IIT - JEE – 2008
FIITJEE Solutions toIIT - JEE – 2008(Paper – 1, Code 7)Time: 3 hoursM. Marks: 246Note: (i) The question paper consists of 3 parts (Part I : Mathematics, Part II : Physics, Part III : Chemistry). Each part has 4sections.(ii) Section I contains 6 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which onlyone is correct.(iii) Section II contains 4 multiple correct answer type questions. Each question has 4 choices (A), (B), (C) and (D), out ofwhich one or more answers are correct.(vi) Section III contains 4 Reasoning type questions. Each question contains STATEMENT–1 and STATEMENT–2.Bubble (A) if both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-1Bubble (B) if both the statements are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT- 1Bubble (C) if STATEMENT-1 is TRUE and STATEMENT-2 is FALSE.Bubble (D) if STATEMENT-1 is FALSE and STATEMENT-2 is TRUE.(iv) Section IV contains 3 sets of Linked Comprehension type questions. Each set consists of a paragraph followed by 3questions. Each question has 4 choices (A), (B), (C) and (D), out of which only one is correct.Marking Scheme:(i) For each question in Section I, you will be awarded 3 Marks if you have darkened only the bubble corresponding to thecorrect answer and zero mark if no bubble is darkened. In all other cases, minus one (– 1) mark will be awarded.(ii) For each question in Section II, you will be awarded 4 Marks if you have darkened all the bubble(s) corresponding tothe correct answer and zero mark for all other cases. It may be noted that there is no negative marking for wronganswer.(iii) For each question in Section III, you will be awarded 3 Marks if you have darkened only the bubble corresponding tothe correct answer and zero mark if no bubble is darkened. In all other cases, minus one (– 1) mark will be awarded.(iv) For each question in Section IV, you will be awarded 4 Marks if you have darkened only the bubble corresponding tothe correct answer and zero mark if no bubble is darkened. In all other cases, minus one (– 1) mark will be awarded.MathematicsPART – ISECTION – IStraight Objective TypeThis section contains 6 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of whichONLY ONE is correct.1.Let a and b be non-zero real numbers. Then, the equation (ax2 by2 c) (x2 – 5xy 6y2) 0 represents(A) four straight lines, when c 0 and a, b are of the same sign(B) two straight lines and a circle, when a b, and c is of sign opposite to that of a(C) two straight lines and a hyperbola, when a and b are of the same sign and c is of sign opposite to that of a(D) a circle and an ellipse, when a and b are of the same sign and c is of sign opposite to that of aFIITJEE Ltd. ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph : 26515949, 26569493, Fax : 26513942
IIT-JEE2008-PAPER-1-2Sol.(B)(ax2 by2 c) (x2 – 5xy 6y2) 0 ax2 by2 c 0 or x2 – 5xy 6y2 0 c x2 y2 iff a b, x – 2y 0 and x – 3y 0 a Hence the given equation represents two straight lines and a circle, when a b and c is of sign opposite to that of a.3 (2 x) ,The total number of local maxima and local minima of the function f (x) 2/3 x ,(A) 0(B) 1(C) 2(D) 32.Sol.(C)Local maximum at x – 1and local minimum at x 0Hence total number of local maxima and localminima is 2. 3 x 1 1 x 2is1–3–2–102(x 1) n; 0 x 2, m and n are integers, m 0, n 0, and let p be the left hand derivative of x – 1 log cos m (x 1)at x 1. If lim g ( x ) p, then3.Let g (x) x 1(A) n 1, m 1(C) n 2, m 2Sol.(B) n 1, m – 1(D) n 2, m n(C)From graph, p – 1 lim g ( x ) 1x 1–x 1 lim g (1 h ) 1x–1h 0 hn lim 1h 0 log cos m h 01 h n 1 n h n 1 n – lim 1 , which holds if n m 2.h 0 m ( tanh ) m h 0 tanh lim1/ 24.If 0 x 1, then(A)2 1 x 2 {x cos ( cot 1 x ) sin ( cot 1 x )} 1 x(B) x1 x2(C) x 1 x 2Sol.is equal to(D)1 x2(C)1/ 221 x 2 ( x cos cot 1 x sin cot 1 x ) 1 1/ 2 2 x1 1 x 2 x coscos 1 sin sin 1 1 22 1 x1 x 2 x 2 1 1 1 x 21 x2 1 x 1 x 2 ( x 2 1 1)1/ 221/ 2 x 1 x2 .FIITJEE Ltd. ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph : 26515949, 26569493, Fax : 26513942
