AITS-CRT-I PCM(S)-JEE(Main)/4 From assroom/integrated School Programs 7 in Top, in Top, 54 in Top, 6 in Top 5 All India Ranks & 4 Students from assroom /Integrated School Programs & 7 Students from All Programs have been Awarded a Rank in JEE (Advanced), FIITJEE ALL INDIA TEST SERIES ANSWERS, HINTS & SOLUTIONS CRT I (Main) JEE(Main)-4 Q. No. PHYSICS CHEMISTRY MATHEMATICS. C D A. B B C. C C C 4. D C B 5. D C A 6. A A B 7. C B C 8. D D D 9. B B A. C A B. C C C. B B B. D B B 4. D A C 5. A B B 6. C B C 7. B B A 8. B D B 9. B B C. B B B. C C C. A C B. D D A 4. B B A 5. B C B 6. C C C 7. D D C 8. B B C 9. C C B. B A C FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, 656949, Fa 6594
AITS-CRT-I PCM(S)-JEE(Main)/4 Physics PART I.. 6 4 5 5.7 ohm metre. r L R r L R ohm metre. 5 7 5 ohm metre 5.7 ohm metre 5.7 m 54 m P 4 P 4 9 4 % t. A N e Ae N e e e e t t. tt vo sin 4. At t,v ˆ v cos i g v ˆ ˆ vk gtj v v v kˆ gt ˆj v cos ˆi 5. f a Z b. (D) t 6. dv v d v k mg m k mg vdv m v m k mg o d 7. I ma I min I I I I FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, 656949, Fa 6594
AITS-CRT-I PCM(S)-JEE(Main)/4 I I if I > I 8. Ic sin5t 5 4 /4 5 /4 IL sin5t 5 4 I sin(5t) /4 5 IIc+IL /4 9. mg sin T m(a) T ma mg a g / & T Net force Result of these forces T ˆj T cos6 ˆj T sin 6 ˆi ˆ Ti Tj ˆ mg i ˆ j ˆ 6 As net force in pulley, mg i ˆ ˆ j 6 force by clamp T 6 T. The circuit can be simplified as:,peak current Ampere. 5.4F.H 5 c sin [5t] volts..4f L 5 5. When u decreases v increases. For plane mirror u decreases. the image in plane mirror moves towards the left.. I 4. sin45 sin45 4 R R I I R R 4. V 4 volts 5 volt 4 4 4 V 4 4 4 4 4.9 volts V V. volt FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, 656949, Fa 6594
AITS-CRT-I PCM(S)-JEE(Main)/4 4 5. In both cases applying com, mv Mv mv V M V : V 6. In steady state :- E E VP VQ R R E E VQ VS R R V V S P R V V E A R Q CE Q B CE 7. V V P - volts. 8. For the two blocks to move together F F min M m g Mg As min, the two blocks move together. Velocity of the system (M+m) is F v at t M F F P Fv F t t. m M 9. P 4r At mct 4r mct t PA. A b T.. After collision, the rod starts rotating about its centre of mass. Now the torque of mg about the centre of mass, hence is constant. FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, 656949, Fa 6594
5 AITS-CRT-I PCM(S)-JEE(Main)/4. elongation in equilibrium ma cos mgsin k () For elongation from equilibrium mgsin macos k net force k (i) k acc m T m k Mg sin ma θ N K( + ) Mg cos 4 r n n g 6 rv T.. V P T g r n n 9 g 4 r n n g r.n.g. 9 5 8r n n n g 7 55. cm.m 4. k. rad 5. T T P gh P r r T ghr. 6. M G GM 4 GM R 5R 5R 7. The apparent frequency initially decreases & then increases in one time period. 8. GMem GMem E R 4R GMem 6R 9. v u gh 4 h h 8m where h is maimum height reached by ball A v u gt t 4 sec, 8 t ' sec u u A A B u' gt ' m / s A(t4sec) U A h 8 m u' B 4 m / s B 4m/sec u B Thus the speed of each ball gets interchanged (for equal mass & elastic collision) u A u B (Tt+t) FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, 656949, Fa 6594
AITS-CRT-I PCM(S)-JEE(Main)/4 6. T T m r T m r r T m r r Thus T T m T T T Chemistry PART II. M wt C 6 H 5.5 M wt C 6 H 5 CH 9 Given:.5 C6H5 So mass% composition reaction of C 6 H 5 is % C 6 H 5.5 4.5 55%. As density increases, volume of gas decreases so intermolecular forces increase and diffisubility decreases.. For 4 n 4, l, m Angular momentum h h h Angular node So total node h h 4. 5. 7. o 5 C 4 4 7 Pyrophosphoric acid H PO H P O H O HO P O P OH H O + P H B P O + O + H O OH B B O + P O + H H O + H p p back bonding between B and O and p d back bonding between P and O is possible. K Cr O H SO CrO KHSO H O 7 4 4 Chromic anhydride Red Crystal O OH 9. NH behaves as a base and E elimination takes place. FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, 656949, Fa 6594
7 AITS-CRT-I PCM(S)-JEE(Main)/4. OH OH OH C CHO NaOH C H. NaSO AgNO AgSO AgS white black ppt. Tc 8a so Tc ; and b 4 N A V b r 7Rb b. AB type structure in which Zn + is tetrahedrally surrounded by S ion, S is present in half of tetradedral acid. 4. It is an etensive property. 5. 6. 7. d B K K t K t A Ke Ke dt K K Fe H O OH H O ksp, Ksp should be high. FeHO 6 H Ka C KaC.5 8. G (net) G G Total e echanged (a + b) G nfe a bfe afe bfe E ae be a b 9. KCrO7 KCrO4 CrO O. O H C C 6 H 5 M PPh H C C 6 H POPh (Wittig reaction) 5. Nitrogen is less electronegative than oygen and on rd reactive site it is sp hybridised. FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, 656949, Fa 6594
