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PAPER-I
MODULE TEST-2 : XI SYNCHRO
Time : 3 hrs.
Date : 03/12/2011
(IIT-JEE PATTERN)
M.M.: 243
PBG
S
-1
TESTET
CODE - A
ANSWERS
Physics: Section I to III
1
(a)
2.
(a)
3.
(c)
4.
(d)
5.
(a)
6.
(d)
7.
(d)
8.
(b,c)
9.
(b,d)
10.
(d)
11.
(c,d)
12. (b,c)
AMITY
13.
(a)
14.
(d)
15.
(b)
16.
(d)
17.
(b)
18. (c)
Chemistry: Section I to III
19.
(a)
20.
(d)
21.
(c)
22.
(b)
23.
(b)
24. (b)
25.
(d)
26.
(a,c)
27.
(a,b,c)
28.
(b,c)
29.
(a,c,d)
30. (a,b,c)
31.
(c)
32.
(a)
33.
(d)
34.
(c)
35.
(a)
36. (d)
37.
(a)
38.
(d)
39.
(b)
40.
(c)
41.
(a)
42. (c)
43.
(b)
44.
(a,b,c)
45.
(b)
46.
(a,b)
47.
(a,b)
48. (a,d)
49.
(c)
50.
(d)
51.
(d)
52.
(c)
53.
(c)
54. (a)
Institute for Competitive Examinations
Mathematics: Section I to III
Section-IV (PCM)
1
(1)
2.
(9)
3.
(3)
4.
(2)
5.
(4)
6.
(4)
7.
(9)
8.
(5)
9.
(1)
10.
(4)
11.
(1)
12.
(4)
[13]
Module Test-2-I (03-12-11) XI Synchro (TC-A)
MODULE TEST-2 : XI SYNCHRO
HINT/SOLUTIONS
PHYSICS
SECTION- I
STRAIGHT OBJECTIVE TYPE
This section contains 7 multiple choice questions numbered 1 to 7. Each question has 4 choice (A), (B), (C)
and (D), out of which ONLY-ONE is correct
1.
A ball of mass m moving with velocity v strikes the bob of a pendulum at rest. The mass of the bob is also m.
If the collision is perfectly inelastic, the height to which the bob will rise is given by
(a)
v2
8g
(b)
v2
2g
(c)
2v 2
g
(d)
v2
g
Hint: (a) Use conservation of linear momentum to find the velocity of combined mass.
m1u1 + m2u2 = m1V+ m2V
Apply conservation of energy,
2.
1
(m + m2)V2 = (m1 + m2)gh
2 1
AMITY
A particle of mass 5m which is at rest explodes into 3 fragments. Two equal fragments each of mass 2m are found
to move with a speed v each in opposite directions. The energy released in the process of explosion is
(a) 2 mv2
(b)
5
mv2
2
(c) 5 mv2
(d) 3 mv2
Hint: (a) From conservation of momentum, find speed of the third particle, 0 = 2mV – 2mV + mV´  V´ =0
Energy released = k – k
Institute
1
1 for Competitive Examinations
= (2m)V + (2m)V
f
2
3.
i
2
2
2
In a vertical plane inside a smooth hollow thin tube, a block of same mass as that of tube is released as shown in
figure. When it is slightly disturbed it moves towards right. By the time the block reaches the right end of the tube,
the displacement of the tube will be (where ‘R’ is the mean radius of tube, assume that the tube remains in vertical
plane)
m
(a)
(c)
2R

(b)
4R

R
2
(d) R
m
R
Hint: (c) Along horizontal direction, shift in the centre of mass of the system is zero.



m x  m2 x2
xm  1 1
m1  m2
4.
A system of two blocks A and B are connected by an inextensible massless string as shown in figure. The pulley
is massless and frictionless. Initially the system is at rest. A bullet of mass ‘m’ moving with a velocity ‘u’ as
shown hits block ‘B’ and gets embedded into it. The impulse imparted by tension force to the block of mass 3m
is
(a)
5 mu
4
(b)
4 mu
5
(c)
2 mu
5
(d)
3 mu
5
m
u
m B
A 3m
[14]
Module Test-2-I (03-12-11) XI Synchro (TC-A)
Hint: (d) Impulse-momentum equations for blocks A, B and the bullet gives the velocity of A just after collision.

p = change in momentum = impulse.
5.
After a perfectly inelastic collision, two objects of the same mass and same initial speeds are found to move
together at half of their initial speeds. The angle between the initial velocities of the objects is
(a) 120°
(b) 60°
(c) 150°
(d) 45°
Hint: (a) Apply conservation of linear momentum.
6.
A uniform rod of mass m is bent into the form of a semicircle of radius R. The moment of inertia of the rod about
an axis passing through A and perpendicular to the plane of the paper is
A
(a)
2
mR 2
3
(b) mR2
(c)
5
mR 2

