Dear Students,
Motivate Yourself
One of the greatest virtues of human beings is their ability to think and act accordingly. The emergence of the techno savvy human from the tree swinging ape has really been a long journey. This transition has taken a span of countless centuries and lots of thinking caps have been involved. Inquisitiveness and aspiration to come out with the best have been the pillars for man's quest for development. Self-motivation is the sheer force, which pulled him apart and distinguished him from his primitive ancestors.
Many times, in our life, when we are reviving old memories we get into a phase of nostalgia. We feel that we could have done better than what we had achieved. But thinking back won’t rewind the tireless worker called time. All we can do is promise ourselves that we will give our very best in the future. But do we really keep up to our mental commitments? I can guess that 90% answers are in the negative. This is because of that creepy careless attitude which is slowly, but surely entering into our mind. We easily forget the pains of yesterday to relish the joys of today. This is the only time in our life, when we can control our fate, by controlling our mind. So it is time to pull up our socks and really motivate ourselves so that we can give our best shot in the future. Self-motivation is the need of the hour. Only we can control and restrict ourselves. It’s up to us, how we use our mental capabilities to the best of our abilities.
Here are some Funda's for self-motivation. Don't just read them digest each one of them and apply them and I bet it will make a better YOU.
• The ultimate motivator is defeat. Once you are defeated, you have nowhere to go except the top.
• Then only thing stopping you is yourself.
• There is no guarantee that tomorrow will come. So do it today. • Intentions don't count, but action's do.
• Don't let who you are, stunt what you want to be. • Success is the greatest motivator.
• Your goals must be clear, but the guidelines must be flexible. Try to include these one liners in your scrapbook or on your favorite poster. You will be sub-consciously tuned to achieve what you want. Also do keep in mind that nothing can control your destiny but you! With Best Wishes for Your Future.
Yours truly
Pramod Maheshwari, B.Tech., IIT Delhi Every effort has been made to avoid errors or
omission in this publication. In spite of this, errors are possible. Any mistake, error or discrepancy noted may be brought to our notice which shall be taken care of in the forthcoming edition, hence any suggestion is welcome. It is notified that neither the publisher nor the author or seller will be responsible for any damage or loss of action to any one, of any kind, in any manner, there from.
• No Portion of the magazine can be published/ reproduced without the written permission of the publisher
• All disputes are subject to the exclusive jurisdiction of the Kota Courts only.
Owned & Published by Pramod Maheshwari, 112, Shakti Nagar, Dadabari, Kota & Printed by Naval Maheshwari, Published & Printed at 112, Shakti Nagar, Dadabari, Kota.
Editor : Pramod Maheshwari
To fly, we have to have resistance
Volume - 5 Issue - 10
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Volume-5 Issue-10
April, 2010 (Monthly Magazine)
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 – 2010 & 2011
Success Tips for the Month • "The way to succeed is to double your
error rate."
• "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
4• President buries 'Time Capsule' on IIT Kanpur campus
• IITs admission criteria set for an overhaul
IITian ON THE PATH OF SUCCESS
6Mr. Krishnamurthy Rengarajan
KNOW IIT-JEE
7Previous IIT-JEE Question
XTRAEDGE TEST SERIES
52 Mock Test IIT-JEE Paper-1 & Paper-2Mock Test AIEEE Mock Test BIT SAT
SOLUTIONS
92Regulars ...
DYNAMIC PHYSICS
148-Challenging Problems [Set# 12] Students’ Forum
Physics Fundamentals
Calorimetry, K.T.G., Heat Transfer Atomic Structure, X-Ray & Radio Activity
CATALYST CHEMISTRY
30 Key ConceptAromatic Hydrocarbon Solubility Product
Understanding: Organic Chemistry
DICEY MATHS
38 Mathematical Challenges Students’ Forum Key Concept Calculus AlgebraStudy Time...
Test Time ...
President buries 'Time Capsule' on IIT Kanpur campus
Kanpur: President Pratibha Patil recently buried a 'Time Capsule' on the Indian Institute of Technology, Kanpur (IIT-K) campus on the occasion of its golden jubilee celebrations and also unveiled a nanosatellite developed by the institute. The capsule, which is made of a special metal, contains pen-drives, chips, images and several other documents related to the landmark achieve-ments of the IIT-K.
Lauding the nanosatellite Jugnu's development team, Patil said it projects the complex nature of tasks that the students there were equipped to handle.
Congratulating IIT-K students and faculty, Patil said that the institute has come a long way in its 50 years of its existence, and also called upon the institute's students and faculty members to develop such devices that can harness energy in efficient ways with minimal negative impact on the environment.
"It (IIT-K) had made an impact on technical education within the country, while its students through their innovations, have played an important role in India, as well as around the world," the President said. Jugnu, developed by a team of 50 IIT students, will help in collection of information related to floods, drought and other natural calamities.
IITs admission criteria set for an overhaul
New Delhi: The admission criteria for admission to the premier engineering institutes of the country - the Indian Institutes of Technology (IITs), is all set for an overhaul.
According to the recommendation of a panel set up under IIT Kaharagpur's Director, Mr. Damodar Acharya, the Joint Entrance Examination (JEE) for undergraduates and Graduate Aptitude Test in Engineering (GATE) for postgraduates may soon not be the only criteria for their admission. The committee proposes to consider the Class XII marks as well.
Presently, the eligibility for a student to sit for a JEE is 60% in class XII after which a cut-off is decided every year for admission into an IIT. Following the consensus of the last meeting of the IIT Council to give more weightage to the school leaving examination, the committee proposes to mull over the issue of factoring in class XII examination result in the cut-off for admission to the IITs.
The IIT Directors in a meeting with the Human Resource Development (HRD) Minister, Kapil Sibal, in Manesar, also expressed their discontent over the existing pattern for GATE as they felt that not enough students with research orientation were being picked up through this exam. Sibal said that each IIT should submit a proposal within a month on one area of expertise in higher
research. Although the admission criteria is becoming stringent with the proposal of IIT-Kharagpur Director's recommendation, the HRD minister retreated with a firm 'No' to IIT-Kanpur Director's proposal for a fee hike. According to the fee-hike proposal, which the IIT-Kanpur's Director framed, it suggested to hike fees by eight times. At present, the B.Tech students pay Rs.50,000 per year as their fee which the committee proposed to increase up to Rs.4 lakh per annum over the period of 10 years with a Rs.35,000 mark-up every year.
"The proposal on fee hike should be discussed as the government was planning to set up a Higher Education Funding Corporation which would address poor students," said a senior official. During the meeting, Sibal also asked the IITs to come up with their plans for the future endeavors in a specialized area in which they want to emerge as global giants by 2020 in four weeks time.
Nanotech is used to treat cancer - IIT-B & Docs
In Mumbai it could have been India’s second nano success if it passes the muster. Only this nano creation is revealed in the healthcare field and it was possible because of partnership between oncologists and scientists of the Indian Institute of Technology-Bombay.
The joint effort of the IIT -B and doctors from Tata Memorial
Hospital in Parel and Apollo Hospital in Hyderabad has the possibility to transform treatment of retinoblastoma—a rare cancer of the retina that mainly affects children under two years of age. They have developed a nano-particle that could conquer the child killer.
Shirin Thakur, Guntur-based teenager was suffering from recurrent retinoblastoma since she was two years old. Last week, she took the third injection of a special mixture into the tissues around her left eye which was made of nano-particles of carboplatin which is commonly used to treat retinoblastoma. While she was standing with her doctor in the Apollo hospital, Hyderabad said, “I have been suffering from attacks of retinoblastoma in my left eye since I was two. Even in the US, they told me there is no hope but to remove my eye.’’ Now, she has “fuzzy’’ vision in the nearly blind eye. “My vision gets better every day.’’
