sol for paper-1

8/12/2019 SOL FOR PAPER-1

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Solutions for Model Grand Test :: Paper - 1

Chemistry

1. Ans. : D

Sol.d[x]

dt = K 1[M] K 2[X]

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

2

2 1 1

K1ln

K K K

=1 10

ln1 1 120.6932 10

= ( )20 2.3 log10 log 20.693 8

t = 5.8 hrs.

2. Ans. : BSol. 4 10 -5 = 4 10 -3 XB (use P = K H . X B)

XB = 10 -2

2H OX = 0.99

2 3H COX = 0.01

M =0.01

10000.99 18

1018

H2CO 3 H+ + HCO 3 Ka 1 >> Ka 2

[H +] = 1CKa =510 10

18

pH = 12 (5 log10 / 18)

= 2.62.

3. Ans. : ASol. A (g) 3B (g) + 2C (g)

initially P O OAt equation P x 3x 2x

2x = 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. argV =8RT M

2OV = 3

8R 560 32 10

HeV = 38R 500

20 10

2COV = 3

8R 440 44 10

HeV = 38R 140

4 10

Maximum vary is of He.

10. Ans. : C

11. Ans. : ACD


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Sol. The complex is ( )2Ag CN

12. Ans. : ABD

13. Ans. : A

14. Ans. : A15. Ans. : C

Sol. A = B = C = D =

16. Ans. : A17. Ans. : D18. Ans. : D

Sol. D =1.07 3600 1

96500

= 0.04 F

= Fe 3+ + e Fe 2+ Fe2+ formed = 0.04 moles. 5 M 25 10 -3 = 0.04 M = 0.32Moles of Fe 3+ left = (0.200 0.25 0.04)

= 0.01M = 0.05

KMnO 4 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. : CSol. f (1+) = 1, f (1) = 0, f (1) = f (1+) f (1)

23. Ans. : CSol. 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+ =

75

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 = a 2(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.

OH OOH

O

CH 3

O

CH 3

O O


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26. Ans. : DSol. = (5 sin 4 cos )2 = 41 sin 2( ) 41.

27. Ans. : C

Sol. 1 2 3 Z Z Z = 1 2 3Z + Z + Z =22 2

31 2

1 2 3

Z Z Z

Z Z Z + + =

1 2 3

1 4 9 Z Z Z

+ + = 1

|9Z 1Z2 + 4Z 3Z1 + Z 2Z3| = |Z 1 Z2 Z3| = |Z 1| |Z2| |Z 3| = 1 . 2 . 3 = 6.

28. Ans. : CSol. Divide by tan 2

(tan + cot )2 + 4a(tan + cot ) + 16 = 0Let X = tan + cot then, X 2 + 4aX + 16 = 0 has two distinct roots if16a 2 64 > 0 (a + 2) (a 2) > 0 a (,2) (2, ) .. (1)Also, for the interval (0, /2), X = tan + cot 2 [ AM GM] The two distinct roots of X 2 + 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 >

52

(3)

From (1), (2) and (3) we get a 5

, 22

29. Ans. : ABCDSol. (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 secA2

=2bc

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

(iv) EF = EF + DF = 2ED = 2AD tanA2

=4bc

b + c sin

A2

30. Ans. : CDSol. Simplifying the given equation to eliminate the denominators

and let f(x) = A 1(x a 2) (x a 3) + A 2(x a 1) (x a 3) + A 3(x a 1) (x a 2)then f(a 1) > 0, f(a 2) < 0 and f(a 3 > 0.

By the theorem on continuous functions we get the result.31. Ans. : CD

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

Mx +M

Any tangent to y 2 = 8x with slope 3M is y =2

3Mx +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 y 2 8x > 0.

32. Ans. : ABCDSol. f(x) = f(1 x)

B

A

CD

E

F

b

A2

A2


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Put x = x +12

then12

f x

+ =

12

f x

Hence,12

f x

+ is an even function and therefore

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 = , we have f (1 / 2) = 0Again, given f (1/4) = 0, f (3/4) = 0 and also f (1/2) = 0But, Rolles theorem, f (x) = 0 at least twice in [0, 1].

Also, ( )1

sin

1/2

1 dtt f t e = ( )1/2

sin

0

dy y f y e (by putting 1 t = y).

33. Ans. : A34. Ans. : A35. Ans. : BSols.

Clearly, O is the mid point of SS and also HH .Thus, diagonals of the quadrilateral HS HS bisect each other, which means it is aparallelogram. SH = S H and HS = H SNow, HS + HS = 2a (sum of the focal distances)

SS = 2ae 1, AA = 2aHH = 2ae 2, BB = 2a

From HOS , (HS )2 = OH 2 + OS 2 2(OH) (OS ) cos(180 ) (HS )2 = a 2e22 + a 2e12 + 2a 2e1e2 cos

HS is maximum when cos = 1, minimum when cos = 1 HS / max = a(e 1 + e 2)

HS / min = a|e 1 e 2|.

36. Ans. : C37. Ans. : C38. Ans. : CSols. Let P 1 begin the letters chain writing to P 2 and P 3

Now, P 2 on receiving a letter from P 1 will write to two others persons other than himself.So, P 2 has (n 1) C 2 choices including a choice in which he may write to P 1.

Probability that P 2 will not write to P 1 is2

21

2

m

nC

C

Similarly, probability that P 3 will not write to P 1 is2

21

2

m

nC

C

.

Thus, the probability that P 1 does not receive a letter at the second state =23

1nn

Now, each of the four persons receiving a letter from P 2 and P 3 will write two letters eachto other persons.

