ALL INDIA TEST SERIES

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1 AITS-CRT-I-PCM (Sol)-JEE(Main)/7 In JEE Advanced 6, FIITJEE Students bag 36 in Top AIR, 75 in Top AIR, 83 in Top 5 AIR. 354 Students from Long Term Classroom/ Integrated School Program & 443 Students from All Programs have qualified in JEE Advanced, 6 FIITJEE ALL INDIA TEST SERIES JEE(Main)-7 ANSWERS, HINTS & SOLUTIONS CONCEPT RECAPITULATION TEST I (Main) Q. No. PHYSICS CHEMISTRY MATHEMATICS. A B B. A C D 3. A C B 4. A D C 5. B A D 6. A B A 7. D D B 8. C C C 9. B C A. A B A. A D D. B A B 3. A B C 4. C C C 5. C B C 6. D B A 7. B D B 8. D B A 9. C D A. C B D. C B C. C C A 3. D D D 4. A C A 5. D B B 6. D D D 7. C A C 8. B B A 9. D C C 3. B B C FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fa 65394

2 AITS-CRT-I-PCM (Sol)-JEE(Main)/7 Physics PART I. A 4t t d 4 V = dt 4t 4t a = SECTION A dv 4 ( )V 3/ dt ( 4t). A Neutron law for cooling Rate of decrease of temperature temperature difference between object and surrounding A 35 f f f 35 5 f = 75 Hz (i) (ii) 4. A v g + KVT v g The ball is dynamic equilibrium. KV T v g v g 5. B k k3 k = 3 k 6. A R = V gv 3 o V V ; R V 3 gv o 7. D Q = U + W (i) and U = nc v (T) (ii) FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fa 65394

3 3 AITS-CRT-I-PCM (Sol)-JEE(Main)/7 8. C The centre of mass = R Linear velocity = V = r. R V R 9. B Sol: Y =. A 4L F. A. = A F YA. 9 F A n =.5 Ad 3 A 7.7 n = 7Hz 5 (i) 4L i L 5 (ii) L 5 3. A Escape velocity = GM R. B I I 4 r cos 45 4r B sin 9 sin A F IBL F f F f ma fr I a r 4. C A > C FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fa 65394

4 AITS-CRT-I-PCM (Sol)-JEE(Main)/ C T g eff g ; eff qe g (with electric field) m 6. D For electron in electric field, magnitude of its momentum may increase, decrease or remain constant w.r.t. initial value. 7. B Q Q.6 Q Q Q = 5, W = Q Q = 3 J 8. D I L and I C will be in opposite phases. I net = I L I C =.6.4 =. A 9. C Variation of moment of inertia with temperature I = I o ( + T) C.O.A.M: I i i = I f f. C Polarization occurs for transverse waves and not for longitudinal waves.. C D W ; d D W = d. C. VAC.8. 5 Also, V AC = (.5) 5 (i) (ii) 3. D = C (critical angle) R tan C FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fa 65394

5 5 AITS-CRT-I-PCM (Sol)-JEE(Main)/7 R cm 4. A = Bv develops across rod PQ. By Lenz s law, plate M get positively charged. 5. D A V = A V dh a gh A dt a A t h dt A h dh gh 6. D E 4n En 6n 4n E n E n E E 4n E E n = 4 = constant 7. C m =.6 amu Q value =.4 MeV t =.8 s 8. B hc hv eV v C & v ' C v ' v ev h(v) 4..5 V 5.9 V FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fa 65394

6 AITS-CRT-I-PCM (Sol)-JEE(Main)/ D The potential drop across R is 4 V while current is ma hence, 3 R = 4. R = 4. m V ma R 3. B The electron density in doped semiconductor, 6 n i (3 ) e n 4.5 = /m 3 h 6 V FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fa 65394

7 7 AITS-CRT-I-PCM (Sol)-JEE(Main)/7 Chemistry PART II SECTION A. B Meq. of FeSO 4 = Meq. of KMnO 4 or, N V = N V or, N =. 5 N =, also M =. C It has unsymmetrical structures. 3. C One mole of NH 3 occupy.4 L at NTP, not at 3 K and atm. 4. D Sulphur forms two sigma bonds with two lone pairs. 5. A BeSO 4 forms comple with water i.e. [Be(H O) 4 ] +. Hence it is most soluble. 6. B The unit of k will be mol -/ L / s. 7. D K a increases only by increasing temperature. 8. C K C Y 4 X C For equilibrium system, S(system) = S(surrounding). B HO HO CH SiCl CH Si OH Silicones 3 3. D py pz 3 3 KP 3 p 3 P equilibrium = p + p y + p z = = 9 atm. A NaOH CO HCOONa FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fa 65394

8 AITS-CRT-I-PCM (Sol)-JEE(Main)/ B SO H SO H O 3 4. C HCOOH is stronger than CH 3 COOH. Hence the solution of HCOOH will contain more no. of ions than that of CH 3 COOH. 5. B Zn + form colourless compounds in solution due to absence of d-d transition. 6. B m 3 m 3 e e n-factor of MgCl.KCl.6 H O = D Due to weak Bi H bond. 8. B Halogen atoms ecept fluorine form compounds with valency, 3, 5 and D Xe undergoes sp 3 hybridization and forms three sigma and three pi-bonds with a lone pair on it.. B CrOCl NaOH Na CrO H O Cl 4 Pb Cl PbCl 4 4 Pb CrO PbCrO. B One Na + is removed and remaining three(effective no. per unit cell) are present.. C The fraction is 3 3. D E = cell E cell > (due to M concentration of ions) 4. C In cyclopentadienyl anion, each carbon is sp hybridized H H H H H FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fa 65394

