Pcm 11th Paper -1 Code a (Jk&Pqrs) 02-01-2011

8/2/2019 Pcm 11th Paper -1 Code a (Jk&Pqrs) 02-01-2011

1/15

11th PQRS/JK (Date: 02-01-2011) Review Test-7/6

Paper-1

Code-A

ANSWER KEY

MATHS

SECTION-3

PART-A

Q.1 D

Q.2 B

Q.3 C

Q.4 BQ.5 D

Q.6 A

Q.7 B

Q.8 C

Q.9 A

Q.10 A,D

Q.11 C,D

Q.12 A,C

PART-B

Q.1 (A) Q,

(B) S,

(C) T,

(D) P

PART-C

Q.1 0001

Q.2 0003

Q.3 0084

Q.4 0050

Q.5 0008

Q.6 5050

PHYSICS

SECTION-1

PART-A

Q.1 D

Q.2 D

Q.3 B

Q.4 BQ.5 A

Q.6 C

Q.7 C

Q.8 B

Q.9 B

Q.10 B,C,D or B,C

Q.11 A,C,D

Q.12 A,B,C

PART-B

Q.1 (A) P,S

(B) Q,T

(C) R,

(D) R

PART-C

Q.1 0008

Q.2 0002

Q.3 0018

Q.4 0066

Q.5 0001

Q.6 0026

CHEMISTRY

SECTION-2

PART-A

Q.1 B

Q.2 D

Q.3 C

Q.4 C

Q.5 B

Q.6 C

Q.7 C

Q.8 B

Q.9 A

Q.10 C,D

Q.11 A,B,C

Q.12 B,C

PART-B

Q.1 (A)Q,R

(B) P,Q

(C) Q,R,T

(D) P,S or P,Q,S

PART-C

Q.1 0007

Q.2 0003

Q.3 0004

Q.4 0014

Q.5 7642

Q.6 2463


2/15

PHYSICS

Code-A Page # 1

PART-A

Q.1

[Sol.

f1

f2

v

]

Q.3

[Sol. P is black so it will abosrb more energy faster. ]

Q.4

[Sol.F

a

f

r F > f ]

Q.5

[Sol. (PE)max is same(KE)

maxfor 1 > (KE)

maxfor 2

A1

< A2

1

> 2

Because KEmax

=22Am

2

1 and m is same.]

Q.6

[Sol. For no toppling

N = f= f

2

h

]

Q.7

[Sol. Adiabatic process

B = P ]

Q.8

[Sol. Adiabatic process, 1

TP= constant, so by differentiating with respect to y..

dydT =

PT

1

dydP ]

Q.9

[Sol. PV = nRT or P =M

RT

dy

P=

M

Rdy

dT]


3/15

PHYSICS

Code-A Page # 2

Q.10

[Sol. a =2xx = A sin (t + ) phase

KE =2

1m2 (A2x2)

PE =2

1kx2 ]

Q.11

[Sol. (a) y is property of material

(b) Breaking stres depends on material, force radius and length.

(c) Strain depends on material, radius and force.

(d) PE =2

1(strain) (stress) (volume)

So, it depends on force, material, radius and length. ]

Q.12

[Sol. (ML2

T3

) = LX(ML3

)y (ML2

T2

)1 = y + z 2 = x3y2z3 =2z or z = 3/2

and y =1/2 x = 7/2P (linear size)7/2 (density of air)1/2 (g density of helicopter)3/2 ]

PART-B

Q.1

[Sol. (i) m >M, (ii) m


4/15

PHYSICS

Code-A Page # 3

Alternative :

A1

A2 t = 0

12cm

12cm

6cm

= t

t = t

t = 0/36

12

=3

t =3

t =6

T= 2 sec. ]

Q.3

[Sol. mgh =2

1mv2 +

2

1I2

20m

13m

h = 7m

v

H

=2

1m(2g (H13)) +

2

1

5

2m(r22)

g(7) = g(H13) +5

1(v2)

= g(H

13) +51 (2g(H

13))

7 = (H13) +5

2(H13)

7 =5

7(H13)

H = 18 m ]

Q.4

[Sol. T' = T

ag

u2

= 2

g

ag

=g

g (1 + ) = (g + a)Reading = m (g + a) = mg(1 + ) = (60) (10) (1 + 20 104 50) = 600 (1 + 0.1) = 660 NewtonApparent weight = 66 kg ]

Q.5

[Sol. = ma

2

sin

m(a)

(pseudo force)

a =

3

m

sin2

ma

2


5/15

PHYSICS

Code-A Page # 4

=2

a3sin

for slightly open is very smallsin

=

2

a3

2

2

dt

d = (2)

= 0 cos (t)

time taken to shut =4

T=

4

/2 =

2

a32

= a6

distance travelled by car = 0 +2

1at2 =

2

1a p2

a6

=

12

2

=12

)2.1)(10(= 1 m ]

Q.6

[Sol. 3kx1

+ kx2

= ma

3k k

m

xx

x2

xx 21

6kx1

= ma

m

kx3= a

a =3

6763= 26 ]


6/15

CHEMISTRY

Code-A Page # 1

PART-A

Q.1

[Sol. N 1s2 2s2 2p3

O 1s2 2s2 2p4

Half filled configurationmore stable, required more energy to remove the electron as compare to

O-atom ]

Q.2

[Sol.

