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PAPER-II
MODULE TEST-3 : XI SYNCHRO
Time : 3 hrs.
Date : 07/01/2012
(IIT-JEE PATTERN)
M.M.: 270
PBG
S
-1
TESTET
CODE - A
ANSWERS
Physics: Section I and II
1
(c)
2.
(a)
3.
(b)
4.
(b)
5.
(c)
6.
(a)
7.
(d)
8.
(b)
9.
(b)
10.
(c)
11.
(c)
12.
(c)
13.
(b)
14.
(a)
15.
(c)
AMITY
Chemistry: Section I and II
16.
(a)
17.
(d)
18.
(c)
19.
(d)
20.
(b)
21.
(d)
22.
(c)
23.
(b)
24.
(b)
25.
(b)
26.
(b)
27.
(a)
28.
(b)
29.
(a)
30.
(c)
Mathematics : Section I and II
31. (d)
32. (b)
33. (a)
34. (d)
35. (d)
Institute
for
Competitive
Examinations
36. (b)
37. (a)
38. (d)
39. (c)
40. (d)
41.
(a)
42.
(c)
43.
(c)
44.
(c)
45.
(a)
Section-III (PCM)
1
(3)
2.
(2)
3.
(5)
4.
(4)
5.
(1)
6.
(9)
7.
(1)
8.
(1)
9.
(2)
10.
(5)
11.
(0)
12.
(0)
2.
A-(r), (s); B-(p), (s); C-(q); D-(p), (q)
Section-IV (PCM)
Physics
1.
A-(r); B-(q); C-(p); D-(s)
3.
A-(s), (t); B-(r); C-(s); D-(p), (q)
Chemistry
4.
A-(q); B-(r); C-(s); D-(p)
5.
A-(q); B-(p); C-(p); D-(p)
6.
A-(p), (q), (s); B-(p), (q); C-(q); D-(r),(s)
Mathematics
7.
A-(q); B-(r); C-(p); D-(s)
9.
A-(r); B-(p); C-(p); D-(q)
8.
A-(r); B-(p); C-(p); D-(p)
[15]
Module Test-3-II (07-01-12) XI Synchro (TC-A)
MODULE TEST-3 : XI SYNCHRO
HINT/SOLUTIONS
PHYSICS
SECTION- I
STRAIGHT OBJECTIVE TYPE
This section contains 10 multiple choice questions numbered 1 to 10. Each question has 4 choice (A), (B),
(C) and (D), out of which ONLY-ONE is correct.
1.
A light cylindrical vessel is kept on a horizontal surface. Its base area is A. A hole of cross sectional area a is made
just at its bottom side. The minimum coefficient of friction necessary to prevent sliding of the vessel due to the
reaction force of the emerging liquid is (a<< A)
(a)
varying
(b) a/A
(c) 2a/A
(d) None of these
Hint: (c) The velocity of efflux of the liquid is given as
v=
2gy

The reaction force of the emerging liquid on the (vessel + liquid content) is equal to
AMITY
F=v

dm
= vav = av2
dt
F = a ( 2gy)2 = 2agy
y
The force of friction f = F

µN = 2agy

µ=
v
µ(Agy) = 2a gy
N
2a
A
Institute
for Competitive Examinations
Hence (c) is correct.
2.
A cube of mass m and density D is suspended from the point P by a
spring of stiffness k. The system is kept inside a beaker filled with a
liquid of density d. The elongation in the spring, assuming D > d, is:
Hint.
(a)
mg 
d
1 

k  D
(c)
mg 
d
1+
k  D 
(b)
mg  D 
1 
k 
d
m
(d) none of these.
(a) The cube is in equilibrium under the following three forces,
(i)
spring force kx, where x = elongation of the spring,
(ii) gravitational force w; weight of the cube = mg
(iii) buoyant force Fb (or upward thrust) imparted by the liquid on the cube given as Fb = Vdg, where V =
volume of the immersed portion of the cube.
For equilibrium of the cube,
x =
kx
kx + Fb = mg
mg  Fb mg  Vdg

