PPER -II PRT I: CHEMISTRY SECTION- I Single Correct Choice Type This section contains 4 multiple choice questions. Each question has 4 choices (), (B), (C) and (D) for its answer, 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 C C C OH H H 4 C CH 5 CH () CH at C-4 (B) H at C-4 (C) CH at C- (D) H at C- Sol: ns [D] The carbocation is also resonance stabilised. C H H C C OH H H C CH CH H C C CH H O CH CH CH H C C CH CH CH O CH H. For a first order reaction P, the temperature (T) dependent rate constant (k) was found to follow the equation log k (000) 6.0. The pre-eponential factor and the activation energy E T a, respectively are ().0 0 6 s and 9. kj mol (B) 6.0 s and 6.6 kj mol (C).0 0 6 s and 6.6 kj mol (D).0 0 6 s and 8. kj mol Sol: ns [D] Ea / RT k e, Ea log k log.0r T Ea log k log, log = 6, = 0.0R T 6 s Ea 000.0R Ea = 000.0 8.4 0 = 8. kj/mole. The spin only magnetic moment value (in Bohr magneton units) of Cr(CO) 6 is () 0 (B).84 (C) 4.90 (D) 5.9 Sol: ns [] 0 There is no unpaired electron in Cr(CO) 6 mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4 - -
IIT-JEE-009 4. The correct stability order of the following resonance structures is H C N N H CN N H CN N H CN N (I) (II) (III) (IV) () (I) (II) (IV) (III) (C) (II) (I) (III) (IV) Sol: ns [B] (B) (I) (III) (II) (IV) (D) (III) (I) (IV) (II) SECTION- II Multiple Correct Choice Type This section contains 5 multiple choice questions. Each question has 4 choices (), (B), (C) and (D) for its answer, out of which ONE OR MORE is/are correct. 5. For the reduction of NO ion in an aqueous solution, E 0 is 0.96V. Values of E 0 for some metal ions are given below V (aq) e V Fe (aq) e Fe E 0 =.9 V E 0 = 0.04 V u (aq) e u E 0 =.40 V Hg (aq) e g E 0 = 0.86 V The pair(s) of metals that is (are) oidized by NO in aqueous solution is (are) () V and Hg (B) Hg and Fe (C) Fe and u (D) Fe and V Sol: ns [, B, D] Because Fe and vanadium has more negative standard reduction potential, therefore stronger reducing agents. 6. In the reaction. X B H 6 [BH (X) ] [BH 4] the amine(s) X is(are) () NH (B) CH NH (C) (CH ) NH (D) (CH ) N Sol: ns [, B, C, D] ll substances have lone pair of electrons hence act as Lewis base. 7. mong the following, the state function(s) is(are) () Internal energy (C) Reversible epansion work (B) Irreversible epansion work (D) Molar enthalpy Sol: ns [, D] 8. The nitrogen oide(s) that contains(s) N N bond(s) is(are) () N O (B) N O (C) N O 4 (D) N O 5 Sol: ns [, B, C] O N N O - N N O O O O N N O O mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4 - -
IIT-JEE-009 9. The correct statement(s) about the following sugars X and Y is(are) CH OH H O H H H OH HO H OH O HOH C O H OH CH OH H O CH OH H OH H H OH HO H H OH H O H CH OH O H OH H OH H H OH X () X is a reducing sugar and Y is a non-reducing sugar (B) X is a non-reducing sugar and Y is a reducing sugar (C) The glucosidic linkages in X and Y are and, respectively (D) The glucosidic linkages in X and Y are and, respectively Y Sol: ns [B, C] SECTION- III Matri - 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, B, C and D, while the statements in Column II are labelled p, q, r, s and t. ny 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 eample: If the correct matches are 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. p q r s t p q r s t B C D p q r s t p q r s t p q r s t 0. Match each of the reactions given in Column I with the corresponding product(s) given in Column II Column I Column II () 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 ) Sol: ns [ p, s]; [B q, s]; [C r, t]; [D q, t] (t) Zn(NO ) mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4 - -
