Page 1 :
JEE Main 2013, , MathonGo, , PART A −CHEMISTRY, 1., , 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:, (1) tertiary alcohol by SN1, (2) secondary alcohol by SN2, (3) tertiary alcohol by SN2, (4) secondary alcohol by SN1, , Sol., , (1), Reaction proceeds through carbocation formation as 30 carbocation is highly stable, hence reaction proceeds, through SN1 with 30 alcohol., , 2., , The first ionization potential of Na is 5.1 eV. The value of electron gain enthalpy of Na+ will be:, (3) –10.2 eV, (1) – 5.1 eV, (3) + 2.55 eV, (4) – 2.55 eV, , Sol., , (1), ∆H =+5.1eV, , , → Na + + e − , here the backward reaction releases same amount of energy and known as, Na ←, , ∆H =−5.1eV, , Electron gain enthalpy., 3., , Sol., , Stability of the species Li2, Li −2 and Li +2 increases in the order of:, (1) Li 2− < Li +2 < Li 2, , (3) Li 2 < Li 2− < Li 2+, , (3) Li −2 < Li 2 < Li 2+, , (4) Li 2 < Li +2 < Li −2, , (1), , Li 2 ( 6 ) = σ1s2 σ* 1s 2 σ2s 2, B.O. =, , 4−2, =1, 2, , Li +2 ( 5 ) = σ1s2 σ* 1s 2 σ2s1, B. O. =, , 3− 2, = 0.5, 2, , Li −2 ( 7 ) = σ1s 2 σ* 1s 2 σ2s 2 σ* 2s1, , 4−3, = 0.5, 2, Li +2 is more stable than Li −2 because Li −2 has more numbers of antibonding electrons., , B.O. =, , 4., , The molarity of a solution obtained by mixing 750 mL of 0.5 (M) HCl with 250 mL of 2(M)HCl will be:, (1) 1.00 M, (2) 1. 75 M, (3) 0.975 M, (4) 0.875 M, , Sol., , (4), M1V1 + M2V2 = MV, M V + M 2 V2 0.5 × 750 + 2 × 250, M= 1 1, =, V, 1000, M = 0. 875, , 5., , Which of the following is the wrong statement?, (1) O3 molecule is bent, (3) Ozone is diamagnetic gas, , Sol., , (2) Ozone is violet-black in solid state, (4) ONCl and ONO– are not isoelectronic, , (All the options are correct statements), O, O, O is bent., (1) Correct, as, , www.mathongo.com
Page 2 :
JEE Main 2013, , 6., , MathonGo, , (2) Correct, as ozone is violet-black solid., (3) Correct, as ozone is diamagnetic., (4) Correct, as ONCl = 32 electrons and ONO− = 24 electron hence are not isoelectronic., All options are correct statements., Four successive members of the first row transition elements are listed below with atomic numbers. Which, one of them is expected to have the highest E 0M3+ /M2+ value?, , (1) Mn(Z = 25), (3) Co(Z = 27), Sol., , (2) Fe(Z = 26), (4) Cr(Z = 24), , (3), E 0Mn +3 / Mn +2 = 1.57 V, , E 0Fe+3 / Fe+2 = 0.77 V, E 0Co+3 / Co+2 = 1.97 V, , E 0Cr+3 /Cr +2 = −0.41 V, , 7., , A solution of (–) –1 – chloro –1 – phenylethane is toluene racemises slowly in the presence of a small, amount of SbCl5, due to the formation of :, (1) carbene, (2) carbocation, (3) free radical, (4) carbanion, , Sol., , (2), , CH, , +, , −, , SbCl5, CH3 , → [ Ph − CH − CH 3 ] [SbCl6 ], Planer structure, , Cl, 8., , The coagulating power of electrolytes having ions Na+, Al3+ and Ba2+ for arsenic sulphide sol increases in, the order:, (2) Ba 2+ < Na + < Al3+, (1) Na + < Ba 2+ < Al3+, 3+, +, 2+, (3) Al < Na < Ba, (4) Al3+ < Ba 2+ < Na +, , Sol., , (1), As2S3 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 1 litre of an aqueous solution of HCl with a pH of 1 to create an, aqueous solution with pH of 2?, (1) 0.9 L, (2) 2.0 L, (3) 9.0 L, (4) 0.1 L, , Sol., , (3), Initial pH = 1, i.e. [H+] = 0.1 mole/litre, New pH = 2, i.e. [H+] = 0.01 mole/litre, In case of dilution: M1V1 = M2V2, 0.1 ×1 =0.01 × V2, V2 = 10 litre., Volume of water added = 9 litre., , 10., , Which one of the following molecules is expected to exhibit diamagnetic behaviour?, (1) N2, (2) O2, (3) S2, (4) C2, , Sol., , (1) & (4) both are correct answers., N2 → Diamagnetic, O2 → Paramagnetic, S2 → Paramagnetic, , www.mathongo.com
Page 3 :
