Part I : (CHEMISTRY) SECTION I (Total Marks : 24) (Single Correct Answer Type) 10/04/2011

Transcription

1 PAPER- IIT-JEE 011 EXAMINATION Part I : (CHEMISTRY) SECTION I (Total Marks : 4) CODE - 9 (Single Correct Answer Type) 10/04/011 This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Ans. [D] + 3 Haematite is F eo Magnetite is Fe 3 O 4 or FeO. FeO3 Ans. [B] Aldohexose 1 / 38

2 PAPER- IIT-JEE 011 EXAMINATION Ans. [B] O R- CH OH H (anhydrous) O OCH R Acetal Ans. [C] NaNO + HCl (dil.) CH 3 NH CH C N Cl Alkaline solution OH OH CH 3 H = N Coloured dye / 38

3 PAPER- IIT-JEE 011 EXAMINATION Ans. [A] T = k f m i = = Ans. [D] E cell = E cell O [Fe ] log 4 + [P ][H ] 4 = [10 ] 1.67 log 4 [.1][10 ] = 1.57 V 3 4 Ans. [A] Cu +, Hg + are group II basic radicals 3 / 38

4 PAPER- IIT-JEE 011 EXAMINATION Ans. [C] (L) : [Co(NH 3 ) 6 ]Cl 3 (M) : Na 3 [Co(Ox) 3 ] (O) : K [Pt(CN) 4 ] (P) : [Zn(H O) 6 ] (NO 3 ) SECTION II (Total Marks : 16) (Multiple Correct Answers Type) This section contains 4 multiple choice questions. Each questions has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct. Ans. [A, C, D] In alkaline solution, KMnO 4 is first reduced to mangnate and then to insoluble MnO Neutral KMnO 4 + HO MnO 4 + KOH + 3[O] I + acidic + MnO4 + 8H + 5e Mn + 4HO Ans. [A, B, D] 4 / 38

5 PAPER- kt C = C e [A] A A0 IIT-JEE 011 EXAMINATION t 1 = = K Ea / A e 0 RT = e A 0 Ea / RT [B] = 4 = log 100 = n = = 8 1 log [D] Ans. [B, C, D] Cu Cl, Cu (CN) and Cu (SCN) are stable Ans. [C, D] Factual * The most appropriate answer to this question is (A,B,C,D) But because of ambiguity in language, IIT has declared (C & D) as correct answer 5 / 38

6 PAPER- IIT-JEE 011 EXAMINATION SECTION III (Total Marks : 4) (Integer Answer Type) This section contains 6 Question. The answer to each of the question is a single-digit integer, ranging from 0 to 9. The total bubble corresponding answer it to be darkened in the ORS. Ans. [7] + [Ag ] = K 1 1 K + K Q K 1 < < K K1 + K K [Ag x = ] = = Ans. [8] CH 3 CH CH CH 3 CH 3 CH 3 Cl / hν * Cl CH CH CH CH CH 3 () CH 3 or * * CH 3 CH CH CH CH 3 Cl (4) CH 3 Cl or CH 3 CH C CH CH 3 (1) CH 3 or CH 3 CH CH CH CH 3 (1) CH Cl Ans. [4] PCl 5 + H O POCl 3 + HCl PCl 5 + H SO POCl 3 + H O + SO Cl 6PCl 5 + P 4 O 10 10POCl 3 6 / 38

7 PAPER- IIT-JEE 011 EXAMINATION PCl 5 + SO POCl 3 + SO Cl Ans. [8] 8 Hexagonal faces Ans. [6] 0.1 V = V = = 6 ml 0.1 Ans. [6] 6 (α H 6) 7 / 38

8 PAPER- IIT-JEE 011 EXAMINATION SECTION IV (Total Marks : 16) (Matrix-Match Type) This section contain 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 ONE or MORE 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 bubbles corresponding to q and r in the ORS Ans. [A p, r, s; B r, s; C t ; D p, q, t] [A] CO (s) CO (g) p, r, s [B] CaCO 3 (s) CaO (s) + CO (g) r, s [C] [D] H H (g) t P white P red p, q, t 8 / 38

9 PAPER- IIT-JEE 011 EXAMINATION Ans. [A r, t, s; B p, s, t; C r, s ; D r, q] Factual * The most appropriate answer to this question is A r, s, t; B p, s, t; C r, s; D q, r But because of ambiguity in language, IIT has declared A r, s, t; B p, s; C r, s; D q & r as correct answer 9 / 38

