FIITJEE ALL INDIA TEST SERIES

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1 FIITJEE LL INDI TEST SERIES PRT TEST II JEE (Main)-8-9 TEST DTE: --8 Time llotted: Hours Maximum Marks: 6 General Instructions: The test consists of total 9 questions. Each subject (PM) has questions. This question paper contains Three Parts. Part-I is Physics, Part-II is hemistry and Part-III is Mathematics. Each part has only one section: Section-. Section- (, 6, 6 9) contains 9 multiple choice questions which have only one correct answer. Each question carries +4 marks for correct answer and mark for wrong answer. FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

2 ITS-PT-II-PM-JEE(Main)/9 Physics PRT I SETIN (ne ptions orrect Type) This section contains multiple choice questions. Each question has four choices,, and, out of which NLY NE option is correct.. n array of point charges and a set of closed Gaussian surfaces S, S, and S are illustrated in the figure. hoose the correct statement: S S +4 + S Flux passing through S is Flux passing through S is Flux passing through S is Flux passing through S is 4 Gauss Law q flux = enclosed. Two electrons enter a region at same point between charged capacitor plates with equal speed v. Electron is moving horizontally to the left while electron s velocity is directed at below the horizontal. Each electron reaches left plate. Which one of the following choice best compares the speeds of electron upon arrival at the left plate? Ignore mutual interaction between electrons, gravity, any fringing effect of the fields and relativistic effects. ssume electric field between plates does not change due to electrons. (v : speed of electron when it reaches left plate. v : speed of electron when it reaches left plate.) +Q v v d Q v v v v FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

3 ITS-PT-II-PM-JEE(Main)/9 v v The answer depends on magnitude of the charge, Q on each plate. Using conservation of mechanical energy for the particle field system as the particle cross the region between the plates, we write KE + PE = where PE = qv. Since the potential difference between the plates will be the same for each charge, the change in PE of the charge field system is the same in each case, meaning the change in KE is the same for each charge. Hence, each charge reaches the other side with the same speed.. Five identical light bulbs are connected in a circuit as shown. ll wires are ideal with no resistance, and the ideal battery has emf V. When the switch S in the circuit is closed, aside from bulb 5, which of the other bulbs glow brighter than before. V 4 5 S nly ulb 4 nly ulb and nly ulb and 4 nly ulb,, and 4 The equivalent circuit before and after the switch is closed for the resistors is shown in the figure. In words, by closing the switch the resistance of the entire circuit goes down since the resistance of the bottom branch drops from R to (/)R. Since there is less resistance in the circuit, there is more current, meaning that there is more current through bulb from Kirchhoff s loop rule with the battery and bulb. ulb is dimmer. Finally, since the resistance of the bottom branch decreased, it now gets a higher amount of current. V V FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

4 ITS-PT-II-PM-JEE(Main)/ n infinite long cylindrical shell of radius 5 cm has a charge of n/m per unit axial length on its surface (uniformly distributed), as shown in the figure. hoose the INRRET statement. cm cm cm 5 cm D The electric field at point at distance of cm from axis of cylinder is N/ The electric field at point at distance of cm from axis of cylinder is 54 N/ The electric field at point at distance of cm from axis of cylinder is 54 N/ The electric field at point D at distance of 5 cm from axis of cylinder is 8 N/ D E N / r r r 5. Eight identical charged particles are located on a circle as shown in the figure-(). Which vector shown in figure-(), best represents the direction of net force acting on the charged particle-4 due to other charged particles. y y x D 45 x Figure Figure D FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

