Iit Paper I Solutions

7/23/2019 Iit Paper I Solutions

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IIT-JEE Model Paper Solutions published on 03-04-2011

in EENADU Pratibha pages.Chemistry-1. Ans. : D

Sol.d[x]

dt = K1[M] – K2[X]

for the equation, time for maximum activity is given by t =( )

2

2 1 1

K1ln

K K K

−

=1 10

ln1 1 12

0.6932 10

−

= ( )20

2.3 log10 log 20.693 8

× −×

t = 5.8 hrs.

2. Ans. : B

Sol. 4 × 10-5

= 4 × 10-3

XB (use P = KH . XB)XB = 10-2

∴ 2H OX = 0.99

2 3H COX = 0.01

∴ M =0.01

10000.99 18

××

≈ 10

18

H2CO3 ƒ H+ + HCO3

– Ka1 >> Ka2

∴ [H+] = 1CKa = 51010

18

−×

pH =1

2

(5 – log10 / 18)

= 2.62.

3. Ans. : A

Sol. A(g) ƒ 3B(g) + 2C(g)

initially P O O

At equation P – x 3x 2x2x = P/4

Kp =( ) ( )

( )

2 3P/4 . 3P/8

7P/8 =

34

10

3P

7.2 x = P/8.

4. Ans. : B

.

5. Ans. : ABC

6. Ans. : ABCD

7. Ans. : C

8. Ans. : A

9. Ans. : D

Sol.arg

V =8RT

πM

2OV =

3

8R 560

π 32 10−

×

× ×

HeV =

3

8R 500

π 20 10−

×

× ×

2COV =3

8R 440

π 44 10−

×

× × HeV =

3

8R 140

π 4 10−

×

× ×

Maximum vary is of He.

10. Ans. : C

11. Ans. : ACD

Sol. The complex is ( )2

Ag CN−



12. Ans. : ABD

13. Ans. : A

14. Ans. : A

15. Ans. : C

Sol. A = B = C = D =

16. Ans. : A

17. Ans. : D

18. Ans. : D

Sol. D =1.07 3600 1

96500

× × = 0.04 F

= Fe3+

+ e– → Fe

2+

Fe2+ formed = 0.04 moles.

∴ 5 × M × 25 × 10-3

= 0.04 ⇒ M = 0.32

Moles of Fe3+ left = (0.200 × 0.25 – 0.04)

= 0.01M = 0.05

KMnO4 is a self indicator.

19. Ans. : A – T, B – PR, C – S, D – Q;

20. Ans. : A – T, B – S, C – R, D – PQ

SOLUTIONS – MATHEMATICS – P-1 – MODEL EXAM – 3 – 03-04-2010

21. Ans. : D

Sol. [ ]10

1

ln dx x ∫ = [ ] [ ] [ ] [ ] [ ] [ ]2 3 4 8 9 10

1 2 3 7 8 9

ln1 dx ln 2 dx ln 3 dx ln 7 dx ln8 dx ln 9 dx+ + + + +∫ ∫ ∫ ∫ ∫ ∫

=

2 3 4 8 9 10

1 2 3 7 8 9

0 dx 0 dx 1 dx ..... 1 dx 2 dx 2 dx+ + + + + +∫ ∫ ∫ ∫ ∫ ∫

= 1 × 5 + 2 × 2 = 9.

22. Ans. : C

Sol. f (1+) = 1, f (1–) = 0, f (1) = f (1+) ≠ f (1–)

23. Ans. : C

Sol. Equation of the plane 3(x – 1) + 0(y – 1) + 4(z – 1) = 0 ⇒ 3x + 4z = 7

Distance from origin =2 2

| 7 |

3 4+ =

7

5

24. Ans. : B

Sol. Foci of2 2

25 9

x y+ = 1 is (4, 0) and (–4, 0).

For the hyperbola, ae = 4, e = 2, a = 2. ⇒ b2 = a2(e2 – 1) = 12.

∴ Equation of the hyperbola is2 2

4 12

x y− = 1.

25. Ans. B

Sol. Since vectors a and b lie on the same plane, ( ) ( ).r a a b− × = 0

⇒ r a b

= 0 is the equation of the plane.

26. Ans. : DSol. ∆ = (5 sin θ – 4 cos θ)

2 = 41 sin

2(θ – α) ≤ 41.

