hints & solution - wordpress.com · 9.total number of geometrical isomers in the given ........

RESONANCE PSOL090111 - 1

Hints & SolutionDATE : 09-01-2011 COURSE : VIJETA (P) & VISHWAAS (F)

PAPER-1PART-I (Chemistry)

1. The standard molar enthalpies of formation of .......................

Sol. + H2 ;

Hº1 = � 28.5 Kcal/mol.

+ 3H2 ;

Hº2 = 3 × Hº1 = � 85.5 Kcal/mol. ...... (1)

Also, + 3H2 ;

Hº3 = Hº�(cyclohexane)

� Hº�(benzene)

...... (2) = � 37 � (11.5) = � 48.5 Kcal/mol.

From (1) and (2)

;

Hº4 = RE = Hº2 � Hº3 = (� 85.5) � (� 48.5) = � 37 Kcal/mol.

2. If two moles of a hypothetical real gas X occupy .......................

Sol. Z = nRT

PV =

30121

2

410

= 0.8

Z < 1, therefore attractive tendencies are dominant among gasmolecules. So, Gas X is more compressible with respect to anideal gas.1 mole of gas X occupies a volume of 22.4 L at a pressure lowerthan that at STP (keeping temperature constant).

3. Which of the following statements is INCOR .......................Sol. K2CO3 and not KOH, can be prepared by Le-Blanc process, but

not by Solvay�s process commercially.

4. A proton beam can undergo diffraction by .......................Sol. For a proton,

(Å) =

V

286.0 0.143 =

V

286.0 V = 4 volts

5. The solubility of Fe(OH)3 in a buffer solution of pH .......................

Sol. Fe(OH)3 Fe3+ + 3OH�

S' 10�8 S' = 4.32 × 10�2 mol/L.Now, Ksp = [Fe3+] [OH�]3

27S4 = 4.32 × 10�2 (10�8)3 S = solubility in pure water. S = 2 × 10�7 mol/L.

Ratio = S

'S = 7

2

102

1032.4

= 216000

6. The density of a mixture of O2 and N2 at NTP is 1.4 .......................Sol. Mmix = 1.4 × 22.4 = 31.36

Let there be 1 mole of gas mixture, containing x mole of O2 and(1 � x) mole of N2.

32 x + 28 (1 � x) = 31.36 x = 0.84

2Op = 2OX × Ptotal= 1

84.0 × 1 = 0.84 atm.

7. Which of the following compounds of Xenon .......................Sol. XeO2F2 has see-saw shape. So, it has non-zero dipole moment.

8. In which of the following processes on an ideal .......................Sol. For Reversible Isothermal expansion, Ssurr. < 0 and for

Irreversible adiabatic process, Isothermal free expansion (againstvacuum) and Adiabatic free expansion (against vacuum),Ssurr. = 0.

9. Total number of geometrical isomers in the given .......................

Sol. Geometrical isomers = 2n � 1 + 21n

2

n = 3 Geometrical isomers = 6

MAJOR TEST-2 (MT-02)TARGET IIT-JEE 2011

10. The total number of isomers for the alkyne with .......................

Sol. enantiomer

enantiomer

enantiomer

enantiomer

11. What percentage of total KMnO4 is being .......................

Sol. 5Fe2+ + KMnO4 + H Mn2+ + Fe3+

vf = 1 vf = 5Moles of extra KMnO4 used by FeSO4

= 5

1 × Moles of FeSO4 =

5

1 × (6 × 1) = 1.2

% of total KMnO4 consumed by FeSO4

= 10

2.1 × 100 = 12%

12. In what molar ratio, is KMnO4 getting consumed .......................

Sol. 5Cu2S + 8KMnO4 + H Mn2+ + Cu2+ + SO2

vf = 8 vf = 5

5CuS + 6KMnO4 + H Mn2+ + Cu2+ + SO2

vf = 6 vf = 5 Moles of Cu2S + Moles of CuS = 7 .......... (1)

&5

8Moles of Cu2S +

5

6Moles of CuS = 10 � 1.2 = 8.8 .... (2)

By (1) & (2)

SCu2n = 1 and CuSn = 6

Moles of KMnO4 consumed by Cu2S = 1 × 5

8

and Moles of KMnO4 consumed by CuS = 6 × 5

6

Therefore, Ratio = 5

8 :

5

36 = 2 : 9

13. Work done by the gas mixture in the above .......................

Sol. mix = 21

21

V2V1

P2P1

CnCn

CnCn

= )2/R3(4)R3(2

)2/R5(4)R4(2

=

2

3

For mixture, equation of process is :

log T = � 2

1 log V + constant

TV1/2 = constant

1mixTV = constantSo, the process is a reversible adiabatic process.Now, TiVi

1/2 = TfVf1/2

300 × 21/2 = Tf × 81/2

Tf = 160 K

W = 1

nR

(Tf � Ti) =

2/1

R6 (160 � 320) = � 1920 R.

14. What can be said about the value of Ssystem for .......................Sol. For a reversible adiabatic process, Ssys. = Ssurr. = Suniv = 0.


15. What would be the ionisation constant (Ka) of .......................

Sol. CH3COOH + NaOH CH3COONa + H2O150 × 0.1 50 × 0.1

pH = pKa + log ]acid[

]salt[(Since acidic buffer)

4.7 = pKa + log 1.0100

1.050

pKa = 5 ka = 10�5

16. What will be the pH of solution, when 'A' and 'C' .......................

Sol. CH3COOH + KOH CH3COOK + H2Om mol 150 × 0.1 150 × 0.1m mol 0 0 15

[CH3COOK] = 300

15 =

20

1

pH = 7 + 2

1 × 5 +

2

1 log

20

1( Salt of WA + SB)

pH = 7 + 2.5 � 2

1 × 1.3

pH = 9.5 � 0.65 = 8.85

17. The number of electrons, having anticlokwise .......................Ans. FalseSol. n = 3, m = 1, therefore, one orbital of 3p subshell and one orbital

of 3d subshell.Zn2+ = 1s2 2s2 2p6 3s2 3p6 3d10

No. of electrons with anticlockwise spin = 1 + 1 = 2.

