NARAYANA IIT ACADEMY

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1 Hints and Solutions : CPT (XI-IC SPARK) Exa Date XI IC SPARK ANSWER KEY EXAM DATE : Physics Cheistry Matheatics. d. d 6. b. c. c 6. d. b. a 6. c 4. b 4. b 64. c 5. b 5. d 65. c 6. d 6. d 66. d 7. c 7. d 67. c 8. a 8. d 68. c 9. a 9. a 69. a 0. d 40. a 70. c. b 4. d 7. a. a 4. b 7. a. d 4. d 7. a 4. d 44. c 74. b 5. b 45. d 75. c 6. a 46. a 76. a 7. b 47. c 77. c 8. c 48. c 78. b 9. b 49. c 79. b 0. b 50. b 80. a. a 5. d 8. b. a 5. c 8. c. c 5. b 8. c 4. c 54. c 84. b 5. a 55. b 85. a 6. b 56. c 86. b 7. a 57. a 87. a 8. c 58. b 88. d 9. b 59. b 89. b 0. c 60. c 90. c

2 Hints and Solutions : CPT (XI-IC SPARK) Exa Date HINTS & SOLUTION PHYSICS. F F = upthrust F = F + upthrust F = (p 0 + ρgh) πr + Vρg = p 0 πr + ρg(πr h + V) Most appropriate option is (4) g = lρ g... i. 4. g = ( l ) lρ g...( ii) l = 0c 00g + 97vg = 000vg and v = A A =.4 ( 0.5) ρ = ρ ( ) ρ ( ρ ρ ) = v ( ρ ρ ) V g V g V v g v 5. We have A ( + ) g. = k.x + Mg a =.4 kg tan ρ σ θ = ρ + σ. 6. P =.6 ρg P = = N/. 7. Using Pascal s law at the botto of two tubes g P0 + + ρ gh = ρ 0 + ρ g(h + h) a Solving h =. a ρ 8. Fro the F.B.D.S of the two blocks, Vρg kx Vρg 0g Vρg + kx = 0g and Vρg = kx x = 0 c N = Conceptual 0. Conceptual. (B) kx

3 Hints and Solutions : CPT (XI-IC SPARK) Exa Date For level horizontal P= hρg P A B = hρg l l PA PB = ρ a0 + ρ a0 hρ g= ρla0. (A) P A = P B (liquid is in equilibriu) xcos0ρ g + ( l x) ρ cos0 g = ( l x) ρ cos0 g + xρ cos0 g x = l/. (D) Let be ass of the liquid. Let h A, h B and h C be the heights of liquid level in three vessels. Then = ρ A h = ρ A h = ρ A h A A B B C c ρ AhA= ρbhb= ρchc= A Pressure at the botto in vessel A, pa = p0 + ρ AghA Siirly, at B and C, pb = p0 + ρbghb pc = p0 + ρc ghc i.e., pa= pb= pc So, the sae force will be exerted by the liquid in all the three vessels. 4. (D) The point A and C are in sae horizontal level Hence P C P A = ρal (refer to fig. in question to identify A, B, C) Now, P B P C = ρgh P B = (P + ρal) = ρgh P B P A = hρg + lρa 5 (B) Volue of block= l. Let h be the height of the block above the surface of ercury. Volue of ercury displace = ( l h) l. Weight of ercury displaced = ( l h) l ρ g. This is equal to the weight of the block which is ρ l g. s Thus ( l h) l ρ g = ρ sl g s h = l ρ which gives 6. (A) ρω 0 g = ρoil.5 g 0 ρoil =.5 ρ ω ρoil 0.8 ρ = ω 7. (B) ρ

Hints and Solutions : CPT (XI-IC SPARK) Exa Date 04..07 V + V 0 initial occupied vol. M ice M stone Final Volue = V + + ρ ρ ω ρ = + V g g ω 0 ice s V M ice s 0 = + ρω ρω stone ρs > ρω Final vol.

