RBS PAPER - 1 : MATHEMATICS, PHYSICS & CHEMISTRY

Transcription

1 RBS Test Booklet Code D PAPER - : MATHEMATICS, PHYSICS & CHEMISTRY Do not open this Test Booklet until you are asked to do so. Read carefully the Instructions on the Back Cover of this Test Booklet. Important Instructions:. Immediately fill in the particulars on this page of the Test Booklet with only Black Ball Point Pen provided in the examination hall.. The Answer Sheet is kept inside this Test Booklet. When you are directed to open the Test Booklet, take out the Answer Sheet and fill in the particulars carefully.. The test is of hours duration. 4. The Test Booklet consists of 90 questions. The maximum marks are There are three parts in the question paper A, B, C consisting of, Mathematics, Physics and Chemistry having 0 questions in each part of equal weightage. Each question is allotted 4 (four) marks for each correct response. 6. Candidates will be awarded marks as stated above in instruction No. 5 for correct response of each question. /4 (one fourth) marks of the total marks allotted to the question (i.e., mark) 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. 7. There is only one correct response for each question. Filling 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 6 above. 8. For writing particulars/ marking responses on Side- and Side- of the Answer Sheet use only Black Ball Point Pen provided in the examination hall. 9. No candidates is allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, any electronic device, etc., except the Admit Card inside the examination room/hall. 0. Rough work is to be done on the space provided for this purpose in the Test Booklet only. This space is given at the bottom of each page and in 4 pages (Pages 0) at the end of the booklet.. n 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.. The CDE for this Booklet is D. Make sure that the CDE printed on Side of the Answer Sheet and also tally the serial number of the Test Booklet and Answer Sheet are the same as that on this booklet. In case of discrepancy, the candidate should immediately report the matter to the invigilator for replacement of both the Test Booklet and the Answer Sheet.. Do not fold or make any stray marks on the Answer Sheet. Name of the Candidate (in Capital letters) : Roll Number : in figures : in words Examination Centre Number : Name of Examination Centre (in Capital letters) : Candidate s Signature :. Invigilator s Signature :. Invigilator s Signature : (Pg. )

2 JEE-MAIN 07 : Question Paper and Solution () Read the following instructions carefully :. The candidates should fill in the required particulars on the Test Booklet and Answer Sheet (Side-) with Black Ball Point Pen.. For writing/marking particulars on Side- of the Answer Sheet, use Black Ball Point Pen only.. The candidates should not write their Roll Numbers anywhere else (except in the specified space) on the Test Booklet/Answer Sheet. 4. ut of the four options given for each question, only one option is the correct answer. 5. For each incorrect response, ¼ (one-fourth) of the total marks allotted to the question (i.e., mark) will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Sheet. 6. Handle the Test Booklet and Answer Sheet with care, as under no circumstances (except for discrepancy in Test Booklet Code and Answer Sheet Code), another set will be provided. 7. The candidates are not allowed to do any rough work or writing work on the Answer Sheet. All calculations/writing works are to be done in the space provided for this purpose in the Test Booklet itself, marked Space for Rough Work. This space is given at the bottom of each page and in 4 pages (Pages 0 ) at the end of the booklet. 8. n completion of the test, the candidates 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. 9. Each candidate must show on demand his/her Admit Card to the Invigilator. 0. No candidate, without special permission of the Superintendent or Invigilator, should leave his/her seat.. The candidates should not leave the Examination Hall without handing over their Answer Sheet to the Invigilator on duty and sign the Attendance Sheet again. Cases where a candidate has not signed the Attendance Sheet second time will be deemed not to have handed over the Answer Sheet and dealt with as an unfair means case. The candidates are also required to put their left hand THUMB impression in the space provided in the Attendance Sheet.. Use of Electronic/Manual Calculator and any Electronic Item like mobile phone, pager etc. is prohibited.. The candidates are governed by all Rules and Regulations of the Examination body with regard to their conduct in the Examination Hall. All cases of unfair means will be dealt with as per Rules and Regulations of the Examination body. 4. No part of the Test Booklet and Answer Sheet shall be detached under any circumstances. 5. Candidates are not allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, electronic device or any other material except the Admit Card inside the examination room/hall. (Pg. )

3 () VIDYALANKAR : JEE-MAIN 07 : Question Paper and Solution.Questions and Solutions. PART- A : MATHEMATICS. If S is the set of distinct values of b for which the following system of linear equations x + y + z =, x + ay + z =, ax + by + z = 0 has no solution, then S is : () an empty set () an infinite set () a finite set containing two or more elements (4) a singleton. (4) = a = (a ) a b x = (a ) y = 0 z = a(a ) If a = = x = y = z = 0 Infinite solution For b = & a = x + y + z = x + y + z = 0 no solution nly one value of b.. The following statement (p q) [(~ p q) q] is : () a tautology () equivalent to ~ p q () equivalent to p ~ q (4) a fallacy. () (p q) [(p q) q] is p q p p q p q [(p q)q] 4 6 T T F T T T T T F F F T F T F T T T T T T F F T T F T T. If 5tan x cos x (). (4) = cos x + 9, then the value of cos 4x is : () 5 5 tan x cos x = cos x sec x cos x = (cos x ) cos x cos x = 4 cos x + 9 () 9 (4) 7 9 (Pg. )

