NATIONAL LEVEL SCIENCE TALENT SEARCH EXAMINATION

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1 UNIIED COUNCIL An ISO 9: 8 Certified Organisation NATIONAL LEVEL SCIENCE TALENT SEARCH EXAMINATION Mathematics. (C) or continuity at, we must have: f () / lim f () lim (+) lim (+) e / lim (+) e Paper Code: UN 4 Solutions for Class : (PCM). (B) The position of the vector of the midpoint P of the line joining the points A and B having position vectors and is given by The correct answer is (c).. (A) Put t, so that d dt and (( ) ( ) ( (+ t)) t t t t + 4 t / I sin + C / + sin ( ) + C Alternative Solution Put t ( + ), so that dt d and ( ) ( ) t t 4. (B) O a b A P B f g gf f g f g 5 ( ) andso r I sin t + C / sin ( ) +C andso (B) We have from the given equation tan (a + b) π (c + d) tan ab cd (a + b) (c + d) tan cot ab cd cd tan (c + d) ( ab) ( cd) (a +b) (c+ d) 4 ab +abcd

2 6. (C) 5. (D) Putting v y/ so that have dv dy + v, we d d (5 ) ( + ) (4 + 5) 6 9 ( ) ( + ) or. 7. (B) The value of determinant of a unit matri of any order is equal to. 8. (A) f () +, (,) f '() 9 Clearly when (, ), then < 9, i.e., 9 < f ' () < V (, ) and so f () is decreasing in the interval (, ). 9. (A) The parametric form of the given equation is t, y t. The equation of any tangent at t is t y + t. Differentiating, we get t y. Putting this value in the equation of tangent, we have y / y + (y /) 4y 4y + y The order of this equation is one.. (D) A {, y}. (B) We have The subsets of A are φ, {}, {y}, {, y} The power set of A { φ, {}, {y}, {, y}} and for n, I e d e ] e n n n I e ( ) ] n e ( ) d ( ) n ni n I () I (e ) e I I ( e) e 5 and I I (e 5) 6 e n d v d v + v v +φ(/ v) d φ(/ v) dv log C (C being constant φ( / v) of integration) But y log C is the general solution d v so log C y v φ (/ v) φ (/v) /v φ (/y) y /. (D) The given vectors are co-planar if k 9 5 5( 5k 8) 6 (5 7) +7 (4+k) 4k + k 8 4. (B) fog (fog) () () 5. (C) f {g()} f ' {g()}. g ' () f ' {g(a)}. g ' (a) f ' (b). f ' (b) The correct answer is (b). i i i i PQ i i i i i i i + i i + + i + + i + i i + + i + +.

3 6. (C) Let S be the area and l be the length of each side of square metal sheet. Then, S d s d l l. d t d t When l cm and d l cm/s, we have: dt ds dt ( ) sq.cm/s 6 sq.cm/s. 7. (A) Put tan θ so that + secθ d sec θ d θ tan a b a b sin,cos a + b a + b or a b sin,cos a + b a + b a a b tan sin ±,cos± b a + b a + b I θ θ dθ θ dθ 4 4 tan θ sin θ sec sec cos a b When sin,cos a + b a + b, + sin θ C we have : f '' () ( + ) / sec θ + C + C tan θ a a b + < a + b a + b b.. a b 8. (C) A is orthogonal A.A T I n A I n A AA T A(I n A T ) A I n A T A I n A [Q I n A (I n A) T ] (I n ) T A T I n A T ] dy du 9. (B) Putting y/ u, we have u +. d d The given differential equation can be written as u + du (u) du (u) u φ φ + d φ (u) d φ (u) This is a point of maimum. a b When sin ±,cos± a + b a + b we have : f '' () a b a b.. a + b >. a + b a + b This is a point of maimum. Maimum value of, φ (u) d du. φ(u) Intergrating, we get log φ (u) log + log k, so φ (u) k, i.e., φ (y/) k.. (A) R is Refleive since (, ) R A. R is not symmetric since (, ) R but (, ) R. R is not Transitive since (, ) R, (, ) R but (, ) R.. (D) f () a sin + b cos f ' () a cos b sin f '' () a sin b cos Now, f ' () a cos b sin a cos b sin a b f () a. + b. a + b a + b a + b Note: we get the minimum value of f () when a b sin,cos a + b a + b Minimum value of a b f () a. + b. ( a + b ) + + a b a b Thus, for a function f () a sin + b cos, the minimum value is ( a + b ) and the maimum value is ( a + b ).

