Solutions to Mock IIT Advanced/Test - 2[Paper-2]/2013

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1 .(D) 5 x 0 4 x Molality molarity 0.8 Vidyamandir Classes Solutions to Mock IIT Advanced/Test - [Paper-]/03 Tf C Hence final freezing point is 3. C [CHEMISTRY].(B) 3.(B) If there were three possible values for spin quantum numbers. For first period (n ) Orbitals & elements 3. For second period (n ) Orbitals 4 ; elements (A) 5.(B) pt 4 0. N HCl, 0. N NaOH Let V ml of base (V + x) ml of acid are mixed. H + added 0. (V + x) + H left 0.(V + x) 0.(V) 0. x. Required of consternation H + at titration point x 4 0 V + x x 3 0 V+ x V + x x 000 % Error V / V + / V x H left 0.x x 00 + H added 0.( V + x) V + x 6.(B) % 5 0.0% 0 0 %. K 7.(A) CrO4 NH X yellow ppt 3 No change Solutions KCrO4 NH Z Re d ppt 3 ppt dissolves Solutions KCrO4 NH Y Red ppt 3 black ppt Solutions X must be Pb + (Pb CrO 4 (yellow) Y must be Hg + (Hs CrO 4 (red), dissolves in NH 3 ) + Hg(NH 3 ) Z must be Ag + (Ag CrO 4, with NH 3 VMC/03/Solutions Mock IIT Advanced/Test - /Paper-

2 O 8.(C) Alanine CH3 CH C OH NH O Glycine CH C OH NH O Phenyl alanine Ph CH CH C OH NH Polypeptide Al Gly Phal O O O HN CH C NH CH C NH CH C OH CH3 CH Ph 9.(C) For an exothermic Reaction H EP ER For H to be negative. E P < E R 0.(A) Prussian blue is Ferri-ferrocyanide Fe 4 [Fe(CN) 6 ] 3..(B) During detection of nitrogen in organic compound, nitrogen present in organic compound is converted to NaCN. So presence of N and C both is necessary condition..(b) 3.(B) 4.(B) Position () in being activated due to lone pair of e from the lower ring but position () is not getting any e density. Note that the N of Right hand ring cannot donate its lone pairs. 5.(ACD) Lead prefers to form di-valent compounds. (Inert pair effect) 6.(AC) 7.(ABC) Toluene doesn t react with HCl but aniline does. Hot conc. H SO 4 dissolves amines but dehydrate alcohols to give alkenes Ceric ammonium nitrate gives Reddish brown ppt with phenols. VMC/03/Solutions 3 Mock IIT Advanced/Test - /Paper-

3 8.(ACD) NO [BF 4 ] NO BF 4 Bond of order NO + > NO BF4 has all σ bonds (pure) But in BF 3 due to back bonding, B F bond length is slightly reduced. 9.(ABCD) + 3MnO4 + 4H MnO4 + MnO + HO (Green) (Purple) (Blackish brown ppt) KMnO 4 forms dark purple (almost black) crystals on crystallization Also, KMnO4 K MnO4 + MnO + O 0.(CD) Solutions to Mock IIT Advanced/Test - [Paper-]/03 [PHYSICS].(C) Component of relative velocity parallel to mirror zero and component of relative velocity perpendicular to the mirror Aω sin 60 3 A ω 3 A k m.(a) By mirror formula v 5cm v ( ) m + Image revolves in a circle of radius ½ cm. Image of a radius is erect particle will revolve in the same direction as the particle. The image will complete the revolution in same time s. π π Velocity of image wr cm. 57 cms VMC/03/Solutions 4 Mock IIT Advanced/Test - /Paper-

4 3.(C) a E. 4π r r C / r. 4π r dr + q 0 πc r a + q E. 4π r 0 ( q πca ) C C E +, is const 0 4π 0 r 0 if πc r a + q E 4π 0 r q C πa 4.(A) We can assume that a part of length l / 6 is cut from the lower portion of side B and put below A. Mass of this part m/6 Length of part l / 6 m l mgl Hence work done g (B) The velocities of the end A and B along the length of rod should be same Hence v cos 30 v cos 30 v v v A B A B Hence Angular velocity of Rod 8 30 vab v sin v ω l l l 6.(A) Absolute pressure in bulb Gauge pressure + Atmospheric pressure P + Pa ρav Using Bernoull s Equation, Pa + P PBC + Where ρ a is the density of Ar. 3 PBC Pa + P v and BC a So equating there two equation for P BC 7.(B) Given ω 3π 8.(D) ω f. 5 π Also x. 0 cm π π π φ x l 8 λ λ 6 cm v fλ cm / s P P ρ gh v P + ρ gh VMC/03/Solutions 5 Mock IIT Advanced/Test - /Paper-

