edt -06 (jee) solutions

SAFE HANDS & IIT-ian's PACE EDT-06 (JEE) Solutions

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1. Answer : Option (1)

2. E =

kx2 =

kx x =

Fx

Answer : Option (2)

3.

F = 2T = 2 (Stress)(Area) = 2 Y (Strain) S = 2Y α ΔT S

Answer : Option (3)

4. Consider any point on the graph e.g. when W = 20N,

Δl = 1x10-4m.

Y =

=

Answer : Option (4)

5. Higher is the Young’s modulus higher us the

elasticity. Elasticity is not about getting stretched

under a force; its about regaining the shape once

the force is removed.

Answer : Option (1)

6.

Bubble will be detached from the surface when

force due to surface tension is equal to the force of

buoyancy on the bubble.

F cosθ =

π R3 ρW g 2π r T cosθ =

π R3 ρW g

Also, from figure, cosθ =

Answer : Option (2)

7. 2T L = mg

Note that we will have to consider two surfaces and

hence force of surface tension is 2T L and not TL.

Answer : Option (3)

8. Answer : Option (4)

9. Smaller is the radius, larger is the pressure.

Answer : Option (2)

10. Stress = B(Strain) ρ g h = B

h =

Also,

= 0.1% = 0.1/100 = 10-3

Answer : Option (3)

11. k =

In c as volume is doubled and length is unchanged,

area must be doubled.

In d as volume is made eight times and length is

doubled, area must be made four times.

Answer : Option (2)

12. Answer : Option (3)

13. Answer : Option (1)

14. If tensile strength point and fracture point are far

apart, material is more ductile.

Answer : Option (2)

15. For 1st case: Δℓ = ℓ1 - ℓ =

For 2nd case: Δℓ = ℓ2 - ℓ =

Solving these two equations, we get, ℓ =

Answer : Option (3)

16. Energy stored = ½ x Stress x Strain x Volume

=

x

x

x Aℓ =

x

x

Answer : Option (1)

17. Stress in both the rods will be same and will be

equal to

.

Δℓ1 =

Δℓ2 =

Δℓ = Δℓ1 + Δℓ2 =

Answer : Option (4)

T T

F

F F θ

θ

SAFE HANDS & IIT-ian's PACE EDT-06 (JEE) Solutions

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18.

Assume that we place mass m at distance d from

point A.

Equating torque about that point will give,

T1 d = T2 (L – d) ………………… (1)

Also, as strain in both the wires are same and their

Young’s moduli are Y and 2Y, Stress in them must be

in ratio 1 : 2.

As area of cross-section for both are same,

2T1 = T2 ………………… (2)

From (1) and (2),

d = 2L/3

Answer : Option (3)

19. Had there been no walls and hence no compressive

force, length of the rod would have been higher. So

there is strain as well as stress in the rod.

Answer : Option (2)

20. Answer : Option (1)

21. Strain = θ =

x

= 1/180C

F = ηA x strain = 25 kN

Answer : Option (3)

22. ρ =

New density = ρ’ =

=

=

Answer : Option (1)

23. Force required to punch a hole = (Ultimate shear

stress)x(Area of cylindrical surface to be punched

out) = ηmax x (2πr t)

where t – thickness of the sheet.

Answer : Option (3)

24. Consider a small element of

length dy at distance y from

the bottom of the rod.

This element support the

weight of part of rod below

it i.e. weight of length y.

Mass of length y = y

Weight of length y = gy

Extension in this elememt = d(Δℓ) = Fℓ/AY =

y dy

Total entension = ∫

= ∫

=

Answer : Option (1)

25. Answer : Option (2)

26. Volume of each new drop is one-eighth of original

drop. Hence, radius of new drop will be R/2.

Also, we will have eight such drops. So total surface

area after splitting will be = 8 x 4π (R/2)2 = 8π R2

Work done = Change in surface energy = (Surface

Tension) x (Change in surface Area)

= T *8πR2 – 4π R2+ = 4π TR2

Note – Drop has only one surface.

Answer : Option (4)

27. As angle of contact is zero, force of surface tension

will be tangential to the needle surface.