4 Hints and Solutions : CPT (XI-IC SPARK) Exa Date V + V 0 initial occupied vol. M ice M stone Final Volue = V + + ρ ρ ω ρ = + V g g ω 0 ice s V M ice s 0 = + ρω ρω stone ρs > ρω Final vol. < initial vol. 8. (C) / 4 D d 4 D d d σ π ρσ g = π ρ g ; = = D σ D σ 9. (B) Effective gravity, geff = g + a ; P = ρ ( g + a) h 0. Due to surface tension, the surface of a liquid behaves like a stretched ebrane.. Conceptual 4 4 Pr P r. Q = π π and Q 8η l = 8η l. Since the tubes are connected in series Q = Q or. 4 4 π Pr π P r = or 8η l 8η l P l r = = = P l r σ ρ a = g ρ, v = ( σ ρ) gh ρ v σ ρ H = = h g ρ 4. Vdg = V d g + V d g V = V + V 5. U= V ρ = 0.5 store water Reading = = kg 6. T = ρ vg ρvg ( ρ ρ ) = vg When lift oves upward

5 Hints and Solutions : CPT (XI-IC SPARK) Exa Date T = ρρ v g + a 7. Conceptual 8. h 000g= 600 g ρvg = ρvg Fnet = ρvg g a = Tie required 0 + at 0. df = ρ gydy d Γ = ρ gydy y Γ net = dt + F = 0 0

6 Hints and Solutions : CPT (XI-IC SPARK) Exa Date CHEMISTRY. Oxide and peroxide of etal not have unpaired electron diaagnetic And it not have vacant d-orbital colourless Super oxide have unpaired electron - paraagnetic Mg N + 6H O Mg OH + NH.. Due to presence of Aoniated electron. 4. Ca + H O Ca OH + H CaH + H O Ca OH + H o 500 C NaNO NaNO + O 5 NaNO Na O N O Burning Mg + O MgO Burning Mg + N Mg N o 800 C Evaporation of aonia dark blue solution produce M. 6NH, which further decoposes on heating. 8. BeH exists in polyeric for (c-e bonding) H H Be Be Be H H 9. Bicarbonate of Li fors only in water. 40. BeO is water insoluble but it is soluble in NaOH and HCl, NHO due to aphoteric nature. 4. CaCO + HO + CO Ca ( HCO ) Carbonated water ( Water soluble) 4. CaCO = Insoluble in water Solubility of bicarbonates of Group I increases down the group. 4. Solubility of fluorides of IIA BeF > BaF > SrF > CaF > MgF (BeF is ost soluble) NaHCO s Na CO + H O + CO 44. NaHCO + HCl NaCl + HO + CO 45. Mg does not produce colour in flae. 46. Na in liq. NH NaNH Standing + H 47. Bead produced with CaSO 4 is colourless Ca(BO ), becauses Ca + is a colourless ion. 48. It is cyclic silicone. 49. Al (SO 4 ) = acidic salt so potash alu produce acidic solution. 50. (BN) n = Inorganic graphite (Graphite like structure) (BN) n = Borazone(Diaond like structure) 5. I III M ( HO) SO 4. M( HO) ( SO4 ) 6 6 ( Alus) 5. Boron faily eleents = ns np + 5. B H NH Roo teperature BH ( NH ) [ BH ] + 6 4

