ANSWER KEY JEE-MAIN ONLINE

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ANSWER KEY JEE-MAIN ONLINE - 5 PHYSICS Date : -4-5. [4]. [4]. [4] 4. [] 5. [] 6. [] 7. [] 8. [] 9. []. []. [4]. []. [] 4. [] 5. [] 6. [] 7. [] 8. [] 9. []. [4]. []. []. [] 4. [4] 5. [] 6. [] 7. [4] 8.

1 ANSWER KEY JEE-MAIN ONLINE - 5 PHYSICS Date : [4]. [4]. [4] 4. [] 5. [] 6. [] 7. [] 8. [] 9. []. []. [4]. []. [] 4. [] 5. [] 6. [] 7. [] 8. [] 9. []. [4]. []. []. [] 4. [4] 5. [] 6. [] 7. [4] 8. [4] 9. [4]. [] CHEMISTRY. []. []. [4] 4. [] 5. [] 6. [] 7. [] 8. [] 9. []. []. [4]. []. [4] 4. [] 5. [] 6. [] 7. [4] 8. [4] 9. [,]. []. []. []. [] 4. [] 5. [] 6. [4] 7. [] 8. [4] 9. []. [] MATHEMATICS. []. []. [] 4. [] 5. [] 6. [4] 7. [4] 8. [] 9. []. []. []. []. [] 4. [] 5. [*] 6. [] 7. [] 8. [4] 9. []. []. []. [4]. [4] 4. [4] 5. [] 6. [] 7. [] 8. [] 9. []. []

2 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS PHYSICS = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =. In propagation of light E and B oscillate in mutually perpendicular directions. E B direction of propagation = +z direction only option (4) satisfies both conditions of () E B E B directed () along the z-axis.. Half life = 5 hrs. =.69.46hr N moles of Na No.of particles (disintegrations) = N N e moles e 4 =. moles.5 no.of particles = 7.4. Amplitude in a damped oscillation is given by A Ae t 4. e m energy A E Ee t where E is initial energy 5 45e 5sec 5 e on taking log both sides ln 5 ln 4 IT M c LT h ML T MLI T If e m c h a b c d a b c MLI T IT M LT ML T by eqating powers, we get a, b, c, d h ce d

3 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS 5. At cm from the magnet on its equitorial plane Bmagnet BM (newtral point) so by equating their magnitude M.6 4 r 5 Tesla 7. M.6 5 Tesla 6. e M Am m IT M c LT h ML T MLI T If e m c h a b c d a b c MLI T IT M LT ML T by eqating powers, we get a, b, c, d d 7. h ce F Rsin X B T R T For the area to be in equilibrium, F T sin I Rsin B T sin & F I Rsin B T IRB 8. When positive terminal of battery is connected to A, current passes through D diode. V current supplied 5 =.4 A When positive terminal is connected to B current passes through D. V current supplied.a SANTACRUZ ANDHERI GOREGAON KANDIVALI (E) KANDIWALI (W) BORIVALI BHAYANDER VASAI POWAI DADAR SION THANE LOKPURAM (THANE) DOMBIVLI KALYAN PANVEL KAMOTHE NERUL SANPADA KHARGHAR

4 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS 9. As collisions are elastic and masses are equal, velocities of colliding particles get exchanged. P in each collision with the supports = mv Time interval between consecutive collisions with one support L nr v F avg P mv mv T L nr / v L nr. When the currents are parallel, I I is positive and the force between then is attractive (i.e. negative) similarly when currents are anti parallel I I is negative and the force between than is repulsive (i.e. positive) so option () satisfies the conditions.. dv E dr dr dxˆi dyˆj ˆ ˆ ˆ ˆ dv 5i j dxi dy j v dv 5dx dy V 5 x y V 5 V volt J / C. For the given situation S V s V O frequency listened by an observer is f. f So, f V V f V Vs f f V f V V V V s s V equating the equation y mx C f m V V s So choice is (A). 4

