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RESONANCE Page # 1
QUESTIONS & SOLUTIONS OFKISHORE VAIGYANIK PROTSAHAN YOJANA (KVPY) 2010
TEST PAPER CLASS - XII
PART-I
One-Mark QuestionsMATHEMATICS
1. Let denote the matrix
0i
i0, where i2 = �1, and let I denote the identity matrix
10
01. Then
I + A + A2 + .............+ A2010 is
(A)
00
00(B)
0i
i0(C)
1i
i1(D)
1�0
01�
Ans. (C)
Sol. I =
10
01 + A + A2 + A3 + ............+ A2010
A =
0i
i0( + A + A2 + A3) + A4 (+ A + A2 + A3) ..... + A2008 (+ A + A2)
A2 =
1�0
01�
A3 =
0i�
i�0= 0 +
0i
i0
A4 = A4n =
+ A + A2 + A3 = 0
2. Suppose the sides of a triangle from a geometric progression with common ratio r. Then r lies in the interval
(A)
251�
,0 (B)
252
,2
51(C)
251
,2
51�(D)
,
252
Ans. (A)
(TEST PAPER HELD ON 31-10-2010)
RESONANCE Page # 2
Sol. Let a, ar, ar2 are sides.
Case 1 : r > 1
a + ar > ar2
r2 � r � 1 < 0
r
251
,2
5�1 r
25�1
,1
Case 2 : r = 1 equilateral triangle
Case 3 : r < 1
Final solution
ar2 + ar > a r
215
,12
1�5,0
r2 + r � 1 > 0 r
251�
,0
3. The number of rectangles that can be obtained by joining four of the twelve vertices of a 12-sided regular
polygon is
(A) 66 (B) 30 (C) 24 (D) 15
Ans. (D)
Sol. Number of diagonals passing through centre = 6
Number of rectangles = 6C2 = 15
4. Let 1, and 2 be the cube roots of unity. The least possible degree of a polynomial, with real coefficients,
having 22, 3 + 4, 3 + 42 and 5 � �2 as roots is
(A) 4 (B) 5 (C) 6 (D) 8
Ans. (B)
Sol. 22, 3 + 4, 3 + 42, 5 � � 2
� 1 � 3 i , 1 + 32 i , 1 � 32 i, 6
where = 2
i31� , 2 =
2i3�1�
1 + 32 i , 1 32� i are conjugate of each other..
Least possible degree = 5
5. A circle touches the parabola y2 = 4x at (1, 2) and also touches its directrix. the y-coordinate of the point of
contact of the circle and the directrix is
(A) 2 (B) 2 (C) 22 (D) 4
Ans. (A)
RESONANCE Page # 3
Sol. Tangent at (2, 1)
y.2 = 2 (x + 1)
y = x + 1
x � y + 1 = 0
equation to circle
(x � 1)2 + (y � 2)2 + (x � y + 1) = 0
putting x = � 1
4 + (y � 1)2 + (� y) = 0
y2 � 4y � y + 8 = 0
D = 0 (+ 4)2 = 32
y = 4
D)4(
y = 4
024 = ± 2
6. Let ABC be an equilateral triangle; let KLMN be a rectangle with K, L on BC, M on AC and N on AB. Suppose
AN / NB = 2 and the area of triangle BKN is 6. The area of the triangle ABC is
(A) 54 (B) 108
(C) 48 (D) not determinable with the above data
Ans. (B)
Sol. BKN = 6
91
APBBKN
3
1
AB
BN
1
2
BN
AN
APB = 19
× 6
= 1
54
ABC = 108
7. Let P be an arbitrary point on the ellipse 1b
y
a
x2
2
2
2
, a > b > = 0. Suppose F1 and F
2 are the foci of the
ellipse. The locus of the centroid of the triangle PF1F
2 as P moves on the ellipse is
(A) a circle (B) an ellipse (C) a parabola (D) a hyperbola
Ans. (B)
RESONANCE Page # 4
Sol. 2
2
2
2
b
y
a
x = 1 , a > b > 0
h = 3x
, k = 3y
2
2
2
2
b
)k3(
a
)h3( = 1 which is ellipse
8. The number of roots of the equation cos7 � sin6 = 1 that lie in the interval [0, 2] is
(A) 2 (B) 3 (C) 4 (D) 8
Ans. (A)
Sol.
1
6
1
7 sin1cos
LH.S is less than or equal 1
while R.H.S is greater or equal to 1
possible when
cos = 1 and sin = 0
= 0, 2
9. The product
(1 + tan 1º) (1 + tan 2º) (1 + tan 3º) ... (1 + tan 45º)
equals
(A) 221 (B) 222 (C) 223 (D) 224
Ans. (C)
Sol. (1 + tan 1º) (1+ tan 2º) .........(1+ tan 45º)
(1 + tan A) (1 + tan B) = 2 if A + B = 45º
Ans = 223
10. Let f : R R be a differentiable function such that f (a) = 0 = f (b) and f�(a) f�(b) > 0 for some a < b. Then the
minimum number of roots of f�(x) = 0 in the interval (a, b) is
(A) 3 (B) 2 (C) 1 (D) 0
Ans. (B)
Sol. f is differentiable function .
f (a) f(b) > 0 f is increasing at x = a and b
or f is decreasing at x = a and b
minimum number of roots of f (a) = 0 in (a, b) is 2,
RESONANCE Page # 5
11. The roots of (x � 41)49 + (x � 49)41 + (x � 2009)2009 = 0 are
(A) all necessarily real
(B) non-real except one positive real root
(C) non-real except three positive real roots
(D) non-real except for three real roots of which exactly one is positive
Ans. (B)
Sol. When x is less than 49 then f(x) has negative value.
