PART- I (PHYSICS) SECTION I. Straight Objective Type

Transcription

1 FIITJEE Solutions to IITJEE-007 (Paper-I, Code-7) Time: 3 hours M. Marks: 43 Note: (i) The question paper consists of 3 parts (Physics, Chemistry and Mathematics). Each part has 4 sections. (ii) Section I contains 9 multiple choice questions which have only one correct answer. Each question carries +3 marks each for correct answer and mark for each wrong answer. (iii) Section II contains 4 questions. Each question contains STATEMENT- (Assertion) and STATEMENT- (Reason). Bubble if both the statements are TRUE and STATEMENT- is the correct eplanation of STATEMENT- Bubble if both the statements are TRUE but STATEMENT- is NT the correct eplanation of STATEMENT- Bubble if STATEMENT- is TRUE and STATEMENT- is FALSE. Bubble if STATEMENT- is FALSE and STATEMENT- is TRUE. carries +3 marks each for correct answer and mark for each wrong answer. (iv) Section III contains paragraphs. Based upon each paragraph, 3 multiple choice questions have to be answered. Each question has only one correct answer and carries +4 marks for correct answer and mark for wrong answer. (v) Section IV contains 3 questions. Each question contains statements given in columns. Statements in the first column have to be matched with statements in the second column and each question carries +6 marks and marks will be awarded if all the four parts are correctly matched. No marks will be given for any wrong match in any question. There is no negative marking. PART- I (PYSICS) SECTIN I Straight bjective Type This section contains 9 multiple choice questions numbered to 9. Each question has 4 choices,, and, out of which only one is correct.. A circuit is connected as shown in the figure with the switch S open. When the switch is closed the total amount of charge that flows from Y to X is 3µF 6µF X 0 54µC 7 µc 8 µc S 3Ω 6Ω Y 7 µc 9V FIITJEE Ltd. ICES ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 006, Ph : , , Fa : 65394

2 IIT-JEE007-PAPER-I- 3µF 6µF X +8µC +8µC S 3Ω 6Ω A Y 9V 3µF 6µF +9µC X +8µC +9µC 3Ω Y 9V +36µC S +7µC Initial charge distribution (when switch S is open) Final charge distribution (when switch S is closed). A long, hollow conducting cylinder is kept coaially inside another long, hollow conducting cylinder of larger radius. Both the cylinders are initially electrically neutral. A potential difference appears between the two cylinders when a charge density is given to the inner cylinder. A potential difference appears between the two cylinders when a charge density is given to the outer cylinder. No potential difference appears between the two cylinders when a uniform line charge is kept along the ais of the cylinders. No potential difference appears between the two cylinders when same charge density is given to both the cylinders. dv = E dr λ and E = πε 0 r where r is distance from the ais of cylindrical charge distribution (r is equal to or greater than radius of cylindrical charge distribution). 3. In the options given below, let E denote the rest mass energy of a nucleus and n a neutron. The correct option is E U E I E Y E(n) E U < E I + E Y + E(n) ( 9 ) > ( 53 ) + ( 39 ) + ( 9 ) ( 53 ) ( 39 ) E( 9 U) < E( 56 Ba) + E( 36Kr) + E(n) ( 9 ) ( 56 ) ( 36 ) 6Ω E U = E Ba + E Kr + E(n) 3. Rest mass energy of U will be greater than the rest mass energy of the nucleus in which it breaks (as conservation of momentum is always followed) 4. In an eperiment to determine the focal length (f) of a concave mirror by the u v method, a student places the object pin A on the principal ais at a distance from the pole P. The student looks at the pin and its inverted image from a distance keeping his/her eye in line with PA. When the student shifts his/her eye towards left, the image appears to the right of the object pin. Then, < f f < < f = f > f Due to paralla 5. The largest wavelength in the ultraviolet region of the hydrogen spectrum is nm. The smallest wavelength in the infrared region of the hydrogen spectrum (to the nearest integer) is 80 nm 83 nm 88 nm 648 nm Transition from to n = 3 will produce smallest wavelength in infrared region. 6. A resistance of Ω is connected across one gap of a metre-bridge (the length of the wire is 00 cm) and an unknown resistance, greater than Ω, is connected across the other gap. When these resistance are interchanged, the balance point shifts by 0 cm. Neglecting any corrections, the unknown resistance is 3 Ω 4 Ω 5 Ω 6 Ω FIITJEE Ltd. ICES ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 006, Ph : , , Fa : 65394

