M. Alvarez Méndez (1986, 13)

+ =

289 324 =

17 18 17N + 68 = 18N + 18 50 = N

3.[B]

4.[A]

5.[B] u = 0 a = constant v0 = u + at = at n' = _



 



 + v

v ₀

n = n + v

t n a

6.[C] LC = 1 MSD – 1VSD

LC = 



 

 − 30

1 29 MSD = 30

1 MSD 720 MSD = 360°

1 MSD = 720

360° = 1/2°

So LC =

30 1 ×

2 1° =

60 1° =

60 ' 60 = 1'

7.[C]

8.[D] Ticker timer is a better device than a stop watch.

9.[C] M = 2

M M₁+ ₂

= 2

62 . 12 30 . 10 +

= 11.46 g 10.[B]

11.[B]

∆Y

= D 2∆D

+ l

∆l

∆Y

= 2 



 



 4 . 0

01 .

0 + 



 



 8 . 0

05 . 0 = 2× 0.025 + 0.0625

∆Y

= 0.05 + 0.0625 = 0.1125 ∆Y = 2× 10¹¹ × 0. 1125 = 0.225 × 10¹¹ So (2 ± 0.2) × 10¹¹ N/m²

12.[B]

13.[B]

^d f

F f

It is only possible if object and image coincide.

14.[D] I = I 0 cos² 2 φ

2 I₀

= I 0 cos² 2 φ

cos 2 φ =

2 1

φ = 90° = λ

π 2 ∆x

2 π =

λ 2π ∆x

∆x = 4 λ ⇒

D d y =

4 λ

y = 4d λD

= ⁷ ₄

10 1 4

1 10 5

−

× =

4 5 × 10^–3 y = 1.25 mm

15.[C] ^→τ = ^→p × ^→E = q (2^→a ) × ^→E

Here 2^→a = (2 – 1) iˆ + (– 1–0) jˆ + (5 – 4) kˆ = iˆ– jˆ + kˆ

→E = 0.20 iˆ V/cm = 20 iˆ V/m

∴ ^→τ = (4 × 10^–6)

[ (

^iˆ^–^jˆ⁺^kˆ

)

^×²⁰^iˆ

]

^{= 8 × 10}^–5

( )

^kˆ⁺^jˆ

Magnitude of torque τ = 8 2 × 10^–5 N-m 16.[D]

– σ + σ + σ c b

a C B

SOLUTION FOR MOCK TEST

PAPER - II _AIEEE

Potential of shell A is, Potential of shell C is, VC =

18.[B] Resistivity of conductors increases with increase in temperature because rate of collisions between free electrones and ions increases with increase of temperature.

However, the resistivity of semiconductors decreases with increase in temperature because more and more covalent bonds are broken at higher temperatures.

L [as cross-section is same]

Now Lc = 2 LA and (volume)c = (volume)P

20.[A] Current flowing through potential wire is – I = 15r r

Potential drop across potential wire is, V = I × 15 r = Magnetic moment of second magnet, M' = (3m)(2 × 2l) = 6M

34.[C] Calcined Gypsum is calcium sulphate

35.[A] Inter particles forces between CH3COCH3 &

CHCl3 are strong H-bonding. Thus solution shows negative deviation. ∆V_mixing = Negative.

36.[C] Addition of inert gas at constant volume does not cause any effect on the equilibrium.

37.[A] For I order Reaction t1/2 is constant

39.[B] Solution is decinormal, that is N/10 & x factor is 1, so conc. = 0.1M

[H⁺] = c.α

= 0.1 ×1.3/100 = 13×10^–4 pH = –log [13×10^–4] = 2.89 40.[A] The no. of atoms in fcc lattice = 4

a = 400 pm = 4×10^–10 m = 4×10^–8cm

d = ₃

a No

M n

× = ₂₃ _–₈ ₃

) 10 4 ( 10 6

60 4

d = 6.23 g/cm³.

41.[B] For any Value of l possible values of m are m = –l to + l

l = 2, m = –2, –1, 0 +1, +2 So option is (B)

42.[B] C : O : H 6gm : 3.01 × 10²³atoms : 2 mole Ratio ½ : ½ : 2 of mole

1 : 1 : 4

COH4 or CH4O

43.[B]

3 2 1

44.[B] NaNO²/HCl gives HNO2 which gives different products with Pri. and Sec. amines.

45. [A] CH3

+ Cl2h→ν CH2Cl +HCl

46.[D] 2 CuSO⁴ + 2 KCN → Cu(CN)2 + K2SO4 2Cu(CN)2 → 2CuCN ↓ + (CN)² ↑

3 KCN + CuCN → K³[Cu(CN)4] 47. [B] Blue print process occurs with the help of Iron Compound.

48.[A] Effective nuclear charge increases therefore ionic radius follow the order.