18. NCl3 does not undergo reaction with water due .......................Ans. FalseSol. NCl3 undergo reaction with water due to H-bonding.

19. Under identical conditons, a gas with lower mole.......................Ans. TrueSol. Under identical conditons, a gas with lower molecular mass

diffuses faster as compared to another gas with higher molecularmass. (According to Graham Law of diffusion)

20. Epoxyethane and methoxymethane are function.......................Ans. False

Sol. & CH3�O�CH3

Epoxyethane MethoxymethaneBoth have different molecular formula.

21. Determine the sum of fully filled and half filled .......................Sol. Mn2+ : 1s2 2s2 2p6 3s2 3p6 3d5

(R=A=0) (R=1,A=0) (R=0,A=1) (R=2,A=0) (R=1,A=1) (R=0,A=2)So, condition in the question corresponds to orbitals of 2p and3d subshells. Total number of half filled and fully filled orbitals = 3 + 5 = 8.

Ans. 8

22. Determine the sum of number of d-p bonds and .......................Sol. For Caro's Acid 2 d-p bonds + 1 peroxy linkage

For Marshall's Acid 4 d-p bonds + 1 peroxy linkage Total = 2 + 1 + 4 + 1 = 8 Ans. 8

23. Calculate the difference between enthalpy .......................

Sol. C7H16() + 11O2(g) 7CO2(g) + 8H2O() ; ng = 7 � 11 = � 4

H � E = (ng)RT

= 1000

30024� = � 2.4 Kcal = 2 (on rounding off) Ans. 2

24. How many of the following compounds do .......................Sol. Compounds of Be and Mg do not impart any colour to Bunsen

flame. Ans. 3

25. Write the total geometrical isomers (only cyclic) .......................

Sol. &

Geometrical isomers = 2 Geometrical isomers = 3Total = 5 Ans. 5

PART-II (Mathematics)26. The complete range of k for ............................Sol. kcos2x � kcosx + 1 0

k(cos2x � cosx) + 1 0

But cos2x � cosx = (cosx � 21

) 2 � 41

� 4

1 cos2x � cosx 2

we have 2k + 1 0 and � 4k

+ 1 0

k � 21

and k 4 � 21 k 4

27. If the equation .............................................Sol. 2x2 + 4x(y � 3) + 7y2 � 2y + t = 0

D = 0 (for one solution ) 16(y �3)2 � 8(7y2 � 2y + t) = 0 5y2 + 10y + t � 18 = 0 again D = 0 (for one solution )100 � 20 (t � 18) = 0t = 23for t = 23 , 5y2 + 10y + 5 = 0(y + 1)2 = 1, y = � 1for y = �1 , 2x2 � 16x + 32 = 0x2 � 8x + 16 = 0 x = 4 x + y = 3

28. If z is a complex number ..............................Sol. We have |z � (1 + i)|2 = 2

(x � 1)2 + (y � 1)2 = 2 Put z = x + iyx2 + y2 = 2(x + y) ..........(i)

Let = h + ik = z2

= iyx2

= 22 yx

)iy�x(2

h = 22 yx

x2

, k =

22 yx

y2�

h � k = 22 yx

)yx(2

= 1 from equation (i)

Locus of the point (h, k) will be (h, k)x � y = 1

29. If Sn =

n

1r234 rr2r

1r2,............................

Sol. Sn =

n

1r234 rr2r

1r2=

n

1r22 )1r(r

1r2=

n

1r22 )1r(

1�

r

1

S10 =

1211

�1 = 121120

30. If a, b, c, d, e are positive.............................Sol. (a,b,c,d,e) = (1, 1, 2, 2, 2), (1, 1, 1, 3, 3), (1, 1, 1, 2, 5)

and their permutations = !2!3!5

+ !2!3!5

+ !3!5

=10+10+20 = 40

31. BC is latus rectum .......................................Sol. Let tangent at P(at2, 2at) make an angle with x-axis, then

tan = t1

Projection of BC on tangent = BC sin

= 2t1

a4

2a2 as ( � 1 t 1)

32. If , and be the eccentric.....................Sol. + + + = 2n

, , be eccentric angles at concylic points thencos( + + + ) = cos 2n= 1cos( + + + ) = cos 2n= 1

33. The locus of P(x, y) such that .......................Sol. The distance between the two points (3, 0) and (0, �4) is 5

Locus of P is the part of line 14y

�3x

34. If sin(sinx + cosx) = cos(cosx � sinx).............

Sol. cos(cosx � sinx) = cos

)xcosx(sin�

2


cosx � sinx = 2n ±

)xcosx(sin�

2(i) Taking +ve sign

cosx � sinx = 2n + 2� sinx � cosx

2cosx = 2n + 2

cosx = n + 4

n = 0, cosx = 4

(only possible values)

sinx = 16�16 2

= 4�16 2

(ii) Taking �ve sign

cosx �sinx = 2n�2

+ sinx + cosx

2sinx=2� 2n

sinx = 4� n sinx =

4

(only possible values)

at 4

> 4�16 2

sinx|max = 4

35. Number of 9 lettered words...........................Sol. 'MEENANSHU' number of letters = 9 (EE = 2, NN = 2)

Number of ways = total words formed � n (A B)

= !2!2!9�

!7�

!2!8.2

= 18.7! � 7.7! = (18 � 7) 7! = 11.7! k = 11

36. a, b, c are....................................................37. Measure of ..................................................Sol.(36, 37)

Given log

aca + log

ba

= log2

log b

ca = log2

a + c = 2b ............(i) A.P. Ans.Also (c � a) ax2 + 2bx + (c + a) = 0 has equal roots D = 04b2 � 4(c2 � a2) = 0b2 = c2 � a2 ..........(ii)c2 = a2 + b2

