sol_major test-6.pdf

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MAJOR # 6

HINTS & SOLUTIONS

CHEMISTRY

Qus. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Ans. 3 1 4 2 1 1 1 1 3 4 3 1 4 2 1 2 2 1 4 4

Qus. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

Ans. 1 4 1 4 3 1 4 3 4 3 3 1 2 2 4 3 3 2 3 3

MATHEMATICS

Qus. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60Ans 4 4 3 2 2 2 3 2 3 2 1 3 4 1 1 2 4 2 4 2

Qus. 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

Ans 3 1 2 1 2 1 2 4 3 4 2 2 3 2 1 3 3 4 1 2

PHYSICS

Ques. 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

Ans. 4 1 1 3 1 3 4 1 1 1 3 3 1 3 1 2 1 4 4 1

Ques. 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120

Ans. 1 2 1 3 3 3 3 2 3 4 2 4 3 3 2 3 1 2 2 2

CHEMISTRY

1. Ans. Ni 2+

, Ti 3+

Hint : Unpaired e– results in colour

2. Ans. + 154.4 kJ

Hint : ∆G = – 2 × 96500 × (– 0.80) J= 154.4 KJ

3. Ans. Reaction of sodium hydride with water

Hint : Reaction of sodium hydride with water is

violent.

4. Ans. C 2 H 5CO 2CH 3

Hint : Ester

5. Ans. 2

K 2 / 3α αα α

====

Hint : K =]CO[

]O][CO[

2

2 / 12

6. Ans.

OHOH

Hint : It is pinacol-pinacolone reaction

7. Ans. ––I

Hint : Good L.G favers rate of reaction.

8. Ans. Homo polymer

Hint : Cellulose is polymer of β-D glucose& starch is polymer of α-D glucose

9. Ans. Distillation

Hint : BP of Acetone → 56°CBP of Methanol → 65°C

10. Ans. Troposphere

Hint : Fact based

11. Ans. van der waals forces

Hint : Layers of carbon atoms in graphite are held

together by van der Waals forces

12. Ans. Zero

Hint : Q Rate is constant w.r.t. time

∴ order of reaction is 013. Ans. Ba

Hint : Ba imparts green colour

14. Ans. 2 atm

Hint : π1 = C1RT1, π2 = C2RT2

or22

11

22

11

2

1

TC

TC

RTC

RTC==

π

π


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π2 = π1 ×11

22

TC

TC

= 3 atm ×

K300C

K400)2 / C(

1

1

×

× = 2 atm

15. Ans. Toluene

Hint :

CH3

CH3–CH2–CH2–CH2–CH3

CH3–C––CH3

CH3

CH3

CH2

Most stable radical BDE is lowest

, , ,

16. Ans. III > I > IV > II

Hint : Order of basic strength ⇒ III > I > IV > II

17. Ans. Nitrolic acid

Hint : R––C H2––NO

O

O=N–OH

– H2O

R–C–NO2

N–OHNitrolic acid

18. Ans. + 1

Hint : [Fe(H2O)5NO] SO4

x + 5(0) + 1 – 2 = 0

Fe = +1

19. Ans. 8

56 3 ×

Hint : Ew =

8

563

3 / 8

56 ×=

20. Ans. 5

Hint :

CH2–OH O–CH3 CH3

,

,

OH

ortho

CH3

OH

meta

CH3

,

OHpera

21. Ans. 1 radial node and 1 angular node

Hint : Radial node = n – l – 1

Angular node = l

22. Ans. O––C 2 H 5

Hint : Ether

23. Ans. a, b and c

Hint : Flocculation value depends on charge.

24. Ans. To attain the five valent state, d orbital

must be used and hydrogen is not

sufficiently electronegative to make d

orbitals contract sufficiently.

