pcm 11th paper -1 code a (jk&pqrs) 02-01-2011
TRANSCRIPT
-
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
-
8/2/2019 Pcm 11th Paper -1 Code a (Jk&Pqrs) 02-01-2011
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]
-
8/2/2019 Pcm 11th Paper -1 Code a (Jk&Pqrs) 02-01-2011
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
-
8/2/2019 Pcm 11th Paper -1 Code a (Jk&Pqrs) 02-01-2011
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
-
8/2/2019 Pcm 11th Paper -1 Code a (Jk&Pqrs) 02-01-2011
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 ]
-
8/2/2019 Pcm 11th Paper -1 Code a (Jk&Pqrs) 02-01-2011
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. ]
-
8/2/2019 Pcm 11th Paper -1 Code a (Jk&Pqrs) 02-01-2011
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 ]
-
8/2/2019 Pcm 11th Paper -1 Code a (Jk&Pqrs) 02-01-2011
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. ]
-
8/2/2019 Pcm 11th Paper -1 Code a (Jk&Pqrs) 02-01-2011
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. ]
-
8/2/2019 Pcm 11th Paper -1 Code a (Jk&Pqrs) 02-01-2011
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,
-
8/2/2019 Pcm 11th Paper -1 Code a (Jk&Pqrs) 02-01-2011
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
-
8/2/2019 Pcm 11th Paper -1 Code a (Jk&Pqrs) 02-01-2011
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
-
8/2/2019 Pcm 11th Paper -1 Code a (Jk&Pqrs) 02-01-2011
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.]
-
8/2/2019 Pcm 11th Paper -1 Code a (Jk&Pqrs) 02-01-2011
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)
-
8/2/2019 Pcm 11th Paper -1 Code a (Jk&Pqrs) 02-01-2011
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.]