Solution 3 (Class XII)

24 

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PART-I (1 Mark)

MATHEMATICS

1. Given : a1, a2, a3 ...AP and a1, a2, a4, a8 ...GP. Let common difference of A.P. = d

a2 = a1 + d a4 = a1 + 3d a8 = a1 + 7d  1 2 a a = 2 4 a a = 4 8 a a = r 1 1 a d a  = aa 3dd 1 1   = aa 37dd 1 1   = r (a1 + d)2 = a 1(a1 + 3d) a12 + d2 + 2 a 1d = a1 2 + 3a 1 d d2 = a 1 d (d  0) d = a1 ....(i) Hence, 1 2 a a = r ; 1 1 a d a  = r 1 1 1 a a a  = r (using (i)) r = 2.

ANSW ER KEY

HINTS & SOLUTIONS (PRACTICE PAPER-3)

Que s. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans. A B C B D B D A A C B B D C A Que s. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Ans. B C B C D B B D D C C D C A D Que s. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 Ans. D C A B B C A A A B C D C C D Que s. 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 Ans. A C B A B A D C D B D A D B B Que s. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 Ans. A C C C B B C D A D D D B A B Que s. 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 Ans. B B D C A C B C B B D C D B C Que s. 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 Ans. D A B C D A C C B B B D A A B Que s. 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 Ans. C A C D B C B A B A D B A C C

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2. k 101 T T 1 – k k  till k = 10 Tk > Tk–1 Let k = 11 T11 < T10  T10 is maximum at k = 10. 3. x = 2 + 3+ 6

x 2

2 =

2 6 3 x2 + 2 – 2 2 = 9 + 6 2 x2 – 7 = 8 2 (x2 – 7)2 = 64 × 2

So, smallest possible value of n is 4. 4. Let the three players are A, B, C.

Now, each player get 0 score after playing 9 games. It happened only when each player wins 3 games and loss 6.

So,

A win 3 games out of 9 9C3

B win 3 games out of remaining 6 6C3

C win 3 games out of remaining 3 3C3

So, required way = 9C 3 × 6C 3 × 3C 3 = ! 6 1 2 3 ! 6 7 8 9       × ! 3 1 2 3 ! 3 4 5 6       ×1 = 1680. 5. B C A (2, 3) (x, y) (4, 0) (2, z) O O is circumcenter  OA = OB = OC = circumradius (2 – 2)2 + (z– 3)2 = (4 – 2)2 + (0 – z)2 z2 + 9 – 6z = 4 + z2 9 – 6z = 4 5 = 6z 6 5 = z  Circumcenter = (z3)2(22)2 = | z – 3 | = 6 3 5  = 6 13 .

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6. P P' (5, 15) (21, 15)L A Mid point of PP’ =         2 15 15 , 2 21 5 L = (13, 15)  Point A will be (13, 0) By property PA + PA’ = 2a PA = (13 5)2 (0 15)2    = 64225 = 289 = 17 cm PA’=

(

13

21

)

2

(

0

15

)

2 = 64225 = 289 = 17 cm  2a = PA + PA’ 2a = 17 + 17 2a = 34 cm

So, length of major axis = 2a = 34 cm. 7. B C P (10, 10) (a, b) (0, 6) 2x + 3y = 18 PB = PC (10 – 0)2 + (10 – 6)2 = (a – 10)2 + (b – 10)2 100 + 16 = a2 + 100 – 20a + b2 + 100 – 20b a2 + b2 – 20a – 20b + 84 = 0 ....(i)

Also (a, b) i.e. on 2x + 3y = 18 2a + 3b = 18

a = 9 –

2 b 3 Using equation (i)

2 2 b 3 9        + b2 – 20        2 b 3 9 – 20b + 84 = 0 81 + 4 b 9 2 – 27b + b2– 180 + 30b – 20b + 84 = 0 4 b 13 2 – 17b – 15 = 0 13b2 – 68b – 60 = 0

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13b2 – 78b + 10b – 60 = 0 13b(b – 6) + 10 (b – 6) = 0 b = 6 or b = 13 10   When b = 6, then a = 9 – 2 6 3 = 0 When b = 13 10  , then a = 9 + 13 2 10 3   = 9 + 26 30 = 13 132  8a + 2b = 8 × 13 132 + 2 × 13 10  =

79

13

20

1056

. 8. cosec2(

 + ) – sin2(– ) + sin2(2– ) = cos2(– )

cosec2(  + ) + sin2(2– ) = 1 ) ( sin ) ( cos2   2  cosec2(  + ) = 1 – sin2(2– ) cosec2(  + ) = cos2(2– )

Minimum value of cosec2(

 + ) is 1 and maximum value of cos2(2– ) is 1.  They will be equal for the value 1.

