07 08 2011 xi pqrs paper i code a

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11th PQRS (Date: 07-08-2011) Review Test-3

PAPER-1

Code-A

ANSWER KEY

PHYSICS

SECTION-1

PART-A

Q.1 B

Q.2 B

Q.3 B

Q.4 C

Q.5 D

Q.6 A

Q.7 D

Q.8 D

Q.9 B

Q.10 D

Q.11 C

Q.12 D

Q.13 A

Q.14 B

Q.15 D

Q.16 B,C

Q.17 B,D

Q.18 A,D

Q.19 A,C,D

Q.20 A,C

Q.21 B

CHEMISTRY

SECTION-2

PART-A

Q.1 C

Q.2 C

Q.3 A

Q.4 A

Q.5 D

Q.6 B

Q.7 C Q.8 D

Q.9 A

Q.10 D

Q.11 B

Q.12 C

Q.13 B

Q.14 B

Q.15 A

Q.16 A,C,D

Q.17 A,D

Q.18 A,B,C,D or ABC

Q.19 A,B,C

Q.20 B,C

Q.21 A,C,D

MATHS

SECTION-3

PART-A

Q.1 A

Q.2 D

Q.3 B

Q.4 D

Q.5 C

Q.6 C

Q.7 D

Q.8 D

Q.9 C

Q.10 C

Q.11 B

Q.12 D

Q.13 A

Q.14 B

Q.15 A

Q.16 A,B,C,D

Q.17 A,B

Q.18 B,C,D

Q.19 A,C

Q.20 A,B

Q.21 A,B,D



PHYSICS

Code-A Page # 1

PART-A

Q.1

[Sol.

vAW vAG

vWGA

B]

Q.2

[Sol. Radius of curvature at point B is minimum. ]

Q.3

[Sol.B A

5×10=50

5×10=50

]

BAABvvv

= ) jˆ(50) jˆ(50 = jˆ100 ]

Q.4

[Sol. Hmax

=2gT

8

1

v

v]

Q.5

[Sol. 2 second later

40 m/s

30 m/s

Relative acceleration = g – g = 0

velocity of separation = 40 – 30 = 10

so reparation keeps on increasing before hitting ground. ]

Q.6

[Sol. K.Emax

=21 m2A2 =

21 m (2f)2A2 = 22mf 2A2

or check dimensionally ]



PHYSICS

Code-A Page # 2

Q.7

[Sol.

F=TT

T

T3T

T

T

3T = ma = 3F,m

F3a ]

Q.8

[Sol.

vA

gh

gvB

VB

2 = VA

2 + 2gh, acceleration for both g downward. ]

Q.9

[Sol. l0

+ 1k N4

l0

+ 2k

N6

l0

= 3l1 – 2l

2]

Q.10

[Sol. 2

2a

Fsp5Fsp

a

36N

Fsp

= 2 × 2a

36N – Fsp

= 5a Fsp

= 16N =100

16m = 16 cm ]

Q.11

[Sol.

N

mg

N – mg = ma N = m(g + a) = 50 (10 + 6) = 800 N ]

Q.12

[Sol.

105)192(a2

1u

65)152(a2

1u

u = 20, a = 10 } 20 +2

1 × 10(2 × 30 – 1) = 315 ]



PHYSICS

Code-A Page # 3

Q.13

[Sol.

v=gt0

v=0

relative acceleration = g – g = 0

relative velocity of separation = gt0so separation keep on increasing ]

Q.14

[Sol. g ]

Q.15

[Sol.

T T

2T sin = W

sin2

WT as 0, T so not possible ]

Q.16

[Sol. = rad/sx

y

(1,0), t=0

3

1t , = t = ×

3

1=

3

,

jˆ

3sin1iˆ

3cos1k ˆv

= jˆ

2iˆ

2

3

ra 2 =

jˆ

3sin1iˆ

3cos1

2=

jˆ

2

3

2

iˆ2]

Q.17

[Sol. R = 20m, u = 210 ,g

2sin2u = R sin 2 = 1, = 45°, u

x= 4 cos = 210 sin 45° = 10

|ux

| relative to observer should become half as vertical motion is same as observed by ground and time

of flight same. Velocity of observer =2

1010 = 5 or 15 ]



PHYSICS

Code-A Page # 4

Q.18

[Sol.

sinv

dt

V sin V

V sin V

tmin

= 90° ]

Q.19[Sol. y

1= y

2max height

t1 2t

2, t

2= 2t

1

x1

= u cos 2

T, x

2= u cos T

2

1

x

x=

2

1,

2

1

t

t=

2

1]