IIT-JEE2008-PAPER-1-35.Consider the two curves C1 : y2 4x, C2 : x2 y2 – 6x 1 0. Then,(A) C1 and C2 touch each other only at one point(B) C1 and C2 touch each other exactly at two points(C) C1 and C2 intersect (but do not touch) at exactly two points(D) C1 and C2 neither intersect nor touch each otherSol.(B)The circle and the parabola touch each other at x 1i.e. at the points (1, 2) and (1, – 2) as shown in thefigure.(1, 2)(3, 0)(1, 0)2 2(1, –2)ˆ cˆ suchˆ b,The edges of a parallelopiped are of unit length and are parallel to non-coplanar unit vectors a,that aˆ bˆ bˆ cˆ cˆ aˆ 1/ 2 . Then the volume of the parallelopiped is6.12(B)3(C)2(D)(A)Sol.12 213(A)Volume aˆ ( bˆ cˆ ) aˆ aˆbˆ aˆcˆ aˆ aˆ bˆ aˆ cˆbˆ bˆ bˆ cˆcˆ bˆ cˆ cˆ1 1/ 2 1/ 211/ 2 1 1/ 2 .21/ 2 1/ 2 1SECTION – IIMultiple Correct Answers TypeThis section contains 4 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D),out of which ONE OR MORE is/are correct. 1 Let f (x) be a non-constant twice differentiable function defined on (– , ) such that f (x) f (1 – x) and f ′ 0. 4 Then 1 (A) f″ (x) vanishes at least twice on [0, 1](B) f ′ 0 2 7.1/ 2(C)Sol.1 f x 2 sin x dx 0 1/ 21/ 2(D) f ( t )esin πt dt 0(A, B, C, D)f (x) f (1 – x)Put x 1/2 x 1 1 f x f x 2 2 Hence f (x 1/2) is an even function or f (x 1/2) sin x is an odd function.Also, f′ (x) – f′ (1 – x) and for x 1/2, we have f ′ (1/ 2 ) 0 .1Also, f (1 t ) e1/ 2sin πt1 f (1 t ) esin πtdt1/ 23/41/41/20dt f ( y ) esin πy dy (obtained by putting, 1 – t y).1/ 2Since f′(1/4) 0, f′ (3/4) 0. Also f′ (1/2) 0 f″ (x) 0 atleast twice in [0, 1] (Rolle’s Theorem)FIITJEE Ltd. ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph : 26515949, 26569493, Fax : 26513942
IIT-JEE2008-PAPER-1-48.A straight line through the vertex P of a triangle PQR intersects the side QR at the point S and the circumcircle of thetriangle PQR at the point T. If S is not the centre of the circumcircle, then112112 (B) (A)PS STPS STQS SRQS SR(C)Sol.114 PS ST QR(D)114 PS ST QR(B, D)PS ST QS SR11 PS ST 1 12PS ST112 PS STQS SRQS SR QS SR2QR QS SR 2 P OQRST12 QS SR QR114 .PS ST QRLet P (x1, y1) and Q (x2, y2), y1 0, y2 0, be the end points of the latus rectum of the ellipse x2 4y2 4. Theequations of parabolas with latus rectum PQ are(B) x 2 2 3 y 3 3(A) x 2 2 3 y 3 39.(C) x 2 2 3 y 3 3Sol.(D) x 2 2 3 y 3 3(B, C)x 2 y2 141b2 a2 (1 e2)3 e 21 1 P 3, and Q 3, (given y1 and y2 less than 0).2 2 Co-ordinates of mid-point of PQ are1 R 0, .2 RQ(x2, y2)P(x2, y2)PQ 2 3 length of latus rectum. 3 1 3 1 two parabola are possible whose vertices are 0, and 0, .2 2 2 2 Hence the equations of the parabolas are x2 2 3y 3 3and x2 2 3y 3 3 .n 1nnTand for n 1, 2, 3, . Then, n2222k 1 n kn kk 0 n kn kn10.Let Sn (A) Sn (C) Tn π3 3π3 3(B) Sn (D) Tn π3 3π3 3FIITJEE Ltd. ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph : 26515949, 26569493, Fax : 26513942
