire to get into an IIT. Close to 400,000 candidates lined up this year. One in 65 made the cut. Twenty years ago, the number of applicants wasn’t as staggering but there were fewer seats. Golfers will tell you that the odds of an amateur pulling off a hole-in-one are 1 in 12,750. Still, that’s a doddle compared to what Buch was up against.
“Hardly anyone in Bhopal even wrote the JEE, let alone got in,” says Buch, 37, a venture capitalist based in San Francisco. “I found it tough to get the right books, like a Russian physics book by IE Irodov. My parents [who were IAS officers] requested the Indian embassy in Moscow to photocopy the book and send it across.”
In June 1989, Buch was declared No. 1 in the IIT JEE exam, arguably the most challenging and competitive exam in the world. Only around 50 Indians have experienced the feeling—the numbness, the ecstasy, the dizziness.
Once every year, JEE toppers appear on television and newspapers carry congratulatory messages. You see mug shots of students, interviews with parents, and advertisements for coaching centres. We spend a lot of time celebrating their success, but rarely do we look further.
What becomes of these brilliant 17-year-olds? What are the challenges they encounter? Do any of them pursue unconventional careers? These were some of the questions
Open set out with while tracking down the very elite group
of JEE toppers.
IT HELPS TO BE NO. 1
During his days in IIT Kanpur, Buch was a long-distance athlete, weightlifter and footballer. He competed in both the 5,000 and 10,000 metres. But in August 1993, a doctor at Delhi’s All India Institute of Medical Sciences diagnosed the 20-year-old with ankylosing spondylitis, a progressively crippling disease without a cure.
Buch suffered inflammation of the eyes and internal organs. “Sometimes it was so hard for me to even sit, stand or sleep,” he recalls. Things got progressively worse over his two-year graduate program at Cornell University. “When I finished in 1995, I was immobilised throughout much of my body. A doctor advised me to stop working and apply for disability payments.”
Buch refused. He moved to San Francisco and started a self-directed rehabilitation programme. He began with long sessions of swimming and gradually started to walk, bike and hike. In 2001, he successfully undertook the Death Ride over five alpine passes on the Sierra Nevada mountain range in California, US. But biking hurt his knees. Searching for a sport that didn’t tax his legs, he discovered surf skiing, one that uses a long, narrow, lightweight kayak with an open cockpit and a foot-pedal controlled rudder. On 17 May, Buch took part in the 2009 Molokai Challenge in Hawaii, a 32-mile surf ski race between Molokai and Oahu, in rough waters swarming with tiger sharks. He finished the race.
“I thought being No 1 in JEE was tough,” says Buch. “But overcoming this disease has been something else. The JEE effort definitely helped with this—I knew the levels of determination I was capable of and refused to give up.”
Success Story
Success Story
This article contains storie/interviews of persons who succeed after graduation from different IITs
Mr. Vineet Buch
B-Tech from IIT-KanpurPHYSICS
1. Two narrow cylindrical pipes A and B have the same length. Pipe A is open at both ends and is filled with a monoatomic gas of molar mass MA. Pipe B is open at one end and closed at the other end, and is filled with a diatomic gas of molar mass MB. Both gases are at the same temperature. [IIT- 2002] (a) If the frequency of the second harmonic of the fundamental mode in pipe A is equal to the frequecy of the third harmonic of the fundamental mode in pipe B, determine the value of MA/MB.
(b) Now the open end of pipe B is also closed (so that the pipe is closed at bout ends). FInd the ratio of the fundamental frequency in pipe A to that in pipe B. Sol. (a) Second harmonic in pipe A = 2 [(v0)A] Third
harmonic of pipe B = 3 [(v0)B] = 2 l 2 v = 3 l 4 v = l 1 A A M RT γ = l 4 3 B B M RT γ A Gas (Monoatomic) MA l B Gas (Diatomic) MB l
Given that second harmonic in pipe A = Third harmonic of pipe B ⇒ l 1 A A M RT γ = l 4 3 B B M RT γ ⇒ B A M M = 189 400 [γ A = 1.67 and γβ = 1.4] (b) (v0)A = A A M RT γ (v0)B = B B M RT γ ∴ B A v v ) ( ) ( 0 0 = B B A A M M × γ γ = 4 3
2. A non-conducting disc of radius a and uniform positive surcface charge density σ is placed on the ground, with its axis vertical. A particle of mass m and positive charge q is dropped, along the axis of the disc, from a height H with zero initial velocity. The particle has
q/m = ε0g/σ.
(a) Find the value of H if the particle just reaches the disc.
(b) Sketch the potential energy of the particle as a function of its height and find its equilibirum position. [IIT- 1999] Sol. (a) Given that : a = radius of disc, σ = surface
charge density, q/m = 4ε0g/σ
The K.E. of the particle, when it reaches the disc can be taken as zero.