5 5 ITS-PT-II-PM-JEE(Main)/9 Since all of the charges around the circle are equal, the forces between them are repulsive. Particle and 5 are at same distance away from particle-4 and repel the particle-4 with the same force in magnitude and resulting force will be along. Likewise for particles and 6 and particle and 7. The force of particle-8 on particle-4 will be along. Hence net force on particle-4 will be along. 6. In the circuit shown, the switch S has been left open for a very long time. ll circuit elements are considered to be ideal. Which one of the following statements best describes the behaviour of the current through the switch S once it is closed? 9 V k S k The current initially is m and decreases to a steady m The current initially is m and increases to a steady m The current initially is 9 m and decreases to a steady m The current initially is 6 m and decreases to a steady m Using Kirchoff loop Rule around the outside of the circuit reveals that the potential difference across the capacitor is 9V since there is no current after a long time. nce the switch is closed, there is a loop that encloses the battery and k resistor resulting in a total of 9 = I I = m through the resistor. For the other branch on the right, there is a capacitor initially charged and a k resistor. 7. n LR series circuit is connected to a sinusoidal voltage of variable angular frequency and the graph between peak current and angular frequency is plotted as shown in the figure. If capacitance of capacitor used is 6 F, then choose the correct alternative. i 6F L R i / ~ 8 (rad/s) L = 6.67 H, R =. k L =.4 H, R = k L = 6.67 H, R =.67 k FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

6 ITS-PT-II-PM-JEE(Main)/9 6 can t be calculated 5 Q 4 L Q 5 R R R R k L 6.67H 4 L ll rods have identical length and cross-sectional area. Find the temperature at junction. The thermal conductivity of E D k 75 rod-f, rod-gh rod-l & rod-d are k and the thermal conductivity of rod-ef, rod-d, rod- & rod-g are k. F k k/ k k/ k k/ k/ H 5 G L Zero 5 x x 75 x 5 x 5 R R R R 4x 5 5 x 4 FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

7 7 ITS-PT-II-PM-JEE(Main)/9 9. n L R circuit is made so that the - input (angular frequency : ) is applied across the combination of L and R. The output is taken across R. The circuit works as a filter circuit. If the time constant of the D (L R) circuit is, then Vout the ratio: equals V in (where V out and V in are peak values of output and input voltage respectively.) V in L R V out The phasor diagram of the circuit, in operation, is shown in the adjacent figure Z R x, where XL The ratio L V V out in R R X L R L Vin = iz ir = V out ix L. solenoid of cross sectional area, N turns and length carries a constant current I. n iron rod of constant permeability and cross sectional area is partially inserted in the solenoid along its axis. alculate the force acting on the iron rod. (Neglect the end effects and assume that when the iron rod is moved slightly through a distance x from its position, the field structure remains the same.) x x N I N I FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

8 ITS-PT-II-PM-JEE(Main)/9 8 4 N I N I The only difference is that a length x of the rod is effectively transferred from the extreme right hand end of the rod (outside the field region) to the uniform field region within the solenoid. The difference in the magnetic energy of the two configurations is U Ux x Ux H dv, (i) v Where V x and H = NI/ = the magnetic field intensity inside the solenoid. Since H is constant, we obtain from (i) N I U x. The force on the rod is U N I Fx x I. conducting rod of length rotates about its one end with angular velocity. Find the potential difference between and (e : charge on electron, m : mass of electron). m m m m e e e e W V q dv = m xdx/e mw xdx dv e d x x FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

9 9 ITS-PT-II-PM-JEE(Main)/9 m dv e m V e V xdx m e. In the circuit shown, find the value of and L if net current from the source is to lead net voltage by a time 4 seconds. 5 mf L H ~ V, 5/ Hz.5 mf, H 5 mf, H 5 mf, H.5 mf, H Phasor diagram : To lead by s 4 4, I I I I I I R V XL V (X X ) R V X I I I L ~ /4 /4 I V I FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