IIT-JEE-009. Match each of the compounds given in Column I with the reaction(s), that they can undergo, given in Column II. Column I Column II Br () O (p) Nucleophilic substitution (B) OH (q) Elimination CHO (C) OH Br (r) Nucleophilic addition (D) NO (s) Esterification with acetic anhydride Sol: ns [ p, q, t]; [B p, s, t]; [C r, s]; [D p] (t) SECTION- IV Integer nswer Type Dehydrogenation This section contains 8 questions. The answer to each of the questions is a single digit integer, ranging from 0 to 9. The appropriate bubbles below the respective question numbers in the ORS have to be darkened. For eample, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and, respectively, then the correct darkening of bubbles will look like the following. X Y Z W 0 0 0 0 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9 mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4-4 -
IIT-JEE-009. The dissociation constant of a substituted benzoic acid at 5 C is.0 0 4. The ph of a 0.0M solution of its sodium salt is Sol: ns [8] ph [pk w pka log C] [4 4 ] 8. t 400K, the root mean square (rms) speed of a gas X (molecular weight = 40) is equal to the most probable speed of gas Y at 60 K. The molecular weight of the gas Y is Sol: ns [4] RT M RT M 400 60 = M = 4 40 M 4. The total number of and particles emitted in the nuclear reaction 4 8 9 9 U Pb is Sol: ns [8] 8 4 4 0 9 U 8 Pb 6 He particles = 6 = = 8 5. The oidation number of Mn in the product of alkaline oidative fusion of MnO is Sol: ns [6] MnO K MnO alkaline oidation 4 Oidation number = 6 6. The coordination number of l in the crystalline state of lcl is Sol: ns [4] The coordination number is 4 7. The number of water molecule(s) directly bonded to the metal atom in CuSO 4.5H O are Sol: ns [4] 8. In a constant volume calorimeter,.5g of a gas with molecular weight 8 was burnt in ecess oygen at 98.0 K. The temperature of the calorimeter was found to increase from 98.0K to 98.45K 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 Sol: ns [9] C v =.6 kj K C p C v = R C p = C v R = (5000 8.4) JK mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4-5 -
IIT-JEE-009 H = C p T 508.4 8 0.45.5 9000Jmole = 9.0 kj mole 9. The total number of cyclic structural as well as stereo isomers possible for a compound with the molecular formula C 5 H 0 are Sol: ns [7] CH CH CH H C CH CH 4 5,-dimethyl cyclo propane also show cis-trans stereoisomerism trans isomer also show optical isomer Total = 7 PRT II : MTHEMTICS SECTION- I Single Correct Choice Type This section contains 8 multiple choice questions. Each question has 4 choice (), (B), (C) and (D), for its answer, out of which ONLY-ONE is correct 0. The normal at a point P on the ellipse 4y = 6 meets the -ais 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 () 5, 7 (B) 5 9, 4 (C), 7 (D), 4 7 Sol: ns [C] Let P(4 cos, sin ) then equation of normal is 4 y Point Q is (cos, 0) cos sin 7 4 Mid point is cos, sin, locus of mid point is y 49 Putting ae 4 we get. The points are,. 7. If the sum of first n terms of an.p. is cn, then the sum of squares of these n terms is. () n(4n ) c 6 (B) n(4n ) c (C) n(4n ) c (D) n(4n ) c 6 mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4-6 -
IIT-JEE-009 Sol: ns [C] Given that n [ a ( n ) d] cn a (n )d = cn a = d and d = c a = c Then required sum = a a 5 a n terms = a [ 5 ] = n(4n ) c... line with positive direction cosines passes through the point P(,, ) and makes equal angles with the coordinate aes. The line meets the plane y z = 9 at point Q. The length of the line segment PQ equals () (B) (C) (D) Sol: ns [C] The direction cosines of the line are,,. Let point Q be,, then 9 4 4 Q is (, 0, ) PQ ( ) (0 ) ( ). The locus of the orthocentre of the triangle formed by the lines ( p) py p( p) = 0, ( q) qy q( q) = 0, and y = 0, where p q, is () a hyperbola (B) a parabola (C) an ellipse (D) a straight line Sol: ns [D] mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4-7 -