PM was asked to fine-tune JEE by Maths Wizard
PM Manmohan singh was given suggestions on the improvement of IIT-JEE pattern by Maths Wizard Anand Kumar. He pleaded that poor aspirants should be given at least three attempts for IIT-JEE exam, as they are usually the late starters.
Every year Anand’s Super-30 offers free accommodation to 30 poor students and provides coaching to them free of cost in order to help them to crack JEE. This noble initiative, has of late reported 100% success rate with all its 30 students making it to IIT’S, without seeking any financial support from government or non-government sources.
In his 15-minute interaction with the PM in New Delhi, Anand enlightened him of his efforts usher in a new awakening in Bihar by sending 182 poor students to IITs within the past seven years. He further informed the PM that he has decided to increase the student intake to 60 from 30. Very soon he would open schools for talented poor children so as to provide them the right momentum at the secondary school level itself. He added that government should run coaching programmes on the pattern of Super-30 for talented students from the countryside and it should not be confined to IIT only.
Audibly elated Anand told TOI over the phone that Pm patted him on his back and also instructed PMO to look into his suggestions.
He said that many problems asked in JEE are of Maths Olympiad level so students from villages find it difficult to solve such problems even if they have sound knowledge of the subject at +2 level. He pleaded that the exam should be designed in such a way that conceptual or analytical problems of the +2 level should be asked in JEE."
IIT Rajasthan moves to state varsity campus
Jaipur: The Indian Institute of Technology, Rajasthan (IIT-R) has been temporarily shifted to the engineering faculty building of Jai Narain Vyas University according to a Memorandum of Understanding (MoU) signed in Jodhpur over the weekend. Students of IIT-R earlier had to rely on IIT-Kanpur and other institutions to attend classes.
IIT-Rajasthan founder-director P. K. Kalra and JNV University Registrar Nirmala Meena, in the
presence of Chief Minister Ashok Gehlot who was on a visit to Jodhpur, signed the MoU.
The CM also inaugurated the work to extend the building. Gehlot pointed out in an official press release that the establishment of IIT in Jodhpur was recommended and headed by the noted economist V.S. Vyas after he studied the facilities and educational standard of various towns in the State.
"The final decision to nod for an IIT in the city was accepted at the central level," added Gehlot. He further announced that the construction of a new building in Rajasthan would require some time, and for the time being the classes would be held in JNV University while those in the extended portion of the building would begin from May his year. Following special efforts made by the state government Mr. Gehlot said that the union government has also nodded to use the newly constructed A.S.K. Hostel in the university for IIT students.
"The establishment of an IIT in the state would definitely boost its higher education and will prove to be a milestone," said Gehlot.
According to the MoU, it approves the functioning of IIT-Rajasthan only on temporary basis for two years in the engineering faculty building.
State Technical Education Minister Mahendrajit Singh Malviya and JNV University Vice-Chancellor Naveen Mathur signed the MoU as witnesses. Among others, Jodhpur MP Chandresh Kumari, Barmer-Jaisalmer MP Harish Chaudhary, Jodhpur Mayor Rameshwar Dadich, Deputy Mayor Niyaz Mohammed and Collector Naveen Mahajan were present on the occasion.
Knowledge is indeed wealth. Who better exemplifies it than Krishnamurthy Rengarajan,IIT-B gold medallist (B Tech dual-degree course). Krishnamurthy's story is that of hard work, sheer grit and determination. His undying passion for learning and excellence has paid off. Coming from a lower middle class background, Krishnamurthy has made his parents proud when he passed with flying colours.
His father, who works as a typist at Bharatiya Vidya Bhavan is overwhelmed by his son's achievement. Rengarajan, who hails from Tamil Nadu, came to Mumbai 28 years ago and settled down in a distant Mumbai suburb of Dombivli. Though the family went through a lot of hardships initially, he made sure that his children were well educated.
"My son always wanted to join the IIT. When people asked him what if you don't get through the entrance examinations, he used to say, `there is no question of me not clearing the test," says his proud father. And, of course, he did top all the five years at IIT, a result of sheer hard work and brilliance, says his mother, barely able to control her excitement. "I am very happy for him," says Radha Rengarajan. Krishnamurthy did his schooling at the Kidland School in Dombivli and pre-degree from V G Vaze College at Mulund. His favourite subject being mathematics it was obvious that he would pursue a degree in engineering. He won the Rakesh Mathur award of Rs 1 lakh (Rs 100,000) during his third year and other scholarships throughout the four years. Here's what Krishnamurthy had to say on his IIT experience.
My IIT experience
The five years I spent at IIT were the best in my life. I will cherish each and every moment here. I loved everything here: the professors are the best one can ever get, the facilities to study and the extra-curricular activities are excellent. I made best of friends and thoroughly enjoyed my college life.
I don't think I will ever get this experience anywhere else.
On studies
Before joining IIT, I used to study for 7 to 8 hours daily. After joining IIT, I used to spend about a couple of hours. I did not go for anything coaching classes. I learnt through Brilliant Tutorial correspondence course and my preparation began after I finished my 10th standard.
Why IIT
IIT is one of the premier institutes in India. I always wanted to get good higher education, so I opted for IIT.
My mantra for success
There is no short cut to success. One has to work very hard, put in a lot of effort, should have a problem-solving mentality and a right approach to every problem. My parents always stood by me, their support has been invaluable and am overwhelmed.
Advice to IIT aspirants
Work hard. You have to spend a lot of time preparing as exams are getting more and more competitive. You must also have problem-solving skills.
Interests
Solving math puzzles, reading books. I used to play cricket, but now I don't get the time.
Next move
Money is the least important thing for Krishnamurthy. So no jobs for the time being. "I have been selected for the scholarship programme at Stanford University for a PhD in operations research. I would like to research on optimising computer networks and operation systems. Quality research is available abroad. After the PhD programme I would like to join any academia of good repute and continue my research activities. Among corporates, I admire Google. It is the one company that reflects perfection, hard work and efficiency."
Will you come back to India?
Of course, I will. The brain drain phenomenon is dying out.
It's the time for reverse brain.
Mr.
Krishnamurthy Rengarajan
IIT-B Gold Medallist
Success Story
PHYSICS
1. A spherical ball of mass m is kept at the highest point
in the space between two fixed, concentric spheres A and B (see figure in solution). The smaller sphere A has a radius R and the space between the two spheres has a width d. The ball has a diameter very slightly less than d. All surface are frictionless. The ball is given a gentle push (towards the right see figure in solution) The angle made by the radius vector of the ball with the upward vertical is denoted by θ [IIT-2002] (a) Express the total normal reaction force exerted by
the sphere on the ball as a function of angle θ.
(b) Let NA and NB denote the magnitudes of the
normal reaction forces on the ball exerted by the sphere A and B, respectively. Sketch the variations of NA and NB as functions of cos θ in the range 0 ≤ θ ≤ π
by drawing two separate graphs in your answer book, taking cos θ on the horizontal axes.
Sol. The ball is moving in a circular motion. The
necessary centripetal force is provided by (mg cos θ – N). Therefore V mgsinθ mgcosθ mg R θ θ C d/2 NA D A B mg cos θ – NA = + 2 d R mv2 …(i)
According to energy conservation 2 1mv2 = mg + 2 d R (1 – cos θ) …(ii) From (i) and (ii) NA = mg (3 cos θ – 2) …(iii)
The above equation shows that as θ increases NA
decreases. At a particular value of θ, NA will become
zero and the ball will lose contact with sphere A. This condition can be found by putting NA = 0 in eq. (iii)
0 = mg (3 cos θ – 2) ∴ θ = cos–1 3 2
The graph between NA and cos θ
From eq (iii) when θ = 0, NA = mg.