JEE Main 2013, , MathonGo, , C2 → Diamagnetic, 11., , Which of the following arrangements does not represent the correct order of the property stated against it ?, (1) Ni2+ < Co2+ < Fe2+ < Mn2+ : ionic size, (2) Co3+ < Fe3+ < Cr3+ < Sc3+ : stability in aqueous solution, (3) Sc < Ti < Cr < Mn : number of oxidation states, (4) V2+ < Cr2+ < Mn2+ < Fe2+ : paramagnetic behaviour, , Sol., , (2) & (4) both are correct answers), The exothermic hydration enthalpies of the given trivalent cations are:, Sc+3 = 3960 kJ/mole, Fe+3 = 4429 kJ/mole, Co+3 = 4653 kJ/mole, Cr+3 = 4563 kJ/mole, Hence Sc+3 is least hydrated; so least stable (not most stable), Fe+2 contains 4 unpaired electrons where as Mn+2 contains 5 unpaired electrons hence (4) is incorrect., , 12., , Experimentally it was found that a metal oxide has formula M0.98O. Metal M, is present as M2+ and M3+ in, its oxide. Fraction of the metal which exists as M3+ would be:, (1) 4.08%, (2) 6.05%, (3) 5.08%, (4) 7.01%, , Sol., , (1), Metal oxide = M0.98O, If ‘x’ ions of M are in +3 state, then, 3x + (0.98 – x) × 2 = 2, x = 0.04, , So the percentage of metal in +3 state would be, , 0.04, × 100 = 4.08%, 0.98, , 13., , A compound with molecular mass 180 is acylated with CH3COCl to get a compound with molecular mass, 390. The number of amino groups present per molecule of the former compound is:, (1) 5, (2) 4, (3) 6, (4) 2, , Sol., , (1), O, , O, , ||, , ||, , → R − NH − C − CH 3 + HCl, R − NH 2 + CH 3 − C − Cl , O, ||, , Each CH 3 − C addition increases the molecular wt. by 42., Total increase in m.wt. = 390 – 180 = 210, 210, Then number of NH2 groups =, =5, 42, 14., , Given, E 0Cr3+ /Cr = −0.74 V; E 0MnO− /Mn 2+ = 1.51 V, 4, , E 0Cr O2− /Cr3+ = 1.33 V; E 0Cl/ Cl− = 1.36 V, 2, , 7, , Based on the data given above, strongest oxidising agent will be:, (1) Cr3+, (2) Mn2+, −, (3) MnO 4, (4) Cl−, Sol., , (3), As per data mentioned, MnO −4 is strongest oxidising agent as it has maximum SRP value., , www.mathongo.com
Page 5 :
JEE Main 2013, , MathonGo, , NH3, H3 N, , NH3, H3N, , Cl, , Cl, , Co, , H3 N, , Co, , Cl, , Cl, , NH3, , Cl, facial, , Cl, meridonial, , Both of these forms are achiral. Hence, [Co(NH3)3Cl3] does not show optical isomerism., 19., , A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant, temperature of 37.00C. As it does so, it absorbs 208J of heat. The values of q and w for the process will be:, (R = 8.314 J/mol K) ( l n 7.5 = 2.01), (1) q = – 208 J, w = – 208 J, (2) q = – 208 J, w = + 208 J, (3) q = + 208 J, w = + 208 J, (4) q = + 208 J, w = – 208 J, , Sol., , (4), Process is isothermal reversible expansion, hence ∆U = 0., ∴ q = −W, As q = +208 J, Hence W = −208 J, , 20., , A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of CO2. The empirical formula of, the hydrocarbon is:, (1) C3H4, (2) C6H5, (3) C7H8, (4) C2H4, , Sol., , (3), , æ yö, y, ® xCO2 + H 2 O, C x H y + ççx + ÷, ÷O2 ¾ ¾, çè, ø, 4÷, 2, Weight (g ), , 3.08 g, , 0.72 g, , moles, , 0.07, , 0.04, , x, 0.07, =, y / 2 0.04, Þ, , 21., , x 7, =, y 8, , The order of stability of the following carbocations:, , CH2, , H2C, , CH CH2 ; H3C, I, , CH2, II, , CH2 ;, III, , is:, (1) II > III > I, (3) III > I > II, Sol., , (2) I > II > III, (4) III > II > I, , (3), Order of stability is III > I > II., , www.mathongo.com
Page 6 :
JEE Main 2013, , MathonGo, , (Stability ∝ extent of delocalization), 22., , Which of the following represents the correct order of increasing first ionization enthalpy for Ca, Ba, S, Se, and Ar?, (1) S < Se < Ca < Ba < Ar, (2) Ba < Ca < Se < S < Ar, (3) Ca < Ba < S < Se < Ar, (4) Ca < S < Ba < Se < Ar, , Sol., , (2), Increasing order of first ionization enthalpy is, Ba < Ca < Se < S < Ar, , 23., , For 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:, (2) C* : C : C = 1:1.128 :1.225, (1) C* : C : C = 1.128 :1.225 :1, (3) C* : C : C = 1:1.125 :1.128, (4) C* : C : C = 1.225 :1.128 :1, , Sol., , (2), , C* =, , 2RT, , C=, M, , 8RT, , C=, πM, , 3RT, M, , 24., , The gas leaked from a storage tank of the Union Carbide plant in Bhopal gas tragedy was:, (1) Methylamine, (2) Ammonia, (3) Phosgene, (4) Methylisocyanate, , Sol., , (4), It was methyl isocyanate (CH3NCO), , 25., , Consider the following reaction:, z, H2O, 2, The values of x, y and z in the reaction are, respectively:, (1) 2, 5 and 8, (2) 2, 5 and 16, (3) 5, 2 and 8, (4) 