IIT-JEE-009 The verte of the are ( p, 0), ( q, 0) and (pq, ( p)( q). Equation of altitude BD is q y 0 ( p) q Putting = pq we get q ( ) ( ) pq y pq p q pq q q B ( p, 0) ( pq, ( p)( q)) O D C ( q, 0) y = 0 SECTION- II Multiple Correct Choice Type This section contains 5 multiple choice questions. Each question has 4 choice (), (B), (C) and (D) for its answer, out of which ONE OR MORE is/are correct. 4. If sin n In d, n 0,,,, then ( )sin () I n = I n (B) 0 m I m 0 (C) 0 Im 0 (D) I n = I n m Sol: ns [, B, C] (sin( n ) sin n) I I d ( )sin n n cos( n ) sin d ( )sin cos( n ) d (say) d I ( ) I cos( n ) d ( ) (i) b b cos( n ) ( ) ( ) d f d f a b d ( ) a a mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4-8 -
IIT-JEE-009 cos( n ) d ( ) (ii) dding (i) and (ii), cos( n ) I ( ) d ( ) cos( n ) d 0 I = 0 I n = I n So I m = I and I m = I 0 0 0 I I 0I m m m I sin d ( ) sin d (iii) b b d g( ) d f ( a b ) d a a d (iv) (iii) and (iv) gives ( ) d I So I = 0 0 I I 0I 0 (s I 0) m 0 0 0 m m s I 0 I, I n I n. mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4-9 -
IIT-JEE-009 5. For the function f ( ) cos,, () for at least one in the interval [, ), f ( ) f () < (B) lim f ( ) (C) for all in the interval [, ), f ( ) f () > (D) f () is strictly decreasing in the interval [, ) Sol: ns [B, C, D] f() = cos f ( ) cos sin 0 s 0, So, f() is increasing lim f ( ) lim cos sin f ( ) sin sin cos = cos 0 So, f () is strictly decreasing in [, ) So, f ( ) < f () Let g() = f( ) f() g() = f ( ) f () < 0 g() is decreasing So, min value of g() occurs at =. lim g( ) lim f ( ) f ( ) lim ( ) cos cos = lim cos cos cos = lim sin sin cos mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4-0 -
IIT-JEE-009 = lim sin sin cos ( ) ( = ( ) lim cos ( ) ( = ( ) lim cos ( ) So, min. value of f( ) f() is f( ) f() > in [, ) 6. n ellipse intersects the hyperbola y = orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the aes of the ellipse are along the coordinate aes, then () Equation of ellipse is y = (B) The foci of ellipse are (±, 0) (C) Equation of ellipse is y = 4 (D) The foci of ellipse are (, 0) Sol: ns [, B] Eccentricity of the ellipse =. So a b a. So equation of ellipse is y = a Solving it with hyperbola, a a, y 6 Ellipse is y = a 4y dy d = 0 dy (Considering point of intersection as ( d y, y )) Hyperbola is y dy y 0 d dy d y Since the curves are orthogonal,. y y = y mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4 - -
IIT-JEE-009 a a 6 a =. So required ellipse is y =. Its foci are (±, 0). 7. The tangent PT and the normal PN to the parabola y = 4a at a point P on it meet its ais at points T and N, respectively. The locus of the centroid of the triangle PTN is a parabola whose () verte is a, 0 (B) directri is = 0 (C) latus rectum is a (D) focus is (a, 0) Sol: ns [, D] Let P (at, at) Equation of PT is ty = at T ( at, 0) Equation of PN is y = t at at N (a at, 0) Centroid of PTN is a at at, Let centroid be (h, k) h k t a a at, k at 9k h a 4a So, locus is 9y = 4a( a) i.e. 4a a y mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4 - -
IIT-JEE-009 Its verte is a,0 Focus is (a, 0) Directri is a Latus rectum is 4 a. 8. For 0, the solution(s) of 6 ( m ) m cosec cosec 4 is(are) m 4 4 () 4 (B) 6 (C) (D) 5 Sol: ns [C, D] 6 m cosec ( m ) cosec = m 4 4 6 m sin / 4 m sin ( m ) sin 4 4 = m sin ( ) 6 m 4 4 m m = sin ( m ) sin 4 4 6 m cot ( m ) cot m 4 4 = 5 cot cot cot cot... cot cot 4 4 4 cot cot cot tan 4 = tan cot = 4 tan 4tan = 0 tan 5,. mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4 - -
IIT-JEE-009 SECTION- III Matri 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, B, C and D, while the statements in Column II are labelled p, q, r, s and t. ny 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 eample: If the correct matches are 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. p q r s t p q r s t B C D p q r s t p q r s t p q r s t 9. Match the statements/epressions given in Column I with the values given in Column II. Column I () Root(s) of the equation sin sin = 6 (B) Points of discontinuity of the function f ( ) cos, where [y] denotes the largest integer less than or equal to y (p) (q) Column II 6 4 (C) Volume of the parallelopiped with its edges represented by the vectors iˆ ˆj, iˆ ˆj and iˆ ˆj kˆ (D) ngle between vectors a and b where a, b and c are (s) unit vectors satisfying a b c 0 (t) Sol: [( q, s), (B p, r, s, t), (C t), (D r)] () sin sin = sin 4 sin ( sin ) = sin 4 sin 4 sin 4 = 4 sin 4 6 sin = 0 Let sin = y y y = 0 (r) mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4-4 -