Which is not possible.
when x > 49
then f(x) = 49 (x � 41)48 + 41 (x � 49)40
+ 2009 (x � 2009)2008
so sing of f(x) does not change
f(x) > 0
non real except one positive root.
12. The figure shown below is the graph of the derivative of some function y = � (x)
Then
(A) f has local minima at x = a, b and a local maximum at x = c
(B) f has local minima at x = b, c and a local maximum at x = a
(C) f has local minima at x = c, a and a local maximum at x = b
(D) the given figure is insufficient to conclude any thing about the local minima and local maxima of f
Ans. (C)
Sol.
f (x) is given smooth curve hence differentiable i- given domain
By first derivative test
x = a, c is point of local maxima
x = b is point of local minima
RESONANCE Page # 6
13. The following figure shows the graph of continuous function y = � (x) on the interval [1, 3]. The points A, B,
C have coordinates (1, 1), (3, 2), (2, 3) respectively, and the lines 1 and
2 are parallel, with
1 being tangent
to the curve at C. If the area under the graph of y = � (x) from x = 1 to x = 3 is 4 square units, then the area
of the shaded region is
(A) 2 (B) 3 (C) 4 (D) 5
Ans. (A)
Sol.
3
i
4dx)x(f
shaded area = Trapezium Area � Area under curve
= 2
)1�3(27
25
� 4 = 2
14. Let
e
1
nn )x(log dx, where n is a non-negative integer..
Then I2011
+ 2011 I2010
is equal to
(A) 1000
+ 999 998
(B) 889
+ 890 891
(C) 100
+ 100 99
(D) 53
+ 54 52
Ans. (C)
RESONANCE Page # 7
Sol. n = dx.)x(log
e
1
n
n = dxx.
x)x(logn
�)x(logxe
i
1�ne
1
n
n = (e � 0) � n
n�1
n + n
n�1 = e
2011
+ 2011 2010
= e
100
+ 100 99
= e
15. Consider the regions A = {(x, y) | x2 + y2 100} and B = {(x, y) | sin (x + y) > 0} in the plane. Then the area
of the region A B is
(A) 10 (B) 100 (C) 100 (D) 50
Ans. (D)
Sol. x2 + y2 100 Area bounded = 100
sin (x + y) > 0
Hence common area of region bounded = 50 as in half the region sin (x + y) > 0 and in other half
sin (x + y) < 0
16. Three vertices are chosen randomly from the seven vertices of a regular 7�sided polygon. The probability that
they form the vertices of an isosceles triangle is
(A) 71
(B) 31
(C) 73
(D) 53
Ans. (D)
Sol. n(S) = 7C3 = 35 n () = 7 + 7 + 7 = 21 P () =
53
3521
17. Let u
= j��i�2 + k� , v = � 3 j� + 2 k� be vectors in R3 and w be a unit vector in the xy-plane. Then the
maximum possible value of |(u
× v ) . w
| is
(A) 5 (B) 12 (C) 13 (D) 17
Ans. (D)
Sol. w).vu(
, vu
= 23�0
11�2
k�j�i�
= )6(�k�)4(j��)32(�i� = 53
greatest when x2 + y2 = 1
fmax.
= 17 Ans. (D)
RESONANCE Page # 8
18. How many six-digit numbers are there in which no digit is repeated, even digits appear at even places, odd
digits appear at odd places and the number is divisible by 4?
(A) 3600 (B) 2700 (C) 2160 (D) 1440
Ans. (D)
Sol.
Number of ways = 4P2 × 4P
2 × 10 = 1440
19. The number of natural number n in the interval [1005, 2010] for which the polynomial
1 + x + x2 + x3 +.......+ xn�1 divides the polynomial 1 + x2 + x4 + x6 + .........+ x2010 is
(A) 0 (B) 100 (C) 503 (D) 1006
Ans. (C)
Sol. 1 + x2 + x4 + .........+ x2010
= 2
10062
x�1
)x(�1
= )x�1)(x1(
)x(�1 21006
= )x�1(
)x�1()x1(
)x1( 10061006
= (1 + x1006) )x1(
)x1()x�1(
)x�1( 503503
= (1 + x1006) (1 + x + x2 + ...............+ x502) (1 � x + x2 � x3 + ............ + x502)
n � 1 = 502 n = 503
20. Let a0 = 0 and a
n = 3 a
n�1 + 1 for n 1. Then the remainder obtained on dividing a
2010 by 11 is
(A) 0 (B) 7 (C) 3 (D) 4
Ans. (A)
Sol. an = 3a
n�1 + 1
= 3(3an�2
+ 1) + 1
= 32 an�2
+ 3 + 1
= 32 (3an�3
+ 1) + 3 + 1
= 33 an�3
+ 32 + 3 + 1
aN = 3n a
0 + 3N�1 + 3N�2 + .... 1
RESONANCE Page # 9
Method - I
a2010
= (1 + 3 + 32 + 33 + 34) + ....... + 32009
= (1 + 3 + 32 + 33 + 34)+ 35 (1 + 3 + 32 + 33 + 34)+ 310 (1 + 3 + 32 + 34 + 35) +.....+32005 (1+ 3 + 32 + 33 + 34)
each (1 + 3 + 32 + 33 + 34) is divisble by
Method-II
a2010
=2
1�)1242(2
1�)243(2
1�32
1�3 40240240252010
= 11 k (k I)
Hence remainder is 0
PHYSICS
21. A pen of mass m is lying on a piece of paper of mass M placed on a rough table. If the coefficients of friction
between the pen and paper, and, the paper and the table are 1 and
2 respectively, then the minimum
horizontal force with which the paper has to be pulled for the pen to start slipping is given by :
(A) (m + M) (1 +
2)g (B) (m
1 + M
2)g
(C) (m 1 + (m + M)
2)g (D) m(
1 +
2)g
Ans. (A)
Sol.
