PART - I : PHYSICS ALLEN. According to problem particle is to land on disc. If we consider a time 't' then x component of displacement is Rwt

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1 IIT-JEE 0 EXAMINATIN TARGET : IIT-JEE 0 (eld on ) PAPER - ( ANSWERS & SLUTINS)

2 IIT-JEE 0 EXAMINATIN (ELD N : ) PART - I : PYSICS PAPER - SECTIN I : Single Correct Answer Type This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which NLY NE is correct.. Consider a disc rotating in the horizontal plane with a constant angular speed w about its centre. The disc has a shaded region on one side of the diameter and an unshaded region on the other side as shown in the figure. When the disc is in the orientation as shown, two pebbles P and Q are simultaneously projected at an angle towards R. The velocity of projection is in the y-z plane and is same for both pebbles with respect to the disc. Assume that (i) they land back on the disc before the disc has completed rotation, (ii) their range is less than half the disc radius, and (iii) w remains 8 constant throughout. Then (A) P lands in the shaded region and Q in the unshaded region. (B) P lands in the unshaded region and Q in the shaded region. (C) Both P and Q land in the unshaded region. (D) Both P and Q land in the shaded region. Ans. (C) y x R P Q w wt Rsinwt Rwt According to problem particle is to land on disc. If we consider a time 't' then x component of displacement is Rwt Rsinwt < Rwt Thus particle P lands in unshaded region. For Q, x-component is very small and y-component equal to P it will also land in unshaded region. Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA-4005 PNE : , Fax : , info@allen.ac.in Website: / 9

3 IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER -. In the given circuit, a charge of +80 mc is given to the upper plate of the 4mF capacitor. Then in the steady state, the charge on the upper plate of the mf capacitor is +80mC 4mF mf mf (A) + mc (B) + 40mC (C) +48mC (D) + 80mC Ans. (C) KCL at x (x 0) + (x 0) 80 = 0 Þ 5x = 80 Þ x =6 v Þ charge on mf = (x 0) (mf) = 48 mc +80mC -80m C mf. Two moles of ideal helium gas are in a rubber balloon at 0 C. The balloon is fully expandable and can be assumed to required no energy in its expansion. The temperature of the gas in the balloon is slowly changed to 5 C. The amount of heat required in raising the temperature is nearly (take R = 8. J/mol.K) (A) 6 J (B) 04 J (C) 4 J (D) 08 J Ans. (D) (x) mf f æ f ö DQ = DU + DW = nrdt + PDV = ç + nrdt è ø æ ö = ç + ( 8. )( 5 ) = 08 J è ø 4. A loop carrying current I lies in the x-y plane as shown in the figure. The unit vector ˆk is coming out of the plane of the paper. The magnetic moment of the current loop is 0 / 9 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA-4005 PNE : , Fax : , info@allen.ac.in Website:

4 Ans. (B) (A) ˆ a Ik (B) IIT-JEE 0 EXAMINATIN (ELD N : ) æp ö ç + a Ik ˆ è ø (C) PAPER - æp ö - ç + a Ik ˆ è ø (D) ( p + ) a Ik ˆ é Magnetic moment = ( )( ) æaö ù æp ö ina = I ê pç + a ú= ç + a Ik ˆ ë èø û è ø 5. Two identical discs of same radius R are rotating about their axes in opposite directions with the same constant angular speed w. The discs are in the same horizontal plane. At time t =0, the points P and Q are facing each other as shown in the figure. The relative speed between the two points P and Q is v r. In one time period (T) of rotation of the discs, v r as a function of time is best represented by (A) (C) Ans. (A) v r R 0 T v r 0 T t t P w (B) (D) w Q v r R 0 T v r 0 T t t vsinq q v q vcosq q vcosq q v vsinq V = Vsinq = Vsin ( wt) rcl Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA-4005 PNE : , Fax : , info@allen.ac.in Website: / 9

5 IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER - 6. A thin uniform cylindrical shell, closed at both ends, is partially filled with water. It is floating vertically in water in half-submerged state. If r C is the relative density of the material of the shell with respect to water, then the correct statement is that the shell is (A) more than half-filled if r C is less than 0.5 (B) more than half-filled if r C is more than.0 (C) half-filled if r C is more than 0.5 (D) less than half-filled if r c is less than 0.5 Ans. (A) V W For floating : mcg+ mwg= FB inside volume=v & container material volume = V C æv VCö V rcvg C + Vg w = ç + g è ø ÞVw V ö = + Cç -rc è ø æ ö if ç r C < è ø V V w > 7. A student is performing the experiment of Resonance Column. The diameter of the column tube is 4 cm. The frequency of the tuning fork is 5 z. The air temperature is 8 C in which the speed of sound is 6 m/s. The zero of the meter scale coincides with the top end of the Resonance column tube. When the first resonance occurs, the reading of the water level in the column is (A) 4.0 cm (B) 5. cm (C) 6.4 cm (D) 7.6 cm Ans. (B) l l v In resonance column experiment = l +e so, l = - e= -0.d 4 4 4f 6 00 cm = -( 0. )( 4 cm ) = = 5. cm An infinitely long hollow conducting cylinder with inner radius R/ and outer radius R carries a uniform current density along its length. The magnitude of the magnetic field, B r as a function of the radial distance r from the axis is best represented by (A) B (B) B R/ R r R/ R r (C) 4 / 9 B R/ R r B (D) R/ R Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA-4005 PNE : , Fax : , info@allen.ac.in Website: r

