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Note:For the benefit of the students, specially the aspiring ones, the question of JEE(Main), 2018 are also given in thisbooklet. Keeping the interest of students studying in class XI, the questions based on topics from class XI have beenmarked with ‘*’, which can be attempted as a test.FIITJEESolutions to JEE(Main)-2018Test Booklet CodeDInstructions as printed on question paper.PAPER – 1 MATHEMATICS, PHYSICS & CHEMISTRYImportant Instructions:1.Immediately fill in the particulars on this page of the Test Booklet with only Black Ball Point Pen provided in theexamination hall.2.The Answer Sheet is kept inside this Test Booklet. When you are directed to open the Test Booklet, take out theAnswer Sheet and fill in the particulars carefully.3.The test is of 3 hours duration.4.The Test Booklet consists of 90 questions. The maximum marks are 360.5.There are three parts in the question paper A, B, C consisting of Mathematics, Physics and Chemistry having 30questions in each part of equal weightage. Each question is allotted 4 (four) marks for each correct response.6.Candidates will be awarded marks as started above in instruction No. 5 for correct response of each question. ¼ (onefourth) marks of the total marks allotted to the question (i.e. 1 mark) will be deducted for indicating incorrect responseof each question. No deduction from that total score will be made if no response is indicated for an item in the answersheet.7.There is only one correct response for each question. Filling up more than one response in any question will be treatedas wrong response and marks for wrong response will be deducted accordingly as per instruction 6 above.8.For writing particulars / marking responses on Side–1 and Side–2 of the Answer Sheet use only Black Ball Point Penprovided in the examination hall.9.No candidate is allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, anyelectronic device, etc. except the Admit Card inside the examination room/hall.10. Rough work is to be done on the space provided for this purpose in the Test Booklet only. This space is given at thebottom of each page and in four pages at the end of the booklet.11. On completion of the test, the candidate must hand over the Answer Sheet to the Invigilator on duty in the Room / Hall.However, the candidates are allowed to take away this Test Booklet with them.12. The CODE for this Booklet is D. Make sure that the CODE printed on Side–2 of the Answer Sheet is same as that onthis Booklet. Also tally the serial number of Test Booklet and Answer Sheet are the same as that on this booklet. In caseof discrepancy, the candidate should immediately report the matter to the Invigilator for replacement of both the TestBooklet and the Answer Sheet.13. Do not fold or make any stray mark on the Answer Sheet.Name of the Candidate (in Capital letters):Roll Number: in figures: in wordsExamination Centre Number :Name of Examination Centre (in Capital letters):Candidate's Signature :1. Invigilator's Signature :2. Invigilator's Signature :FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com.

JEE-MAIN-MPC-2018-2PART A – MATHEMATICS1.If the curves y2 6x, 9x2 by2 16 intersect each other at right angles, then the value of b is :9(1)(2) 627(3)(4) 42Sol.(1)Let the point of intersection be (x1, y1) finding slope of both the curves at point of intersectionfor y2 6x, 9x2 by2 16dy62y 6 , m1 dx2y118x1dy 0 , m2 dx2by1m1m2 –1 6 18x1 1 2y1 2by1 18x 2by 6 18 x1 1 4b 6x1 2.Sol. b 92 Let u be a vector coplanar with the vectors a 2iˆ 3 ˆj kˆ and b ˆj kˆ . If u is perpendicular to a and 2u b 24 , then u is equal to :(1) 84(2) 336(3) 315(4) 256(2) u 1a 2 b u a 1a a 2 a b0 1 4 9 1 2 3 1 2 7 1 u b 1a b 2 b b24 1(2) (–7 1)(2)24 –12 1 1 –2 2 14 u 2 2iˆ 3jˆ kˆ 14 ˆj kˆ 4iˆ 8jˆ 16kˆ 2u 3363. 1 2 15 For each t R, let [t] be the greatest integer less than or equal to t. Then lim x . x 0 x x x (1) does not exist (in R)(2) is equal to 0(3) is equal to 15(4) is equal to 120FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com.

