2 Answer key & Solution

10. avg

2V 2m / s

1� �

11.

dd/2 h

V

12. dv adt�� � = Area under a-t curve

1

11 10 55m / s2

� � � �

13.dv

a vdx

� ���� �� ��� �

dv

dx is negative and v is decreasing

a

x

14.x

y x tan 1R

� ���� � � �� ��� � ..........(i)

5xy 16x 1

64

� ���� � �� ��� � ............(ii)

Compare equation (i) and (ii)

64R 12.8m

5� �

15.

u

2gh v

2v u 2gh� �

16. HA = H

B = H

C

2yu

H2g

� ��� �� � �� ���� �

so that (uy)

A = (u

y)

B = (u

y)

C

yA B C

2uT so T T T

g� � �

yuu

sin�

�

A B CQ� � � � so that A B Cu u u� �

xR u� �

C B A x C x B x AR R R (u ) (u ) (u )� � � � � �

17. R (range) is same for two angle . so angle are � and

90º - �.

1

2u sinT

g

��

2

2u sin(90º )T

g

���

2

1 2 2

4u sin cosT T

g

� ��

1 2

2T T R

g�

1 2T T R�

18. y = ax – bx2 compare with 2

2 2

gxy x tan

2u cos� ��

�tan� = a � � = tan–1a

22

gu

2b cos�

�2 2 2u sin tan

H2g 4b

� �� �

2aH

4b�

19.1 3 3

tan tan4 4

� � ���� � � ���� ��� �

R 4

H tan�

�

R 16

H 3�

20. t = gx hy

[t] = [gx] [hy]

[M0L0T1] = [M0LxT–2x] [Ly]

[M0L0T1] = [M0Lx+y T–2x]

x+y = 0

3 Answer key & Solution

– 2x = 1 � 1

x2

��

1y

2�

21. a = 2t + 4t2

v t

o o

dv adt�� �t

2

o

v (2t 4t )dt� ��3

2 4tv t

3� �

6x 6

o o

dx vdt�� �6

2 36

o

4x t t dt

3

� ���� � �� ��� ��x

6 = 504 m

22. in down stream

m w m w

2(v v )25 v v

25� � � � �� .........(i)

in up stream.

m w m w(v v ) 40 v v40

� � � � � ��

�

Equation (i) and (ii)

m2v25 40

� �� �

m

13v

400�

�

in still water

m

tv

��

400t sec

13�

23.2u sin 2

Rg

�� , for R

max , sin2� = 1

2� = 90º ������45º

24.21

kE mv2

at highest point v is minimum.so kE is mimimum at highest point

25. Pressure = [M1 L–1 T–2]

energy density = 1 1 2E

[M L T ]v

� �� �� � �� �� �

Torque = [M1 L2 T–2]work = [M1 L2 T–2]Linear momentum = [M1 L1 T–1]

1 2 21 2 1

1

E [M L T ]E h h [M L T ]

T

��

�� �� � � ��

]

strees = �. dimensionlessso [stress] = [�]

26. n

qS u (2n 1)

2� � �

4

10S 0 [2 4 1]

2� � � �

S4 = 5× 7 = 35 m

27. Acceleration = dv

dt

dvslop of v t graph

dt

� �� �� �� �� �

28.

1 1 1x% ( a%) ( b%) ( c%) 4( d%)

2 2 2� � � � � � � � �

1 1 1x% 2 4 2 4 0.5

2 2 2� � � � � � � � �

x% 6%� �

29.x

y x tan 1R

� �� �� � �� �� �

ay a tan 1

a b

� �� �� � �� ��� �

ay a tan

a b

� �� �� �� ��� �

aby tan

a b

� ���� ��� ��� ��

30.dx

vdt

� [x sin kt)��

dv ( sin kt)

dt� �

v = –k cos kt

31.R p dv

C J / mole Ky 1 n dt

� ���� � ��� ��� ��

4 Answer key & Solution

32. v = 4t3 – 2t

3dx4t 2t

dt� �

� �2 t

3

0 0

dx 4t 2t dt� �� �

t 2 sec�

2dva 12t 2

dt� � �

a (at t = 2 sec) = 22 m/s2

33. A V B

3L

4

L

4

v

2

<V> = 1 2

L

t t�

L3L L4V 2V

��

= 0.80V

34. Slope of x – t graph represent the velocity.

35. x = 5 sin 10t

y = 5 cos 10t

x

dxv 50cos10t

dt� � 2 2

x yv v v� �

y

dyv 50 sin10t

dt� ��

= 50 m/s

36. v = u + atv

0 = 0 + at (2n)

as u (2n 1)