10 PAPER- IIT-JEE 011 EXAMINATION SECTION III (Total Marks : 4) (Integer Answer Type) This section contains 6 Question. The answer to each of the question is a single-digit integer, ranging from 0 to 9. The total bubble corresponding answer it to be darkened in the ORS. Ans. [7] + [Ag ] = K 1 1 K + K Q K 1 < < K K1 + K K [Ag x = ] = = Ans. [8] CH 3 CH CH CH 3 CH 3 CH 3 Cl / hν * Cl CH CH CH CH CH 3 () CH 3 or * * CH 3 CH CH CH CH 3 (4) Cl CH 3 Cl or CH 3 CH C CH CH 3 (1) CH 3 or CH 3 CH CH CH CH 3 (1) CH Cl Ans. [4] PCl 5 + H O POCl 3 + HCl PCl 5 + H SO POCl 3 + H O + SO Cl 6PCl 5 + P 4 O 10 10POCl 3 PCl 5 + SO POCl 3 + SO Cl 10 / 38

11 PAPER- IIT-JEE 011 EXAMINATION Ans. [8] 8 Hexagonal faces Ans. [6] 0.1 V = V = = 6 ml 0.1 Ans. [6] 6 (α H 6) 11 / 38

12 PAPER- IIT-JEE 011 EXAMINATION SECTION IV (Total Marks : 16) (Matrix-Match Type) This section contain 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 ONE or MORE 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 bubbles corresponding to q and r in the ORS Ans. [A p, r, s; B r, s; C t ; D p, q, t] [A] CO (s) CO (g) p, r, s [B] CaCO 3 (s) CaO (s) + CO (g) r, s [C] [D] H H (g) t P white P red p, q, t 1 / 38

13 PAPER- IIT-JEE 011 EXAMINATION Ans. [A r, t, s; B p, s, t; C r, s ; D r, q] Factual * The most appropriate answer to this question is A r, s, t; B p, s, t; C r, s; D q, r But because of ambiguity in language, IIT has declared A r, s, t; B p, s; C r, s; D q & r as correct answer 13 / 38

14 PAPER- IIT-JEE 011 EXAMINATION Code : 9 10/04/011 Ans.[C] θ θ e When θ > θ C, no ray will transmit T = 0, T + R = 100 % and R > 0 T θ c 14 / 38

15 PAPER- IIT-JEE 011 EXAMINATION Ans.[A] In order to have net force zero, the mean position will be shifted towards right but the time period will remain unaffected. Ans.[C] Pitch = 0.5 mm divisions on the = 50 circular scale 0. 5 least count of screw gauge = 50 main scale, reading =.5 mm circular scale reading = 0 ρ ρ reading =.5 mm + (0 0.01) mm ρ = m = m = 0.01 =.5 mm + 0. mm =.7 mm m 4π D 3 D + 3 D 3 ρ 0.01 %error = 100 = % = 3.1. ρ.7 15 / 38

16 PAPER- IIT-JEE 011 EXAMINATION Ans.[D] m 1 = 0.01 kg m = 0. kg v H = 5m Ball Bullet 0 m 100 m T = H g = 1 sec Let v 1 & v be velocity of bullet & ball respectively just after collision. v 1 = 0 v = 0 & v 1 = 100 From conservation of momentum 0.01 v = ( ) + (0. 0) 0.01 v = = 5 v = 5 10 = 500 m/sec. 16 / 38

17 PAPER- IIT-JEE 011 EXAMINATION Ans.[C] Electric lines of force for induced electric field is closed loop. Ans.[B] A θ 30º 10º A B Here φ = π + θ A cos 30º = B sin θ B sin θ = 3A Solving above, B = A and θ = 60º = 3 π. and A sin 30º + B cosθ = A B cos θ = A Hence φ = 40º = 3 4π 17 / 38

18 PAPER- IIT-JEE 011 EXAMINATION Ans.[A] y dr a r x b No. of turns per unit thickness = N b a magnetic field at centre due to element = db = B = µ 0 i r µ 0 in (b a) N b a b a dr r dr µ = 0iN b ln (b a) a µ 0 (dn)i r 18 / 38