10 ITS-PT-II-PM-JEE(Main)/9. metallic charged ball, having charge q is immersed into the liquid of dielectric constant K and low resistivity. Find the time constant of disappearance of charge on the ball K K K K s we know that E q J 4 Kr I s J 4 r 4 Kr K I q q K K I q q 4. n external force is applied to move a square loop of dimension 5 cm 5 cm and resistance at a constant speed across a region of uniform magnetic field =. Tesla. The side of the square loop makes an angle = 45 with the boundary of the field region, as shown in figure. t t =, the loop is completely inside the field region, with its right vertex at the boundary. Total charge flown through loop when loop is completely outside the magnetic region. l v.5 milli coulomb. milli coulomb.4 milli coulomb milli coulomb a Q R R 5. Two bar magnets having same geometry with magnetic moments.5 M and.5 M are placed in such a way that their similar poles are on the same side, and its time period of oscillations is T. Now if the polarity of one of the magnets is reversed, keeping other quantities same, then the time period of oscillation is T. hoose the correct option. FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

11 ITS-PT-II-PM-JEE(Main)/9 T T T T T T 4 T 4 T T M.5M.5M T M T M.5M.5M 6. uniform charged (thin) non-conducting rod is located on the central axis a distance b from the center of an uniformly charged non-conducting disk. The length of the rod is L and has a linear charge density. The disk has radius a and a surface charge density. The electrostatic force on the rod due to disc is z z=b+l z=b a F L a b b L a k ˆ F 4 b a L k ˆ F 8 b a L k ˆ L F kˆ L a The electric field created at any point along the central axis is given by z Ez kˆ a z FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

12 ITS-PT-II-PM-JEE(Main)/9 reaking up the rod into an infinite number of infinitesimally small point charges dq, we have that the net force on each tiny charge is df dqe z. Summing up all these contributions, and using the fact that dq = dz gives bl z F dq kˆ z ˆk dz a z b a z L a b L a b k ˆ 7. cycle followed by an engine (using one mole of an ideal gas in a cylinder with a piston) is shown in figure. Heat exchanged by the engine, with the surroundings for each section of the cycle is ( v = (/) R). : constant volume : constant pressure D : adiabatic D : constant pressure P =P P =P D P D V =V V V D V 5 Q V P P, Q / P V V, QD Q V P P, Q 5 / P V V, QD Q V P P, Q 5 / P V V 5, QD Q V P P, Q 5 / P V V, QD, Q 5 / P V V D D, Q 5 / P V V D D, Q / P V V D D, Q 5 / P V V D D Q U W Q U RT T V P P / P V V P V V 5 / P V V Q Q 5 / P V V D D FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

13 ITS-PT-II-PM-JEE(Main)/9 8. The figure shows the variation of average power versus angular frequency of source voltage of two series. LR-ircuits connected to same voltage source (the circuit- has resistance R, capacitance, inductance L and quality factor Q while circuit- has resistance R, capacitance, inductance L and quality factor Q respectively. hoose the correct option: P av ircuit- ircuit- Q Q and R R, LL Q Q and R R, L L Q Q and R R, L L Q Q and R R, L / L / Quality Factor = Q L Q Q Q R V Pav R max R R L L 9. The figure shows a non-conducting (thin) disk with a hole. The radius of the disk is b and the radius of the hole is a. total charge Q is uniformly distributed on its surface. ssuming that the electric potential at infinity is zero, what is the electric potential at the center of the disk? a b Q b a Q b a Q b a The potential produced by a charged disk of radius R, at a distance z from it, along its central axis was FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

14 ITS-PT-II-PM-JEE(Main)/9 4 z R z () y superposition, we can think of the potential created by disk with the hole as the sum of two disks, with the same but opposite surface densities: V z z b z z a z z b z a Since we are only interested at the center, we have V b a Finally The total area of the disk with the hole is b Q a Q V b a b a. hoose the INRRET Statement kq b a b ab a b a, thus kq b a The paramagnetic material displays greater magnetization when cooled, for the same magnetizing field. The diamagnetic property of material is almost independent temperature. If a toroid uses bismuth for its core, the magnetic field in the core will be slightly greater than when core is empty. The permeability of a ferromagnetic material depends on the magnetic field. n cooling, the tendency of thermal agitations to disrupt the alignment of magnetic dipoles decreases in case of paramagnetic materials. These they display greater amount of magnetism. on placing a sample of diamagnetic material in magnetic field, the magnetization (Induced dipole moment) is opposite direction of magnetizing field. Thus it does not affected by temperature. The field in core will be slightly less than when core is empty as ismuth is a diamagnetic material ( = r ni). FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