(d) 2mR2
R
Hint: (d) IA = IC + m(R)2 = mR2 + mR2
7.
AMITY
Three rings, each of mass m and radius r, are so placed that they touch each other. Find the moment of inertia
about the axis as shown in figure.
(a) 5 mr2
(b)
5 2
mr
7
(c) 7 mr2
(d)
7 2
mr
2
Institute for Competitive Examinations
Hint: (d) Parallel axes theorem provides the necessary result.
I=
1
1
mR2 + ( mR2 + mR2)2
2
2
SECTION- II
MULTIPLE CORRECT ANSWERS TYPE
This section contains 5 multiple choice questions numbered 8 to 12. Each question has 4 choice (A), (B),
(C) and (D), out of which ONE OR MORE is/are correct
8.
A particle moving with kinetic energy E makes an head on elastic collision with an identical particle at rest. During
the collision
(a)
elastic potential energy of the system is always zero
(b) maximum elastic potential energy of the system is E/2
(c)
minimum kinetic energy of the system is E/2
(d) all of these
Hint: (b,c)
Maximum potential energy and minimum kinetic energy of the system are obtained simultaneously.
This is achived when both more will equal velocity.
From COM, m1 u1 + m2u2 = m1V1 + m2V2
u
1
1
 u1 
 mu1 + 0 = 2mV  V = 1 ; E = mu12, E´ = (2m)  
2
2
2
2
[15]
2
Module Test-2-I (03-12-11) XI Synchro (TC-A)
9.
A steel ball of mass 2m suffers one-dimensional collision with a row of three balls, each of mass m. If mass 2m has
collided with velocity v and the three balls numbered 1, 2, 3 were initially at rest, then after the collision, which of
the following statement is/are correct (e = 1 for all collisions)?
(a) Balls 1, 2 and 3 would start moving to the right, each with velocity
2v
3
2
m
v 1
m
2m
3
m
4v
3
2v
(c) Balls 1 and 2 will come to rest and ball 3 will start moving to the right with velocity
3
(d) Ball of mass 2m will move with velocity v/3 after collision.
(b) Balls 1 and 2 will come to rest and ball 3 will start moving to the right with velocity
Hint: (b,d)
For 2 m and m, find velocity of each just after collision, with the help of COM and e(coefficient of
restitution)
For 1 & 2 collision and 2 & 3 collision, momentum and velocity get interchanged.
10.
Which of the following is not the moments of inertia of a thin square plate ABCD
4
A
B
AMITY
of uniform thickness about an axis passing thorugh the centre O and perpendicular
2
to the plate?
1
O
(a) I1 + I2
(b) I3 + I4
(c) I1 + I3
(d) I1 + I2 +I3 + I4
D
3
C
Hint: (d) From perpendicular axes theorem,
I =I +I =I +I
Institute
for Competitive Examinations
z
11.
1
2
3
4
A solid sphere (mass = m, radius = R) rolling on a horizontal rough surface
collides elastically with a smooth vertical wall, as shown in the figure. Which
of the following statement is false?
(a) After collision, the velocity of the centre of mass gets reversed.
v
(b) The moment of inertia of the sphere about an axis passing through P
may be (7/5) mR2.
P
(c) Angular momentum of the sphere about any stationary point on the horizontal surface is conserved.
(d) Just after collision the point of contact with the horizontal surface remains at rest.
Hint: (c,d)
12.
Elastic collision ensures the magnitude of speed remains constant.
2
7
Ip = mR2 + mR2 = mr2
5
5
Any stationary point indicates that frictional force may pass through it or not.
After collision, rolling with not take place.
A spool of wire rests on a horizontal surface as shown in figure. As the wire is pulled, the spool does not slip at
contact point P. On separate trials, each one of the forces F1, F2, F3 and F4 is applied to the spool. For each one
F3 F2
of these forces the spool
(a)
will rotate anticlockwise if F1 is applied
F4
(b) will not rotate if F2 is applied
(c)
F1
will rotate anticlockwise if F3 is applied
(d) will rotate clockwise if F4 is applied.
P
[16]
Module Test-2-I (03-12-11) XI Synchro (TC-A)
Hint: (b,c)
Point of contact P is at rest. Taking torque of all the forces about point P we see that:
torque is clockwise if F1 is applied i.e., spool will rotate clockwise
torque of F2 is zero i.e. spool will not rotate if F2 is applied.
torque of F3 or F4 is anticlockwise i.e. spool will rotate anticlockwise if only F3 or F4 is applied.
SECTION- III
LINKED COMPREHENSION TYPE
This section contains 2 Paragraphs. Based upon each paragraph, 3 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
13.
The coefficient of restitution of balls is
(a)
2
3
(b)
1
3
(c)
1
2
(d) 1
14.
The loss is kinetic energy is
15.
AMITY
(a)
1 2
mu
2
(b)
1 2
mu
4
(c)
1 2
mu
3
(d)
1 2
mu
6
(c)
mu
6
(d)
mu
2
Impulse on the each sphere by the hitting sphere is
(a)
mu
2
(b)
mu
3
Institute for Competitive Examinations
v
Hint: 13. (a), 14. (d), 15. (b)
mu = 2mv cos 30°
....(i)
v2  v1
e = u u
1
2
....(ii)
u
Impulse = change in momentum
v
Comprehension-II
16.
If the sphere after colliding with the wall rolls without slipping in opposite
direction,then coefficient of friction µ is
1
2
(a)
(b)
2
3
17.
1
3
(d)
1
4
What is linear impulse imparted by the wall on the sphere during impact?
(a)
18.
(c)
32 N-s
(b) 16 N-s
(c)
4 5 N-s
(d) 15 2 N-s
What is the angular impulse imparted by wall during impact?
(a) 20 N-m-sec
(b) 15 N-m-sec
(c) 4 N-m-sec
(d) 10 N-m-sec
Hint: 16. (d), 17. (b), 18. (c)
Apply equations of linear impulse and angular impulse on the sphere
V
Linear impulse = change in linear momentum
Angular impulse = change in angular momentum
[17]
Module Test-2-I (03-12-11) XI Synchro (TC-A)
CHEMISTRY
STRAIGHT OBJECTIVE TYPE
This section contains 7 multiple choice questions numbered 19 to 25. Each question has 4 choice (A), (B),
19.
Distribution of fraction of molecules with velocity is represented in
figure. Velocity corresponding to point ‘x’ is
(a)
2 RT
M
(c)
8 RT
M
(b)
3RT
M
(d)
8 RT
M
Fraction of molecules
(C) and (D), out of which ONLY-ONE is correct
‘X’
Velocity
Hint: (a)
20.
At relatively high pressure the vander waal’s equation of state reduced to
AMITY
(a)
PV = RT –
a
V
(b) PV =
a RT
V2
(c) P = RT–
a
V2
(d) PV = RT + Pb