Potential due to a charged disc at any axial point situated at a distance x from 0.
V(x) = 0 2ε σ ] – [ a2+x2 x ] Hence, V(H) 0 2ε σ ] – [ a2+H2 H and V(O) = 0 2ε σa
According to law of conservation of energy, Loss of gravitation potential energy = gain in electric potential energy H (m,q) a O H mgH = qDV = q[V(0) – V(H)] mgH = g[a – { (a2+H2)– H}] 0 2ε σ …(1)
From the given relatuion : 0
2ε σq
= 2 mg (given) Putting this is equation (1), we get,
MgH = 2mg[a – { (a2+H2)–H }] or H = 2[a + H – (a2+H2)] or H = 2a + 2H – 2 (a2+H2) or 2 (a2+H2) = H + 2a or 4a2 + 4H2 = H2 + 4a2 + 4aH or 3H2 + 4aH or H = 3 4a [Q H = O is not valid]
KNOW IIT-JEE
By Previous Exam Questions(b) Total potential energy of the particle at height h U(x) = mgx + qV(x) = mgx + ( – ) 2 2 2 0 x x a q + ε σ = mgx + 2mg [ a2+x2 –x] = mg [2 a2+x2)–x] …(2) For equilibrium : dx dU = 0 This gives : x = 3 a
From equation (2), graph between U(x) and x is as shown above.
O a / 3 H = 4a/3 X U
2 mga 3mga
3. A wheel of radius R having charge Q, uniformly distributed on the rim of the wheel is free to rotate about a light horizontal rod. The rod is suspended by ligh inextensible strings and a magnetic field B is applied as shown in the figure. The initial tensions in the strings are T0. If the breaking tension of the stringas are
2
3T0 , find the maximum angular velocity ω
0 with which the wheel can be rotated. [IIT-2003]
T0 T0
d
B ω0
Sol. From above figure, when the ring is not rotating wt. of ring = Tension in string mg = 2T0
∴ T0 =
2
mg …(1)
When the ring is rotating, we can treat it as a current carrying loop. The magnetic mement of this loop
M = iA = T Q × πr2 = π 2 Q ω × πr2
This current carrying loop will create its own magnetic field which will interact with the given vertical magnetic field in such a way that the tensions in the strings will become unequal. Let the tension in the string be T1 and T2.
For translational equilibrium
T1 + T2 = mg …(2) For rotational equilibrium
Torque acting on the ring about the centre of ring → τ = M × → B → t = M × B × sin 90º = π 2 Q ω × πr2 × B = 2 2 Br Qω
For rotational equilibrium, the torque about the centre of ring should be zero.
∴ T1 × 2 D – T 2 × 2 D = 2 2 Br Qω ⇒ T1 – T2 = D Br Qω 2 …(3) On solving (2) and (3) we get
T1 = 2 mg + D Br Q 2 2 ω
But the maximam tension is
2 3T0 ∴ 2 3T0 = T 0 + D Br Q 2 2 max ω = 2 0 mg T Q ∴ ωmax = 02 BQr DT
4. An object is moving with velocity 0.01 m/s towards a convex lens of focal length 0.3 m. Find the magnitude of rate of separation of image from the lens when the object is at a distance of 0.4m From the lens. Also calculated the magnitude of the rate of change of the lateral magnification. [IIT-2004] Sol. f = 0.3 m, u = – 0.4 m
Using lens formula v 1 – 4 . 0 – 1 = 3 . 0 1 ⇒ v = 1.2 m
Now we have v 1 – u 1 = f 1 , differentiating w.r.t. t we have – 2 v 1 dt dv + 12 u dt du = 0 given dt du = 0.01 m/s ⇒ dt dv = 2 2 ) 4 . 0 ( ) 120 ( × 0.01 = 0.09 m/s
So, rate of seperation of the image (w.r.t. the lens) = 0.09 m/s Now, m = u v ⇒ dt dm = 2 u dt vdu – dt udv 2 ) 4 . 0 ( ) 01 . 0 )( 2 . 1 ( – ) 09 . 0 )( 4 . 0 ( = – 0.35
So magnitude of the rate of change of lateral magnification = 0.35.