IIT-JEE-009 y y y = 0 y(y ) (y ) = 0 (y )(y ) = 0 y = sin = sin = ± or or or, 4 So (q, s) y sin sin (B) 6 f ( ) cos Let t f () = [t] cos[t] The points to be checked for discontinuity are t,,,, 4 nd clearly f () = [t] cos [t] is discontinuous at t,,,,,, 6 B (p, r, s, t). (C) Value of parallelopited [ i j iˆ ˆj iˆ ˆj kˆ ] 0 0 = ( ) = ns. C t. (D) a b c 0 a b c a b c a b c ( a b c ) mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4-5 -
IIT-JEE-009 a b cos = cos = cos D r. 0. Match the statements/epressions in Column I with the values given in Column II. Column I () The number of solutions of the equation e sin cos = 0 (p) in the interval 0, (B) Value(s) of k for which the planes k 4y z = 0, (q) 4 ky z = 0 and y z = 0 intersect in a straight line (r) (C) Value(s) of k for which = 4k (s) 4 has integer solution(s) (t) 5 (D) If y = y and y(0) = then value(s) of y(ln ) Sol: ns [ p, B q, s, C q, r, s, t, D r] () Let f() = e sin cos f () = e sin e sin cos sin > 0 f() is increasing. Maimum one intersection with ais. (P) Column II (B) Planes k 4y z = 0, 4 ky z = 0 and y z = 0 intersect in a straight line. The normal vector of all three planes are coplanar. k 4 4 k 0 k =, 4 B (, 5) (C) Given, = 4k Let, y = and y = 4k Minimum value of y is 6 and graph goes to. So for y = y, k can take values,, 4, 5 c q, r, s, t mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4-6 -
IIT-JEE-009 dy (D) y d I.F. = e sin d = e So, solution ye e d c ye e c Now, y(0) = = c c = So, y e = e y = e y(ln ) = e ln = 4 = D (r). SECTION- IV Integer nswer Type This section contains 8 questions. The answer to each of the questions is a single digit integer, ranging from 0 to 9. The appropriate bubbles below the respective question numbers in the ORS have to be darkened. For eample, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and, respectively, then the correct darkening of bubbles will look like the following. X Y Z W 0 0 0 0 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9. Let BC and BC be two non-congruent triangles with sides B = 4, C = C = and angle B = 0. The absolute value of the difference between the areas of these triangles is Sol: Sol: ns [4] mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4-7 -
IIT-JEE-009 a c b cos B ac a cos0 6 ( ) a 4 4 a 8 0 4 6 a a (C C ) ( ) ( ) 4 Difference of rea 4 4 sin 0 4.. Let (, y, z) be points with integer coordinates satisfying the system of homogeneous equations : y z = 0 z = 0 y z = 0 Then the number of such points for which y z 00 is Sol: ns [7] It can be written as 0 0 y 0 z 0 X = 0 But = 0 So, system has infinite number of solutions. Let = k z = k and y = 0 Now, y z 00 k 0 9k 00 0k 00 0 k 0 So, there are 7 value of k. mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4-8 -
IIT-JEE-009 p( ). Let p() be a polynomial of degree 4 having etremum at =, and lim. 0 Then the value of p() is Sol: ns [0] Let p() = a 4 b c d e So, p () = 4a b c d Now, p() has etremum at =, So, p () = 0 and p () = 0 4a b c d = 0 and a b 4c d = 0... (i) p( ) Now, lim 0 d e lim a b c 0 lim d e c 0 lim c d e 0 Now, it must be of the form (%) so e = 0 c d lim 0 c d lim 0 gain, it must be of the form (%) so d = 0 c lim 0 c = c = So, c =, d = 0, e = 0 Now, from (i) 4a b = 0 a b 4 = 0 mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4-9 -
IIT-JEE-009 So, a, b 4 p( ) 4 4 p() 6 8 4 4 8 4 0 4 4. Let f : R R be a continuous function which satisfies f ( ) f ( t) dt. 0 Then the value of f (ln 5) is Sol: ns [0] f ( ) f ( t) dt 0 f () = f() df ( ) d f ( ) df ( ) f ( ) d ln f() = c f() = e c f(0) = 0 f() = e (Let e c = ) f(0) = e 0 = 0 f() = 0 f(ln 5) = 0 5. The smallest value of k, for which both the roots of the equation 8k 6 (k k ) = 0 are real, distinct and have values at least 4, is Sol: ns [] 8k 6(k k ) = 0 For real and distinct D > 0 64k 64(k k ) > 0 k > 0 k > mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4-0 -
