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) 6.. (D) 6. 4. 64. 5. 65. 6. 66. 7. 67. 8. 68. 9. 69. 40. 70. 4. 7. 4. 7. 4. 7. 44. 74. 45. 75. 46. 76. 47. (D) 77. 48. 78. 49. 79. 50. 80. 5. (D) 8. 5. 8. 5. 8. 54. 84. 55. 85. 56. 86. 57. 87. 58. (D) 88. 59. (D) 89. 60. 90. Exam Date :06--07 Mathematics (D) (D) (D) (D) Page No.
CPT-08_XI (LJ) 06..07. T m( a g). B Here, a=0, = + = 6. + 9.8 = 7 HINTS & SOLUTION PHYSICS From F.B.D. T= M g, T N = µ, = ( + ) N M m g M Mg = µ ( M + m) g m = M µ. A Since the block is at rest, static friction acts which is equal to the applied force,i.e mg sin θ = 9.8 0.5 = 9.8N 4. vmax= µ rg 5. C f = 0 N; f = 5 N; F f f = 8 a;75 0 5 = 8a A B A B 40 = 8 a; a = 5 m / s 6. A The normal force between block B and the floor is N = 8 + g = 00N The maximum possible friction is f = µ N = 0.5 00 = 50N max As the applied force 0N is less than f max, the blocks will not move and the friction force between blocks A and B will be zero. 7. g mg= ( m) a; a= ; mg= ma; a= g 8. When lift moves up or down with uniform speed. The apparent weight=real weith W=mg 9. A Page No.
CPT-08_XI (LJ) 06..07 0 cos0 0 = 5 N 0 mg sin 0 = 5N Therefore no need of friction f=0 0. D F kx = M a kx = F M a... i kx = M a... ii F M a = M a F M a a = M. A h = g sin θ t sin θ For block A, t For block B, h = g From these equn., we have sin θ = 4sin θ sin θ = θ = 0. D Since, there is no friction between P and wall, there will be no upward force that can support its weight.. B This is a limiting case where friction between M and ground is µn and acceleration is zero. Note that there will be no normal force and hence no friction between blocks M and m. From F.B.D of block M, Mg + T = N and T + T = µ N T = µ ( Mg + T) T µ Mg = µ From F.B.D of block m, T = mg µ Mg µ M 0.4 0 mg = m = = =.5kg µ µ 0.4 Page No.
CPT-08_XI (LJ) 06..07 4. A v u = as In Case of smooth plane, v 0 = g sin θ s (i) 5. In Case of rough plane, v = θ µ θ 0 g sin cos s n On dividing () by (), we get.() = µ cot θ µ = tan θ n n mv 0 0 00 mv + Mv = v = = = = 0.5 m / s M 8 4 6. A At t = s, F = = 4N, µ mg = 0. 4 0 = 8N Since F < µ mg Friction force, f = F = 4N 7. C Mv From F.B.D, N Mg = R MV N = + Mg R 8. A The coin will just slip when mω r > µ N or, mω r > µ mg s µ sg or, r > ω s Hence, if ω is doubled, r becomes one fourth, i.e, cm Page No.4
CPT-08_XI (LJ) 06..07 9. C 0. The force acting on the bob are shown in F.B.D. From F.B.D, we have mv Tsin rg = θ v 0 And mg = Tcos θ tan θ = = = θ = 45 rg 0 0 dp dp x y Fx =, Fy =, F = Fx + Fy dt dt. = + = ; Keq = K K 4K 4K K eq. Force of friction between the two will be maximum i.e., µ mg µ mg Retardation of A is aa= = µ g and acceleration of B is a m Acceleration of B relative to A is µ g ab = a A B + aa = = g 4. B µ mg µ g = = m 4. f = µ m g = 0.6 0 = 6N f = µ m g = 0.5 0 = 0N Fnet = F F = 5 4 = N; As, Fnet < f + f The system will remain at rest and the values of friction forces on the blocks will be f = + N, f = + 0 N, f + f = N Page No.5
CPT-08_XI (LJ) 06..07 x + x = l Differentiating with respect to time, We get v +v =0 Again differentiating w.r.to to time a + a = 0 a = a, a = a ( x x ) ( x x ) + = l x + x x = l Diffetntiting w.r.t to time v +v -v =0 Again differentiating w.r.t. time a +a -a=0 a = a ; a = a 5. A mg sinθ = 5N max B A f = µ mg cosθ = 6.98 N; f > mg sinθ T = 0 max 6. ( θ µ k θ ) ( sinθ µ cosθ ) F = mg sin + cos, F = mg k tanθ tanθ F= F ; µ k= ; = ; θ= 0 7. h = R R cos θ ; µ = tanθ 8. C tan θ = µ, tanθ= θ= 60 9. C J= F. dt 4 0 0. D 0 + t dt = 0 = 6 N = W cos θ; N sinθ = T W sinθ cosθ = T or T = W = T = W 4 Page No.6
