IIT - JEE 2013 (Advanced) CODE : O

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1 IIT - JEE 0 (Advanced) CODE : O

2 () Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution

3 IIT JEE 0 Advanced : Question Paper & Solution (Paper I) () PART I : PHYSICS SECTION I : (Only One option correct Type) This section contains 0 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. x. The work done on a particle of mass m by a force, ˆ y ˆ K i+ j / / ( x y ) (x + y ) + (K being a constant of appropriate dimension), when the particle is taken from the point (a, 0) to the point (0, a) along a circular path of radius a about the origin in the x-y plane is (A) K π (B) K π (C) K π (D) 0 a a a. (D) k F r F is conservative r ˆ i ˆ F K x + i U (x + y ) y (x + y ) U K [potential energy] (x + y ) U(0, a) U(a, 0) W 0. Two rectangular block, having identical dimensions, can be arranged either in configuration I or in configuration II as shown in the figure. One of the block has thermal conductivity k and the other k. The temperature difference between the ends along the x- axis is the same in both the configurations. It takes 9 s to transport a certain amount of heat from the hot end to the cold end in the configuration I. The time to transport the same amount of heat in the configuration II is (A).0 s (B).0 s (C) 4.5 s (D) 6.0 s. (A) Thermal resistance of individual block are R and R. For config ; R R; For config ; R Q Ratio of heat currents R Q R 9 Q Q t Q t t t 9s sec Q 9 R

4 (4) Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution. two non-reactive monoatomic ideal gases have their atomic masses in the ratio :. The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature, is 4 :. The ratio of their densities is (A) : 4 (B) : (C) 6 : 9 (D) 8 : 9. (D) PV PV P Mρ P Mρ ρ 4 8/9 ρ 4. A particle of mass m is projected from the ground with an initial speed u 0 at an angle α with the horizontal. At the highest point of its trajectory, it makes a completely inelastic collision with another identical particle, which was thrown vertically upward from the ground with the same initial speed u 0. The angle that the composite system makes with the horizontal immediately after the collision is π π π π (A) (B) +α (C) α (D) (A) u H sin α g Speed the particle that was thrown vertically at the highest point be v. u0 sin α v u0 g v u 0 cos α g py tan θ p x 5. A pulse of light of duration 00 ns is absorbed completely by a small object initially at rest. Power of the pulse is 0 mw and the speed of light is 0 8 ms. The final momentum of the object is (A) kg ms (B) kg ms (C) kg ms (D) kg ms 5. (B) P h/d or E PC P In the Young s double slit experiment using a monochromatic light of wavelength λ, the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is λ λ λ λ (A) ( n + ) (B) (n + ) (C) (n + ) (D) (n + ) (B) I δ 4I0 cos If I I 0 mu 0 cos α θ mu 0 cos α 4

5 IIT JEE 0 Advanced : Question Paper & Solution (Paper I) (5) cos δ cos δ δ π ± nπ± 4 π Δ x π nπ± Δ x n ± λ λ The image of an object, formed by a plano-convex lens at a distance of 8 m behind the lens, is real and is one-third the size of the object. The wavelength of light inside the lens is times the wavelength in free space. The radius of the curved surface of the lens is (A) m (B) m (C) m (D) 6 m 7. (C) v 8 m, μ m v v m v u f f u f v m f f 8 f f 6 f R μ R 6 R ( ) 8. One end of a horizontal thick copper wire of length L and radius R is welded to an end of another horizontal thin copper wire of length L and radius R. When the arrangement is stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is (A) 0.5 (B) 0.50 (C).00 (D) (C) F R L L R F F F 4A A Δ Δ L L 4Δ Δ Δ Δ Δ Δ 9. A ray of light travelling in the direction ( i j ˆ ) reflection, it travels along the direction ( i j ˆ ) + is incident on a plane mirror. After. The angle of incidence is (A) 0 (B) 45 (C) 60 (D) 75 5

6 (6) Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution 9. (A) Angle between IandR is ( i ˆj)( i ˆ IR. + j) cos δ I R ( ) ( ) 4 δ 0 80 i 0 i 60 i 0 0. The diameter of a cylinder is measured using a Vernier caliper with no zero error. It is found that the zero of the Vernier scale lies between 5.0 cm and 5.5 cm of the main scale. The Vernier scale has 50 divisions equivalent to.45 cm. The 4 th division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is (A) 5. cm (B) 5.4 cm (C) 5.6 cm (D) 5.48 cm 0. (B) M.S.R. 5. cm V. C cm. 50 L.C cm V.S.R. 4 d (5.) cm cm SECTION II : (One or more options correct Type) This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct.. In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance C. The switch S is pressed first to fully charge the capacitor C and then released. The switch S is then pressed to charge the capacitor C. After some time, S is released and then S is pressed. After some time, (A) the charge on the upper plate of C is CV 0. (B) the charge on the upper plate of C is CV 0. (C) the charge on the upper plate of C is 0. (D) the charge on the upper plate of C is CV 0. I δ R. (B), (D) After S is released : +CV 0 CV 0 C C 6

7 After S is released IIT JEE 0 Advanced : Question Paper & Solution (Paper I) (7) +CV 0 CV 0 +CV 0 CV 0 C C After S is released +CV 0 CV 0 CV 0 +CV 0 C C. A particle of mass M and positive charge Q, moving with a constant velocity u 4i ˆ ms, enters a region of uniform static magnetic field normal to the x-y plane. The region of the magnetic field extends from x 0 to x L for all values of y. After passing through this region, the particle emerges on the other side after 0 milliseconds with a velocity u ( i ˆ + ˆj) ms. The correct statement(s) is (are) (A) The direction of the magnetic field is z direction. (B) The direction of the magnetic field is +z direction. (C) The magnitude of the magnetic field 50 π M units. Q (D) The magnitude of the magnetic field is 00 π M units. Q. (A), (C) Magnetic field is clearly in z axis. vy tan θ vx mv r qb π θ 6 rsinθ L mv mv sin θ L B sinθ qb ql r vt Time of travel θ t r v θ vt sin θ L θ mv m π 50πm B. θ q v t q q 7

8 (8) Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution. A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, y(x, t) (0.0 m) sin [(6.8 m )x]. cos[(68 s )t]. Assuming π.4, the correct statement(s) is (are) (A) The number of nodes is 5. (B) The length of the string is 0.5 m. (C) The maximum displacement of the midpoint of the string, from its equilibrium position is 0.0 m. (D) The fundamental frequency is 00 Hz.. (B), (C) y 0.0 sin [(6.8 m ) x] cos [(68 s ) t] 0.0 sin (0π x) cos (00 πt) π k 0 π λ m 0cm K 0 ω ω 00 π ν 00s π ω v 0 m/s k in 5 th harmonic. 5v 5 ν λ 5cm (B) 6 nodes A is wrong. Mid-point is antinode. sin (0 πx) y 0.0 cos (00 π t) y max 0.0 m (C) ν v hz. (D) is wrong A solid sphere of radius R and density ρ is attached to one end of a mass-less spring of force constant k. The other end of the spring is connected to another solid sphere of radius R and density ρ. The complete arrangement is placed in a liquid of density ρ and is allowed to reach equilibrium. The correct statement(s) is (are) 4πR ρg (A) the net elongation of the spring is k 8πR ρ g (B) the net elongation of the spring is. k (C) the light sphere is partially submerged. (D) the light sphere is completely submerged. 4. (A), (D) B ρ vg mg ρ vg Kx + B mg Kx ρ vg Kx B m mg 8

9 IIT JEE 0 Advanced : Question Paper & Solution (Paper I) (9) x 4 Rg ρ π ρvg K K B mg + Kx ρ vg + ρ vg B ρ vg ρ Δvg ρ vg v Δv Sphere is completely submerged 4πR ρg K 5. Two non-conducting solid spheres of radii R and R, having uniform volume charge densities ρ and ρ respectively, touch each other. The net electric field at a distance R from the centre of the smaller sphere, along the line joining the centres of the spheres, is ρ zero. The ratio ρ (A) 4 5. (B), (D) Q ρ V Q 8ρ V can be (B) A R P R B ρ V 8 ρ V (C) 5 5 B Kx mg (D) 4 ρ Case I : If P is between A and B then + V e ρ ρ Case II : If P is left of A then ve ρ Case II Case I +ve ve R R P R Q Q Q Q (R) 4 πε0(r) 4 πε0(5r) 4 πε0(r) 4 πε0(r) Q 4 Q Q Q ρ V 4 Q 8 ρ V 5 Q ρ ρ V ρ 5 8ρ V ρ ρ 4 ρ 5 ρ 9

