Narayana IIT Academy

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IDIA Sec: Sr. II_IZ Jee-Advanced Date: 9-01-18 ime: 09:00 AM to 1:00 oon 011_P1 Model Max.

1 IDIA Sec: Sr. II_IZ Jee-Advanced Date: ime: 09:00 AM to 1:00 oon 011_P1 Model Max.Marks: 40 KEY SEE CEMISRY 1 B A 3 A 4 A 5 B 6 D 7 B 8 ABCD 9 BD 10 ABC 11 BC 1 C 13 B 14 C 15 C 16 B PYSICS 4 D 5 C 6 C 7 A 8 D 9 B 30 B 31 ACD 3 ACD 33 ABC 34 CD 35 D 36 B 37 C 38 D 39 A MAS 47 C 48 B 49 C 50 D 51 C 5 A 53 A 54 AB 55 ABCD 56 AC 57 ABCD 58 C 59 C 60 A 61 B 6 C

2 01. BF3 O 3BO3 F _Sr.II_IZ_JEE-Adv_(011_P1)_GA-8_Key & Sol s CEMISRY Cl3 O 3 OCl SCl4 O SO3 Cl PCl5 O 3PO4 Cl 0. When I is oxidized by MnO 4 in alkaline medium I converts into IO KMnO KO K MnO O O K MnO O MnO 3KO 0 KMnO O MnO KO 3 O alkaline 4 KI 3O KIO3 ence KMnO KI O KO MnO KIO E hc and 1500P P P hc E 10 J P 1 A 10 m 1 1 E 8.8 ev 1 P 8.8 E ev P Maximum when electron jumps from P to n =1 E 8.8eV A 8.8 Minimum when electron jumps from P = to n = 1 3 E A Sec: Sr.II_IZ Page

3 O _Sr.II_IZ_JEE-Adv_(011_P1)_GA-8_Key & Sol s 5. Indicated position is metal to C O and para to hydrocarbon 06. his is Beckmann rearrangement. Group anti to O migrates intramolecularly with 7. retension of configuration Cl Cl O 11. C3CO C3 I C 3 a MeI Me 1. S 6O3 SO4 6O O B 3 3 S a SO a S O SO S E C D 13. S 6O3 SO4 6O O B 3 3 S a SO a S O SO S E C D Sec: Sr.II_IZ Page 3

4 S 6O3 SO4 6O O B 3 3 S a SO a S O SO S E Ksp 4S 3 [ O ] S 10 Ag Cl 10 m Ag 0.05 Ag I 410 m C D m I 10 Ag 7 M AgO KI AgI KO m moles 0.v 0.3v f f 0.3v conc K v 0.1v Conc I v 0.v ConceO v K m v 0.v _Sr.II_IZ_JEE-Adv_(011_P1)_GA-8_Key & Sol s 18. CER Page O. 160 (Part I ) E q p S 0 t sur S S ve, G ve sys total sys Since the value of o G is positive, the indicated reaction cannot be spontaneous ,, S and P 4 (red)/ O produce I on reaction with I Greenwood, pp.810 Sec: Sr.II_IZ Page 4

5 _Sr.II_IZ_JEE-Adv_(011_P1)_GA-8_Key & Sol s MnO 3 O Y X =, Y = 1. Specific rotaion of R-enantiomer = 48 Eanantiomeric encess with 5% S isomer 50% Specific rotaion of mixture x 4 x 4. Conceptual 3. Option f will not perform as required PYSICS 4. Let distances of the pulleys and block from the fixed wall are l, l1 & l as Sec: Sr.II_IZ Page 5

6 _Sr.II_IZ_JEE-Adv_(011_P1)_GA-8_Key & Sol s Also as the force F is gradually increased, the block and pulleys are in equilibrium. From FBD of block and pulleys F F F 1 +F =F and F =F 1 hence F1 and F 3 3 F F Correspondingly x1 and x 3k 3k l l x and l l l l l x Also x From the above; 1 x l 3 5. Both men do equal amount of work as for both As the second man does the same work in lesser time, he delivers more power when the particle is above z=0 plane its radius is and when it is below z=0 plane its radius is herefore, when it crosses z=0 plane for the first time its coordinates are x= -r 1 and y=0. When it crosses z=0 plane for the second time its coordinates are x= -r 1 and y= - r. hird time x=-4r 1 and y= -r cohesive force between water molecules is much weaker than adhesive force between water and glass molecules. Whereas cohesive force between mercury molecules is much stronger than adhesive force between mercury and glass molecules 30. Apply lens formulae let in the current circuit at any later time t is as shown Using Kirchhoff s law Sec: Sr.II_IZ Page 6