49.[B] xy2 xy + y t = 0 : 600 0 0 teq^m : 600-P P P Now : 600 – P + P + P = 800

P = 200 mm Hg

100 mm Hg

400 200 200 )

Pxy (

) Py )(

Pxy Kp (

× =

50.[C] Cell reaction : Zn + Cu⁺² → Zn⁺²+ Cu Cell emf:

Ecell = E°cell – n 059 .

0 log _^











+ + 2 2

Cu Zn

So doubling the conc. of ions.

E_cell remains same.

51.[A] pKa = –log Ka = –log (1.8×10^–5) = 4.7447 [CH3COOH] =

L 4 mol . 5 0 . 0 60

12 =

× [CH3COONa] =

5 . 0 4 . 16

82 = 0.4 L mol

Now, pH = pKa + log 

 acid

salt

= 4.7447 + log _^₀⁰_.^.₄⁴_^ = 4.7447

52.[B] FeC₂O4 → Fe⁺³ + CO2

+2 +3 +3 +4 Increase = 1

Increase= 1 Total Increase in O.N. = 3 So valence factor of FeC2O4 = 3 KMnO4 →Mn⁺²

+7 +2 v.f. (KMnO4) = 5 gm E KMnO4 = gm E of FeC2O4

Mole × v.f. = Mole × v.f.

1× 5 = x × 3 ⇒ x = 5/3 53.[A] ∆G = ∆H – T ∆S

∆G = Θ , ∆G < 0 (Spontaneous process) ∆G = ∆H – T ∆S

= ∆E + P ∆V – T ∆S (∆G)E,V = 0 + 0 – T ∆S

(∆S)_E,V = ⊕ ⇒ ∆G = Θ (Spontaneous process) 54.[B]

n Carbocatio 2

CH3 – H C – CH3

CHC| 3 3– H CH

2O H 3 – CH – OH| CH – CH3

CH| 3 C – 3C H

° + +

→



1, 2-Methyl shift

) stable More (

n Carbocatio 3

CH – CHCH| 3 – CH| 3 C – C CH H

– CH| 3 C CHC| 3 – C

H₃ ₃ ^–^H ₃ ₃

 +

= ← ⁺

55.[C] Factual Q.

56.[D]

OH ^Zndust__→ ^H^C^–^Cl

AlCl

3 3



 →

 CH3

KMnO4

alk → COOH

57.[D]

58.[A] 4 HCl + O2 → 2 H₂O + 2 Cl2

Chlorine is in the form of cloud.

59.[A] Coordination no. = 6 Oxidation no. = 3 no. of d electron = 6 no. of Unpaired d electron = 0

60.[C] Resonance structure should have same number of electron pairs.

MATHEMATICS

61.[A]

60º O

A 30º

Let the height of the tower is h 1000 = h cot 30º – hcot 60º

⇒ h =

º 60 cot º 30 cot

1000

− h = 500 3m

62.[D] since variance is independent of change in origin. Hence variance of observations 101, 102, ... 200 is same as variance of 151, 152, ...250.

∴ V_A= VB

⇒

B A

V V = 1

63.[C] centroid of triangle is given by











 + + −

3 3 b ,a 3

3 c²

if centroid lie on y axis ⇒ abscissae = 0 c² + 3 = 0 ⇒ no real c exist if centroid lie on x axis ⇒ ordinate = 0

⇒ a + b – 3 = 0 ⇒ a + b = 3

64.[C] f(x) = sinx –cosx – kx + b f'(x) = cosx + sinx – k f'(x) = 2 sin(π/4 + x) – k

if f(x) is decreases for all x ∴ f'(x) is –ve i. e. k > max. of 2 sin(π/4 + x) i. e. k > 2

65.[A] _x^lim_→₀











 −

2 x

x x cos e ²

= _x^lim_→₀ _^









 −

− +

2 2

x x cos 1 x

1 e ²

1 + _x^lim_→₀ ²₂ x

2 / x sin

2 = 1 +

2 1 =

2 3

66.[A] Second determinant has been obtained from the first by the operation

C1 → C₁ + 2C2 – 3C3. so its value remains unchanged

67.[C] given 3sinx – 4sin³x – k = 0

⇒ 3sinx – 4sin³x = k

⇒ sin3x = k

...(i)

angle A and B satisfy the equation (i)

∴ sin 3A = sin3B = k ⇒ sin3A = sin3B But A > B ⇒ A B

Now, sin3A = sin(π – 3B) 3A = π – 3B

A + B = 3

π ⇒ C = 3 2π

68.[B] given statement (p q) ∨ ~ r → (p ∧ r) (F ↔ F) ∨ F → (F ∧ T) T ∨ F → F

T → F = F 69.[C]