; where V = (m/D)
k
k
mg 
d
x = k  1  D 


m
Fb
mg
Hence (a) is correct.
[16]
Module Test-3-II (07-01-12) XI Synchro (TC-A)
3.
A small block having density less than that of liquid is suspended inside the
liquid by a string as shown in the figure. The tension in the string is 20 N
when the container is at rest. If container start accelerating up with acceleration
a = 2ms–2
2 m/s2 the tension in the string will be. (g = 10 m/s2)
(a)
30 N
(b) 24 N
(c)
35 N
(d) 22 N
Hint: (b)

a
T  To  1  
g

4.
The curve in figure represents potential energy (U) in between two atoms in a diatomic molecules as a function of
U
distance x between atoms. The atoms are:
(a)
attracted when x lies between A and B and repelled when x lies between B and C
A
(b) attracted when x lies between B and C and repelled
when x lies between A and B
B
C
AMITY
x
(c) attracted when they reach B
(d) repelled when they reach B
Hint: (b)
F 
du
dx
Institute for Competitive Examinations
= – (slope of potential energy curve)
Two wires of the same material (Young’s modulus Y) and same length L but radii R
L, 2R
5.
and 2R respectively are joined end to end and a weight w is suspended from the
combination as shown in the figure. The elastic potential energy in the system is
(a)
3w2 L
4R 2Y
(b)
3w2 L
8R 2Y
(c)
5w2 L
8R 2Y
(d)
w2 L
R 2Y
L, R
w
Hint: (c)
Elongation in the wires is given by
F = K1y1
F = K2y2
Y1 A1
K1 = L
1
K2 
Y2 A2
L2
Net energy store d is
1
1
K1 y12  K 2 y22
2
2
[17]
Module Test-3-II (07-01-12) XI Synchro (TC-A)
6.
In a steady flow of fluid the velocity
(a) at each point of space remains constant with time
(b) varies from point to point with time
(c) remains constant only with time but not in space
(d) varies in both space and time
Hint: (a) Conceptual.
7.
A satellite of mass m revolves around the earth of radius R at a height x from its surface. If g is the acceleration due
to gravity on the surface of the earth, the orbital speed of the satellite is
(a) gx
Hint: (d) V0 =
8.
(b)
gR
Rx
(c)
gR 2
Rx
1/2
(d)
 gR 2 


R x
GM
, GM = gR2
xR
AMITY
Consider a big container filled with air is in a space where gravity is absent. An apple A and a balloons filled with
helium one inside the container as shown in figure. The container is accelerated towards right. (density of helium
is less than density of air)
(a)
apple moves towards right and balloon towards left with respect to container
(b) apple moves towards left and balloon towards right with respect
to container
both moves towards
right with
respect to container
Institute
for
Competitive
Examinations
A
B
(c)
a
(d) both moves towards left with respect to container
Hint: (b) Conceptual
9.
The rms speed of oxygen at room temperature is about 500 m/s. The rms speed of hydrogen at same temperature
is about
(a)
1100 m/s
Hint: (b) Vrms =
10.
(b) 2000 m/s
(c) 1000 m/s
(d) 1500 ms
3RT
M
A spherical liquid drop of radius R is divided into eight equal droplets. If the surface tension is T, then the work
done in this process will be
(a)
2R2 T
(b) 3R2 T
(c) 4R2 T
(d) 2RT 2
Hint: (c) work done = T × change in surface area
SECTION- II
LINKED COMPREHENSION TYPE
This section contains 2 Paragraphs. Based upon each paragraph, 3 and 2 multiple choice questions have
to be answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.
[18]
Module Test-3-II (07-01-12) XI Synchro (TC-A)
Comprehension-I
A viscous liquid of coefficient of viscosity ‘’ is flowing through a tube cylindrical tube. The velocity at a
distance ‘r’ from the axis is given by V  Vo 1 

r  where R is the radius of cross section of the cylindrical

R
tube.
11.
Volume of the liquid coming out of the cross-section of tube
(a)
2
VoR
Vo R 2
2
(b)
Vo R 2
3
(d) none of these
(c) 2lV0
(d) none of these
(c)
Hint: (c) Volume coming out per second is area × velocity.
12.
Friction force applied by liquid on the tube
(a)
4lV0
(b) lV0
Hint: (c) Viscous force = |A
13.
dv
|
dr
AMITY
Kinetic energy of the liquid flowing the tube
(a)
R 2V02l
6
(b)
1
R 2V02l
12
(c)
1 2 2
R V0 l
3
(d) none of these
Hint: (b)
Institute for Competitive
Examinations
dr
r
2
r R
1
r

