JEE-Advanced-2013 Paper 1 Code 0

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1 JEE-Advanced-03 Paper Code 0 June, 03 9:00 AM Noon Detailed solutions to the paper by Vidyamandir Classes VMC/Solutions Paper - Code 0 JEE Advanced - 03

2 Part : PHYSICS Section : (Only One option correct Type) This section contains 0 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct x y. The work done on a particle of mass m by a force, K ˆi + ˆj 3 3 (K being a constant of ( x + y ) ( x + y ) appropriate dimensions), when the particle is taken from the point (a, 0) to the point (0, a) along a circular path of radius a about the origin in the x-y plane is : Kπ (A) (B) a W = F.dr = F dx+ F dy (D) ( x y ) xdx W = K + ydy 3 / 3 / ( x + y ) ( x + y ) ( 0, a ) = K d = = 0 ( ) x + y x + y Kπ Kπ (C) (D) 0 a a 0,a a, 0 a, 0. Two rectangular blocks, having identical dimensions, can be arranged either in configuration I or in configuration II as shown in the figure. One of the blocks has thermal conductivity κ and the other κ. The temperature difference between the ends along the x-axis is the same in both the configurations. It takes 9s to transport a certain amount of heat from the hot end to the cold end in the configuration I. The time to transport the same amount of heat in the configuration II is : (A) (A).0s (B) 3.0s (C) 4.5s (D) 6.0s dq T T H = = Q = t dt R R Since Q and T is same. t R t = R l l R = + KA KA R R 3l = KA KA KA 3KA = + = R l l l l = 3KA VMC/Solutions Paper - Code 0 JEE Advanced - 03 (i) (configuration ) (configuration )

3 Putting in (i) 9 t = 3l / KA l / 3KA t = sec 3. Two non-reactive monoatomic ideal gases have their atomic masses in the ratio : 3. The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature, is 4 : 3. The ratio of their densities is : (A) : 4 (B) : (C) 6 : 9 (D) 8 : 9 (D) M M P P = 3 4 = 3 Using PM = ρrt P M P M ρ P M 4 8 = = = = ρ ρ ρ P M A particle of mass m is projected from the ground with an initial speed u 0 at an angle α with the horizontal. At the highest point of its trajectory, it makes a completely inelastic collision with another identical particle, which was thrown vertically upward from the ground with the same initial speed u 0. The angle that the composite system makes with the horizontal immediately after the collision is : (A) π 4 (B) π + α (C) π α (D) π 4 (A) From energy conservation both will have equal speeds at the point of collision. V= u 0 cosα For perfectly inelastic collision, to conserve linear momentum Final velocity of combined mass will be along the Resultant θ = A pulse of light of duration 00 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30 mw and the speed of light is ms. The final momentum of the object is : (A) (C) kg ms (B) kg ms (D) kg ms kg ms (B) 3 E power t 30 0 P = = = C C 3 0 = 7 0 kgm / s 9 6. In the Young's double slit experiment using a monochromatic light of wavelength λ, the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is : VMC/Solutions 3 Paper - Code 0 JEE Advanced - 03

4 λ Vidyamandir Classes λ 4 λ 8 λ 6 (A) ( n+ ) (B) ( n+ ) (C) ( n+ ) (D) ( n+ ) (B) I = 4I cos 0 φ I = 4I cos cos φ 0 0 φ = = ( n + ) φ π 4 π π λ p.d. = ( n+ ) p.d = ( n + ) λ The image of an object, formed by a plano-convex lens at a distance of 8 m behind the lens, is real and is one-third the size of the object. The wavelength of light inside the lens is 3 times the wavelength in free space. The radius of the curved surface of the lens is : (A) m (B) m (C) 3m (D) 6m (C) V 8 = = u = 4 u 3 u 3 O I = = f = 6 v u f f 8 4 λ0 λ0 λ = λ 0 = µ 3 µ µ = 3 / µ = f R 3 / = 6 R R = 3 8. One end of a horizontal thick copper wire of length L and radius R is welded to an end of another horizontal thin copper wire of length L and radius R. When the arrangement is stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is : (A) 0.5 (B) 0.50 (C).00 (D) 4.00 (C) Force in both the wires will be same. Fl Fl x= = Ay π r y x l r l r L R = =. x = = r l l r L R x (ratio of elongation of thin to thick wire) x = VMC/Solutions 4 Paper - Code 0 JEE Advanced - 03