10 (0) Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution SECTION III : (Integer value correct Type) This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive). 6. A bob of mass m, suspended by a string of length l, is given a minimum velocity required to complete a full circle in the vertical plane. At the highest point, it collides elastically with another bob of mass m suspended by a string of length l, which is initially at rest. Both the strings are mass-less and inextensible. If the second bob, after collision acquires the minimum speed required to complete a full circle in the vertical plane, the ratio is 6. [5] m g m 5g 5 5 5g m m 7. A particle of mass 0. kg is moving in one dimension under a force that delivers a constant power 0.5 W to the particle. If the initial speed (in ms ) of the particle is zero, the speed (in ms ) after 5 s is 7. [5] d P mv dt d dt ν P m ν t dν P ν dt m ; P ν dv dt m 0 0 ν P t m ν P 0.5 t 5 5 m 0. ν 5 m/s ν 5 8. The work functions of Silver and Sodium are 4.6 and. ev, respectively. The ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium is 8. [] V s h φ ν e e Slope h for both silver and Na e Ratio of slope 0

11 IIT JEE 0 Advanced : Question Paper & Solution (Paper I) () 9. A freshly prepared sample of a radioisotope of half-life 86 s has activity 0 disintegrations per second. Given that n 0.69, the fraction of the initial number of nuclei (expressed in nearest integer percentage) that will decay in the first 80 s after preparation of the sample is 9. [4] t N Ne λ 0 N e λt N0 N λt e N0 ΔN λt ( e ) N0 λ t << λt e λ t ΔN λt N0 n t t / Δ N 00 4 ΔN % 4 % N 0 0. A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 0 rad s about its own axis, which is vertical. Two uniform circular rings, each of mass 6.5 kg and radius 0. m, are gently placed symmetrically on the disc in such a manner that they are touching each other along the axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest relative to the disc and the system rotates about the original axis. The new angular velocity (in rad s ) of the system is 0. [8] R 0.4 M 50 km ω 0 m R 6.5 kg r 0. mr ω ( mrr ) mr + ω ω ω ( ) ω ω 8 R

12 () Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution PART II : CHEMISTRY SECTION I : (Only One option correct Type) This section contains 0 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.. Consider the following complex ions, P, Q and R. P [FeF 6 ], Q [V(H O) 6 ] + and R [Fe(H O) 6 ] + The correct order of the complex ions, according to their spin only magnetic moment values (in B.M.) is (A) R < Q < P (B) Q < R < P (C) R < P < Q (D) Q < P < R. (B) P FeF Fe Ar d 5 [ ] 6 i.e. [ ] Magnetic moment n( n+ ) 5 BM BM ( ) V + [ ] Q V H O + 6 Ar d Magnetic moment 5 BM ( ) i.e. Fe + 6 [ ] Magnetic moment 4( 4+ ) R Fe H O + P > R > Q 4 BM 6 Ar d. The arrangement of X ions around A + ion in solid AX is given in the figure (not drawn to scale). If the radius of X is 50 pm, the radius of A + is (A) 04 pm (B) 5 pm (C) 8 pm (D) 57 pm. (A) + + ra ra 0.44 rx rx r A pm 04 pm. Sulfide ores are common for the metals (A) Ag, Cu and Pb (B) Ag, Cu and Sn (C) Ag, Mg and Pb (D) Al, Cu and Pb. (A) Factual 4. The standard enthalpies of formation of CO (g), H O ( ) and glucose (s) at 5 C are 400 kj/mol, 00 kj/mol and 00 kj/mol, respectively. The standard enthalpy of combustion per gram of glucose at 5 C is (A) kj (B) 900 kj (C) 6. kj (D) + 6. kj 4. (C) CH 6 O6 + O 6CO + 6HO CO 400 kj/ mol

13 IIT JEE 0 Advanced : Question Paper & Solution (Paper I) () H O 00 kj / mol Glucose 00 kj / mol ΔH ΔH P ΔH R 6ΔHf(CO) + 6Hf(HO) Δ Hf( C6H Ο6) (6 400) + 6( 00) ( 00) kj / mol 900 ΔH 6. kj / gm Upon treatment with ammoniacal H S, the metal ion that precipitates as a sulfide is (A) Fe(III) (B) Al(III) (C) Mg(II) (D) Zn(II) 5. (D) White ppt of zinc sulfide by adding ammoniacal H S to Zn(II). ZnS + HCl ZnCl + HS ZnCl + NaOH Zn(OH) + NaCl Zn(OH) + NaOH NaZnO + H O Na ZnO H S ZnS NaOH + + white ppt. 6. Methylene blue, from its aqueous solution is adsorbed on activated charcoal at 5 C. for this process, the correct statement is (A) The adsorption requires activation at 5 C. (B) The adsorption is accompanied by a decrease in enthalpy. (C) The adsorption increases with increase of temperature. (D) The adsorption is irreversible. 6. (B) Factual 7. KI in acetone, undergoes S N reaction with each of P, Q, R and S. The rates of the reaction vary as (A) P > Q > R > S (B) S > P > R > Q (C) P > R > Q > S (D) R > P > S > Q 7. (B) S N reactions are through back attack of attacking Nu and not on the basis of stability of carbocation. 8. In the reaction, P+ Q R + S the time taken for 75% reaction of P is twice the time taken for 50% reaction of P. The concentration of Q varies with reaction time as shown in the figure. The overall order of the reaction is (A) (B) (C) 0 (D)

14 (4) Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution 8. (D) For P, t 75% t 50% I st order wrt P For Q, [Q] [Q 0 ] kt. i.e. zero order. i.e. overall order of a reaction 9. Concentrated nitric acid, upon long standing, turns yellow brown due to the formation of (A) NO (B) NO (C) N O (D) N O 4 9. (B) Sunlight 4HNO 4NO + H O + O 0. The compound that does NOT liberate CO, on treatment with aqueous sodium bicarbonate solution, is (A) Benzoic acid (B) Benzenesulphonic acid (C) Salicylic acid (D) Carbolic acid (Phenol) 0. (D) Compound with COOH gr or SO H group can liberate CO with NaHCO solution. SECTION II : (One or more options correct Type) This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct.. The initial rate of hydrolysis of methyl acetate (M) by a weak acid (HA, M) is /00 th of that of a strong acid (HX, M), at 5 C. the K a of HA is (A) 0 4 (B) 0 5 (C) 0 6 (D) 0. (A) Ka Cα (0.0) 0 4. The hyperconjugative stabilities of tert butyl cation and butene, respectively, are due to (A) σ p (empty) and σ π * electron delocalisations. (B) σ σ * and σ π electron delocalisations. (C) σ p (filled) and σ π electron delocalisations. (D) p(filled) σ * and σ π * electron delocalisations.. (A) H + CH C CH σ p(empty) delocalisation CH CH CH CH (σ π * electron) delocalization.. The pair(s) of coordination complexes / ions exhibiting the same kind of isomerism is (are) (A) [Cr(NH ) 5 Cl]Cl and [Cr(NH ) 4 Cl ]Cl (B) [Co(NH ) 4 Cl ] + and [Pt(NH ) (H O)Cl] + (C) [CoBr Cl ] and [PtBr Cl ] (D) [Pt(NH ) (NO )]Cl and [Pt(NH ) Cl]Br. (B), (D) Factual. 4

15 IIT JEE 0 Advanced : Question Paper & Solution (Paper I) (5) 4. Among P, Q, R and S, the aromatic compound(s) is /are (A) P (B) Q (C) R (D) S 4. (A), (B), (C), (D) N H All are Aromatic except (R) 5 OH (P) (Q) (R) (S) 5. Benzene and naphthalene from an ideal solution at room temperature. For this process, the true statement(s) is(are) (A) ΔG is positive (B) ΔS system is positive (C) ΔS surroundings 0 (D) ΔH 0 5. (B), (C), (D) Conditions for Ideal Solution. SECTION III : (Integer value correct Type) This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive). 6. The atomic masses of He and Ne are 4 and 0 a.m.u, respectively. The value of the de Broglie wavelength of He gas at 7 C is M times that of the de Broglie wavelength of Ne at 77 C. M is 6. [5] λ He M Ne.TNe λ M.T 4 00 Ne He He 7. EDTA 4 is ethylenediaminetetraacetate ion. The total number of N Co O bond angles in [Co(EDTA)] complex ion is 7. [8] Total no. of N Co O bond angle in [Co (EDTA)] 8

16 (6) Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution 8. The total number of carboxylic acid groups in the product P is 8. [] O O C... O. + HO Heat C O O Total no. of carboxylic group O C C O O HO 9. A tetrapeptide has COOH group on alanine. This produces glycine (Gly), valine (Val), phenyl alanine (Phe) and alanine(ala), on complete hydrolysis. For this tetrapeptide, the number of possible sequences (primary structures) with NH group attached to a chiral center is 9. [4] HO HO O O O C C O Glycine CH NH COOH Alanine CH CH NH COOH Valine (CH ) CH CH NH COOH NH phenyl alanine Ph CH CH COOH No. of possible sequences (primary structures) with NH group attached to a chiral center is The total number of lone pairs of electrons in melamine is 40. [6].... N NH N.. N. NH... Lone Pairs 6... NH Melamine 6

