Vidyamandir Classes SOLUTIONS. Joint Entrance Exam IITJEE-2018 Paper Code - B. 8th April AM PM. VMC/JEE Mains Solutions

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1 SOLUTIONS Joint Entrance Exam IITJEE-08 Paper Code - B 8th April AM.0 PM VMC/JEE Mains-08 Solutions

2 Joint Entrance Exam JEE Mains 08 PART-A PHYSICS.() For collision with deuterium: mv o mv mv (Conservation of momentum )... () v v v ( e )... () v By () and () v mv mv 8 Pd 0.89 mv 9 For collision with carbon Nucleus mv 0 mv mv (Conservation of momentum )... () v v v ( e )... () By () and () v v mv m v 48 Pc 0.8 mv 69.() Change in momentum of a single molecule. u P0 m0 Total change in momentum per second P np0 n. m0u F nm Pressure 0u A A Substituting values: P.5 0 N / m..() P B V V V V P B V r Also, (As V r r P mg r B Ka V 4r r ) VMC/JEE Mains-08 Solutions

3 4.(4) Equivalent circuit is 5.() Where, K U r r eq V V r r veq. r r 0 VAB Veq =.55 volts 0 req du K F r r dr r K r K. E. = mv mv (As Force towards center ) r r K mv r Total energy = KE + PE 6.(4) For m to be at rest T 5g For m & m to be at rest f ( N) f T 5g f 0.5( m m) g m. kg K K = Zero r r Amongst the options minimum mass that can be kept for no motion is 7. kg 7.() For Series limit of Lyman : n and n P RcZ For Series limit of Pfund: n 5 and n L P RcZ 5 5 VMC/JEE Mains-08 Solutions

4 8.() When an unpolarized light of intensity I passes through a polarizer for the st time, intensity of output is I (irrespective of orientation of polarizer) So, i.e., polarizers A and B have axes parallel to each other. Now let the axis of C make an angle with A, and with B. I 4 I cos 8 Solving, 45 9 (). RZ n n n RZ n Since n is very large, using binomial n RZ n n RZ RZ n B r n h n A As n n n n 4 mze n 0 ()..(4) Voltage across Si diode in forward bias is 0.7 volts. Hence voltage across 00 resister is 0.7 =.V mv r qb. I =.5 ma 00 mk qb r r e e m K eb 4m K eb p VMC/JEE Mains-08 4 Solutions

5 r p.() qi p m K eb Comparing (), () and () q f CV r r r e KCV qinduced q f qi ( K ) CV 0.() Quality factor Q 0 L R p nc 4.() Overall bandwidth use for transmission 0% of C 5.() Number of telephonic channel c f l Y Total bandwidth Channel bandwidth () Hz 5kHz 7 MR MR 55 I0 6 M ( R) MR 4 MR O is the centre of mass of the system. Applying parallel axis theorem between O & P IP I0 7 M ( R) MR 6MR MR kq 7.(4) A kq C B kq VB b b c ( )(4 a ) ( )(4 b ) ( )(4 c ) k b b c a b 4 c 4 0 b b a b c 0 b 8.(4) Let potential difference per unit length of potentiometer wire be x. In case-i (5)( x) (i) In case-ii i r 5 ir (40)( x) VMC/JEE Mains-08 5 Solutions

6 40 r 5 r x 5 40x r 5 From (i) & (ii) r 5 5x 40x 5 5 r 5 r (ii) 9.() From wave equations : In air:, k c In medium: kc, k k k, k c c 0 r 0 r 0 r 0 r Medium and air are non-magnetic ; r r 4 4 r r r r c c c c 0.() Angular width of central maxima ; (where a is slit width and is wavelength) a (i) a In YDSE, fringe width.(4).().(4) D [where d is slit separation and D is distance of screen from slits) d D d 0 f T k m a 6 D a d d 60 R M 9M R R I M 9MR MR MR I 4MR k m m It is given that final total kinetic energy has increased, so some internal energy of the system must have been converted into kinetic energy. VMC/JEE Mains-08 6 Solutions

7 mv mv.5 mv0 v v.5 v0... (i) Since, there is no external force, momentum can be conserved mv mv mv0 v v v0... (ii) From (i) & (ii) v v0 v.5 v0 v v0v 0.5v0 0 Relative velocity v v Difference of roots D v0 a 4 (). 0I B ; m I r r 0I B ; m I r r 5 (). m r m r B r B r M M V L M L M L r r r r Maximum % error in density.5% % 4.5% 6.() Let the resistances in left and right slot be r and 000 r respectively Initial: r(00 x) (000 r)( x)...() After interchanging: (000 r)[00 ( x 0)] r( x 0) (000 r)(0 x) r( x 0)...() From (): 00r rx 000x rx r 0x r r From (): (000 r) 0 r (4) P Vrms Irms cos ; 00 0 P Wattless current, (000 r)(00 r) r 00r r r r 00r I I rms sin r VMC/JEE Mains-08 7 Solutions

