PART - I (CHEMISTRY) ALLEN. oxidation State of Fe = III oxidation State of Fe = II, III OH H OH H OH RCH2OH

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1 IIT-JEE 0 EXAMINATIN (ELD N : ) PART - I (CEMISTRY) PAPER - SECTIN I : (Total Marks : 4) (Single Correct Answer Type) This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which NLY NE is correct.. xidation states of the metal in the minerals haematite and magnetite, respectively, are (A) II, III in haematite and III in magnetite (B) II, III in haematite and II in magnetite (C) II in haematite and II, III in magnetite (D) III in haematite and II, III in magnetite. Ans. (D) aematite Fe 3 Magnetite Fe 3 4 (Fe. Fe 3 ). The following carohydrate is (A) a ketohexose (C) an a-furanose. Ans. (B) -pyranose (Aldohexose) 3. The major product of the following reaction is oxidation State of Fe = III oxidation State of Fe = II, III (B) an aldohexose (D) an a-pyranose (A) a hemiacetal (C) an ether 3. Ans.(B) RC Å ( anhydrous) (B) an acetal (D) an ester Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA PNE : , Fax : , info@allen.ac.in Wesite: / 9

2 IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER - R C (An acetal) C R Å + Å R C Å CR + CR 4. Amongst the compounds given, the one that would form a rilliant coloured dye on treatment with NaN in dil. Cl followed y addition to an alkaline solution of -naphthol is - (A) 4. Ans. (C) N(C) 3 (B) NC 3 (C) 3 C Cl + NaN + C 3 N Me N Cl N (D) alkaline N=N range coloured dye Me C N 5. The freezing point (in C) of a solution containing 0. g of K 3 [Fe(CN) 6 ] (Mol. Wt. 39) in 00 g of water (K f =.86 K kg mol ) is - (A).3 0 (B) (C) (D) Ans. (A) DT f = m K f i = 0./ = 0.06 =.3 0 T f = 0 DT f =.3 0 C / 9 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA PNE : , Fax : , info@allen.ac.in Wesite:

3 6. Consider the following cell reaction : Fe (s) + (g) +4 + (aq) Fe+ (aq.) + (l) IIT-JEE 0 EXAMINATIN (ELD N : ) E =.67 V At[Fe + ] = 0 3 M, P( ) = 0. atm and p = 3, the cell potential at 5 C is - (A).47 V (B).77 V (C).87 V (D).57 V 6. Ans. (D) Fe (s) + (g) +4 + (aq) ¾ Fe+ (aq.) + (l) PAPER - E = E [Fe ] log + n P.[ ] 0.06 (0 ) =.67 log (0 ) 0.06 E =.67 log0 7 4 = = =.565 V ~.57 V Passing S gas into a mixture of Mn +, Ni +, Cu + and g + ions in an acidified aqueous solution precipitates (A) CuS and gs (C) MnS and NiS 7. Ans. (A) (B) MnS and CuS (D) NiS and gs CuS and gs will e precipitated when S in passed through aqueous solution containing Mn +, Ni +, Cu + and g + ions in acidic medium. 8. Among the following complexes (K P) K 3 [Fe(CN) 6 ] (K), [Co(N 3 ) 6 ]Cl 3 (L), Na 3 [Co(oxalate) 3 ] (M), [Ni( ) 6 ]Cl (N), K [Pt(CN) 4 ] () and [Zn( ) 6 ] (N 3 ) (P) The diamagnetic complex are - (A) K, L, M, N (C) L, M,, P 8. Ans. (C) Diamanetic complexes are (L) [Co(N 3 ) 6 ]Cl 3 ; (M) Na 3 [Co(oxalate) 3 ] (B) K, M,, P (D) L, M, N, () K [Pt(CN) 4 ] ; (P) [Zn( ) 6 ] (N 3 ) Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA PNE : , Fax : , info@allen.ac.in Wesite: 3 / 9

4 IIT-JEE 0 EXAMINATIN (ELD N : ) SECTIN II : (Total Marks : 6) (Multiple Correct Answer Type) PAPER - This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which NE or MRE may e correct. 9. Reduction of the metal centre in aqueous permanganate ion involves - (A) 3 electrons in neutral medium (C) 3 electrons in alkaline medium 9. Ans. (A,C,D) 7+ Mn 4 acidic strong alkaline Mn + Mn 6+ Faintly alkaline or neutral Mn For the first order reaction N 5 (g) ¾ 4N (g) + (g) (B) 5 electrons in neutral medium (D) 5 electrons in acidic medium (A) the concentration of the reactant decreases exponentially with time (B) the half-life of the reaction decreases with increasing temperature. (C) the half-life of the reaction depends on the initial concentration of the reactant. (D) the reaction proceeds to 99.6% completion in eight half-life duration. 0. Ans. (A,B,D) A t = A 0 e kt for option (D) / / 99.6% t99.6 ln 50 = = ; 8 t ln ln = ln t 50 t 0.4. The equilirium Cu I Cu o + Cu II in aqueous medium at 5 C shifts towards the left in the presence of (A) N 3 (B) Cl (C) SCN (D) CN 4 / 9 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA PNE : , Fax : , info@allen.ac.in Wesite:

5 . Ans. (B,C,D) IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER - The eq n. Cu I Cu o + Cu II shifts towards left in the presence of Cl, SCN, and CN due to formation of CuCl(ppt). [Cu(SCN) 4 ] 3 and [Cu(CN) 4 ] 3 ions respectively.. The correct functional group X and the reagent/reaction conditions Y in the following scheme are (i) Y X (C ) 4 X (ii) C (C ) 4 C (A) X = CC 3, Y = /Ni/heat (B) X = CN, Y = /Ni/heat (C) X = CN, Y = Br /Na (D) X = CN, Y = /Ni/heat Ans.(A,B,C,D) X (C ) X 4 (i) y (ii) C (C ) C CC (C ) CC NC (C ) CN 4 NC (C ) CN 4 NC (C ) CN 4 4 Condensation polymer condensation polymer Br /Na /Ni D /Ni D /Ni D (C ) 6 exane-,6-diol C (C ) 4 C D C (C ) 4 C N (C ) 6 N D examethylene diamine C (C ) 4 C N (C ) 4 N D Tetramethylene diamine C (C ) 4 C N (C ) 6 N D examethylene diamine Condensation polyester Condensation polyamide Condensation polyamide Condensation polyamide Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA PNE : , Fax : , info@allen.ac.in Wesite: 5 / 9

6 IIT-JEE 0 EXAMINATIN (ELD N : ) SECTIN III : (Total Marks : 4) (Integer Answer Type) PAPER - This Section contains 6 questions. The answer to each of the questions is a single-digit integer, ranging from 0 to 9. The ule corresponding to the correct answer is to e darkened in the RS. 3. In L saturated solution of AgCl [K sp (AgCl) = ], 0. mol of CuCl [K sp (CuCl) = ] is added. The resultant concentration of Ag + in the solution is.6 0 x. The value of x is. 3. Ans.(7) AgCl Ag + + Cl a a + CuCl Cu + + Cl sp ( + a) K sp(agcl) a(a + ) = = a K (CuCl) ( + a) = [Ag] + [Cu ] Since a<<< Q = 0 6 Þ = 0 3 and a = so x = 7 a = =.6 0 x = + 6 = The maximum numer of isomers (including stereoisomers) that are possile on mono-chlorination of the following compounds, is C 3 C C3C C C 3 4. Ans.(8) C 3 C C C C C 3 3 C 3 monochlorination Cl hn C3C C CCCl * + Cl * C 3 C C C C3 * C 3 + Cl C C C C C 3 3 ( isomers) (4 isomers) + C 3 C Cl C C C C C / 9 C 3 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA PNE : , Fax : , info@allen.ac.in Wesite:

7 IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER - 5. Among the following, the numer of compounds that can react with PCl 5 to give PCl 3 is, C, S,, S 4, P Ans. (4) S,, S 4, P 4 0 PCl 5 + ¾ PCl 3 + Cl 6PCl 5 + P 4 0 ¾ 0PCl 3 S 4 + PCl 5 ¾ PCl 3 + S Cl + S + PCl 5 ¾ PCl 3 + SCl 6. The numer of hexagonal faces that present in a truncated octahedron is. 6. Ans. (8) In a truncated octahedron, there are 4 faces (8 regular hexagonal and 6 square), 36 edges and 4 vertices. 7. The volume (in ml) of 0.M AgN 3 required for complete precipitation of chloride ions present in 30 ml of 0.0M solution of [Cr( ) 5 Cl]Cl, as silver chloride is close to. 7. Ans.(6) millimoles of AgN 3 = millimoles of (Cr( ) 5 Cl)Cl 0. V = = 0.6 V = 6 ml 8. The total numer of contriuting structures showing hyperconjugation (involving C onds) for the following carocation is. C 3 Å CC3 8. Ans. (6) The contriuting hyperconjugating structures (involving C onds) are = 6 No. of a = 6 C + C C C 3 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA PNE : , Fax : , info@allen.ac.in Wesite: 7 / 9

8 IIT-JEE 0 EXAMINATIN (ELD N : ) SECTIN IV : (Total Marks : 6) (Matrix-Match Type) PAPER - This Section contains questions. Each question has four statements (A, B, C and D) given in Column I and five statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching with NE or MRE statement(s) given in Column II. For example, if for a given question, statement B matches with the statements given in q and r, then for the particular question, against statement B, darken the ules corresponding to q and r in the RS. 9. Match the transformations in Column-I with appropriate option in Column-II Column-I Column-II (A) C (s) C (g) (p) phase transition (B) CaC 3 (g) Ca(s) + C (g) (q) allotropic change (C) (g) (r) D is positive (D) P (white, solid) P (red,solid) (s) DS is positive 9. Ans. (A) (p, r, s) ; (B) (r, s) ; (C) (t) ; (D) (q, t) C (s) ¾ C (g) phase transition D > 0, DS > 0 (t) DS is negative 8 / 9 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA PNE : , Fax : , info@allen.ac.in Wesite:

9 IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER - 0. Match the reactions in Column-I with appropriate types of step/reactive intermediate involved in these reactions as given in Column-II Column-I Column-II C 3 (A) aq Na ¾ ¾ ¾ ¾ (p) Nucleophilic sustitution C3MgI (B) CCC Cl ¾ ¾ ¾ ¾ 8 S4 (C) C C C ¾ ¾ ¾ (D) C C C C(C ) 3 S4 ¾ ¾ ¾ C 3 8 C 3 C 3 0. Ans. (A) (r,s,t) ; (B) (p, s, t) ; (C) (r, s) ; (D) (q, r) (q) Electrophilic sustitution (r) Dehydration (s) Nucleophilic addition (t) Caranion Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA PNE : , Fax : , info@allen.ac.in Wesite: 9 / 9

10 IIT-JEE 0 EXAMINATIN (ELD N : ) PART -II (PYSICS) PAPER - SECTIN I : (Total Marks : 4) (Single Correct Answer Type) This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which NLY NE is correct.. A light ray traveling in glass medium is incident on glass-air interface at an angle of incidence q. The reflected (R) and transmitted (T) intensities, oth as function of q, are plotted. The correct sketch is (A) (C) 00% Intensity 00% Intensity T R 0 q 90% T R 0 q 90% (B) (D) 00% Intensity 00% Intensity T R 0 q 90% T R 0 q 90% Ans. (C) When q =0, maximum light is transmitted. At q > q C (critical angle), no further light is transmitted. A wooden lock performs SM on a frictionless surface with frequency, n 0. The lock carries a charge r +Q on its surface. If now a uniform electric field E is switched-on as shown, then the SM of the lock will e (A) of the same frequency and with shifted mean position (B) of the same frequency and with the same mean position (C) of changed frequency and with shifted mean position (D) of changed frequency and with the same mean position Ans. (A) Time period of spring lock system depends on spring constant and mass of lock. n applying electric field only the equilirium position gets shifted. E +Q \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ 0 / 9 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA PNE : , Fax : , info@allen.ac.in Wesite:

11 IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER - 3. The density of a solid all is to e determined in an experiment. The diameter of the all is measured with a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale. The reading on the main scale is.5 mm and that on the circular scale is 0 divisions. If the measured mass of the all has a relative error of %, the relative percentage error in the density is (A) 0.9% (B).4% (C) 3.% (D) 4.% Ans. (C) P = M 4 3 DP DM Drö 00 = ç + 00 P è M r ø 3 p r, æ 3 Dr = least count = 0.0 Þ r =.7 DP æ 0.0 ö 00 = % + ç 3 00 = 3.% P è 070ø 4. A all of mass 0. kg rests on a vertical post of height 5m. A ullet of mass 0.0 kg, traveling with a velocity V m/s in a horizontal direction, hits the centre of the all. After the collision, the all and ullet travel independently. The all hits the ground at a distance of 0 m and the ullet at a distance of 00 m from the foot of the post. The initial velocity V of the ullet is V m/s (A) 50 m/s (B) 50 m/s (C) 400 m/s (D) 500 m/s Ans. (D) 0.0 V = 0. u u Time of flight t = s; Range for all = u t Þ 0 = u Þ u = 0 m/s Þ V = 500 m/s 5. Which of the field patterns given elow is valid for electric field as well as for magnetic field? (A) (B) (C) (D) Ans. (C) Magnetic field lines and induced electric field lines always from closed loop. Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA PNE : , Fax : , info@allen.ac.in Wesite: / 9

12 IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER - 6. A point mass is sujected to two simultaneous sinusoidal displacements in x-direction, x ( ) t = Asin wt and () æ pö x t = Asin ç w+ t 3 mass to a complete rest. The values of B and f are (A) Ans. (B) x (t) 3 A, p (B) 4 = w +f rings the è ø. Adding a third sinusoidal displacement x ( ) 3 t Bsin ( t ) 4 A, p (C) A, p (D) A, p 6 3 A p/3 A x (t) x (t) + x (t) + x 3 (t) = 0 x 3 (t) has to e such that resultant is zero. So it should make 4 p from x (t) anticlockwise A long insulated copper wire is closely wound as a spiral of N turns. The spiral has inner radius a and outer radius. The spiral lies in the X-Y plane and a steady current I flows through the wire. The Z- component of the magnetic field at the center of the spiral is (A) (B) (C) m 0NI ln æ ç ö ( - a) èaø m 0NI æ a n + ö l ç ( -a) è-aø m0ni æö lnç èaø 0 (D) m NI æ a n + ö l ç è - aø Ans. (A) I a Y X Taking an elemental strip of radius x and width dx. Y Area of strip = px dx Numer of turns through area = N dx - a I a x X ò N æö m0 Idx m0niln ( a) ç è db aø = - ò = x ( -a) a / 9 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA PNE : , Fax : , info@allen.ac.in Wesite:

13 IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER - 8. A satellite is moving with a constant speed V in a circular orit aout the earth. An oject of mass m is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of its ejection, the kinetic energy of the oject is (A) mv 3 (B) mv (C) mv (D) mv Ans. (B) KE of oject = mv when it moves with satellite ; PE of oject = mv At the time of ejection KE + PE = 0 to make it escape from gravitational pull. KE = mv. SECTIN II : (Total Marks : 6) (Multiple Correct Answer Type) This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which NE or MRE may e correct. 9. Two solid spheres A and B of equal volumes ut of different densities d A and d B are connected y a string. They are fully immersed in a fluid of density d F. They get arranged into an equilirium state as shown in the figure with a tension in the string. The arrangement is possile only if A B (A) d A < d F (B) d B > d F (C) d A > d F (D) d A + d B = d F Ans. (ABD) F Buoyant = (m a + m B )g ; vd F g = v (d A +d B )g d A + d B = d F. Therefore a,, d 30. Which of the following statement(s) is/are correct? (A) If the electric field due to a point charge varies as r.5 instead of r, then the Gauss law will still e valid. (B) The Gauss law can e used to calculate the field distriution around an electric dipole. (C) If the electric field etween two point charges is zero somewhere, then the sign of the two charges is the same. (D) The work done y the external force in moving a unit positive charge from point A at potential V A to point B at potential V B is (V B V A ) Ans. (CD) The field distriution for a dipole can not e calculated y using Gauss law only, therefore (C,D) Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA PNE : , Fax : , info@allen.ac.in Wesite: 3 / 9

14 IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER - 3. A series R-C circuit is connected to AC voltage source. Consider two cases ; (A) when C is without a dielectric medium and (B) when C is filled with dielectric of constant 4. The current I R through the resistor and voltage V C across the capacitor are compared in the two cases. Which of the following is/are true? (A) I Ans. (BC) A R > I (B) I B R A R < I (C) V B R A C > V (D) V B C A C < V B C X C decreases therefore impedence decreases and current increases. I B > I A As I B increases the voltage across R increases therefore V C decreases. 3. A thin ring of mass kg and radius 0.5 m is rolling without slipping on a horizontal plane with velocity m/s. A small all of mass 0. kg, moving with velocity 0 m/s in the opposite direction, hits the ring at a height of 0.75 m and goes vertically up with velocity 0 m/s. Immediately after the collision 0.75m 0m/s 0m/s m/s (A) the ring has pure rotation aout its stationary CM (B) the ring comes to a complete stop (C) friction etween the ring and the ground is to the left (D) there is no friction etween the ring and the ground Ans. (AC) Since momentum of all and ring has same magnitude ut they are opposite in direction and final momentum of all after the collision in horizontal direction is zero, therefore the ring has pure rotation aout its stationary CM just after collision (assuming non impulsive friction). SECTIN III : (Total Marks : 4) (Integer Answer Type) This Section contains 6 questions. The answer to each of the questions is a single-digit integer, ranging from 0 to 9. The ule corresponding to the correct answer is to e darkened in the RS. 33. Two atteries of different emfs and different internal resistances are connected as shown. The voltage across AB in volts is 6V W A B Ans.(5) 3V W v AB = 4 / = = Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA PNE : , Fax : , info@allen.ac.in Wesite:

15 IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER A series R-C comination is connected to an AC voltage of angular frequency w=500 radian/s. If the impedance of the R-C circuit is R.5, the time constant (in millisecond) of the circuit is Ans. (4).5 R = R + æ ö ç è wcø 0.5 R = æ ö ç è wcø ; 0.5 R = ; C = 500 C 50R ; RC = sec 50 t = 4 millisecond; t = A train is moving along a straight line with a constant acceleration a. A oy standing in the train throws a all forward with a speed of 0 m/s, at an angle of 60 to the horizontal. The oy has to move forward y.5 m inside the train to catch the all ack at the initial height. The acceleration of the train, in m/s, is Ans. (5) With respect to train : Velocity : Acceleration : v y T = 5 3 = = g 0.5 = 5t at Þ a =5 m/s Water (with refractive index = 4 ) in a tank is 8 cm deep. il of 3 refractive index 7 lies on water making a convex surface of radius 4 of curvature R = 6 cm as shown. Consider oil to act as a thin lens. An oject S is placed 4 cm aove water surface. The location of its image is at x cm aove the ottom of the tank. Then x is Ans. () First refraction: m =, u = 4, m = 7 4, R = + 6, m m m -m - = v u R After solving v = Now for second refraction : h = = 6 ( /6) So, from ottom 8 6 = So, x = R=6cm Denser h I S m=.0 m=7/4 m=4/3 Rarer R=6cm S m=.0 m=7/4 m=4/3 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA PNE : , Fax : , info@allen.ac.in Wesite: 5 / 9

16 IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER A lock of mass 0.8 kg is attached to a spring of force-constant N/m. The coefficient of friction etween the lock and the floor is 0.. Initially the lock is at rest and the spring is un-stretched. An impulse is given to the lock as shown in the figure. The lock slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the lock in m/s is V= N/0. Then N is Ans. (4) -mmgx - kx = 0- mv.44 4 v = = Þ N = Ans. (7) 38. A silver sphere of radius cm and work function 4.7 ev is suspended from an insulating thread in freespace. It is under continuous illumination of 00 nm wavelength light. As photoelectrons are emitted, the sphere gets charged and acquires a potential. The maximum numer of photoelectrons emitted from the sphere is A 0 z (where < A < 0). The value of Z is hc ( NeK ) -f= ev = e l R \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ - ( ) æ40 ö - 9 N ç = è 00 ø / = N.6 6 / 9 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA PNE : , Fax : , info@allen.ac.in Wesite:

17 IIT-JEE 0 EXAMINATIN (ELD N : ) SECTIN IV : (Total Marks : 6) (Matrix-Match Type) PAPER - This Section contains questions. Each question has four statements (A, B, C and D) given in Column I and five statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching with NE or MRE statement(s) given in Column II. For example, if for a given question, statement B matches with the statements given in q and r, then for the particular question, against statement B, darken the ules corresponding to q and r in the RS. 39. ne mole of a monatomic ideal gas is taken through a cycle ABCDA as shown in P-V diagram. Column II gives the characteristics involved in the cycle. Match them with each of the processes given in Column I. Column I 3P P P B C A 0 V 3V D 9V V Column II (A) Process A B (p) Internal energy decreases. (B) Process B C (q) Internal energy increases. (C) Process C D (r) eat is lost. (D) Process D A (s) eat is gained. Ans. (A) p,r,t (B) p,r (C) q,s (D) r,t (t) Work is done on the gas. For (A) : For (B) : In process AB (isoaric compression) Work is negative, DU is negative, DQ is negative BC process (Isochoric) 3P P B A Work is zero, DU is negative, DQ is negative For (C) : CD Process (Isoaric expansion) P C D Work is negative, DU is positive, DQ is positive 0 V 3V 9V V For (D) : DA Process (V = decreases Isothermal) Work is negative, DU is zero, DQ is negative Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA PNE : , Fax : , info@allen.ac.in Wesite: 7 / 9

18 IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER Column I shows four systems, each of the same length L, for producing standing waves. The lowest possile natural frequency of a system is called its fundamental frequency, whose wavelength is denoted as l f. Match each system with statements given in Column II descriing the nature and wavelength of the standing waves. Column I Column II (A) Pipe closed at one end (p) Longitudinal waves 0 L (B) Pipe open at oth ends (q) Transverse Waves 0 L (C) Stretched wire clamped at oth ends (r) l f =L 0 L (D) Stretched wire clamped at oth ends (s) l f = L and at mid-point (t) l f = 4L 0 L/ Ans. (A) p,t (B) p,s (C) q,s (D) q,r For (A) : For (B) : L Sound wave is longitudinal wave l F = L Þ l 4 F = 4L Sound wave is longitudinal wave l F = L Þ l F = L For (C) : String wave is transverse l F = L Þ l F = L For (D) : l F = L Þ so (q & r) 8 / 9 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA PNE : , Fax : , info@allen.ac.in Wesite:

19 IIT-JEE 0 EXAMINATIN (ELD N : ) PART - III (MATEMATICS) PAPER - SECTIN I : (Total Marks : 4) (Single Correct Answer Type) This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which NLY NE is correct. x y 4. Let P(6, 3) e a point on the hyperola - =. If the normal at the point P intersects the a x-axis at (9,0), then the eccentricity of the hyperola is - Sol. 5 (A) Ans. (B) (B) 3 Equation of normal at P(6, 3) on It intersects x-axis at (9, 0) x a y - = is (C) (D) 3 ax y + = ae Þ a ae 6 = Þ 3 e = 4. Let (x,y) e any point on the paraola y = 4x. Let P e the point that divides the line segment from (0,0) to (x,y) in the ratio : 3. Then the locus of P is- (A) x = y (B) y = x (C) y = x (D) x = y Sol. Ans. (C) Let P e (h, k) æx yö on using section formula P ç, è4 4ø \ x y h = and k = 4 4 Þ x = 4h and y = 4k Q (x, y) lies on y = 4x \ 6k = 6h Þ k = h Locus of point P is y = x. 43. Let f(x) = x and g(x) = sinx for all xîr. Then the set of all x satisfying (ƒ o g o g o ƒ) (x) = (g o g o ƒ) (x), where (ƒ o g) (x) = ƒ(g(x)), is- (A) ± n p,n Î{ 0,,,... } (B) ± n p, n Î{,,... } Sol. (C) p + n p, n Î{..., -, -,0,,,...} Ans. (A) Given ƒ(x) = x ; g(x) = sinx ƒogogoƒ(x) = sin (sinx ) and gogoƒ(x) = sin(sinx ) given ƒogogoƒ(x) = gogoƒ(x) Þ sin (sinx ) = sin(sinx ) Þ sin(sinx ) = 0 or (rejected) (D) n p, n Î{..., -,-,0,,,...} sin(sinx ) = 0 Þ x = np Þ x =± n p; x Î{ 0,,, 3,... } Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA PNE : , Fax : , info@allen.ac.in Wesite: 9 / 9

