Narayana IIT Academy

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1 INDIA Sec: Jr. IIT_N10(CHAINA) JEE-MAIN Date: Time: 08:00 AM to 11:00 AM CTM- Ma.Marks: 60 KEY SHEET MATHS PHYSICS CHEMISTRY

2 1: Outside of ABC but inside of ABC _Jr.IIT_N10(CHAINA)_Jee-Main_CTM-0_Key&Sol s SOLUTIONS MATHS D = A + C B = ( 6, 7 ) 4. The image of B w.r.t the line AI lies on AC.Similarly the image of AI w.r.t the line BI lies on BC.Bysolvity AC and BC we get C, 0 5. Solve y 4. 5y 7 we get PI as ( -1, ) clearly ( -1, ) satisfy y 1 0, y, 60, y Ref equation of line 0 m tan 60 m 0, 1 m 7. P 1, 1 ; y 4 r , (5, 4) Sec: Jr.IIT_N10 (CHAINA) Page

3 8. By using Manelaw s theorem _Jr.IIT_N10(CHAINA)_Jee-Main_CTM-0_Key&Sol s 9. GD 5 AG 5 A 1, 6 Slope of AB is m Slope of BC is 1 tan m m m, m 10. Let equation of AB ; 7 y 4 = 0 m1 7 Equation of AC is + y + 1 = 0 m 1 Slope of BC is m m m m m 1 1 mm 1 m m 1 tan, Solving + 5y 1 = 0, 5 + y + 7 =0 we get 1 A, a1 a b1 b : f f Sec: Jr.IIT_N10 (CHAINA) Page

4 _Jr.IIT_N10(CHAINA)_Jee-Main_CTM-0_Key&Sol s 14. Range of f() is [ 0, 4] 15. Min value of f() holds in [50, 51] 16. Fundamental period is 17. Equation of line passing through P(, 1) is y 1 = m ( ) m y m 1 0 m 1 its distance form origin is f m m 1 say f m is maimum iff m = -. Then the line is y 1 = - ( -) + y = 10 Required area = sq units 18. Let the equation of the line passing through (0, 0) is y = m. Which meets + y = at A m A, m m and y =m meets y + =0 in B. Where m B, m 1 m 1 Let ( h, k) both mid point of AB k k mh m h and m m 1 h 1 1 m m 1 h 1 1 h k k 1 h h Locus of ( h, k ) is 1 1 y y 1 Sec: Jr.IIT_N10 (CHAINA) Page 4

5 y y _Jr.IIT_N10(CHAINA)_Jee-Main_CTM-0_Key&Sol s y y y y 4 y y y y y y y Let both side of triangle DEF a' be the side of triangle ABC cos0 BD BD DC cot 60 0 DC BD DC a Area of DEF. 4 1 Areaq of ABC a 4 AB 0. tan OA AC AB tan OA OA tan tan slope of L1 is m Slope of L is m Sec: Jr.IIT_N10 (CHAINA) Page 5

6 _Jr.IIT_N10(CHAINA)_Jee-Main_CTM-0_Key&Sol s h h m m m b b a m m m b 4 h a ab h ab 9 8 b a b 1: P(, y) moves such that sum of its distances from given lines is, then locus of P is a reactangle of sides, sin / cos / d 8 and area is d sin sin Where is angle between given lines 1 Clearly sin required area = 8 16 sq. units 1. Given family passing through intersection of 4y 6 0 y 0 y 1 y 0..1 y 4 0. Sec: Jr.IIT_N10 (CHAINA) Page 6

7 _Jr.IIT_N10(CHAINA)_Jee-Main_CTM-0_Key&Sol s Concurrent point ( 1, ) = A Let Q = (, ) Let P(, y) is its image AQ = AP AP AQ y 1 1! y y y y 4 0. Let = cos t, t 0, So, we have 1 sin t Then inequality becomes, sin t + cos t a The maimum value of f(t) = sin t + cos t on the interval 0, is. Hence the range of a is set of all real numbers not eceeding & 1 For,, and 1 m where m N 1 m 5. " 0" ; (i) the given inequality is when 1 Which has solution [1,) (ii) ln when 0, 1 Which has solution ( 0, 1) Sec: Jr.IIT_N10 (CHAINA) Page 7

