From Long Term Classroom Programs and Medium / Short Classroom Program 4 in Top 0, 0 in Top 0, 4 in Top 00, 75 in Top 00, 59 in Top 500 Ranks & 54 t o t a l s e l e c t i o n s i n I I T - J E E 0 FIITJEE ALL INDIA TEST SERIES JEE (Main)-0 ANSWERS, INTS & SLUTINS CRT(Set IV) ANSWERS KEY PYSICS CEMISTRY MATEMATICS Q. No. ANSWER ANSWER ANSWER. C B D. C D C. A A B 4. C D A 5. D B D 6. A D B 7. A C D 8. D B B 9. C D C 0. A D C. A C C. C B B. A C A 4. A C A 5. D A A 6. B B B 7. A C A 8. B B A 9. C B D 0. D C A. B D C. B A B. B A D 4. B C A 5. D D D 6. D D B 7. B D B 8. B C D 9. B B D 0. A A D FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 4606000, 656949, Fa 6594
Physics PART I AITS-CRT(Set-IV)-PCM(S)-JEE(Main)/. Since V L = V C, so it is resonance. V and i = R SECTIN A. Energy lost in the collisions = mu rrel ( e). 4. The equivalent circuit is V V I = R R and current through ammeter is halved I ammeter eq I V = = R V R R A 4. q q V = ; E = 4πε d 4πε d 0 0 5. initially the current in the circuit is zero, and there is no potential drop across the internal resistance. di di L = L =ε, at t = 0. dt dt At any time t, di di L = L =ε ir dt dt 4 6. (. 0 0 0 0 ) (0 0 ) F = Buoyant force = weight F = 0 N in down ward direction 7. % error in V 6% % error in i % maimum percentage error in value of R is 6%. R = (0 ±.). 8. Kinetic energy is maimum at centre and potential energy is minimum. 0.. C F =. 5 9 dv a = dt t v dv dv kv =, dt kdt = v 0 v0 FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 4606000, 656949, Fa 6594
AITS-CRT(Set-IV)-PCM(S)-JEE(Main)/ kt + v0 v v = + = 0 kv v 0 t. When an incident ray I is reflected by a mirror whose normal is N, the reflected ray is given by the following : (I.N)N R = + I (N.N) Using this epression twice, we get the result, j ˆ+ kˆ ˆi R =. θ =60º Let A = acceleration of (5M) (horizontal rightward) FBD of (5M) along X-ais T+ Tcosθ Nsinθ= 5MA N = Mg cosθ= Mg cos 60 = Mg T+ T Mg = 5MA T Mg = 0MA (i) FBD of (M) along X-ais Let a = acceleration of M T + MA = Mg (ii) FBD of M along incline Mg sin θ - MA cos θ - T = Ma Nsinθ y T (horizontal) T a MA T Tcosθ A 4. Mg MA T = Ma Mg MA T = Ma (iii) (i), (ii) and (iii) A = 0 y B B res B 5. λ min for n =. 7. The relative speed between the two image is zero I u I u u FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 6, Ph: 4606000, 655949, Fa: 6594.
4 AITS-CRT(Set-IV)-PCM(S)-JEE(Main)/ 8. mg du = A. d LA y U = l mga d LAY 0 mgl U =. 6AY 9. The torque due to the normal reaction (N) and friction (f) about the C.M. topple the body in the forward direction.. A node is a point where there is no vibration. There is vibration at other points where nodes are not formed.. W = F.dr bdr =.d r dr R b πrb = dr = R = π bj. R. i i C C L L L L 5. This balanced wheat stone bridge 4 4 8 + 4 8+ 4 8 = ; = 4 6 4. = 8Ω Ω 0 T 4Ω 4Ω α P G 8Ω 8Ω 4Ω 7. Use Kirchoff s current law. 8. Using A v = A v (i) and Bernoulli s principle we get P + ρ v = P + ρ v..(ii) From (i) and (ii) (P P ) v = A ρ(a A ) (P P ) Rate of flow = A v = AA ρ(a A ) But P P = P FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 4606000, 656949, Fa 6594
AITS-CRT(Set-IV)-PCM(S)-JEE(Main)/ 5 P rate of flow of glycerine = AA ρ(a A ). 9. Surface integral = EdS = Eds cos π = EdS m 0 E ds Gm0 = ds = 4π m 0 G r 0. In adiabatic compression, the temperature always increases and since PV = nrt, the quantity PV also increases. FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 6, Ph: 4606000, 655949, Fa: 6594.
