Rt Solutions-18!12!2011 XIII VXY Paper I Code A

7/30/2019 Rt Solutions-18!12!2011 XIII VXY Paper I Code A

1/17

13th VXY (Date: 18-12-2011) Review Test-5

PAPER-1

Code-A

ANSWER KEY

CHEMISTRY

SECTION-2

PART-A

Q.1 D

Q.2 D

Q.3 C

Q.4 A

Q.5 A

[Only V-Group Batch]

Q.6 C

[Only XY Batch]

Q.6 A

Q.7 C

Q.8 B

Q.9 A

Q.10 A,B,C,D

Q.11 A,C,D


Q.12 A,B,C,D

[Only XY Batch]

Q.12 A,D

PART-B

(Only V-Group Batch)

Q.1 (A) P,Q,R,S,T

(B) P,Q,R,S

(C) P,S

(D) P,R,S,T

(Only XY Batch)

Q.1 (A) Q,T

(B) P,T

(C) P,S

(D) R,S

PART-C

Q.1 3006


Q.2 2345

[Only XY Batch]

Q.2 9890

Q.3 3049

Q.4 0004

PHYSICS

SECTION-1

PART-A

Q.1 C

Q.2 B

Q.3 D

Q.4 A

Q.5 B

Q.6 A

Q.7 D

Q.8 BQ.9 A

Q.10 A,B

Q.11 C

Q.12 B,C,D

PART-B

Q.1 (A) P,Q

(B) P,S

(C) P,S

(D) R

PART-C

Q.1 0500

Q.2 0006

Q.3 0001

Q.4 0005

MATHS

SECTION-3

PART-A

Q.1 D

Q.2 B

Q.3 B

Q.4 D

Q.5 D

Q.6 B

Q.7 B

Q.8 DQ.9 D

Q.10 B,C,D

Q.11 B,C

Q.12 B,D

PART-B

Q.1 (A) S

(B) P,Q

(C) P

(D) Q

PART-C

Q.1 0002

Q.2 0006

Q.3 0002

Q.4 0010


2/17

PHYSICS

Code-A Page # 1

PART-A

Q.1

[Sol. P =R

q2

R =A

P

R8

q'P

2

R' =4A/

)2( l= 8R ]

Q.2

[Sol. T2 =GM

4 2r3

G =sR

3

4T

4

32

2

R

3= 2

T

3

]

Q.3

[Sol. Let T be final temp.

V1S(TT

1) + V

2S(TT

2) = 0

T =21

2211

VV

VTVT

]

Q.4

[Sol. msdt

dT=A (T4T

04)

dtdT

m1

Lesser mass cools faster and hollow sphere is having less mass. ]

Q.5

[Sol. PV = nRTV

n=

RT

P

Since temperature in all are same.

One having more pressure has heightest particle density. ]

Q.6[Sol. P T3

P = k3

nR

PV

P2V3 = const. PV3/2 = const. Y = 3/2 ]


3/17

PHYSICS

Code-A Page # 2

Paragraph for question nos. 7 to 9

[Sol. Q.No. 7 & Q.No. 8

i

R

27 iR

)27R(i = ..... (1)

R

3

ni

(ni)i

i

iR = 3(n1)i R = 3(n1) ...... (2)

From (1) and (2)

(R + 27) = R

13

R R = 9 and n = 4 ]

Q.9

[Sol.

378V

40A

8 A

9 27 330

32A 40A 10A

9

9]

Q.10

[Sol. g =

Rrr

GM

RrR

GMr

2

3

(A) r, r2

< R 2

1

FF

=2

1

rr

(B) r, r2

> R 2

1

F

F= 2

1

22

r

r]

Q.11

[Sol.

3F 5F

30V

56.25+q1

56.25+q

30V

7F

q = 56.25 Ci

q

q2

3

q25.56

7

q12

..... (1)


4/17

PHYSICS

Code-A Page # 3

5

qq25.56

7

q212

= 30 ..... (2)

q2

= 70 ; q1

=26.25

q = 43.75 ]

Q.12

[Sol.