IIT-JEE2008-PAPER-1-5Sol.(A, D)n11Sn lim Sn lim 2n n n1 k / n (k / n)k 11 dx 1 x x2 30Now, Tn π3 3π3n 11nk 00k 1as h f ( kh ) f ( x ) dx h f ( kh )SECTION – IIIReasoning TypeThis section contains 4 reasoning type questions. Each question has 4 choices (A), (B), (C) and (D), out of whichONLY ONE is correct.11.Consider the system of equations ax by 0, cx dy 0, where a, b, c, d {0, 1}.STATEMENT – 1: The probability that the system of equations has a unique solution is 3/8.andSTATEMENT – 2: The probability that the system of equations has a solution is 1.(A) Statement-1 is True, Statement -2 is True; Statement-2 is a correct explanation for Statement-1(B) Statement -1 is True, Statement -2 is True; Statement-2 is NOT a correct explanation for Statement-1(C) Statement -1 is True, Statement -2 is False(D) Statement -1 is False, Statement -2 is TrueSol.(B)For unique solutiona b 0 where a, b, c, d {0, 1}c dTotal cases 16.Favorable cases 6 (Either ad 1, bc 0 or ad 0, bc 1).Probability that system of equations has unique solution is6 3 and system of equations has either unique solution16 8or infinite solutions so that probability for system to have a solution is 1.12.Consider the system of equationsx 2y 3z 1 x y 2z kx 3y 4z 1STATEMENT -1 : The system of equations has no solution for k 3.and1 3 1STATEMENT -2 : The determinant 1 2 k 0 , for k 3.1 4 1(A) Statement-1 is True, Statement -2 is True; Statement-2 is a correct explanation for Statement-1(B) Statement -1 is True, Statement -2 is True; Statement-2 is NOT a correct explanation for Statement-1(C) Statement -1 is True, Statement -2 is False(D) Statement -1 is False, Statement -2 is TrueFIITJEE Ltd. ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph : 26515949, 26569493, Fax : 26513942
IIT-JEE2008-PAPER-1-6Sol.(A)1 2 3D 1 1 2 01 3 4 1 2 3and D1 k 1 2 (3 k) 0 if k 31 3 41 1 3D 2 1 k 2 (k 3) 0 , if k 31 1 41 2 1D3 1 1 k (k 3) 0 , if k 31 3 1 system of equations has no solution for k 3.13.Let f and g be real valued functions defined on interval ( 1, 1) such that g″(x) is continuous, g(0) 0, g′(0) 0, g″(0) 0, and f(x) g(x) sinx.STATEMENT -1 : lim[g(x)cot x g(0)cosecx] f ′′(0) .x 0andSTATEMENT -2 : f′(0) g(0).Sol.14.(A) Statement-1 is True, Statement -2 is True; Statement-2 is a correct explanation for Statement-1(B) Statement -1 is True, Statement -2 is True; Statement-2 is NOT a correct explanation for Statement-1(C) Statement -1 is True, Statement -2 is False(D) Statement -1 is False, Statement -2 is True(B)f′(x) g(x)cosx sinx.g′(x) f′(0) g(0)f″ (x) 2g′ (x) cos x – g (x) sin x sin x g″ (x) f″ (0) 2g′ (0) 0g ( x ) cos x g ( 0 )g′ ( x ) cos x g ( x ) sin x lim g′ ( 0 ) 0 f″ (0).But lim [ g ( x ) cot x g ( 0 ) cosec x ] limx 0x 0x 0sin xcos xConsider three planesP1 : x y z 1P2 : x y z 1P3 : x 3y 3z 2.Let L1, L2, L3 be the lines of intersection of the planes P2 and P3, P3 and P1, and P1 and P2, respectively.STATEMENT -1 : At least two of the lines L1, L2 and L3 are non-parallel.andSTATEMENT -2 : The three planes do not have a common point.Sol.(A) Statement-1 is True, Statement -2 is True; Statement-2 is a correct explanation for Statement-1(B) Statement -1 is True, Statement -2 is True; Statement-2 is NOT a correct explanation for Statement-1(C) Statement -1 is True, Statement -2 is False(D) Statement -1 is False, Statement -2 is True(D)The direction cosines of each of the lines L1, L2, L3 are proportional to (0, 1, 1).FIITJEE Ltd. ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph : 26515949, 26569493, Fax : 26513942