4 4 cos x cos x 5 = 4 cos cos x x cos x 5 cos 4 x = 4 cos 4 x + 7 cos x 9 cos 4 x + cos x 5 = 0 Put, cos x = m 9m + m 5 = 0 9m + 5m m 5 = 0 m (m + 5) (m + 5) = 0 (m + 5) (m ) = 0 m = 5 or m = cos x = or cos 5 x = But, cos 5 x JEE-MAIN 07 : Question Paper and Solution (4) cos x = Now, cos 4x = cos x = (cos x) = ( cos x ) = = = = 9 = 9 = For three events A, B and C, P (Exactly one of A or B occurs) = P (Exactly one of B or C occurs) = P (Exactly one of C or A occurs) = and P (All the three events occur simultaneously = () Then the probability that at least one of the events occurs is : 7 () () 7 7 () 6 64 P(exactly one of A or B occurs) = 4 P(A) + P(B) P(A B) = 4 (4) 6 P(exactly one of B & C occurs) = 4 P(B) + P(C) P(B C) = 4 P(exactly one of C or A occurs) = 4 P(C) + P(A) P(A C) = 4 P(A B C) = 6 Ad. [P(A) + P(B) + (C) P(A B) P(B C) P(C A)] = 8. () Probability of at least one event = P(A B C) = P(A) + P(B) + P(C) P(A B) P(B C) P(C A) + P(A B C) = = (Pg. 4)

5 (5) VIDYALANKAR : JEE-MAIN 07 : Question Paper and Solution 5. Let be a complex number such that + = z, where z =. If then k is equal to : () z () z () (4) 5. () i i i w = w =,w + w + w = 0, w = k = w w w w 7 w w w w = (w w 4 ) (w w ) + (w w) = w w 4 w + w + w w = w w w 4 = w w w = (w w) k = w i i w = i i i = i z (w + = z = i) 7 = k, 6. Let k be an integer such that the triangle with vertices (k, k), (5, k) and (k, ) has area 8 sq. units. Then the orthocenter of this triangle is at the point : (), (), (), (4), (4) = (k 0 k ) ( 5k k k) 8 5k k0 56 5k + k + 0 = 56 5k + k + 0 = 56 5k + k 46 = 0 5k + k 0k 46 = 0 k (5k + ) (5k + ) = 0 k =, 5 k = as integer A (, 6), B (5, ), C (, ). As BC perpendicular to xaxis eqn. of altitude AD is x = For BE, y = (x 5) C(, ) D B (5,) E A(, 6) Solving orthocentre, (Pg. 5)

6 JEE-MAIN 07 : Question Paper and Solution (6) 7. Twenty meters of wire is available for fencing off a flowerbed in the form of a circular sector. Then the maximum area (in sq. m) of the flowerbed is : ().5 () 0 () 5 (4) 0 7. () r + s = 0 (length of wire) Now s = r 0r r + r = 0 = s r r A = r 0r r = r r A = 0 r r da = 0 r = 0 r = 5 dr da = < 0 Area is maximum dr 0 r Maximum Area = r 5 r 00 = = 5 0 = 5 sq.m The area (in sq. units) of the region 8. (4) () 59 Area = = 0 0 () xdy ydx PQR ydy xdx 0 0 x, y : x0, x y, x 4yand y x is : () 7 (4) 5 = = = 5 9. If the image of the point P (,, ) in the plane, x + y 4z + = 0, measured parallel to the line, x y z is Q, then PQ is equal to : 4 5 () 5 () 4 () 4 (4) () Line PM (Parallel to given line) is x y z r(let) P(,,) 4 5 x y z M (r +, 4r, 5r + ) 4 5 M satisfy plane, x + y 4z + = 0 r + + r 6 0r + = 0 M 6r + 6 = 0 r = So, M = (,, 8) PM = So, PQ = 4 (0,) (0,) S ( ) ( ) (8 ) = Q(,) R P(,) (,0) x = 4y y x (Pg. 6)

7 (7) VIDYALANKAR : JEE-MAIN 07 : Question Paper and Solution 0. If for 9 () 9x 0. () x 0, 4, the derivative of 6x x tan 9x is y = tan () x x x x = tan tan dy dx = x 9x 9 g(x) = 9x x x 9x () = = tan x x. Let x x tan = 9 x 9x = x 9x 9 x 9x x.gx, then g(x) equals : dy. If sinx y cos x 0 and y(0) =, then y is equal to : dx () () () (4) 4. () dy ycos x cosx 0 dx sinx sin x dy cos x cos x y dx sin x sin x cosx dx sin x I.F. = e sin x y. ( + sin x) = c + ( cos x)dx y ( + sin x) = c sin x Given y (0) = ( + 0) = c 0 c = Soln. y( sin x) = sin x y(/) = (4) 9x. Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP = AB. If BPC =, then tan is equal to : () 9 () 6 7 () 4. () AP = AB tan = AB AP AB AC tan = AP AB 4 C B A (4) 4 9 P (Pg. 7)