2mg NA mg cos θ1 2mg cos θ –1 5mg NB When θ = cos–1 3 2 ; N A = 0
The graph is a straight line as shown. When θ > cos–1 3 2 NB – (– mg cos θ) = 2 d R mv2 + ⇒ NB + mg cos θ = + 2 d R mv2 …(iv)
Using energy conservation 2 mv 2 1 = mg θ + − + cos 2 d R 2 d R + 2 d R mv2 = 2 mg [1 – cos θ] …(v)
From (iv) and (v) we get
NB + mg cos θ = 2 mg – 2mg cos θ NB – mg (2 – 3 cos θ) When cos θ = 3 2, N B = 0 When cos θ = – 1, NB = 5 mg
Therefore the graph is as shown.
2. A cylindrical block of length 0.4 m and area of cross section 0.04 m2 is placed coaxially on a a thin metal
disc of mass 0.4 Kg and the same cross section. The upper face of the cylinder is maintained at a constant temperature of 400 K and the initial temperature of the disc is 300 K. If the thermal conductivity of the materical of all cylinder is 10 watt/kg. K, how long will it take for the temperature of the disc to increases to 350 K? Assume, for purpose of calculation, the thermal conductivity of the disc to be very high and
KNOW IIT-JEE
the system to be thermally insulated except for the upper face of the cylinder. [IIT-1992] Sol. Initially the temperature at the open end is 400 K and
the temperature at the metal disc-cylinder interface is 300 K. Metal disc Insulation Cylinder H Open end θ
As heat passes through the cylinder and reaches the metal disc, the temperature of metal disc rises. Since the conductivity (thermal) of metal disc is very high so temperature of the whole disc will rise and along with that the temperature of the other end of the cylinder (metal disc-cylinder interface) also rises simultaneously. Let at any instant of time, the temperature of the metal disc-cylinder interface is θ. At this instant the rate of heat crossing the cylinder.
dt dQ = l ) – 400 ( KA θ ...(i) where K = thermal conductivity
A = area of cross section of cylinder l = length of cylinder
The same amount of heat is received by the metal disc. Therefore dt dQ = mc dt dθ ...(ii) m = mass of disc
c = specified heat of metal disc From (i) and (ii)
mc dt dθ = l ) – 400 ( KA θ θ − θ 400 d × KA mcl = dt On integrating At t = 0, θ = 300
∫
t 0dt = KA mcl∫
−θ θ 350 300(400 ) d At t = t, θ = 350 ⇒ t = KA mcl − [log (400 – θ) 350 300 ] = 04 . 0 10 303 . 2 4 . 0 600 4 . 0 × × × × − log10 300 400 350 400 − − = 166.38 sec.3. A particle of mass m and charge q is moving in a region where uniform, constant electric and magnetic fields E and B are present. E and B are parallel to each other. At time t = 0 the velocity v0 of the particle
is perpendicular to E. (Assume that its speed is always <<c, the speed of light in vaccum.) Find the velocity v of the particle at time t. You must express your answer in terms of t, q, m, the vectors v0, E and
B and their magnitudes v0, E and B. [IIT-1998]
Sol. Because the forces due to parallel electric and magnetic fields on a charged particle moving perpendicular to the fields will be at right angles to each other (electric force being along the direction of
→
E while magnetic force perpendicular to the plane containing v and B ) so magnetic force will not affect the motion of charged particle in the direction of electric field and vice-versa. So the problem is equivalent to superposition of two independent motions as shown in the adjoining figures.
Y X Z Fm v0 k jˆ iˆ B E Fe
So for motion of the particle under electric field alone, ay = m qE i.e., dt dvy = m qE or
∫
y 0 dv = y∫
y 0 m qE dt i.e., vy = m qE t ...(1) While at the same instant, the charged particle under the action of magnetic field will describe a circle in the x-z plane withr = qB mv0 i.e., ω = r v0 = m qB X v0 v0cos θ v0 θ θ Z v0 sin θ
So angular position of the particle at time t in the x-z plane will be given by
θ = ωt = m qB
t
and therefore in accordance with figure vx = v0 cosθ = v0 cos ωt = v0 cos
t m qB ...(2)
and vz = v0 sinθ = v0 sin ωt = v0sin
t m qB ...(3) So in the light of equations (1), (2) and (3), we get
→ v= iˆvx + jˆvy + kˆvz = iˆv0 cos t m qB + jˆ t m qE + v 0 sin t m qB kˆ
But because here, iˆ = 0 0 v v→ , jˆ = E E → = B B → and kˆ = B v B v 0 0× So, v = →0 → 0 0 v v v 0 cos m qBt + → E E m qEt + B v B v 0 0× v 0sin m qBt
4. A long solenoid of radius 'a' and number of turns per unit length n is enclosed by cylindrical shell of radius R, thickness d(d<<R) and length L. A variable current i = i0 sin ωt flows through the coil. If the
resitivity of the material of cylindrical shell is ρ, find the induced current in the shell. [IIT-2005] Sol. The magnetic field in the solenoid is given by
B = µ0ni
a
L
d R
⇒ B = µ0 n i0 sin ωt [Q i = i0 sin ωt given]
The magnetic flux linked with the solenoid φ = →B.→A = BA cos 90º
= (µ0 n i0 sin ωt) (πa2)
∴ The rate of change of magnetic flux through the solenoid
dt dφ
= π µ0 n a2 i0 ω cos ωt
The same rate of change of flux is linked with the cylindrical shell. By the principle of electromagetic induction, the induced emf produced in the cylinderical shell is I TOP VIEW e = – dt dφ = – πµ0 n a2 i0 ω cos ωt ...(i)
The resistance offered by the cylindrical shell to the flow of induced current I will be
R = ρ A l Here, l = 2πR, A = L × d ∴ R = ρ Ld R 2π ...(ii) The induced current I will be
I = R | e | = R 2 Ld ] t cos i na µ [ 0 2 0 π × ρ × ω ω π I = R 2 Ld t cos i na µ0 2 0 ρ × ω ω π
5. A beam of light has three wavelength 4144 Å, 4972 Å and 6216 Å with a total intensity of 3.6 × 10–3
Wm–2 equally distributed amongst the three
wavelengths. The beam falls normally on an area 1.0 cm2 of a clean metallic surface of work function
2.3 eV. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photo electrons liberated in two seconds. [IIT-1989]
Sol. Work function
= 2.3 eV = 2.3 × 1.6 × 10–19J = 3.68 × 10–19 J
Energy of a photon of wave length (4144 Å)
= λ hc = 10 8 34 10 4144 10 3 10 6 . 6 − − × × × × = 4.78 × 10–3 × 10–16 = 4.78 × 10–19 J
Energy of a photon of wave length 4972 Å
= 34 10 8 10 4972 10 3 10 6 . 6 − − × × × × = 3.98 × 10–19J
Energy of a photon of wavelength 6216 Å
= 10 8 34 10 6216 10 3 10 6 . 6 − − × × × × = 3.18 × 10–19 J
This means that light of wavelength 4144 Å and 4972Å are capable of ejection of electrons from the metal surface
The energy incident on 1 cm2 area of metal per
second = 3.6 × 10–3 × 10–4 = 3.6 × 10–7 J
The energy/s of wavelength 4972 Å = 1.2 × 10–7 J
No of photon incident of wavelength 4972 Å
E = λ hc n ∴ n = hc Eλ ⇒ n = 7 34 810 10 3 10 6 . 6 10 4972 10 2 . 1 × × × × × × − − − = 301.33 × 109 = 3.01 × 1011
Similarly number of photon incident of wavelength 4144 Å. n = 7 34 810 10 3 10 6 . 6 10 4144 10 2 . 1 × × × × × × − − − = 2.51 × 1011
⇒ Total number of photons capable of ejection of electrons per second
= 3.01 × 1011 + 2.51 × 1011 = 5.52 × 1011
∴ Total number of photoelectrons ejected in two seconds ≈ 11 × 1011.
