Joint Entrance Exam/JEE Mains 2015 Code A

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1 Joint Entrance Exam/JEE Mains 05 Code A PART-A PHYSICS.() Δy urt 0 0 t 0t (As long as in air) When one reaches ground graph will be a parabola opening downwards..() L t n g n Lt g Δ g Δ Δ L t g L t %.() For equation of A f 0 For equilibrium of B f 00 f f 0.() From Conservation of momentum 5.() mv mv V V m V mv m V % energy loss m V mv 5mV / 00 mv % 9 xdm xcm M h x r dx 0 h M 00 6.() For Max. volume R a R a a, M M πr M VMC/JEE Mains Solutions/JEE-05

2 M ' a M R Icube 6 6 MR 9 7.() 8.() 9.() M Vc V full V cut M cut 8 GM R GM / 8 = R R R / = GM R l gt T π l0 g π Also, aδl Y Fl0 gtm l ' π Aπ g Tm T 0 Mg gt π A Tm Mg T P T (Given) PV nrt (Ideal gas) T V π R T R 0.() We have assumed temperature to be in Kelvin Sequentially keeping in contact with reservoirs dq dt ΔS ms T T dt dt n T T Sequentially keeping in contact with 8 reservoirs T T 00 dt dt dt Δs... n T T T 00 T T 7 Where T T T... T7 are temperature after each st, nd, rd. reservoirs have supplied heat to body. mean free path.() Time( t) = molecular speed We know that: Hence So, V V T mean free path volume Molecular speed T volume τ T Q VMC/JEE Mains Solutions/JEE-05

3 .() Q V T const Q V T const We know that, for adiabatic Process : TV PE K x γ const γ Q TV T V γ Q γ Q KE K ( A x ) Hence ()..() Δf 0 f 0 0 f = 0 f Δf f 00 0 %.() At P, the field due to upper part is E & due to lower point is E Hence net field will be in the direction of E net Hence (). 5.(,) For solid sphere. 0 V 0 Q Vin R r R r 8πε R R V out Q V πε 0 r Q Given V0 πε 0 R V V,V V in 0 0 out 0 R r R = 0 5V0 V0 R R R R R V 0 V0R R R R E E E net V0 V0R 8R R R, R R Hence R R R Also R > R R VMC/JEE Mains Solutions/JEE-05

4 Q Q 6.() Q C CE Q E C C Q CE CE, whose graph will be as shown in () C C E E at C, Q E at C, Q E 7.() V ρl neavd A 5 ρ ρ Ωm 8.() from kirhhoffs law 6 i i i 0... (i) 9 i i i (ii) i 0. AQ to P 9.() As the solenoids are coaxial net force by the solenoid on other should be zero (by symmetry) F F 0 0.() μ sinθ 0 I L T π(l sin θ) T cosθ (λ L) g μ tan θ 0 I L πl sin θ λlg πlλg sin θ tan θ I μ0 I sinθ πλgl μ 0 cosθ.() For stable equilibrium magnetic moment should be in the direction of external field and for unstable equilibrium magnetic moment should be opposite to external field. 5.() I 0 0. A Rt I I L 0 e K closed for long VMC/JEE Mains Solutions/JEE-05

5 I e I ma e 50 K open K ma.() P 0 E C πr P E π 0 Cr 9 0. E 90 E.5 V / m 8 0.() For transmission at AC r r C sin μ Also, sinθ μ sinr = μ sina r θ A A r r sin sin A sin 5.() As μ is increasing in the upward direction, hence speed is decreasing as we go up. So the rays at the bottom will be ahead of the rays at the top, hence the wavefront bends as shown. And the waves move to the wavefront. Hence the beam will bend upwards. B C 6.(). Δθ d x. 5( Δθ) 0m () z n V r n z K mv V Kqq r E. 6 z n 8.() (A) (ii) (B) (i) (C) (iii) 9.() Resultant frequencies are c m, c and c m. VMC/JEE Mains 5 Solutions/JEE-05

