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IIT JEE Exam 2021 : Pace : AITS MAINS 3 (SOLVED)

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AITS 16 (Main) Physics PART (A) : PHYSICS SINGLE CORRECT ANSWER TYPE This section contains 20 multiple choice questions. Each question has 4 options (A), (B), (C) and (D) for its answer, out of which ONLY ONE option can be correct. Marking scheme: +4 for correct answer, 0 if not attempted and 1 in all other cases. 1. A box is put on a scale which is adjusted to read zero when the box is empty. A stream of pebbles is then poured into the base from a height h above its bottom at a rate of n pebbles. Each pebble has a mass m. If the pebbles collide with the box such that they immediately come to rest after collision then the scale reading at time t after the pebbles begin to fill the box is [neglect piling up of pebbles] (B) mn 2gh gt (C) mngt (A) mnt 1. (D) Zero (B) Scale reading will be the sum of weight and force due to their impulse m mnt g Scale reading v t 2ghmn mngt mn 2. A particle of mass m is suspended from a ceiling through a string of length L .The particle moves in a horizontal circle of radius r .The speed (v) of the particle and the tension (T) in the string is (A) v (C) v 2. 2gh gt 2r g L 2 r 2 12 3r g L 2 r 2 12 and T 3mgL L 2 r 2 12 and (B) v (D) v r g L 2 r 2 14 r g L 2 r 2 14 and T and T L mgL 2 L r2 12 mgL 2 r2 12 (D) The situation is shown in the diagram .The angle mode by the string with the vertical is given by r sin L The forces on the particle are (1) The tension T along the string and (2) The weight mg vertically downward CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 1 AITS 16 (Main) Physics The particle is moving in circle with constant speed v .Thus the radial acceleration towards the centre has magnitude v2 Resolving the forces along the radial direction and applying Newton s r second law, mv 2 r As there is no acceleration in the vertical direction, net force in the vertical direction is zero, T cos mg (iii) Dividing (ii) and (iii) T sin tan v2 rg Or V rg tan mg cos Using (i), T V 3. r g L r 2 2 14 and T L mgL 2 r2 12 Graph of position of image vs position of a point object from a convex lens is shown in the figure. The focal length of the lens is ------ (A) 0.50 0.05 cm (B) 5.00 0.05 cm (C) 0.50 0.10 cm (D) 5.00 0.10 cm 3. (B) 2F 10 t 5cm 1 1 1 r v u df dv du f 2 v2 u 2 dv du df f 2 2 2 u v 1 1 ,dv 10 10 (as there are 10 division in 1cm scale least count is 0.1 .i.e, considered as error) u 10, v 10,du So 1 1 df 25 10 10 2 10 10 2 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 2 AITS 16 (Main) Physics 2 25 10 100 so f 5.00 0.05 cm df 0.05 4. In the given circuit the R.M.S. value of voltages across the capacitor C, inductor L and resistor R1 are 12V,10V and 5V respectively .Then the peak voltage across R2 is (A) 7 2V 4. (B) 69 V (C) 138 V (D) None of these (C) In a parallel combination, each branch should have equal potential but in AC circuit in a branch Veff VR2 VC VL 2 VC2 VR21 VL2 VR22 12 2 52 10 2 VR22 V R2 rms 69 Vpeak 2Vrms 5. V R2 peak 138V The potential energy of particle varies as U x E 0 for 0 x 1 0 for x 1 for 0 x 1 , the deBroglie wavelength is 1 and for x 1 1the de-Broglie wavelength is 2 . The total energy of the particle is 2E 0 . Find 1 2 5. (A) 3 (B) (C) For 0 x 1, U E 0 7 (C) 2 Kinetic energy K1 total energy U 2E 0 E 0 E 0 1 (D) 5 h 2mE 0 For x 1, U 0 kinetic energy K 2 total energy U 2E 0 2 h 2m 2E 0 . (ii) From Eqs. (i) and (ii), we have 1 2 2 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 3 AITS 16 (Main) Physics 6. A sphere of radius 0.1m and mass 8 kg is attached to the lower end of a steel wire of length 5.0 m & diameter 10 3m.The wire is suspended from 5.22m high ceiling of a room. When the sphere is made to swing like a simple pendulum it just grazes the floor at its lowest point. Calculate the velocity of the sphere at the lowest position. Y for steel 1.994 1011 Nm 2 (A) 7.7 ms 1 (B) 4.4 ms 1 (C) 2.2 ms 1 (D) 8.8ms 1 6. (D) As the length of the wire is 5 m and diameter 2 0.1 0.2 m and at the lowest point it grazes the floor which is at a distance 5.22m from the root, the increase in the length of the wire at lowest point L 5.22 5 0.2 0.02m So the tension in the wire (due to elasticity) 1.994 1011 5 10 4 0.02 YA T L L 5 199.4 N 2 And as the equation of circular motion of a mass tied to a string a vertical plane is mv 2 T mg cos r So at the lowest point mv 2 T mg as 0 r But here r 5 0.02 0.1 5.12m 7. So 8 v 2 199.4 8 9.8 5.12 i.e, v 2 121 5.12 / 8 77.44 So, v 8.8 ms 1 An observer can see through a pinhole the top end of a thin rod of height h placed as shown in the figure. The beaker height is 3h and its radius h. when the beaker is filled with a liquid up to height 2h, he can see the lower end of the rod. Then refractive index of the liquid is CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 