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. 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 3 4 2 3 3 a r π × = 2 2 3 4 2 3 3 a a π × = 2 3 π = 0.74 = 74%
∴ The percentage of void space = 100 – 74 = 26%
7. (The standard reduction potential of Ag+/Ag electrode at 298 K is 0.799V. Given that for AgI, Ksp = 8.7 × 10–17, evaluate the potential of Ag+/Ag electrode in a saturated solution of AgI. Also calculate the standard reduction potential of I–electrode. [IIT-1994] Sol. In the saturated solution of AgI, the half cell
reactions are
At anode : Ag → Ag+ + e– At cathode : AgI + e– → Ag + I– Cell reaction AgI → Ag+ + I–
On applying Nernst equation Ecell = Eºcell –
n 0591 . 0 log [Ag+] [I–] For electrode Ag+ + e– → Ag ∴ EAg+/Ag= EºAg+/Ag– n 0591 . 0 log ] Ag [ 1 + Ksp of AgI = [Ag+] [I–] Q [Ag+] = [I–] ∴ Ksp of AgI = [Ag+]2 ∴ [Ag+] of AgI = sp K of AgI [Ag+] = 8.7×10−17 = 9.3 × 10–9 M So EAg+/Ag = 0.799 – 1 0591 . 0 log 9 10 3 . 9 1 − × = + 0.799 + 0.0591 log 9.3 – 0.0591 × 9 log 10 = + 0.799 + 0.0591 × 0.9785 – 0.0591 × 9 = 0.325 V
For above cell reaction Ecell = Eºcell –
n 0591 . 0 log [Ag+] [I–] = Eºcell – n 0591 . 0 log (Ksp of AgI) At equilibrium Ecell = 0 ∴ Eºcell = 1 0591 . 0 log(8.7 × 10–17) = –0.95 volt Eºcell = Eºcathode + Eºanode
–0.95 = –0.799 + EºAg/AgI/I–
(In form of cell reaction) EºAg/AgI/I– = – 0.95 + 0.799 = –0.151 V or EºI–/AgI/Ag = + 0.151 V
8. An organic compound A, C6H10O, on reaction with CH3MgBr followed by acid treatment gives compound B. The compound B on ozonolysis gives compound C, which in presence of a base gives 1-acetyl cyclopentene D. The compound B on reaction with HBr gives compound E. Write the structures of A, B, C and E. Show how D is formed from C.
[IIT-2000] Sol. The given reactions are as follows.
O CH3MgBr OMgBr CH3 H+ –H2O CH3 HBr CH3 Br (A) (B) (E) CH3 O O COCH3 O Base COCH3 (D) (C)
The conversion of C into D may involve the following mechanism. COCH3 (C) CH2 O –BH+ B+ COCH3 HC O COCH3 HC O– –B BH+ COCH3 OH –BH+ +B COCH3 OH – –OH– COCH3 (D) 9. A colourless solid (A) on heating gives a white solid
(B) and a colourless gas (C). (B) gives off reddish-brown fumes on treating with H2SO4. On treating with NH4Cl, (B) gives a colourless gas (D) and a residue (E). The compound (A) on heating with (NH4)2SO4 gives a colourless gas (F) and white residue (G). Both (E) and (G) impart bright yellow colour to Bunsen flame. The gas (C) forms white powder with strongly heated Mg metal which on hydrolysis produces Mg(OH)2. The gas (D) on heating with Ca gives a compound which on hydrolysis produces NH3. Identify compounds (A) to (G) giving chemical equations involved.
Sol. The given information is as follows : (i) A →Heat B + C Colourless Solid Colourless
Solid gas
(ii) B + H2SO4 →∆ Reddish brown gas (iii) B + NH4Cl →∆ D + E
Colourless gas (iv) A + (NH4)2SO4 →∆ F + G
olourless gas White Residue (v) E and G imparts yellow colour to the flame.
(vi) C + Mg →Heat White powder →H2O Mg(OH) 2 (vii) D + Ca →Heat Compound H →2O NH
3
Information of (v) indicates that (E) and (G) and also (A) are the salts of sodium because Na+ ions give yellow coloured flame. Observations of (ii) indicate that the anion associated with Na+ in (A) may be NO3–. Thus, the compound (A) is NaNO3.
The reactions involved are as follows : (i) 2NaNO3 →∆ 2NaNO2 + O2 ↑
(A) (B) (C)
(ii) 2NaNO2 + H2SO4 → Na2SO4 + 2HNO2 (B) Dil.
3HNO2 → HNO3 + H2O + 2NO↑ 2NO + O2 → 2NO2 ↑
Reddish brown
Fumes
(iii) NaNO2 + NH4Cl → NaCl + N2 ↑ + 2H2O (B) (E) (D) (iv) 2NaNO3 + (NH4)2SO4 →∆ Na2SO4 + 2NH3 (A) (G) (F) 2HNO3 (v) O2 + 2Mg →∆ 2MgO →H2O Mg(OH)2 (C) (vi) N2 + 3Ca →∆ Ca3N2 (D) Ca3N2 + 6H2O → 3Ca(OH)2 + 2NH3 ↑ Hence, (A) is NaNO3, (B) is NaNO2, (C) is O2, (D) is N2, (E) is NaCl, (F) is NH3 and (G) is Na2SO4.
