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1 PRACTICE TEST-4 KISHORE VAIGYANIK PROTSAHAN YOJANA (KVPY) 7 STREAM (SA)_ DATE : -9-7 ANSWER KEY PART- A. (C). (D) 3. (D) 4. (B) 5. (D) 6. (C) 7. (D) 8. (B) 9. (D). (B). (B). (B) 3. (A) 4. (D) 5. (C) 6. (B) 7. (A) 8. (A) 9. (D). (C). (D). (B) 3. (A) 4. (C) 5. (B) 6. (D) 7. (D) 8. (D) 9. (A) 3. (A) 3. (D) 3. (A) 33. (A) 34. (D) 35. (C) 36. (D) 37. (B) 38. (C) 39. (D) 4. (D) 4. (B) 4. (A) 43. (D) 44. (D) 45. (A) 46. (C) 47. (B) 48. (C) 49. (B) 5. (C) 5. (D) 5. (B) 53. (C) 54. (B) 55. (D) 56. (C) 57. (D) 58. (D) 59. (A) 6. (B) PART- B 6. (A) 6. (A) 63. (B) 64. (D) 65. (D) 66. (A) 67. (C) 68. (C) 69. (C) 7. (B) 7. (B) 7. (A) 73. (D) 74. (D) 75. (C) 76. (B) 77. (A) 78. (B) 79. (A) 8. (D) PAGE -
2 HINTS & SOLUTIONS PART- A MATHEMATICS. Note that this sum can be zeo only when n is of the fom 4k o 4K + 3 othewise sum is non-zeo. y x( 5, 5) 9 pais y x <3 8pais 55 pais y x <7 8pais y 3 x <7 pais 3. {D, E, F} {6, 4, } o {8, 6, 4} o {,, 4} {G, H, I, J} {7, 5, 3, } o {9, 7, 5, 3} {A, B, C} {, 8, 9} o {, 8, } o {,, } A + B + C 9 A 8, B, C 4. y k a(x b) eflecting about y k make it y k a(x b) thee (a + b + c + d + e + f) k 5. In witing fist hunded natual numbes each non-zeo digit is witten times, hence E() ( ) ( ) 5 () ()! 5.(5)! f(x) 5 5 g(x) M, N 5 P, Q 8. (a + bx + cx ) n eplacing x by x n x we get n b c a x x n x (ax + bx + c) n n x Hence coefficient of x coefficient of x n a c n PAGE -
3 9. Root the equation ae positive (as sum & poduct both ae positive) Let p, p, q N and g.c.d(p,q), be a oot of this equation q So we have ap bpq + aq p divides a and q also divides a (p, q) (, a), (a, ), (a, a) (, ) p q a o (as is not a oot) a Hence b a a + a b a + b is even (D). th n(n ) n(n ) tem of the sequence is and est all ae 's hence 34 n(n + ) 468 Hence Till fist (5) th tems thee ae 49 numbe of 's and 76 numbe of 's post this till (34) th tems all ae 's. Hence sum 49 + (85) a. S, also a S 4 ( ) (If <, S <,hence > & of couse < ).. Requied aea 4 (aea in fist quadant) {by symmety} In fist quadant equation is x + y Aea 4 Hence equied ae ( + ) 3. Given sum is coefficient of x in ( x) x 4. ab Given sum is coefficient of x in Hence if it is zeo then is odd. q a b b a (ab ) 9 ab 4 4 (x ) x (p + ) (q + ) ( + ) (pq + pq + q + p) + (p + q + ) + Hence equied (p + ) (q + ) ( + ),which is maximum when p, q, ae as close to each othe as possible two if them 3 & one is 4. Hence equied PAGE - 3
4 PHYSICS 6. Released enegy MeV. 7. Q measues acceleation of P to be zeo. Q measues velocity of P, i.e., to be constant. Hence Q obseves P to move along staight line. Fo P and Q to collide Q should obseve P to move along line PQ. Hence PQ should not otate. 8. x A cos wt (as it stats fom est at t ) A a A cos w...() A (a + b) Acosw...() Solving () and () fo A we get A a 3a b 9. Resultant Displacement y y + y fo y to be zeo (x 3t) (x + 3t 6) on solving (x 3 ) (t ) Theefoe, at x 3, esultant displacement is zeo fo all values of t.. v A cos37º v A 5 m/s and v A cos37º v B cos53º v B m/s. PAGE - 4
5 . tansitional k.e. otational k.e. mv I V Impulse eq. Fdt mv...() Fhdt mr...(3) 5 R dividing () and (3) h...(4) 5 V 5 R... () Solving () and (4) h R 5. Sping on the left of the block ae in seies, hence thee equivalent is K (K) (K) K K K Spings on the ight of the block ae in paallel, hence thee equivalent is K K + K 3K Now again both K and K ae in paallel K eq. K + K K + 3K 4K Hence, fequency is f 3. As p V T V 3 K eq M 4K M i.e. if tempeatue inceases, volume also inceases hence w.d. will be positive. 4. Fom (i) A and C both ae chaged, eithe positively o negatively. Fom (ii) Both D and E has no chage and fom (iii), A is positively chaged. Theefoe fom (i), B is negatively chaged. 5. Figue shows one of the legs of the mosquito landing upon the wate suface. Theefoe, T. a 8 W weight of the mosquito. 6. Given cicuit is equivalent to solve it can be calculated 7. It is clea fom the figue that acceleation does not change sign, i.e., does not change in diection. Only the magnitude of acceleation fist inceases and then deceases. Velocity keeps on inceasing. Hence displacement also keeps on inceasing. PAGE - 5