K   (2r dr ( lV02 1    
2
 R
r 0
Comprehension-II
O
A uniform metal sphere of radius a and mass M is surrounded by a thin
uniform spherical shell of equal mass and radius 4a. The centre of the
shell C fall on the surface of the inner sphere, the centre of sphere is O.
14.
C a
4a
P1
Find the gravitational field intensity at point P1
(a)
GM
16a 2
(b)
GM
8a 2
(c)
P2
GM
2a 2
(d)
a
a
GM
4a 2
Hint: (a) Field due to a shell at interior point is zero and for exterior points it behaves as point mass.
15.
Find the gravitational field intensity at point P2
(a)
21GM
900a 2
(b)
61GM
450a 2
(c)
61GM
900a 2
(d)
61GM
1800a 2
Hint: (c) Field due to a shell at interior point is zero and for exterior points it behaves as point mass.
[19]
Module Test-3-II (07-01-12) XI Synchro (TC-A)
CHEMISTRY
STRAIGHT OBJECTIVE TYPE
This section contains 10 multiple choice questions numbered 16 to 25. Each question has 4 choice (A), (B),
(C) and (D), out of which ONLY-ONE is correct
16.
In CO32 & H 2 CO3 , the resonance stabilisation would be
(a)
greater in CO32  than H 2 CO3
(b) greater in H 2 CO3 and CO23 
(c)
equal in H 2 CO3 and CO32 
(d) No stabilisation in either
Hint: (a) Because Canonical structures are equally stable. But in case of H2CO3. Charge separation is present
O
O
O
C
C
C
O
O
O
O
O
O
(Equally stable)
17.
AMITY
Which of the following structure cannot represent resonance forms for N2O?
(a)
...N = N = ..O.
(b)
(c)
(d)
.N.. = N = O.. .
Hint: (d) In nitrogen and oxygen when octet is complete, all electrons will be present in paired form.
18.
Consider the following species
Institute
for Competitive
Examinations
(II) CH OH
(III) OH
(IV) CH COO
C H O
(I)


6
5

3
3
(V) CH3  S
Decreasing order of nucleophilicity in an aqueous solution will be
(a)
IV > II > III > I > V
(b) IV > III > II > I > V
(c)
V > III > I > IV > II
(d) III > IV > V > I > II

In aqueous solution, OH– is max. hydroated CH 3  S is less hydrated, so it has more site for the attack
Hint: (c)
on carbon. OH– is more nucleophilic then C6H5–O– because ini latter case –ve charge is stabilized by
resonance C6H5 – O– is more nucleophilic then CH3COO– because it is less hydrated.
19.
Consider the following compounds:
(I)
(II)
C
NC
(III)
CN
O
Which of the above compounds exhibit aromaticity?
(a)
I & II only
(b) II & III only
(c) I & III only
[20]
(d) I, II & III
Module Test-3-II (07-01-12) XI Synchro (TC-A)
Hint: (d)
C
NC
CN
O
(Aromatic character is present in these compounds)
20.
An acid H2A has dissociation constants K1 & K2 are 0.10 & 10–7 respectively. What is the concentration of [H+] in
a solution prepared by dissolving .2 moles of the acid in water to give 1 litre solution?
(a)
0.20 M
(b) 0.10 M
(c) 0.15 M
(d) 017 M