IIT-JEE2008-PAPER-1-7SECTION IVLinked Comprehension TypeThis section contains 3 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be answered. Eachquestion has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.Paragraph for Question Nos. 15 to 17Consider the functions defined implicitly by the equation y3 3y x 0 on various intervals in the real line.If x ( , 2) (2, ), the equation implicitly defines a unique real valued differentiable function y f(x).If x ( 2, 2), the equation implicitly defines a unique real valued differentiable function y g(x) satisfying g(0) 0.15.Sol.()((A)4 27332(B) 4 27332(C)4 2733(D) 4 2733(B)Differentiating the given equation, we get3y2y′ 3y′ 1 01 y′ 10 2 21Differentiation again we get 6yy′2 3y2y″ 3y″ 0()()6.2 24 2 3 2 .4(21)73 f ′′ 10 2 16.The area of the region bounded by the curves y f(x), the x-axis, and the lines x a and x b, where a b 2, isb(A) aabx()dx bf (b) af (a))dx bf (b) af (a)3 ( f (x) ) 12b(C) Sol.)If f 10 2 2 2 , then f ′′ 10 2 a(3 ( f (x) ) 12(D) ax)dx bf (b) af (a))dx bf (b) af (a)3 ( f (x) ) 1bx((B) (2x3 ( f (x) ) 12(A)The required area bbaa f (x)dx xf (x)b xf ′(x)dxab bf (b) af (a) ax3 (f (x) 2 1) dx .117. g′(x)dx 1(A) 2g( 1)(C) 2g(1)Sol.(B) 0(D) 2g(1)(D)We have y′ ( (13 1 f (x)2))which is even1Hence g′(x) g(1) g( 1) 2g(1). 1FIITJEE Ltd. ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph : 26515949, 26569493, Fax : 26513942
IIT-JEE2008-PAPER-1-8Paragraph for Question Nos. 18 to 20A circle C of radius 1 is inscribed in an equilateral triangle PQR. The points of contact of C with the sides PQ, QR, RP 3 3 3 , . Further, itare D, E, F, respectively. The line PQ is given by the equation 3x y 6 0 and the point D is 2 2 is given that the origin and the centre of C are on the same side of the line PQ.18.The equation of circle C is((A) x 2 3((C) x 3Sol.))22( ( y 1) 12( ( y 1) 12(D) x 3)2)22 ( y 1) 1(B) x 2 31 y 12 2(D)Q(3, 3 3 3 , 2 2 ) 3 3 3 , 2 2 EDC60 R60 Fx 3Equation of CD is()3, 1()2 ( y 1) 1 .2Points E and F are given by 3 3 (A) , , 3, 0 2 2 3 1 , ,(B) 2 2 3 3 3 1 , , , (C) 2 2 2 2 3 3 3 1 , (D) , , 2 2 2 2 (Sol.3x y 6 03 33y 2 2 11322Equation of the circle is x 319.)Px 3x C (3, 0)(A)Since the radius of the circle is 1 and CEquation of CE is(3, 1) , coordinates of F ((3, 0)3, 0 )x 3 y 1 113 22 3 3 , . E 2 2 FIITJEE Ltd. ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph : 26515949, 26569493, Fax : 26513942
IIT-JEE2008-PAPER-1-920.Equation of the sides QR, RP are22x 1, y x 1(A) y 33(C) y Sol.(B) y 33x 1, y x 122(D)Equation of QR is y 3 1x, y 03(D) y 3x, y 0(3 x 3) y 3xEquation of RP is y 0.Paragraph for Question Nos. 21 to 23Let A, B, C be three sets of complex numbers as defined belowA {z : Imz 1}B {z : z 2 i 3}{}C z : Re ( (1 i ) z ) 2 .21.The number of elements in the set A B C is(A) 0(C) 2(B) 1(D) Sol.(B)A Set of points on and above the line y 1 in the Argand plane.B Set of points on the circle (x 2)2 (y 1)2 32C Re(1 i) z Re((1 i) (x iy) x y 2Hence (A B C) has only one point of intersection.22.Let z be any point in A B C. Then, z 1 i z 5 i lies between2(A) 25 and 29(C) 35 and 39Sol.23.(B) 30 and 34(D) 40 and 44(C)The points (– 1, 1) and (5, 1) are the extremities of a diameter of the given circle .Hence z 1 – i 2 z 5 – i 2 36.Let z be any point in A B C and let w be any point satisfying w 2 i 3 . Then, z w 3 lies between(A) 6 and 3(C) 6 and 6Sol.2(B) 3 and 6(D) 3 and 9(D)z w z wand z w Distance between z and wz is fixed. Hence distance between z and w would be maximum for diametrically opposite points. z w 6 6 z w 6 3 z w 3 9.FIITJEE Ltd. ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph : 26515949, 26569493, Fax : 26513942