19 PAPER- IIT-JEE 011 EXAMINATION Ans.[B] V M r mv GmMe = r r If K.E. of mass m = was k then from r = GM V e...(1) E = K GmM e = 0 K = m r GM e = mv r 19 / 38

20 PAPER- Ans.[A,B,D] IIT-JEE 011 EXAMINATION A B Vd f g Vd A g Vd f g Vd B g system will be in equilibrium with tension in string only if d f > d A and d B > d f. If both A & B are considered as a system then Vd f g = V (d A + d b )g d A + d B = d f Ans.[C,D] Ans.[B,C] R C ~ R 4C Z 1 = R 1 + ωc ~ z 1 > z V A C A IK = ; ωc Z = R A B I R < IR B B IK VC = ; 4ωC 1 + 4ωC B C V < V A C 0 / 38

21 PAPER- IIT-JEE 011 EXAMINATION Ans.[A,C] or [A] As no data is given about nature of horizontal plane. * The most appropriate answer to this question is (A,C), but because of ambiguity in language, IIT has declared [(A, C), (A)] as correct answer 1 / 38

22 PAPER- IIT-JEE 011 EXAMINATION Ans.[5] V A V B = = = = 5V Ans.[4] Z = R 1. 5 τ = RC R + R + 1 = Z 500C 1 = R C 1 = 0.5 R 500C 1 = 0.5R 500C 1 = RC = RC sec = RC RC = 4 mill second. Ans.[5] T = = 3 sec x = 10 cos 60 (T) = 5 3 m In frame of train, 5 3 = 1 a ( 3 ) (a : acceleration of train) a = 5 m/sec / 38

23 PAPER- IIT-JEE 011 EXAMINATION Ans.[] µ v 3 µ 1 u = µ µ R µ 3 R µ = 4 3V Ans. = (18 16) cm = cm = = V = 16 cm 3V 4 8 3V 1 Ans.[4] k = N/m m = 0.18 kg µ = 0.1 Using W E theorem 1 m(u) 1 = K (x) + µmg (x) 1 (0.18) u 1 = u = 0.4 m/sec m/sec. 3 / 38

24 PAPER- IIT-JEE 011 EXAMINATION 140 Ans.[7] Energy of photon ev = 6. ev 00 Maximum KE of a electron = 6. ev 4.7 ev When potential on surface of sphere becomes equal to 1.5V q = 1.5 V q = 1.5 (4π ε 0 ) r 4π 0 r No. of photoelectron emitted n = 1.5 (4πε )r 0 19 = / 38

25 PAPER- IIT-JEE 011 EXAMINATION Ans. (A) p,r,t; (B) p,r; (C) q,s; (D) r,t Process AB : (Pressure is constant) T If T A = T T B = 3 So U = Negative [Q U = nc v T] W = nr T = Negative Q = U + W = Negative Process BC : (Volume is constant) T T If T B = then TC = 3 9 U = nc v T = Negative W = Zero Q = Negative Process C D : (Pressure is constant) T If T C = then TD = T 9 U = nc v T = positive W = positive Q = positive Process D A : T D = T andt A = T Hence process is isothermal U = 0 W = negative Q = negative 5 / 38

26 PAPER- IIT-JEE 011 EXAMINATION Ans. (A) p,t; (B) p,s; (C) q,s; (D) q,r (A) λf = L 4 λ f = 4L (C) Stretched wire clamped at both ends (B) Longitudinal waves (D) λf = L λ f = L λf = L λ f λ + f = L λ f = L 6 / 38

27 PAPER- IIT-JEE 011 EXAMINATION Part III : (MATHEMATICS) SECTION I (Total Marks : 4) (Single Correct Answer Type) Code-9 10/04/011 This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Ans. [B] Equation of the normal at (6, 3) is a x + 6 b y = a + b 3 it passes through (9, 0) so 9a = a + b 6 b a = Now b = a (e 1) e 1 = 1 e = 3 e = 3 Ans. [C] t t h =, k = P(h, k) (t, t) (0, 0) t = 4h, t = k so 4k = 4h 7 / 38