H 2 A 
 H  HA
Hint: (b)
C  C
C
C
Ka 
21.
C 2
1 
1
2

2 1 
or
or 22 +  – 1 = 0
Which one of the following is the correct statement? The odd electron of methyl radical occupies
AMITY
(a)
One of the sp2 hybridised orbitals of C
(b) One of the sp3 hybridised orbitals of C
(c)
One of the sp hybridised orbitals of C
(d) The non-hybridised p-orbitals of C
Hint: (d) CH3 is sp2 hybridised.
22.
Among the given compounds, the one which is least basic is
Institute
for Competitive
Examinations
(b)
(c)
(d)
N
N
(a)
N
N
H
Hint: (c) Due to aromatic compound have pair is busy in conjugation.
23.
Which one of the following molecules has all the effects namely inductive, mesomeric and hyperconjugative?
(a)
HC  C – C  C – Cl
(b) CH  CH  CH 2  CH 3
|
Cl
(c)
CH2 = CH – Cl
(d) All of these
Hint: (b)
24.
Which of the following compounds can be optically active?
CH 3
Br
(I)
CH3  CH 2  CH  CH3
|
OH
(II)
Cl
H
(III)
CH3
C=C=C
H
CH3
(IV)
CH 3
(a)
I & III
(b) I, II & III
(c) I & II
[21]
CH3
(d) All of these
Module Test-3-II (07-01-12) XI Synchro (TC-A)
Hint: (b) In first and second case chiral carbon is present. In third case structure is non-planar.
25.
Which of the following will act as buffer solution?
(I)
100 ml of .1 M NaOH + 100 ml of .1 M CH3COOH
(II) 100 ml of .1 M NaOH + 200 ml of .1 M CH3COOH
(III) 200 ml of .1 M NH3 + 100 ml of .1 M HCl
(IV) 100 ml of .1 M HCl + 100 ml of .1 M CH3COONa
(a)
I, II & III
(b) II & III
(c) II, III & IV
(d) I & IV
Hint: (b) In case II, buffer CH3COOH + CH3COONa is formed. In case III, buffer NH3 + NH4Cl is formed.
SECTION- II
LINKED COMPREHENSION TYPE
This section contains 2 Paragraphs. Based upon each paragraph, 3 and 2 multiple choice questions have
to be answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.
AMITY
Comprehension-I
Double bond equivalents or DBE 
 n (V  2)  1
2
where n = no. of different kinds of atoms present in the molecule.
Institute for Competitive Examinations
V = valency of each atom
DBE implies the sum of double bonds and rings present in the molecule.
26.
Give the number of acyclic structures of C3H6O excluding tautomerism
(a)
3
(b) 4
(c) 5
(d) 6
O
||
Hint: (b) CH3  C  CH 3 , CH 3  CH 2  CHO, CH 2  CH  O  CH 3 & CH2 = CH – CH2 – OH
27.
Give the no. of cyclic structures of C3H6O
(a)
3
(b) 2
(c) 5
(d) 6
OH
O
O
Hint: (a)
28.
Give the no. of cyclic structures of C5H10.
(a)
4
(b) 5
(c) 6
(d) 7
Hint: (b)
[22]
Module Test-3-II (07-01-12) XI Synchro (TC-A)
Comprehension-II
The phenomenon of resonance said to occur whenever for a molecule we can write two or more lewis structures
which differ in the position of electrons but not in relative position of atoms. Various Lewis structures drawn are
called Canonical structures.
29. Which of the following statements is incorrect?
(a)
are not resonating structure
4
3
2
1
(b) In CH 2  CH  CH  O nucleophilic centre position is 4 while electrophilic centre is position 1
(c)
SnCl4 acts as an electrophile
(d) Triplet carbene will have 2 unpaired electrons.
Hint: (a) Delocalization of lone pair is present in this structure (Rule of resonance).
30.
What is the decreasing order of stability of the ions given below?


(I)
CH3  CH  CH 3
(a)
I > II > III
O
||
(III) CH 3  CH  C  CH 3

(II) CH3  CH  O  CH 3
AMITY
(b) III > II > I
(c) II > I > III
(d) III > I > II
Hint: (c) II case  Resonance is taking place
I case  six hyperconjugate structure
III case  three hyperconjugative structure and resonance structures cannot be drawn.
MATHEMATICS
Institute for Competitive
Examinations
SECTION- I
STRAIGHT OBJECTIVE TYPE
This section contains 10 multiple choice questions numbered 31 to 40. Each question has 4 choice (A), (B),
(C) and (D), out of which ONLY-ONE is correct
31.
x2 y 2

 1 and its director circle is/are
Number of common tangents to the hyperbola
9
4
(a)
1
(b) 2
(c) 3
(d) none of these
Hint: (d) Equation of director circle is x2 + y2 = 5
Equation of tangent to
y = mx +

9m 2  4
1  m2
x2 y 2

 1 is
9
4
9m 2  4
 5

9m2 – 4 = 5 + 5m|

4m2 = – 9
No such ‘m’ possible.
[23]
Module Test-3-II (07-01-12) XI Synchro (TC-A)
32.
6 boys and 4 girls are to be seated on 10 chairs arranged as shown.
The number of ways in which they can be seated such that all girls never sit together is
(a)
10! – 7! × 4!
(b) 10! – 6! × 4!
(c)
10!
(d) 7! × 4!
Hint: (b) Total no. of ways of arrangement = 10!
If all girls sit together then total number of arrangements = 6! × 4!
10 – 6! × 4! are the total required ways.
33.
Number of roots of the equation x  7 x  18  0 is /are
(a)
1
Hint: (a) Let x = t
t2 – 7t – 18 = 0
(b) 2

t  [0, )

