MODEL SOLUTIONS TO IIT JEE 2012

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1 MDEL SLUTINS T IIT JEE Paper II Code PAT I B A D C B C A D 9 C D B C D A D A, C A, C A, B, C B, D A, B π. A a a kˆ a π kˆ π M Ι A a Ιkˆ Section I. Let V m volume of material of shell V w volume of water in shell V a volume of air in shell utside volume of shell, V V m V w V a (i) ρ m density of shell material ρ w density of water ρ a density of air (treated as zero) W weight of floating body V mρ mg V wρ wg (ii) B weight of water displaced V ( Vm Vw Va ) ρ wg ρwg (iii) (iii) [ (ii) [Q Body floats] V ] m Vw Va ρwg V mρ mg V wρ wg Vm Vw Va ρm m ρw [ ] V Vw ρ m V m Vm Va V w V w V w ρw ρ V w m V m Vm Va ρw ρ If m >., V w V a ηv m (where < η) ρw V w < V a V w is less than half the volume of vessel. ρ If m <., V w V a ηv m ρw V w > V a V w is greater than half the volume of vessel.. For r, B For B < r µ Ι r r πr i.e. non-linear increase with decreasing slope db dr For r >, B r

2 . Let the disc rotate for time t, after the body is projected from P. The position P on the disc comes to P, such that PP (curved path) ωt. But the projected part has an initial velocity vî. ωî. Hence, in time t its -displacement is PP ωt. bviously lengthwise curved path PP and PP (straight) are same. But P is to the right of P because straight line motion. Also because projectile motion (Given range PP < ) it will land somewhere in P in the unshaded area. bviously Q lands in unshaded region. v. (L e) υ 6 L e.6 m 6. cm L 6. cm. (e.d). cm Y Q.7 P P P P ω.8 X Since there are three particles after disintegration of neutron (which was initially at rest), if the electron is at rest, the linear momentum of proton will be equal and opposite to the linear momentum of anti-neutrino. This linear momentum of proton will be more than the case when the electron was having maimum velocity and the anti-neutrino was at rest. So the KE of proton will be more than its KE when electron was having maimum KE. Consequently, when electron is at rest, the KE of anti-neutrino will be maimum but slightly less than.8 6 ev.. When electron is at rest (possible), its KE is zero, which is the minimum value. If the rest mass of anti-neutrino is not zero, the velocity attained by electron cannot be equal to the maimum velocity attained when rest mass of anti-neutrino is zero. Hence the maimum KE of electron will be less than.8 6 ev. i.e. K.8 6 ev. n c v v n c. For meta-materials, the refracted ray emerges on the same side of the normal as the incident ray.. By the time the CM completes one rotation, all points on the disc has also completed one rotation ω is same for all points ( ω of CM). QC 6. Q ( C C ) 8 8 µc ( ) 7. In one rotation, there are two cycles of relative velocity, with initial relative velocity zero. 8. Constant pressure heating of a monoatomic gas Q nc p θ ( ) 8 J. v P aω; v C a v Q a v Q v C a P ω ω ; ' vq vc ω ω ' ω C Q Section II Ais of rotation is vertical. 9. When anti-neutrino is at rest; electron will have maimum velocity and maimum kinetic energy.

3 Section III l. t a g sin θ a ; Ι P > Ι Q Ι mr a p < a q t P > t Q mgh mv Ι mr K Q > K P V Q > V P ω Q > ω P 6. The net flu linked with coil is zero, whether current is constant or current is changing. 7. Z L (. ) Ω Z C 6 i L. A ; ZL. i C A V i. V i. A. V ; il ic.6 A V Ω 8. Due to q and q, q, q E K along D Due to q and q, E K along D E E E 6K along D V at is zero (as net charge is zero) P is the equatorial line of three dipoles V P zero; V ST is not same. 9. v esp g ρ (Q same ρ) P, Q (from surface areas) P 9 / [ P ( P) 9 ] (v esc) > (v esc) Q > (v esc) ( P v ) esc Also / 9 ( vesc ) P ( vesc ) P ( v ) esc Q. v ω î (Considering instantaneous centre of rotation) ω ω v P sin î cos kˆ ω ω î kˆ ω vp v vp ω î ω î ωkˆ ω kˆ

4 PAT II C C B B D D A A 9 B D C A C A B, D A, C A, C, D A, C, D A, C A, B, C Section I. Paramagnetic Ni comple is tetrahedral Diamagnetic comple is square planar H CH CH C CH CH - H. P NaH H PH NaH P NaH P PH Na HP idation number of P in PH is idation number of P in Na HP is. idising agent educing agent Zn dust. β-keto acids undergo decarboylation readily on heating. C H (g) C (g) H (g) H kj 8 kj mol F 6. Xe F sp d hybridised See-Saw 7. CH CH C CH CH CH C CN CH CN H CH CH C CH CH Note: In presence of conc. H S, cyanide may get hydrolysed to give amide (ption D). 8. T b K b m. m.76 M M 9. P P W M P M W 76 P P 6 P 7 Section IΙ 9. M M e M (.M) e M M (.M) M E cell.9 log Ksp / (For MX M X S S K sp S ).9.9 log Ksp / K sp / K sp