8 JEE-MAIN 07 : Question Paper and Solution (8) tan ( + ) = tan tan tan tan tan = 4.tan 4 4 tan + 8 tan = 4 tan 4 tan 9 tan = tan = 9. If A = 4, then adj (A + A) is equal to : () 6 5 () 84 7 () 6 7. () A 6 9 = 4 4 A A = = 84 5 Adj. (A A) = 84 7 (4) For any three positive real numbers a, b and c, 9 (5a + b ) + 5 (c ac) = 5b (a + c). Then : () b, c and a are in G.P. () b, c and a are in A.P. () a, b and c are in A.P. (4) a, b and c are in G.P. 4. () 5a + 9b + 5c 75ac 45ab 5bc = 0 450a + 8b + 50c 50ac 90ab 0bc = 0 (5a 5c) + (5a b) + (b 5c) = 0 5a = 5c 5a = b b = 5c 5a = b = 5c a b c 5 b, c, a in A.P. 5. The distance of the point (,, 7) from the plane passing through the point (,, ) having normal perpendicular to both the lines x y z 4 and x y z 7 is : () () () (4) () Plane passing (,,) is a(x ) + b(y + ) + c(z + ) = 0 (Pg. 8)

9 (9) VIDYALANKAR : JEE-MAIN 07 : Question Paper and Solution Normal of line containg both the line is ˆi ˆj kˆ 5i ˆ 7j ˆ kˆ From given condition normal of plane parallel to 5i ˆ 7j ˆ kˆ Plane is 5(x ) + 7(y + ) + (z + ) = 0 5x + 7y + z + 5 = 0 Distance = 5 7 ( 7) n 6. Let I n = tan xdx, n then the ordered pair (a, b) is equal to : (), (), () I n = = I 6 + I 4 = n tan x.(sec x) dx n n tan x.sec xdx tan xdx n tan x I n = I n n n tan x I n + I n = C n n = 6 5 tan x 5 C Equate a = 5, b = 0. If I 4 + I 6 = a tan 5 x + bx 5 + C, where C is a constant of integration, (), 5 (4), 5 (a, b) =, The eccentricity of an ellipse whose centre is at the origin is. If one of its directrices is x = 4, then the equation of the normal to it at, is : () y x = () 4x y = () 4x + y = 7 (4) x + y = 4 7. () e a x = e 4 = 4 = a a = a x = 4 (Pg. 9)

10 e b = a 4 = b 4 b 4 4 b = JEE-MAIN 07 : Question Paper and Solution (0) x y ellipse = 4 x y dy = 0 4 dx dy x x dx 4 y 4y m tangent = 4 Normal at, is y = (x ) y = 4x 4 4x y = 0 8. A hyperbola passes through the point P, and has foci at (, 0). Then the tangent to this 8. () hyperbola at P also passes through the point : (), (), (), (4), S (, 0) S (, 0) foci S ( ae, 0) ae = 4 a e = 4 a b a = 4 a + b = 4 a = 4 b Hyperbola x y = x y a b It passes P, = a b = 4 b b b + b = b (4 b ) b 4 + b = 0 (b + 4) (b ) = 0 b = a = (Pg. 0)

11 () VIDYALANKAR : JEE-MAIN 07 : Question Paper and Solution y Tangent at, is x = bviously, satisfy 9. The function f :R, defined as f(x) = x, is : x () invertible () injective but not surjective () surjective but not injective (4) neither injective nor surjective 9. () bviously from graph not oneone. f(x) = f(x) x ( x ) bviously range < 0 x R,. 0. cot x cos x lim equals : x () 0. () 4 Let L = x cot x cos x lim x x () 6 () 8 Put, x = = x x = (4) 4 As x, 0 cot cos L = lim 0 sin sin tan sin cos = lim = lim 0 0 = sin sin cos lim cos 0 sin cos = lim = 0 cos = 6 = 6 sin sin 4 lim lim lim 6 cos (Pg. )

12 JEE-MAIN 07 : Question Paper and Solution (). Let a i ˆ ˆj kˆ and b ˆi ˆj. Let c be a vector such that c a a b c = and the angle between c and a b be 0. Then a. c is equal to : () 5 8 () () 5 (4) 8. () a b = i ˆ j ˆ kˆ c a = (c a) (c a) = 9 a b = c a a c = 9 Given : (a b) c = a c = 9 a b c sin 0 = a c = c = c = =,. The normal to the curve y (x ) (x ) = x + 6 at the point where the curve intersects the yaxis passes through the point : (), (), (), (4),. () yaxis at (0, ) y (x 5x + 6) = x + 6 x 6 y = x 5x 6 dy dx = x 5x 6 x 6 x 5 x 5x 6 dy dx 0, = 6 = Normal at (0, ) is y = (x 0) y + x = It passes,. If two different numbers are taken from the set {0,,,,, 0}; then the probability that their sum as well as absolute difference are both multiple of 4, is : 6 () () () 4 7 (4) () {0,,,, 4, 5, 6, 7, 8, 9, 0} 0 n(s) C 55 Let A be the event that the sum as well as absolute difference of two different numbers taken from given get are both multiple of = 8 divisible by divisible by = divisible by 4 (Pg. )