5, 2 and 16, ® xMn 2+ + 2yCO 2 +, xMnO-4 + yC2 O 24- + zH+ ¾ ¾, , Sol., , (2), , → 2Mn +2 + 10CO 2 + 8H 2O, 2MnO 4− + 5C2O4−2 + 16H + , x = 2, y = 5, z = 16, 26., , Which of the following exists as covalent crystals in the solid state?, (1) Silicon, (2) Sulphur, (3) Phosphorous, (4) Iodine, , Sol., , (1), Silicon (Si) – covalent solid, Sulphur (S8) – molecular solid, Phosphorous (P4) – Molecular solid, Iodine (I2) – Molecular solid, , 27., , Compound (A), C8H9Br, gives a white precipitate when warmed with alcoholic AgNO3. Oxidation of (A), gives a acid (B), C8H6O4. (B) easily forms anhydride on heating. Identify the compound (A)., , www.mathongo.com
Page 8 :
JEE Main 2013, , MathonGo, , O, CH3CH2, , C, , O, NH 3, , OH ¾ ¾ ¾, ® CH3CH2, , (A), , C, (B), , ONH4, , ∆ / (- H 2 O), O, , CH3CH2, , C NH2, (C), , Br2 / KOH, , CH3CH2 NH2, 30., , In which of the following pairs of molecules/ions, both the species are not likely to exist?, (2) H 22+ , He 2, (1) H-2 , He22(3) H-2 , He22+, , Sol., , (4) H +2 , He 22-, , (2), Bond order of H 22+ and He2 is zero, thus their existence is not possible., , www.mathongo.com
Page 9 :
JEE Main 2013, , MathonGo, , PART B − MATHEMATICS, 31., , The circle passing through (1, −2) and touching the axis of x at (3, 0) also passes through the point, (1) (2, −5), (2) (5, −2), (3) (−2, 5), (4) (−5, 2), , Sol., , (2), (x − 3)2 + y2 + λy = 0, The circle passes through (1, − 2), ⇒ 4 + 4 − 2λ = 0 ⇒ λ = 4, (x − 3)2 + y2 + 4y = 0 ⇒ Clearly (5, − 2) satisfies., , 32., , 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, p2 + q 2, p 2 + q 2 cos θ, (2) 2, (1), p cos θ + q sin θ, p cos θ + q 2 sin θ, (3), , Sol., , (p, , 2, , + q 2 ) sin θ, , ( p cos θ + q sin θ ), , ⇒ AB =, , ⇒ AB =, , + q 2 ) sin θ, , p cos θ + q sin θ, , A, , p + q sin θ, , (p, , 2, , + q 2 ) sin θ, , ( p cos θ + q sin θ ), , α, , p, , 2, , sin θ cos α + cos θ sin α, , B, , π −(θ + α), , p + q sin θ, sin θ ⋅ q, cos θ ⋅ p, +, 2, 2, p +q, p2 + q 2, 2, , =, , p2 + q 2, , 2, , θ, D, , α, q, , C, , ., , Given : A circle, 2x2 + 2y2 = 5 and a parabola, y2 = 4 5 x., Statement − I : An equation of a common tangent to these curves is y = x + 5 ., Statement − II : If the line, y = mx +, , 0., (1), (2), (3), (4), Sol., , 2, , (4), Using sine rule in triangle ABD, AB, BD, =, sin θ sin ( θ + α ), 2, , 33., , (4), , 2, , (p, , 5, (m ≠ 0) is their common tangent, then m satisfies m4 − 3m2 + 2 =, m, , Statement - I is True; Statement -II is true; Statement-II is not a correct explanation for Statement-I, Statement -I is True; Statement -II is False., Statement -I is False; Statement -II is True, Statement -I is True; Statement -II is True; Statement-II is a correct explanation for Statement-I, , (1), , 5, (m ≠ 0)., m, Now, its distance from the centre of the circle must be equal to the radius of the circle., , Let the tangent to the parabola be y = mx +, 5, =, m, , 5, , 1 + m 2 ⇒ (1 + m2) m2 = 2 ⇒ m4 + m2 − 2 = 0., 2, ⇒ (m2 − 1) (m2 + 2) = 0 ⇒ m = ± 1, So, the common tangents are y = x + 5 and y = − x − 5 ., , So,, , 34., , A ray of light along x + 3y = 3 gets reflected upon reaching x-axis, the equation of the reflected rays is, , www.mathongo.com
Page 10 :
JEE Main 2013, , Sol., , MathonGo, , (1), , 3y = x − 3, , (2) y = 3x − 3, , (3), , 3y = x − 1, , (4) y = x + 3, , (1), , Slope of the incident ray is −, , 1, , ., , 3, , So, the slope of the reflected ray must be, , 1, , ., , 3, , The point of incidence is, , (, , ), , 3, 0 . So, the equation of reflected ray is y =, , 1, 3, , (x − 3) ., , 35., , All the students of a class performed poorly in Mathematics. The teacher decided to give grace marks of 10, to each of the students. Which of the following statistical measures will not change even after the grace, marks were given ?, (1) median, (2) mode, (3) variance, (4) mean, , Sol., , (3), Variance is not changed by the change of origin., Alternate Solution:, σ=, σ1 =, , ∑ x−x, , 2, , for y = x + 10 ⇒ y = x + 10, , n, , ∑ y + 10 − y − 10, n, , ∑ y−y, , 2, , =, , 2, , n, , =σ., , 36., , If x, y, z are in A.P. and tan−1x, tan−1y and tan−1z are also in A.P., then, (1) 2x = 3y = 6z, (2) 6x = 3y = 2z, (3) 6x = 4y = 3z, (4) x = y = z, , Sol., , (4), If x, y, z are in A.P., 2y = x + z, and tan−1x, tan−1y, tan−1z are in A.P., 2 tan−1 y = tan−1x + tan−1z ⇒ x = y = z., Note: If y = 0, then none of the options is appropriate., , 37., , If, , ∫ f ( x ) dx = Ψ ( x ) , then ∫ x f ( x ) dx, 5, , 3, , 1 3, x Ψ ( x 3 ) − 3∫ x 3 Ψ ( x 3 ) dx + C, 3, 1, (3) x 3 Ψ ( x 3 ) − ∫ x 3 Ψ ( x 3 ) dx + C, , 3, , (1), , Sol., , is equal to, 1 3, x Ψ ( x 3 ) − ∫ x 2 Ψ ( x 3 ) dx + C, 3, 1, (4) x 3 Ψ ( x 3 ) − ∫ x 2 Ψ ( x 3 ) dx + C, , 3, , (2), , (2), ∫ f ( x ) dx = ψ ( x ), , Let x3 = t, 3x2dx = dt, then, =, , ∫ x f ( x ) dx, 5, , 3, , {, , =, , 1, tf ( t ) dt, 3∫, , }, , 1, 1, t f ( t )dt − ∫ 1 ⋅ ∫ f ( t ) dt dt = x 3 ψ ( x 3 ) − ∫ x 2 ψ ( x 3 ) dx + C ., , 3 ∫, 3, , www.mathongo.com
Page 11 :
JEE Main 2013, , 38., , Sol., , MathonGo, , (1) x 2 + y 2 − 6y + 7 = 0, , x 2 y2, +, = 1 , and having centre at (0, 3) is, 16 9, (2) x 2 + y 2 − 6y − 5 = 0, , (3) x 2 + y 2 − 6y + 5 = 0, , (4) x 2 + y 2 − 6y − 7 = 0, , The equation of the circle passing through the foci of the ellipse, , (4), foci ≡ (± ae, 0), We have a2e2 = a2 − b2 = 7, , Equation of circle (x − 0)2 + (y − 3)2 =, , (, , 7 −0, , ), , 2, , + ( 0 − 3), , 2, , ⇒ x2 + y2 − 6y − 7 = 0., , 39., , The x-coordinate of the incentre of the triangle that has the coordinates of mid points of its sides as (0, 1), (1, 1) and (1, 0) is, (2) 1 + 2, (1) 2 − 2, (3) 1 − 2, , Sol., , (4) 2 + 2, , (1), , C(0, 2), , ax + bx 2 + cx 3, x-coordinate = 1, a+b+c, , =, , 2× 2 + 2 2 ×0 + 2× 0, , (1, 1), , (0, 1), , 2+2+2 2, 2, =, =, = 2− 2 ., 4+2 2, 2+ 2, Alternate Solution:, x-coordinate = r = (s − a) tan A/2, 4+2 2, , π, = , − 2 2 tan = 2 − 2 ., 2, 4, , , , 4, , B(2, 0), A(0, 0), , (1, 0), , x, , 40., , The intercepts on x-axis made by tangents to the curve, y = ∫ t dt , x ∈ R, which are parallel to the line, 0, , y = 2x, are equal to, (1) ± 2, (3) ± 4, Sol., , (4), dy, = x =2 ⇒x=±2⇒y=, dx, , (2) ± 3, (4) ± 1, , 2, , ∫ t dt = 2 for x = 2, 0, , −2, , and y =, , ∫, , t dt = −2 for x = − 2, , 0, , ∴ tangents are y − 2 = 2 (x − 2) ⇒ y = 2x − 2, and y + 2 = 2 (x + 2) ⇒ y = 2x + 2, Putting y = 0, we get x = 1 and − 1., 41., , The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, ….. , is, 7, 7, (1) ( 99 − 10−20 ), (2), (179 + 10−20 ), 9, 81, 7, 7, (4), (3) ( 99 + 10−20 ), (179 − 10−20 ), 9, 81, , Sol., , (2), tr = 0.777 ….. r times, , www.mathongo.com
Page 12 :
JEE Main 2013, , MathonGo, , = 7 (10–1 + 10–2 + 10–3 + ….. + 10–r), 7, = (1 − 10− r ), 9, 20, 20, 7, 7, 1, 7(, , S20 =, t r = 20 − 10− r = 20 − (1 − 10−20 ) =, 179 + 10−20 ), , 9, 9, 9, 81, , r =1, r =1, , , , ∑, , ∑, , 42., , Consider :, Statement − I : (p ∧ ~ q) ∧ (~ p ∧ q) is a fallacy., Statement − II : (p → q) ↔ (~ q → ~ p) is a tautology., (1), Statement - I is True; Statement -II is true; Statement-II is not a correct explanation for Statement-I, (2), Statement -I is True; Statement -II is False., (3), Statement -I is False; Statement -II is True, (4), Statement -I is True; Statement -II is True; Statement-II is a correct explanation for Statement-I, , Sol., , (1), S1:, , p, T, T, F, F, , q, T, F, T, F, , ~p, F, F, T, T, , ~q, F, T, F, T, , p^~q, F, T, F, F, , ~p^q, F, F, T, F, , (p^~q)^(~p^q), F, F, F, F, Fallacy, , S2:, p, T, T, F, F, , q, T, F, T, F, , ~p, F, F, T, T, , ~q, F, T, F, T, , p⇒q, T, F, T, T, , ~q⇒~p, T, F, T, T, , (p ⇒ q) ⇔ (~ q ⇒ ~ p), T, T, T, T, Tautology, , S2 is not an explanation of S1, 43., , The area (in square units) bounded by the curves y = x , 2y − x + 3 = 0 , x-axis, and lying in the first, quadrant is, (1) 36, (2) 18, 27, (3), (4) 9, 4, , Sol., , (4), 2 x = x −3, 4x = x2 – 6x + 9, x2 – 10x + 9, x = 9, x = 1, , 3, , (3, 0), , 3, , 3, , −0, , , x, , 2, ∫ ( 2y + 3) − y dy, 0, 3, , 2, y3 , y + 3y − = 9 + 9 – 9 = 9, 3 0, , , 44., , Sol., , tan A, cot A, can be written as, +, 1 − cot A 1 − tan A, (1) secA cosecA + 1, (2) tanA + cotA, (3) secA + cosecA, (4) sinA cosA + 1, , The expression, , (1), , www.mathongo.com, , 9
