EXAMINATION PAPER-IIT-JEE 2009
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1 EXAMINATION PAPER-IIT-JEE 9 (QUESTION & SOLUTIONS) PAPER II Part I (CHEMISTRY) SECTION I Straight Objective Type //9 This section contains multiple choice questions. Each question has choices (A), (B), (C) and (D), out of which ONLY ONE is correct.. In the following carbocation, H/CH that is most likely to migrate to the positively charged carbon is - H H + 5 H C C C C CH HO H CH (A) CH at C- (B) H at C- (C) CH at C- (D) H at C- [Ans.D] Group will migrate from C because of formation of more stable carbocation. Migrating nature of H is more than CH.. The spin only magnetic moment value (in Bohr magneton units) of Cr(CO) is - (A) (B).8 (C).9 (D) 5.9 [Ans. A] Cr d 5 s d t g eg CO is a strong field ligand. CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com
2 . The correct stability order of the following resonance structures is - + H C=N=N H C N=N + + H C N N + H C N=N (I) (II) (III) (IV) (A) (I) > (II) > (IV) > (III) (C) (II) > (I) > (III) > (IV) (B) (I) > (III) > (II) > (IV) (D) (III) > (I) > (IV) > (II) [Ans. B] Explanation : () The species with incomplete octet are less stable. Therefore II & IV are less stable than I & III. () In II & IV comparison is done on the basis of E.N of atom carrying (+) charge. () I is more stable than III because ve charge is on more E.N. atoms i.e., N.. For a first order reaction A P, the temperature (T) dependent rate constant(k) was found to follow the equation logk = () T +.. The pre-exponential factor A and the activation energy Ea,respectively, are - (A). s and 9. kj mol (B). s and. kj mol (C). s and. kj mol (D). s and 8. kj mol [Ans. D] k = Ae Ea/RT or, lnk = lna Ea/RT or, log k = log A Ea.R T Given, log k = () T + log A = A = sec. E a =.R E a =. 8. = 8. kj mol. CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com
3 SECTION II Multiple Correct Answers Type This section contains 5 multiple correct answer(s) type questions. Each question has choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct. 5. The nitrogen oxide(s) that contain(s) N-N bond(s) is (are) - (A) N O (B) N O (C) N O (D) N O 5 A N O N N O B N O O O N N O asymmetrical form O O C N O N N O O O O D N O 5 N O N O O O N N = O O symmetrical form [Ans. A,B,C]. The correct statement(s) about the following sugars X and Y is (are) - H HO CH OH O H OH H H OH H HOH C O O H HO OH H H CH OH H HO CH OH O H OH H H OH O H H CH OH O H H HO OH H H OH X (A) X is a reducing sugar and Y is a non-reducing sugar (B) X is a non-raducing sugar and Y is a reducing sugar (C) The glucosidic linkages in X and Y are α and β, respectively (D) The glucosidic linkages in X and Y are β and α, respectively Y [Ans. B,C] CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com
4 In 'X' there is no free C = O group and α-linkages are present while in 'Y' free C = O and β-linkage is present. 7. In the reaction X + B H [BH (X) ] + [BH ] the amine(s) X is (are)- (A) NH (B) CH NH (C) (CH ) NH (D) (CH ) N [Ans.A] B H + NH excess NH B H. NH low temperature ionic compound can be represented as [BH (NH ) ] + [BH ] 8. Among the following, the state function(s) is (are)- (A) Internal energy (B) Irreversible expansion work (C) Reversible expansion work (D) Molar enthalpy State functions are independent of path [Ans.A,D] 9. For the reduction of given below - V + (aq) + e V Fe + (aq) + e Fe Au + (aq) + e Au NO ion in an aqueous solution, E is +.9V. Values of E for some metal ions are E =.9V E =.V E = +.V Hg + (aq) + e Hg E = +.8V The pair(s) of metals that is (are) oxidized by (A) V and Hg (C) Fe and Au (B) Hg and Fe (D) Fe and V NO in aqueous solution is (are) [Ans.A,B,D] V + NO V + + NO E cell =.9 (.9) =.5 V spontaneous Fe + NO Fe + + NO E cell =.9 (.) = V spontaneous cell Hg + NO Hg + + NO E =.9 (.8) =. V spontaneous CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com