4 AITS 16 (Main) (A) 7. Physics 5 2 (B) 5 2 (C) 3 2 (D) 3 2 (B) PQ QR 2h i 45o ST RT, h KM MN So KS h 2 2h 2 h 5 sin r 8. h 5 1 5 sin i sin 45o 5 sin r 1/ 5 2 A cubical block of side moving with a velocity v on a horizontal smooth plane as shown. It hits a ridge at point O and sticks to it (collision is perfectly inelastic). The angular speed of the block after it hits O is (A) 8. h 3v 4a (B) 3v 2a (C) 3v 2a (D) Zero (A) CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 5 AITS 16 (Main) Physics Net torque about O is zero .Therefore angular momentum (L) about point O will be conserved Or, L1 Lf a Mv 0 com Mr 2 2 2 a 2 2 Ma 2 M w Ma 3 6 2 3v / 4a 9. One mole of monoatomic ideal gas undergoes the process A B, as in the given P V diagram. The specific heat for this process is (A) 9. 3R 2 (B) 15R 7 (C) 30R 7 (D) 20R 7 (B) Q U W form 1st low Here U nC v T Temperature at A,TA 2P0 V0 R 16P0 V0 R 3 14P0 V0 So, U R 21P0 V0 2 R 1 W 2P0 3V0 3V0 2P0 2 [Area under P V diagram] 9P0 V0 At B, TB So Q 21P0 V0 9P0 V0 30P0 V0 Q nC T C Q 30P0 V0 15 R n T 14P0 V0 7 R CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 6 AITS 16 (Main) 10. 10. Physics If x, v and a denote the displacement the velocity and the acceleration of a particle executing simple harmonic motion of time period T, the which of the following do not change with time? aT (A) (B) aT 2 v (C) a 2 T 2 2 2 v2 (D) aT v (A) In option aT LT 2 T M 0 L0 T o v LT 1 So according to the result above relation does not depend on time. 11. 11. A 0.5kg block slides from A on horizontal track with an initial speed of 3m s 1 towards a weightless horizontal spring of length 1m of and force constant 2N m 1 .The part AB of the track is friction less and the part BC has the coefficient of static and kinetic friction as 0.22 and 0.20 respectively. If the distance AB and BD are 2 m and 2.14m respectively the total distance through which the block moves before it comes to rest completely is (A) 2.5 m (C) At D.KE (B) 4.42 m (C) 4.24 m (D) 2.44 m 1 mv 2 k mg BD 2 1 0.5 32 0.2 0.5 10 2.14 2 2.25 2.14 0.11J 1 0.11 kx 2 R mgx x 2 x 0.11 0 2 x 0.1m Total distance covered 2 2.14 0.1 4.24 m 12. 12. A sample consisting of hydrogen atoms in the ground state is excited by monochromatic radiation of energy 12.75eV. If we were to observe the emission spectrum of this sample, then the number of spectral lines observed will be (A) 3 (B) 6 (C) 10 (D) 15 (B) The energy of an electron of hydrogen atom in the ground state is E 0 13.6eV The energy of incoming photon 12.75eV The energy of an electron in the excited state is 13.6 2 13.6 12.75 0.85eV n 13.6 n 22 16 0.85 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 7 AITS 16 (Main) Physics n2 4 n n 1 No of spectral lines n 2 n1 2 1 2 13. A ball is thrown horizontally from a height h above a staircase as shown in the figure. If the coefficient of restitution for any collision between the ball and the staircase is e, then the value of h for which the ball will bounce the same height above each step is (A) 13. d 1 e2 (B) d 1 e2 (C) de 2 1 e2 (D) de 2 1 e2 (A) Vertical velocity of the ball just before each collision is v app 2gh The height up to which the ball bounces after each collision is h' h d Vertical velocity of the ball just after the collision is vsep evapp e 2gh Using the principle of conservation of mechanical energy we get 2 1 m e 2gh mg h d 2 1 2 e 2gh g h d 2 e2 h h d d h 1 e 2 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 8 AITS 16 (Main) h 14. 14. d 1 e2 The figure below shows the top view of a hollow cylindrical room having a frictionless floor .A block of mass m is projected with a speed v on the smooth horizontal floor, along the wall of the room, the average normal force exerted by the cylindrical walls on the block, during its motion from A to B is mv2 (A) R (D) | P | 2mv t 15. 15. Physics R mv2 (B) R F V 2mv 2 (D) R 2mv 2mv 2 F R /V R A fully charged capacitor C with initial charge Q 0 is connected to an ideal inductor of self inductance L at t = 0 .The time at which energy is stored equally between the electric and the magnetic fields is LC (A) (B) 2 LC (C) LC (D) LC 4 (A) The charge on the capacitor at nay instant of time is Q Q 0 cos wt i dQ wQ0 sin wt dt Q0 cos wt 2 1 Li wQ0 sin wt 2 2C 1 w LC t 2 tan wt 1 wt tan1 1 16. 2mv 2 (C) R 4 LC 4 4 60g of ice at 00C is added to 200g of water initially at 700C in a calorimeter of unknown water equivalent W. If the final temperature of the mixture is 400C ,then the value of W is [Take latent heat of fusion of ice 80 L f 80 cal g 1 and specific heat capacity of water s 1 cal g 1 o C 1 ] CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 9 AITS 16 (Main) 16. 17. 