JEE-MAIN-MPC-2018-3Sol.(4) 1 2 15 lim x . x 0 x x x 1 215 1 2 15 lim x . . x 0x x x x x x 1 15 lim 1 2 . 15 x . 120x 0 x x 4.If L1 is the line of intersection of the planes 2x 2y 3z 2 0, x y z 1 0 and L2 is the line ofintersection of the planes x 2y z 3 0, 3x y 2z 1 0, then the distance of the origin from theplane, containing the lines L1 and L2 , is :11(1)(2)24 211(3)(4)3 22 2Sol.(3)Vectors along the given lines L1, L2 areˆi ˆj kˆn̂ 2 2 3 iˆ ˆjL1 011 1 1ˆiˆjkˆand n̂ 2 1 2 1 3iˆ 5jˆ 7kˆ3 1 2Putting y 0 in 1st two equation2x 3z 22x 2z 2z 4Point on the plane is (–5, 0, 4) and normal vector of required plane isˆi ˆjkˆ1 1 0 7iˆ 7ˆj 8kˆ3 5 7Hence, equation of plane is –7x 7y – 8z – 3 031Perpendicular distance is 162 3 2 25.The value ofsin 2 x 1 2xdx is : 2 4 (3)2(1)(2) 8(4) 4 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com.L2 0

JEE-MAIN-MPC-2018-4Sol.(1) 2sin 2 x 1 2Let I xdx 2 2 I sin 2 x 1 2 xdx 2 2 sin 2I 22xdx 2sin 2 xdx 20 2 2I 1 cos 2x dx 20 I 46.Let g(x) cos x2, f (x) x , and , ( ) be the roots of the quadratic equation 18x2 9 x 2 0.Then the area (in sq. units) bounded by the curve y (gof ) (x) and the lines x , x and y 0, is :11(1)(2)2 13 12211(3)3 1(4)3 222 Sol. Y3 1Required area cos x dx 2 y cos x6O 6 3Sol. (2) 3*7. X 1 If sum of all the solutions of the equation 8cos x cos x cos x 1 in [0, ] is k , then k is 6 6 2 equal to :202(1)(2)93138(3)(4)99(3) 1 8cos x cos x cos x 166 2 1 8cos x cos 2 sin 2 x 162 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com.

JEE-MAIN-MPC-2018-5 6 cos x – 8 cos x sin2 x – 4 cos x 1 2 cos x – 8 cos x (1 – cos2 x) 11 2 cos 3x 1 cos 3x 2 2n 3x 2n x 339 5 7 x ,,9 9 913 13k k 998.Let f x x 2 1x2and g x x of h(x) is :(1) 2 2(3) 3Sol.f x 1, x R { 1, 0, 1}. If h(x) , then the local minimum valuexg x (2) 3(4) 2 2(1)f(x) x2 1xg(x) x –21x1x2h(x) 1g x x xf x x2 21 x 2x 1 x x 1 2 h(x) x x x 1x 1 2 h x 1 2 1 2 x 1 x x 1 h x 0 at x 2x1x 2 point of local minimaxHence, local minimum value is 2 2FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com.

JEE-MAIN-MPC-2018-69.The integral 1(1)31 cot xsin 2 x cos 2 x sin 332325x cos x sin x sin x cos x cos x C 1(3)53 1 tan x (2) C(4) 2dx is equal to :1 3 1 tan 3 x11 cot 3 x C C(where C is a constant of integration)Sol.(3)sin 2 x cos 2 x sin 5x cos3 x sin 2 x sin 3 x cos 2 x cos5 xsin 2 x cos 2 x sin3x cos3 xtan 2 x sec 2 x 1 tan x 32 2 2dxdxdxput 1 tan3x t1 dt1 c23 t3t1 c3 1 tan 3 x 10.A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed andthis ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn atrandom from the bag, then the probability that this drawn ball is red, is :33(1)(2)41021(3)(4)55Sol.(3)Initially 4 Red balls and 6-Black balls4 6 6 42Required probability 10 12 10 125*11.Let the orthocentre and centroid of a triangle be A( 3, 5) and B(3, 3) respectively. If C is the circumcentreof this triangle, then the radius of the circle having line segment AC as diameter, is :(1)3 52(3) 2 10(2)(4) 31052FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com.