5 9. A ray of light travelling in the direction ( ˆi 3ˆj ) the direction ( ˆi 3ˆj ). The angle of incidence is : + is incident on a plane mirror. After reflection, it travels along (A) 30 (B) 45 (C) 60 (D) 75 a.b (A) Cosθ = = a b θ = 0 80 θ i= = The diameter of a cylinder is measured using a Vernier callipers with no zero errors. It is found that the zero of the (B) Vernier scale lies between 5.0 cm and 5.5 cm of the main scale. The Vernier scale has 50 divisions equivalent to.45cm. The 4th division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is : (A) 5. cm (B) 5.4 cm (D) 5.36 cm (D) 5.48 cm. 45 L.C = MSD VSD = cm cm 50 = = cm D = 5. 0 cm cm = 5. 0 cm cm = 5. 4 cm VMC/Solutions 5 Paper - Code 0 JEE Advanced - 03

6 Section : (One or more options correct Type) This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct. In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance C. The switch S is pressed first to fully charge the capacitor C and then released. The switch S is then pressed to charge the capacitor C. After some time, S is released and then S 3 is pressed. After some time, (A) the charge on the upper plate of C is CV 0 (B) the charge on the upper plate of C is CV 0 (C) the charge on the upper plate of C is 0 (D) the charge on the upper plate of C is CV0 (BD). A particle of mass M and positive charge Q, moving with a constant velocity u = 4ˆ i ms, enters a region of uniform static magnetic field normal to the x y plane. The region of the magnetic field extends from x = 0 to x = L for all values of y. After passing through this region, the particle emerges on the other side after 0 milliseconds with a velocity u ( 3ˆ i ˆ = + j ) ms. The correct statement(s) is (are) (A) (B) The direction of the magnetic field is -z direction. The direction of the magnetic field is +z direction (C) The magnitude of the magnetic field 50 π M 3Q VMC/Solutions 6 Paper - Code 0 JEE Advanced - 03 units. (D) The magnitude of the magnetic field is 00 π M units. 3Q (AC) u = 4ˆ i u = ( 3 ˆi + ˆj ) Tanθ = θ = 30 3 B should be along θ M t = QB π.m = 6 QB z direction. π M 50π M B= 00= units 6Q 3Q

7 3. A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, y( x,t ) = (0. 0 m) sin [(6. 8 m ) x] cos[68 s ) t ]. Assuming π = 3. 4, the correct statements(s) is (are) (A) The number of nodes is 5. (B) The length of the string is 0.5 m. (C) The maximum displacement of the midpoint of the string, from its equilibrium position is 0.0 m. (D) The fundamental frequency is 00 Hz. y m m x s t For 5 th harmonic λ 5 = l π π π k = 6.8= λ= m λ λ π l= = = m= 0.5 m (BC) = ( 0.0 ) sin ( 6.8 ) cos ( 68 ) Number of nodes = 6 At midpoint of string there is an antinode y max at midpoint = 0.0 m f 0 v ω / k 68/ = = = = = 0 Hz l l A solid sphere of radius R and density ρ is attached to one end of a mass-less spring of force constant k. The other end of the spring is connected to another solid sphere of radius R and density 3ρ. the complete arrangement is placed in a liquid of density ρ and is allowed to reach equilibrium. The correct statement(s) is (are) (A) (B) (C) (D) the net elongation of the spring is the net elongation of the spring is 3 4π R ρg 3k 3 8π R ρg 3k the light sphere is partially submerged. the light sphere is completely submerged. (AD) In equilibrium let elongation of spring is x 3 ρvg 4ρπ R g Then kx= ρvg x= = k 3k In equilibrium net force of weight and spring force on light sphere in downward direction is kx+ ρvg = ρvg = FB So it is fully submerged. 5. Two non-conducting solid spheres of radii R and R, having uniform volume charge densities ρ and ρ respectively, touch each other, The net electric field at a distance R from the centre of the smaller sphere, along the line joining the centres of the spheres, is zero. The ratio VMC/Solutions 7 Paper - Code 0 JEE Advanced - 03 ρ ρ can be.