ABC is right angled at C C = 90º

38. The values of 'a' .........................................39. The equation of circle....................................Sol.(38, 39)

Line x + 2y + a = 0Circle x2 + y2 = 4

Solve x2 +

2

2xa

= 4

5x2 + 2ax + (a2 � 16) = 0 will have two distinct root then4a2 > 20(a2 � 16)

a2 < 20 � 52 < a < 52Equation of circle passing through intersection points of circlex2 + y2 � 4x � 2y + 1 = 0with line 12x � 6y � 41 = 0(x2 + y2 � 4x � 2y + 1) + (12x � 6y � 41) = 0 .......(i)Equation of circle passing through intersection point of circlex2 + y2 = 4 with line x + 2y + a = 0 is given by(x2 + y2 � 4) + µ( x + 2y + a) = 0 .......(ii)compare (i) and (ii)

� 4 + 12 = µ� 2 � 6 = 2µ1 � 41 = � 4 + µa a = 2

39. From Q.38 put the value of = 51

in (i)

then equation of circle passing through points A,B,C,D is 5x2 + 5y2 � 8x � 16y � 36 = 0

40. They are all consecutive ...............................41. They form an A.P. ........................................

Sol.(40, 41)Three consecutive numbers {1,2,3}, {2,3,4},.....{16,17,18} 16 casesNo two are consecutive the number of ways= 18 � 3 + 1C3 = 16C3 = 560Total ways = 560 + 16 = 576

41. If a < b < c and a, b, c in A.P. a + c = 2b, even number both a and c are odd or both even The number of ways = 2 . 9C2 = 72

42. Number of ways in which ..............................Sol. 7C4 . 1! = 35

43. Number of ways in which ..............................Sol. 5C0 + 5C1 + 5C2 + 5C3 + 5C4 + 5C5 = 25 = 32

44. Equation x4 � 2x2 .........................................

Sol. x4 + 1 = 2x2 sin2 2x

x2 + 2x

1= 2sin2

2x

Now x2 + 2x

1 2

x2 + 2x

1= 2 x2 = 1 x = ±1

45. Equation tan ...............................................

Sol. tan

6x = 2tanx

)6/tan(.xtan�1)6/tan(xtan

= 2 tanx

3

y�1

3

1y

= 2y Let tanx = y

y�3

1y3 = 2y

2y2 � y3 + 1 = 0 its D < 0 so no solution

46. If the variable line .........................................Sol. C1 : (x � 1)2 + (y � 1)2 = 1

C2 : (x � 8)2 + (y � 1)2 = 4The given line L : 3x � 4y + k = 0 will lie between these circlesif centres of the circles lie on opposite sides of the lineL (1, 1) . L(8,1)(k � 1)(k + 20) < 0 k (�20, 1) ......(1)Also the line L will neither touch nor intersect the circle iflength of perpendicular drawn from centre to L >corresponding radius.

For C1 5

|k1.4�1.3| > 1

|k � 1| > 5k � 1 < �5 or k � 1 > 5k < � 4 or k > 6 .........(2)

and for C2 5

|k1.4�8.3| > 2 |k + 20| > 10

k + 20 < �10 or k + 20 > 10k < �30 or k > �10 .......(3)from (1), (2),(3) k (�10, �4)

a = �10, b = �4 (b � a) = �4 + 10 = 6

47. If the sum of all the solutions ........................Sol. 3 cot2 + 10 cot + 3 = 0

cot = �3 or � 3

1

cot = � 3 two solution one in 2nd and other in 4thquadrant say 1 and + 1.

similarly cot = � 31

two solution 2 and + 2.

sum = 2[ + (1 + 2)] = 2[ + cot�1(�3) + cot�1

31

� ]

= 2[ + � cot�13 + � cot�1

31

] = 2[3 �(cot�13 + tan�13)]

= 2[3 � 2

] = 5


48. If is the imaginary cube ............................Sol. We have |a + b| = 1

|a + b|2 = 1

(a + b) )ba( = 1 a2 + ab (+ ) + b2 = 1(as 1 + + 2 = 0) a2 � ab + b2 = 1 (a � b)2 + ab = 1 (a � b)2 = 0 and ab = 1 then (1, 1), (�1, �1)

(a � b)2 = 1 and ab = 0 then(0, 1), (1, 0), (0, �1), (�1, 0)Hence (0, 1), (1, 0), (0, �1), (�1, 0) (1, 1), (�1, �1)i.e. 6 ordered pairs.

49. The line 3x + 6y = k intersect .......................

Sol. We have k

y6x3 = 1 ..........(1)

2x2 + 2xy + 3y2 � 1 = 0 .....(2)Now homogenising (2) with the help of (1), we get

2x2 + 3xy + 3y2 �

2

ky6x3

= 0

k2(2x2 + 3xy + 3y2) � (3x + 6y)2 = 0Now coefficient of x2 + coefficient of y2 = 0(2k2 � 9) + (3k2 � 36) = 05k2 = 45k2 = 9 k = 3, � 3 sum = 3 + (� 3) = 0

50. Let A be the set of all 3 × 3 skew ...................Sol. In a skew symmetric matrix diagonal elements are zero.

Also aij + aji = 0Hence number of matrices = 2 × 2 × 2 = 8

PART-III (Physics)51. A tube of length open at only one end is ...................................Sol. According to given condition,

e

2 4

v13

=

e2

2 2

v7

e = 24

So, r = 6e10

r = 725

53. A non�uniform rope of length hangs ...................................

Sol. vy = y

yT

Ty = g dyey

0

y0

Ty = 0(e

y �1).g

vy = ye

gg vy

2 = g(1 � e�y).