20 IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER Let ƒ:[,] [0, ) e a continuous function such that ƒ(x) = ƒ( x) for all x Î [,]. Let R = ò xƒ(x)dx, Sol. and R e the area of the region ounded y y = ƒ(x), x=, x=, and the x-axis. Then - (A) R = R (B) R = 3R (C) R = R (D) 3R = R Ans. (C) - R = ò ƒ(x)dx, - R =ò - xƒ(x)dx 45. If Sol. (A) x 0 ò = ( -x)ƒ( -x)dx = - ò - ( -x)ƒ(x)dx ò = ƒ(x)dx - ò - - xƒ(x)dx or R = R R Þ R = R æ ö ç Qòƒ(x)dx = òƒ(a + -x)dx è a a ø (given ƒ(x) = ƒ( x)) lim é ë + xln( + ) ù û x = sin q, > 0 and q Î ( p,p], then the value of q is- p ± 4 (B) Ans. (D) x 0 p ± (C) 3 p ± (D) 6 xln(+ ) lim x x 0 x limé ë+ xln(+ ) ù û = e = + ence + = sin q Þ æ ö sin q= ç + ³ è ø p ± \ sin p q = Þ sinq=± Þ q = ± 46. The circle passing through the point (,0) and touching the y-axis at (0,) also passes through the point - Sol. (A) æ 3 ç - ö,0 è ø (B) æ 5 ç - ö, è ø Ans.(D) Family of circle which touches y-axis at (0,) is x + (y ) + lx = 0 Passing through (,0) Þ + 4 l = 0 Þ l = 5 \ x + y + 5x 4y + 4 = 0 which satisfy the point ( 4,0). (C) æ 3 5 ç - ö, è ø (D) ( 4,0) 0 / 9 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA PNE : , Fax : , info@allen.ac.in Wesite:

21 IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER - é a ù ê w ú 47. Let w¹ e a cue root of unity and S e the set of all non-singular matrices of the form ê c ú êëw w ú, û where each of a, and c is either w or w. Then the numer of distinct matrices in the set S is- (A) (B) 6 (C) 4 (D) 8 Sol. Ans.(A) a w w c w = cw a(w w c) = ( cw) aw( cw)=( cw) ( aw) for non singular matrix c ¹ & a ¹ w w Þ c ¹ w, a ¹ w Þ a & c must e w & can e w or w \ total matrices = 48. A value of for which the equations Sol. have one root in common is - x + x = 0 x + x + = 0, (A) - (B) -i 3 (C) i 5 (D) Ans.(B) x -x = = Þ + x= -...(i) + & x = - from (i) & (ii)...(ii) æ+ ö + ç = è-ø - Þ ( + ) ( ) = ( + ) Þ = + + Þ 3 3 = 0 Þ ( + 3) = 0 Þ = 0, =± 3i Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA PNE : , Fax : , info@allen.ac.in Wesite: / 9

22 IIT-JEE 0 EXAMINATIN (ELD N : ) SECTIN II : (Total Marks : 6) (Multiple Correct Answer Type) PAPER - This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which NE or MRE may e correct. ì p p ï - x -, x - ï ï p 49. If ƒ(x) = í - cosx, - < x 0 then - ï ï x-, 0< x ï î lnx, x> Sol. (A) ƒ(x) is continuous at p x =- (B) ƒ(x) is not differentiale at x = 0 (C) ƒ(x) is differentiale at x = (D) ƒ(x) is differentiale at x =- 3 Ans.(A,B,C,D) - æ p ö ƒ ç - = 0 è ø, + æ p ö ƒ ç - = 0 è ø ì p ï - x ï p ïsinx - < x 0 ƒ'(x) =í ï 0< x ï ï x > ïî x ƒ'(0 ) = 0, ƒ'(0 + ) = \ not differentiale at x = 0 ƒ'( ) =, ƒ'( + ) = \ differentiale at x = as 3 æ p ö - Î ç -,0 è ø ƒ'(x) = sinx which is differentiale at x = Let L e a normal to the paraola y = 4x. If L passes through the point (9,6), then L is given y- (A) y x + 3 =0 (B) y + 3x 33 = 0 (C) y + x 5 = 0 (D) y x + = 0 Sol. Ans. (A,B,D) Equation of normal is y = mx m m 3 It passes through the point (9, 6) then 6 = 9m m m 3 Þ m 3 7m + 6 = 0 Þ (m )(m )(m + 3) = 0 Þ m =,, 3 Equations of normals are y x + 3 = 0, y + 3x 33 = 0 & y x + = 0 / 9 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA PNE : , Fax : , info@allen.ac.in Wesite:

23 IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER - 5. Let E and F e two independent events. The proaility that exactly one of them occurs is 5 Sol. proaility of none of them occurring is T, then - (A) P(E) = 4 5, P(F) = 3 5 (C) P(E) =, P(F) = 5 5 Ans. (A,D) Let P(E) = x & P(F) = y According to given condition x( y) + y( x) = 5 Þ x + y xy = 5 Also, ( x)( y) = 5 Þ x + y xy = 3 5 from (i) & (ii) xy = 5, x + y = (i) Solving this x = 4 5, y = 3 5 or 3 x =, 5...(ii) and the. If P(T) denotes the proaility of occurrence of the event 5 4 y = 5 - x 5. Let ƒ : (0,) R e defined y ƒ(x) = - x, where is a constant such that 0 < <. Then Sol. (B) P(E) =,P(F) = 5 5 (D) P(E) = 3,P(F) = (A) ƒ is not invertile on (0,) (B) ƒ ¹ ƒ on (0,) and ƒ'() = ƒ'(0) (C) ƒ = ƒ on (0,) and ƒ'() = ƒ'(0) Ans. (A) ƒ : (0,) R -x ƒ(x) = Î (0,) - x - Þ ƒ'(x) = (x -) Þ ƒ'(x) < 0 " x Î (0, ) hence ƒ(x) is decreasing function hence its range (, ) Þ Þ co-domain ¹ range ƒ(x) is non-invertile function (D) ƒ is differentiale on (0,) Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA PNE : , Fax : , info@allen.ac.in Wesite: 3 / 9