CHEMISTRY
6. From the following data, form the reaction between
A and B. [IIT-1994]
Initial rate (mol L–1s–1)
[A] mol L–1 mol L[B]–1 300 K 320 K 2.5 ×10–4 3.0 ×10–5 5.0 ×10–4 2.0 × 10–3 5.0 × 10–4 6.0 × 10–4 4.0 × 10–3 – 1.0 × 10–3 6.0 × 10–5 1.6 × 10–2 – Calculate
(a) the order of reaction with respect to A and with respect to B,
(b) the rate constant at 300 K, (c) the energy of activation, (d) the pre exponential factor.
Sol. Rate of reaction = k[A]l [B]m
where l and m are the order of reaction with respect to A and B respectively. From the given data, we obtain following expressions :
5.0 × 10–4 = k[2.5 × 10–4]l [3.0 × 10–5]m ...(i)
4.0 × 10–3 = k[5.0 × 10–4] l [6.0 × 10–5]m ...(ii)
1.6 × 10–2 = k[1.0 × 10–3]l [6.0 × 10–5]m ..(iii)
From eq. (ii) and eq. (iii), we get
2 3 10 6 . 1 10 0 . 4 − − × × = l × × − − 3 4 10 0 . 1 10 0 . 5 or 0.25 = (0.5)l or (0.5)2 = (0.5) l or l = 2
From eq. (i) and eq. (ii), we get
3 4 10 0 . 1 10 0 . 5 − − × × = m 5 5 2 4 4 10 0 . 6 10 0 . 3 10 0 . 5 10 5 . 2 × × × × − − − − or 8 1 = 4 1 × m 2 1 or 2 1 = m 2 1 or m = 1 (b) At T1 = 300 K, k1 = 2 1 ] B [ ] A [ reaction of Rate = ] 10 0 . 3 [ ] 10 5 . 2 [ 10 0 . 5 5 2 4 4 − − − × × × = 2.67 × 108 L2 mol–2 s–1 (c) At T2 = 320 K, k2 = 2 1 ] B [ ] A [ reaction of Rate = ] 10 0 . 3 [ ] 10 5 . 2 [ 10 0 . 2 5 2 4 3 − − − × × × = 1.067 × 109 L2 mol–2 s–1 We know, 2.303 log 1 2 k k = − 2 1 1 2 a T T T T R E or 2.303 log 8 9 10 67 . 2 10 067 . 1 × × = × − 300 320 300 320 314 . 8 Ea or 2.303 × 0.6017 = ×300 320 20 314 . 8 Ea or Ea = 20 300 320 314 . 8 6017 . 0 303 . 2 × × × × = 55.3 kJ mol–1
(d) According to Arrhenius equation, k = Ae−Ea/RT or 2.303 log k = 2.303 log A – RT Ea At 300 K, 2.303 log (2.67 × 108) = 2.303 log A – 300 314 . 8 10 3 . 55 3 × × or 2.303 × 8.4265 = 2.303 log A – 22.17 or logA = 303 . 2 17 . 22 4062 . 19 + = 303 . 2 5762 . 41 = 18.0531 A = Antilog 18.0531 = 1.13 × 1018 s–1
7. 1 g of a mixture containing equal number of moles of carbonates of two alkali metals, required 44.4 ml of 0.5 N HCl for complete reaction. The atomic weight of one metal is 7, find the atomic weight of other metal. Also calculate the amount of sulphate formed on quantitative conversion of 1.0 g of the mixture in
two sulphates. [IIT-1972]
Sol. Let, Mass of one alkali metal carbonate M2CO3 = xg
Then, mass of other alkali metal carbonate M2´CO3
= (1 – x)g Step 1. Equivalent mass of M2CO3 = 2 mass Molecular = 2 74 = 37 (at mass of M = 7) Meq. of M2CO3 = 37 x ×1000 Moles of M2CO3 = 74 x Step 2
Equivalent mass of M2´CO3 =
2 mass Molecular = 2 60 m 2 + ( m = atomic mass of M´) Meq. of M2´CO3 = 60 m 2 ) x 1 ( 2 + − × 1000 Moles of M2´CO3 = 60 m 2 x 1 + − Step 3. Meq. of HCl = NHCl × VHCl = 0.5 × 44.4
Step 4. According to the question,
Moles of M2CO3 = Moles of M2´CO3
or 74 x = 60 m 2 x 1 + − ...(i)
And Meq. of M2CO3 + Meq . of M2´CO3 = Meq. of HCl or 37 x × 1000 + 60 m 2 ) x 1 ( 2 + − × 1000 = 0.5 × 44.4 ...(ii) Solving eq. (i) and (ii), we get
m = 23 and x = 0.41
∴ Mass of M2CO3 = x = 0.41 g
and Mass of M2´CO3 = 1 – x = 0.59 g
Step 5. Equivalent mass of M2SO4 = 2 mass Molecular = 2 110 = 55 Meq. of M2SO4 = 55 WM2SO4 × 1000
But, Meq. of M2SO4 = Meq. of M2CO3
∴ 55 WM2SO4 × 1000 = 37 41 . 0 × 1000 or WM2SO4 = 0.6095 g Step 6.
Equivalent mass of M2´SO4 =
2 mass Molecular = 2 142 = 71 Meq. of M2´SO4 = 71 WM2´SO4 × 1000 But, Meq. of M2´SO4 = Meq. of M2´CO3
∴ 71 WM2´SO4 × 1000 = 106 59 . 0 2× × 1000 or WM2´SO4= 0.7904 g
∴ Total mass of sulphates = WM2SO4 + WM2´SO4
= 0.6095 + 0.7904
= 1.3999 g
8. 0.9 g of a solid organic compound (molecular mass 90), containing carbon, hydrogen and oxygen, was heated with oxygen corresponding to a volume of 224 ml at STP. After combustion the total volume of the gases was 560 ml at STP. On treatment with potassium hydroxide, the volume decreased to 112 ml. Determine the molecular formula of the
compound. [IIT-1972]
Sol. Given that,
Mass of solid organic compound = 0.9 g Molecular mass of organic compound = 90 ∴ No. of moles of organic compound available
= 90 9 . 0 = 0.01 Volume of O2 taken = 224 ml Volume of O2 used = 224 – 112 = 112 ml 22400 ml O2 = 1 mol. ∴ 112 ml O2 = 22400 112 = 0.005 mol ∴ At STP, no. of moles of O2 used =
22400 112
= 0.005 mol
Volume of CO2 obtained = 560 – 112 = 448 ml
∴ At STP, no. of moles of CO2 used =
22400 448 = 0.02 mol 0.01 mol organic compound yields = 0.02 mol CO2.
∴ 1 mol organic compound yields
= 2 mol CO2 or 2 mol C
∴ The molecular formula of organic compound is C2HyOz. The reaction is : C2HyOz + 2 1 O2 → 2CO2 + 2 y H2O
Equating no. of oxygen atoms, z + 1 = 4 + 2 y or z = 3 + 2 y Molecular mass of C2HyOz = 2 × 12 + y × 1 + z × 16 Hence, 2 × 12 + y × 1 + + 2 y 3 × 16 = 90 or y = 2 and z = 3 + 2 y = 4
∴ The molecular formula of organic compound is C2H2O4.