5 SECTION III Matrix - Match Type This section contains questions. Each question contains statements given in two columns, which have to be matched. The statements in Column I are labeled A, B, C and D, while the statements in Column II are labeled p, q, r, s and t. any given statement in Column I can have correct matching with ONE OR MORE statements (s) in column II. The appropriate bubbled corresponding to the answers to these questions have to be darkened as illustrated in the following example : If the correct matches are A p, s and t; B q and r; C p and q; and D s and t; then the correct darkening of bubbles will look like the following. A B C D p q r s t q r p s t r s t p q r. Match each of reactions given in Column I with the corresponding product(s) given in ColumnII. Column I Column II (A ) Cu + dil HNO (p) NO (B) Cu + conc HNO (q) NO (C) Zn + dil HNO (r) N O (D) Zn + conc HNO (s) Cu(NO ) (t) Zn(NO ) [Ans. (A) p, s ; (B) q, s ; (C) p, t ; (D) q, t] Cu + 8HNO NO + Cu (NO ) + H O Dilute Cu + HNO NO + Cu (NO ) + H O Conc. Zn + HNO N O + Zn(NO ) + 5H O Dilute Zn + HNO NO + Zn (NO ) + H O Conc. CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com 5
6 . Match each of reactions given in Column I with the corresponding product(s) given in ColumnII. Column I (A) Br (p) Nucleopilic substitution Column II O (B) OH (q) Elimination (C) CHO (r) Nucleophilic addition (D) OH Br (s) Esterification with acetic anhydride NO (t) Dehydrogenation [Ans. (A) p, q; (B) p, s, t; (C) r,s; (D) p] SECTION IV Integer Answer Type This section contains 8 questions. The answer to each of the questions is a single-digit integer, ranging from to 9. The appropriate bubbles below the respective question numbers in the ORS have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are,, 9 and, respectively, then the correct darkening of bubbles will look like the following : X Y Z W CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com
7 . The coordination number of Al in the crystalline state of AlCl is [Ans. ] Cl Cl Al Cl Cl Al Cl Cl. The oxidation number of Mn in the product of alkaline oxidative fusion of MnO is Fuse in MnO + KNO K MnO +NO [Ans. ]. The total number of α and β particles emitted in the nuclear reaction 9U 8 Pb is [Ans. 8] α & β emission occurs. Six alpha emission decreases mass number by and atomic number by. β emission increases mass number by The dissociation constant of a substituted benzoic acid at 5ºC is.. The ph of a.m solution of its sodium salt is [Ans. 8] ph = 7 + [pka + log C] = 7 + [ + log ] = 8. The number of water molecule(s) directly bonded to the metal centre in CuSO. 5H O is [Ans. ] [Cu (H O) ] SO. H O 7. At K, the root mean square (rms) speed of a gas X (molecular weight = ) is equal to the most probable speed of gas Y at K. The molecular weight of the gas Y is - [Ans. ] C rms x = C m y R R = M or = M or M = CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com 7
8 8. The total number of cyclic structural as well as stereo isomers possible for a compound with the molecular formula C 5 H is- [Ans. 7] 5 structural isomers ( stereo isomers are possible, these are given below ) Me Me Me Me Enantiomers Stereo isomers = Me Me 9. In a constant volume calorimeter,.5 g of a gas with molecular weight 8 was burnt in excess oxygen at 98. K. The temperature of the calorimeter was found to increase from 98. K to 98.5 K due to the combustion process. Given that the heat capacity of the calorimeter is.5 kj K, the numerical value for the enthalpy of combustion of the gas in kj mol is- [Ans. 9] Q = C T =.5.5 =.5 kj U = = 9 kj H = U + nr T = kJ. CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com 8
9 Part II (MATHEMATICS) SECTION I Single Correct Choice Type This section contains multiple choice questions. Each question has choices (A), (B), (C) and (D), for its answer, out of which ONLY ONE is correct.. The normal at a point P on the ellipse x + y = meets the x- axis at Q. If M is the mid point of the line segment PQ, then the locus of M intersects the latus rectums of the given ellipse at the points : (A) ± 5, ± 7 (B) ± (C) ±, ± (D) ± 7 [C] x y + = 5 P ( cos θ, sin θ) x y x + y = y =. y y (sin θ) M N = = = tan θ (M N = slope of normal) x cosθ y sin θ = tan θ (x cos θ) Q ( cos θ cos θ, ) Q ( cos θ, ) M (h, k) (7/ cos θ, sin θ) x + y = 9, ± 9, ± 7 x = ae x = ae Q b = a ( e ) e = x =, x = (equation of latus rectum). + y = y = 9 9 y = ± 7 CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com 9