8 (A) 4 (B) 3 5 (C) 3 5 (D) 4 (BD) For E P = 0 3 ( R ) 4 ρ π Q ρr 3 ρr = = 4πε 0( R) 3ε 0 4πε 0( R) 3ε 0 ρ = 4 ρ Also for E Q = 0 Q Q Q 4 ρ 3 + = 0 = = 4πε 0( R) 4πε 0( 5R) Q 5 ρ 5 Section 3: (Integer value correct Type) This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive) VMC/Solutions 8 Paper - Code 0 JEE Advanced - 03

9 6. A bob of mass m, suspended by a string of length l, is given a minimum velocity required to complete a full circle in the vertical plane. At the highest point, it collides elastically with another bob of mass m suspended by a string of length l, which is initially at rest. Both the strings are mass-less and inextensible. If the second bob, after collision acquires the minimum speed required to complete a full circle in the vertical plane, the ratio (5) u = 5 gl At highest point v = gl (from energy conservation) After collision (identical mass and elastic collision) u = v = gl To complete vertical circle l gl = 5gl = 5 l l is l 7. A particle of mass 0. kg is moving in one dimension under a force that delivers a constant power 0.5 W to the particle. If the initial speed (5) P= 0.5 W W = Pdt= Pt= 0.5 5=.5 J K = W 0. mv W v.5 = = v = 5 v= 5 m / s (in ms ) of the particle is zero, the speed (in ms ) after 5 s is 8. The work functions of Silver and Sodium are 4.6 and.3 ev, respectively. The ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium is () h υ = w + ev 0 C h w V = υ C e e Slope of V C versus υ is Hence ratio = 0 h Constant e = 9. A freshly prepared sample of a radioisotope of half-life 386 s has activity 0 3 disintegrations per second. Given that ln = 0.693, the fraction of the initial number of nuclei (expressed in nearest integer percentage) that will decay in the first 80 s after preparation of the sample is T = = 386 λ= = = / λ λt N = N e (4) 3 f 0 λt N N N ( e ) 0 0 = = = N N 0 0 e λt VMC/Solutions 9 Paper - Code 0 JEE Advanced - 03

10 λt= 80= = e = (0.04) = 0.96 f = ( e ) = 0.96= 0.04 x Using e x for x<< Percentage fraction = 4 0. A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 0 rad s about its own axis, which is vertical. Two uniform circular rings, each of mass 6.5 kg and radius 0. m, are gently placed symmetrically on the disc in such a manner that they are touching each other along the axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest relative to the disc and the system rotates about the original axis. The new angular velocity (in rad s ) of the system is (8) From conservation of angular momentum Iω= ( I+ I ) ω MR MR R R ω= + m + m MR MR ω= + mr ω = ω ω = 8 rad/s ω Part : CHEMISTRY Section : (Only One option correct Type) VMC/Solutions 0 Paper - Code 0 JEE Advanced - 03

11 This section contains 0 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Consider the following complex ions, P, Q and R [ ] ( ) ( ) P = FeF, Q= V H O and R= Fe H O. The correct order of the complex ions, according to their spin-only magnetic moment values (in B.M.) is. (A) R < Q < P (B) Q < R < P (C) R < P < Q (D) Q < P < R 3 (B) P = [FeF 6] Fe 4s 3d n= 5 Q = [V(HO) 6] V 4s 3d n= 3 R = [Fe(HO) 6] Fe 4s 3d n= 4 Magnetic moment will be in order Q < R < P. The arrangement of X ions around A + ion in solid AX is given in the figure (not drawn to scale). If the radius of X is 50 pm, the radius of A + is. (A) 04 pm (B) 5 pm (C) 83 pm (D) 57 pm (A) Using radius ratio for octahedral void (FCC) + r 0.44 r = r + = = Sulfide ores are common for the metals (A) Ag, Cu and Pb (B) Ag, Cu and Sn (C) Ag, Mg and Pb (D) Al, Cu and Pb (A) Ag S Cu S, CuFeS PbS Silver glance Copper glance & Copper pyrite Galena 4. The standard enthalpies of formation of CO (g), H O(l) and glucose (s) at 5 C 400 kj / mol, 300 kj / mol and 300 kj / mol, respectively. The standard enthalpy of combustion per gram of glucose at 5 C is. VMC/Solutions Paper - Code 0 JEE Advanced - 03 are

