ANSWER KEY JEE-MAIN ONLINE

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

2 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS PHYSICS = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =. F f ma and fr I for Rolling a R F f ma...i F mr fr a. R f f ma...ii so, f ma ma f ma. I 5A I A I 5 A t.sec Emf = 5v emf di L dt 5 L. 5 L 5 L.67 H. speed v m / sec v a R v constant a soa R R

3 . Emwave RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS f Hz E 7 V / m we know E B C 8 so 7 B 9 T 8 8 now for.5 6 m so 8 B 9 sin t kˆ k F m r k is some constant so mv km v r constant r r T v T r straight line r 6. Time = 5 min = sec. Volume = 5 ltr. = 5 m Diameter = cm Area of tap = M Velocity of water = R w Vd V At 5 m /sec 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 7. For cylinderical T P R no option 8. de Kh d h / E Kh d h /...(i) I d h / h tan d hsec d I h tan hsec d h sec sin d h cos h I h h h Put it in equation (i) h E h a a h a a o E h a ch C A

5 9. C e a h c RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS y z a Capacitance C M L A T e AT c LT h M L T a L y z a M L A T AT L M L T LT Compare = z = M L A T A T M L T y a L LT y T L LT a So e a u hc a y. Using perpendicular ais theorem. Ma I a I m f. angular magnfication on m f e 5 5 so, tan tan 5 tan tan 5 β 6 SANTACRUZ ANDHERI GOREGAON KANDIVALI (E) KANDIWALI (W) BORIVALI BHAYANDER VASAI POWAI DADAR SION THANE LOKPURAM (THANE) DOMBIVLI KALYAN PANVEL KAMOTHE NERUL SANPADA KHARGHAR

6 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS. Vd E I neav d V and V E l I so E V and I V d so I V I V. Electron concentration in n-region is more as compared to that in p-region.. sin t sin t at t = & velocity both has to be zero 5. Py d h to go straight otherwise diffraction will occur 6. Circuit a i i e for b t / RC i i e t / L / R so (b) (a) 6

7 7. Energy of proton = RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS mv qv v qv m R v magnetic force = q v B mv R d mv R qb d dqb dqb m sin R mv m qv sin Bd q mv 8. momentum P me and E ev h p h me M.7Å so value =.56 + error = =.59 cm 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

8 nh. mvr... i RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS and mv qvb R mv so qb...ii R qb nh mv mv. mv qbnh. m mv nhqb m hqb E n m. a sin wt y a sin wt y asin wt cos wt y a y a a a. On outer surface there will be no charge. So Q On inner surface total charge will be zero but charge distribution will be there so Q 8

9 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS. If block to come to rest after m. u So, d a a g m.5.5 / sec V u.5 u m / sec by momentum conservation 5 5 V kg =.5 M approimately 5 V V m / sec V so to get V m / sec V gh 8 h m. The block comes to rest means its velocity at that point was m/sec. So that point K. E. mv..9.5j K. / at displacement P. E. T. E. K. E. T. E. So T. E..5 =.6 J 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 5. f ' 8 m/sec 8 = 856 Hz f = 8 Hz 6. P a b d c V 7. l cm r cm N 5turms i 5A 55 M AM 8. mv F t l distance t v velocity t v F v and K. E. T mv T V T F T

11 9. u cm v 5cm f? v u f 5 f R 5 5 R RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS R 6cm 5. 5 i Amp 5 V volt v i 5v.6 V CHEMISTRY = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =. Metal oide calculation is the process of heating ore in abscence of air e.g. Calcium carbonate gives CaCO CaO+ CO as calculation.. Only temporary hardness which is due to HCO (Bicarbonate) ions is removed by Boiling. N O NO. g g t t equilibrium where Degree of disociation. Mol.wt Miture Now, As per Ideal gas Equation PV nrt PM miture drt M M NO NO PM mi d. g/l RT 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. Starch is a miture of amylose & amylopetin poly saccharides and monomer is glucose. 5. Cis Ni HO NH & facial Ni H O NH have optical isomers. 6. Fluorine is the most electronegative element & has least tendency to form double bonds. M 7. Mili equivalents of H SO 6 M Mili equivalents of NaOH Mili equivalents of NH. N V NH W % of Nitrogen.. 8. Drugs which relieve pain is called analgesics. Non -narcoties :- aspirin (acetyl salicylic acid) paracetamol (parahydrous acetanilide) 9. In XeOF, Xe is sp d, hybridised having 5 Bond pairs & lone pair SHAPE: Square Pyramidal

13 . Basicity + M group RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS M group ( L.P is not involved in conjugation it is more basic ). The sequence of filling electrons in sith period : i.e. 6s f 5d 6 p ns n f n d np. (Thermal stability of Ionic character hydrodies) or Lattice energy Mg OH < Ca OH < Sr OH < Ba OH. Weight of Hydrated BaCl = 6g Weight of Anhydrous BaCl = 5g Loss in mass 9g Assuming BaCl. H O as hydrate Mass of HO = 9g 9 Moles of HO = =.5 8 Grass molecular let of BaCl = 8 9 % of HO in this hydrated BaCl = =.75% on solving This percentage is present in BaCl.HO 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. Stable due to aromatic 5. Polymer Use Polystyrene Manufacture of toys Glyptal Paints and lacquars. P.V.C Rain Coats Bakelite Computer discs s 5p valence shell configuration of group 6 element present in period 5. Tellurium (Te). 8. After adsorption these is decrease in the residual forces due to bond formation. G, H & S all are negative in the case of adsorption, 9. EMF of galvanic cell.volt If Eet from cathode to anode as polarity is changed. EMF then electron flows steadily from anode to cathode while If Eet EMF there electron flows. Fe, radical gives blood red with SCN Cl radical gives chromyl chloride Test.