10 . A line with positive direction cosines passes through the point P(,, ) and makes equal angles with the coordinate axes. The line meets the plane x + y + z = 9 at point Q. The length of the line segment PQ equals : (A) (B) (C) (D) [C] x y + z = = = λ P (,, ) Q ( + λ, + λ, + λ) ( + λ) + (λ ) + (λ + ) = 9 + λ + λ + λ + = 9 λ = λ = Q (,, ) PQ + + =. The locus of the orthocentre of the triangle formed by the lines ( + p)x py + p( + p) = ( + q)x qy + q ( + q) =, and y =, where p q, is : (A) a hyperbola (B) a parabola (C) an ellipse (D) a straight line [D] Intersection points of given lines are ( p, ), ( q, ), [pq, (p +) (q + )] respectively. A(pq, (p +) (q + ) E B ( p, ) D C ( q, ) Now equation of altitudes AD and BE are x = pq, and qx + (q + )y + pq = their point of intersection is (pq, pq) so, h = pq, k = pq so locus is h = k h + k = x + y = which is a straight line. CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com
11 . If the sum of first n terms of an A.P. is cn, then the sum of squares of these n terms is : n(n )c n(n + )c (A) (B) n(n )c (C) [C] T n = S n S n = cn c(n ) = cn cn + cn c = cn c T n = c (n ) = c (n n +) T n = c [ n n + n] = c n(n + )(n + ) n(n + ) + n = = nc [8n + n + n + ] nc [8n ] + n(n )c (D) = nc (n ) SECTION II Multiple Correct Choice Type This section contains 5 Multiple choice questions. Each question has choices (A), (B), (C) and (D), for its answer, out of which ONE OR MORE is/are correct.. The tangent PT and the normal PN to the parabola y = ax at a point P on it meet its axis at points T and N, respectively. The locus of the centroid of the triangle PTN is a parabola whose : a (A) vertex is, (C) latus rectum is a (B) directrix is x = (D) focus is (a, ) CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com
12 [A, D] P(at, at) T( at, ) N(a + at, ) ty = x + at y = tx + at + at h = at at + a + at, k = at + + h = a ( + t ), t = k a h = a 9k + a ah = 8a + 9k 9y = ax 8a y a = x a a vertex, directrix x = a a Latus rectum, Focus (a, ) 5. For < θ <, the solution (s) of m= (m ) m cos ec θ + cosec θ + = is (are) : (A) (B) (C) 5 (D) CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com
13 [C, D] sin( / ) sin( / ) m (m ) sin θ + θ + = (m ) m sin θ + sin θ + m (m ) m cot θ + cot θ + m= sin / cot θ cot θ + + cot θ + cot θ cot θ + cot θ + = cot θ cot θ + = sin / cot θ + tan θ = sin θ + cos θ = sin θcosθ sin θ = 5 θ =,. For the function f(x) = x cos x, x, (A) for at least one x in the interval [, ), f(x + ) f(x) < (B) lim f '(x) = x (C) for all x in the interval [, ), f(x + ) f(x) > (D) f '(x) is strictly decreasing in the interval [, ) [B,C, D] f(x) = x cos x f (x) = cos + x (B) lim f (x) = x = + = x sin (/ x) x lim cos (/x) + lim x sin (/ x) x CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com
14 cos(/ x) (D) f (x) = x Q for x [, ) f (x) is negative. so f (x) is decreasing for x [, ) using LMVT f (x + ) f (x) = f (x) x + x = cos + sin x x x Q f (x) < so f (x) is decreasing f (x) > f (x) > lim f (x) x f (x + ) f (x) so > x + x f(x + ) f(x) > f(x + ) f(x) > 7. An ellipse intersects the hyperbola x y = orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then : (A) Equation of ellipse is x + y = (B) The foci of ellipse are (±, ) (C) Equation of ellipse is x + y = (D) The foci of ellipse are (±, ) [A, B] Since both conic cuts orthogonally so foci coincides Now foci of hyperbola is (±, ) foci of ellipse (±, ) eccentricity of rectangular hyperbola is eccentricity of ellipse = equation of ellipse is x + y = CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com
