NARAYANA I I T / N E E T A C A D E M Y. C o m m o n P r a c t i c e T e s t 1 3 XI-IC SPARK Date: PHYSICS CHEMISTRY MATHEMATICS
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1 . (C). (C) 3. (A) 4. (B) 5. (C) 6. (A) 7. (B) 8. (A) 9. (B) 0. (B). (D). (D) 3. (A) 4. (C) 5. (D) NARAYANA I I T / N E E T A C A D E M Y XIS-IC-IIT-SPARK (8..7) C o m m o n P r a c t i c e T e s t 3 XI-IC SPARK Date: 8..7 ANSWER PHYSICS CHEMISTRY MATHEMATICS 6. (A) 3. (C) 46. (A) 6. (C) 7. (A) 3. (B) 47. (B) 6. (A) 8. (D) 33. (A) 48. (D) 63. (D) 9. (A) 34. (A) 49. (B) 64. (D) 0. (D) 35. (C) 50. (A) 65. (A). (B) 36. (B) 5. (B) 66. (D). (A) 37. (C) 5. (C) 67. (D) 3. (D) 38. (B) 53. (B) 68. (B) 4. (C) 39. (B) 54. (B) 69. (A) 5. (C) 40. (B) 55. (B) 70. (D) 6. (B) 4. (C) 56. (A) 7. (B) 7. (D) 4. (B) 57. (C) 7. (A) 8. (D) 43. (C) 58. (B) 73. (C) 9. (A) 44. (A) 59. (A) 74. (A) 30. (D) 45. (A) 60. (C) 75. (C) (Hint & SolutionA PART A : PHYSICS 76. (A) 77. (C) 78. (B) 79. (A) 80. (A) 8. (A) 8. (A) 83. (C) 84. (A) 85. (C) 86. (A) 87. (B) 88. (A) 89. (A) 90. (A). (C) In this question, we will have to assume that temperature of enclosed air about water is constant (or pv=constant) p p0 gh p0 A 500 H p A 300 Solving these two equations, we get H=06 mm Level fall=(06 00) mm=6 mm 4. (B) From equation of continuity (Av=constant) v... (i) 4 4 Here, v is the velocity of water with which water comes out of the syringe (Horizontally). Solving Eq. (i), we get V=4 m/s The path of water after leaving the syringe will be a parabola. Substituting proper values in equation of trajectory. NARAYANA IIT/NEET ACADEMY ()
2 XIS-IC-IIT-SPARK (8..7) gx y x tan u cos According to question, we have, o.5 R tan 0 0R o 4 cos 0 (R=horizontal range) Solving this equation, we get R=m. 7. (B) Stress = F/A, F =Mg 8. (A) Strain = l / l 9. (B) Net pressure =pressure due to atmosphere 0. F Y (B) l A l. (D) P0 g h P 0 v P0 gh g h P 0 v v gh. v gh. (D) Applying Bernoulli's equation at C and D, we have P0 0 g3.6 P0 v 0 v 6 m / s Volume blown per unit time av r v 96 Similarly, at A and C, PA v g3.6.8 P0 v 0 PA N / m 5 3. (A) The meniscus between two plates is cylinidrical in shape. Pressure at A(the lowest point of the meniscus ) p p T / r A 0 Pressure at B=Pressure at C=p 0 =Pressure at A gh T T T P p gh, h r gr gd B 0 Alternative method : Force upward T cos T o 0 Gravitational pull = (Volume Density) g=lhdg T lt=lhdg h. d g 4. (C) Let P 0 be atmospheric pressure and P the pressure of air within the sealed tube. NARAYANA IIT/NEET ACADEMY ()
3 XIS-IC-IIT-SPARK (8..7) Therefore equilibrium pressure just below the meniscus should be equal to atmospheric pressure because levels of water inside and outside the tube is same. T T i.e., P p0 or P p0 r r If L=0.m is the length of tube and x the Length of immersed part, then from Boyle's law T pv pv ; P0 La P0 r L x a Where a is the cross-sectional area of tube, T i.e., P0 L P0 L x r T P0 L P0 L x L x r 4TL x P d 4T o 5. (D) Energy released n 4 a 4 b b Now, n ra b or n 3 3 a b a 3 3 Now, b 4a 4b 4b a 4 b v 3 4b / 3 3, Therefore, energy released is b a 6 or v a b. 6. (A) Let the mass of the needle be m. As the liquid surface is distorted, the surface tension forces acing on both sides of the needle make an angle, say, with vertical. Since the forces acting on the needle are F, F and mg, resolving the forces vertically for its equilibrium, we have NARAYANA IIT/NEET ACADEMY (3)