25. Ans. NO 2 , brown gas

26. Ans. C N

O Me

Me

Hint :

O

C OHPCl5

O

C Cl

Me2NH

C NMe

Me

O. .

27. Ans. Histidine and arginine

Hint : Fact based

28. Ans. Analgesics

Hint : These are Narcotic Analgesics

29. Ans. 0.045 M

Hint : CH3COOH CH3COO– + H

+

C

C (1 – α) Cα Cα

K =α

α=

α

α⋅α

–1

C

)–1(C

CC 2 = C α2

or C = K/ α2 =2

5–

)02.0(

108.1 × = 0.045 M

30. Ans. 1/4m

Hint : PV = nRT =M

m RT

2

1P × V =

M

m′ R ×

3

2T

∴ M

m

3

4

M

m ′= ⇒ m′ =

4

3m

∴ mass escaped = m –4

3m =

4

1m


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31. Ans. H 2O > H 2S > H 2Se > H 2Te(Order of volatility)

Hint : Correct order is :H2O < H2Te < H2Se < H2S

32. Ans. –1758 kJ Hint : Sum of bond energies of reactants

= 4 × C – H + 2 × C ≡ C + 5 × O = O= 4 × 414 + 2 × 812 + 5 × 494= 1656 + 1624 + 2470 = 5750 kJ

Sum of bond energies of products= 8 × C = O + 4 × O – H= 8 × 707 + 4 × 463= 5656 + 1852 = 7508 kJ

Enthalpy of combustion,

∆H = 5750 – 7508 = – 1758 kJ

33. Ans. Ph–CH 2 –C–CH 3|

| Br

CH 3Hint :

Ph–CH=C–CH3

CH3HBr/peroxide

Br Ph–CH–C–CH3

CH3

Br

HBr

Stable radical

Ph–CH2–C–CH3

CH3

Br

34. Ans. H 5C 6

H C=C

H

CH(CH 3 ) 2Hint :

C6H5–CH2–CH–CH–CH3

OH

CH3

conc. H2SO4

C6H5C=C

H

H

CH(CH3)2

Anti elimination

takes place

35. Ans. In all the reactions, the product (C) is

aniline Hint :

MeO C–NH2

OBr2 /OH

MeO NH2

36. Ans. The overall reaction involves the

elimination of (CO) as CO 2 Hint : In R.D.S formation of acylnitrene take place

37. Ans. Zr and Hf have about the same radius Hint : Due to lanthanide contraction

38. Ans. C 2 H 6

Hint : CnH2n + 2 +2

1n3 +O2 → nCO2 + (n + 1) H2O

4

7

V

V

2

2

CO

O

= = 4

7

)n(2

1n3

=

+

12n + 4 = 14 n

2n = 4 ⇒ n = 2C2H6

39. Ans. (II) and (IV) Hint : Molecular formula same and show E and

Z-isomer.

40. Ans. cis-[Pt(en) 2Cl 2 ] +2 , [Ni(en) 2 ]

MATHEMATICS

41. Ans. All of these

Hint: Ref. (a, b) R (a, b)

Sym. (a, b) R (c, d) → (c, d) R (a, b)

Tra.

(a, b) R (c, d) & (c, d) R (e, f) → (a, b) R (e, f)

a + d = b + c

c + f = d + e

a + f = b + e

42. Ans. None

Hint: (xn–1) = (x – 1) (x – ω) …..(x – ωn–1)

⇒ ln(xn –1) = ln(x–1) + ln(x – ω) …+ ln(x–ωn–1)

dx

d

⇒ 1x

nxn

1n

−

−

=1nx

1....

x

1

1x

1−ω−

+ω−

+−

(put x = 2)

12

2.nn

1n

−

−

= 1+1n2

1....

