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1 IIJEE Solutions to JEE(MAIN)- XYZ PAPER : CHEMISRY, MAHEMAICS & PHYSICS est Booklet Code Read carefully the Instructions on the Back Cover of this est Booklet. Q Important Instructions:. Immediately fill in the particulars on this page of the est Booklet with Blue/Black Ball Point Pen. Use of pencil is strictly prohibited.. he test is of hours duration.. he est Booklet consists of 9 questions. he maximum marks are here are three parts in the question paper A, B, C consisting of Chemistry, Mathematics and Physics having questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response. 5. Candidates will be awarded marks as stated above in instruction No. 4 for correct response of each question. (/4) (one fourth) marks will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer sheet. 6. here is only one correct response for each question. illing up more than one response in any question will be treated as wrong response and marks for wrong response will be deducted accordingly as per instruction 5 above. IIJEE Ltd., IIJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , ax 6594

2 JEE (MAIN)--CMP - PAR A CHEMISRY. An unknown alcohol is treated with the Lucas reagent to determine whether the alcohol is primary, secondary or tertiary. Which alcohol reacts fastest and by what mechanism: () tertiary alcohol by S N () secondary alcohol by S N () tertiary alcohol by S N (4) secondary alcohol by S N Reaction proceeds through carbocation formation as carbocation is highly stable, hence reaction proceeds through S N with alcohol.. he first ionization potential of Na is 5. ev. he value of electron gain enthalpy of Na + will be: () 5. ev (). ev () +.55 ev (4).55 ev H =+ 5.eV + Na Na + e, here the backward reaction releases same amount of energy and known as H= 5.eV Electron gain enthalpy.. Stability of the species Li, Li and Li + increases in the order of: () + Li Li Li () + Li Li Li + < < () Li < Li < Li + < < (4) Li < Li < Li * ( ) =σ σ σ Li 6 s s s B.. = 4 = + ( ) * Li 5 =σs σs σ s B.. = =.5 * * ( ) =σ σ σ σ Li 7 s s s s 4 B.. = =.5 Li + is more stable than Li because Li has more numbers of antibonding electrons. 4. he molarity of a solution obtained by mixing 75 ml of.5 (M) HCl with 5 ml of (M)HCl will be: (). M (). 75 M ().975 M (4).875 M M V + M V = MV MV + MV M = = V M = Which of the following is the wrong statement? () molecule is bent () zone is violet-black in solid state () zone is diamagnetic gas (4) NCl and N are not isoelectronic Sol. (All the options are correct statements) () Correct, as is bent. IIJEE Ltd., IIJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , ax 6594

3 JEE (MAIN)--CMP- () Correct, as ozone is violet-black solid. () Correct, as ozone is diamagnetic. (4) Correct, as NCl = electrons and N = 4 electron hence are not isoelectronic. All options are correct statements. 6. our successive members of the first row transition elements are listed below with atomic numbers. Which Sol. () E M /M one of them is expected to have the highest E + + value? () Mn(Z = 5) () e(z = 6) () Co(Z = 7) (4) Cr(Z = 4) E E + + = Mn /Mn + + = e / e + + = Co /Co + + = Cr /Cr.57 V.77 V.97 V E.4V 7. A solution of ( ) chloro phenylethane is toluene racemises slowly in the presence of a small amount of SbCl 5, due to the formation of : () carbene () carbocation () free radical (4) carbanion Sol. () CH Cl SbCl CH 5 [ Ph CH CH ] [ SbCl ] Planer structure he coagulating power of electrolytes having ions Na +, Al + and Ba + for arsenic sulphide sol increases in the order: () () Na < Ba < Al () Al < Na < Ba (4) Ba < Na < Al Al < Ba < Na As S is an anionic sol (negative sol) hence coagulation will depend upon coagulating power of cation, which is directly proportional to the valency of cation (Hardy-Schulze rule). 9. How many litres of water must be added to litre of an aqueous solution of HCl with a ph of to create an aqueous solution with ph of? ().9 L (). L () 9. L (4). L Sol. () Initial ph =, i.e. [H + ] =. mole/litre New ph =, i.e. [H + ] =. mole/litre In case of dilution: M V = M V. =. V V = litre. Volume of water added = 9 litre.. Which one of the following molecules is expected to exhibit diamagnetic behaviour? () N () () S (4) C Sol. () & (4) both are correct answers. N Diamagnetic Paramagnetic S Paramagnetic IIJEE Ltd., IIJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , ax 6594

