TEST - 7 (Paper-I) ANSWERS

Transcription

1 Test - 7 (Paper-I) (Answers & ints) All India Aakash Test Series for JEE (Main)-0 TEST - 7 (Paper-I) ANSWES PYSICS CEMISTY MATEMATICS. (). () 3. (). (). (). () 7. (3) 8. () 9. (3) 0. (). (3). () 3. (). (3). (). () 7. () 8. (3) 9. (3) 0. (3). (). (3) 3. (3). (3). (). () 7. () 8. () 9. () 30. () 3. () 3. () 33. (3) 3. () 3. () 3. () 37. () 38. () 39. () 0. (). (). (3) 3. (). (). (). () 7. () 8. () 9. () 0. (3). (). (3) 3. (). (). (). () 7. () 8. () 9. () 0. (). (3). () 3. (). (). (). (3) 7. () 8. () 9. () 70. () 7. () 7. () 73. () 7. () 7. () 7. () 77. (3) 78. () 79. () 80. (3) 8. () 8. (3) 83. () 8. () 8. () 8. () 87. (3) 88. () 89. () 90. (3) /9

2 All India Aakash Test Series for JEE (Main)-0. Answer (). Answer () h real h app μ PAT - A (PYSICS) Test - 7 (Paper-I) (Answers & ints) 0 0 dz 0 3 ln cm z Answer () The plane of the mirror and incident ray must be perpendicular to each other. 3. Answer () 80 f 0 cm Sun rays fall normally to the mirror which converges them to focus. For water surface μω μω v u μ 0 v cm μ 3 ω. Answer () f, f μ μ μ μ F eq ( μ μ) F eq. Answer () μ μ μ μ + 3 v 3( d) d 3d v 7. Answer (3) y vi x v 0 d μ 3 v 0 (0) 9 v 0 (0) cm/s 8. Answer () λ ( Z ) 0º 30º 30º 30º 0º Z 9 9. Answer (3) Z Answer () μ 3 Dλ Dλ d d sin0 3 sinr r 30 Angle between reflected and refracted ray 90. /9 80 λ λ nm

3 Test - 7 (Paper-I) (Answers & ints). Answer (3) P y All India Aakash Test Series for JEE (Main)-0 I I + I 9 net I. Answer () S 3λ S D S P S P 3λcosθ λ cosθ 3 D y y D. Answer () d mm, D m DΔx y d 3 0 Δx 3 0 θ D D + y 0 Δ x n λ For n, λ 00 nm 3. Answer () Each photon shares half the energy So E 3 MeV A0 A0 A t 0 0 Price 09 s 0. Answer () 7. Answer () 8. Answer (3) At closest approach there velocity will be equal to v. m P.u m P v + m P.v v u K e P P + P +. mu m u m v r Ke r m u P. 9. Answer (3) 0 E. ev 7 KE of emitted electrons. 3. ev. λ db 0 Å 0. Answer (3) Now, hc 3 0 λ λ m 8.37 fm. Answer (3) 90 θ n θ θ n Due to reflection from water surface, wave suffers an additional phase change of π. From Snell's law, n 3 π Δφ π n sinθ n n sin(90 θ) n 3... (i) Net Δφ π + π π Waves are reaching in phase n cosθ n 3 Squaring and adding,... (ii) n n + n 3 3/9

4 All India Aakash Test Series for JEE (Main)-0. Answer () E E ev First excitation potential 0. V. Answer (3) ΔV i v, ΔI b 0 0 A, ΔI C 0 3 A g m 3. Answer (3) d 3 Δ IC 0 0. Ω 3 Δ V 0 0 i h d h d h h h Test - 7 (Paper-I) (Answers & ints) h h 0 m 0 m Increase in height m. Answer (3) Separation f 0 + f + f e 80 + (3.) + 99 cm. Answer () λ For first diffraction minima, sinθ d λ sin30. cm. Answer () 7. Answer () 8. Answer () 9. Answer () 30. Answer () PAT - B (CEMISTY) 3. Answer () C Conc. C C Equivalent mass of benzaldehyde M + M M 3. Answer () + is most reactive due to small size of ring which increases strain in ring and make it to have low E a and high reactivity. 33. Answer (3) In reaction carbanion formation takes place which is betterly stablize in case of third iodination 3. Answer () A N (Syn) B (Major) : N : (Minor) (Anti) 3. Answer () * dil Δ (A) Loss of marked as + makes (A) aromatic. 3. Answer () C C C C X reduction LA xidation (KMn ) C /9

5 Test - 7 (Paper-I) (Answers & ints) All India Aakash Test Series for JEE (Main)-0 3. Answer () Y C reduction LA C CN dil + C CN C C 3 C 3 xidation (KMn ) C dil + C C C C (Terphthalic acid) 37. Answer () 38. Answer () 39. Answer () Br (i) Mg (ii) Dry ether (iii) (A) Formic acid is stronger acid than Benzoic acid 0. Answer () Δ eaction goes through cyclic intermediate. Answer () Cl has Cl as good leaving group hence its rate of reaction is maximum. Answer (3) dil S Since () group is not present hence it does not respond Victor Mayer test.. Answer () Gabriel phthalimide synthesis follows S N pathway hence it is fastest for C 3 Br.. Answer () ArS Cl N N S Ar N S Ar (Soluble). Answer () Electron withdrawer on diazonium salt and other strongly activated benzene ring favours the coupling reaction. 7. Answer () Quaternary ammonium hydroxide is an ionic hydroxide hence it is as strong base as K. 8. Answer () 9. Answer () 0. Answer (3). Answer () In highly acidic medium all N groups get protonated.. Answer (3) 3. Answer () If C C linkage is there than for both glucose and galactose anomeric carbon atoms are in acetal linkage hence produces non reducing sugar. /9