In addition, for needle, we will have two needle-

water interface.

2Tℓ = mg m =

Answer : Option (1)

28. If radius is doubled, the length of contact and hence

the force due to surface tension will also be

doubled. So weight of the liquid in the capillary will

also double and so will the mass.

Answer : Option (1)

29. Answer : Option (1)

30. Answer : Option (2)

A B

Y 2Y T1 T2

y

dy

SAFE HANDS & IIT-ian's PACE EDT-06 (JEE) Solutions

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CHEMISTRY SOLUTION 31. (4)

AB + CD ⇌ AD + CD mole at 0t 1 1 0 0

Mole at equ.

4

31

4

31 ⇌

4

3

4

3

0.25 0.25 ⇌ 0.75 0.75

90625.0

5625.0

25.025.0

75.075.0

cK

32. (1)

33. (1)H2O = H+ OH-, is endothermic, as tempreture increases, Kw increases, which also changes the pH scale. 34.

Solution: (1) At start, 2

5PCl ⇌ 0

20

3 ClPCl

At equilibrium. 100

602

100

402

100

402

Volume of cantainer = 2 litre

2100

6022100

402

2100

402

cK = 0.266

35. (3)

36. (1)

Solution: (1) Dissociation constant of 910 HA

HA ⇌ AH

CKH a ][19 1010

510

510][ H

5pH

14 pOHpH

pHpOH 14 514 ; 9pOH

37. (3)

38. (3)

04.0101.0

10

2.0

10

2.0

][

][][

5

23

PCl

ClPClKc

39. (1)

COOHCH 3 ⇌ HCOOCH 3

Initial 1 0 0

At equilibrium )1( C C C

)1(01.0 01.0 01.0

)1(01.0 01.0

)1.001.0( [In presence of HClM1.0 ]

][

]][[

3

3

COOHCH

HCOOCHK

;

)1(01.0

)1.001.0(01.0106.1 5

On solving, %016.0 40. (3)

OH 2 ⇌ ][ H ][ OH ; HCl ⇌ ][ H

][ Cl

Total 87 1010][][][

2

HClOH HHH

]101[10 17

10

1110][ 7 H

10

1110log]log[ 7HpH ;

98.6pH

41.(2)

34

2243 23)( POCaPOCa

MPOCa )103.3(2

3][

2

3][ 73

42

M71095.4

234

32 ].[][ POCaK sp

2737 )103.3()1095.4( 321032.1

42. (4) pH of a solution of a salt of weak base

with strong acid

)log(2

1CpKpK bw

)1.0log108.1log14(2

1 5 1.5

43. (4)

44. (2) 510 aK ; 6pH

pH = ][

][loglog

acid

saltKa or 6 =

][

][log10log 5

acid

salt

or 6 = ][

][log10log5

acid

salt or 6 =

][

][log5

acid

salt

or 156][

][log

acid

salt or

1

10

][

][

acid

salt

45. (2)

Solubility of 342 )(SOAl

342 )(SOAl ⇌ 4

32 SOAl

324

23 ][][ SOAlK sp

46. (2)of common ion effect

47. (2)The solubility of 4BaSO in g/litre is given 31033.2

SAFE HANDS & IIT-ian's PACE EDT-06 (JEE) Solutions

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in mole/litre.wtm

Wn

.

5101 233

1033.2 3

Because 4BaSO is a compound

252 ]101[ SKsp 10101

48. (2) )(24 sCOONHNH ⇌

PP

gCOgNH

3

12

3

23 )()(2

185.1)2(27

4

27

4

3

1

3

2 332

2

23

PPPPPK CONHp

49. (4) )6.01(32

SO ⇌ )3.0(

2)6.0(22 OSO

675.04.04.0

3.06.06.0

][

][][

3

22

2

SO

OSOKc .