Q.20

[Sol. g sin

g

xy

g cos

V32 = V

12 – 2g sin x so V

3< V

1

V42 = V

22 2g cos (y) = V

22 – 2g cos (0) = V

22

V4

= V2

]

Q.21[Sol. x = (t – 1) (t – 2) x(0) = 2 0

3t2dt

dx 3

dt

dx

0t

(-ve)

2dt

xd2

2

speed =dt

dx

speed decreasing from t = 0 to t = 3/2 then increasing. ]



Code-A Page # 1

CHEMISTRY

PART-A

Q.1

[Sol. Zeff

= Z –

1s2 2s2 2p6 3s2

Zeff

= 12 – (2 + 8 × 0.85 + 0.35)

= 2.85 unit Ans.]

Q.2

[Sol. Molecular mass of Na2SO4.xH2O = (142 + 18x)After heating, Na

2SO

4will form and it's molecular mass = 142

(142 + 18x)g Na2SO

4·xH

2O gives loss of 18x g due to evaporation of water

100 g Na2SO

4·xH

2O gives loss of g

)x18142(

100x18

= 47 gm

x = 6.9958

7 (nearest integer) ]

Q.3

[Sol. Specific heat capacity = 1.68 J/g-k

=2.468.1 cal/g-k = 0.4 cal/g-k

Specific heat capacity × atomic weight 6.4

Atomic weight =4.0

4.6= 16 ]

Q.4

[Sol. Let x gm KCl is added in 60g H2O

Total weight of solution = (60 + x)

Percent weight of KCl =

x60

100x

= 20

x = 0.2 (60 + x)

x =8

120= 15g ]

Q.5

[Sol. IE2

and IE3

are very high compared to IE1

, so it will prefer to show valency of 1 and form AF, A2O and

A3N ]

Q.6

[Sol. Let the volume of each be 1 lit. (arbitravily chosen) so each solution contains 0.2 mole H+, 0.2 mole Cl –

and 0.4 mol K+ , 0.4 mole OH – before mixing aAfter mixing :

HCl + KOH KCl + H2O

0.2 0.4

mole of H+ 0

mole of OH – = 0.4 – 0.2 = 0.2

mole of K+ = 0.4

mole of Cl – = 0.2

Volume of solution = 2 lit



Code-A Page # 2

CHEMISTRY

So molarity of [K+] =2

4.0= 0.2 M

solutionof volumeTotal

soluteof moleM

molarity of [Cl – ] =2

2.0= 0.1 M

molarity of [OH – ] =2

2.0= 0.1 M ]

Q.7

[Sol. Lets weight of SO3

in 0.5 gm sample = x

Weight of H2SO

4in 0.5 gm sample = 0.5 – x

H2SO

4+ 2NaOH Na

2SO

4+ 2H

2O

98

x5.0

SO3

+ 2NaOH Na2SO

4+ H

2O

80

x

1000

7.264.0

98

x5.0

80

x2

x = 0.103

% SO3

= 1005.0

103.0 = 20.6 ]

Q.8

[Sol. m mole of Na2S

2O

3consumed = 20 × 0.3 = 6

Corresponding m mole of I2

=26 = 3 m mole

Corresponding m mole of H2O

2=

60

1003 = 5 m mole

25 ml H2O

2solution contains 5 m mole = 5 × 10 – 3 mole

Molarity =025.0

1053

=5

1

Volume strength =5

1 × 11.2 = 2.24 ]

Q.9

[Sol. (NH4)2SO

4+ NaOH Na

2SO

4+ 2NH

3+ 2H

2O

NH3

+ HCl NH4Cl

HCl + NaOH NaCl + H2O

Actual m moles of HCl used for reacting NH3

= 100 × 0.4 – 40 × 0.3

= 28 m mole

m mole of N-atom = 28 m mole



Code-A Page # 3

CHEMISTRY

mass of Nitrogen atom = 28 × 10 – 3 × 14 g = 0.392 g

Percent N is fertilizer = 10092.3

392.0 = 10% ]

Q.10

[Sol. Fe +2

1O

2 FeO

(1 – x)

2x1 (1 – x)

Fe +4

3O

2

2

1Fe

2O

3

x4

3x

2

x

2

x1+ x

4

3= 0.65

x = 5

3

FeO : Fe2O

3= (1 – x) :

2

x

=5

2:10

3

= 4 : 3 ]

Q.12

[Sol. Let x mole HCOOH & y mole H2C

2O

4is taken.