24 IIT-JEE 0 EXAMINATIN (ELD N : ) SECTIN III : (Total Marks : 4) (Integer Answer Type) PAPER - This Section contains 6 questions. The answer to each of the questions is a single-digit integer, ranging from 0 to 9. The ule corresponding to the correct answer is to e darkened in the RS. 53. Let y'(x) + y(x)g'(x) = g(x)g'(x), y(0) = 0, x Î R, where ƒ'(x) denotes dƒ(x) and g(x) is a given non-constant dx differentiale function on R with g(0) = g() = 0. Then the value of y() is Sol. Ans. 0 Given y(0) = 0, g(0) = g() = 0 Let y'(x) + y(x). g'(x) = g(x) g'(x) Þ y'(x) + (y(x) g(x)) g'(x) = 0 y'(x) Þ + y(x) = g'(x) g(x) Þ I.F. = òd(g(x)) g(x) e = e Þ g(x) g(x) g(x) y(x).e = g(x).e - e + c Þ dy(x) y(x) g(x) dg(x) + = g(x) g(x) y(x).e e g(x).dg(x) = ò put x = 0 Þ 0 = 0 + c Þ c = Þ y(). e g() = g()e g() e g() Þ y() = 0 e 0 + Þ y() = 0 r r Let a =-ˆi- k, ˆ =- ˆi+ ˆ r j and c = ˆi + ˆj + 3k ˆ e three given vectors. If r is a vector such that r r r r = c and rr r r.a= 0, then the value of r. r is Sol. Ans. 9 r a=-ˆi-kˆ r =- ˆi+ ˆj r c = ˆi + j ˆ+ 3kˆ r r r r = c Taking cross product y a r r r r r r r (r ) a = (c ) a Þ rrr r rr rrr r rr (r.a) -(.a)r = (c.a)) -(.a)c Þ r 0 = (--3)(- ˆi + ˆj) -()(i ˆ+ j ˆ+ 3k) ˆ r = - 3i ˆ+ 6j ˆ+ 3kˆ r r r. = 3+ 6= i /3 Let w= e p, and a,, c, x, y, z e non-zero complex numers such that a + + c = x a + w + cw = y a + w + cw = z. x + y + z Then the value of a + + c Sol. Ans. 3 Comment : If w = e ip/3 then a,,c is x + y + z a + + c is not always an integer, infect its value depends upon 4 / 9 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA PNE : , Fax : , info@allen.ac.in Wesite:

25 Þ Let w = e i p/3 x = (a + + c) ( a + + c) IIT-JEE 0 EXAMINATIN (ELD N : ) = a + + c + a+ ac+ a + c+ ca+ c y = (a + w + cw ) ( a + w + cw) PAPER - = a + + c + aw + acw+ aw+ cw + caw + cw z = (a+ w + c w )(a+ w+ c w ) = a + + c + aw+ acw + aw + cw+ caw+ cw \ x + y + z = 3 ( a + + c ) x + y + z Þ = 3 a + + c 56. Let M e 3 3 matrix satisfying é0ù é-ù é ù é ù M ê ú = ê ú ê - ú = ê ú ê ú ê ú, M ê ú ê ú êë úû êë úû êë úû êë- ú and û Then the sum of the diagonal entries of M is Sol. Ans. 9 Let éa cù M= ê ú ê x y z ú êël m núû according to question éa cùé0ù é-ù ê úê ú = ê ú ê x y z úê ú ê ú êël m núê ûë0úû êë 3 úû éù é0ù M = ê ú ê 0 ú ê ú = ê ú êë úû êëúû Þ =, y =, m = 3...() Þ éa cùé ù é ù ê x y z úê ú ê ú ê úê - ú = ê ú êël m núê ûë 0 úû êë-úû a = x y = l m = from () a = 0 x = 3 l = éa cùéù é0 ù ê úê ú = ê ú ê x y z úê ú ê 0 ú êël m núê ûëúû êëúû Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA PNE : , Fax : , info@allen.ac.in Wesite: 5 / 9

