INDIA Sec: Sr. IIT_IZ RPTA-6 Date:8--7 Time: : PM to 5: PM 6_P Max.Marks: 86 KEY SHEET PHYSICS D D B C 5 A 6 B 7 C 8 ABC 9 BD AD ABD BCD ABCD ABD 5 A 6 A 7 A 8 D CHEMISTRY 9 D C D C D D 5 ABD 6 ABD 7 ABCD 8 ABCD 9 BCD BD ABCD AC C B 5 B 6 B MATHS 7 C 8 D 9 A A C C AD ABCD 5 BC 6 ABD 7 C 8 BCD 9 AC 5 AC 5 B 5 A 5 C 5 A
8--7_Sr.IIT_IZ_JEE-Adv_(6_P)_RPTA-6_Key&Sol s SOLUTIONS PHYSICS. Mass distribution is volumetric charge distribution is areal. Potential is zero on the sphere of radius with centre at (,, )., and form isolated system 8 i. sin d d sin Maximum value of d is m for 9. By varying d and, we get many 5. A possibilities of P. There are two possibilities of P on the line x y = cube cube Flux due to single wire from whole cube 8 similarly four wires out of twelve will have same contribution and eight will have zero face 8 6. The peak value of the current is I R v v R C when the angular frequency is changed to The new peak value is I v v v I I C R R R 7. Conceptual 8. Zero magnetic field will be recorded at Q when particle is at A and B such that AQ and BQ are tangent to circle. Sec: Sr. IIT_IZ Page
A 8--7_Sr.IIT_IZ_JEE-Adv_(6_P)_RPTA-6_Key&Sol s C o 6 o Q B zero magnetic field will be recorded at Q at time T t T T, T, T, T... 6 6 6 6 where 5 7 t,,,,... T 9. For rotation equilibrium, taking torque about the point from where normal force passes. a a lab lab mgx la B x mg For the block not to topple a mg x l ab q A B. Initially, V k R R and A B R R q q V k. During charging, more current flows in A whereas in steady state it is same. In steady state i v R across capacitor 8 Energy = cv c. Conceptual. Z q tan. Charge flowing in a circuit = Passage I : 5 6 : Let VA volts R Net current entering node C = Sec: Sr. IIT_IZ Page
VC VC VC 65 V C 7 V 7 PD across AC V 7 P. G V / m 6 di 7. L VR dt di i and Li dt 8--7_Sr.IIT_IZ_JEE-Adv_(6_P)_RPTA-6_Key&Sol s 8. When S is closed battery current and hence current through L charges L will again oppose di dt 9. HgI HgI CHEMISTRY Scarlet red Yellow. Conceptual. ) BaCl KI noppt ) BaCl CrO BaCrO (insolubile in CHCOOH ) yellow ppt. CuS has low Ksp precipitated first due low solubility product Ksp. A) Fe Fe CN Fe Fe CN Fe FeCN 6 6 6 Prussion blue Fe Fe CN Fe Fe CN 6 6 prussion blue B) Brown colour solution Fe K Fe CN K 6 Fe Fe CN K 6 Fe K Fe CN 6 6 white ppt K Fe CN K Sec: Sr. IIT_IZ Page
8--7_Sr.IIT_IZ_JEE-Adv_(6_P)_RPTA-6_Key&Sol s Prussion blue C) Fe KSCN no clour ateon Fe KSCN Blood red colour. The green precipitate Ni OH is soluble is eases of NH 6 6 deep blue N OH NH N NH OH 5. Roastiong ZnS ZnO SO a FeS O Fe O 8SO 6. A) Only Cu forms soluble complex not Bi B) Only Ag forms soluble complex Ag NH not Hg and C) CH and Cd both formed soluble complex with NH, Cu NH. 7. Conceptual Cd NH 8. Cl, Br, NO and ClO liberated in these reactions liberate iodine from KI and turns starch to blue. 9. Conceptual. Both K and NH produces yellow ppt. All produced yellow precipitate PbCrO AgI CdS. Conceptual &. OH 5 & 6.,, Cu NO NaOH Cu NaNO CuO H O Black ppt Zn NaNO 7NaOH Na ZnO H O NH CuO H SO CuSO H O C 6 CuSO K Fe CN u Fe CN K 6 SO chocolate ppt Pb Sec: Sr. IIT_IZ Page 5
8--7_Sr.IIT_IZ_JEE-Adv_(6_P)_RPTA-6_Key&Sol s 7. (X) is NH Cl It liberates CrOCl (A) with KCrO 7 and conc. H SO (Chromyl chloride test) CrO Cl NaOH Na CrO NaCl H O Na CrO Pb PbCrO yellow ppt Na NH Cl NaOH NH NaCl H O NH I Hg O Hg NH KHgl Reddish brown ppt n n 7. n n 5 MATHS 8. the possibilities after the first two throws are: /6/6 we have a total of 6 without a double and we stop. 6/66/6 we don't have a total of 6 or a double and we stop. 6/66/6 we have a double and must continue. We need to look more carefully. /6/6 we have a double, we throw more dice and have a /6/6 chance of getting a total of 6. /6/6 we have double, we throw more and have a /6/6 chance of getting 6. /6/6 we have a double or more, we throw again but we can't get 6. So, overall a successful outcome has probability of /6+/6 /6+/6 /6=8/96/6+/6 /6+/6 /6=8/96. 9.,6,...96,7,...97,8,,...98 Sec: Sr. IIT_IZ Page 6
,9,...99 8--7_Sr.IIT_IZ_JEE-Adv_(6_P)_RPTA-6_Key&Sol s 5,,5... x -y = (x-y) (x+y) (x +y ) n(e) = 5 ( C )+6( C )( C ) = 5.. conceptual. m m.. n 9m n m n m m n m.. m 8n n n n 9n. x + y = xy xy 75 x (-x) x -x + 75 x x 5 5 5 x 5 x p( E). P( E F) P( E) P( F) P( E F) P( E) P( F) 7 ( P( E))( P( F)) P( E) P( F) P( E), P( F) OR P( E), P( F) P( A B). 6 6 P( B C) P a is even and b is odd Sec: Sr. IIT_IZ Page 7
8--7_Sr.IIT_IZ_JEE-Adv_(6_P)_RPTA-6_Key&Sol s P( A C) P a is odd and b is even P( A B C) P 5. -(-m)(-p)(-c) = (or) m + p + c - mp - mc - pc + mpc = / mp(-c) + pc(-m) + mc(-p) + mpc = mp + mc + pc-mpc =/ mp(-c) + pc(-m) + mc(-p) =/ mp + mc + pc -mpc =/5 7 mpc, mp mc pc 5 5 P P P,, 6. Conceptual 7. conceptual. 8. conceptual 9. conceptual 5. 5. Probability that lock will opens at the (k-m) trail n n n... n n n n ( k ) n. 5. Required Probability = Probability of (k-) successive failure then one success at the K th trail k n. n n 5. Probability that 'Y' wins a match in th game =.c.b + 6.c.ba = c b(b + a) 5. Let E n = even that 'x' wins in the n th game n probability that 'x' wins n PEn P E n n n n n n. a b n n a b c n n a n b a c. ( n( n ) b ) n n a b b a c. a c a c a a c a c a a c Sec: Sr. IIT_IZ Page 8