Narayana IIT Academy

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1 IDIA Sec: Sr. IIT_IZ Jee-Advanced Date: 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 A-PQRS B-PQRST C-PQRS D-PQRS A-QRT B-PRT C-ST D-PRT PYSICS 1 A B C 4 B 5 A 6 D 7 B 8 C 9 ABC BD 1 BC AC A-PR A-PQR B-QR C-RS 4 B-QST C-PQRT D-RT D-PQRT MATS 41 B 4 A 4 B 44 C 45 A 46 D 47 D 48 C 49 BC 5 BC 51 BD 5 AC A-Q B-S C-R D-S 6 A-Q B-P C-R D-S

2 8-1-18_Sr.IIT_IZ_JEE-Adv_(11_P)_GTA-8_Key & Sol s SOLUTIOS CEMISTRY. OPh SO4 E reaction is not possible. O O Br 4. Br Fe CPh P CPh Q O 5. C C COO Br red P C Br COO + C COO C C 6. For Li +, 4 th excited state = p de G 8. G nfe dt nft 1. Sol : 5 O O E oxidized = E reduced = s x 1 y s Sec: Sr.IIT_IZ Page

3 8-1-18_Sr.IIT_IZ_JEE-Adv_(11_P)_GTA-8_Key & Sol s C a Br C C C C liq. C 4 Br Br C C C Et C C Et a D liq. i D D D D C C C C -Pd BaSO 4 C C Et Et Et C C C -Pd BaSO 4 cold.dil. alk.kmno 4 O O O O. C C 1. A PYSICS d dt = t for d dt to be minimum; d dt = d dt = t d dt = t (t ) = or =. B t t The peak value of the current is. Sec: Sr.IIT_IZ Page

4 I = R v v 1 R C _Sr.IIT_IZ_JEE-Adv_(11_P)_GTA-8_Key & Sol s when the angular frequency is changed to The new peak value is. (C) v v v I ' ' I 4R R R I C Q A Q' (r r ) Q 4r = B In the circular motion around the Earth, the centripetal force on the satellite is a gravitational force. Therefore, v = GM /R, where M is the mass of the Earth, R is the radius of the orbit of satellite and G is the universal gravitational constant. Therefore, the kinetic energy increases with the decrease in the radius of the orbit. The gravitational potential energy is negative and decreases with the decrease in radius. 5. A The slab does not contribute to deviation. For minimum deviation by prism, r1 = r = as shown in figure. sin i = sin or i = 45 Minimum deviation = i A = 9 6 = 6. D Sec: Sr.IIT_IZ Page 4

5 idcos db = 4 a cos _Sr.IIT_IZ_JEE-Adv_(11_P)_GTA-8_Key & Sol s = a tan d = a sec d 4 a db = icos d (d in radians) icos = d 4 a 18 (d in degrees) 7. B 8. C = icos d (d in degrees) 7a As the jar is pushed down, due to increase in hydrostatic pressure volume of gas trapped decreases. ence net bouyant force decreases. Since angular velocity is constant, acceleration of centre of mass of disc is zero. ence the magnitude of acceleration of point S is wx where w is angular speed of disc and x is the distance of S from centre. Therefore the graph is 9. A, B, C Zero magnetic field will be recorded at Q when particle is at A and B such that AQ and BQ are tangent to circle. zero magnetic field will be recorded at Q at time T, T T, T T, T T... where T = t = ,,,,... Sec: Sr.IIT_IZ Page 5

6 8-1-18_Sr.IIT_IZ_JEE-Adv_(11_P)_GTA-8_Key & Sol s. (B,D) 1. B, C x = sin 1 t + 8 cos 5 t = sin 1 t + 8[1 cos1t] x = 4 + sin 1 t + 4 cos 1 t (x 4) = 5 sin (1t +) Amplitude = 5 units Maximum displacement = 9 units. 4 tan Applying Gauss theorem to volume containing cuboid indicated by ABCD E q EA enc A k A or qenc = k Electrostatic energy stored in dielectric medium = 1 E k k At = E At k.. (A,C). Energy available for excitation = E/ if E <.4 ev excitation will not take place. For E <.4 ev collistion is perfectly elastic and neutron must come to rest. v = 5 V 1 e = 1 GM R Applying energy conservation GMm 1 GMm mv R (R h) v = GM GM R R h 1 GM. = 1 1 GM 4 R R R h Sec: Sr.IIT_IZ Page 6

7 1 4R = h R(R h) _Sr.IIT_IZ_JEE-Adv_(11_P)_GTA-8_Key & Sol s R + h = 4h h = R/ 4. 1 Capacitor is directly connected with 1 V cell in parallel to it Q = CV = 1 1 = 1 µc. 5. Tension in rod at a distance x from right edge is T = F( x L ) T dx YA net extension in rod = L = F L YA 6. 5 The plate is free to rotate about vertical axis yy. Let v, vcm and be the velocity of particle, velocity of centre of mass of plate and angular velocity of plate just after collision. From conservation of angular momentum about vertical axis passing though O is mu a = mv a + ma...(1) since the collision is elastic, the equation of coefficient of restitution is vcm e = u v = 1...() But vcm = a...() solving equation (1), () and () we get = 1 u 7 a Sec: Sr.IIT_IZ Page 7