13 () VIDYALANKAR : JEE-MAIN 07 : Question Paper and Solution 0 8 divisible by = divisible by = 4 n(a) 6 P(A). n(s) A man X has 7 friends, 4 of them are ladies and are men. His wife Y also has 7 friends, of them are ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X and Y together can throw a party inviting ladies and men, so that friends of each of X and Y are in this party, is : () 485 () 468 () 469 (4) () Man (7 friends ) Wife (7 friends) 4 Ladies Men Ladies 4 Men friends Party Invites Ladies and Men (G ) 4 Ladies + Men Ladies + 4 Men (G ) () () Ladies and Men G G L M 4 C C M L C C L and M L and M 4 4 C C C C L and M L and M 4 4 C C C C Req. no. = 4 C C C C C C C C C C C 4 C 5. The value of 4 4 = = = 485 C 0 C C 0 C C 0 C C4 0 C 4.. C0 0 C0 () is : () 0 () 0 9 (4) 0 0 (Pg. )

14 5. (4) C C... C0 0 C 0 C... 0 C0 0 0 = ( ) ( ) = 0 0 As C C... C0 JEE-MAIN 07 : Question Paper and Solution (4) C0 C C... C0 C... C0 C 0 0 C C... C 6. A box contains 5 green and 0 yellow balls. If 0 balls are randomly drawn, onebyone, with replacement, then the variance of the number of green balls drawn is : 6 () 6 () 4 (4) 5 () 5 6. () 5 G, 0 Y. P(Green ball in each step) = P(yellow ball in each step) = 5 Variance = npq = 0 = Let a, b, c R. If f(x) = ax + bx + c is such that a + b + c = and f (x + y) = f(x) + f(y) + xy, x, y R, then 0 fn is equal to : n () 0 () 65 () 90 (4) () f(x)= ax + bx + c f()= a + b + c = f( + ) = f() + f()= + = 7 f( + ) = f() + f() + f()= = f( + ) = f() + f() + = + + = 8 S n = t n _ S n = _ + _ 7 + _ + + _ t n _ + t n 0 = ( n terms) t n = n t n = n [6 (n )] = n [n ] 0 f (n) = n = n 5 n 0 0 n n = 55 = 0 5n 0 () () 0 () 5. 6 = [ ] (Pg. 4)

15 (5) VIDYALANKAR : JEE-MAIN 07 : Question Paper and Solution 8. The radius of a circle, having minimum area, which touches the curve y = 4 x and the lines, y = x is : () () () 4 (4) 4 8. () Let centre of circle (0, k) As touches x y = 0 line 0 k r = k = r (radius) Circle is (x 0) + (y k) = As touches y = 4 x It passes (0, 4) 0 + (4 k) k = k 6k + = 0 k = 8 4 Radius for minimum radius = = 4 4 k (as symmetric curves) = 4 (0, 4) y = x 9. If for a positive integer n, the quadratic equation, xx x x... x n x n 0n has two consecutive integral solutions, then n is equal to : () () 9 () 0 (4) 9. (4) Let the roots be, + ( + ) + ( + ) ( + ) + ( + (n ) ( + n) = 0n and ( + ) ( + ) + ( + ) ( + ) + + ( + n) ( + n + ) = 0n Subtract, ( + ) ( + n) ( + n + ) = 0 + (n + ) n (n + ) = 0 n n (n + ) = 0 = (n + ) n = So n must be odd n = 9, = 4 n =, = 5 Put, = 4, (4) () + () () + () () + () = = 60 0n (Pg. 5)

16 JEE-MAIN 07 : Question Paper and Solution (6) Put, = 5, (5) (4) + (4) () = = 0 = 0n n = 0. The integral /4 /4 dx is equal to: cosx () () () 4 (4) 0. () I = I = 4 dx () cos x 4 4 dx () cosx 4 By adding, () + () I = I = 4 dx cos x 4 4 cosec xdx = 4 PART- B : PHYSICS 4 4 cot x = cot cot 4 4 = [ ] =. A radioactive nucleus A with half time T, decays into a nucleus B. At t = 0, there is no nucleus B. At sometimes t, the ratio of the number of B to that of A is 0.. Then t is given by : T T log log. () t () t () log(.) log. t T (4) t T log(.) log. () T A B t0 N00 t NA = N0 et NB = N0 e t A t NB e 0. t N e t = n. T n. t n e t. (Pg. 6)

17 (7) VIDYALANKAR : JEE-MAIN 07 : Question Paper and Solution. The following, observations were taken for determining surface tension t of water by capillary method : diameter of capillary, D =.5 0 m rise of water, h =.45 0 m Using g = 9.80 m/s r h g and the simplified relation T 0 N/m, the possible error in surface tension is closest to : () 0 % () 0.5% ().5 % (4).4 %. () rgh T = 0 dt dr dg dh T r g h % dt % T An electron beam is accelerated by a potential different V to hit a metallic target to produce Xrays. It produces continuous as well as characteristic Xrays. If min is the smallest possible wavelength of Xray in the spectrum, the variation of log min with log V correctly represented in: () () () (4). () hc KEmax ev min min V log min = log C log V graph is a straight line with slope = intercept = +ve log min log V 4. The moment of inertia of a uniform cylinder of length and radius R about its perpendicular bisector is I. What is the ratio of I/R such that the moment of inertia is minimum? () 4. () M dm dx L dm R di dm x 4 () () (4) (Pg. 7) x d