27. Ans. : C

OH O

OH

O

CH3

O

CH3

O O



Sol. 1 2 3 Z Z Z = 1 2 3Z + Z + Z =

22 231 2

1 2 3

Z Z Z

Z Z Z + + =

1 2 3

1 4 9

Z Z Z + + = 1

⇒ |9Z1Z2 + 4Z3Z1 + Z2Z3| = |Z1 Z2 Z3| = |Z1| |Z2| |Z3| = 1 . 2 . 3 = 6.

28. Ans. : C

Sol. Divide by tan2 θ

(tan θ + cot θ)

2

+ 4a(tan θ + cot θ) + 16 = 0Let X = tan θ + cot θ then, X

2 + 4aX + 16 = 0 has two distinct roots if

16a2 – 64 > 0 ⇒ (a + 2) (a – 2) > 0 ⇒ a ∈ (–∞,–2) ∪ (2, ∞) ….. (1)

Also, for the interval (0, π /2), X = tan θ + cot θ ≥ 2 [ AM ≥ GM]

∴ The two distinct roots of X2 + 4aX + 16 = 0 are each greater than or equal to 2.

∴ Sum of roots = –4a > 4 ⇒ a < –1 ….. (2)

and f(2) > 0 ⇒ 4 + 8a + 16 > 0 ⇒ a >5

2

− …… (3)

From (1), (2) and (3) we get a ∈ 5

, 22

− −

29. Ans. : ABCD

Sol. (i) By geometry, in ∆AEF, AE = AF = AD sec A/2∴ ∆AEF is isosceles.

(ii) Area of ∆ABC = Area of ∆ABD + Area of ∆ADC

⇒ 1

bc sin A2

=1 A

b . AD sin2 2

+1 A

c . AD sin2 2

⇒ AD =2bc

b + c cos

A

2.

(iii) From (i) AE = AD secA

2=

2bc

b + c = H.M. of b and c.

(iv) EF = EF + DF = 2ED = 2AD tanA

2

=4bc

b + c

sinA

2

30. Ans. : CD

Sol. Simplifying the given equation to eliminate the denominators

and let f(x) = A1(x – a2) (x – a3) + A2(x – a1) (x – a3) + A3(x – a1) (x – a2)

then f(a1) > 0, f(a2) < 0 and f(a3 > 0.By the theorem on continuous functions we get the result.

31. Ans. : CD

Sol. Any tangent to y2 = 8x with slope M is y =2

Mx +M

Any tangent to y2 = 8x with slope 3M is y = 23Mx +3M

P(h, k) lies on both the tangents.BY eliminating m, we get 3k 2 = 32h.

The points should satisfy the above equation and also y2 – 8x > 0.

32. Ans. : ABCD

Sol. f(x) = f(1 – x)

Put x = x +1

2 then

1

2 f x

+

=

1

2 f x

−

Hence,1

2 f x

+

is an even function and therefore

B

A

CD

E

F

b

A

2A

2



1

2 f x

+

sin x is an odd function. ∴

1/2

1

1sin x dx

2 f x

−

+

∫ = 0

Also, f ′(x) = – f ′(1 – x) and for x = ½, we have f ′(1 / 2) = 0

Again, given f ′(1/4) = 0, ∴ f ′(3/4) = 0 and also f ′(1/2) = 0

But, Rolle’s theorem, f ″(x) = 0 at least twice in [0, 1].

Also, ( )1

sin

1/2

1 dtt f t e π −∫ = ( )

1/2sin

0

dy y f y e π

∫ (by putting 1 – t = y).

33. Ans. : A34. Ans. : A

35. Ans. : B

Sols.

Clearly, O is the mid point of SS′ and also HH′.

Thus, diagonals of the quadrilateral HS′H′S bisect each other, which means it is aparallelogram.

∴ SH = S′H′ and HS′ = H′S

Now, HS + HS′ = 2a (sum of the focal distances)

SS′ = 2ae1, AA′ = 2a

HH′ = 2ae2, BB′ = 2a

From ∆HOS′, (HS′)2 = OH2 + OS′2 – 2(OH) (OS′) cos(180 – θ)

⇒ (HS′)

2

= a

2

e2

2

+ a

2

e1

2

+ 2a

2

e1e2 cos θ HS′ is maximum when cos θ = 1, minimum when cos θ = – 1

∴ HS′ / max = a(e1 + e2)

HS′ / min = a|e1 – e2|.

36. Ans. : C

37. Ans. : C38. Ans. : C

Sols. Let P1 begin the letters chain writing to P2 and P3

Now, P2 on receiving a letter from P1 will write to two others persons other than himself.So, P2 has (n – 1) C2 choices including a choice in which he may write to P1.

