Narayana II Academy INDIA Sec: Sr. II_IZ (Incoming) CU- Date:-5-8 ime: 7: AM to : AM 6_P Max.Marks: 86 KEY SHEE PHYSICS C A C 4 D 5 A 6 AC 7 AD 8 AD 9 ACD AC ACD ACD C 4 5 5 5 6 7 6 8 CHEMISRY 9 C A D A 4 AD 5 CD 6 AC 7 8 A 9 C AC AC 4 7 5 6 9 MAHS 7 A 8 9 A 4 C 4 A 4 CD 4 AC 44 CD 45 AC 46 CD 47 A 48 D 49 CD 5 5 5 6 5 8 5 54 7
Narayana II Academy ' Q r r. Q A.5 Q 4 r -5-8_Sr.II_IZ(Incom)_JEE-Adv_(6_P)_CU-_Key&Sol s SLUINS PHYSICS. Slab does not contribute to deviation sin i r r i 45 sin r Minimum deviation = i A. Since angular velocity is constant, acceleration of centre of mass of disc is zero. Acceleration = x for the point S 4. Force on the container by escaping liquid = S gh. Acceleration of the container = Sgh g Sh g Sh Sh g Shg 5. H U ; a is same for A,, C a USin U U Sin 9 H A ; H ; HC H H A HC a a a 6. ft Acceleration of block =f/m. If slipping stops in time t then v at. m For the plank ; F f p Fv and w pt Fvt fvt mv v Heat produced = w KEgain mv mv mv q R q R A 7. k... ; qa q k... R R From equation () and (), qa q. When is earthed A A q q q q Sec: Sr. II_IZ Page
Narayana II Academy -5-8_Sr.II_IZ(Incom)_JEE-Adv_(6_P)_CU-_Key&Sol s qa qa A 4 R R. ut q qa in eq 8 R We get, qa 4 R 8. A A 6.5 6 Req ; I 7.5mA 7.5 5 R eq 7.5 5 ; 4 5 9 5 P ; P : ; P 54mw R R RL R R RL When R and R are interchanged,.5.5.5 Req 6 L 4.5.5 6.5 P 6mw.5 L L RL u ' ' cm cm ext mul mu L u 9. mu m const F mu mu Pf mu mu u u. de Wdy A Y ky W Y ky W kl Integrate e log log A k Ak Y L W kl Strain energy = Fe W log Ak Y. 6 4 dv F r ma r dt Sec: Sr. II_IZ Page
Narayana II Academy 6 F 4 6 r r r dv dt -5-8_Sr.II_IZ(Incom)_JEE-Adv_(6_P)_CU-_Key&Sol s 9 dv 9 t F dt ln F 6 r r r 6 yd. Path difference ; D Phase difference path difference I I cos. Conceptual 4. From angular momentum conservation, mu mv... 5. For elastic collision, ut cm a.... cm e... u u From equation (), (), and (),. 7a R U ; W Area under P- graph R a a ma Equation of straight line is P P P... At P Pt, P P... For the process, from initial state of P R to the state of P,,, heat given to system is Q R P P Sec: Sr. II_IZ Page 4
Narayana II Academy -5-8_Sr.II_IZ(Incom)_JEE-Adv_(6_P)_CU-_Key&Sol s P P P P P P P P P P... P 5 5 From equations (), (), (), Q P P 4 his process switches from endothermic to exothermic as dq d dq negative i.e. d. Solving, we get 5 8 changes from positive to 6. 7. 8. Y F F Y Y and A A 4S 4 P 4S 4 4 5R P R R 5R 4 96S 6nS P R R x 6 6 x 6 x x 6 x i 9. conceptual. key: CHEMISRY For fluorides, m.p. decreases down the group. Ionic mobilities in aqueous solution increase down the group.. key: A Hint: SN attack at primary carbon no change in the confugiration. key: D Hint: e is most acidic and c is least between d & f, f forms carbanion which goes into conjugation with ring so more stable. Along with conjugation with oxygen which is more EN. key: A Sec: Sr. II_IZ Page 5