17 IIT JEE 0 Advanced : Question Paper & Solution (Paper I) (7) PART III : MATHEMATICS SECTION I : (Only One option correct Type) This section contains 0 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. 4. For a > b > c > 0, the distance between (, ) and the point of intersection of the lines ax + by + c 0 and bx + ay + c 0 is less than.then (A) a + b c > 0 (B) a b + c < 0 (C) a b + c > 0 (D) a + b c < 0 4. (A) Given equations are ax + by + c 0 bx + ay + c 0 Now, x y bc ac bc ac a x ( b a ) c c a b a+ b (b a)c c y a b a+ b Now c c a+ b a+ b (a + b + c) < (a + b) (a + b+ c) < a+ b (a + b + c) < a + b a + b c > 0 b < 4. The area enclosed by the curves y sin x + cos x and y cos x sin x over the interval π 0, is (A) 4( ) (B) ( ) (C) ( + ) (D) ( + ) 4. (B) π y sin x+ 4 π y sin x 4 Area of shaded part π/4 π π sin x+ sin x dx π/ π π + sin x+ sin x dx 4 4 π/4 π 4 7

18 (8) Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution π/4 π/ π π π π sin x + + sin x dx + sin x + sin x π/ The number of points in (, ), for which x x sin x cos x 0, is (A) 6 (B) 4 (C) (D) 0 4. (C) f(x) x x sin x cos x f (x) x sin x x cos x + sin x x ( cos x) > 0 x> 0 So, f (x) 0 x 0 For x 0, f(x) Hence two solutions < 0 x < 0 (0, ) n 44. The value of cot cot + k is n k (A) (B) 5 (C) (D) (B) n cot cot + k n k ( n+ ) cot cot + n cot n cot ( n + n + ) n (n + ) n cot tan cot tan n + n(n+ ) n + (n + ).n cot ( tan (n + ) tan n) cot tan 4 tan n 4 cot tan + 4. cot 5 cot 5 π 45. A curve passes through the point,. Let the slope of the curve at each point (x, y) 6 be y sec y +, x > 0. Then the equation of the curve is x x y (A) sin y log x + (B) cosec log x + x x y y (C) sec log x + (D) cos log x + x x 8

19 45. (A) dy dx y sec y + x x let IIT JEE 0 Advanced : Question Paper & Solution (Paper I) (9) v + x dv dx v + sec v y x v y vx dy dx v + x dv dx dx cos v dv x sin v lnx + lnc π Given it passes (, ) 6 Sin 6 π ln + lnc lnc So, curve is sin v lnx+ sin y x lnx Let f :, (the set of all real numbers) be a positive, non-constant and differentiable function such that f (x) < f (x) and f. Then the value of / f(x)dxlies in the interval (A)( e, e) (B) (e, e ) e (C),e 46. (D) f (x) < f(x) f(x) < f(x) f(x) dx < dx f(x) n{ f(x) } < x+ c f(x) < e x + c f(/) < e + c < e + c.(f(/)) ) e 0 < e + c 0 < + c c >.(i) (D) 0, e 9

20 (0) Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution / / / / / x + c f(x)dx< e dx x + c f(x)dx< e / + + f(x)dx< e e + c f(x)dx< e [ e ] Maximum value of c c / And minimum value of f(x) dx is greater than zero because it is positive, nonconstant function. f(x)dx is [e ] / 47. Let PR i ˆ+ ˆj kˆ and SQ ˆ i j ˆ 4k ˆ determine diagonals of a parallelogram PQRS and PT ˆi + j ˆ+ kˆ be another vector. Then the volume of the parallelepiped determined by the vectors PT, PQ and PS is (A) 5 (B) 0 (C) 0 (D) (C) PR i ˆ+ ˆj kˆ and SQ ˆ i ˆ j 4k ˆ are diagonals of a parallelogram. PT ˆ i+ ˆ j+ k ˆ is another vector. Volume of the parallelepiped by the vector PT, PQ & PS a+ b i ˆ+ ˆj kˆ S a R a b ˆi ˆj 4kˆ b a i ˆ ˆj kˆ b b ˆi+ ˆj+ kˆ P Q a Volume of parallelepiped formed by PT, PQ & RS is given by scalar triple product of ˆi+ ˆj+ kˆ, i ˆ ˆj kˆ & ˆ i+ ˆ j+ k ˆ Volume (5) + ( 5) + (5) Perpendiculars are drawn from points on the line x + y + z to the plane x + y + z. The feet of perpendiculars lie on the line (A) x y z (B) x y z (C) x y z (D) x y z

21 IIT JEE 0 Advanced : Question Paper & Solution (Paper I) () 48. (D) Any point on x + y + z is (r, r, r). Let P(x, y) be feet of perpendicular then, x x y y z z ax+ by+ cz+ d a b c a + b + c x (r ) y ( r ) ) z r 4r 6 So, x r 4r 6 r ; 7r 5r + 6 y ; z r 7r 5r + 6 Obviously,, satisfy x y z Four persons independently solve a certain problem correctly with probabilities,,,. Then the probability that the problem is solved correctly by at least one of them is (A) 5 (B) (C) (D) (A) Required Probability P(No one solves the problem correctly) Let complex numbers α and α lie on circles (x x 0) + (y y 0 ) r and (x x 0 ) + (y y 0 ) 4r, respectively. If z 0 x 0 + iy 0 satisfies the equation z 0 r +, then α (A) (B) (C) (D) (C) α Z 0 r () and Z0 r () α () (α Z 0 ) ( α Z0 ) r α α Z0 Z0α+ α Z 0 + Z 0 α α + () α Z0 r α ( Z 0 ).( Z0 ) r + r (, Z 0 r α α 4r α α α Z 0 α Z 0 + α r + 4 r α α Z 0 + α Z 0 + α () (4) α + () 7r α (4) r +) r α + 7 r α α 7

22 () Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution SECTION II : (One or more options correct Type) This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct. 5. A line passing through the origin is perpendicular to the lines ˆ ˆ ˆ :(+ t)i + ( + t)j + (4+ t)k, < t < ˆ ˆ ˆ :(+ s)i + (+ s)j + ( + s)k, < s < Then, the coordinate (s) of the point (s) on at a distance of 7 from the point of intersection of and is (are) (A),, (B) (,,0) (C) (,,) (D),, (B), (D) ˆ ˆ ˆ ˆ ˆ ˆ : r (i j+ 4k) + t(i + j+ k) ˆ ˆ ˆ ˆ ˆ ˆ : r (i + j+ k) + s(i + j+ k) Vector perpendicular to ˆ i+ j ˆ + k ˆ and i ˆ+ j ˆ+ kˆ is ˆi ˆj kˆ i ˆ+ j ˆ kˆ So, line is r k( i ˆ+ j ˆ k) ˆ For point of intersection of and k ( i ˆ+ j ˆ k) ˆ i ˆ ˆj + 4kˆ + t (i ˆ+ j ˆ+ k) ˆ ( k )i ˆ+ (k+ )j ˆ+ ( k 4)kˆ t(i ˆ+ j ˆ+ k) ˆ k t (i) and k + t (ii) k 4 t (iii) k + (ii) (iii) k 4 k + k 4 k From (i), ( ) t t. So, : r i ˆ j ˆ+ kˆ So, point of intersection of and is A (,, ). The line can be written as x y z s Any point on (s +, s +, s + ) B Given, AB 7 (s + ) + (s + 6) + s 7 9s + 8s (s + ) (9s + 0) 0 s or s 0 9 So, point B (,, 0) and 7 7 8,, 9 9 9

23 IIT JEE 0 Advanced : Question Paper & Solution (Paper I) () 5. Let f(x) x sin πx, x > 0. Then for all natural numbers n, f' ( x) vanishes at (A) a unique point in the interval n, n + (B) a unique point in the interval n +,n+ (C) a unique point in the interval (n, n + ) (D) two points in the interval (n, n + ) 5. (B), (C) f(x) x sinπx f (x) sinπx + (x) π(cosπx) sinπx + xπ cosπx 0 sinπx πx cos πx tanπx πx y 0 x 5. Let S n 4n k k(k+ ) ( ) k. Then S n can take value(s) (A) 056 (B) 088 (C) 0 (D) 5. (A), (D) S n 4n k k(k+ ) ( ).k ( ). + ( ). + ( ) 6. + ( ) upto 4n terms (4n) n (n + terms) n n times 6n + 4n 54. For matrices M and N, which of the following statement(s) is (are) NOT correct? (A) N T M N is symmetric or skew symmetric, according as M is symmetric or skew symmetric (B) M N N M is skew symmetric for all symmetric matrices M and N (C) M N is symmetric for all symmetric matrices M and N (D) (adj M) (adj N) adj (M N) for all invertible matrices M and N

