Vidyamandir Classes SOLUTIONS JEE Entrance Examination - Advanced Paper-2

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1 SOLUTIONS 8-JEE Entrance Eamination - Advanced Paper- PART-I PHYSICS.(ABD).(ACD) K dk t dk t dt dt t t t K mv v t v t ( B is correct) m dv a F constant (A is correct) dt m As the force is constant, its work depends only on net displacement and not on path. So it is conservative (D is correct) d t d t dt dt m m t Viscous drag force between layers is t t m dv u Fd A A d h This is also the force eerted by liquid on the plate and the floor of tank. ACD are correct (C is incorrect) 3.(AB) Flu through shell is enc (Rsin 6 ) 3R (A is correct) Field due to infinite charged wire is perpendicular to the wire E at all points (B is correct) z The shell is non-conducting (D is incorrect) 4.(D) For AC: v f ( f ) v f VMC Paper- JEE Entrance Eam-8 Advanced

2 v ( f ) f m hi () u f f For DE: v ( f ) f ( ) v f f v f ( f ) / f f m hi ( ) f u ( f ) y-coordinate of image of line BC is same for all points The same can be seen from ray diagram also Note: We have used paraial rays approimation despite the object not being very small. 5.(AC) 6.(ABC or AC) 7.(6.3 or 6.3) Th Pb N He N e Conserving mass number: 3 N (4) N 5 (A is correct) Conserving atomic number: 9 8 N () N ( ) N (C is correct) Note: anti-neutrino will also be emitted along with particles. v 5 (n ) (.57 e) (n ) v 4(.57 e) v 5 (n 3) (.839 e) (n 3) v 4(.839 e) (ii) - (i) ( ) v v (.33) 33m/s (A is correct) v m 66.4 cm (C is correct) f 5 Neglecting end correction to find the mode for length 5.7 cm. (i). (ii) (33) 5 (n ) n 3 (first overtone) (D is incorrect) 4(.57) 3(33) For e: 5 e.9m 4(.57 e).9 cm (The absolute value of end correction may be considered as.9) I I mv v.5 m/s m.4 / t v v e d v e dt t t/ d v e dt / t t/ t [ e ] t / v v ( e ) ( / ) At t, v ( e ) 4(.5)(.37) = 6.3 m 8.(3 or 3. or 3.) VMC Paper- JEE Entrance Eam-8 Advanced

3 u sin (45 ) () u 48 vy After collision: tan 3 v 3v v Also 48 mv mu v 4 y v v 4 y y y 3v v 4 v 6 H 9.( or. or.) v y 6 3 g m F qe qe sin t qe a sin t m v t qe dv m sin t dt qe v ( cos t) m.(5.55 or 5.56) v y qe cost m vma 3 3 t qe ()() m/s m ( )( ) If the maimum deflection in galvanometer is for current I g, then torque on coil for this current is equal to torsional torque due to twist of. rad. ( NIg A)( B) C For conversion into ammeter of range - A : I g 4 ( )(.).A 4 (5)( )(.) (.) R (.9) R R s g S (3 or 3. or 3.) Elongation in the steel wire due to additional load is Mg (.)()() 4 l m mm AY 6 3. ( ) 4 For Vernier calliper : VSD 9MSD VSD.9 MSD ) Least count MSD VSD.MSD.mm The reading of Vernier calliper will be.3 mm. So the 3 rd Vernier scale division will coincide with a main scale division..(9 or 9. or 9.) r r /3 /3 /3 T TV TV TV T (8 V ) T 4V T 5K 4 3 U ncvt () 8 (5 ) 9J Decrease in internal energy 9J VMC Paper- 3 JEE Entrance Eam-8 Advanced

4 3.(4 or 4. or 4.) Since the frequency of light is just above the threshold frequency, so energy of incident photons is only slightly greater than work function E 6.5eV No. of incident photons per second: n P E Emission efficiency is % Number of emitted electrons = no. of incident photons (per second) = n Momentum with which an electron strikes the anode mk mev Electrons strike normally and are absorbed Change in momentum mev 4.(3 or 3. or 3.) Force = momentum change per second = (Number of electrons striking per second change in momentum of electron) P 3 9 mev (9 )(.6 )(5) 9 E N Z Z Z Z (B) Z Z Z 3 For point charge: E kq d. P is a point on ais of dipole 3. For infinite line charge: 4. E P E k d k k l k d l d l d l 5. For infinite sheet of charge E kl d The correct match is: P-5, Q-3, R-,4, S- d l kp k ( Q l E ) 3 3 d d 6.(B) For a satellite is an orbit of radius R, m GM mv v GM R R R GM L mvr m R m GMR R v v R 4 R L m R L m R 4 GM K K mv m m R (4) 8 R K m R VMC Paper- 4 JEE Entrance Eam-8 Advanced