10. An alkyl halide X, of formula C6H13Cl on treatment with potassium t-butoxide gives two isomeric alkenes Y and Z(C6H12). Both alkenes on hydrogenation give 2, 3-dimethyl butane. Predict the structures of X, Y
and Z. [IIT-1996]
Sol. The alkyl halide X, on dehydrohalogenation gives two isomeric alkenes.
X13 6H Cl C HCl – ; butoxide t K ∆ − − → 12 6H C Z Y+
Both, Y and Z have the same molecular formula C6H12(CnH2n). Since, both Y and Z absorb one mol of H2 to give same alkane 2, 3-dimethyl butane, hence they should have the skeleton of this alkane.
Y and Z (C6H12) Ni H2 → CH3 – CH – CH – CH3 CH3 CH3 2,3-dimethyl butane
The above alkane can be prepared from two alkenes CH3 – C = C – CH3 CH3CH3 2,3-dimethyl butene-2 (Y) and CH3 – CH – C = CH2 CH3 CH3 2,3-dimethyl butene-1 (Z)
The hydrogenation of Y and Z is shown below :
CH3 – C = C – CH3 CH3CH3 (Y) H2 Ni CH3 – CH – CH – CH3 CH3 CH3 CH3 – CH – C = CH2 CH3 CH3 (Z) H2 Ni CH3 – CH – CH – CH3 CH3 CH3 Both, Y and Z can be obtained from following alkyl halide :
CH3 – C – CH – CH3 CH3 CH3 2-chloro-2,3-dimethyl butane (X) K-t-butoxide ∆; –HCl CH2 = C — CH – CH3 CH3 CH3 Cl + CH3 – C = C – CH3 CH3CH3 (Z) 20% (Y) 80% Hence, X, CH3 – C – CH – CH3 CH3 CH3 Cl Y, CH3 – C = C – CH3 CH3 CH3 Z, CH3 – CH – C = CH2 CH3 CH3
MATHEMATICS
11. The curve y = ax3 + bx2 + cx + 5, touches the x-axis at
P(–2, 0) and cuts the y axis at a point Q, where its gradient is 3. Find a, b, c. [IIT-1994] Sol. It is given that y = ax3 + bx2 + cx + 5 touches x-axis at
P(–2, 0) which implies that x-axis is tangent at (–2, 0) and the curve is also passes through (–2, 0). The curve cuts y-axis at (0, 5) and gradient at this point is given 3 therefore at (0, 5) slope of the tangent is 3.
Now, dx dy
= 3ax2 + 2bx + c
since x-axis is tangent at (–2, 0) therefore 2 − = x dx dy = 0 ⇒ 0 = 3a(–2)2 + 2b(–2) + c ⇒ 0 = 12a – 4b + c ...(1) again slope of tangent at (0, 5) is 3 ⇒ ) 5 , 0 ( dx dy = 3 ⇒ 3 = 3a(0)2 + 2b(0) + c ⇒ 3 = c ...(2)
Since, the curve passes through (–2, 0), we get 0 = a(–2)3 + b(–2)2 + c(–2) + 5
0 = – 8a + 4b – 2c + 5 ...(3) from (1) and (2), we get
12a – 4b = –3 ...(4)
from (3) and (2), we get
– 8a + 4b = 1 ...(5)
adding (4) and (5), we get 4a = –2 ⇒ a = –1/2 Putting a = –1/2 in (4), we get 12(–1/2) – 4b = –3 ⇒ – 6 – 4b = –3 ⇒ – 3 = 4b ⇒ b = –3/4 Hence, a = –1/2, b = –3/4 and c = 3
12. In a triangle ABC, the median to the side BC is of length
3 6 11
1
− and it divides the angle A into
angles 30º and 45º. Find the length of the side BC. [IIT-1985] Sol. Let AD be the median to the base BC = a of ∆ABC
let ∠ADC = θ then
+ 2 2 a a cot θ = 2 acot 30º – 2 a cot 45º ⇒ cot θ = 2 1 3−
Applying sine rule in ∆ADC, we get
C D a/2 a/2 B A 45º 30º θ ) º 45 sin( AD − θ − π = sin45º DC ⇒ ) º 45 sin( AD + θ = 1/ 2 2 / a ⇒ AD = 2
a {sin 45º cosθ + cos45ºsinθ}
⇒ AD = θ+ θ 2 sin cos 2 a = 2 a(cos θ + sin θ) ⇒ 3 6 11 1 − = − + − − 3 2 8 2 3 2 8 1 3 2 a ⇒ a = 3 6 11 ) 1 3 ( 3 2 8 2 − + − ⇒ a = 3 6 11 ) 1 3 ( 3 2 8 2 2 − + −
⇒ a = ) 3 6 11 )( 3 2 4 ( 3 2 8 2 − + − ⇒ a = 2 36 3 22 3 24 44 3 2 8 − + − − = 2 3 2 8 3 2 8 − − = 2
13. Without using tables, prove that (sin 12º) (sin 48º) (sin 54º) =
8 1
[IIT-1982] Sol. (sin 12º) (sin 48º) (sin 54º)
=
2
1(2 sin 12º sin 48º) sin 54º
=
2 1
{cos (36º) – cos (60º)}sin 54º
= 2 1 − 2 1 º 36 cos sin 54º = 4 1
{2 cos 36º sin 54º – sin 54º}
=
4
1(sin 90º + sin 18º – sin 54º)
= 4 1 + − − + 4 1 5 4 1 5 1 = 4 1 − − − + 4 1 5 1 5 1 = 4 1 − 2 1 1 = 8 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 nn C 6 ) 3 2 . 3 3 ( − + ×6 3 15. 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 af(x)dx = = − − = −∫
( ) , ( ) ( ) 2 ) ( ) ( , 0 0 f x dx 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 .1. Two capacitors C1 and C2, can be charged to a potential V/2 each by having
V S1 S2 R R C2 C1 O
(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.