24 (4) Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution 54. (C), (D) (C) is wrong as (MN) T N T M T NM MN (D) is wrong as adj (MN) adj (N) adj (M) for all invertible M and N. 55. A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8 : 5 is converted into an open rectangular box by folding after removing squares of equal area from all four corners. If the total area of removed squares is 00, the resulting box has maximum volume. Then the lengths of the sides of the rectangular sheet are (A) 4 (B) (C) 45 (D) (A), (C) Let the sides be 8C and 5C where C is a parameter given (8C + 5 C) K (constant) C K 46 Now, 5C x Volume is given by V (8C x) (5C x) x x x x x or V 4x 46x C + 0C x dv dx x 9 xc + 0 C 8C x () area of all squares 4x 00 x 5 dv 0 for maximum volume at x 5 dx using x 5 in () we get 0 C 460 C C C C or 5/6 but C 5/6 and x 5 makes sides negative. Thus C is the only value of the parameter. Hence sides are 4, 45. SECTION III : (Integer value correct Type) This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive). { } 56. Consider the set of eight vectors V ai + bj + ck:a, b,c {,}. Three non coplanar vectors can be chosen from V in p ways. Then p is, 56. [5] Total number of vectors are, ˆi+ ˆj+ kˆ ˆi+ ˆj kˆ ˆ i+ ˆ j+ k ˆ ˆ i+ ˆ j k ˆ ˆi ˆj+ kˆ ˆi ˆj kˆ ˆ i ˆ j+ k ˆ ˆ i ˆ j k ˆ Total no. of ways choosing non coplanar vectors ( 4 C ) ( 4 ) ( 4 C + C C + C) P P 5 4

25 IIT JEE 0 Advanced : Question Paper & Solution (Paper I) (5) 57. Of the three independent events E, E and E, the probability that only E occurs is α, only E occurs is β and only E occurs is γ. Let the probability p that none of events E, E or E occurs satisfy the equations (α β) p αβ and (β γ) p βγ. All the given probabilities are assumed to lie in the interval (0, ). Pr obability of occurrence of E Then Pr obability of occurrence of E 57. [6] P (E ) x P (E ) y P (E ) z Then x ( y) ( z) α y ( z) ( x) β z ( x) ( y) γ ( x) ( y) ( z) P Now, (α β) P αβ [x ( y) ( z) y ( z) ( x)] ( x) ( y) ( z) xy ( y) ( z) ( x) x ( y) y ( x) xy x xy y + xy xy x y x y () Also, [y ( z) ( x) z ( x) ( y)] ( x) ( y) ( z) yz ( z) ( y) ( x) y ( z) z ( y) yz y y z 0 z x z The coefficients of three consecutive terms of ( + x) n + 5 are in the ratio 5 : 0 : 4. Then n 58. [6] Let the coefficients of three consecutive terms be n + 5 Cr, n + 5 Cr+ and n + 5 Cr+ Given n + 5 Cr : n + 5 Cr+ : n + 5 Cr+ : : 5 : 0 : 4 Solving n + 5 Cr : n + 5 Cr+ : : 5 : 0 We get r n + () likewise solving n+ 5 Cr+ 0 we get n+ 5 Cr+ 4 r 5n + 6 () solving () & () we get n A pack contains n cards numbered from to n. Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 4. If the smaller of the numbers on the removed cards is k, then k 0 5

26 (6) Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution 59. [5] n(n + ) The sum of first n cards is, S. if two consecutive cards numbered k and k + are removed then s n(n + ) (k + k + ) 4 (given) n(n + ) k, which is satisfied for n 50 and k 5 in the given limits k 5 and k 0 5. x y 60. A vertical line passing through the point (h, 0) intersects the ellipse + at the 4 points P and Q. Let the tangents to the ellipse at P and Q meet at the point R. max min If Δ(h) area of the triangle PQR, Δ Δ( h) and, ( h) / h Δ / h Δ then Δ Δ 60. [9] Let, line x h For P and Q points h y + P h, 4 h 4 h y 4 (h, 0) R y (4 h ) 4 Q h, 4 h y ± 4 h Tangent at P, y 4 h xh + 4. So, R, 0 4 h / 4 Δ P Q R ( 4 h ) h h 4 h (4 h ) 4 h h h Δ 6 4 h < 0 h h Δ is decreasing / 4 / 4 5 So, Δ max Δ (h) at h h 4 Δ / min Δ (h) (at h ) So, / Δ Δ 5 /

27 IIT JEE 0 Advanced : Question Paper & Solution (Paper I) (7) 7

28 IIT - JEE 0 (Advanced) CODE : O

29 () Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution

30 IIT JEE 0 Advanced : Question Paper & Solution (Paper II) () PART I : PHYSICS SECTION : (One or more options correct Type) This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct.. The figure below shows the variation of specific heat capacity (C) of solid as a function of temperature (T). The temperature is increased continuously from 0 to 500 K at a constant rate. Ignoring any volume change, the following statement(s) is(are) correct to a reasonable approximation. (A) the rate at which heat is absorbed in the range 0 00 K varies linearly with temperature T. (B) heat absorbed in increasing the temperature from 0 00 K is less than the heat required for increasing the temperature from K. (C) there is no change in the rate of heat absorption in the range K. (D) the rate of heat absorption increases in the range K. C (A), (B), (C), (D) T (K) dq mc dt dq m (at + b) in the range (0 00 k) dt option (A) is correct. Considering (0 00 k), the graph to be linear using the reasonable approximation given in the question. During (400k 500k) C is the maximum. option (B) is correct. Option (C) is correct as C is constant in the range (400k 500k) and option (D) is also correct.. The radius of the orbit of an electron in a Hydrogen-like atom is 4.5 a 0, where a 0 is the Bohr radius. Its orbital angular momentum is h. It is given that h is Planck constant and π R is Rydberg constant. The possible wavelength(s), when the atom de-excites, is (are) (A) (B) (C) (D) R 6R 5R R. (A), (C) r n n a0 z 4.5a 0 n 9 z n a0 z z and z L nh π L h π n

31 (4) Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution Wavelength is given by λ Rz nf n i λ 4R n f n i For transition, n λ 9 R 4R 9 For transition n 6 λ 4R 5 4R R (A) (C). Using the expression d sin θ λ, one calculates the values of d by measuring the corresponding angles θ in the range 0 to 90º. The wavelength λ is exactly known and the error in θ is constant for all values of θ. As θ increases from 0º. (A) the absolute error in d remains constant. (B) the absolute error in d increases. (C) the fractional error in d remains constant. (D) the fractional error in d decreases.. (D) d sin θ λ δ (d sin θ) δλ But δλ 0 ( sin θ) δd + d cos θ δθ 0 δd + cot θ δθ 0 d δd cot θ (dθ) d δθ constant absolute error in d is δd d cot θ (δθ) λ cos θ δd d θ sin θ at θ increase from 0 to 90, δd decreases. Hence (B) is not true. By the same fractional error decreases. 4. Two non-conducting spheres of radii R and R and carrying uniform volume charge densities +ρ and ρ, respectively, are placed such that they partially overlap, as shown in the figure. At all points in the overlapping region, (A) the electrostatic field is zero. (B) the electrostatic potential is constant. (C) the electrostatic field is constant in magnitude. (D) the electrostatic field has same direction. 4

32 IIT JEE 0 Advanced : Question Paper & Solution (Paper II) (5) 4. (C), (D) ρ E OP 0 ρ E PO 0 Net field at P E E+ E ρ [OP + PO ] 0 ρ E OO 0 O E is constant (C) And option (D) is also correct. E P E O 5. A steady current I flows along an infinitely long hollow cylindrical conductor of radius R. This cylinder is placed coaxially inside an infinite solenoid of radius R. The solenoid has n turns per unit length and carries a steady current I. Consider a point P at a distance r from the common axis. The correct statement(s) is (are) (A) In the region 0 < r < R, the magnetic field is non-zero. (B) In the region R < r < R, the magnetic field is along the common axis. (C) In the region R < r < R, the magnetic field is tangential to the circle of radius r, centered on the axis. (D) In the region r > R, the magnetic field is non-zero. 5. (A), (D) n N L Bd μ 0 I μ B 0I πr B μ 0 ni along axis of cylinder along axis of solenoid (i) 0 < r < R, magnetic field At this point, there will be field only due to solenoid. (ii) R < r < R μ0 B I by the rod. (Along the tangent to the circular part around the cylinder) πr B solenoid μ 0 ni along axis 6. Two vehicles, each moving with speed u on the same horizontal straight road, are approaching each other. Wind blows along the road with velocity w. One of these vehicles blows a whistle of frequency f. An observer in the other vehicle hears the frequency of the whistle to be f. The speed of sound in still air is V. The correct statements(s) is (are) (A) If the wind blows from the observer to the source, f > f. (B) If the wind blows from the source to the observer, f > f. (C) If the wind blows from observer to the source, f < f. (D) If the wind blows from the source to the observer, f < f. 5