15 8. If I n = then : ( + sin nx x )sin x dx, n =,,,, (A) I n = I n+ (B) I m = = (C) m= I n = I n = m I m = (D) I n = I n + sin n x x ( + )sin x Adding I n = or I n = I n+ = Consider I n+ I n = x sin nx dx x ( + )sin x sin nx dx sin x sin (nx) dx sin x sin(n + )x dx sin x sin (n + )x sin nx dx sin x cos (n + )x sin x = dx = sin x I n+ = I n () again I m+ = I + I 5 + I I m= using () = I = I = I m m= = because + I m = sin mx dx = use f(a x) = f(x) sin x Also I n+ I n as by trial I I CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com 5
16 SECTION III Matrix - Match Type This section contains questions. Each question contains statements given in two columns which have to be matched. The statements in Column I are labelled A, B, C and D, while the statements in Column II are labelled p,q, r, s and t. Any given statement in Column I can have correct matching with ONE OR MORE statement (s) in Column II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following example: If the correct matches are A -p, s and t; B -q and r; C -p and q; and D s and t; then the correct darkening of bubbles will look line the following. A B C D p q r s t p q r s t p p q q r r s s t t p q r s t 9. Match the statements/expressions given in Column I with the values given in Column II. Column I Column II (A) Root(s) of the equation sin θ + sin θ = (B) Points of discontinuity of the function x x f(x) = cos where[y] denotes the largest integer less than or equal to y (C) Volume of the parallelepiped with its edges represented by the vectors î + ĵ, î + ĵ and î + ĵ+ kˆ (D) Angle between vectors a and b where a, b and c are unit vectors satisfying, (p) (q) (r) (s) a + b + c = (t) A q, s; B p, r, s, t; C t; D r CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com
17 (A) sin θ + sin θ = sin θ + sin θ cos θ = sin θ + sin θ ( sin θ) = sin θ sin θ + = ( sin θ ) (sin θ ) = sin θ = or sin θ = θ =, (B) x f(x) = cos x at x = x x = [] = & = = at x = x x = [] = & = [] = So point of discontinuity x =,,, (C) Volume of parallelepiped = = (D) c r = a r b r c r = ( a r b r ).( a r b r ) = a r + b r + a r. b r cos θ cos θ = θ = CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com 7
18 . Match the statements/expressions given in Column I with the values given in Column II. Column I Column II (A) The number of solutions of the (p) equation xe sinx cos x = in the interval, (B) Value (s) of k for which the planes (q) kx + y + z =, x + ky + z = and x + y + z = intersect in a straight line (C) Value(s) of k for which (r) x + x + x + + x + = k has integer solution (s) (D) If y' = y + and y() =, then value (s) of y(ln ) (s) (t) 5 A p; B q, s; C q, r, s,t; D r (A) f(x) = xe sinx cos x f (x) = e sinx + xe sinx cos x + sin x > in (, /) f(x) increasing in x (, /) y = x f() = < f(/) = e > solution (B) k k = k (k ) ( ) + (8 k) = k (k ) (k ) = (k ) (k ) = k =, CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com 8
19 (C) If x x = k x = k (k =,,, 5) If x < x + x + x + = k x + = k; x = k = (Not possible) x < x + x + x + + x + = k = k (Not possible) x < x + x x + x + = k x + = k (Not possible) x < x + x x x = k x = k x = k k = (D) dy dy = y + dx = y + dx + c ln (y + ) = x + c y() = ln = c ln (y + ) = x + ln y + = e x y (ln ) = e ln = SECTION IV Integer Answer Type This section contains 8 questions. The answer to each of the questions is a single-digit integer, ranging from to 9. The appropriate bubbles below the respective question numbers in the ORS have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are,, 9 and, respectively, then the correct darkening of bubbles will look like the following : X Y Z W CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com 9
20 . The smallest value of k, for which both the roots of the equation x 8kx + (k k + ) = are real, distinct and have values at least, is : [] D > k (k k + ) > k k + k > k > B > k > k > A f() k + k k + k 8 k + k k + (k ) (k ) k, k so k [, ) so smallest integer value of k is. x. Let f: R R be a continuous function which satisfies f(x) = [] f (t)dt. Then the value of f(ln 5) is : f (x) = f(x) f (x) = f (x) f (x) dx f (x) = dx ln [f(x)] = x + c f(x) = e x+c f(x) = Ae x as f() = (Given) So A = f(x) = f(ln 5) = CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com