8 8.(4) The (), () and () graphs can represent the motion of a ball that is thrown in vertically upward direction. Initially speed decreases, becomes zero and then on the return trip, speed increases. Slope of graph in option (4) does not explain it. 9.() For mono atomic gas 0.() Using TV constant F n R 5 (00) V T V 00 T 89 K U n RT kj k mv F n R R R R Now T R V ( n) R k V n mr n ( n) V R VMC/JEE Mains-08 8 Solutions

9 PART-B MATHEMATICS.() y 7 x 6 x y 7 x y 5 x y 5 0 Also, centre of the circle is 8, 6 and the radius is 64 6 c c 5.(4) x y z 0 x y z 0 So the equation of plane is 7x 7y 8z c c 95.() Now, distance from origin equal to x x 0 and x, (where and are non-real cube roots of unity) () Equation of PQ, 4x 0 y 6 VMC/JEE Mains-08 9 Solutions

10 y Area of TPQ () yy' 6 6 y' y y 8x by y' 0 8x 9x 7x 7x y' b by by by y 9 y 6x b 6.(4) k k k x ky z 0..(i) x ky z 0.(ii) x 4y z 0..(iii) On solving (i) and (ii) x 5z 0..(iv) On solving (iii) and (iv) 4y z 5 z z xz 0 y z 4 7.() x x x Case-I: x x x 6 x 6 0 x 4 x 0 x 0, 6 As x 9 x 6 Case-II: x x 6 x 6 x 6 0 x 8 x 0 x x 6 0 x 6, 4 8.(4) As, x x 4 There are exactly two elements in the given set. 8 cos x cos sin x 6 8cos x cos x 4 8cos x 4cos x 4 cos x 4 cos x cos x VMC/JEE Mains-08 0 Solutions

11 cosx cosx x 0, x,, Sum =. 9 9.(4) 4 6 Total probability () Let g x x t x g' x 0 x t R 0 ; t 0, x x f x x x t, Let h t h x x f g x f x t t g x t t t t h' t t Local minimum value occurs at t Local minimum value h 4.(4) Since Set A is, a 5 4 < a < 6 VMC/JEE Mains-08 Solutions

12 and b 5 4 < b < 6 Now B is (a 6) (b 5) 9 4 It can be seen that all vertices of rectangle lie inside the ellipse, therefore A B 4.() ( p q) ( p q) p q ( p q) p q ~ p T F F F F T F F F F F T F T T F F T F T 4.(4) The equation of tangent at P y 6 (x 6) A ( 6, 0) The normal is y 6 (x 6) B (4, 0) Since APB AB is the diameter. Center of the circle C (4, 0) Slope of PB m 44.() 4 m m Slope of CP m tan mm x 4 x x x x 4 x (A Bx)(x A) x x x 4 Put x = A A 4 Put x = (A B)( A) (9 4) ( 6 4) (4 6) VMC/JEE Mains-08 Solutions

13 5 0 0 ( 4 B)5 ( 4 B) B = 5 45.() Let x y x y x y C0 x 5 5 Cx 4 y... 5 C5 y 5 5 C0 x 5 5C x 4 y... 5C5 y C 0x Cx y C 4xy C0x 5C x y C 4xy x x x x x x x x x x x x 0x 0x 5x 5x 0x () a a5 a 9...a49 46 a 4d (i) a9 a4 66 a 5d.(ii) From (i) and (ii) d and a Now, a a...a m 7 8 r 40m r 47.() Let, R ( h, k) P (0, k) Q ( h,0) Equation of line would be, k h hk x y... (i) h k h k Locus of (h, k) is y x xy / sin x 48.() Given dx x / 49.() 7 r 7 r 40m m m 4 x / / sin x (sin x) f ( x) f ( x) sin x x x sin x dx sin x dx g( x) cos x f ( x) x g( f ( x) ) cos x Given, x, 6 Area = 8x 9x 0 (6 x )( x ) 0 / cos x dx / 6 VMC/JEE Mains-08 Solutions

14 50.() 5 lim x... x0 x x x 5 5 lim x... x0 x x x x x x 5 = lim (... 5) lim x... x0 x0 x x x Now 0 x x R 0 5.() Variance = 45 () Variance 5.(4) sin x cos x dx 5 5 (sin x cos x sin x sin x cos x cos x) 6 tan x sec x 5 tan x tan x tan x Put tan x t t ( t ) dt (t ) (t ) t y dy t dt dx dt sec x dx dy C y (y) C (tan x ) 5.() Doubtful points for differentiability are 0 and At x = 0 h h ( e ) sin h 0 f (0 ) lim ho h lim ho h ( h) ( e ) sin h h sinh h lim and lim e 0 h0 h ho f (0 ) 0 0 h h ( e ) sin h 0 f (0 ) lim ho h h ( h) ( e ) sin h lim h0 h sinh h lim and lim e 0 h0 h ho f (0 ) 0 0 VMC/JEE Mains-08 4 Solutions