an 2 
Hint: (d)  P  2  (V – nb) = nRT
V 

a 

 P  2  (V – b) = RT
V 

Institute for Competitive Examinations
at high pressure
a
can be neglected but b can’t be neglected
V2
P(V – b) = RT
PV – Pb = RT
PV = RT + Pb
21.
NH3 gas is liquefied more easily than N2. Therefore
(a)
Vanderwaal’s constant a and b of NH3 > N2 (b) Vanderwaal’s constant a and b of N2 > NH3
(c)
a(NH3) > a (N2) but b(NH3) < b(N2)
(d) a(NH3) < a (N2) but b(NH3) > b(N2)
Hint: (c) NH3 has greater intermolecular forces of attraction than N2. Thus a(NH3) > a(N2) but b(NH3) < b(N2)
22.
One mole of N2O4(g) at 300 K is kept in a closed container under one atmosphere. It is heated to 600 K when 20%
by mass of N2O4(g) decomposes to NO2(g). The resultant pressure is
(a)
Hint: (b)
1.2 atm
(b) 2.4 atm
(c) 2.0 atm
(d) 1.0 atm
 2NO
N2O4 
2
1
0
–0.2
0.4
1–0.2
0.4
total moles = 0.8 + 0.4 = 1.2
p nT (at cost V)
[18]
Module Test-2-I (03-12-11) XI Synchro (TC-A)
P1
P
 2
n1T1 n2T2
P2 =
23.
1  1.2  600
= 2.4 atm.
1  300
 C + D.
In the study of the reaction A + B 
A and B were mixed in a vessel kept at TK. The initial concentration of A was twice the initial concentration of B.
After the equilibrium was reached the equilibrium concentration of C was three times the equilibrium concentration
of B. Calculate equilibrium constant.
(a)
2.0
Hint: (b)
(b) 1.8
A
+
B


(c) 4.0
C
+
(d) 2.8
D
2
1
0
0
–x
–x
+x
+x
2– x
1– x
x
x
AMITY
At equilibrium [C] = 3[B]
x = 3(1 – x)
4x = 3
x=
3
4
3 3