2

1−ω−

+ω−

⇒ S = 112

2nn

1n

−−

−

43. Ans. 4

1−

Hint: Now |α – β| is same⇒ S1

2 – 4P1 = S2

2 – 4P2

b12 – 4C1 = b2

2 – 4C2

[f(x)] min =4

1

a4

−=

∆−⇒

4

1

4

C4b 12

1 =−

⇒ b12 – 4C1 = 1

[g(x)] min occurs at2

7

2

b2 =−

⇒ b2 = –7


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Now [g(x)] min =a4

∆− = –

4

1

4

c4b 22

2 −=

−

44. Ans. –7

Hint: Also b2 = –7

45 Ans. x + n +1

Hint: [ ])n2x(...)4x()2x(n

1'x n21 ++++++=

=

++

2

)1n(n.

n

2x

Mean = x + (n +1)

46. Ans. A(–α αα α , – β ββ β )

Hint: A(α, β)–1

=

αα

α−αββ

ββ

β

1000cosesine

0sinecose

e

1

= A(–α, –β)47. Ans. [4 1]

Hint: cofactor of a21, in matrix

−−

120

211

201

= a = 4

Cofactor of a33 in matrix

−−

120

211

201

= b = 1

[4 1]

48. Ans. bba

aa

ba

b

||||

||

||||

||

++

+

Hint: Since→a and

→b are equally inclined to

→c

→c must be of the form t

+

→

→

→

→

|b|

b

|a|

a

Now,→

→→

→

→

→→

→

++

+b

|b||a|

|a|a

|b||a|

|b|

=

+

+→

→

→

→

→→

→→

|b|

|b|

|a|

|a|

|b||a|

|b||a|

= t

+

→

→

→

→

|b|

b

|a|

a

49. Ans. right angled

Hint: A( â ); B( b̂ ); C( ĉ )

⇒ ĉ.db̂.dâ.d→→→

== & )ĉb̂(d +λ=→

….(1)

⇒ )ĉ.âb̂.â( +λ = )ĉ.b̂1( +λ

⇒ 1+ ĉ.b̂ = ĉ.âb̂.â +

⇒ 1– 0ĉ.âĉ.b̂b̂.â =−+

⇒ 0ĉ).âb̂(b̂.ââ.â =−+−

⇒ 0)ĉâ).(b̂â( =−− ⇒ )ĉâ()b̂â( −⊥−

i. e. the triangle is right angled

50. Ans. < 1, 1, 2 >

Hint:

+=

=+

jin

y x

ˆˆ

3

1 …….(1)

Equation of the plane through A(1, 0, 0) is → ⇒ a(x–1) + b(y–0) + c(z–0) = 0 ….(2)

Qplane (2) is also passing through the point

B(0, 1, 0), then –a + b = 0 …….(3)

∴ cos4

π =

011cba

)0(c)1(b)1(a

222 ++++

++

QAngle between planes (1) & (2) is4

π

⇒ 222 cba ++ = a + b

⇒ 2ab = c2 ⇒ 2a2 = c2

⇒ c = a2

Hence a : b : c = a : a : a2

= 1 : 1 : 2

51. Ans.

3

2,

3

4,

3

2

Hint: )k ˆ ĵî()k ˆ ĵî(r −+−λ+++=r

1

1z

1

1y

1

1x

−

−=

−=

−

− = λ ……(1)