 +  =

2

...(i) 2–  = 0 ...(ii)

By adding (i) & (ii) 3 = 2   = 6   = 3   sin(– ) = sin ( 6  – 3  ) = – sin ( 6  ) = 2 1  . 9. sinx + siny = 5 7 ....(1) cos x + cosy = 5 1 ....(2) By (1)2 + (2)2 we get

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sinx siny + cosx cosy = 0 cos(x – y) = 0

 x – y = 90º

By (1) × (2) we get

sinx cosx + sinx cosy + siny cosx + siny cosy = 25

7

sin(90 + y)cosx + sin(x + y) + sin(x – 90) cos y =

25 7

cosy cosx + sin(x + y) – cosx cosy =

25 7 sin(x + y) = 25 7 . 10. 6 x y = 0 1 (1, 1) y = sin x

Clearly, curve meet each other twice in 2– 3

4– 5 6– 7 8– 9 10– 11  Total 10 Times. 11. f(x) is differentiable on R. So, it will be contincous on R. Continuity at x = 0 LHL – 0 x lim  x x sin 2 Put x = 0 – h, then h  0 0 hlim h ) h 0 sin( 2   0 hlim h h h h sin   = 0 RHL  0 x lim x2 + ax + b Put x = 0 + h, then h  0 0 hlim h 2 + ah + b = b Value of f(x) at x = 0 f(0) = b.

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 f(x) is contineous at x = 0  LHL = RHL = f(0) 0 = b = b b = 0 Differentiability at x = 0 LHD 0 hlim

h

f

h

f

)

(

0

)

0

(

0 hlim h b h sinh2    0 hlim 2 2 h h sin = 1 RHD 0 hlim h ) 0 ( f ) h 0 ( f   0 hlim h b – b ah h2  0 hlim h ) a h ( h  = a.  f(x)is differentiable at x = 0, LHD = RHD a = 1.

12. Let point p(x1, y1) is on the curve y2 = 4x.

 y1 2 = 4x 1 x1 = 4 y12 PA = 2 1 2 1 0) (y 3) x (    AP2 = x 1 2 + y 1 2 – 6y1 + 9 AP2 = x 1 2 + y 1 2 – 6y1 + 9 Let AP = z z2 = x 1 2 + y 1 2 – 6y1 + 9 z2 = 2 2 1 4 y         + y12 – 6y1 + 9 z2 = 16 y14 + y12 – 6y1 + 9 Diff. w.r.t. y1 2z 1 dy dz = 16 y 4 13 + 2y1– 6 2z 1 dy dz = 4 y13 + 2y 1– 6 = 4 24 y 8 y13 1 2z 1 dy dz = (y1– 2) (y12 + 2y1 + 12)

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For the critical points 1 dy dz = 0 (y1– 2)(y1 2 + 2y 1 + 12) = 0 y1 = 2  y12 = 4x1  (2)2 = 4x1  x1 = 1. 2 2 1 dy dz         + 2z 2 1 2 dy z d = (y12 + 2y 1 + 12) + (y1– 2) ( 2y1 + 2) = y12 + 2y 1 + 12 + 2y1 2 – 4y1 + 2y1– 4 = 3y12 + 8. when y1 = 2 and 1 dy dz = 0 2 1 2 dy z d > 0  z is min at (1, 2) Minimum distance = (10)2(23)2 = 11 = 2 .

13. We can find the answer through option as the sum of weight of packet taken from trucks is 1022870 gm and its unit digit is 0. The truck that have heavier bags have unit digit 0. So, the truck have lighter bags in which the sum of weight of bags must have unit digit 0.

So, according to option D. i.e. truck no. 2, 8 Track 2 have 21 bags and total weight = 21

× 999 gm = ...8 gm

Truck have 27 bags and total weight = 27

× 999 = 128 × 999 gm = ...2 gm

So, the unit digit of the weight contain by truck 2, 8 together is 0.