4 JEE (MAIN)--CMP -4 C Diamagnetic. Which of the following arrangements does not represent the correct order of the property stated against it? () Ni + < Co + < e + < Mn + : ionic size () Co + < e + < Cr + < Sc + : stability in aqueous solution () Sc < i < Cr < Mn : number of oxidation states (4) V + < Cr + < Mn + < e + : paramagnetic behaviour Sol. () & (4) both are correct answers) he exothermic hydration enthalpies of the given trivalent cations are: Sc + = 96 kj/mole e + = 449 kj/mole Co + = 465 kj/mole Cr + = 456 kj/mole Hence Sc + is least hydrated; so least stable (not most stable) e + contains 4 unpaired electrons where as Mn + contains 5 unpaired electrons hence (4) is incorrect.. Experimentally it was found that a metal oxide has formula M.98. Metal M, is present as M + and M + in its oxide. raction of the metal which exists as M + would be: () 4.8% () 6.5% () 5.8% (4) 7.% Metal oxide = M.98 If x ions of M are in + state, then x + (.98 x) = x =.4 So the percentage of metal in + state would be.4 4.8%.98 =. A compound with molecular mass 8 is acylated with CH CCl to get a compound with molecular mass 9. he number of amino groups present per molecule of the former compound is: () 5 () 4 () 6 (4) R NH + CH C Cl R NH C CH + HCl Each CH C addition increases the molecular wt. by 4. otal increase in m.wt. = 9 8 = hen number of NH groups = 5 4 = 4. Given E + =.74 V; E + =.5V Cr /Cr Mn 4 /Mn E =. V; E =.6 V + Cr 7 /Cr Cl/Cl Based on the data given above, strongest oxidising agent will be: () Cr + () Mn + () Mn 4 (4) Cl Sol. () As per data mentioned Mn is strongest oxidising agent as it has maximum SRP value. 4 IIJEE Ltd., IIJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , ax 6594

5 JEE (MAIN)--CMP-5 5. Arrange the following compounds in order of decreasing acidity: H H H H Cl (I) CH (II) N (III) CH (IV) () I > II > III > IV () III > I > II > IV () IV > III > I > II (4) II > IV > I > III Sol. () Correct order of acidic strength is III > I > II > IV 6. he rate of a reaction doubles when its temperature changes from K to K. Activation energy of such a reaction will be: (R = 8.4 JK mol and log =.) () 48.6 kj mol () 58.5 kj mol () 6.5 kj mol (4) 5.6 kj mol As per Arrhenius equation: k E æ a ö ln =- k R ç - çè ø E æ a ö. log =- ç - 8.4çè ø E a = 5.6 kj/mole 7. Synthesis of each molecule of glucose in photosynthesis involves: () molecules of AP () 8 molecules of AP () 6 molecules of AP (4) 8 molecules of AP Light reaction H + NADP + 8ADP 6 + 8AP + NADPH 6C + NADPH + 8AP C H + NADP + 8ADP + 6H Dark reaction 6 6 Net reaction: 6C + 6H C6H Which of the following complex species is not expected to exhibit optical isomerism? () Co( en) Cl () ( )( ) + + Co en NH Cl Sol. () [Co(NH ) Cl ] exists in two forms (facial and meridonial) () Co( NH ) (4) Co( en) Cl + IIJEE Ltd., IIJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , ax 6594

6 JEE (MAIN)--CMP -6 NH NH H N Cl H N Cl Co Co H N Cl Cl NH Cl facial Cl meridonial Both of these forms are achiral. Hence, [Co(NH ) Cl ] does not show optical isomerism. 9. A piston filled with.4 mol of an ideal gas expands reversibly from 5. ml to 75 ml at a constant temperature of 7. C. As it does so, it absorbs 8J of heat. he values of q and w for the process will be: (R = 8.4 J/mol K) (l n 7.5 =.) () q = 8 J, w = 8 J () q = 8 J, w = + 8 J () q = + 8 J, w = + 8 J (4) q = + 8 J, w = 8 J Process is isothermal reversible expansion, hence U =. q = W As q = +8 J Hence W = 8 J. A gaseous hydrocarbon gives upon combustion.7 g of water and.8 g of C. he empirical formula of the hydrocarbon is: () C H 4 () C 6 H 5 () C 7 H 8 (4) C H 4 Sol. () æ yö y CH x y + ç x+ ¾ ¾ xc + H çè 4ø Weight() g.8 g.7 g moles.7.4 x.7 = y/.4 x 7 Þ = y 8. he order of stability of the following carbocations: CH H C CH CH ; H C CH CH ; I II III is: () II > III > I () I > II > III () III > I > II (4) III > II > I Sol. () rder of stability is III > I > II. IIJEE Ltd., IIJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , ax 6594