17. (A) 70g (B) 80g (C) 40g (D) 20g (C) The principle of calorimetry states that for an isolated system, The heat gained by the cold body (Here it is ice ) = The heat lost by the hot body (Here it is water + calorimeter) m ice L f m ice s T m water W 8 T 60 80 60 1 40 200 W 1 30 480 240 600 3W W 40g Two uniform rods of equal length but different masses are rigidly joint to form an L shaped body which is then pivoted as shown. If in equilibrium, the body is in the shown configuration, ratio M/m will be (A) 2 (B) 3 (D) Net torque about O should be zero Hence 1 1 mg sin 60o Mg sin 30o 2 2 18. 18. Physics (C) 2 (D) 3 M sin 60o 3 m sin 30o What is the modulation index of an FM signal having a carrier swing of 100kHz when the modulating single has a frequency of 8 k HZ? (A) 6.25 (B) 6.23 (C) 6.55 (D) 6.33 (A) Given :Carrier Swing =100kHz f a 8kHz Form the defining equation, f mf fa First determining f Carrier Swing f 2 100 103 2 50kHz CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 10 AITS 16 (Main) Physics Now substituting into the equation for m f 50 103 8 103 6.25 mf 19. 19. In YDSE, the slits have different widths. As a result, the amplitude of waves from two slits is A and 2A respectively .If I0 be the maximum intensity of the interference pattern, then the intensity of the I pattern at a point where the phase difference between waves is is given by 0 5 4 cos .Where P P is in integer .Find the value of P? (A) 9 (2) 3 (3) 6 (4) 1 (A) As the amplitudes are A and 2A then the ratio of intensities is 1:4 I max I0 I1 I2 2 I1 I 2 I 4I 2 2I I0 9 Intensity at any point : I0 9I I I ' I1 I2 2 I1 I2 cos I ' I 4I 2 I 4I cos I ' 5I 4I cos ; I ' I 5 4 cos I0 I 5 4 cos 0 5 4 cos 9 P P 9 I' 20. The transverse displacement of a wave on string is given by x, t e ______ (A) Wave moving along + x axis with speed (B) Wave moving along x axis with ax 2 bt 2 2 abxt , this represent a a b b a (C) Standing wave of frequency 20. b 1 (D) Standing wave of frequency b (B) General equation of travelling wave is y f ax bt f x vt y f ax bt f x vt y e ax 2 bt 2 2 abxt e 2 ax bt CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 11 AITS 16 (Main) Physics So, the wave will be travelling along x- axis with speed b a NUMERICAL VALUE TYPE This section contains 10 questions. Attempt any 5 questions out of 10. Each question is numerical value type. For each question, enter the correct numerical value (in decimal notation (e.g. 6.25, 7.00, 7, 0.33, .30, 30.27, 127.30). Marking scheme: +4 for correct answer, 0 if not attempted and 0 in all other cases. 21. In the circuit shown in the figure (A) R 3 is a variable resistance,. As the value R 3 is changed the current I through the cell varies as shown in the figure (B). As R 3 , the current I 6A . What is the value of the sum of the resistance R1 and R 2 in ohms? 21. (4) For R 3 0, I 9A (from the graph). In this situation the circuit can be drawn as shown in figure (A) [Current through R 2 will be because We have a short circuit R 3 0 across R 2 ] Here I 36 9A R1 R 2 R 1 R 2 4 22. 22. A gas mixture consists of 2 moles of oxygen and 4 moles of argon at temperature T. Neglecting all vibrational modes we calculate the total internal energy of the system to be xRT .What is the value of x? (11) 1 The internal energy of n moles of a gas is U nFRT 2 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 12 AITS 16 (Main) Physics Where F = number of degrees of freedom The internal energy of 2moles of oxygen at temperature T is 1 U1 2 5RT 5RT (F = 5 for oxygen molecule) 2 The total internal energy of 4 moles at argon at temperature T is 1 U 2 4 3RT 6RT (F = 3 for argon molecule) 2 Total internal energy U1 U 2 11RT x 11 23. 23. A long straight wire carrying a current of 30A is placed in an external uniform magnetic field of induction 4 10 4 T (The direction of the current is parallel to magnetic field). The magnitude of the resultant magnetic induction at a point 2.0cm away from the wire is n 10 4 T . What is the value of n? (5) Magnetic field due to wire 0 I 4 10 7 30 3 10 4 T 2 2 r 2 2 10 This magnetic field will be perpendicular to external magnetic field Net magnetic field B B B2 B02 3 10 4 10 4 2 4 2 5 10 4 T 24. 24. A ball is projected from the ground at an angle of 450with the horizontal surface. It reaches a maximum height of 120m and return to the ground. Upon hitting the ground for the first time it loses half of its kinetic energy .Immediately after the bounce, the velocity of the ball makes an angle 300with the horizontal surface .The maximum height it reaches after the bounce (in meters) is (30) H1 u 2 sin 2 45o 120 2g u2 120 4g . (i) When half of kinetic energy is lost v u 2 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 13 AITS 16 (Main) Physics 2 u 2 o sin 30 u2 2 H2 (ii) 2g 16g Form (i) and (ii) H H 2 1 30m on 30.00 4 25. In Young s double slit experiment the separation between the slits is d 0.25cm and the distance of the screen from the slits is D =100cm.When the wavelength of light used in the experiment is o 6000 A the intensity at a distance x 4 10 5 m from the