4 4
K = 5 1  1.8

4 4
Institute for Competitive Examinations
24.
The heat of neutralisation of four acid a, b, c and d when neutralised against a common base are 13.1, 9.4,
12.4 KCal respectively. The weakest among these acid is
(a)
a
(b) b
(c) c
(d) d
Hint: (b) Enthalpy of neutralisation  strength of acid
25.
The dissociation energy of CH4 and C2H6 are 360 and 620 Kcal/mol respectively. The bond energy of C – C bond
is
(a)
260 kcal/mol
(b) 180 kcal/mol
(c) 360 kcal/mol
(d) 80 kcal/mol
Hint: (d) Let the bond energy of C – H is x kcal and C – C bond is y kcal
4x = 360
x = 90

6x + y = 620
6 × 90 + y = 620
y = 80 kcal
[19]
Module Test-2-I (03-12-11) XI Synchro (TC-A)
SECTION- II
MULTIPLE CORRECT ANSWERS TYPE
This section contains 5 multiple choice questions numbered 26 to 30. Each question has 4 choice (A), (B),
(C) and (D), out of which ONE OR MORE is/are correct
26.
Which is an irreversible process
(a)
Mixing two gases by diffusion
(b) Evaporation of water at 373 K and 1 atm pressure
(c)
Dissolution of NaCl in water
(d) all of these
Hint: (a,c)
27.
Which of the following is correct about the equilibrium?
(a)
G = 0
(b) Equilibrium constant is independent of initial concentration of reactants
(c)
Catalyst has no effect on equilibrium state
(d) reaction stops at equilibrium
AMITY
Hint: (a,c)
28.
For the following reaction
 2AB(g), K
A2(g) + B2(g) 
1
1
1
 AB(g), K
A2(g) + B2(g) 
2
2
2
Institute for Competitive Examinations
 A (g) + B (g), K
2AB(g) 
2
2
3

AB (g) 
1
1
A2(g) + B2(g), K4
2
2
then which of the following relations is/are correct
(a)
K1 × K2 = 1
(b)
K1 × K4 = 1
(c)
K 3 × K2 =1
(d) none of these
Hint: (b,c)
29.
Which of the following statement is correct on the basis of charle’s law?
(a)
The volume of ideal gas can never be zero
(b) The pressure of ideal gas can be zero
(c)
At zero pressure, all molecular motion ceases in a gas and it does not exist any pressure on the wall of the
container
(d) It is not possible to attain absolute zero.
Hint: (a,c,d)
30.
For the adiabatic expansion of an ideal gas
(a)
PV = constant
(b) TV –1 = constant
(c) T P1 –  = constant
(d) none of these
Hint: (a,b,c)
[20]
Module Test-2-I (03-12-11) XI Synchro (TC-A)
SECTION- III
LINKED COMPREHENSION TYPE
This section contains 2 Paragraphs. Based upon each paragraph, 3 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
31.
How much heat flows into the system along path ADB if the work done is 10 J
(a)
80 J
(b) –50 J
(c) 60 J
(d) 30 J
Hint: (c)
32.
When the system is returned from state B to A along the curved path, workdone on the system is 20 J. Does the
system absorber liberate heat and how much?
(a)
system liberates, –70 J
(b) system absorbs, 70 J
(c)
system absorbs, 90 J
(d) system liberate, –90 J
AMITY
Hint: (a)
33.
If HD – HA = 40 J, heat absorbed in the process DB is
(a)
+50 J
(b) +30 J
(c) +60 J
(d) +10 J
Hint: (d)
Institute for Competitive Examinations
Comprehension-II
34.
In a particular experiment, a gas undergoes adiabatic expansion satisfying the equation VT3 = constant. The
ratio of specific heats  is:
(a)
4
(b) 3
(c)
5
3
(d)
4
3
35.
In an adiabatic expansion of air (assume it a mixture of N2 and O2), the volume increases by 5%. The percentage
change in pressure is:
(a) 7%
(b) 6%
(c) 4%
(d) 3%
36.
In a thermodynamic process helium gas occupies the law
T
= constant. The heat given to a gas when temperature
P 2/ 5
of m moles of He is raised from T to 2T is:
(a)
8RT
(b) 4RT
(c) 16 RT
(d) zero
Hint: 34. (c); 35. (a); 36. (d)
34.
 VT3 = constant
or V1/3. T = constant
Also, V - 1 . T = constant

35.
 1 
1
5
or  
3
3
PV = constant

105V 
P1  
  constant
 100 
[21]
Module Test-2-I (03-12-11) XI Synchro (TC-A)


P1 105 

1
P 100 
or
P1 100 

P 105 
or
P1 100 

P 105 


P1 = 0.93 P
7% decrease in P.

7/5
T
 constant
P2 / 5
36.