9. (a) Write the intermediate steps for each of the following reactions. (i) C6H5CHOHC ≡ CH → + O H3 C 6H5CH=CHCHO (ii) OH H+ O CH3 (b) Show the steps to carry out the following transformations :
(i) Ethylbenzene → benzene
(ii) Ethylbenzene → 2-phenylpropionic acid [IIT-1998]
Sol. (a) (i)
C6H5CH(OH)C ≡ CH H + C6H5CH – C ≡ CH OH2+ –H2O C6H5CH = C = CH C6H5CH – C ≡ CH ⊕ ⊕ OH– C6H5CH = C = CHOH C6H5CH = CHCHO Tautomerism
(ii) OH H+ CH 3 OH CH3 CH3 O O H⊕ ⊕ (b) (i) C2H5 alk. KMnO4 Ethyl benzene COOH NaOH Benzoic acid –H2O COONa Sodium benzoate NaOH +CaO Benzene + Na2CaO3 (ii) CH2CH3 Br2/hv CHBrCH3 Mg Ether CH3CH – MgBr CO2 CH3CHCOOMgBr CH3CHCOOH H2O/H+ Mg + Br OH 2-Phenylpropionic acid
10. Interpret the non-linear shape of H2S molecule and
non-planar shape of PCl3 using valence shell electron
pair repulsion (VSEPR) theory. (Atomic numbers : H = 1, P = 15, S = 16, Cl = 17.) [IIT-1998] Sol. In H2S, no. of hybrid orbitals =
2
1(6 + 2 – 0 + 0) = 4 Hence here sulphur is sp3 hybridised, so
16S = 1s2, 2s22p6, 4 4 3 4 4 2 1 ion hybridisat sp 1 z 1 y 2 x 2 3 p 3 p 3 p 3 s 3 or S H H H H S
Due to repulsion between lp - lp; the geometry of H2S is distorted from tetrahedral to V-shape.
In PCl3, no. of hybrid orbitals =
2 1
[5 + 3 – 0 + 0] = 4 Hence, here P shows sp3-hybridisation
15P = 1s2, 2s22p6, 4 4 3 4 4 2 1 ion hybridisat sp 1 z 1 y 1 x 2 3 p 3 p 3 p 3 s 3 P or P Cl Cl Cl Cl Cl Cl
Thus due to repulsion between lp – bp, geometry is distorted from tetrahedral to pyramidal.T
MATHEMATICS
11. 7 relatives of a man comprises 4 ladies and 3 gentlemen; his wife has also 7 relatives; 3 of them are ladies and 4 gentlemen. In how many ways can they invite a dinner party 3 ladies and 3 gentlemen so that there are 3 of man’s relative and 3 of the wife’s relatives ? [IIT-1985] Sol. The possible cases are :
Case I : A man invites 3 ladies and women invites 3 gentlemen
⇒ 4C
3.4C3 = 16
Case II : A man invites (2 ladies, 1 gentleman) and women invites (2 gentlemen, 1 lady)
⇒ (4C
2.3C1).(3C1.4C2) = 324
Case III : A man invites (1 lady, 2 gentlemen) and women invites (2 ladies, 1 gentleman)
⇒ (4C
1.3C2).(3C2.4C1) = 144
Case IV : A man invites (3 gentlemen) and women invites (3 ladies)
⇒ 3C
3.3C3 = 1
∴ Total number of ways = 16 + 324 + 144 + 1 = 485
12. Let n be a positive integer and (1 + x + x2)n = a
0 + a1x + ... + a2nx2n
Show that a02 – a12 + ... + a2n2 = an. [IIT-1994] Sol. (1 + x + x2)n = a 0 + a1x + .... + a2nx2n ...(1) Replacing x by –1/x. we obtain n 2 x 1 x 1 1 − + = a 0 – x a1 + 2 2 x a – 33 x a +...+ 22nn x a ..(2) Now, a02 – a12 + a22 – a32 + ... + a2n2 = coefficient of
the term independent of x in [a0 + a1x + a2x2 + ... + a2nx2n] − + − + n 2 n 2 2 2 1 0 x a ... x a x a a
= coefficient of the term independent of x in (1 + x + x2)n n 2 x 1 x 1 1 − + But, R.H.S. = (1 + x + x2)n n 2 x 1 x 1 1 − + = 2n n 2 n 2 x ) 1 x x ( ) x x 1 ( + + − +
= 2n n 2 2 2 x ] x ) 1 x [( + − = 2 2n4 2 n x ) x x x 2 1 ( + + − = 2n n 4 2 x ) x x 2 1 ( + + Thus, a02 – a12 + a22 – a32 + ... a2n2
= coefficient of the term independent of x in
n 2 x 1 (1 + x2 + x4)n = coefficient of x2n in (1 + x2 + x4)n = coefficient of tn in (1 + t + t2)n = a n
13. Solve for x the following equation : log(2x + 3) (6x2 + 23x + 21) = 4 – log(3x + 7) (4x2 + 12x + 9) [IIT-1987] Sol. log(2x + 3) (6x2 + 23x + 21) = 4 – log(3x + 7) (4x2 + 12x + 9) ⇒ log(2x + 3)(2x + 3) . (3x + 7) = 4 – log(3x + 7)(2x + 3)2 ⇒ 1 + log(2x + 3) )(3x + 7) = 4 – 2log(3x + 7) (2x + 3) Put log(2x + 3) (3x + 7) = y ⇒ y + y 2 – 3 = 0 ⇒ y2 – 3y + 2 = 0 ⇒ (y – 1) (y – 2) = 0 ⇒ y = 1 or y = 2 ⇒ log(2x + 3) (3x + 7) = 1 or log(2x + 3)(3x + 7) = 2 ⇒ 3x + 7 = 2x + 3 or (3x + 7) = (2x + 3)2 ⇒ x = – 4 or 3x + 7 = 4x2 + 12x + 9 4x2 + 9x + 2 = 0 4x2 + 8x + x + 2 = 0 (4x + 1) (x + 2) = 0 x = – 2, –1/4 ∴ x = – 2, –4, –1/4
But, log exists only when, 6x2 + 23x + 21 > 0,
4x2 + 12x + 9 > 0,
2x + 3 > 0 and 3x + 7 > 0 ⇒ x > –3/2
∴ x = –1/4 is the only solution.
