KISHORE VAIGYANIK PROTSAHAN YOJANA Date : Duration : 3 Hours Max. Marks : 100 STREAM - SA

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1 KISHORE VAIGYANIK PROTSAHAN YOJANA Date : Duration : 3 Hours Max. Marks : 100 STREAM - SA GENERAL INSTRUCTIONS The Test Booklet consists of 80 questions. There are Two parts in the question paper. The distribution of marks subjectwise in each part is as under for each correct response. MARKING SCHEME : PART-I : MATHEMATICS Question No. 1 to 15 consist of ONE (1) mark for each correct response. PHYSICS Question No. 16 to 30 consist of ONE (1) mark for each correct response. CHEMISTRY Question No. 31 to 45 consist of ONE (1) mark for each correct response. BIOLOGY Question No. 46 to 60 consist of ONE (1) mark for each correct response. PART-II : MATHEMATICS Question No. 61 to 65 consist of TWO () marks for each correct response. PHYSICS Question No. 66 to 70 consist of TWO () marks for each correct response. CHEMISTRY Question No. 71 to 75 consist of TWO () marks for each correct response. BIOLOGY Question No. 76 to 80 consist of TWO () marks for each correct response.

2 1. r be a root r + r (1) now (r+) (r+3) (r+4) (r+5) (r + 5r +6) (r + 9r + 0) (3r) (7r + 14) using (i) 1 (r + r) -16 using (i) PART-I One Mark Questions MATHEMATICS. Given f(x) + (x+ 1 ) f (1-x) 1...(1) but x 0 f (0) + 1 f (1) 1 f (0) + f (1)...() put x 1 in (1) f (1) + 3 f (0) 1 f (1) +3 f (0)...(3) Solving () & (3) we have F(0) & f (1) - f(0) + f (1) ( n) n(n+ 1) n nn ( + 1)(n+ 1) 6 n(n+ 1) n + 1 1n + 6n 1( n 1) + 6( n+ 1) + 6 n + 1 n + 1 I + 6 n + 1 If the given terms is an intiger, then n 1,,5 Sum 8 6 must be an integer n X ab or x 10 a + b y ba or y 10 b + a Now x - y (10a +b) - (10 b + a) 99 (a - b ) 3 x 11(a + b) (a b) (1) According of Q (a + b)(a b) 11 and a b 1 a + b 11 and a b 1 PAGE -

3 a 6, b 5 Hence x 65 y 56 and m 33 x + y + m HCF x 1 p(x) x 5x + a x 5x + 4 (x 1) (x 4)...(1) and q(x) x 3x + b x 3x + (x 1) (x )...() k(x) (x 1) (x ) (x 4) Hence (x 1) + R(x) (x 1) + (x 1) (x ) (x ) (x 4) (x 1) (x 3) Hence sum of roots C B A O area of sector r 1 π π OACB θ 3 6 area of shaded region π area of 6 OAB π Hence area of line Area of semi-circle area of shaded region 1 1 π 3 π π π π. 4 4 AI b+ c Q... (1) IF a BI a+ c 3 Q...() ID b PAGE - 3

4 Hence 9. CI a+ c Q IE c 1 a+ b c...(3) () a + c 3b using to a + a + b 3b using (3) 3a b 3 b a...(4) Now again (3) c a+ b 3 a+ a 5 c + a a+ a AI b+ c 4 11 IF a a 4 B 10. RP RA 10 α RS 10 α + β... (1) Also SQ SD 10 β RS 10 β + α...() (1) and () α β, Hence RS AOB is equilatrual ( AOB OAB OBA 60 ). OBC is right angled isosceles ( OBC 90 ) 3. ABC is isosceles ( BAC BCA 15 ) 4. OAC 60 CAB 45 PAGE - 4

5 5. AOF is right angled isosceles ( AOF 90, OFA 45 ) 6. BOF 90 AOB OBC is right angled isosceles ( BOC 45 ) BOF 30 BOC Let total seats 100 on first day, Ticket price 00 sneots ful 60% Revenue R On second day Tricked price 00 0% of Scents full % of Revenue R % Increase is Revenue % 1. year Population x According to Q x - 39 k (60) & 63 kr 63 x x x x x 84 & - 40 Ans(B) 39 (60)(63) 0 R R1 100 R 1 PAGE - 5

6 13. N ab ab ab 1< a 9 0 < b 9 ab, I N 10 5 a b a + 10 b + 10a + b ( )(10 a b) ( )( )(10 a b) (10 a+ b)... (1) then 10a + b P 1 P p 1,p prime and 10 10a+ b 99 a b 10a + b Ans(C) 14. Let house no are αα,, +, α + 4, α + 6, α + 8, α + 10,... α + 10 a α a (1) House no. will be (+) α a 10 > 0 α > 10 α 1 as a is each too...() n Now S [ α + ( n 1) d] n n ( 1)() n( α + ( n 1)) [ α n ] na ( 10+ n 1) na ( 11 + n) n + na ( 11) (11 ) ( 11) 680 a ± a + n...(3) Q n 6 (11 ) ( 11) 680 a ± a a...(4) 4 From () and (4) 1 a 3 Now checking through (3) for a 1, 14,...; we have a 18, n 10 and S n 170 Hence options Ans(C) PAGE - 6

7 a + 840a + 10a + 4a + 7a + a a + 840a + 10a + 4a + 7a + a Let a a3 a4 1 a 5 0 a 6 4 a 7 Ans(B) slope is increasing at point R No Bnoycncy force in vaccum PHYSICS T F mg F tan θ ( F same) mg 1 tanθα m m m Case-1 v Case- gh U + ke w f B H 45 1 gh mgh+ m µ mgh 9 8 µ 9 PAGE - 7