6 Ans. (D) B = 0 for r < R/ IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER - B æm0jr m0 I ö = ç - è r ø for R r R = B B m I r 0 = r ³ R B B R/ R B r Section-II : Paragraph Type This section contains 6 multiple choice questions relating to three paragraphs with two questions on each paragraph. Each question has four choices (A), (B), (C) and (D) out of which NLY NE is correct. Paragraph for Questions 9 and 0 Most materials have the refractive index, n>. So, when a light ray from air enters a naturally occurring material, then by Snell's law, sinq n sinq = n, it is understood that the refracted ray bends towards the normal. But it never emerges on the same side of the normal as the incident ray. According to æcö electromagnetism, the refractive index of the medium is given by the relation, n = ç =± em r r è vø, where c is the speed of electromagnetic waves in vacuum, v its speed in the medium, e r and m r are the relative permittivity and permeability of the medium respectively. In normal materials, both e r and m r are positive, implying positive n for the medium. When both e r and m r are negative, one must choose the negative root of n. Such negative refractive index materials can now be artificially prepared and are called meta-materials. They exhibit significantly different optical behaviour, without violating any physical laws. Since n is negative, it results in a change in the direction of propagation of the refracted light. owever, similar to normal materials, the frequency of light remains unchanged upon refraction even in meta-materials. 9. For light incident from air on a meta-material, the appropriate ray diagram is Air q Air q (A) Meta-material q (B) Meta-material q Air q Air q (C) Meta-material q (D) Meta-material q Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA-4005 PNE : , Fax : , info@allen.ac.in Website: 5 / 9

7 Ans. (C) Snell's law : IIT-JEE 0 EXAMINATIN (ELD N : ) sinq n sinq = n so, sinq = n sinq n sinq For Air & meta material sinq ( ) =, q - n is ( ve) 0. Choose the correct statement. (A) The speed of light in the meta-material is v = c n PAPER - Ans. (B) (B) The speed of light in the meta-material is v = c n (C) The speed of light in the meta-material is v = c. (D) The wavelength of the light in the meta-material (l m ) is given by l m = l air n, where l air is the wavelength of the light in air. C By definition of n : V = n Paragraph for Questions and The general motion of a rigid body can be considered to be a combination of (i) a motion of its centre of mass about an axis, and (ii) its motion about an instantaneous axis passing through the centre of mass. These axes need not be stationary. Consider, for example, a thin uniform disc welded (rigidly fixed) horizontally at its rim to a massless stick, as shown in the figure. When the disc-stick system is rotated about the origin on a horizontal frictionless plane with angular speed w, the motion at any instant can be taken as a combination of (i) a rotation of the centre of mass of the disc about the z- axis, and (ii) a rotation of the disc through an instantaneous vertical axis passing through its centre of mass (as is seen from the changed orientation of points P and Q). Both these motions have the same angular speed w in this case. Q P z w P Q y x Now consider two similar systems as shown in the figure : case (A) the disc with its face vertical and parallel to x-z plane; Case (B) the disc with its face making an angle of 45 with x-y plane and its horizontal diameter parallel to x-axis. In both the cases, the disc is welded at point P, and the systems are rotated with constant angular speed w about the z-axis. z w Q z w Q P y P y x case (a) x case (b) 6 / 9 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA-4005 PNE : , Fax : , info@allen.ac.in Website:

8 IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER -. Which of the following statements regarding the angular speed about the instantaneous axis (passing through the centre of mass) is correct? (A) It is w for both the cases. (B) It is w for case (a); and w for case (b). (C) It is w for case (a); and w for case (b). (D) It is w for both the cases. Ans. (D) Since body is rigid therefore w is same about each point.. Which of the following statements about the instantaneous axis (passing through the centre of mass) is correct? (A) It is vertical for both the cases (a) and (b). (B) It is vertical for case (a); and is at 45 to the x-z plane and lies in the plane of the disc for case (b). (C) It is horizontal for case (a); and is at 45 to the x-z plane and is normal to the plane of the disc for case (b). (D) It is vertical for case (a); and is at 45 to the x-z plane and is normal to the plane of the disc for case (b). Ans. (A) Since body is rigid therefore all of its part rotate about same axis. Paragraph for Questions and 4 The b-decay process, discovered around 900, is basically the decay of a neutron (n). In the laboratory, a proton (p) and an electron (e ) are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has a continuous spectrum. Considering a three-body decay process, i.e. n p + e + v r e, around 90, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino ( v r e ) massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is ev. The kinetic energy carried by the proton is only the recoil energy.. If the anti-neutrino had a mass of ev/c (where c is the speed of light) instead of zero mass, what should be the range of the kinetic energy, K, of the electron? (A) 6 0 K ev (B) 6.0eV K ev (C) 6.0eV K < ev (D) 6 0 K < ev Ans. (D) Electron kinetic energy has continuous spectrum. Q-value of this reaction is very close to 0.8 MeV. 6 So 0 K < ev 4. What is the maximum energy of the anti-neutrino? (A) zero (B) much less than ev (C) Nearly ev (D) Much larger than ev Ans. (C) Sum of kinetic energy of electron and energy of neutrino remains nearly constant and proton carries away with negligible energy due to heavy mass. to be Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA-4005 PNE : , Fax : , info@allen.ac.in Website: 7 / 9