7 JEE (MAIN)--CMP-7 (Stability extent of delocalization). Which of the following represents the correct order of increasing first ionization enthalpy for Ca, Ba, S, Se and Ar? () S < Se < Ca < Ba < Ar () Ba < Ca < Se < S < Ar () Ca < Ba < S < Se < Ar (4) Ca < S < Ba < Se < Ar Sol. () Increasing order of first ionization enthalpy is Ba < Ca < Se < S < Ar. or gaseous state, if most probable speed is denoated by C*, average speed by C and mean square speed by C, then for a large number of molecules the ratios of these speeds are: () C* : C : C =.8 :.5 : () C* : C : C = :.8 :.5 () C* : C : C = :.5 :.8 (4) C* : C : C =.5 :.8 : Sol. () R C* =, M 8R C =, πm C = R M 4. he gas leaked from a storage tank of the Union Carbide plant in Bhopal gas tragedy was: () Methylamine () Ammonia () Phosgene (4) Methylisocyanate It was methyl isocyanate (CH NC) 5. Consider the following reaction: z xmn4 + yc4 + zh ¾ ¾ xmn + yc + H he values of x, y and z in the reaction are, respectively: (), 5 and 8 (), 5 and 6 () 5, and 8 (4) 5, and 6 Sol. () Mn 5C 6H Mn C 8H x =, y = 5, z = 6 6. Which of the following exists as covalent crystals in the solid state? () Silicon () Sulphur () Phosphorous (4) Iodine Silicon (Si) covalent solid Sulphur (S 8 ) molecular solid Phosphorous (P 4 ) Molecular solid Iodine (I ) Molecular solid 7. Compound (A), C 8 H 9 Br, gives a white precipitate when warmed with alcoholic AgN. xidation of (A) gives a acid (B), C 8 H 6 4. (B) easily forms anhydride on heating. Identify the compound (A). IIJEE Ltd., IIJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , ax 6594

8 JEE (MAIN)--CMP -8 () C H 5 () CH Br Br () CH Br CH (4) CH Br CH CH Sol. () (A) CH Br AgN( alc. ) CH [] CH CH (B) H + CH + AgBr ( pale yellow ) CH pthalic anhydride 8. Energy of an electron is given by æ - 8 Z ö ç E=-.78 J ç n. Wavelength of light required to excite an çè ø electron in an hydrogen atom from level n = to n = will be (h = Js and c =. 8 ms ) ().86 7 m () m () m (4).4 7 m hc 8 E = =.78 Z λ 7 λ=.4 m 9. An organic compound A upon reacting with NH gives B. n heating B gives C. C in presence of KH reacts with Br to give CH CH NH. A is () CH CH CH CH () H C CH CH Sol. () CH () CH CH CH (4) CH CH IIJEE Ltd., IIJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , ax 6594

9 JEE (MAIN)--CMP-9 CH CH C (A) NH H ¾¾¾ CH CH C (B) /(- H) NH 4 CH CH C (C) NH Br / KH CH CH NH. In which of the following pairs of molecules/ions, both the species are not likely to exist? () H, He () H, He () H, He (4) H, He Sol. () Bond order of H + and He is zero, thus their existence is not possible. IIJEE Ltd., IIJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , ax 6594