A

10V 2.510

i = 4A

Maximum

A

10V 2.5

i = 2A

Minimum

]

PART-B

Q.1

[Sol. (A) .const

P

T1y

y

P

T

y

1y

TP

= const. R

(B) E = 2/322)Rx(

kQx

E

x

(C)U

VR.O.R.I.

R.O.V.I.

(D) Speed decreases as infinity does not exist. ]

PART-C

Q.1

[Sol. E =y2 0

sin =

y2 0

22 hy

h

Y

Z

h

h(y,0)

=h8

0

y = h ]


5/17

PHYSICS

Code-A Page # 4

Q.2

[Sol. T = (lx)g ...... (1)T = xg ...... (2)

l

x=

1

]

Q.3

[Sol. More the resistance less will be power dissipated.

One should connecte the battery is 6 resistor.

2

11V

]

Q.4

[Sol.

Kd

Id2

2

2m12

1 (2) =K

K =12

1ml2

2

T

2

= 5 ]


6/17

CHEMISTRY

Code-A Page # 1

PART-A

Q.1

[Sol. 2C (graphite) + O2

(g) 2CO (g) rH =110.5 2

CO2(g) C(graphite) + O

2(g)

rH = 393.5

C(diamond) C (graphite) rH = 2

__________________________________________________

CO2

(g) + C(diamond) 2CO(g) rH = + 170.5 kJ/mol Ans. ]

Q.2

[Sol. CH2

= CHCH2I

)excess(HI

II||CHCHCH 23 2I

CH3CH=CH

2

HI

I|

CHCHCH 33 ]

Q.3

[Sol. A : Cr3+ in [Cr(H2O)

6]3+

Hyb : d sp

= 3.89 B.M.

2 3

eff

3d 4s 4p

B : Cu+ in [Cu(CN)4]3

Hyb : sp

= 0

3

eff

3d 4s 4p

C : V+++ in [V (H2O)

6]3+

Hyb : d sp

= 2.83 B.M.

2 3

eff

3d 4s 4p

D : Co++

in [Co(H2O)6]++

Hyb : sp d= 3.89 B.M.

3 2

eff

3d 4s 4p

]


7/17

CHEMISTRY

Code-A Page # 2

Q.4

[Sol. A CO2 5600 counts 2800 1400

B CO2 1400 counts

2 half lives. ]

Q.5

[Sol. NO2

will showI, butOMe will show +M.

NO2 NO2 OMeNO2

NO2

CH CO

CH

CH

O O

O

O

+ OH +OH

CH OH2

]


Q.6

[Sol. H3PO2 H3PO4 + PH3

H3PO

3

H3PO

4+ PH

3

H3PO

4

P4O

10+ H

2O

HNO2

NO + HNO3

H3PO

4doesn't undergoes disproportionation ]

[Only XY Batch]

Q.6

[Sol. (A) CCl ClCl

H

(B)

C

C

O OH

Cl ClCl

H H

Chloral (CCl3CHO) Chloral hydrate CCl

3CHO . H

2O

(C)

H

H

O

O

C (D)

HO

Cl

Ortho hydroxy benzaldehyde ortho chlorophenol ]


8/17

CHEMISTRY

Code-A Page # 3


[Sol.(7) M = 0.1

M =2.11

strengthvolume

0.1 =2.11

V volume strength = 1.12 Ans.

(8) [Sol. H2O

2+ 2KI I

2

nf= 2 n

f=1 n

f=2

I2

+ hypo 2I

nf= 2 n

f= 1

Eq. H2O

2= Eq. of I

2= Eq. Hypo .....(1)

Eq. Hypo = 200 0.1 = 20

KIO3

+ KI I2n

f= 5 n

f= 1 n

f= 5/3

5 mm of KIO3

=3

5 mm of I

2....(2)

from equation (1) 2 mm of I2 = 20mm of I

2= 10

mm of KIO3

=3

52=

3

10Ans.