T F

T F F F

T T

T T T F

F F

F T F T

F F

F T T T

~ q ) q p ( p

~ q p

~ q p q

p ∨ ∧ ∨ ↔ ∧

Hence neither tautology nor contradiction 70.[D] Given sequence can be written as

2 5,

13 20,

9 10,

20, ...

or 8

20, 13 20 ,

18 20,

23 20 ...

which is a H.P.

n^thterm of corresponding A.P.

an =

71.[B] Let any point P divides line joining A(–2, 4, 7) & (3, –5, 8) in ratio λ : 1 is 4 units above x-axis

⇒ 13 – p² = 4 ⇒ p = ± 3 also lies in I^stquadrant ⇒ p < 0

⇒ p = – 3

75.[C] If lines x² + 2λx + 2y² = 0

& (1 + λ) x² – 8xy + y² = 0 are equally inclined

⇒ Their bisectors eqⁿmust be same

76.[D] given circle in standard form is x² + y² + 2 

77.[C]

∫

− 4

) x (

f dx = 4

⇒

∫

− 2

) x (

f dx +

∫

⁴

) x (

f dx = 4

⇒

∫

− 2

) x (

f dx = 4 –

∫

⁴

) x (

f dx ...(i)

Q ⁽³^–^f(x)⁾

∫

2 dx = 7 (given)

⇒

∫

⁴

dx 3 –

∫

⁴

) x (

f dx = 7

⇒ 3[x]⁴₂= 7 +

∫

⁴

) x ( f dx

⇒ 3 × 2 = 7 +

∫

⁴

) x (

f dx ⇒

∫

⁴

) x (

f dx = – 1 ...(ii)

put this value from (ii) in (i), we get

−

∫

) x (

f dx = 4 – (–1) = 5

⇒ ⁻¹

∫

) x (

f dx = – 5

78.[A] lines – 2x – y + 6 = 0 & 4x – 2y + 7 = 0 (make c1 & c2 +ve) now a1a2 + b1b2 = – 8 + 2 = – 6 < 0

bisector by +ve is acute and contains origin eqⁿ given by

5 6 y x 2 − +

− = +

5 2

) 7 y 2 x 4

( − +

– 4x – 2y + 12 = 4x – 2y + 7

8x = 5

79.[C]

∫

[ ]

⁻

) 1

x ( g

f f'g(x). g'(x) dx put f g(x) = t ⇒ f'g(x) g'(x) dx = dt

∴ required integral

[ ]

x ( g f log

log (f g(2)) – log (f g(1)) = 0 {Q g(1) = g(2)}

80.[C] f(x) =

∫

^x²₁⁺₊^sin_x²²^x ^sec²^{x dx}

∫

^^_^^x²⁺₁¹₊⁻_x^cos² ²^x^^_^^sec²^{x dx}

∫

^^_^¹⁻^cos₁₊²_x²^x^^_^^sec²^{x dx}

∫

^_^^sec²^x⁻₁₊¹_x²^_^^dx

= tan x – tan^–1x + c

∴ f(0) = 0 ⇒ tan0 –tan^–10 + c = 0

⇒ c = 0

∴ f(x) = tanx – tan^–1x

∴ f(1) = tan1 – tan^–11

= tan1 – π/4

81.[B] we know x –[x] = {x}

∴ Domain of {x} is R & {x} ∈ [0, 1)

but in f(x), {x} is in denominator & it should not be equal to zero

∴ {x} 0 ⇒ x I

and domain of sec^–1x is R –(–1, 1)

∴ domain of f(x) is R – (–1, 1) – I

82.[C] Q sin^–1x is defined for |x| ≤ 1 and sec^–1x is defined for |x| ≥ 1 therefore

both defined for |x| = 1 ⇒ x = {1,–1}

∴ D_f = {–1, 1}

further f(–1) = sin^–1(–1) + sec^–1(–1) = – π/2 + π = π/2

and f(1) = sin^–1(1) + sec^–1(1) = π/2 + 0 = π/2 Hence Rf = {π/2}

83.[B] Let A(z1), B(z2) and C(z3) be the vertices of the triangle then

|z1| = |z2| = |z3|

⇒ |OA| = |OB| = |OC|

O being the origin

⇒ O is circumcentre of the triangle, Also, the triangle is equilateral, therefore circumcentre coincide with the centroid

