NARAYANA IIT ACADEMY ANSWER KEY IITJEE MAINS MODEL

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1 CPT-08_XI (LJ) XI STUD (LJ) physics (D).. (D) (D) (D) NARAYANA IIT ACADEMY ANSWER KEY IITJEE MAINS MODEL Chemistry. 6.. (D) 6.. (D) (D) (D) (D) (D) Exam Date : Mathematics (D) (D) (D) (D) Page No.

2 CPT-08_XI (LJ) T m( a g). B Here, a=0, = + = = 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 = = 9.8N 4. vmax= µ rg 5. C f = 0 N; f = 5 N; F f f = 8 a; = 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 = = 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.

3 CPT-08_XI (LJ) 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 mg = m = = =.5kg µ µ 0.4 Page No.

4 CPT-08_XI (LJ) 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 mv + Mv = v = = = = 0.5 m / s M A At t = s, F = = 4N, µ mg = = 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

5 CPT-08_XI (LJ) 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 = = 6N f = µ m g = = 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

6 CPT-08_XI (LJ) 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= C J= F. dt 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

7 CPT-08_XI (LJ) 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 = = 0.6 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 Va = = =. 0 m 4N 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

8 CPT-08_XI (LJ) 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 = y = x Slope of line parallel to the above equation m = ( y ) ( x ) x y = 7y = 4x + 8 4x + 7y 9 = 0 + = ( x ) + = 0 x = not possible. Page No.8

9 CPT-08_XI (LJ) 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

10 CPT-08_XI (LJ) = p 4q p + 4q = + 4q + 4q = q + 7. B α, α α + = α 5 k k Also, α = = α 5 5 k = A α, 6α 4 4 α + 6α = 7α = α = p p p 8 Product α 6α = p 8 6 = p p p 4 p = = C a b c = = = λ & λ= A 4a + 5b + 6c = 0 5 a + b + c = 0 6 Comparing from ax + by + c = 0 5 x =, y = 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 +, =, Page No.0

11 CPT-08_XI (LJ) 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 = = = C a- > 0 & D < 0 Page No.

12 CPT-08_XI (LJ) 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 = 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 P = = = C π m = tan = 4 y = m x a y = x 4 x y 4 = B y = mx + c 4 = m + c (i) = m + c.(ii) Subtracting the above two equation 4m = 6 m = + c a + b =, Page No.

ANSWER KEY IITJEE MAINS MODEL

CPT-09_XI (LJ) 3..07 XI STUD (LJ) NARAYANA IIT ACADEMY ANSWER KEY IITJEE MAINS MODEL Exam Date :3--07 physics.. 3. 4. 5. 6. 7. (D) 8. 9. 0... (D) 3. 4. 5. 6. 7. (D) 8. (D) 9. 0... 3. 4. 5. 6. 7. 8. 9.