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1 INDIA Sec: S. IIT_IZ JEE-MAIN Date: 4--8 Time: 7: AM to : AM GTM-5 Max.Maks: 6 KEY SHEET MATHS PHYSICS CHEMISTRY

2 4--8_S.IIT_IZ_Ph-I_JEE-MAIN_GTM-5_Key&Sol s SOLUTIONS MATHS. Since lines foms a tiangle, 4 cicles touching then ae in cicle and ex cicles d, a, b, c d. a b c. Since le ABC, is equilateal I G and C G A B. Equation of plane is x y z 4 x y z 4 5 d Distance b / w paallel planes is 6 d Paallel line though P,, is Lines on plane x y z 9 t x y z t, 6 Q,, 9 any point Q t, t, 6t P S / 5. Aea S maximum when f 6. a b k a b In nbd of, sin x x. so sin x x and x sin x f x x 8. g( f x x g ( f x. f x g f x 9. f x x x 4 x 4 x x g 5 x x x x dx dx x dx x x x x log x c. Diffeentiate. f x. f x x. f x f x x Sec: S.IIT_IZ Page

3 . Aea x f x c, Put x f c c 6 x 6 f x so f ydx ydx ydx. It is satisfied only when dy dx 4--8_S.IIT_IZ_Ph-I_JEE-MAIN_GTM-5_Key&Sol s a b c. Given homozeonous equations has a non zeo solution b c a c a b 4. x x c 4. x c. x... c. x c. x... Coefficient of 4 k k x c c.c c x x x x x x k. 5 can be distibuted in 4 4 c 6c ways can be distibuted in 4 ways 5 can be distibuted in 4 ways Numbe of ways Since a, b, c, d {,}. 4 n s 6 a b To have unique solution has 6 posibilities c d Pobability S n n n.. n n n n 8. Sum of positive tems can neve be zeo 9. Total aea of given egion. tanx. x a Multiply and divide with sin Sec: S.IIT_IZ Page

4 4--8_S.IIT_IZ_Ph-I_JEE-MAIN_GTM-5_Key&Sol s 4 6 sin cos sin.cos sin.cos sin 7 h. In le PAC, sin h PAsin PA A P C Bh In le PAB, sin PA cos ce PA h PA sin cos ec sin. ap b bp c cp d b c d p a b c 4. p q p q 5. At x is multiple of y then y need not be multiple of x 6. Any vecto along bisecto is a a b b 7. a. b a b let a i j 6k b xi yj zk 4 7. x y z x y z 4 x 8. x x x x x x x x 6 x x x, 4 a. When sides of a tapezium ae equal to a then maximum aea Sec: S.IIT_IZ Page 4

5 FL 4MgL. y Al d l Whee M. kg exat, g 9.8 ms exact L m exact, l.8mm.8 m 4--8_S.IIT_IZ_Ph-I_JEE-MAIN_GTM-5_Key&Sol s PHYSICS l.5 mm, d.4mm.4 m d. mm Substituting the values of M,g,L,d and l in Eq.() we get Y. Nm Fom Eq. () the popotionate uncetainty in Y is given by Y M g L d l Y M g L d l Since the values of M, g and L ae exact, M, g and L. Hence Y d l.mm.5mm Y d l.4mm.8mm Y.5Y.5..5 Nm Since the value of Y is coect only up to the fist decimal place, the value of Y must be ounded off to the fist decimal place. Thus Y. Nm. Theefoe, the esult of the expeiment is Y Y.. Nm. Fom Boh s quantization condition, m nh nh Whee is the speed of the electon, which fom is given by m h Fo gound state, n =. Hence m Cuent aound the obit is (hee T = time peiod) e I But T T Sec: S.IIT_IZ Page 5

