NARAYANA I I T / N E E T A C A D E M Y

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CODE Phase Test IV NARAYANA I I T / N E E T A C A D E M Y PHASE TEST - IV XII-NEW-REG BATCHES :: PAPER I & II :: Date: 8.0.8 PAPER I KEY PHYSICS CHEMISTRY MATHEMATICS. (7). (). (8) 4. (4) 5. (6) 6.

1 CODE Phase Test IV NARAYANA I I T / N E E T A C A D E M Y PHASE TEST - IV XII-NEW-REG BATCHES :: PAPER I & II :: Date: PAPER I KEY PHYSICS CHEMISTRY MATHEMATICS. (7). (). (8) 4. (4) 5. (6) 6. () 7. (5) 8. (9) 9. (A), (C), (D) 0. (B), (C). (D). (A), (B), (C). (A), (D) 4. (A) (B) (C) (D) 5. (A), (B) 6. (A), (C) 7. (A) (C) (D) 8. (C), (D) 9. A-Q, B-P, C-S, D-R 0. A Q, B R, C S, D P. (). (). (6) 4. () 5. () 6. (7) 7. (5) 8. (5) 9. (A), (C), (D) 0. (A), (B). (A) (B) (C), (D). (C), (D). (A), (B), (D) 4. (A) (C) 5. (B), (D) 6. (A), (C) 7. (B) 8. (A) 9. (A) 0. (A). (). (). (5) 4. (6) 5. () 6. (5) 7. (6) 8. () 9. (B), (D) 0. (B), (D). (A), (B), (C). (C), (D). (C), (D) 4. (B),(C) 5. (BD) 6. (BD) 7. (B), (C), (D) 8. (A), (B), (D) 9. (A PQS), (B P Q R), (C QR), (D QRS) 40. (A Q), (B P), (C S), (D Q,S) 4. () 4. () 4. (4) 44. () 45. () 46. () 47. () 48. (8) 49. (B) 50. (A) 5. (A), (C) 5. (A), (C) 5. (ABD) 54. (ABC) 55. (A), (B), (C) 56. (A), (B), (C) 57. (A), (D) 58. (B), (C) 59. A-Q, B-R, C-S, D-P 60. A S, B S, C-Q, D-P PAPER II KEY PHYSICS CHEMISTRY MATHEMATICS. () 4. (4). (5) 4. (). () 4. () 4. () 44. () 5. (4) 45. () 6. () 46. (6) 7. (4) 47. () 8. (0) 48. (5) 9. (A), (B), (C) 49. (B) 0. (B), (D) 50. (BD). (C), (D) 5. (B), (C), (D). (B) (C) 5. (A), (C). (ABCD) 5. (A), (C) 4. (C), (D) 54. (B), (C), (D) 5. (A), (B), (D) 55. (A) (B), (C) (D) 6. (A), (B), (C) 56. (B), (C), (D) 7. (B) 57. (A) 8. (D) 58. (B) 9. (A) 59. (D) 40. (A) 60. (B)

2 . (7) n T n T v 40 l l n T 40 0x l Solving x = 7. () Given that P = av b or PV b = a HINTS & SOLUTION PAPER - I PHYSICS (PAPER I) Comparing with PV r = constant, we have r = b R We know that C= C v r R R R Here, C = 0 = C v O or b b P 0.0. (8) P 0.0 or 4. (4) The bob will execute SHM about the stationary axis passing through AB. l' l T ; Where ' ;g ' g cos g ' sin 5. (6) Initially ice will absorb heat to raise it's temperature to 0 o C then it's melting takes place If m = Initial mass of ice, m' = Mass of ice that melts and m w = Initial mass of water By Law of mixture Heat gain by ice = Heat loss by water m c ( 0) m' L = mwc w [0] 0.5(0) m ' 80 = 5 0 m ' = kg So final mass of water = Initial mass of water + Mass of ice that melts = 5 + = 6 kg. 6. () No beats are observed by motorcyclist. So apparent frequency observed by him must be equal,

3 =. f = f 0 v 0 v v = m/s N = 7. (5) In mixture of gases in average translational kinetic energy of each molecule is same. V Helium rms m Argon V Argon m 4 n = 5 8. (9) rms Helium 9. (ACD) There should be a node at x = 0 and antinode at x = m. Also, v 00m / s. k y 0at x 0 And y A at x m. Only (a) (c) and (d) satisfy the condition. 0. (B,C) When the particle of mass m at O is pushed by y in the direction of A, the spring A will be compressed by y while springs B and C will be stretched by, so that the total restoring force on the mass m along OA, where Extension in A and (B and C) are y and respectively and related by. So, choice () and () are correct and choice () and (4) are wrong.