14. cos( x)cos([2x] )dx 1 0

  = cos( x)cos0dx 2 / 1 0

 + cos( x)cos dx 1 2 / 1

  = cos( x)dx 2 / 1 0

cos( x)dx 1 2 / 1

 =

x

sin

1/2 0 –

x

sin

1 2 / 1 =        1 –        1 0 =  2 .

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15. IN =        n x ) nx cos( 10 1 0 +

1 0 9 n dx x ) nx cos( 10 = 0 +               

1 0 8 1 0 n dx x ) nx sin( n 9 n 10

n

x

)

nx

sin(

9

=          

1 0 8 2 sin(nx)x dx n 9 10 =        

1 0 10 sin(nx)dx n ! 10 = 0 as Denom  16. y = x2 & y = 1 – x2

Point of intersections of graphs x2 = 1 – x2 2x2 = 1 x = ± 2 1  Point of intersections =        2 1 , 2 1 and         2 1 , 2 1 . Area under graph :

=

   2 / 1 2 / 1 2 2 (1 x) x =

  2 / 1 2 / 1 2 1 x 2 = 2 / 1 2 / 1 3 x 3 x 2           = 2 × 2 1 2 6 2  = 2 2 6 6 2 = 2 3 4 = 3 2 2 . 17. a3i4kand b  = 5j12k

 

4 5 ) 3 ( a  2 2  and b (5)

 

12 13 2 2    

Therefore, a vector which bisects the angle is 13

3i4k

+ 5 (5j 12k

   ) = 39i 25j 8k      .

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19. Let M 2x1.3x2.5x3...  , N 2y1.3y2.5y3...  xi & yi  w

m / d d =

N / d d         1 – 2 1 – 2x1         1 – 3 1 – 3x2         1 – 5 1 – 5x3 ... =                1 – 3 1 – 3 1 – 2 1 – 2y1 y2 ...

N / d M / d d / 1 d / 1 = .... 1 – 3 / 1 1 – 3 1 1 – 2 / 1 1 – 2 1 .... 1 – 3 / 1 1 – 3 1 1 – 2 / 1 1 – 2 1 2 1 2 1 y y x x                                                                                  2 1 2 1 2 1 2 1 y y y y x x x x 3 2 )... 1 – 3 )( 1 – 2 ( ... 3 2 )... 1 – 3 )( 1 – 2 ( = 1 M N  . 20. C 2 m m A from elementonlyone

C C m n m n mC n m 2 1 0        (1 + n) m

PHYSICS

27. Sphere is hollow so potential inside sphere will be same as that on surface.

28. Heat supplied Q = du + W (at contat pressure)

PV = RT PdV = RdT dT = R PdV Q = CVdT + PdV Q = CV R PdV + PdV

Work done at constant pressue, W W = PdV PdV PdV R PdV C W Q V   1 R CV W Q 

 (For diatomic gos, CV = R

2 5 ) 1 R 2 R 5 W Q   2 7 W Q   7 2 Q W 

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29. series balmer For 3 1 2 1 R 1 series yman l For 2 1 1 1 R 1 2 2 2 2                     ( 36 4 9 9 1 4 1    ) 36 / 5 4 / 3      = 5 27 5 9 1 3   27 5      30. q A q B unchanged charge divides 2 q and 2 q Than, on touching 2 q sphere to q Charge divides 4 q 3 2 q 2 / q   force between 4 q 3 2 q R   f= 2 2 R 8 q 3 K f= 8 3 × F 8 3 R kq 2 2 

31. Intially block enters in the magnetic filed rate of change in flux will be constant so costant current will produce, when it mass in side the magnetic field there is no change in magnetic flux, current I = 0, when it use the filed the rate of the change in flux will be again constant between in decrecaing order so contant current will induced on opposite.

32. No change in moment of inertia

34.

E A

O +q

–q

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38.

m

a

Total force in upward direction m × (g + a) because mass m is stationary on inclined plane and whole

system is accelerated with acceleatration a in upward. 39. Force of positve charge = Electric force + Magnetic force

F = (qE + qVB)

This force is in upward direction so no any particle will pass through the hole. 40. Potential energy at H height = Kinetic energy at the lowest point of circular path.

mgH = 2 1

mv2

To complete the circular motion minmum velocity at lowest point will be V = 5gR

mgH = 2 1 m (5gR) H = 2 5 R

CHEMISTRY

41. According to Graham’s law

Rate of diffusion  Molarmass

1

due to highest molar mass of CO2 rate of diffusion is slowest.