6 e I Using () in (), we get eh I 4 m 4--8_S.IIT_IZ_Ph-I_JEE-MAIN_GTM-5_Key&Sol s Magnetic dipole moment IA I eh 4 m eh 4 m t dn. N Ne Ne dt dn Ne dt t t Slope of the gaph is, which is Now T / ln ln ln, which is choice(b) 4., a Given a 5% of nm nm 5. Mean angula deviations poduced by cown and flint glass pisms espectively ae d A and d A Thei dispesive powes ae and Thei angula dispesions espectively ae D A and D A When the pisms ae combined, the dispesion by the combination will be zeo if Sec: S.IIT_IZ Page 6

D D o A A O d d, 4--8_S.IIT_IZ_Ph-I_JEE-MAIN_GTM-5_Key&Sol s o A A 6. 8 9.5 ms k 6 8 Refactive index 8 c n 8.5 Also n, Fo a non-magnetic medium Theefoe n n 4 Dielectic constant k 4 E whee R t / 7.

7 D D o A A O d d, 4--8_S.IIT_IZ_Ph-I_JEE-MAIN_GTM-5_Key&Sol s o A A ms k 6 8 Refactive index 8 c n 8.5 Also n, Fo a non-magnetic medium Theefoe n n 4 Dielectic constant k 4 E whee R t / 7. i e L R di E E e e dt R L t / t/ di t / Induced emf is e L Ee dt t / Fom (), e E e E e i R E R R O e E ir e Using this in () we have E Equation () shows that the gaph of e against i is a staight line with negative slope and positive intecept. So the coect gaph is (d) Q 8. Chage pe unit aea of the disc is. Divide the disc into a lage numbe of R concentic cicula elements as shown in the figue Aea of element d d d Sec: S.IIT_IZ Page 7

8 Chage on the element is dq 4--8_S.IIT_IZ_Ph-I_JEE-MAIN_GTM-5_Key&Sol s Q Q d d d R R The magnetic at O due to the element dq dq Qd Q d db 4 4 R R R The magnetic field at O due to the complete disc is R Q B db d v 9. Q R Thus B R The cuent element SR is paallel to B. theefoe. Hence foce exeted on am SR B I a sin. The cuent element QP is antipaallel to B. theefoe, 8. Hence, foce exeted on am QP B I a sin 8. Fom Fleming s left hand ule, the foce exeted on am PS is F opposite foces constitute a couple which exet a toque = F PQ B I b a This toque will otate the coil in the clockwise sense. B I b but diected out of the page. These equal and E 4. By Ohm s law, the cuent density J and electic field R ae elated as J E ; ( = conductivity) E J ax b J Since the coss-sectional aea is constant, J is constant Now dv E dx dv E dx Sec: S.IIT_IZ Page 8

9 V Integating x dv ax b J dx V 4--8_S.IIT_IZ_Ph-I_JEE-MAIN_GTM-5_Key&Sol s ax ax V V J bx o V V V J bx At x, V Eq () is the equation of a symmetic paabola (as in the case of the tajectoy of a pojectile). 4. a b d E a d Fom gusss s law q E. ds E 4 q a 4 q 8 a q Chage density q 8 a 4 4 6a 4. Since y x, t e ax bt abxt ax bt i ax bt e The eal pat of,, cos y x t is R y x t ax bt Fo a tavelling along x diection,, cos y x t A kx t Compaing () and (), k a and b k b a and i, the given equation can be witten as 4. Let n be the numbe of moles of the gas in the box. The kinetic enegy of the gas n M. When the box is suddenly stopped, this enegy is used up in changing Sec: S.IIT_IZ Page 9