50. (4)

xaa4

HSNH ⇌

xx5.0atm5.0

)(2)(3

gg SHNH

Total pressure 84.025.0 x

i.e., 17.0x

SHNHp PPK23

. )17.0().67.0( =0.1139

51. (3) 52. (1)

G° = – 2.303 RT log K

= – 2.303 2 300 cals 10 log10303.2

13

= – 600 Kcals

53. (4) Solution: 2HI(g) H2(g) + I2(g)

1– 2

2

Where is the degree of dissociations

50

11

synthesis

dissK

K

21

22

50

1

25

1

212

25 = 2–2

2+5 2 = 2

= 252

2

= 0.22

(4) 54. (3)Since CH3SH is acid, CH3S

– will be its conjugate base. 55. (3)Relative strength of bases can be shown by their conjugated acids. Conjugate acid of

OH is OH2 which is a weak acid conjugate acid

of COOCH 3 is COOHCH3 which is stronger than

OH2 . While conjugate acid of Cl is HCl which

is strongest out of there. so the order of relative

strength of bases is ClCOOCHOH 3 .

56. (1) S8(g) 4S2(g) Initial partial 1 atm 0 pressure Equ. par. (1- 0.29) 4 × 0.29 pressure = 0.71 atm = 1.16 atm

Now, KP = 8

2

S

4S

P

P =

71.0

)16.1( 4

= 2.55 atm3

57. (2) Le Chatlier’sprinciple 58. (1) PCl3 + Cl2 PCl5

0.20 0.10 0.40 moles/litre If 0.20 mole of Cl2 is added then at

equilibrium 0.20 – x 0.30 – x 0.40 + x

20 = )x–30.0)(x–20.0(

x40.0

or x = 0.08 [PCl5] = 0.4 + 0.08 = 0.48 moles

59.

(3) pH = 12 so [OH–] = 10–2 M

Now Ba(OH)2(s) Ba2+ + 2OH– ,

5 × 10–3 M10–2 M

so Ksp = [Ba2+] [OH–]2

= (5 × 10–3) (10–2)2

= 5 × 10–7

60. (1) For the reaction,

CaCO3(g) ⇌ CaO(s) + CO2(g)

Kp = 2COP and KC = [CO2]

( [CaCO3] = 1 and [CaO] = 1 for solids] According to Arrhenius equation we have

RTH rAeK/

Taking logarithm, we have

)303.2(

loglogRT

HAK

or

p

This is an equation of straight line. When log Kp is plotted against 1 / T. we get a straight line.

The intercept of this line = log A, slope = –

∆H°r / 2.303 R Knowing the value of slope from the plot

and universal gas constant R, ∆Hr° can be calculated.

(Equation of straight line : Y = mx + C. Here,

ATR

HK

or

p log1

303.2log

Y m x C

2log COp

1 / T

(2

loglog COp pK )

SAFE HANDS & IIT-ian's PACE EDT-06 (JEE) Solutions

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1. (c) Accordingly, 22

2

11

3

2

baC

baC

T

Tnn

nn

......(i)

3

33

212

3

4

3

baC

baC

T

Tnn

nn

......(ii)

(i) = (ii) )1)(2)(3(2

)2)(3(6

)1(

2

nnn

nn

nn

n

5)1(3)1(2 nnn .

2. (c) .....15.10.5

5.3.1

10.5

3.1

5

11 S

..!3

)2)(1(

!2

)1(

!11)1( 32

x

nnnx

nnnxx n

5

1nx and

10.5

3.1

!2

)1( 2

xnn

2

1n and

5

2x

3

5

5

3

5

21

2/12/1

S .

3. (a)

n

rr

r

r

r

rrnr mC

032

termsupto....2

7

2

3

2

1)1(

r

r

rn

n

r

r

r

n

r

rnr CC

200 2

3.)1(

2

1.)1(

+ .....2

7)1(

30

r

rn

r

rnr C

....8

71

4

31

2

11

nnn

up to m terms.

....8

1

4

1

2

1

nnnupto m terms

)12(2

12

2

11

2

11

2

1

nmn

mn

n

nmn

4. (c) Middle term in expansion of 22

44 )()1( xCx

Middle term in expansion of 33

66 )()1( xCx

According to question, 33

622

4 CC

10/3 .

