ANSWER KEY IITJEE MAINS MODEL

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1 CPT-09_XI (LJ) XI STUD (LJ) NARAYANA IIT ACADEMY ANSWER KEY IITJEE MAINS MODEL Exam Date :3--07 physics (D) (D) (D) 8. (D) Chemistry (D) (D) (D) (D) (D) (D) Mathematics (D) Page No.

2 . B MgX max = 0 + kx max. A l 3. By conservation of energy mg ( h h ) = mv where, h = 00 Increase in P.E. mg ( cos ) h 4. A = 0 V = 40 m / s 5. Tension, mv T = + mg cosθ r For T, θ= 30 and T, θ= 60 So, T >T 6. C HINTS & SOLUTION PHYSICS o = θ = cos 60 = J 3 W = Fdx = 7 x + 3x dx = 7x x x + = 35 J d s F = ma = m = 3N dt t Now, W= Fds= W= 3 dt= 3J Work done by centripetal force is always zero because force and instantaneous displacement are always perpendicular. 9. A By conservation of energy m v + m v = m u + m u v = 4 m / s KE = mv = 88J 0. C Page No.

3 ( ) Fl sin 45 = mgl cos 45 F = mg 0 0. Work done does not depend on time.. D 3. The sliding shall start at lower surface first if F > g [ ] 4. B h 8 Work done = Weight h / = mg = 4 0 = 60J 5. C m v v = =.5 0 J 6. B From v = u + at, 0 = 0 + a 0, a = m/s From s = ut + at s = = 00m 4 W= F s= ma s= 50 00= 0 J 7. (i) When block is to slide down + mg sinθ = µ mg cos θ...( i) (ii) when block is to slide up mg sinθ + µ mg cosθ = 7 ii 8. D 9. A Work done = increase in K.E. = From (i) & (ii) 4 µ = 3 3 Limiting friction = µ mg= 7.84N Static friction on Body = applied friction =.5N 0. For block a = g sin θ µ, g cosθ For block a = g sinθ µ g cosθ Since a, > a so both blocks move separately. A ar = rω = υω. C P.E. w.r.t. lowest point = Mg (L) From conservation of energy, Mv = Mg L i.e. v= gl 3. C Page No.3

4 W= F.S = 6i ˆ + j ˆ 3k ˆ. i ˆ 3j ˆ + xk ˆ i.e. 0 = ( 6 ) + ( 3) + ( 3 x) i.e. x = Rough Sheet 4. A 5. B N = F sinθ + w and F cosθ = f = µ N ( sin ) F cosθ = µ F θ + w given µ = tanφ wsinφ F = cos ( θ + φ ) 6. C 7. A 8. A m= m 0 kg / m = 0 kg Effective height h= m 3 W = mgh = 0 0 = 5000J 9. The force is constant and hence conservation W =W 30. FBD of block mg cos mg cosθ N = masinθ for N = 0 a = g cotθ 3 B 3. C Glucose is not a polymer 33. A Both NO3 and 34. D CHEMISTRY CO3 have 3 electrons and central atom in each is 35. C 36. B Increasing charge on cation polarization increases 37. D 38. B Vanderwaal s forces α molecular weight. 39. B 40. B sp hybridization. Page No.4

5 4. A 4. D Vanderwaals forces are physical forces hence weakest. 43. B Bond order increases stability increases 44. B ( = ) Li : s, s EA ve ; Be : s,s p ( EA = + ve) ; 45. C 46. D 47. B 48. A 49. D 50. C Oxygen cannot expand its octet because of non availability of d-orbitals. 5. B C N sp hybridization linear structure 5. C % ionic character = µ observe 00 µ B B. O. in O = 54. D 55. C 56. A 57. C 58. B Size of cation decreases lattice energy increases 59. B 60. B O contains one unpaired electron hence it is paramagnetic MATHEMATICS 6. C aa + bb = 3 < 0, cc = 30> 0 ax + by + c ax + by + c Use = a + b a + b 6. A 63. B Let The equation of line is Page No.5

6 x y + = a b = a b... ( i) a + b = 7... ii 64. A 65. C Algebric sum of the distances from the three non collinear points to variable line is zero then the line passing through centroid of the triangle formed by this points. 66. C ( h, k) = ( 5, ) xy + x 5y = 0 ( x + 5)( y ) + ( x + 5) 5( y ) = 0 xy = 67. C Area of an equilateral triangle is h where h is the height of the triangle B (h, k) = (, ) x = X +, y = Y + Y 8X = 0 a = 69. A perpendicular bisector of AB is x + 3y + 0 = 0... y =... Solve above equations 70. C x = a + bsecθ and y = b + a tan θ x a = sec θ and y b = tan θ b a sec θ tan θ = x a y b = b a 7. B Opposite vertex of square are 7. B ( a cos α,sin α ) &( asin α, a cosα ) Page No.6

7 73. A 74. C ( 3+ k) x + ( 4 + k) y + ( 5 3k) = 0 Coefficient of x = 0 ( 3a + ) x ( a + 3) y + 9 a = 0 ( 3x y ) a + x 3y + 9 = 0 3x y = 0 x 3y + 9 = 0 x y = = x y = = 8 7 x = 3, y = 4 ( x, y) ( 3,4) Write the image of in X-axis and write the equation through that point and ( 4,0) 75. A 76. C aa + bb < 0 a x + b y + c a x + b y + c use = a + b a + b 77. A Slope of line is equation is 7 y = ( x + ) x + 7 y = B Y = B PA = PB 80. B 8. B 8. B Let C ( λ, λ ) 8 λ = 0 λ = ± 5, λ = 7, 3 C ( 7,5 ),( 3, 5) Page No.7

8 83. A x co α + y sin α = p 84. B ( x cosα + ysin α) cos α + ( xsinα + y cos α) sin α = p x cos α 4sin α cos α + x sin α + 4sin α cos α = p x cos α + sin α = p x = p ( x + y + ) + k ( x y ) = 0 ( + k) x + ( k) y + ( k) = 0 k + m = k, m = 3 m m = 4k + = k = 5 3k B (,4 ) Letα α be the point 4α α 0 = ± 5 α = 3, α = 7, 86. C aa + bb = 4 > 0 a x + b y + c a x + b y + c = a + b a + b use 87. C x + y 3x 6y 5 = x 9 = D = k 0 k = C Verify LpL p< 0 for options 90. B Use x = Xcos θ Ysin θ y = Xsin θ + Ycosθ Page No.8

NARAYANA IIT ACADEMY ANSWER KEY IITJEE MAINS MODEL

CPT-08_XI (LJ) 06..07 XI STUD (LJ) physics... 4. 5. 6. 7. 8. 9. 0. (D).. (D). 4. 5. 6. 7. 8. 9. 0.... (D) 4. 5. 6. 7. 8. 9. 0. (D) NARAYANA IIT ACADEMY ANSWER KEY IITJEE MAINS MODEL Chemistry. 6.. (D)