central maximum is p I0 where I0 is q the intensity of central maxima an P and q are the smallest integers .What is the value of p + q? 25. (7) Path diff xd D Path diff 4 10 5 0.25 10 2 1 Path diff. 1 10 7 path diff 2 Phase diff 1 10 7 2 6 10 7 2 Phase diff 6 Path diff 60o 3 Path diff R 1 2 2 1 2 cos 60o R 1 2 2 3 0 4 p 3, q 4 1 2 R 3 R 26. p q 7 A block of mass 1 kg is released from the top of a rough incline having 1 . The initial speed of 3 the block is 2ms 1 just before compressing the block .The inclined plane has un known length and has a spring of spring constant k =1N m 1 connected at the base as in the figure .Find the maximum compression of spring (answer in metre) CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 14 AITS 16 (Main) 26. Physics (2) ma mg sin 30o 1 mg cos 60o 3 a 0 By energy conservation w mg w fr w spring K.E w mg w fr 1 1 m kx 2 0 mv 2 x v 2m 2 2 k 27. 27. In a sample initially, there are an equal number of atoms of two radioactive isotopes A and B .Three days later ,the number of atoms of A is twice that of B .The half-life of B is 1.5 days .What is the half life ( in days ) of isotope A? (3) t 0 A B N0 N0 t 0 3 days 2N 2N N 0 0.5 t 0 T2 N N 0 0.5 0 T2 t N 1 1 t r2 2 0.5 0 r1 3 2 0.5 1 0.5 r1 28. 1 3 2 days r1 T1 3 A totally reflecting small plane mirror placed horizontally faces a parallel beam of light as shown in the figure. The mass of the mirror is 200 g. Assume that there is no absorption in the lens and that CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 15 AITS 16 (Main) Physics 20% of the light emitted by the source goes through the lens .The power of the source needs to support the weight of the mirror is P 108 watt, where P is [take g =10ms 2] 28. (15) 2P C 200 10 3 10 104 3 P' 2 p P 1500 106 15 108 W 0.2 mg 29. A uniform rod of mass m and length l is hinged at one of its end with the ceiling and another end of the rod is attached with a thread which is attached with the horizontal ceiling at point P .If one end of the rod is slightly displaced horizontally and perpendicular to the rod and released .If the time period 2 sin of small oscillation is 2 . Find x. xg 29. (3) Consider a small angular displacement Torque about the rotational axis 1 mg sin 2 I 1 m 1sin 2 3 m 12 sin 2 3 1 m sin 2 mg sin 2 3 3g 2 sin T 2 2 sin 3g CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 16 AITS 16 (Main) Physics 30. The figure shows a square loop PQRS with an edge length a .The resistance of the wire PQR is r and 2 0i that of PSR is 2r.The value of the magnetic field at the centre of the loop is found to be , k a then what is the value of k? 30. (3) The magnetic field due to a straight current carrying wire is I B 0 sin sin 4 d 2i 0 3 BPQR 2 sin 45o sin 45o 4 2 i 0 3 BPSR 2 sin 45o sin 45o 4 2 Bnet BPQR BPSR Bnet k 3 2 0i 3 a CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 17 AITS 16 (Main) Chemistry PART (B) : CHEMISTRY SINGLE CORRECT ANSWER TYPE This section contains 20 multiple choice questions. Each question has 4 options (A), (B), (C) and (D) for its answer, out of which ONLY ONE option can be correct. Marking scheme: +4 for correct answer, 0 if not attempted and 1 in all other cases. 31. 31. A metallic element crystallizes into a lattice containing a sequence of hexagonal layers of ABABAB .. What percentage by volume of this lattice is empty space? (A) 26% (B) 74% (C) 50% (D) 85% (A) Number of atoms in hcp is 6 per units cell 4 6 R 3 3 Packing fraction Base area c Base area 6 3 4r 2 4 2 C 4r 3 2 0.74 3 Void % 1 0.74 100 26 Packing fraction 32. 32. Consider the following sequence of reactions The final product C is (A) (B) (C) (D) (D) CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI# 1 AITS 16 (Main) 33. Chemistry The degree of dissociation of PCl5 g for the equilibrium PCl5 g PCl3 g Cl2 g is approximately related to the pressure at equilibrium (P) by the relation 1 (A) P 33. (B) 1 P (C) 1 P2 (D) 1 P4 (B) P 1 1 P Kp 1 P 1 2 P 2 P 1 2 1 Neglecting 2 K p 2P Kp P CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI# 2 AITS 16 (Main) 34. The increasing order of basicity of the following compounds is (1) (2) (3) 34. Chemistry (4) (A) 4 2 1 3 (B) 1 2 3 4 (C) 2 1 3 4 (D) 2 1 4 3 (D) 3 is most basic because its conjugate acid is stabilized by equivalent resonance. Out of 1 and 4, 4 is more basic due to +1 effect. 2 is less basic than 1 and 4 because N atom is sp2 hybridized (more s%). 