42. Moles of H2 = 2 3 , Moles of O2 = 32 4

Kinetic energy of n moles of gas = 2 3 nRT so, oxygen of energy Kinetic hydrogen of energy Kinetic = RT n 2 3 RT n 2 3 2 1 = 2 1 n n = 32 / 4 2 / 3 = 12 : 1

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44. ClF3 sp3d hybridisation,

but due to presence of two lp on central atom Cl, according to VSEPR theory shape is ‘T’

45. HCO3 + H+   H2CO3 Bronsted base HCO3   H+ + CO3 2– Bronsted acid

47. Isoelectronic means same no. of electrons CO has 6 + 8 = 14 electrons

CN– has 6 + 7 + 1 = 14 electrons

48. CO2, due to sp hybridisation bond angle = 180º

49. Diethyl ether, because it is inert towards the Grignard reagent 50. CH3– CH2– CH2– CHO + CH3– CH2– CH2– MgBr CH 3– CH – CH – C – CH – CH – CH2 2 2 2 3 H OMgBr H O3 + CH 3– CH – CH – C – CH – CH – CH2 2 2 2 3 H OH

Achiral Secondary alcohol

51. [Ni (PPh3)2 Cl2]  dsp2 hybridisation, because PPh3 is strong ligand hence pairing of electrons

takes place [NiCl4]2–

 sp3 hybridisation, because Cl– is weak ligand hence pairing of electrons is not

takes place 52. 16H + 2MnO + 5COO + – –

4

COO –

53. Suppose equilibrium constant for the following reaction is K1

N2 + 3H2 2NH3 ; K1 = 3 2 2 2 3 ] H ][ N [ ] NH [ --- (i) and equilibrium constant for the following reaction is K2

2 1 N2 + 2 3 H2 NH3 ; K2 = 3/2 2 2 / 1 2 3 ] H [ ] N [ ] NH [ --- (ii) square the both side of equation (ii)

K22 = 3 2 2 2 3 ] H ][ N [ ] NH [ K22 = k

1 [by equation (i)

K2 = k1

K2 = 41 [ K1 = 41]

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54. Suppose reaction is 2A Product

according to rate law Rate R = k [A]2 or 2 1 R R = 2 2 2 1 ] A [ ] A [ according to question 2 1 R R = 2 1 2 1 2 ] A [ ] A [             2 ] A [ ] A [ 1 2   2 1 R R = 4 R2 = 4 R1 55. HCO3–   H+ + CO32– Conjugate base NH3 H+ + NH2– Conjugate base

56. (II) & (IV)

Because both have close system of conjugated double bond and follow Huckel’s (4n+2)  e– rule.

57. N• •

H

p of N takes part in resonance with conjugated double bonds, so it is not easily available on N for the

protonation.

N

• •

p is not taken part in resonance so easily available for the protonation.

N

• •

H

O

due to high E.N. of O availability of p on N decreases.

N

• •

H

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58. Gauche conformer.

because angle between same groups is 60º

59. Suppose initial quantity = No after 75% completion of the reaction remaining quantity N = No× 100 25 = 4 No T = K 303 . 2 log       N No T = K 303 . 2 log        4 / N N o o T = K 386 . 1 --- (i)  T1/2 = K 693 . 0 K = 30 693 . 0 --- (ii) so by equation (i) and (ii) T = 30 / 693 . 0 386 . 1 T = 60 min. 60. Concentration of H+ ions in H 2SO4 solution = 2 × 0.1 = 0.2 M

So no. of moles of H+ ions in 10 ml H 2SO4 solution = 1000 10 2 . 0  = 0.002

concentration of OH– ions in 0.1 KOH solution

= 1 × 0.1

= 0.1 M

So no. of moles of OH– ions in 10 ml KOH

solution = 1000 10 1 . 0  = 0.001

after mixing remaining moles of H+ ions = 0.002 – 0.001

= 0.001 so concentration of H+ ions in mixture of

solutions = 10 10 001 . 0  × 1000 = 0.05 M

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PART-II (2 Mark)