Chemistry 6 PART II AITS-CRT(Set-IV)-PCM(S)-JEE(Main)/. M C + SECTIN A E G = nfe ;n = ; and = nf E + T T P. cell cell 4. 5. [ ] d = K M + K M K dt n applying ssa for intermediate o K[ M][ ] = K[ ][ ][ M] + K[ ][ ] [ ][ ] [ K M ] = K [ M ][ ] + K [ ] d [ ] KK [ M][ ] Then we get = dt K[ M][ ] + K[ ] And if K [ ] >> K [ ][M] d [ ] KK [ M][ ] Then = = K[ M][ ] dt K[ ] ence K = K (for overall process) So in the given condition. E = E (for overall process) a a [ ][ ] [ ][ ][ ] [ ][ ] S + ( CC ) NaAc S + Ac C C C S + CC C Ac Ac S Ac Ac S S + Ac ( RDS) S S + CC CC 6. (A) Bulky groups are stabilized at anti position. (B) θ N θ 60 o hence lone pair resides in almost sp hybrid orbital, i.e. in an orbital with % in S-character. While in N lone pair of N-occupy sp hybrid orbital so its donation is more easy. 7. + (C) At high p : C C +, so two anionic groups stabilised at anti position. B B B FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 4606000, 656949, Fa 6594
AITS-CRT(Set-IV)-PCM(S)-JEE(Main)/ 7 8. 0. More stabilised group or ion is better leaving group.. S S S. Probability density ψ = (at nucleous r = 0) π s a0. According to Werner s theory, only those ions are precipitated which are attached to the metal atoms with ionic bonds and are present outside the co-ordination sphere. 4. It is cyanide process of etraction of Ag and Au. 5. [Co(en) Cl ] + shows geometrical as well as optical isomerism. 6. Si Si N Trigonal planar. Lone pair is no longer available to donate to any lewis acid. Si N C Pyramidal, l.p. is localized on N. C C 4 8. b = 4 NA π r put the values of b, N A and π and get r = 0.944 nm. 0. This is because acidity increases with increase in the size of borane. In larger borane, the charge formed upon deprotonation can be better delocalized over a large anion with many boron atoms than over a small one. T S = ncv n ;CV Ar = R T. ( ) FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 6, Ph: 4606000, 655949, Fa: 6594.
8 AITS-CRT(Set-IV)-PCM(S)-JEE(Main)/. Me Me o ( ) ( ) Me more stable Me ( less stable), methyl shift Me o ( ) + Me Me Me. Diels-Alder reaction is given by conjugated (S)-cis and (S)-trans conformations. 4. o 490 C C C 5. In (I) lone pair is delocalized to activate the ring while in (III) lone pair is not delocalized. Among I and II, N is more basic than sulphur because the lone pair of N resides in a orbital of high p- character. 6. For E Z priority order is: 5 4 Priority for R S configuration > = bond > R-grp. > C R 4 But least prior group is attached to solid line so result is S. 8. C g( Ac),ether C NaB4,Et g(ac) 5 6 C methylcycloheanol 9. β-ketoacid decarboylate at a faster rate. 0. The is abstracted from the aial position due to stereo electronic factors. FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 4606000, 656949, Fa 6594
AITS-CRT(Set-IV)-PCM(S)-JEE(Main)/ Mathematics 9 PART III SECTIN A. R R = {(, y);, 0 y 5} dom R R = [, ] Range R R = [0, 5] [0, 4] (0, 0) R R and (0, 5) R R (, 4) 0 is related to 0 as well as 5. ence R R does not define a function. (0, 5) (, 4). z + z z + z. For real λ f() = + b + c ( + b) + c b (c b ) f() min = c b g() c + b ( + c) + c + b (b + c ) ( + c) (b + c ) g() ma = b + c By the given condition c b > b + c c > b 4. Since cos (n! π) will be a proper fraction between and + (ecluding 0 and ) and (It is) 0 as m. 5. y = 8 dy y = 0 d y = dy d 5 At point, = = m dy 5 d Also 9 + 5y = 5 8 + 50y dy 0 d = d 5y = dy 9 5 At point, d 5 = = m dy mm = π so θ = 90º = FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 6, Ph: 4606000, 655949, Fa: 6594.