12 (A) kj (B) 900 kj (C) 6.kJ (D) + 6. kj (C) C6HO6 + 6O 6CO + 6HO As we know To calculate H= Hf (Product) Hf (Reactants) = 6H + 6H H = 6 ( 400) + 6( 300) ( 300) f (CO ) f (HO) f (C6H O 6 ) = = 900 kj H for one gram. H per gram = 900 = 6. kj Upon treatment with ammoniacal H S, the metal ion that precipitates as a sulfide is. (A) Fe (III) (B) Al (III) (C) Mg (II) (D) Zn (II) + + (D) HS+ Zn ZnS+ H 6. Methylene blue, from its aqueous solution, is absorbed on activated charcoal at 5 C. For this process, the correct statement is. (A) The adsorption requires activation at 5 C (B) The adsorption is accompanied by a decrease in enthalpy (C) The adsorption increases with increase of temperature (D) The adsorption is irreversible (B) Methylene blue on activated charcoal is an example of physiosorption. Since entropy decreases adsorption therefore enthalpy has to be negative to make G< KI in acetone, undergoes S N reaction with each of P, Q, R and S. The rates of the reaction vary as. (A) P > Q > R > S (B) S > P > R > Q (C) P > R > Q > S (D) R > P > S > Q (B) Rate of S N : 8. In the reaction, P+ Q R + S the time taken for 75% reaction of P is twice the time taken for 50% reaction of P. The concentration of Q varies with reaction time as shown in the figure. The overall order of the reaction is. VMC/Solutions Paper - Code 0 JEE Advanced - 03

13 (A) (B) 3 (C) 0 (D) (D) Since time taken for 75% reaction of P is twice the time taken for 50% reaction of P. Hence reaction is first order w.r.t. P. Analysis of graph of Q w.r.t. time shows, reaction of Q is of zero order. Hence overall order of the reaction is one. 9. Concentrated nitric acid, upon long standing, turns yellow-brown due to the formation of (A) NO (B) NO (C) N O (D) N O 4 (B) HNO 3 on standing decomposes as HNO3 NO + O + HO 30. The compound that does NOT liberate CO, on treatment with aqueous sodium bicarbonate solution, is. (A) Benzoic acid (B) Benzenesulphonic acid (C) Salicylic acid (D) Carbolic acid (Phenol) (D) Phenol is too weak to liberate CO with NaHCO 3. Section : (One or more options correct Type) This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct VMC/Solutions 3 Paper - Code 0 JEE Advanced - 03

14 3. The initial rate ofr hydrolysis of methyl acetate (M) by a weak acid (HA, M) is /00 th of that of a strong acid (HX, M), at 5 o C. The K a of HA is (A) 0 4 (B) 0 5 (C) 0 6 (D) 0 3 (A) R= K[RCOOR][HO] Clearly But So R w Kw = = (Given) [Since (RCOOR) is common] Rs Ks 00 Ea / Rt K= Ae [ E a will not change as catalyst is same (H + )] Kw Aw = =....(i) Ks As 00 + Also, A depends on no. of particles i.e. A [H ] So for HA + HO H3O + + A C Cα Cα Cα In equation (i) Cα Cα = α= 0.0 ( C = M) and 4 Ka = = 0 00 α 3. The hyperconjugative stabilities of tert-butyl cation and -butene, respectively, are due to (A) σ p ( empty) and σ π * electron deloaclisations. (B) σ σ * and σ π electron deloaclisations. (C) σ p ( filled ) and σ π electron deloaclisations. (D) p ( filled) σ * and σ π * electron deloaclisations. (A) Hyperconjugation in tert-butyl carbocation refers to delocalisation of σ -electrons over the empty p-orbitals of C +. In -butene, it refers to delocalisation of σ -electrons over the π * orbitals of C = C. 33. The pair(s) of coordination complexes/ions exhibiting the same kind of isomerism is(are) Cr( NH ) Cl Cl and Cr( NH ) Cl Cl (A) [ 3 5 ] [ 3 4 ] (B) [ Co( NH3) 4Cl] and [ Pt( NH3) ( H O) Cl] (C) [ ] [ ] CoBr Cl and PtBr Cl (D) [ ( ) ( )] [ ( ) ] + + Pt NH NO Cl and Pt NH Cl Br (BD)(B) [Co(NH 3) 4Cl ] and [Pt(NH 3) (HO) Cl] Both can show geometrical isomerism. (D) [Pt(NH 3) 3 NO 3]Cl and [Pt(NH 3) 3Cl] Br Both can show ionisation isomerism. 34. Among P, Q, R AND S, the aromatic compound(s) is/are VMC/Solutions 4 Paper - Code 0 JEE Advanced - 03