JEE-MAIN-MPC-2018-7Sol.(4)Orthocentre A (–3, 5) centroid B (3, 3)21A(–3, 5)B(3, 3)C(6, 2)AC 3 10Radius 3 105 322*12.If the tangent at (1, 7) to the curve x 2 y 6 touches the circle x2 y2 16x 12y c 0 then the value ofc is :(1) 95(2) 195(3) 185(4) 85Sol.(1)Equation tangent at (1, 7) 2x – y 5 0perpendicular (–8, –6) to line2 8 6 5 82 62 c5 5 82 6 2 cc 95.*13.If , C are the distinct roots, of the equation x2 x 1 0, then 101 107 is equal to :(1) 2(2) 1(3) 0(4) 1Sol.(4)x2 – x 1 0Roots (– , – 2), – , – 2101 101 107 2107 1*14.PQR is a triangular park with PQ PR 200m. A T.V. tower stands at the mid-point of QR. If the anglesof elevation of the top of the tower at P, Q and R are respectively 45 , 30 and 30 , then the height of thetower (in m) is :(1) 50 2(2) 100(3) 50(4) 100 3Sol.(2)Let ST h (height of tower)PT ST hST tan 30ºQTPS45º200 m200 mQT h 3Now PT2 QT2 20024h2 2002h 100Qh30º30ºTFIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com.R

JEE-MAIN-MPC-2018-89*15.If9 x 5 9 and x 5 iii 12 45 , then the standard deviation of the 9 items x1, x2 , , x9 is :i 1(1) 3(3) 4Sol.(4)xi – 5 yi *16.(2) 9(4) 2 y92i y i9 22 45 9 29 9 (x 1) is :(1) 2(3) 0Sol.5 5The sum of the co-efficients of all odd degree terms in the expansion of x x 3 1 x x 3 1 ,(2) 1(4) 1(1)2 2 5 C0 x5 5 C2 x 3 x 3 1 5 C4 x x 3 1 2 x 10 x x 5 x x 1 2x 56363Sum of coefficient of odd powers 2(1 – 10 10) 2.*17.Tangents are drawn to the hyperbola 4x2 y2 36 at the points P and Q. If these tangents intersect at thepoint T(0, 3) then the area (in sq. units) of PTQ is :(1) 36 5(2) 45 5(3) 54 3(4) 60 3Sol.(2)Chord of contact is y – 124x2 y2 36 x2 451 6 5 15 45 52*18.From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected andarranged in a row on a shelf so that the dictionary is always in the middle. The number of sucharrangements is :(1) at least 750 but less than 1000(2) at least 1000(3) less than 500(4) at least 500 but less than 750Sol.(2)Dictionary can be chosen in 3C1 3 waysNovels can be arranged in 6C4 4! 360 waysN 3 360 1080FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com.

JEE-MAIN-MPC-2018-919.If the system of linear equationsx ky 3z 03x ky 2z 02x 4y 3z 0has a non-zero solution (x, y, z), thenxzy2is equal to :(2) 10(4) 30(1) 30(3) 10Sol.(3)For non–zero solution D 01 k 33 k 2 02 4 3 1(–3k 8) – k(–9 4) 3(12 – 2k) 0 –3k 8 5k 36 – 6k 0 – 4k 44 0k 11hence equations are x 11y 3z 03x 11y – 2z 0and 2x 4y – 3z 0let z t5tthen we get, x t and y 22 5 t t xz 2 thus, 2 2 10yt420.x 4 2xIf 2 x x 42x2x2x (A Bx) (x A)2, then the ordered pair (A, B) is equal to :x 42x(2) ( 4, 5)(4) ( 4, 5)(1) (4, 5)(3) ( 4, 3)Sol.(4)12x2x(5x – 4) 1 x 4 2x1 2x x 412x2x (5x – 4) 0 (x 4)000 x 4 (5x – 4)(x 4)2 (A Bx)(x – A)2 A – 4, B 5FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com.

JEE-MAIN-MPC-2018-10*21.Two sets A and B are as under :A {(a, b) R R : a 5 1 and b 5 1} ;B {(a, b) R R : 4(a 6)2 9(b 5)2 36}. Then :(1) neither A B nor B A(2) B A(3) A B(4) A B (an empty set)Sol.(3)A a, b : a 4, 6 , b 4, 6 B a, b : 4 a 6 a 6 9 A B2 b 5 2bb 6 2 9 b 5 36b 42(6, 3) 14(6, 7)aa 4 a 6*22.Tangent and normal are drawn at P(16, 16) on the parabola y2 16x, which intersect the axis of theparabola at A and B, respectively. If C is the centre of the circle through the points P, A and B and CPB , then a value of tan is :41(1)(2)32(3) 2(4) 3Sol.(3)Focus will be centre of Circumcircle of PABB : (24, 0)C : (4, 0)16 4mCP , mPB –212 34 2tan 3 281 3P(16, 16) AC(4, 0)B23.Let S {t R : f (x) x . (e x 1) sin x is not differentiable at t}. Then the set S is equal to :(1) {0, }(2) (an empty set)(3) {0}(4) { }Sol.(2) x S {t : R : f(x) x – e 1 sin x is not diff. at t} x f(x) x – (e – 1) sin x at x 0, h e h 1 sin h f (0 ) lim h 0h h e h 1 sin h 0 lim h 0 h h e h 1 sin h 0f (0 ) lim h 0 h – h e h 1 sin h f ( ) lim h 0h h e h 1 sinh 0 hlim 0 h FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com.

