rt solutions-09!10!2011 xiii vxy paper i code b

7/30/2019 Rt Solutions-09!10!2011 XIII VXY Paper I Code B

1/16

13th VXY (Date: 09-10-2011) Review Test-3

PAPER-1

Code-B

ANSWER KEY

CHEMISTRY

SECTION-2

PART-A

Q.1 A

Q.2 D

[Only V-Group Batch]

Q.3 B

[Only XY Batch]

Q.3 B

Q.4 D

Q.5 A

Q.6 C

Q.7 C

Q.8 C

Q.9 A

Q.10 A

Q.11 C


Q.12 A

[Only XY Batch]

Q.12 A

Q.13 A

Q.14 A,C,D

Q.15 A,B,C,D


Q.16 A,B,C,D

[Only XY Batch]

Q.16 A,B,C

Q.17 A,B,C,D

PART-C

Q.1 0007

Q.2 0006

Q.3 0102

Q.4 0012 or 0021

MATHS

SECTION-1

PART-A

Q.1 B

Q.2 D

Q.3 C

Q.4 B

Q.5 A

Q.6 A

Q.7 A

Q.8 B

Q.9 C

Q.10 A

Q.11 D

Q.12 B or C

Q.13 B

Q.14 A,B,C

Q.15 A,B,D

Q.16 A,B,D

Q.17 A,B,C,D

PART-C

Q.1 0005

Q.2 0021

Q.3 0010

Q.4 0018

PHYSICS

SECTION-3

PART-A

Q.1 B

Q.2 D

Q.3 B

Q.4 B

Q.5 A

Q.6 C

Q.7 A

Q.8 A

Q.9 D

Q.10 B

Q.11 A

Q.12 B

Q.13 D

Q.14 A,B

Q.15 A,B,D

Q.16 B,D

Q.17 B,C

PART-C

Q.1 0023

Q.2 0005

Q.3 0028

Q.4 0008


2/16

MATHEMATICS

Code-B Page # 1

PART-A

Q.1

[Sol. Given, Tn

=4n

n4

So, Sn

=

1n224

n4)4n4n(

n=

1n222

)n2(2n

n=

1n22

)n22n()n22n(

n

=

1n22

22

)n22n()n22n(

)n22n()n22n(

4

1=

2n2n

1

2n2n

1

4

122

T1

=

5

11

4

1

T2

=

10

1

2

1

4

1

T3 = 171

51

41 ....

and so on

Sum =8

3. Ans.]

Q.2

[Sol. As f is continuous )1x(fLim0x

= f (1) = f (1) = 1

Also, )x(gLim0x

= )x(gLim0x

= 3

Now, verify alternatives. ]

Q.3

[Sol. f '(1+) =h

)1(f)h1(fLim

0h

=h

h

1sinh

Lim0h

=

h

1sinLim

0h= non-existent

f is not differentiable at x = 1.

For x = 0, since (x1) and sin

1x1

both are differentiable at x = 0

(x1) sin

1x1

is differentiable at x = 0. Ans.

Note: f ' (0+) = f ' (0) = cos 1 sin 1. ]


3/16


4/16

MATHEMATICS

Code-B Page # 3

Q.8, .9, 10

[Sol. P(x) = x59x4 + px327x2 + qx + r

P(x) is divisible by x2

q = r = 0P(x) = x59x4 + px327x2

2x

)x(P= x39x2 + px27 = 0

, , R+

+ + = 9, = 27

A.M. of, , =3

= 3

G.M. of , , = ()1/3 = (27)1/3 = 3 A.M. = G.M. = = 3 = 9 = 3 = = p = + + = 27

(i) p + q + r = 27 + 0 + 0 = 27

(ii) 1 = 2, + 3 = 6, + 7 = 10 S

n= 2 + 6 + 10 + ....... to n terms = 2 (1 + 3 + 5 + ...... to n terms) = 2n2

Now,

2n 1nn SS

1=

2n 22 )1n(2n2

1=

2n )1n(n1

2

1=

2n n

1

1n

1

2

1

=

.......3

1

2

1

2

1

1

1

2

1=

2

. Ans.

(iii)

n32n)27(

1........