(9) N1V

1= N

2V

2

V 0.1 2 = 0.1 200

V = 100 ml ]

Q.10

[Sol. (A)

H

exp

ring

ringexp.

OH

OH2

(B) H

OH2

OH

(C) H

then same as (A)

(D) H

OH2

OH

]


9/17

CHEMISTRY

Code-A Page # 4

Q.11

[Sol. (A) Due to high value of Zeff

in Ir+3, which belongs to IIIrd transition series (P < 0)

(B) Due to higher oxidation state of Cobalt in3

62

III

])OH(Co[ , P < 0, also H2O lies towards

stronger ligand side in spectrochemical series.

(C) CO is a strong field ligand hence P <

(D) Due to high value of Zeff

of Pd2+ , which belongs to 2nd transition series , (P < ) ]


Q.12

[Sol.

N N

+ N=NNH

NHN=N

Me

Me

H O3+

OH OH NH2

NH2

Me Me

+ + +H O3

+

]

[Only XY Batch]

Q.13

[Sol.

NH2

2HNO

N N

2N

H

~H

ringexp.

H O2

HO

Possible products

OH

and

HO

]


10/17

CHEMISTRY

Code-A Page # 5

PART-C


Q.2

[Sol. (a)

Me Me

CMe3 CMe3

+ E

E

(2)

(b)

NO2 NO2

+ E

E

(3)

(c)

NHCMe NHCMe

O O

+ E

E

(4)

(d)

O O

O O

C CMe MeMe Me

+ E

E

(5) ]

[Only XY Batch]

Q.2

[Sol. (a)

O||

CHCCH 33 +

OD

Me

H CH CH2

OHCross aldol

O||

CHCCH 23

OD

Me

H CH

C

H+


11/17

CHEMISTRY

Code-A Page # 6

Products:

OCH

|||CHCCHCCH

|OH

3

323 (1) H

D O

Me

CH2CHCH CCH2 3

OH

*(2)

*CH

C

CH3

CH3 CHO

OH Me

D

H (2) H H

D D

Me Me

CH2

CH

CH

OH

**

CHO

(4)

Total 9

(b) 8

(c) Other than first all are diastereomers therefore 9

(d) No enantiomers are formed therefore 0

9890 ]

Q.3

[Sol. t1/2 = 30 minin 90 min, total 3 half lives

Volume of H2O2 consumed = 28 + 2

28+

2

14= 49 ml Ans. ]

Q.4

[Sol. C5H

10O

5

ROH + CH CCl3

O O

ROCCH3

MW1 + 43

MW + 42

C5H

10O

5MW = 150

Remaining MW = 318150

= 168

4 42 4OH

Structure

CHO

CH OH2

OH

OH

OH

H

H

H ]


12/17

MATHEMATICS

Code-A Page # 1

PART-A

1. Using L Hospital's rule, we get

L =4

1

x

xsin4

1

0xLimit

)xsin2(2

xsinx

0xLimit

x2cos22

xsinx

0xLimit

2

]

2. The equation of plane containing A, B and C is

212

011 1z1y1x

= 0

A(1,1,1)

M( , 2, 1)

C(3,0,1)

B(2,2,1)

P(x,yz)

2x + 2y3z + 3 = 0 ........(1)Clearly,M (, 2, 1) satisfy equation (1), so

2 + 43 + 3 = 0 = 2. Ans.]

3. Given x33x =c

Let y = x33x and y =c

Now,dx

dy= 3x23

1 1O

y =

c

2

2

x

y(1,2)

(1, 2)

x = 1 or1 (Graph of y = x33x is as shown)Clearly, for 3 distinct roots, we have

c (2, 2) number of integers = {1, 0, 1} 3 Ans.