T = constant  P2/5
PV = RT
PV = R  constant  P2/5

P
 V = constant
P2 / 5
or
P3/5  V = constant


5


AMITY
or
PV5/3 = constant    3 for He 
or PV = constant
Thus process is adiabatic  Q = 0
MATHEMATICS
SECTION- I
STRAIGHT OBJECTIVE TYPE
Institute for Competitive
Examinations
This section contains 7 multiple choice questions numbered 37 to 43. Each question has 4 choice (A), (B),
(C) and (D), out of which ONLY-ONE is correct
37.
The circle pasing through the distinct points (1, t), (t, 1) and (t, t) for all values of t, passes through the point
(a)
(1, 1)
(b) (–1, –1)
(c) (1, –1)
(d) (–1, 1)
Hint: (a) Let equation of the circle passing through the given points be
then
and
x2 + y2 + 2gx + 2fy + c = 0
.....(i)
1 + t2 + 2g + 2f t+ c = 0
.....(ii)
t2 + 1 + 2gt + 2f + c = 0
.....(iii)
t2 + t2 + 2gt + 2ft + c = 0
....(iv)
from (ii) and (iii), we have
2g (1 – t) + 2f (t –1) = 0  2(t –1) (f – g) = 0
That is, f = g, because t  1, otherwise the three given points would be the same. Again, From (iii) and
(iv), we get
and
t2 + 1 + 2gt + 2g + c = 0
....(v)
2t2 + 2gt + 2gt + c = 0
....(vi)
subtracting these, we get
t2 –1 + 2g (t –1) = 0  2g = – (t + 1)
[22]
Module Test-2-I (03-12-11) XI Synchro (TC-A)
Finally, from (vi), we get c = –2t2 + 2t (t + 1) = 2t, so that the equation of the circle (i) can be written as
x2 + y2 – (t + 1)x – (t + 1)y + 2t = 0
Which passes through (1, 1) for all values of t while the points given by (b), (c) and (d) do not satisfy
this equation.
38.
The line 9x + y – 28 = 0 is the chord of contact of the point P (h, k) with respect to the circle 2x2 + 2y2 – 3x +
5y – 7 = 0, for
(a)
P(3, –1)
(b) P(3, 1)
(c) P(–3, 1)
(d) no position of P
Hint: (d) An equation of the chord of contact of P with respect to the given circle is

2xh + 2yk –
3
5
(x + h) +
(y + k) – 7 = 0
2
2
3

x  2h   
2


5 3
5

y  2k    h  k  7  0
2
2
2


which should be same as the given line
9x + y – 28 = 0
comparing (i) and (ii) we get
AMITY
4h  3 4k  5 3h  5k  14


18
2
2  28
then h – 9k = 12 and 3h –117k =126
solving for h, k we get h = 3, k = –1.
Institute
Competitive
so, the point Pfor
is (3, –1).
But we note the S(3, –1) = 2 × Examinations
9 + 2(–1) –3(3) + 5(–1) – 7 < 0 where
2
s = 2x2 + 2y2 – 3x + 5y –7
Showing that P lies inside the given circle and then cannot be the required point. Thus the given line
cannot be the chord of contact of the given circle for any point P.
39.
A line is drawn through the point P(3, 11) to cut the circle x2 + y2 = 9 at A and B. Then PA · PB is equal to
(a)
9
(b) 121
(c) 205
(d) 139
Hint: (b) From geometry we know PA · PB = (PT)2 where PT is the length of the tangent from P to the circle.
Hence PA · PB = (3)2 + (11)2 – 9 = 112 = 121.
40.
If the line x cos  + y sin  = p represents the common chord APQB of
the circles x2+y2=a2 and x2+y2= b2(a > b) as shown in the figure, then
O
AP is equal to
(a)
a 2  p 2  b2  p 2
(b)
a 2  p 2  b2  p 2
(c)
a 2  p 2  b2  p 2
(d)
a 2  p 2  b2  p 2
A
P
p
L
Q
B
Hint: (c) The given circles are concentric with centre at (0, 0) and the length of the perpendicular from (0, 0) on the
given line is p, Let OL = p
then
AL =
(OA) 2  (OL) 2  a 2  p 2
[23]
Module Test-2-I (03-12-11) XI Synchro (TC-A)
41.
and
PL =
(OP )2  (OL ) 2  b2  p 2

AP =
a 2  p 2  b2  p 2
The normal at an end of a latus rectum of the ellipse x2/a2 + y2/b2 = 1 passes through an end of the minor axis if
(a)
e4 + e2 = 1
(b) e3 + e2 = 1
(c) e2 + e = 1
(d) e3 + e = 1