5 R R T R R v GM GM R The correct match is: P-3, Q-, R-4, S- T T 3/ 3/ R R (C) P-V curve of adiabatic process is steeper than isothermal process. So, the processes corresponding to the graph are I. Adiabatic II. Isobaric III. Isochoric IV. Isothermal So correct match is: P-3, Q-4, R-, S- 8.(A) P. r t iˆ t ˆj v iˆ ˆj a F P-,,3,4,5 Q. r cost iˆ sin t ˆj v sin t iˆ cos t ˆj ˆ a cost i sin t ˆj a r force is towards centre 3 a v ( ) sin t cost Work is done by the force and kinetic energy of particle will not be conserved. Q-,5 R. r cost iˆ sin t ˆj S. Uniform circular motion R-,3,4,5 ˆ r t i t ˆj v iˆ t ˆj a ˆj force is not passing through origin S-5 PART-II CHEMISTRY [Co en NH H O].(ABD) If en is replaced by CN we will get Co CN NH 3 H O VMC Paper- 5 JEE Entrance Eam-8 Advanced

6 .(BD) 3.(D) 3 COenNH3 HO As en is strong ligand so pairing will occurs and hence it will NOT be paramagnetic. Also as NH 3 is strong ligand then H O. So it will absorb longer wavelength. Both Cu salts characteristic green colour in flame test Cu will be precipitated by passing HS gas in acidic medium while precipitated by passing HS in alk. medium. the E Cu and M / M Mn will be Mn will be precipitated on passing HS gas in basic medium because metals value is positive only for Cu. Cu / Cu Mn / Mn E.34V, E.8V 4.(D) For L - Glucopyramose VMC Paper- 6 JEE Entrance Eam-8 Advanced

7 For -anomer, both anomeric OH and CHOH should be on same side For L-sugar, CHOH group should be on downward side. 5.(AD) Ag Bg Cg First order reaction A g B g C g t A t t A A Pt (P P ) t (Pt P ) A A ln kt ln kt A A t ln P P P t kt P t P ln kt 3P P ln P ln 3P P kt t ln 3Po Pt ln Po kt (so slope k (-ve)) y C m t/ 3 n3; t / 3 is independent of [A] ; k is constant. K 6.(AC) As the concentration of product decreases by increasing temperature H. as given As we know G RT ln K ln k ln k T T Also and T T P A G RT ln K G RT ln K G G G RT ln k RT ln k eq eq K as Keq T can't be -ve VMC Paper- 7 JEE Entrance Eam-8 Advanced

8 7.(6.) ln K G T ln k G ln K G T ln k G T T G G Since G ln k as ln k T T So, G G because G increases on increase in temperature as G on increases Temp. G H T S and H So S 8.(6.47) PbS 3O PbO SO 3PbS 3O 3Pb 3SO PbS PbO 3Pb SO mole of O mole of Pb w 3 7 w gm 6.47 kg 3 9.(6.) Eq. of MnCl Eq.of oalic acid.(7.) w w 6 mg 9 5 VMC Paper- 8 JEE Entrance Eam-8 Advanced

9 .(495.) C6H5COCH3 3NaOBr C6H5COONa CHBr3 NaOH ; (6% yield) 6 mol mol C H COOH NH C H CONH H O ; (5% yield) 6 mol 6 5 mol C6H5CONH Br 4KOH C6H5NH KBr KCO3 HO ; (5% yield) mol mol VMC Paper- 9 JEE Entrance Eam-8 Advanced