IIT-JEE2008-PAPER-1-10PhysicsPART - IIUseful Data:Plank’s constant h 4.1 10 15 eV.sVelocity of light c 3 108 m/sSECTION – IStraight Objective TypeThis section contains 6 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of whichONLY ONE is correct.24.Two beams of red and violet colours are made to pass separately through a prism (angle of the prism is 60 ). In theposition of minimum deviation, the angle of refraction will be(A) 30 for both the colours(B) greater for the violet colour(C) greater for the red colourSol.(D) equal but not 30 for both the colours(A)At minimum deviation for any wavelengthr1 r2 A/2, Because r1 r2 A25.Which one of the following statements is WRONG in the context of X-rays generated from a X-ray tube?(A) Wavelength of characteristic X-rays decreases when the atomic number of the target increases.(B) Cut-off wavelength of the continuous X-rays depends on the atomic number of the target(C) Intensity of the characteristic X-rays depends on the electrical power given to the X-ray tube(D) Cut-off wavelength of the continuous X-rays depends on the energy of the electrons in the X-ray tubeSol.(B)λcutoff hceV(independent of atomic number)26.An ideal gas is expanding such that PT2 constant. The coefficient of volume expansion of the gas is12(A)(B)TT34(D)(C)TTSol.(C)γ 1 dV V dT PT2 constantnRT 2T constantV3 γ TFIITJEE Ltd. ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph : 26515949, 26569493, Fax : 26513942
IIT-JEE2008-PAPER-1-1127.A spherically symmetric gravitational system of particles has a mass density ρ for r Rρ 0 0 for r Rwhere ρ0 is a constant. A test mass can undergo circular motion under the influence of the gravitational field ofparticles. Its speed V as a function of distance r (0 r ) from the centre of the system is represented by(A) V(B) VR(C)(D)RSol.rVr(C)4v2,G πρ0r 3rrRrr RHence, v r 43 G 3 πρ0 R v 2 ,r2 r 1Hence v r28.RVr RStudents I, II and III perform an experiment for measuring the acceleration due to gravity (g) using a simple pendulum.They use different lengths of the pendulum and /or record time for different number of oscillations. The observationsare shown in the table.Least count for length 0.1 cmLeast count for time 0.1 sStudentIIILength of thependulum (cm)64.064.0Number ofoscillations (n)84III20.04Total time for(n) oscillations (s)128.064.036.0Timeperiod (s)16.016.09.0 g If EI, EII and EIII are the percentage errors in g, i.e., 100 for students I, II and III, respectively,g (A) EI 0(C) EI EIISol.(B) EI is minimum(D EII is maximum(B) A g 4π 2 2 T g A T 2AgT E A t 2 , greater the value of t, lesser the errorAtHence, fractional error in the Ist observation is minimumFIITJEE Ltd. ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph : 26515949, 26569493, Fax : 26513942
IIT-JEE2008-PAPER-1-1229.Figure shows three resistor configurations R1, R2 and R3 connected to 3V battery. If the power dissipated by theconfiguration R1, R2 and R3 is P1, P2 and P3, respectively, 1(A) P1 P2 P3(C) P2 P1 P3Sol.1Ω1Ω1Ω1Ω1Ω1Ω1ΩR2R3(B) P1 P3 P2(D) P3 P2 P1(C)V2RR1 1 Ω, R2 1/2 Ω, R3 2 Ω P2 P1 P3P SECTION – IIMultiple Correct Answers TypeThis section contains 4 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of whichONE OR MORE is/are correct.30.In a Young’s double slit experiment, the separation between the two slits is d and the wavelength of the light is λ. Theintensity of light falling on slit 1 is four times the intensity of light falling on slit 2. Choose the correct choice(s).(A) If d λ, the screen will contain only one maximum(B) If λ d 2λ, at least one more maximum (besides the central maximum) will be observed on the screen(C) If the intensity of light falling on slit 1 is reduced so that it becomes equal to that of slit 2, the intensities of theobserved dark and bright fringes will increase(D) If the intensity of light falling on slit 2 is increased so that it becomes equal to that of slit 1, the intensities of theobserved dark and bright fringes will increaseSol.