Page 13 :
JEE Main 2013, , MathonGo, , 1, cot 2 A, 1 − cot 3 A, cos ec 2 A + cot A, −, =, =, = 1 + sec A cosec A, cot A (1 − cot A ) (1 − cot A ) cot A (1 − cot A ), cot A, 45., , The real number k for which the equation, 2x3 + 3x + k = 0 has two distinct real roots in [0, 1], (1) lies between 2 and 3, (2) lies between −1 and 0, (3) does not exist, (4) lies between 1 and 2, , Sol., , (3), If 2x3 + 3x + k = 0 has 2 distinct real roots in [0, 1], then f′ (x) will change sign, but f′(x) = 6x2 + 3 > 0, So no value of k exists., , 46., , lim, , x tan 4x, , is equal to, , 1, 2, , (2) 1, , (3) 2, , (4) −, , (1), , Sol., , (1 − cos 2x ) ( 3 + cos x ), , x →0, , 1, 4, , (3), lim, , x →0, , (1 − cos 2x ), ( 3 + cos x ), x ( tan 4x ), 2, , 1, sin x 1 4x , lim 2 , ⋅ , ( 3 + cos x ) = 2 ×1× × 1× ( 3 + 1) = 2 ., x →0, 4, x, 4, tan, 4x, , , , , 47., , Let Tn be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If, Tn+1 − Tn = 10, then the value of n is, (1) 5, (2) 10, (3) 8, (4) 7, , Sol., , (1), n +1, , C3 − n C3 = 10 ⇒ nC2 = 10 ⇒ n = 5., , 48., , At present, a firm is manufacturing 2000 items. It is estimated that the rate of change of production P w.r.t., dP, additional number of workers x is given by, = 100 − 12 x . If the firm employs 25 more workers, then, dx, the new level of production of items is, (1) 3000, (2) 3500, (3) 4500, (4) 2500, , Sol., , (2), P, , ∫, , 25, , dP =, , 2000, , ∫ (100 − 12, , x ) dx, , 0, , (P – 2000) = 25 × 100 –, , 12 × 2, ( 25 )3/ 2, 3, , P = 3500., π /3, , 49., , Statement − I : The value of the integral, , ∫ 1+, , π /6, , Statement − II :, , (1), (2), , b, , b, , a, , a, , dx, , is equal to, , tan x, , π, ., 6, , ∫ f ( x ) dx = ∫ f ( a + b − x ) dx ., , Statement - I is True; Statement -II is true; Statement-II is not a correct explanation for Statement-I, Statement -I is True; Statement -II is False., , www.mathongo.com
Page 14 :
JEE Main 2013, , MathonGo, , (3), (4), Sol., , Statement -I is False; Statement -II is True, Statement -I is True; Statement -II is True; Statement-II is a correct explanation for Statement-I, , (3), π/3, , I=, , ∫ 1+, , dx, , ∫ 1+, , tan x, , π/6, π/3, , I=, , π/6, , tan x, , tan x, , dx, , π, 6, π, I= ., 12, , 2I =, , 50., , Sol., , 1 α 3 , If P = 1 3 3 is the adjoint of a 3 × 3 matrix A and |A| = 4, then α is equal to, 2 4 4 , (1) 11, (2) 5, (3) 0, (4) 4, (1), , 1 α 3 , P = 1 3 3 , 2 4 4 , |Adj A| = |A|2, |Adj A| = 16, 1 (12 − 12) − α (4 − 6) + 3 (4 − 6) = 16., 2α − 6 = 16., 2α = 22., α = 11., , 51., , The number of values of k, for which the system of equations, (k + 1)x + 8y = 4k, kx + (k + 3)y = 3k − 1, has no solution, is, (1) 1, (2) 2, (3) 3, (4) infinite, , Sol., , (1), For no solution, k +1, 8, 4k, … (1), =, ≠, k, k + 3 3k − 1, ⇒ (k + 1) (k + 3) − 8k = 0, or k2 − 4k + 3 = 0 ⇒ k = 1, 3, But for k = 1, equation (1) is not satisfied, Hence k = 3., , 52., , If y = sec(tan−1x), then, (1), (3), , 1, 2, , dy, at x = 1 is equal to, dx, , (2) 1, 2, , (4), , 1, 2, , www.mathongo.com
Page 15 :
JEE Main 2013, , MathonGo, , Sol., , (4), y = sec (tan−1x), dy, 1, = sec ( tan −1 x ) tan ( tan −1 x ) ⋅, dx, 1+ x2, dy, 1, 1, = 2 × 1× =, ., dx x =1, 2, 2, , 53., , If the lines, , Sol., , (2), 1 −1 −1, , x −2 y −3 z −4, x −1 y − 4 z − 5, =, =, and, =, =, are coplanar, then k can have, 1, 1, −k, k, 2, 1, (1) exactly one value, (2) exactly two values, (3) exactly three values, (4) any value, , 1, , 1, , −k = 0, , k, , 2, , 1, , 1 (1 + 2k) + 1 (1 + k2) − 1 (2 − k) = 0, k2 + 1 + 2k + 1 − 2 + k = 0, k2 + 3k = 0, (k) (k + 3) = 0, 2 values of k., 54., , Let A and B be two sets containing 2 elements and 4 elements respectively. The number of subsets of A × B, having 3 or more elements is, (1) 220, (2) 219, (3) 211, (4) 256, , Sol., , (2), A × B will have 8 elements., 28 − 8C0 − 8C1 − 8C2 = 256 − 1 − 8 − 28 = 219., , 55., , Sol., , uuur, uuur, If the vectors AB = 3iˆ + 4kˆ and AC = 5iˆ − 2ˆj + 4kˆ are the sides of a triangle ABC, then the length of the, median through A is, (1) 72, (2) 33, (3) 45, (4) 18, C, , (2), uuur uuur, uuuur AB + AC, AM =, 2, uuuur, AM = 4iˆ − ˆj + 4kˆ, uuuur, AM = 16 + 16 + 1 = 33, , M, , A, , B, , 56., , A multiple choice examination has 5 questions. Each question has three alternative answers of which, exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is, 13, 11, (1) 5, (2) 5, 3, 3, 10, 17, (3) 5, (4) 5, 3, 3, , Sol., , (2), P (correct answer) = 1/3, , www.mathongo.com
Page 16 :
JEE Main 2013, , MathonGo, , 4, , 1, , 1 2, C 4 + 5 C5, 3 3, 5× 2, 1, 11, + 5 = 5 ., 5, ( 3) ( 3) 3, , 5, , 57., , 1, , 3, , 5, , 1+ z , If z is a complex number of unit modulus and argument θ, then arg , equals, 1+ z , π, (2) θ, (1) − θ, 2, (3) π − θ, (4) − θ, , Sol., , (2), |z| = 1 ⇒ zz = 1, 1+ z 1+ z, =, =z., 1, 1+ z, 1+, z, , 58., , If the equations x2 + 2x + 3 = 0 and ax2 + bx + c = 0, a, b, c ∈ R, have a common root, then, a : b : c is, (1) 3 : 2 : 1, (2) 1 : 3 : 2, (3) 3 : 1 : 2, (4) 1 : 2 : 3, , Sol., , (4), For equation x2 + 2x + 3 = 0, both roots are imaginary., Since a, b, c ∈ R., If one root is common then both roots are common, a b c, Hence, = =, 1 2 3, a : b : c = 1 : 2 : 3., , 59., , Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is, 5, 7, (2), (1), 2, 2, 9, 3, (3), (4), 2, 2, , Sol., , (2), 4x + 2y + 4z = 16, 4x + 2y + 4z = − 5, 21, 21 7, dmin =, =, = ., 36 6 2, 10, , 60., , Sol., , x +1, x −1 , , The term independent of x in expansion of 2/3, −, 1/3, 1/ 2 , x − x +1 x − x , (1) 120, (2) 210, (3) 310, (4) 4, , (2), , ( x1/3 + 1)( x 2/3 − x1/3 + 1), 1, , −, ⋅, 2/3, 1/3, , x − x +1, x, , , Tr+1 = (− 1)r 10Cr, 10, C4 = 210., , x, , 20 − 5r, 6, , (, , )(, , ), , (, , 10, , x −1 , , , x −1, , , x +1, , ), , = (x1/3 − x−1/2)10, , ⇒r=4, , www.mathongo.com, , is
Page 17 :
JEE Main 2013, , MathonGo, , PART C − PHYSICS, 61., , In an LCR circuit as shown below both switches are open initially. Now switch S1 is closed, S2 kept open., (q is charge on the capacitor and τ = RC is capacitive time constant). Which of the following statement is, correct?, V, R, S1, C, S2, L, , (1) At t = τ, q = CV/2, (2) At t = 2τ, q = CV(1 − e−2), τ, (3) At t = , q = CV(1 − e−1 ), 2, (4) Work done by the battery is half of the energy dissipated in the resistor., Sol., , (2), Charge on the capacitor at any time ‘t’ is, , (, , q = CV 1 − e− t / τ, at t = 2τ, , (, , q = CV 1 − e −2, , ), , ), , 62., , A diode detector is used to detect an amplitude modulated wave of 60% modulation by using a condenser of, capacity 250 pico farad in parallel with a load resistance 100 kilo ohm. Find the maximum modulated, frequency which could be detected by it., (1) 10.62 kHz, (2) 5.31 MHz, (3) 5.31 kHz, (4) 10.62 MHz, , Sol., , (3), , 1, 1, =, = 6.37 kHz, 2πRC 2 × 3.14 × 100 × 103 × 250 × 10−12, fC = cut off frequency, As we know that fm fC, ∴ (3) is correct, Note: The maximum frequency of modulation must be less than fm, where, fC =, , 1 − m2, m, m ⇒ modulation index, , f m = fC, , 63., , The supply voltage to a room is 120 V. The resistance of the lead wires is 6 Ω. A 60 W bulb is already, switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel, to the bulb?, (1) 2.9 Volt, (2) 13.3 Volt, (3) 10.04 Volt, (4) zero volt, , Sol., , (3), , Heater, , 120 × 120, = 240Ω, 60, 120 × 120, Resistance of Heater =, = 60Ω, 240, , Resistance of bulb =, , Bulb, , 6Ω, , 120 V, , www.mathongo.com
Page 18 :