3 IIT-JEE007-PAPER-I-3 = = 80 Solving (i) and (ii) = 3Ω (i) (ii) G A ray of light travelling in water is incident on its surface open to air. The angle of incidence is θ, which is less than the critical angle. Then there will be only a reflected ray and no refracted ray only a refracted ray and no reflected ray a reflected ray and a refracted ray and the angle between them would be less than 80 θ a reflected ray and a refracted ray and the angle between them would be greater than 80 θ θ Refracted ray air water θ θ Incident ray Reflected ray 8. Two particle of mass m each are tied at the ends of a light string of length a. The whole system is kept on a frictionless horizontal surface with the string held tight so that each mass is at a distance a from the center P (as shown in the figure). Now, the mid-point of the string is pulled vertically upwards with a small but constant force F. As a result, the particles move towards each other on the surface. The magnitude of acceleration, when the separation between them becomes is F a F m a m a m a P F a m F m a F a m T sin θ = F T cos θ = ma F tanθ= ma F A = m a N a T θ mg P F T m 9. Consider a neutral conducting sphere. A positive point charge is placed outside the sphere. The net charge on the sphere is then, negative and distributed uniformly over the surface of the sphere negative and appears only at the point on the sphere closest to the point charge negative and distributed non-uniformly over the entire surface of the sphere zero SECTIN II Assertion - Reason Type This section contains 4 questions numbered 0 to 3. Each question contains STATEMENT- (Assertion) and STATEMENT- (Reason). Each question has 4 choices,, and out of which NLY NE is correct. FIITJEE Ltd. ICES ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 006, Ph : , , Fa : 65394

4 IIT-JEE007-PAPER-I-4 0. STATEMENT- The formula connecting u, v and f for a spherical mirror is valid only for mirrors whose sizes are very small compared to their radii of curvature. STATEMENT- Laws of reflection are strictly valid for plane surfaces, but not for large spherical surfaces. Statement- is True, Statement- is True; Statement - is a correct eplanation for Statement-. Statement- is True, Statement- is True; Statement - is NT a correct eplanation for Statement-. Statement - is True, Statement- is False. Statement - is False, Statement- is True.. STATEMENT- If the accelerating potential in an X-ray tube is increased, the wavelengths of the characteristic X-rays do not change. STATEMENT - When an electron beam strikes the target in an X-ray tube, part of the kinetic energy is converted into X-ray energy. Statement- is True, Statement- is True; Statement - is a correct eplanation for Statement-. Statement- is True, Statement- is True; Statement - is NT a correct eplanation for Statement-. Statement - is True, Statement- is False. Statement - is False, Statement- is True.. STATEMENT- A block of mass m starts moving on a rough horizontal surface with a velocity v. It stops due to friction between the block and the surface after moving through a certain distance. The surface is now tilted to an angle of 30 0 with the horizontal and the same block is made to go up on the surface with the same initial velocity v. The decrease in the mechanical energy in the second situation is smaller than that in the first situation. STATEMENT- The coefficient of friction between the block and the surface decreases with the increase in the angle of inclination. Statement - is True, Statement- is True; Statement- is a correct eplanation for statement-. Statement - is True, Statement- is True; Statement- is NT a correct eplanation for statement-. Statement - is True, Statement- is False. Statement - is False, Statement- is True. 3. STATEMENT- In an elastic collision between two bodies, the relative speed of the bodies after collision is equal to the relative speed before the collision. STATEMENT- In an elastic collision, the linear momentum of the system is conserved. Statement - is True, Statement- is True; Statement - is a correct eplanation for Statement-. Statement - is True, Statement- is True; Statement - is NT a correct eplanation for Statement-. Statement - is True, Statement- is False. Statement - is False, Statement- is True. SECTIN III Linked Comprehension Type This section contains paragraphs P 4-6 and P 7-9. Based upon each paragraph, 3 multiple choice questions have to be answered. Each question has 4 choices,, and, out of which NLY NE is correct. FIITJEE Ltd. ICES ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 006, Ph : , , Fa : 65394