JEE-MAIN-MPC-2018-11 h e h 1 sinh f ( ) lim 0h 0 h f(x) is diff. for all x R–*24.The Boolean expression (p q) ( p q) is equivalent to :(1) q(2) p(3) p(4) qSol.(2)( p q) ( p q) p ( q q) pAlternate solutionpq pp q p q (p q)TTTFFFTFTFFFFTTTTFFFFTFT (p q) ( p q)FFTT*25.A straight line through a fixed point (2, 3) intersects the coordinate axes at distinct points P and Q. If O isthe origin and the rectangle OPRQ is completed, then the locus of R is :(1) 3x 2y 6xy(2) 3x 2y 6(3) 2x 3y xy(4) 3x 2y xySol.(4)Let the point R(a, b)yx ythen equation of line be 1a b2 3 1a b2 3Locus of R is 1x yi.e. 2y 3x xyR(a, b)Q(0, b)O*26.Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series12 2.22 32 2.42 52 2.62 .If B 2A 100 , then is equal to :(1) 496(2) 232(3) 248(4) 464Sol.(3)A (12 22 32 202) (22 42 202) (12 22 32 202) 22(12 22 32 102)20 21 41 4 10 11 21 66 4410B (12 22 32 402) (22 42 62 402) 12 22 32 402 4(12 22 . 202)40 41 81 4 20 21 41 66 33620B – 2A 33620 – 8820 24800 100 P(a, 0)FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com.x

JEE-MAIN-MPC-2018-12 248Alternate solutionSn n n 1 22, n is even.412212 40 22 100 400 62 248 B – 2A 40 27.Sol.Let y y(x) be the solution of the differential equation sin xdy y cos x 4 x , x (0, ). If y 0 ,dx 2 then y is equal to : 6 4(1) 29 8 2(3) 9 3 24(2)9 38(4) 29(4)d(y sinx) 4xdxx2 c2y sinx 2x2 cgiven y ( /2) 0c – 2 /2Thus y sin x 2x2 – 2/28now y( /6) 29 y sinx 428.The length of the projection of the line segment joining the points (5, 1, 4) and (4, 1, 3) on the plane,x y z 7 is :(1)2332(3)3Sol.2(2)1(4)3(1)D.R’s of AB (1, 0, 1)D.R’s of normal to plane (1, 1, 1)1 1 cos 2 3 2 B(4, –1, 3)2 sin 3A(5, –1, 4)AB 2Length of projection AB cos 23 P:x y z 7FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com.

JEE-MAIN-MPC-2018-13*29.Let S {x R : x 0 and 2x 3 x x 6 6 0 }. Then S :(1) contains exactly four elements(3) contains exactly one elementSol.(2) is an empty set(4) contains exactly two elements(4)2 x 3 x x 6 6 0let 0 x 3 6 2 x x x 2 x 4 (Ignoring x 2 6 x 6 0x 6)2letx 3 2 x 6 x 4 x 16 (Ignoring 6 x 6 0x 0)12*30.Let a1, a2, a3, . , a49 be in A.P. such that a4 k 12 416 and a9 a43 66. If a12 a22 . a17 140m,k 0then m is equal to :(1) 33(3) 68Sol.(2) 66(4) 34(4)12 a4k 1 416 (let common difference of A.P. be d)k 012 a1 4 kd 416k 0 a1 24d 32Given a9 a43 66 a1 25d 33 d 1, a1 817 17a 2r r 1 7 r 2 4760 140mr 1 m 34FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com.