5 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS. B A C By triangle rule A C B B A C B A C B sin B A B sin again B cos A cos B A so, B A B A F V E D 4. 4V 4 R A 9V B V C If current in 4 is zero then v BCDE VEB VBC VCD VDE 4 VCD VCD volt again VA 9 VD V V 5V A D SANTACRUZ ANDHERI GOREGAON KANDIVALI (E) KANDIWALI (W) BORIVALI BHAYANDER VASAI POWAI DADAR SION THANE LOKPURAM (THANE) DOMBIVLI KALYAN PANVEL KAMOTHE NERUL SANPADA KHARGHAR 4 4

6 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS 5. Let block is floating with disolve depth h. Then about equilibrium h M g F Block up AH g Ah g B L When block depressed by distance x then F F' M g Net up Block AH x L g AH B g from equation () F Net Ax L g F Ax g Net L d x AHBlock Ax Lg dt d x Lg x dt H Block Lg H B For simple pendulum g H Equating L B cm 9 6

7 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS 6. O This combinations will behave like a mirror of power. P P P eq L M Peq f F eq f so the behaviour will be like a mirror of focal length f a/ I a O Using mirror equation V U f eq 9 a f 4 a f a f L n I 7. ni L R S I L V i V L Voltage drop across zener diode is V so voltage drop across L V V V n I R R RS i L L S S Vi VL n I L SANTACRUZ ANDHERI GOREGAON KANDIVALI (E) KANDIWALI (W) BORIVALI BHAYANDER VASAI POWAI DADAR SION THANE LOKPURAM (THANE) DOMBIVLI KALYAN PANVEL KAMOTHE NERUL SANPADA KHARGHAR 6 6 KS

8 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS 8. Let A is derived quantity which is derived by three fundamental quantities S, and h eg By using property of homogeneity. x S A h eg y z m T A m L T S LT h m L T eg x y z m L T m L T m LT m L T x z, x y z & x y z x y z x y z y y, x, z S so, A h 9. Potential V(r) due to large planet of radius R is given by out V r r > R GM r R V s r r = R GM R GM r V R R in V(r) r<r r=r r>r r r < R 8

9 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS. Due to quarter ring electric field intensity is k E sin R when So, due to each quarter section, field intensity is E E k k E sin R 4 R so Net E Net E R / k R L R k k 4 R R 4 R so, E 4 4 Net 4 R L L In a potentiometer, the null point will fluctuate due to varying current & voltage. In the moving magnet / coil galvanometer, the dial will be unsteady due to varying current through it. In hot wire voltmeter, the principle of heat due to current is used to measure the voltage. P avg V R RMS V RMs R P avg hot wire voltmeter SANTACRUZ ANDHERI GOREGAON KANDIVALI (E) KANDIWALI (W) BORIVALI BHAYANDER VASAI POWAI DADAR SION THANE LOKPURAM (THANE) DOMBIVLI KALYAN PANVEL KAMOTHE NERUL SANPADA KHARGHAR 8 8

10 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS. A x dx B x COM L L dx x dx L bx ax dx 7 L L L bx a dx L bl al 7 L L bl al a b 7 b a a b. mv r r r K.E. r P.E r dr r P.E r r T.E. 5 T.E. r 6

11 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS 4. As current leads voltage thus VL il VB ir i V C i C Since power factor has to be made Effective capacitance has to be increased thus connecting in parallel. C' L C R cos i i L C C' C C' L C' C L C' LC L 5. I 4Icos I 4I max in parallel Imax Now, I 4I cos cos x x 4 d D y y D 4 4d y 4 SANTACRUZ ANDHERI GOREGAON KANDIVALI (E) KANDIWALI (W) BORIVALI BHAYANDER VASAI POWAI DADAR SION THANE LOKPURAM (THANE) DOMBIVLI KALYAN PANVEL KAMOTHE NERUL SANPADA KHARGHAR