t = 9, t = – 2
(c) 4
O
t = 9 is only possible root
34.
(d) none
AMITY
If both the root of x2 + x + p = 0 are negative, then
(a)
p<0
(b) p < 1/4
(c) p > 1/4
(d) none of these
Hint: (d) 1 – 4p  0 or p  1/4
 p   0, 1 
 4 
p>0
Institute for Competitive Examinations
35.
Locus of a point p(x, y), which follows the equation
(a)
parabola
(b) hyperbola
( x  2) 2  ( y  3) 2  ( x  3) 2  ( y  2) 2  5 2 is a
(c) circle
(d) none of these
Hint: (d) Distance between A(2, 3) and B(–3, –2) is 52.
Locus represents the line joining (2, 3) and (–3, –2) except the segment AB.
36.
If f(x) = ax2 + bx + c such that b2 < 4ac and a – 2b + 4c < 0 then value of (5a – b + 2c) is
(a)
>0
(b) < 0
(c) = 0
(d) can’t say
Hint: (b) D < 0 and f(–½) < 0
 f(x) < 0  x 
Now 5a – b + 2c = f(1) + f(2) < 0
37.
The common tangents of an auxillary circle ( of a hyperbola) with its corresponding rectangular hyperbola and
with its conjugate hyperbola form a
(a)
square
(b) trapezium
(c)
a parallelogram with non-equal adjacent sides
(d) none of these
[24]
Module Test-3-II (07-01-12) XI Synchro (TC-A)
Hint:(a)
Let the rectangular hyperbola be x2 – y2 = a2
...(i)
Then equation of its conjugate hyperbola is x2 – y2 = – a2
...(ii)
and equation of its auxillary circle is x2 + y2 = a2
...(iii)
Equation of common tangents of (i) and (iii) rare x   a
and equation of common tangents of (ii) and (iii) are y   a
Hence, then form a square.
38.
If a, b, c are 3 positive consecutive integers, (where a < b < c) then number of possible equations ax2 + bx + c = 0 with
real roots is/are
(a)
1
(b) 2
(c) infinite
(d) none of these
Hint: (d) Let the numbers be a, a + 1, a + 2
Then D = (a +1)2 – 4a (a + 2)  0
 a2 + 1 + 2a – 4a2 – 8a  0
 3a2 + 6a – 1  0

2
2 
, 1 

 a   1 
3
3

 No such ‘a’ possible.
39.
AMITY
a  I+
Using all the alphabets of the word “DALHOUSIE”, the number of words which can be formed such that only
vowels occupy even places are
(a)
5! × 6P4
(b) 4! × 6P5
(c) 4! × 5!
(d) none of these
Institute for Competitive Examinations
Hint: (c) 4 vowels can occurly even places is 4! ways
Rest 4 places can be filled by either vowel or consonants is 5! ways.
Total number of ways = 5! × 4!
40.
Hyperbola xy = c2 and circle x2 + y2 = r2 intersect at 4 points pi (xi , yi ), i =1, 2, 3, 4 then value of x1 × x2 × x3×x4 =
(a)
c2
(b) r2
(c) –1
(d) none of these
Hint: (a) Solving xy = c2 and x2 + y2 = r2 simultaneously we get
c4
 r2