28 PAPER- k = h hence required locus is y = x IIT-JEE 011 EXAMINATION Ans. [A] gof(x) = gf(x) = g(x ) = sin x go (gof(x)) = g(sin x ) = sin (sin x ) fo(gogof(x)) = f(sin (sin x )) = (sin(sin x )) (sin (sin x )) = sin (sin x ) sin (sin x ) (sin (sin x ) 1) = 0 sin (sin x ) = 0 or sin (sin x ) = 1 sin x = nπ sin x = nπ + π At n = 0 At n = 0 sin x = 0 x = nπ x = ± n π ; n {0, 1,,.} sin x = π Not possible Ans. [C] R 1 = x f(x) dx (i) 1 R 1 = ( 1 x)f(1 x)dx 1 = ( 1 x)f(x)dx (ii) 1 (i) + (ii) R 1 = 1 f (x)dx = R R 1 = R 8 / 38

29 PAPER- IIT-JEE 011 EXAMINATION Ans. [D] x lim 0 (1 + x ln(1 + b )] 1/x = b sin θ b > 0; θ ( π, π) lim x 0 e [1 + xl n(1 + b ln(1+ b ) = b sin θ 1 + b = b sin θ sin 1 θ = b + b )] 1 xln(1+ b ) ln(1+ b ) = b sin θ 1 RHS = b + as b > 0 b But LHS = sin θ Only possibility sin θ = sin θ = 1 π θ = ± Ans. [D] (h 0) + ( ) = (h + 1) + ( 0) h = h h + 4 (h, ) (0, ) h = 5 Equation of circle is ( 1, 0) 5 x + + (y ) 5 = 0 9 / 38

30 PAPER- IIT-JEE 011 EXAMINATION x x + y + 4 4y = 4 x + y + 5x 4y + 4 = from given points only point ( 4, 0) satisfies this equation. Ans. [A] 1 a b ω 1 c 0 ω ω 1 (1 ωc) a (ω ω c) + b(ω ω ) 0 1 ωc aω + acω 0 (1 ωc) aω (1 ωc) 0 (1 ωc) (1 aω) 0 c ω & a ω & b = ω or ω (a, b, c) (ω, ω, ω) or (ω, ω, ω) Ans. [B] x + bx 1 = 0 (i) x + x + b = 0 (i) (ii) we get x = Put this value in (i) b + 1 b 1 b + 1 b b 1 = 0 b 1 b 1 b 3 + 3b = 0 b(b + 3) = 0 b = 0 or b = ± i 3 (ii) 30 / 38

31 PAPER- IIT-JEE 011 EXAMINATION SECTION II (Total Marks : 16) (Multiple Correct Answers Type) This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONE OR MORE may be correct. Ans. [A, B, C, D] At x = π π LHL = 0, RHL = 0, f = 0 So f(x) is continuous at x = At x = 0 π LHD = 0; RHD = 1 So f(x) is not differentiable at x = 0 At x = 1 LHD = 1, RHD = 1 So f(x) is differentiable at x = 1 π in, 0 ; f(x) = cos x so f(x) is differentiable at x = 3 31 / 38

32 PAPER- IIT-JEE 011 EXAMINATION Ans. [A. B, D] y = mx m m 3 It passes through (9, 6) 6 = 9m m m 3 m 3 7m + 6 = 0 (m 1) (m ) (m + 3) = 0 m = 3, 1, Hence equations will be y = x 3, y = x 1 and y = 3x + 33 Ans. [A, D] 11 P(E) (1 P(F)) + (1 P(E)) P(F) = 5 11 P(E) + P(F) P (E) P(F) = 5 (1) (1 P(E)) (1 P(F)) = 5 1 P(E) P(F) + P(E) P(F) = 5 3 P(E) + P(F) P(E) P(F) = 5 () From (1) & () 1 P(E) P(F) = 5 and P(E) + P(F) = 5 7 so either P(E) = 5 4, P(F) = 5 3 and P(E) = 5 3, P(F) = / 38

33 PAPER- IIT-JEE 011 EXAMINATION Ans. [A, B] f : (0, 1) R b x f(x) = 1 bx b (0, 1) b 1 f (x) = = ( ) ve (1 bx) So f(x) is monotonically decreasing for x (0, 1) so for x (0, 1) f(x) (f(1), f(0)) f(x) ( 1, b) so f(x) is not onto. so f(x) is not invertible function. * The most appropriate answer to this question is (A, B), but because of ambiguity in language, IIT has declared (A) as correct answer. SECTION III (Total Marks : 15) (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 bubble corresponding to the correct answer is to be darkened in the ORS Ans. [0] dy + y = g dg I. F. = 1. dg = g y.e g = ge g. dg ye g = ge g e g + c y = g 1 + ce g = ge g e g. dg 33 / 38