4 . (D). (ABC) From For st resonance, so length of water column is 95 cm. For nd resonance,, so length of water column is (0-75) cm = 45 cm rd resonance wont be established as for that the length of air column required is, length of tube. Separation between consecutive nodes is, 4 m 4 m m 4 m. (A), (D) T k and T k k T k T When these two springs are connected in series and the same mass m is attached at lower end and then for series combination k k k By substituting the values of k,k T T T 4 m 4 m 4 m

5 Time period of the system T T T When these two springs are connected in parallel and the same mass m is attached at lower end and then for parallel combination k k k By substituting the values of k,k Time period of the system T T 4 m T T T T 4 4. (A), (B), (C), (D) y 0 sin 60t x 4 m 4 m T T 4 We have A = 0 m, = 60 rad/s and k = m - v = /k = 0 m/s ; f = / = 0/ Hz and = /k = (m) The sign of (60t) and (x) are of same sign, and thus wave should travel in negative x- direction. 5. x a b c (A), (B) Maximum relative error in x is x a b c x Hence 00% %, Hence options (A) and (B) are correct. x 6. (A), (C) N Let O be the mean position and a be the acceleration at a displacement x from O. At position I, N mg = ma. II a N 0 x mg At position II, mg N = ma O For N = 0 (loss of contact), g = a = x. N g I a Loss of contact will occur for amplitude x max at the highest point of the motion. n R 7. (A), (C), (D) PV = Nrt RT T V, P 8. (C), (D) According to Wien s displacement law m T = constant. As > >, T > T > T. Total area under the graph gives total energy emitted by all wavelengths in unit time which is proportional to T 4 (Stefan s law). 9. (A) (Q); (B) (P); (C) (S); (D) (R) 0. (A) (Q); (B) (R); (C) (S); (D) (P) CHEMISTRY (PAPER I). () Diborane possesses four BH covalent bonds and two three centred (two electrons) BHB or hydrogen bride bond. These bonds are known as banana bonds.. () Salts (NH 4 ) SO 4, FeCl and NH 4 NO give acidic solutions on hydrolysis.. (5) K w = K a K b K 0 K 50 4 w b 6 Ka 0 6. (5) (i) CH CH CH CH CH 9 mg (ii) CH CH CH CH CH CH 5

6 (iii) CH CH CH CH CH CH CH CH CH CH CH (iv) CH CH C CH (v) CH CH CH 7. (6) No. of geometrical isomers for compounds with symmetry having odd stereocentres is n n given by. 8. () Dicarboxylic acid with molecular formula C4H 4O 4 are. (C), (D) NaNO (s) NaNO (s) + O (g) H > 0 NaNO and NaNO are in solid state, changing their amount has no effect on equilibrium. Increasing temperature will favour forward reaction due to endothermic nature of reaction. Also, increasing pressure will favour backward reaction in which some O (g) will combine with NaNO (s) forming NaNO. 8. (ABD) Only compound (C) do not show geometrical isomerism. 9. (A) PQS; (B) PQR; (C) QR; (D) QRS 40. (A) Q; (B) P; (C) S; (D) Q, S Refer reactions. BH and AlCl are electron deficient. MATHEMATICS (PAPER I) 4. () cos sin a b a sin x b cos x a b cos sin cos sin 4. ( ) 6 cot cot cot 6

7 6cos sin sin sin sin 6 cos sin sin sin cos sin sin sin sin sin 4. (4) AC AC' b, B = 0 o, AB = c = 4 c a b Using cos B = ac cos 0 o 4 a 8 = a 4 a 8 0, a a 4, a a = 8.a.4 a a 4, a a a a 4aa o o Now, acsin 0 a csin 0 a a 4 sin 44. () f n cos cos 4n sin = sin n sin n = sin n n sin n sin n = sin n sin n sin n cos n cos n sin n = cot(n ) cot(n + ) 45. () sin x = and cos x = x = n + ( ) n, n Z 46. () tan = cot 5 cos 6 = 4cos cos = 0 cos = 0 or,sin cos 4 sin + sin = 0 sin + sin sin = 0, sin 0 and sin cos 0 and sin 47. () AD csin B DP Rcos BcosC r 48. (8) We have a cot A R R 49. (B) By projection formula b cos C c cos B a. cosa cosb cosc Take a b c b c a a c b a b c Now use, cos A, cos B and cosc bc ac ab C B 50. (A) bcos ccos b ( cos C) c ( cos B) b c (bcosc ccosb) (using projection formula) 7