55. A monkey of mass 'm' climbs�up on ...................................

Sol. Let 2a

is the acceleration of centre of circular object

T + f = 8m2a

..............(i)

(T � f) R = 8mR2 .R2a

..............(ii)

From equation (i) & (ii),f = 0

Then, T � mg =

a

2g

m

4ma = mg + 2

mg � ma a =

10g3

So,acceleration of monkey with respect to ground=10

g32g =

5g

So, = 21

5g

t2 So, t = g

10.

56. A metal block of area 0.10 m2 ...................................Sol. Shear force F = T = mg = 0.020 × 10 = 0.2 N

Shear stress on the fluid = AF

= 1.0

2.0 = 2

Strain rate =

= 31030.0

090.0

= rate nstrai

stress =

)090.0()1030.0(2 3

= 320

× 10�3 Pa s.

57. Extension in a uniform rod of length ...................................

Sol. Tension at point P in the rod isT =

mx g

Extension of width dx length of rod is

= 2rY

Tdx

=

2rY

mg

xdx

Total extension in rod =

0

2rY

mg xdx = Yr2

gm2

58. 0.056 kg of nitrogen is enclosed in a ...................................Sol. As gas is enclosed in the cylinder, V = constant

(Q)v = n CVT

n = 28

10056.0 3 = 2 mol.

and as nitrogen is diatomic, CV =

25

R

1

2

1mPs

2mPs

TT

)V(

)V( = 2 T2 = 4T1

T = 3T1 = 900

(Q)V = 2 × 2

5 × 2 × 900 (Q)V = 9000 cal

59. A gas has molar heat capacity C = 24.9 J ...................................Sol. For one mole of an ideal gas

PV = RT and P2 T = k

P = Tk

and V = k

R

P

RT T3/2

dV = dT.Tk

R

2

3

so dW = dT.T.k

R23

.Tk

PdvdTT

T

= 23

RdT

NowdQ = dU + dw

CdT = CVdT + 23

RdT

CV = C � 23

R CV = 2R3

2R3

�3.8R9.24

60. A small ball of mass m is released from ...................................

Sol. Nmm2N

=

R

v.m

2

2N

+ N = R

)gh2.(m

2N3

= R

mgh2 N =

R3mgh4

.

61. Beat frequency observed by the driver...................................62. Consider the sound wave observed ...............................61 & 62.

(61) f1i = f1r = fvv

v

c

f2i = f2r = fvv

v

cNow, for driver

fdr1 = r1c f

v

vv and fdr2 = r2

c fv

vv


So, beat frequency = | fdr1 � fdr2 |

= r2c

r1c f

v

vvf

v

vv

= f

)vv)(vv()vv()vv(

cc

2c

2c

= f v

vv42

c

= f

v

v4 c

.

(62) 1 = f

vv c2 =

f

vv c

1 � 2 = f

v2 c 1 + 2 =

fv2

21

21

=

vvc .

63. Minimum value of impulse P so that disc ...................................64. Find the friction force on sphere just ...................................63 & 64.

(63) For Pmin

P(R) = 23

mR2.

= mR3P2

By energy conservation,

21

.23

mR2. 22

2

Rm9

P4 =

2mgR

P = gR23

m

(64) mgR23

= 2

3mR2.

= R 3

g

f2mg3

= m(R)

fmg23

= 3

mg f =

23

mg � 3

mg

f = 32

mg.

65. Ratio of maximum temperature and ...................................

Sol. After collision velocity of (particle + piston) = 2v

So, K.E. = 2

1(2m)

4v2

= 4

mv2

As, Q = 0, U = �W

So,4

mv2

= 2R3

T T = R6

mv 2

So, Tmax = R6

mv

R

mv 22

= 67

Rmv2

and Tmin = R6

mvR

mv 22

= R

mv65 2

So, min

max

T

T =

57

.

66. Assume whole system is placed in gravity ...................................Sol. When particle strikes elastically, particle comes to rest instanta-

neously and piston moves with velocity v.

So,21

mv2 = 23

RT T = R3

mv2

So, Tmax = R

mv34 2

and Tmin = R

mv2

So,min

max

T

T =

34

.

71. A uniform string of length , fixed at both ...................................

Sol. Total energy E =

0

2dmv21

=

0

22xdmA

21

=

0

222 .kxsinA.dxm

2

1=

41

mA22

= 2f = 2.

T2v3

= mT3

Energy = 41

ma2. 2

29

.

mT

Energy = 4

9

Ta 22

So, energy between two consecutive nodes =

Ta43 22

.

72. A tall glass jar has a small hole of radius ...................................Sol. Water will begin to enter the jar through the hole when the

hydrostatic pressure exceeds the allowed excess pressure of2T/r on the concave side of the water surface at the hole. So letdepth is x, then

xdg = 2T/r

x = 3�1006.0101000

06.02dgr

T2

= 0.2 m = 20 cm =

10

x= 2

73. 4 gms of steam at 100°C is added to 20 ...................................Sol. Heat released by steam inconversion to water at 100°C is

Q1 = mL = 4 × 540 = 2160 cal.Heat required to raise temperature of water from 46°C t100°C is Q2 = mS = 20 × 1 × 54 = 1080 J

Q1 > Q2 and 2

1

Q

Q = 2

Hence all steam is not converted to water only half steamshall be converted to water

Final mass of water = 20 + 2 = 22 gm

74. A uniform rectangular plate of mass 1 kg is ...................................

Sol. (K.E.) of plate w.r.to A = 21

IA2

A = 34

kg m2

(K.E.)A = 21

× 34

× 4 = 38

J.

75. Initially a small block is in equilibrium and ...................................Sol. From given conditions

10 = ut + 21

.gt2 u = 5 m/s

So, block held with spring will have velocity 5 m/s in verticallyupward direction.Now, v2 = 2(A2 � x2)

25 = 2

10

22

k

mgA

5 = A2 � 4

A = 3m.

PAPER-2PART-I (Chemistry)

1. Select the correct option for the vapour pressure .......................Sol. A liquid having lesser boiling point has greater vapour pressure

at a particular temperature.