33 (6) Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution 6. (A), (B) W u SW u w S O u OW u + w u u v+ u+ w v+ u+ w f f f v (u w) v u+ w f Clearly f > f irrespective of sign of w. 7. Two bodies, each of mass M, are kept fixed with a separation L. A particle of mass m is projected from the midpoint of the line joining their centres, perpendicular to the line. The gravitational constant is G. The correct statement(s) is (are) (A) The minimum initial velocity of the mass m to escape the gravitational field of the two GM bodies is 4 L. (B) The minimum initial velocity of the mass m to escape the gravitational field of the two GM bodies is L. (C) The minimum initial velocity of the mass m to escape the gravitational field of the two GM bodies is L. (D) The energy of the mass m remains constant. 7. (B), (D) (D) is clearly correct, since gravitational field is conservative. GMm + mve 0 L v e GM L 8. A particle of mass m is attached to one end of a mass-less spring of force constant k, lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time t 0 with an initial velocity u 0. When the speed of the particle is 0.5 u 0, it collides elastically with a rigid wall. After this collision, (A) the speed of the particle when it returns to its equilibrium position is u 0. (B) the time at which the particle passes through the equilibrium position for the first time m is t π. k (C) the time at which the maximum compression of the spring occurs is t 4 π m. k (D) the time at which the particle passes through the equilibrium position for the second time is t 5 π m. k 8. (A), (D) The impulse of the wall merely reverses the direction of motion is thout changing particle speed. Hence (A) 6

34 IIT JEE 0 Advanced : Question Paper & Solution (Paper II) (7) Displacement from mean position be x u0 x w sin (ωt) x u 0 cos (ωt) u0 when particle speed is cos (ωt) π ωt [for first time] π t ω Time after which it passes the mean position t π π m u K Hence (B) is incorrect. Time after which particle crosses mean position for second time Hence (D). SECTION : (Paragraph Type) π π + ω ω 5 π ω This section contains 4 paragraphs each describing theory, experiment, data etc. Eight questions relate to four paragraphs with two questions on each paragraph. Each question of a paragraph has only one correct answer among the four choices (A), (B), (C) and (D).. Paragraph for Questions 9 and 0 A small block of mass kg is released from rest at the top of a rough track. The track is a circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figure below, is 50 J. (Take the acceleration due to gravity, g 0 ms ). 9. The speed of the block when it reaches the point Q is (A) 5 ms (B) 0 ms (C) 0 ms (D) 0 ms 9. (B) Kinetic energy of block at Q mgr sin 0 50 J mv J r 50 0 ms 7

35 (8) Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution 0. The magnitude of the normal reaction that acts on the block at the point Q is (A) 7.5 N (B) 8.6 N (C).5 N (D).5 N 0. (A) N mg cos 60 mv r 0 mv N mg cos 60 + r N N 40 mg Paragraph for Questions and A thermal power plant produces electric power of 600 kw at 4000 V, which is to be transported to a place 0 km away from the power plant for consumers usage. It can be transported either directly with a cable of large current carrying capacity or by using a combination of step-up and step-down transformers at the two ends. The drawback of the direct transmission is the large energy dissipation. In the method using transformers, the dissipation is much smaller. In this method, a step-up transformer is used at the plant side so that the current is reduced to a smaller value. At the consumers end, a step-down transformer is used to supply power to the consumers at the specified lower voltage. It is reasonable to assume that the power cable is purely resistive and the transformers are ideal with a power factor unity. All the currents and voltages mentioned are rms values.. If the direct transmission method with a cable of resistance 0.4Ω km is used, the power dissipation (in%) during transmission is (A) 0 (B) 0 (C) 40 (D) 50. (B) P 600 kw V 4000 V I A 4000 Power Loss I R (50) kw % Power Loss %. In the method using the transformers, assume that the ratio of the number of turns in the primary to that in the secondary in the step-up transformer is :0. If the power to the consumers has to be supplied at 00 V, the ratio of the number of turns in the primary to that in the secondary in the step-down transformer is (A) 00 : (B) 50 : (C) 00 : (D) 50 :. (A) Step up transformer : V 40,000 V Step down transform 40,000 V 00 turns ratio : 8

36 IIT JEE 0 Advanced : Question Paper & Solution (Paper II) (9) Paragraph for Questions and 4 A point charge Q is moving in a circular orbit of radius R in the x-y plane with an angular velocity ω. This can be considered as equivalent to a loop carrying a steady current Q ω π. A uniform magnetic field along the positive z-axis is now switched on, which increases at a constant rate from 0 to B in one second. Assume that the radius of the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around a closed loop. It is known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a proportionality constant γ.. The magnitude of the induced electric field in the orbit at any instant of time during the time interval of the magnetic field change is (A) BR 4 (B) BR (C) BR. (B) Magnitude of induced emf is given by A db π πr dt πr R B BR (D) BR 4. The change in the magnetic dipole moment associated with the orbit, at the end of the time interval of the magnetic field change, is (A) γbqr BQR BQR (B) γ (C) γ (D) γbqr 4. (B) ω f ω i α t qer () MR qe () MR E BR Δω q BR MR qb M μ γl γiω Δμ qb γ MR ( Δω ) γmr M γqbr Paragraph for Questions 5 and 6 The mass of a nucleus A Z X is less than the sum of the masses of (A-Z) number of neutrons and Z number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass M can break into two light nuclei of masses m and m only if (m + m ) < M. Also two light nuclei of masses m and m 4 can undergo complete fusion and form a heavy nucleus of mass M only if (m + m 4 ) > M. The masses of some neutral atoms are given in the table below : H u Η.040 u H u 4 He u 6 Li 6.05 u 7 Li u 70 0 Zn u 8 4 Se u 5 64 Gd u 06 8 Pb u 09 8 Bi u 0 84 Po u ( u 9 MeV/c ) 9

37 (0) Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution 5. The correct statement is (A)The nucleus 6 Li can emit an alpha particle. (B) The nucleus 0 84 Po can emit a proton. (C) Deuteron and alpha particle can undergo complete fusion. (D)The nuclei 70 0 Zn and (C) 4 6Li H + He M 6.05 m.040 m m + m m + m > M (not possible) 0 09 (B) 84 Po 8 Bi + P M m m m + m M < m + m (not possible) (C) 4 6 H+ H Li (m + m ) > M Possible (D) 0Zn + 4Se 64 Gd m m m + m M m + m < M (not possible) Se can undergo complete fusion. 6. The kinetic energy (in kev) of the alpha particle, when the nucleus 0 84 Po at rest undergoes alpha decay, is (A) 59 (B) 54 (C) 5707 (D) (A) Po 8 Pb + He M m m m + m M (m+ m ) E Δ mc ( )9 54 K+ K E 0

38 IIT JEE 0 Advanced : Question Paper & Solution (Paper II) () P m P + E m P μ E P Em m m + m p Em E(06) 0 k E m m+ m k (54) SECTION : (Matching List Type) This section contains 4 multiple choice questions. Each question has matching lists. The codes for the lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct 7. One mole of a monatomic ideal gas is taken along two cyclic processes E F G E and E F H E as shown in the PV diagram. The processes involved are purely isochoric, isobaric, isothermal or adiabatic. Match the paths in List I with the magnitudes of the work done in List II and select the correct answer using the codes given below the lists. List I List II P. G E. 60 P 0 V 0 n Q. G H. 6 P 0 V 0 R. F H. 4 P 0 V 0 S. F G 4. P 0 V 0 Codes : P Q R S (A) 4 (B) 4 (C) 4 (D) 4 7. (A) PV F F PGVG P0V 0 PV 0 G

39 () Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution V V PV PV G 0 γ γ F F PHVH γ γ 0 0 PV 0 H γ V H V 0 ( ) 5 / V0 8V0 WGE P0 VE VG P0 V0 V0 P0V 0 4. W P 8V V 4P V () (P) ( ) ( ) (Q) ( ) GH ( ) PV H H PV P F F 0 8V0 P0V0 (R) WFH γ 5 4P0V 0 6P0V 0 ( ) VG V0 (A) WFG PFVF n P0 V0 n V V P0V 0 n P V n ( ) F 0 60P0V 0 n () (P) 4; (Q) (); (R) (); (S) () 8. Match List I of the nuclear processes with List II containing parent nucleus and one of the end products of each process and then select the correct answer using the codes given below the lists: List I List II P. Alpha decay. 5 5 O N Q. β + decay. R. Fission. S. Proton emission 4. Codes : P Q R S (A) 4 (B) 4 (C) 4 (D) 4 8. (C) P Q R S U Th Bi Pb Pu La +...