4 7. (A) F Fcos Fcos mg 0 y This gives Tl cos Then m g Fcos m where F=Tl g For m to be maximum, cos Tl Hence, mmax. g or A t H/ / dt y dy 0 H t a A a g A or t a g g y H H/ H A H or t Similarly, H a g. (ii) A t 0 / dt y dy 0 H/ a g A H or t a g.. (iii) From Eqs. (ii) and (iii), we get t or t t t 8. (D) (a) Applying Bernoulli's equation between points () and () P v gh P v gh Since, area of reservoir >> area of pipe v 0, also P P atmospheric pressure So, v g h h =.7 m/s XIS-IC-IIT-SPARK (8..7) (b) The minimum, pressure in the bend will be at A. Therefore, applying Bernoulli's equation between () and (A) P v gh PA va gh A Again, v 0 and from conservation of mass va v P P g h h v Therefore, substituting the values, we have or A A NARAYANA IIT/NEET ACADEMY (4)
5 PA XIS-IC-IIT-SPARK (8..7) N / m 9. (A) As mass of the air is conserved, n n n (as PV=nRT) P V P V PV RT RT RT As temperature is constant, P V P V PV Solving, this we get T T T 4S 4 3 4S 4 3 P0 r P0 r r 3 r 3 4S 4 3 P0 r r 3 P r r r S 4 r r r (D) v g 0 h.. (i) Component of its velocity parallel to the plane is v cos 30. Let the stream strikes the plane after time t. Then Further or or o vcot 30 t g. o o 0 vcos 30 g sin 30 T o v cot 30 x vt 3y g o v cot 30 v g 3 h gt g 3v g v cot 30 3 h g g o 3 v h g 5 v g or 50 h h h = 8.33 m. h. (B) F v A NARAYANA IIT/NEET ACADEMY (5)
6 or F Here, v is the velocity of liquid, with which it comes out of the hole. Av. (i) XIS-IC-IIT-SPARK (8..7) Further V= Ax.. (ii) V t sv.. (iii) and W = F.x.(iv) From the above four equations, V W Av A 3 V V.V. s t s t. (A) From continuity equation, va vb v0 p A gh p B 0 B p p gh (i) A Now let us make pressure equation from manometer. Hg pa g h H gh pb Putting pb pa gh we get h=0. R 3. (D) Radius of meniscus r cos P due to spherical surface r P p 4. (C) Viscous force F area So let F ka B atm 0 cos R F k A A (i) and T k A (ii) Dividing Eq.(ii) by Eq. (i) we get, T A F0 A F A A T A A 0 H H 5. (C) h Top.5H Since this point lies in the tank. So hole should be made at this point. 6. (B) Force from right hand side liquid on left hand side liquid. (i) Due to surface tension force = RT(towards right) (ii) Due to liquid pressure force NARAYANA IIT/NEET ACADEMY (6)