F � 1mg �
2(m + M)g = M.
mmg1
F = 1mg +
2(M + m)g +
1Mg
= 1g(M + m) +
2g(m + M)
= (1g +
2g) (M + m)
22. Two masses m1 and m
2 connected by a spring of spring constant k rest on a frictionless surface. If the
masses are pulled apart and let go, the time period of oscillation is :
(A) T =
21
21
mm
mm
k1
2 (B) T =
21
21
mm
mm k2 (C) T =
k
m2 1 (D) T =
k
m2 2
Ans. (A)
Sol. T = k1
mm
mm2
21
21
we can use concept of reduce mass (A).
RESONANCE Page # 10
23. A bead of mass m is attached to the mid�point of a taut, weightless string of length and placed on a
frictionless horizontal table.
Under a small transverse displacement x, as shown, if the tension in the string is T, then the frequency of
oscillation is :
(A) mT2
21
(B) mT4
21
(C) mT4
21
(D) mT2
21
Ans. (B)
Sol. 2T cos = � ma ; 22
x4
xmT2
= a
If x < < < < 2
2
xmT2
= a a = m
Tx4
= mT4
T = T4
m2
.
24. A comet (assumed to be in an elliptical orbit around the sun) is at a distance of 0.4 AU from the sun at the
perihelion. If the time period of the comet is 125 years, what is the aphelion distance ? AU : Astronomical
Unit.
(A) 50 AU (B) 25 AU (C) 49.6 AU (D) 24.6 AU
Ans. (C)
Sol. AU = mean distance between Sun and Earth
and Time period = 1y
T2 r3
(1 yr)2 (1A0)3
(125)2 3
a
2
4.0r
25 =
2
4.0ra
ra = 50 � 0.4
= 49.6.
RESONANCE Page # 11
25. The circuit shown consists of a switch (S), a battery (B) of emf E, a resistance R, and an inductor L.
The current in the circuit at the instant the switch is closed is :
(A) E/R (B) E/(R(1 � e)) (C) (D) 0
Ans. (D)
Sol. zero (D)
26. Consider a uniform spherical volume charge distribution of radius R. Which of the following graphs correctly
represents the magnitude of the electric field E at a distance r from the center of the sphere ?
(A) (B) (C) (D)
Ans. (A)
Sol. 3R
KQrE r inside
2r
KQE 2r
1 outside.
27. A charge +q is placed somewhere inside the cavity of a thick conducting spherical shell of inner radius R1
and outer radius R2. A charge +Q is placed at a distance r > R
2 from the center of the shell. Then the electric
field in the hollow cavity.
(A) depends on both +q and +Q (B) is zero
(C) is only that due to +Q (D) is only that due to +q
Ans. (D)
Sol. There is no net effect of outside charge.
So, (D).
28. The following travelling electromagnetic wave
Ex = 0, E
y = E
0sin(kx + t), E
z = �2E
0sin(kx + t) is :
(A) elliptically polarized (B) linearly polarized
(C) circularly polarized (D) unpolarized
Ans. (C)
RESONANCE Page # 12
29. A point source of light is placed at the bottom of a vessel which is filled with water of refractive index to a
height h. If a floating opaque disc has to be placed exactly above it so that the source is invisible from above,
the radius of the disc should be :
(A) 1
h
(B)
1
h2
(C) 1
h2
(D) 1
h2
Ans. (B)
Sol. tanC = hr
from snells law
sinC =
1
r = 1
h2
30. Three transparent media of refractive indices 1,
2,
3, respectively, are stacked as shown. A ray of light
follows the path shown. No light enters the third medium.
Then :
(A) 1 <
2 <
3(B)
2 <
1 <
3(C)
1 <
3 <
2(D)
3 <
1 <
2
Ans. (B)
Sol. 2 >
3
1sin i
1 =
2 × sin r
1 <
2
From condition TIR
2 >
3and
1 >
3
So, 2 >
1 >
3 .
31. A nucleus has a half�life of 30 minute. At 3PM its decay rate was measured as 120,000 counts/sec. What
will be the decay rate at 5 PM ?
(A) 120,000 counts/sec. (B) 60,000 counts/sec.
(C) 30,000 counts/sec. (D) 7,500 counts/sec.
Ans. (D)
Sol. A = T/t0
2
AA = 42
120000=
222210100120
= 7500 (D).