8 Note : ln _Jr.IIT_N10(CHAINA)_Jee-Main_CTM-0_Key&Sol s Solution set :0,. 6. f 6 7 cos 6 7 cos 6 6 has period 1 6 and cos has period. Then period of f Hence, the period is LCM of and The equation a a 1 5 a... a a 1 5 a a a 1 5 a 1so no solution for 1 8. We have y a 1 y a y 0 As R, so discriminant 0 y 4 a y 0, y R y 4y 4 4a 4, y R y 4 a 1, y R 4a 4 0 a 1 (But a 1, think?) Hence, a, cossin 0 true for all R; log 0 for 0, 1 0. Clearly f ' e 1 0 ' f ' 4 5 f (0, e ](0, e] ' ' f is increasing is into f is one-one range of PHYSICS Sec: Jr.IIT_N10 (CHAINA) Page 8

9 _Jr.IIT_N10(CHAINA)_Jee-Main_CTM-0_Key&Sol s 1. Angle of friction is the angle between contact force the normal force. N mg cos F C f mg sin. f mg sin tan tan N mg cos 1 f mg sin g sin 10 g cos m g. B 5 mb mb. kg 4. L g g L or L 1 5. N mg N ma a g 10 5 m / s m Acceleration of block is ' g ' Speed V as V or S g 7. As mg cos mg sin T 0 g g tan 1 a V / R Net force is centripetal r 40. If f1 f 0 for the same tension 41. simultaneously m N V 1 T r r as it is not possible f 1 and f cannot be zero 4. mg mv N mg R mv N mg R Sec: Jr.IIT_N10 (CHAINA) Page 9

10 N _Jr.IIT_N10(CHAINA)_Jee-Main_CTM-0_Key&Sol s V 4. net 44. mg mv mg N R N 0 V0 Rg F 6j ˆ r i ˆ 4j ˆ 4j ˆ net F r 4 J T a = g/ mg mg T mg / 4 T mg / 4 T S T d mg d 4 V m 90 f 47. f mg sin S Vt f s f Scos 90 f Ssin mg sin Vt sin mgvt sin P 1 k t Sec: Jr.IIT_N10 (CHAINA) Page 10

11 mv 1 or V kt mv 1 t _Jr.IIT_N10(CHAINA)_Jee-Main_CTM-0_Key&Sol s dv k a dt 48. As displacement is along +ve -direction, initially it is moving along +-direction and work done is +ve upto 1m So mv mu J mgh P 10 J / s t t sec dm dm P 10 P gh 0 kg / sec dt dt gh 1010 dm m t kg 100 litres dt u 4 du 4 0 d at d y d stable equilibrium V gh 5Rg 5R h 5. To complete the circle V 5lg 54. u 5Rg V V Rg g R a a a r g g g 10 t a r V a t 55. While it is moving up kinetic energy decreases, at the given moment only it has maimum kinetic energy. 56. From the energy conservation mg h 57. V g R h Sec: Jr.IIT_N10 (CHAINA) Page 11 a 0 b c m 11 m b c m1 m a b a b

12 mg sin 60 mgsin 0 g 1 g 58. a 1 m 4 mai ˆ maj ˆ a acm ˆ i ˆ a g j or acm 1 m Position of COM follows the original trajectory. So _Jr.IIT_N10(CHAINA)_Jee-Main_CTM-0_Key&Sol s cm Range 60. Let be the displacement of COM of shell, then cm 0 gives 4m m 10R 4 10R or R u sin g SOLUTIONS: CHEMISTRY 61. P 0 P s loss in weight of water chamber and P s loss in wt. of solution chamber. P0 Ps n w M P N m W (1) or 6. As we know m = m180 o P T = P A X S + P B X A o o o = PA XA PB (1 XA) o o = P PA B X A + P o B But from question, P T = 180 X A + 90 (1) () Equating equation (1) and () o A o B P P = 180 P o B 90 (when A = 0 and B = 1, P = P 0 B or P 0 B = = 90) o P A = = 70 o P A 70 = : o P 90 B () 6. HA H + + A - Initially conc. C 0 0 Sec: Jr.IIT_N10 (CHAINA) Page 1