7 XIS-IC-IIT-SPARK (8..7) Net force is xh x 0 p gh R.x dx 0 p0rh R gh (towards left) p0rh R gh RT. 7. (D) The bubble will detach if, Buoyant force Surface tension force sin T dl T r sin 4 R 3 w g T dlsin w R g T r sin 3 r sin R 4 wr g wg Solving, r R 3T 3T 8. (D) Using geometry b b cos R R cos Using pressure equation along the path p S 0 hg p0 R Substituting the value of R, we get S S h cos Rg bg. r g T s L 9. (A) Terminal velocity v and viscous force F= 6 rv T 9 Rate of production of heat (power): as viscous force is the only dissipative force. Hence. dq Fv T 6 rv T vt 6 rvt dt r g 8g 6r s L s L r 9 7 dq or r 5 dt. 30. (D) Decrease in surface energy=heat required in vaporization. 5 NARAYANA IIT/NEET ACADEMY (7)
8 T ds Ldm T 4 rdr L4 r dr T r. L PART B : CHEMISTRY 33. (A) (i) OH show +R effect (ii) OH show i effect (iii) NO show R effect (iv) NO show i effect 37. (C) It has more no. of bonds than other. 39. (B) All are acceptor i power - NO CN F 40. (B) in (ii) second is aromatic while first is antiaromatic 4. (C) electron participated in conjugation (resonance) are counted. 4. (B) Nitrogen with double bond is SP hybridized. 43. (C) SP 3 nitrogen is more basic than SP nitrogen. 50. (A) 3 8 XIS-IC-IIT-SPARK (8..7) (C) 3 CH C H 5 6. (C) Let P x, y be the point of contact PART C : MATHEMATICS dy dy y 4a and x 4a dx dx. For the tangency of curves, 4a y required locus. 6. (A) Locus of (h, k) is y a b x a b Which is equation of parabola. y 63. (D) Solve x, x y We get y Put in y ax a a a x 0 a a 64. (D) mm tt 4Equation of PQ is yt t x att 0 y = 0 x= 4a t t t t 65. (A) tan, tan tt tt t t 0 4ttor t t 66. (D) h k 4h tan or a a a, tan h a k 4ah x 4a xy 4a, which is the NARAYANA IIT/NEET ACADEMY (8)
9 XIS-IC-IIT-SPARK (8..7) 67. (D) 68. (B) 69. (A) Circumcircle of ABC passes through focus Putting y = 0, in equation of circumcircle, we get either,0 of the given parabola. 4k 4 = or 4k 4 4k 4 3 NARAYANA IIT/NEET ACADEMY (9)
10 XIS-IC-IIT-SPARK (8..7) 70. k t t (D) Slope of AB slope of BC = - k 4 t 4 t 4 7. (B) Slopes m,3m4m ;3m, Eliminates m we have (A) All the circles described on focal chords of a parabola contains same constant term hence the common chords will pass through origin. 73. (C) Family of lines passes thorough focus hence latus rectum will makes shortest intercept. p 74. (A) Solving the equation x y p and y px we get 3p p 3p x px 0 x or x 4 3p But x is not acceptable hence the required points are p p, p or, p 75. (C) 76. (A) r or r 3 and r or r 5 r or r 5 Also r r 6 r 6r 5 r 5 x y... i (C) Given ellipse is Equation of a circle centered at (,0) can be written as x y r... ii The abscissa of the intersection points of the circle and the ellipse is given by the equation 6 x x r 4 4 x x 6 x 4r, i.e. 3x 3y 6x 8 0 i.e. x y 3 x y (B) 4 / 4 PA PB a 4 NARAYANA IIT/NEET ACADEMY (0)
11 XIS-IC-IIT-SPARK (8..7) 79. (A) 80. (A) NARAYANA IIT/NEET ACADEMY ()
12 XIS-IC-IIT-SPARK (8..7) 8. (A) 8. (A) 83. (C) NARAYANA IIT/NEET ACADEMY ()
13 XIS-IC-IIT-SPARK (8..7) 84. (A) 85. (C) 86. (A) 5 ab a5a6 5 3,e, a 0, b NARAYANA IIT/NEET ACADEMY (3)
14 XIS-IC-IIT-SPARK (8..7) 87. (B) 88. (A) NARAYANA IIT/NEET ACADEMY (4)
15 XIS-IC-IIT-SPARK (8..7) 89. (A) 90. (A) The given ratio is b a b 3 3 Now e e a 4 4 NARAYANA IIT/NEET ACADEMY (5)
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