35. A solid compound X on heating gives CO2 gas and a residue .The residue mixed with water forms Y On passing an excess of CO2 through Y in water a clear solution Z is obtained .On boiling Z compound X is reformed .The compounds X is (A) Ca HCO3 2 (B) CaCO3 (C) Na 2 CO3 (D) K 2 CO3 35. (B) The give compound X must be CaCO3 . It can be explained by following reaction CaCO3 CaO CO 2 CaO H 2 O Ca OH 2 36. The product of the reaction given below is (A) (B) CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI# 3 AITS 16 (Main) Chemistry (C) (D) 36. (C) 37. The major product formed in the following reaction is (A) (B) (C) (D) 37. (A) 38. A certain metal which irradiated by light r 3.2 1016 Hz emits photoelectrons with twice kinetic energy as did photoelectrons when the same metal is irradiated by light r 2.0 1016 Hz . Then V0 of metal is 38. (A) 1.2 1014 Hz (B) KE 1 hv1 hv 0 (B) 8 1015 Hz (C) 1.2 1016 Hz (D) 4 1012 Hz KE 2 hv 2 hv 0 As KE 1 2 KE 2 hv1 hv 0 2 hv 2 hv 0 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI# 4 AITS 16 (Main) Chemistry Or hv 0 2hv 2 hv1 Or v 0 2v1 v1 2 2 1016 3.2 1016 0.8 1016 Hz 8 1015 Hz 39. Suppose that gold is being plated on to another metal in an electrolytic cell. The half-cell reaction producing the Au (s) is AuCl 4 3e Au s 4Cl . If a 0.30A current runs for 15.00 minute, what mass of Au (s) will be 39. plated, assume all the electrons are used in the reduction of AuCl 4 ? The Faraday constant is 96485 C/mol and molar mass of Au is 197g. (A) 0.184 g (B) 0.551 g (C) 1.84 g (D) 0.613 g (A) Mass of Au deposited =Number of Faraday passed Eq.mass 0.30 15 60 197 0.184g 96485 3 40. The value of n in the molecular formula Be n Al 2Si6 O18 is 40. (A) 1 (C) For Be n Al 2Si 6 O18 (B) 2 (C) 3 (D) 4 According to charge balance in a molecule 2n 6 24 36 0 This gives, n 3 41. Which is the correct IUPAC name of this compound? 41. (A) 3, 4 diethyl 5, 5 dimethyl 1,4 hexadiene (B) 3, 4 diethyl 5 methyl 1,4 hexadiene (C) 3, 4 diethyl 2 methyl 2,5 hexadiene (D) None of these (B) 3, 4 diethyl 5 methyl 1,4 hexadiene CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI# 5 AITS 16 (Main) 42. 42. Chemistry The enthalpy of vaporization of liquid water using the date: 1 (i) H 2 g O 2 g H 2 O 1 2 H 285.77kJ / mol 1 (ii) H 2 g O 2 g H 2 O g 2 H 241.8kJ / mol in KJ / mol is (A) + 43.93 (B) 43.93 (C) + 527.61 (A) Enthalpy of vaporization of H 2 O is (D) 527.61 H 2O H 2 O g i.e, ii i Enthaply of vaporization of liquid water 285.77 241.84 43.96 kJ mol 1 43. 43. 44. 44. The correct decreasing order of electropositive character among the following elements is: Fe,Sc, Rb, Br, Te, F, Ca (A) Fe Sc Rb Br Te F Ca (C) Rb Ca Sc Fe Br Te F (D) (B) Ca Rb Cs Fe Te F Br (D) Rb Ca Sc Fe Te Br F KMnO4 reacts with KI, in basic medium to form I2 and MnO 2 . When 250ml of 0.1M KI solution reacts with 250mL of 0.02M KMnO4 in basic medium, what is the number of moles of I 2 formed? (A) 0.015 (B) (B) 0.0075 (C) 0.005 (D) 0.01 Number of milli equivalents of MnO 4 0.02 3 250 15 Number of milli equivalents of I2 0.1 1 250 25 Thus, here MnO 4 is limiting reagent Number of milli equivalent of I 2 formed = Number of milli equivalent of MnO 4 15 15 0.015 1000 0.015 0.0075 Number of moles of I2 formed 2 Or number of equivalent of I2 formed CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI# 6 AITS 16 (Main) 45. 45. 46. 46. Chemistry The solubility of sulphates of alkaline earth metals in water decreases down the group because (A) The lattice energy of sulphates of group II decrease down the group (B) The lattice energy of sulphates of group II increase down the group (C) Both hydration and lattice energies decreases down the group (D) The decrese in hydration energy is more than the decreases in lattice energy (D) Both lattice energy and hydration energy decreases down the group of alkaline earth sulphates but hydration energy decreases more rapidly than lattice energy When 2.5 g of a sample of Mohr s salt reacts completely with 50 mL of purity of the sample of Mohr s salt is: (A) 78.4 (B) 70 (A) meq. FeSO4 ( NH 4 ) SO4 6 H 2O (C) 37 N KMnO4 solution. The % 10 (D) 40 = meq of KMnO4 (n = 1) W 1 1000 0.1 50; W 1.96 g 392 Hence, % of Mohr s salt 196 100 78.4% 2.5 47. 47. 48. Rearrangement of an oxime to an amide in the presence of a strong acid is called (A) Curtius rearrangement (B) Fries rearrangement (C) Beckmann s rearrangement (D) Aldol condensation (C) Which of the following is the strongest base: (A) (B) CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI# 7 AITS 16 (Main) Chemistry (C) 48. (D) (D) Thus the order of basic strength of conjugate base will be opposite Hence, the strongest conjugate base is 49. Identify the product Z in the following sequence of reactions Na C2 H5 OH K 2 Cr2 O 7 HNO2 CH 3 CN X Y Z H2O H 2SO4 (A) CH 3CHO (B) CH 3CONH 2 (C) CH 3COOH (D) CH 3CH 2 NHOH 49. (C) 50. An electrolysis of a oxytungsten complex ion using 1.10 A for 40 min produces 0.838 % of tungsten. What is the charge on tungsten in the material? (Atomic mass of W = 184) (A) 6 (B) 2 (C) 4 (D) 1 (A) W I t 0.838 n ; n M 96500 184 40 60 1.0 96500 n 6 50. NUMERICAL VALUE TYPE This section contains 5 questions. Each question is numerical value type. For each question, enter the correct numerical value (in decimal notation, truncated/rounded-off to second decimal place. (e.g. 6.25, 7.00, 7, 0.33, .30, 