21 . Let p(x) be a polynomial of degree having extremum at x =, and of p() is : [] Let P(x) = ax + bx + cx + dx + e Q x lim P(x) + = x x lim p(x) + =. Then the value x so lim P(x) = x x x lim d = e = & c = so P(x) = ax + bx + x Q P (x) = ax + bx + x Q P () = P () = so a + b + = & a + b + = + On solving a = & b = P() = a + 8b + c = ax d e + bx + c + + x = x. Let ABC and ABC' be two non-congruent triangles with sides AB =, AC = AC' = and angle B = º. The absolute value of the difference between the areas of these triangles is : [] A B θ C D C AD = sin = difference of areas = ACC = sin θ θ = 5 side CC = ( cos 5 ) = ACC =.. = CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com
22 5. The centres of two circles C and C each of unit radius are at a distance of units from each other. Let P be the mid point of the line segment joining the centres of C and C and C be a circle touching circles C and C externally. If a common tangent to C and C passing through P is also a common tangent to C and C, then the radius of the circle C is : [8] Let the coordinate system is as follows C (, k) B C (, ) (, ) A (, ) equation of AB is y = m(x ) Q AB is tangent to C so m = ± But m should be positive m = So equation of AB = = x Q C & C are touching each other externally. So CC = r + r 9 + k = (r + ) () Q AB is tangent to circle C so k + r r = k = () 8 + So solving () and () r = 8 C. Let (x, y, z) be points with integer coordinates satisfying the system of homogeneous equations : x y z =, x + z =, x + y + z =. Then the number of such points for which x + y + z is : [7] Q x y z =, x z =, x y z = On solving these three y = z = x so x + y + z x + + 9x x x =,,, so total no. of different points possible are 7 (,, ), (,, ), (,, ), (,, ), (,, ), (,, ), (,, ) CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com
23 x 7. If the function f(x) = x + e and g(x) = f (x), then the value of g'() is : [] g(x) = f (x) f(g(x)) = x g(f(x)) = x g (f(x)) f (x) = g (f(x)) = f (x) put x = g () = f () f (x) = x + e x/ f () = g () = 8. The maximum value of the function f(x) = x 5x + x 8 on the set A = {x x + 9x} is : [7] f(x) = x 5x + x 8 f (x) = (x ) (x ) Q A = {x x + 9x } x 5 so f(x) is increasing for x [, ) so (f (x)) max at x [, 5] is f(5) so (f(x)) max = f(5) = 7 CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com
24 Part III (PHYSICS) SECTION I Single Correct Choice Type This section contains multiple choice questions. Each question has choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct. 9. Photoelectric effect experiments are performed using three different metal plates p, q and r having work functions φ p =. ev, φ q =.5 ev and φ r =. ev, respectively. A light beam containing wavelengths of 55 nm, 5 nm and 5 nm with equal intensities illuminates each of the plates. The correct I-V graph for the experiment is : [Take hc = ev nm] (A) (C) I V p q r r q p E 55 nm = =.5 ev 55 I V (B) (D) p r q q r p I I V V Ans.[A] E 5 nm = =.75 ev 5 E 5 = =.5 ev 5 ev s = E φ For metal p, V s =.5 V s =.5 V For metal q, V s =.5.5 =. V For metal r, V s =.5 =.5 V hence option 'A' is correct. Also I p > I q > I r CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com
25 . A piece of wire is bent in the shape of a parabola y = kx (y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x- axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is a (A) gk (B) a gk dy a = g tanθ where tan θ = = kx dx a = kx. g a x = kg a (C) gk (D) a gk Ans.[B]. A uniform rod of length L and mass M is pivoted at the centre. Its two ends are attached to two springs of equal spring constants k. The springs are fixed to rigid supports as shown in the figure, and the rod is free to oscillate in the horizontal plane. The rod is gently pushed through a small angle θ in one direction and released. The frequency of oscillation is (A) k M θ L k θ (B) k M (C) k M (D) k M Ans.[C] O L k θ Take τ torque about 'O' L L ML. k θ. =. α k α =. θ M f = k M CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com 5