18 L/ L/ MR M I dx x dx 4L L L/ L/ MR M L L I L 4L L M I R L For I to be minimum (R + L ) should be minimum f = R +L df dr 6R L 0 dl dl dr L () dl R Again volume V = R L = constant R drl + R dl = 0 LdR + RdL = 0 dr R () dl L From () and () L R R L L L R R JEE-MAIN 07 : Question Paper and Solution (8) 5. A slender uniform rod of mass M and length is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes at angle with the vertical is : () g cos () g sin 5. () Torque about origin = I L ML mg sin gsin = L () g sin (4) g cos 6. C p and C are specific heats at constant pressure and constant volume respectively. It is observed that, C C a for hydrogen gas p z C p C = b for nitrogen gas The correct relation between a and b is : () a = 8 b () a b () a = b (4) a = 4 b 4 mg x (Pg. 8)

19 (9) VIDYALANKAR : JEE-MAIN 07 : Question Paper and Solution 6. (4) Q Molar specific heat C = nt Q Specific heat C = mt C' m M (Molecular weight) C n C = C' M ' ' C C R P V R CP CV M for Hydrogen M = R a = CP CV for nitrogen gas M = 8 R b = CP CV 8 a 4 b 7. A copper ball of mass 00 gm is at a temperature T. It is dropped in a copper calorimeter of mass 00 gm, filled with 70 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75C. T is given by : (Given : room temperature = 0C, specific heat of copper = 0. cal/gmc) () 85C () 800C () 885C (4) 50C 7. () Heat released by ball = Heat absorbed by 'Calorimeter + water' system (T 75) = 70 (75 0) (75 0) (T 75) = T 75 = 80 T = 885C 8. In amplitude modulation, sinusoidal carrier frequency used is denoted by c and the signal frequency is denoted by m. The bandwidth ( m ) of the signal is such that m << c. Which of the following frequencies is not contained in the modulated wave? () c m () m () c (4) m + c 8. () 9. The temperature of an open room of volume 0 m increases from 7C to 7C due to the sunshine. The atmospheric pressure in the room remains 0 5 Pa. If n i and n f are the number of molecules in the room before and after heating, then n f n i will be : () ().6 0 ().8 0 (4) () PV 0 0 ni RT PV 0 0 nf RT (Pg. 9)

20 PV nf ni = 4.5 R T T Number of molecules = = JEE-MAIN 07 : Question Paper and Solution (0) 40. In a Young s double slit experiment, slits are separated by 0.5 mm, and the screen is placed 50 cm away. A beam of light consisting of two wavelengths, 650 nm and 50 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is : () 5.6 mm ().56 mm () 7.8 mm (4) 9.75 mm 40. () d = 0.5 mm D = 50 cm = 650 nm and = 50 nm D Fringe width = d = 950 m.95 mm = 560 m.56 mm 0.50 where both waves coincide y = L.C.M. of and y = 7.8 mm y = 4 = 5 = 7.8 mm 4. A particle A of mass m and initial velocity collides with a particle B of mass m which is at rest. The collision is head on, and elastic. The ratio of the debroglie wavelengths A and B after the collision is : A () A () A () A (4) B B B B 4. () m v m/ Rest A before B From conservation of linear momentum m v mv = mv A + v B B v = v A + also, v = v B v A vb v = (add) 4v v vb andva h v A mv A h B m 4v vb A v A B v B (Pg. 0)

21 () VIDYALANKAR : JEE-MAIN 07 : Question Paper and Solution 4. A magnetic needle of magnetic moment Am and moment of inertia kg m is performing simple harmonic oscillations in a magnetic field of 0.0 T. Time taken for 0 complete oscillation is : () 8.76 s () 6.65 s () 8.89 s (4) 6.98 s 4. () Time period is given by I 7.50 T = MB for 0 oscillation = = 6.65 S 6 4. An electric dipole has a fixed dipole moment p, which makes angle with respect to xaxis. When subjected to an electric field E ˆ Ei, it experiences a torque T ˆ k. When subjected to another electric field E Eˆ j it experiences a torque T T. The angle is : () 90 () 0 () 45 (4) (4) T pe pcos ˆi psin ˆj Eiˆ T psin E kˆ In the second situation T p E p cos ˆi psin ˆ j Ej ˆ p cos E kˆ T T as pcos E p sin tan = = In a coil of resistance 00, a current is induced by changing the magnetic flux through it as shown in the figure. The magnitude of change in flux through the coil is : () 75 Wb () 00 Wb () 5 Wb (4) 50 Wb 44. (4) Equation of current in the coil i = 0 t i = 0 t + 0 for the coil d ir dt d = R i dt = R ( 0t + 0) dt 0.5 = R 0 0t 0 dt 00 0t 0t 0.5 = 000 t t 000 (0.5 0) (0.5 0) Magnitude of change in flux = 50 Wb (Pg. )