(A) & (B)For at least one maxima, sin θ λ/dIf λ d, sin θ 1 and y If λ d 2d, sin θ exists and y is finite31.GGThe balls, having linear momenta p1 piˆ and p 2 piˆ , undergo a collision in free space. There is no external forceGGacting on the balls. Let p1' and p '2 be their final momenta. The following option(s) is (are) NOT ALLOWED for anynon-zero value of p, a1, a2, b1, b2, c1 and c2.(A) pG 1' a1ˆi b1ˆj c1kˆGp '2 a 2ˆi b 2ˆj(C) pG 1' a1ˆi b1ˆj c1kˆGp ' a ˆi b ˆj c kˆ2Sol.221(B)(D)Gp1' c1kˆGp '2 c 2 kˆGp1' a1ˆi b1ˆjGp ' a ˆi b ˆj221(A) & (D)G G GG GP P1 P2 P1′ P2′Fext 0GP 0FIITJEE Ltd. ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph : 26515949, 26569493, Fax : 26513942
IIT-JEE2008-PAPER-1-1332.Assume that the nuclear binding energy per nucleon (B/A) versus mass number (A) is as shown in the figure. Use thisplot to choose the correct choice(s) given below.Figure: B/A86420100200A(A) Fusion of two nuclei with mass numbers lying in the range of 1 A 50 will release energy(B) Fusion of two nuclei with mass numbers lying in the range of 51 A 100 will release energy(C) Fission of a nucleus lying in the mass range of 100 A 200 will release energy when broken into two equalfragments(D) Fission of a nucleus lying in the mass range of 200 A 260 will release energy when broken into two equalfragmentsSol.(B) & (D)If (BE)final (BE)initial 0Energy will be released.33.A particle of mass m and charge q, moving with velocity V enters Region II normal to the boundary as shown in thefigure. Region II has a uniform magnetic field B perpendicular to the plane of the paper. The length of the Region II isA. Choose the correct choice(s).Region IFigure:VRegion II Region III AqAB(A) The particle enters Region III only if its velocity V mqAB(B) The particle enters Region III only if its velocity V mqABm(D) Time spent in Region II is same for any velocity V as long as the particle returns to Region I(C) Path length of the particle in Region II is maximum when velocity V Sol.(A), (C) & (D)SECTION – IIIReasoning TypeThis section contains 4 reasoning type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE iscorrect.34.STATEMENT-1The stream of water flowing at high speed from a garden hose pipe tends to spread like a fountain when held verticallyup, but tends to narrow down when held vertically down.andSTATEMENT-2In any steady flow of an incompressible fluid, the volume flow rate of the fluid remains constant.(A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1(B) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation forSTATEMENT-1(C) STATEMENT -1 is True, STATEMENT-2 is False(D) STATEMENT -1 is False, STATEMENT-2 is TrueSol.(A)FIITJEE Ltd. ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph : 26515949, 26569493, Fax : 26513942
IIT-JEE2008-PAPER-1-1435.STATEMENT-1Two cylinders, one hollow (metal) and the other solid (wood) with the same mass and identical dimensions aresimultaneously allowed to roll without slipping down an inclined plane from the same height. The hollow cylinder willreach the bottom of the inclined plane first.andSTATEMENT-2By the principle of conservation of energy, the total kinetic energies of both the cylinders are identical when they reachthe bottom of the incline.