2. The nodal plane in the -bond of ethene is .......................

Sol.

This -bond is formed by lateral overlapping of two p-orbitals oftwo C-atoms. These p-orbitals are parallel to the plane containingethene molecule, therefore molecular plane does not have any electron density and nodal plane lies along the molecular plane.

3. In which of the following cases upon mixing, .......................Sol. A buffer solution always contains either a weak acid and its salt

or a weak base and its salt.4. For the equilibrium .......................Sol. Value of Kp decreases exponentially with increase in temperature,

according to Vant Hoff's equation for exothermic reaction.

5. Suppose you are given two oleum samples .......................

Sol. %SO3 in sample A = 981801

801

× 100 45%


and %SO3 in sample B = 50%Greater the % of SO3 in the oleum sample, greater is the %strength of oleum sample.

6. An unknown gaseous hydrocarbon and oxygen .......................

Sol. CxHy +

4

yx O2

xCO2 + 2

yH2O ()

i a 7a 0

� 0 7a �

4

yx a xa

Given : 7a �

4

yx a = xa

8x + y = 28Only C3H4 hydrocarbon satisfies the above relation.

9. 20 mL of pure formic acid (HCOOH) liquid is .......................

Sol. (Vmix)actual = mix

mixd

m=

mix

OHHCOOH

d

mm2

=8/9

4012015.1 = 56 mL

(Vmix)expected = 20 + 40 = 60 mL

% contraction in volume = 60

5660 × 100 = 6.66%

XHCOOH =

1848

462015.146

2015.1

= 49

9

% w/v = .solV

wtHCOOH × 100 =

56

23 × 100 = 41.07%

10. Consider the following equilibrium .......................Sol. As pressure increases, equilibrium shifts towards lesser number

of gaseous moles, i.e. forward direction.So, moles of H2 and CO decrease while moles of CH3OH willincreases.Also, as Pequi. increases, so partial pressure and henceconcentration of all the species will increase.

( P = V

nRT = CRT)

11. Which of the following statements is/are correct .......................Sol. (B) Mineral Carnallite (KCl.MgCl2.6H2O) contains the metal

Magnesium.(D) Slaked lime is used for removing temporary hardness ofwater.

12. Which of the following statements is/are .......................Sol. (A) Lothar Meyer curve is the 'atomic volume' vs 'atomic mass'

curve, in which alkali metals occupy the crests of curve.(C) If IE1 for He-atom is 10 eV/atom, then the energy required forthe conversion He (g) He2+ (g) + 2e� is 10 + 13.6 (22)= 64.4 eV/atom.(D) The element having atomic number 29 (Cu), having oneelectron in outermost shell, belongs to d-block of Modern periodictable.

13. Which of the following statements is/are true for .......................Sol. (C) Dichromate titrations require an external indicator for endpoint

detection (Diphenyl amine).(D) In iodometric titrations, generally KI is used as an reducingagent.

15. (A ) PCl3F2 and PCl2F3 .......................(B) C6H6 and B3N3H6 .......................(C) ICl2

� and I3� .......................

(D) CO2 and CN22� .......................

Sol. (A ) PCl3F2

= zero (non-polar)Hybridisation of central atom = sp3d.

Shape of molecule about central atom / ion = trigonalbipyramidal.

PCl2F3

zero (polar)Hybridisation of central atom = sp3d.

Shape of molecule about central atom / ion = trigonalbipyramidal.

(B) C6H6

= zero (non-polar)Hybridisation of central atom = sp2.

Shape of molecule about central atom / ion = trigonal plannar.

B3N3H6

= zero (non-polar)Hybridisation of central atom = sp2.

Shape of molecule about central atom / ion = trigonal plannar.

(C) ICl2�


Shape of molecule about central atom / ion = linear.

I3�


Shape of molecule about central atom / ion = linear.(D) CO2 O=C=O

= zero (non-polar)Hybridisation of central atom = sp.

Shape of molecule about central atom / ion = linear.CN2

�2 N=C=N

= zero (non-polar)Hybridisation of central atom = sp.

Shape of molecule about central atom / ion = linear.

17. If the pH of a 0.2 M solution of a weak base BOH .......................Sol. For solution of weak base BOH,

pOH = 2

1(pKb � log C)

14 � 10.3 = 2

1(pKb � log C)

pKb = 6.7Now, for making the pH of the resulting solution 7.3 (pOH = 14 �7.3 = 6.7 = pKb), the solution should be at half equivalence point.

For eq. point, volume of HCl required = 4.0

2.020 = 10 mL

For half eq. point, volume of HCl required = 2

10 = 5 mL

Ans. 5

18. For the given reaction at � 23ºC .......................

Sol. For the given reaction : Kp = ( OH2p )9 = 10�9

OH2p = 0.1 atm

OH2n required for consumption of solid completely = 0.5 × 9 =

0.45 (according to stoichiometry of reaction).

Also, OH2n required to maintain the above equilibrium

= RT

Vp OH2 = 240

121

101.0

= 0.05

( OH2n )total required = 0.45 + 0.04 = 0.5 moles

Mass of H2O to be added = 0.5 × 18 = 9 gramAns. 9

19. How many of the following compounds contain .......................Sol. KO2, NO2, ClO2, NO, KO3 are paramagnetic (contain unpaired

electrons).Ans. 5

20. How many of the following compounds can .......................Sol. SO2 , H2O2 , HNO2 , H3PO2 can act as both oxidising agent and

reducing agent due to intermediate O.S. of S(+4), O(�1), N(+3)and P(+1). Ans. 4


21. Write the number of acyclic ester have molecular .......................Sol. C4H6O2

D.U. = 2, ,

, ,

Ans. 5

PART-II (Mathematics)22. If (a2 + b2 + c2 + ab + bc + ca)......................Sol. We have a2 + b2 + c2 + ab + bc + ca 0

( a + b)2 + (b + c)2 + (c + a)2 0 a + b = 0 b + c = 0 c + a = 0 a = b = c = 0

Now

410

041

104

= 4(16 � 0) + 1(1 � 0) = 65

23. Let 1a

1+

2a1

+ ..................................