Any point lies on the line (1) is

P(–λ + 1, λ+1, –λ+1) & origin (0,0,0)0

∴ OP = 222 )1()1()1( λ−+λ++λ−

⇒ D = (OP)2 = 2(1–λ)2+(1+λ)2

⇒ λd

dD = – 4(1–λ) +2(1+λ) = 0

λ=3

1 ∴ point P

3

2,

3

4,

3

2


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52. Ans. I 1 = I 2 = I 3

Hint: I1 = ∫π

+

2 /

0

2

2

xcos1

xcosdx

Apply prob. ⇒ I1 = I2

I1 = ∫π

+

2 /

0

2

2

xsin1

xsindx

I1 + I2 = ∫π

++

+

2 /

0

2

2

2

2

dxxsin1

xsin

xcos1

xcos

2I1 = ∫π

+

+2 /

0

22

22

xsinxcos2

xcosxsin21dx

I1 = dxxcosxsin24

xcosxsin212 /

022

22

∫

π

+

+ = I3

∴ I1 = I353. Ans. None

Hint: ∫π

π

3 /

4 /

dxxcosecxcos

∫π

π

3 /

4 /

xdxcot = 3 / 4 / |]xsin|[lnππ

= ln

2

1ln

2

3+

= ln2

3

54. Ans. π ππ π

Hint: f'(x) =)x(f

xsin2

1 2−

∫ ∫−

= dx2

xsin21dx)x('f )x(f

2

c x x f

+=

2

2sin

2

1

2

)( 2

f(x) = 'c2

x2sin+ period = π

55. Ans. even

Hint: the given function can be rewritten as,

f(x) =

≥−


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x =2

0cos

p+

α , y =2

sin

p0

α+

cos α = x2

p, sin α = y2

p

cos2α + sin2α = 1

2

2

)x2(

p +

2

2

)y2(

p = 1

22 y

1

x

1+ =

2p

4

60. Ans. ( h2 – r

2) x – 2 rhy = 0

Hint: (x – r)2 + (y – h)

2 = r

2

(0,0)

Equation of tangent in slope form

y – h = m(x – r) ± r2

m1+

is passing (0, 0)

– h = m (– r) ± r 2m1+

(mr – h)2

= r2

(1 + m2

)m

2r

2 – 2mrh + h

2 = m

2r

2 + r

2

2mrh = h2 – r

2

m =hr2

rh 22 −

y = xhr2

rh 22

−

61. Ans. ( ∑ ∑∑ ∑ xi , ∑ ∑∑ ∑ yi )

Hint:A(x1, y1)

C(x3, y1)(x2, y2)

H(x, y)

(0, 0) 1 2

GO

∑∑ ii y

3

1,x

3

1

ix

3

1∑ =

3

0x +

iy3

1∑ =

3

0y +

y = ∑yi

(∑xi, ∑yi)

62. Ans. 8

9

Hint: ∫−

−

+2

1

2

44

2dx

x x

x2=4y

2

(–2,0)

–1

(0,2)

O

=

2

1

32

432

241

−

×−

+ x x x

=

+−−

8−

3

1

2

3

4

1

36

4

1

=8

9

24

27

246

5==

7+

63. Ans. f '(x) = 0, ∀∀∀∀ x ∈∈∈∈ R

Hint: x = y = z, 3f(x) + f(x)3 = 14

⇒ f(x) = 2

⇒ f '(x) = 0

64. Ans. 20

974 ±

Hint: If sin(α + β) =4

3then cos(α + β) =

4

7

cos(α – β) =5

4 then sin (α – β) = ±

5

3

cos2α = cos(α + β + α – β)


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=

±−

5

3

4

3

5

4.

4

7

⇒ cos2α–

±

20

974

65. Ans. 2

Hint: Let xn

=π

sin3x = sin2x

sin3x – sin2x = 0

2cos 02

sin2

5=

x x

02

5cos =

x

x = (2k +1)5π

5)12(

π+=

πk

n

n=

+12

5

k ⇒ n= –1, –5, + 5, + 1

only 2 values satisfies given equation

66. Ans.1

Hint:m

mm x )) / (cos(lim

∞→

=

m

m m

x

−+

∞→1cos1lim

=

)2 / (sin2

)2 / (sin2

1

2

2

2

2

2sin21lim

m xm

m x

m m

x

−−

∞→

−

= e0 = 1

67. Ans. π /2 – tan–1(9)

Hint :

−

4

3,

3

5dx

dy= tan θ = 2

2

5 + 4 = 9

θ = tan–1 9.

∴ angle with y axis =2

π– tan

–1 (9)

68. Ans. π

Hint : f ' (x) = ex [cos – sinx] = y [let]

∴

dx

dy= – 2 e

x sin x = 0

x = 0, π in [0, 2π]

2

2

dx

yd = – 2e

x[cos x – sin x]

0x

2

2

dx

yd

=

= – 2 [Max.]