8 For min deviatom i e 60 A r r 1 A r 1 r S t 3. For minimum devation i should lie between 40 to F F 0.5 divorcing lens 4. In option B it will not move, in option C & D path will be straight line. 5. kq µ i E d Q / Q Q µ f kq k( Q) k Q + + d d d kq E d PAGE - 8

9 6. Useing lenz's law upper face first become North pole then south pole S N anticolume In SHM particle comes times at every position in 1 oscillation, so actual histogram may be option (A) but since at it random snap shots so it should be option (D) CHEMISTRY CO & N are isoelectronic Molecule No.of Electron CO N HYDRAZINE N H H N H LP BP 5 N Cs ( ) + O( g) CO( g) moles 1mole 1mole 1mole weight 1gm 3gm 44gm 1gm of C require 1 mole of O H H.4gm of C will recquire mole of O volume of.4/1 mole O at STP 4.48 litre litre PAGE - 9

10 34. Nonpolar substance will have high R f value as solvent is nonpolar therefore option (A) will have high R f value as it have low dipole moment r n RH n Z r H e + 53 n Z approx NH4 Cl acidic Salt ( PH < 7) NaCl Neutral Salt (PH 7) CH3COONa Basic salt (PH > 7) 39. average speed α 1 M V V He O 3 4 M M O He NH Cl NaNO NaCl N HO CH3 O CH CH CH3 CH3 CH O CH CH3 CH 3 CH 3 O CH CH 3 PAGE - 10

11 Alkaline KMnO 4 (Syn) Oxidation OH OH 44. I,II & IV compound form H bond III do not form H Bond G RTlnK eq 45. As we move from left to right in period ionisaton energy increases. BIOLOGY 46. (B) 47. (A) 48. (B) 49. (D) 50. (B) 51. (D) 5. (C) 53 (A) 54. (B) 55. (A) 56. (D) 57. (B) 58. (B) 59. (C) 60. (D) 61. a + b + c 0, a,b,c R 0 a +b +c + (ab + bc +ca) 0 q a +b +c, r a 4 + b 4 +c 4 r q - (a b +b c +c a ) r q -[(ab + bc + ca) - abc(a + b + c)] r q - (q / 4) r q / ANS - B Similarly & n 0 + ANS - A PART-II Two Mark Questions MATHEMATICS 63. x + y a...(1) x y + 4 x 1 y 1 x a - y put in equation ()...() PAGE - 11

12 a y y + 4 a y 1 y 1 (a -y) (y-1) + y (a-y-1) 4 (a-y-1) (y-1) y (-a) +y(a -a) + 4a - 4 -a 0 D 0 (a - a) - 4( - a)(4a - 4 -a ) 0 (a - a) - 4(a - )(a - 4a + 4) 0 a (a - ) - 4(a - )(a - ) 0 (a - ) [a - 4a + 8] 0 {D < 0 +ive} (a - ) 0 ( ) a R Q a [1,014] ANS -D A ( 0, 5) P 64. C ( 0,0) B ( 7,0) Equation of line AB is x y α Let P [ α, 5(1 )] 7 on solving 16(PA) 9(PB) P[ 7, 5 ] 3 3 Let BP : PC λ :1 then λ BP:PC : 1 ANS - (A) 65. abc + ab + bc + ca + a + b + c 9 ab (c + 1) +b (c + 1) + a(c + 1) + c (a + 1) (b +1) (c+1) 30 9 a 1 0 b, c 9 a,b,c I Then no of sol. 18 ANS - (D) PAGE - 1

13 PHYSICS 66. Finaly com at p X am AX 1 1+AX A +A 1 (a-b) (a-b) a (a-b) + ba ( b)( a b+ b/) a( a-b) +( a-b) b A a a 1 0 b b a 1+ 5 b 67. Weight F 0 4πr t ρ W g + 4/3πr 3 ρ Ne g 4/3 πr 3 ρ air g t 3.5 um 68. Sol Heat lost heat gas 0.05 x 900 x ( ) 1 x 400 x (T - 30) T r i r also 45 + r > C 45 - r > C 90 > C Q µ > Ans (D) PAGE - 13

14 P Q R G R E R C R A R 1 ma 8 ma 3 ma 1 ma R R R R 13 ma 5 ma ma 34 ma X H F P B Using KCL At point A Current is 3mA At point C Current is 8 ma At point E Current is 1 ma At point G Current through GH is 34 ma V PQ V GH i R GH 34 V CHEMISTRY H( g) + O( g) HOl () 0. mole 0.1 mole 0. mole moles of gas remainting 9.7 at consiant (T) & (V) n n p p & p p NH 3+HSO4 (NH )SO 10ml M 4 4 mmol of NH3 millimole of HSO 4 0 mmol NH3 mmol of N 40 WN g % of N PAGE - 14

15 L of H produced by 0.1 eqvivalent of metal L a of H will be produced by eqvivalents 1.15 No of gram eqvivalent of metal Eqvivalent weight x x 74. x 1.16 CaO + C CaC + CO CaC + HO HC CH + Ca( OH) 75. RedHot 3HC CH Fe BIOLOGY 76. (C) 77. (C) 78. (B) 79 (A) 80. (A) PAGE - 15

V(STP) No.of mequivalents = n. Analytical chemistry

V(STP) No.of mequivalents = n. Analytical chemistry Analytical chemistry 1-qualitative analysis. Is concerned with the identification of ions molecules elements and compounds present in sample 2- Quantitative analysis :- Is concerned with the Determination