JEE-MAIN-MPC-2018-14PART B – PHYSICSALL THE GRAPHS/DIAGRAMS GIVEN ARESCHEMATIC AND NOT DRAWN TO SCALE31.Three concentric metal shells A, B and C of respective radii a, b and c (a b c) have surface chargedensities , – and respectively. The potential of shell B is : b 2 c2 a 2 b2(1) a (2) c 0 c 0 a (3)Sol. a 2 b2 c 0 b (4) b 2 c2 a 0 b (3)VB VB I VB II VB III VB K4 a K4 b K4 c bbc222 a 2 b2 c 0 b cabIIIIII*32.Sol.Seven identical circular planar disks, each of mass M and radius R arewelded symmetrically as shown. The moment of inertia of the arrangementabout the axis normal to the plane and passing through the point P is :18119(1)(2)MR 2MR 2225573(3)(4)MR 2MR 222PO(1)I I cm 7M 3R 2 MR 2 MR 2 81 MR 22 2 6 M 2R 7M 3R 2 2 2*33.From a uniform circular disc of radius R and mass 9M, a small discRof radiusis removed as shown in the figure. The moment of3inertia of the remaining disc about an axis perpendicular to the planeof the disc and passing through centre of disc is :37(1)(2) 4 MR2MR 2940(3)(4) 10 MR2MR 292R3RFIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com.

JEE-MAIN-MPC-2018-15Sol.(2)I 34.229MR 2 M R 2R 2 M 4 MR2233 The reading of the ammeter for a silicon diode in the given circuitis :(1) 13.5 mA(2) 0(3) 15 mA(4) 11.5 mA200 3VSol.(4)I 0.7 V3 0.7A 11.5 mA200200 3V35.Unpolarized light of intensity I passes through an ideal polarizer A. Another identical polarizer B is placedIbehind A. The intensity of light beyond B is found to be . Now another identical polarizer C is placed2Ibetween A and B. The intensity beyond B is now found to be . The angle between polarizer A and C is :8(1) 60 (2) 0 (3) 30 (4) 45 Sol.(4)2I I cos 2 8 21 cos 2 45 36.For an RLC circuit driven with voltage of amplitude vm and frequency 0 resonance. The quality factor, Q is given by :CR(1) 0(3) 0 RL(2) 0 LR(4)R 0 C 1LCthe current exhibitsFIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com.

JEE-MAIN-MPC-2018-16Sol.(2) 1 2 1 L 0 L1 [ here 02 ] 0R CRLCAlternate solution 0 Lis the only dimensionless quantity, hence must be the quality factor.RQ *37.Two masses m1 5 kg and m2 10 kg, connected by an inextensiblestring over a frictionless pulley, are moving as shown in the figure. Thecoefficient of friction of horizontal surface is 0.15. The minimumweight m that should be put on top of m2 to stop the motion is :(1) 10.3 kg(2) 18.3 kg(3) 27.3 kg(4) 43.3 kgmTm2Tm1m1gSol.(3)T 5g 10 m g 5g50 15m 23.33 kgThe minimum value from the options, satisfying the above condition is, m 27.3 kg10 m *38.In a collinear collision, a particle with an initial speed v0 strikes a stationary particle of the same mass. Ifthe final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relativevelocity between the two particles, after collision, is :vv(1) 0(2) 042v(3) 2 v0(4) 02Sol.(3)kf 1.5 kiv12 v 22 1.5v 20From conservation of momentumv1 v2 v0from (i) and (ii)2v1v2 –0.5 v02v0mv1restmBefore collisionmv2mAfter collisionSo, v 2 v1 v 22 v12 2v1 v 2 1.5v 02 0.5v 02 2 v 0*39.A particle is moving with a uniform speed in a circular orbit of radius R in a central force inverselyproportional to the nth power of R. If the period of rotation of the particle is T, then :(1) T Rn/2(2) T R3/2 for any n.n(3) T R 2 1(4) T R(n 1)/2FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com.