15 5 ITS-PT-II-PM-JEE(Main)/9, so permeability of ferromagnetic H material is dependent on the applied magnetic field. s shown in graph for & H, is larger for smaller value of H, thus permeability is greater for lower fields. c b H f a d e. If gm water at is mixed with gm ice at. Let in final state system consists of m gm ice and m gm water. hoose the correct option. Specific heat capacity of ice is.5 cal/g/. Specific heat capacity of water is cal/g/. Latent heat of fusion of water is a 8 cal/g. Latent heat of vaporisation of water is a 54 cal/g. m = gm, m = gm, = 5 m = 87.5 gm, m =.5 gm, = m =.5 gm, m = 87.5 gm, = m = gm, m = gm, = ll ice will not melt. It means in final state, the system will contain both ice and water at. Let x gm of ice melts due excess heat. x.5gm 8 Water in final state =.5 gm ice in final state = 87.5 gm Water gm Water at Q = = cal Ice Ice at Water at = / = cal = 8 = 8 cal Second method : Let whole system is at in the state of water so. Heat constant = Q H H = 7 cal It means the heat available will force the water to freeze to compensate the 7 cal (negative heat). Let x gm of water to fulfil the condition so 7 x = 87.5 gm 8 Water =.5 gm FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

16 ITS-PT-II-PM-JEE(Main)/9 6. conducting rod of mass = kg, length = m suspended in uniform magnetic field Tk ˆ. The rod is attached with two different string as shown in the figure. Tension in each string is N. Find the value of current in the wire. (g = m/s ). Non-conducting string ˆ ˆ k k mp 5 mp.5 mp.5 mp T sin i mg T i mg i i i = 5. Non-conducting string ˆ ˆ k k T i mg T. n ideal mutual inductor is made from a primary coil of inductance 5mH and a secondary coil of inductance mh. Find the value of the Mutual inductance ( pproximately). ssume no flux leakage. 8 mh mh 4 mh 7 mh D M Lp Ls 5 5 H = 7 mh FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

17 7 ITS-PT-II-PM-JEE(Main)/9 4. non-conducting sphere has mass m = 88 g and radius r =. cm. flat compact coil of wire with 5 turns is wrapped tightly around it, with each turn concentric with the sphere, as shown. The sphere is placed on a rough inclined plane that slopes downward to the left, making an angle θ with the horizontal, so that the coil is parallel to the inclined plane. uniform magnetic field of.5 T vertically upward exists in the region of the sphere. Take ; g m / s. 7 The current in the coil, that will enable the sphere to rest in equilibrium on the inclined plane is.5. The current in the coil, that will enable the sphere to rest in equilibrium on the inclined plane is.5. The current in the coil, that will enable the sphere to rest in equilibrium on the inclined plane is.8. The current in the coil, that will enable the sphere to rest in equilibrium on the inclined plane is.6. The diagram shows a free-body diagram for the sphere on the incline. We will use the following notation: f is the force of static friction; is the external magnetic filed given as =.5 T; I is the unknown current in the coil; µ is the magnetic moment of the current in the coil; m=.88 kg is the mass of the sphere; g = m/s is the free fall acceleration; r =. m is the radius of the sphere; and N = 5 is the number of turns in the coil. We will use everywhere three significant digits as given in the problem statement. The sphere is in translational equilibrium, thus fs mg sin.. () The sphere is in rotational equilibrium. If torques are taken about the center of the sphere, the magnetic field produces a clockwise torque of magnitude µ sinθ, and the frictional force a counter-clockwise torque of magnitude f s R. Thus: f s R µsinθ =. () mg I f s From (): fs= Mg sinθ. Substituting this in () and cancelling out sinθ, one obtains mgr mg 88 µ = MgR I.8 N r N r 5 / 7..5 The current must be counter-clockwise as seen from above. The result does not depend upon the angle of inclination θ. FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