10 JEE (MAIN)--CMP - PAR B MAHEMAICS. he circle passing through (, ) and touching the axis of x at (, ) also passes through the point () (, 5) () (5, ) () (, 5) (4) ( 5, ) Sol. () (x ) + y + λy = he circle passes through (, ) λ = λ = 4 (x ) + y + 4y = Clearly (5, ) satisfies.. ABCD is a trapezium such that AB and CD are parallel and BC CD. If ADB = θ, BC = p and CD = q, then AB is equal to p + q cosθ p + q () () pcosθ+ qsinθ p cosθ+ q sinθ () ( ) p + q sinθ ( pcosθ+ qsinθ) p + q sinθ ( ) (4) pcosθ+ qsinθ Using sine rule in triangle ABD AB BD = sin θ sin θ+α AB = AB = ( ) p + q sinθ p + q sinθ = sin θcos α+ cos θsin α sin θ q cos θ p + p + q p + q ( ) p + q sinθ θ+ θ. ( pcos qsin ) A π (θ + α) p + q θ α D q α B C p. Given : A circle, x + y = 5 and a parabola, y = 4 5 x. Statement I : An equation of a common tangent to these curves is y = x Statement II : If the line, y = mx+ (m ) is their common tangent, then m satisfies m 4 m + = m. () Statement - I is rue; Statement -II is true; Statement-II is not a correct explanation for Statement-I () Statement -I is rue; Statement -II is alse. () Statement -I is alse; Statement -II is rue (4) Statement -I is rue; Statement -II is rue; Statement-II is a correct explanation for Statement-I 5 Let the tangent to the parabola be y = mx + (m ). m Now, its distance from the centre of the circle must be equal to the radius of the circle. 5 5 So, m = + ( + m ) m = m 4 + m =. m (m ) (m + ) = m = ± So, the common tangents are y = x + 5 and y = x A ray of light along x+ y = gets reflected upon reaching x-axis, the equation of the reflected rays is IIJEE Ltd., IIJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , ax 6594

11 JEE (MAIN)--CMP- () y = x () y = x () y = x (4) y = x+ Slope of the incident ray is. So, the slope of the reflected ray must be. he point of incidence is (, ). So, the equation of reflected ray is y ( x ) =. 5. All the students of a class performed poorly in Mathematics. he teacher decided to give grace marks of to each of the students. Which of the following statistical measures will not change even after the grace marks were given? () median () mode () variance (4) mean Sol. () Variance is not changed by the change of origin. Alternate Solution: x x σ= for y = x + y= x + n y+ y y y. σ = = = σ n n 6. If x, y, z are in A.P. and tan x, tan y and tan z are also in A.P., then () x = y = 6z () 6x = y = z () 6x = 4y = z (4) x = y = z If x, y, z are in A.P. y = x + z and tan x, tan y, tan z are in A.P. tan y = tan x + tan z x = y = z. Note: If y =, then none of the options is appropriate., then xf 5 ( x ) 7. If f ( x) dx =Ψ( x) x x x x dx C x Ψ Ψ dx + C dxis equal to x Ψ x x Ψ x dx + C x x x x dx C Ψ Ψ + () Ψ( ) Ψ ( ) + () ( ) ( ) () ( ) ( ) (4) ( ) ( ) Sol. () f x dx =ψ ( ) ( x) Let x = t x dx = dt dx= tf () t dt t f t dt { f t dt } dt 5 then xf( x ) = ( ) ( ) = () () x ψ x x ψ x dx + C. IIJEE Ltd., IIJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , ax 6594

12 JEE (MAIN)--CMP - 8. he equation of the circle passing through the foci of the ellipse () () x + y 6y+ 7 = () x + y 6y+ 5= (4) foci (± ae, ) We have a e = a b = 7 Equation of circle (x ) + (y ) = ( 7 ) + ( ) x + y 6y 7 =. x y + =, and having centre at (, ) is 6 9 x + y 6y 5= x + y 6y 7 = 9. he x-coordinate of the incentre of the triangle that has the coordinates of mid points of its sides as (, ) (, ) and (, ) is () () + () (4) + x-coordinate = ax + bx + cx a+ b+ c = = = 4+ + =. Alternate Solution: x-coordinate = r = (s a) tan A/ 4+ π = tan 4 =. C(, ) (, ) (, ) A(, ) (, ) B(, ) 4. he intercepts on x-axis made by tangents to the curve, y = x, are equal to () ± () ± () ± 4 (4) ± dy x dx = = x = ± y = tdt= for x = and y = tdt= for x = tangents are y = (x ) y = x and y + = (x + ) y = x + Putting y =, we get x = and. 4. he sum of first terms of the sequence.7,.77,.777,.., is 7 () ( 99 7 ) () ( 99 7 ) Sol. () t r = r times () ( ) + (4) ( ) x y = t dt, x R, which are parallel to the line IIJEE Ltd., IIJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , ax 6594