x2
x1 × x2 × x3 × x4 = c4
x2 

x4 – r2x2 + c4 = 0
SECTION- II
LINKED COMPREHENSION TYPE
This section contains 2 Paragraphs. Based upon each paragraph, 3 and 2 multiple choice questions have
to be answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.
Comprehension-I
Any equation of the form a(f(x))2 + b(f(x)) + c = 0 can be transformed to a quadratic equation by substitiuling
f(x) = t, where t  [Minimum of f(x), Maximum of f(x)]
41.
For the equation 4x – 3 × 2x – 5 = 0 to have real solutions, the roots of equation t2 – 3x – 5 = 0 should
(a)
>0
(b)  (0, 2)
(c)  R
(d) none of these
Hint: (a) Let 2x = t
 t  (0, )
2
Roots of t – at + b = 0 should lie in (0, )
[25]
Module Test-3-II (07-01-12) XI Synchro (TC-A)
42.
Equation sec2x – 5secx + 2 = 0 has real solution then roots of t2 – 5t + 2 = 0 should belong to
(a)
(– , 0)
(b) (– , –1]
(c) (– , –1]  [1, )
(d) none of these
Hint: (c) Let t = sec x
 t (– , –1]  [1, )
2
Roots of t – at + b = 0 should lie in [– , –1]  [1, )
43.
Number of real solutions of (log10x)2 – 3 log10x – 2 = 0 is/are
(a)
0
(b) 1
(c) 2
(d) none of these
Hint: (c) Let log10x = t
 t(–, )
2
Roots of t – 3t – 2 = 0 are real
 Equation has = real roots
Comprehension-II
Consider the set of digits 1, 2, 3.....,9.
44.
Number of nine digit numbers which can be formed which are divisible by 4 are
(a) 10 × 7!
(b) 8 × 7!
(c) 16 × 7!
(d) none of these
AMITY
Hint: (c) 2 × 5 × 7! + 2 × 3 × 7! = 16 × 7!
45.
Number of nine digit numbers which can be formed which are divisible by 9 are
(a) 9!
(b) 8!
(c) 3 × 8!
(d) none of these
Hint: (a) 9!
SECTION- III
INTEGER ANSWER TYPE
This section contains 12 questions. The answer to each of the question is a single
digit integer, ranging from 0 to 9. The bubble corresponding to the correct answer is
to be darkened in the ORS. The appropriate bubbles corresponding to the answers
to these questions have to be darkened as illustrated in the following example:
If answer of question number (1) is 8, then the correct darkening of bubbles will
look like the following.
Institute for Competitive Examinations
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
PHYSICS
1.
F=mg/2
Consider the following situation as shown in figure. The force
mg
F is equal to
. If the area of cross section of string is A
2
and its Young’s modulus Y. The stress developed in the string
x mg
is 4 AY . Find ‘x’. The string is light end there is no friction.
m
m
Hint: (3) Tension developed in string creates stress in wire.
N
T
mg/2
T
mg
mg
[26]
Module Test-3-II (07-01-12) XI Synchro (TC-A)
2.
Consider an ice cube of edge 1 cm. kept in a gravity free hall. The surface area of water when the ice melts is
[(18) x]1/ 3 cm2 specific gravity ice is 0.9 and surface tension of water is 0.075 J/m2.
Hint: (2) In gravity free space.
Ice cube after melting will acquire splerical shape.
3.
The weight of an object is 80 N on surface of earth of radius R. If the object is taken to a height of (2 –1)R from
earth surface. The weight of object is 8x N.
Hint: (5) g ' 
4.
g
h

1  
 R
2
Two rain drops are falling with terminal velocity 5 m/s. Drops combine to form a big drop in air. The terminal
velocity of new drop is 5(x)1/3. Find ‘x’
Hint: (4) Terminal velocity is directly propertional to square of radius of drop.
AMITY
V1  r1 
 
V2  r2 
2
4 3
4

r2  2    r13 
3
3

CHEMISTRY
5.
Institute for Competitive Examinations
100 cc of HCl of pH value 1 is mixed with 100 ml of distilled water. What will be final pH of the resultant solution.
(Give the final answer according to rounding off rules) (Hint: log5 = .6990)
Hint: (1) Final [H+] =
.01
moles/L
.2
pH = 1.3
after rounding off pH = 1.
6.
An aqueous solution contains 10% NH3 by mass and has a density of 0.99 g/cc. Ka for NH4+ = 5 × 10–10. [H+]
conc. will be x × 10–13 M. Here what will be the value of x?
Hint: (9) 10% NH3 by means 10g NH3/100 g solution
D = .99 g/cc
99
M
17
Apply Ka × Kb = 10–14
Molarity of NH3 
Kb = 2 × 10–5
 NH + + OH–
NH4OH 
4
At equilibrium
C – C
–
C
C
+
After calculation [OH ], calculate [H ]
[27]
Module Test-3-II (07-01-12) XI Synchro (TC-A)
7.
Calculate the % hydrolysis in .003 M aqueous solution of NaOCN. Ka for HOCN = 3.33 × 10–4 M. % of
hydrolysis = x × 10–2. Here x will be
Kw
Hint: (1) K h  K
a
K h  C 2
  104
8.
Then % =
C
 100
C
Solubility of Pb(OH)2 in water is 6.7 × 10–6 M. The solubility of Pb (OH)2 in a buffer solution of pH = 8 will be x
× 10–3 (after rounding off.). Here x will be
Hint: (1)
 Pb+2(aq) + 2OH–(aq)
Pb(OH)2(s) 
In buffer
x
10–6
Assume x is new conc. of Pb+2 in buffer solution from solubility you can calculate the value of Ksp.
AMITY
Apply Ksp = x × (10–6)2
Here, x can be calculated
MATHEMATICS
9.
,   are roots of ax2 + bx + c = 0,  > 0,  < 0 and 1  |  |  3 and 4  |  |  5, then maximum value of
Institute for Competitive Examinations
b
2a
is
Hint: (2) 
b 