1

2

3

5

4

6

7

8

6-ethyl-2, 2, 3, 7-tetramethyl octane. ]

Q.3

[Sol. 98 gram 80 gram

H2SO

4+ 2NaOH Na

2SO

4+ 2H

2O

H3PO4 + NaOH Na2HPO4 + H2O98 gram 40 gram

40 gram NaOH reacts with 49 gram H2SO

4& 98 gram H

3PO

4to form sulphate & dihydrogen

phosphate.

wt. ratio

H2SO

4: H

3PO

4

49 98

1 :2 Ans. ]

Q.4

[Sol. N

H

it is not homocyclic since hetero atom is present within the ring. ]

Q.5

[Sol. H O2

H2

T1

= T

P1

= 830

30 = 800 mm

P1

= 2

T2

= 0.9 T

2

2

1

1

T

P

T

P or

T9.0

P

T

800 2

P2

= 720 mm

Ptotal

at 2nd temp. = 720 + 45 = 745 mm Ans. ]


7/15

CHEMISTRY

Code-A Page # 2

Q.6

[Sol.Vem2

h

8

Ve2'm2

h

8

Ve2'm = m e V

m' = 4m ]

Q.7

[Sol.V

150

V = 24 volt Energy of free electron = 24 eV,

E = 13.6 (Z)2

1612

= 1.51 eV

total energy released = 24(1.51) = 25.51 ]

Q.8

[Sol. (n2)

62 = 4 ]

Q.9

[Sol. Min. energy transition for 6

5 ]

Q.10

[Sol. SiF2Cl

2 sp3 not sp3d ]

Q.11

[Sol. Molecular formulaC4H

6

D. B. E. 2

Possibility 2bond

or 1 bond & 1 ring

or 2 ring ]

PART-B

Q.1[Sol. (D) Assuming only translational kinetic energy Ans. P,S

Assuming translational, rotational as well as vibrational K.E. Ans. P,Q,S ]


8/15

CHEMISTRY

Code-A Page # 3

PART-C

Q.1

[Sol. CH3Cl sp3 tetrahedral

C

H

H H

Cl

One according toabove structure

6along the faces

Total = 6 + 1 = 7 ]

Q.2

[Sol. wsol.

= 1140

wB

= 210

wA

+ wsolvent

= 930

10nA

+ OH2

n18 = 930

19nA

+ OH2

n = 7

342nA

OH2n18 = 126 352 n

A= 1056 ; n

A= 3 ]

Q.3

[Sol Zr+3 [Ar] 5s0 4d1

1s2

2s2

3s2

2p6

3p6

4p6

3d10

4d1

5s0

4s2 LHS to the 4d all electrons

contributes one

Z = Z

= 40 36 1 = 4

1

eff

]

Q.4

[Sol. Total number of SS bond = (n1)

n : total number of S-atom

S15O62

S = 15SS bonds are 151 = 14 ]

Q.5

[Sol. ]


9/15

CHEMISTRY

Code-A Page # 4

Q.6

[Sol. PGL PVm

= RT

Vm

=11

K300

P

RT

VG

= R3011

K3001.1 = 30 0.0821

VG 2.463 lit. = 2463 Ans. ]


10/15

MATHEMATICS

Code-A Page # 1

PART-A

Q.1

[Sol. Clearly, min.(2x2ax + 2) and max. (b 1 + 2x x2) will occur at the vertices of the parabolas

y = 2x2ax + 2 and y = b1 + 2xx2 respectively, so that min. (2x2ax + 2) = 28

a2

and

max. (b1 + 2xx2) = b

Hence, 2

8

a2

> b 16a2 > 8b a

2 + 8b

16 < 0 a2

4 (4

2b) < 0.

Now, discriminant of 2x2 + ax + (2b) = 0, is a24 (42b), which is less than zero.

Hence, all roots of the equation 2x2 + ax + 2b = 0 are imaginary. Ans.]