5 IIT-JEE007-PAPER-I-5 P 4 6 : Paragraph for Question Nos. 4 to 6 A fied thermally conducting cylinder has a radius R and height L 0. The cylinder is open at its bottom and has a small hole at its top. A piston of mass M is held at a distance L from the top surface, as shown in the figure. The atmospheric pressure is P 0. L R 4. The piston is now pulled out slowly and held at a distance L from the top. The pressure in the cylinder between its top and the piston will then be P0 P 0 P0 Mg P0 Mg + π R π R Piston 5. While the piston is at a distance L from the top, the hole at the top is sealed. The piston is then released, to a position where it can stay in equilibrium. In this condition, the distance of the piston from the top is P0πR P ( L) 0πR Mg π RP0 + Mg ( L) RP π 0 P 0π R + Mg P0 πr ( L) πrp ( L) 0 πrp0 Mg L 0 Mg + P(πR ) = P 0 πr P 0 (LπR ) = P(πR ) P0πR L πrp0 Mg = ( ) (P `V = P V for isothermal process) 6. The piston is taken completely out of the cylinder. The hole at the top is sealed. A water tank is brought below the cylinder and put in a position so that the water surface in the tank is at the same level as the top of the cylinder as shown in the figure. The density of the water is ρ. In equilibrium, the height of the water column in the cylinder satisfies ρg(l 0 ) + P 0 (L 0 ) + L 0 P 0 = 0 ρg(l 0 ) P 0 (L 0 ) L 0 P 0 = 0 ρg(l 0 ) + P 0 (L 0 ) L 0 P 0 = 0 ρg(l 0 ) P 0 (L 0 ) + L 0 P 0 = 0 L 0 πr P 0 L 0 = P(L 0 )πr P = P 0 + ρg(l 0 ) Solving (i) & (ii), we get the answer.... (i)... (ii) P 7 9 : Paragraph for Question Nos. 7 to 9 Two discs A and B are mounted coaially on a vertical ale. The discs have moments of inertia I and I respectively about the common ais. Disc A is imparted an initial angular velocity ω using the entire potential energy of a spring compressed by a distance. Disc B is imparted an angular velocity ω by a spring having the same spring constant and compressed by a distance. Both the discs rotate in the clockwise direction. 7. The ratio of / is k ( ) = I ω FIITJEE Ltd. ICES ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 006, Ph : , , Fa : 65394

6 IIT-JEE007-PAPER-I-6 k = ω = ( I)( ) 8. When disc B is brought in contact with disc A, they acquire a common angular velocity in time t. The average frictional torque on one disc by the other during this period is Iω 9Iω 3t t 9Iω 3Iω 4t t Applying conservation of angular momentum I ( ω ) + I( ω) 4ω ω= =... (i) 3I 3 τ ω=ω+ t... (ii) I From () & (ii), τ = I ω 3t 9. The loss of kinetic energy during the above process is Iω Iω 4 Iω 3 Iω 6 SECTIN IV Matri-Match Type This section contains 3 questions. Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in column I have to be matched with statements (p, q, r, s) in column II. The answers to these questions have to be appropriately bubbled as illustrated in the following eample. If the correct match are A-p, A-s, B-r, C-p, C-q and D-s, then the correctly bubbled 4 4 matri should be as follows: p q r s A B C D p q r s p q r s p q r s p q r s 0. Some physical quantities are given in Column I and some possible SI units in which these quantities may be epressed are given in Column II. Match the physical quantities in Column I with the units in Column II and indicate your answer by darkening appropriate bubbles in the 4 4 matri given in the RS. Column I Column II GMeM s G universal gravitational constant, M s mass of the Sun M e mass of the earth, (p) (volt) (coulomb) (metre) FIITJEE Ltd. ICES ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 006, Ph : , , Fa : 65394

7 IIT-JEE007-PAPER-I-7 3RT M ; R universal gas constant, T absolute temperature, M molar mass F (r) (meter) qb ; F force, q charge, B magnetic field GM R e e, G universal gravitational constant, M e mass of the earth, R e radius of the earth (q) (kilogram) (metre) 3 (second) (second) (s) (farad) (volt) (kg) A (p) & (q), B (r) & (s), C (r) & (s), D (r) & (s). Some laws/processes are given in Column I. Match these with the physical phenomena given in Column II and indicate your answer by darkening appropriate bubbles in the 4 4 matri given in the RS. Column I Column II Transition between two atomic energy levels Electron emission from a material (p) Characteristic X-rays (q) Photoelectric effect Mosley s law (r) ydrogen spectrum Change of photon energy into kinetic energy of electrons (s) β-decay A (p) & (r), B (q) & (s), C (p), D (q). olumn I gives certain situations in which a straight metallic wire of resistance R is used and Column II gives some resulting effects. Match the statements in Column I with the statements in Column II and indicate your answer by darkening appropriate bubbles in the 4 4 matri given in the RS. Column I Column II A charged capacitor is connected to the ends of the wire (p) A constant current flows through the wire The wire is moved perpendicular to its length with a constant velocity in a uniform magnetic field perpendicular to the plane of motion The wire is placed in a constant electric field that has a direction along the length of the wire. A battery of constant emf is connected to the ends of the wire (q) Thermal energy is generated in the wire (r) A constant potential difference develops between the ends of the wire (s) Charges of constant magnitude appear at the ends of the wire A (q), B (r) & (s), C (r) & (s), D (p), (q) & (r) FIITJEE Ltd. ICES ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 006, Ph : , , Fa : 65394