⇒ origin is centroid

⇒ 3

z z z₁+ ₂+ ₃

= 0

⇒ z1 + z2 + z3 = 0 84.[B] Total students n = 100 Average marks x = 72 Total boys n1 = 70

average marks of boys x₁ = 75 Total girls n2 = 30

now x =

n x n x n₁ ₁+ ₁ ₂

∴ x₂=

2 1 1

n x n x n −

x₂ =

30 75 70 72 100× − ×

= 65

85.[C] Letters of word 'STATISTICS' are 1A, 2I, 1C, 3S, 3T total = 10 Letters of word 'ASSISTANT' are 2A, 1I, 1N, 3S, 2T total =9 common letters are A, I, S & T probability of choosing A is =

10 1 ×

9 2 =

90 2

probability of choosing I = 10

2 × 9 1 =

90 2

probability of choosing S = 10

3 × 9 3=

90 9

probability of choosing T = 10

3 × 9 2 =

90 6

total =

90 2 +

90 9 +

90 6 =

90 19

86.[D] A = { – 2, –1, 0, 1, 2} ⇒ n(A) = 5 B = {0, 1, 2, 3} ⇒ n(B) = 4 C = {1, 2} ⇒ n(C) = 2

D = {(1, 7),(2, 6),(3, 5),(4, 4),(5, 3)(6, 2),(7, 1),}

⇒ n(D) = 7

(A ∪ B ∪ C) = {– 2, – 1, 0, 1, 2, 3}

⇒ n(A ∪ B ∪ C ) = 6

n(D) = 7

n(B ∪ C) = 4 87.[B] this is of the form

f'(y) dx

dy + f(y). f(x) = Q(x) put tany = z

∴ sec²y dx dy =

dx dz

∴dx

dz + 2xz = x³

∴ _e∫^pdx₌_e∫²^xdx₌_e^x²

∴ solⁿis ze^x²=

∫

^e^x²^.^x³^dx

= 2

∫

^e^x²^.^x² ^{2x dx}

= 2

∫

^e^t^.^tdt⁼₂¹^e^t (t – 1) + c

∴ tany e^x² = 2 1 _x²

e (x² – 1) + c

⇒ tany = ce⁻^x² + 2

1 (x² – 1)

88.[D] The given curves are y²= 8x ...(1)

and xy = – 1 ...(2)

any tangent to (1) is y = mx +

m 2 ...(3)









⇒

∴

2 a 8 a

4 m

mx a y is

ax 2 4 y to gent Qtan

we shall find m so that it touches (2)

∴ from (2) & (3) m 1 mx 2

x =−



 



 + ⇒ m²x² + 2x + m = 0 ...(4)

(3) touches (2) if quadratic (4) has equal roots

∴ D = 0

⇒ 4 – 4m³= 0 ⇒ m³ = 1 ⇒ m = 1

∴ required common tangent is y = x + 2 89.[B] h(x) = (f(x))² + (g(x))²

∴ h'(x) = 2f(x)f'(x) + 2g(x) g'(x)

= 2f(x)g(x) + 2g(x)g'(x) {Q f'(x) = g(x)}

Also f'(x) = g(x)

∴ f"(x) = g'(x) ⇒ – f(x) = g'(x)

∴ h'(x) = 2f(x) g(x) – 2g(x)f(x)

h'(x) = 0

∴ h(x) = constant for all x but h(5) = 11 ∴ h(x) = 1 for all x

∴ h(10) = 11

90.[A] for m1 : Let I & N are assumed single letter in IN order Now total no. of letter are IN T E G E R = 6 letter

these 6 letter can be arranged in row =

6 × 2= 720 ways

∴ Letter of words INTEGER can be arranged in a row =

7 = 2520 ways

Now no. of ways in which IN are not together is m1 = 2520 – 720 = 1800

Now for m2 : I- - - - - R → rest five can arrange

= 2

5 = 60 ways

∴

2 1

m m =

60 1800 = 30

PHYSICS

1.[B] In case of damped vibrations, amplitude decreases exponentially with time

∴ A = A₀e^–bt or

(Where y is the vertical distance of particle from x axis)

Here m,v and y all are fixed so ^→L. remains constant.

5. [B] Using, weight of floating body = weight of liquid displaced.

we get V ρ g =  (buoyant forces of mercury and oil act in opposite direction)

Then, ρ = given work done

∴ 10 + WCA = 5 or WCA = –5 J

For zero p.d. the fall of potential should be equal to in emf.

SOLUTION FOR MOCK TEST

In document kRCni~ ENELAULA DE DE (página 183-200)

[ (

)

]

( )

SOLUTION FOR MOCK TEST

PAPER - II AIEEE

3 2 1

MATHEMATICS

∫

∫

∫

∫

∫

∫

∫

∫

∫

∫

∫

∫

∫

∫

[ ]

[ ]

∫

∫

∫

∫

∫

∫

∫

PHYSICS

SOLUTION FOR MOCK TEST

PAPER - II _AIEEE