18 ITS-PT-II-PM-JEE(Main)/ Two circular loops and have their planes parallel to each other, as shown in figure. Loop has a current flowing in the counterclockwise direction, viewed from above. I If the current in loop decreases with time, the two loops attract each other. If the current in loop increases with time, the two loops attract each other If current in loop decreases current in is clockwise. If current in loop increases, current in will be anticlockwise. pply Lenz s law 6. If magnetic dip in India is found to be 8. ne of the option, that is closer to the magnetic dip in ritain, is ritain is closer to magnetic north. 7. In the system of three charged particles, each charge particle is in equilibrium under their electrostatic forces only. hoose the INRRET option The particle must be collinear. ll the charges cannot have the same magnitude. ll the charges cannot have the same sign. The equilibrium is stable. D asic concept 8. In a moving coil galvanometer the number of turns N =, area of the coil = 4 m and the magnetic field strength =.T. (The resistance per unit length of the wire of coil is ). hoose the INRRET option To increase its voltage sensitivity by 5% we increase number of turns to 6 To increase its voltage sensitivity by 5% we increase area to 9 m FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

19 9 ITS-PT-II-PM-JEE(Main)/9 To increase its voltage sensitivity by 5% we increase magnetic field to.45 T To increase its voltage sensitivity by 5% we change the material of wire such that its specific resistance would be / rd of the specific resistance of the present wire. IN = I N R V IR () N R N where is a constant. a V SV d. dv 9. non conducting solid cylinder of infinite length having uniform charge density and radius of cylinder is 5R. Find the flux passing through the surface D as shown in figure. 8R R D R 6R R R FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

20 ITS-PT-II-PM-JEE(Main)/9 8R R R q in R 5R R 8R. conducting rod of length 5 cm is free to slide on two parallel conducting bars as shown in the figure. In addition, two resistors 5 and are connected across the ends of the bars. There is a uniform magnetic field. tesla pointing inside the plane of paper. Suppose an external agent pulls the bar towards the left at a constant speed 5 ms. R v R Sliding rod urrent through 5 resistor is 5 m urrent through resistor is m Force applied by agent to maintain constant 5 ms velocity is N Power delivered by external agent is 5 watt. v ir. 5 i 5 4 FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

21 ITS-PT-II-PM-JEE(Main)/9 hemistry PRT II SETIN Straight bjective Type This section contains multiple choice questions. Each question has four choices,, and out of which NLY NE is correct.. Hydrolysis of mustard gas give Hl and a diol. Which of the following intermediate will form during the hydrolysis l S H l S S + H S l l S l Fast -l l S + Mustard gas H H S + l H S H H H H S H. H H S 4 Major product formed in the above reaction is: FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

22 ITS-PT-II-PM-JEE(Main)/9 H HS4 H Ring closure H. Number of monochloro structural isomers form when undergoes monchlorination l / h Products Monochlorination FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

23 ITS-PT-II-PM-JEE(Main)/ Which of the following is correct? n achiral molecule which is superimposable on its mirror image cannot exist as two enantiomers. hanging the configuration of a molecule always means that bonds are broken. oth and None of these 5. Which of the following correct for E elimination in chloro cyclohexanes: H l H l Rate of E elimination in is faster than. Rate of E elimination in is faster than. can undergo E elimination only if undergoes ring inversion. oth and are correct. 6. Which of the following molecule showing the correct direction of dipole moment: N H FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

24 ITS-PT-II-PM-JEE(Main)/9 4 N N N l N H 7. Predict the major product formed when the following reaction occurs in dilute base: H/ FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