0 AITS-CRT(Set-IV)-PCM(S)-JEE(Main)/ d 6. + +α + +β Put + = z d = dz dz (z +α )(z +β) dz z + ( α+β )z+αβ dz α+β α β z + ( α+β) α+β α β log z + z c + + ( α+β) log z + (z +α )(z +β ) + c 7. Since S. D. Range = (b a) Variance (b a) Var.(X) (b a) or (b a) Var.(X) 8. sin + sin y + sin z = π sin + sin y = π sin z sin y + y = π sin z y + y = z 9. z = z z = z zz = (z )(z ) z+ z =..(i) also z = z + z = z + zz = (z + )(z + ) z+ z =.(ii) By (i) and (ii) z+ z = 0. The system of linear equation will have a non zero solution if Now operate c c and c c, then epand.. B = CAC B = (CAC ) (CAC ) = CA C B = B B = (CA C )(CAC ) = CA (C C)AC = CA.AC = CA C a (a+ ) (a+ ) a (a+ ) (a+ ) = 0 FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 4606000, 656949, Fa 6594
AITS-CRT(Set-IV)-PCM(S)-JEE(Main)/. The number of students answering at least r questions incorrectly = n r The number of students answering eactly r questions (where r (n )) Incorrectly = n r n (r + ) The number of students answering all questions wrongly º = Thus the total number of wrong answer is.( n n ) + ( n n ) n = 4095 n = 4096 n =. b = a + c..(i) pr q = p+ r..(ii) and b q = (ap) (cr) put b from (i) and q from (ii) in equation (iii) then on simplification we get required result...(iii) 4. lim f() = lim (p[ + ] q[ ]) = p() q( ) = p + q lim f() = lim p[ + ] q[ ] = p q(0) = p + + f() is constant at = p + q = p p = q p q = 0 5. Since f () = g().( a) f () > 0 if g(a) > 0 and < 0 if g(a) < 0 f is increasing in the nbd of a if g(a) > 0 and f is decreasing in the nbd of a if g(a) < 0 6. f() = + b + c f() = + b + c f() = 8 + 4b + c By Role s Theorem f() = f() b + c + 7 = 0 f () = + b + c 4 f' = 0 By Rolle s theorem 8b + c + 6 = 0 By (i) and (ii) b c = = 5 8 b = 5 c = 8...(i)..(ii) 7. Put a =, b =, c = 0 0 d π π = = ( + 4)( + 9) ( + )( + 0)(0 + ) 60 FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 6, Ph: 4606000, 655949, Fa: 6594.
AITS-CRT(Set-IV)-PCM(S)-JEE(Main)/ 8. Area of ABC c d ( )d 0 0 (c+ ) d d 0 0 0 d π..(i) 4 Area inside the circle and outside the ellipse π π 4 π (4 ) + 4 C (0, ) A B X 9. Put y z = dy = dz + z d d dz + z = z+φ d z dz =φ d z dz d = + log c φ z Since log c = = y z = dz z φ z = z φ z y φ = z = z y φ = y FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 4606000, 656949, Fa 6594
AITS-CRT(Set-IV)-PCM(S)-JEE(Main)/ 0. In right angled PAC θ CA g + f c tan = = PA S θ tan Now cosθ= θ + tan P (, y ) θ/ θ/ A B C ( g, f). Equation of the tangent at the verte is y + = 0 Equation of the ais of the parabola is + y + k = 0 Since it passes through (0, 0) 0 + 0 + k = 0 k = 0 Equation of ais is + y = 0..(i)..(ii)..(iii) Verte Focus Z A S Where z is a point on directri Solving (i) and (iii) we get A, z is (, ) Now directri is y + c = 0..(iv) But equation (iv) is passes through z + c = 0 c = So directri is y + = 0..(v) Using PS = PM PS = PM. Equation of a tangent to the ellipse cosθ ysinθ + = a b a b A,0 B 0, cosθ sinθ Let P is the mid point of AB, P (h, k) a h = cosθ b k = sinθ Eliminate θ by (i) and (ii). PQ = b tan θ Q = P = a sec θ+ b tan θ since Q = P = PQ sin θ= (e ) sin θ < (e ) < a y + = is b..(i).(ii) Y B (0, 0) P A X P(asec θ, btan θ) Q(asec θ, btan θ) FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi - 6, Ph: 4606000, 655949, Fa: 6594.
4 AITS-CRT(Set-IV)-PCM(S)-JEE(Main)/ e > 4. Since a is perpendicular to b Let V = a+ yb+ z(a b).. (A) Given a.b = 0, V.a = 0, V.b =, [V a b] = By equation (i) V.a = a = a = 0 = 0.. (i) Again V.b = y b = yb y =.. (ii) b Again V.(a b) = z a b = z a b z =.. (iii) a b Put (i)/(ii)/(iii) in equation (A). 5. Probability that problem is not solved by st = = Probability that problem is not solved by nd = = Probability that problem is not solved by rd = = 4 4 Probability that problem is not solved by any one of the three =.. = 4 4 ence the required Probability = 4 4 6. Mean = np = 4 Variance = npq = n = 6, q =,p = 4 4 So P(X ) = P(X = 0) = 4 7. sin (θ + φ) = sin θ cos φ + cos θ.sin φ 6 5 4 = 5 + 5 56 = 65 8. cos θ = sin α π cosθ= cos α π θ= nπ± α 9. h = tan 60º h = tan 0º. h h 60º 0º A B C 0. Put b = ar, c = ar, d = ar, e = ar 4, f = ar 5. FIITJEE Ltd., FIITJEE ouse, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph 4606000, 656949, Fa 6594