(A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1(B) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation forSTATEMENT-1(C) STATEMENT -1 is True, STATEMENT-2 is False(D) STATEMENT -1 is False, STATEMENT-2 is TrueSol.(D)a 36.mgR 2 sin θI cm mR 2STATEMENT-1In a Meter Bridge experiment, null point for an unknown resistance is measured. Now, the unknown resistance is putinside an enclosure maintained at a higher temperature. The null point can be obtained at the same point as before bydecreasing the value of the standard resistance.andSTATEMENT-2Resistance of a metal increases with increase in temperature.(A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1(B) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation forSTATEMENT-1(C) STATEMENT -1 is True, STATEMENT-2 is False(D) STATEMENT -1 is False, STATEMENT-2 is TrueSol.(D)Runknown 37.R (100 A )ASTATEMENT-1An astronaut in an orbiting space station above the Earth experiences weightlessness.andSTATEMENT-2An object moving around the Earth under the influence of Earth’s gravitational force is in a state of ‘free-fall’.(A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1(B) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation forSTATEMENT-1(C) STATEMENT -1 is True, STATEMENT-2 is False(D) STATEMENT -1 is False, STATEMENT-2 is TrueSol.(A)FIITJEE Ltd. ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph : 26515949, 26569493, Fax : 26513942
IIT-JEE2008-PAPER-1-15SECTION – IVLinked Comprehension TypeThis section contains 3 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be answered. Each questionhas 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.Paragraph for Question Nos. 38 to 40In a mixture of H He gas (He is singly ionized He atom), H atoms and He ions are excited to their respective firstexcited states. Subsequently, H atoms transfer their total excitation energy to He ions (by collisions). Assume that theBohr model of atom is exactly valid.38.Sol.39.Sol.40.Sol.The quantum number n of the state finally populated in He ions is(A) 2(B) 3(C) 4(C)(D) 5The wavelength of light emitted in the visible region by He ions after collisions with H atoms is(A) 6.5 10 7 m(B) 5.6 10 7 m(C) 4.8 10 7 m(D) 4.0 10 7 m(C)hc[ λ: visible region]E4 E3 λThe ratio of the kinetic energy of the n 2 electron for the H atom to that of He ion is11(B)(C) 1(A)42(A)KE Z2/n22KE H ZH ZHe 1 2 4KE He 2(D) 2Paragraph for Question Nos. 41 to 435 A small spherical monoatomic ideal gas bubble γ is trapped inside a liquid of density ρA (see figure). Assume3 that the bubble does not exchange any heat with the liquid. The bubble contains n moles of gas. The temperature of thegas when the bubble is at the bottom is T0, the height of the liquid is H and the atmospheric pressure is P0 (Neglectsurface tension).Figure:P0LiquidHy41.Sol.42.As the bubble moves upwards, besides the buoyancy force the following forces are acting on it(A) Only the force of gravity(B) The force due to gravity and the force due to the pressure of the liquid(C) The force due to gravity, the force due to the pressure of the liquid and the force due to viscosity of the liquid(D) The force due to gravity and the force due to viscosity of the liquid(D)Buoyant force is resultant of pressure-force of liquid.When the gas bubble is at a height y from the bottom, its temperature is2/5 P ρA g(H y) (B) T0 0 P0 ρA gH 3/ 5 P ρA g(H y) (D) T0 0 P0 ρA gH P ρA gH (A) T0 0 P0 ρA gy P ρA gH (C) T0 0 P0 ρA gy 2/53/ 5FIITJEE Ltd. ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph : 26515949, 26569493, Fax : 26513942