AITS-CRT-I PCM(S)-JEE(Main)/4 8. O O O C CH () NaOI COOH CO O. As metallic character increases down the group so acidic nature of oide decreases. 6. - NH - Ni 4- and Ni 4- NH NH NH Cis Trans 7. 4 As O 5H O AsO H 4e 9. BeO C BeC CO. Only lithium nitride with formula LiN in first group and Mg with Mg N. Mathematics PART III. Since the equations are consistent D (a ) (a ) (a ) (a ) (a ) (a ) Put u (a + ), v a +, w a + u v, v wz, w u u + v + w a + 6 u v w u v w (u v) (v w) (w u) (u + v + w) () () () (a + 6) i.e. a.. 4 n +, so 4 567 m +,89 k, so 89 p. Hence 4 567 + 89 n 4 is even, So 4 567 4q. 89 4r +, so 89 4t +. Hence 4 567 + 89 4l +. So it is 9 +. FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, 656949, Fa 6594
9 AITS-CRT-I PCM(S)-JEE(Main)/4 c 4. Here, tan A + tan B and tan A. tan B ab tan A. tan B tan A cot B A 9 B A + B 9 C 9 sin A a c, sin B b c sin A + sin B + sin C a c b c 5. I I I d 4 5 e d I 5 I. d 4 5 e 6. Since f() and g() are one-one and onto and are also the mirror images of each other with respect to the line y. It clearly indicates that h() f() + g() will be a constant function and will always be equal to 4. 7. Equation of the curve passing through the points A, B, C, D is ( + 4y 4)(4 + y 4) y If it represents a parabola then the quadratic terms must form a perfect square. or 49. r 8. Locus is a hyperbola with eccentricity r r r 9. In an equilateral triangle r R Also eradius r 4R sin A cos B cos C 4R r, R, r are in A.P... R. lim / lim t t / t t dt form dt 4 4 lim 4 5 lim 5 8.. (m + )(n ) n, so if p divides n, then it also divides n and hence also (n ) n(n ) n, and hence also (n ) (n ). So n, ± or ±. But n must be an integer, so,, do not work. gives n and hence m, which is a solution. gives n and hence m, which is a solution. FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, 656949, Fa 6594
AITS-CRT-I PCM(S)-JEE(Main)/4. Two lines can meet in a point 6 C 5 line and circle meet in two points ( 6 C 4 C ) 48. Two circles meet in points 4 C total number of points 48 + 5 + 75. 7. + cos+ cos 6 cos cos cos sin cos, sin 7, and,,,.. is the common value which satisfies both n + (6n + ) 5. 6. ( ) d 4 ( )co t dt (t )cot t Put t d dt du u Let cot t u dt du ln u + c ( t ) ln cot c 7 7 7 7 9 7 co t d FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, 656949, Fa 6594
AITS-CRT-I PCM(S)-JEE(Main)/4 7. Let be the period of sin + cosa then sin( + ) + cosa( + ) sin + cosa for all In this identity, putting and, we het sin + cosa and sin + cos solving these equation, we get sin and cosa hence n and a m. Where m, n are non zero integers hence a m n or a m n (since ) 8. A, A A 4 A n then A(I A) n A(I + na) A na A. 9. sin A + sin (5 A) + sin A sin (45 A) + sin45 sin(a 45) +.sin A 45 < and > < A < 5 < A < 7 45 < A 45 < 5 sina 45 sin A sin B.. The curve y c and the circle + y touches each other so c + 4 + c will have equal roots so ( ) 4c 4c c 4 c. 4 Roots of the equations are 4 y. early distance between the points of contact units. Sum of 7 digits a multiple of 9 Sum of number,,, 4, 5, 6, 7, 8, 9 45 so two left should also have sum of 9. The pair of left numbers is (, 8), (, 7), (, 6), (4, 5). with each pair left number of 7 digit number 7! So with all 4 pairs 4 7!. FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, 656949, Fa 6594
AITS-CRT-I PCM(S)-JEE(Main)/4. Equation of Ist line is y ( ) y and equation of IInd line is + y. early point of intersection is (, ).. The centre c of the circle (5, 7) and the radius PC 5 7 5 5 5 5 q PA 5 5 and p PB + 5 G.M of p and q pq 5 5 5 5 P ( 7, ) A C (5, 7) B 69 5. 4. a, b, c are in G.P b ac and c a + b c a + ac a a c c a c cos B 5 cos B 5 (B is acute) B a c A C b cos A sin B cos B 5 5. y tan From graph it is clear that it will have two real roots, y / / 6. The given equation is p + q r r. Its roots are given,, so we have q r + + ; + + ; r p p p. D. 7. For local maima at R + and minima at certain R the roots of f () are of opposite sign f () < b <. FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, 656949, Fa 6594
AITS-CRT-I PCM(S)-JEE(Main)/4 8. Drawing the curves y log e and y So, number of solutions (, ) (, ) (, ) (, ) 9. Putting c c + z c ac ac f(c)d f(c z)dz c c So, the required value c ac c d c(a ). z z z z z z ( z + z + z ) (z z z z z z z z z z z z ) 6 (zz zz zz zz zz zz ) () Now z + z + z z + z + z + zz zz zz zz zz zz (z z z z z z z z z z z z ) () from () and () maimum value 6 + 9 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, 656949, Fa 6594