RESONANCE Page # 13
32. A block is resting on a shelf that is undergoing vertical simple harmonic oscillations with an amplitude of 2.5
cm. What is the minimum frequency of oscillation of the shelf for which the book will lose contact with the
shelf ? (Assume that g = 10 m/s2)
(A) 20 Hz (B) 3.18 Hz (C) 125.6 Hz (D) 10 Hz
Ans. (D)
Sol. 2 A = g
Ag
f = 21
Ag
= 2105.2
8.921
= 3.18
33. A vander Waal's gas obeys the equation of state
2
2
V
anP (V � nb) = nRT. Its internal energy is given by
U = CT � Van2
. The equation of a quasistatic adiabat for this gas is given by :
(A) TC/nRV = constant (B) T(C+nR)/nRV = constant
(C) TC/nR (V � nb) = constant (D) P(C+nR)/nR (V � nb) = constant
Ans. (A)
Sol. dQ = 0
�du = dw
�nCvdt = Pdv
du = C dT + 2
2
v
andv
cdT + dvv
an2
2
= Pdv
P = 2
2
v
annbv
nRT
cdT + dvv
an2
2
= dvv
annbv
nRT2
2
nRC
nbvdv
TdT
RESONANCE Page # 14
nRC
nT = n (v � nb) + C
n nR/CT � n (v � nb) = C
nbvT nR/C
= ec = K
nR/CT = Kv � Knb
Kv nR/CT = � Knb = constant
v nR/CT = constant.
34. An ideal gas is made to undergo a cycle depicted by the PV diagram alongside. The curved line from A to B
is an adiabat.
Then :
(A) The efficiency of this cycle is given by unity as no heat is released during the cycle
(B) Heat is absorbed in the upper part of the straight line path and released in the lower part
(C) If T1 and T
2 are the maximum and minimum temperatures reached during the cycle, then the efficiency is
given by 1 � 1
2
TT
(D) The cycle can only be carried out in the reverse of the direction shown in the figure
Ans. (B)
35. A bus driving along at 39.6 kmph is approaching a person who is standing at the bus stop, while honking
repeatedly at an interval of 30 seconds. If the speed of sound is 330 ms�1, at what interval will the person hear
the horn.
(A) 31 sec. (B) 29 sec.
(C) 30 sec. (D) The interval will depend on the distance of the bus from the passenger
Ans. (B)
Sol.
RESONANCE Page # 15
v0 = 39.6 kmph = 39.6 ×
185
= 2.2 × 5 = 111
t1 =
330d
t2 =
3303011d
= 330
d �
3303011
t = t1 � t
2 = 1
time interval = 30 � 1 = 29 second
Ans. (B)
36. Velocity of sound measured at a given temperature in oxygen and hydrogen is in the ratio :
(A) 1 : 4 (B) 4 : 1 (C) 1 : 1 (D) 32 : 1
Ans. (A)
Sol. v = MRT
41
322
M
M
v
v
0
H
H
0
37. In Young's double slit experiment, the distance between the two slits is 0.1 mm, the distance between the
slits and the screen is 1m and the wavelength of the light used is 600 nm. The intensity at a point on the
screen is 75% of the maximum intensity. What is the smallest distance of this point from the central fringe?
(A) 1.0 mm (B) 2.0 mm (C) 0.5 mm (D) 1.5 mm
Ans. (A)
Sol. d = 0.1 mm
D = 1m
= 600 nm
given IR = 75% of maximum = 75% of 4 = 3
0
R = 3 = + + 2 cos
= 3
x = Ddy
=
2 ×
3
= 6
y = d6
D = 3
9
101.06
106001
y = 1 meter
RESONANCE Page # 16
38. Two masses m1 and m
2 are connected by a massless spring of spring constant k and unstretched length .
The masses are placed on a frictionless straight channel � which we consider our x-axis. They are initially at
rest at x = 0 and x = , respectively. At t = 0, a velocity of v0 is suddenly imparted to the first particle. At a later
time t, the center of mass of the two masses is at :
(A) x = 21
2
mmm
(B) x = 21
02
21
1
mm
tvm
mmm
(C) x = 21
02
21
2
mm
tvm
mmm
(D) x = 21
01
21
2
mm
tvm
mmm
Ans. (D)
Sol.
vcom
= 21
01
mm
0vm
x
i =
21
21
mm)(m)0(m
xcm
= xi + v
com t
xcm
= 21
01
21
2
mm
tvm
mm)(m
39. A charged particle of charge q and mass m, gets deflected through an angle upon passing through a square
region of side a which contains a uniform magnetic field B normal to its plane. Assuming that the particle
entered the square at right angles to one side, what is the speed of the particle ?
(A) cotamqB
(B) tanamqB
(C) 2cota
mqB
(D) 2tana
mqB
Ans.
Sol. According to given information
sin = Ra
RESONANCE Page # 17
CHEMISTRY
41. The number of isomers of Co(diethylene triamine) Cl3 is.
(A) 2 (B) 3 (C) 4 (D) 5
Ans. (B)
Sol. [Co(dien)Cl3]
dien = NH2� CH
2 � CH
2 � NH � CH
2 � CH
2 � NH
2
[Co(dien)Cl3] have two geometrical isomers cis and trans.
cis form is optically active but trans form is optical inactive.
42. Among the following, the -acid ligand is :
(A) F� (B) NH3
(C) CN� (D) I�
Ans. (C)
Sol. The ligands which allow back bonding to sufficient extent are called acid acceptors.
Example CO, CN�, NO etc.