9 IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER - Section III : Multiple Correct Answer (s) Type This section contains 6 multiple questions. Each question has four choices (A)_, (B), (C) and (D) out of which NE or MRE are correct. 5. A current carrying infinitely long wire is kept along the diameter of a circular wire loop, without touching it. The correct statement(s) is (are) (A) The emf induced in the loop is zero if the current is constant. (B) The emf induced in the loop is infinite if the current is constant. (C) The emf induced in the loop is zero if the current decreases at a steady rate. (D) The emf induced in the loop is finite if the current decreases at a steady rate. Ans. (A,C) Since flux of wire on the loop is zero therefore emf will not be induced. 6. In the given circuit, the AC source has w = 00 rad/s. Considering the inductor and capacitor to be ideal, the correct choice (s) is(are) 00mF W I 0V (A) The current through the circuit, I is 0. A. (B) The current through the circuit, i is 0. ÖA. (C) The voltage across 00W resistor = 0ÖV. (D) The voltage across 50W resistor = 0V. Ans. (A,C) 0 0 I = = = z A at 45 leading 00 5 I 0 0 = = = A at 45 lagging z 50 5 I = I + I = A» 0.A 0 VR = I R = 00 = 0 V 5 50W I X C=00mF I X =50R L R =00W 0V R =50W B I B 8 / 9 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA-4005 PNE : , Fax : , info@allen.ac.in Website:

10 IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER - 7. Six point charges are kept at the vertices of a regular hexagon of side L and centre, as shown in q figure. Given that K = 4p Î, which of the following statement(s) is(are) correct? L 0 +q F L P E -q A +q S B +q (A) The electric field at is 6 K along D. (B) The potential at is zero. R C -q (C) The potential at all points on the line PR is same. (D) The potential at all points on the line ST is same. Ans. (A,B,C) T D -q Resultant is 6 K along D. For every point of PR there are two charges of opposite sign and equal magnitude at equal distance. 8. Two spherical planets P and Q have the same uniform density r, masses M P and M Q, and surface areas A and 4A, respectively. A spherical planet R also has uniform density r and its mass is (M P + M Q ). The escape velocities from the planets P, Q and R, are V P, V Q and V R, respectively. Then Ans. (B,D) (A) V Q > V R > V P (B) V R > V Q > V P (C) V R /V P = (D) V P /V Q = / 4 p Escape velocity = GM R µ µ R R [since density of each planet is same] Area Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA-4005 PNE : , Fax : , info@allen.ac.in Website: 9 / 9

11 IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER - 9. The figure shows a system consisting of (i) a ring of outer radius R rolling clockwise without slipping on a horizontal surface with angular speed w and (ii) an inner disc of radius R rotating anti-clockwise with angular speed w/. The ring and disc are separated by frictionless ball bearing. The system is in the x-z plane. The point P on the inner disc is at a distance R from the origin, where P makes an angle of 0 with the horizontal. Then with respect to the horizontal surface, z w (A) the point has a linear velocity Rwiˆ w/ R R P 0 R (B) the point P has a linear velocity Rwi ˆ + Rwkˆ 4 4 (C) the point P has a linear velocity Rwi ˆ - Rwkˆ 4 4 æ ö (D) the point P has a linear velocity ç - Rwiˆ+ Rwkˆ è 4 ø 4 Ans. (AB) Velocity of the centre = v = rw, ˆ Rw sin 0 ˆ Rw v cos 0 ˆ P = Rwi - i + k v= Rwiˆ Rw/ 0 Rw ˆ ˆ Rwk vp = i Two solid cylinders P and Q of same mass and same radius start rolling down a fixed inclined plane from the same height at the same time. Cylinder P has most of its mass concentrated near its surface, while Q has most of its mass concentrated near the axis. Which statement(s) is(are) correct? (A) Both cylinders P and Q reach the ground at the same time. (B) Cylinder P has larger acceleration than cylinder Q. (C) Both cylinders reach the ground with same translational kinetic energy. (D) Cylinder Q reaches the ground with larger angular speed. Ans. (D) x P w Rw 0 / 9 q g sinq a = (where k is radius of curvature); k P > k Q k + R Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA-4005 PNE : , Fax : , info@allen.ac.in Website:

12 IIT-JEE 0 EXAMINATIN (ELD N : ) PART - II : CEMISTRY PAPER - SECTIN I : Single Correct Answer Type This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which NLY NE is correct.. The compound that undergoes decarboxylation most readily under mild condition is : (A) C C C (B) C (C) C C (D) C C Ans. (B) It is example of b-keto acid which undergoes decarboxylation most readily.. The shape of Xe F molecule is Ans. (D) (A) Trigonal bipyramidal (C) tetrahedral F Xe F See-saw (B) Square planar (D) see-saw. For a dilute solution containing.5 g of a non-volatile non-electrolyte solute in 00 g of water, the Ans. (A) elevation in boiling point at atm pressure is C. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of g) of the solution is (take K b =0.76 K kg mol ) (A) 74 (B) 740 (C) 76 (D) 78 DT b = K b m...(i) 760 PS 8 = m [for a dilute solution]...(ii) By eq. (i) / (ii) = P 8 S P S = 74 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA-4005 PNE : , Fax : , info@allen.ac.in Website: / 9

13 IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER - 4. NiCl {P(C 5 ) (C 6 5 )} exhibits temperature dependent magnetic behavior (paramagnetic/diamagnetic). The coordination geometries of Ni + in the paramagnetic and diamagnetic states are respectively : (A) tetrahedral and tetrahedral (B) square planar and square planar (C) tetrahedral and square planar (D) square planar and tetrahedral Ans. (C) For tetrahedral compound For square planar compound + 8 Ni = d + 8 Ni = d d xy d yz d zx d x y d 8 d x y d z Two unpaired electron. ence paramagnetic d xy d z d xz d yz No unpaired electron ence dimagnetic 5. The major product of the given reaction sequence is : Ans. (B) C C C C CN (A) C C = C C C (C) C C C C C G 95% S eat 4 (B) C C= C CN C (D) C C= C C N C C C C C CN C C C C CN C C=C CN+ C 95% S4 (acts as dehydrating reagent) / 9 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA-4005 PNE : , Fax : , info@allen.ac.in Website:

14 IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER - 6. The reaction of white phosphorus with aqueous Na gives phosphine along with another phosphorus containing compound. The reaction type ; the oxidation states of phosphorus in phosphine and the other product are respectively (A) redox reaction ; and 5 (B) redox reaction ; + and +5 (C) disproportionation reaction ; and +5 (D) disproportionation reaction ; and + Ans. N PTIN IS CRRECT P 4 + Na (aq.) ¾ ¾ P - + xidation state of P in P = Na P + xidation state of P in Na P = + ence, (No option) is correct 7. Using the data provided, calculate the multiple bond energy (kj mol ) of a C º C bond in C. Ans. (D) That energy is (take the bond energy of a C bond as 50 kj mol.) C(s) + (g) ¾ C (g) D = 5 kj mol C(s) ¾ C(g) D = 40 kj mol (g) ¾ (g) D = 0 kj mol (A) 65 (B) 87 (C) 865 (D) 85 C(s) + (g) ¾ C º C (g) D = 5 kj/mole Þ 5 = D sub C(s) + B.E. B.E. C B.E. CºC 5 = B.E. CºC 5 = (B.E.) CºC B.E. CºC = 85 kj/mole 8. In the cyanide extraction process of silver from argentite ore, the oxidizing and reducing agents used are : Ans. (B) (A) and C respectively. (C) N and Zn dust respectively. (B) and Zn dust respectively. (D) N and C respectively. NaCN Ag S ¾¾¾¾¾ [Ag(CN) in presenceof air ] Ag S + 4NaCN Na [Ag(CN) ] + Na S Na S + + ¾ ¾ Na S 4 + S + Na ence, oxidising agent is. [Ag(CN )] + Zn ¾ ¾ Ag + [Zn (CN) 4 ] ence, reducing agent is 'Zn'. Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA-4005 PNE : , Fax : , info@allen.ac.in Website: / 9

15 IIT-JEE 0 EXAMINATIN (ELD N : ) SECTIN-II : Paragraph Type PAPER - This section contains 6 multiple choice questions relating to three paragraphs with two questions on each paragraph. Each question has four choices (A), (B), (C) and (D), out of which NLY NE is correct. Paragraph for Question 9 and 0 Bleaching powder and bleach solution are produced on a large scale and used in several house-hold products. The effectiveness of bleach solution is often measured by iodometry. 9. Bleaching powder contains a salt of an oxoacid as one of its components. The anhydride of that oxoacid is : Ans. (A) (A) Cl (B) Cl 7 (C) Cl (D) Cl 6 Bleaching powder is Ca(Cl)Cl. So, the associated oxo acid is Cl. ence, the anhydride of this acid is Cl. (Cl ¾ ¾ ¾ - Cl ) 0. 5 ml of household bleach solution was mixed with 0 ml of 0.50 M KI and 0 ml of 4 N acetic Ans. (C) acid. In the titration of the liberated iodine, 48 ml of 0.5 N Na S was used to reach the end point. The molarity of the household bleach solution is (A) 0.48 M (B) 0.96 M (C) 0.4 M (D) 0.04 M Eq. of CaCl = Eq. of Na S 5 M = [ + + CaCl + e ¾ Ca Cl ] (nf = ) M = 0.4 Paragraph for Question and In the following reaction sequence, the compound J is an intermediate. (C C) I C CNa J (i), Pd/C (ii) SCl (iii) anhyd. AlCl J (C 9 8 ) gives effervescence on treatment with NaC and a positive Baeyer s test.. The compound K is : K (A) (B) (C) (D) Ans. (C) 4 / 9 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA-4005 PNE : , Fax : , info@allen.ac.in Website:

16 . The compound I is IIT-JEE 0 EXAMINATIN (ELD N : ) C PAPER - (A) (B) (C) (D) Ans. (A) C=C C I (C C) CCNa J (C) 9 8 J gives effervescence on treatment with NaC due to presence of C group and positive Baeyer's test due to presence of unsaturation. C=C C C C C (Pd) SCl C Cl AlCl K (Intramoleculer Fridel craft reaction) Paragraph for Question and 4 The electrochemical cell shown below is a concentration cell. M M + (saturated solution of a sparingly soluble salt, MX ) M + (0.00 mol dm ) M The emf of the cell depends on the difference in concentrations of M + ions at the two electrodes. The emf of the cell at 98 K is 0.059V.. The value of DG (kj mol ) for the given cell is (take If = C mol ) (A) 5.7 (B) 5.7 (C).4 (D).4. Ans. (D) Q DG = nfe DG = Joule =.4 kj/mole 4. The solubility product (K sp ; mol dm 9 ) of MX at 98 K based on the information available for the given concentration cell is (take.0 R 98/F = V) (A) 0 5 (B) (C) 0 (D) 0 Ans. (B) = [M ] log [M ] - = [M ++ ] = 0 5 = s K sp = 4s = 4 (0 5 ) = Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA-4005 PNE : , Fax : , info@allen.ac.in Website: 5 / 9

17 IIT-JEE 0 EXAMINATIN (ELD N : ) SECTIN-III : Multiple Correct Answser(s) Type PAPER - This section contains 5 multiple choice question. Each question has four choices (A), (B), (C) and (D) out of which NE or MRE are correct. 5. The given graphs / data I, II, III and IV represent general trends observed for different physisorption and chemisorption processes under mild conditions of temperature and pressure. Which of the following choice(s) about I, II, III and IV is (are) correct? Amount of gas adsorbed Amount of gas adsorbed T P (I) (III) P constant 00K 50K (A) I is physisorption and II is chemisorption (C) IV is chemisorption and II is chemisorption Ans. (A,C) Amount of gas absorbed Potential energy (II) T (IV) P constant 0 Distance of molecule from the surface D uds=50kjmol (B) I is physisorption and III is chemisorption (D) IV is chemisorption and III is chemisorption 6. The reversible expansion of an ideal gas under adiabatic and isothermal conditions is shown in the figure. Which of the following statement(s) is (are) correct? (P,V,T) P adiabatic isothermal (P,V,T) (P,V,T) V (A) T = T (B) T > T (C) w isothermal > w adiabatic (D) DU isothermal > DU adiabatic 6 / 9 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA-4005 PNE : , Fax : , info@allen.ac.in Website:

18 Ans. (A,D) IIT-JEE 0 EXAMINATIN (ELD N : ) (A) Since process is isothermal. (B) In adiabatic expansion temperature decreases. (C) Although W isothermal,exp > W adiabatic,exp Since expansion work is ve, so option is wrong. (D) Q DU isothermal = 0 & DU adiabatic < 0 Þ DU isothermal > DU adiabatic 7. For the given aqueous reactions, which of the statement(s) is (are) true? dilute S4 excess KI + K [Fe(CN) 6] brownish-yellow solution PAPER - ZnS 4 white precipitate + brownish-yellow filtrate NaS colourless solution (A) The first reaction is a redox reaction. (B) White precipitate is Zn [Fe(CN) 6 ]. (C) Addition of filtrate to starch solution gives blue colour. (D) White precipitate is soluble in Na solution. Ans. (A,C,D) I + [Fe(CN) 6 ] ¾ ¾ [Fe(CN) 6 ] 4 + I (Redox reaction) I + I ¾ ¾ I (Brownish yellow filtrate or solution). [Fe(CN) 6 ] 4 + Zn ¾ ¾ Zn [Fe(CN) 6] or K Zn [Fe(CN) 6] and I + S ¾ ¾ I + S 4 6 white ppt. (soluble in Na) I I + I ; starch + I ¾ ¾ deep blue solution. ence, answers are A,C, D 8. With reference to the scheme given, which of the given statement(s) about T,U,V and W is (are) correct? C T LiAl 4 (excess) Å Cr / (CC) V U W (A) T is soluble in hot aqueous Na (B) U is optically active (C) Molecular formula of W is C (D) V gives effervescence on treatment with aqueous NaC Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA-4005 PNE : , Fax : , info@allen.ac.in Website: 7 / 9

19 Ans. (A,C,D) IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER - C (T) hot aq. Na C Na + (Soluble in aq. Na) C LiAl 4 C Cr / + C (U) optically (inactive) (No Chiral centre) C C C (V) (due to presence of C group it gives effervescence with aq. NaC ) excess (C C) C C C C C C C Moleculor formula C Which of the given statement(s) about N,,P and Q with respect to M is (are) correct? Cl C Cl C Cl C M N Cl C C Cl P Cl Q (A) M and N are non-mirror image stereoisomers (B) M and are identical (C) M and P are enantiomers (D) M and Q are identical 8 / 9 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA-4005 PNE : , Fax : , info@allen.ac.in Website:

20 Ans. (A,B,C) IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER - (M) C Cl Cl C (N) () (P) (Q) Cl C C Cl C Cl C Cl Cl Cl Cl C C C Cl C MN M MP MQ Diastereomers (non mirror image stereo isomers) Identical Enantiomer Diastereomer 40. With respect to graphite and diamond, which of the statement(s) given below is (are) correct? Ans. (B,D) (A) Graphite is harder than diamond. (B) Graphite has higher electrical conductivity than diamond. (C) Graphite has higher thermal conductivity than diamond. (D) Graphite has higher C C bond order than diamond. Due to resonance the bond order is greater than in case of graphite. Diamond has higher thermal conductivity than graphite. Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA-4005 PNE : , Fax : , info@allen.ac.in Website: 9 / 9

21 IIT-JEE 0 EXAMINATIN (ELD N : ) PART - III : MATEMATICS SECTIN I : Single Correct Answer Type PAPER - This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which NLY NE is correct. 4. Four fair dice D, D, D and D 4, each having six faces numbered,,, 4, 5 and 6 are rolled simultaneously. The probability that D 4 shows a number appearing on one of D, D and D is - (A) 9 6 Ans. (A) 6 C = (B) 08 6 (C) 5 6 (D) If P is a matrix such that P T = P + I, where P T is the transpose of P and I is the identity matrix, éxù é0ù then there exists a column matrix X= ê y ú ¹ ê 0 ú such that ê ú ê ú êëzúû êë0úû (A) é0ù PX= ê 0 ú ê ú êë0 úû Ans. (D) P T = P + I Þ P = P T + I Þ P = (P + I) + I Þ P = 4P + I Þ P = I Þ PX = X 4. Let a(a) and b(a) be the roots of the equation ( ) ( ) ( ) 6 (B) PX = X (C) PX = X (D) PX = X + a - x + + a - x+ + a - = 0 where a >. Then lim a (a) and + lim b (a) are + a 0 a 0 (A) Ans. (B) 5 - and (B) - and (C) ææ aö ö ææ aö ö æ a ö ç ç ç ç ç 6 èè ø ø èè ø ø è ø æx x ö aç + + = 0 Þ x + x + = 0 è 6ø Þ x =-,- Þ lim a (a) a 0 + and lim b ( a) + are a x x+ + - = and (D) - and 9 - and 0 / 9 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA-4005 PNE : , Fax : , info@allen.ac.in Website:

22 IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER The equation of a plane passing through the line of intersection of the planes x + y + z = and x y + z = and at a distance from the point (,, ) is (A) 5x y + z = 7 (B) x+ y = - (C) x + y + z = (D) x- y = - Ans. (A) Let required plane be (x + y + z ) + l(x y + z ) = 0 Q plane is at a distance Þ l( ) ( +l ) + ( -l ) + ( +l) from the point (,, ). ( ) = ( +l ) + (-l) + ( +l) Þ l = Þ l = l + l 4l + 6l Þ l=- æ 7 ö \ required plane is (x + y + z ) + ç - ( x- y+ z- ) = 0 è ø Þ 5x y + z = Let a, a, a,... be in harmonic progression with a = 5 and a 0 = 5. The least positive integer n for which a n < 0 is (A) (B) (C) 4 (D) 5 Ans. (D) a,a,a...be in.p Þ,,... be in A.P. a a a in A.P. T = = and T0 = = a 5 a0 5 \ T 0 = T + 9d 4 = + 9d Þ d = T n = T + (n )d < 0 Þ (n - - ).4 < Þ < n 4 Þ Þ 99 n 4 < Þ least positive integer n is 5. 4(n -) < Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA-4005 PNE : , Fax : , info@allen.ac.in Website: / 9

23 IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER If a r and b r r r r are vectors such that a + b = 9 and a (i ˆ+ j ˆ+ 4k) ˆ = (i ˆ+ j ˆ+ 4k) ˆ r b, then a possible r r value of (a + b).( - 7i ˆ+ j ˆ+ k) ˆ is (A) 0 (B) (C) 4 (D) 8 Ans. (C) r r (a + b) (i ˆ+ j ˆ+ 4k) ˆ = 0 Þ r r a + b =l (i ˆ+ j ˆ+ 4k) ˆ r r a + b = 9 Þ l = r r a + b = (i ˆ+ j ˆ+ 4k) ˆ r r (a + b).( 7i ˆ+ ˆj + k) ˆ = = The value of the integral (A) 0 Ans. (B) p/ p/ ò x cos x dx + p/ p/ = (B) p/ p / ò x cos x dx = ò I p/ 0 p+ x p ò / æ ö ç x + ln cos xdx p-x -p/è ø is p - (C) 4 æp+ xö ò l nç p-x cosxdx è ø x cos x dx æ p/ ö p/ = (x sin x) 0 - xsin xdx ç ò è 0 ø = æ p/ p æ öö p/ ç - (x cos x) 0 + cos x dx 4 ç ò è è øø p/ æp ö = ç - cos x dx 4 ò è ø æp ö p = ç - = -4 è 4 ø 0 0 II p p + 4 (D) 48. Let PQR be a triangle of area D with a =, b = 7 and 5 c =, where a, b and c are the lengths of the sides of the triangle opposite to the angles at P, Q and R respectively. Then sin P - sin P equals sinp+ sinp (A) 4D (B) 45 4D æ ö (C) ç è4d ø (D) æ 45 ö ç è4d ø / 9 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA-4005 PNE : , Fax : , info@allen.ac.in Website:

24 Ans. (C) IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER - sinp-sinpcosp sinp+ sinpcosp = (- cos P) ( + cosp) = tan A = D s (s- a) = ( (s -b) (s -c) ) D s = 4 = ææ 7öæ 5öö 4-4- ç ç è ç øè ø ç è D ø = æ ö ç è4dø SECTIN II : Paragraph Type This section contains 6 multiple choice questions relating to three paragraphs with two questions on each paragraph. Each question has four choices (A), (B), (C) and (D), out of which NLY NE is correct. Paragraph for Question 49 and 50 Let a n denotes the number of all n-digit positive integers formed by the digits 0, or both such that no consecutive digits in them are 0. Let b n = the number of such n-digit integers ending with digit and c n = the number of such n-digit integers ending with digit The value of b 6 is (A) 7 (B) 8 (C) 9 (D) 50. Which of the following is correct? (A) a 7 = a 6 + a 5 (B) c 7 ¹ c 6 + c 5 (C) b 7 ¹ b 6 + c 6 (D) a 7 = c 7 + b 6 Solution for Q.49 & Q.50 For a n The first digit should be For b n ( n- Places) Last digit is. so b n is equal to number of ways of a n (i.e. remaining (n ) places) b n = a n For c n Last digit is 0 so second last digit must be So c n = a n b n + c n = a n So a n = a n + a n Similarly b n = b n + b n 49. Ans.(B) a =, a = So a =, a 4 = 5 a 5 = 8 Þ b 6 = a 5 = 8 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA-4005 PNE : , Fax : , info@allen.ac.in Website: / 9

25 50. Ans.(A) a n = a n + a n put n = 7 a 7 = a 6 + a 5 c n = c n + c n So put n = 7 c 7 = c 6 + c 5 IIT-JEE 0 EXAMINATIN (ELD N : ) (A) is correct (B) is incorrect PAPER - b n = b n + b n put n = 7 b 7 = b 6 + b 5 a 7 = a 6 + a 5 while (D) says a 7 = a 5 + a 5 (C) is incorrect (D) is incorrect Paragraph for Question 5 and 5 A tangent PT is drawn to the circle x + y = 4 at the point P(, ). A straight line L, perpendicular to PT is a tangent to the circle (x ) + y =. 5. A common tangent of the two circles is Ans. (D) (A) x = 4 (B) y = (C) x+ y= 4 (D) x+ y = 6-0 h = = 6 - equation of tangents from (6, 0) : y 0 = m(x 6) Þ y mx + 6m = 0 use p = r 6m + m = k= (0,0) (,0) (,0) (h,0) Þ 6m = 4 + 4m m = 4 m = /8 Þ m =± at m =- equation of tangent will be x+ y = A possible equation of L is (A) x- y = (B) x+ y = (C) x- y =- (D) x+ y = 5 4 / 9 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA-4005 PNE : , Fax : , info@allen.ac.in Website:

26 Ans. (A) IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER - Equation of tangent at P will be x+ y = 4 Slope of line L will be Let equation of L be : x y= + c Þ x- y+ c= 0 Now this L is tangent to nd circle So + c =± Þ using c =- or c =- c =- x y= - Þ x- y =. ence (A) 5 Paragraph for Question 5 and 54 Let ƒ(x) = ( x) sin x + x for all x Î IR, and let 5. Consider the statements : P : There exists some x Î IR such that ƒ(x) + x = ( + x ) Q : There exists some x Î IR such that ƒ(x) + = x( + x) Then (A) both P and Q are true (C) P is false and Q is true Ans (C) ƒ(x) = ( x) sin x + x P : ƒ(x) + x = ( + x ) Þ ( x) sin x + x + x = + x Þ ( x) sin x x + x = 0 ( - ) ( x) cos x + = 0 which is not possible. \ P is false. Q : ƒ(x) + = x( + x) x + ( x) sin x + = x + x ( x) sin x x + = 0. Let h(x) = ( x) sin x x +, clearly h() = and h(x) = (x x + )sin x x + é æ ö ù = x êç - +.sin x x - + x x x ú ë è ø û \ h(x) as x. \ By intermediate value theorem h(x) = 0 has a root which is greater than. ence Q is true. x æ t ö g(x) = ç -l nt ƒ(t)dt for all x Î (, ). òè t+ ø (B) P is true and Q is false (D) both P and Q are false Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA-4005 PNE : , Fax : , info@allen.ac.in Website: 5 / 9