15 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS. Chemical pollutants in photo chemical smog are Nitrogen oides (NO and NO ), volatlie organic compounds, Ozone (O ), peroyacetyl nitrile.. Hell - volhard- Zelinsky Reaction. + Co ion is precipitated by HS & NHOH present in group IV cationic Analysis. At high pressure real gas particles are easily compressed which was later gives by van-der waal s equations as per kinetic theory. 5. First order reaction rate = KNO5 N O NO + O 5 g g g t pressure 5 t min pressure 5 p p p Total pressure 5 p p p 5 p 87.5 mm Hg P.5 mm Hg P 5& P 5 for N O reactant t K log log min 5 6 min On solving.5 mm Hg 5 p P 8.75 mm Hg Total pressure 5 p 6.5 mm Hg 6. Total V.P. of solution P X P X A A B B Total V.P. of solution torr torr 8 torr Mole fraction of Benzene in vapour form SANTACRUZ ANDHERI GOREGAON KANDIVALI (E) KANDIWALI (W) BORIVALI BHAYANDER VASAI POWAI DADAR SION THANE LOKPURAM (THANE) DOMBIVLI KALYAN PANVEL KAMOTHE NERUL SANPADA KHARGHAR

16 7. RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS 8. Ligand donats electrons to metal. In methan there is no electron to donate it is stable with octet configuration. 9. In CH BE 6 KJ/mol BE 9 KJ/mol C H C H CH6 BE 6 BE 6 KJ/mol In CC CH BE CC 8 KJ/mol hc Now, E BE nm 8 6. CC 8 J/molecule. Assertion is correct but reason is Incorrect. Bonding MO involves contructive interference. MATHEMATICS = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =. Let average wage of Night shift worker is Conduction for Rolls theorem f() = f( ) and f ' c and b b c 6

17 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS. Solving y y Points of intersection (, ) and (, ) Area = d d (,) (, ) 8. A, B, C 89, Slope of AB Slope of BC So, lies on same line 5. Question incomplete ! 5! Hence, Option is not matching f tan tan 8. f f 5 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 9. ; i 9 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS. hyperbola is y 9 foci, e e e equation of ellipse a y b, a foci of ellipse a b a. b b equation of ellipse y 8

19 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS. aa a a a d sum of last four term = 78 a a a a n n n n 78 Their mean.5 a.5.5 n = 9 Median = n p 5% nc 5% n P' C ' 65% C 5% n P n P C np nc np C % 5%,. / lnt f dt r t let t z dt dz z ln z dz z z ln t f dz r z ln z ln 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 n n n. C r C r C r 7 By solving line get r = 6 so, if is 7 th term. 5. a b angle between a and b is 6. a b is r to plane containing a and b c a b a b c a b a n a b n. cos 6 sin 6 a b sin 6. n n n r c 9 c 7 7 / 55/ c 55 so, h sin 5 6. Equation of line L y (,) y... line y... solving there two /5, 5

21 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS 7. sin A sin B sin 5 sin 5 cos5 sin5 8. OP sin OB cos 6 Area = OP OB sin cos sin A P (o,h) B least value sin 5 9. e sin Lim sin cos Lim e sin cos. r r 6 = 778 r6. Given that y equation of a choral of the circle then y...i y so the equation of the circle is y y 9 y y 9 y y 9 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. d / 5/ d / put E dt d. Given that cos t t sin t d so, sint t cost sin t dt d cost t sin t cost dt dy t sin t...ii d dy sin t t d t sint dy tan t d dy d t / so the slopw of the normal is and at l / and y / to the equation of normal is / / y y / y so the distacne from the origin is

23 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS y z. L y z y 7 L so point P(,, ) [on line L ] point on other line [on line L ] Q (,, ) [on line L ] Q (,,) PQ iˆ ˆj kˆ vector II to line L iˆ ˆj kˆ iˆ ˆj kˆ vector to r 8 L iˆ ˆj kˆ n iˆ ˆj kˆ distance = PQ. mˆ 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 5. Given that A A A I A A I A I A A I I A A A A 6. Required probability is C C C... C C 7. y 6 y 6 y centre of circle (, ) y 6 7 8, The centre positive of the statement is If i will come, then it is not raining

25 RAO IIT ACADEMY / JEE - MAIN / ONLINE EXAM - 5 / SOLUTIONS 9. Radius CP y 5 5 C(,) d dt (,) (, ) P (,) +y= dt t / t lt t / c. dy d 9 dy d 9 dy 9 d dy 9 d y d y d d y log c Given that y = () = log z c y log log y 8 8 log log = SANTACRUZ ANDHERI GOREGAON KANDIVALI (E) KANDIWALI (W) BORIVALI BHAYANDER VASAI POWAI DADAR SION THANE LOKPURAM (THANE) DOMBIVLI KALYAN PANVEL KAMOTHE NERUL SANPADA KHARGHAR

ANSWER KEY JEE-MAIN ONLINE

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.