∴ Probability that P2 will not write to P1 is2 21

2

m

nC

C

−−

Similarly, probability that P3 will not write to P1 is2

21

2

m

n

C

C

−

−.

Thus, the probability that P1 does not receive a letter at the second state =

23

1

n

n

−

−

Now, each of the four persons receiving a letter from P2 and P3 will write two letters each

to other persons.

Probability that none of these four persons write letters to P1 is

43

1

n

n

−

− . This is the third

stage .

By probability of none of the 8 recipients of the letters write letters to P1 in the 4th

stage is8

3

1

n

n

−

− .

So, on probability of none of the 2m – 1

recipients of the 2m – 1

recipients of the letters at (m

– 1)th stage write letters to P1 in the Mth

stage is (0), (4) f f

123

1

m

n

n

−

−

− .

∴ Probability of P1 not receiving a letter in first ‘M’ stages

=

2 3 12 2 2 23 3 3 3

.....1 1 1 1

m

n n n n

n n n n

−

− − − −

− − − −

A′

B′

H′

S′ ASO

H

B

Let the angle between

the major axes of the

two ellipses be ‘θ



=

2 3 12 2 2 ..... 23

1

m

n

n

−+ + + +−

−

=

2 23

1

m

n

n

−−

−

.

39. Ans. :Sol. 1) Same ax + by + c = 0 is a focal chord of the parabola the angle between the tangents at

A and B is 90o.

ii) Equation of a circle touching y = x at (2, 2) is (x – 2)2 + (y – 2)

2 + λ(y – x) = 0

Since this touches the y-axis also (λ– 4)2 = 32 ⇒ λ = 4 ±4 2 .

∴ Min radius =| |

2

λ = 4 – 2 2

iii) 1

2

arg z z

z z

−

− =

4

π represents the major arc of a circle chord subtending 45

o on

circumference, and 90o at the centre.

r2 + r2 =2

1 2 z z

2r2 = 62 ⇒ r = 3 2

iv)

( )a b c× × =

( )a b c× ×

( ) ( ). .a c b a b c/ −/ // = ( ) ( ). .a c b b c a/ −/ //

⇒ ( ) ( ). .a b c b c a=

⇒ a and c are collinear vectors

∴ angle between them is zero.

40. Ans. : A – P, B – S, C – S, D – Q.Sol. A. k = 2

y – 1 =1

1 x − ⇒ dy

dx =( )

2

1

1 x

−

− ⇒

2

2

dy

dx =( )

3

2

1 x

−

−

∴ ( )2

21

d y y

dx− =

4

2

( 1) x − =

2

2 dy

dx

B. λ = 3Max. distance = sum of the radii + distance between the centraes = 3

C. k = 3

f ′(x) = 3x2 – 6x – 1 = 0 at x = 1 ±

2

3

in [0, 4], x = 1 +2

3 is a point of minimum.

Thus, max f(x) = max. (0), (4) f f = 15.

D. Ans. : 0

F(x) = (cos2 x – 3)2 – 4 and 4 ≤ (cos2 x – 3)2 ≤ 9Thus, min. of f(x) is zero.

SOLUTIONS – PHYSICS – MODEL EXAM. – 3 – 03-04-2010

41. Ans. : C

Sol. y = 05

sin cos2

x y t

l

π ω 5

sin2

x

l

π =1

2±

x = 10 cm, 50 cm, 70 cm, 110 cm, 130 cm.

42. Ans. : A

Sol. 0 1

t V

i e R

τ −

= −

where τ = L

R

charge flown =

0

i dt

τ

∫ = 0

0

T

t V

t e

R

τ

τ −

+

= ( )10 1V

e R

τ τ − + −

= 0

Re

V τ = 0

2R e

V L.

A′

A/2

S′ A

45o

Or r

Z2(3, 5) Z1(9, 5)



43. Ans. A

Sol. 0

2

Q

C = ∈L

( )

20

2 2

Q

C = U0 Q0 = 02 CU

∴ ∈L = 0U

C

44. Ans. : B

Sol. f = m2

a

mx″ - f = ma

f =2

sin3

mAt

ω ω −

max friction force =2

3

mAω

45. Ans. : B

Sol. Centre of mass is the inertial frame of reference.2

21 33

2 2

Gmmv

r

−

=

23Gm

d

−

v =1 1

22

Gmr d

−

Relative velocity = 2 cos30ov = 3v

=1 1

62

Gmr d

−

46. Ans. : A

Sol. The electric field between A and B, and that between B and C =0

3

2

Q

A∈

VA – VB =0

3

2

Qd

A∈ VB – VC =

0

3(2 )