14. Let f[(x + y)/2] = {f(x) + f(y)} / 2 for all real x and y, if f´(0) exists and equals –1 and f(0) = 1, find f(2).
[ΙΙΤ−1992] Sol. f + 2 y x = 2 ) y ( f ) x ( f + ∀ x, y ∈ R (given) Putting y = 0, we get f 2 x = 2 ) 0 ( f ) x ( f + = 2 1 [1 + f(x)] [Q f(0) = 1] ⇒ 2f(x/2) = f(x) + 1 ⇒ f(x) = 2f(x/2) – 1 ∀ x, y ∈ R ...(1) Since f´(0) = –1, we get ⇒ h ) 0 ( f ) h 0 ( f lim 0 h − + → = – 1 ⇒ h 1 ) h ( f lim 0 h − →
Now, let x ∈ R then applying formula of differentiability. f´(x) = h ) x ( f ) h x ( f lim 0 h − + → = h ) x ( f 2 h 2 x 2 f lim 0 h − + → = h ) x ( f 2 ) h 2 ( f ) x 2 ( f lim 0 h − + → = h ) x ( f 1 2 h 2 f 2 1 2 x 2 f 2 2 1 lim 0 h − − + − → [using equation (1)] = h ) x ( f } 1 ) h ( f 2 1 – ) x ( f 2 { 2 1 lim 0 h − − + → = h 1 ) h ( f lim 0 h − → = –1 Therefore f´(x) = – ∀ x ∈ R ⇒
∫
f´(x)dx =∫
−1dx ⇒ f(x) = – x + k where k is a constant. But f(0) = 1, therefore f(0) = – 0 + k ⇒ 1 = k ⇒ f(x) = 1 – x ∀ x ∈ R ⇒ f(2) = – 115. If (a + bx)ey/x = x, then prove that
x3 2 2 dx y d = 2 y dx dy x − [IIT-1983]
Sol. (a + bx).ey/x = x ...(1)
Differentiating both sides, we get
(a + bx).ey/x. − 2 x y dx dy x + bey/x = 1 ⇒ x. 2 x y dx dy x − + beb/x = 1, (using (1)) or dx dy – x y + bey/x = 1,
Again differentiation both sides,
2 2 dx y d – 2 x y dx dy x − + bey/x. − 2 x y dx dy x = 0 ..(2) from (2), 2 2 dx y d – 2 x y dx dy x − − x y dx dy = 0 ⇒ x3 2 2 dx y d = y 2 dx dy x −
1. Four infinite thin current carrying sheets are placed in YZ plane. The 2D view of the arrangement is as shown in fig. Direction of current has also been shown in the figure. The linear current density. i.e. current per unit width in the four sheets are I, 2I, 3I and 4I, respectively.
I Y X II III IV a a a a
The magnetic field as a function of x is best represented by (A) B X µ0I a 2µ0I 5µ0I 2a 3a 4a 5a (B) B X µ0I a 3µ0I –µ0I 2a 3a 4a 5a (C) B X +µ0I a 4µ0I –µ0I 2a 3a 4a 5a (D) B X +µ0I a 2µ0I –µ0I 2a 3a 4a 5a
2. Match the column
Column – I Column – II
(A) a charge particle is (P) Velocity of the moving in uniform particle may be electric and magnetic constant fields in gravity free
space
(B) a charge particle is (Q) Path of the particle moving in uniform may be straight line electric, magnetic
and gravitational
fields
(C) a charge particle is (R) Path of the particle moving in uniform may be circular magnetic and
gravitational fields (where electric field is zero)
(D) A charge particle is (S) Path of the particle moving in only may be helical uniform electric field
(T) None
3. Magnetic flux in a circular coil of resistance 10Ω changes with time as shown in fig. Cross indicates a direction perpendicular to paper inwards.
Match the following : 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 given in same issue
φ(Magnetic flux) 10 2 t(s) –10 6 8 10 14 16 × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × Column – I Column – II
(A) At 1s, induced current is (P) Clockwise (B) At 5s, induced current is (Q) Anticlockwise (C) At 9s, induced current is (R) Zero
(D) At 15s, induced current is (S) 2A
(T) None
4. A conducting rod of length l is moved at constant velocity v0 on two parallel, conducting, smooth, fixed
rails, which are placed in a uniform constant magnetic field B perpendicular to the plane of the rails as shown in figure. A resistance R is connected between the two ends of the rail. Then which of the following is/are correct ?
⊗v0 R ⊗B
(A) The thermal power dissipated in the resistor is equal to the rate of work done by an external person pulling the rod
(B) If applied external force is doubled, then a part of the external power increases the velocity of the rod
(C) Lenz’s law is not satisfied if the rod is accelerated by an external force
(D) If resistance R is doubled, then power required to maintain the constant velocity v0 becomes half
5. The x-z plane separates two media A & B of refractive indices µ1 = 1.5 & µ2 = 2. A ray of light
travels from A to B. Its directions in the two media
are given by unit vectors µ→1=a∧i+b∧,jµ→2 =c∧i+b∧j.
Then (A) 3 4 c a = (B) 4 3 c a = (C) 3 4 d b = (D) 4 3 d b =
6. Two converging lenses of the same focal length f are separated by distance 2f. The axis of the second lens is inclined at angle θ=60º with respect to the axis of the first lens. A parallel paraxial beam of light is incident from left side of the lens. Then
2f
60º
(A) Final image after all possible refraction will formed at optical centre of first lens
(B) Final image after all possible refraction will formed at optical centre of second lens
(C) Final image after all possible refraction will formed at distance f from second lens
(D) Final image after all possible refraction will formed at distance f from first lens
7. If Cv for an ideal gas is given by Cv = (3 + 2T)R,
where T is absolute temperature of gas, then the equation of adiabatic process for this gas is
(A) VT2 = constant
(B) VT3e-2T = C
(C) VT2e2T = constant
(D) VT3e2T = constant
8. The pressure of one mole of ideal gas varies according to the law P = P0 – αV2 where P0 & α are
positive constant constants. The highest temperature that gas may attain -
(A) 2 / 1 0 0 3 P R 3 P 2 α (B) 2 / 1 0 0 3 P R 2 P 3 α (C) 2 / 1 0 0 3 P R P α (D) 2 / 1 0 0 P R P α
1.[C] The given circuit as an R-L-C series circuit when frequency of the source varies the impedance of the R-L-C series circuit varies and correspondingly the current in the circuit get varied
Impedance variation and current variation are shown in figure. R = Z minimum f1 f=fr f2 f ∆f=f2-f1 I I0 f1 f=fr f2 f ∆f=f2-f1 2 I I= 0
At frequency f1 XC > XL Power factor – leading
nature of circuit is capacitance At frequency f2 XL > XC Power factor – leading
nature of circuit is inducting At frequency f1 and frequency f2 impedance Z= 2.R
Because of the fact –
As ...(i) R V Zhm V I0 = = Z V 2 I0 = = ± ⇒ Z V 2 2 / V = ⇒ 2 R and Z V 2 RV = f1 < f < f2
2.[A] ∆f=f2 −f1 = B and width of R-L-C series circuit
L / R . 2 1 π = 3[C] At frequency f1 current . Amp 2 10 2 20 10 200 . 2 1 R V . 2 1 2 I0 = = = = ±
Watt less current φ = µ Isin I . Amp 10 2 1 . 2 10 Iµ = = º 45 Asφ= because 2 1 2 R R Z / R cosφ= = =
4[A] At frequency current 20Amp. 10 200 R V Zm V I= = = =
Potential difference across capacitor VC = IC.XC = 1.XC = 20.XC Charge on capacitor QC = C.VC = C. 20XC = C. (2v) C 1 ω π = π = π = ω = 5 1 ) 50 ( 2 20 f 2 20 20 Coulomb ) 5 ( π −1 = cb ) 5 ( π −1 = 5.[C] Longest wavelength m 59 . 0 600 5 8 . 0 350 f v v s max = × × = + = λ 6.[B] f v v v f s max = − 350 0.8 5 607Hz 350 = × − = 7.[A] 345.5/346.0 × 600 = 599 Hz 8.[A] 2 2 2 2 2 dt y d v 1 dx y d & dt dy v 1 dx dy=− =+
Solution
Physics Challenging Problems
Set # 11
1.[C] 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 II III (a) 2 J 0 µ (b) 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 Bresultant =µ0 −µ0 −µ0 +µ0 For a < x < 2a, J 2 ) J 4 ( 2 ) J 3 ( 2 ) J 2 ( 2 J Bresultant =µ0 −µ0 −µ0 +µ0 =−µ0 For 2a < x < 3a, 0 2 ) J 4 ( 2 ) J 3 ( 2 ) J 2 ( 2 J Bresultant =µ0 +µ0 −µ0 −µ0 = So, the required curve is
B
X O a 2a 3a 4a 5a
2. A → P,Q,S B → P,Q,R,S
C → P,Q,R,S D → Q
(A) 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.