10 4--8_S.IIT_IZ_Ph-I_JEE-MAIN_GTM-5_Key&Sol s the intenal enegy, as a esult of which the tempeatue of the gas ises. The change in intenal enegy is given by U nc T Fo a diatomic gas 5 C R Hence 5R n T n M o mv T 5R 44. Initially the tempeatue of the metal piece falls apidly with time because is high. The fall in tempeatue. 45. Fo an ideal gas, pv nrt tempeatue is non=linea. Finally the piece attains the oom. Diffeentiating, we have since T is kept constant dv p dp dv V V o dp p dv Hence theefoe, choice (a) coectly epesents the gaph of vesus V dp p p. 46. Let us find the foce exeted on wate in the ight side of section ABCD. This foce F = foce f, due to pessue on the left side of section ABCD + foce f due to suface tension. Now f = pessue at the cente X aea of ABCD h P g Rh diected towads ight f = suface tension X length AB T R T Diected towad left h F P g Rh RT P Rh Rgh RT 47. If the gavitational field is zeo at a point at a distance x fom M, then GM GM x x x M m x M 4m o Which gives x, theefoe x. the gavitational potential at GM GM U x x Gm G 4m 9Gm / / x is Sec: S.IIT_IZ Page

11 4--8_S.IIT_IZ_Ph-I_JEE-MAIN_GTM-5_Key&Sol s 48. Let OC RC and let V c be the velocity of the cente of mass of the disc. The linea momentum of the cente of mass is P c Mv c. If L c is angula momentum of the disc about C, then the angula momentum about oigin O is L Lc Rc Pc Magnitude L Ic Rc Mvc sin MR MRc vc sin Ic MR MR MR R R sin c R and vc R MR 49. F x x du d dx dx x Acceleations x F A m x Fom Eq.() it follows that A when x which gives x m and x 4m. Also A is minimum ms at x m. 5. Given h m, R.5m and hoizontal distance s m. When the sting beaks, the stone is pojected in the hoizontal diection, which means that thee is no initial vetical velocity. Fom s ut gt, h gt (i) we have u Sec: S.IIT_IZ Page

12 4--8_S.IIT_IZ_Ph-I_JEE-MAIN_GTM-5_Key&Sol s Whee is the velocity of the stone in the hoizontal diection which is the same as its velocity in cicula motion Eliminating t fom (i) and (ii) we get Now, centipetal acceleation is a c gs ms R hr.5 5. Suface aea of bubble of adius 4. Suface ae of bubble of adius. Theefoe, incease in suface aea 6 4. since a 4 6 bubble has two suface, the total incease in suface ae 4 Enegy spent = wok done = 4 5. Suface aea of one face of the thin plate L L L Total suface aea is A L A L L.%.% A L L 5. The tension in the sting is given by mv T mg l Whee l is the length of the sting. Velocity v of the bob is maximum when it passes though the mean position; then the tension T is also maximum. Now A l l max m m A l m g l g max l lg l m m m Tmax mg lg mg l m m 54. The given netwok of esistances is a balanced Wheatstone s bidge. Theefoe, no cuent flows though the 6 esisto and it can be omitted. The equivalent esistance of the netwok is that of a paallel combination of R R R R R and R R R 4R 6R which is R R R 6R R R R R R 6R The powe deliveed will be maximum if the esistance of the R o R 4 R Sec: S.IIT_IZ Page

13 4--8_S.IIT_IZ_Ph-I_JEE-MAIN_GTM-5_Key&Sol s 55. Feomagnetic substances have a vey high (of the ode of o moe) and positive susceptibility x. Now x 56. The esonant angula fequency is LC / Theefoe, the esonant fequency is V 5 5 Hz 57. Foce of fiction mg. Theefoe, etadation a mg / m g. Also. But p m. ax o am x m p gm x p o gm x 59. The given logic cicuit yields an OR gate Theefoe am x p But a g. Theefoe, IC I E 8. A abd IC 7.8 A. Now.975 I E But 6. (4) 6. () 6. I I C B, Theefoe I. log k 5 = B IC. log k CHEMISTRY 4 A A. k =.97 k ; k k =.97. = (4) 65. () 66. () 67. () 68. () Sec: S.IIT_IZ Page