Probability that none of these four persons write letters to P 1 is43

1nn

. This is the third

stage .By probability of none of the 8 recipients of the letters write letters to P 1 in the 4 th stage is

83

1

n

n

.

A

B

H

S ASO

H

B

Let the angle betweenthe major axes of the

two ellipses be


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So, on probability of none of the 2 m 1 recipients of the 2 m 1 recipients of the letters at (m

1)th stage write letters to P 1 in the M th stage is { }(0), (4) f f 123

1

m

nn

.

Probability of P 1 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

=

2 3 12 2 2 ..... 231

m

nn

+ + + +

=2 23

1

m

nn

.

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 90 o.ii) Equation of a circle touching y = x at (2, 2) is (x 2) 2 + (y 2) 2 + (y x) = 0Since this touches the y-axis also ( 4) 2 = 32 = 4 4 2 .

Min radius =| |

2

= 4 2 2

iii) 12

arg z z z z

=

4

represents the major arc of a circle chord subtending 45 o on

circumference, and 90 o at the centre.

r2 + r 2 =2

1 2 z z

2r2 = 6 2 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

dydx

=( )2

1

1 x

2

2dy

dx =

( )32

1 x

( ) 221 d y ydx

= 42

( 1) x = 22 dy

dx

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

C. k = 3

f (x) = 3x 2 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. : 0F(x) = (cos 2 x 3) 2 4 and 4 (cos 2 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

5

sin2

xl

=

12

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

42. Ans. : A

45 o

Or r

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


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Sol. 0 1t

V i e

R

= where =

L R

charge flown =0

i dt

= 00

T t

V t e

R

+

= ( )10 1V e R + =

0

ReV = 02R e

V L .

43. Ans. A

Sol. 02Q

C = L ( )

20

2 2Q

C = U 0 Q0 = 02 CU

L = 0U C

44. Ans. : BSol. f = m

2a

mx - f = ma

f =2

sin3

mAt

max friction force =2

3mA

45. Ans. : B

Sol. Centre of mass is the inertial frame of reference. 221 332 2

Gmmvr

= 23Gm

d

v =1 1

22

Gmr d

Relative velocity = 2 cos30 ov = 3v

=1 1

62

Gmr d

46. Ans. : A

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

32

Q A

VA V B =0

32

Qd A

VB V C =0

3(2 )

2Q

d A

VA V C =0

92

Qd A

47. Ans. : B

Sol. W ab = 0 Q ab = ( )03

22 R

N T

Wbc = ( )03 ln 3 Nr T Qbc = ( )03 ln 3 Nr T W ca = ( )0 03 NR T T = 02 NRT

Qca = ( )0 05

32 R

n T T

= 05 NRT

Efficiency =( )( )

0

0

3ln 3 2

3 3ln 3

NRT

NRT

+

=( )

3ln 3 23 1 ln 3

+

48. Ans. : B

Sol. Required = ( )2 24

R R R R

R+ + = 3R = 30 .

mg

G

a

N f f


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49. Ans. : ABDSol. 0V ir i R = 0

0 03V ir V = 0 iR = 02V , i R = 3V 0

i i = 05V R

aV ir = V b Vb V a = 2V 0

UL =

2051

2V

L R

=

20

225 LV

R

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

50. Ans. : ACD

Sol. U C U A = NCr(T C T A) = ( )C C A ACr

P V P V R

= 0 03

2P V

QBC = ( ) p C B NC T T = ( )P C C B BC

P V P V R

= 0 05

2P V

WABC = 0 05

2

P V QABC = 4P 0V0 C = 4R

51. Ans. : ABCD

Sol.2

00

1 M P r

= + = 4 0R

( )2

20 0

0

1 z I r R

= + = 4 0R

3

( )( )2

0

0

11 cosG x R R M

= + = 0

y = ( )( )2

00

11 sin

v R R

M

+ = 2

R

.

52. Ans. : ABCD

Sol.8 3

3 6 1i

+ =

+ + =

12

A from to + in cell 2

P2 = 812

= 4W P 1 = 312

= 1.5

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

= ( )21

3 1 62

+ + = 2.5 W.

53. Ans. : CSol. 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 = 43 sin 30 = 2

3

cos =23

sin =59

=5

3

55. Ans. : CSol. length of the shadow on the water surface = cot

=2

5

56. Ans. : D

3m

60 o

3 3 m


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Sol. ( )24 E R =0

Ze+

E = 204

Ze

R independent of a.

57. Ans. : B

Sol. For a = 0 ( )d R r R

= ( )20

4 R

Ze r dr =

d = 33 Ze

R

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

= constant a = R

So that E(4 r2) =

3

0

43

r

E =03

r

.

59. Ans. : A Q, B T, C R, D P

Sol. L X L = 1

C X C =

1 LC

X X C

= ( )2 LC

X LC

X = = ( )29 n LC = 9

XL = 9R Z = ( )2 28 R R+ = 65 R1

cos65

= 00 9V

I R

=

0 sin9 2 Rvv t =

= 0 cos9

V t

0659 2ad

V v sub t

= where

1cos

65 =

= 1065 1

cos cos9 65

V t

+

08 sin9bd

V v t = ( )0 sin

9cd

vv t = + . = 0 sin

9

vt

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 4m l

v

=

2

212 2 4ml m l

v

22 3l v = 03v . (2)

e = 1 1 24l

v v

+ + = 0v .. (3)

solving equations (1), (2) and (3) 01815v

v = 02 15v

v =

GV =( ) ( )0 02

32

mv m

m

+ = 0

3v

I0 XL

I0 XC

I0 R


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sol for paper-1

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