JEE-MAIN-MPC-2018-17Sol.(4)F v 1Rn1n 1R 22 RT vT RT R1 n 12n 1240.Two batteries with e.m.f 12 V and 13 V are connected in parallel across a load resistor of 10 . Theinternal resistances of the two batteries are 1 and 2 respectively. The voltage across the load liesbetween :(1) 11.7 V and 11.8 V(2) 11.6 V and 11.7 V(3) 11.5 V and 11.6 V(4) 11.4 V and 11.5 VSol.(3)The circuit may be drawn as shownin the figure. iri eq 11.56 V1 ri12 V1 13 V2 0V10 41.In an a.c. circuit, the instantaneous e.m.f. and current are given bye 100 sin 30 t i 20 sin 30 t 4 In one cycle of a.c., the average power consumed by the circuit and the wattless current are, respectively :(1) 50, 0(2) 50, 10100050(3)(4), 10,022Sol.(3) 0 I0 cos 100 20 cos 4 1000P watt22220 1Wattless current Irms sin 10A2 2e0 45 i0FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com.

JEE-MAIN-MPC-2018-1842.An EM wave from air enters a medium. The electric fields are z E1 E 01 xˆ cos 2 t in air and c E 2 E 02 xˆ cos k 2z ct in medium, where the wave number k and frequency v refer to their valuesin air. The medium is non-magnetic. If r1 and r2 refer to relative permittivities of air and mediumrespectively, which of the following options is correct ? r r111(1) (2) 4 r2 r22(3)Sol. r1 r2(4) 2 r1 r2 14(4) 0 0 r rvairc22 2 vmedc/2 0 0 r r1 r1 r2 11443.A telephonic communication service is working at carrier frequency of 10 GHz. Only 10% of it is utilizedfor transmission. How many telephonic channels can be transmitted simultaneously if each channelrequires a bandwidth of 5 kHz ?(1) 2 106(2) 2 1034(3) 2 10(4) 2 105Sol.(4)No. of telephonic channels that can be transmitted simultaneously 0.1 10 109 2 1055 103*44.A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. Thedensity of granite is 2.7 103 kg/m3 and its Young’s modulus is 9.27 1010 Pa. What will be thefundamental frequency of the longitudinal vibrations ?(1) 7.5 kHz(2) 5 kHz(3) 2.5 kHz(4) 10 kHzSol.(2)Y 5.86 103 m/s For fundamental mode, 1.2 mv fundamental frequency 4.88 kHz 5 kHz Wave velocity (v) *45.It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of itsenergy is pd; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc. Thevalues of pd and pc are respectively :(1) (0, 1)(2) ( 89, 28)(3) ( 28, 89)(4) (0, 0)FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com.

JEE-MAIN-MPC-2018-19Sol.(2)Before collisionAfter collisionK0Neutron (1) Neucleus (A) (1)K1(A)K2K 0 K1 AK 2 (from conservation of momentum)and K 0 K1 K 2 (for elastic collision)So after solvingKK1 A 1 (1 A) 1 2K0K0For Deuterium, A 2, 1 For Carbon, A 12, 1 K1 0.89K0K1 0.28K046.The density of a material in the shape of a cube is determined by measuring three sides of the cube and itsmass. If the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximumerror in determining the density is:(1) 6%(2) 2.5%(3) 3.5%(4) 4.5%Sol.(4) m3 a 100 100 100 ma 4.5%*47.Two moles of an ideal monoatomic gas occupies a volume V at 27 C. The gas expands adiabatically to avolume 2 V. Calculate (a) the final temperature of the gas and (b) change in its internal energy.(1) (a) 195 K(b) 2.7 kJ(2) (a) 189 K(b) 2.7 kJ(3) (a) 195 K(b) –2.7 kJ(4) (a) 189 K(b) – 2.7 kJSol.(4)T1 V1 1 T2 V2 1 T2 189 K U n Cv T – 2.7 kJ*48.A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in acylindrical container. A massless piston of area a floats on the surface of the liquid, covering entire crosssection of cylindrical container. When a mass m is placed on the surface of the piston to compress the dr liquid, the fractional decrement in the radius of the sphere, , is: r mgKa(1)(2)Kamg(3)Ka3mg(4)mg3KaFIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com.

JEE-MAIN-MPC-2018-20Sol.(4) V 3 r VrPK V / V r mg r 3Ka49.A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material5of dielectric constant K is inserted between the plates, the magnitude of the induced charge will be :3(1) 0.9 n C(2) 1.2 n C(3) 0.3 n C(4) 2.4 n CSol.(2)Q KCVQinduced Q 1 1/ K Qinduced 1.2 n C50.The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of theloop is B1 . When the dipole moment is doubled by keeping the current constant, the magnetic field at thecentre of the loop is B2 . The ratio(1)(3)Sol.1B1is:B2(2) 22(4)32(4)m I r 22m I r 2 r 2r IBr B 0 1 22 rB2 r51.An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let n , g be the de Broglie wavelength of the electron in the nth state and the ground state respectively. Let n be the wavelength of the emitted photon in the transition from the nth state to the ground state. For largen, (A, B are constants)B(1) 2n (2) n A 2 n(3) n A B nSol.(4) 2n A B 2n(2)2 r n n n 2 r 2 r0 n 2 2 r0 nnn (1)FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com.