2
Hint: (a) Let an end of a latus rectum be ae, b 1  e , then the equation of the normal at this end is
x  ae y  b 1  e 2

ae / a 2 b 1  e 2 / b2
It will pass through the end (0, –b) if

b 2 1  1  e 2
2
a 
1  e2
 or b
2
a2

1  e2
1  1  e2
or (1  e 2 )[1  1  e 2 ]  1  e 2
AMITY
or
42.
1  e 2  1  e 2  1 or e 4  e2  1
A normal to the parabola y2 = 4ax with slope m touches the rectangular hyperbola x2 – y2 = a2
(a)
m6 + 4m4 – 3m2 + 1 = 0
(b) m6 – 4m4 + 3m2 – 1 = 0
(c)
m6 + 4m4 + 3m2 + 1 = 0
(d) m6 – 4m4 – 3m2 + 1 = 0
Institute
for Competitive Examinations
rectangular hyperbola x – y = a if (–2 am – am )
Hint: (c) Equation of the normal to the parabola y2 = 4ax with a slope m is y = mx – 2am – am3. It touches the
2
2
2
3 2
a2m2 (2 + m2)2 = a2(m2 – 1)
m2 (m4 + 4m2 + 4] = m2 – 1
m6 + 4m4 + 3m2 + 1 = 0
43.
Equation of the locus of the pole with respect to the ellipse
x2 y 2

  where
a 4 b4
(b) 2 = 1/a2
circle is the curve
(a)
2 = a2
(c) 2 = b2
x2 y2

 1, of any tangent line to the auxiliary
a 2 b2
(d) 2 = 1/b2
Hint: (b) Equation of the auxiliary circle is
x2 + y2 = a2
...(i)
Let (h, k) be the pole, then the equation of the polar of (h, k) with respect to the ellipse
x2 y2

 1 is
a 2 b2
Since (ii) is a tangent to circle (i)
1
2
 a 
 h   k 
 2   2 
a  b 
Locus of (h, k) is
h2 k 2
1
 4  2
4
a
b
a
x2 y2
1


a 4 b4 a 2
[24]
Module Test-2-I (03-12-11) XI Synchro (TC-A)
SECTION- II
MULTIPLE CORRECT ANSWERS TYPE
This section contains 5 multiple choice questions numbered 44 to 48. Each question has 4 choice (A), (B),
(C) and (D), out of which ONE OR MORE is/are correct
44.
A line L1 intersects x and y axes at P and Q respectively. Another line L2, perpendicular to L1, cuts x and y axis at
R and S respectively. The locus of the point of intersection of the lines PS and QR is a circle passing through the
(a)
origin
(b) point P
Hint: (a),(b),(c) Let the equation of L1 be
(c) point Q
(d) point R
x y
  1 , so that the coordinates of P and Q
a b
are (a, 0) and (0, b), respectively. Now, the slope of L1 is 
therefore, that of L2 is
Q(0, b)
b
and
a
L2
a
. Let the equation of L2 be
b
P(a, 0)
R
0
L1
x
y

1
S
bk ak
Then the coordinates of R and S are (bk, 0) and (0, – ak), respectively. Also, the equation of PS is
AMITY
x y
x y

 1 , and that of RQ is
  1 . Therefore, eliminating k, we get the required locus as
a ak
bk b
xa b y

y
x
Institute
for
Examinations
 x(x –a) + y(y
– b) =Competitive
0
 x2 + y2 – ax – by = 0
which is a circle passing through the origin, points P and Q.
45.
5
, which subtends
An equation of a circle through the origin, making an intercept of 10 on the line y  2 x 
2
an angle of 45° at the origin is
(a)
x2 + y2 – 4x – 2y = 0
(b) x2 + y2 – 2x – 4y = 0
(c)
x2 + y2 + 4x + 2y = 0
(d) x3 + y2 + 2x + 4y = 0
Hint: (b) Let an equation of the circle through the origin be
x2 + y2 + 2gx + 2fy = 0
5
10 made by the line y  2 x  2 on this circle. Since it is given
that PQ subtends an angle of 45° at the origin (a point on the circle). It subtends a right angle at the
centre (–g, –f) of the circle.
Let PQ be the intercept of length
 PQ2 = CP2 + CQ2 = 2r2
 10 = 2r2  r  5
So that g2 + f2 = 5
(i)
C (- g, - f )
Let CL be the perpendicular from C on PQ.
r
r
r/ 2
Then CL  LP  10 / 2
P
[25]
L
Q
y = 2x + 5 / 2
Module Test-2-I (03-12-11) XI Synchro (TC-A)

 f  2g  5 / 2
10

2
1 4
 2 g  f  
5
5

2
2
 2 g  f  0 or 2 g  f  5 2
2 g  f  5 2 gives imaginary values of g and f from (i). So 2g – f = 0  2g and the required
equations are
x2 + y2 – 2x – 4y = 0 and x2 + y2 + 2x + 4y = 0
46.
The Cartesian equation of the curve whose parametric equation is x = 2t – 3 and y = 4t2 – 1 is given by
(a)
(x + 3)2 – y – 1 = 0
(b) x2 + 6x – y + 8 = 0
(c)
(y + 1)2 + x + 3 = 0
(d) y2 + 6x – 2y + 4 = 0
Hint: (a), (b) Eliminating t from the given equations, we get
AMITY
2
 x  3
y  4
 1  y  1  ( x  3) 2