15 (A) P (B) Q (C) R (D) S (ABCD) (A) (B) (C) 0 5 C (NH 4 ) CO3 NH3 + CO + HO (D) 35. Benzene and naphthalene form an ideal solution at room temperature. For this process, the true statement(s) is (are) (a) G is positive (b) S system is positive (c) S = 0 surroundings (d) H = 0 (BCD) Benzene and Naphthalene form an ideal solution at room temperature. H= 0 For mixing of liquids, Ssystem is positive. (Entropy increase for mixing). qsys SSurr = = 0 Since for ideal solution Hmix = 0 T ( ) Section 3: (Integer value correct Type) VMC/Solutions 5 Paper - Code 0 JEE Advanced - 03

16 This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive) 36. The atomic masses of He and Ne are 4 and 0 a.m.u., respectively. The value of the de Broglie wavelength of He gas at C is M times that of the de Broglie wavelength of Ne at 77 0 C. M is h (5) As we know λ= m( KE) As we know KE avg =3/ K b T λhe mnetne = M = = = 5 m T 4 00 λ Ne He He So M = EDTA is ethylenediaminetetraacetate ion. The total number of N-Co-O bond angles in [ Co EDTA ] complex ion is ( ) (8) Co Total number of N Co O bond angles = (4 ) = The total number of carboxylic acid groups in the product P is () 39. A tetrapeptide has COOH group on alanine. This produces glycine (Gly), valine (Val), phenyl alanine (Phe) and alanine (Ala), on complete hydrolysis. For this tetrapeptide, the number of possible sequences (primary structures) with NH group attached to a chiral centre is VMC/Solutions 6 Paper - Code 0 JEE Advanced - 03

17 O (4) (gly/valine/ph-al) NH C C OH CH3 Primary structure If glycine is at left corner, then NH will be at if valine at corner O gly NH C H C NH alanine Phenyl alanine i pr If Phenyl alanine is at left corner O gly NH C H C NH alanine Phenyl alanine Ph 4 possible structures. arrangements arrangements 40. The total number of lone-pairs of electrons in melamine is (6) The structure of Melamine is Hence number of lone pairs = 6 VMC/Solutions 7 Paper - Code 0 JEE Advanced - 03

18 Part 3 : MATHS Section : (Only One option correct Type) This section contains 0 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct 4. For a> b> c > 0, the distance between (, ) and the point of intersection of the lines ax+ by+ c= 0 and bx + ay + c = 0 is less than. Then (A) a+ b c> 0 (B) a b+ c< 0 (C) a b+ c> 0 (D) a+ b c< 0 π 4. The area enclosed by the curves y= sin x+ cos x and y = cos x sin x over the interval 0, is (A) 4( ) (B) ( ) (C) ( + ) (D) ( + ) 43. The number of points in (, ), for which x x sin x cos x = 0, is (A) 6 (B) 4 (C) (D) 0 VMC/Solutions 8 Paper - Code 0 JEE Advanced - 03

19 44. The value of cot (A) (C) cot + n= k= n k is: (B) (D) π y y 45. A curve passes through the point,. 6 Let the slope of the curve at each point x, y be + sec,x 0 x > x. Then the equation of the curve is (A) (C) y sin = log x + x y sec = log x + x y (B) cos ec = log x + x (D) y cos = log x + x VMC/Solutions 9 Paper - Code 0 JEE Advanced - 03

20 46. Let f :, R (the set of all real numbers) be a positive, non-constant and differentiable function such that f '(x) < f ( x ) and f =. Then the value of f ( x )dx lies in the interval / (A) (e, e) (B) ( e, e ) (C) e,e (D) e 0, 47. Let PR= 3 ˆ i+ ˆ j ˆ k and SQ= ˆ i 3 ˆ j 4ˆ k determine diagonals of a parallelogram PQRS and PT = ˆ i+ ˆ j+ 3ˆ k be another vector. Then the volume of the parallelepiped determined by the vectors PT,PQ (A) 5 (B) 0 (C) 0 (D) 30 and PS is : VMC/Solutions 0 Paper - Code 0 JEE Advanced - 03