7 Hints and Solutions : CPT (XI-IC SPARK) Exa Date Disproportionation 54. Ca ( OH) + Cl Ca ( OCl) + CaCl or CaOCl ( Bleaching powder) 55. ( Aq. ) HBO + NaOH Na B OH HBO + CH5OH CH5 BO + HO ethyl borate ( Green flae) 57. Al + O AlO Protective layer 58. Inert pair effect explain stability of + state in lower eleents of carbon faily. 59. Al Cl + H O [Al(H O) 6 ]Cl 60. Conceptual MATHEMATICS 6 The given point (-a, a) lies on the directrix x= a of the parabola y = 4ax. Thus, the tangents are at right angle. 6. The coordinates of the focus of the parabola y = 4ax are (a, 0). The line x-y-a=0 passes through this point. Therefore it is a focal chord of the parabola. Hence the tangents intersect right angle. 6. No of ways: C = C = = 45ways 8 y = k x, k copare with Y = 4AX k 8 8 k A =, X = x,y = y Directrix is X = A or x = 4 k k 4 8 k k x =, coparing with x= we get, = k + 4k = 0 k = 8, 4 k 4 4k x + y x y x + y (x y) = t +, = t = + 4 (x y) (x + y) + 4 = 0, since second degree ter for a perfect square it represent a parabola. 66. Least length of focal chord=length of latus rectu =9 No. of focal chords=0 67. Focus (, ) and vertex (, ) Equation of the parabola is (y + ) = 4 4(x + ) (y ) = 6(x + ) (y ) = 6(x + ) 68. Let the parabola be x = ay + by + c...() since it passes through the point (,), (,), (, ) = a + b + c a + b + c =...() = 4a + b + c 4a + b + c =...() = 9a + b + c 9a + b + c =...(4) Subtracting equation () fro equation () a + b =...(5) 5a + b =...(6) 5 On solving a =,b = and c = 0 Hence parabola is 5y + x y + 0 = Line 4x + y + K = 0 and y = 8x y = 4x K a It will intersect if C< 70. Chord is 4y=x+8 and y 4 K y x 4 K = + =,C = and 4a=8 a = K K 9 < > K > 4 = 8x

8 Hints and Solutions : CPT (XI-IC SPARK) Exa Date y 4y = Hence point of interaction are (8,8) and ( 8 y y + 64 = 0 y = 8, 8 8, Length of coon chord= = Focus (, -4) and directrix x+y+7=0 Equation of parabola is x + y + 7 (x ) + y + 4 = ( + + ) x y 7 x + y 6x + 8y + 5 = x + y x + 6y + 50 = x + y xy + 4y + 4x x + y xy 6x + y + = 0 7. Equation of tangent to It will be tangent to x y = + x x = 4x is = y if y = x + x = x + x = x + 4 x x = 0 = 4 = = Fro equation () y = x y = x 4 x + y + 4 = y = 4x is y = x + It will be tangent to x + y = x + x + + x = 8 x ( + ) + + 6x 8 = 0 7. Equation of tangent to 6 6 x + x + = 8 x + x + + x = 8 6 ( + ) x + x + 8 = 0 ( + ) x. + x + ( 6 8 ) = = 8 + = ( + )( ) 4 = = 0 + = 0 = ± Equation of tangent is y = ± x ± 6 x + y + 6 = If y = x + c is noral to the parabola Fro given condition, y = x y y = 4..x a = And x + y = k y = ( ) x + k = And c= k c = k = ( ) ( ) = 9 = 4ax. then c = a a.