27 IIT-JEE 0 EXAMINATIN (ELD N : ) 54. Which of the following is true? (A) g is increasing on (, ) (B) g is decreasing on (, ) (C) g is increasing on (,) and decreasing on (, ) (D) g is decreasing on (,) and increasing on (, ) Ans. (B) g( x) x ( - ) ( t+ ) æ t ö = ç -lnt ƒ(t)dt òè ø PAPER - æ(x -) ö g' ( x) = ç -lnx ƒ(x) è x+ ø ƒ(x) > 0 " xîr Suppose. (x -) h(x) = -lnx x+ æ 4 ö h(x) = - ç + lnx èx+ ø 4 h'(x) = (x + ) - x ( ) h'x ( x-) ( + ) =- x x h'(x) < 0 So h(x) is decreasing so h(x) < h(). " x > h(x) < 0 " x > So g'(x) = h(x) ƒ(x) g'(x) < 0 " x > g(x) is decreasing in (, ). SECTIN III : Multiple Correct Answer(s) Type This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which NE or MRE are correct. x ƒ(x) e t t dt ò for all x Î(0, ), then - t 55. If = ( - )( - ) 0 (A) ƒ has a local maximum at x = (B) ƒ is decreasing on (,) (C) there exists some c Î (0, ) such that ƒ"(c) = 0 (D) ƒ has a local minimum at x = 6 / 9 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA-4005 PNE : , Fax : , info@allen.ac.in Website:

28 Ans. (A,B,C,D) ƒ(x) = x ò 0 t e (t -)(t -)dt IIT-JEE 0 EXAMINATIN (ELD N : ) + + PAPER - x Þ ƒ'(x) = e (x -)(x -) \ ƒ'() = ƒ'() = 0 Þ ƒ''(c) = 0 for same c Î (, ) (by Rolle's theorem) 56. For every integer n, let a n and b n be real numbers. Let function ƒ : IR IR be given by [ ] ( ) ì ïan + sin px, for xî n,n + ƒ(x) = í, for all integers n. ïîb n + cos px, for x Î n -,n If ƒ is continuous, then which of the following holds(s) for all n? (A) a n b n = 0 (B) a n b n = (C) a n b n+ = (D) a n b n = Ans. (B,D) For ƒ to be continuous : ƒ(n ) = ƒ(n + ). Þ Þ b n + cosnp = a n + sinnp b n + = a n Þ a n b n = (\ B is correct) Also ébn + cos px (n -, n) ê an + sin p x [n, n + ] ƒ(x) = ê ê bn+ + cos p x (n +, n + ) ê êëa n+ sin p x [n +, n + ] Again ƒ((n + ) ) = ƒ((n + ) + ) Þ a n = b n+ Þ a n b n+ = Þ a n b n = (\ D is correct) 57. If the straight lines x - y + = = z and x + y + = = z are coplanar, then the plane(s) containing k 5 k these two lines is(are) (A) y + z = (B) y + z = (C) y z = (D) y z = Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA-4005 PNE : , Fax : , info@allen.ac.in Website: 7 / 9

29 Ans. (B,C) (,, 0); (,, 0) For coplanarity of lines IIT-JEE 0 EXAMINATIN (ELD N : ) PAPER k = 0 Þ (k 4) = 0 5 k Þ k = ± for k = r Normal vector n = ˆj-kˆ \ Required plane : y z = l Q Passes through (,, 0) Þ l = \ y z = for k = r n = ˆj+ kˆ \ Required plane : y + z = l Q Passes through (,, 0) Þ l = \ y + z = é 4 4ù ê 7 ú 58. If the adjoint of a matrix P is ê ú, then the possible value(s) of the determinant of P is (are)- êë úû (A) (B) (C) (D) Ans. (A,D) adjp = Þ P = 4 Þ P = ± æ pö æp pö 59. Let ƒ : (,) IR be such that ƒ(cos4 q ) = for qîç 0, Èç,. Then the value(s) of - sec q è 4ø è4 ø æ ö ƒç è ø is (are)- (A) - (B) + (C) - (D) + 8 / 9 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA-4005 PNE : , Fax : , info@allen.ac.in Website:

30 Ans. (A,B) Q æ 0, p ö æ p, p ö qîç Èç è 4ø è4 ø æ pö æp ö Þ qîç 0, Èç, p è ø è ø IIT-JEE 0 EXAMINATIN (ELD N : ) + cosq Now ƒ(cos 4 q ) = = = + -sec q cos q cos q Let cos4q= Þ Þ Þ cos q- = cosq=± From (i) æö ƒç = ± èø Þ (A, B) are correct...(i) 60. Let X and Y be two events such that P( X Y ) =,P(Y X) = and P( X Y) following is(are) correct? (A) P( X Y) È = (B) X and Y are independent (C) X and Y are not independent (D) P( X Y) Ans. (A,B) P(X Ç Y) = P(X), P(Y/X) Þ P(X) = Also P(X Ç Y) = P(Y).P(X/Y) Þ P(Y) = Þ P(X Ç Y) = P(X).P(Y) Þ X,Y are independent P(X È Y) = P(X) + P(Y) P(X Ç Y) = + - = 6 P(X C Ç Y) = P(Y) P(X Ç Y) = - = 6 6 Þ (A, B) are correct Ç = C PAPER - Ç =. Which of the 6 Corporate ffice : CAREER INSTITUTE, SANKALP, CP-6, INDRA VIAR, KTA-4005 PNE : , Fax : , info@allen.ac.in Website: 9 / 9