+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + A C B
(A) e.m.f. induced in any one rod is 16V
(B) e.m.f. induced in the complete ∆ABC is 48 3V
(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 mininum value of n. (n, d are natural numbers) 1 2 3 4 2d 5 6 d d d d
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 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
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 b
c a
Q
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
Regents Physics
You Should Know
Nuclear Physics :
• Alpha particles are the same as helium nuclei and have the symbol .
• The atomic number is equal to the number of protons (2 for alpha)
• Deuterium ( ) is an isotope of hydrogen ( )
• The number of nucleons is equal to protons + neutrons (4 for alpha)
• Only charged particles can be accelerated in a particle accelerator such as a cyclotron or Van Der Graaf generator.
• Natural radiation is alpha ( ), beta ( ) and gamma (high energy x-rays)
• A loss of a beta particle results in an increase in atomic number.
• All nuclei weigh less than their parts. This mass defect is converted into binding energy. (E=mc2)
• Isotopes have different neutron numbers and atomic masses but the same number of protons (atomic numbers).
• Geiger counters, photographic plates, cloud and bubble chambers are all used to detect or observe radiation.
• Rutherford discovered the positive nucleus using his famous gold-foil experiment.
• Fusion requires that hydrogen be combined to make helium.
• Fission requires that a neutron causes uranium to be split into middle size atoms and produce extra neutrons.
• Radioactive half-lives can not be changed by heat or pressure.
• One AMU of mass is equal to 931 meV of energy (E = mc2).
1. Option [C] is correct
Magnetic field due to infinite current carrying sheet is given by ,
2 J
B=µ0 where J is linear current density. 2 J 0 µ I IV (b) (a) 2 J 0 µ 2 J 0 µ 2 J 0 µ
Fig. (a) and (b) represent the direction of magnetic field due to current carrying sheets. For x < a,
2 ) J 4 ( 2 ) J 3 ( 2 ) J 2 ( J 2 J B 0 0 0 0 t tan resul µ + µ − µ − µ = For a < x < 2a, J 2 ) J 4 ( 2 ) J 3 ( 2 ) J 2 ( 2 J B 0 0 0 0 0 t tan resul =−µ µ + µ − µ − µ = For 2a < x < 3a, 0 2 ) J 4 ( 2 ) J 3 ( 2 ) J 2 ( 2 J B 0 0 0 0 t tan resul = µ − µ − µ + µ =
So, the required curve is
2. A → P,Q,S ; B → P,Q,R,S C → P,Q,R,S ; D → Q
i. Velocity of the particle may be constant, if forces of electric and magnetic fields balance each other. Then, path of particle will be straight line. Also, path of particle may be helical if magnetic and electric fields are in same direction. But path of particle cannot be circular. Path can be circular if only magnetic field is present, or if some other forces is present which can cancel the effect of electric field.
ii. Here, all the possibilities are possible depending upon the combinations of the three fields. iii. This situation is similar to part (i)
iv. In a uniform electric field, path can be only straight line or parabolic.
3. A → Q B → R
C → P D → Q
i. At t = 1s, flux is increasing in the inward direction, hence induced e.m.f. will be in anticlockwise direction.
ii. At t = 5s, there is no change in flux, so induced e.m.f. is zero
iii. At t = 9s, flux is increasing in upward direction hence induced e.m.f. will be in clockwise direction.
iv. At t = 15s, flux is decreasing in upward direction, so induced e.m.f. will be in anticlockwise direction.
4. Option [A,B,D] is correct
Rate of work done by external agent is
de/dt = BIL.dx/dt = BILv and thermal power dissipated in resistor = eI = (BvL) I clearly both are equal, hence (A).