12 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS 6. For metals, there is no free motion but rather oscillation about mean position. Thus these have K.E. & P.E., which are almost equal. i.e. P.E avg K.E avg RT T.E. K.E. P.E. T.E. RT per mole R specific heat C M 8.4 C 7 J C 95 kg K F F + a 7. b F F V for upper link Ceq 6 F 5 F F F F Q Q C upper lower on closing switch charge on F is C & that on F is 5C qi q j 5 5C charge 5Cflowsfrom b to a

13 8. At Brewster s angle RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS tan i The reflected light is completely polarized, where as refracted light has both components to electric field. Thus, the reflected ray will have lesser intensity compared to refracted ray. I reflected I 9..8m.6 r.6 7.m /s R.8kˆ.6i ˆ m V 7.ˆj m / s L m R V L 5.76i ˆ 4.kˆ L 4.4kg m s. h m also mr nh r n r n n for n = 4, the de Broglie wavelength is four times that of ground state. SANTACRUZ ANDHERI GOREGAON KANDIVALI (E) KANDIWALI (W) BORIVALI BHAYANDER VASAI POWAI DADAR SION THANE LOKPURAM (THANE) DOMBIVLI KALYAN PANVEL KAMOTHE NERUL SANPADA KHARGHAR

14 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS CHEMISTRY = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =. A B C ABC given : 6. g of A, 6. atoms of B and.6 mole of C yields 4.8 gm of compound ABC. Atomic mass of A 6 amu Atomic mass of C 8 amu Mole of 6 A. mole 6 6. Mole of B 6. Mole of C.6 mole So accroding to reaction A B C ABC.6 C is limiting reagent which consumed. mole So. mole of C formed. mole of ABC. So Mole of ABC wt molecular wt 4.8. of AB C Molecular wt So Molecular wt. of ABC 4 So atomic mass of A Atomic massof B atomic mass of C 4 So atomic mass of B 5amu 6 B 8 4. dipole moment q d d (distance) 8.67Å.67 cm 8.8D.8 esu cm.8 q d So fractional charge Particle chagre Total charge q Q =.5

15 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS. Cl + HO HCl+ HClO 4. FeF 6 oxidiation state of Fe Fe Ar d, 5 F is weak field Ligand 5. Lactic acid is formed in muscles during vigorous exercise. This is due to anaerobic respiration. Glucose Lactic acid + energy C H O C H O energy H 4P O6 7. Gas deviate the most from its ideal behaviour at high pressure and law temeperature. 8. BeCl according to Fajan s rule, covalent nature small of cation. 9. Ef 8 kj/mole ; Eb kj/mole given, A g B g H Ef Eb 4 kg/mole we know that Ef Eb H Ef Eb 4 Eb Eb 4 Eb kg/mole Ef 8 kg/mole SANTACRUZ ANDHERI GOREGAON KANDIVALI (E) KANDIWALI (W) BORIVALI BHAYANDER VASAI POWAI DADAR SION THANE LOKPURAM (THANE) DOMBIVLI KALYAN PANVEL KAMOTHE NERUL SANPADA KHARGHAR 4

16 . C 6 H 4 isomers are RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS. Neoprene is polymer of chloroprene, nch CH C CH CH CH C CH,4addition polymerisation Cl Cl n. Calamine is an ore of zinc.. Order of de brog lie wavelength visible photon > Thremal electron> Thermal neutran 4. Higher the reduction potential easier to reduce and oxidise the other species with lower reduction potential. 5. Dihydrogen is infalmmable gas. 6. Na Cr O 7 is more soluble than KCrO 7 7. Aldol condensation reaction + 8. A pink coloured salt terms blue on heating in the presence of CO 9. CH D-Cl + CH + H D CH + Cl D Both & are formed in equal amounts. H Cl CH D H H Cl CH D 6