13 _Jr.IIT_N10(CHAINA)_Jee-Main_CTM-0_Key&Sol s on. After dissociation C(1 ) C C Here, i = 1 + (-1)= 1+ T f =i m K f 0.06 = (1+) = 0.9 K a = [ H ][ A ] [ HA] = C C C(1 ) C 1 = K a = 10 Ans: 4 64: For 1M NaClsolution (i C) = 1M M Urea solution (i C) = M (C = Molarity) 1M CaCl solution (i C) = 1M = M 1.5 M KCl solution (i C) = 1.5M = M Similarly for 1.5 M AlCl and 1.5M Na SO 4 The value of i C be different and.5 M KCl and 1M Al (SO 4 ) has 5 M each. So, the each pair of solution A,B and D have same concentration so, all pairs of solution be isotonic, ecept Ans: 65. () (T f ) cal = K f m (T f ) cal = = 1.86 i = ( Tf ) ( T ) f obs cal = H O = = 1 + (n 1) Hence = 1, n = Two species will be produced from single species which is only possible for [Pt(H O) Cl ]Cl, Sec: Jr.IIT_N10 (CHAINA) Page 1

14 _Jr.IIT_N10(CHAINA)_Jee-Main_CTM-0_Key&Sol s 66. (4) Blue colour formation due to formation of ferric ferrocyanide, Fe 4 [Fe(CN) 6 ]. Though the osmotic pressure on side X much greater than the osmotic pressure of side Y, hence, solvent molecules from side X passed to side Y. Therefore no passage of K + or [Fe(CN) 6 ] 4-, Fe + and Cl will occur through the semipermeable membrane. Hence, there is no formation of blue colour. 67. () True statements of Henry s Law 68. (). Negative deviation is when H-bonding is established 69..[4] T (C 0 ) 1 n or T = const. (C 0 ) 1 n or T (C 0 ) n 1 = const. or T (C 0 ) n 1 = const. n 1 or T ( C 0 ) = const But from question n 1 = 1 70.[1] n = N O 5 4 NO + O Initial pressure 0.1 atm0 0 pressure at time, t (0.1 ) atm atm / atm Total pressure = (0.1 ) + + / = / = 0.15 (from Question) = 0.1/ Now, from Ist order kinetics,. 0 t= K P log 0.0 = log P =54 sec. 71. (1) For zero order reaction [A] t = [A] 0 kt for completion of reaction [A] t =0and t = [ A] 0 & A k 0 = a Sec: Jr.IIT_N10 (CHAINA) Page 14

15 t = k a Correct answer is (A) _Jr.IIT_N10(CHAINA)_Jee-Main_CTM-0_Key&Sol s 7 (1). Slow step must have the highest Ea value and being eothermic product(s) energy must be less then the reactant(s) energy. 7. (). Let the order of reaction w.r.t. [A] be and that w.r.t. [B] b y then rate equation r = k[a] [B] y From (i) & (ii) data = = = 1 10 Now from (i) & (ii) = From (i) data 0.1 = 0.1 y Rate equation r = k[a] 0.05 = k (0.1) k = = = 0.5 s () rate epression () d[cro dt 7 it is given that y y y = 0 ] 1 d[cr dt d[cr O7 ] = mole L 1 s 1 dt d [Cr so dt = mole L 1 s 1 ] ] = mole L 1 s [NF and H + O] are pyramidal while [NO and BF ] are trigonal planar () Sec: Jr.IIT_N10 (CHAINA) Page 15

16 81. A A t 4 ; 1 Ay 1 Ayt _Jr.IIT_N10(CHAINA)_Jee-Main_CTM-0_Key&Sol s 1 A A y y ln t A A 1 1 t 4 y ln If t hrs t y t ln ln T y 50, T50, y T 50, y 0 min 8. Since X has valence electrons, it will lose three electrons to acquire noble gas configuration. Similarly Y will gain electrons to acquire octet in its outermost shell. () I: BrF 4 + : Trigonal bipyramidal structure and see saw shape II : True statement III : Br and Br are of different shape and structure. IV: In PCl 5, hybrid orbital of P is sp d z Polarization in the molecule increases with increases of charge and decrease in size of the cation. O O HNO : N OH Sec: Jr.IIT_N10 (CHAINA) Page 16