2� � �

as (2n 1)

2� �

0v(2n 1)

4n� �

37. ˆ ˆu 3i 4j� ��

ˆ ˆa 0.4i 0.3j� ��

t = 10 sec

ˆ ˆ ˆ ˆv u at (3i 4j) (0.4i 0.3j)10� � � � � �� � �

ˆ ˆv 7i 7j� ��

| v | 7 2 unit��

38. x – t graph slope represent the velocity slope is nega-

tive in AB.

39. x = t3 – 3t2 + 3t+3

2dxv 3t 6t 3

dt

dva 6t 6

dtif v = 0 then t = 1

1/2

3V m / s

4

2V 3

direction no change

40. 2 2 2 2 2 22 1 x 9 2 [2 (2x 1) 1 ]

2x 2,

341. Area of ink blot (A) = (3t2 + 7)

dA6t

dt�

t 5

dA6 5 30

dt �

� ��� � � ��� ��� �cm2/s

42. Given data y = 8x – x3

Findout x 2

dy??

dx �

� ��� ��� ��� �

Approach using differentiation

2dy8 3x

dx� �

2

x 2

dy8 3(2)

dx �

� ��� � ��� ��� �

= 8 – 3 × 4 = 8 – 12 = 4

43. Given data � = t2 + 3t –5

Findout ?? at t 5

d

dt� = 2t + 3� = 2 × 5 + 3� = 13 rad/s

44. Given data 120º

10N

8N

Findout1 2 2 1| F F | or | F F | ??� � ��� �� �� ��

Approach using 2 2A B A BF F 2F F cos� � � �

5 Answer key & Solution

60º

8

102F�

2 210 8 2 8 10 cos 60� � � � � �

180 64� �

244�

45.

60º

120º 70º

110º

A

B

C

D

Approach �

Approach of angle b/w two vectors�

AB = 70º

�CD

= 120º�

BC = 110º

46. Given F = 10

ˆ ˆF (3i 4j)� �

Approach ˆ ˆF.(3i 4j) 0� ��

checking option

(A) ˆ ˆ| ui 3j | 10� �

(B) � � � �ˆ ˆ ˆ ˆ5 2i 5 2 j . 3i 4 j 10� � �

(C) � � � �ˆ ˆ ˆ ˆ8i 6 j . 3i 4 j 24 24 0� � � � �

(D) � � � �ˆ ˆ ˆ ˆ3i 4 j . 8i 6 j 24 24 0� � � � �

47. ˆ ˆ ˆA 3i 4 j 12k��

ˆ ˆ ˆB 2i j (x)k��

� = 23ºcos23º = 12/13

A.B ABcos�����

83x

6048. Given data

10(at+3]

Approach [at] = [3] = [M0 L0 T0]

[a] [T1] = [M0 L0 T0]

[a] = [T–1]

49. Given datab

v att c

� ��

Approach [v] = [at] = [b] [b] [b]

[t c] [t] [c]� �

�

[b][v]

[t]�

[L1T–1] [T1] = [b][L1] = [b][M0L1T0] = [b]

50. Given data

old unit

Length � metre

new unit

Length � x

findout 1m2 (old unit) = �����������(new unit)

Approach using unit analysis

2 22

2

1x m1m

x�

2

1

x� (new unit)

51. Density of air = 0.001293 gm/ml

1 mole of air at STP have 22.4 lit volume

� Mass of 1 mole = 22400 × 0.001293 = 28.96 gm

Vapour density = Molar mass

2 =

28.96

2 = 14.48.

52.V.D. of x

V.D. of y = Molar mass of x

Molar mass of y

2 = 4

Molar mass of y

Molar mass of y = 2 gm/mole.