13 JEE (MAIN)--CMP- = 7 ( r ) 7 = ( r ) 9 7 r 7 S = tr = 9 r= r= 9 9 = ( ) 7 79 = ( + ) 8 4. Consider : Statement I : (p ~ q) (~ p q) is a fallacy. Statement II : (p q) (~ q ~ p) is a tautology. () Statement - I is rue; Statement -II is true; Statement-II is not a correct explanation for Statement-I () Statement -I is rue; Statement -II is alse. () Statement -I is alse; Statement -II is rue (4) Statement -I is rue; Statement -II is rue; Statement-II is a correct explanation for Statement-I S: p q ~p ~q p^~q ~p^q (p^~q)^(~p^q) allacy S: p q ~ p ~ q p q ~ q ~ p (p q) (~ q ~ p) S is not an explanation of S autology 4. he area (in square units) bounded by the curves y = x, y x+ =, x-axis, and lying in the first quadrant is () 6 () 8 () 7 4 (4) 9 x = x 4x = x 6x + 9 x x + 9 x = 9, x = ( ) y + y dy y y + y = = tan A cot A he expression + cota tana can be written as () seca coseca + () tana + cota () seca + coseca (4) sina cosa +, x (, ) 9 IIJEE Ltd., IIJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , ax 6594

14 JEE (MAIN)--CMP -4 cot A cot A cosec A+ cota = = cot A cot A cot A cot A cot A cot A ( ) ( ) ( ) = + sec A cosec A 45. he real number k for which the equation, x + x + k = has two distinct real roots in [, ] () lies between and () lies between and () does not exist (4) lies between and Sol. () If x + x + k = has distinct real roots in [, ], then f (x) will change sign but f (x) = 6x + > So no value of k exists. 46. lim x ( cosx) ( + cosx) xtan4x is equal to () () () (4) 4 Sol. () ( cosx) lim + cos x x x tan4x ( ) ( ) sin x 4x lim + cos x x x 4 tan4x ( ) + =. 4 = ( ) 47. Let n be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If n+ n =, then the value of n is () 5 () () 8 (4) 7 n+ n C C = n C = n = At present, a firm is manufacturing items. It is estimated that the rate of change of production P w.r.t. Sol. () additional number of workers x is given by dp = x. If the firm employs 5 more workers, then dx the new level of production of items is () () 5 () 45 (4) 5 P 5 ( ) dp = x dx (P ) = 5 ( 5) / P = Statement I : he value of the integral b π/ dx π is equal to. tanx 6 π /6 + Statement II : f ( x) dx = f ( a+ b x) dx. a b a () Statement - I is rue; Statement -II is true; Statement-II is not a correct explanation for Statement-I () Statement -I is rue; Statement -II is alse. IIJEE Ltd., IIJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , ax 6594

15 JEE (MAIN)--CMP-5 Sol. () () Statement -I is alse; Statement -II is rue (4) Statement -I is rue; Statement -II is rue; Statement-II is a correct explanation for Statement-I I = π/ π/6 π/ π /6 dx + tanx tan x I = dx + tanx π I = 6 π I =. α 5. If P = is the adjoint of a matrix A and A = 4, then α is equal to 4 4 () () 5 () (4) 4 α P = 4 4 Adj A = A Adj A = 6 ( ) α (4 6) + (4 6) = 6. α 6 = 6. α =. α =. 5. he number of values of k, for which the system of equations (k + )x + 8y = 4k kx + (k + )y = k has no solution, is () () () (4) infinite or no solution k+ 8 4k = () k k+ k (k + ) (k + ) 8k = or k 4k + = k =, But for k =, equation () is not satisfied Hence k =. 5. If y = sec(tan x), then dy dx at x = is equal to () () () (4) IIJEE Ltd., IIJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , ax 6594

16 JEE (MAIN)--CMP -6 y = sec (tan x) dy = sec ( tan x) tan ( tan x) dx + x dy = =. dx x= 5. x y z 4 x y 4 z 5 If the lines = = and = = are coplanar, then k can have k k () exactly one value () exactly two values () exactly three values (4) any value Sol. () k = k ( + k) + ( + k ) ( k) = k + + k + + k = k + k = (k) (k + ) = values of k. 54. Let A and B be two sets containing elements and 4 elements respectively. he number of subsets of A B having or more elements is () () 9 () (4) 56 Sol. () A B will have 8 elements. 8 8 C 8 C 8 C = = uuur uuur If the vectors AB = i ˆ+ 4kˆ and AC = 5i ˆ j ˆ+ 4kˆ are the sides of a triangle ABC, then the length of the median through A is () 7 () () 45 (4) 8 Sol. () uuur uuur uuuur AB + AC AM = uuuur AM = 4i ˆ ˆj + 4kˆ uuuur AM = = A C M B 56. A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. he probability that a student will get 4 or more correct answers just by guessing is () () () (4) 5 5 Sol. () P (correct answer) = / IIJEE Ltd., IIJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , ax 6594