6 All India Aakash Test Series for JEE (Main)-0 Test - 7 (Paper-I) (Answers & ints). Answer () Carbohydrate with minimum number of carbon atoms is C. Answer (). Answer () 7. Answer () C 8. Answer () 9. Answer () 0. Answer () PAT - C (MATEMATICS). Answer (3) Mean x + y x + y... (i) Variance. / x + y (mean). x + y... (ii) From (i) and (ii) we get x, y 7. Answer () L and W can be filled at places in ways n(s) Now 3 L's and W can be arranged at places in ways, hence n (A) 7 P 3. Answer () a b Δ c d 3 Δ ad bc ad bc (, ) 0 ( ) 0 (, ) ( ) ( ) equired probability Answer () Either the number ends with (, 3) or (7, 9) So, possibilities are Probability. Answer () numbers iˆ j k 3.(3 iˆ+ jˆ) DC. Let θ D C a. Answer (3) Suppose a, b and c are the components a + b + c (3) a b c λ 3 9λ + λ + 3λ 3 λ 8 λ ±9 Since a 3λ < 0 as the line makes an obtuse angle with x-axis, λ 9 Components are 7, 8,

7 Test - 7 (Paper-I) (Answers & ints) 7. Answer () ˆ ˆ ˆ ( p i) ( p i).( p i) All India Aakash Test Series for JEE (Main) Answer () α + β + α. β ( α + β α. β) c b [where p aiˆ b jˆ c kˆ] So, the given expression is equal to ( a + b + c ) p 3 8. Answer () α. β 7. Answer () f( x) dy yf ( x) y yf '( x) f ( x) dy 0 y ( a b).( a c) ( a b). u is where u a c a.( b u) a.[ b ( a c)] a.[( b. c) a ( a. b) c] f( x) d 0 y 7. Answer () put y sin x v or, f( x) d x 0 y a.[( b. c). a] [ a. b 0] a ( b. c) Also ( a b).( a ( b c)) 0 9. Answer () equired probability 9 ut of 03, 3, 3, 03, 3, 3, 83, all are prime no except 3 Divisible by 3 Then dy dv y x+ y x sin cos So, the given equation becomes dv sin x cos x dv sin x cosx, Now integrate v sin x + c i.e. y sin x sin x + c π put y, x c 0 So, y sin x 73. Answer () Put x sinθ and y sinφ cosθ + cosφ a(sinθ sinφ) θ+φ θ φ cos cos θ+φ θ φ a cos sin 8 φ cot a sin x sin y cot a. y 0 x 7/9

8 All India Aakash Test Series for JEE (Main)-0 7. Answer () dy x sin y x sin + y dy x y cos sin y x cosec. dy cos n integrating both sides y x log tan c sin 7. Answer () f '( x) 3( f x) f '( x) 3 f( x) Integrating we have 3x + C log f( x) Put x 0 C log log f(x) log + 3 x f( x). e f(). e 3 x 7. Answer () A (A X) A B (A.X)A (A.A)x A B A A x A B ence x A A B A 77. Answer (3) xy Let m dy for required family of curves at (x, y) dy Let m for the hyperbola xy mm dy dy x x dy x 3 x y + c Test - 7 (Paper-I) (Answers & ints) 78. Answer () First if we arrange the given values in ascending order of magnitude. 7 x, x 3, x, x, x, x +, x +, x+ Median x + x x M xi M,,,,,,, MD M. ( ) 79. Answer () PA ( B) PAPB ( ). ( ) ( P(A)) P(B) ( xy ) y xy (i) PA ( B) PAPB ( ). ( ) x( y) x xy (ii) From the equation (i) and (ii) x + y 30 30y y 30y (y )(y ) 0 P(B) or 8/9

9 Test - 7 (Paper-I) (Answers & ints) 80. Answer (3) Prime numbers in to 0, 3,,, 7,, 3, 7, 9 The no's in to 00 having sum of their digits is prime to Total 37 no's Probability Answer () c b a ziˆ xkˆ 8. Answer (3) There are 8-octants, so, the probability will be Answer () y + dy y y x + c y 3 + C C 3 y 9 x + y xy 3y 3 + 9y 3y 3 + 9y xy 8. Answer () 9 All India Aakash Test Series for JEE (Main)-0 8. Answer () ( L. L )( L. L ) ( L. L )( L. L ) (. )( cos θ) L L ( L L) 8. Answer () 87. Answer (3) P(A B) is true because A and B are exhaustive But P(A B) P(A) + P(B) P(A B) is true for exhaustive events A and B. 88. Answer () a is a sphere x + y is a circle 89. Answer () Aobc (,, )& Baoc (,, ) & (0, 0, 0) A.o + B.b + C.c 0 () A.a + B.o + C.c 0 () Solving () and (), we get A B C λ bc ac ab A λ bc, B λ ac, C λab equired equation is bcx + acy abz 0 put a, b, x we get x + y z 0 Statement - & Statement - are correct and statement is correct explanation of statement Answer (3) 9/9