26 . The mass M shown in the figure oscillates in simple harmonic motion with amplitude A. The amplitude of the point P is k k P M (A) ka k (B) ka k (C) ka k + k (D) ka k + k Ans.[D] x + x = A and k x = k x x + kx k = A x = k A (k + k ) SECTION II Multiple Correct Choice Type This section contains 5 multiple choice questions. Each question has choices (A), (B), (C) and (D) for its answer, out of which ONE OR MORE is/are correct.. Under the influence of the Coulomb field of charge +Q, a charge q is moving around it in an elliptical orbital. Find out the correct statement(s). (A) The angular momentum of the charge q is constant (B) The linear momentum of the charge q is constant (C) The angular velocity of the charge q is constant (D) The linear speed of the charge q is constant Ans.[A] Torque of coulombic force on ' q' about 'Q' is zero. * Angular momentum of without specifying a point about which it is to be calculated, do not have significance. CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com
27 . A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure, A is the point of contact, B is the centre of the sphere and C is its topmost point. Then C B A (A) (B) V r C V r B = V r B V r A r r r (C) = (D) V V = V A V r C V r r r A = (VB VC ) r r r r VC VA VB VC V = B V = V (Let) C V = V C A B Ans.[B, C] 5. The figure shows the P V plot of an ideal gas taken through a cycle ABCDA. The part ABC is a semi-circle and CDA is half of an ellipse. Then P A D B C V (A) the process during the path A B is isothermal (B) heat flows out of the gas during the path B C D (C) work done during the path A B C is zero (D) positive work is done by the gas in the cycle ABCDA (B) Q = U + W In BCD : W is negative Pf Vf Pi Vi U = ( γ ) = ve (D) Cycle is clockwise. Ans.[B, D] CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com 7
28 . A student performed the experiment to measure the speed of sound in air using resonance air-column method. Two resonances in the air-column were obtained by lowering the water level. The resonance with the shorter air-column is the first resonance and that with the longer air-column is the second resonance. Then - (A) the intensity of the sound heard at the first resonance was more than that at the second resonance (B) the prongs of the tuning fork were kept in a horizontal plane above the resonance tube (C) the amplitude of vibration of the ends of the prongs is typically around cm (D) the length of the air-column at the first resonance was somewhat shorter than / th of the wavelength of the sound in air Ans.[A,D] 7. Two metallic rings A and B, identical in shape and size but having different resistivities ρ A and ρ B, are kept on top of two identical solenoids as shown in the figure. When current I is switched on in both the solenoids in identical manner, the rings A and B jump to heights h A and h B, respectively, with h A > h B. The possible relation(s) between their resistivities and their masses m A and m B is (are) A B (A) ρ A > ρ B and m A = m B (C) ρ A > ρ B and m A > m B (B) ρ A < ρ B and m A = m B (D) ρ A < ρ B and m A < m B Ans.[B,D] B = µ ni φ = µ ni A where A = Area of ring µ ni A e = t CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com 8
29 induced current = I = µ nia (area) t ρ L Force due to induced current in ring is For short duration is t Impulse = F t Impulse induced current m v ρ v ρm so maximum height depend on v A h max ρ m SECTION III Matrix-Match Type This section contains questions. Each question contains statements given in two columns, which have to be matched. The statements in Column I are labelled A, B, C and D, while the statements in Column II are labelled p, q, r, s and t. Any given statement in Column I can have correct matching with ONE OR MORE statement(s) in Column II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following example. If the correct matches are A p, s and t; B q and r; C p and q; and D s and t; then the correct darkening of bubbles will look like the following. A B C D p q r s t p q r s t p p q q r r s s t t p q r s t CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com 9