If applied external force is doubled, the rod will experience a net force and hence acceleration. As a result velocity increase, hence (B).
Since, I = e/R
On doubling R, current and hence required power become half.
Since, P = BILv Hence (D)
5. Option [A] is correct
) j ( 2 ) j ( 5 . 1 µ→1×∧ = µ→2×∧ ] j ) j d i c [( 2 j ) j b i a ( 5 . 1 ∧+ ∧ ×∧= ∧+ ∧ ×∧ ∧ ∧ =2ck k a 5 . 1 3 4 5 . 1 20 c a= =
Solution
Physics Challenging Problems
Set # 9
6. Option [A] is correct f sin60º f f cos60º I2 f – + I2 x u = -f cos60º f = +f º 60 cos f 1 v 1 f 1 − − = f 2 v 1 f 1 + = v 1 f 2 f 1 = − v = -f º 60 cos xf = x º 60 cos f = x = 2f
∴ final image will formed at optical centre of first
lens. 7. Option [C] is correct Cv = (3 + 2T)R dQ = dU + PdV adiabatic process dQ = 0 0 = Rn (3 + 2T)dT + PdV dV V nRT dT ) T 2 3 ( Rn 0= + +
∫
∫
+ = − dT T T 2 3 V dV -log V = 3 logT + 2T + C -logV – logT3 = 2T + C log VT3 = 2T + C VT3 = e2T VT3e-2T = C 8. Option [A] is correct2 0 V P P= −α PV = RT 2 0 V P V RT α − = R V R V P T= 0 −α 3 0 dVdT = 0 R V 3 R P 2 0 − α = α = 3 P V 0 Now put V in T.
WHAT ARE EARTHQUAKES?
Earthquakes like hurricanes are not only super destructive forces but continue to remain a mystery in terms of how to predict and anticipate them. To understand the level of destruction associated with earthquakes you really need to look at some examples of the past.
If we go back to the 27th July 1976 in Tangshan, China, a huge earthquake racked up an official death toll of 255,000 people. In addition to this an estimated 690,000 were also injured, whole families, industries and areas were wiped out in the blink of a second. The scale of destruction is hard to imagine but earthquakes of all scales continue to happen all the time.
So what exactly are they ? Well the earths outer layer is made up of a thin crust divided into a number of plates. The edges of these plates are referred to as boundaries and it’s at these boundaries that the plates collide, slide and rub against each other. Over time when the pressure at the plate edges gets too much, something has to give which results in the sudden and often violent tremblings we know as earthquakes.
The strength of an earthquake is measured using a machine called a seismograph. It records the trembling of the ground and scientists are able to measure the exact power of the quake via a scale known as the richter scale. The numbers range from 1-10 with 1 being a minor earthquake (happen multiple times per day and in most case we don’t even feel them) and 7-10 being the stronger quakes (happen around once every 10-20 years). There’s a lot to learn about earthquakes so hopefully we’ll release some more cool facts in the coming months.
1. A trolley initially at rest with a solid cylinder placed on its bed such that cylinder axis makes angle θ with direction of motion of trolley as shown in Figure starts to move forward with constant acceleration a. If initial distance of mid point of cylinder axis from rear edge of trolley bed is d, calculate the distance s which the trolley goes before the cylinder rolls off the edge of its horizontal bed. Assume dimensions of cylinder to be very small in comparison to other dimensions. Neglect slipping.
d θ
Calculate also, frictional force acting on the cylinder. Sol. Since, axis of cylinder is inclined at angle θ with the direction of motion of trolley, therefore components of acceleration a of trolley are acosθ along axis of cylinder and asinθ normal to axis of the cylinder. Cylinder rolls backward due to this normal component asinθ.
Let mass and radius of cylinder be m and r respectively and let angular acceleration of cylinder be α.
Due to angular acceleration, cylinder axis has acceleration relative to trolley bed, which will be equal to rα normal to cylinder axis. But component of acceleration of trolley normal to cylinder axis is
asinθ. Therefore, net acceleration of cylinder axis is (asinθ – rα) normal to axis.
Consider free body diagram of the cylinder as shown figure
Note : There are two components of friction (i) F1 (normal to cylinder axis) and
(ii) F2 (along cylinder axis). F2 prevents cylinder from sliding along axis or acosθ component of acceleration of cylinder along axis is due to F2.
mg
l.α F1
N m(a.sin θ –rα)
∴ F2 = ma cos θ
F2 is not shown in the free body diagram because in this diagram forces action normal to cylinder axis are shown.