22 JEE-MAIN 07 : Question Paper and Solution () 45. A time dependent force F = 6t acts on a particle of mass kg. If the particle starts from rest, the work done by the force during the first sec. will be : () 8 J () 4.5 J () J (4) 9 J 45. () a = F 6t 6t m dv 6t dt dv = 6t dt v = t 6 tdt 0 ds t dt ds = t dt w = Fds = i 8 4 6t t dt 8 t dt t Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths, r = /, is given by : 4 () r () r () r (4) r () = hc E hc hc E ( E) E hc hc 4 E E E () / () gives r =.. ().. () In the above circuit the current in each resistance is : () 0 A () A () 0.5 A (4) 0.5 A 47. () Points at two ends of resistors are at the same potential. Current through them = 0. (Pg. )

23 () VIDYALANKAR : JEE-MAIN 07 : Question Paper and Solution 48. A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time? () () () (4) 48. (4) At the highest point direction or sign of velocity changes so only correct answer can be (4). 49. A capacitance of F is required in an electrical circuit across a potential difference of.0 kv. A large number of F capacitors are available which can withstand a potential difference of not more than 00 V. The minimum number of capacitors required to achieve this is : () () () 6 (4) () 50V 50V 50V 50V 8 such rows and 4 capacitors in each row. 000V Capacitance of one row = F 4 Total capacitance = F and number of capacitance = In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be : r () CE () CE (r r) r r () CE (4) CE (r r) (r r ) 50. (4) E In steady state E i r r Er Potential difference across the capacitor = C r r CEr and charge on the capacitor =. r r r r i = 0 i i (Pg. )

24 JEE-MAIN 07 : Question Paper and Solution (4) 5. In a common emitter amplifier circuit using an npn transistor, the phase difference between the input and the output voltages will be : () 80 () 45 () 90 (4) 5 5. () The input and output voltage and 80 apart in phase. 5. Which of the following statements is false? () Kirchhoff s second law represents energy conservation. () Wheatstone bridge is the most sensitive when all the four resistances are of the same order of magnitude. () In a balanced Wheatstone bridge if the cell and the galvanometer are exchanged, the null point is disturbed. (4) A rheostat can be used as a potential divider. 5. () 5. A particle is executing simple harmonic motion with a time period T. At time t = 0, it is at its position of equilibrium. The kinetic energytime graph of the particle will look like : () () () (4) 5. () The correct graph is (). as K.E. and P.E. both oscillate with double the frequency of SHM. 54. An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency 0 GHz. What is the frequency of the microwave measured by the observer? (speed of light = 0 8 ms ) () 5. GHz () 0. GHz (). GHz (4) 7. GHz 54. (4) Doppler's effect in light If observer is moving towards source f f 0, where, = v c f = f 0 f 0 7. GHz 55. A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains same, the stress in the leg will change by a factor of : () () density = constant volume = constant () 9 () 9 (4) 8 (Pg. 4)

25 (5) VIDYALANKAR : JEE-MAIN 07 : Question Paper and Solution Area length = constant da dl 0 A L da dl 9 A L Area decreases by a factor of 9. Stress = Mg A A Stress increase by a factor of When a current of 5 ma is passed through a galvanometer having a coil of resistance 5, it shows full scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range 0 0 V is : () () () (4) () i0 5mA i 0 R R 5 g v i R R 0 0 g v0 0 R = Rg 5 i0 50 R = 0 5 = = 985 G 57. The variation of acceleration due to gravity g with distance d from centre of the earth is best represented by (R = Earth s radius) : () () () (4) 57. () According due to gravity is same as gravitation field due to earth. (I) for d R g GMd E = d R (II) for d R E = GM d d d 58. An external pressure P is applied on a cube at 0C so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by : () PK () P K () P K (4) PK (Pg. 5)

26 JEE-MAIN 07 : Question Paper and Solution (6) 58. () Volume strain due to excess pressure V P.. () V K Increase in volume due to increase in temperature. V T T.. () V from equation () & () P T K Increase in temperature P T K 59. A diverging lens with magnitude of focal length 5 cm is placed at a distance of 5 cm from a converging lens of magnitude of focal length 0 cm. A beam of parallel light falls on the diverging lens. The final image formed is : () real and at a distance of 6 cm from the convergent lens. () real and at a distance of 40 cm from convergent lens. () virtual and at a distance of 40 cm from convergent lens. (4) real and at a distance of 40 cm from the divergent lens. 59. () for converging lens u = (5 + 5) = 40 cm. v u f 5 cm v f u v = + 40 cm 60. A body of mass m = 0 kg is moving in a medium and experiences a frictional force F = Its initial speed is 0 = 0 ms. If, after 0 s, its energy is m 0, the value of k will be : 8 () 0 kg m s () 0 kg m () 0 kg s (4) 0 4 kg m 60. (4) m = 0 kg, fr = kv, v 0 = 0 m/s, after t = 0 sec KE = mv0 8 v0 final velocity = 5 m / s fr = kv Acceleration v 0 / 0 v v m 0 0 dv kv dt m dv k dt 0k v v m 0 0 m 0 k = 0 0 v 00 0 v 0 / 0k v m 4 v0. k (Pg. 6)