30 8. Column II gives certain systems undergoing a process. Column I suggests changes in some of the parameters related to the system. Match the statements in Column I to the appropriate process(es) from Column II. Column I Column II (A) The energy of the system is increased (p) System : A capacitor, initially uncharged Process : It is connected to a battery (B) Mechanical energy is provided to the (q) System : A gas in an adiabatic system, which is converted into energy of random motion of its parts container fitted with an adiabatic piston Process : The gas is compressed by pushing the piston (C) Internal energy of the system is (r) System : A gas in a rigid container converted into its mechanical energy Process : The gas gets cooled due to colder atmosphere surrounding it (D) Mass of the system is decreased (s) System : A heavy nucleus, initially at rest Process : The nucleus fissions into two fragments of nearly equal masses and some neutrons are emitted (t) System : A resistive wire loop Process : The loop is placed in a time varying magnetic field perpendicular to its plane A p,q,s,t ; B q ; C s, D s CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com
31 9. Column I shows four situations of standard Young's double slit arrangement with the screen placed far away from the slits S and S. In each of these cases S P = S P, S P S P = λ/ and S P S P = λ/, where λ is the wavelength of the light used. In the cases B, C and D, a transparent sheet of refractive index µ and thickness t is pasted on slit S. The thicknesses of the sheets are different in different cases. The phase difference between the light waves reaching a point P on the screen from the two slits is denoted by δ(p) and the intensity by I(P). Match each situation given in Column I with the statement(s) in Column II valid for that situation. Column I Column II (A) S P P P (p) δ(p ) = S (B) (µ )t = λ/ S P P P (q) δ(p ) = S (C) (µ )t = λ/ S P P P (r) I(P ) = S (D) (µ )t = λ/ S P P P (s) I(P ) > I(P ) S (t) I(P ) > I(P ) A p,s ; B q ; C t ; D r, s t CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com
32 (A) At P δ(p ) = I (P ) = I At P λ δ(p ) = I(P ) = I I(P ) > I(P ) At P λ δ(p ) = φ = λ λ = (B) δ(p ) = λ I(P ) = I cos = I I(P ) > I(P ) I(P ) = I δ(p ) = I(P ) = I δ(p ) = λ λ = λ φ = λ λ = I(P ) = I ( + cos ) = I ( + ) = I + I I(P ) > I(P ) CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com
33 (C) δ(p ) = λ I(P ) = λ λ λ δ(p ) = = I(P ) = I λ λ λ δ(p ) = = λ φ = = λ I(P ) = I cos = I λ (D) δ(p ) = I(P ) = I λ λ δ(p ) = I(P ) = λ λ δ(p ) = = λ I(P ) I(P ) > I(P ) 9λ λ = 5λ = SECTION IV Integer Answer Type This section contains 8 questions. The answer to each of the questions is a single-digit integer, ranging from to 9. The appropriate bubbles below the respective question numbers in the ORS have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are,, 9 and, respectively, then the correct darkening of bubbles will look like the following : X Y Z W CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com
34 5. A solid sphere of radius R has a charge Q distributed in its volume with a charge density ρ = κr a, where κ and a are constants and r is the distance from its centre. If the electric field at r = R is 8 times that at r = R, find the value of a. Ans. [] R/ for r R R / ρ Q in = r dr R / = k r R / a+ dr a = + k r dr k (a + ) + a R / = [ r ] Q in = k R (a + ) a+ k R R (a + ) E r = = R For r R k a+ Q in = R (a + ) k a R (a + ) E(r = R) = (R) + a+ CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com