2


(x + 3)2 – (y + 1) = 0 or x2 + 6x – y + 8 = 0
47.
The slope of a tangent to the parabola y2 = 9x which passes through the point (4, 10) is
(a)
9/4
(b) 1/4
(c) 3/4
(d) 1/3
Institute for Competitive Examinations
Hint: (a), (b) The equation of the parabola is of the form y2 = 4ax, where a = 9/4. Equation of a tangent with slope
m is
y  mx 
a
9
 y  mx 
m
4m
Since it passes through (4, 10)
10  4m 
9
 16m 2  40m  9  0
4m
(4m – 1) (4m – 9) = 0 m = 1/4 or 9/4
48.
The equation of a directrix of the ellipse (x2/16) + (y2/25) = 1 is
(a)
y = 25/3
Hint: (a,d)
(b) x = 3
Here directrix is y = 
(c) x = –3
(d) y = – 25/3
b
b
b2
25



e
3
b2  a 2
b2  a 2
b
SECTION- III
LINKED COMPREHENSION TYPE
This section contains 2 Paragraphs. Based upon each paragraph, 3 multiple choice questions have to be
answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.
[26]
Module Test-2-I (03-12-11) XI Synchro (TC-A)
Comprehension-I
49.
The equation represents a family of circles passing through two fixed points whose coordinates are
(a)
50.
51.
(–2, 0) and (2, 0) (b) (2, 0) and (–4, 0)
(c) (4, 0) and (–2, 0)
(d) (–4, 0) and (–2, 0)
Equation of a circle C of this family, tangents so which at these fixed points intersects on the line x + 2y + 5 = 0 is
(a)
x2 + y2 – 2x – 8y – 8 = 0
(b) x2 + y2 – 2x + 6y – 8 = 0
(c)
x2 + y2 – 2x + 8y – 8 = 0
(d) x2 + y2 – 2x – 6y – 8 = 0
If the chord joining the fixed points subtends an angle  at the centre of the circle C, then  =
(a)
/6
(b) /4
(c) /3
(d) /2
Hint: 49. (c); 50. (d); 51. (d)
Equation of the given circle can be written as (x2 + y2 – 2x – 8) –2a (y) = 0 which represents a family of
circles passing through the intersection of the circle x2 + y2 – 2x – 8 = 0 and the line y = 0.
The circle and the line intersect at the points, P (–2, 0) and Q (4, 0).
Let the tangents at P and Q to a member of this family intersect at (h, k) then PQ is the chord of contact
AMITY
of (h, k) and its equation is
hx + ky – (x + h) – a(y + k) – 8 = 0
or
x(h – 1) + y(k – a) – (h + ak + 8) = 0
Comparing this with equation y = 0 of PQ, we get
h = 1, h + ak = 8 = 0
Since (h, k) lies on the given line x + 2y + 5 = 0
1 + 2k + 5 = 0 k = –3
Institute
for Competitive Examinations
and
1 – 3a + 8 = 0 a = 3
Hence the equation of the required member C of the family is x2 + y2 – 2x – 6y – 8 = 0
Centre of the circle C or R(1/3)
Slope of PR = 1 and of QR = –1
= /2
Comprehension-II
52.
Area of the rectangle formed by joining the vertices of the latera recta of the two parabolas P1 and P2 is
(a)
2a2 sq. units
(b) 4a2 sq. units
(c) 8a2 sq. units
(d) 16a2 sq. units
Hint: (c)
Required area 4a(2a) = 8a2
53.
If a = 4, then equation of the ellipse having the line joining the focci of the parabolas P1 and P2 as the major axis
and eccentricity equal to 1/2 is
(a)
x2 y2

1
4
3
(b)
x2 y2

1
3
4
x2 y2

1
(c)
16 12
[27]
x2 y2

1
(d)
12 16
Module Test-2-I (03-12-11) XI Synchro (TC-A)
 1
Hint: (c) 2a = 8 a = 4, b2 = a2 (1 – e2) = 16 1    12
 4
Ellipse is
54.
x2 y 2
x2 y 2


1

1
i.e.
a 2 b2
16 12
Equation of the tangent at the point on the parabola P1, where the line L meets the parabola is
(a)
x – 2y 4a = 0
(b) x + 2y – 4a = 0
(c) x + 2y – 8a = 0
(d) x – 2y + 8a = 0
SECTION- IV
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
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
look like the following.
AMITY
PHYSICS
1.
Two blocks of the masses 10 kg and 20 kg are placed on the x-axis. The first mass is moved on the axis by a
distance 2 cm. By what distance (in m) should the second mass be moved to keep the position of centre of mass
unchanged?