8 b c a b c a C B k bcos ccos (where k = a + b + c, given) 5. (A), (C) From the given condition ( sin A) = ( sin B) = ( sin A) sin A sin A + = 0 sin A = /, sin A = A = /6, / But A is acute angle. Similarly, sin B = sin(/6) = ½ B = /4 5. (A), (C) sin = cos cos cos n n n So, its value can be 0 or. 5. (ABD) Let BD =x, CE = x+, AF = x+ So, BD = BF = x 8 CE = CD = x + AF = AE = x + BC = a = CD + BD = x + AC = b = AE + EC = x + AB = c = AF + FB = x + a b c s x and r s s s a s b s c s x x 4 0 x 6 x 4 x 4 (As x can t be negative integer) So, a = 9, b =, c = 0 So perimeter of ABC i) AB CA c b 0 ii) rs iii) 54. (ABC) 5 0 abc 9 0 R s A C 0 A C A 90 & C 0 0 0

9 A A B C r s atan 4Rsin sin sin s a b c bc 4R R r a. a b c 55. (ABC) r, R 4, 5 abc r s 5; R abc 80 s 4 5 r r r r 4R 6 7, r r r r r r r c 56. (ABC) cosc c 48 c Also, A 0 sin A As C 60, A 0 B 90 ac 44 4 Now, ac r s a b c a b c Now verify alternatives 57. (AD) We have, r(r r rr rr ) and r rr (s a)(s b)(s c) (s a)(s b)(s c) x r(r r r r r r )x rr r 0 It satisfied by x= So, one root is x= and other root is therefore r r r 58. (BC) We have sin C cos C sin(b C) cos(b C) [sin(b C) sin C] [cos C cos(b C)] sin(b C) cos B sin(b C)sin B sin(80 A) cos B sin(80 A)sin B sin A(cos B sin B) 9

10 sin A cos B sin B sin A sin B, is possible only when sina= and sin B 4 4 A 90 and B 45, then C 45 r, r 59. (A-Q) (B-R) (C-S) (D-P) Applying r a r b r r c 60. (A) (S); (B) (S); (C) (Q); (D) (P) (A) We have, sin x 4sin x k 0 sinx k[0 k ] Since A, B both satisfy this equation sin A sin B A B A B C 80 (A B) (B) Since the sides are positive x > or else c will be negative a x x x x x b Also, a > c. Hence, a is the greatest side, so A is the greatest angle. b c a cos A bc A 0 (C) We have, (a + b + c)(b + c a) = bc (a b c)(a b c a) bc s(s a) bc 4s(s a) bc s(s a) s(s a) bc 4 bc A cos cos0 or A 0 A (D) We have, a b c c (a b ) c c (a b ) a b 0 4 c c (a b ) (a b ) a b a b c (a b c ) a b ab cosc C ( C is acute) 4 PAPER - II PHYSICS (PAPER II) ncv n CV. ( R) CV n n 5 R R R. () cos C 0

11 . (6) 4. (m) 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 gx values in equation of trajectory. y x tan u cos 0R According to question, we have.5 R tan 4 cos 5. () (R = horizontal range) Solving this equation, we get R = m. 6. (7)

12 7. (5) Phase difference between the two waves is 90º. Amplitudes are added by vector method. 5 4 Answer is (5) Distance between the successive nodes, v T / d f f Substituting the values we get, D = 5cm 9. (A,C,D) Conceptual 0. (A, B)

13 . (A, B, C, D) Since both the gases are contained in the same vessel, temperature of both the gases is same. Average KE per molecule of a diatomic gas is 5/ KT. Hence, average KE per molecule of both the gases is same. Therefore, a is correct RT vrms V rms, Hence, M v M M rms 6 4, Hence, 'b ' is correct m m and 6M M Hence, ; Hence d is correct 8 Partially pressure exerted by a gas is RT P N V P N Hence, 8 P N Therefore, c is also correct. (C) (D). (A) (B) (D) During process A and B, pressure and volume both are decreasing. Therefore, temperature and hence internal energy of the gas will decrease T PV or VA B negative. Further, WA B is negative. In process B to C, pressure of the gas is constant while volume is increasing. Hence, temperature should increase or UB C = Positive. During C to A volume is constant while pressure is increasing. Therefore, temperature and hence internal energy of the gas should increase of UC A = Positive. During process CAB, volume of the gas is decreasing. Hence, work done by the gas is negative.

4. (A, C) 5. (B) (D) (A) p-v graph is not rectangular hyperbola. Therefore, process A-B is not isothermal. (B) In process BCD, product of pv (therefore temperature and internal energy) is decreasing.