5. (c) 11649 nn = 116)481( nn

116)48(....)48()48(1 221 nCCC n

nnnn

= nn

nnn CCCnn )48(...)48()48()1648( 33

22

= ..8.6.8.6.6.[864 244

33

22

2 CCCn nnn]8.6. 2 nn

nnC

Hence, 11649 nn is divisible by 64.

6. (c) In the expansion of (1+x)2n, the general term

nkCkn 20,2

As given for 22

32,2,1 r

nr

n CCnr

SAFE HANDS & IIT-ian's PACE EDT-06 (JEE) Solutions

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Either 23 rr

or )2(23 rnr , )( rnn

rn CC

1r or 1212 rnrn , )1( r .

7. (b) 2/1)21( x can be expanded if 1|2| x i.e. if 2

1|| x , i.e.

2

1,

2

1if ..

2

1

2

1if xeix

8. (c) We have

7

)13(log)5/1(

79log

12

12

2

12

x

x

7

5/11

1

)13(

179

x

x

5

5/11

571

57

6)13(

179

x

xCT

)13(

1)79(

1

15

7

x

xC

Now 84)13(

)79(84

1

1

57

6

x

x

CT

)13(479 11 xx

02712027)3(123 22 yyxx

(Where xy 3 )

2,19,339,3 xy x

9(a) The numerator is of the form

333 )()(3 babaabba

33 25)718( rN

242

6151

66 2.32.33 CCDr 66

655

6424

6333

6 22.32.32.3 CCCC

This is clearly the expansion of 366 )25(5)23(

1)25(

)25(3

3

r

r

D

N

10(b) In the expansion of 5510/15/1 )( xy , the general term is

10/5/115510/1555/1551 .)()( rr

rrr

rr xyCxyCT .

This 1rT will be independent of radicals if the exponents 5/r and 10/r are integers, for 550 r which is possible

only when 50,40,30,20,10,0r .

There are six terms viz. 51413121111 ,,,,, TTTTTT which are independent of radicals.

11(b) We have 452 )1()1( xx

= ...)( 42

521

50

5 xCxCC )( 44

433

422

41

40

4 xCxCxCxCC

So coefficient of 5x in ])1()1[( 452 xx

= .60.. 15

34

14

25 CCCC

12(b) We know that

2

)1(.....32

12

3

1

2

0

1

nn

C

Cn

C

C

C

C

C

C

n

n

SAFE HANDS & IIT-ian's PACE EDT-06 (JEE) Solutions

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Putting n=15, then 1202

)115(15

.

13(c) Putting x=1 in (1+x– 3x2 )2163.

we get sum of the coefficients as 1)1()311( 21632163 .

14(b) By hypothesis, 12240962 12 nn

Since n is even, hence greatest coefficient

9246.5.4.3.2.1

7.8.9.10.11.126

122/ CCn

n .

15(a) In the expansion of

11

2 1

bx

ax , the general term is r

r

rr

r

rrr x

baC

bxaxCT 322111111211

1

11)(

For x7, we must have 22 – 3r = 7 r = 5, and the coefficient of x7 =5

6

511

5

5115

11 1.

b

aC

baC

Similarly, in the expansion of ,1

11

2

bx

ax the general term is r

r

rr

rr xb

aCT 311

1111

1 .)1(

For x–7 we must have, 11 – 3r = –7 r = 6, and the coefficient of 7x is 6

5

511

6

5

611

b

aC

b

aC .

As given, 16

5

511

5

6

511 ab

b

aC

b

aC

16(a) Words starting from A are !5 = 120

Words starting from I are 5 ! = 120

Words starting from KA are 4 ! = 24

Words starting from KI are 4 ! = 24

Words starting from KN are 4 ! = 24

Words starting from KRA are 3 ! = 6

Words starting from KRIA are 2 ! = 2

Words starting from KRIN are 2 ! = 2

Words starting from KRISA are 1 ! = 1

Words starting from KRISNA are 1 ! = 1

Hence rank of the word KRISNA is 324.

17(c) 341

31 CCC nnn 34 CC nn

714

31

3

4

nn

C

Cn

n

.

18(a)

19(c)

20(b)

21(a)

22(c)

23(a)

24(a)

25(b)

26(c)

27(d)

28(a)

29(a)

30(a)