27 (7) VIDYALANKAR : JEE-MAIN 07 : Question Paper and Solution PART- C : CHEMISTRY 6. gram of a carbonate (M C ) on treatment with excess HCl produces mole of C. The molar mass of M C in g mol is : () 84. () 8.6 ().86 (4) () M C HCl MClH C mol M M M = 84. g / mol Given, C (graphite) + (g) C (g) ; o r H = 9.5 kj mol H (g) (g) H (l) ; r H o = 85.8 kj mol C (g) + H (l) CH 4 (g) + (g) ; r H o = kj mol Based on the above thermochemical equations, the value of r H o at 98 K for the reaction C H (g) CH (g) will be : (graphite) 4 () kj mol () 74.8 kj mol () 44.0 kj mol (4) kj mol 6. () C (graphite) + (g) C (g) ; H (g) (g) H (l) ; o rh = 9.5 kj mol H o = 85.8 kj mol o C (g) + H (l) CH 4 (g) + (g) ; H = kj mol C H (g) CH (g) H =? (graphite) 4 o H = o o o H H H = 74.8 kj/mol 6. The freezing point of benzene decreases by 0.45C when 0. g of acetic acid is added to 0 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be : (K f for benzene = 5. K kg mol ) () 80.4% () 74.6% () 94.6% (4) 64.6% 6. () CH CH (CH CH) t = 0 0 t = t / i = T f = i K f m 0.45 = 0. / 60 i (5.) 0 0 i = 0.5 = / = i.e % (Pg. 7)

28 JEE-MAIN 07 : Question Paper and Solution (8) 64. The most abundant elements by mass in the body of a healthy human adult are : xygen (6.4%); Carbon (.9%), Hydrogen (0.0%); and Nitrogen (.6%). The weight which a 75 kg person would gain if all H atoms are replaced by H atoms is : () 7.5 kg () 7.5 kg () 0 kg (4) 5 kg 64. () Weight of hydrogen atoms = 0 75 = 7.5 kg 00 If all H atoms are replaced by H i.e. increases in weight by 7.5 kg. 65. U is equal to : () Isobaric work () Adiabatic work () Isothermal work (4) Isochoric work 65. () Using I st law of thermodynamics U = q + w For adiabatic process, q = 0 U = w U is equal to adiabatic work. 66. The formation of which of the following polymers involves hydrolysis reaction? () Bakelite () Nylon 6, 6 () Terylene (4) Nylon (4).. N H NHH H H NH Polymerisation Nylon-6 H C (CH ) 5 NH Cl /Cl Cr /Cr Cr 7 /Cr Mn 4 /Mn 67. Given E.6V, E 0.74V E.V, E.5V. H (Hydrolysis) Among the following, the strongest reducing agent is : () Mn + () Cr + () Cl (4) Cr 67. (4) Strongest reducing agent i.e. least value of reduction potential o Cr /Cr E = 0.74 V 68. The Tyndall effect is observed only when following conditions are satisfied: (a) The diameter of the dispersed particles is much smaller than the wavelength of the light used. (b) The diameter of the dispersed particle is not much smaller than the wavelength of the light used. (c) The refractive indices of the dispersed phase and dispersion medium are almost similar in magnitude. (d) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude. () (b) and (d) () (a) and (c) () (b) and (c) (4) (a) and (d) (Pg. 8)

29 (9) VIDYALANKAR : JEE-MAIN 07 : Question Paper and Solution 68. () Tyndall effect is observed only when the following two conditions are satisfied. (i) The diameter of the dispersed particles is not much smaller than the wavelength of the light used ; and (ii) The refractive indices of the dispersed phase and the dispersion medium differ greatly in magnitude. 69. In the following reactions, Zn is respectively acting as a/an : (a) Zn + Na Na Zn (b) Zn + C ZnC () base and base () acid and acid () acid and base (4) base and acid. 69. () ZnNa Na Zn Acid Base Base Zn C ZnC Acid 70. Which of the following compounds will behave as a reducing sugar in an aqueous KH solution? () () () (4) 70. (4) HCH H H CH H CCH KH HCH H H Hemi-Acetal is reducing sugar CH H H CH CK 7. The major product obtained in the following reaction is : () C 6 H 5 CH = CHC 6 H 5 () (+)C 6 H 5 CH( t Bu)CH C 6 H 5 () ()C 6 H 5 CH( t Bu)CH C 6 H 5 (4) ()C 6 H 5 CH( t Bu)CH C 6 H 5 (Pg. 9)