Sol.1a�a

1�a2

r2r

r

=

)a)(a(

a22

rr

2r

=

)a(1

r +

)a(

12

r .....(i)

so

n

1r ra1

= i

n

1r ra

1 = � i

n

1r2

ra

1 = � i

so

n

1r r2r

r

1a�a

1�a2 = 0 from(i)

24. Number of chords represented.......................Sol. All chord will be diameter of circle so infinite chords are

possible.Centre of circle is (�2, � 3) and radius is 3.

25. In ABC, angle A is 120º, BC + CA ..............

Sol. Given a + b = 20 and c + a = 21now , a2 = b2 + c2 � 2bc cos 120º

a2 = (20 � a)2 + (21 � a)2 + 2

)a�21)(a�20(2

a2 = (21 � a)2 + (20 � a)[(20 � a) + (21 � a)]a2 = (21 � a)2 + (20 � a)(41 � 2a)a2 = 441 + a2 � 42a + 820 � 81a + 2a2

2a2 � 123a + 1261 = 0(2a � 97)(a � 13) = 0

a = 13 or a = 2

97= 48

2

1(not possible)

a = 13

26. The number of real points at which................

Sol. Equation of asymptotes is y � 2 = ± 32

(x � 1)

3y � 2x = 4 and 3y + 2x = 8as given line is parallel to one of the asymptotesso it will cut at one point only

27. Let z = x + iy then locus of moving................

Sol. Given z

z1 is real

z

z1 =

zz1

z + 2z = z + z2

)z�z( + )z�z( )zz( = 0

)z�z( )1zz( = 0

zz (z 0) or z + 1z = 0

y = 0 or x = �21

but excluding origin.

28. If x 2,0 , log0.1 sin 2x + log10 ...............

Sol. log0.1 sin2x + log0.1 secx = log0.1 3

sin2x. secx = 3

sinx = 23

x = 3

, 32

, but cosx > 0

x = 3

29. The number of integers between ....................Sol. The number are of the form p9, pq4, where p, q are primes

3.24 = 48, 5.24 = 80, 7.24 = 112, 11.24 = 176, 2.34 = 162Hence number of integers is 5.

30. If |z � 3| = min {|z � 1|, |z � 5|}........................

Sol.

31. Circle x2 + y2 + 2gx + 2fy + c = 0 ..................Sol. centre (� g, � f)

radius = c�fg 22

distance of centre from origin radius of circleg2 + f2 g2 + f2 � c c 0circle should be cut distinct points on both axis.x2 + y2 + 2gx + 2fy + c = 0put y = 0 x2 + 2gx + c = 0D > 0 4g2 � 4c > 0 g2 > csimilarly put x = 0 f 2 > c

32. If p,q,r are positive rational.............................Sol. f(x) = (p + q � 2r) x2 + (q + r � 2p) x + (r + p � 2q)

Given p > 0, q > 0, r > 0 and f(x) has a root in (�1, 0)also f(1) = 0 one root of f(x) is 1.

f(0) < 0 r + p � 2q < 0

qpr

< 2 AA

Also product of roots = r2�qpq2�pr

= .1

since p,q,r are rational is rational both roots are rational (B)Also r + p < 2q(r + p)2 < 4q2

4q2 > (p + r)2

4q2 � 4pr > (p � r)2 roots of px2 + 2qx + r = 0 are real and distinctDiscriminant of px2 + 2qx + r = 0 a positive quantity

33. Let P(x) = x3 � 8x2 + cx � d ..........................Sol. We have x1 + x2 + x3 = 8

x1 x2 x3 = dx1 x2 + x2 x3 + x3 x1 = c


Possible roots are 1,2,5 or 1,3,4 d = 10 or d = 12 c = 2 + 10 + 5 = 17 or 3 + 12 + 4 = 19

34. For any odd integer n 1, if the.....................Sol. S = 13 � 23 + 33 � 43 +......�(n � 1)3 + n3

= (13 + 23 + 33 +....+ n3 ) � 2(23 + 43+......+ (n � 1)3)

=

2

2)1n(n

� 24

333

21�n

...21

= 4

)1n(n 22 � 16

42

1n2

1�n22

= 4

)1n(n 22

� 4

)1�n( 22

= 4

)1n( 2[n2 � (n � 1)2]

= 4

)1�n2()1n( 2

for n = 7 S = 413.64

= 208

for n = 5 S = 4

9.36= 81

for n = 9 S = 4

17.100= 425

for n = 11 S = 36.21 = 756

35. In ABC, if r = 1, R = 3, s = 5, then ..............

Sol. r = s

s = 5, a + b + c = 10, = r.s = 1.5 = 5

= R4

abc abc = 60

2 = s(s � a) (s � b) (s � c)5 = (5 � a) (5 � b) (5 � c)5 = 125 � 25(a + b + c) + 5 (ab + bc + ca) � abcab + bc + ca = 38a2 + b2 + c2 = (a + b + c)2 � 2(38) = 24

Also r1 + r2 + r3 � r = 4R

r1 + r2 + r3 = r + 4R = 13

36. (A) P, Q, R are three points.....................Sol. (A) Area of triangle formed with tangents at P,Q,R

= 2

1(area of PQR) =

21

.6 = 3

(B) 201122 = 20092.42 = 20092)16(

= 16 × 16 × 16 × .......... end with 6

(C) we have x4 � 2x3 � 3x2 + 2x + 2 = 0(x4 � 3x2 + 2) � 2x (x2 � 1) = 0(x2 � 2)(x2 � 1) � 2x (x2 � 1) = 0(x2 � 1)(x2 � 2x � 2) = 0

x = 1 or � 1 x = 2

842 = 1 ± 3

sum of integral solution is = 1 + (�1) = 0

(D) Let H.P. a1

, da

1

,

d2a1

, d3a

1

,........

then a1

= 5

2 and

da1

= 1312

a = 25

, d = 12

7�

now nth term of H.P. = d)1�n(a1

=

n7�3712

so the nth term is largest positive value when (37 � 7n) hasleast positive value , which occurs for n = 5

Hence t5 = 57�37

12

= 6

37. (A) If |z| = |z � 2| = |z � 1 � i 3 | ................