Q.2

[Sol. Let required line be, (y0) = m (x6), m > 0

mxy6m = 0 .....(1)As distance of above line from N(1, 3) is 5, so

2m1

m63m

= 5 (3 + 5m)2 = 25(1+m2) 9 + 30m = 25 m =

30

16=

15

8. Ans.]

Q.3

[Sol. Case-I: ;!2!3!7

!10

Case-II: 0 0 1 1 1 1 1 1 2 2 ; !2!2!2!2!6

!10

Hence total =

!2!2!2!2!6

!10

!2!3!7

!10!10 = 31560!10 = 375 10! k = 375 Ans.]

Q.4

[Sol. We have 12

zz

2

zz

| x | + | y | = 1

Also, |zi| + |z + i| = 2

A line segment between

(0, 1) and (0,

1).So, number of solution is 2

i.e., z = i and i ]

Q.5

[Sol. Clearly, equation of required circle, is(x + 1)2 + (y1)2 + (x + y) = 0

x2 + y2 + ( + 2)x + (2)y + 2 = 0 ..........(1)As circle (1) intersects the circle x2 + y2 + 6x4y + 18 = 0 orthogonally,


11/15

MATHEMATICS

Code-A Page # 2

so using orthogonality condition, we get

2

2

223

2

2= 18 + 2

(3 + 6)(24) = 20 = 10So, putting = 10 in equation, we get

x2 + y2 + 12x + 8y + 2 = 0.

Clearly, radius = 246 22 = 21636 = 50 = 25 Ans.]Q.6

[Sol. We have

Q =

n

0r1r

r

)3cos(

)3sin(=

3cos

sin+

9cos

3sin+

27cos

9sin+ ............... +

)3cos(

)3sin(1n

n

As,

3cos

sin=

3coscos2

cossin2=

3coscos2

2sin

=21

3coscos

3sin =21 (tan 3tan )

Q =2

1[(tan 3tan ) + (tan 9tan 3) + ........... + (tan 3n + 1tan3n)]

Q =2

P P = 2Q. Ans.]

[PARAGRAPH TYPE]

Q.7 to Q.9 has four choices (A), (B), (C), (D) out of which ONLY ONE is correct.

Paragraph for question no. 7 to 9

[Sol. Given C: x22x cos + cos 2 + cos +2

1

(i) x =a2

b=

2

cos2 = cos

As vertex lies on y = x

cos = cos22cos2 + (2cos21) + cos +2

1

0 = cos221 cos2 =

21 =

47,

45,

43,

4 4 values Ans.

(ii) If C touches l then discriminant = 0

x2(2cos + 1)x + 2cos2 + cos 2

1= 0 has equal roots

So, (2cos + 1)2 = 4

2

1coscos2 2

4cos2 + 4cos + 1 = 8cos2 + 4cos 2


12/15

MATHEMATICS

Code-A Page # 3

4cos2 = 3 cos2 =4

3

=6

11,

6

7,

6

5,

6

sum =

6

24= 4Ans.

(iii) If y = x intersects at A (x1, x

1) and B (x

2, x

2) (x , x )

1 1

A

y=

xB(x , x )

2 2

X

Y

O (0, 0)

l

C

(AB)2 = 2 (x1x

2)2

= 2[(x1

+ x2)24x

1x

2] .......(1)

Solving, we get

x22x cos + cos 2 + cos +2

1= x

(2cos + 1)x + 2cos2 + cos 2

1= 0

Now, sum of roots = x1

+ x2

= 2cos + 1

Product of roots = x1x

2= 2cos2 + cos

2

1

Hence (AB)2 = 2

2

1coscos241cos2 22

= 2[4cos2 + 1 + 4cos 8cos24cos + 2] = 2 [34cos2] = 68cos2

So, AB is max. if =2

or

2

3(i.e., cos2 = 0)

(AB)max.

= 6 = L

Hence, 6 = 2 Ans.]

Q.10

[Sol. We have

2xcos

1xcos

2

2

, 1 + tan22y 1, 2 3 + sin 3z 4

So, the only possibility is cos2x +xcos

12 = 2, 1 + tan

22y = 1, 3 + sin 3z = 2.

cosx = 1 x = n, n I

Also, tan 2y = 0 y =2

m, m I and sin 3z =1 z = (4k1)

2

, k I. Ans.]

Q.11

[Sol. We have

x1

+ x2

+ ........ + x100

=2

100(x

1+ x

100) = 1 ............(1)

and x2

+ x4

+ ....... + x100

=2

50(x

1+ d + x

100) = 1 ............(2)

Solving (1) and (2), we get

d =50

3


13/15

MATHEMATICS

Code-A Page # 4

So, x1

=50

149

Also, x100

=25

74

Now, sum of infinite G.P. =

50

31

25

47

=

47

94= 2. Ans.]

Q.12

[Sol. Let the slope of tangent at B be m and it makes 45 with 3x5y = 1, so tan 45 =

5

m31

5

3m

m = 4,

4

1. Ans.]

PART-B

Q.1

[Sol.