CPT-08_XI (LJ) 06..07 CHEMISTRY. D 4. C b = 4x Vol. occupied by the molecule 4 = 4 π r N A Average velocity is given as C = 8RT πm The ratio of velocities at two different temperatures for the same gas is given as C T = C T Substituting the values, the ratio is 0. 00 = = 0.6 T The temperature 0 T = 00 = 00K = 97 C 5. A RMS velocity of (C ) is given as C= Given temperature T= 66K RT RT C= and C = M M The ratio of RMS velocities RT M C TM 66 M = = = C T M 08 M RMS Speed of XY molecules = C C = = 4ms 6. B Excluded volume = b = 4 ( Va N0 ) Where V a is the actual volume of helium atom. b.9 0 7 Va = = =. 0 m 4N 0 4 6.0 0 Volume of helium atom = 4 πr Radius of helium atom, 40. C EαT 0 r =. 0 m 44. C RT P = ;[ y = mx] V b 45. A Boyle s temperature = a/rb Page No.7
CPT-08_XI (LJ) 06..07 50. C V 5. A mp = RT M In HO, hybridization is sp and the shape is angular due to bp & lp 54. C 55. C 56. C All of them have total four bond pairs with zero lp. But X e F 4 have four bond pair and two lone pair so it have square planer structure. 58. D Bp, LP MATHEMATICS 6. A OA=OB so triangle OAB is isoceles. 6. B / / x + x = 0 / Put x = t 6. (D) ax + bx + c = a(x + α)(x + β) changing x to, we get x a + b + c = a + α + β x x x x a + bx + cx = a( + αx) ( + βx) 64. B x x y + = 0 + y = 65. α β α + β + = + β α αβ α β + = β α b a π π x cos + ysin = = ( α + β ) ( b / a = ) αβ α β + 0 β α b / a α β b + + = β α a b a 66. C 67. A 4x + 7y = 9 4 9 y = x + 7 4 Slope of line parallel to the above equation m = ( y ) ( x ) 4 7 4 x y = 7y = 4x + 8 4x + 7y 9 = 0 + = 7 9 9 4 7 ( x ) + = 0 x = not possible. Page No.8
CPT-08_XI (LJ) 06..07 68. C m + n b q = = b r = q c mn c r 69. A p= 7 & p = 4q 70. A Slope of OD slope of BC = β + = = α β+ = = β+ = 4α + 4 α = 4α + β = () Slope of OE slope of AC= β = α 6 β = β = 4α 4 4 α 4α β = On solving 8α = α = Form eqn(i) 4 + β = 6+ β = 5 5 α, β, β = 5 β = 7. B x 4x + is minimum at b 4 x = = = a 7. C Page No.9
CPT-08_XI (LJ) 06..07 = p 4q p + 4q = + 4q + 4q = q + 7. B α, α α + = α 5 k k Also, α = = α 5 5 k = 5 74. A α, 6α 4 4 α + 6α = 7α = α = p p p 8 Product α 6α = p 8 6 = p p p 4 p = = 8 75. C a b c = = = λ & λ= 5 76. A 4a + 5b + 6c = 0 5 a + b + c = 0 6 Comparing from ax + by + c = 0 5 x =, y = 6 77. A α = β α + β + γ = p γ = p Since γ is a root of the given equation, so it satisfies the equation i.e. γ pγ + qγ r = 0 p p + pq r = 0 r = pq 78. D Since triangle is equilateral so In centre and centroid coincide In centre= + 0 +, + 0 + 0 =, Page No.0
CPT-08_XI (LJ) 06..07 79. C The quadratic equation: x + αx + β = 0.(i) and x + px + q = 0 (ii) Let m be the common root of the equations. m m q β = = m = αq βq q β p α α p 80. C p, x + y = 0 x y + = = + x = and y = x =, y = ( x, y) = (, ) 8. D Since given points from a right angled triangle Orthocenter (l,m) 8. B Given: first quadratic equation: x 5x + 6 = 0 and its roots = α and β Second quadratic equation: x + px + q = 0 and its roots = ( α + β ) and α + β = 5, αβ = 6, α + β = 7 αβ 8. B Quadratic is positive always, so discriminant < 0 4a 4 0 a < 0 a + a 0 < 0 a + 5 a < 0 5 < a < 84. C 8x + 6y + 5 = 0 5 4x + y + = 0 4x + y 5 = 0 Distance b/w parallel lines 5 + 5 55 = = = 6 + 9 5 85. C a- > 0 & D < 0 Page No.
CPT-08_XI (LJ) 06..07 86. B ( x ) (, y =,a, x, y =,b, x, y c, ) x + x + x y + y + y ( x, y ) =, on x axis y = 0 a + b = 0 87. D f x = x + bx + b b + c = x b + c b c b min f ( x) = c b g ( x) = x cx c + c + b = x + c + c + b c + b max g ( x) = c + b Now min f ( x) > max g ( x) c b > c + b c > b c > b 88. A + 5 7 7 5 P = = = 44 + 5 89. C π m = tan = 4 y = m x a y = x 4 x y 4 = 0 90. B y = mx + c 4 = m + c (i) = m + c.(ii) Subtracting the above two equation 4m = 6 m = + c a + b =, Page No.