6 8. BE m.c [( ] u.c MeV 8.4 MeV. 9. v Ah dv v dh da dh A h t s t s A 5 /ºC. 3. Let S, L be the density of silve and liquid. Also m and V be the mass and volume of silve block. Tension in sting mg bouyant foce T S Vg L Vg ( S L ) Vg Also V m s T s S s mg (.7) N. CHEMISTRY 3. (4a + 96)g X 4 O 6 4a g X 4a g X 4 O 6 has g X 4a 96 4a 5.7 4a 96 a (B) All molecules do not have same speed thee is maxwell distibution of molecula speeds. (C) Vapou pessue is constant at constant tempeatue fo a substance. (D) Z V m,eal V,ideal m so V m. eal <.4 L m V,eal (.4L) (at STP) 34. E 3 E + E Theefoe, 3 Also, hc 3 hc hc + Thus, 3 So, 3 PAGE - 6
7 35. K.E. ev h h mev mev 36. NO sp NO 3 sp NO sp NO sp 37. A + B C + D t t eq x x + x + x ( x) ( x) x x + x x x 9 x 9 [D] + 9 [D] Li + O (g) Li O (excess) Na + O (g) Na O (excess) K + O (g) KO (excess) 39. Alcohols with fomula C4HO ae - n-butaniol o sec-butanol popan--ol 4. 3-Bomocyclohexa-,5-diene-,4-dicaboxylic acid 4. Only Ag +, Zn + and 3 Ga has the configuation s p 6 d (pseudo inet gas), Sn + has 8 + electon configuation, wheeas all othe K +, Mg +, S, O, N 3 and Cl has octet configuation. PAGE - 7
8 4. M f M V % w/v n L solute added V L M GMM solute.5..5 M %. Stength (g/l) M GMM solute g/l Mole faction of uea cannot be calculated as final density of the solution is not given. 43. Both OH and O(CH 3 ) has sp 3 hybidised "O" atom. Both OH and O(CH 3 ) has two lone pais but bond angle COC in O(CH 3 ) is geate than OH due to steic epulsion between two CH 3 goups. 45. CaC + H 3 O + Ca(OH) + C H (alkyne) Mg C + H O Mg(OH) + CH 4 Al 4 C 3 + H 3 O + Al(OH) 3 + CH 4 Be C + H 3 O+ Be(OH) + CH 4 PART- B MATHEMATICS 6. Let b n a n b n (b n ) (a n ) ( + b ) ( + b ) ( + b ).. i ( + b ) ( + b ) ( + b 4 )... n (b ) b (whee n tends to infinity) 3 6. Poduct of imaginay oots must also be intege (A) 63. y 4 x y 4 x So 4 x + z z 4 x 3 x 4 x 4 x 3 x + x 7 3 (4x x + 3 x) 3 7x (3 x) x 3x + 9 3x x 7x 4x x + 9 (x 3) x 3, y 5, z 5 3 xyz PAGE - 8
9 64. (a + b), (b + c), (c + d) ae in A.P. i.e. b + c (a + c) + (b + d) c b d (a b + c) If a + c b a, b, c, d ae in A.P. Numbe ae (34, 345,...,6789) o (357, 468, 3579) o 9 numbes If a + c b & c b d c 6 c 6 b a 5 & b + d (not possible) c 7 b a 6 & b + d 4 Numbes ae 47 c 8 b a 7 & b + d 6 Numbes ae 568, 358, 48 c 9 b a 8 & b + d 8 Numbes ae 889, 679, 469, 593 So total numbe f(x) ([k[x] k{x}] k[x]) (whee x [x] + {x}) k 4 (k[x] [k{x}] k[x]) (whee [x + m] [x] + m, m z) k 4 k [k{x}] [{x}] + [3{x}] + [4{x}] Note that as {x}, 4,, 4 3,, 3,, 3, 3, 3 4, 3, 4 Values of f(x) is,,, 4, 5, 6 Hence numbe of intege in ange is The equation of staight line AB is P mv + c () PHYSICS whee m is the slope and c is the intecept. Fo points A and B, we have and P mv + c P m(v ) + c P These equations give m and c 3 P V. nrt Now. PV nrt P Using this in eq () we have V () nr T (mv cv) T will be maximum if dt dv and d T dv. Diffeentiating eq () with espect to V and putting dt, we get dv m V + c c which gives V m T max c (3P ) 9 P V 4nRm 4nR P / V 4nR Thus, the coect choice is (A). PAGE - 9
10 67. T ma 3T a 3 mg g KE m a cos t 7 ; PE m a sin t KE PE m a (cos t sin t) m a cos t Angula fequency The time peiod T s 69. Fo sphee : T 4 S m C. Fo cube : d dt d T 4. S m.c dt d dt d dt sphee cube m sphee cube V m V [S 6a (4 )] 6 7. Toque about CM : F b. 4 I ( ) () () (g). 4 g 4I '' will be same fo all points. PAGE -
11 CHEMISTRY 7. [HS ] 8 a c K M 7. Hº xn Hº f [N H 4 ] [(94 + 8) ( )] Hº 338 kj 73. (A) º amine evolved N gas with NaNO /HCl. (B) Aomatic aldehyde does not give Fehling solution test. (C) OH gives iodofom test. (D) Since cabon is not pesent in H N NH, it does not gives lassigne's test. 74. H : s He : s, s Li : s, s, s Be : s, s, s, s B : s, s, s, s, p x p y C : s, s, s, s, p x p y N : s, s, s, s, p x p y, p z O : s, s, s, s, p z, p x p y, *p x p y F : s, s, s, s, p z, p x p y, *p x p y 75. Species lone pai on cental atom TeB : + BF : SNF 3 : XeF 3 : 3 Total Numbe of lone pai 6 N SNF 3 : F S F F PAGE -
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