43. The bond order in O2
2� is
(A) 2 (B) 3 (C) 1.5 (D) 1
Ans. (D)
Sol. O2
2� : Total electron = 18 ; Bond order = 1.
Peroxide (O22�) : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p
z)2 (2p2
x = 2p2
y ) (*2p
x2 = *2p2
y)
Bond order =
2
N�N AB =
28�10
= 1.
RESONANCE Page # 18
44. The energy of a photon of wavelength = 1 meter is (Plank's constant = 6.626 10�34 Js, speed of light = 3 108
ms�1)
(A) 1.988 10�25 J (B) 1.988 10�30 J (C) 1.988 10�28 (D) 1.988 10�31
Ans. (A)
Sol. E =
hc =
110310262.6 834
= 1.988 × 10�25 J.
45. The concentration of a substance undergoing a chemical reaction becomes one half of its original value after
time t, regardless of the initial concentration. The reaction is an example of a
(A) zero order reaction (B) second order reaction
(C) first order reaction (D) third order reaction
Ans. (B)
Sol. t1/2
(half-life) is independent of initial concentration so reaction is first order.
46. The shape of the molecule CIF3 is :
(A) trigonal planar (B) T-shaped (C) pyramidal (D) Y-shaped
Ans. (C)
Sol. ClF3
F
Cl
F
87.5º
87.5º
F
F
....
F nearly 'T' shaped.
47. Friedal�Crafts acylation is :
(A) -acylation of a carbonyl compound (B) acylation of phenols to generate esters
(C) acylation of aliphatic olefins (D) acylation of aromatic nucleus
Ans. (D)
Sol. 3AlCl
O||
Cl�C�R
Freidal craft acylation
RESONANCE Page # 19
48. The order of acidity of compounds I�IV, is
I II III IV
(A) I < III < II < IV (B) III < I < II < IV (C) IV < I < II < III (D) II < IV < III < I
Ans. (A)
Sol. > > >
Stability conjugate base order
> > >
49. The most stable conformation for n-butane is
(A) (B) (C) (D)
Ans. (A)
Sol. anti conformation is most stable
50. In the nuclear reaction
Th23490 Pa234
91 + X
X is :
(A) e01 (B) e0
1 (C) H11 (D) H2
1
Ans. (A)
Sol. ePaTh 01�
23491
23490
RESONANCE Page # 20
51. A concentrated solution of copper sulphate, which is dark blue in colour, is mixed at room temperature with
a dilute solution of copper sulphate, which is light blue. For this process
(A) Entropy change is positive, but enthalpy change is negative.
(B) Entropy and enthalpy changes are both positive.
(C) Entropy change is positive and enthalpy does not change.
(D) Entropy change is negative and enthalpy change is positive.
Ans. (A)
Sol. Entropy change is positive, but enthalpy change is negative.
52. Increasing the temperature increacses the rate of reaction but does not increase the :
(A) number of collisions (B) activation energy
(C) average energy of collisions (D) average velocity of the reactant molecules
Ans. (B)
Sol. On increasing temperature, change in activation energy is not significant.
53. In metallic solids, the number of atoms for the face centred and the body-centered cubic unit cells, are,
respectively.
(A) 2,4 (B) 2,2 (C) 4,2 (D) 4,4
Ans. (C)
Sol. For FCC unit cell number of atom per unit cell
Z = 8[corner] × 81
+ 6 [face center] ×21
= 4.
For body center unit cell number of atom per unit cell
Z = 8[corner] × 81
+ 1 [body center] × 1 = 2.
54. From equation 1 and 2
CO2 CO +
21
O2
[1cK = 9.1 10�12 at 1000°C] (eq 1)
H2O H
2 +
21
O2
[2cK = 7.1 10�12 at 1000°C] (eq 2)
(A) 0.78 (B) 2.0 (C) 16.2 (D) 1.28
Ans. (D)
RESONANCE Page # 21
Sol. CO2 CO +
21
O2 1CK = 9.1 × 10�12 .....(i)
H2O H
2 +
21
O2 2CK = 7.1 × 10�12 .....(ii)
Target eq.
CO2 + H
2 CO + H
2O K
C = ?
Target eq.
[Eq. (i) � Eq(ii)] KC =
2
1
C
C
K
K = 12
12
101.7
101.9
= 1.28.
55. For a first order reaction R P, the rate constant is k. If the initial concentration of R is [R0], the concentra-
tion of R at any time 't' is given by the expression.
(A) [R0] ekt (B) [R
0] (e�kt) (C) [R
0] e�kt (D) [R
0] (�ekt)
Ans. (B)
Sol. R P
t=0 R0
0
t=t (R0 � x) x
Rate = � dt
]R[d = K[R]1
= dtdx
= K[R0 � x] =
x
0 0dx.
)x�R(1
= t
0
dt.K
= ln
x�R
R
0
0= Kt
[R] = (R0 � x] = R
0e�Kt
56. The correct structure of PCl3F
2 is
(A) (B) (C) (D)
Ans. (C)
Sol. PCl3F
2 :
RESONANCE Page # 22
57. The enantiomeric pair among the following four structures is
I II
III IV
(A) I & II (B) I & IV (C) II & III (D) II & IV
Ans. (B)
Sol. (I) & (IV) are enantiomeric pair
I IV
58. Consider the reaction : 2NO2 (g) 2NO (g) + O
2(g). In the figure below, identify the curves X,Y and Z
associated with the three species in the reaction
(A) X = NO, Y = O2, Z = NO
2(B) X = O
2, Y = NO, Z = NO
2
(C) X = NO2, Y = NO , Z = O
2(D) X = O
2, Y= NO
2, Z = NO
Ans. (A)
RESONANCE Page # 23
Sol. 2NO2 (g) 2NO (g) + O
2 (g)
NO2 is reactant so its concentration is decrease with time while NO and O
2 are product so their concentration
increases with time.