Mole of CO formed = (x + y)mole of CO

2formed = y

Potash absorb CO2

so only CO will remain = (x + y)

Volume contraction = (x + 2y) – (x + y) =6

1( x + 2y)

1

4

y

x ]

Q.13

[Sol. XA moles of solute present in 1 mole of solutionXB

= XA

, XA

= 1 – XA

m = 17

1000

)X1(

X

A

A

m =A

A

X1

)X(82.58

]



Code-A Page # 4

CHEMISTRY

Q.14

[Sol. 3 – x = 0.208 2 / 1)10081( – 115

= 109 – 115208.0 = 25208.0

= 0.208 × 5 = 1.04

x = 3 – 1.04 = 1.96 Ans. ]

Q.15

[Sol. )AHCO(n2

1n

3BOH2

)AHCO(n2

1n

3BCO2

22 COOHmm = 6.2 g

Let M is the molar mass of AHCO3

2.644M

8.16

2

118

M

8.16

2

1 g

M = 842.6

8.1631

2.6

)229(8.16 ]

Q.17

[Sol. (C) M =100

4215.02512.007.010 = 1.0

100

10 M

(D) M =100

15.02512.02007.055

M =100

75.32485.3 = 0.1 M ]

Q.19

[Sol. Ionic mobility sizeHydrated

1]

Q.20

[Sol. Mass fraction =260

10= 0.0385

Mole fraction =

18

250

5.58

105.58 / 10

=89.1317.0

17.0

=06.1417.0 = 0.012

Molality = 1000260

5.58 / 10 = 68.01000

250

17.0

Molarity = 6538.01000260

5.58 / 10 ]



MATHEMATICS

Code-A Page # 1

PART-A

Q.1

[Sol. Let the roots be 2k and 3k.

Now, sum of roots = 5k = 2 – a ......(1)

and product of roots = 6k 2 = a + 2 ......(2)

Now proceed.

Q.2

[Sol. E = (sec 3A + cosec 3A)2 = 22626

Hence, E = 262 = 24. Ans.]

Q.5

[Sol. Given, cos A + cos B + cos C = a .......(1)

and sin A + sin B + sin C = b .......(2)

Square and add, we get the result.]

Q.7

[Sol.

10

1r

1r 3r82 =

10

1r

10

1r

10

1r

1r 13r82 =2

)110()10(8

12

1210

– 3 (10)

= 1024 – 1 + 440 – 30 = 1433. Ans.]

Q.8

[Sol. f (1) < 0

and f (2) < 0

21 f(x)

x-axis]

Q.10

[Sol. Using A.M. G.M. inx

z,

z

y,

y

x]

Q.14

[Sol. Let a1

be the 1st term and d is the common difference.

a7

= 9 = a1

+ 6d (given) a1

= 9 – 6d

a1a

2a

7= a

1(a

1+ d) (a

1+ 6d) = 9(9 – 6d) (9 – 5d) = 27

upwardopeningparabola

2)27d33d10(

Now proceed. ]

Q.15

[Sol. a1

+ a3

+ a5

+ ....... + a19

+ a21

=2

n[2a

1+ (n

1 – 1) 2d] =

2

11[2a

1+ 10 · 2d]

= 11 [a1

+ 10d] = 11 × 33 = 363 Ans.]



MATHEMATICS

Code-A Page # 2

Q.16

[Sol. Given 0 < , , <2

.......(1)

Also, 2cot = cot + cot .......(2)

Now proceed.

Q.17

[Sol. Given, + = –

a, = band 2 + 2 = – A, 2 2 = B

Now, 2 + 2 = ( + )2 – 2 a2 – 2b = – A

So, 2ab2A

Also, 2 2 = B (b)2 = B 2bB . Ans.]

Q.18

[Sol. Let first term and common difference of A.P. be a and d respectively.

Given, a3

+ a9

= 8 (a + 2d) + (a + 8d) = 8 a + 5d = 4 .........(1)

Also, S7 = 14

2

7(2a + 6d) = 14 a + 3d = 2 ..........(2)

From (1) and (2), we get

d = 1 and a = – 1. Now verify alternatives.Ans.]

Q.21

[Sol. Put x = 1, (a + 2b – 3c) + b + 2c – 3a + 2a – 3b = 0

so x= 1, is its one root.

Also, other root = Product of roots = c3b2a

b3a2c

.

Now proceed. ]

07 08 2011 xi pqrs paper i code a

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