)27(

1

)27(

1

27

1Lim { q = r = 0}

= 32 )27(

1

)27(

1

27

1 + ....... =

27

11

27

1

=

26

1. Ans.]

Q.11

[Sol. S-1: Given, p, q, r are in A.P.,

So p + r = 2q .........(1)

Also, (q

p), (r

q), p are in G.P.,So, (rq)2 = (qp) p .........(2)

d2 = ap as d 0, where d is the common difference of A.P. p = d q = 2dand r = 3d

p : q : r = 1 : 2 : 3 S

1is false and S-2 is obviously true. Ans.]


5/16

MATHEMATICS

Code-B Page # 4

Q.12

[Sol. Given, ax2ax + 1 > 0 for atleast one x R.If a 0, then clearly ax2ax + 1 > 0 is true for atleast one x R.If a < 0, then graph of y = ax2ax + 1 is parabola opening downward.

But Disc. = a24a > 0 (As < 0)

ax2ax + 1 > 0 is true for atleast one xR then aR.Hence a R. Ans.]

Q.13

[Sol. Clearly S - 1 and S-2 are true but S-2 is not explaining S - 1B is correctNow, for S-2, we have

(gof) (x) = g(f(x)) = 1 xR. Ans. ]

Q.14

[Sol. Let A = B = , C = +

Now, A + B + C =

4

3 3 =

4

=

4

.

Also, sin sin () sin ( + ) =225

12 sin () sin ( + ) =

25

12

2 sin () sin ( + ) =25

24cos 2cos 2 =

25

24

4

,As

So,cos 2 =25

24= cos (CA)

As, cos 2 = 25

24

tan2

=

2cos1

2cos1

=

25

241

25

241

tan = 7

1

.

Now, tan A = tan () =

tantan1

tantan=

3

4

7

11

7

11

or4

3.

4

,As

Also, cos A cos C = cos () cos ( + ) = cos2sin2

= cos2

4

2

2cos1=

2

1

2

1

2

1

25

24=

25

12.

Also,

2

ACtan

2 =)AC(cos1

)AC(cos1

=

25

241

25

241

=

49

1

2

ACcot

2 = 49.

Also, (A + C) =224

3

sin (A + C) + cosec (A + C) = 1 + 1 = 2.

Now verify alternatives. ]


6/16

MATHEMATICS

Code-B Page # 5

Q.15

[Sol.

(A) We know that every continuous function in closed interval [a, b] is bounded. So, Given statement is

explained by extreme value theorem.

(B) Given f(x) = x22x + 3 = (x1)2 + 2

vertex of f(x) = (1, 2)

Clearly, f(x) has neither a maximum nor a minimum in interval (1, 3).

(C) Let f(x) = tan4x + (x1) sin x + 5cos x

Clearly, f(x) is continuous function in [0, 1].

Also, f(0) = 5 and f(1) = 1 + 05 =4 f(0) f(1) < 0So, f(x) = 0 has atleast one root in (0, 1) (Using intermediate value theorem)

(D) Given, 12a + 6b + 4c + 3d = 0

a +2

b+

3

c+

4

d= 0 ........(1)

Now, I =

1

0

32dx)dxcxbxa(

=

1

0

432

3

dx

3

cx

2

bxax

= a +

2

b+

3

c+

4

d= 0 (Using equation (1))

f(x) must have atleast one root in (0, 1). (Using Property of definite integral. )So, the given statement is true. Ans.]

Q.16

[Sol. Given, f '(t) = et (cos2tsin 2t)

Integrate both sides with respect to t, we getf(t) = et cos2t + C

But, f(0) = 1 C = 0So, f(t) = et cos2t.

Clearly,

[As, < t < 0 so 0 < et < 1 and 0 cos2t 1 0 et cos2t 1 f is bounded in (, 0).]Note that f is neither odd nor even function.

Also, f(t) = et et cos2t = et cos2t = 1 t = 0, , 2 [0, 2]. So, 3 solutions exist.

Also, t1

0t

)t(fLim

(1 form) = eL, where L =

t

1tcoseLim

2t

0t

(

0

0form)

=1

)t2sint(coseLim

2t

0t

= et (Using L'hospital rule)]


7/16

MATHEMATICS

Code-B Page # 6

Q.17

[Sol. Let a be the first term and d be the common difference.