Institute
for
Competitive
Examinations

m x  m x
Hint: (1) Shift in the centre of mass = 0  x 
1
cm
2.
1
2
2
m1  m2
At the instant shown, the truck is traveling to the right at
3 m/s, while the pipe is rolling counterclockwise at
 = 8 rad/s without slipping at B. Determine the speed
(in m/sec) of the pipe’s center G.
3





Hint: (9) vB  vBG  vG 3iˆ  8  iˆ  vG  vG  (3  12)iˆ
2
3.
Two particles A and B move towards each other under a mutual force of attraction while the centre of mass of the
system moves with 3 m/sec. At the instant when the speed of A is v and the speed of B is 2v, what is the speed (in
m/sec) of the centre of mass of the system?


Hint: (3) Iext = 0  acm  0  vcm  0
4.
An inextensible and flexible carpet of length 8m is wound together and is placed
A
on a long inclined surface (as shown in the figure) with one of its ends A
fixed on the incline. When it is allowed to roll down starting from zero initial
velocity, determine the time (in sec) taken by the carpet to unwind completely.
Hint: (2) a =
37º
g sin 
1 2
at
2
2 ; S = ut +
2
1 k / R
[28]
Module Test-2-I (03-12-11) XI Synchro (TC-A)
CHEMISTRY
5.
 2CO(g) and the equilibrium
If 50% of CO2 converts to CO at the following equilibrium: C(s) + CO2(g) 
pressure is 3 atm, calculate KP
 2CO(g)
Hint: (4) C(s) + CO2(g) 
6.
O
+P
+P
PCO = 2
PCO2 =1
P=2
KP =
22
4
1
Rate of diffusion of CH4 is 2 times that of an unknown gas x. Find the molar mass of x.
AMITY
Hint: (4)
7.
P
–0.5P
P – 0.5 P
P – 0.5 P + P = 3
1.5 P = 3
rCH 4
rx

Mx
16
When one mole of each CH3COOH and C2H5OH are allowed to react, 3/4th of the reactants changes in to product
by the time equilibrium, find the equilibrium constant for this reaction.
Institute for Competitive Examinations
Hint: (9)
8.
5 mole of an ideal gas was allowed to expand isothermally and reversibly at 300 K from initial volume to final
volume (20 lit.). During this work done by the gas is 1.386 Kcal, find the initial volume of the gas.
V2
Hint: (5) w = –2.303 nRT log V
1
MATHEMATICS
9.
The minimum number of normals that can be drawn through any point in the Cartesian plane to the parabola y2 =
4ax is .....
Hint: (1) The minimum number of normals is 1 as a cubic equation must have at least one real root.
10.
If two distinct chords of a parabola y2 = 4ax, passing through (a, 2a) are bisected on the line x + y = 1, then length
of latus rectum can be less than
Hint: (4) Let the point where the chord is bisector be (, 1 – ). Now equation of the chord is
y(1 – ) – 2ax = (1 –)2
Putting x = a and y = 2a we get 2a(1 – ) = (1 – )2
i.e. a(a – 1) < 0
i.e. 0 < a < 1
i.e. length of LR < 4.
[29]
Module Test-2-I (03-12-11) XI Synchro (TC-A)
11.
The circle x2 + y2 + 4x – 7y + 12 = 0 cuts an intercept on y-axis equal to ..... units.
Hint: (1) The length of interception on y-axis is given by
2
7
2 f 2  c  2    12  1
2
12
.
If a chord of the circle x2 + y2 = 8 make equal intercept of length a on the coordinate axes, then a < ....
Hint: (4) A chord with equal intercept length a is given by x + y = a.
Since a chord intersects a circle at two distinct points.
The D of x2 + (a – x)2 = 8 is positive.
2x2 – 2ax + a2 – 8 = 0
4a2 – 8(a2 – 8) > 0
–4a2 + 64 > 0
AMITY
a2 – 16 < 0
–4 < a < 4
Institute for Competitive Examinations
[30]
Module Test-2-I (03-12-11) XI Synchro (TC-A)