Formation of moles of NO is double than O2.
So X = NO, Y = O2 and Z = NO
2.
59. The aromatic carbocation among the following is
(A) (B) (C) (D)
Ans. (C)
Sol. � Cyclic
� Planar
� Complete conjugation
� (4n + 2) pe�
� Tropylium carbocation
60. Cyclohexene is reacted with bromine in CCl4 in the dark. The product of the reaction is
(A) (B) (C) (D)
Ans. (A)
Sol. Anti addition takes place.
BIOLOGY
61. Ribouncleic Acid (RNA) that catalyze enzymatic reactions are called ribozymes. Which one of the following
acts as a ribozyme?
(A) Ribosome (B) Amylase (C) tRNA (D) Riboflavin
Ans. (B)
62. In 1670, Robert Boyle conducted an experiment wherein he placed a viper (a poisonous snake) in a chamber
and rapidly reduced the pressure in that chamber. Which of the following would be true?
(A) Gas bubble developed in the tissues of the snake
(B) The basal metabolic rate of the snake increased tremendously
(C) The venom of the snake was found to decrease in potency
(D) The venom of the snake was found to increase in potency
Ans. (A)
RESONANCE Page # 24
63. Bacteria can survive by absorbing soluble nutrients via their outer body surface, but animals cannot, because
(A) Bacteria cannot ingest particles but animals can
(B) Bacteria have cell walls and animals do not
(C) Animals have too small a surface area per unit volume as compared to bacteria
(D) Animals cannot metabolize soluble nutrients
Ans. (B)
64. A horse has 64 chromosomes and a donkey has 62. Mules result from crossing a horse and donkey. State
which of the following in INCORRECT?
(A) Mules can have either 64, 63 or 62 chromosomes
(B) Mules are infertile
(C) Mules have well defined gender (male/female)
(D) Mules have 63 chromosomes
Ans. (C)
65. If the total number of photons falling per unit area of a leaf per minute is kept constant, then which of the
following will result in maximum photosynthesis?
(A) Shining green light
(B) Shining sunlight
(C) Shining blue light
(D) Shining ultraviolet light
Ans. (B)
66. Path-finding by ants is by means of
(A) Visually observing landmarks
(B) Visually observing other ants
(C) Chemical signals between ants
(D) Using the earth's magnetic field
Ans. (C)
67. Sometimes urea is fed to ruminates to improve their health. It works by
(A) Helping growth of gut microbes that break down cellulose
(B) Killing harmful microorganisms in their gut
(C) Increasing salt content in the gut
(D) Directly stimulating blood cell proliferation
Ans. (A)
RESONANCE Page # 25
68. If you compare adults of two herbivore species of different sizes, but form the same geographical area, the
amount of faeces produced per kg body weight would be
(A) More in the smaller one than the larger one
(B) More in the larger one than the smaller one
(C) Roughly the same amount in both
(D) Not possible to possible to predict which would be more
Ans. (A)
69. Fruit wrapped in paper ripens faster than when kept in open air because
(A) Heat of respiration is retained better
(B) A chemical in the paper helps fruit ripening
(C) A volatile substance produced by the fruit is retained better and helps in ripening
(D) The fruit is cut off from the ambient oxygen which is an inhibitor is fruit ripening.
Ans. (C)
70. When a person is suffering form high fever, it is sometimes observed that the skin has a reddish tinge. Why
does this happen?
(A) Red colour of the skin radiates more heat
(B) fever causes the release of a red pigment in the skin
(C) There is more blood circulation to the skin to keep the body warm
(D) There is more blood circulation to the skin to release heat from the body
Ans. (D)
71. Bacteriochlorophylls are photosynthetic pigments found in phototrophic bacteria. Their function is distinct
from the plant chlorophylls in that they
(A) do not produce oxygen
(B) do not conduct photosynthesis
(C) absorb only blue light
(D) function without a light source
Ans. (A)