10 .( 4.6 ) Cu O (g) Cu O(s), G 78kJ H (g) O (g) HO(g), G 78kJ 3 Cu(s) H O(g) Cu O(s) H (g), G kj Eq.(i) Eq.(ii) = Eq.(iii) G G RT nq ; G (for min) 8 P 5 n P H 5 P H 5 n 3.(85.) 8 n P H PH A B AB HO E E RT ; H RT ( % of bar) ab af E af / RT Af 4 A kf A f e Keq E a b b / RT b b RT K 4e RT eq 4.(.6) eq k A e Keq 4e n P n H G RT n K 85J / mol ; The absolute value of n n A(s) B (aq) A (aq) B(s) M M... (ii)... (iii)... (i) VMC Paper- JEE Entrance Eam-8 Advanced G 85J mol n A K 4 n B G G RT n Q G RT n K G RT n 4 5.(C) G G TS ; H G G RTn 4 S R n4 T T S.6 J K mol FeF ; sp d 3 Ni CO ; sp 4

11 3 TiHO Cl 3 3 ; d sp 3 3 Vidyamandir Classes Ni CN ; dsp 4 Cr NH 3 6 ; d sp 3 FeCl 4 ; sp 6.(D) 7.(B) 8.(D) P CH3COOH NaOH CH3COONa HO.... (Acidic Buffer) No effect of dilution on [H ] Q CH3COOH NaOH CH3COONa HO (salt of W/A and S/B)... [H ] K K C w a w a ; [H ] old K K Kw Ka ; [H ] new [H ] old C C R NH3 HCl NH4Cl (salt of W/B and S/A)... VMC Paper- JEE Entrance Eam-8 Advanced

12 [H ] K C w w ; [H ] old Kb Kb Ni OH (s) Ni (aq) OH (aq) S No effect of dilution on [H ] K C K w C ; [H ] new [H ] old K as solution remain saturated. b PART-III MATHEMATICS.(D) n fn ( ) tan ( j)( j ) j ( j) ( j ) Tj tan ( j)( j ) Tj tan ( j) tan ( j ) T tan ( ) tan T tan ( ) tan ( ) Tn tan ( n) tan ( n ) fn ( ) tan n ( n) tan tan n (A) (B) n tan tan n n tan ( fn ( )) ( n) tan ( fn ()) n 5 n( n )( n ) tan f j () 6 j f ' n ( ) ( n) f ' n () n f ' n () n [ f ' n ()]sec ( fn ()) n n [ f ' j ()][sec f j ()] j n n (C) lim tan( fn ( )) lim tan tan lim n n (D) lim sec fn ( ) lim tan ( fn ( ) Since = is not in a domain.(bd) VMC Paper- JEE Entrance Eam-8 Advanced

13 PMQ 8 ( ) 9 PMQ 9 Locus of M : ( )( ) ( y 7)( y 5) E : y y 39 E ecept P and Q 3.(AD) Locus of S : T S h ky ( y k) 39 h k k 39 h k ( k ) h k k E : y y As E does not contain some points, E won t contain some points For atleast one solution, must be infinite solution ( ) 3 (A) (B) (C) b 5 b 4 3 b3 b 7b 3 b3 : S (Unique solution) b b b b 3 5 b 6 3( b b 3 b 3) b3 3 which will not be zero for each b, b, b 3 in S No solution y 5z b (i) 4y z b (ii) y 5z b3 (iii) (ii) + (i) b b (iii) + (i) VMC Paper- 3 JEE Entrance Eam-8 Advanced

14 4.(AC) (D) b b3 y 5 z : b : b b : b b3 For a solution b b b b3 b b b b3 For every b, b, b3 S infinite solution will not eist unique solution 4 5 Equation of tangent to y 4 y m m Applying p = r for circle m Equation of common tangents are y y Q (, ) For ellipse OQ a b b e a b Length of latus-ractum a y / Required area d ( ) 4 5.(ACD) s [ sz t z r ] 6.(BCD) t [ sz tz r ] s z t z rs tr rs tr z t s For unique solution s t For no solution s t and rs tr For infinite solutions s t and rs tr f ( )sin t f ( t)sin lim sin t t Applying L hospital rule VMC Paper- 4 JEE Entrance Eam-8 Advanced