5 mv mv mv + mv and v v ev and e ( 3 gh Hence height rises by tennis ball ) 9h g Max height reach by tennis ball above ground 9h + d 9.(B) Optical path difference at P, p µ SP+ ( µ g µ ) t µ SP ( g ) µ S P S P + µ µ t µ xd Thus p ( µ g µ ) For maximum 0 p xd + t as SP SP D D ( µ g µ ) td ( 4 ) x µ d At point O, x 0 Thus T 4sec T td ( 0 ) T d dx 6Dt 30.(A) Speed of central maxima v dt 0 T d Central maxima is at O at time t 4 sec 6 6Dt v 3 0 m / s 36d Fringe pattern (and so central maxima) is moving downwards. v 3 m v m 3v + m m 3.(D) Time constant of larger circuit t RC 0Ω 0 mf 00ms Thus current as function of time is : i ( 00V ) ( 0Ω) t e 00ms 0 e At t 00 ms, we obtain i ( 0A)( e ) 3.(A) According to question only long wire nearest the small loop produces an appreciable magnetic field through the small loop. c + a µ 0 µ 0 dr l π r π c ib ib a φb n + c dφ µ 0b a di ε ln dt π + c dt 7 ( π ) ε ln3 i. 50 e R π 5 t µ 0b a 0 l n e 00ms π + c 0. t 00ms 7 0ln3 0ln3 7 7 i 0 0 A A e 5 e R a + b 5 5 Ω VMC/03/Solutions 6 Mock IIT Advanced/Test - /Paper-

6 33.(C) From Bernoulli s theorem for level A and For highest level, CD PA P + ρv + ρ gh....(i) Since the tube has uniform cross section and water is incompressible v v gh....(ii) B p P gh gh p p ρ g h + h A ρ ρ atm The minimum value of p 0 ( h ) p 0 3 7m ρ g 0 0 atm h max 3 π R db / dt 34.(A) Induced electric field at point p π R R db E towards right dt ee er db Acceleration of electron a toward left m m dt 35.(ABD) 36.(CD) R ε0 A ε0 A C, C d d C C V AV d d Extra charge flown ε0 Work done by battery: Wb Vx charge flown ε0 AV d d Change in P.E. of capacitor ( ) hh 5 U C C V ε0 AV d d R would maximum If hh maximum Hence (A) hh (B) hh (C) hh (D) hh (BCD) F f m a 5a f ma 7 fr m R α 6 a+ Rα a a 38.(ABCD) Current from E will increase or decrease acc. to as jockey is shifted left or right respectively. So potential gradient in the potentiometer wire will increase or decrease respectively so balancing length (corresponding to E ) will decrease or increase respectively. And pot. grad. is also increased if E is increased so l is decreased if S is closed then p.d across E is decreased so l is also decreased and if E is increased l is increased m 39.(BCD) Here F > µ mg k + M, so slipping will occur between blocks Form m : f µ mg ma a. m / s k r For M : mgµ MA A 0. 4 m / s a / 5m / s k VMC/03/Solutions 7 Mock IIT Advanced/Test - /Paper-

7 d d 40.(ACD) Li L i dt dt di di L, i Li dt dt i L i L i i di V L di / dt V L dt and V L V / Li W / L i L i 4 W L i V 4 L V 4 W 4 W Solutions to Mock IIT Advanced/Test - [Paper-]/03 4.(D) a > < x< a 0< loga x< loga x 0 a a a x a e x a dx x dx [MATHEMATICS] a e 4.(C) lx + my 0 is rotated about line of intersection with the plane z 0 (x-y plane) lx + my + kz 0, k R (By family of intersection of plane) lx+ my+ nz 0 is making an angle α with lx + my 0 cosα n ˆi. nˆ 43.(B) 3 l + m cosα l + m + n l + m l + m + n sec α l + m n tan α l + m f x x + x + 00x+ 7 sin x Equation cosα n± l + m tanα l + m + n l + m + n ( y f )( y f ( 3) ) + ( y f )( y f ( 3) ) + 3( y f )( y f ) 0 g y ( y f )( y f ) ( y f )( y f ) ( y f )( y f ) Let f x 3x + x+ 00> 0 x R f (x) strictly increasing function x R g(f ()) > 0, g(f ()) > 0, g(f (3)) > 0 44.(B) 4 lines intersect each other in 4 C 6 points 4 circles intersect each other in 4 P Every lines cuts 4 circles into 8 points. Therefore, 4 lines cut four circles into 3 points. Therefore, the maximum no. of points (B) x y+ z 0 x y+ z 0 yx y+ z 0 if D 0 has many solution VMC/03/Solutions 8 Mock IIT Advanced/Test - /Paper-