IIT-JEE-009 Now, Put, k = 8k k 6 6 7 5, So, k =. 6. The centres of two circles C and C each of unit radius are at a distance of 6 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 eternally. 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 Sol: ns [9] sin 4 C C P sin r r r r 9. 7. The maimum value of the function f() = 5 6 48 on the set = { 0 9} is Sol: ns [7] = { 0 9} 0 9 0 ( 5)( 4) 0 [4, 5] f () = 5 6 48 f () = 6 0 6 = 0 f () = 6( )( ) For [4, 5] f () > 0 f () is increasing function. mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4 - -
IIT-JEE-009 t = 5 f () will be maimum. f (5) = 50 75 80 48 = 40 47 = 7 f () ma = 7. 8. If the function f ( ) e and g() = f (), then the value of g () is Sol: ns [] fog () = d fog( ) f ( g( )) g( ) d g( ) g( ) ( g( )) e g() g () ( g()) e Since f (0) = f () = g() = 0 g(). 0 e PRT III : PHYSICS SECTION- I Single Correct Choice Type This section contains 8 multiple choice questions. Each question has 4 choices (), (B), (C) and (D) for its answer, out of which ONLY-ONE is correct 9. Photoelectric effect eperiments are performed using three different metal plates p, q and r having work functions p. 0 ev, q. 5 ev and r. 0 ev, respectively. light beam containing wavelengths of 550 nm, 450 nm and 50 nm with equal intensities illuminates each of the plates. The correct I-V graph for the eperiment is [Take hc = 40 ev nm] () I p q r (B) p q r I V V (C) I r q p V (D) r q p I V mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4 - -
IIT-JEE-009 Sol: ns [] ll wavelengths are below threshould wavelength for metal P, hence its intensity will be maimum. 40. The mass M shown in the figure oscillates in simple harmonic motion with amplitude. The amplitude of the point P is () k k (B) k k k P k M (C) k k k (D) k k k Sol: ns [D] and k = k = k k = k = k k k k 4. piece of wire is bent in the shape of a parabola y = k (y-ais 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 -ais 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-ais is () a gk (B) a gk (C) a gk (D) a 4gk Sol: ns [B] F.B.D. of bead relative to wire is as shown N sin = ma N y N cos = mg ma tan = a mg g a also tan = dy d = k k = a g = a kg ma mg N mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4 - -
IIT-JEE-009 4. 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 fied 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 in one direction and released. The frequency of oscillation is () k M (B) k M (C) 6k M (D) 4k M Sol: ns [C] torque about O L L is 0 = k. 0 = k L O 0 = ML = 6k M f = 6k M SECTION- II MULTIPLE CORRECT CHOICE TYPE This section contains 5 multiple choice questions. Each question has 4 choices (), (B), (C) and (D), for its answer out of which ONE OR MORE is/are correct. 4. Two metallic rings and B, identical in shape and size but having different resistivities and 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 and B jump to heights h and h B, respectively, with h > h B. The possible relation (s) between their resistivities and their masses m and m B is (are) B () (C) B and m = m B (B) B B and m > m B (D) B and m = m B and m < m B mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4-4 -
IIT-JEE-009 Sol: ns [B, D] Induced current i Impulse = mv i mv mv = constant...(i) where v is initial velocity of rings if m = m B v = v B B for h > h B, < B lso if < B and m < m B v > v B (from (i) h > h B. 44. The figure shows the P-V plot of an ideal gas taken through a cycle BCD. The part BC is a semi circle and CD is half of an ellipse. Then, P () the process during the path B is isothermal (B) heat flows out of the gas during the path B C D D B (C) work done during the path B C is zero C (D) positive work is done by the gas in the cycle BCD 0 V Sol: ns [B, D] B is not isothermal in C D, W = ve U is negative Q is negative i.e. heat flows out of gas. W B C is not zero. net work done is positive 45. student performed the eperiment 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, () 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 mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4-5 -