For horizontal forces,
F1 = m (a sin θ – rα) …(1) F1r = I α where I = 2 2 mr ∴ F1 = 2 1 mrα …(2)
Form equation (1) and (2),
rα =
3
2 a sin θ
The cylinder will roll off the edge of trolley bed when its centre of mass reaches the edge. Since. cylinder axis is inclined at an angle 'θ' with direction of motion of trolley, therefore, its centre of mass follows a straight line path relative to the trolley bed, and that straight line is normal to cylinder axis. Hence, displacement of centre of mass of the cylinde, relative to trolley is equal to (d. cosec θ).
considereing motion of cylinder relative to the trolley, u = 0, acceleration = rα = 3 2 a sin θ, s = d cosec θ, t = ? Using, s = ut + 2 1at2, or t = θ 2 sin 3 a d
Now considering motion of trolley during this interval ofd time,
u = 0, acceleration a , t = θ 2 sin 3 a d , s = ? Using, s = ut + 2 1at2, s = 2 3d cosec2 θ Ans. F1 = 2 1 m. rα = 3 1ma sin θ
Total frictional force acting on the cylinder is
F = F12+F22 =
3 1
ma sin2θ+9cos2θ Ans.
Expert’s Solution for Question asked by IIT-JEE Aspirants
Students' Forum
2. A particle of mass m is placed on centre of curvature of a fixed, uniform semi-circular ring of radius R and
M as shown in Figure. Claculate m M
R
(i) interaction force between the ring and the particle and
(ii) work required to displace the particle from centre of curvature to infinity.
Sol. To calculate, interaction force, consider two equal are lengths R dθ each of the semi-circular ring as shown in figure Rdθ x dθ dθ θ θ Rdθ
Mass of each arc, dM =
R M
π Rdθ = π θ
md
Gravitational force exerted by each arc on the particle, dF = 2 R GmdM = 2 R GMm π dθ
Since, force exerted by each arc is directly towards the arc, therefore, resultant of these two forces is along negative x-axis and the resultant force = dF1 cos θ
= 2 2
R GMm
π cos θ dθ
Total force on the particle is
F = 2 2 R GMm π
∫
π = θ = θ θ θ 2 / 0 cos d or F = 2 2 R GMm π Ans. (i)Work done during displacement of particle from centre of the curvature to infinity is used to increase gravitational potential energy of the system.
Initial gravitational potential energy of particle with each arc is dU = – R dM Gm. = – R GMm π dθ ∴ Total initial potential energy,
U1 = – R GMm π
∫
π π = θ θ 2 / 2 / – d or U1 = – R GMmWhen separation between particle and semicircular ring becomes large, potential energy becomes U2 = 0 ∴ Work done = U2 – U1 =
R GMm
Ans.(ii)
3. A long round conductor of radius a is made of a material whose thermal conductivity depends on distance r from axis of the conductor as K = cr2 , where c is a constant. Calculate
(i) thermal resistance per unit length of such a conductor and
(ii) temperature gradient if rate of heat flow through the rod is H.
Sol. Since, thermal conductivity of material of the conductor depends upon distance from its axis, therefore, conductivity at every point of a co-axial cylindrical surface will be the same. To calculate thermal resistance of the given conductor, it may be assumed to be composed of thin co-axial cylindrical shells which are in parallel combination with each other.
Consider a thin co-axial cylindrical shell of radius x, radial thickness dx and of unit length as shown in figure
Its cross sectional area, A = 2πx.dx
Thermal conductivity K = cx2 and length l = 1 m ∴ Its thermal resistance, dR =
KA l = dx . x 2 ) cx ( 2 π l or dR = dx cx . 2 1 3 π
Since, such cylindrical shells are in parallel with each other, therefore, equivalent resistance R per unit length is given by R 1 =
∫
dR 1 =∫
= = π a x x dx cx 0 3. 2 or R = 24 ca π Ans.(i) Since, temperature gradient is temperature difference per unit length, therefore, temperature gradient = rate of heat flow × resistance per unit lengthor dt dθ = H × R = 2 4 ca H π Ans. (ii) 4. Switch S of circuit shown in Figure is in position 1
for a long time. At instant t = 0, it is thrown from position 1 to 2. Calculate thermal power P1(t) and P2(t) generated across resistance R1 and R2 respectively.
C R2 R1 2 S 1 + – E
Sol. Since, initially the switch was in position 1 for a long time, therefore, initially the capacitor was fully charged or potential difference across capacitor at
t = 0 was equal to emf E fo the battery.
∴ Initial charge on capacitor, q0 = CE When switch is thrown to position 2, capacitor starts to discharge through resistance R1 and R2. To calculate thermal power P1(t) and P2(t) generated across R1 and R2 respectively, current I at time t through the circuit must be known.
Let at instant t, charge remaining on the capacitor be
q and let current through the circuit be I.