25 5 ITS-PT-II-PM-JEE(Main)/9 H H 8. Propanol is more volatile as compared to glycerol because of Less extent of hydrogen bonding High molar mass of propanol Hybridization ll of the above 9. The IUP name of the following compound is: yclopentanedioic anhydride Pentanedicarboxylic anhydride Pentanedioic anhydride Dicyclopentane anhydride FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

26 ITS-PT-II-PM-JEE(Main)/ Predict the product of the following reaction: Rg I gi : lkyl iodide arboxylic acid Ester lcohol 4. Which of the following compound will form when acetic anhydride reacts with o-fluoroanisole in presence of dry ll. H H F H F H H F H F FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

27 7 ITS-PT-II-PM-JEE(Main)/9 4. Which of the following species is most stable? N N N N D 4. The hybridization of - and - carbon atoms marked in the following structure of benzyne intermediate? sp and sp sp and sp sp and sp None of these 44. Predict the major product formed in the following reaction: Ni H o H 6 FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

28 ITS-PT-II-PM-JEE(Main)/9 8 No reaction 45. i I /Red P lcohol ii gn ompound does not react with HN and gives a colourless solution with NaH. Degree of the alcohol taken initially None of these 46. Predict the product of the following reaction: H H Lactone nhydride lcohol lkene 47. The correct order of basic strength of the following compound is: N H N H N H 4 N > > > 4 > > > 4 4 > > > > > 4 > D FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

29 9 ITS-PT-II-PM-JEE(Main)/9 48. Predict the product of the following reaction? i PhP HI iiuli iii H H H 49. Which mechanism the following reaction will follows: Sl 5 Pyridine 5 H H H l S S N S N S N i None of the above 5. In the following reaction g H /H Pl H H The compound is: H H H H l H H l l H l FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

30 ITS-PT-II-PM-JEE(Main)/9 5. Which of the following product will form in the following reaction: H H H H H H 5. The possible product of the following reaction: i HMgr H H H ii H H H H H H H H H H 4 D 4 i H, Hl, HH KN ii H ompound formed in the above reaction is: NH l 5. Glycine lanine Threonine Serine 54. The only two pk a values of an amino acid are. and 9.6. The isoelectric point of the amino acid is: 7. FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

31 ITS-PT-II-PM-JEE(Main)/ Which of the following products will form when cellulose undergoes complete hydrolysis: D-fructose D-ribose L-glucose D-glucose D 56. Neoprene is prepared by polymerization of: -hloro-,-butadiene utadiene crylonitrile Styrene 57. n organic compound X gives gas on reaction with NaH. The compound X is: Picric acid H H S H ll of these D FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

32 ITS-PT-II-PM-JEE(Main)/9 58. The degree of unsaturation in the following compound is: Predict the compound X in the following reaction: NH r NaH X Na Nar H NH H N H NH 6. ompound on reductive ozonolysis will give: i ii Zn/H H H FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

33 ITS-PT-II-PM-JEE(Main)/9 FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

34 ITS-PT-II-PM-JEE(Main)/9 4 Mathematics PRT III SETIN Straight bjective Type This section contains multiple choice questions. Each question has 4 choices,, and, out of which only NE is correct 6. Normals having slope m, m, m ; (m < ) are drawn at points P(a, b), Q, R respectively on the curve y 6y 6x + 7 = so as to intersect at point S(9, 6), then a + b is equal to Roots of the equation 4m 7m + = are m, m, m solving m 9 a 4 4, b 8 a =, b = 5 6. The locus of the centroid of triangle PSQ, where PQ is any chord of the parabola y = 8(x + ) subtending right angle at the vertex and S be its focus is also a parabola whose latus rectum is equal to 4 8 FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