40 IIT JEE 0 Advanced : Question Paper & Solution (Paper II) () 9. A right angled prism of refractive index μ is placed in a rectangular block of refractive index μ, which is surrounded by a medium of refractive index μ, as shown in the figure. A ray of light 'e' enters the rectangular block at normal incidence. Depending upon the relationships between μ, μ and μ, it takes one of the four possible paths 'ef', 'eg', 'eh' or 'ei'. Match the paths in List I with conditions of refractive indices in List II and select the correct answer using the codes given below the lists: List I List II P. e f. μ > μ Q. e g. μ > μ and μ > μ R. e h. μ μ S. e i 4. μ < μ < μ and μ > μ Codes : P Q R S (A) 4 (B) 4 (C) 4 (D) 4 9. (D) (P) e f for towards normal μ > μ For always from normal μ < μ i.e., () (Q) e g no deviation any where μ μ μ, i.e., () (R) e h away from normal. μ < μ away from normal. μ < μ And also at no I.R. μ sin 45 < μ μ < μ μ <μ < μ i.e., (4) (S) e i TIR at

41 (4) Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution μ sin 45 > μ i.e., μ > μ (P) ; (Q) ; (R) (4); (S) (). 0. Match List I with List II and select the correct answer using the codes given below the lists: List I List II P. Boltzmann constant. [ML T ] Q. Coefficient of viscosity. [ML T ] R. Planck constant. [MLT T ] S. Thermal conductivity 4. [ML T K ] Codes : P Q R S (A) 4 (B) 4 (C) 4 (D) 4 0. (C) (P) u [u] K B T [ KBT] [ ML T ] [ KB] [ K] [K B ] [ ML T K ] 4 (Q) F 6π rη V [η] rv LLT ( ) F MLT [ ] ML T (R) E hν [h] ν T (S) Q KA( T T) t [ ML T ] [K] L [ ] [ L ][ K] E ML T [ ML T ] ( ) [ MLT K ] ( ) (P) 4; (Q) (); (R) (); (S) () 4

42 IIT JEE 0 Advanced : Question Paper & Solution (Paper II) (5) PART II : CHEMISTRY SECTION : (One or more options correct Type) This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct.. The carbon-based reduction method is NOT used for the extraction of (A) tin from SnO (B) iron from Fe O (C) aluminium from Al O (D) magnesium from MgCO. CaCO. (C), (D) The oxides of the less electropositive methods are reduced by strongly heating them with coke.. The thermal dissociation equilibrium of CaCO (s) is studied under different conditions. CaCO (s) CaO(s) + CO (g) For this equilibrium, the correct statement(s) is (are) (A) ΔH is dependent on T (B) K is independent of the initial amount of CaCO (C) K is dependent on the pressure of CO at a given T (C) ΔH is independent of the catalyst, if any. (A), (B), (D) K Equilibrium constant is independent of the partial pressure of CO. It is independent of the catalyst.. The correct statement(s) about O is(are) (A) O O bond lengths are equal (B) Thermal decomposition of O is endothermic (C) O is diamagnetic in nature (D) O has a bent structure. (A), (C), (D) Δ O O + 68 k cal i.e. Exothermic 4. In the nuclear transmutation Be + X Be + Y (X, Y) is (are) (A) (γ, n) (B) (p, D) (C) (n, D) (D) (γ, p) 4. (A), (B) Be + γ Be + n (A) (B) (C) (D) 9 8 4Be 4Be H Be H Be D Be + n Be + D (False) + γ + (False) 5

43 (6) Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution 5. The major product(s) of the following reaction is(are) (A) P (B) Q (C) R (D) S 5. (B) OH Br OH Br aq. Br SO H Br (Q) 6. After completion of the reactions (I and II), the organic compound(s) in the reaction mixtures is(are) NaOH (A) Reaction I : (B) Reaction I : (C) Reaction I : (D) Reaction I : 6. (C) O P and Reaction II : P U, acetone and Reaction II : Q, acetone T, U, acetone and Reaction II : P R, acetone and Reaction II : S, acetone Br (mol) ( m o l) NaOH CH C CH (T) + (U) Br (mol) CHCOOH (p) 6

44 IIT JEE 0 Advanced : Question Paper & Solution (Paper II) (7) 7. The K sp of Ag CrO 4 is. 0 at 98 J. The solubility (in mol / L) of Ag CrO 4 in a 0. M AgNO solution is (A). 0 (B). 0 0 (C). 0 (D) (B) + AgCrO4(s) Ag (aq) + CrO 4 (aq) s + 0. s 0. K (0.) s. 0 sp 0 s. 0 M 8. In the following reaction, the product(s) formed is (are) (A) P (major) (B) Q (minor) (C) R (minor) (D) S (major) 8. (B), (D) O.. CCl O H CCl O CHCl OH CH OH CHO CH O CHO CH HO CH major O CH O HC CHCl (minor) 7

45 (8) Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution SECTION : (Paragraph Type) This section contains 4 paragraphs each describing theory, experiment, data etc. Eight questions relate to four paragraphs with two questions on each paragraph. Each question of a paragraph has only one correct answer among the four choices (A), (B), (C) and (D).. Using the following Paragraph, Solve Q. No. 9 and 0 An aqueous solution of a mixture of two inorganic salts, when treated with dilute HCl, gave a precipitate (P) and a filtrate (Q). The precipitate P was found to dissolve in hot water. The filtrate (Q) remained unchanged, when treated with H S in a dilute mineral acid medium. However, it gave a precipitate (R) with H S in an ammoniacal medium. The precipitate R gave a coloured solution (S), when treated with H O in an aqueous NaOH medium. 9. The precipitate P contains (A) Pb + (B) Hg + (C) Ag + (D) Hg + 9. (A) Pb + + HCl PbCl soluble in hot water (p) 0. The coloured solution S contains (A) Fe (SO 4 ) (B) CuSO 4 (C) ZnSO 4 (D) Na CrO 4 Pb + + HCl PbCl soluble in hot water 0. (D) (p) Cr(OH) + 4NaOH + [O] Na CrO + 5H O (R) 4 (S) Using the following Paragraph, Solve Q. No. and P and Q are isomers of dicarboxylic acid C 4 H 4 O 4. Both decolorize Br / H O. On heating, P forms the cyclic anhydride. Upon treatment with dilute alkaline KMnO 4, P as well as Q could produce one or more than one from S, T and U.. Compounds formed from P and Q are, respectively (A) Optically active S and optically active pair (T, U) (B) Optically inactive S and optically inactive pair (T, U) (C) Optically active pair (T, U) and optically active S (D) Optically inactive pair (T, U) and optically inactive S. 8

46 IIT JEE 0 Advanced : Question Paper & Solution (Paper II) (9). (B) H H C C (P) (cis) H COOH COOH C COOH Δ HO O H C C O C C H O cyclic anhydride HOOC C (Q) H dil. KMnO 4 P H H COOH OH OH COOH (S) (meso) COOH COOH dil. KMnO 4 H OH HO H Q + HO H H OH COOH COOH (T) (U). In the following reaction sequences V and W are, respectively (A) (B) (C) (D) 9

47 (0) Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution. (A) O (Q) H /Ni Δ CH CH C C O (V) + (V) O anhy. AlCl HPO 4 O C CH CH COOH Zn-Hg/conc.HCl CH CH CH COOH (W) Using the following Paragraph, Solve Q. No. and 4 O A fixed mass m of a gas is subjected to transformation of states from K to L to M to N and back to K as shown in the figure. The succeeding operations that enable this transformation of states are (A) Heating, cooling, heating, cooling (C) Cooling, heating cooling, heating (C) Heating, cooling, cooling, heating (D) Cooling, heating, heating, cooling. (C) Factual 4. The pair of isochoric processes among the transformation of states is (A) K to L and L to M (B) L to M and N to K (C) L to M and M to N (D) M to N and N to K 4. (B) Factual Using the following Paragraph, Solve Q. No. 5 and 6 The reactions of Cl gas with cold dilute and hot concentrated NaOH in water give sodium salts of two (different) oxoacids of chlorine, P and Q respectively. The Cl gas reacts with SO gas, in presence of charcoal, to give a product R. R reacts with white phosphorus to give a compound S. On hydrolysis, S gives an oxoacid of phosphorus, T. 0