7 _Sr.II_IZ_JEE-Adv_(011_P1)_GA-8_Key & Sol s And Also and q is increasing from Solving, we get ence 33. If A= 40 0, then angle of incidence at PR is less than critical angle for all values of i. therefore option(a) is correct If A =80 0, then angle of incidence at PR is less than critical angle for some values of i and greater than critical angle for other values of i. so (B) is also correct. If A= 9 0, then for all values of i, angle of incidence at PR is greater than critical angle and therefore (C) is also correct has two components i.e. And he vector diagram is as shown Similarly, acceleration of A is resultant of 3 vectors as shown below Sec: Sr.II_IZ Page 7

8 _Sr.II_IZ_JEE-Adv_(011_P1)_GA-8_Key & Sol s 35. EMF is induced only in the part AB. ence equivalent circuit is shown 36. Resistance also current density So, R=R 1 +R Solving we get and ow let α, -α and α be surface charge densities at interface of plates and dielectric; and interface of the dielectrics hen So, Where Q= charge of stationary nucleus, q=charge of orbiting particle and m=mass of orbiting particle hen r 0 = r 0 /9, v 0 =3v 0 and E 0 =7E EMF is induced only in the part AB. ence equivalent circuit is shown 41. breaking stress depends only on the nature of the material, so Breaking tension α r Sec: Sr.II_IZ Page 8

9 4. Let coordinates of point P(x,y) (1) _Sr.II_IZ_JEE-Adv_(011_P1)_GA-8_Key & Sol s (),(3) Integrating LS from 0 to x max and RS from v 0 to 0, we get 43. From momentum conservation v 1 = v Loss in KE = first excitation energy of - atom 44. umber of fringes shifted = herefore, now at O there is a minimum And 4n 4 MAEMAICS S dx n r1 r r 10 0 x 3 x n n z a ab,a ab,a ab where is a complex cube root of 1 z1 z a 1 a 3 Sec: Sr.II_IZ Page 9

10 49. Roots are 1, cos,sin number of for which _Sr.II_IZ_JEE-Adv_(011_P1)_GA-8_Key & Sol s sin 1,cos 1,cos sin,cos sin 1are respectively he plane ax by cz d 0 the ratio,,, in 0, divides the segment joining ax by cz d ax by cz d 51. Let tr A x,tr B y given x y 1, x y 3 x y A x,y,z,b x,y,z in Given dy tan x y cos x dx cos x sin x cos x cos x If dx cos x1 cos x 1 cos x cos x e solution of given d.e is 1 cos x 4 when 3 3 x,y 6 8 cos x y sin x c, tan x C 0 Y cos x 53. Required probability C C3 10 C 5 s s a b c ss a b bc bc c a b c b c a b bc b c a b b c a b 55. he feet of from 3 A, S, to given tangents, are 5 5 and B 1 1, tangent at vertex is equation of AB 4x 6y 5 0. Given tangents as per equation of perpendicular. he point of intersection of tangents, lies on directvix. Distance from focus to tangent at vertex is a and 4a is latus rectum. Also semi latus rectum is M of focal segments. 56. Given b ac, x a b, y b c Sec: Sr.II_IZ Page 10

11 _Sr.II_IZ_JEE-Adv_(011_P1)_GA-8_Key & Sol s a c a b c c a b a c x y a b b c a b b c a c b a c b 1 1 x y a b b c ab b bc b a b c b Let g x x 14x 4x,and hx 3 p graph of g x is Identify p such that h x intersect at required number of points (1,11) -3 1 (,8) (-3,-117) 58. x y z a b c a b c A BC A B and 1 1 C A B A B A B A B C A B C A B A B A B A B C A B C A B ow C A B C A B C A B C A B Adding the two result, given C AC A (v1y),y Sec: Sr.II_IZ Page 11

12 Required area= _Sr.II_IZ_JEE-Adv_(011_P1)_GA-8_Key & Sol s = = m he 10 th digit is Put x r cos,y r sin then r r sin 1 1 sin r 4 3 cos sin max r replace x by x, 65. Put f x 1 x f x x f x 1V 1v 1v f x f x f x f x f 0 0 x t n 4 n4 n I t 1 t t... t 1 3t 5t... n 3 t n 1 t dt n n1 3 5 n1 t t t...t d t t t... t I n n 66. Using Baye s theorem p 14 p Sec: Sr.II_IZ Page 1

Narayana IIT Academy

Narayana IIT Academy IDIA Sec: Sr. IIT_IZ Jee-Advanced Date: 8-1-18 Time: : PM to 5: PM 11_P Model Max.Marks: 4 KEY SEET CEMISTRY 1 B C B 4 B 5 B 6 B 7 B 8 B 9 ABCD 1 ABCD 11 A 1 ABCD 1 1 14 15 16 4 17 6 18 19 A-PQRS B-PQRST