5 . G nfe 96.9 J. kj mol. CaCl HAc CaAc H Cl KI Cl KCl I Na S I NaI Na S 6 Number of moles of CaCl number of moles of Na S 8. Molarity of (CaCl ) bleach solution 8... CaCl is the salt of HCl and HCl with Ca(H) The anhydride of HCl is Cl HCl Cl H. CH (I) (CH C) CH CNa CH CH CH (i) H / Pd - C Section IΙΙ. Graphite has higher electrical conductivity than diamond (A). Graphite has higher C C bond order than diamond. 6. Graphs I and III represent physisorption and graphs II and IV represent chemisorption 7. T T, since isothermal T < T, since adiabatic epansion U isothermal U adiabatic epansion ve 8. KI K Fe(CN) 6 K Fe(CN) 6 KI ZnS Zn Fe(CN) 6 KI 9. Filter residue, Zn Fe(CN) 6 filtrate, KI (brown) Zn Fe(CN) 6 8NaH Na [Zn(H) ] KI starch blue colour (T) LiAlH Soluble Na [Fe(CN) 6] CH H CH H (ii) SCl CH anhy.alcl CH Cl C Cr (V) CH CH CH CH ecess (CH C) CH CCH CH CCH C (W) (K). (A), (B), (C). The compound (I) is benzaldehyde

6 PAT III A C C A B D D B 9 B C A B D A A, B A, B, C, D B, D B, C A, D A, B Section I. Eq. of plane y λ ( yz-) (λ) ( λ)y (λ)z λ -----() Distance from (,, ) to Eq () is Clearly distance from (,, ) to Choice (A) is only satisfies. ( a b) 9. a b a. b 9 Let c i j k Given a c c b ( b c) ( a b) c a b is parallel to c a b λ ( i j k) \ (λ) ( 7) (λ) () (λ) ( 6)λ λ 9λ 6λ 9 λ ± Choice (C) sinp sin p sinp sin p ( sinp) cos p ( sinp) cos p { } { } 9 cosp sinp sin p sinp sin S 7 S a S b P S c 6 6 Choice (C). D, D, D, D Total number of choices 6 D D D D D D 6 6 Total cases Total cases 6 6 Favorable cases Probability 6 6 Choice (A) π. cos π π π log cos π π Q

7 π cos d π sin d sin ( ) π ( cos ) π ( sin ) π π π π 6. P T P Ι () Taking transpore T P P Ι () T T P P P P ( ) P T P P is symmetric P T P Ι P P Ι P Ι PX X Choice (D) cos d) 7. A, A, Corresponding A. P A, A 9d d 9 99 n A n 9 9 n th term of H.P 99 n 99 n < n > 99 N > Min value Choice (D) 8. αβ 6 a as a a 6 a a lim(αβ) lim a ( a) ( a) So Choice (C) Section II 9. f () for all ( ) and g () cos f() ( ) Consider cos cos n Derivative ( ) ( ) ( ) < < in {(, )} g() is decreasing in (, ) Choice (B). P ( ) sin ( ) ( ) sin ( ) ( ) Sin > y ( ) no eists P is false Q ( ) sin \ Sin ( ) If ( ) ie 6 which could be simplified\ Q is True. a a : a : a : a all ones : one zero : 8 two zeros : We note that for a positive integer, a n a n a n Hence Choice (A)

8 . b 6 all ones one zero two zeros Total 8 cases Choice (B). is not tangents to the first circle y is not a tangent to the nd circle Choice (D) y is not a tangent to the nd circle y 6 is tangent to both. L is of the form y k and y is tangent to nd equation Choice (A) Section III. P() P(Y X) P (X Y) P(X) 6 P(Y) 6 P(X) P(Y) P(X Y), X, Y are 6 independent. 6 (A) and (B) are true & (C) is false X C, Y are independent P(X C Y) 6 Choice (A) & (B) t 6. f () e ( t )( t ) f e ( )( ) f () e ( )( ) e ( ) e ( ) e [ ( )( ) ] dt e ( ) f (), It is obvious that f () at atleast one c (, ) (, ) f () > (, ) f () < (, ) f () > Choice (A), (B), (C) & (D) 7. b n cosπ (n, n ) f() a n sinπ for [n-, n ] b n cosπ for (n, n] a n sinπ for {n, n] b n cosπ for (n, n) a n sinπ for (n, n) At n b n a n b n At n a n b n a n b n At n b n a n a n b n at n a n b n a n b n Thus (B) & (D) 8. Since the line are Coplanar k k (k ) k k ± Equation plane is ± y y z and y z Choice B and C 9. adjp IP n adja P P ± A and D are true cos θ 6. f(cosθ) cos θ cos θ cos θ f ( cos θ ) f ( ) Where ± cos ± f ± Choice (A) and (B) z ± θ

MODEL SOLUTIONS TO IIT JEE 2012

MODEL SOLUTIONS TO IIT JEE 2012 MDEL SLUTINS T IIT JEE Paper II Code PAT I 7 8 B D D A B C A D 9 B C B C D A 7 8 9 D A, C A, C A, B, C B, D A, B. A a a kˆ a kˆ M Ι A a Ιkˆ. ρ cg wρg ρg ρ cg w ρg Section I w ρc w ρ w ρ c If ρ c