4 . (C) a b a b sin θ where θ is the angle between a and b a or b or sin θ Also, a. b a b cos θ where θ is the angle between a and b a or b or cos θ Since sin θ and cos θ are simultaneously not possible for value of θ, so we have : a b a.b either a or b. (B) or continuity at, we must have f () f ( + ) f ( ). f () (a + b) (a + b) f ( + ) lim h f ( + h) lim h {b( + h) + a ( + h) + c} 4. (C) Let A be of the order m n. Since A + B is defined, so B is of the same order as A. i.e., B is of the order m n. Also, since AB is defined, so the number of columns of A number of rows of B i.e., n m. A is of order m m and B is of order m m. Hence A, B are square matrices of the same order. 5. (D) The two curves represent parabolas with vertices at (, ) and (, ). They intersect at (, ) and (, ), so that requried area is area of OPMQO (area of OPMO). O (, ) P Q M (, ) lim h {bh + (b +a) h + (b + a + c)} b + a + c f () f ( + ) (a + b) b + a + c c. Now, f() being differentiable at, we have R f ' () L f ' (), f ( + h) f () Rf () lim h h b( + h) + a(+ h) + c (a+ b) lim h h bh + (b + a)h + c lim h h orm,otherwise lim bh + (b + a) lim (b + a) h f ( h) f () L f ' () lim h h a( - h) + b - (a + b) lim h h ah - ah lim orm h -h lim (-ah+ a) a. h Rf ' ()Lf '() b+a a a b Hence, a b, c The correct answer is (B) 4 d + d. ( ) / / 4. / (D) If we denote cos by y, then since cos π y π () Also since π π / sin ( ) / π/ sin sin(y) π/ ( ) π / y π / () rom () and () we find y π / y π / 4 cos π / 4 which holds if /.

5 7. (C) R a A a R A R d A R dl 8. (C) f () 4 + π + sin 9 9π f ' () 4 + cos 8π f '' () sin Now, f ' () 4 9π + cos. (D) + t [] t [] t dt +.89 dt t t.89 + t ( ) + (.7 ).4, sgn,, >,, <, >, sgn, < π The doubtful point about continuity and differentiability is f (), At π, we have: f '' () f ( + ) lim h f ( + h) lim h h f '' π 6 + > π. f ( ) lim h f ( h) lim h ( h ) lim h h π is a point of minima. Least (minimum) value of f () π f 6 π+ 6 π π 9. (A) Let R be a relation defined on a set A containing 6 elements. Let A {a, b, c, d, e, f} Then, R must contain atleast 6 elements (ordered pairs) to be an equivalence relation. i.e., we must have R {(a, a), (b, b), (c, c), (d, d), (e, e), (f, f)}. (B) Put t or tord d t t I.89 [] t dt t [] t [] t.89 [] t + + dt dt dt t t t. (A) I f() is continuous at also. f ( + h) f () Now Rf () lim h h h lim limh h h h f ( h) f () Lf () lim h h ( h) h h lim lim h h Rf () Lf (). Hence, f() is derivable at. The correct answer is (D). + sin sin d sin cos d sin + C sin 5

6 . (D) dy / d e + 4y e e 4y e 4y dy e d 4y e e + C 4 Putting, we have /4 / C C 7/ 4y e e 7 4y 7 e 4e Hence (B) f is one-one since f( ) f( ) (B) 6. (C) +. Let y f() y y + y y ( y) + y. y Clearly, is not defined when y. (, ) has no pre-image in (, ). Hence, f is not onto. a b a b a b b c b c b c [C C C +C ] (ac b ) b ac. Now, b ac b ac a, b, c are in G.P. π π sin π / 4 sin a π / 4 8. (B) f( ) y where y f() + + y + y ( + y) y y + y f (y) y + y f (y) f() + 9. (D) We can write I a a a d a sin + a a π a [Q y f() f (y)] 4. (C) Given that (a+b) m m C A m + m C A m B+ m C A m B m C m B m... (i) Putting m in (i) we get: (A + B) C A + C A B + C B A +AB+B... (ii) But we have: (A + B) (A + B).(A + B) A +AB+BA+B... (iii) rom (ii) and (iii) we get: AB AB + BA AB BA. a ( ) a ( / ) 7. (C) s r 8r + 5 velocity v ds 6t 8 dt Now, the body will stop when velocity becoems zero. 8 4 v 6t 8 t sec sec. 6 6