JEE Main 2013, , MathonGo, , 120, × 240, 246, 120, Voltage across bulb after heater is switched on, V2 =, × 48, 54, Decrease in the voltage is V1 − V2 = 10.04 (approximately), Note: Here supply voltage is taken as rated voltage., , Voltage across bulb before heater is switched on, V1 =, , 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. The extension x0 of the spring when it is in equilibrium is:, Mg LAσ , Mg LAσ , (2), (1), 1 −, , 1 −, , k , M , k , 2M , Mg, Mg LAσ , (3), (4), 1 +, , k, k , M , (Here k is spring constant), , Sol., , (2), At equilibrium ΣF = 0, AL , kx 0 + , σg − Mg = 0, 2, , LAσ , x 0 = Mg 1 −, 2M , , , 65., , Mg, , Two 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 0 = is placed at the origin. If charge q0 is given a small displacement (y a) along the y-axis, the net, 2, force acting on the particle is proportional to:, 1, (1) −y, (2), y, (3) −, , Sol., , Buoyant force, , kx0, , 1, y, , (4) y, , (4), Fnet = 2Fcos θ, , =2, , =, , k ⋅q ⋅q / 2, , (, , a 2 + y2, , kq 2 y, a3, , ), , F, θ θ, q/2, , F, 2, , ⋅, , y, , y, , 2, , a + y2, q, , a, , a, , q, , (y a), , 66., , A beam of unpolarised light of intensity I0 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. The intensity of, the emergent light is:, (1) I0/2, (2) I0/4, (4) I0, (3) I0/8, , Sol., , (2), , 67., , The anode voltage of a photocell is kept fixed. The wavelength λ of the light falling on the cathode is, gradually changed. The plate current I of the photocell varies as follows:, , www.mathongo.com
Page 19 :
JEE Main 2013, , MathonGo, , (1), , (2), I, , I, , O, , O, , λ, , (3), I, , I, , O, , Sol., 68., , Sol., 69., , O, , λ, , λ, , (3), Two coherent point sources S1 and S2 are separated by a small distance ‘d’ as, shown. The fringes obtained on the screen will be:, (1) straight lines, (2) semi-circles, (3) concentric circles, (4) points, , d, S1, , S2, , Screen, , D, , (3), A metallic rod of length ‘l’ is tied to a string of length 2l 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:, , 3Bωl 2, 2, 5Bωl 2, (3), 2, , 4Bωl 2, 2, 2Bωl 2, (4), 2, , (1), , Sol., , λ, , (4), , (2), , (3), de = B(ωx) ⋅ dx, 3L, , 2L, , e = Bω ∫ xdx, , ω, L, , x, , 2L, , =, , 5BωL2, 2, , 70., , In a hydrogen like atom electron makes transition from an energy level with quantum number n to another, with quantum number (n – 1). If n >> 1, the frequency of radiation emitted is proportional to, 1, 1, (1) 2, (2) 3/ 2, n, n, 1, 1, (3) 3, (4), n, n, , Sol., , (3), , 1, 1 , ν∝, − 2, 2, n , (n − 1), (2n − 1), ∝ 2, n (n − 1) 2, , ∝, , 1, n3, , (since n, , 1), , www.mathongo.com
Page 21 :
JEE Main 2013, , MathonGo, , ∆Q =, , 3 P0 V0 5 2P0 V0 13, R, + R, =, P0 V0, 2 R 2 R 2, , 75., , A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1 %., What is the fundamental frequency of steel if density and elasticity of steel are 7.7 × 103 kg/m3 and, 2.2 × 1011N/m2 respectively?, (1) 178.2 Hz, (2) 200.5 Hz, (3) 770 Hz, (4) 188.5 Hz, , Sol., , (1), , Fundamental frequency f =, , 1 T, 2l µ, , =, , 1, T, 2l Aρ, , =, , 1 stress, 1, 2.2 × 1011 ×10−2, =, ., 2l, 2 × 1.5, ρ, 7.7 × 103, , 76., , This question has statement I and statement II. Of 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: To increase the range of ammeter, additional shunt needs to be used across it., (1) Statement – I is true, Statement – II is true, Statement – II is not the correct explanation of Statement–I., (2) Statement – I is true, statement – II is false., (3) 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., , (3), , For Ammeter, S =, , Ig G, I − Ig, , 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 2R?, GmM, 2GmM, (2), (1), 2R, 3R, GmM, 5GmM, (3), (4), 3R, 6R, , Sol., , (4), , GMm, 6R, GMm, T. Ei = −, R, , T. Ef = −, , ∆W = T.Ef – T.Ei =, , 78., , 5GMm, 6R, , (, , ), , A projectile is given an initial velocity of iˆ + 2 ˆj m/s, where iˆ is along the ground and ĵ is along the, 2, , vertical. If g = 10 m/s , the equation of its trajectory is:, (2) 4 y = 2 x − 5 x 2, (1) y = 2 x − 5 x 2, (3) 4 y = 2 x − 25 x 2, Sol., , (4) y = x − 5 x 2, , (1), , www.mathongo.com
Page 23 :