21 48. Perpendiculars are drawn from points on the line perpendicular lie on the line. (A) (C) x + y + = = z 3 x y z = = (B) x y z = = (D) to the plane x+ y+ z = 3. The feet of x y z = = 3 5 x y z = = Four persons independently solve a certain problem correctly with probabilities, 3,,. Then the probability that the problem is solved correctly by at least one of them is (A) (C) (B) (D) VMC/Solutions Paper - Code 0 JEE Advanced - 03

22 50. Let complex numbers α and respectively. If (A) (C) 7 α lie on circles ( x x0 ) + ( y y0 ) = r and ( x x0 ) + ( y y0 ) = 4r, z = x + iy. satisfies the equation z = r +, then 0 (B) (D) 3 α = VMC/Solutions Paper - Code 0 JEE Advanced - 03

23 Section : (One or more options correct Type) This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct 5. A line l passing through the origin is perpendicular to the lines l : 3+ t iˆ + + t ˆj + 4+ t kˆ, < t< ( ) ( ) ( ) ( ) ˆ ( ) ˆ ( ) l : 3+ s i + 3+ s j+ + s kˆ, < s< Then, the coordinate(s) of the point(s) on l at a distance of 7 from the point of intersection of l and l is (are) (A) 7 7 5,, (B) (,, 0) (C) (,, ) (D) 7 7 8,, Let f ( x) x sin π x, x 0 = >. Then for all natural numbers n, f ( x) (A) a unique point in the interval n, n+ (B) a unique point in the interval n+, n+ n, n+ (C) a unique point in the interval ( ) (D) two points in the interval ( n, n+ ) vanishes at VMC/Solutions 3 Paper - Code 0 JEE Advanced - 03

24 4n 53. Let S = ( ) n k= k ( k+ ) k. Then S n can take value(s) (A) 056 (B) 088 (C) 0 (D) For 3x3 matrices M and N, which of the following statement(s) is(are) NOT correct? (A) N T M N is symmetric or skew symmetric, according as M is symmetric or skew symmetric (B) M N N M is skew symmetric for all symmetric matrices M and N (C) M N is symmetric for all symmetric matrices M and N (D) (adj M) (adj N) = adj (M N) for all invertible matrices M and N VMC/Solutions 4 Paper - Code 0 JEE Advanced - 03

25 55. A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8:5 is converted into an open rectangular box by folding after removing squares of equal area from all four corners. If the total area of removed squares is 00, the resulting box has maximum volume. Then the lengths of the sides of the rectangular sheet are (A) 4 (B) 3 (C) 45 (D) 60 VMC/Solutions 5 Paper - Code 0 JEE Advanced - 03

26 Section 3: (Integer value correct Type) This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive) { } 56. Consider the set of eight vectors V ai bj ckˆ a b c { } chosen from V in p ways. Then p is = ˆ+ ˆ + :,,,. Three non-coplanar vectors can be 57. Of the independent events E, E and E 3, the probability that only E occurs is α, only E occurs is β and only E 3 occurs is γ. Let the probability p that none of events E, E or E 3 occurs satisfy the equations ( α β ) p= αβ and ( β 3 γ ) p= βγ. All the given probabilities are assumed to lie in the interval (0,). Then Probability of occurrence of E = Probability of occurrence of E 3 VMC/Solutions 6 Paper - Code 0 JEE Advanced - 03

27 5 58. The coefficients of three consecutive terms of ( + x ) n+ are in the ratio 5 : 0 : 4. Then n = 59. A pack contain n cards numbered from to n. Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 4. If the smaller of the numbers on the removed cards is k, then k 0 = x y 60. A vertical line passing through the point (h, 0) intersects the ellipse + = at the points P and Q. Let the 4 3 tangents to the ellipse at P and Q meet at the point R. If ( h) =area of the triangle PQR, = max ( h) and = min ( h), then / h = / h VMC/Solutions 7 Paper - Code 0 JEE Advanced - 03

28 JEE Advanced 03 Answer Key Paper - Code 0 PHYSICS CHEMISTRY MATHS D B 4 A A A 4 B 3 D 3 A 43 C 4 A 4 C 44 B 5 B 5 D 45 A 6 B 6 B 46 D 7 C 7 B 47 C 8 C 8 D 48 D 9 A 9 B 49 A 0 B 30 D 50 C B,D 3 A 5 B,D A,C 3 A 5 B,C 3 B,C 33 B,D 53 A,D 4 A,D 34 A,B,C,D 54 C,D 5 B,D 35 B,C,D 55 A,C VMC/Solutions 8 Paper - Code 0 JEE Advanced - 03