MATHEMATICS

81. p(x) = a0 + a1x + ...+ anxn p(0) = 7 a0 = 7 p(1) = a0 + a1 + a2 + ...+ an = 9 p(– 1) = a0– a1 + a2 ... = 1 p(2) = a0 + 2a1 + 4a2 + ... = 13 p(– 2) = a0– 2a1 + 4a2 ... = – 15 p(1) + p(– 1) = 2[a0 + a2 + ...] = 10 a0 + a2 + a4 = 5 ...(1) 7 + a2 + a4 = 5 a2 + a4 = – 2 ...(2) p(2) + p(– 2) = 13 – 15 2(a0 + 4a2 + ...) = – 2 a0 + 4a2 + 16a4 = – 1 4a2 + 16a4 = – 8 ...(2) p(3) = 25 a0 + 3a1 + 9a2 + ... = 25

a0 + 3a1 + 9a2 + 27a3 + 81a4 + 243a5 = 25 ...(3)

From (1) and (2) 4a + 4a = 2 4 – 8 4a + 16a = 2 4 – 8 – – + a = 0 and a = 4 2 – 2  Smallest possible value of n is 3. 82.

 abc 2 0 ) b – a (  ab 1 a2 

|a – b| < c ... (1) [Triangle inequalities] |b – c| < a ... (2) |c – a| < b ... (3)

Squaring and adding

a2 + b2 + c2 < 2ab + 2bc + 2ca 2 ab a2    So, b  [1,2). 83. y = |x3|4 – 5 – 6 – 1 3 7 12 O – x 6 x – 4 2 x x – 1 2 – 5 When, x < – 1 y = | 3 – x – 4 | – 5 y = – x – 1 – 5 y = – x – 6 When –1  x < 3 y = | 3 – x – 4 | – 5 = | – x – 1| – 5 = x + 1 – 5 = x – 4 When, 3  x < 7 y = |x – 7| – 5 y = 7 – x – 5 y = 2 – x

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When, x  7

y = | x – 7| – 5

= x – 7 – 5

= x – 12.

Area bounded region

=

    1 6 dx ) 6 x ( +

  3 1 dx ) 4 x ( +

 7 3 dx ) x 2 ( +

 12 7 dx ) 12 x ( = 1 6 2 x 6 2 x             + 3 1 2 x 4 2 x           + 7 3 2 2 x x 2          + 12 7 2 x 12 2 x          =         6 2 1 –         36 2 36 +       12 2 9 –       4 2 1 +        2 49 14        2 9 6 +       144 2 144 –       84 2 49 = 2 11 – 18 – 2 15 – 2 9 – 2 21 – 2 3 – 72 + 2 119 = 2 130 – 2 48 – 90 = 65 – 24 – 90 = 49 sq. unit. 84. b b a b a A B D C cos = ab 2 a b a2 2 2 = a 2 b ....(i) cos(180º – ) = 2 2 2 2 b 2 a b b   – cos = 2 2 2 b 2 a b 2  cos = 2 2 2 b 2 b 2 a  ....(ii) From (i) & (ii)

2 2 2 b 2 b 2 a  =

a

b

2

3 b a       – 2       b a – 1 = 0

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Let b a = x x3 – 2x – 1 = 0 (x + 1)(x2 – x –1) = 0 x = – 1 or x = 2 ) 1 )( 1 ( 4 1 1   = 2 5 1  x cannot be negative  x = 2 1 ) 1 5 (  85. an =

2

a

1

n1 a1 = 2 a 1 0 a1 = 2 cos 1  a1 =

2

2

cos

2

2

a1 = 2 cos a2 =

2

a

1

1 = 2 2 cos 1  =

2

4

cos

2

2

= cos 2 2  -an = cos n 2    nlim 4 n(1 – an)   nlim 4 n        n 2 cos 1

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  nlim 4 n × 2sin2 n 1

2

  nlim 1 n 1 n 1 n 2 n 2 2 2 2 sin 2 2         × n1 n 1 2 2        nlim 2 2  × 1 n 1 n 1 n 2 2 2 2 sin        = 2 2  . 86. f(x) = (sin x)sinx

f(x) =

e

sinxlogsinx

Minimum value of sinxlog(sinx) is 0.

Maximum value of (sin x)sinx is e0 = 1.

Maximum value of sinxlog(sinx) is –

e

1

.

Minimum value of (sin x)sinx is e

1

e

 . 87.