(B) Here, all the possibilities are possible depending upon the combinations of the three fields.
(C) This situation is similar to part (i)
(D) In a uniform electric field, path can be only
3. A → Q B → R C → P D → Q
(A) At t = 1s, flux is increasing in the inward direction, hence induced e.m.f. will be in anticlockwise direction.
(B) At t = 5s, there is no change in flux, so induced e.m.f. is zero
(C) At t = 9s, flux is increasing in upward direction hence induced e.m.f. will be in clockwise direction.
(D) At t = 15s, flux is decreasing in upward direction, so induced e.m.f. will be in anticlockwise direction.
4.[A, B, D]
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.[A] xz n1sini = n2sinr ^j ) 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 # 12
6.[A] I2 + – f f cos60º I1 f sin60º f 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.[C] 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.[A] 2 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.• Saturn’s rings are made up of particles of ice, dust and rock. Some particles are as small as grains of sand while others are much larger than skyscrapers.
• Jupiter is larger than 1,000 Earths.
• The Great Red Spot on Jupiter is a hurricane-like storm system that was first detected in the early 1600’s.
• Comet Hale-Bopp is putting out approximately 250 tons of gas and dust per second. This is about 50 times more than most comets produce.
• The Sun looks 1600 times fainter from Pluto than it does from the Earth.
• There is a supermassive black hole right in the middle of the Milky Way galaxy that is 4 million times the mass of the Sun.
• Halley’s Comet appears about every 76 years.
• The orbits of most asteroids lie partially between the orbits of Mars and Jupiter.
• Asteroids and comets are believed to be ancient remnants of the formation of our Solar System (More than 4 billion years ago!).
• Comets are bodies of ice, rock and organic compounds that can be several miles in diameter.
• The most dangerous asteroids, those capable of causing major regional or global disasters, usually impact the Earth only once every 100,000 years on average.
• Some large asteroids even have their own moon.
• Near-Earth asteriods have orbits that cross the Earth’s orbit. These could potentially impact the Earth.
• There are over 20 million observable meteors per day.
• Only one or two meteorites per day reach the surface of Earth.
• The largest found meteorite was found in Hoba, Namibia. It weighed 60 tons.
1. A beam of length L, breadth b and thickness d when loaded by a weight Mg in the middle, a depression e is produced in it. By measuring this depression e, the value of Young's modulus of the material of beam can be calculated by using the expression
Y = e d b 4 L g M 3 3
Following are the values of different physical quantities obtained in one set of observations on this experiment :
M = 1000 gms, L = 200 cm,
b = 2.54 cm, d = 0.620 cm, e = 0.1764 cm.
If M is measured by spring balance, L by metre scale, b by vernier calipers, d by screw gauge and e by spherometer, then what will be the maximum possible percentage errors in Y ?
Sol. Given that Y =
e d b 4 L g M 3 3
Taking log on both sides of above equation, we get log Y = log M + log g + 3 log L – log 4
– log b – 3 log d – log e Differentiating above equations, we have :
Y Y ∆ = M M ∆ + 3 L L ∆ – b b ∆ – 3 d d ∆ – e e ∆
In order to calculate maximum possible error, we shall convert negative sign into positive sign.
∴ Y Y ∆ = M M ∆ + 3 L L ∆ + b b ∆ + 3 d d ∆ + e e ∆ Now, least counts of the different measuring instruments used in the experiment are as under : Least count of spring balance = 5 gm i.e. ∆M = 5gm Least count of metre scale = 0.1 cm i.e. ∆L = 0.1 cm Least count of vernier callipers = 0.01 cm
i.e. ∆b = 0.001 cm
Least count of screw gauge = 0.001 cm i.e. ∆d = 0.001 cm
Least count of spherometer = 0.005 cm i.e ∆e = 0.005 cm ∴ Y Y ∆ = 1000 5 + 200 1 . 0 3× + 54 . 2 01 . 0 + 62 . 0 001 . 0 3× + 1764 . 0 005 . 0 = 0.005 + 0.0015 + 0.00393 + 0.00484 + 0.02834 = 0.0436 or 4.36%
Hence the maximum possible percentage error is 4.36%.
2. A small glass ball is released from rest from the top of a smooth incline plane of constant base b. find the angle of inclination of the plane for minimum time of motion of the glass ball.
A
B C
b θ
Sol. Let the angle of inclination be θ. If, the glass ball reaches the bottom B of the inclined plane after a time, say t, the equation of motion along the plane is given as AB = (VA)t + 2 1(g sin θ)t2 A B C b θ g L = b sec θ g sin θ
Since the glass ball is released from rest VA = 0,
hence AB =
2
1 (g sin θ)t2 ...(1)
Putting AB = (BC) sec θ = b sec θ, in equation (1), we obtain t = θ θcos sin g b 2 = θ 2 sin g b 4
For t to be minimum, sin 2θ is maximum ∴ sin 2θ|max = 1 or, 2θ = 2 π or, θ = 4 π
3. Two particles, both of mass m, attract each other with the force F rr(r) = – rˆ
r2 α
where α is a positive constant. At a certain moment (t = 0), the distance between the particles is R, and their velocities are
− = = xˆ v 2 v xˆ v v 0 2 0 1
Expert’s Solution for Question asked by IIT-JEE Aspirants
Students' Forum
Assuming the two-particle system reaches a minimum of kinetic energy at a certain moment and at a certain finite distance between the particles (in the laboratory frame), find the distance between the particles at that moment and the value of that minimal kinetic energy.
Sol. For the sake of convenience, we will first solve the problem in the frame of the centre of mass. Then, we will transform the results into the laboratory frame. We will determine the velocity of the center of mass by : cm vr = xˆ m 2 mv 2 mv0− 0 = – v xˆ 2 1 0 ...(1)
This velocity remains constant since there are no external forces acting on the whole system. In the velocity of the centre of mass, the velocities of the particles are : − = − = = − = xˆ v 2 3 v v u xˆ v 2 3 v v u 0 cm 2 2 0 cm 1 1 r r r r r r ...(2)
By definition of a centre of mass frame, the total momentum of the particles is zero. The kinetic energy in this system at t = 0 is :
K´ = 2 1 mu12 + 2 1 mu22 = 4 9 mv02 ...(3)
Therefore, the total energy in the center of mass frame at t = 0 is E´ = K´ + U´ = 4 9 mv02 – R α ...(4) where we define R ≡ r(t = 0). Note that since the force is conservative, we have Fr = – ∇u. The scalar function is u = – α/r.