72. Athletes often experience muscle cramps. Which of the following statements is true about muscle cramps?
(A) Muscle cramp is caused due to conversion of pyruvic acid into lactic acid in the cytoplasm
(B) Muscle cramp is caused due to conversion of pyruvic acid into lactic acid in the mitochondria
(C) Muscle cramp is caused due to nonconversion of glucose to pyruvate in the cytoplasm
(D) Muscle cramp is caused due to conversion of pyruvic acid into ethanol in the cytoplasm
Ans. (A)
73. A couple went to a doctor and reported that both of them are "carries" for a particular disorder, their first childis suffering from that disorder and that they are expecting their second child. What is the probability that thenew child would be affected by the same disorder?(A) 100% (B) 50% (C) 25% (D) 75%
Ans. (C)
RESONANCE Page # 26
74. Of the following combinations of cell biological processes which one is associated with embryogenesis?(A) Mitosis and Meiosis (B) Mitosis and Differentiation(C) Meiosis and Differentiation (D) Differentiation and Reprogramming
Ans. (D)
75. Conversion of the Bt protoxin produced by Bacillus thuringinesis to its active form in the gut of the insects ismediated by(A) acidic pH of the gut (B) alkaline pH of the gut(C) lipid modification of the protein (D) cleavage by chymotrypsin
Ans. (A)
76. If you dip a sack full of paddy seeds in water overnight and then keep it out for a couple of days, it feels warm.What generates the heat?(A) Imbibition(B) Exothermic reaction between water and seed coats(C) Friction amount seeds due to swelling(D) Respiration
Ans. (D)
77. Restriction endonucleases are enzymes that cleave DNA molecules into smaller fragments. Which type ofbond do they act on?(A) N-glycosidic Bond(B) Phosphodiester bond(C) Hydrogen bond(D) Disulfide bond
Ans. (B)
78. The fluid part of blood flows in and out of capillaries in tissues to exchange nutrients and waste materials.Under which of the following conditions will fluid flow out from the capillaries into the surrounding tissue?(A) When arterial blood pressure exceeds blood osmotic pressure(B) When arterial blood pressure is less than blood osmotic pressure(C) When arterial blood pressure is equal to blood osmotic pressure(D) Arterial blood pressure and blood osmotic pressure have nothing to do with the outflow of fluid fromcapillaries
Ans. (A)
79. The distance between two consecutive DNA base pairs is 0.34 nm. If the length of a chromosome is 1 mmthe number of base pairs in the chromosome is approximately(A) 3 million (B) 1.5 million (C) 30 million (D) 6 million
Ans. (A)
80. Estimate the order of the speed of propagation of an action potential or nerve impulse(A) nm/s (B) micron/s (C) cm/s (D) m/s
Ans. (D)
RESONANCE Page # 27
Two-Marks QuestionsPART - II
MATHEMATICS
81. Arrange the expansion of
n
4/12/1
x2
1x
in decreasing powers of x. Suppose the coefficients of the first
three terms form an arithmetic progression. Then the number of terms in the expansion having integer
powers of x is
(A) 1 (B) 2 (C) 3 (D) more than 3
Ans. (C)
Sol.n
4/12/1
x2
1x
=
n
4/32/n
x2
11x
= xn/2
.........
x2
1c....
x2
1c
x2
1cc
4/3rn
2
4/32n
4/31n
0n
nc0 , 2
2n
1n
2
c,
2c
are in AP
1, 2n
, 8
)1�n(n are in APAP
n = 1 + 8
)1�n(n
8(n � 1) � n (n � 1) = 0
(n � 1) (8 � n) = 0 n = 1 or n = 8
but n = 1 i.e. not posible
n = 8
Now8
4/12/1
x2
1x
=
r
4/1
8
0r
r�82/1r
8
x2
1)x(c
= 4r3
�4r8
0rr
8 x21
c
required number of terms = 3 (r = 0, 4, 8) ans (C)
RESONANCE Page # 28
82. Let r be a real number and n N be such that the polynomial 2x2 + 2x +1 divides the polynomial (x + 1)n�r.
Then (n, r) can be
(A) (4000, 41000) (B)
10004
1,4000 (C)
1000
1000
4
1,4 (D)
40001
,4000
Ans. (B)
Sol. Roots of 2x2 + 2x = 1 = 0 are , 2
i1
It will satisfies the (x + 1)n � r = 0
n
12
i1�
= r
n
12
i1�
= r,, r R
n
4in
e2
1
= r
n
4i
e
will be real only if n is multiple of 4.
For n = 4000
r = 4000
2
1
= 10004
1
83. Suppose a, b are real numbers such that ab 0. Which of the following four figures represents the curve
(y � ax � b) (bx2 + ay2 � ab) = 0 ?
(A) Fig.1 (B) Fig.2 (C) Fig.3 (D) Fig.4
Ans. (B)
Sol. y = ax + b and 1by
ax 22
fig. 1 and 3 incorrect (a < 0 b the equation of line y = ax + b)
Now for a > 0 and b < 0 fig. 2 is correct fig. 2
RESONANCE Page # 29
84. Among all cyclic quadrilaterals inscribed in a circle of radius R with one of its angles equal to 120º, consider
the one with maximum possible area. Its area is
(A) 2R2 (B) 2 R2 (C) 2R3 (D) 2R32
Ans. (C)
Sol. Area of cyclic quadratic and is maximum only if area of ABC and area of ADC is maximum
Area of ABC is maximum if ABC is equilateral .
AC = 2R sin 60º = 3 R.
maximum Area of ABC = 2)R3(43
= 2R433
maximum area of ADC = 21
(AC) (DM)
= 21
º60cot
2R3
)R3(
= 4R3 2
area of ABCD = 222 R3R43
R433
85. The following figure shows the graph of a differentiable function y = � (x) on the interval [a, b] (not containing
0).
Let g(x) = � (x) / x. Which of the following is a possible graph of y = g (x)?
RESONANCE Page # 30
(A) Fig.1 (B) Fig.2 (C) Fig.3 (D) Fig.4
Ans. (B)
Sol.
f(x) is monotonic increasing in [a, )
f(x) > 0 x [a,)
f(x) is monotonic decreasing in (, b] f(x) < 0 x (, b]
Now g(x) = x
)x(f
g(x) = 2x
)x(f�)x(xf
h(x) = xf(x) � f(x)
h(x) = f(x) + xf(x) � f(x)
h(x) = xf < 0 { x > 0 and f(x) < 0 concavity}
h(x) is M.D.