JEE-MAIN-MPC-2018-211 1 1 R 2 2 n 1 n 1 1 1 R n 2 1 1 1 n 1 2 R n n n 1 From equation (1) and (2) n (2)1 4 2 r02 B 1 2 A 2R n n*52.The mass of a hydrogen molecule is 3.32 10 27 kg. If 1023 hydrogen molecules strike, per second, a fixedwall of area 2 cm 2 at an angle of 45 to the normal, and rebound elastically with a speed of 103 m/s, thenthe pressure on the wall is nearly:(1) 4.70 10 2 N / m 2(2) 2.35 103 N / m 2(3) 4.70 103 N / m 2(4) 2.35 10 2 N / m 2Sol.(2)P 2mv cos 45 nA2 3.32 10 27 103 (P Pressure, A Area)1 10 2322 10 4 2.35 10 N / m 23*53.All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it epositiontimetimeSol.(3)In graph ‘3’ initial slope is zero which is not possible, since initial velocity is non zero in all other threegraphs.54.An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits ofradii re, rp, r respectively in a uniform magnetic field B. The relation between re, rp, r is:(1) re r rp(2) re rp r (3) re rp r (4) re rp r FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com.

JEE-MAIN-MPC-2018-22Sol.(3)r 2Km(K Kinetic energy)qBr mqSo re rp r 55.On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. Theresistance of their series combination is 1 k . How much was the resistance on the left slot beforeinterchanging the resistances?(1) 910 (2) 990 (3) 505 (4) 550 Sol.(4)R1 R 2 100 and (i)R2 10 R1 100 10 (ii)From (i) and (ii) 55 cm.R1 55 R 2 45and R1 R 2 1000 from (i) and (iii)R1 550 (i) (iii)56.In a potentiometer experiment, it is found that no current passes through the galvanometer when theterminals of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by aresistance of 5 , a balance is found when the cell is connected across 40 cm of the wire. Find the internalresistance of the cell.(1) 2.5 (2) 1 (3) 1.5 (4) 2 Sol.(3)When the cell is shunted by a resistance of 5 i 5 r ir 40Now 52r3 5 r 13 r 1.5 57.If the series limit frequency of the Lyman series is L, then the series limit frequency of the Pfund series is:(1) L/25(2) 25 L(3) 16 L(4) L/16FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942website: www.fiitjee.com.

JEE-MAIN-MPC-2018-23Sol.(1)Series limit frequency of the Lyman series is given by 1 1 L RcZ 2 2 2 1 L RcZ 2 (1)1 1Series limit frequency of the Pfund series P RcZ 2 2 2 5 P L2558.The angular width of the central maximum in a single slit diffraction pattern is 60 . The width of the slit is1 m. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it,Young’s fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observedfringe width is 1 cm, what is slit separation distance?(i.e. distance between the centres of each slit.)(1) 100 m(2) 25 m(3) 50 m(4) 75 mSol.(2) b 1 10 6 5 10 7 mb22 D D 5 10 7 0.5Fringe width, d d 1 10 2d 25 msin 30 *59.A particle is moving in a circular path of radius a under the action of an attractive potential U total energy is:3 k(1) 22ak(3)2a 2Sol.(2) k. Its2 r2k4a 2(4) Zero(4)kdUk Fr 322rdrrSince the particle moves in a circular path of radius ‘a’mv 2 kk 3 mv 2 2 (1)aaa1kKinetic Energy mv 2 222akkTotal Energy 2 2 02a2aGiven U 60.A silver atom in a solid oscillate

FIITJEE Ltd., FIITJEE House, 29 -A, Kalu Sarai, Sarvapriya Vihar, New Delhi 110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com. Note: For the benefit of the students, specially the aspiring ones, the question of JEE(Main), 2018 are also given in this booklet. Keeping the interest of students studying in class XI, the questions based on topics

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