2

Qd

A∈ VA – VC =

0

9

2

Qd

A∈

47. Ans. : B

Sol. Wab = 0 Qab = ( )0

32

2

R N T

Wbc = ( )0

3 ln 3 Nr T Qbc = ( )0

3 ln 3 Nr T

Wca = ( )0 03 NR T T − = 02 NRT −

Qca = ( )0 0

53

2

Rn T T

−

= 05 NRT −

Efficiency η =( )

( )0

0

3ln 3 2

3 3ln 3

NRT

NRT

−

+

=( )

3ln 3 2

3 1 ln 3

−

+

48. Ans. : B

Sol. Required =( )2 2

4

R R R R

R+ + = 3R = 30 Ω.

49. Ans. : ABD

Sol. 0V ir i R′− − = 0

0 03V ir V − − = 0 iR = 02V − , i R′ = 3V0

i i′− = 05V

R

− aV ir − = Vb Vb – Va = 2V0

UL =

2051

2

V

L R

=

20

2

25 LV

R

When S is opened + 0 05 3 L V V ∈ − + = 0 L∈ = 2V0

mg

G

α

a

N f

f



50. Ans. : ACD

Sol. UC – UA = NCr(TC – TA) = ( )C C A A

Cr P V P V

R− = 0 03

2

P V

QBC = ( ) p C B NC T T − = ( )PC C B B

C P V P V

R− = 0 05

2

P V

WABC = 0 05

2

P V QABC = 4P0V0 C = 4R

51. Ans. : ABCD

Sol.

2

0

0

1 M P r

π θ

αθ π

= +

∫ = 4πρ0R

( )2

20 0

0

1 z I r R

π θ

ρ αθ π

= +

∫ = 4πρ0R

3

( )( )2

0

0

11 cosG x R R

M

π θ

ρ αθ θ π

= +

∫ = 0

y = ( )( )2

0

0

11 sin

v R R

M

π ρ αθ θ

π +

∫ =2

R

π − .

52. Ans. : ABCD

Sol.8 3

3 6 1i

+ −=

+ + =

1

2A from – to + in cell 2

P2 = 81

2

= 4W P1 = 31

2

= 1.5

Cell 1 is charging. Cell 2 is discharging.Power dissipated in the circuit

= ( )2

13 1 6

2

+ +

= 2.5 W.

53. Ans. : C

Sol. At t = 300 s the height of water

= length of the pole= 300(1) = 300 cm = 3 m

54. Ans. : B

Sol. ( )01 sin( )a θ − =

4

3 sin 30 =

2

3

cosθ =2

3 sinθ =

5

9 =

5

3

55. Ans. : C

Sol. length of the shadow on the water surface = cot θ

=2

5

56. Ans. : D

Sol. ( )24 E Rπ =0

Ze+

∈ E =

204

Ze

Rπ ∈ independent of a.

57. Ans. : B

Sol. For a = 0 ( )d

R r R

ρ = − ∴ ( )2

0

4

R

Ze r dr π ρ = ∫

∴ d =3

3 Ze

Rπ

58. Ans. : ASol. For E to linearly dependent on r

ρ = constant a = R

3m

θ

60o

3 3 m



So that E(4πr2) =

3

0

4

3r ρ π

∈ E =

03

r ρ

∈.

59. Ans. : A – Q, B – T, C – R, D – P

Sol. L X Lω = 1

C X C ω

=

1 L

C

X X C ω

= ( )2 L

C

X LC X

ω = = ( )29 n LC ω = 9

XL = 9R Z = ( )2 28 R R+ = 65 R

1cos

65θ = 0

09

V I

R=

0 sin9 2

R

vv t

π ω

= −

= 0 cos

9

V t ω −

065

9 2ad

V v sub t

π ω θ

= − −

where

1cos

65θ =

= 1065 1cos cos

9 65

V t ω −

− +

08sin

9bd

V v t ω = ( )0 sin

9cd

vv t ω π = + . = 0 sin

9

vt ω −

60. Ans. : A – Q, B – P, C – R, D – S.

Sol. 0 1 22 2

M m

v mv v= −

1 22v v− = 0v ……. (1)

02 4

m lv

=2

212 2 4

ml m lvω

−

22 3l vω − = 03v ……. (2)

e = 1 1 24

lv v

ω + + = 0v ….. (3)

solving equations (1), (2) and (3) 01

8

15

vv = 0

215

vv =

GV =( ) ( )0 0

23

2

m v m

m

+ = 0

3

v

I0 XL

I0 XC

I0 R

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