7 Physics 4. (C) In air : β. mm. m; λ 6 Å 6 7 m; λ Now, β D d D β. or 7 dλ 6 In liquid: Let λ be the wavelength of light in liquid and β be the corresponding fringe width. Then, λ λ µ Now, 6. 7 λ 7 7 D. 6 β d 6..5 m.5 mm. 4. (C) requency of emitted radiation v Z V H ZH V Z Li Li v Li 9v H 9v di di 4. (A) v L 4t L dt dt L 4 i 4t dt di i A Energy stored, U Li 5 J 44. (C) Maimum magnetic field is at the surface and is given by µ π i BR i µ B, R ( / π ) ( ) A 45. (A) Dipole moment p l q 6 8 cm. Electric intensity, E NC θ Torque, τ pe sin θ 8 sin 46. (B) Here, E V; r Ω ; I.5 A Let R be the resistance of the resistor. Then, E I R+ r E or R r 7Ω I.5 Now, the terminal voltage of the battery potential difference across R IR V 47. (D) Image formed by objective (I ) is at second focus of it because objective is focussed for distant objects. Therefore, O e P I P f o 6 cm urther I should lie at first focus of eyepiece because final image is formed at infinity. P I f e Given P P 6 cm f + f e 6... () urther angular magnification is given as 5. Therefore, f f 5... () e Solving Eqs. () and (), we get f cm and f e 6 cm. 48. (A) The energy of a photon of wavelength 5 nm is hc 4 ev nm.48 ev λ 5 nm f e The energy of a photon of wavelength 7 nm is hc 4 ev nm.77 ev λ 7 nm The energy difference between the states involved in the transition should, therefore, be between.77 ev and.48 ev. n 4, E.85 ev n, E.5 ev n, E.4 ev n, E.6 ev 8 5 Nm 7

8 The given figure shows some of the energies of hydrogen states. It is clear that only those transitions which end at n may emit photons of energy between.77 ev and.48 ev. Out of these only n n falls in the proper range. The energy of the photon emitted in the transition n to n is E (.4.5) ev.9 ev. The wavelength is hc λ E 4 ev nm 654 nm..9ev 49. (C) Here, R ohm; X L 4 ohm Therefore, impedance of the LR-circuit, ( ) ( ) Z R + X + 4 L ohm R Power factor, cosφ.6 Z 5 Now, E volt; I ampere P E I cosφ av v v 5. (D) Let the potential be zero at the point P, at a distance from the charge q. q 4µC 4 6 C, q 6 C Ap, PB ( )m Potential due to q at P is V A Potential due to q at P is. V B 4πε ( ) q Total potential at P is zero q q. +. 4πε 4πε q q ( ) ( ) ( ).667 m q. 4πε E I.6 cosφ 66 watt 5. (D) The energy density is u av ε E (8.85 C /N m ) (5 N/C). 8 J/m The volume of the cylinder is V cm 5 cm 5 4 m. The energy contained in this volume is U (. 8 J/m ) (5 4 m ) 5.5 J. 5. (C) Every current carrying loop is a magnetic dipole. If it lies in the plane of paper and current in it is in clockwise direction magnetic field at all points lying within the loop is perpendicular to paper inwards and at points outside the loop magnetic field is perpendicular to paper in outward direction. or θ < 8, the centre O lies outside the loop and current is clockwise. Therefore, magnetic field is perpendicular to paper in outward direction. S i R P i Q 8 The electric potential is zero at a point distance.667 m from the 4 µc charge. 5. (B) Amount left after years A A A /5 t/t / A A 4 Probability of decay A A / 4 % A 75% (A) When current flows in any of the coils, the flu linked with the other coil will be maimum in the first case. Therefore, mutual inductance will be maimum in case (a). 55. (A) or real image u u, v u, f cm Substituting in + v u f We get u u or u cm

9 or vitual image u u, v u, f cm u u or u cm Distance between two positions of the object are u u or cm cm cm. 56. (A) hc The energy of the photon is E λ 4 ev nm 589 nm. ev. Hence, the minimum energy required to create a hole-electron pair is. ev. 57. (D) Total magnetic flu passing through whole of the X-Y plane will be zero, because magnetic lines form a closed loop. So as many lines will move in Z direction and same will return to +Z direction from the X-Y plane. 58. (B) Radius of the loop r cm. m Total flu φ. 5 Nm /C But φ E. πr 5 φ. 6 E 4.4 N / C π r.4. ( ) 59. (B) Here, work function, ω. ev J Wavelength of the incident radiation, λ 5 Å 5 m If photoelectrons are emitted with maimum velocity v ma, then mvma hv ω hc ω λ J If e is the charge on electron and V is the stopping potential, then e V mvma.5 9 J 6. (C) Here, I v.5 ma.5 A, ω rad s, R kω 4 Ω and C.5 µ.5 6 Now, X C know Ω ω C.5 6 Therefore, r.m.s. voltage across the capacitor, V C I v X C V Also, Z R XC ( ) ( ). 4 Ω 6. (C) or a given length and area of crosssection, the resistance of a material is directly proportional to its specific resistance. Since, the specific resistance is least for silver, it is the best conductor of the three given materials. 6. (C) In any nuclear reaction mass number and atomic number should remain conserved. Reaction (C) satisfies this condition. Also for 9 9 Np, neutron to proton ratio is greater than.5 which makes it unstable. 6. (A) l m, m. Kg, I A (a) The tension in the wire is zero when the magnetic force on the wire BIl is equal and opposite to its weight mg. mg. 9.8 B.96T Il (b) When the current is reversed, the magnetic force BI l and mg act vertically downwards. The tension in the wire. T BI l +mg mg N. 64. (D) λ 6 Å 6 m, D m, β.5 mm.5 m λ D 6 β.5 d. m. or V.5 e V 9.6 9

