The torque on the system about O is zero throughout since the net external force on the system is zero.a) 0b) vL sinωtc) vL cosωt d) vL

2. INSTRUCTIONS A. General 1. This booklet is your Question Paper. Do not break the seals of this boolket before being instructed to do so by the invigilators. 2. The question paper CODE is printed on the right hand top corner of this page and on the back page (Page No. 28) of this booklet. 3. Blank spaces and blank pages are provided in this booklet for your rough work. No additional sheets will be provided for rough work. 4. Blank papers, clipboards , log tables, slide rules, calculators, cameras, cellular phones, pagers and electronic gadgets are NOT allowed inside the examination hall. 5. Answer to the questions and personal details are to be filled on a two-part carbon-less paper, which is provided separately. You should not separate these parts. The invigilator will separate them at the end of examination. The upper sheet is a machine-gradable Ob jective Response Sheet (ORS) which will be taken back up the invigilator. You will be allowed to take away the bottom sheet at the end of the examination. 6. Using a black ball point pen, darken the bubbles on the upper original sheet. Apply sufficinet pressure so that the impression is created on the bottom sheet. 7. DO NOT TAMPER WITH / MUTILATE THE ORS OR THE BOOKLET. 8. On breaking the seals of the booklet check that it contains 28 pages and all the 60 questions and corresponding answer choices are legible. Read carefully the instructions printed at the beginining of each section. B. Filling the Right Part of the ORS 9. The ORS has CODES printed on its left and right parts. 10. Check that the same CODE is printed on the ORS and on this boolet. IF IT NOT THEN ASK FOR A CHANGE OF THE BOOKLET. Sign at the place provided on the ORS affirming that you have verified that all the codes are same. 11. Write your name , Registration Number and the name of examination centre and sign with pen in the boxes provided on the right part of the ORS. Do not write any of this information anywhere else. Darken the appropriate bubble UNDER each digit of your Registration Number in such a way that the impression is created on the bottom sheet. Also darken the paper CODE given on the right side of ORS (R4) C Question Paper Format The questions paper consists of 3 parts (Physics, Chemistry and Mathematics). Each part consist of three sections. 12. Section I contains 10 multiple choice questions. Each question has four choices. (A), (B), (C) and (D) out of which ONLY ONE is correct. 13. Section II contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct. 14. Section III contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive). D.Marking Scheme 15. For each question in Section I, you will be awarded 3 marks if you darken the bubble corresponding to the correct answer ONLY and zero marks if no bubbles are darkened. In all ohter cases, minus one (–1) mark will be awarded in this section. 16. For each question in Section II, you will be awareded 4 marks if you darken ALL the bubble(s) corresponding to the correct answer(s) ONLY. In all other cases zero (0) marks will be awarded. No negative marks will be awarded for incorrect answers in this section. 17. For each question in Section III, you will be awarded 4 marks if you darken the bubble corresponding to the correct answer ONLY. In all other cases zero (0) will be awarded. No negative marks will be awarded for incorrect answers in this section.

3. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 3 Sri Chaitanya IIT Academy PHYSICS SECTION – I ( SINGLE CORRECT CHOICE TYPE ) This section contains 10 multiple choicse questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct 1. A small block is connected to one end of massless spring of un–stretched length 4.9 m. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by 0.2 m and released from rest at t= 0. It then executes simple harmonic motion with angular frequency / 3 rad s    . Simultaneously at t = 0, a small pebble is projected with speed v from point P at an angle of 450 as shown in the figure. Point P is at a horizontal distance to 10 m from O. If the pebble hits the block at t = 1s, the value of v is (take g = 10 m/s2 ) a) 50 /m s b) 51 /m s c) 52 /m s d) 53 /m s Ans :A Equation of SHM cosy A t At 1t  0.2cos (1) 0.2cos 0.1 3 y m      Block is at a distance 5m  Range of pebble is R =5m 2 2sin90 50 10 V R V     50 /V m s 

4. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 4 Sri Chaitanya IIT Academy 2. In the determination of Young’s modulus 2 4MLg Y ld       by using Searle’s method, a wire of length L = 2 m and diameter d = 0.5 mm is used. For a load M = 2.5 kg, an extension l = 0.25 mm in the length of the wire is observed. Quantities d and l are measured using a screw gauge and a micrometer, respectively. They have the same pitch of 0.5 mm. The number of divisions on their circular scale is 100. The contributions to the maximum probable error of the Y measurement. a) due to the errors in the measurements of d and l are the same b) due to the error in the measurement of d is twice that due to the error in the measurement of l. c) due to the error in the measurement of l is twice that due to the error in the measurement of d. d) due to the error in the measurement of d is four times that due to the error in the measurement of l. Ans :A 2 2 2 4 d d d d v              4 0.25 l l l l      3. A small mass m is attached to a massless string whose other end is fixed at P as shown in the figure. The mass is undergoingcircular motion in the x–y plane with centre at O and constant angular speed  . If the angular momentum of the system, calculated about O and P are de- noted by 0L  and PL  respectively, then

5. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 5 Sri Chaitanya IIT Academy a) 0L  and PL  do not vary with time b) 0L  varieswithtime while PL  remainsconstant c) 0L  remains constant while PL  varies with time d) 0L  and PL  both vary with time Ans :C Torque about O is zero angular momentumwill be constant about ‘O’. Torque about Pis non zero. Angular momentum about P varies with time.  0L m r v   PL m OP v  direction of PL changes with OP and PL OP 4. Young’s double slit experiment is carried out by using green, red and blue light, one color at a time. The fringe widths recorded are ,G R  and B , respectively. Then a) G B R    b) B G R    c) R B G    d) R G B    Ans :D D d       R G B    R G B   

6. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 6 Sri Chaitanya IIT Academy 5. Consider a thin spherical shell of radius R with its centre at the origin, carrying uniform positive surface charge density. The variation of the magnitude of the electric field  E r  and the electric potential V(r) with the distance r from the centre, is best represented by whichgraph ? a) b) c) d) Ans : D From centre to surface E = 0, then 1 E r  From centre to surface V = k, then 1 V r  6. A bi–convex lens is formed with two thin plano–convex lenses as shown in the figure. Re- fractive index n of the first lens is 1.5 and that of the second lens is 1.2. Both the curved surfaces are of the same radius of curvature R = 14 cm. For this bi–convex lens, for an object distance of 40 cm, the image distance will be a) – 280. 0 cm b) 40.0 cm c) 21. 5 cm d) 13.3 cm

7. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 7 Sri Chaitanya IIT Academy Ans :B 1 2 1 1 1 F f f   =    1 21 1 R     = 0.5 0.2 0.7 1 14 14 20    20F cm 40R cm  1 1 1 1 1 1 1 1 1 2 1 1 20 40 20 40 40 40F v u v v             40v cm  7. Two large vertical and parallel metal plates having as separation of 1 cm are connected to a DC voltage source of potential difference X.Aproton is released at rest midway between the two plates. It is found to move at 450 to the vertical JUST after release. Then X is nearly a) 5 1 10 V  b) 7 1 10 V  c) 9 1 10 V  d) 10 1 10 V  Ans :C mg Eq 45 mg Eq pVq d   9 10 mgd V V approx q    8. Three very large plates of same area are kept parallel and close to each other. They are con- sidered as ideal black surfaces and have very high thermal conductivity. The first and third plates are maintained at temperatures 2T and 3T respectively. The temperature of the middle (i.e. second) plate under steady state condition is a) 1 465 2 T       b) 1 497 4 T       c) 1 497 2 T       d)   1 497 T Ans : C    4 4 4 4 1 2 2 3A T T A T T        1 44 44 4 2 2 2 97 2 3 2 T T T T T T           

8. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 8 Sri Chaitanya IIT Academy 9. A thin uniform rod, pivoted at O, is rotating in the horizontal plane with constant angular speed  , as shown in the figure. At time t = 0, a small insect starts from O and moves with constant speed v with respect to the rod towards the other end. It reaches the end of the rod at t = T and stops. The angular speed of the system remains  throughout. The magnitude of the torque of the torque    on the system about O, as a function of time is best represented by which plot ? a) b) c) d) Ans :B   dL d dI I dt dt dt      constantrodI  2 red d I mr dt     = 0 2 dr m r dt         v r = vt 2 2 2mr v m v t   t  when the insect reaches the end it stops v = 0 0 

9. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 9 Sri Chaitanya IIT Academy 10. A mixture of 2 moles of helium gas (atomic mass = 4amu) and 1 mole of argon gas (atomic mass = 40 amu) is kept at 300 K in a container. The ratio of the rms speeds     helium argon rms rms v v        is a) 0.32 b) 0.45 c) 2.24 d) 3.16 Ans : D rms T V M  1 2 2 1 40 10 4 He Ar VV M V V M     SECTION – II ( MULTIPLE CORRECT ANSWER TYPE) This section contains 5 multiple choicse questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONE or More are correct. 11. A small block of mass of 0.1 kg lies on a fixed inclined plane PQ which makes an angle  with the horizontal. Ahorizontal force of 1 N acts on the block through its center of mass as shown in the figure. The block remains stationary if (take 2 10 /g m s ) a) 0 45  b) 0 45  and a frictional force acts on the block towards P. c) 0 45  and a frictional force acts on the block towards Q. d) 0 45  and a frictional force acts on the block towards Q. Ans :A(or)AC

10. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 10 Sri Chaitanya IIT Academy 1cos sinmg  45   when 45  friction force is upwards If friction force sufficient block will be in state of rest 12. Consider the motion of a positive point charge in a region where there are simulataneous uniform electric and magnetic fields 0 ˆE E j  and 0 ˆB B j  . At time t = 0 , this charge has velcoity v  in the x – y plane, making an angle  with the x – axis. Which of the following option(s) is(are) correct for time t > 0 ? a) If 0 0  , the charge moves in a circular path in the x – z plane b) If 0 0  , the charge undergoes helical motion with constant pitch along the y – axis. c)If 0 10  ,thechargeundergoeshelicalmotionwithitspitchincreasingwithtime,alongthey–axis d) If 0 90  , the charge undergoes linear but accelerated motion along the y–axis Ans :C,D when 90  0mF  when 10   path is helical and with increasing pitch.

11. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 11 Sri Chaitanya IIT Academy 13. For the resistance network shown in the figure, choose the correct option(s). a) The current through PQ is zero b) 1 3I A c) The potential at S is less than that at Q. d) 2 2I A Ans:A,B,C,D 1 12 3 4 I   2 12 3 18 I   2A P QV V S QalsoV V  14. A person blows into open – end of a long pipe. A s a result, a high – pressure pulse of air travels down the pipe.When this pulse reaches the other end of the pipe, a) a high – pressure pulse starts travelling up the pipe, if the other end of the pipe is open b) a low – pressure pulse starts travelling up the pipe, if the other end of the pipe is open c) a low – pressure pulse starts travelling up the pipe, if the other end of the pipe is closed d) a high – pressure pulse starts travelling up the pipe, if the other end of the pipe is closed Ans : B, D

12. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 12 Sri Chaitanya IIT Academy Reflected pressure pulse Incident pressure pulse Open end At open end pressure should not change pressure wave will get phase change as  Reflected pressure pulse Incident pressure pulse Closed end At enclosed end pressure wave will get reflected without phase change 15. A cubical region of side a has its centre at the origin. It encloses three fixed point charges, q at  0, / 4,0 , 3a q  at (0, 0, 0) and q at  0, / 4,0a . Choose the correct option(s). a) The net electric flux crossing the plane / 2x a  is equal to the net electric flux crossing the plane / 2x a  b)The net electric fluxcrossingtheplane / 2y a  is morethan the net electricflux crossing the plane / 2y a  . c) The net electric flux crossing the entire region is 0 q  . d) The net electric flux crossing the plane / 2z a  is equal to the net electric flux crossing the plane / 2x a 

13. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 13 Sri Chaitanya IIT Academy Ans:ACD  Net flux passing the plane 2 a x   and 2 a x   is same  Net flux through cube is 0 q   Flux through plane 2 a z  and 2 a x  is same SECTION – III ( INTEGER ANSWER TYPE) This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive). 16. A lamina is made by removing a small disc of diameter 2R from a bigger disc of uniform mass density and radius 2R, as shown in the figure. The moment of inertia of this lamina about axes passing through O and P is 0I and PI , respectively. Both these axes are perpendicular to the plane of the lamina. The ratio 0 PI I to the nearest integer is Ans :3        2 2 2 2 2 2 0 1 1 2 2 2 2 I R R R R R R            4 43 8 2 R R   4 213 13 2 2 R MR       2 2 2 21 1 4 2 4 2 5 2 2 PI M R M R MR M R               2 2 211 8 16 2 MR MR MR         2 211 37 24 2 2 MR MR       

14. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 14 Sri Chaitanya IIT Academy 17. Acircular wire loop of radius R is placed in the x - y plane centered at the origin O.Asquare loop of side  a a R having two turns is placed with its center at 3z R along the axis of the circular wire loop, as shown in figure. The plane of the square loop makes an angle of 45with respect to the z - axis. If the mutual inductance between the loops is given by 2 0 /2 2p a R  , then the value of p is Ans : 7   2 20 3/22 2 2 1 4 2 3 iR B a i R R           (n = 2) 2 2 20 0 9/2 1 2 22 a M R a      18. A cylindrical cavity of diameter a exists inside a cylinder of diameter 2a as shown in the figure. Both the cylinder and the cavity are infinetely long.Auniformcurrent density J flows along the length. If the magnitude of the magnetic field at the point P is given by 0 12 N aJ , then the value of N is

15. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 15 Sri Chaitanya IIT Academy Ans : 5 0 1 2 ja B   2 20 0 2 2 3 12 2 2 a j ja B a a                   1 2 0 1 1 2 12 B B ja         0 5 12 ja  19. A proton is fired from very far away towards a nucleus with charge 120Q e , where e is the electronic charge. It makes a closest approach of 10 fm to the nucleus. The de Broglie wave- length (in units of fm) of the proton at its start is: (take the proton mass,   27 5 / 3 10 ;pm kg   15 / 4.2 10 . /H e j S c   ; 9 0 1 9 10 / ; 4 m F    15 1 10fm m  ) Ans :7 2 2 9 15 120 9 10 10 10 2 p e P m     9 2 27 2 15 5 9 10 120 2 10 3 10 10 e P              17 15 3 2 10 3 2 10 10 e e P          15 4.2 10 6 10 h p     

16. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 16 Sri Chaitanya IIT Academy 20. An infinitely long solid cylinder of radius R has a uniform volume charge density  . It has a spherical cavity of radius R/2 with its centre on the axis on the cylinder, as shown in the figure. The magnitude of the electric field at the point P, which is at the distance 2R from the axis of the cylinder, is given by the expression 0 23 16 R k   . The value of k is Ans: 6   3 2 2 4 2 3 8 2 2 p R K k R E R R      4 1 1 3 32 pE k R        0 1 1 1 4 24 R        0 23 4 24 p R E          24 4 6 16 K   

17. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 17 Sri Chaitanya IIT Academy CHEMISRY SECTION – I ( SINGLE CORRECT CHOICE TYPE ) This section contains 10 multiple choicse questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct 21. The number of aldol reaction(s) that occurs in the given transformation is . . 3 4 conc aq NaOH CH CHO CHO  OH OH HO OH a) 1 b) 2 c) 3 d) 4 Ans : C 3 2 OH CH CHO CH CHO 2 2 2H C H CH CHO O CH CH CHO        O H 2 2 OH CH CHO HO CH CH CHO    H - C - H O H 'HCHO Cannizaro s NaOH reaction CH CH  CHO OH OH CHO HO OH OH OH OH OHHO Number of aldol reaction = 3

18. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 18 Sri Chaitanya IIT Academy 22. A compound M P XP has cubic close packing (ccp) arrangement of X. Its unit cell structure is shown below. The empirical formula of the compound is a) MX b) MX2 c) M2 X d) M5 X14 Ans : B Corner Face center 1 6 3 2   1 8 1 8   4 X : Edge centre Body center 1 1 1  1 4 1 4   2 M : 4 2X M 4 2X M 2MX 23. The carbonyl functional group (–COOH) is present in a) picric acid b) barbituric acid c) ascorbic acid d) aspirin Ans : D Acetyle salicylic acid =Asprin COOH OCOCH3

19. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 19 Sri Chaitanya IIT Academy 24. The colour of light absorbed by an aqueous solution of CuSO4 is : a) organe-red b) blue-green c) yellow d) violet Ans : A Orange - red is absorbed Based up on Munshell wagon wheel V B G O R Y 25. The number of optically active products obtained from the complete ozonolysis of the given compound is : CH3 - CH = CH - C - CH = CH - C - CH = CH - CH3 HCH3 H CH3 a) 0 b) 1 c) 2 d) 4 Ans : A Asthe ozonolysis products do not have chiral centres, number of opticallyactive products = 0 26. As per IUPAC nomenclautre, the name of the complex    2 3 34 2 Co H O NH Cl   is a) Tetraaqudiaminecobalt (III) chloride b) Tetraaquadiamminecobalt (III) chloride c) Diammineteraaquacobalt (III) chloride d) Diamminetetraaquacobalt (III) chloride Ans : D    2 3 34 2 Co H O NH Cl   Diammineteraaquacobalt (III) chloride 27. Which ordering of compound is according to the decreasingorder of the oxidation state of nitrogen ? a) 3 4 2, , ,HNO NO NH Cl N b) 3 2 4, , ,HNO NO N NH Cl c) 3 4 2, , ,HNO NH Cl NO N d) 3 4 2, , ,NO HNO NH Cl N

20. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 20 Sri Chaitanya IIT Academy Ans : B 3 5HNO   2NO   2 0N  4 3NH Cl   Decreasing order of O -S = 3 2 4, , ,HNO NO N NH Cl 28. For one mole of a van der Waals gas when b = 0 and T = 300 K, the PV vs. I/V plot is shown below. The value of the van der Waals constant a (atm. liter2 mol–2 ) is a) 1.0 b) 4.5 c) 1.5 d) 3.0 Ans : C   2 2 n a P v nb nRT v         ’ n = 1, b = 0  2 q p v b RT v         a pv RT v   a pv RT v    2 221.6 20.1 1.5 . 3 2 y c mx m a slop atm t mole        

21. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 21 Sri Chaitanya IIT Academy 29. In allene  3 4C H , the type(s) of hybridisation of the carbon atoms is (are) a) sp and sp3 b) sp and sp2 c) only sp2 d) sp2 and sp3 Ans : B 2 2H C C CH  2 2 sp sp sp 30. The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [a0 is Bohr radius] a) 2 2 2 04 h ma b) 2 2 2 016 h ma c) 2 2 2 032 h ma d) 2 2 2 064 h ma Ans : C mvr= 2 nh  v= 2 nh mr for second Bohr’s orbit 2r =4 0a V= 04 h ma 2 2 2 2 2 016 h V m a  K.E 2 2 2 2 0 1 2 32 h mv ma  

22. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 22 Sri Chaitanya IIT Academy SECTION – II ( MULTIPLE CORRECT ANSWER TYPE) This section contains 5 multiple choicse questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONE or More are correct. 31. Which of the following molecules, in prue form, is (are) unstable at room temperature ? a) b) c) O d) O Ans : BC cyclobutadiene and cyclopentadien are anti aromatic. 32. Which of the following hydrogen halides react(s) with  3AgNO aq to give a precipitate that dissolves in  2 2 3Na S O aq ? a) HCl b) HF c) HBr d) HI Ans : ACD 3 3AgNO HX AgX HNO   aq aq  2 2 3 3 2 3 2 2AgX Na S O Na Ag S O NaX    aq aq s AgF is soluble 33. Choose the correct reason(s) for the stability of the lyophobic colloidal particles > a) Preferential adsorption of ions on their surface from the solution b) Preferential adsorption of solvent of their surface the solution c)Attraction between different particles having opposite charges on their surface d) Potential difference between the fixed layer and the diffused layer of opposite charges around the coolidal particles Ans :AD

23. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 23 Sri Chaitanya IIT Academy 34. For an ideal gas, consider only P-V work in going from an initial state X to the final state Z. The can be reached by either of the two paths shown in the figure. Which of the following choice(s) is (are) correct ? [Take S as change in entropy and w as work done] a) x z x y y zS S S       b) x z x y y zW w w    c) x y z x yw w   d) x y z x yS S     Ans :AC Based up on the given graph x y x y y zS S S       x y z x yW W   35. Identify the binary mixtrue(s) that can be separated into individual compounds, by differential extraction,a s shown in the given scheme. a) 6 5 6 5C H OH and C H COOH b) 6 5 6 5 2C H COOH and C H CH OH c) 6 5 2 6 5C H CH OH and C H OH d) 6 5 2 6 5 2C H CH OH and C H CH COOH Ans : BD Benzyl alcohol is insoluble in both NaOH and 3NaHCO ,

24. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 24 Sri Chaitanya IIT Academy SECTION – III ( INTEGER ANSWER TYPE) This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive). 36. When the following aldohexose exists in its B-configuration, the total number C CH2 CHOH CHOH CHO CH2OH OHH of stereoisomrs in its pyranose form is : Ans : 8 As the configration at C-5 is fixed, (D-configration), then only 4 stero isomers are possible solution 37. 29.2% (w/w) HCl stock solution has a densityh of 1.25 g mL–1 . The molecular weight of HCl is 36.5 g mol–1 . The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is : Ans : 8 Molarity of stock solution = 10 d x GMW   = 10 1.25 29.2 36.5   =10M 1 1 2 2 1 1 10 200 0.4 8 ( ) V M V M V V ML approx      38. The periodic table consists of 18 groups. An isotope of copper, on bombardment with pro- tons, undergoes a nuclear reaction yielding element X as shown below. To which group,element X belongs int he periodic table ? Ans : 8 63 1 1 4 1 52 29 1 0 2 1 266 2CU H n He H X     26 Fe  8 th group

25. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 25 Sri Chaitanya IIT Academy 39. An organic compound undergoes first-order decomposition. The time taken for its decomposition to 1/8 and 1/10 of its initial concentration are t1/8 and t1/10 respectively. What is the Ans : 9 1 8 1 10 2.303 log 8 2.303 log 10 a t ak a t ak   1 8 1 10 log8 log10 t t  1 8 1 10 3log 2 t t   1 8 1 10 10 9 t t   40. The substituents R1 and R2 for nine peptides are listed inthe table given below. How many of these peptides are positively charged at pH = 7.0 ? Ans : 4 At pH below pI(Isoelectric point)Aminoacid exist as Cation.As BasicAminoacids of pH around 9-8, these are expected to exists as cation of pH=7IV,VI,VIII,IX.

26. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 26 Sri Chaitanya IIT Academy MATHEMATICS SECTION – I ( SINGLE CORRECT CHOICE TYPE ) This section contains 10 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct 41. Let z be a complex number such that the imaginary part of z is nonzero and and 2 1a z z   is real. Then a connot take the value a) –1 b) 1 3 c) 1 2 d) 3 4 Ans. D Let Z x iy   0y  2 1a z z   becomes     2 1a x iy x iy        2 2 1 2 0x y x i xy y a i       Comparingreal & imaginary parts 2 2 1x y x a    ––––––(1) 2 0xy y  –––––––(2) From (2) 1 2 x   as 0y  Put value of x in (1) 2 21 1 1 2 2 y a           21 1 4 2 y a   2 3 0 4 y a   3 4 a 

27. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 27 Sri Chaitanya IIT Academy 42. If 2 1 lim 4 1x x x ax b x          , then a) a = 1, b = 4 b) a = 1, b = –4 c) a = 2, b = –3 d) a = 2, b = 3 Ans. B 2 1 lim 4 1x x x ax b x          2 1 1 1 4 1x x a x a b b lim x         For existense of limit coefficient of 2 0x  1 – a = 0 a = 1  1 1 lim 4 1x x a b b x       1 4a b   b = – 4 43. Let ijP a    be a 3 3 matrix and let ijQ b    , where 2i j ij ijb a  for 1 , 3i j  . If the determinant of P is 2, then the determinant of the matrix Q is a) 210 b) 211 c) 212 d) 213 Ans. D as 2i j ij ijb a  11 11 12 12 13 134 , 8 , 16b a b a b a   21 2 22 22 23 238 , 16 , 32b a b a b a   31 31 32 32 33 3316 , 32 , 64b a b a b a   11 12 13 21 22 23 31 32 33 2 a a a A a a a a a a         

28. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 28 Sri Chaitanya IIT Academy 11 12 13 21 22 23 31 32 33 4 8 16 8 16 32 16 32 64 a a a B a a a a a a          11 12 13 21 22 23 31 32 33 4 8 16 2 2 2 4 4 4 a a a a a a a a a            11 12 13 21 22 23 31 32 33 4 8 16 2 4 a a a a a a a a a              4 8 16 2 4 2      13 2B  44. The ellipse 2 2 1 : 1 9 4 x y E   is inscribed in a rectangle R whose sides are parallel to the coor- dinates axes. Another ellipse E2 passing through the point (0, 4) circumscribes the rectangle R. The eccentricity of the ellipse E2 is a) 2 2 b) 3 2 c) 1 2 d) 3 4 Ans. C Given Ellipse 2 2 1 9 4 x y   As ellipse is bounded by &x a y b    Sides of rectangle are 3 & 2x y       3,2 , 3,2C D  D (0,2) (3,0) (0,-2) (-3,0) A B C    3, 2 , 3, 2A B   As new ellipse Circumscribe the rectangle & passes through (0 ,4).

29. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 29 Sri Chaitanya IIT Academy Let new ellipse is 2 2 2 2 1 x y a b   Passes through   2 2 16 0,4 1 16b b     Passes through  3,2 2 9 4 1 16a   2 12a  So, Vertical ellipse 2 2 1 12 16 x y    2 2 2 1a b e   2 12 16 1 e  2 12 1 1 16 4 e    1 2 e  45. The function    : 0,3 1,29f  , defined by   3 2 2 15 36 1f x x x x    is a) one-one and onto b) onto but not one-one c) one-one but not onto d) neither one-one nor onto Ans. B    : 0,3 1,29f    3 2 2 15 36 1f x x x x     ' 2 6 30 36f x x x     6 2 3x x      '' 6 2 5f x x  2x  is pt of maxima.

30. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 30 Sri Chaitanya IIT Academy 3x  is pt of minima.  & 2 29 0f    3 0f  so curve is n f is not one one 2 3 Min value  0 1f  Maxvalue  2 29f  so n f is Onto. 46. The locus of the mid-point of the chord of contact of tangents drawn from points lying on the straight line 4 5 20x y  to the circle 2 2 9x y  is a)  2 2 20 36 45 0x y x y    b)  2 2 20 36 45 0x y x y    c)  2 2 36 20 45 0x y x y    d)  2 2 36 20 45 0x y x y    Ans. A Let    1 1, & ,P h k Q x y As P lies of 4x-5y=20 4 20 , 5 h P h        P A B Q Equation of chord of contact 1 0AB S  4 20 9 5 h xh y         5 4 20 45xh y h   ------------------(1) Equation of chord AB whose mid point is  1 1,Q x y

31. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 31 Sri Chaitanya IIT Academy 2 2 1 1 1 1xx yy x y   ---------------(2) Compare (1) & (2) 2 2 1 1 1 1 5 4 20 45h h x y x y     1 1 2 2 2 2 1 1 1 1 9 45 &4 20 x y h h x y x y      On eliminating h 1 1 2 2 2 2 1 1 1 1 49 45 20 x y x y x y      2 2 1 1 1 120 36 45x y x y   Locus is  2 2 20 36 45 0x y x y    47. The point P is the intersection of the straight line joining the points Q (2, 3, 5) and R (1, –1, 4) with the plane 5 4 1x y z   . If S is the foot of the perpendicular drawn from the point T (2, 1, 4) to QR, then the length of the line segement PS is a) 1 2 b) 2 c) 2 d) 2 2 Ans. A Equation of 1 1 4 1 4 1 x y z QR         Dr’s of  1,4,1QR Co-ordinates of  1 , 1 4 ,4P       As P is point of Intersection of QR & given plane, so      5 1 4 1 4 4 1         1 3   T(2,1,4) Q S R

32. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 32 Sri Chaitanya IIT Academy 4 1 13 , , 3 3 3 P        Let  1 , 1 4 ,4S       Dr’s of  1,4 2,TS     TS QR So,    1 1 4 4 2 0       1 2   3 9 ,1, 2 2 S        2 2 2 4 3 1 13 9 1 3 2 3 3 2 PS                         1 2  48. Let   2 cos , 0 0, 0 x x f x x x       , x IR , then f is a) differentiable both at x = 0 and at x = 2 b) differentiable at x = 0 but not differentiable at x = 2 c) not differentiable at x = 0 but differentiable at x = 2 d) differentiable neither at x = 0 nor at x = 2 Ans. B   2 cos ; 0 0 ; 0 x x f x x x       At x=0

33. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 33 Sri Chaitanya IIT Academy     2 0 0 cos 00 0 lim lim h h hf h f hRHD h h        0 lim cos 0 h h h              2 0 0 cos 0 0 0 lim lim h h h f h f h LHD h h                0 lim cos 0 h h h              Differentiable at x=0 At x=2       2 0 0 2 cos 4 cos 2 2 2 2 lim lim h h h f h f h RHD h h                    2 0 0 2 cos 4 cos 2 2 2 2 lim lim h h h f h f h LHD h h                Not Differentiable at x=2. 49. The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball is a) 75 b) 150 c) 210 d) 243 Ans. B 5 Balls to 3 persons, so that all persons get atleast one ball. Case-I Totalways 2 2 1 2 1 2 1 2 2  5 3 1 2 2 1 3C C C   

34. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 34 Sri Chaitanya IIT Academy 10 3 1 3 90     Case-II Totalways  5 2 1 3 1 1 3C C C    11 1 13 3 1 31 10 2 1 3 60     Total ways = 90 + 60 = 150. 50. The integral   2 9/ 2 sec sec tan x dx x x  equals (for some arbitrary constant K) a)     2 11/2 1 1 1 sec tan 11 7sec tan x x K x x          b)     2 11/2 1 1 1 sec tan 11 7sec tan x x K x x         c)     2 11/ 2 1 1 1 sec tan 11 7sec tan x x K x x          d)     2 11/2 1 1 1 sec tan 11 7sec tan x x K x x         Ans. C   9 2 sec sec tan x dx x x  Let sec tanx x t  -------------(1) So,  2 sec tan secx x x dx dt  sec dt x dx t  ------------(2)

35. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 35 Sri Chaitanya IIT Academy 1 sec tanx x t   ----------(3) From (1) & (3) 1 1 sec 2 x t t        So, Integral becomes   9 2 sec .sec sec tan x x dx x x  9 2 1 1 2 dtt tt t        2 13 2 1 1 2 t dt t   9 13 2 2 1 2 t dt t dt         7 11 2 2 1 7 112 22 t t K            7 11 2 2 7 11 t t K            7 11 22 1 1 11 sec tan7 sec tan x xx x         2 11 2 1 1 1 sec tan 11 7sec tan x x K x x          

36. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 36 Sri Chaitanya IIT Academy SECTION – II ( MULTIPLE CORRECT CHOICE TYPE ) This sectioncontains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out ofwhich ONE OR MORE is/ are correct 51. A ship is fitted with three engines E1 , E2 and E3 . The engines function independently of each other with respective probabilities 1 1 , 2 4 and 1 4 . For the ship to be operational at least two of its engines must function. Let X denote the event that the ship is operational and let X1 , X2 and X3 denote respectively the events that the engines E1 , E2 and E3 are functioning. Which of the following is (are) true ? a) 1 3 | 16 c P x X    b) P [Exactly two engines of the ship are functioning | X] 7 8  c)  2 5 | 16 P X X  d)  1 7 | 16 P X X  Ans. B,D A)     1 1 1 1 12 4 4 1 8 4 c P X X P X       1 1 3 1 1 1 1 2 2 2 4 4 2 4 4 4 P X                     B)   1 1 3 1 1 1 2 72 4 4 2 4 4 / 1 8 4 P Exactlytwoengines of the ship are functioning X             C)       2 2 2 1 1 3 1 1 1 2 52 4 4 2 4 4 / 1 8 4 P X X P X X P X                    D)       1 1 1 1 1 3 1 1 1 2 72 4 4 2 4 4 / 1 16 2 P X X P X X P X             

37. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 37 Sri Chaitanya IIT Academy 52. Tangents are drawn to the hyperbola 2 2 1 9 4 x y   , parallel to the straight line 2 1x y  . The points of contact of the tangents on the hyperbola are a) 9 1 , 2 2 2       b) 9 1 , 2 2 2        c)  3 3, 2 2 d)  3 3,2 2 Ans. A, B 2 2 1 9 4 x y   Tangent to the hyperbola is of the form 2y x    2 9 4 4 36 4 32      4 2   Tangent is : 2 4 2 0x y   Point of contact is 2 2 9 1 9 1 , , , , 2 2 2 2 2 2 a l b m n n                     53. Let S be the area of the region enclosed by 2 x y e  , y = 0, x = 0, and x = 1. Then a) 1 S e  b) 1 1S e   c) 1 1 1 4 S e        d) 1 1 1 1 2 2 S e         Ans. A, B, D 2 1 1 0 0 1 1x x I e dx e dx e        1 1I e   2 1 1x e e    2 1 1 1 1 0 0 0 x e dx e dx dx      2 1 0 1x e dx e   

38. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 38 Sri Chaitanya IIT Academy 54. If y (x) satisfies the differential equation ' tan 2 secy y x x x  and y (0) = 0, then a) 2 4 8 2 y        b) 2 ' 4 18 y        c) 2 3 9 y        d) 2 4 2 ' 3 3 3 3 y          Ans. A,D tan 2 sec dy y x x x dx   tan . cos dx I F e x   General solution is    cos 2 sec cosy x x x x dx  2 x c  2 sec secy x x c x   0,0 0 0 0c c    . 2 secy x x   2 2 4 16 y        2 8 2   2 2 3 9 y         2 sec tan 2 sec dy x x x x x dx   .   2 1 2 3 2 2 3 9 3 y                2 2 4 33 3    

39. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 39 Sri Chaitanya IIT Academy 55. Let  ,  0,2  be such that   2 2cos 1 sin sin tan cot cos 1 2 2                ,  tan 2 0   and 3 1 sin 2     . Then  cannot satisfy a) 0 2    b) 4 2 3     c) 4 3 3 2     d) 3 2 2     Ans. A,C, D   3 tan 2 0, 1 sin 2         3 5 1 , 0 cos 2 3 2               ______ (1)    2 2cos 1 sin sin tan / 2 cot / 2 cos 1           1 sin cos 2         1 sin 1 2      (from (1)) 13 17 6 6       2 7 3 6      (from (1)) cannot satisfied byA, C, D

40. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 40 Sri Chaitanya IIT Academy SECTION –III (INTEGERANSWERTYPE) This sectioncontains5 questions . Theanswer to eachofthe questions is asingle digit integer, rangingfrom0 to 9. The appropriate bubbles below the respective question numbers in the ORS have to be darkened. 56. Let p (x) be a real polynomial of least degree which has a local maximum at x = 1 and a local minimum at x =3. If p (1) = 6 and p (3) = 2, then  ' 0p is Ans. 9   3 2 P x ax bx cx d     1 6 6P a b c d       3 2 27 9 3 2P a b c d       1 1 0 3 2 0P a b c      1 3 0 27 6 0P a b c     1, 6, 9a b c     1 2 3 2P x ax bx c    1 0 9P c  57. If ,a b  and c  are unit vectors satisfying 2 2 2 9a b b c c a           ,then 2 5 5a b c    is Ans. 3 sol. 1a b c     2 2 2 9a b b c c a       3 . 2 a b b c c a        3 2 . . 0a b b c c a      22 2 2 . . . 0a b c a b b c c a       2 0a b c  

41. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 41 Sri Chaitanya IIT Academy  0a b c   ––––––(1) Takingdot product with (1) by , &a b c then 1 1 1 . , . , . 2 2 2 a b b c c a       2 5 5 3a b c    58. The value of 3 2 1 1 1 1 6 log 4 4 4 ... 3 2 3 2 3 2 3 2            is Ans. 4 sol. 1 4 3 2 x x  3 2 4x x  squaring on both sides 2 18 4x x  2 18 4 0x x   2 18 9 8 4 0x x x       9 2 1 4 2 1 0x x x    .   2 1 9 4 0x x   1 4 , 2 9 x   4 9 x   0x   Required answer is 2 3/2 3 6 log 2         = 6 – 2 = 4 59. Let S be the focus of the parabola 2 8y x and let PQ be the common chord of the circle 2 2 2 4 0x y x y    and the given parabola. The area of the triangle PQS is Ans. 4

42. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 42 Sri Chaitanya IIT Academy sol. P Q S (0, 0) (2,4) (2, 0) 2 (0,4) 2 8 2 4 2 0x x x x       0,2 0,0 , 2,4x P Q     area of SPQ = 1 2 4 2   = 4 60. Let :f IR IR be defined as   2 1f x x x   . The total number of points at which f attains either a local maximum or a local minimum is Ans. 5 sol.   2 1, 1f x x x x      2 1 , 1 0x x x       2 1 , 0 1x x x     2 1, 1x x x     ' 2 1, 1f x x x    2 1, 1 0x x      0-1 1 x y 1 2 , 0 1x x    1 2 , 1x x   differentiable at 1 1 , 2 2 x   , not differentiable at –1, 0, 1 The number of points = 5

43. Sri Chaitanya IIT Academy., A.P. IIT-JEE 2012 Paper - 1 along with Key & Solutions Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081

45. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 3 Sri Chaitanya IIT Academy PHYSICS SECTION – I ( SINGLE CORRECT CHOICE TYPE ) This section contains 10 multiple choicse questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct 1. A small block is connected to one end of massless spring of un–stretched length 4.9 m. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by 0.2 m and released from rest at t= 0. It then executes simple harmonic motion with angular frequency / 3 rad s    . Simultaneously at t = 0, a small pebble is projected with speed v from point P at an angle of 450 as shown in the figure. Point P is at a horizontal distance to 10 m from O. If the pebble hits the block at t = 1s, the value of v is (take g = 10 m/s2 ) a) 50 /m s b) 51 /m s c) 52 /m s d) 53 /m s Ans :A Equation of SHM cosy A t At 1t  0.2cos (1) 0.2cos 0.1 3 y m      Block is at a distance 5m  Range of pebble is R =5m 2 2sin90 50 10 V R V     50 /V m s 

46. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 4 Sri Chaitanya IIT Academy 2. In the determination of Young’s modulus 2 4MLg Y ld       by using Searle’s method, a wire of length L = 2 m and diameter d = 0.5 mm is used. For a load M = 2.5 kg, an extension l = 0.25 mm in the length of the wire is observed. Quantities d and l are measured using a screw gauge and a micrometer, respectively. They have the same pitch of 0.5 mm. The number of divisions on their circular scale is 100. The contributions to the maximum probable error of the Y measurement. a) due to the errors in the measurements of d and l are the same b) due to the error in the measurement of d is twice that due to the error in the measurement of l. c) due to the error in the measurement of l is twice that due to the error in the measurement of d. d) due to the error in the measurement of d is four times that due to the error in the measurement of l. Ans :A 2 2 2 4 d d d d v              4 0.25 l l l l      3. A small mass m is attached to a massless string whose other end is fixed at P as shown in the figure. The mass is undergoingcircular motion in the x–y plane with centre at O and constant angular speed  . If the angular momentum of the system, calculated about O and P are de- noted by 0L  and PL  respectively, then

47. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 5 Sri Chaitanya IIT Academy a) 0L  and PL  do not vary with time b) 0L  varieswithtime while PL  remainsconstant c) 0L  remains constant while PL  varies with time d) 0L  and PL  both vary with time Ans :C Torque about O is zero angular momentumwill be constant about ‘O’. Torque about Pis non zero. Angular momentum about P varies with time.  0L m r v   PL m OP v  direction of PL changes with OP and PL OP 4. Young’s double slit experiment is carried out by using green, red and blue light, one color at a time. The fringe widths recorded are ,G R  and B , respectively. Then a) G B R    b) B G R    c) R B G    d) R G B    Ans :D D d       R G B    R G B   

48. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 6 Sri Chaitanya IIT Academy 5. Consider a thin spherical shell of radius R with its centre at the origin, carrying uniform positive surface charge density. The variation of the magnitude of the electric field  E r  and the electric potential V(r) with the distance r from the centre, is best represented by whichgraph ? a) b) c) d) Ans : D From centre to surface E = 0, then 1 E r  From centre to surface V = k, then 1 V r  6. A bi–convex lens is formed with two thin plano–convex lenses as shown in the figure. Re- fractive index n of the first lens is 1.5 and that of the second lens is 1.2. Both the curved surfaces are of the same radius of curvature R = 14 cm. For this bi–convex lens, for an object distance of 40 cm, the image distance will be a) – 280. 0 cm b) 40.0 cm c) 21. 5 cm d) 13.3 cm

49. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 7 Sri Chaitanya IIT Academy Ans :B 1 2 1 1 1 F f f   =    1 21 1 R     = 0.5 0.2 0.7 1 14 14 20    20F cm 40R cm  1 1 1 1 1 1 1 1 1 2 1 1 20 40 20 40 40 40F v u v v             40v cm  7. Two large vertical and parallel metal plates having as separation of 1 cm are connected to a DC voltage source of potential difference X.Aproton is released at rest midway between the two plates. It is found to move at 450 to the vertical JUST after release. Then X is nearly a) 5 1 10 V  b) 7 1 10 V  c) 9 1 10 V  d) 10 1 10 V  Ans :C mg Eq 45 mg Eq pVq d   9 10 mgd V V approx q    8. Three very large plates of same area are kept parallel and close to each other. They are con- sidered as ideal black surfaces and have very high thermal conductivity. The first and third plates are maintained at temperatures 2T and 3T respectively. The temperature of the middle (i.e. second) plate under steady state condition is a) 1 465 2 T       b) 1 497 4 T       c) 1 497 2 T       d)   1 497 T Ans : C    4 4 4 4 1 2 2 3A T T A T T        1 44 44 4 2 2 2 97 2 3 2 T T T T T T           

50. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 8 Sri Chaitanya IIT Academy 9. A thin uniform rod, pivoted at O, is rotating in the horizontal plane with constant angular speed  , as shown in the figure. At time t = 0, a small insect starts from O and moves with constant speed v with respect to the rod towards the other end. It reaches the end of the rod at t = T and stops. The angular speed of the system remains  throughout. The magnitude of the torque of the torque    on the system about O, as a function of time is best represented by which plot ? a) b) c) d) Ans :B   dL d dI I dt dt dt      constantrodI  2 red d I mr dt     = 0 2 dr m r dt         v r = vt 2 2 2mr v m v t   t  when the insect reaches the end it stops v = 0 0 

51. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 9 Sri Chaitanya IIT Academy 10. A mixture of 2 moles of helium gas (atomic mass = 4amu) and 1 mole of argon gas (atomic mass = 40 amu) is kept at 300 K in a container. The ratio of the rms speeds     helium argon rms rms v v        is a) 0.32 b) 0.45 c) 2.24 d) 3.16 Ans : D rms T V M  1 2 2 1 40 10 4 He Ar VV M V V M     SECTION – II ( MULTIPLE CORRECT ANSWER TYPE) This section contains 5 multiple choicse questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONE or More are correct. 11. A small block of mass of 0.1 kg lies on a fixed inclined plane PQ which makes an angle  with the horizontal. Ahorizontal force of 1 N acts on the block through its center of mass as shown in the figure. The block remains stationary if (take 2 10 /g m s ) a) 0 45  b) 0 45  and a frictional force acts on the block towards P. c) 0 45  and a frictional force acts on the block towards Q. d) 0 45  and a frictional force acts on the block towards Q. Ans :A(or)AC

52. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 10 Sri Chaitanya IIT Academy 1cos sinmg  45   when 45  friction force is upwards If friction force sufficient block will be in state of rest 12. Consider the motion of a positive point charge in a region where there are simulataneous uniform electric and magnetic fields 0 ˆE E j  and 0 ˆB B j  . At time t = 0 , this charge has velcoity v  in the x – y plane, making an angle  with the x – axis. Which of the following option(s) is(are) correct for time t > 0 ? a) If 0 0  , the charge moves in a circular path in the x – z plane b) If 0 0  , the charge undergoes helical motion with constant pitch along the y – axis. c)If 0 10  ,thechargeundergoeshelicalmotionwithitspitchincreasingwithtime,alongthey–axis d) If 0 90  , the charge undergoes linear but accelerated motion along the y–axis Ans :C,D when 90  0mF  when 10   path is helical and with increasing pitch.

53. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 11 Sri Chaitanya IIT Academy 13. For the resistance network shown in the figure, choose the correct option(s). a) The current through PQ is zero b) 1 3I A c) The potential at S is less than that at Q. d) 2 2I A Ans:A,B,C,D 1 12 3 4 I   2 12 3 18 I   2A P QV V S QalsoV V  14. A person blows into open – end of a long pipe. A s a result, a high – pressure pulse of air travels down the pipe.When this pulse reaches the other end of the pipe, a) a high – pressure pulse starts travelling up the pipe, if the other end of the pipe is open b) a low – pressure pulse starts travelling up the pipe, if the other end of the pipe is open c) a low – pressure pulse starts travelling up the pipe, if the other end of the pipe is closed d) a high – pressure pulse starts travelling up the pipe, if the other end of the pipe is closed Ans : B, D

54. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 12 Sri Chaitanya IIT Academy Reflected pressure pulse Incident pressure pulse Open end At open end pressure should not change pressure wave will get phase change as  Reflected pressure pulse Incident pressure pulse Closed end At enclosed end pressure wave will get reflected without phase change 15. A cubical region of side a has its centre at the origin. It encloses three fixed point charges, q at  0, / 4,0 , 3a q  at (0, 0, 0) and q at  0, / 4,0a . Choose the correct option(s). a) The net electric flux crossing the plane / 2x a  is equal to the net electric flux crossing the plane / 2x a  b)The net electric fluxcrossingtheplane / 2y a  is morethan the net electricflux crossing the plane / 2y a  . c) The net electric flux crossing the entire region is 0 q  . d) The net electric flux crossing the plane / 2z a  is equal to the net electric flux crossing the plane / 2x a 

55. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 13 Sri Chaitanya IIT Academy Ans:ACD  Net flux passing the plane 2 a x   and 2 a x   is same  Net flux through cube is 0 q   Flux through plane 2 a z  and 2 a x  is same SECTION – III ( INTEGER ANSWER TYPE) This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive). 16. A lamina is made by removing a small disc of diameter 2R from a bigger disc of uniform mass density and radius 2R, as shown in the figure. The moment of inertia of this lamina about axes passing through O and P is 0I and PI , respectively. Both these axes are perpendicular to the plane of the lamina. The ratio 0 PI I to the nearest integer is Ans :3        2 2 2 2 2 2 0 1 1 2 2 2 2 I R R R R R R            4 43 8 2 R R   4 213 13 2 2 R MR       2 2 2 21 1 4 2 4 2 5 2 2 PI M R M R MR M R               2 2 211 8 16 2 MR MR MR         2 211 37 24 2 2 MR MR       

56. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 14 Sri Chaitanya IIT Academy 17. Acircular wire loop of radius R is placed in the x - y plane centered at the origin O.Asquare loop of side  a a R having two turns is placed with its center at 3z R along the axis of the circular wire loop, as shown in figure. The plane of the square loop makes an angle of 45with respect to the z - axis. If the mutual inductance between the loops is given by 2 0 /2 2p a R  , then the value of p is Ans : 7   2 20 3/22 2 2 1 4 2 3 iR B a i R R           (n = 2) 2 2 20 0 9/2 1 2 22 a M R a      18. A cylindrical cavity of diameter a exists inside a cylinder of diameter 2a as shown in the figure. Both the cylinder and the cavity are infinetely long.Auniformcurrent density J flows along the length. If the magnitude of the magnetic field at the point P is given by 0 12 N aJ , then the value of N is

57. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 15 Sri Chaitanya IIT Academy Ans : 5 0 1 2 ja B   2 20 0 2 2 3 12 2 2 a j ja B a a                   1 2 0 1 1 2 12 B B ja         0 5 12 ja  19. A proton is fired from very far away towards a nucleus with charge 120Q e , where e is the electronic charge. It makes a closest approach of 10 fm to the nucleus. The de Broglie wave- length (in units of fm) of the proton at its start is: (take the proton mass,   27 5 / 3 10 ;pm kg   15 / 4.2 10 . /H e j S c   ; 9 0 1 9 10 / ; 4 m F    15 1 10fm m  ) Ans :7 2 2 9 15 120 9 10 10 10 2 p e P m     9 2 27 2 15 5 9 10 120 2 10 3 10 10 e P              17 15 3 2 10 3 2 10 10 e e P          15 4.2 10 6 10 h p     

58. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 16 Sri Chaitanya IIT Academy 20. An infinitely long solid cylinder of radius R has a uniform volume charge density  . It has a spherical cavity of radius R/2 with its centre on the axis on the cylinder, as shown in the figure. The magnitude of the electric field at the point P, which is at the distance 2R from the axis of the cylinder, is given by the expression 0 23 16 R k   . The value of k is Ans: 6   3 2 2 4 2 3 8 2 2 p R K k R E R R      4 1 1 3 32 pE k R        0 1 1 1 4 24 R        0 23 4 24 p R E          24 4 6 16 K   

59. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 17 Sri Chaitanya IIT Academy CHEMISRY SECTION – I ( SINGLE CORRECT CHOICE TYPE ) This section contains 10 multiple choicse questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct 21. The number of aldol reaction(s) that occurs in the given transformation is . . 3 4 conc aq NaOH CH CHO CHO  OH OH HO OH a) 1 b) 2 c) 3 d) 4 Ans : C 3 2 OH CH CHO CH CHO 2 2 2H C H CH CHO O CH CH CHO        O H 2 2 OH CH CHO HO CH CH CHO    H - C - H O H 'HCHO Cannizaro s NaOH reaction CH CH  CHO OH OH CHO HO OH OH OH OH OHHO Number of aldol reaction = 3

60. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 18 Sri Chaitanya IIT Academy 22. A compound M P XP has cubic close packing (ccp) arrangement of X. Its unit cell structure is shown below. The empirical formula of the compound is a) MX b) MX2 c) M2 X d) M5 X14 Ans : B Corner Face center 1 6 3 2   1 8 1 8   4 X : Edge centre Body center 1 1 1  1 4 1 4   2 M : 4 2X M 4 2X M 2MX 23. The carbonyl functional group (–COOH) is present in a) picric acid b) barbituric acid c) ascorbic acid d) aspirin Ans : D Acetyle salicylic acid =Asprin COOH OCOCH3

61. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 19 Sri Chaitanya IIT Academy 24. The colour of light absorbed by an aqueous solution of CuSO4 is : a) organe-red b) blue-green c) yellow d) violet Ans : A Orange - red is absorbed Based up on Munshell wagon wheel V B G O R Y 25. The number of optically active products obtained from the complete ozonolysis of the given compound is : CH3 - CH = CH - C - CH = CH - C - CH = CH - CH3 HCH3 H CH3 a) 0 b) 1 c) 2 d) 4 Ans : A Asthe ozonolysis products do not have chiral centres, number of opticallyactive products = 0 26. As per IUPAC nomenclautre, the name of the complex    2 3 34 2 Co H O NH Cl   is a) Tetraaqudiaminecobalt (III) chloride b) Tetraaquadiamminecobalt (III) chloride c) Diammineteraaquacobalt (III) chloride d) Diamminetetraaquacobalt (III) chloride Ans : D    2 3 34 2 Co H O NH Cl   Diammineteraaquacobalt (III) chloride 27. Which ordering of compound is according to the decreasingorder of the oxidation state of nitrogen ? a) 3 4 2, , ,HNO NO NH Cl N b) 3 2 4, , ,HNO NO N NH Cl c) 3 4 2, , ,HNO NH Cl NO N d) 3 4 2, , ,NO HNO NH Cl N

62. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 20 Sri Chaitanya IIT Academy Ans : B 3 5HNO   2NO   2 0N  4 3NH Cl   Decreasing order of O -S = 3 2 4, , ,HNO NO N NH Cl 28. For one mole of a van der Waals gas when b = 0 and T = 300 K, the PV vs. I/V plot is shown below. The value of the van der Waals constant a (atm. liter2 mol–2 ) is a) 1.0 b) 4.5 c) 1.5 d) 3.0 Ans : C   2 2 n a P v nb nRT v         ’ n = 1, b = 0  2 q p v b RT v         a pv RT v   a pv RT v    2 221.6 20.1 1.5 . 3 2 y c mx m a slop atm t mole        

63. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 21 Sri Chaitanya IIT Academy 29. In allene  3 4C H , the type(s) of hybridisation of the carbon atoms is (are) a) sp and sp3 b) sp and sp2 c) only sp2 d) sp2 and sp3 Ans : B 2 2H C C CH  2 2 sp sp sp 30. The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [a0 is Bohr radius] a) 2 2 2 04 h ma b) 2 2 2 016 h ma c) 2 2 2 032 h ma d) 2 2 2 064 h ma Ans : C mvr= 2 nh  v= 2 nh mr for second Bohr’s orbit 2r =4 0a V= 04 h ma 2 2 2 2 2 016 h V m a  K.E 2 2 2 2 0 1 2 32 h mv ma  

64. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 22 Sri Chaitanya IIT Academy SECTION – II ( MULTIPLE CORRECT ANSWER TYPE) This section contains 5 multiple choicse questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONE or More are correct. 31. Which of the following molecules, in prue form, is (are) unstable at room temperature ? a) b) c) O d) O Ans : BC cyclobutadiene and cyclopentadien are anti aromatic. 32. Which of the following hydrogen halides react(s) with  3AgNO aq to give a precipitate that dissolves in  2 2 3Na S O aq ? a) HCl b) HF c) HBr d) HI Ans : ACD 3 3AgNO HX AgX HNO   aq aq  2 2 3 3 2 3 2 2AgX Na S O Na Ag S O NaX    aq aq s AgF is soluble 33. Choose the correct reason(s) for the stability of the lyophobic colloidal particles > a) Preferential adsorption of ions on their surface from the solution b) Preferential adsorption of solvent of their surface the solution c)Attraction between different particles having opposite charges on their surface d) Potential difference between the fixed layer and the diffused layer of opposite charges around the coolidal particles Ans :AD

65. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 23 Sri Chaitanya IIT Academy 34. For an ideal gas, consider only P-V work in going from an initial state X to the final state Z. The can be reached by either of the two paths shown in the figure. Which of the following choice(s) is (are) correct ? [Take S as change in entropy and w as work done] a) x z x y y zS S S       b) x z x y y zW w w    c) x y z x yw w   d) x y z x yS S     Ans :AC Based up on the given graph x y x y y zS S S       x y z x yW W   35. Identify the binary mixtrue(s) that can be separated into individual compounds, by differential extraction,a s shown in the given scheme. a) 6 5 6 5C H OH and C H COOH b) 6 5 6 5 2C H COOH and C H CH OH c) 6 5 2 6 5C H CH OH and C H OH d) 6 5 2 6 5 2C H CH OH and C H CH COOH Ans : BD Benzyl alcohol is insoluble in both NaOH and 3NaHCO ,

66. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 24 Sri Chaitanya IIT Academy SECTION – III ( INTEGER ANSWER TYPE) This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive). 36. When the following aldohexose exists in its B-configuration, the total number C CH2 CHOH CHOH CHO CH2OH OHH of stereoisomrs in its pyranose form is : Ans : 8 As the configration at C-5 is fixed, (D-configration), then only 4 stero isomers are possible solution 37. 29.2% (w/w) HCl stock solution has a densityh of 1.25 g mL–1 . The molecular weight of HCl is 36.5 g mol–1 . The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is : Ans : 8 Molarity of stock solution = 10 d x GMW   = 10 1.25 29.2 36.5   =10M 1 1 2 2 1 1 10 200 0.4 8 ( ) V M V M V V ML approx      38. The periodic table consists of 18 groups. An isotope of copper, on bombardment with pro- tons, undergoes a nuclear reaction yielding element X as shown below. To which group,element X belongs int he periodic table ? Ans : 8 63 1 1 4 1 52 29 1 0 2 1 266 2CU H n He H X     26 Fe  8 th group

67. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 25 Sri Chaitanya IIT Academy 39. An organic compound undergoes first-order decomposition. The time taken for its decomposition to 1/8 and 1/10 of its initial concentration are t1/8 and t1/10 respectively. What is the Ans : 9 1 8 1 10 2.303 log 8 2.303 log 10 a t ak a t ak   1 8 1 10 log8 log10 t t  1 8 1 10 3log 2 t t   1 8 1 10 10 9 t t   40. The substituents R1 and R2 for nine peptides are listed inthe table given below. How many of these peptides are positively charged at pH = 7.0 ? Ans : 4 At pH below pI(Isoelectric point)Aminoacid exist as Cation.As BasicAminoacids of pH around 9-8, these are expected to exists as cation of pH=7IV,VI,VIII,IX.

68. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 26 Sri Chaitanya IIT Academy MATHEMATICS SECTION – I ( SINGLE CORRECT CHOICE TYPE ) This section contains 10 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct 41. Let z be a complex number such that the imaginary part of z is nonzero and and 2 1a z z   is real. Then a connot take the value a) –1 b) 1 3 c) 1 2 d) 3 4 Ans. D Let Z x iy   0y  2 1a z z   becomes     2 1a x iy x iy        2 2 1 2 0x y x i xy y a i       Comparingreal & imaginary parts 2 2 1x y x a    ––––––(1) 2 0xy y  –––––––(2) From (2) 1 2 x   as 0y  Put value of x in (1) 2 21 1 1 2 2 y a           21 1 4 2 y a   2 3 0 4 y a   3 4 a 

69. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 27 Sri Chaitanya IIT Academy 42. If 2 1 lim 4 1x x x ax b x          , then a) a = 1, b = 4 b) a = 1, b = –4 c) a = 2, b = –3 d) a = 2, b = 3 Ans. B 2 1 lim 4 1x x x ax b x          2 1 1 1 4 1x x a x a b b lim x         For existense of limit coefficient of 2 0x  1 – a = 0 a = 1  1 1 lim 4 1x x a b b x       1 4a b   b = – 4 43. Let ijP a    be a 3 3 matrix and let ijQ b    , where 2i j ij ijb a  for 1 , 3i j  . If the determinant of P is 2, then the determinant of the matrix Q is a) 210 b) 211 c) 212 d) 213 Ans. D as 2i j ij ijb a  11 11 12 12 13 134 , 8 , 16b a b a b a   21 2 22 22 23 238 , 16 , 32b a b a b a   31 31 32 32 33 3316 , 32 , 64b a b a b a   11 12 13 21 22 23 31 32 33 2 a a a A a a a a a a         

70. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 28 Sri Chaitanya IIT Academy 11 12 13 21 22 23 31 32 33 4 8 16 8 16 32 16 32 64 a a a B a a a a a a          11 12 13 21 22 23 31 32 33 4 8 16 2 2 2 4 4 4 a a a a a a a a a            11 12 13 21 22 23 31 32 33 4 8 16 2 4 a a a a a a a a a              4 8 16 2 4 2      13 2B  44. The ellipse 2 2 1 : 1 9 4 x y E   is inscribed in a rectangle R whose sides are parallel to the coor- dinates axes. Another ellipse E2 passing through the point (0, 4) circumscribes the rectangle R. The eccentricity of the ellipse E2 is a) 2 2 b) 3 2 c) 1 2 d) 3 4 Ans. C Given Ellipse 2 2 1 9 4 x y   As ellipse is bounded by &x a y b    Sides of rectangle are 3 & 2x y       3,2 , 3,2C D  D (0,2) (3,0) (0,-2) (-3,0) A B C    3, 2 , 3, 2A B   As new ellipse Circumscribe the rectangle & passes through (0 ,4).

71. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 29 Sri Chaitanya IIT Academy Let new ellipse is 2 2 2 2 1 x y a b   Passes through   2 2 16 0,4 1 16b b     Passes through  3,2 2 9 4 1 16a   2 12a  So, Vertical ellipse 2 2 1 12 16 x y    2 2 2 1a b e   2 12 16 1 e  2 12 1 1 16 4 e    1 2 e  45. The function    : 0,3 1,29f  , defined by   3 2 2 15 36 1f x x x x    is a) one-one and onto b) onto but not one-one c) one-one but not onto d) neither one-one nor onto Ans. B    : 0,3 1,29f    3 2 2 15 36 1f x x x x     ' 2 6 30 36f x x x     6 2 3x x      '' 6 2 5f x x  2x  is pt of maxima.

72. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 30 Sri Chaitanya IIT Academy 3x  is pt of minima.  & 2 29 0f    3 0f  so curve is n f is not one one 2 3 Min value  0 1f  Maxvalue  2 29f  so n f is Onto. 46. The locus of the mid-point of the chord of contact of tangents drawn from points lying on the straight line 4 5 20x y  to the circle 2 2 9x y  is a)  2 2 20 36 45 0x y x y    b)  2 2 20 36 45 0x y x y    c)  2 2 36 20 45 0x y x y    d)  2 2 36 20 45 0x y x y    Ans. A Let    1 1, & ,P h k Q x y As P lies of 4x-5y=20 4 20 , 5 h P h        P A B Q Equation of chord of contact 1 0AB S  4 20 9 5 h xh y         5 4 20 45xh y h   ------------------(1) Equation of chord AB whose mid point is  1 1,Q x y

73. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 31 Sri Chaitanya IIT Academy 2 2 1 1 1 1xx yy x y   ---------------(2) Compare (1) & (2) 2 2 1 1 1 1 5 4 20 45h h x y x y     1 1 2 2 2 2 1 1 1 1 9 45 &4 20 x y h h x y x y      On eliminating h 1 1 2 2 2 2 1 1 1 1 49 45 20 x y x y x y      2 2 1 1 1 120 36 45x y x y   Locus is  2 2 20 36 45 0x y x y    47. The point P is the intersection of the straight line joining the points Q (2, 3, 5) and R (1, –1, 4) with the plane 5 4 1x y z   . If S is the foot of the perpendicular drawn from the point T (2, 1, 4) to QR, then the length of the line segement PS is a) 1 2 b) 2 c) 2 d) 2 2 Ans. A Equation of 1 1 4 1 4 1 x y z QR         Dr’s of  1,4,1QR Co-ordinates of  1 , 1 4 ,4P       As P is point of Intersection of QR & given plane, so      5 1 4 1 4 4 1         1 3   T(2,1,4) Q S R

74. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 32 Sri Chaitanya IIT Academy 4 1 13 , , 3 3 3 P        Let  1 , 1 4 ,4S       Dr’s of  1,4 2,TS     TS QR So,    1 1 4 4 2 0       1 2   3 9 ,1, 2 2 S        2 2 2 4 3 1 13 9 1 3 2 3 3 2 PS                         1 2  48. Let   2 cos , 0 0, 0 x x f x x x       , x IR , then f is a) differentiable both at x = 0 and at x = 2 b) differentiable at x = 0 but not differentiable at x = 2 c) not differentiable at x = 0 but differentiable at x = 2 d) differentiable neither at x = 0 nor at x = 2 Ans. B   2 cos ; 0 0 ; 0 x x f x x x       At x=0

75. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 33 Sri Chaitanya IIT Academy     2 0 0 cos 00 0 lim lim h h hf h f hRHD h h        0 lim cos 0 h h h              2 0 0 cos 0 0 0 lim lim h h h f h f h LHD h h                0 lim cos 0 h h h              Differentiable at x=0 At x=2       2 0 0 2 cos 4 cos 2 2 2 2 lim lim h h h f h f h RHD h h                    2 0 0 2 cos 4 cos 2 2 2 2 lim lim h h h f h f h LHD h h                Not Differentiable at x=2. 49. The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball is a) 75 b) 150 c) 210 d) 243 Ans. B 5 Balls to 3 persons, so that all persons get atleast one ball. Case-I Totalways 2 2 1 2 1 2 1 2 2  5 3 1 2 2 1 3C C C   

76. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 34 Sri Chaitanya IIT Academy 10 3 1 3 90     Case-II Totalways  5 2 1 3 1 1 3C C C    11 1 13 3 1 31 10 2 1 3 60     Total ways = 90 + 60 = 150. 50. The integral   2 9/ 2 sec sec tan x dx x x  equals (for some arbitrary constant K) a)     2 11/2 1 1 1 sec tan 11 7sec tan x x K x x          b)     2 11/2 1 1 1 sec tan 11 7sec tan x x K x x         c)     2 11/ 2 1 1 1 sec tan 11 7sec tan x x K x x          d)     2 11/2 1 1 1 sec tan 11 7sec tan x x K x x         Ans. C   9 2 sec sec tan x dx x x  Let sec tanx x t  -------------(1) So,  2 sec tan secx x x dx dt  sec dt x dx t  ------------(2)

77. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 35 Sri Chaitanya IIT Academy 1 sec tanx x t   ----------(3) From (1) & (3) 1 1 sec 2 x t t        So, Integral becomes   9 2 sec .sec sec tan x x dx x x  9 2 1 1 2 dtt tt t        2 13 2 1 1 2 t dt t   9 13 2 2 1 2 t dt t dt         7 11 2 2 1 7 112 22 t t K            7 11 2 2 7 11 t t K            7 11 22 1 1 11 sec tan7 sec tan x xx x         2 11 2 1 1 1 sec tan 11 7sec tan x x K x x          

78. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 36 Sri Chaitanya IIT Academy SECTION – II ( MULTIPLE CORRECT CHOICE TYPE ) This sectioncontains 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out ofwhich ONE OR MORE is/ are correct 51. A ship is fitted with three engines E1 , E2 and E3 . The engines function independently of each other with respective probabilities 1 1 , 2 4 and 1 4 . For the ship to be operational at least two of its engines must function. Let X denote the event that the ship is operational and let X1 , X2 and X3 denote respectively the events that the engines E1 , E2 and E3 are functioning. Which of the following is (are) true ? a) 1 3 | 16 c P x X    b) P [Exactly two engines of the ship are functioning | X] 7 8  c)  2 5 | 16 P X X  d)  1 7 | 16 P X X  Ans. B,D A)     1 1 1 1 12 4 4 1 8 4 c P X X P X       1 1 3 1 1 1 1 2 2 2 4 4 2 4 4 4 P X                     B)   1 1 3 1 1 1 2 72 4 4 2 4 4 / 1 8 4 P Exactlytwoengines of the ship are functioning X             C)       2 2 2 1 1 3 1 1 1 2 52 4 4 2 4 4 / 1 8 4 P X X P X X P X                    D)       1 1 1 1 1 3 1 1 1 2 72 4 4 2 4 4 / 1 16 2 P X X P X X P X             

79. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 37 Sri Chaitanya IIT Academy 52. Tangents are drawn to the hyperbola 2 2 1 9 4 x y   , parallel to the straight line 2 1x y  . The points of contact of the tangents on the hyperbola are a) 9 1 , 2 2 2       b) 9 1 , 2 2 2        c)  3 3, 2 2 d)  3 3,2 2 Ans. A, B 2 2 1 9 4 x y   Tangent to the hyperbola is of the form 2y x    2 9 4 4 36 4 32      4 2   Tangent is : 2 4 2 0x y   Point of contact is 2 2 9 1 9 1 , , , , 2 2 2 2 2 2 a l b m n n                     53. Let S be the area of the region enclosed by 2 x y e  , y = 0, x = 0, and x = 1. Then a) 1 S e  b) 1 1S e   c) 1 1 1 4 S e        d) 1 1 1 1 2 2 S e         Ans. A, B, D 2 1 1 0 0 1 1x x I e dx e dx e        1 1I e   2 1 1x e e    2 1 1 1 1 0 0 0 x e dx e dx dx      2 1 0 1x e dx e   

80. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 38 Sri Chaitanya IIT Academy 54. If y (x) satisfies the differential equation ' tan 2 secy y x x x  and y (0) = 0, then a) 2 4 8 2 y        b) 2 ' 4 18 y        c) 2 3 9 y        d) 2 4 2 ' 3 3 3 3 y          Ans. A,D tan 2 sec dy y x x x dx   tan . cos dx I F e x   General solution is    cos 2 sec cosy x x x x dx  2 x c  2 sec secy x x c x   0,0 0 0 0c c    . 2 secy x x   2 2 4 16 y        2 8 2   2 2 3 9 y         2 sec tan 2 sec dy x x x x x dx   .   2 1 2 3 2 2 3 9 3 y                2 2 4 33 3    

81. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 39 Sri Chaitanya IIT Academy 55. Let  ,  0,2  be such that   2 2cos 1 sin sin tan cot cos 1 2 2                ,  tan 2 0   and 3 1 sin 2     . Then  cannot satisfy a) 0 2    b) 4 2 3     c) 4 3 3 2     d) 3 2 2     Ans. A,C, D   3 tan 2 0, 1 sin 2         3 5 1 , 0 cos 2 3 2               ______ (1)    2 2cos 1 sin sin tan / 2 cot / 2 cos 1           1 sin cos 2         1 sin 1 2      (from (1)) 13 17 6 6       2 7 3 6      (from (1)) cannot satisfied byA, C, D

82. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 40 Sri Chaitanya IIT Academy SECTION –III (INTEGERANSWERTYPE) This sectioncontains5 questions . Theanswer to eachofthe questions is asingle digit integer, rangingfrom0 to 9. The appropriate bubbles below the respective question numbers in the ORS have to be darkened. 56. Let p (x) be a real polynomial of least degree which has a local maximum at x = 1 and a local minimum at x =3. If p (1) = 6 and p (3) = 2, then  ' 0p is Ans. 9   3 2 P x ax bx cx d     1 6 6P a b c d       3 2 27 9 3 2P a b c d       1 1 0 3 2 0P a b c      1 3 0 27 6 0P a b c     1, 6, 9a b c     1 2 3 2P x ax bx c    1 0 9P c  57. If ,a b  and c  are unit vectors satisfying 2 2 2 9a b b c c a           ,then 2 5 5a b c    is Ans. 3 sol. 1a b c     2 2 2 9a b b c c a       3 . 2 a b b c c a        3 2 . . 0a b b c c a      22 2 2 . . . 0a b c a b b c c a       2 0a b c  

83. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 41 Sri Chaitanya IIT Academy  0a b c   ––––––(1) Takingdot product with (1) by , &a b c then 1 1 1 . , . , . 2 2 2 a b b c c a       2 5 5 3a b c    58. The value of 3 2 1 1 1 1 6 log 4 4 4 ... 3 2 3 2 3 2 3 2            is Ans. 4 sol. 1 4 3 2 x x  3 2 4x x  squaring on both sides 2 18 4x x  2 18 4 0x x   2 18 9 8 4 0x x x       9 2 1 4 2 1 0x x x    .   2 1 9 4 0x x   1 4 , 2 9 x   4 9 x   0x   Required answer is 2 3/2 3 6 log 2         = 6 – 2 = 4 59. Let S be the focus of the parabola 2 8y x and let PQ be the common chord of the circle 2 2 2 4 0x y x y    and the given parabola. The area of the triangle PQS is Ans. 4

84. Sri Chaitanya IIT Academy PAPER-1 KEY & SOLUTIONS Sri Chaitanya IIT Academy Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081 # 42 Sri Chaitanya IIT Academy sol. P Q S (0, 0) (2,4) (2, 0) 2 (0,4) 2 8 2 4 2 0x x x x       0,2 0,0 , 2,4x P Q     area of SPQ = 1 2 4 2   = 4 60. Let :f IR IR be defined as   2 1f x x x   . The total number of points at which f attains either a local maximum or a local minimum is Ans. 5 sol.   2 1, 1f x x x x      2 1 , 1 0x x x       2 1 , 0 1x x x     2 1, 1x x x     ' 2 1, 1f x x x    2 1, 1 0x x      0-1 1 x y 1 2 , 0 1x x    1 2 , 1x x   differentiable at 1 1 , 2 2 x   , not differentiable at –1, 0, 1 The number of points = 5

85. Sri Chaitanya IIT Academy., A.P. IIT-JEE 2012 Paper - 1 along with Key & Solutions Admin Off : Plot No : 304, Kasetty Heights. Ayyappa Society - Madhapur , Pin : 500081

The torque on the system about O is zero throughout since the net external force on the system is zero.a) 0b) vL sinωtc) vL cosωt d) vL

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The torque on the system about O is zero throughout since the net external force on the system is zero.a) 0b) vL sinωtc) vL cosωt d) vL