17 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS NH. A) Na FeSO4 NaCN Fe FeCN 4 6 Blue ppt B) S Violet colour Na FeCl C) NH C NH NaSCN Fe FeSCN 4 6 Thiourea Red colour. Ice water Volume of Ice is more than compare to water so on increase the pressure reaction shift in the forward direction.. Molar mass of acetic acid in benzene using freezing point depreasion is affected by association.. Addition of phosphate fertiliser to water bodies caues enhamced growth of algae. 4. ) O H and O are fuctional isomers ) O CH CH C CH CH and O CH C CH CH CH are positional iosmers ) O and O H are fuctional isomers 4) O and O H are fuctional isomers 5. - COCl 4 reaction CO H O 4Cl CO Cl 6H O pink 6 4 pink Blue SANTACRUZ ANDHERI GOREGAON KANDIVALI (E) KANDIWALI (W) BORIVALI BHAYANDER VASAI POWAI DADAR SION THANE LOKPURAM (THANE) DOMBIVLI KALYAN PANVEL KAMOTHE NERUL SANPADA KHARGHAR 6

18 6. RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS CH CH N CH Br 5 will form micelles in aqueous solution at lowest molar concentration. 7. Sucralose contains chlorine as it is trichloroderivative of sucrose. Sucralose 8. A B C R K AB Rate... According to condition R K A B Rate... equation eq eq R R R R 9. Incorrect formula is XCl. Dipolemoment e transfer (or) e delocalisation. NH In e transfer is maximum. NO 8

19 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS MATHEMATICS = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =. m tan 6 m m m m or m equation of required line is y x i.e., y x. f x 5 x x 5 f ' x x. f, f f.4 x, log t t t.6.4 dt g t C Differentiating both sides log t t t g t g ' log g t t t log 5 g t 4. In an equilateral triangle incentre & curcumcentre all same & R r Now, r R equation of circumcircle is x y 6 x y x y 4 SANTACRUZ ANDHERI GOREGAON KANDIVALI (E) KANDIWALI (W) BORIVALI BHAYANDER VASAI POWAI DADAR SION THANE LOKPURAM (THANE) DOMBIVLI KALYAN PANVEL KAMOTHE NERUL SANPADA KHARGHAR 8

20 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS 5. cos cos... sin cos... Dividing () by (), tan tan given & sin cos sin cos sin cos cos 6. symbolic form P ~ R Q ~ P ~ R Q ~ Q P ~ R Using Demerojans law 7. 5 adj A 5 5 adj A 5 adj A 5 A 5 A 5

21 8. a b c a b 4c RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS a b c For a point on pout z x y x y 4 By solving we get x 5, y, z Point is 5,, equation line is x 5 y z a a b Shortest distance = b 9. lim f x f x e x lim x x sin x log x k 4 k x 4 x k 4 4k k. End points of bouble ordinate can be taken as t, t & t, t 4 x t t & y t t x t & y t according to given condition. i.e., eliminatry t x t & t y x y i.e., x 9 4 y i.e., 9y 4x SANTACRUZ ANDHERI GOREGAON KANDIVALI (E) KANDIWALI (W) BORIVALI BHAYANDER VASAI POWAI DADAR SION THANE LOKPURAM (THANE) DOMBIVLI KALYAN PANVEL KAMOTHE NERUL SANPADA KHARGHAR

22 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS. b ae a b ae a b e a b also e e e e a. np npq q, p, n 4 p x p x p x 4 4 C Put x both side 4 a a Required probility 7 NOTE : Don t consider equilateral triangle Consider only cases of isosceles t riangle, each case occuring thrice,,,,,,,,,, 4,,5 4, 4, 4, 4, 4,4, 4, 4,5 4, 4,6 5,5, 5,5, 5,5, 5,5, 4 5,5,6 6,6, 6,6, 6,6, 6,6,4 6,6,5 out of which 6,6,5 has maximum area. Hence requaired probility is 6 NOTE : If we consider equilateral triangle, there are 6 occurrences of non equilateral isoscales triangles and 6 occurences of equilateral triangle out of which 6,6,6 has maximum area. so the required probability would have been 69 and not 7.