π=

x2

2

dx

yd= + 2e

π[Min.]

∴ Min. of f ' (x) = slope is at x = π

69. Ans. a straight line with slope 2

Hint : f " (x) = – f (x) & f ' (x) = g (x)

& h' (x) = (f (x))2 + (g (x))

2

⇒ h ' (x) = 2 f (x). f ' (x) + 2 g (x) g (x)= 2 f (x) . g (x) – 2 g (x). f (x)

h " (x) = 0 ⇒ h ' (x) = const ⇒ slope constant& by h (0) = 2, h (1) = 4 seems it is

a st. line with slope constant.

70. Ans. a = 4b

Hint : Min. value of a tan

2

x + b cot

2

x = ab2 Max. value of a sin

2 x + b cos

2x = a

⇒ a = ab2

⇒ a2 – 4 ab = 0a(a – 4b) = 0

a = 4b 71. Ans. 2

Hint: tan4

π =

α+α

α−α

cossin

sincos

ab

ab

⇒ (bsin α + acos α) = ± (b cos α – a sinα)⇒ b2sin2α + a2cos2α + 2absinα cosα

= b2cos2α + a2sin2α –2absinα. cos α ⇒ (b2 – a2)sin2α + (a2–b2)cos2α = – 4ab sinα cosα =(a

2–b

2)cos2α = – 2ab sin2α .......(1)

0cossin

sincos

α−α

−αα p

pba

⇒ a(–p cosα) – bp sinα –p = 0⇒ –p(acosα + bsinα + 1) = 0⇒ (acosα + bsinα)2 = 1⇒ a2cos2α + b2sin2α + 2ab sinα cosα = 1


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⇒ a2cos2α + b2sin2α+2

)sin)(cos( 2222 α−α− ab=1

⇒

2

)sin)(cosba( 2222 α+α+ = 1

⇒ a2 + b2 = 2

72. Ans. G.P.

Hint: P( 21at , 2at1) Q(22at , 2at2)

T ≡ (at1t2 a(t1+t2))also S(a, 0)`

SP = 21at + a

SQ = 22at + a

ST = )tt(a)atat( 2122

21 ++−

= a2

21

2

21 )tt()1tt( ++−

= a 2122

2121

221 tt2tttt21)tt( +++−+

=22

21

22

21 tt1tta +++

= )1t)(1t(a22

21 ++

= ))(( SQSP

73. Ans. a + 1006 d

Hint:

[a–a+a–a+a…+ a] +d[0–1+2–3+4 …. +2012]2013 times

= a –d [(–1)+(–1) + …. +(–1)] 1006 times

a + 1006d

74. Ans. 3

1

Hint: λ =ternsnnextof sum

ternsnfirstof sum

λ =]d)1n()nda(2[

2

n

]d)1n(a2[2

n

−++

−+ =

nd3)da2(nd)da2(

+− +−

λ will be constant iff 2a – d = 0d = 2a

⇒ λ =nd3

nd =

3

1

75. Ans. 4

Hint: (1–x)5.

44

x1

)x1(1

−

−

= (1–x). (1–x4)

4

(1– x)[1– 4C1.x

4+

4C2x

8–

4C3x

12….]

= ……(–1) (–4C3).x

13……

= ….. +4. x13

……….

76. Ans. Zero

Hint: f '(0) = 0

011

000

101

100

202

101

100

000

010

=++

01

0

1

)0('f

x

)x(f lim

0x===

→(use L Hospital rule)