17 JEE (MAIN)--CMP C 4 + C = ( ) ( ) 57. If z is a complex number of unit modulus and argument θ, then + z arg + z equals π () θ () θ () π θ (4) θ Sol. () z = zz = + z + z = = z. + z + z 58. If the equations x + x + = and ax + bx + c =, a, b, c R, have a common root, then a : b : c is () : : () : : () : : (4) : : or equation x + x + = both roots are imaginary. Since a, b, c R. If one root is common then both roots are common Hence, a = b = c a : b : c = : :. 59. Distance between two parallel planes x + y + z = 8 and 4x + y + 4z + 5 = is () 5 () 7 () 9 (4) Sol. () 4x + y + 4z = 6 4x + y + 4z = 5 7 d min = 6 = 6 =. 6. x+ x he term independent of x in expansion of / / / x x + x x () () () (4) 4 is Sol. () / / / ( x + )( x x + ) ( x + ) ( x ) / / x x + x ( x ) r+ = ( ) r C r C 4 =. 5r 6 x r = 4 = (x / x / ) IIJEE Ltd., IIJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , ax 6594

18 JEE (MAIN)--CMP -8 PAR C PHYSICS 6. In an LCR circuit as shown below both switches are open initially. Now switch S is closed, S kept open. (q is charge on the capacitor and τ = RC is capacitive time constant). Which of the following statement is correct? V R S C S L () At t = τ, q = CV/ () At t = τ, q = CV( e ) τ () At t =,q= CV( e ) (4) Work done by the battery is half of the energy dissipated in the resistor. Sol. () Charge on the capacitor at any time t is t/ ( ) q = CV e τ at t = τ q = CV( e ) 6. A diode detector is used to detect an amplitude modulated wave of 6% modulation by using a condenser of capacity 5 pico farad in parallel with a load resistance kilo ohm. ind the maximum modulated frequency which could be detected by it. ().6 khz () 5. MHz () 5. khz (4).6 MHz Sol. () fc = = = 6.7 khz π RC.4 5 f C = cut off frequency As we know that f m f C () is correct Note: he maximum frequency of modulation must be less than f m, where m fm = fc m m modulation index 6. he supply voltage to a room is V. he resistance of the lead wires is 6 Ω. A 6 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 4 W heater is switched on in parallel to the bulb? ().9 Volt (). Volt ().4 Volt (4) zero volt Sol. () Resistance of bulb = = 4Ω 6 Resistance of Heater = = 6Ω 4 6 Ω Heater Bulb V IIJEE Ltd., IIJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , ax 6594

19 JEE (MAIN)--CMP-9 Voltage across bulb before heater is switched on, V = 4 46 Voltage across bulb after heater is switched on, V = Decrease in the voltage is V V =.4 (approximately) Note: Here supply voltage is taken as rated voltage. 64. A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half submerged in a liquid of density σ at equilibrium position. he extension x of the spring when it is in equilibrium is: Mg LAσ Mg LAσ () () k M k M Mg LAσ () + (4) Mg k M k (Here k is spring constant) Sol. () At equilibrium Σ = AL kx + σg Mg = LAσ x = Mg M kx Mg Buoyant force 65. wo charges, each equal to q, are kept at x = a and x = a on the x-axis. A particle of mass m and charge q q = is placed at the origin. If charge q is given a small displacement (y a) along the y-axis, the net force acting on the particle is proportional to: () y () y () (4) y y net = cosθ k q q/ = y ( ) a + y a + y kq y = (y a) a q a θ θ q/ y a q 66. A beam of unpolarised light of intensity I is passed through a polaroid A and then through another polaroid B which is oriented so that its principal plane makes an angle of 45 relative to that of A. he intensity of the emergent light is: () I / () I /4 () I /8 (4) I Sol. () 67. he anode voltage of a photocell is kept fixed. he wavelength λ of the light falling on the cathode is gradually changed. he plate current I of the photocell varies as follows: IIJEE Ltd., IIJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , ax 6594