30.27, 127.30). Marking scheme: +4 for correct answer, 0 if not attempted and 0 in all other cases. CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI# 8 AITS 16 (Main) 51. Chemistry All the energy released from the reaction X Y, r G o 193kJ mol 1 is used for oxidizing M as M M 3 2e , E o 0.25V Under standard conditions the number of moles of M+ 51. oxidized when one mole of X is converted to Y is F 96500C mol 1 (4) G o nFEo n 193 103 8 96500 0.25 So No of moles of M oxidized by 1 moles of X 52. In Borax Na 2 B4 O 7 .10H 2O 8 4 2 if number of sp2 hybridized B-atoms are X and number of sp3 hybridized B-atom are Y .What is the value of sum X + Y? 52. (4) In Borax 2B atoms are sp 2 and 2 B atoms are sp3 hydridised X 2, Y 2 X Y 4 53. The number of hydroxyl group (s) in Q is 53. (4) 54. In neutral or faintly alkaline solution 8 moles of permanganate anion quantitatively oxidize thiosulphate anions to produce X moles of a Sulphur containing product the magnitude of X is (6) 8MnO 4 3S2 O 32 H 2 O 8MnO 2 6SO 24 2OH 54. 55. The work done (in Cal) in adiabatic compression of 2 moles of an ideal monotomic gas by the constant external pressure 2 atm and an initial temperature of 300K is : R 2cal / mol K CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI# 9 AITS 16 (Main) 55. Chemistry (720) To calculate the work done we need to know the final temperature nRT1 T1 300k, P1 1atm, n 2 mole, P2 2atm, V1 P1 w Y nC v T2 T1 P2 V2 V1 For monatomic gas 3 Cv R 2 3 nR T2 T1 P2 V2 V1 2 3 T T nR T2 T1 P2 nR 2 nR 1 2 P2 P1 3 P T2 T1 T2 2 T1 2 P1 ST2 3 P2 T1 2 2 P1 T2 0.4 1.5 2 300 420 K 56. 56. The half-life of a radioactive isotope is three hours .If the initial mass of the isotope were 256g the mass of it in grams remaining undecayed after 18 hours would be: (4) t1 2 3 hours, n T t1/2 18 3 6 6 1 N N0 2 1 1 256 4g 8 8 57. 57. How many of the following element exclusively occur in combined state? Gold, iron, zinc, aluminum, platinum, sodium, magnesium (5) Gold and platinum occur in the free state while iron, zinc, aluminum, sodium and magnesium occur in combined state 58. Consider following reactions 58. The sum of molecular masses of gas A and B is ____________ u. (46) Gas A is H 2 gas B is CO 2 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI# 10 AITS 16 (Main) Chemistry 59. The number of unpaired electrons in CoF6 59. (4) number of unpaired electron 4 60. For the reaction N 2 O 4 2NO 2 g the degree of dissociation of N 2 O 4 is 0.2 at 1 atm. Then the K p 3 are of 2NO2 N 2 O4 is 60. (6) N 2 O 2 g 2NO 2 g At t 0 at 1 1 1 1 Mole fraction PNO Kp 2 PN2O4 Kp 2 2 1 2 4 2 p 1 2 Kp 4 0.2 1 0.2 2 2 0.16 0.96 For 2NO 2 N 2O 4 ; K P 0.96 6 0.16 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI# 11 AITS 16 (Main) Mathematics PART (C) : MATHEMATICS SINGLE CORRECT ANSWER TYPE This section contains 20 multiple choice questions. Each question has 4 options (A), (B), (C) and (D) for its answer, out of which ONLY ONE option can be correct. Marking scheme: +4 for correct answer, 0 if not attempted and 1 in all other cases. 61. The compound statement p q p ~ q is logically equivalent to 61. (A) p q (C) (B) p q (C) tautology (D) contradiction We know that - p q p q given statement is p q p q Which is a tautology as p p t 62. 62. 63. 63. The number of times the digit 0 is used in writing the numbers from 1to1000 is equal to (A) 189 (B) 300 (C) 192 (D) 270 (C) Consider the number from 000,001,002, .998,999. Since, each digit has an equal likelihood of coming and digits used are 3 1000 . Hence, 0 appears 300 times But, 000,001,002, .,098,099; Have some unwanted zeroes, we subtract these zero s i.e 3 1 2 9 1 90 111 Thus total times 0 appears 300 111 189 Also add 3 zero s in 1000 189 3 192 1 1 1 Let A 1 1 0 , A1 be a matrix formed by the cofactors of the elements of the matrix A and 0 1 1 A 2 be matrix formed from cofactors of matrix A1 and so on, then the value of A10 is (A) 310 (D) (B) 320 (C) 9 (D) 31024 A 1 1 0 1 1 1 3 2 A1 A (by property) 2 4 2 3 A 2 A1 A A3 A2 A Similarly, A10 A 210 10 32 31024 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 1 AITS 16 (Main) 64. 64. Mathematics The area bounded by the parabola 4y 3x 2 , the line 2y 3x 12 and the y-axis in 1st quadrant is (A) 10 sq.units (B) 20 sq.units (C) 30 sq. units (D) 36 sq. units (B) The intersection point is A A 4,12 by solving 3x 2 2 3x 12 4 3x 12 3x 2 Thus the required are dx 2 4 0 4 4 3 x3 x 2 6x 4 0 4 3 16 24 16 12 8 20sq.units 4 65. The mean and variance of a random variable X having a binomial probability distribution are 6 and 3 respectively then the probability P X 2 is (A) 65. 10 4096 (B) 4083 4096 (C) 3 1024 (D) 13 2048 (B) np 6, npq 3 1 1 q , p , n 12 2 2 P x 2 1 p X 0 P X 2 1 P X 0 P X 1 12 1 11 1 1 1 1 12 C 0 12 C1 2 2 2 13 4083 1 12 2 4096 9 66. 66. 