15 f (0 ) f (0 ) 0 Similarly f ( ) f ( ) 0 Hence f ( x ) is differentiable x R dy y cot x 4x cosec x d y sin x 4xdx dx 54.() Integrating both sides we get: y sin x x c Also, y 0 c y sin x x 8 y () u ( a b) 0; u a 0 and u b 4. Let b ( b aˆ ) aˆ ( b uˆ ) uˆ ˆ ˆ b ( b a) ( b u) ( b uˆ ) b ( b aˆ ) uˆ (4) 7 uˆ u 6 56.() 57.() x 5 y z 4 P 5,, 4 P is foot of perpendicular from A to plane P,, x 4 y z Q 4,, Q is foot of perpendicular from B to plane 6 7 h x x h 00 h h 0 Q,, 4 6 PQ VMC/JEE Mains-08 5 Solutions

16 4 h (00) 4h h () 6 C4 C 4! (4) 60.() A A B a a a 6 b 5 b 4 b AC (6 ) Diameter AC Radius 5 PART-C CHEMISTRY 6.() I is - sp d hybridised - linear shape 6.(4) CH COOK is a salt of a weak acid and a strong base Most basic 6.() VMC/JEE Mains-08 6 Solutions

17 64.() Amidines, are stronger organic bases. 65.() Methyl orange is used for titration of strong acid and weak base. 66.() 67.() CxHyO z has z oxygen atom y y CxHy x O x CO HO 4 y O atoms required for combustion x 4 y z x 4 y z x 4 68.() During reduction HO HO During oxidation HO O 69.() Option () is correct [NCERT Class XII Part-II, Page No.-40] VMC/JEE Mains-08 7 Solutions

18 70.() BH6 O BO HO 7.66 nbh n O required = HO H O n-factor for O = 4 Number of equivalent = 4 F 96500C 7.() G H T S i t t s h. hr RT nk H TS H S nk RT R H Slope is R Since H is ve 7.() n 7.() r k A Slope is positive. n k (i) n 0.5 k (ii) From (i) and (ii) n Option () is correct [NCERT Class XII Part-II, Page No 405] 74.(4) Case - I Case - II VMC/JEE Mains-08 8 Solutions

19 Two isomers (fac and mer) are produced if reactant complex ion is a cis isomer. Only one isomer (fac) is formed if reactant complex ion is a trans isomer. 75 (4). 76 (). OH OH + NaOH H O + CO COOH (X) OH OCOCH COOH + (CO CO) O HSO4 Cat COOH + CH COOH (X) Aspirin [NCERT class XII part II/Page No. 0] 77.() 0 BaSO 4(s) Ba (aq) SO 4 (aq) Ksp 0 NaSO4 Na SO4 Conc. of For final solution M V M V i i f f SO4 in final solution 0 Ba SO M C C. 0 M 9 Ba 0 M 78.(4) Kjeldahl method is not applicable to compounds containing nitrogen in nitro (NO ) and azo (N = N ) groups and nitrogen present in the ring (pyridine) as nitrogen of these compounds does not change to ammonium sulphate. [NCERT Class XI part II/Page No. 58] NaOH Al NaOH Al OH NaAlO (M) white gel ppt (X) 79.() AlO Al OH AlO is used in chromatography as an absorbent. (Refer NCERT Class XIth/Part-II, Page-5) VMC/JEE Mains-08 9 Solutions

20 80.(4) HCl H Cl 0.M 0. M HS H HS HS H S K K HS H S K K K. 0 0 H S K H 0.M HS 0. S () (Refer NCERT Class XIth Part-II, Page-407), HS 0. 0 S 0 M The F ions make the enamel on teeth much harder by converting hydroxyapatite, into much harder fluorapatite i.e. Ca PO 8.() NH4NO N HO NH4 SO 4 NH HSO4 Ba N Ba N NH4 Cr O 7 N 4HO CrO 8.() NCERT Class XII/Part-II, Page No () Cr H O Cl 6 x 0 0 x Cr C6H6 x 0 0 x 0 K Cr CN O O NH x x 6 4 CaF 85.() Pressure of cation in interstitial sites is Frenkel defect. 86.() C6H 6( ) 5 O (g) 6CO (g) HO( ) n(g) H U n(g) RT kj / mol () BCl and AlCl are e deficient and thus act as Lewis acid Ca PO4 Ca OH, VMC/JEE Mains-08 0 Solutions

21 88.() KCl exist as K and Cl 89.() Depression in freezing pt Tf i Kf m Less the value of i, Higher the value of freezing pt. 90.() For () i = (min) H does not exist as Bond order is zero Electronic configuration of H : s * s B.O 0 VMC/JEE Mains-08 Solutions

Joint Entrance Exam Mains-2018

Joint Entrance Exam Mains-2018 Important Instructions: Joint Entrance Exam Mains-08 Paper Code - B 8 th April 08 9.0 AM.0 PM CHMISTRY, MATHEMATICS & PHYSICS. Immediately fill in the particulars on this page of the Test Booklet with