8 IIT-JEE007-PAPER-I-8 PART- II (CEMISTRY) SECTIN I Straight bjective Type This section contains 9 multiple choice questions numbered 3 to 3. Each question has 4 choices,, and, out of which NLY NE is correct. 3. The number of structural isomers for C 6 4 is C C C C C C C 3 3 C C C C C 3 3 C C C C C 3 C 3 3 C C C C 3 C 3 C 3 C 3 C 3 3 C C C C 3 ence is correct. C 3 4. In the following reaction, N conc. N3 conc. S4 X the structure of the major product X is N N N N N N N N N C conc. N3 conc. S4 N N C Due to presence of lone pair of electron on nitrogen atom, it will activate the ring and it will stabilize intermediate cation at o and p positions. ence is correct. FIITJEE Ltd. ICES ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 006, Ph : , , Fa : 65394

9 IIT-JEE007-PAPER-I-9 5. When 0 g of naphthoic acid (C 8 ) is dissolved in 50 g of benzene (K f =.7 K kg mol ), a freezing point depression of K is observed. The van t off factor (i) is T f = K f molality i =.7 i 7 50 i = 0.5 ence is correct. 6. Among the following, the paramagnetic compound is Na 3 N K = σs σ*s, σs σ*s, σp, z π p =πp, y π *p =π *py Number of unpaired electrons = 0. N = N Number of unpaired electrons = 0 Number of unpaired electrons = 0 = σs, σ*s σs, σ* s, σp z, π p =πp y, π * p =π * p y Number of unpaired electrons = Thus is paramagnetic. ence is correct. 7. The value of log 0 K for a reaction A B is r 98K r 98K (Given : = 54.07kJ mol, S = 0JK mol and R= 8.34 JK mol ; = 5705) G = T S = = J mol = 5705 log 0 K log 0 K = 0 ence is correct 8. The species having bond order different from that in C is N N + CN N N (6 electron system) Bond order =. N, CN and N are isoelectronic with C therefore all have same bond order (= 3) ence is correct. 9. The percentage of p-character in the orbitals forming P P bonds in P 4 is P is sp 3 hybridized in P 4. P P P P ence is correct. FIITJEE Ltd. ICES ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 006, Ph : , , Fa : 65394

10 IIT-JEE007-PAPER-I Etraction of zinc from zinc blende is achieved by electrolytic reduction roasting followed by reduction with another metal roasting followed by reduction with carbon roasting followed by self-reduction ption is correct. 3. The reagent(s) for the following conversion, Br Br? is/are alcoholic K alcoholic K followed by NaN aqueous K followed by NaN Zn/C 3 C C Alc. K NaN C C C C Br Br Br Because C = C Br has partial C Br double bond character, it requires more stronger base to remove Br. ence is correct. SECTIN II Assertion-Reason Type This section contains 4 questions numbered 3 to 35. Each question contains STATEMENT- (Assertion) and STATEMENT- (Reason). Each question has 4 choices,, and out of which NLY NE is correct. 3. STATEMENT-: p-ydroybenzoic acid has a lower boiling point than o-hydroybenzoic acid. STATEMENT-: o-ydroybenzoic acid has intramolecular hydrogen bonding. Statement- is True, Statement- is True; Statement- is a correct eplanation for Statement-. Statement- is True, Statement- is True; Statement- is NT a correct eplanation for Statement-. Statement- is True, Statement - is False. Statement- is False, Statement- is True. C More stabilized by intramolecular hydrogen bonding C More stronger intermolecular forces increases the boiling point. C ence is correct. FIITJEE Ltd. ICES ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 006, Ph : , , Fa : 65394

11 IIT-JEE007-PAPER-I- 33. STATEMENT-: Micelles are formed by surfactant molecules above the critical micellar concentration (CMC). STATEMENT-: The conductivity of a solution having surfactant molecules decreases sharply at the CMC. Statement- is True, Statement- is True; Statement- is a correct eplanation for Statement-. Statement- is True, Statement- is True; Statement- is NT a correct eplanation for Statement-. Statement- is True, Statement- is False. Statement- is False, Statement- is True. The formation of micelles takes places only above a particular temperature called Kraft temperature (T k ) and above a particular concentration called critical micelle concentration (CMC). Each micelle contains at least 00 molecules. Therefore conductivity of the solution decreases sharply at the CMC. ence is correct. 34. STATEMENT-: Boron always forms covalent bond. STATEMENT-: The small size of B 3+ favours formation of covalent bond. Statement- is True, Statement- is True; Statement- is a correct eplanation for Statement-. Statement- is True, Statement- is True; Statement- is NT a correct eplanation for Statement-. Statement- is True, Statement- is True. Statement- is False, Statement- is True. According to Fajan s rule small cations having high charge density always have tendency to form covalent bond. ence is correct 35. STATEMENT-: In water, orthoboric acid behaves as a weak monobasic acid. STATEMENT-: In water, orthoboric acid acts as a proton donor. Statement- is True, Statement- is True; Statement- is a correct eplanation for Statement-. Statement- is True, Statement- is True; Statement- is NT a correct eplanation for Statement-. Statement-is True, Statement- is False. Statement- is False, Statement- is True. 3 B 3 (orthoboric acid) is a weak lewis acid. B 3 3 B ( ) It does not donate proton rather it acceptors form water. ence is correct SECTIN III Linked Comprehension Type This section contains paragraphs C and C Based upon each paragraph, 3 multiple choice questions have to be answered. Each question has 4 choices,, and, out of which NLY NE is correct. C : Paragraph for question Nos 36 to 38 Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules (approimately ) are present in a few grams of any chemical compound varying with their atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following eample illustrates a typical case, involving chemical/electrochemical reaction, which requires a clear understanding of the mole concept. A 4.0 molar aqueous solution of NaCl is prepared and 500 ml of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes (atomic mass: Na = 3, g = 00; Faraday=96500 coulombs) 36. The total number of moles of chlorine gas evolved is FIITJEE Ltd. ICES ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 006, Ph : , , Fa : 65394