9 Hints and Solutions : CPT (XI-IC SPARK) Exa Date Any tangent to y = 4x is o the for y = x +,( a = ) and this touches the circle () + 0 x + y = 9. If = + [ Centre of the circle is (, 0) and radius is ]. + = ± + + = ± = = = = ± If the tangent touches the parabola and circle above the x-axis, then slope should be positive. = and the equation is y = x + Or y = x Equation of focal chord is y 0 = ( x 4) x y 4 = 0 x 6 + y = And circle is Centre ( 6,0) and r= It will be tangent to the circle = = ±, { } = (C) The first digit can be chosen in 9 ways( other than zero), the second can be chosen in 9 ways (any digit other than the first digit), the third digit can be chosen in 9 ways (any digit other then the second digit) and so on. Hence required nuber of nubers is (n ties) = 9 n. 78. (B) 60 = 5 Now a divisor will be of the for (4n + ) if divisor is fored with the help of (4n + ) type nuber or by (4n + ) types nuber taken even ties. Hence divisors are, 5,,, 5, 5, i.e., (B) First, 6 distinct digits can be selected in 0 C 6 ways. Now the position of sallest digit in the is fixed i.e. position 4. Of the reaining 5 digits, two digits can be selected in 5 C ways. These two digits can be placed to the right of 4 th position in one way only. The reaining three digits to the left of 4 th position are in the required order autoatically. So n(s) = 0 C 6 5 C = 0 0 = (A) Here x x x = 5 Let nuber of two s given to each of x, x, x be a, b, c. Then a + b + c =, a, b, c 0 The nuber of integral solutions of this equations is equal to coefficient of x in ( x) - i.e. 4 C i.e. the available two s can be distributed aong x, x and x in 4 C = 6 ways. Siilarly, the available three can be distributed aong x, x, x in C = ways. (= coefficient of x in ( x) - ) Total nuber of ways = 4 C C C = 6 = 54 ways. 9! 8. (B) Total nuber of perutations =!

10 Hints and Solutions : CPT (XI-IC SPARK) Exa Date Nuber of those containing HIN = 7! 7! Nuber of those containing DUS =! Nuber of those containing TAN = 7! Nuber of those containing HIN and DUS = 5! Nuber of those containing HIN and TAN = 5! Nuber of those containing TAN and DUS = 5! Nuber of those containing HIN, DUS and TAN =! 9! 7! Required nuber = 7! + 7! + + 5!! = 6994.! 8. C M, T, A, H, E, I, C, S. Nuber of words in which both M are together + Nuber of words in which both T are together Nuber of words in which both T and both M are together = required nuber of words Required nuber of words = 0!!! + 0!!! 9! 5 9! + 5 9! 9! 9 9! = =.!!! 8. C Three digit odd nubers using only and only 5 are. Three digit odd nubers using, 4 and 5, are 4. Three digit odd nubers using 4, 5 and 6 are. Three digit odd nubers using two and one 6 are Three digit odd nubers using two 6 and one are. So total three digit nubers =. 84. B Total nuber of ways of posting the letters = 6 C 6 + C = (A) First key will be tried for at the ost (n-) locks. Second key will be tried for at the ost (n-) locks and so on. Thus the axiu nuber of trials needed n( n ) = (n ) + (n ) + = = n C 86. B Let n be the nuber of sides hence nuber of diagonals = n C n = 44 n( n ) n = 44 n n n = 88 n n 88 = 0 (n ) (n+8) = 0 n = 87. A 6! 70 Required nuber of ways = = = 0!! ( D)Nuber of groups having 4 boys and girl = 4 C 4. C = Nuber of groups having boys and girls = 4 C. C = Nuber of groups having boys and girls = 4 C. C = 6 So, total nuber of dolls = = (B) Let odd natural nubers be a, b, c where a, b, c are natural nubers a + b + c = 5 a + b + c = 7 a, b, c... () No. of solutions of () is coefficient of x 4 in (-x) - = 6 C = x 5 = C If all are different, then we have 4 p = 4 words. If two are sae and one different, we have = 9 words so total no of words = 4+9 =.

ELEC NCERT. 1. Which cell will measure standard electrode potential of copper electrode? (g,0.1 bar) H + (aq.,1 M) Cu 2+ (aq.

ELEC NCERT. 1. Which cell will measure standard electrode potential of copper electrode? (g,0.1 bar) H + (aq.,1 M) Cu 2+ (aq. I. Multiple Choice Questions (Type-I) 1. Which cell will easure standard electrode potential of copper electrode? Pt (s) H 2 (g,0.1 bar) H + (aq.,1 M) Cu 2+ (aq.,1m) Cu Pt(s) H 2 (g, 1 bar) H + (aq.,1