48 IIT JEE 0 Advanced : Question Paper & Solution (Paper II) () 5. P and Q respectively, are the sodium salts of (A) hypochlorus and chloric acids (B) hypoclorus and chlorus acids (C) chloric and perchloric acids (D) chloric and hypochlorus acids 5. (A) Cold NaOH + Cl NaCl + NaOCl(P) + H O hot 6NaOH + Cl 5NaCl + NaClO (Q) + H O 6. R, S and T, respectively are (A) SO Cl, PCl 5 and H PO 4 (B) SO Cl, PCl and H PO (C) SOCl, PCl and H PO (D) SOCl, PCl 5 and H PO 4 6. (A) SO + Cl SOCl (R) P4 + 0SOCl 4PCl5+ 0 SO PCl5 + 4HO HPO4 + 5HCl SECTION : (Matching List Type) This section contains 4 multiple choice questions. Each question has matching lists. The codes for the lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct 7. An aqueous solution of X is added slowly to an aqueous solution of Y as shown in List I. The variation in conductivity of these reactions is given in List II. Match List I with List II and select the correct answer using the code given below the lists : List I List II P. (CH 5) Ν + CHCOOH. Conductivity decreases and then increases X Y Q. KI (0.M) + AgNO (0.0M). Conductivity decreases and then does not X Y change much R. CHCΟΟΗ + KOH. Conductivity increases and then does not X Y change much S. NaOH + HI 4. Conductivity does not change much and then X Y increases Codes : P Q R S (A) 4 (B) 4 (C) 4 (D) 4 7. (A) Factual 8. The standard reduction potential data at 5 C is given below. E (Fe +, Fe + ) V; E (Fe +, Fe) 0.44 V E (Cu +, Cu) V E (Cu +, Cu) V E [O (g) + 4H + + 4e H O] +. V;

49 () Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution E [O (g) + H O + 4e 4OH ] V E (Cr +, Cr) 0.74 V ; E (Cr +, Cr) 0.9 V Match E of the redox pair in List I with the values given in List II and select the correct answer using the code given below the lists : List I P. E (Fe +, Fe). 0.8 V Q. + E (4HO 4H + 4OH ). 0.4 V R. E (Cu + + Cu Cu + ) V S. E (Cr +, Cr + ) V Codes : List II P Q R S (A) 4 (B) 4 (C) 4 (D) 4 8. (D) Use formula nf E nfe + nfe E ne + ne n 9. The unbalanced chemical reactions given in List I show missing reagent or condition (?) which are provided in List II. Match List I with List II and select the correct answer using the code given below the lists : List I P. PbO + H SO? PbSO + O + other product. NO 4 4 Q.? NaSO + HO NaHSO4 + other product. I R.? N H N + other product. Warm 4 S.? XeF Xe + other product 4. Cl Codes : P Q R S (A) 4 (B) 4 (C) 4 (D) 4 9. (D) P PbO + H SO warm PbSO + O + H O 4 4 Cl 4 4 I NO Q Na S O + H O NaHSo + NaCl + HCl R N H N + 4HI S XeF Xe + NOF List II

50 IIT JEE 0 Advanced : Question Paper & Solution (Paper II) () 40. Match the chemical conversions in List I with the appropriate reagents in List II and select the correct answer using the code given below the lists : List I List II P.. (i) Hg(OAc),; (ii) NaBH 4 Q.. NaOEt R.. Et Br S. 4. (i) BH ; (ii) H O / NaOH Codes : P Q R S (A) 4 (B) 4 (C) 4 (D) (A) P, Q, R, S 4 PART III : MATHEMATICS SECTION : (One or more options correct Type) This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct. 4. Let ω be a complex cube root of unity with ω and P [p ij ] be a n n matrix with p ij ω i + j. Then P 0, when n (A) 57 (B) 55 (C) 58 (D) (B), (C), (D) P O only when n is multiple of. Ex : ω ω ω ω ω ω ω ω ω ω ω ω P 0, when n 55, 56, The function f(x) x + x + x + x has a local minimum or a local maximum at x (A) (B) (C) (D)

51 (4) Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution 4. (A), (B) Given, f(x) x + x + x + x Sign scheme of x + x is For x f (x) 4 x ve +ve / y 8 8/ / ve x x For x < x > f(x) x + y 8 8/ Now, f(x) 4 x, / x x x +, x < x > y 8 8/ / x The point of maxima or minima are,, 0. + i 4. Let w and P {w n : n,,, }. Further H z :Rez> and H z :Rez <, where is the set of all complex numbers. If z P H, z P H and O represents the origin, then z O z π π (A) (B) (C) π (D) 5 π (C), (D) i π π ω + cos + i sin 6 6 ω n nπ nπ cos + i sin (by De Moivre s Theorem) 6 6 nπ If z P H, then cos 6 > which is true for n 0,,. 4

52 IIT JEE 0 Advanced : Question Paper & Solution (Paper II) (5) If z P H, then, nπ cos 6 <, which is true for n 5, 6, 7., (, 0) (, 0),, z oz can be 5 π or. π If x 4 x, then x log (A) (B) log log 44. (A), (B), (C) x 4 x xlog (x ) log 4 x log x x ( log ) x (B) log x 4 x x log4 x x (C) log4 x 4 x x log (x ) log 4 x (x ) log 4 (x 9). log.log x (A).log (C) log 4 log (D) log 45. Two lines L y z y z :x 5, and L :x α, α α are coplanar. Then α can take values (s) (A) (B) (C) (D) (A), (D) L : x 5 y 0 α Z Line L passes (x, y, z ) (5, 0, 0) L : x α y z 0 α Line L passes (x, y, z ) (α, 0, 0) As Line co-planar Shortest distance 0 5

53 (6) Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution x x y y z z m n 0 m n α α 0 (α 5) (α 4) (α ) 0 0 α α, 4, In a triangle PQR, P is the largest angle and cos P. Further the incircle of the triangle touches the sides PQ, QR and RP at N, L and M respectively, such that the lengths of PN, QL and RM are consecutive even integers. Then possible length(s) of the side(s) of the triangle is (are) (A)6 (B) 8 (C) 4 (D) 46. (B), (D) (4x + 4) + (4x + ) (4x + 6) cos p (4x + 4) (4x + ) 6x 6 6(x + ) (x + ) x x + x + x x + x + x x 4 0 x 4 or. Hence x 4 So, possible lengths of sides are 8,. 47. For a R (the set of all real numbers), a, lim x Then a a a a ( )... n 60 a ( n + ) ( na + ) + ( na + ) +...( na + n) (A) 5 (B) 7 (C) 47. (B), (D) Let, L n a a a n lim n na (na )... (na n) a ( + ) (( + ) ) a a a a n n n n n lim n(n + ) ( n+ ) n a+ n a a r n n n r n lim lim n a n+ n ( n+ ) n na+ n a x + Q ( ) x x + P (D) a a xdx n 0 a + 7 n+ n na + x x + 4 x + 4 R 6

54 L IIT JEE 0 Advanced : Question Paper & Solution (Paper II) (7) Using limit of a sum we get n a r n n a xdx r lim n a n a+ a n+ n+ na + ( ) (a + ) ( a + ) (a + ) (a + ) 0 a + a a 7 or a Answer is (B), (D) 0 a+ a n n lim n a Given L Circle(s) touching x axis at a distance from the origin and having an intercept of length 7 on y axis is (are) (A) x + y 6x + 8y (B) x + y 6x + 7y (C) x + y 6x 8y (D) x + y 6x 7y (A), (C) Let centre, C (, α) and BC α c 7 α c 7 () Agai α α + 9 c c 9 () From (), α 9 7 α ± 4 Eq is x + y 6x ± 8y C B 0 A e α SECTION : (Paragraph Type) This section contains 4 paragraphs each describing theory, experiment, data etc. Eight questions relate to four paragraphs with two questions on each paragraph. Each question of a paragraph has only one correct answer among the four choices (A), (B), (C) and (D).. Paragraph for Question 49 and 50 Let f : [0, ] R (the set of all real numbers) be a function. Suppose the function f is twice x f'' x f' x + f x e,x 0,. differentiable, f(0) f() 0 and satisfies ( ) ( ) ( ) [ ] 49. Which of the following is true for 0 < x <? (A) 0< f( x) < (B) f ( x) 49. (D) 4 < f x < 0 < < (C) < f( x) < (D) ( ) 7

55 (8) Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution 50. If the function e x f(x) assumes its minimum in the interval [0, ] at following is true? 4 4 (C) f' ( x) f( x ),0 x (C) Let g(x) e x f(x) f' x > f x,0< x< 4 f' x < f x, < x< 4 (A) f' ( x) < f( x ), < x< (B) ( ) ( ) If x 4 < < < (D) ( ) ( ) is point of minima, then g (x) e x (f (x) f(x)) x, which of the 4 e x (f (x) f(x)) < 0 f (x) < f(x) x 0, 4 x 0, 4 Paragraph for Question 5 and 5 Let PQ be a focal chord of the parabola y 4ax. The tangents to the parabola at P and Q meet at a point lying on the line y x + a, a > Length of chord PQ is (A) 7a (B) 5a (C) a (D) a 5. (B) P + T Given point T lies on y x + a a(t + t ) at t + a t + t + (as focal chord t t ) t + t ( t t ) ( ) t + t 4t t 5 + So, PQ a ( t t ) ( at at ) a ( t t ) 4( t t ) + a (( ) 5) 4( 5) 8 + a a 5. If chord PQ subtends an angle θ at the vertex of y 4ax, then tan θ (A) 7 (B) 7 (C) 5 (D) 5. (D) Slope of OP at at t m Slope of OQ t m m m tanθ + m m as θ is nd quadrant. t t + t t ( ( + t )) T at t,a t t t tt (,at) P at (,at) Q at 5