26 l + m + n = Þ n = Þ n = 7 Now a + y + n = = 9 IIT-JEE 0 EXAMINATIN (ELD N : ) 57. The numer of distinct real roots of x 4 4x 3 + x + x = 0 is Sol. Ans. Let ƒ(x) = x 4 4x 3 + x + x Þ Q ƒ'(x) = 4x 3 x + 4x ƒ"(x) = x 4x + 4 = (x x + ) > 0 ƒ'(x) is strictly increasing function ƒ'(x) is cuic polynomial hence numer of roots of ƒ'(x) = 0 is Þ Numer of maximum roots of ƒ(x) = 0 are Now ƒ(0) =, ƒ() = 9, ƒ( ) = 5 Þ ƒ(x) has exactly distinct real roots. 58. The straight line x 3y = divides the circular region x + y 6 into two parts. If ìæ 3ö æ5 3öæ öæ öü S= íç,, ç, ç,,- ç,, ý, îè 4ø è 4øè4 4øè8 4øþ then the numer of point(s) in S lying inside the smaller part is Sol. Ans. If the point lies inside the smaller part, then origin and point should give opposite signs w.r.t. line & point should lie inside the circle. for origin : = ( ve) for (, 3 4 ): = 3 (+ve); point lies inside the circle 4 (0, 6) (0, /3) y PAPER - x 3y =0 x (/,0) ( 6,0) for ( 5, 3 4 ): = 7 (+ve) ; point lies outside the circle 4 For æ ö ç,- è4 4ø : 4 3 æ ö ç - è 4 ø = 4 (+ve) ; point lies inside the circle æ ö For ç, è8 4ø : 8 3 æ ö ç è4 ø = - 3 ( ve) ; point lies inside the circle. \ points lie inside smaller part. 6 / 9 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA PNE : , Fax : , info@allen.ac.in Wesite:

27 IIT-JEE 0 EXAMINATIN (ELD N : ) SECTIN IV : (Total Marks : 6) (Matrix-Match Type) PAPER - This Section contains questions. Each question has four statements (A, B, C and D) given in Column I and five statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching with NE or MRE statement(s) given in Column II. For example, if for a given question, statement B matches with the statements given in q and r, then for the particular question, against statement B, darken the ules corresponding to q and r in the RS. 59. Match the statements given in Column I with the values given in Column II Sol. Column I r r (A) If a = ˆj + 3k, ˆ = - ˆj + 3kˆ and r c = 3kˆ form (B) (C) (D) a triangle, then the internal angle of the triangle etween a r and r is If ò (ƒ(x) - 3x)dx = a -, then the value of ƒ æp ö ç è6 ø is (q) p 3 a The value of p ln3 56 / ò 7/6 sec( px)dx is æ ö The maximum value of Argç for (s) p è - z ø z =, z ¹ is given y (p) Column II Ans. (A) (q); (B) (p); (C) (s); (D) (t) (A) c q a c = = = 3 a r r r a + - c cos q = r r = = Þ a.. q= p 3 (r) (t) p 6 p 3 p (B) ò (ƒ(x) 3x)dx = a = a ò a (-x)dx Þ ò(ƒ(x) - x)dx = 0 Þ one of the possile solution of this equation is a æpö p ƒ(x) = x Þ ƒç = è 6 ø 6 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA PNE : , Fax : , info@allen.ac.in Wesite: 7 / 9

28 IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER - (C) p ln3 5/6 ò 7/6 (secpx)dx p ln3 p l = [ n secp x+ tanpx ] 5/6 7/6 p = ln l n3 5p 5p sec + tan 6 6 7p 7p sec + tan 6 6 p = l n3 ln 3 = p (D) Let Þ æ ö q= Argç è - z ø z æ0-ö q= Arg ç which is shown in adjacent diagram. è z - ø Þ Maximum value of q is approaching to p 60. Match the statements given in Column I with the intervals/union of intervals given in Column II Column I (A) The set ì æ iz íre ç ö : z is a complex numer, z =, z ¹± ü ý î è-z ø þ (B) (C) is x- -æ 8(3) ö The domain of the function ƒ(x) = sin ç (x -) è-3 ø tan q (0,0) Column II (p) (-,-) È (, ) is (q) ( -,0) È (0, ) If ƒ( q ) = -tan q tan q, then the set (r) [, ) - -tan q q (,0) ì pü íƒ( q ):0 q< ý î þ is (s) ( -, - ] È [, ) Sol. (D) If ƒ(x) = x 3/ (3x 0), x ³ 0, then ƒ(x) is increasing in (t) (-,0] È [, ) Ans. (A) (s); (B) (t); (C) (r); (D) (r) (A) Let z = cosq + isinq æ i(cosq+ isin q) ö æ cosqi -sin q ö Reç = Re - q+ q ç q- q q è (cos i sin ) ø èsin i cos sin ø æ ö - = Reç - = è sin qø sin q \ Set will e (-, -] È [, ) 8 / 9 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA PNE : , Fax : , info@allen.ac.in Wesite:

29 IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER - (B) (x-) (x -) -3 x ¹ Þ x x x (3-3)(3+ 3) 3 x = t \ t > 0 (C) 8t (3 - t)(t + 3) ³- 8t (3 - t)(t + 3) Taking intersection, x Î( -, 0] È [, ) Þ tî( 0,3 ) È [9, ) Þ x Î( -,) È [, ) Þ t Î(0,] È (3, ) Þ x Î( -, 0] È (, ) tan ƒ( q ) = -tan q tan q C C + C 3 Þ - -tanq tanq ƒ( q ) = 0 tan q 0 -tanq Þ ƒ(q) = sec q Þ ƒ( q) Î [, ) (D) ƒ(x) = 3x 5/ 0x 3/ ƒ'(x) = / / x - x > 0 Þ 5 x(x- ) ³ 0 Þ x ³ Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA PNE : , Fax : , info@allen.ac.in Wesite: 9 / 9