35 R E r = = E(r = R) 8 k R (a + ) R a+ = 8 k a+ R (a + ) (R) a = a+ = 5. A steady current I goes through a wire loop PQR having shape of a right angle triangle with PQ = x, PR = x and QR = 5x. If the magnitude of the magnetic field at P due to this loop is Q µ I k, find the value of k. 8x Ans. [7] x 5º M 5x 7º P x R PM sin7º = x x PM = x = 5 5 B P = µ I [ sin 5º + sin 7º] x 5 µ = I 5 + (x) 5 5 = µ I 5 8x k = µ I = 7 8x CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com 5
36 5. Three objects A, B and C are kept in a straight line on a frictionless horizontal surface. These have masses m, m and m, respectively. The object A moves towards B with a speed 9 m/s and makes an elastic collision with it. Thereafter, B makes completely inelastic collision with C. All motions occur on the same straight line. Find the final speed (in m/s) of the object C. 9m/s m m m A B C v v Collision between A and B, m 9 = mv + mv 9 = v + v...() e = = ( v v ) 9 v v = 9...() from () and () v = m/s m/s m m m A B C Ans. [] m B m C m = m v v = m/s 5. Two soap bubbles A and B are kept in a closed chamber where the air is maintained at pressure 8 N/m. The radii of bubbles A and B are cm and cm, respectively. Surface tension of the soap-water used to make bubbles is. N/m. Find the ratio n B /n A, where n A and n B are the number of moles of air in bubbles A and B, respectively. [Neglect the effect of gravity.] Ans. [] CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com
37 cm A T P A = P atm + r A P A = N/m T P B = P atm + r B P A = N/m For ideal gas PV = nrt P A V A = n A RT PA VA n A = RT PB VB n B = RT n n B A = PBV P V A B A = cm B. = = 8 +. (.) (.) = () = 5. A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses. kg and.7 kg. Taking g = m/s, find the work done (in joules) by the string on the block of mass. kg during the first second after the system is released from rest. Ans. [8] CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com 7
38 a T. kg T.7 kg a a = = (m m)g (m + m ) (.7.)g (.7 +.) a = g S = at g = () T =.(g + a) = g =.(g + g ). g = =. T =.8 N Work done by tension =.8 W = 8 J 55. A cylindrical vessel of height 5 mm has an orifice (small hole) at its bottom. The orifice is initially closed and water is filled in it up to height H. Now the top is completely sealed with a cap and the orifice at the bottom is opened. Some water comes from the orifice and the water level in the vessel becomes steady with height of water column being mm. Find the fall in height (in mm) of water level due to opening of the orifice. [Take atmospheric pressure =. 5 N/m, density of water = kg/m and g = m/s. Neglect any effect of surface tension.] Ans. [] CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com 8
39 P P=P atm 5 mm H mm Final pressure inside two cylinder, P f = P atm ρgh P f = 5. P f = 5. 5 P f =.98 5 By Boyle's Law for gas P i V i = P f V f 5 (5 H) =.98 5 (5 H) = 9 H = 5 9 H = mm Fall in height = = mm P atm 5. A cm long string, having a mass of. g, is fixed at both the ends. The tension in the string is.5 N. The string is set into vibrations using an external vibrator of frequency Hz. Find the separation (in cm) between the successive nodes on the string. Ans. [5] x= x=cm T =.5 N T.5. v = = =. = m/s m v = λf v λ = = =. f Separation between successive node λ = = = 5 cm CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com 9
40 57. A metal rod AB of length x has its one end A in ice at C and the other end B in water at C. If a point P on the rod is maintained at, then it is found that equal amounts of water and ice evaporate and melt per unit time. The latent heat of evaporation of water is 5 cal/g and latent heat of melting of ice is 8 cal/g. If the point P is at a distance of λx from the ice end A, find the value of λ. [Neglect any heat loss to the surrounding.] Ans. [9] ºC ºC ºC A λx B ice water x For ice k A m 8 = λx For water...() m 5 = k A (x λx)...() eq.()/eq.() 8 ( λ) = 5 λ λ = 7 ( λ) λ = 9 9λ λ = 9 λ = 9 CORPORATE OFFICE, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: 7-59 Website : info@careerpointgroup.com
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