Narayana II Academy Sec: Sr. II_IZ Page 6-5-8_Sr.II_IZ(Incom)_JEE-Adv_(6_P)_CU-_Key&Sol s Na C into NaHC Hint:Since, phenolphthalein indicates only conversion of hence, x ml, of HCl will be further required to convert NaHC toh C. So, total volume of HCl required to convert Na C into H C x x x ml 4. key: A,,D Hint: 6NaH + Cl 5. KEY: CD NaCl + 5NaCl + H 5 n f Hint: i. Since t / is independent of concentration of A at constant ph means that the order w r t [A] = x ii. [ ] x Let r H r when ph x r r when ph r hus, r x x Hence order w r t H Hence the rate So dx k AH dt dx dt x r k A H dx k A H dt H r k A r hus r r is the correct answer. r k A H... i r k A H... ii r k A H r 4 4 r 4r
Narayana II Academy 6. Cannizaro reaction 7. key: Hint: Rearrangement 8. conceptual 9. key: C. KEY: AC -5-8_Sr.II_IZ(Incom)_JEE-Adv_(6_P)_CU-_Key&Sol s H i) ollens ii) H H CH R CH Mg Pr r CH CH H H H H S H / HF 4 H 5 H Hint: CH C CH H 5 H H. Key:- AC a P b R R a P v v Pdv v v R a dv dv v b v v. key: H Hint: orax H H 4 H 4 H H 4H. key: Hint: W 5 7.5 4. Given energy of ideal mixing Sec: Sr. II_IZ Page 7
Narayana II Academy maxg nr X ln X X ln X A A -5-8_Sr.II_IZ(Incom)_JEE-Adv_(6_P)_CU-_Key&Sol s X A X mixg 484 ln ln 48.4 ln 6.96KJ 5. conceptual 6. key: 9 NaC. H after effloresces. MAHS 7. Perimeter of Region R = perimeter of trapezium ACD where A = (, ), = (, ), C = (4, ) and D = (, 4) 8. Conceptual. 9. Solving the given equations, we get 4. hx b a x a x a b D b a a for maximum area e e e e e x x, x, x, x, x x x y e - x 4. g x hx is not differentiable at x,, f g x f g x 8 and f() = 5 g(5) = 4. log f x r r log x r r Differentiate and then put x = 4. f x sin 4xdx Sec: Sr. II_IZ Page 8
Narayana II Academy... 4 44. ;... etc. K 9 47, K 6 f x x 45. 46. (A) 47. AA A A A A Need not be symmetric () A A A A Need not be skew symmetric adj A (C) A adj A A Now, c adj A adja c adj A adja adja adja c skew, symmetric matrix (D) A & A A A A 5 Let c Now 5 5 5 5 5 5 C A A -5-8_Sr.II_IZ(Incom)_JEE-Adv_(6_P)_CU-_Key&Sol s s s a s b s c. s s a s s c s s a s c s b s s b a c b a,b,c are in A.P Also a c ac Also A tan tan A C tan tan tan 48. PQR ab sin sin It is maximum when 49. f x x 8x 5 x 5x f is in (5, 6), in (, 5) Sec: Sr. II_IZ Page 9
Narayana II Academy x x x x -5-8_Sr.II_IZ(Incom)_JEE-Adv_(6_P)_CU-_Key&Sol s 9 5 6, 5 g x 9,5 x 6 x 8, x 6 5. Let the vertices of the heptagon be at the 7 th roots of unity, then 6 a 5 b 4 c 5. f x x 5. ACD is a rectangle. A t, C t, t t, t t, t t Area t t 5. x f x has roots,,,..., 9 9! 5 x f x A x x... x 9 A, f 54. Conceptual Sec: Sr. II_IZ Page