IIT-JEE-009 (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 /4th of the wavelength of the sound in air. Sol: ns [, D] 46. Under the influence of the Coulomb field of charge Q, a charge q is moving around it in an elliptical orbit. Find out the correct statement (s) () 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 Sol: ns [] s electrostatic field due to charge Q is central hence angular momentum is constant. Due to elliptical orbit linear speed will change and linear momentum will also change. 47. sphere is rolling without slipping on a fied horizontal plane surface. In the figure is the point of contact, B is the centre of the sphere and C is its topmost point. Then C B V 0 () V V V V C (B) B C VC VB VB V (C) V V V V (D) V V 4 V Sol: ns [B, C] C B C C B V C V = V 0 V C V B = V 0 V B V = V 0 V C V B = V B V V C V = V B V C V C V = (V B V ) as V = 0 C B V 0 V C V = V B mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4-6 -
IIT-JEE-009 SECTION- III MTRIX-MTCH 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, B, C and D, while the statements in Column II are labelled p, q, r, s and t. ny 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 eample: If the correct matches are 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. p q r s t p q r s t B C D p q r s t p q r s t p q r s t 48. Column II gives certain systems undergoing a process. Column I suggests changes in some of the parameters related to the system. Match the systems in Column I to the appropriate process (es) from Column II. Column-I Column-II () The energy of the system is increased (p) System: capacitor, initially uncharged. (B) Mechanical energy is provided to the system, (q) which is converted into energy of random motion of its parts (C) Internal energy of the system is converted into (r) its mechanical energy Process: It is connected to a battery System: gas in an adiabatic container fitted with an adiabatic piston Process: The gas is compressed by pushing the piston System: gas in a rigid container Process: The gas gets cooled due to colder atmosphere surrounding it. (D) Mass of the system is decreased (s) System: heavy nucleus, initially at rest Sol: ns [] (p, q, s, t); [B] (q); [C] (s); [D] (s) (t) Process: The nucleus fissions into two fragments of nearly equal masses and some neutrons are emitted System: resistive wire loop. Process: The loop is placed in a time varying magnetic field perpendicular to its plane. mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4-7 -
IIT-JEE-009 49. 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 0 = S P 0, S P S P =/4 and S P S P =/, where is the wavelength of the light used. In the cases B, C and D, a transparent sheet of refractive inde 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 (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 () S P P P 0 (p) ( P) 0 S (B) ( ) t / 4 S P P P 0 (q) ( P ) 0 S (C) ( ) t / S P P P 0 (r) I ( P ) 0 S (D) ( ) t / 4 S P P P 0 (s) I P ) I ( ) ( 0 P S (t) I P ) I ( ) ( P Sol: ns [] (p, s); [B] (q); [C] (t); [D] (r, s, t) mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4-8 -
IIT-JEE-009 SECTION- IV Integer nswer Type This section contains 8 questions. The answer to each of the questions is a single digit integer, ranging from 0 to 9. The appropriate bubbles below the respective question numbers in the ORS have to be darkened. For eample, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and, respectively, then the correct darkening of bubbles will look like the following. X Y Z W 0 0 0 0 4 4 4 4 5 5 5 5 6 6 6 6 7 7 7 7 8 8 8 8 9 9 9 9 50. Two soap bubbles and B are kept in a closed chamber where the air is maintained at pressure 8 N/m. The radii of bubbles and B are cm and 4 cm, respectively. Surface tension of the soap water used to make bubbles is 0.04 N/m. Find the ratio n B /n, where n and n B are the number of moles of air in bubbles and B, respectively. [Neglect the effect of gravity] Sol: r = 0.0 cm R = 0.04 cm T = 0.04 N/m P = P 0 4T r = 8 4 0.04 0.0 = 6 r P 0 = 8 N/m R P B = P 0 4T R = 8 4 0.04 0.04 = mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4-9 -