6 bt A t Ae m in Damped SHM max 0.() As Similarly Q t Q e max 0 Rt Q L max Q 0 e L R A B Rt L LA LB LA LB R R Then A correspond to L And B correspond to L Solutions JEE Mains-05 PART-B CHEMISTRY.() Mole of g resin Mol of 06 Ca uptake mol of 06 C8H7SONa Ca ( C8H7SO ) Ca Na Ca per g of resin.() r a.().7 o o r.9 A.86 A z En.6 ev n z = Atomic Number For hydrogen (z = ).6 En ev n Energy for possible excited states is given by :.() Fion-dipole r.6 n, E. ev 5.() NO(g) O (g) NO (g) () o G NO 86.6 kj / mol or J / mol f G o f (O ) 0 VMC/JEE Mains 6 Solutions/JEE-05

7 G o f (NO )? o G (for the above reaction ) Writing for the reaction: G o reaction G o o f (NO ) G f NO 0 = RT ln k R(98) ln(.6 0 ) R 98 ln.60 G o f (NO ) G o f (NO ) 0.5[ R(98) ln(.6 0 )] P 6.() A n x B o B P n A A 7.(). / M B / 58 MB 6 g mol o G 9.J RT ln K eq Keq e Q Qc Keq reaction goes backward 8.() Cu (aq) e Cu(s) Faraday gives mole of Cu. Weight of deposited Cu = 6.5 g 9.() Probability of collision of more than three particles simultaneously at certain orientation is very low. 0.() Initial millimoles of acetic acid After filtration, millimoles of acetic acid left Adsorbed millimoles of acetic acid. 0.9 Wt. of acetic acid adsorbed mg. Wt. of acetic acid adsorbed per gm of charcoal 5 / 8.() Order of ionic radii in N > O > F Hence correct order is.7,.0 and.6 VMC/JEE Mains 7 Solutions/JEE-05

8 .() Equation at cathode Al (melt) e Al( l) At Anode C( s) O ( melt) CO( g) e C( s) O ( melt) CO ( g) e In the metallurgy of aluminium, purified AlO is mixed with NaAlF 6 or CaF which lowers the melting point of the mix and brings conductivity. The fused matrix is electrolyzed. Na AlF 6 alone is not the electrolyte.. () HO HO O O O e O e O So H O can act as both reducing as well as oxidizing agent..() For alkaline earth metal sulphates solubility decreases down the group. So only for the soluble BeSO, hydration energy is greater than its lattice enthalpy. 5.() For inter-halogen compounds (X X ) bond energy is lesser than general X X bond {except for F - F}, and also inter-halogen bond is polar. So they are more reactive. 6.() TiCl Ziegler-Natta polymerization PdCl Wacker process CuCl Deacon s process VO5 Contact process 7.() Down the group in inert gases inter-particle forces increase, so boiling point increases. 8.() Complex is a Mabcd type square planar complex, so it has possible geometrical isomers. 9.() KMnO is coloured due to charge transfer from oxide to Mn +7 i.e from ligand to metal (L M). 50.() Nitrogen and oxygen do not react at ambient temperatures. In order to produce various oxides of nitrogen, they have to undergo endothermic reaction at high temperatures. Generally the range of temperature to react N and o o O to form oxides is 00 C to 800 C. 5.() Mass of organic compound = 50 mg Mass of AgBr = mg. Mole of Br = mole of AgBr Mass of Br % of Bromine in organic compound = %. 50 VMC/JEE Mains 8 Solutions/JEE-05

9 5.() Ph CH CH = CH CH -phenyl--butene exhibits geometrical isomerism 5.(). O. H O/Zn 5-keto--methyl hexanal. 5.() Alkyl fluorides are generally formed by swartz Reaction. AgF/ HgF / CoF R X R F ppt. X Cl / Br 55.() 56.() 57.() Glyptal is a polyester of glycerol and pthalic acid used in paints and lacquers. 58.() Vitamin C is soluble in water 59.() Phenelzine is used as an anti-depressant and is not an antacid. 60.() Zn on reaction with K [Fe(CN) 6] produce bluish white ppt. due to formation of Zn [Fe(CN) 6]. All other substances are yellow coloured. VMC/JEE Mains 9 Solutions/JEE-05