56 IIT JEE 0 Advanced : Question Paper & Solution (Paper II) (9) Paragraph for Question 5 and 54 Let S S S S, where z + i S { z C: z < 4}, S z C:Im > 0 i S z C:Rez> 0 { } and 5. Area of S (A) 0 π (B) 0 π (C) 6 π (D) π 5. (B) S {z z < 4} (0, 4) B y x C ( 4, 0) (4, 0) A D (0, 4) y x z + i S z c:im > 0 i Let z x + iy z + i (x + iy) + i i i ( ) (x ) + i y + ( + i) i (+ i) { (x ) (y + ) } + i { (x ) + (y + } z + i x+ y Im i 4 z + i Im > 0 i x+ y > 0 4 x + y > 0 4 O x 9

57 (0) Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution S {z c, Re(z) > 0} x > 0 S S S y (0, 4) x + y 4 O π/ (4, 0) x y + x 0 (, ) Area of sector is π (4) 5 ππ /6 π π min 54. z S i z (A) 54. (C) min i z z s (B) + (C) i z z (i ) z ( + i) PM (Perpendicular distance is minimum distance) y (0, 4) (D) + x O M (4, 0) x P(, ) (, ) y 0

58 IIT JEE 0 Advanced : Question Paper & Solution (Paper II) () Paragraph for Question 55 and 56 A box B contains white ball, red balls and black balls. Another box B contains white balls, red balls and 4 black balls. A third box B contains white balls, 4 red balls and 5 black balls. 55. If ball is drawn from each of the boxes B, B and B, the probability that all drawn balls are of the same colour is (A) 8 (B) 90 (C) 558 (D) (A) W R B W R 4B W 4R 5B B B B E one white from each box E one red from each box E one black from each box E, E, E are mutually exclusive events P(E) P(E ) + P(E ) + P(E ) If balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these balls are drawn from box B is (A) 6 (B) 6 (C) 65 (D) (D) P(B ) P E B 5 P(B ) P E B 6 P(B ) B P E P E B

59 () Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution SECTION : (Matching List Type) This section contains 4 multiple choice questions. Each question has matching lists. The codes for the lists have choices (A), (B), (C) and (D) out of which ONLY ONE is correct 57. Match List I with List II and select the correct answer using the code given below the lists : P Q R S List I List II Volume of parallelepiped determined by vectors a,b and c is. Then the volume of the parallelepiped determined by vectors. 00 a ( b,b ) ( c) and( c a) is Volume of parallelepiped determined by vectors a,b and c is 5. Then the volume of the parallelepiped determined by vectors. 0 a+ b,b+ c andc+ a is ( ) ( ) ( ) Area of a triangle with adjacent sides determined by vectors aandb is 0. Then the area of the triangle with adjacent sides. 4 a + b and a b is determined by vectors ( ) ( ) Area of a parallelogram with adjacent sides determined by vectors aandb is 0. Then the area of the parallelogram with adjacent a+ b anda is sides determined by vectors ( ) Codes : P Q R S (A) 4 (B) 4 (C) 4 (D) (C) (P) Given volume of parallelepiped formed by a,b,c is equal to [abc] () a b b c and c a is given by volume of parallelepiped formed by ( ), ( ) v a ( b,b ) ( c,c ) a or v 6 [ abc] v 4 p (Q) given [ abc] 5 volume of parallelepiped formed by a+ b, b c c+ a is given by ( ) +, and ( ) v a ( + b,b ) + c,c ( + a) v 6 a+ b,b+ c, c+ a v [ abc] v 60 Q 4

60 (R) given ( a b ) IIT JEE 0 Advanced : Question Paper & Solution (Paper II) () 0 a b 40 () Area of triangle with adjacent sides a b a b is given as ( + ) and ( ) Δ ( a + b ) ( a b ) 5a b 00 R Δ ( ) (S) given a b 0 area of parallelogram with sides a+ b and a is given as area ( a+ b) a b a 0 S P Q R S 4 Answer is (C) x y z+ x 4 y+ z+ 58. Consider the lines L :, L : and the planes P : 7x + y + z, P : x + 5y 6z 4. Let ax + by + cz d be the equation of the plane passing through the point of intersection of lines L and L, and perpendicular to planes P and P. Match List I with List II and select the correct answer using the code given below the lists : List I List II P a. Q b. R c. S d 4. Codes : P Q R S (A) 4 (B) 4 (C) 4 (D) (A) Let L : x and L : x 4 z + y t y + z + s any point on L is M (t +, t, t ) and any point on L is M (s + 4, s, s )

61 (4) Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution at the point on integration t + s + 4 () t s () t s () Solving (), () and () we get t and s M M (5,, ) Also if the plane is perpendicular to P and P then 7a + b + c 0 (4) and a + 5b 6c 0 (5) b a; c a hence the plane becomes, ax ay az d (6) using M or M in (6) we get 5a + 6a + a d d a the plane becomes ax ay az a or x y z (a 0) (P) a P (Q) b Q (R) c R 4 (S) d S P Q R S Match List I with List II and select the correct answer using the code given below the lists : P Q R S List I ( ) + ( ) ( ) + ( ) List II cos tan y y sin tan y 4 5 y + y takes value. cot sin y tan sin y If cos x + cos y + cos z 0 sin x + sin y + sin z then possible x y. value of cos is π If cos x cos x + sin x sin x sec x cos x sin x sec x 4 π. + cos + x cos x 4 then possible value of sec x is ( ) If ( ) ( ) cot sin x sin tan x 6, x 0, then possible value of x is 4. 4

62 IIT JEE 0 Advanced : Question Paper & Solution (Paper II) (5) Codes : P Q R S (A) 4 (B) 4 (C) 4 (D) (B) (P) Now, cos( tan y) + y sin ( tan y) + y cot ( sin y) + tan ( sin y) 4 y y cos cos + y sin sin + y + y y y y cot cot tan tan + y y / y + + y + y 4 y y + y y + y y P 4 (Q) From the given equations we have cos z (cos x + cos y) () sin z (sin x + sin y) () Squaring and adding () and () we get cos x + cos y + cos x cos y + sin x + sin y + sin x sin y + cos (x y) cos (x y) x y cos x y cos Q (R) The given equation can be simplified as π sin sin xcosx 4 sin x sec x (cos x sin x) sinxcosx sin x sec x (cos x sin x) sin x cos x sin x ( cos x sin x) / + y 4 / 5

63 (6) Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution cos x 0 sin x ( cos x sin x) cos x sin x sin x ( cos x sin x) 0 cos x + sin x sin x (cos x sin x) 0 π sin x ( cos x sin x) sin x π x 4 sec x R (S) Now, ( ) cot sin x ( cot sin x ) ( ) and sin tan ( x 6 ) cot cot x x sin sin x x x 6 + 6x () ( ) ( ) sin tan x 6 x 6 + 6x () From () and () we get x x x 6 + 6x ( x 0) Solving this we x 5 or x 5 S P Q R S A line L : y mx + meets y axis at E (0, ) and the arc of the parabola y 6x, 0 y 6 at the point F(x 0, y 0 ). The tangent to the parabola at F(x 0, y 0 ) intersects the y axis at G(0, y ). The slope m of the line L is chosen such that the area of the triangle EFG has a local maximum. Match List I and List II and select the correct answer using the code given below the lists : 6

64 IIT JEE 0 Advanced : Question Paper & Solution (Paper II) (7) List I List II P m. Q Maximum area of ΔEFG is. 4 R y 0. S y 4. Codes : P Q R S (A) (A) (B) 4 (C) 4 (D) 4 8x0 (0, y ) 0, y 0 E G F(x 0, y 0 ) (, 4) y 6x Tangent at P(x 0, y 0 ) to y 6x is yy 0 4(x + x 0 ) 8x0 S0, G 0, y 0 As F (x 0, y 0 ) lies on y 6x y 0 6x 0 () Δ EG x 0 8x0 y0 y 0 6 y [y 8x ] [6y 0 y 0 ] () 64 7

65 (8) Vidyalankar : IIT JEE 0 Advanced : Question Paper & Solution Δ [y 0 y ] 0 64 Δ 0 y 0 (4 y 0 ) 0 y 0 0 or y 0 4 Δ y0 4 [ 6y 0 ] < 0 64 local maximum when y 0 4 So, put in () y 0 4, Δ max [6 4 4 ] 64 [6 4] 4 From (), y 0 6x 0 6 6x 0 x 0 8x y 0 y y Line GF, 4 x 0 y x + So, slope m Hence, (A) option correct

66 IIT JEE 0 Advanced : Question Paper & Solution (Paper II) (9) 9