1 3 The coefficient of the term independent of x in the expansion of 1 x 2x x 2 is 3x 2 1 19 17 1 (A) (B) (C) (D) 3 54 54 4 (C) 3 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 2 AITS 16 (Main) Mathematics coefficient of the term independent of x is the coefficient of x 0 in 9 9 1 3 2 1 3 2 1 3 3 2 x x x 2x x 3x 3x 3x 2 2 2 9 9 1 3 Now, in x 2 , 3x 2 9 r 3 Tr 1 9 C r 2 (i) coefficient of r 1 18 3r x 3 x 0 18 3r 0 r 6 3 6 3 1 Coefficient of x 0 is 9 C 6 2 3 (ii) Coefficient of 19 x 1 18 3r 1 r (not possible) 3 1 (iii) Coefficient of x 18 3r 1 r 7 2 3 1 Coefficient of x 3 is 9 C7 2 3 Hence the required coefficient is 3 9 2 3 1 3 1 1 9 C 6 1 0 2 9 C7 2 3 2 3 9 8 7 1 1 2 3 2 8 273 7 7 2 17 18 27 54 3 1 cos 10 d 67. The value of 67. 5 5 (A) tan 8 c 8 (B) 1 cos 13 10 is equal to (where, C is the constant of integration) 8 5 5 (B) tan + C 8 2 3 10 8 5 5 (C) tan 16 2 5 5 16 (D) tan c 8 2 3 5 2 sin 1 cos 2 d 13 1 cos d 13 5 210 cos 2 3 10 3 1 5 tan sec 2 d 2 2 2 Let, tan 1 t sec2 d dt 2 2 2 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 3 AITS 16 (Main) Mathematics 8 3 5 t5 t dt C 8 5 8 5 5 tan C 8 2 68. 68. In an increasing geometric progression the sum of the first and the last term is 99 and their product is 288. If sum of all the terms of GP is 189, then the number of terms in the progression is equal to (A) 5 (B) 6 (C) 7 (D) 8 (B) G.P. is increasing i.e. r 1 Given a ar n 1 99, ar n 2 288 and a 1 r n 1 r 189 a 1 r n 1 99 and a 2 r n 1 288 a 2 288 99a a 2 99a 288 0 As a 3,96 r n 1 32 for a 3 Now, a 1 r n 189 3 1 r.r n 1 1 r 1 r 1 r.32 63 1 32r 63 1 r 63r 31r 62 r 2 Now, 2 n 1 32 n 6 189 69. The number of roots of the equation sin 1 x cos 1 x sin 1 5x 3 is/are 69. (A) 3 (B) (B) 1 (C) 2 (D) 0 70. If the graph of the function y a b x 2 2 a b 2c x 1 is strictly a b above the x-axis then (A) b c a (B) a c b (C) b a c (D) c b a (C) 71. The value of 70. 2 (A) 71. 2 2 cos x dx is equal to 1 ex (B) (C) 0 (D) 2 (C) CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 4 AITS 16 (Main) Let Mathematics 2 cos x dx 1 ex 2 cos x 2 cos x I dx 1 ex 1 e x 0 1 ex I 2 cos x x x 1 e 1 e 0 dx I 2 cos x dx 0 I 0 72. Let the tangents PQ and PR are drawn to y 2 4ax from any point O on the line x 4a 0 . The angle subtended by the chord of contact QR at the vertex of the parabola y 2 4ax is (A) 72. 4 (B) 3 (C) 2 (D) 6 (C) Let, Q at 12, 2at1 and R at 22 , 2at 2 , then P at1t 2 ,a t1 t 2 which lies on a 4a 0 at1t 2 4a t1t 2 4 Now the vertex of the parabola y 2 4ax is the origin O 0, 0 slope of OQ 2 2 & slope of OR t1 t2 2 2 4 4 1 t1 t 2 t1 t 2 4 OQ is perpendicular to OR QOR 90 73. 2 Given two independent events, if the probability that both the events occur is that exactly one of them occurs is 73. 8 , then probability 49 26 and the probability of more probable of the two events is 49 then 14 is equal to (A) 2 (B) 4 (C) Let A and B are two independent events 8 P A B P A .P B (1) 49 (C) 8 (D) 7 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 5 AITS 16 (Main) Mathematics P A BC P B A C 26 49 P A P B 2P A .P B 26 49 42 6 49 7 From (1) and (2) we get, 4 2 P A , P B or 7 7 2 4 P A , P B 7 7 2 P A P B Hence the probability of the more probable event is 74. 74. 4 7 The number of integral value (s) of k such that the system of equation kx 2y z x, ky z z 3x and 2x kz 2y z has non-trivial solution, is/are (A) 0 (B) 1 (C) 2 (D) 3 (D) Rearranging all the equations, we get k 1 x 2y z 0 3x ky 2x 0 2x 2y k 1 z 0 For a no-trivial solution k 1 3 2 k 1 2 0 2 2 k 1 k 1 k 2 k 4 2 3k 1 1 6 2k 0 k 3 k 2 4k k 2 k 4 6k 2 6 2k 0 k 3 9k 0 k 0, 3, 3 75. 75. The solution of the differential equation cos x (A) y c sin x cos x (C) y tan x c (B) dy y tan x sec x dx tan x dx I.F e e logsec x sec x dy ysin x 1 is (where, C is an arbitrary constant) dx (B) y sin x c cos x (D) y sin x sin x c Hence the solution is, y.sec x sec 2 xdx CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 6 AITS 16 (Main) Mathematics y sec x tan x C y sin x c cos x 76. 76. 40 1 The coefficient of x 20 in the expansion of 1 x 2 x 2 2 2 x (A) 30 C10 (B) 30 C15 (C) 30 C 25 5 is (D) 30 C 20 (C) 1 2 1 x 2 2.x. x x 1 x 2 40 1 x x x 2 1 x 1 x 2 40 1 x 2 40 1 x 1 x 2 10 2 x 10 5 5 10 x10 1 x 2 30 Coefficient of x 20 in x10 1 x 2 30 Coefficient of x10 in 1 x 2 30 So, the required coefficient is 1 x 2 30 30 r 0 30 C r x 2 2r 10 r 5 r 30 C5 30 C 25 77. Let the vertices of a triangle are A 3 2sin , 4 2cos , B 3 2cos , 4 2sin and 77. C 3 2sin , 4 2 cos , then the distance between the centroid and the circumcenter of ABC is 2 3 1 1 (A) units (B) units (C) units (D) units 3 2 2 3 (A) A,B,C the lie on the circle x 3 y 4 2 2 2 78. The value of lim 78. (A) 1 (C) x lim x lim x sin 2 cos 2 x tan sec2 x 2 is equal to (C) 2 (B) 2 (D) 0 lim sin 2 2 sin x tan tan x 1 tan x sin 2 1 sin 2 x 2 2 x 2 sin 2 sin 2 x tan tan 2 x CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 7 AITS 16 (Main) Mathematics sin 2 sin 2 x 2 sin 2 x tan 2 x 1 2 1 2 x 2 sin 2 x tan 2 x tan tan 2 x lim 79. 79. 