12 IIT-JEE007-PAPER-I- + NaCl Na + Cl At anode: Cl Cl Moles of Cl = in 500 ml. Therefore mole of Cl evolves. ence is correct. 37. If the cathode is a g electrode, the maimum weight (g) of amalgam formed from this solution is Na g (amalgam) formed = moles at cathode. ence is correct. 38. The total charge (coulombs) required for complete electrolysis is moles of electrons ( Faraday) are required. F = F = ence is correct. C 39-4 : Paragraph for Question Nos. 39 to 4 The noble gases have closed-shell electronic configuration and are monoatomic gases under normal conditions. The low boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions. The direct reaction of enon with fluorine leads to a series of compounds with oidation numbers +, +4 and +6. XeF 4 reacts violently with water to give Xe 3. The compounds of enon ehibit rich stereochemistry and their geometries can be deduced considering the total number of electron pairs in the valence shell. 39. Argon is used in arc welding of its low reactivity with metal ability to lower the melting point of metal flammability high calorific value Argon is used mainly to provide an inert atmosphere in high temperature metallurgical (arc welding of metals/alloys) etraction. ence is correct. 40. The structure of Xe 3 is linear pyramidal Xe sp 3 hybridized pyramidal structure. ence is correct. 4. XeF 4 and XeF 6 are epected to be oidizing unreactive 6XeF4 + 4Xe + Xe3 + 4F + 3 XeF6 + 3 Xe3 + 6F ence is correct. planar T-shaped reducing strongly basic FIITJEE Ltd. ICES ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 006, Ph : , , Fa : 65394

13 IIT-JEE007-PAPER-I-3 SECTIN IV Matri-Match Type This section contains 3 questions. Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column-I have to be matched with statements (p, q, r, s) in Column-II. The answers to these questions have to be appropriately bubbled as illustrated in the following eample. If the correct matches are A-p, A-s, B-q, B-r, C-p, C-q and D-s, then the correctly bubbled 4 4 matri should be as follows: A B C D p q r s p q r s p q r s p q r s p q r s 4. Match the complees in Column-I with their properties listed in Column-II. Indicate your answer by darkening the appropriate bubbles of the 4 4 matri given in the RS. Column-I Column-II [Co(N 3 ) 4 ( ) ]Cl (p) geometrical isomers [Pt(N 3 ) Cl ] (q) paramagnetic [Co( ) 5 Cl]Cl (r) diamagnetic [Ni( ) 6 ]Cl (s) metal ion with + oidation state A p, q, s B p, r, s C q, s D q, s + N N N 3 Co Co 3 N N 3 3 N N 3 (trans) (cis) Co + = 3d 7 (Paramagnetic) [Pt(N 3 ) Cl ] is square planar. Cl Cl 3 N 3 N + + Pt Pt 3 N Cl Cl N 3 (cis) (trans) Pt + = 5d 8 4s 0 (diamagnetic) + + Co Co Cl Co + = 3d 7 (paramagnetic) Ni + = 3d 8 (weak field ligand, paramagnetic) FIITJEE Ltd. ICES ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 006, Ph : , , Fa : 65394