77. Ans. 3

Hint: 0 × 5! + 0 × 4! + 3!/3! +!2

!2.1 + 0.1! + 1

= 3

78. Ans. 4

Hint: L. C. M of {2, 5, 6, 8} is 120 min so all 4

bells be ringing simultaneously in every 120min so

in 8 hr they ringing 4 times

79. Ans. 335

88

Hint: P

R

E 1

R

R R R

R G

E1 E2 E3

=

+

+

3

3

2

2

1

1

1

1

E

RP)E(P

E

RP)E(P

E

RP)E(P

E

RP).E(P

=335

88

10

3

12

5

11

3

12

4

13

4

12

313

4

12

3

=×+×+×

×

80. Ans. 92

63

Hint: p =2

1,

2

1=q

required prob. =9C4 p

4×q

5 × p

=9C4 × p

5q

5 =

92

63


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PHYSICS

81. Ans. None

Hint :

0

inqdr.E

∈

=→→

∫

E. 4πr2 =0

inq

∈

r

dr

=0

r

0

2drr4

ε

π×ρ∫

E. 4πr2 =0

1

∈ drr4cr 2

r

0

2 π∫

E. 4πr2 =0

c4

∈

π

5

r5

E =0

3

5

cr

∈

∴ At r =2

R, E =

05

c

∈

3

2

R

=

0

3

40

cR

ε

82. Ans. 2 d

4

π ππ π

φ φφ φ

Hint : φmax = E S

φ = E.4

d2π

E =2d

4

π

φ

83. Ans. increases with increase in refractive index of

a prism

Hint : δ = A (µ–1)

84. Ans. 91 Ω ΩΩ Ω

Hint : Use 100 =S1000

S1000

−

85. Ans. high ρ ρρ ρ , low α αα α

86. Ans. Small temperature difference 30° to 35°C

between hot liquid and surrounding

87. Ans. 24 Ω ΩΩ Ω

Hint :30

20 =

R

16

R =20

3016× = 24 Ω

88. Ans.

16

9 I

Hint : I1 = I0 cos2 30° =

4

I3 0

and I2 = I1 cos230° =

4

I3 0 ×4

3

=16

I9 0

89. Ans. When a current flows through a solenoid, it

has a tendency to increase its radius if no

external magnetic field exists in the space

Hint : Solonoid is wounded over a cylinder hence

magnetic field can be induced in it

90. Ans. 120°

Hint :

× ×× × ×

r × ××× ××

120°P

2

r

O

P M

rθ

O

2

r

M

cosθ =2

1 or θ = 60°

Deviation = 120°

91. Ans. the reading of the voltmeter is 50 V

Hint : Reading of volt meter = 50V

200 Ω 200 Ω

100V

92. Ans. 3

2v

Hint : v =R

GM

v' =

2

RR

GM

+ =

R3

GM2 v' =

3

2v


10/12


93. Ans. mg and T only

94. Ans. 0.5

Hint : a =

37

49

+

= 4.9 m/s2

Now Fps = µ MgM (a) = µ Mg

95. Ans. decreases

Hint : dw → +vedU → –veT↓ , PV↓

96. Ans. 10 gm

Hint : H.C = 42 J/C = 10 cal/C

weq = 10 gm

97. Ans. A →→→→Q ; B→→→→ T ; C →→→→ R,S ; D→→→→ R, S ; D→→→→ P

Hint : charle's law V ∝ T A → Q

(B)2

1 KT B → T

(C) Adiabatic Pressure dW = –dU

P ∝ Tγ / γ –1 C → R,S

(D) Work done PdV = n R d T

D → P

98. Ans. y = 0.1 sin

+ 4 4t

π ππ π

Hint : KE at mean position =2

1 mω2 a2 = 8 × 10–3

or ω =2

1

2

3

ma

1082

×× −

=2

1

2

3

)1.0(1.0

1082

×

×× − = 4

Equation of SHM is,

y = a sin (ωt + θ)= 0.1 sin

π+

4t4

99. Ans. None of these

100. Ans. 21

200

Hint :

2O

He

V

V =

325

RT7

43

RT5

×

× =

743

3255

××

×× =

21

200

101. Ans. 1 : 1

Hint : Let2

1

M

M =

1

2

M1' = M1 ×

4

2

1

M2' = M2 ×

3

2

1

∴'