34 PAPER- Q y(0) = 0 & g(0) = 0 at x = 0 0 = Ce 0 C = 1 y = g 1 + e g at x = y() = e 0 = 0 IIT-JEE 011 EXAMINATION Ans. [9] a r = î kˆ, b r = î + ĵ, c r = î + ĵ + 3 kˆ r r r r r r r r r ( r c) b = 0 r c = λb r = c + λb r r Q r a = 0 r r r r a. c + λb. a = 0 r r a. c λ = r r = 4 b. a r r r r r r. b = c. b + λ b = 9 Ans. [ * ] wrong question if ω = i π / 3 e then ans is 3. If ω = e i π / 3 then no integral solution is possible. Ans. [9] a b c Let M = d e f g h i 0 1 Q M 1 = b = 1, e =, h = / 38

35 PAPER- IIT-JEE 011 EXAMINATION 1 M 1 = 0 1 M 1 = a = 0, d = 3, g = c = 1, f = 5, i = 7 1 So a + e + i = = 9 Ans. [] Let f(x) = x 4 4x 3 + 1x + x 1 Let α, β, γ, δ are the root of equation. αβγδ = 1 so the equation has at least two real roots...(i) f'(x) = 4x 3 1x + 4x + 1 f"(x) = 1x 4x + 4 = 1((x + 1) + 1) so f"(x) > 0 so f'(x) = 0 has only one real roots so f(x) = 0 has at most two real roots. from (i) & (ii) f(x) = 0 has exactly two real roots..(ii) Ans. [] 1,0 (, 3) Pont (x 1, y 1 ) lies inside the region if x 1 + y1 6 0 & x 1 3y P 1, True > 0 True P, False / 38

36 PAPER- IIT-JEE 011 EXAMINATION P 3, True > 0 True P 4, True > 0 False 8 4 So P 1 & P 3 lies in the interval SECTION IV(Total marks : 8) (Integer Answer Type) 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 Column I can have correct matching with ONE or MORE 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 bubbles corresponds to q and r in the ORS. Ans. [A q ; B p, q, r, s, t or p ; C s ; D t] (A) 36 / 38

37 PAPER- IIT-JEE 011 EXAMINATION c r br θ a r r r a b 1 cosθ = r r = a b π θ = 3 (B) b a ( f ( x) 3( x)) dx = a b differentiating w.r.t (b). f(b) 3b = b f ( b) = b π π So f = ; if a = b then any value of f(x) is possible 6 6 π (C) I = ln3 5 / 6 sec πx dx 7 / 6 π I = ln sec πx + tanπx πln3 π I = l 3 = π ln3 n (D) z = 1 5 / 6 7 / 6 z = cos θ + i sin θ. θ ( π. π] and θ 0. Arg 1 (1 z) = 1 Arg = 1 cosθ i sinθ θ i cot 1 Arg + = π θ so maximum value is π. * The most appropriate answer to this question is A q; B p or p, q, r, s & t; C s; D s But because of ambiguity in language, IIT has declared A q; B p or p, q, r, s & t; C s; D t as correct answer 37 / 38

38 PAPER- IIT-JEE 011 EXAMINATION Ans. [A p, r, s ; B t ; C r ; D r] (A) Let z = cos θ + i sin θ iz i(cos+ i sinθ) so = = cosecθ 1 z 1 cosθ i sinθ iz so Re = cosec θ (, 1] [1, ) 1 z 8 3 (B) 1 3 x x 8t 1 9 t 8 3 = 9 3 x x θ (n + 1) π x Let 3 x = t So f(x) = sin t = sin x t 1 on solving x (, 0] [, ) {1} (C) f(θ) = sec θ so f(θ) [, ) (D) f(x) = 3x 5/ 10x 3/ * The most appropriate answer to this question is A q; B p; C s; D s But because of ambiguity in language, IIT has declared A s; B t; C r; D r as correct answer 15 x f'(x) = ( x ) So f(x) is increasing for f'(x) 0 x [, ) 38 / 38