14 IIT-JEE007-PAPER-I Match gases under specified conditions listed in Column-I with their properties/laws in Column-II. Indicate your answer by darkening the appropriate bubsbles of the 4 4 matri given in the RS. Column-I Column-II hydrogen gas (P = 00 atm, T = 73K) (p) compressibility factor hydrogen gas (P 0, T = 73K) (q) attractive forces are dominant C (P = atm, T = 73K) (r) PV = nrt real gas with very large molar volume (s) P(V nb) = nrt A p, s B r C p, q D p, s PVm Z = at high pressure and low temperature. RT an Equation P+ ( V nb) V = nrt reduces to P(V nb) = nrt. For hydrogen gas value of Z = at P = 0 and it increase continuously on increasing pressure. C molecules have larger attractive forces, under normal conditions. PVm Z =, at very large molar volume Z. RT 44. Match the chemical substances in Column-I with type of polymers/type of bonds in Column-II. Indicate your answer by darkening the appropriate bubbles of the 4 4 matri given in the RS. Column-I Column-II cellulose (p) natural polymer nylon-6, 6 (q) synthetic polymer protein (r) amide linkage sucrose (s) glycoside linkage A p, s, B q, r; C p, r D s Cellulose Nylon 6, 6 (Glycoside linkage) C N N C (Amide linakge) Protein R C N C N C N R' (Amide linkage) Sucrose C C (Glycoside linkage) FIITJEE Ltd. ICES ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 006, Ph : , , Fa : 65394

15 IIT-JEE007-PAPER-I-5 PART III (MATEMATICS) SECTIN I Straight bjective Type This section contains 9 multiple choice questions numbered 45 to 53. Each question has 4 choices,, and, out of which NLY NE is correct. 45. A hyperbola, having the transverse ais of length sinθ, is confocal with the ellipse 3 + 4y =. Then its equation is cosec θ y sec θ = sec θ y cosec θ = sin θ y cos θ = cos θ y sin θ = The given ellipse is y + = 4 3 a =, b = 3 3 = 4 ( e ) e = so that ae = ence the eccentricity e, of the hyperbola is given by = e sinθ e = cosecθ b = sin θ(cosec θ ) = cos θ y ence the hyperbola is = or cosec θ y sec θ = sin θ cos θ 46. The tangent to the curve y = e drawn at the point (c, e c ) intersects the line joining the points (c, e c ) and (c +, e c+ ) on the left of = c on the right of = c at no point at all points Slope of the line joining the points (c, e c ) and (c +, e c + ) is equal to e c+ c e c > e tangent to the curve y = e will intersect the given line to the left of the line = c. Alternative The equation of the tangent to the curve y = e at (c, e c ) is y e c = e c ( c) () Equation of the line joining the given points is c c e(e e ) y e = [ (c )] () Eliminating y from () and (), we get [ (c )] [ (e e )] = e e+ e or c= < 0 < c. (e e ) the line () and () meet on the left of the line = c. A y (c, e c ) B 47. A man walks a distance of 3 units from the origin towards the north-east (N 45 E) direction. From there, he walks a distance of 4 units towards the north-west (N 45 W) direction to reach a point P. Then the position of P in the Argand plane is 3e iπ/4 + 4i (3 4i) e iπ/4 (4 + 3i) e iπ/4 (3 + 4i) e iπ/4 FIITJEE Ltd. ICES ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 006, Ph : , , Fa : 65394

16 IIT-JEE007-PAPER-I-6 Let A = 3, so that the comple number associated with A is 3e iπ/4. If z is the comple number associated with P, i π /4 z 3e 4 π i / 4i then = e = i π /4 0 3e 3 3 3z 9e iπ/4 = ie iπ/4 z = (3 + 4i)e iπ/4. P y 4 3 π/4 3e iπ/4 A 48. Let f() be differentiable on the interval (0, ) such that f() =, and ( ) tf f() t lim = t t for each > 0. Then f() is tf() f(t) lim = t t f () f() + = 0 f() = c + also f() = 3 c = 3. ence f() = The number of solutions of the pair of equations sin θ cosθ = 0 cos θ 3 sinθ = 0 in the interval [0, π] is zero two one four sin θ cosθ = 0 sin θ = 4 also cos θ = 3sinθ sinθ = two solutions in [0, π]. 50. Let α, β be the roots of the equation α p + r = 0 and, β be the roots of the equation q + r = 0. Then the value of r is ( p q )( q p ) ( q p )( p q ) 9 9 ( q p )( q p ) ( p q )( q p ) 9 9 FIITJEE Ltd. ICES ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 006, Ph : , , Fa : 65394

17 IIT-JEE007-PAPER-I-7 The equation p + r = 0 has roots (α, β) and the equation α q + r = 0 has roots,β. α r = αβ and α + β = p and + β= q q p (p q) β= and α= 3 3 αβ = r = (q p) (p q) The number of distinct real values of λ, for which the vectors λ î + ĵ+ˆk, î λ ĵ+ˆk and î + ĵ λ ˆk are coplanar, is zero one two three λ λ λ = 0 λ 6 3λ = 0 ( + λ ) (λ ) = 0 λ = ±. 5. ne Indian and four American men and their wives are to be seated randomly around a circular table. Then the conditional probability that the Indian man is seated adjacent to his wife given that each American man is seated adjacent to his wife is / /3 /5 /5 Let E = event when each American man is seated adjacent to his wife A = event when Indian man is seated adjacent to his wife Now, n(a E) = (4!) (!) 5 Even when each American man is seated adjacent to his wife Again n(e) = (5!) (!) 4 5 P A n(a E) (4!) (!) = = E n(e) (5!) (!) 4 =. 5 Alternative Fiing four American couples and one Indian man in between any two couples; we have 5 different ways in which his wife can be seated, of which cases are favorable. required probability = lim π 4 sec f() t dt π 6 8 f ( ) π f π equals f ( ) π 4f() FIITJEE Ltd. ICES ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 006, Ph : , , Fa : 65394