(A) C6

can be seated any where in the row. For C5

there are two options one just before C6

and one just

after C6. similarly for C

4also there are 2 options Total 25 = 32 Ans.

(B)

!31C3

6

places3select

6C4 2! = 12030 = 90 Ans.

Alternatively : 6C4 3 2! = 15 3 2 = 90 Ans.

(C) Give one to each of the 6 children. Now 4 remaining can be given as follows:

(i) 1 marble to each of 4 children 6C4= 15 ways

(ii) 2 marbles to each of two children 6C2= 15

(iii) 3 marbles to one and 1 to other of five = 6C1

5C1

= 30

(iv) 1 marbles to two and 2 marbles to 1 child 6C2 4C

1= 60

So, total ways = 120 Ans.

(D) Select 3 out of six in 6C3ways for one column and 3 rejected children on the other column can be

arranged only in 1 way

Hence number of ways 6C3 1 = 20 ways Ans.]


14/15

MATHEMATICS

Code-A Page # 5

PART-C

Q.1

[Sol. Chord of contact with mid-points (h, k) is hx + ky = h2 + k2 ...(1) Also chord of contact

of (x1, y

1) is xx

1+ yy

1= 2 ...(2)

On comparing (1) & (2), we get

x1

=22 kh

h2

, y

1=

22 kh

k2

As (x1, y1) lies on 3x + 4y = 10 6h + 8k = 10(h2

+ k2

),

x2 + y2 y5

4x

5

3 = 0, which is a circle with centre P

10

4,

10

3

Clearly, distance of M

10

2,

10

11from P

10

4,

10

3

=

10

2

10

4

10

3

10

112

=100

36

100

64 = 1. Ans.]

Q.2

[Sol. Let cos = t, so we get 2t16logtlogt4

2

2

2

t

4log

t16logtlog

2

22

2

2

tlog2

tlog4tlog

2

22

2

Let log2

t = z, so z2 +

z2

z4= 2 z(z22z3) = 0 z = 0,1, 3.

t = 1,21 , 8,

So, cos = 1 or cos =2

1 = 2n

3

or = 2m, n, m I.

Hence, number of solutions are 3

0,

3

5,

3,e.i . Ans.]

Q.3

[Sol. Let a2

= x2

+ 6, a3

= x3

+ 2, a4

= x4

+ 1.

So, a1 + a2 + a3 + a4 = 15 a1 + x2 + x3 + x4 = 9, where a1, x2, x3, x4 0So, using begger's method, we get number of solutions = 9C

3= 84. Ans.]

Q.4

[Sol. If a, x, y, z, b are in A.P., then b = 5th term = a + 4d (d = common difference)

d =4

ab

xyz = (a + d) (a + 2d) (a + 3d) = 55 (given)


15/15

MATHEMATICS

Code-A Page # 6

4

b3a

4

b2a2

4

a3b= 55

(a + 3b) (a + b) (3a + b) = 55 32 .......(1)Again, if a, x, y, z, b are in H.P., then common difference of the associated A.P. is

a

1

b

1

4

1

i.e.,

ab4

ba

x

1

=

ab4

ba

a

1

x = b3a

ab4

y

1=

ab4

ba2

a

1 y =

b2a2

ab4

=

ba

ab2

andz

1=

ab4

ba3

a

1 z =

ba3

ab4

so, xyz =

ba3

ab4

ba

ab2

b3a

ab4=

55

343(given)

3255ba32

33

=

55

343 a3 b3 = 343 ab = 7 a = 7, b = 1 or a = 1, b = 7

So, sum = (a2 + b2) = 50 Ans.]

Q.5

[Sol. AM = MB

Now, APB = 90

AM = PM = MB = 22 )2k()1h( P(1,2)

C(3,4)B

M(h,k)

A

90

In CMB, CB2 = CM2 + MB2

36 = (h3)2 + (k4)2 + (h1)2 + (k2)2

36 = 2h2 + 2k2

8h

12k + 30 Locus of (h, k) is

x2 + y24x6y3 = 0

So, (a + b + c) = 2 + 3 + 3 = 8. Ans.]

Q.6

[Sol. We have 11x2x2

Case I: If x 2

1 | 2x(2x1)| = 1, which is true given equation is satisfied x

2

1.

Case-II : If x < 2

1

)1x2(x2 = 1 | 4x1| = 1 4x1 = 1 x = 0, 2

1

x = 0.

2

1xReject

So, solution set is x {0}

,

2

1.

Hence, sum = 1 + 2 + .......... + 100 =2

101100= 50 101 = 5050. Ans.]

Documents

Pcm 11th Paper -1 Code a (Jk&Pqrs) 02-01-2011