8 0 t 5 t t 0 f ( x) dx f ( x) dx (D) y 4( x+ ) 47.(A) In f x dx f x dx t + 5 x 0 0 dt dx f ( x) dx f ( t 5) dx + ( + 5) normals are possible from (b, k) iff h > a. (See Illustration in Conic) ABC y 4x x> x+ > x > 0 ( K, 0) ; K > f x f x dx dx 0 t 0 0 b+ c c+ a a+ b k 3 b c k + c+ a k a+ b 3k Find b + c a cos A bc a + c b cos B ac a + b c cosc ab And then find the ratio as 7 : 9 : 5 48.(A) r a λ a + µ b + γ a b r a a λ a. a + µ b. a + γ a b a 0 λ a, a b 0 Also, r a b λ a b + µ b b + γ a b b µ r a b γ a b b Also, or ( ) ( ) γ ( ) ( ) a+ b+ c 36k a+ b+ c 8k a 7k, b 6k, c 5k r b a a b r a b b b b a γ r b r b a γ a or VMC/03/Solutions 9 Mock IIT Advanced/Test - /Paper-

9 49.(A) n + n n fn x x + x ( n )! ( n )! ( n )! Vidyamandir Classes n fn x x + x! ( ) n x. e x + xe x n n! n n n x fn( x) x d dx ( x + x) e dx 0 0 n x e x + x dx e x e 0 dx 0 4n 50.(A) Value ( + 4n 4n 4n 4n 4n C0+ + C Cn) + ( + Cn+ + C n Cn) 4n+ 4n+ 4n+ 4n+ 4n 4n+ ( C0 + C Cn) + Cn + Cn 5.(D) Let f ( x) sin x and g( x) 5.(A) Here, f ( a) f ( a) 53.(A) S : S : S 3 : cos x, Also sin x 0 for x 0, π, then by cauchy s theorem f β f α f θ sinβ sinα cosθ cotθ g β g α g θ cosβ cosα sinθ a a e e + because + a a + e + e f ( a) I ( z) g ( z) z ( dz), (Putting x z ) f( a) f a f a { } ( ) ( ) ( ) ( ) x g x x dx x g x x. dx I I I I f( a) f( a) True by definition False (because by the given condition, at least one point may lie on the plane) True (Standard result) S 4 : True shortest distance 54.(D) Area ( l ) x 5 π nx + tan x dx xlnx x dx+ x tan x dx ln ln5+ tan + x 4 P A B P A P B and P( A B) P( A ) P( B) 6 5 If P(B) x then P( A) ( x) and P( A ) x P( A) + P( A ) 5x+ 4( x) 30x( x) 6 x 5x 55.(AB) or 30x 9x or ( x )( x ) 56.(BD) A plane parallel to x+ y+ z 3 is x+ y+ z k A point on x+ y+ z 3 is (,, ). So or + + k or 3 k 4 or 3 k ± 4 or k, 7 the planes are x+ y+ z and x+ y+ z 7 4 x, 6 5 VMC/03/Solutions 0 Mock IIT Advanced/Test - /Paper-

10 3 4 or (AD) sin( B C) cos( B C) t 9 A cot, (absurd) t 4, where + t 5 B C t tan B C b c A A tan ± cot cot 3 b+ c 3 A π A π 4 a k + k 5k a : c 5 : 58.(ACD) c λ a + µ b + λ a b b c b ( λ a + µ b + γ + a b) µ { a b 0, b b, b a b 0} c c c λ a + µ b + γ ( a b) + λµ a b + λγ ( a a b) + µγ b a b or c c λ + µ + γ c c c λ a + µ b + γ a b Also, λ c a + µ c b + γ c a b λ( λ a + µ b + γ a b) a + µµ + γ a b c λ + µ + γ a b c....(ii) (i), (ii) γ γ a b c. γ a b c a + b + a b c a c + b c + a b c λ + µ + γ Also,....(i) 59.(AC) Area ( x sin x ) dx ( 3 x sin x ) dx ( cos ) ( cos cos) + cos + cos (BC) S S n terms n n S ( ) n n + n 4 8 n VMC/03/Solutions Mock IIT Advanced/Test - /Paper-