23 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS 5. a b a b cos...() a b c a b c a b a b c c a b a b...() Now DB AB a b a a a. b c a b a a b c Hence none of the answers is correct. sin x 6. f x dt x Differentiating both sides w.r.t.x f sin x cos x x y f 7. v u gh 48 h h The greatest height 64 6 meters SANTACRUZ ANDHERI GOREGAON KANDIVALI (E) KANDIWALI (W) BORIVALI BHAYANDER VASAI POWAI DADAR SION THANE LOKPURAM (THANE) DOMBIVLI KALYAN PANVEL KAMOTHE NERUL SANPADA KHARGHAR

24 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS 8. a x x x x a x x x x a x x a x x ax x a or For leal & unequal roots D 4a a, a 8 r r 9. The general term is second bracket is C x 8 Total exponent of x is 6 r Term indenpendent of. x y Differentiating we have x x x exp. of x exp. of x exp. of 5 x C C dy dy cos y xcos y sin y dx dx x y r. dy dx dx dy equation of normal is y i.e., y x h y a 9x a y a tan x y h a tan a 9x a a 9x h x a tan 9 h tan h cos asin 9cos x 4

25 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS. AM r Using pythogoras theorem is r r 4 r r r 4 r r r CAM r 5 dy dx. y x y y x y dy y x y dx dy dx dx x y dy y I.F. dy y ny e e y y x y dy c y y solution is x y c y x, y c x y y put y x SANTACRUZ ANDHERI GOREGAON KANDIVALI (E) KANDIWALI (W) BORIVALI BHAYANDER VASAI POWAI DADAR SION THANE LOKPURAM (THANE) DOMBIVLI KALYAN PANVEL KAMOTHE NERUL SANPADA KHARGHAR 4

26 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS 4. n C n 54 n n n 54 n n 8 n 5. 5 n k n n n n k k k 6 6 k 55 6 function is symmetrical about x 6. f x f x function is symmetrical about x 4 & f 4 x f 4 x f x is periodic with period. 5 5 f x dx f x dx 5 5 f x dx A,, & B,, all in the plane AB i ˆ ˆ j k ˆ is in the plane Vector normal of plane i ˆ kˆ i ˆ 5 ˆj 4kˆ equation of plane is 5iˆ ˆj kˆ r iˆ ˆj kˆ iˆ ˆj kˆ 5 5x y z 6

27 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS 8. ar ar 6 & a ar ar ar r r a r r r 6 r, 6 ar r a 5 T7 ar Let Z r cos isin 5 r5 cos5 isin 5 6 Let Z 5 Im Z sin 5 Im Z sin 5 5 sin sin 5 5 dz d 5 4 sin 5cos 5 sin 5 5sin cos sin 5 4 5sin sin cos 5 cos sin 5 sin or sin 4 n n or 4 Z 4 min 4. Selection of three element in A such that f x y 7 C Now for remaining 4 elements in A we have elements in B 7 4 Total number of onto function C C 7 = C 4 4 SANTACRUZ ANDHERI GOREGAON KANDIVALI (E) KANDIWALI (W) BORIVALI BHAYANDER VASAI POWAI DADAR SION THANE LOKPURAM (THANE) DOMBIVLI KALYAN PANVEL KAMOTHE NERUL SANPADA KHARGHAR 6

ANSWER KEY JEE-MAIN ONLINE

ANSWER KEY JEE-MAIN ONLINE - 5 PHYSICS. []. []. []. [*] 5. [] 6. [] 7. [*] 8. [] 9. []. []. []. []. []. [] 5. [] 6. [] 7. [] 8. [] 9. []. []. []. []. [*]. [] 5. [] 6. [] 7. [] 8. [] 9. []. [] Date : --5