Clearly,T5

= a + 4d = 17 .....(1)

Also, S10

=2

10(2a + 9d) ......(2)

As, the sum of the terms in the sequence lies in the interval (180, 190), so

180 0

t2t1 =aLet h(t) = t2t1

Clearly, if f(x) = g(x) has four distinct real roots, so4

5


9/16

CHEMISTRY

Code-B Page # 1

PART-A

Q.2

[Sol. Covalent character Polarisation power Size of cation

Polarisation Size of anion

Polarisation Power of Pseudo inert gas cation (having 18 electrons in valence shell) is Maximum. ]

Q.4

[Sol. (A) At constant P and T

G < 0H TS < 0

H < TS

S isve H isve

(D) P-V work will be zero is free expansion. ]

Q.5

[Sol. AsCHO group isM, the +M group will decrease CC bond length. ]

Q.6

[Sol. ]I[]H[]NO[kdt

dNO

2

1

dt

]H[d

4

1

dt

dNO

2

12

2

dt

dNO2= 2 k

0X = k

1[X]

dt

]H[d

= 4 k0

X = k2

X

dt

dNO= 2k

0X = k

3[X]

K0

=2

k

4

k

2

k 321

k1

=2

k1

= k3

]

Q.7

[Sol. For Hydrocarbons containing same C

HOC stability

1]

Paragraph for question nos. 8 to 10

[Sol (8)SO2Cl

2+ H

2O H

2SO

4(acid) + HCl (acid)

PCl3

+ H2O H

3PO

3(acid) + HCl (acid)

NCl3

+ H2O NH

3(base) + HOCl (acid)

CCl4

+ H2O X (does not hydrolysed at ordinary condition)


10/16

CHEMISTRY

Code-B Page # 2

(9) NCl3

hydrolysed at ordinary temperature

(10) OHHClI11

HOI + HCl ]


Q.12

[Sol. Due to transference of electron from 3d to 4d orbital; Zeff

becomes decreases so unpaired electron is

easily removed & [Co(NH3)6]2+ is easily oxidised to [Co(NH

3)6]3+.

[Co(NH3)6]2+

Co2+ = [Ar]3d 4s 4p 4d

e

Hybridisation is d sp2 3

]

[Only XY Batch]

Q.12

[Sol. Due to the absence of lone pair of electron, acetylene does not form H-bond. ]

Q.14

[Sol. (A) )s(OH,natomizatio 2

H =

HO,BE)s(OH,ationlimsubH2H

2

(C) )s(OH,natomizatio 2

H =2

HHH

OO,BE

HH,BE)s(OH,formation 2

(D) )s(OH,natomizatio 2

H =

)(OH,natomizatio)s(OH,fusion 22HH

l ]


Q.16

[Sol. Due to the -acid behaviour of CO molecule in metal-carbonyls, CO bond length is greater than free

CO molecule. ]

[Only XY Batch]

Q.16

[Sol. Due to the presence of vacant d-orbital in Si and Ge; p-d back bonding is present in (A), (B) and (C)

compound so they are planar. ]


11/16

CHEMISTRY

Code-B Page # 3

PART-C

Q.1 to Q.2 are "Integer Type" questions. (The answer to each of the questions are upto 4 digits) [2 4 = 8]

Q.1

[Sol. 2N2O

5(g) 4NO

2(g) + O

2(g)

t = 0 P0

t = 4 min P02y 4y y

t = 2P0

P0/2

P0 + x = 6502.5 P

0+ x = 1550

1.5 P0

= 900 k =4

6ln

4

1

200600

600ln

4

1

P0

= 600 =2

3ln

4

1=

4

4.0= 0.1

y = 100

t1/2

=1.0

7.0

k

2ln =7 Ans.]

Q.2

[Sol. CCC = C CCC = C CCC = C

CCC = C CCC = CCl

H Cl

Cl Cl

Cl

H

* *+ +

+

(E + Z)

]

Q.3 to Q.4 are "Integer Type" questions. (The answer to each of the questions are upto 4 digits) [2 6 = 12]

Q.3

Sol. C(graphite) + H2O(g) H

2(g) + CO(g)

4 2.5 2.5 fG 10 mole

RfPf)G()G(G

= 130(230)

= 230130 = 100 kJmol1

G = G + 2.303 RT logQ

= 100 + 2.303RT ln

OH

PH

2

CO2

P

P

= 100 + 2.303RT

1010

4

1010

5.2ln

2


12/16

CHEMISTRY

Code-B Page # 4

= 100 + 10ln4

25.6

= 100 + 20 log4

5

= 100 + 20(log 52log2)

= 100 + 20 (0.70.6)

= 100 + 20 0.1

= 100 + 2 = 102 kJ Ans.]