15 7.(.) (A) (B) f ( )cos t f '( t)sin lim sin t f ( ) cos f '( ) sin sin dy y cot sin d y sin sin sin d C y c sin f 6 C = f ( ) sin f g( ) f ( ) 6 3 g '( ) f '( ) 3 g ''( ) f ''( ) (Wrong) g ''( ) sin cos g '''( ) cos 3sin 4 (cos ) 3(sin ) g ''( ) is decreasing function g ''( ) g ''() g ''( ) g '( ) is decreasing function g '( ) g '() g '( ) g( ) is decreasing function g( ) g () g( ) 4 ( ) 6 4 f ( ) 6 f for all (, ] (C) f ( ) sin f () f ( ) f '( ) for some (, ) [Rolle s Theorem] (D) f ''( ) f ( ) cos f '' f d ( 3) / ( ) t d t dt ( ) VMC Paper- 5 JEE Entrance Eam-8 Advanced

16 ( 3) t dt t ( 3 ) ( 3 ) ( 3) t ( 3) / 8.(4.) For maimum value of determinant we can make Diagonal entries minimum and non-diagonal entries maimum ma 4 9.(9.) α : No. of one-one functions from X Y 7 C 5 5! β : No. of on to function from Y X 7! 7! 5! 5! 3! 4!!!3!! (,,,, 3) (,,,, ).(.4) ( ) 9 5! dy (5 )(5 ) d y y 5y ln 5y 5y e 5y C 5y lim e lim 5y f () C = lim y lim f ( ) (.) f ( y) f ( ) f '( y) f '( ) f ( y ) f () Put =, y = f () f () f '() f '() f () f '() f '() Now put y = f ( ) f ( ) f '() f '( ) f () f ( ) f ( ) f '( ) dy y d dy d y ln y C C f ( ) e / f (4).(8.) Equation of OP e log ( ) e e y a b c 3 d 4 e VMC Paper- 6 JEE Entrance Eam-8 Advanced

17 y z T(,, ) will satisfy in plane T,, Q(3, 3, ) Q lies on z-ais. 3, 3. Distance of point P from -ais = P (3, 3, 4) R(3, 3, 4) PR 8 3.(.5) S,, ˆ ˆ p SP i j k ˆ ˆ ˆ q SQ i j k ˆ ˆ ˆ r SR i j k ˆ ˆ ˆ t ST i j k ˆ iˆ ˆj p q iˆ ˆj r t kˆ ( p q) ( r t ) ( p q) ( r t ).5 n n n 4.(646.) r. C. r r Cr Cr r r VMC Paper- 7 JEE Entrance Eam-8 Advanced

18 5.(A) : E n n n Cr Cr r. 9 C (, ) (, ) E sin ln ln e e e e e e e e,, e e e,, e e f ( ) ln Domain of f: (, ) (, ) Range of f: (, ) g( ) sin ln e Domain of g:,, e e Range of g:, P-4, Q-, R-, S- 6.(C) P, Q, R 4, S C3 C n n n Cr Cnr r 9. C 9 α = Eactly + Eactly 4 + Eactly 6 + Eactly 8 + Eactly C C C C C3 C3 C4 C4 C5 C M, α 3 = Total ways No girl Eactly girl C C C C 4 M is included + M G is included + G are ecluded is included C3 C3 C C 34 G is included C C C C C3 C C4 C4 C3 C M and G are ecluded P 4, Q 6, R 5, S VMC Paper- 8 JEE Entrance Eam-8 Advanced

19 7.(B) tan3 b a a 3b Area of LMN 4 3 b a ab b, a 3 8.(D) Length of conjugate ais = b = 4 b Eccentricity a 3 Distance between focii = ae = 8 b 4 Length of latus rectum a 3 P 4, Q 3, R, S ( ) sin f e h sin e f ' ( ) lim h h h h sin e e lim h h h e lim h h e h h h continuous at = Limit does not eist Non differentiable at h = sin ; f ( ) tan ; sin h sin h f ( ) lim lim h tan h h tan h sin( h) sin h f( ) lim lim h tan ( h) h tan h Discontinuous at = f 3 ( ) [sin (ln ( )] sin (ln ( )) ln ( ) / e / e / f3( ) [, e ] f3 ( ) Continuous at = Differentiable at = Derivative is continuous at = sin ; f4 ( ) ; lim sin f () Continuous at = VMC Paper- 9 JEE Entrance Eam-8 Advanced

20 h sin f ' 4 ( ) lim h lim hsin h h h h h sin f ' 4( ) lim h h h f ' 4( ) sin cos lim f ' 4( ) does not eist Derivative is discontinuous at = P, Q, R 4, S 3 Differentiable at h = VMC Paper- JEE Entrance Eam-8 Advanced

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