18 IIT-JEE007-PAPER-I-8 lim π 4 sec f(t)dt π 6 0 form 0 f (sec )secsec tan Let L= lim π 4 f () 8f () L = =. π/4 π SECTIN II Assertion Reason Type This section contains 4 questions numbered 54 to 57. Each question contains STATEMENT (Assertion) and STATEMENT - (Reason). Each question has 4 choices,, and out of which NLY NE is correct. 54. Let the vectors PQ, QR, RS, ST, TU and UP represent the sides of a regular heagon. PQ RS+ ST 0. STATEMENT - : ( ) STATEMENT - : PQ RS = 0 and PQ ST 0. Statement - is True, Statement - is True; Statement- is a correct eplanation for Statement- Statement - is True, Statement - is True; Statement- is NT a correct eplanation for Statement- Statement - is True, Statement - is False Statement - is False, Statement - is True Since PQ / TR TR is resultant of SR and ST T S vector. PQ ( RS + ST) 0. U R But for statement, we have PQ RS = 0 which is not possible as PQ / RS. P Q ence, statement is true and statement is false. 55. Let F() be an indefinite integral of sin. STATEMENT - : The function F() satisfies F( + π) = F() for all real. STATEMENT - : sin ( + π) = sin for all real. Statement - is True, Statement - is True; Statement- is a correct eplanation for Statement- Statement - is True, Statement - is True; Statement- is NT a correct eplanation for Statement- Statement - is True, Statement - is False Statement - is False, Statement - is True F() = cos sin d = d F() = ( sin) + c. 4 Since, F( + π) F(). ence statement is false. But statement is true as sin is periodic with period π. FIITJEE Ltd. ICES ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 006, Ph : , , Fa : 65394

19 IIT-JEE007-PAPER-I Let,,, n be mutually eclusive and ehaustive events with P( i ) > 0, i =,,, n. Let E be any other event with 0 < P(E) <. STATEMENT - : P( i E) > P(E i ). P( i ) for i =,,, n n STATEMENT - : ( ) i= P = i Statement - is True, Statement - is true; Statement- is a correct eplanation for Statement- Statement - is True, Statement - is True; Statement- is NT a correct eplanation for Statement- Statement - is True, Statement - is False Statement - is False, Statement - is True Statement : i E If P( i E) = 0 for some i, then P = P = 0 E i If P( i E) 0 for i =, n, then i P(i E) P( i) P = E P( i ) P(E) E P P( i) i E = > P P( i) [as 0 < P(E) < ] P(E) i ence statement may not always be true. Statement : Clearly n = S (sample space) P( ) + P( ) + + P( n ) =. 57. Tangents are drawn from the point (7, 7) to the circle + y = 69. STATEMENT - : The tangents are mutually perpendicular. STATEMENT - : The locus of the points from which mutually perpendicular tangents can be drawn to the given circle is + y = 338 Statement - is True, Statement - is true; Statement- is a correct eplanation for Statement- Statement - is True, Statement - is true; Statement- is NT a correct eplanation for Statement- Statement - is True, Statement - is False Statement - is False, Statement - is True Since the tangents are perpendicular locus of perpendicular tangents to circle + y = 69 is a director circle having equation + y = 338. SECTIN III Linked Comprehension Type This section contains paragraphs M and M Based upon each paragraph, 3 multiple choice questions have to be answered. Each question has 4 choice,, and, out of which NLY NE is correct. M : Paragraph for question Nos. 58 to 60 Consider the circle + y = 9 and the parabola y = 8. They intersect at P and Q in the first and the fourth quadrants, respectively. Tangents to the circle at P and Q intersect the -ais at R and tangents to the parabola at P and Q intersect the -ais at S. 58. The ratio of the areas of the triangles PQS and PQR is : : : 4 : 8 FIITJEE Ltd. ICES ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 006, Ph : , , Fa : 65394