14 4. (A, C) 5. (B) (D) (A) p-v graph is not rectangular hyperbola. Therefore, process A-B is not isothermal. (B) In process BCD, product of pv (therefore temperature and internal energy) is decreasing. Further, volume is decreasing. Hence, work done is also negative. Hence Q will be negative or heat will flow out of the gas. (C) W ABC = positive (D) For clockwise cycle on p-v diagram with P on y-axis, net work done is positive. dv dv 6. (A, C) s ; F A N dh dh 7. (B) 8. (A) For the passengers in train A there is no relative motion between source and observer, as both are moving with velocity 0 m/s. Therefore, there is no change in observed frequencies and correspondingly there in no change in their intensities. Therefore, the correct option is (A). 9. (A) In one second number of maximas is called the beat frequency. Hence, 00 9 fb f f 4Hz 0. (A) Speed of wave v k 00 9 or v or 00m / s CHEMISTRY (PAPER II). () HO ion is bonded with one water molecules through hydrogen bond 9. (ABC) Compound (D) cannot show geometrical isomerism. 0. (BD) Compounds (B) and (D) has -hydrogen. Hence, they will show keto-enol tautomerism.. (CD) Only compound (C) and (D) can show cis-trans isomerism.. (BC) Unsymmetric trans-alkene and symmetric cis-alkene have dipole moment. 4. (C, D) d-block metals form metallic hydride 5. (A, B, D) Aqueous solution of base changes red litmus into blue. Na O changes red litmus into white due to its bleaching nature. MATHEMATICS (PAPER II) 4. (4) sin B sin A sin C SinA sin(a B) sin A sin A cos B cos A sin B 4

15 sin B( cos A) sin A( cos B) sin B sin A B sin A B sin A tan tan k (Suppose) cos B cos A cos A cos A k k sin A k k cos A sin A k cos A k sin(a ), where tan k k k k 4k k k 0 k k B Maximum value of tan is c b a 4. () k sin0 sin5 sin 55 4 Now, c a k sin 0 k sin 55 k (sin0 sin 55 )(sin0 sin 55 ) k sin cos cos sin k sin65 sin 55 c a ab B 4. () Since, BP BR BPR BRP 90 B 5 a c 0 C 55 b A B B PR BPsin RG BP sin cos C Similarly, QG PC sin cos B sin cos RG BP QG PC C sin cos RG BR RG CQ. QG CQ QG BR r AD AE () r BF BD 0 B B BF BD r cot r cot AD AE A A r cot r cot 5

16 45. () Applying C & D, C cos A B sin r r r s s s 6 r s a s b s c s a s b s c Minimum value = s a s b s c 46. (6) sin 6 x 0 sin x sin 6x 0 x 0,, 6 tan x tan x 0 x, x,, () sin C and C is given to be obtuse. C a b ab cosc cos 4 5 r r. s (5) let sin x a, cos x b,sin x c a b c a b c a b 0or b c 0or c a 0 sin x cos x or sin x cos x or sin x total solution 5 s s a s b s c (B) Using, a b c cos C ab x x x x x x x 6

17 x x x x x x x x x x x x x x x 0 x and x But x c is negative x is the only solution. 50. (BD) Let a straight line through the vertex P of a given PQR intersects the side QR at the point S and the circumcircle of PQR at the point T. Points P, Q, R, T are concyclic, then PS.ST=QS.SR Now, PS ST PS.ST [ AM>GM] and PS ST PS.ST QS.SR Also, SQ QR SQ.SR QR SQ.SR SQ.SR QR 4 SQ.SR QR 4. PS ST QS.SR QR 5. (BCD) a=4, b=8, 5. (AC) o C 60 a b c cosc c 4 ab x 4x 0 x x 0 x o o 0 A, B o C 0 o 5. (AC) B 60º a c sin c sin A sin A C c a 54. (BCD) sin B 7

18 X XY Y x y x y x y x y 4 cos sin sin cos 4 cos sin sin cos 4 sin cos sin cos cos sin 4sin cos 4sin cos 4 cos sin x y x y xy x y xy cos sin X 4XY Y x y A= and B=- 4 cos x x sin x cos x (BCD) cos x cos x cos x 0 Least difference = (D) 60. (B) a b 4 0 4b 4 c a a b c ********** 8

Objective Mathematics

Objective Mathematics . In BC, if angles, B, C are in geometric seq- uence with common ratio, then is : b c a (a) (c) 0 (d) 6. If the angles of a triangle are in the ratio 4 : :, then the ratio of the longest side to the perimeter