35 5 ITS-PT-II-PM-JEE(Main)/9 t t = 4 If (h, k) is the centroid of PSQ, h 4 k Then t t, t t 4 8 k h 4 Latus rectum = 8 S Q(t ) P(t ) 6. hyperbola having foci (4, ) and (4, 5) has x + y 7 = as one of its tangent, then the point of contact of this tangent is 9 5, (, 6) (, 7) (, 5) Image of (4, ) about the tangent x + y 7 = is P(8, ) Equation of line passing through P(8, ) and (4, 5) is y + x 4 = Solving this with tangent, we get point of contact 64. The equation of circle touching the parabola y = x at the point (, ) and having its centre on the line y + x = is (x ) + (y + ) = x y x y (x 4) + (y + 4) = 5 Normal to the parabola y = x at (, ) is 4y x + 4 = Let centre of circle is (, ) which lies on the normal 4 5 Radius is 7 5 FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

36 ITS-PT-II-PM-JEE(Main)/ x y If the ellipse 6 b, b > and the hyperbola value of b is x y intersect orthogonally, then the Foci of the two curves are coincident 66. x y If the line joining the foci of the hyperbola S a b does not subtend a right angle at x y any point on the hyperbola S, then number of integral values of 4e is/are (e is 4a b eccentricity of S = ) 4 ircle having foci of the hyperbola S as the extremities of diameter should not intersect the hyperbola S at real points a + b < 4a b < a e < 7/4 4 < 4e < ne of the sides of a triangle is divided into segments of 4 and 6 units by the point of tangency of the inscribed circle which has radius units, then the largest side of triangle is 4 4 FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

37 7 ITS-PT-II-PM-JEE(Main)/9 D tan, tan 5 tan x 5 cot x = 5 lternate: Use = rs x 4 x Let S n = n ; n N, then 64S n is always less than (n ) 4 (n + ) 4 (n + ) 4 (n + ) 4 64 n 6 n n 6n n line is drawn from ( 4, ) to intersect the curve, then the maximum value of the slope of line is P Q x y at P and Q above x-axis. If P and Q are the roots of the equation (r cos 4) + (r sin ) = 8 r (cos + sin ) 8 cos r + 8 = 8cos P Q 8 cos tan FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

38 ITS-PT-II-PM-JEE(Main)/ If two distinct chords drawn from the point sin, to the circle x + y = sin x + which lies is y, ( is a parameter) are bisected by x-axis, then the exhaustive set in n, n ; n I n, n ; n I 6 6 n, n ; n I 4 4 (n, (n + )); n I sin h sin h h sin h 4sin 4 For distinct real roots D > sin > n, n ; n I 4 4 sin, (h, ) sin, 7. Let be (, 5), a point on the line y = x and point on x-axis such that + + is minimum, then the coordinates of is (5, ), 5, 4, Image of (, 5) about x-axis is P(, 5) and about y = x is Q(5, ) Equation of line PQ is y + x 5 = 5 Putting y =, x FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

39 9 ITS-PT-II-PM-JEE(Main)/9 x y 7. line L intersects the coordinate axes at points and. nother line L 8 perpendicular to L intersects the coordinate axes at and D. The locus of circumcentre of D is 5x 4y = 9 5x 4y = 8 4x 5y = 9 4x 5y = 8 ircumcentre will lie on the perpendicular bisector of 7. The value of 4 cot 4cot 6cot 4 cot is D tan cot 4 tan tan 4 tan 4 tan tan 4 tan 4tan 6 tan 4 tan ; The equation of two sides of a triangle are x + 4y =, x + y 8 =. If the circumcentre is (, ), then the centroid of the triangle is 7, 4, 6, (, 6) FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

40 ITS-PT-II-PM-JEE(Main)/9 4 Foot of perpendicular from (, ) on the line x + y 8 = is (, 4) Hence, P is (, 8) which is orthocentre 4 entroid is, P(, 8) x + y 8 = Q( 4, 6) (, ) x + 4y = R(4, ) 75. The value of tan sec cosec tan cot cot cot cot, < < is 6 tan tan cot cot tan cot tan cot cot cot cot = cot tan cot cot 76. If the incircle of a triangle passes through the circumcentre, then value of (cos + cos + cos ) is 9 4 Distance between circumcentre and incentre = R Rr r R Rr r R 77. The minimum value of the expression (t + ) + (t 4), (t, R) is FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