IIT-JEE-009 n n P V P P B B B 4 6 r 4 R = 6 r R 6 nb n = 6 ns.: 6 5. cylindrical vessel of height 500 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 out from the orifice and the water level in the vessel becomes steady with height of water column being 00 mm. Find the fall in height ( in mm) of water level due to opening of the orifice. [Take atmospheric pressure =.0 0 5 N/m, density of water = 000 kg/m and g = 0 m/s. Neglect any effect of surface tension] Sol: In container, final air pressure, P f = P a hg = 0 5 0. 0 0 = 0.98 0 5 N/m initial air pressure = 0 5 N/m using P i V i = P f V f 0 5 h = 0.98 0 5 00 h = 94 mm fall in height of liquid = 500 (94 00) = 6 mm ns.: 6 5. Three object, B and C are kept in a straight line on a frictionless horizontal surface. These have masses m, m and m, respectively. The object moves towards B with a speed 9 m/s and makes an elastic collision with it. Thereafter, B makes completely inelastic collision with C. ll motions occur on the same straight line. Find the final speed (in m/s) of the object C. m m m B C Sol: From COM between and B 9 ms 9 m = m. v B mv B C 9 = v B v m m m =0 mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4-0 -
IIT-JEE-009 lso, 9 =.(v B v ) v B = 6 from COM between B & C m 6 = m v C v C = 4 m/s ns: 4 5. steady current I goes through a wire loop PQR having shape of a right angle triangle with PQ =, PR = 4 and QR = 5. If the magnitude of the magnetic field at P due to this loop is 0 k I 48, find the value of k. Sol: Due to PQ and RP, B at point P is zero. R l 4 5 sin 5 4 l 90 l Q P 5 oi BP (sin sin(90 ) 4l oi 4 5 5 4 5 o I 7 o I 7 4( ) 48 ns: 7 a 54. solid sphere of radius R has a charge Q distributed in its volume with a charge density r, where and a are constants and r is the distance from its centre. If the electric field at times that at r = R, find the value of a. R r is 8 Q Sol: t r =R, E = 4 o R where Q is total charge dq = r a. 4 dr = 4 r. k. r dr r a q 4k r dr n 0 a r 4k a 4 k a a r R E R 4k q 4 r 4 ( R / ) a ( r R /) 0 o a (ccordint o Gauss Law) mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4 - -
IIT-JEE-009 Q E (r = R) = 4 R 0 given; 4 k R a 4k ( R ) a.. a 8 4 R 4 ( R / ) 0 0 a a = 8 a = ns : 55. light inetensible string that goes over a smooth fied pulley as shown in the figure connects two blocks of masses 0.6 kg and 0.7 kg. Taking g = 0 m/s, find the work done (in joules) by the string on the block of mass 0.6 kg during the first second after the system is released from rest. Sol: Let m = 0.6 mg mg g a m S a g 6 0 6 lso T mg = g g m T = m g 4 T mg ns. 8 V = T.S 4 g 4 0.60 0 mg 8 J 6 6 56. metal rod B of length 0 has its one end in ice at 0 C and the other end B in water at 00 C. If a point P on the rod is maintained at 400 C, 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 540 cal/g and latent heat of melting of ice is 80 cal/g. If the point P is at a distance of from the ice end, find the value of. [Neglect any heat loss to the surrounding] Sol: k(00) dm H 540 (0 ) dt k(400) dm ; H 80 dt Dividing 00 0 400 540 80 9 00 C B H P 400 C H 0 C ns.: 9 mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4 - -
IIT-JEE-009 57. 0 cm long string, having a mass of.0 g, is fied at both the ends. The tension in the string is 0.5N. The string is set into vibrations using an eternal vibrator of frequency 00Hz. Find the separation (in cm) between the successive nodes on the string. Sol: T = 0.5 N; 0 50 0.0 00 V v 0.5 ; 00 00 50 0 m 0cm = 5 ns. : 5 mity Institute for Competitive Eaminations : Phones: 464/44, 557///4, 950-489/4 - -