Applying Kirchhoff's voltage law on the mesh in the circuit of figure C R2 R1 I q + – I C q – IR 2 – IR1 = 0 or I = C R R q ) ( 1+ 2 ...(1)
Since, the capacitor is discharging, therfore,
I = – dt dq ∴ From equation (1), q dq = – C R R dt ) ( 1+ 2 ...(2)
Knowing that at t = 0, q = q0 = CE, integrating equation (2),
∫
= = ? q CE q q dq = –∫
= + t t R R C dt 0( 1 2) ∴ log CE q = – C R R t ) ( 1+ 2 or q=CEe–t/(R1+R2)C But I = – dt dq , therefore, I = ) (R1 R2 E + C R R t e– /( 1+ 2) ...(3) Hence, thermal power across R1 isP1 = I2R1 or P1 = C R R R E ) ( 1 2 1 2 + C R R t e–2/( 1+ 2) Ans. Similarly, thermal power across R2, P2 = I2R2
or P2 = e t R R C R R R E –2/( ) 2 2 1 2 2 2 1 ) ( + + Ans.
5. Two plane mirrors, a source S of light, emitting mono-chromatic rays of wavelength λ and a screen are arranged as shown in figure. If angle θ is very small, calculate fringe width of interference pattern formed on screen by reflected rays.
b a S Screen θ θ
Sol. Since, interference is due to reflected rays, therefore, images S1 and S2 of the source S behave like two coherent sources as shown in figure
b a S O θ θ M N R d D
Distance of source S from each mirror = a cos θ ∴ SS1 = SS2 = 2 × a cos θ
Distance between S1 and S2, d = SS1 sin θ + SS2 sin θ = 4a cos θ sin θ
But θ is very small, therefore cos θ ≈ 1 and sin θ ≈ θ ∴ d = 4aθ
Distance RS = SS1 cos θ = 2a.cos2θ ≈ 2a ∴ Distance of screen from two coherent sources S1 and S2 is
D = RO = RS + SO
or D = (2a + b)
Now the arrangement is similar to Young's double slit arrangement. ∴ Fringe width, ω = d Dλ = θ λ + a b a 4 ) 2 ( Ans.
Matter Waves :
Planck's quantum theory : Wave-particle duality - Planck gave quantum theory while explaining the
radiation spectrum of a black body. According to Planck's theory, energy is always exchanged in integral multiples of a quanta of light or photon. Each photon has an energy E that depends only
on the frequency ν of electromagnetic radiation and is given by :
E = hν ...(1) where h = 6.6 × 10–34 joule-sec, is Planck's
constant. In any interaction, the photon either gives up all of its energy or none of it.
From Einstein's mass-energy equivalence principle, we have
E = mc2 ...(2) Using equations (1) and (2), we get ;
mc2 = hν or m = 2
c hν
...(3) where m represents the mass of a photon in
motion. The velocity v of a photon is equal to that of light, i.e., v = c.
According to theory of relativity, the rest mass m0 of a photon is given by : m0 = 2 2 c v 1 m − Here, m = 2 c hν and v = c Hence, m0= 0 ....(4)
i.e., rest mass of photon is zero, i.e., energy of photon is totally kinetic.
The momentum p of each photon is given by :
p = mc = 2 c hν × c = c hν = ν / c h = λ h ...(5)
The left hand side of the above equation involves the particle aspect of photons (momentum) while the right hand side involves the wave aspect (wavelength) and the Planck's constant is the bridge between the two sides. This shows that electromagnetic radiation exhibits a
wave-particle duality. In certain circumstances, it behaves like a wave, while in other circumstances it behaves like a particle.
The wave-particle is not the sole monopoly of e.m. waves. Even a material particle in motion according to de Broglie will have a wavelength. The de Broglie wavelength λ of the matter waves is also given by : λ = mv h = p h = mK h 2
where K is the kinetic energy of the particle. If a particle of mass m kg and charge q coulomb
is accelerated from rest through a potential difference of V volt. Then
2 1mv2 = qV or mv = 2mqV Hence, λ = mqV h 2 = V 34 . 12 Å Photoelectric effect :
When light of suitable frequency (electromagnetic radiation) is allowed to fall on a metal surface, electrons are emitted from the surface. These electrons are known as photoelectrons and the effect is known as photoelectric effect. Photoelectric effect, light energy is converted into electrical energy.
Laws of photolectric effect :
The kinetic energy of the emitted electron is independent of intensity of incident radiation. But the photoelectric current increases with the increase of intensity of incident radiation.
The kinetic energy of the emitted electron depends on the frequency of the incident radiation. It increases with the increase of frequency of incident radiation.
If the frequency of the incident radiation is less than a certain value, then photoelectric emission is not possible. This frequency is known as threshold frequency. This threshold frequency varies from emitter to emitter, i.e., depends on the material.
There is no time lag between the arrival of light and the emission of photoelectrons, i.e., it is an instantaneous phenomenon.