14 4--8_S.IIT_IZ_Ph-I_JEE-MAIN_GTM-5_Key&Sol s 69. (4) The equied eaction can be obtained by adding the following equation 7. () 7. (4) 7. () FeO(s) Fe(s) + O (g), G C(s) + O (g) CO(g), G FeO(s) + C(s) Fe(s) + CO(g), G G = G + G Fo this eaction to happen, G must be negative, which is possible only at a T > 7 K. Numbe of moles of substance =. Rate of cooling dq =.4 J min dt Duing feezing (t) = (4 ) = minutes. So, heat lost duing feezing = dq = dq t dt =.4 J So, entopy at 4 K = dt dq.4 = 4 = JK mol. 7. () 74. () 75. () 76. () 77. (4) Sol. If thee ae 6 atoms of an element having mass numbe M, atom each of mass numbe M + and M +4, the mean mass numbe is given by 78. () 6M (M ) (M 4) M.75 8 d M 5 Sol..7 kgm Pa dp RT p Sec: S.IIT_IZ Page 4

15 79. () 4--8_S.IIT_IZ_Ph-I_JEE-MAIN_GTM-5_Key&Sol s Sol. P : P : A A 8. () 8. () B X : P : B Total vapou pessue of solution P P X (P )(X ) A A B A A P 5P X A P P X P X A A B A and vapou of (A), P A = PY A ; whee Y A = mole faction of A in vapou pessue. A A A A 5PA XA P P X Y P 5 The mole pecent of A in the vapou in equilibium with solution Sol. 4 I Ce Ce I n n Equivalents of I equivalents of Ce 4+ Equivalents of I in 5 ml.5 Equivalents of I in ml moles of I in ml 4 mass of I in ml 7.54g / lite 8. (4) Sol. Fenkel defect involves the self intestitial type of impefection. So, thee would be no change in density of the cystal lattice. 8. () Sol. Half cell eduction eaction of the given electodes ae Fist electode: AmO e AmO Second electode: 4 4 Thid electode: Am e Am AmO 4H e Am H O Thus it is evident that the half cell eaction of only seconday electode involves H + ion so its eduction potential will change with vaying ph values. Rest two electodes have eduction potential value independent of ph. Sec: S.IIT_IZ Page 5

16 84. () 4--8_S.IIT_IZ_Ph-I_JEE-MAIN_GTM-5_Key&Sol s Sol. Amount of heat equied to lowe the tempeatue of 5 g of wate fom to C J Numbe of moles of ice needed to melt absob this heat Since each ice cube contains one mole of wate. So at least 7 ice cubes ae equied. 85. () 86. (4) Sol. 87. () 88. () Sol. MnO4 KCO O KMnO4 CO KMnO4 Cl KMnO 4 KCl 89. () Sol. Intehalogen compound like B - Cl, I-CI ae halogenating agent fo aomatic compound. Electophilic attack occus though less electonegative halogen. So ICl is a good iodinating agent in pesence of an acid. The given compound has the highest and equal pi electon cloud density at the dotted cicle cente due to stong electon withdawing natue of NO and stong electon supplying natue of NH goup. So electophilic addition of iodine will occu on both these points to give the poduct. 9. () Sol. -OMe and CH ae electon eleasing goups so they deceases the acidic stength fom espective p- and o- positions. Futhemoe deuteium is also moe electon eleasing than potium. So it also deceases the acidic stength. Sec: S.IIT_IZ Page 6

PART- A 1. (C) 2. (D) 3. (D) 4. (B) 5. (D) 6. (C) 7. (D) 8. (B) 9. (D) 10. (B) 11. (B) 12. (B) 13. (A) 14. (D)

PART- A 1. (C) 2. (D) 3. (D) 4. (B) 5. (D) 6. (C) 7. (D) 8. (B) 9. (D) 10. (B) 11. (B) 12. (B) 13. (A) 14. (D) PRACTICE TEST-4 KISHORE VAIGYANIK PROTSAHAN YOJANA (KVPY) 7 STREAM (SA)_ DATE : -9-7 ANSWER KEY PART- A. (C). (D) 3. (D) 4. (B) 5. (D) 6. (C) 7. (D) 8. (B) 9. (D). (B). (B). (B) 3. (A) 4. (D) 5. (C) 6.