53. M. wt. of NaNO3 = 85

70 mg of Na+ are present in 1 mL

50 ml of solution contains 50 × 70 = 3500 mg = 3.5 g

Na+ ion

23 g of Na+ are present in 85 g of NaNO3

3.5 g of Na+ are present in 85

23 × 3.5 = 12.934 g of

NaNO3

54. 2 Ag + S � Ag2 S

2 × 108 g of Ag reacts with 32 g of sulphur

10 g of Ag reacts with 216

32 × 10 =

216

320 > 1 g

6 Answer key & Solution

It means ‘S’ is limiting reagent

32 g of S reacts to form 216 + 32 = 248 g of Ag2S

1 g of S reacts to form = 32

248 = 7.75 g

Alternately

neq

of Ag = 108

10 = 0.0925

neq

of S = 1

16 = 0.0625

(neq

= number of equivalents)

Since neq

of S is less than neq

of Ag

� 0.0625 eq of Ag will react with 0.0625 eq of S to

form 0.0625 eq of Ag2S

Hence , amount of Ag2S = n

eq× Eq. wt. of Ag

2S

= 0.0626 × 124 = 7.75 g

55. m = M 1000

(1000 d M M.Wt.) where ‘m’ is molality, M is

molarity.

= 2

2

10 1000

(1000 1.1 10 106)

= 10

1100 1.6 =

10

1099.4 = 9.00 × 10– 3

[Take 1099.4 = 1100]

56. 1 mole of electrons weighs = 9.1 × 10–31 kg × 6.023

×1023

= 54 × 10–8 kg

= 54 × 10– 8 × 1000 = 54 × 10– 5 g

= 54 × 10– 5 g × 103 mg

= 54 × 10– 2 mg

= 0.54 mg

(1) is not correct because charge on 1 e is 1.6 × 10–19 C

and not on 1 mole of electrons.(3) and (4) ruled out as

explained above.

57. 2SO2 (g) + O

2 (g) 2SO

3 (g)

At t = 0 10 16 0

At equilibrium (10 - 2x) (16 - x) 2x = 8

2x=8

x = 4

Thus, number of moles of SO2 unreacted is 10 - 2x =2

� Number of moles of O2 unreacted

= 16 -x =12

58. CaCl2 + Na

2CO

3 ��� CaCO

3 + 2 NaCl

CaCO3 ����

CaO + CO2

Mole of CaCl2 = mole of CaCO

3 = mole of CaO =

1.62

56

Mass of CaCl2 =

1.62

56 Molar mass of CaCl

2

= 1.62

56 × 111 gm.

% of CaCl2 =

3.21

10 × 100 = 32.1 %.

59. KI is limiting reagent

� 3 mole of KI will give 33 mole of NO2 according

to stoicheometry.

60. An orbital can have maximum two electrons.

61. En � Z2 � Z doubled � E

n becomes four times.

Rn � 1/Z � Z doubled � R

n becomes half.

vn � Z � Z doubled � v

n becomes two times.

62. Number of values of � = total number of subshells = n.

Value of � = 0,1,2........(n – 1).

� = 2 � m = – 2, – 1, 0, + 1 , + 2 (5 values)

m = –�� to + � through zero.

63. Eabsorbed

= Eemitted

�hc

300 =

hc

496 +

hc.

� � = 759 nm.

64. KE = – TE = 3.4 eV.

� � = 150

KE =

150

3.4Å.

65. �e =

e e

h

2m KE = p

h

2 1/1837m 16E

, �p =

p p

h

2m KE = p

h

2m 4E .

�� = h

2m KE = p

h

2 4m E

� �e > �

p = �� .

66. Since some visible quanta were observed along with other

quanta, electrons must have made transition from some

higher state to n = 2 and then from n = 2 to n = 1.

� Transition from 2 � 1 is compulsory, because

electron from n = 2 will finally fall into n = 1.

67. �p =

p p

h

2m q V � �

a =

h

2m q V

�a =

p p

h

2 4m 2q V

��

�p =

8

1 =

2 2

1.

7 Answer key & Solution

68. n(n 2) = 4.9

� No. of unpaired electrons, n = 4.

25Mn : [Ar]4s23d5

For having 4 unpaired electrons, a Mn atom should

lose 3 electrons (2 from 4s and 1 from 3d).

� a = +3.