2a  2 

b
1
  
2a 2
Maximum value of
10.
b
possible when  = 1,  = – 5
2a
9 
Number of roots of 2sin2x – 5sinx + 1 = 0 in 0,
are
 2 
Hint: (5) Roots of 2t2 – 5t +1 = 0 are
11.
t
5  2s  8 5  17

4
4
t
5  17
4
or sin x 
5  17
4
 9 
has 5 solutions in 0,
 2 
Number of points at which a hyperbola and its conjugate intersect is/are
Hint: (0) Hyperbola and its conjugate never touch or intersect each other.
12.
If a normal at p(t1) to the curve xy = c2, meet it again at v(t2) such that pv subtends 90° at the centre of the curve,
then number of such points p; possible is/are
[28]
Module Test-3-II (07-01-12) XI Synchro (TC-A)
Hint: (0) Normal at P(t1) intersect at V(t2) then, t13 t2 = –1 ...(i)
P(ct1, c /t1 )
PV subtends 90° at the origin
c / t1 c / t2
 ct  ct  1
1
2
t12
2 2
 1
 t1 t2 = – 1

t16
(c/t2)
V(ct2)
from (i)
 t14 = – 1 has no selection.
SECTION- IV
MATRIX-MATCH TYPE
This section contains 9 questions. Each question contains statements given in two columns which have to be
matched. Statement (A, B, C, D) in Column I have to be matched with statements (p, q, r, s) in
Column II.
Physics
1.
Consider three masses ‘M’ each are placed on the vertices of an equilateral triangle of side ‘l’ m.
AMITY
A.
Gravitational field intensity at confroid
(p)
2 GM
3 l2
B.
Gravitational field intensity at mid point of any side
(q)
3 3
C.
Gravitational potential at centroid
(r)
zero
GM
l
2GM 
1 
Institute
for Competitive Examinations
Gravitational potential at mid point of any side
(s) 
2

l
3
D.

(t)


3GM
2l
Hint: A-(r); B-(q); C-(p); D-(s)
Gravitational field due to
Point mass is
GM
x2
Gravitational potential is
2.
GM
x
Water is filled in a container upto height H1 and the container is kept on a stand of height H2
A.
For maximum range hole is made in container at H2 from ground
B.
For maximum range hole is made at
C.
Range is same if the hole made at height (H1 – x) or x from the
H1  H 2
from bottom
2
(p) H1 > H2
(q) H2 = 0
(r)
H1 <H2
bottom of container
D.
Horizontal range first increases and then decreases as the depth of the
(s) H1 = H2
hole increases from surface of water
(t)
None of these
Hint: A-(r), (s); B-(p), (s); C-(q); D-(p), (q)
[29]
Module Test-3-II (07-01-12) XI Synchro (TC-A)
Torcelli theroem according to Horizontal range for a container. Rept on stand of height H2 and water level at
height H1 from ground is R  2 ( H 1  H 2  h ) h
3.
A.
Archimedes’ principle
(p) floating of needle on surface of water
B.
Bernoulli’s theorem
(q) dancing of camphor
C.
Stokes’ Law
(r)
working of atomizer
D.
Surface tension
(s)
terminal velocity of rain drop
(t)
floating of ice cube on water
Hint: A-(s, t); B-(r); C-(s); D-(p, q)
Conceptual
Chemistry
4.
A.
Solubility of Hg2Cl2 in 0.1 M NaCl solution
(p) 10 Ksp
B.
Solubility of PbI2 in .01 M KI solution
(q) 100 Ksp
C.
Solubility of Ag2CrO4 in 0.25 M K2CrO4 solution
(r)
D.
Solubility of calcium oxalate in 0.1 M oxalic acid solution
(s)
AMITY
10000 Ksp
K sp
Hint: A-(q); B-(r); C-(s); D-(p)
A:
 Hg 22  (aq)  2Cl  (aq)

x
.1 M
Hg 2 Cl 2 (s)
Institute
K = x × (.1) for Competitive Examinations
2
sp
x = 100 Ksp
B:
5.
 Pb 2 (aq)  2I  (aq)
PbI 2 (s) 
x
.01 M
Ksp = x × (.01)2
x = 10000 Ksp
Further options you can solve easily like this
Stability order
A.