2 2 2 If A and A aA bI 0 then a 2b is equal to (where I is an identity matrix and O null 9 4 matrix of order 2 respectively) (A) 27 (B) 26 (C) 24 (D) 12 (B) 2 2 2 2 22 12 A2 9 4 9 4 54 34 2a 2a b 0 aA , bI 9a 4a 0 b A 2 aA bI 12 2a 22 2a b O 34 4a b 91 54 a 6 22 2a b 0 b 10 a 2b 26 80. The length of the perpendicular (in units) from the point 80. x 2 y 7 z 3 lies in the interval 1 2 1 3 (A) 1, (B) 2,3 2 (C) (C) 0, 2 1, 2, 4 on the straight line (D) 4,5 Let point A is 1, 2, 4 and any general point on the lines is B 2, 2 7, 3 AB 1 i 2 5 j 1 k And a vector parallel to the line is AB i 2 j k AB.b 0 1 4 10 1 0 2 The length of the perpendicular AB 2 1 4 5 2 1 3 2 2 un 3 NUMERICAL VALUE TYPE CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 8 AITS 16 (Main) Mathematics This section contains 10 questions. Attempt any 5 questions out of 10. Each question is numerical value type. For each question, enter the correct numerical value (in decimal notation (e.g. 6.25, 7.00, 7, 0.33, .30, 30.27, 127.30). Marking scheme: +4 for correct answer, 0 if not attempted and 0 in all other cases. 2 81. 81. 4i 3 i 2 If r cos i sin , then cos sin is equal to where,i 1 2i 1 (1.4) 2 4i3 i 4i i 2i 1 2i 1 25 1 4i 1 4i 3 4i 25 3 4i 25 2 3 4i 3 4i 3 4i 3 42 3 4 5 i r cos i sin 5 5 5i 2 2 2 3 4 r 5, cos ,sin 5 5 7 cos sin 1.4 5 82. The distance between the focus and directrix of the parabola 82. (12) Given the equation of parabola is 3x y 2 48 x 3y is 2 3x y x 3y 24 2 2 length of the latus rectum 4 24 distance between the focus and directrix is equal to 2a 12 83. 83. If the direction ratios of a line are 1 , 2 , 4 and if it makes an angle of 60 with the y axis, then the sum of the value of is (7) The direction ratios of the line are 1 , 2 , 4 Direction ratios of the y-axis are 0,1, 0 So cos 60 2 1 2 1 2 16 2 2 4 2 16 16 2 2 2 21 2 2 14 5 0 Hence the sum of the values of 7 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 9 AITS 16 (Main) 84. 84. Mathematics 5 12 2 2 If sin 1 sin 1 sin 1 cos 1 then the value of x is equal to x x x x (13) 5 12 sin 1 and sin 1 are both when x 12 . Therefore x x 25 12 cos 1 1 2 sin 1 x x 2 2 1 1 2 cos x 2 sin x 25 12 sin 1 sin 1 1 2 x x On comparing, we get 12 25 1 2 x x Therefore 144 x 2 25 or x 13 85. 85. The volume of the greatest cone obtained by rotating a right- angle of hypotenuse 2 units about a side k is cubic units, then the value of k is equal to 9 3 (16) The volume generated 1 1 V r 2 h 4 x 2 x 3 3 dV 0 For maximum volume, dx 2 4 3x 2 0 x 3 Vmax Also, 4 2 16 4 3 3 3 9 3 d 2V 0 dx 2 86. The sum of the y- intercepts of the tangents drawn from the point 86. x 2 y2 1 3 2 (6) 2, 1 to the hyperbola CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 10 AITS 16 (Main) Mathematics Let the equation of the tangent form 2, 1 is y mx 3m 2 2 1 2m 3m 2 2 2m 1 3m 2 2 2 m 1, 3 So, the equation of the tangents are y 1 x 2 & y 1 3 x 2 y intercepts of the tangent are 1 & 5 the sum of y- intercepts 6 87. 87. In a class tournament where the participants were to play one game with one another, two of the class players fell ill, having played 3 games each. If the total number of games played is 27, then the number of participants at the beginning was (9) Suppose the two player did not play at all such that the remaining n 2 players played matches Since these two players played 3 matches each hence that total number of matches is n 2 C 2 3 3 27 n 2 C2 21 n 2 n 2 C2 n 2 n 3 7 6 C2 21 2 2 n 2 7 n 9 2 88. The value of x 4 4 x 2 dx x 2 4 x dx is equal to 0 2 2 0 88. (2) 2 x 4 4 x 2 dx 0 1 2 2 2x 4 x dx x 3 0 I 2 II 1 2 dx Using integration by parts we get 3 4 x 2 2 x 3 3 2 3 2 2 2 4 x 2 3x 3 0 2 3 2 1 0 0 2x 2 4 x 2 2 dx 0 2 3 x 2 4 x 2 2 dx 2 0 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 11 AITS 16 (Main) Mathematics x 2 4 x 2 4 x 2 dx 2 0 2 x 4 4 x 2 dx 4 0 2 0 x 2 0 89. 4 x 2 dx 2 x 4 4 x 2 dx x 2 4 x 2 dx 2 Let four circles having radii r1 5units , r2 5units, r3 8 units and r4 units r4 5 are mutually touching each other externally then the value of 89. 2 is equal to r4 (2.25) Let, the centers of circles with radii r1 , r2 , r3 , r4 are C1 , C2 , C3 and C4 respectively. The foot of C3 and C1C 2 is E, then C3E 12 8 r4 4 r4 Now C1C2 r4 90. 90. 2 C 4 E 52 5 r4 4 r4 52 2 2 2 8 2 9 2.25 9 r4 4 x 1 y 2 z 3 measured parallel to the plane 2 3 3 3x 5y 2z 5 is k, then the value of k 2 is equal to (4573) If the distance of point P 3, 2,6 from the line Coordinates of the point Q are 2 1,3 2, 4 3 PQ is perpendicular to the normal vector of the plane, PQ 2 2 i 3 j 4 3 k Normal vector 3i 5j 2k CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 12 AITS 16 (Main) Mathematics Hence 3 2 2 5 3 2 4 3 0 6 6 15 8 6 0 12 0 So, the coordinate of the point Q are 23, 34, 45 Hence, PQ 26 2 36 51 2 2 4573 CENTERS : MUMBAI /DELHI /AKOLA /LUCKNOW /NASHIK /PUNE /NAGPUR /BOKARO /DUBAI # 13

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