IIT-JEE2008-PAPER-1-16Sol.(B)P11 γ T1γ P21 γ T2γP1 P0 ρA gH,AT1 T0P2 P0 ρA g(H y)43.The buoyancy force acting on the gas bubble is (Assume R is the universal gas constant)(P ρA gH) 2 / 5ρA nRgT0(A) ρA nRgT0 0(B)(P0 ρA gy) 7 / 5(P0 ρA gH) 2 / 5 [P0 ρA g(H y)]3/ 5(C) ρA nRgT0Sol.(P0 ρA gH)3/ 5(P0 ρA gy)8/ 5(D)ρA nRgT0(P0 ρA gH)3/ 5[P0 ρA g(H y)]2 / 5(B)ρA Vg Buoyancy force ρA g P ρA g ( H y ) T2 T0 0 P ρA gH P2 P0 ρA g(H y)nRT2P225Paragraph for Question Nos. 44 to 46A small block of mass M moves on a frictionless surface of an inclined plane, as shown in figure. The angle of theincline suddenly changes from 60 to 30 at point B. The block is initially at rest at A. Assume that collisions betweenthe block and the incline are totally inelastic (g 10 m/s2).MFigure:Av60 B30 3m44.Sol.C3 3mThe speed of the block at point B immediately after it strikes the second incline is(A) 60 m/s(B) 45 m/s(C) 30 m/sA(B)Along the plane velocity just before collision3mv 2g ( 3) 60 m/s60 BAlong the plane velocity just after collisionvB v cos 30 45 m/sv3mSol.46.Sol.15 m/s30 30 3m45.(D)C3 3mThe speed of the block at point C, immediately before it leaves the second incline is(A) 120 m/s(B) 105 m/s(C) 90 m/s(B)1mg (3) m ( v C2 v 2B ) vC 105 m/s2(D)75 m/sIf collision between the block and the incline is completely elastic, then the vertical (upward) component of thevelocity of the block at point B, immediately after it strikes the second incline is(A) 30 m/s(B) 15 m/s(C) 0(D) 15 m/sv sin 30 (C)30 vy v sin 30 cos 30 v cos 30 cos 60 v cos 30 v cos 30 30 vy 060 v sin 30 30 Before collision30 After collisionFIITJEE Ltd. ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph : 26515949, 26569493, Fax : 26513942
IIT-JEE2008-PAPER-1-17ChemistryPART – IIISECTION – IStraight Objective TypeThis section contains 6 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of whichONLY ONE is correct.47.Sol.Native silver metal forms a water soluble complex with a dilute aqueous solution of NaCN in the presence of(A) nitrogen(B) oxygen(C) carbon dioxide(D) argon(B)Ag dissociates in a solution of NaCN in the presence of air, and forms sodium argentocyanide.4Ag 8NaCN 2H 2 O O 2 4Na Ag ( CN )2 4NaOH48.2.5 mL of22M HCl in water at 25 C. TheM weak monoacidic base ( K b 1 10 12 at 250 C ) is titrated with155(concentration of H at equivalence point is K w 1 10 14 at 250 C(A) 3.7 10 13 M(B) 3.2 10 7 M 2Sol.)(D) 2.7 10 2 M(C) 3.2 10 M(D)BOH HCl BCl H 2 OCB C (1 h )ZZX BOH H 2 O YZZCh H Ch25 7.5 mlVolume of HCl used 2 /1522.5 5 0.1 MConcen
IIT-JEE2008-PAPER-1-3 FIITJEE Ltd. ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 110016, Ph : 26515949, 26569493, Fax : 26513942 5. Consider the two curves C1: y 2 4x, C 2: x 2 y2 – 6x 1 0. Then, (A) C1 and C2 touch each other only at one point (B) C1 and C2 touch each other exactly at two points (C) C1
Atomic Structure for JEE Main and Advanced (IIT-JEE) m Jupiter (XI) 3 Prepared By: Er. Vineet Loomba (IIT Roorkee) (iv) Some were even scattered in the opposite direction at an angle of 180 [Rutherford was very much surprised by it and remarked that “It was as incredible as if you fired a 15 inch shell at a piece of
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Main Report and Tables A 6 JEE 2011 – Results The Joint Entrance Examination for the year 2011 (JEE 2011) was conducted on Sunday, April 10 . by post to respective zonal JEE offices from where the admit card was issued. The last date for receiving required documents at respective zonal IITs was announced in the Counselling Brochure as 10 .
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