Sol. (A) |z| = |z � 2| = |z � 1 � i 3 |

x = 3

120 = 1 , y =

3300

= 3

1

z = 1 + i. 3

1 |z � 1| =

3

1

(B) Let aa�1 3

= bb�1 3

= cc�1 3

= k xx�1 3

= k

a + b + c = 0 abc = 1Hence a3 + b3 + c3 = 3abc = 3

(C) A2 =

2�4�

12

2�4�

12 =

00

00

I + 2A + 3A2 +.... = I + 2A

=

10

01+

4�8�

24 =

3�8�

25

det. ( I + 2A + 3A2 +.... ) = 3�8�

25 = 1

(D) Maximum number of intersection points

38. Let p be the product of the ............................Sol. x4 � 4x3 + 6x2 � 4x + 1 = 2009

(x � 1)4 = 2009

(x � 1)2 = ± 2009

x2 � 2x + 1 + 417 = 0as we want only the product of non-real roots

product of root = 1 + 417 = 1 + 7(6.42) = 45.95

9p

= 5

39. If a1, a2, a3, a4, 5, 4, a6, a7, a8, a9 ................

Sol. D =

987

6

321

aaa

a45

aaa

Since an = n

20, d =

201

D =

920

820

720

620

520

420

320

220

20

= 74)20( 3

97

87

1

32

54

1

31

21

1

Now R1 R1 � R2 and R2 R2 � R3

= 74)20( 3

97

87

1

91�

403�

0

31�

103�

0

= 2150

= qp

p = 50, q = 21

29

q�p = 1

40. Let S be the sum of all possible.....................Sol. For idempotent matrix A2 = A

A�1A2 = A�1 AA = I (Since A is non-singular) non sinugular idempotent matrix is always a unit matrix.

2 � 3 = 1 = ± 2

m2 � 8 = 1 m = ± 3n2 � 15 = 1 n = ± 4Now p = q = r = 0Now S = (�4) + (�3) + (�2) + 0 + 2 + 3 + 4 = 0


41. Suppose a parabola y = x2 � ax � 1...............Sol. y = x2 � ax � 1

x = 2

4aa 2

= 2

4aa 2 , =

24a�a 2

Equation of family of circles through A and B is(x � )(x � ) + y2 + y = 0As it passes through C(0, �1)

+ 1 � = 0 (But = �1) = 0Equation of circle through A, B and C is(x � )(x � ) + y2 = 0It cuts the y-axis when x = 0, so y2= 0(put = �1)

y2 = 1 y = 1 or �1

Hence t = 1 Ans.

42. Number of words between DANGER ..............Sol. Rank of the word DANGER is 135

Rank of the word GARDEN is 379Number of words between then given words= 379 � 135 � 1 = 243 = 35 m + n = 8

PART-III (Physics)43. Initially spring is in natural length a...................................Sol. Just after collision velocity of (block B + particle) = v/2

At the time of maximum extension both A and B block willhave same velocity (say v')Conserving angular momentum about centre of pulley ofsystem of (A + B + particle)

v' = 4v

By energy conservation,

21

× 2m ×

2

2v

= 2 ×

21

× 2m ×

2

4v

+

21

× 10 × AA2 � 2mgA

On solving,A = 5m.

44. A particle is projected from a long horizontal ...................................Sol. From the given condition,

gcossinu2 2

(1 + e + e2 + .........) = 10

(1 + e + e2 + .........) = 3

2

e11

= 3

2 e =

23

1 .

45. A uniform disc of mass m is attached to a ...................................

Sol. k .23

R.2x3

= 23

mR2 Ra

m2kx3

= a

So, T = k3m2

2 .

46. There are two identical small holes of area ...................................

Sol.

Thrust force F = F1 � F2 = av12 � av2

2

= a(2gh1) � a(2gh2) = 2ag(h1 � h2) = 2agh.

47. Consider a U�shaped frame with a sliding ...................................Sol. For wire to be in equilibrium,

mg = 2S m = gS2

For mass less than gS2

it will move upward

So amongst the options g4S

is the minimum mass.

48. If is density of the material of a uniform ...................................

Sol. = A

x..dx.m

0

2

mA2

= 2

2 =

2

2

=

21

.

49. The specific heat of many solids at low ...................................

Sol. Q = T

0

SdT = T

0

dT.T 3

= 23

T2 = 23

× 900 = 3450 J.

50. Steel and brass strips are welded together ...................................Sol. Let length of the strip is 0 ,

Here, L0(1 + bT) =

2d

R

and L0(1 + ST) =

2d

R

Dividing, T1

T1

S

b

=

2/dR2/dR

R = T)(

d

sb

b � s = TR

d

b � s = 51

101 2

= 2 × 10�3 /ºC.

52. During an experiment, an ideal gas is found ...................................

Sol. Equation of process

2P = constant = C .... (1)

Equation of State TMRP

.... (2)

From 1 and 2 PT = constant C is false, D is true.

As -changes to 2

P changes to 2

Pfrom equation (1) A is false.

Hence T changes to T2 . B is true.

53. Initially distance between two small blocks ...................................Sol. x1 + x2 = 1

3x1 = 2x2 ........(i)

x1 = 52

m,x2 = 53

m

By momentum conservation and energy conservation,

21

× 10 × (1)2 = 21

× 3 × v12 +

21

× 2 × v22 ........(ii)

v1 = 3

2m/sec. , v2 = 3 m/sec.