Q.4

[Sol. Due to the presence of vacant d-orbital in si and Ge; (A) and (B) compound shows p-d back

bonding and they are planar about nitrogen.

HPO3

+ H2O H

3PO

4(X)

]O[ H3PO

5(Z')

2 H3PO

4(X)

OH2 H4P

2O

7(Y)

]O[ H4P

2O

8(Z)

P

O

HOOH

OOH

(1 bond)

P P

O OOO

OH OH

(2 bond)

HO OH ]


13/16

PHYSICS

Code-B Page # 1

PART-A

Q.1

[Sol. N250 = 25

mgNmg = ma

N = 275

Reading = 275 + 50 = 325 N ]

Q.2[Sol. ctf = 0

f = ct

when f = mg sliding

ctmg = mdt

dv

Parabolic graph ]

Q.3

[Sol. Tension will be same in both parts.

Tcos = mg + Tcos

mgT

T

> Ans. ]

Q.5

[Sol. VC

= 1 3R

B

2

1VPC

= 31R

2 2R = 0

31

= 22

3 t

2=

t

2

= 3 ]

Q.6

[Sol.f

1

x

1

v

1

f

1

x

1

v

1

x

1

f

1

v

1

(a) v f(b) diverging lens will diverge rays further.(c) v x ]

Q.7

[Sol. amax

=2A35.28 = (4.2)2 A

A =2.42.4

28.35

= 2 m ]


14/16

PHYSICS

Code-B Page # 2

Q.8

[Sol. L = I1

+ I2

=3

)6.0(4 2

7.0

2

7.0

2=

7.0

44.1 =

7

4.14 ]

Q.9

[Sol. L + L = 0

74.14 + 18 X4 = 0

=1807

4.14

=

35

4aC ]

Q.10

[Sol. 0

+ =7.0

20

0

+354 =

14

w0

=14

35

=

70

85

70

3]

Q.14

[Sol. (A) K =2

1v2

rel

(B) amM = am

aM =

F mM

)mM(

m

F

= am aM

(C) 02 = u22d mM

)mM(F F =

)mM(d2

mMu2

]

Q.15

[Sol. + c

= 60A

B C60

30

c

1 sin 30 = sin sin c = 1

c = sin

1

1

sin1

2

1+ sin1

1=

3

=3

7

2

1= sin (60

c)


15/16

PHYSICS

Code-B Page # 3

1 = sin c

2

1=

2

3cot

c

2

1

cot c

=3

2]

Q.16

[Sol.

F = mg(rightwards)

rest

R

P

N

F

Mg

Torque about P = 0

Angular momentumabout P = conserved ]

Q.17

[Sol. K = 0fff3

3

2

2

1

1 ]

PART-C

Q.1

[Sol. 1

=10030

065.0

2

=

10030

035.0

l = l1a

1T + (30 l

1)

2T

0.058 = l1

3000

065.0+ (30l

1)

3000

035.0

1.74 = 0.065 l1

+ 1.050.035l,

0.69 = +0.03l1

l1

= 23 cm ; l2

= 7 cm ]

Q.2

[Sol. F + f = ma

FR

fR = Ia/Racm

f

F

F = ma6

5

M5

F6a ]


16/16

PHYSICS

Code-B Page # 4

Q.3

[Sol. 152 0.5T 0.2 = I

0.2

0.5

F = 152N

T

300

30g 0.8

T300 = 0.8 30

T = 324

76324 0.2 = I0.2 = 0.8 = 411.2 = I 4

I = 2.8 kg-m2. ]

Q.4

[Sol. Let the distance of the lens from the object be l

when a real image is formed on the screen. Then

l100

1

l

1=

23

1

On solving, we get l = (50 10 2 ) cm.

Now, if the lens performs SHM and a real image is formed after a fixed time gap, then this time

gap must be one-fourth of the time period.

P

Screen

50 cm

100 cm

Phase difference between the two positions of real image must be2

. As the two positions are

symmetrically located about the origin, phase difference of any of these positions from origin must be4

.

10 2 cm = A sin 4

A = 20 cm

To achieve this velocity at the mean position,

0= A= A

m

K

Required impulse p = m0

= A m/K = 8 kg m/s. ]

rt solutions-09!10!2011 xiii vxy paper i code b

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