MODEL SOLUTIONS TO IIT JEE 2012

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1 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 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 <., w <. w <.. Upto, B From to, B is almost linear (increasing). From r >, B r. LHS is the shaded portion. So for anticlockwise rotation by, Q lands in the unshaded part and P lands in the shaded part. v. (L e) υ L e. m. cm L. cm. (e.d). cm QC. 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 Section II 9. Energy of anti-neutrino is much less than.8 e.

2 . Electron will have linear momentum KE is never zero for electron. Since the neutrino has some mass, KE of electron cannot reach.8 e. 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).. v P aω; v C a v Q a v Q v C ω ω ; ' vq vc ω ω ' ω Ais of rotation is vertical. Section III l. t a g sin θ a ; Ι P > Ι Q Ι mr a P a p < a q t P > t Q mgh mv Ι mr K Q > K P Q > P ω Q > ω P C. The net flu linked with coil is zero, whether current is constant or current is changing. Q 7. Z L (. ) Ω Z C i L. A ; ZL. i C A i. i. A. ; il ic. A 8. Due to q and q, q, q E K along D Due to q and q, E K along D E E E K along D at is zero (as net charge is zero) P is the equatorial line of three dipoles P zero; 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 Ω (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

3 PAT II 7 8 C C B B D D D 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. 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 H CH CH C CNH CH 8. T b K b m. m.7 M M 9. P P W M P M W 7 P P P 7 Section IΙ - H CH CH C CNH CH F. Xe F sp d hybridised See-Saw 7. CH CH C CH CH CH C CN CH CN 9% H S 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. G nfe 9.9 J. kj mol

4 . CaCl HAc CaAc H Cl KI Cl KCl I Na S I NaI Na S 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 Cl CH C (i) H / Pd - C (ii) SCl CH anhy.alcl CH CH Section IΙΙ. Graphite has higher electrical conductivity than diamond (A). Graphite has higher C C bond order than diamond.. Graphs I and III represent physisorption and graphs II and I represent chemisorption 7. T T, since isothermal T < T, since adiabatic epansion U isothermal U adiabatic epansion ve 8. KI K Fe(CN) K Fe(CN) KI ZnS Zn Fe(CN) KI 9. Filter residue, Zn Fe(CN) filtrate, KI (brown) Zn Fe(CN) 8NaH Na [Zn(H) ] KI starch blue colour (T) Cr LiAlH () CH Soluble Na [Fe(CN) ] CH CH H CH H C (K). The compound (I) is benzaldehyde. (A), (B), (C) ecess (CH C) (W) CH CCH CH CCH

5 PAT III 7 8 A C C A B D D B 9 B C A A D A A, B A, B, 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) ( )λ λ 9λ λ 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 Choice (C). D, D, D, D Total number of choices D D D D D D Total cases Total cases Favorable cases Probability Choice (A). cos log cos Q

6 cos d sin d sin ( ) ( cos ) ( sin ). 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 a as a a 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 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)

7 . b all ones one zero two zeros Choice (A). 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 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) P(Y) P(X) P(Y) P(X Y), X, Y are independent. (A) and (B) are true & (C) is false X C, Y are independent P(X C Y) Choice (A) & (B) t. f () e ( t )( t ) f e ( )( ) f () e ( )( ) e ( ) e ( ) e [ ( )( ) ] dt e ( ) f (), ( ) f () > (, ) f () < (, ) f () > Choice (A) (B) & (D) 7. b n cos (n, n ) f() a n sin for [n-, n ] b n cos for (n, n] a n sin for {n, n] b n cos for (n, n) a n sin 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. 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 6 7 8 B A D C B C A D 9 C D B C D A 6 7 8 9 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