20 IIT-JEE007-PAPER-I-0 Coordinates of P and Q are (, + ) and (, ). Y Area of PQR = = Area of PQS = 4 4 = Ratio of area of triangle PQS and PQR is : 4. ( 3, 0) S (, 0) (, 0) P (, ) R (9, 0) X Q (, ) 59. The radius of the circumcircle of the triangle PRS is Equation of circumcircle of PRS is ( + ) ( 9) + y + λy = 0 It will pass through (, ), then λ. = 0 8 λ= = Equation of circumcircle is + y 8 + y 9= 0. ence its radius is 3 3. Alternative Let PSR = θ sin θ= 3 PR = 6 = R sinθ R = The radius of the incircle of the triangle PQR is 4 3 8/3 Radius of incircle is r = s as = s= = 8 6 r = =. 8 M 6 63 : Paragraph for Question Nos. 6 to 63 Let V r denote the sum of the first r terms of an arithmetic progression (A.P.) whose first term is r and the common difference is (r ). Let T r = V r+ V r and Q r = T r+ T r for r =,, 6. The sum V + V + + V n is n ( n )( 3n n ) n ( n n ) n n + 3n + n + n 3 n ( )( ) + ( ) FIITJEE Ltd. ICES ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 006, Ph : , , Fa : 65394

21 IIT-JEE007-PAPER-I- V r = r [r + (r )(r )] = (r3 r + r) n(n + )(3n + n + ). Vr = 6. T r is always an odd number a prime number V r + V r = (r + ) 3 r 3 [(r + ) r ] + () = 3r + r + T r = 3r + r = (r + )(3r ) which is a composite number. 63. Which one of the following is a correct statement? Q, Q, Q 3, are in A.P. with common difference 5 Q, Q, Q 3, are in A.P. with common difference 6 Q, Q, Q 3, are in A.P. with common difference Q = Q = Q 3 = T r = 3r + r T r + = 3(r + ) + (r + ) Q r = T r + T r = 3[r + ] + [] Q r = 6r + 5 Q r + = 6(r + ) + 5 Common difference = Q r + Q r = 6. an even number a composite number SECTIN IV Matri-Match Type This section contains 3 questions. Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (p, q, r, s) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following eample. If the correct matches are A p, A s, B q, B r, C p, C q and D s, then the correctly bubbled 4 4 matri should be as follows: p q r s A p q r s B p q r s C p q r s D p q r s 64. Consider the following linear equations a + by + cz = 0 b + cy + az = 0 c + ay + bz = 0 Match the conditions / epressions in Column I with statements in Column II and indicate your answers by darkening the appropriate bubbles in 4 4 matri given in the RS. Column I Column II a + b + c 0 and a + b + c = ab + bc + ca (p) the equations represent planes meeting only at a single point. a + b + c = 0 and a + b + c ab + bc + ca (q) the equations represent the line = y = z. a + b + c 0 and a + b + c ab + bc + ca a + b + c = 0 and a + b + c = ab + bc + ca (r) the equations represent identical planes. (s) the equations represent the whole of the three dimensional space. FIITJEE Ltd. ICES ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 006, Ph : , , Fa : 65394

22 IIT-JEE007-PAPER-I- A r B q C p D s = a b c b c a c a b = (a + b + c)[(a b) + (b c) + (c a) ]... If a + b + c 0 and a + b + c = ab + bc + ca = 0 and a = b = c 0 the equations represent identical planes. a + b + c = 0 and a + b + c ab + bc + ca = 0 the equations have infinitely many solutions. a + by = (a + b)z b + cy = (b + c)z (b ac)y = (b ac)z y = z a + by + cy = 0 a = ay = y = z. a + b + c 0 and a + b + c ab + bc + ca 0 the equation represent planes meeting at only one point.. a + b + c = 0 and a + b + c = ab + bc + ca a = b = c = 0 the equation represent whole of the three dimensional space. 65. Match the integrals in Column I with the values in Column II and indicate your answer by darkening the appropriate bubbles in the 4 4 matri given in the RS. Column I Column II d (p) + log 3 d (q) 0 3 d (r) d (s) log 3 π 3 π A s B s C p D r. d + π =. 0 d π =.. 3 = d ln 3 d π = 3 FIITJEE Ltd. ICES ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 006, Ph : , , Fa : 65394

23 IIT-JEE007-PAPER-I In the following [] denotes the greatest integer less than or equal to. Match the functions in Column I with the properties Column II and indicate your answer by darkening the appropriate bubbles in the 4 4 matri given in the RS. Column I Column II (p) continuous in (, ) (q) differentiable in (, ) + [] (r) strictly increasing in (, ) + + (s) not differentiable at least at one point in (, ) A p, q, r B p, s C r, s D p, q. is continuous, differentiable and strictly increasing in (, ).. is continuous in (, ) and not differentiable at = [] is strictly increasing in (, ) and discontinuous at = 0 not differentiable at = = in (, ) the function is continuous and differentiable in (, ). FIITJEE Ltd. ICES ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 006, Ph : , , Fa : 65394