4. For real roots D 0k24(k2 + 2k4) 0 3k28k + 16 0 3k2 + 8k16 0 3k2 + 12k4k16 0

3k (k + 4)4(k + 4) 0 k

3

4,4 .

Also, 2 + 2 = ( + )22= k22(k2 + 2k4) = k24k + 8 = 12 (k + 2)2Maximum value is 12 when k = 2. Ans.]

5. 1c

z

b

y

a

x2

2

2

2

2

2

.....(1)

1c

z

b

y

a

x2

2

2

2

2

2

......(2)

1c

z

b

y

a

x2

2

2

2

2

2

.....(3)

(1) + (2) + (3) 3c

z

b

y

a

x2

2

2

2

2

2

.....(4)


13/17

MATHEMATICS

Code-A Page # 2

From (1) and (4), 1c

z2

2

z = c

Hence, x = a, y = b, z = c.

So, number of solution (x, y, z) = 2 2 2 = 8. Ans.

6. I =

6

0dxxcos

xsin1

=

6

022 dx

2

xsin2

xcos

2

xsin

2

xcos

=

6

0

2

xsin

2

xcos

dx

=

6

0

42xcos2

dx

=

6

0

dx42

xsec

2

1=

6

02

x

4tan

2

x

4secn2

l

=

12

32n2 l = 3468n2 l . Ans.]


Sol. We have

L1

:1

4z

1

7y

3

6x

= r

1(say) ........(1)

and L2

:4

2z

2

9y

3

x

= r

2(say) .......(2)

Clearly, any point P on line L1

is (3r1

+ 6, r1

+ 7, r1

+ 4) and any point Q on line L1

is

(3r2, 2r

29, 4r

2+ 2).

So, direction ratios of PQ are 2rr4,16rr2,6r3r3 121212 .The points on the lines (1) and (2) which are nearest to each other will lie on line of shortest distance

which will be perpendicular to both the given lines. If PQ is the line of shortest distance, then

3(3r23r

16)1(2r

2+ r

116) + 1(4r

2r

12) = 0

i.e., 7r211r

14 = 0 ........(3)

and 3(3r23r

16) + 2(2r

2+ r

116) + 4(4r

2r

12) = 0

i.e., 29r2

+ 7r122 = 0 .......(4)

On solving ((3) and (4) ,We get r

1=1, r

2= 1

Point P(3, 8, 3) and Q(3,7, 6)

7. The distance between P and Q = 222 368733 = 922536 = 303 .

8. Equaton of plane OPQ, is

673

383

0z0y0x

= 0

P(3,8,3) Q(3,7,6)

O(0, 0, 0)

P(x, y, z)

23x9y + z = 0 Ans.


14/17

MATHEMATICS

Code-A Page # 3

9. The value of parallelopiped formed by OQ,OP and OR where O is origin and R (1,1, 0)

= |011

673

383

||OROQOP|

= | 3(0 + 6)8(06) + 3(3 + 7) | = 304818 = 96. Ans.]

10. We have

(BTAB)T = BTAT (BT)T = BTATB = BTAB, iff A is symmetric.

BTAB is symmetric iff A is symmetric.Also, (BTAB)T = BTATB = BTAB, iff A is skew-symmetric matrix. ]

11.

2 O(0,0)

x=2x

y

9

4,2A

y=1

+

x=1

B(2,4)

+

Graph of f(x) =3)1x(

)4x( 3

Now, verify alternatives. Ans.]

12. Let the line be

1b

y

a

x , a, b > 0

where 1b

4

a

1

a

11

b

4

a

1a

b

4

b =1a

a4

Now, 2A = ab = a

1a

a4

= 1a

)11a(4 2

=

1a

1

)1a(4

da

dA2 =

2)1a(

114 = 0

B(0,8)

A(2,0)Ox

y

a1 = 1 or1

a = 2; b = 8, m =4


15/17

MATHEMATICS

Code-A Page # 4

A]min

=2

16= 8

Equation is (y4) = 4(x1) y4 = 4x + 4 4x + y = 8 Ans.