JEE Main 2013, , MathonGo, , Patm + mg – (P0 +dP) A = m, , d2x, dt 2, , ...(2), , Process is adiabatic ⇒ PV γ = C ⇒ − dP =, 1, 2π, , Using 1, 2, 3 me get f =, , 82., , γ PdV, V, , A γ P0, MV0, 2, , A charge Q is uniformly distributed over a long rod AB of length L as shown in the figure. The electric, potential at the point O lying at a distance L from the end A is:, B, , A, , O, L, , 3Q, , (1), , 4πε 0 L, , Q ln 2, 4πε 0 L, , (3), , Sol., , L, , (2), , Q, 4πε 0 L ln 2, , (4), , Q, 8πε 0 L, , (3), O, x, , V =∫, , dx, , x =2 L, , x=L, , Q ln 2, k Q, dx =, 4πε 0 L, x L , , 83., , A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm. The centre of, the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of, 2.0 A flows through the smaller loop, then the flux linked with bigger loop is, (1) 6 × 10–11 weber, (2) 3.3 × 10–11 weber, –9, (3) 6.6 × 10 weber, (4) 9.1 × 10–11 weber, , Sol., , (4), 2, , 1, r, , R, , d, , Let M12 be the coefficient of mutual induction between loops, φ1 = M12 i2, , ⇒, , µ0 i2 R 2, 2(d 2 + R 2 )3/2, , ⇒ M 12 =, , π r 2 = M 12 i2, , µ 0 R 2π r 2, 2(d 2 + R 2 )3/ 2, , φ2 = M 12 i1 ⇒ φ2 = 9.1×10−11 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 0 the graph between the temperature T of the metal and time t will be closest to :, (1), (2), T, T, θ0, O, , θ0, t, , O, , www.mathongo.com, , t
Page 24 :
JEE Main 2013, , (3), , MathonGo, , (4), , T, , T, , θ0, O, , t, , O, , t, , Sol., , (2), The temperature goes on decreasing with time (non-linearly) The rate of decrease will be more initially, which is depicted in the second graph., , 85., , The I – V characteristic of an LED is, (1), B, G, Y, R, , (2), , I, , O, , (3), , V, , V, , V, , R, Y, G, B, , O, , (4), , Red, Yellow, Green, Blue, , O, , (R)(Y)(G)(B), I, , I, , O, , V, , Sol., , (4), For LED, in forward bias, intensity increases with voltage., , 86., , This question has Statement I and Statement II. Of 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, 1, , m , mass M. If the maximum energy loss possible is given as f mv 2 then f = , ., 2, , M +m, Statement – II : Maximum energy loss occurs when the particles get stuck together as a result of the, collision., (1) Statement – I is true, Statement – II is true, Statement – II is not a correct explanation of Statement – I., (2) Statement – I is true, Statement – II is false., (3) 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., , (3), Loss of energy is maximum when collision is inelastic as in an inelastic collision there will be maximum, deformation., 1 Mm 2, KE in COM frame is , Vrel, 2 M +m, 1 Mm 2, KEi = , KE f = 0 (Q Vrel = 0 ), V, 2 M +m, 1 Mm 2, Hence loss in energy is , V, 2 M +m, M, ⇒ f =, M +m, , www.mathongo.com
Page 25 :
JEE Main 2013, , MathonGo, , 87., , The amplitude of a damped oscillator decreases to 0.9 times its original magnitude is 5s. In another 10s it, will decrease to α times its original magnitude, where α equals., (1) 0.81, (2) 0.729, (3) 0.6, (4) 0.7, , Sol., , (2), A = A0 e − kt, , ⇒ 0.9 A0 = A0 e−5 k, and α A0 = A0 e −15 k, solving ⇒ α = 0.729, 88., , Diameter of plano-convex lens is 6 cm and thickness at the centre is 3 mm. If speed of light in material of, lens is 2 × 108 m/s, the focal length of the lens is :, (1) 20 cm, (2) 30 cm, (3) 10 cm, (4) 15 cm, , Sol., , (2), , R, , R-t, , d, t, , R 2 = d 2 + ( R − t )2, 2, , t, R 2 − d 2 = R 2 1 − , R, 2, d, 2t, 1− 2 = 1−, R, R, (3)2, 90, R=, =, = 15 cm, 2 × (0 ⋅ 3) 6, 1, 1, 1 , = ( µ − 1) −, , f, R, R, 1, 2 , 1 3 1 , = − 1 , f 2 15 , f = 30 cm, 89., , The magnetic field in a travelling electromagnetic wave has a peak value of 20 nT. The peak value of, electric field strength is :, (1) 6 V/m, (2) 9 V/m, (3) 12 V/m, (4) 3 V/m, , Sol., , (1), Ε 0 = CB0, = 3 × 108 × 20 × 10–9, = 6 V/m, , 90., , Two short bar magnets of length 1 cm each have magnetic moments 1.20 Am2 and 1.00 Am2 respectively., They are placed on a horizontal table parallel to each other with their N poles pointing towards the South., They have a common magnetic equator and are separated by a distance of 20.0 cm. The value of the, resultant horizontal magnetic induction at the mid - point O of the line joining their centres is close to, (Horizontal component of earth’s magnetic induction is 3.6 × 10–5 Wb/m2), , www.mathongo.com