2

'1

M

M =

2

1

M

M ×

2

1 =

1

1

102. Ans. 6.5 MeV

Hint : 92U236

→ 92U235

+ 0n1

236.123 amu 235.1214 amu 1.008665 amu

∆m = 236.123 – 236.130∆m = 0.007 amu∴ Energy = ∆mc2 = 0.007 × 931 = 6.5 MeV

103. Ans. A →→→→ U; B →→→→ R; C →→→→ P; D →→→→ S

104. Ans. rate of temp. falling of A is half that of B

Hint : Rate of temperature falling ∝ m

A

105. Ans. R = xy

Hint : For series connection x = nR. For parallel

connection y =n

R. Therefore xy = nR ×

n

R = R

2

106. Ans. greater of less than its emf depending on the

direction of the current through the batteryHint : Terminal voltage V = E – I r, where E is the

emf and r the internal resistance of the battery. If

the current is flowing from the +ve to the negative

terminal in the external circuit V < E. However, ifthe current is flowing in the opposite direction, a

voltmeter across the battery would show V > E.

107. Ans. Momentum of inertia and moment of forceHint : Do yourself

108. Ans. 3

2

Hint : hascending = hdescending

2

0v + ta =

2

'v0 + td


11/12


where v = velocity with which object is projected

v' = velocity with which object is returns

ta = time in ascending motion

td = time in descending motion

∴d

a

tt =

V'V =

h)2g(2

h)2g(2

+− =

128 =

32

109. Ans. 12 : 1Hint : Angular speed of hour hand,

ω1 =)6012(

2

×

π

Angular speed of minute hand ω2 =60

2π

∴1

2

ω

ω =

60

6012× =

1

12

110. Ans. 288 J

Hint : 4v + 8 × 6 = 0 or v = – 12 m/s

K.E. =2

1 × 4 × (12)

2 = 288 J

111. Ans. h/4

Hint : At maximum height, the energy is entirely

potential (mgh). This is equal to the initial kinetic

energy. Let h' be the height at which kinetic energy

is 75% of its initial value. At this height, the

potential energy must be 25% its maximum

potential energy. Thus, at this pointm g h' = (0.25) m g h

or h' = h/4

112. Ans. β ββ β > 4

3 α αα α

Hint : y = x tan θ θ

−22

2

cosu2

gx

β = α tan θ –θα× 2

2

cosg22

gx

⇒ α tan2

θ – 4α tanθ + 4β + α = 0It tanθ is real 16α2 ≥ 4 α (4β + α)

or β ≤4

3α

So it not possible

α>β

4

3

113. Ans. The balance A will read less than 2 kg and B

will read more than 5 kg

Hint : Force of upthrust will be there on mass

m shown in figure, so A weighs less than 2 kg.

Balance will show sum of load of beaker and

reaction of upthrust so it reads more than 5 kg.

114. Ans. v m t F ∆∆∆∆∆∆∆∆ =

Hint : vmtF ∆=∆ ⇒ t

vmF

∆=

By doing so time of change in momentum

increases and impulsive force on knees decreases.

115. Ans. ω ωω ω L =C

1

ω ωω ω

116. Ans. X = 1, Y = 0

Hint : P = YX +

O = Y.X

R = Y.X)YX( ++

117. Ans. 1.1 × 10 –9

m

Hint : λ =mv

h

118. Ans. 4

1 amp

Hint :15

55

V = I R

5 = 20 I

I =4

1

119. Ans. tan –1 (3/2)

Hint :'

tan

tan

δ

δ = cos α

tan δ = cos α × tan δ' = cos 30º × tan 60º

=2

3 × 3 =

2

3

δ = tan–1 2

3


12/12


120. Ans.π ππ π

µ µµ µ

2

Iv0 loge

+

r

L r

Hint : v

dx

r

x

L

dε = B v d x

= ∫+

π

µLr

r

0 dxx2

vI

E =π

µ

2

Iv0 ln

+

r

Lr

sol_major test-6.pdf

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