41 4 ITS-PT-II-PM-JEE(Main)/9 8 Slope of tangent t t = 4 P(, ) PQ Q P ( + t, t) y = x In, = 6, then the maximum value of sin + sin is 5 sin sin 6 = sin cos If 4x 4x 7 8cos y ; x R, then the range of sin y + cos y + is x x 9, 4, 4 9, 4, 4 FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

42 ITS-PT-II-PM-JEE(Main)/9 4 8cos y 4 x x cos y x y 8. Two tangents to the hyperbola having slopes m and m cuts the coordinate axes at 8 four concyclic points. If m and m satisfy the equation 5 + k =, then the value of k is 5 4 m m = 8. parabola is drawn through two given points (, ) and (, ) such that its directrix always touch the circle x y 6, then locus of focus of the parabola is x + 4y = 48 4x + y = 48 x + 4y = 6 4x + y = 6 Let focus be S(h, k), then (h ) + k = 4(cos )... () (h + ) + k = 4(cos + )... () h cos 4 h k 4 4 h h k 6 8. The locus of foot of perpendicular from (, ) on each member of the family of lines ( + t)x + ( t)y + t = ; t R is x + y + x y + = x + y x y + = FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

43 4 ITS-PT-II-PM-JEE(Main)/9 x + y x y 4 = none of these Given family passes through the point (, ) k k h h x(x ) + (y )(y ) = x + y x y + = (h, k) (, ) 8. onsider point P with ordinate t lying on the curve (, ) minimum distance from point P to the line y x =, then x y 4 ; t N. If S n represents the lim t S is t n P t, t t t S n = n 5 lim t S t t t t = lim t 5 = 5 (y rationalizing) 84. In, if medians from and are mutually perpendicular, then the possible value of cot + cot is 5 FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

44 ITS-PT-II-PM-JEE(Main)/ D y y tan tan x x y y x x x y cot y y xy x x y x Similarly cot xy x y cot cot xy x y x y D E 85. If 89 n cot 9k, then k is given by sin n k sinnk sink n ; n I n ; n I n ; n I 6 n ; n I 6 89 = sin n k sinnk n = k 89 sin n k nk sink sin n k sinnk 89 cot k cot 9k cot nk cot n k sink sink sink n cosk ; n I sin k k n 86. Let two tangents x 4y + = and x y = of a parabola intersect the tangent at vertex at points P(, 5) and Q(, ) respectively, then the length of latus rectum is FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

45 45 ITS-PT-II-PM-JEE(Main)/ Focus is (6, ) and equation of tangent at vertex is 5x + y 5 = LR = From point P(8, t); t being the parameter, tangents are drawn to the ellipse touch the curve at and. The locus of the image of P about is x y so as to 6 x = 4 x + y = x = 8 x + y = 4 h 8, k t h = Let the equation of incircle of be 4(x + y ) = 5 which touches the sides,, and at D, E and F respectively. If D, E and F are in P with common difference.5, then the circumradius of is s = x x x d r.5 s x x = 5 Sides are 7.5,, R x x d D x E x d F x + d x + d FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594

46 ITS-PT-II-PM-JEE(Main)/ The number of solutions of the equation sin cosx cos sinx 5 6 ; x [, ] is/are D sin (cos x) = sin (sin x) cos x sin x = cosx 4 n x ; n I 9. The lower one-fifth portion of a vertical tower subtends an angle tan at a point in the 5 horizontal plane through its foot and at a distance m from the foot. If the angle subtended by the upper four fifth portion of tower at point is, then is 5 tan 7 7 tan 7 7 tan 7 tan 7 h tan 5 5 h = m Hence tan( + ) = tan 5 tan tan 5 7 lso, tan h 5 h 5 FIITJEE Ltd., FIITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fax 6594