By elastic collision,

Impulse = 2mBv1 = 34 N.sec.Now, on maximum extension both block move with same velocity,Let at that time velocity is v.By energy conservation,

21

× 2 × 3 + 21

× 3 × 34

= 21

× 10 × x2max +

21

× 5 × v2

By momentum conservation,

3232 = 5v

v = 5

34m/sec. x =

51

m = 20 cm.

55. In the figure shown cross�section ...................................Sol. For case�1 :

gh = 21v1

2 v1 = gh2 t1 = g

h10

For case�2 :

g2h + 2gh = 21

zv22

v2 = gh4 t2 = gh6

For case�3 :

g2h + 2g2h + 3gh = 21

3v32

v3 = gh6 t3 = gh2

x1 = v1t1 = 20 x2 = 24 x3 = 12

x1 : x2 : x3 = 12:24:20

= 6:12:10 = 3:6:5 .

56. An athlete of mass 70 kg participates ...................................Sol. For minimum time,

amax = g

1000 = 21gt2

t = 120 second v = gt = 6

100

So, work done by man = 21

× 70 ×

2

6

100

= 9.7 kJ.

57. (A) m1 and m2 are the masses ...................................Sol. (D) Figure shows the forces acting on the two blocks . As we

are looking for the maximum value of M / m, the equilibrium islimiting . Hence , the frictional forces are equal to times thecorresponding normal forces.

Equilibrium of the block m givesT =

N1 and N1 = mg

which givesT = mg ......(i)

Next, consider the equilibrium of the block M. Taking componentsparallel to the incline

T + N2 = Mg sin Taking components normal to the incline

= Mg cos These give

T = Mg(sin � cos ) .......(ii)

From (i) and (ii) , mg = Mg(sin � cos )

or, M/m =

cossin

If tan < , (sin � cos ) < 0 and the system will not slidefor any value of M/m.

59. Assuming the xylem tissues ...................................Sol. gh r2 = 2rS cos

r = ghcosS2

=

101010

5.0123

= 10�6 m.

60. S1 and S2 are two stationary sources ...................................Sol. For maximum intensity,

p = n

=

v =

1440360

= 41

m

Path differences corresponding to maxima are 0,0,25,0.50,0.75.............So, in one complete revolution detector detects maximumintensity

= 8 × 4 = 32 times.

61. In the situation shown coefficient of ...................................Sol. Let everything moves together

Then,

a = 1212

= 1 m/s2

But fAB maximum = 15NSo, sliding occurs.Now, see if B and C move together.

a = 9

815 =

97

m/s2

So, friction acting between B and C is 97

× 5 m/s2 .

62. P is a point moving with constant speed ...................................Sol. Velocity of approach of P and O is

� dt

dx = v cos 60° = 5 m/s

It can be seen that velocity of approach is always constant.

P reaches O after = 5

100 = 20 sec.

63. A sound wave of wavelength 20 cm ...................................Sol. Sound level in dB is

B = 10 log10

0

If B1 and B2 are the sound levels and I1 and I2 are the intenaities inthe two cases

B1 � B2 = 10 log10

1

2

1

2

= 100

So1

2

01

02

S

S

= 10

S02 = 10 So1

and S01 =

21020

105.1

1023

BK

P 2�

5

3�0

= 10Å

So So2 = 100Å


CODE - 0 PAPER-1ANSWER KEY

Q .N o . 1 2 3 4 5 6 7 8 9 1 0

A n s . A D D B E A C E B C

Q . N o . 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 2 0

A n s . B C A C B D F F T F

Q . N o . 2 1 2 2 2 3 2 4 2 5 2 6 2 7 2 8 2 9 3 0

A n s . 8 8 2 3 5 C A E A C

Q . N o . 3 1 3 2 3 3 3 4 3 5 3 6 3 7 3 8 3 9 4 0

A n s . B A D D C A D A A C

Q . N o . 4 1 4 2 4 3 4 4 4 5 4 6 4 7 4 8 4 9 5 0

A n s . D T T F T 6 5 6 0 8

Q . N o . 5 1 5 2 5 3 5 4 5 5 5 6 5 7 5 8 5 9 6 0

A n s . B B A E C B D B A B

Q . N o . 6 1 6 2 6 3 6 4 6 5 6 6 6 7 6 8 6 9 7 0

A n s . D A A B C A T T F F

Q . N o . 7 1 7 2 7 3 7 4 7 5

A n s . 6 2 4 8 3

PAPER-2CODE - 0

Q.No. 1 2 3 4 5 6 7 8 9 10

Ans. B A D B B A C D BC BCD

Q. No. 11 12 13 14 15 16 17 18 19 20

Ans. AC ACD AB ABCD(A - p,q,r) ; (B - p,q,r,s,t) ;

(C - p,r,s,t) ; (D - p,q,r,s,t).(A - q) ; (B - r) ; (C - s) ; (D - p)

5 9 5 4

Q. No. 21 22 23 24 25 26 27 28 29 30

Ans. 5 A A D A B C B A AB

Q. No. 31 32 33 34 35 36 37 38 39 40

Ans. ABD ABC CD ABC ABCD(A) - q ; (B) - p ;(C) - t ; (D) - p

(A) - q ; (B) - q ;(C) - q ; (D) - t

5 1 0

Q. No. 41 42 43 44 45 46 47 48 49 50

Ans. 1 8 B A B C D A C B

Q. No. 51 52 53 54 55 56 57 58 59 60

Ans. AB BD ACD ABC AD AC(A) � p, r ; (B) � q ;

(C) � s ; (D) � p,r,t

(A) � p,q,r,s,t ; (B) � q,r,s,t

; (C) � r,s,t ; (D) � p,q,r,t 1 4

Q. No. 61 62 63

Ans. 5 4 5

hints & solution - wordpress.com · 9.total number of geometrical isomers in the given ........

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