10 1 dx x 1 =

log

x

101 = log10 = 2.303 B = 1 + 2 1 + ....+ 9 1 = 1 + 0.5 + 0.33 + 0.25 + 0.20 + 0.16 + 0.14 + 0.12 + 0.11 2.81 C = 2 1 + ....+ 9 1 + 10 1  2.81 – 1 + 0.1 =1.91

So, C < A < B and B – A  0.51, A – C = 2.303 – 1.91  0.40. So, B – A > A – C.

88. r r r r 60º 60º r r A B C D

As we want the distance between two point is at least r. Now when the point A, B are at distance r.Then the angle made arc BA is 60º.

Now as chord AB come closer to centre the length of chord AB is increased that is it is greater than r and the angle is also increases i.e. from 60º to 180º and now when chord AB move way from centre then the

length of chord AB decreases , when chord AB reach CD the length of AB equal to r and the angle chang from 180º to 60º

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So, the angle required for desired conditions = 2(180 – 60) = 240

Total angle for all around the circle = 360º

So, required propability = 360 240

= 3 2 89. Let aN be pth digit no.

So 1 – 5 ) 5 ( 4 P–1 < N  1 – 5 ) 1 – 5 ( 4 P 2 × 10P–1 < aN 9 8 (10P –1) 5P–1 – 1 < N  5P–1 min N N log a log         = ) 1 – 5 ln( ) 10 2 ln( P 1 P  = 5 ln 5 – 10 ) 1 – 5 ( ) 10 (ln 10 P 1 – P P 1 – P  = 5 ln 10 ln = 10 5 log max N N log a log         = ) 1 – 5 ln( ) 1 – 10 ( 9 8 ln 1 P P  = (10 1)5 ln5 9 8 ) 1 – 5 ( 10 ln 10 9 8 1 P P 1 P P    = log105

By sandwitch therom limit is log105 .

90. S =

  i j n 1 | j – i | = |1–1| + |1 – 2| + --- |1–n| = 2 n ) 1 – n ( |2 – 2| + --- |2 – n| = 2 ) 1 – n )( 2 – n  ... S =

  N 1 r 2 ) 1 r – n )( r – n ( =n1C3

PHYSICS

91. Mass of sphere of radius r

m = 3 3 R Mr               R3 3 4 M Xcm = m M ) r R ( m 0 M                    m R 3 / 4 M r 3 4 3 3 Xcm =          3 3 3 3 R M R M ) r R ( R Mr

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= ) r rR R )( r R ( ) r R ( r R r R R ) r R ( r 2 2 3 3 3 3 3 3                Xcm = 2 2 3 r rR R r  

92. For planeo convex lens

R 1 ) 1 ( F 1 L    FL = 1 R  

Refraction through less

u 1 R 1 V 1    

This v must be centre of mirror

– u 1 1 r R 1 R 1           u =  R 93. In cyclic process u = 0 u =W = Area of loop = (P1– P2) (V2– V1)

94. On comparing both the figures

x = R 6 ) x R ( ) R 6 )( x R (    x2 + xR – 6R2 = 0 x = 2 ) R 6 4 ( R x 2  2  x = 2R 2 R 5 R   

95. Torque about point A

R sin mg MR MR 5 2 2 2           f A O = R 7 sin g 5 

Applying Newton’s law,

Ma = Mgsin – Mgcos gcosgsin– 7 sin g 5  gcos 7 sin g 2  = tan–1 2 7 Velocity of sound = M RT  T Velocity of sound 

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96.

H F

a/2 A

mg

For height greater than H Balancing torque about point A F × H =

2 a

mg ...(1)

For height less than H

F = gh ...(2) From (1) and (2)  = H 2 a 97. mV /r2 mg N                 sin r mv cos mg sin mg cos r MV2 2 sin           r V g 2 = cos           r V g 2 tan = rg V rg V 2 2     98. 3 5 4 S1 S2 Path difference = 5 – 4 = 1 m

For constructive interference n1= 1m

For destructive interference For n = 1, 1= 1m, 2= 2m m 1 2 ) 1 n 2 ( 2   

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99. Two small blocks slide without losing contact with the surface along two frictionless tracks 1 and 2, starting at the same time with same initial speed v. Track 1 is perfectly horizontal, while track 2 has a dip in the middle, as shown.