69. Energy of one photon = 12400

3100 = 4ev = 4 x 96

= 384 kJ mol–1

� % of energy converted to K.E. = 384 288

384 =

96

384 x 100 = 25%

70. H2C

2O

4 readily participates in oxidation-reduction and

acid-base reactions both (due to 2 acidic H and C in

+ 3 oxidation state).

71. O3 will oxidise H2O2 into oxygen, hence radioacitve

oxygen of H2O2 will go only in oxygen, not in water.

Half reactions : O3 + 2H+ + 2e– ��� O2 + H2O

; H2O2 ��� O2 + 2H+ + 2e–

72. ��� + H2O +

� v.f. of dichloroacetic acid = 2 (4 – 2) + 2 (0 – (–

1)) = 6

m.eq. of dichloroacetic acid = m.eq. of oxidising agent

= 600

� m.moles of dichloroacetic acid = 600

6 = 100

CHCl2COOH + NH

3 ��� CHCl

2COONH

4

From reaction, m.moles of NH3 = m.moles of

dichloroacetic acid = 100

� Moles of NH3 =

100

1000 = 0.1

73. meq. of KMnO4 = meq of H2O2

30 × 12

1 = 20 × N� � N� =

2012

30

� =

8

1N

� strength = N� × equivalent mass = 8

1 × 17 = 2.12

g/L.

74. ; ;

75. No. of meq of H+ = 10 × 1 + 20 × 2 = 50

[� H2SO

4 , N = 2 M]

No. of meq of OH– = 30 × 1 = 30

No. of meq of H+ left unreacted = 50 – 30 = 20 meq

76.

oxidation number (N1 = +1, N2 = –2, N3 = 0)

77. Cl2 + OH– ��� CI– + ClO

3– + H

2O ;

v.f. of Cl2 =

2 10

2 10 =

5

3

� Eq. wt. of Cl2 =

71

5 / 3 = 42.6

78. �EN = 3.0 – 1.2 = 1.8

�EN > 1.7 (ionic)

79. Lattice energy AlBr3 > CaO (more change)

80. Polarisability � charge on anion

� size of anion

F– < Cl– < Br– < I–

81. Atability µ 1/polarisation

�� I– large size, more polarisation

� Cu+2 small size, more polarization

82.S

O O 2�, 2�, one lone pair

83. O = P

OO

P O

O P O

P = OO

O

O

Total sigma bond =

16

84. Cl

O O O = C = O

+O = N = O I – I – I

85. NO2+

+O = N = O Bond angle = 180º

86. Xe

F

F

3 lp

XeF

2 lp

F

F

F Xe

F

1 lp

F

F

FF

F

87. S1 = SnCl

2

Sn =

(ground state)

due to lone pair repulsion bond angle is less than

120º

S2 = XeF

7+ = sp3d3

(All bond length not equal)

S3 Sf

4

8 Answer key & Solution

SF

FF

F

Due to lone pair – bond pair repuloion bond angle

are less than 0180º & 90º

88. SF

FO O = S

O

O Cl

OOO

XeO

OO

89. IO2F

2–

IO

OF

F

– S

F

FF

F

(See-saw Shape)

90. H2CO

3 = sp2 (polar)

S1F

4 = sp3 (non polar)

BF3 = sp2 (non polar)

HClO2 = sp3 (polar)

91. Borax = sp2, sp3

Diborane = sp3

Borazole = sp2

92. (i) Bond length � 1/Bond order

(i) B.L. NO– > NO > NO+

B.O. � 2 2.5 3.0

(ii) Bond energy µ 1/B.O.

H2 > H

2+ > He+

B.O. 1 0.5 0.5

(iii) O22– O2 O

2+2

Unpaired e– �� 1 2 0

paramajnetic charclar = O2 > O

22– = O

22+

93. Z = 118

It is 18 group element

so is is noble gas

94. halogen show maximum oxidation state (+7)

95. Ionic radiums µ negative charge

Mg+2 < Na+ < F– < O2–

96. There is large energy jump in 2nd and 3rd IE so

two valence e–

97. Na IE

EA Na+

IE of Na = EA of Na+

98. Amoniated solution of metal is blue, paramagnatic,

conductor.

99. NaH + H2O � NaOH + H

2 (Basic solution)

100. Na + Al2O

3 � Na

2O

(x) + Al

Na2O + H

2O � NaOH 2CO Na

2CO

3(y)