XII-S (NEW) - Paper-I

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1 XII-S (NEW) / CPT /9/08/6 NARAYANA I I T / P M T A C A D E M Y CPT T EST 5 XII-S (NEW) - Paper-I (JEE ADVANCE MDEL 03-P) ANSWER KEY PHYSICS CHEMISTRY MATHEMATICS Q.N. KEY Q.N. KEY Q.N. KEY. (A) (B) 4 (B). (A) (C) 4 (A) 3. (D) 3 (C) 43 (A) 4. (A) 4 (B) 44 (C) 5. (B) 5 (A) 45 (B) 6. (A) 6 (C) 46 (D) 7. (A) 7 (D) 47 (D) 8. (C) 8 (D) 48 (A) 9. (A) 9 (A) 49 (A) 0. (B) 30 (B) 50 (C). (C,D) 3 (A,C,) 5 (A,C). (B,C) 3 (A,C,D) 5 (A,B,C,D) 3. (B,C) 33 (A,B,C) 53 (A,D) 4. (A,B) 34 (A,C) 54 (A,C) 5. (C,D) 35 (A,C,D) 55 (A,C)

2 . (A) f v u A. (A) Hints and Solutions PART - I : PHYSICS PHYSICS 3. (D) f R P 4. (A) f R R D 5. (B) Frequency does not change during refraction, but wavelength and speed decreases in a denser (here water) medium. 6. (A) r i and r 90 r 90 i sin i sin i Now, R R D D R D r sin i sin c D R sin i or sin c tan i cosi c sin tan i 7. (A) n sin i n3 sin i3 sin 45º sin i3 sin i sin i sin 90 i sin i3 or i3 30º 8. (C) For convex lens, v 0 This image I is therefore, ( ) cm or 40 cm towards left of plane mirror. Therefore, second image I (by the plane mirror) will be formed 40 cm behind the mirror. 9. (A) F f f 0 0 F 5cm v v 5cm v m u I m cm cm NARAYANA IIT/PMT ACADEMY

3 0. R R.5... ii 0.5 R R Here, = refractive index of medium or liquid. Dividing Eq. (i) by Eq. (ii) we get, 5 or 5.5 / 8 0. (B).5... i. (C,D) For total internal reflection to take place; Angle of incidence, I > critical angle, q c r sin I > sin q c or sin45 o > /n r n or n or n.44 Therefore, possible values of n can be.5 or.6 in the given options.. (B,C) First for lens ;V 30cm 30 V 5 For plane mirror object is real at distance 45 cm from mirror its image is at distance 45 cm behind the mirror. Again for lens image from mirror act as real object at distance 60 cm from lens. Lens makes its real image at distance 60 cm. 3. (B, C) 4. (A, B) For real image Let u x, then v x x x 0 Solving we get x = 30 cm. For virtual image, Let u = -x, then v = + x x x 0 or x = 0 cm 5. (C, D) Inverted and real image is formed by concave mirror Let u x, then v x / x 3f Erect and virtual image kis formed by convex mirror. x Let u x then v NARAYANA IIT/PMT ACADEMY 3

4 x / x f x = +f 6. 9 cm/s dy Given, 4cm / s dt Distance of bird as observed by fish 4 Z y x y x 3 dz dy 4 dx dt dt 3 dt dz Given 6cm / s dt Substituting in Eq. (i) we get, dx 9cm / s dt 7. (6) A m sin Using n A sin Given m A sin A 3 A sin A A sin cos A sin Solving this equation we get, A = 60º. D 8. (5) P p p ; M f 9. () ray passing through centre of curvature is only corret. 0. (6) NARAYANA IIT/PMT ACADEMY 4

5 Insect can see the image of source S in the mirror, so far as it remains in field of view of image overlapping with the road. Shaded portion is the field of view, which overlaps with the road upto length PQ. By geometry we can see that, PQ = 3 AB = 60 cm distance 60 t 6s speed 0 CHEMISTRY. (B) ne is oxidise and another is reduce. (C) 3. (C) 4. (B) MgBr H 5. 3 (A) i R' MgBr ii H R C H R C H R C H R' R' 6. (C) Carbocyl group +, 4-Dinistro phenyl hydrazine Coloured product 7. H _ H (D) CH 3CCH 3 CH3 C CH C CH3 CH3 Aldol 8. (D) -Hydrogen containing carbonyl compounds show aldol condensation. 9. (A) nly CH 3 CCH 3 can form carbanion (Claisen-Schmidt reaction) 30. (B) a b + H NNH C NH H CH = NNH C NH (major) Because lone pair of b takes part in resonance 3. (A, C) 3. (A, C, D) 33. (A, B, C) First three reagents produce corresponding alcohol 34. (A, C) -hydrogen containing aldehyde and ketones. 35. (A, C, D) NaHS 3 gives only nucleophilic addition reaction. 36. () 37. (3) 38. (5) 39. (5) A is cinnamic acid CH = CH - CH 6HCH 4NH CH N 6H 40. () MATHEMATICS 4. (B) Component of a perpendicular to the direction of 4. (A) R C B 0, R C B,R C B a b a a b b NARAYANA IIT/PMT ACADEMY 5

6 C.A R.A 0 A.B x y z x y z (A) x y z 0 This plane is perpendicular to plane x y z =. 0 equation of plane is 5x + 4y + z 0 = (C) Dividing with x 3. dy y 3cos x 3 3 dx x x 45. (B) a b c 0 a b c a b b c c a (D) 47. (D) k 3 0 k 5 or k = 5/ 3 k 48. (A) The lengths of edges are a = 5 = 3, b = 9 3 = 6, c = 7 5 = Length of the diagonal = 49. (A) a b 3a c a b 3c 0 a b 3c a b c 7 b 3c a b 3c a b 9 c 6 b c a 6 cos 60º (C) a b 3c a b a b 3 cos 60º 5. (A, C) The diagonals are given by AB BC 4i j 4k, AB BC i j These vectors have magnitude 6 and respectively, and their dot product is. Therefore the angle between them is 3 cos cos or (A,B,C,D) 53. (A, D) Equation of tangent at P(x, y) is NARAYANA IIT/PMT ACADEMY 6

7 dy y Y y X x, Q x, 0, dx dy dx y Area of PQ x y dy dx But area given is a y y a xy a xy dy dy dx dx dx dx x a y xy a, dy dy y y Using concept of linear differential equation P x, y Q a a x cy, x cy y y 54. (A, C) The given 55. (A, C) CNCEPTUAL 56. (7) Area AB AC (8) Plane must pass through 3 5 7,, or,3, dy dx 58. () y x NARAYANA IIT/PMT ACADEMY 7

8 log y log x log c e e e y cx it passese through (, ) c = y = x 59. () AB AC AD 0 dy cos x 60. (3) dx y sin x sin ln y sin x c y x k Now y(0) = ( + ) ( + 0) = k k = 4 Therefore (y + ) ( + sin x) = y. Hence. sin x y 3 NARAYANA IIT/PMT ACADEMY 8