30 7. () Br Ph H (+) Ph t BuK Ph CH CH Ph JEE-MAIN 07 : Question Paper and Solution (0) 7. Which of the following species is not paramagnetic? () C () () B (4) N 7. () C is diamagnetic, B and N are paramagnetic 7. n treatment of 00 ml of 0. M solution of CoCl.6H with excess AgN ;. 0 ions are precipitated. The complex is : () [Co(H ) Cl ].H () [Co(H ) 6 ]Cl () [Co(H ) 5 Cl]Cl.H (4) [Co(H ) 4 Cl ]Cl.H 7. ().0 No of moles of AgCl formed = = 0.0 mol 60 No of moles of CoCl 6H = = 0.0 mol No of chlorine atoms outside the co-ordination sphere = = i.e. [Co(H ) 5 Cl] Cl H 74. pk a of a weak acid (HA) and pk b of a weak base (BH) are. and.4, respectively. The ph of their salt (AB) solution is : () 6.9 () 7.0 ().0 (4) () ph = pk w pk a pk b = 4..4 = The increasing order of the reactivity of the following halides for the S N reaction is : () (II) < (I) < (III) () (I) < (III) < (II) () (II) < (III) < (I) (4) (III) < (II) < (I) 75. () Rate of S N Reaction Stability of Carbocation II < I < III 76. Both lithium and magnesium display several similar properties due to the diagonal relationship; however, the one which is incorrect, is : () both form soluble bicarbonates () both form nitrides () nitrates of both Li and Mg yield N and on heating (4) both form basic carbonates 76. (4) (Pg. 0)

31 () VIDYALANKAR : JEE-MAIN 07 : Question Paper and Solution 77. The correct sequence of reagents for the following conversion will be : () CH MgBr, H + /CH H, [Ag(NH ) ] + H () CH MgBr, [Ag(NH ) ] + H, H + /CH H () [Ag(NH ) ] + H, CH MgBr, H + /CH H (4) [Ag(NH ) ] + H, H + /CH H, CH MgBr 77. (4) C H H (Ag(NH ) H CH H H + CH MgBr (excess) H C H C H C CH H C CH CH 78. The products obtained when chlorine gas reacts with cold and dilute aqueous NaH are : () Cl and Cl () Cl and Cl () Cl and Cl (4) Cl and Cl 78. () Cl + NaH (Cold and dilute) NaCl + NaCl + H 79. Which of the following compounds will form significant amount of meta product during mononitration reaction? () () () (4) 79. () Aniline gives significant amount of meta product Methylpent--ene on reaction with HBr in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is : () Zero () Two () Four (4) Six 80. () CH CH H C CH C CH CH HBr Peroxide H C CH CH CH CH Br (Pg. ) = 4

32 JEE-MAIN 07 : Question Paper and Solution () 8. Two reactions R and R have identical preexponential factors. Activation energy of R exceeds that of R by 0 kj mol. If k and k are rate constants for reactions R and R respectively at 00 K, then ln(k /k ) is equal to : (R = 8.4 J mol K ) () () 6 () 4 (4) 8 8. () K = K = (E a) = Ae Ae E a /RT E a /RT (E ) 0 kj / mol a k k = Ae Ae E a /RT (Ea E a ) e E a /RT e = e 4 K n K = 4 8. Which of the following molecules is least resonance stabilized? () () () (4) 8. () (), () & (4) are aromatic 8. The group having isoelectronic species is : 8. (4) (), F, Na, Mg + (), F, Na, Mg + (), F, Na +, Mg + (4), F, Na +, Mg +, F, Na + and Mg + are Isoelectronic species contains 0 e. 84. The radius of the second Bohr orbit for hydrogen atom is : (Planck s Const. h = Js; mass of electron = kg; charge of electron e = C; permittivity of vacuum 0 = kg m A ) () 4.76 Å () 0.59 Å (). Å (4).65 Å 84. () r n = 0.59n z Å = (0.59 4) Å = (0.59 4) Å =. Å (Pg. )

33 () VIDYALANKAR : JEE-MAIN 07 : Question Paper and Solution 85. The major product obtained in the following reaction is : () () () (4) 85. (4) H H C H DIBALH 86. Which of the following reactions is an example of a redox reaction? () XeF + PF 5 [XeF] + PF 6 () XeF 6 + H XeF 4 + HF () XeF 6 + H Xe F + 4HF (4) XeF 4 + F XeF (4) XeF 4 + F XeF 6 + Xenon is oxidized from +4 to +6 whereas oxygen is reduced from + to 0. C H 87. A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is a, the closest approach between two atoms in metallic crystal will be: () a () a () a (4) a 87. () For FCC, 4r = a The closest approach between two atoms in metallic crystal = r = a a (Pg. )

34 JEE-MAIN 07 : Question Paper and Solution (4) 88. Sodium salt of an organic acid X produces effervescence with conc. H S 4. X reacts with the acidified aqueous CaCl solution to give a white precipitate which decolourises acidic solution of KMn 4. X is : () HCNa () CH CNa () Na C 4 (4) C 6 H 5 CNa 88. () NaC 4 conc HS4 NaS 4 H C C Na C CaCl CaC +NaCl 4 4 white ppt 89. A water sample has ppm level concentration of following anions F = 0; S4 00; N 50 The anion/anions that make/makes the water sample unsuitable for drinking is/are : () both S4 and N () only F () only S4 (4) only N 89. () 90. Which of the following upon treatment with tert-buna followed by addition of bromine water, fails to decolourize the colour of bromine? () () () (4) 90. (4) Due to absence of Beta-hydrogen in compound-4, no alkene is formed. Alkene and alkyne de-colourises Br -water. (Pg. 4)