Also, r =s

=

2

6882

822

1

=

175

8

= 175

Now verify alternatives. ]

PART-B

1.

(A) Sn

= 33.

2

n[2 + (n1) (n1)] = 33n (n1)2 = 64

Hence n = 9. Ans.]

(B) Domain of f(x) is x [1, 1]. Also f(x) is decreasing in its domain.

At x =1 fmax

(x =1) = cos1(1) + cot1(1) = +4

3 =4

7

At x = 1 fmin.

(x =1) = cos1(1) + cot1(1) = 0 +4

=

4

4

f(x)

4

7 0.8 f(x) 5.25 (approx.)

(C) As 1 + sin 0,R so the given equation becomes cos = 1 + sin cos sin = 1.

Clearly, = 0, 2,2

3 . Ans.

(D) D = 1 + 4a = (Odd No)2

because of integral roots.

let odd no = 2 + 1 then 1 + 4a = (2 + 1)2 = 42 + 4 + 1a = (+ 1) ; so a = 6, 12, 20, 30.Hence number of possible values of a are four. Ans.]

PART-C

1. Clearly, g'(x) = f(x) =

2x1,x

1x3,)2x(

3

2

3

g'(x) is a continuous function. ++1 0

Sign scheme of g''(x)


16/17

MATHEMATICS

Code-A Page # 5

g'(x) has a local maximum at x =1 and local minimum at x = 0.

x=2x= 3 (2,0) x= 1 O(0,0)

Graph of g'(x) = f(x)

x

y

Hence, g'(x) has two extremum points. ]

2. We have

z2y0x4

z0yx2

z3y0x3

=

y34

z1

y28

Now, on comparing we get system of equations as3x + 3z = 8 + 2y 3x2y + 3z = 8 ........(1)2x + y = 1 + z 2x + yz = 1 ........(2)4x + 2z = 4 + 3y 4x3y + 2z = 4 .........(3)

On solving (1), (2) and (3), we getx = 1, y = 2, z = 3

Hence, (x + y + z) = 1 + 2 + 3 = 6. Ans.]

3. Given, f (x) = 2x3 + 3(13a)x2 + 6(a2a)x + b

f ' (x) = 6[x2 + (13a)x + a(a1)] .......(1)As, f (x) has positive point of local maximum, so the equation f ' (x) = 0 must have both roots positive

and distinct roots.

D > 0 (13a)2 > 4a(a1) 16a + 9a2 > 4a24a 5a22a + 1 > 0, true a R. ......(2)

Also, sum of roots > 0 (13a) > 0 3a > 1 a >3

1......(3)

and, product of roots > 0 a(a1) > 0 a < 0 or a > 1 .......(4)Hence, (2) (3) (4)

a (1, )So, the smallest integral value of 'a' equals 2.]


17/17

MATHEMATICS

Code-A Page # 6

4. Given, 7 sin 3x2 (3 sin 3x4 sin3 3x) = (1 + tan2) + 4 (1 + cot2)Put, sin 3x = y, we get

y (8y2 + 1) = 9 + (tan 2 cot )2

L.H.S. min occurs minimum value

when y =1 and minimum value =9. = 9

maximum occurs when y = 1 and maximum value = 9 when tan2= 2Hence maximum of L.H.S = minimum value of R.H.S.

sin 3x = 1 = sin2

x =

6

(minimum positive root).

or sin 3x =

2

3sin x =

2

(maximum negative root).

Sum =3

2

6

3

26

.

Hence,

15 )rootnegative(maximumroot)positive(minimum =

15

3

2= 10 Ans.]

Documents

Rt Solutions-18!12!2011 XIII VXY Paper I Code A