(p) True
F3 C  CH 2  CH 3  CH 2
CH 2
CH2
>
B.
(q) False
OCH3
CH3
O
>
OH
C.
(Keto-enol tautomerism)
D.
>
[30]
Module Test-3-II (07-01-12) XI Synchro (TC-A)
Hint: A-(q); B-(p); C-(p); D-(p)

A:
CF3  CH 2 is unstable due to the electron withdrawing nature of CF3 group.
B:
is more stable because in conjugation oxygen will donate its lone pair for further stabilization
OCH3
of +ve charge.
C:
C = O (polar bond) is more stable as compared to
D:
6.
C=C
(non-polar bond)
Due to less angle strain in cyclobutane as compared to cyclopropane.
..
..
A.
CH3 – O – CH – CH 3
(p) Stable due to resonance
B.
F3C 
(q) Destablished due to inductive effect
C.
CH3
|
CH 3  C 
|
CH3
(r)
AMITY
Stabilised by hyperconjugation
(s) A secondary
carbocation
Institute
Examinations
CH  CH  CH for Competitive

D.
3
3
Hint: A-(p),(q),(s); B-(p), (q); C-(q); D-(r),(s)
A:


CH3  O  CH  CH 3  CH 3  O  CH  CH 3
Inductive effect is created by oxygen.
Hyperconjugation effect is created by CH3 group. attached to carbocation.
B:
Inductive effect is created by three F atoms.
C:
Mathematics
7.
Let 0 < a < b
A.
Roots of the equation x2 + (x – a) (x – b) = 0 lie in
(p) (– , a)  (b, )
B.
Roots of the equation (x – a) (x – b) + (x – a) + (x + b) = 0 lie in
(q) either (– , a) or (b )
C.
Roots of the equation (x – a) (x – b) – x2 = 0 lie in
(r)
D.
Roots of the equation x(x – a) +(x – b) = 0 lie in
(s) (– , 0)  (a, b)
(–, a)  (a, b)
Hint: A-(q); B-(r); C-(p); D-(s)
[31]
Module Test-3-II (07-01-12) XI Synchro (TC-A)
A:
Let f(x) = x2 + (x –a) (x –b)
f(a) = a2 > 0
f(b) = b2; > 0
B:
 either both are roots in (– , a) or (b, ) or (a, b)
f(x) = (x –a) (x – b) + (x – a) + (x – b)
then f(x) = a – b ; < 0
f(b) = b – a; > 0
Roots are in (–, a)  (a, b)
C:
Let f(x) = (x – a) (x – b) – x2
f(a) = – a2; < 0
f(b) = – b2; < 0
f(0) = ab; > 0
Roots are in (– , a)  (b, )
D:
Let f(x) = x(x – a) + (x – b)
f(0) = – b; < 0
f(a) = a – b; < 0
AMITY
f(b) = b(b – a); > 0
Roots are in (–, 0) (a, b)
8.
Number of common tangents to xy = 8 and x2 + y2 = r2 are, if
A.
r=1
(p) 0
B.
r=3
(q) 1
r=4
(s)
Institute
for Competitive
Examinations
r = 22
(r) 2
C.
D.
4
Hint: A-(r); B-(p); C-(r); D-(p)
Equation of tangent of xy = 8, is y = mx + 42–m
If it has to be tangent to x2 + y2 = r2, then
4 2m
1  m2
r

– 8m = r2 (1 + m2)
 r2m2 + 8m + r2 = 0
Has to need roots of 64 – 4r4  0

r4  16
0r2
For r = 1; 2 tangnets
r = 22, 3, 4 no common tangent.
9.
If roots of a quadratic equation represents semi transverse axis and semi conjugate axis length; of a hyperbola,
then equation and corresponding eccentricity are
A.
x2 – 5x + 6 = 0
(p) 2
B.
x2 – 42x + 8 = 0
(q) 5/4
C.
x2 – 2ax + a2 = 0, a > 0
(r)
D.
x2 – 7x + 12 = 0
(s)
[32]
13
3
2
Module Test-3-II (07-01-12) XI Synchro (TC-A)
Hint: A-(r); B-(p); C-(p); D-(q)
A:
a = 3, b = 2

e2 = 1 
B:
a = b = 22

e = 2
C:
a=b

e = 2
D:
a = 4, b = 3

e2  1 
4
a

e
9
16

e
13
3
5
4

AMITY
Institute for Competitive Examinations
[33]
Module Test-3-II (07-01-12) XI Synchro (TC-A)