9 9 AITS-CRT-I-PCM (Sol)-JEE(Main)/7 5. B Fructose contains 3 chiral carbon atoms. 6. D 7. A Aliphatic amines are stronger base than aromatic amines 8. B In anthoproteic test, electrophylic nitration takes place. 9. C CH CHO NH4Cl MeCH NH CN HCl MeCH NH COOH 3 KCN H O 3. B The configuration of carbon atoms at 3, 4, 5 positions should be same. FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fa 65394

10 AITS-CRT-I-PCM (Sol)-JEE(Main)/7 Mathematics PART III. B 5 Tr cot 3r tan 36r 5 tan 4 36r 9 3 tan 9 4r r r tan 3 3 r. r 3 3 tan r tan r 3 3 S n tan n tan n 8n 3 tan 8n 3 cot. D Equation of the curve is y 3. B f f ' ln ln e lne ln e ln e Now, consider the function g' y ln y g' y when y e gy SECTION A y ln y gy is an increasing function when e e lne ln FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fa 65394

11 AITS-CRT-I-PCM (Sol)-JEE(Main)/7 lne ln e f ' f is decreasing is, 4. C When cot sin No possible value of. When cot cot cot sin cos 3 5,,, Number of solution = 5. D S, S 6 6. A y OR 7. B 8. C y 3 y y y y f has minimum value at. compared to 4. f will be smaller than f (4) because is closer to as FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fa 65394

12 AITS-CRT-I-PCM (Sol)-JEE(Main)/7 M M M3 M which ever is less is most consistent. 9. A z i arg z i argz iargz i 6 6 y y tan tan 6 y y y 4 3 y y y 4 4 centre 3,, radius = 4. A o o o o o o sin 8 4 sin 6 8 sin S... sin 8 sin 4 sin6 sin 8 sin 89 sin 496 o o o o o o o o S cot 4 cot 89. D 3 3 f h 3h f f ' h 3h. 3h lim lim h h f h h f f ' h h. 4h 4. 4 (Using L Hospital Rule). B sin3 I d 3 3. C 4 a 4 a y 9 f 4 For four distinct solutions a, 9 a, 8 a 8, 3 O 3 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fa 65394

13 3 AITS-CRT-I-PCM (Sol)-JEE(Main)/7 4. C f " f ' Integrating Inf ' c, f ' c f ' e f e k, f k f e Area e d e e y O 5. C Using the property that equal chords subtends equal angles at centre of circle, then problem can be converted to the diagram in adjoining figure AB 4, AC,BC 3 ABC / cos / cos cos / cos A O 4 3 C B 6. A cos cos cos 7. B cos t, and t, log 5t log t 3 log 3 5 ; t and t 5 log 5t t t t 3 t or t (rejected) cos cos 8. A dy 4 e e d d dy g' y e 7 When y then 6 dy d 3 5 7/6 FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fa 65394

14 AITS-CRT-I-PCM (Sol)-JEE(Main)/ A z z z z z z z z z z z z z. D Number of ways = (Number of one-one function from set A = {s, s.s 6 } to A such that f s s 6! D 6! i i 6. C The given plane passes through a and is parallel to the vector b a and c. So it is normal to b a c. Hence, its equation is r a. b a c or r. b c c a a b c a b c The length of the perpendicular from the origin to this plane is b c c a. A n cos n sinn A sinn cos n n A Lim n n 3. D We have T ln 3e 3 e e e 3 e e 3 e e e ln 3 lne e e ln ln ln ln e e 4 4 T 4 4. A D.R s of PQ are, y, z y z y z 3 Also y 3z 4 y z 3,y,z is a point on the two planes y 3z 4 y z 3 line of intersection of these two planes will be the required locus which is y 5 z d FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fa 65394

15 5 AITS-CRT-I-PCM (Sol)-JEE(Main)/7 5. B 3 4y 3 y having focus 3, 5 and point of intersection of lines given by the equation 3y y 3 7y can be obtained by solving equations 4 y 3 and 6 7 (partial differentiation method), i.e centre (point of intersection of normals). Parabola is of the form y 3 3 Hence required circle is Or y y 6 i.e. 4 4y 8 8y D Asymptotes equations are 7. C dy y y n d Put y t 8. A Number of ways y k 9. C The epression can be interpreted as the square of distance between the points tan A, cot A and cosb, sinb The minimum value of this distance is the minimum distance between the curves y and y 3. C For every selection of 6 numbers without loss of generality there will be 5 favorable cases out of every cases. FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph 466, , Fa 65394

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FIITJEE JEE(Main) AITS-FT-II-PCM-Sol.-JEE(Main)/7 In JEE Advanced 06, FIITJEE Students bag 6 in Top 00 AIR, 75 in Top 00 AIR, 8 in Top 500 AIR. 54 Students from Long Term Classroom/ Integrated School Program & 44 Students