8 _Sr.IIT_IZ_JEE-Adv_(11_P)_GTA-8_Key & Sol s Q = U + W where U is the change in the internal energy of the gas; and W is work, done by the gas. For one mole of the monatomic ideal gas U = /R T. Work equals the area under the graph P vs. V The equation of straight line is given as p p V p V.(i) p p V p at point 1, 1 1 V (ii) Therefore, for the process from the initial state with P1V1 = / RT1 to the state with P,V,T the heat given to system is Q = (/) R (T T1) + (1/) (P + P1) (V V1) = (PV P 1V1) + 1 (PV + P 1V + PV1 P1V1)... (iii) = PV + 1 P 1V 1 PV 1 P1V1 from equation (i), (ii) and (iii), we get P 5 5 V1 V P V P V1 V 4 V Q = The process switches from endothermic to exothermic as dq dv changes from positive to 8. (8) negative, that is at dq dv =. Solving we get V = 5 8 V W Q Q 1 U f Ui, C' C C C C' 9 9Q C Q C 9E E 8E n 8 Sec: Sr.IIT_IZ Page 8

9 9. (A) (p), (r); (B) (q), (r); (C) (r), (s); (D) (r), (t) A is instantaneous centre of rotation. mgx I A a cm y mgxy I A mgx mg sin and ma cm f cos min ma cm f I A I cm mr I A mg ma f ma _Sr.IIT_IZ_JEE-Adv_(11_P)_GTA-8_Key & Sol s cm cm R Y x y f a cm COM x R y mg A For semicircular homogenous ring R x, I A mr 4 1, y R = g R a cm g mg 1 5 f mg =.98 min f 5 4 For semicircular homogenous disc 4R, I x A mr, y R Sec: Sr.IIT_IZ Page 9

10 8g 9 R _Sr.IIT_IZ_JEE-Adv_(11_P)_GTA-8_Key & Sol s a cm 8 g mg mg f =.1 9 f 1 min 119 For homogenous hollow hemisphere R 5 x, I A mr, y R g 1R g a cm mg 1 mg f = f 6 min 17 For homogenous hollow hemisphere R 7 x, I A mr, y R g 56R 15 7 a cm g mg mg f = min f (A) (p), (q), (r) ; (B) (q), (s), (t) ; (C) (p), (q), (r), (t) ; (D) (p), (q), (r), (t) Sec: Sr.IIT_IZ Page 1

11 8-1-18_Sr.IIT_IZ_JEE-Adv_(11_P)_GTA-8_Key & Sol s MATS 41. a,b,c are in G.P. In a, In b, In c are in A.P. Let In a,ln b,ln c,, are in A.P. and,, are in A.P or But Common difference 4. Conceptual sin1x sin x 4. sin x sin 4 x... sin x sin11x sin1x cosx cos4 x... cosx cos11x sin x Squaring and adding, sin 1x 1.cos sin x 1nm1 m n x k k i k i k k i k i k k k i i Sec: Sr.IIT_IZ Page 11

12 45. Conceptual _Sr.IIT_IZ_JEE-Adv_(11_P)_GTA-8_Key & Sol s 46. There exists no x value for f(x) = and SP S ' P a f f 48. First adding second row to first, third row to second.upto 1 th roe to 99 th row. 49. Then subtracting first column from second and so on, we obtain det A f x K f x ' f x Kf x f x f x ' ' kx kx e f x e f x 1 1 kx kx, Let g x e f x g x e f x g x. g x one of g x, g x is decreasing Let 1 g1 x be increasing but g1 g1 x f x Let g x be decreasing but g g x f x,1 f x x 5. Replacing x by x x 1 1 and 1 x respectively, x 1 f x tan x tan tan 1 x x f x f 1 x tan x tan tan 1 x tan x 1 x So 51. Conceptual P P 1 P n n1 n1 1 x 1 1 x tan tan x x 1 Sec: Sr.IIT_IZ Page 1

13 8-1-18_Sr.IIT_IZ_JEE-Adv_(11_P)_GTA-8_Key & Sol s 1 Pn. 4 4 Pn 1 1 Pn Pn n1 1 Pn P So P n k1 1 k1 n n1 1 n k. S. x dx k. Permuting summation & integral, 1 n k1 k1 1 n k. S. x dx k 1 n 1 log 1 x n dx log 1 x dx 1. x iy A ib x iy i x y 54. Z Z Z Z iz 1 C id x iy i x iy 1 Z iz 1 So, n A B nc D. 55. Conceptual 56. Conceptual 57. A B tan A tan A i Sec: Sr.IIT_IZ Page 1

14 tan A tan A 1 tan A 5 tan A 5 f x x 5x 5 f ' x x 1x x, Let P x A x x x P x A x x x ow P ' x Ax x As ' P x divides P Considering x, So _Sr.IIT_IZ_JEE-Adv_(11_P)_GTA-8_Key & Sol s P ' ( x) & Px must have two common roots. & as the common roots, we get i & i So, Conceptual 6. Conceptual. Sec: Sr.IIT_IZ Page 14