1

2 V

V

Finish Start

Which block reaches the finish line first ? [Hint : Use velocity-time graph to solve]

(A) Block on track 1 reaches the finish line first (B*) Block on track to reaches the finish line first (C) Both blocks reach the finish line at the same time

(D) It depends on the length of the dip in the second track, relative to the total length of the tracks. 100. Q = u + PV (1st law of thermodynamics ) mv = u + 1.01 ×10 5 ( 3 10 1 8 . 1 / 1 1  ) u 20.8 × 105 J kg–1

CHEMISTRY

101. Millimoles of NH4OH in 10 ml of 0.1 M NH4OH solution = 0.1 × 10 = 1 millimoles of NH4Cl in 10 ml of 1M NH4Cl solution = 1 × 10 = 10 pOH = pKb + log ] Base [ ] Salt [ 6 = pKb + log 10 10 / 1 10 10 / 10   [ pOH = 14 – pH = 14 – 8 = 6] 6 = pKb + log10 pKb = 6 – 1 pKb = 5

102. 2C4H10 + 13O2 8CO2 + 10H2O H = – 2658 kJ/mol

Butane present in cylinder = 11.6 kg = 11600 g =

58 11600

mol

 Combustion of 1 mol of C4H10 gives = 2658 kJ energy  Combustion of 11600/58 mol of C4H10 gives =

58 11600 2658

kJ = 531600 kJ energy energy consumes in 1 day = 15000 kJ

so 531600 kJ energy will be consumed in = 15000 531600  35 days 103. W = d × V = 0.879 × 50 = 43.95 , Kf = 5.12 Kg kg mol –1 = 5120K g mol–1 Tf = mW w Kf o 5.51 – 5.03 = 95 . 43 m 643 . 0 5120   m = 096 . 21 16 . 3292 m = 156 g mol–1

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104. Zn + 2Ag+   Zn2+ + 2Ag (0.04M) (0.28M) Ecell = Eº – n 059 . 0 log ] Zn [ ] Ag [ ] Ag ][ Zn [ 2 2 2   Ecell = 2.57 – 2 059 . 0 log 1 ) 04 . 0 ( 1 28 . 0 2 2   { [Ag] = [Zn] = 1}

by solving the equation we get Ecell  2.50 V 105. Co + Con. HCl 2+ CoCl4 2– HOH(excess) [Co(H O) ]2 6 2+ Pink 106. ln k = T 11067 – + 31.33 2.303 log k = T 11067 – + 31.33

suppose k1 and T1 are rate constant and temperature in case-I and k2 and T2 are rate constant and temperature in case-II So, 2.303 log k1 = – 1 T 11067 + 31.33 --- (1) 2.303 log k2 = – 2 T 11067 + 31.33 --- (2)

by subracting equation (2) from equation (1) 2.303 log 2 1 k k = –11067        2 1 T 1 – T 1 2.303 log 1 1 k 2 k = –11067        2 T 1 – 298 1 [ k2 = 2k1] 2.303 × (–0.3010) = 2 T 11067 298 11067 – 

by solving the above equation we get T2  303.7 K

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107. K1 = ] Br ][ CuCl [ ] Cl ][ Br CuCl [ – – 2 4 – – 2 3 --- (i) K2 = ] Br ][ Br CuCl [ ] Cl ][ Br CuCl [ – – 2 3 – – 2 2 2 --- (ii) K3 = ] Br ][ Br CuCl [ ] Cl ][ CuClBr [ – – 2 2 2 – – 2 3 --- (iii) K4 = ] Br ][ CuClBr [ ] Cl ][ CuBr [ – – 2 3 – – 2 4 --- (iv) equilibrium constant for given equation

K = 2– – 3 4 3 – – 2 3 ] Br ][ CuCl [ ] Cl ][ CuClBr [ --- (v)

by multiplying the right hand side of equation (i), (ii) and (iii) we get right hand side of equation (v) it means K = K1K2K3 108. OH Br in CS2 2 –HBr Br NaOH –H O2 OH Br ONa Me–I –NaI Br OMe (X) (Y) 109. 90234Th82206Pbx24Hey01e By comparing mass no.

234 = 206 + 4x + oy x = 7

By comparing nuclear charge 90 = 82 + 2x 1y y = 82 + 2 × 7 – 90 y = 6 110. O AlCl3 I2/NaOH OH OH II O Me II O Me II O OH Haloform reaction fries rearrangement

Figure

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References

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