20 JEE (MAIN)--CMP - () I () I () I λ (4) I λ λ λ Sol. () 68. wo coherent point sources S and S are separated by a small distance d as shown. he fringes obtained on the screen will be: () straight lines () semi-circles () concentric circles (4) points S d S D Screen Sol. () 69. A metallic rod of length l is tied to a string of length l and made to rotate with angular speed ω on a horizontal table with one end of the string fixed. If there is a vertical magnetic field B in the region, the e.m.f. induced across the ends of the rod is: () () Bωl 5Bωl () (4) 4Bωl Bωl Sol. () de = B( ωx) dx L e B xdx = = ω L 5BωL L x L ω 7. In a hydrogen like atom electron makes transition from an energy level with quantum number n to another with quantum number (n ). If n >>, the frequency of radiation emitted is proportional to () () / n n () (4) n n Sol. () ν (n ) n (n ) n (n ) (since n ) n IIJEE Ltd., IIJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , ax 6594

21 JEE (MAIN)--CMP- 7. Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? he surface tension is, density of liquid is ρ and L is its latent heat of vaporization. () / ρ L () /ρl () /ρl (4) ρl/ Sol. () ρ4πr RL = 4π[R (R - R) ] ρr RL = [R R + R R - R ] ρr RL = R R ( R is very small) R = ρ L. R 7. he graph between angle of deviation (δ) and angle of incidence (i) for a triangular prism is represented by: () δ () δ i i () δ (4) δ Sol. () i i 7. Let [ε ] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, = time and A = electric current, then: () [ε ] = [M - L - 4 A ] () [ε ] = [M - L - A - ] () [ε ] = [M - L - A] (4) [ε ] = [M - L - A] 7. q 4πε r = [A ] 4 ε = = [M L A ] [ML L ] 74. p A B p p D C v v v he above p-v diagram represents the thermodynamic cycle of an engine, operating with an ideal monoatomic gas. he amount of heat extracted from the source in a single cycle is () pv () pv () 4p v (4) p v Sol. () Heat is extracted from the source in path DA and AB is IIJEE Ltd., IIJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , ax 6594

22 JEE (MAIN)--CMP - PV 5 PV Q = R R R + R = PV 75. A sonometer wire of length.5 m is made of steel. he tension in it produces an elastic strain of %. What is the fundamental frequency of steel if density and elasticity of steel are 7.7 kg/m and. N/m respectively? () 78. Hz ().5 Hz () 77 Hz (4) 88.5 Hz undamental frequency f = l µ = l Aρ = stress. = l ρ his question has statement I and statement II. f the four choices given after the statements, choose the one that best describes the two statements. Statement- I: Higher the range, greater is the resistance of ammeter. Statement- II: o increase the range of ammeter, additional shunt needs to be used across it. () Statement I is true, Statement II is true, Statement II is not the correct explanation of Statement I. () Statement I is true, statement II is false. () Statement I is false, Statement II is true (4) Statement I is true, Statement II is true, Statement II is the correct explanation of statement- I. Sol. () or Ammeter, S = IG g I I g So for I to increase, S should decrease, so additional S can be connected across it. 77. What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of R? () GmM () GmM R R () GmM R. E f = GMm 6R. E i = GMm R W =.E f.e i = 5GMm 6R 78. A projectile is given an initial velocity of ( iˆ+ ˆ) vertical. If g = m/s, the equation of its trajectory is: () () y = x 5x () 4y = x 5x (4) (4) 5GmM 6R j m/s, where î is along the ground and ĵ is along the 4y = x 5x y = x 5x IIJEE Ltd., IIJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , ax 6594

23 JEE (MAIN)--CMP- x = t y = t 5t Equation of trajectory is y = x 5x 79. wo capacitors C and C are charged to V and V respectively. It is found that by connecting them together the potential on each one can be made zero. hen : () C = 5C () C+ 5C = () 9C = 4C (4) 5C = C C = C 6C = C C = 5C 8. A hoop of radius r and mass m rotating with an angular velocity ω is placed on a rough horizontal surface. he initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases to slip? () rω () () rω (4) Sol. () rom conservation of angular momentum about any fix point on the surface mr ω = mr ω = ω V CM ω r = ω 8. An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. he piston and cylinder have equal cross sectional area A. When the piston is in equilibrium, the volume of the gas is V and its pressure is P. he piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency: () VMP π A γ MV () π AγP () (4) rω rω 4 π A γ P MV Aγ P π VM Sol. () P atm A P A mg BD of piston at equilibrium P atm A + mg = P A...() P atm A mg (P + dp) A BD of piston when piston is pushed down a distance x IIJEE Ltd., IIJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , ax 6594