10 65. (C) C µ, C 5 µ, C 7 µ (i) The effective capacitance when connected in series C s? (ii) C C C C 5 7 s C s µ The effective capacitance when connected in parallel C p? C p C + C + C µ Chemistry 66. (A) Al + + e Al. One mole requires 965 C One millimole requires 96.5 C 9.65 A time in sec 96.5 Time s (A) The decreasing order of the boiling points of given hydrides is SbH > NH > AsH > PH 68. (A) The coordination number of the central atom or ion is determined by the number of sigma bonds between the ligands and the central atom/ion. 69. (A) 5 5 Ethyl isocyanide Ethyla min e H CHN C + HO CHNH + HCOOH Methanoic acid 7. (B) CH CH CH COOH > CH CH CH CH OH > CH CH CH CHO (a) The OH group of the COOH group in an acid is more strongly polarized due to the presence of carbonyl oygen adjacent to it. (b) Alcohol molecules can form H-bonds through their OH groups. The presence of carbonyl group in acids permits the formation of H-bonds through OH and the carbonyl oygen. Thus, a carboylic acid molecule can enter into hydrogen bonding at two different points. 7. (D) The helical structure of proteins is stabilized by NH ---- CO hydrogen bonds. 7. (A) Amount of NaCl. g 58.5 g/ mol No. of unit cells in. g of NaCl 6. mol. g g mol (A) Xe 6 has seven electron pairs (6 bonding pairs and one lone pair) and would, thus, have a distorted octahedral structure as found eperimentally in the gas phase. Xe 74. (C) Due to stabilization of COOH by resonance, electrophilicity of C-atom of C O group of COOH decreases and hence carboylic acids do not give characteristic reactions (such as cyanohydrin formation, reaction with NH OH, etc) of aldehydes/ ketones. 75. (A) According to Henry s law for a gas at different concentrations C C k, k P P C P C P Given that C.5 g/l P 58 torr, P 8 torr, C? So, C C.7 g/l 76. (A) RCl and RBr can be prepared by free radical halogenation of alkanes while R and RI cannot be prepared. With, the reaction is not only eplosive but also brings cleavage of C-C bonds while with I the reaction is too slow to be of any practical value. 77. (D) In a bessemer converter, copper pyrites are oidised to eo and Cu O. eo is slagged off. Cu O reacts with Cu S left unoidised to give Cu. Cu S + O Cu O + SO Cu O + Cu S 6 Cu + SO 78. (C) Zinc ( d ) configuration shows only + o. state. 79. (D) Ketones are less reactive because of statements (B) and (C). 8. (B) Argentite. Ag S is concentrated by leaching with NACN solution in the presence of air. Ag S + 8 NaCN + O + H O 4 Na[Ag(CN) ] + 4 NaOH+S 8. (C) Both optical and geometrical isomerisms are the types of streoisomerism.

11 8. (C) Rate k [NO] [O ] New rate k ([NO]) ([O ]) k [NO] [O ].8 New rate 8 Rate 8. (D) Low temperature and high pressure favours the conversion of SO to SO in contact process for the manufacture of H SO (C) The IUPAC name of the given compound is, -dimethyl--cycloheanol. CH HO 85. (C) HO HC 6 CH 5 CH 4CH C CH CH C 6 H 5 NH + CH COCl C 6 HNHCOCH 5 + HCl. Ace tan ilide 86. (A) The two structures are nonsuperimposable mirror images of each other and hence are enantiomers. 87. (B) As equal number of Na + and Cl ions are missing from their lattice site, it is Schottky defect. 88. (C) Since, compound A undergoes Cannizzaro reaction, it must be formaldehyde. Since, compound B undergoes iodoform test, it must be a methyl ketone, i.e., - pentanone. 89. (B) Adsorption is an eothermic process. So, it decreases with an increase in temperature. 9. (C) In H, the molecules are associated through hydrogen bonding resulting in increase of b.p. of the liquid.