The advantage of using the center of mass frame is evident when one inspects the moment of arrival at a minimal distance, t0. At that moment, in this system,
the two particles stop and reverse their directions. The kinetic energy, therefore, vanishes at t0 in the
center of mass frame, or,
K´(t0) = K'min = 0 ...(5) Hence, E´(t0) = – min R α = E´ ...(6) Plugging in the value of E´, we find :
Rmax = R mv 9 4 R 4 2 0 − α α ...(7)
We now transform the centre of mass frame to the laboratory frame. Since Rmax is the relative distance
between the two particles, it is unchanged by the transformation. Recall that distance is an invariant quantity of the transformations of displacement and / or rotation. Therefore, Kmin – max R α = E(0) = E = 2 1 mv02 + 2 1 m(2v0)2 – R α ....(8)
The kinetic energy in the laboratory frame is, therefore : Kmin = 2 5mv 02 – R α + R 4 R mv 9 4 02 α − α α = − 4 9 2 5 mv 02 = 4 1 mv 02 ...(9)
Another way of finding the minimal kinetic energy is by using the following formula :
K = K´ + Mv2cm 2 1
...(10) where K´ is the kinetic energy in the centre of mass frame, K is the energy in the laboratory frame, and M is the total mass of the system. In our case,
K = 0 + (2m)Mv2cm 2 1 = 2 0 Mv 4 1 ...(11)
Note : Generally, in transforming from system S to system S´ with relative velocity Vr , the kinetic energy is transformed as : K´ = K – M 2 p2 + 2 M p V 2 M −r r ..(12) where M is the total mass and pris the total momentum in S. K´ is minimal in the center of mass frame if we choose Vr= vrcm= M pr . We obtain : K´ = K = M 2 p2 ...(13)
4. A hot body is being cooled in air according to Newton's law of cooling, the rate of fall of temperature being K times the difference of its temperature with respect to that of surroundings. Calculate the time after which the body will lose half the maximum heat it can lose. The time is to be counted from the instant t = 0.
Sol. According to Newton's law of cooling, we have
dt dθ
= – K(θ – θ0)
where θ0 is the temperature of the surrounding and θ
is the temperature of the body at time t. Suppose θ = θ1 at time t = 0. Then,
∫
θ θ θ−θ θ 1 0 d = –K∫
0tdt or, log 1 0 0 θ − θ θ − θ = – Kt or, θ – θ0 = (θ1 – θ0)e–Kt ...(1)The body continues to lose heat till its temperature becomes equal to that of the surroundings. The loss of heat in this entire period is dQm = ms(θ1 – θ0).
This is the maximum heat the body can lose. If the body loses half this heat, the decreases in its temperature will be ms 2 dQm = 2 0 1−θ θ
If the body loses this heat in time t1, the temperature at t1 will be θ1 – 2 0 1−θ θ = 2 0 1+θ θ
Putting these values of time and temperature in (1) :
2 0 1+θ θ – θ0 = (θ1 – θ0)e−Kt1 or, e−Kt1 = 2 1 or t 1 = K 2 log
5. In a certain region surrounding the origin of the coordinates, →B = 5 × 10–4→kT and →E = kˆ V/m. A
proton enters the fields at the origin with an initial velocity v→0 = 2.5 × 105iˆ m/s. Describe the proton's
motion and give its position after three complete revolutions.
Sol. The z-component of the force →F is a constant electrical force. It produces a constant acceleration along z-axis given as
Z = 2 1at2 = 2 1 m eE t2 B E z x y
The other component of the force F is a magnetic force which provides the necessary centripetal force to keep the proton in a circular path of radius r(say). The period of revolution of the proton
T = 0 v r 2π As Fcp = r mv20 = ev 0B ∴ v0 = m eBr Hence, T = eB m 2π
Since the particle (proton) moves in circular path having a period of revolution T in x, y plane and moves along z-axis with a constant acceleration a = eE/m, the path of the proton is a helix.
After three revolutions, putting t = 3T, we obtain
z = m eE 2 1 (3T)2 = m eET 2 9 2
Putting T = 2πm/eB, we get z = 22 eB Em 18π = 62 4 2 27 ) 10 5 ( 10 6 . 1 10 66 . 1 5 ) 7 / 22 ( 18 − − − × × × × × × = 37m
Interesting Science Facts
• The Universe contains over 100 billion galaxies. • Wounds infested with maggots heal quickly and
without spread of gangrene or other infection.
• More germs are transferred shaking hands than
kissing.
• The longest glacier in Antarctica, the Almbert
glacier, is 250 miles long and 40 miles wide.
• The fastest speed a falling raindrop can hit you is
18mph.
• A salmon-rich, low cholesterol diet means that
Inuits rarely suffer from heart disease.
• Inbreeding causes 3 out of every 10 Dalmation
dogs to suffer from hearing disability.
• The world’s smallest winged insect, the Tanzanian
parasitic wasp, is smaller than the eye of a housefly.
• If the Sun were the size of a beach ball then
Jupiter would be the size of a golf ball and the Earth would be as small as a pea.
• It would take over an hour for a heavy object to
sink 6.7 miles down to the deepest part of the ocean.
• There are more living organisms on the skin of
each human than there are humans on the surface of the earth.
• The grey whale migrates 12,500 miles from the
Artic to Mexico and back every year.
• Quasars emit more energy than 100 giant galaxies. • Quasars are the most distant objects in the
Universe.
• The Saturn V rocket which carried man to the
Moon develops power equivalent to fifty 747 jumbo jets.
• Koalas sleep an average of 22 hours a day, two
hours more than the sloth.
• Light would take .13 seconds to travel around the
Earth.
• Neutron stars are so dense that a teaspoonful
would weigh more than all the people on Earth.
• One in every 2000 babies is born with a tooth. • Every hour the Universe expands by a billion
miles in all directions.
• Somewhere in the flicker of a badly tuned TV set
is the background radiation from the Big Bang.
• The temperature in Antarctica plummets as low
as -35 degrees Celsius.
• Space debris travels through space at over 18,000
Calorimetry :
The specific heat capacity of a material is the amount of heat required to raise the temperature of 1 kg of it by 1 K. This leads to the relation
Q = ms θ
where Q = heat supplied, m = mass, θ = rise in temperature.
The relative specific heat capacity of a material is the ratio of its specific heat capacity to the specific heat capacity of water (4200 J kg–1K–1).
Heat capacity or thermal capacity of a body is the amount of heat required to raise its temperature by 1 K. [Unit : J K–1]
Thus heat capacity = Q/θ = ms Also dt dθ = ms 1 × dt dQ
i.e., the rate of heating (or cooling) of a body depends inversely on its heat capacity.
The water equivalent of a body is that mass of water which has the same heat capacity as the body itself. [Unit : g or kg] This is given by
W =
w s
s m×
where m = mass of body, s = specific heat capacity of the body, sw = specific heat capacity of water.
Principle of Calorimetry : The heat lost by one system = the heat gained by another system. Or, the net heat lost or gainsed by an isolated system is zero. It system with masses m1, m2, ...., specific heat
capacities s1, s2, ...., and initial temperatures θ1, θ2, ....
are mixed and attain an equilibrium temperature θ then θ = ´ ms ms Σ θ Σ
, for equal masses θ = s s
Σ θ Σ
Newton's law of cooling :
The rate of loss of heat from a body in an environment of constant temperature is proportional to the difference between its temperature and that of the surroundings.
If θ = temperature of the surroundings then
– ms
dt
dθ = C´(θ – θ
0)
where C´ is a constant that depends on the nature and extent of the surface exposed. Simplifying
dt dθ = –C(θ – θ 0) where C = ms ´ C = constant
Kinetic theory of gases :
The pressure of an ideal gas is given by p = 3 1
µnC2
where µ = mass of each molecule, n = number of molecules per unit volume and C is the root square speed of molecules. p = 3 1 ρC2 or pV = 3 1 mC2
where ρ is the density of the gas and m = mass of the gas.
Root Mean Square Speed of Molecules : This is defined as C = N C ... C C C12+ 22+ 23+ + 2N
where N = total number of molecules. It can be obtained through these relations
C = ρ p 3 = M RT 3
Total Energy of an ideal gas (E) :
This is equal to the sum of the kinetic energies of all the molecules. It is assumed that the molecules do not have any potential energy. This follows from the assumption that these molecules do not exert any force on each other.
E = 2 1 mC2 = 2 3 M m RT = 2 3 pV
Thus, the energy per unit mass of gas = 2 1C2
The energy per unit volume = 2 3p
The energy per mole = 2 3 pV = 2 3 RT