24 JEE (MAIN)--CMP -4 d x P atm + mg (P +dp) A = m...() dt Process is adiabatic PV γ γ = C dp = PdV V Using,, me get f = π A γ P MV 8. A charge Q is uniformly distributed over a long rod AB of length L as shown in the figure. he electric potential at the point lying at a distance L from the end A is: A B L L Q Q () () 4πε L 4πε L ln Sol. () Q ln () 4πε L (4) Q 8πε L x dx = ln = x L k Q Q V = x= L dx x L 4πε L 8. A circular loop of radius. cm lies parallel to a much bigger circular loop of radius cm. he centre of the small loop is on the axis of the bigger loop. he distance between their centres is 5 cm. If a current of. A flows through the smaller loop, then the flux linked with bigger loop is () 6 weber (). weber () weber (4) 9. weber r R d Let M be the coefficient of mutual induction between loops φ = M i M ir / µ ( d + R ) π r µ R πr = / ( d + R ) = M i φ = Mi φ = 9. weber 84. If a piece of metal is heated to temperature θ and then allowed to cool in a room which is at temperature θ the graph between the temperature of the metal and time t will be closest to : () () θ θ t t IIJEE Ltd., IIJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , ax 6594

25 JEE (MAIN)--CMP-5 () (4) θ t t Sol. () he temperature goes on decreasing with time (non-linearly) he rate of decrease will be more initially which is depicted in the second graph. 85. he I V characteristic of an LED is () B G Y R () () V V (4) R Y G B I I I Red Yellow Green Blue V (R)(Y)(G)(B) or LED, in forward bias, intensity increases with voltage. V 86. his question has Statement I and Statement II. f the four choices given after the Statements, choose the one that best describes the two Statements. Statement I : A point particle of mass m moving with speed v collides with stationary point particle of m mass M. If the maximum energy loss possible is given as f mv then f = M + m. Statement II : Maximum energy loss occurs when the particles get stuck together as a result of the collision. () Statement I is true, Statement II is true, Statement II is not a correct explanation of Statement I. () Statement I is true, Statement II is false. () Statement I is false, Statement II is true (4) Statement I is true, Statement II is true, Statement II is a correct explanation of Statement I. Sol. () Loss of energy is maximum when collision is inelastic as in an inelastic collision there will be maximum deformation. Mm KE in CM frame is V M + m rel Mm KE i = V KE f = ( QV rel = ) M + m Mm Hence loss in energy is V M + m M f = M + m IIJEE Ltd., IIJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , ax 6594

26 JEE (MAIN)--CMP he amplitude of a damped oscillator decreases to.9 times its original magnitude is 5s. In another s it will decrease to α times its original magnitude, where α equals. ().8 ().79 ().6 (4).7 Sol. () A= A e kt 5k =.9A A e 5k and α A = Ae solving α = Diameter of plano-convex lens is 6 cm and thickness at the centre is mm. If speed of light in material of lens is 8 m/s, the focal length of the lens is : () cm () cm () cm (4) 5 cm Sol. () R R-t d t R = d + ( R t ) t R d = R R d t = R R () 9 R = = = 5 cm ( ) 6 = ( µ ) f R R = f 5 f = cm 89. he magnetic field in a travelling electromagnetic wave has a peak value of n. he peak value of electric field strength is : () 6 V/m () 9 V/m () V/m (4) V/m Ε =CB = 8 9 = 6 V/m 9. wo short bar magnets of length cm each have magnetic moments. Am and. Am respectively. hey are placed on a horizontal table parallel to each other with their N poles pointing towards the South. hey have a common magnetic equator and are separated by a distance of. cm. he value of the resultant horizontal magnetic induction at the mid - point of the line joining their centres is close to (Horizontal component of earth s magnetic induction is.6 5 Wb/m ) IIJEE Ltd., IIJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , ax 6594

27 JEE (MAIN)--CMP-7 ().56 4 Wb/m ().5 4 Wb/m () Wb/m (4).6 5 Wb/m B = B + B + B net M M H = µ M 4 + µ M 4 + B πx πx H = µ ( M ) 4 + M + B π x H 7 5 =. +.6 =.56 4 Wb/m IIJEE Ltd., IIJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , ax 6594