10 Solutions JEE Mains-05 PART-C MATHEMATICS 6.() n A n B B 8 n A Number of subsets having atleast three elements C0 C C () z z, z zz z z zz z z z z z z z z z z z z z 0 z 6.() 6.() n n n 6 0 n n n 6 0 Subtract above two equations we get: an 6an an 0 A a b T A A 9I a a b b a b 0 a b.. (i) a b 0 a b.. (ii) b, a an an an VMC/JEE Mains 0 Solutions/JEE-05

11 65.() (), 5, 6, 7, 8 67.() 0,, digit numbers 7 ( & 5 can t come at st place) 5 digit numbers numbers x C0 C x C x C x C x C 50 x x C0 C x C x... C50 x x x C0 C..x... C50 On adding 68.() 69.() Putting x = on both sides n m, G, G, G, n in G.P. n r G n, G n, G G G G n n n m n nn... n n tn... n n n t n n 70.() lim x 0 cos x cos x xtan x sin x lim tan x x 0 x. x x VMC/JEE Mains Solutions/JEE-05

12 7.() g g g k m k g g m k m 8m m 5m 8 m, k 5 5 k m Vidyamandir Classes 7.() x xy y 0..(i) Differentiating w.r.t. x dy dy x y x 6y 0 dx dx dy x y x y 0 dx dy x y dx x y dx x y dy x y At (, ) dx dy Equation of normal at (, ) y x Or y x y x.(ii) Substituting value of y from (ii) in (i) We get x x x x x 0 x x x x x 0 x 6x 0 x x x x 0 0 x, y which is in th quadrant. 7.() f x ax bx cx dx e x f lim 0 x x x f lim x 0 x ax bx cx dx e lim x 0 x d 0, e 0, c f x ax bx x f x ax bx x VMC/JEE Mains Solutions/JEE-05

13 a b (i) f a b 8 0 8a b (ii) Solving (i) and (ii) a a, b f x x x x f () dx dx x x x x Let x t 5 x dx dt dt t x I C t C x C C t x 75.() I dx I 76.() x x 6x 0x x y,, 8 y y y y 9 dy y / / / dy y 77.() dx xlog x I.F. dx x log x e log x Required solution is y log x log x dx y log x x log x C As x = is in domain, replace x = y 0 (0 ) C 0 C C y log x xlog x y e VMC/JEE Mains Solutions/JEE-05

14 78.() No. of points on L (, 9) (9,) = 9 Similarly points on other lines would be 8, 7,.., total points = () Let P, Let point of intersection of x y 0 and x y 0 be Q Q, PQ Since PAR QAR AQ Point Q is always lies at distance of unit from point A. 80.() Equation of circles are: x y 5..(i) C,,r 5 And x y 9 8 C, 9, r 8..(ii) CC 9 = = r r No. of common tangents = VMC/JEE Mains Solutions/JEE-05

15 8.() x y [Equation of tangent x y ] Area of quadrilateral () 8.() t h t t t k Eliminating t h k x y x y z k x k y k satisfies x y z 6 z k k k k 6 Point of intersection k Distance is 5,, k 8.() Equation of family of planes containing line x 5y z and x y z 5 is x y z k x y z xk y k z k k (i) (i) is parallel to x y 6z.(ii) k k 5 k 6 Substituting value of k in (i), we get k 6 k 5 k x y 5 z 5 0 7x y z 9 0 x y 6z () a b c b c a If a.c 0, then b.c b c a.c b b.c a b c a VMC/JEE Mains 5 Solutions/JEE-05

16 86. b c cos b c cos sin 87.() Total = 6 6 = 56 New after removing one 6 Total = 0 & no. of entries = 5 New total after adding,, 5 = 0 + = 5 No. of entries = 5 + = 8 Mean = () Let height of tower be PQ = h. In PAQ, h tan0 AQ h AQ In PBQ, BQ h AB h h In PCQ, tan60 CQ CQ h 89.() BC = h h : h : x tan y tan x tan x, x AB:BC = x tan z x x tan tan tan z For 90.() ~ ~ s ( ~ r s z tan x, x, x x, tan tan x x tan y tan x tan x tan y tan x x x y x s ~ ~ r s s r ~ s s r s ~ s s r s r F VMC/JEE Mains 6 Solutions/JEE-05