So, at x = a it may be M.I. followed by M.D. till x = b
or
M.D. through out a
But can�t be M.I. followed by M.D.
So (B) is correct answer.
RESONANCE Page # 31
86. Let V1 be the volume of a given right circular cone with O as the centre of the base and A as its apex. Let V
2
be the maximum volume of the right circular cone inscribed in the given cone whose apex is O and whose
base is parallel to the base of the given cone. Then the ratio V2 / V
1 is
(A) 253
(B) 94
(C) 274
(D) 278
Ans. (C)
Sol.
Let radius of base of given cone is r and length
as h V1 =
31
r2h ........(1)
Let radius of base 1 and R1 and height H. For the cane with apex O
V2 =
31
R2H ..........(2)
NowH�h
hRr i �
rR
hH
H = h
r
R�1
V =
r
R�1hR
3
1 2
rR2
�R2h31
dRdv 2
= 0 R = 3r2
For maximum volume
R � 3r2
V2 =
r
R�1hR
3
1 2
=
32
�19r4
h31 2
= h814 2 .......(2)
274
1VV2
RESONANCE Page # 32
87. Let � : R R be a continuous function satisfying
x
0
dt)t�(x)x�( , for all x R. Then the number of
elements in the set S = {x R � (x) = 0 } is
(A) 1 (B) 2 (C) 3 (D) 4
Ans. (A)
Sol. f(x) = 1 + f(x) dxdy
= 1 + y
n (1 + y) = x + c
y = ex + c � 1
y = ex � 1
f(x) = ex � 1 ..... (1)
Now f(x) = x + x
0
dt)t(f
ex � 1 = x + et � (t)0
x
ex � 1 = x + ex � x �
= 1
f(x) = ex � 1
for f(x) = 0 ex � 1 = 0
x = 0
88. The value of
2
0
1� )x(coscos|,�x|min dx is
(A) 4
2
(B) 2
2
(C) 8
2(D) 2
Ans. (B)
Sol.
A = 21
.2
+ 21
.2
2
2
RESONANCE Page # 33
89. Let ABC be a triangle and P be a point inside ABC such that 0PC3PB2PA . The ratio of the area of
triangle ABC to that of APC is
(A) 2 (B) 23
(C) 35
(D) 3
Ans. (D)
Sol. Let P is origin and p.v. of A, B, C are a, b, c respectively
Given 0PC3PB2PA
0c3b2a ...... (i)
Area of ABC = 21
|a
× b
+ b
× c
+ c
× a
|
= 21
|a
×
2a�c3�
+
2ac3�
× c
+ c
× a
|
= 21
|23
( c
× a
) � 0 � 0 + 21
( c
× a
) + c
× a
|
= 23
|c
× a
| ...... (ii)
Area of APC = 21
|c
× a
|
APCABC
=
13
90. Suppose m, n are positive integers such that 6m + 2m+n 3n + 2n = 332. The value of the expression m2 + mn +
n2 is
(A) 7 (B) 13 (C) 19 (D) 21
Ans. (C)
Sol. 6m + 2m+n. 3m + 2n = 332 = 4 × 83
2m�2. 3m + 2m+n�2 3m + 2n�2 = 83 .......(1)
83 is prime number so that this is possible only if
m = 2
by (1) 32 + 2n 32 + 2n�2 = 83
2n. 9 + 2n�2 = 74
2n�2 (36 + 1) = 74
2n�2 = 21
n = 3
Hence. m2 + mn + n2
= 4 + 6 + 9 = 19
RESONANCE Page # 34
PHYSICS91. A ball is dropped vertically from a height of h onto a hard surface. If the ball rebounds from the surface with a
fraction r of the speed with which it strikes the latter on each impact, what is the net distance travelled by theball up to the 10th impact ?
(A) r1
r1h2
10
(B) 2
20
r1
r1h
(C) h
r1
r1h2
2
22
(D) h
r1
r1h2
2
20
Ans. (D)
Sol. h10 upto........gVr
gVr
gV th
20
420
220
= g
V20
[1 + r2 + r4 + .............upto 10th] � h
= hr1
)r(1h2
2
102
= hr1
r1h2
2
20
.
92. A certain planet completes one rotation about its axis in time T. The weight of an object placed at the equatoron the planet's surface is a fraction f (f is close to unity) of its weight recorded at a latitude of 60º. The density
of the planet (assumed to be a uniform perfect sphere) is given by :
(A) 2GT4
3f1f4
(B) 2GT4
3f1f4
(C) 2GT4
3f1f34
(D) 2GT4
3f1f24
Ans. (A)
Sol. 2r
GMm =
rmv2
v = r
GM
T = V
r2T =
GMr
23
T = rGM
r2 =
GMr
23
geff
= g � 2 Re cos2
f(g � 2 Re cos260º) = (g � 2Re cos20)
fg � g = 2R
1
4f
g[f � 1] = 4R2
[f � 4] R =
)4f(
)1f(g42
= 2R
Gm(f � 1) =
4R2
(f � 4)
)4f(
)1f(GM42
= R3
)4f(4
)1f(GM4T2
2
= R3
)4f()1f(
= R3 =
3R34
M
=
)1f(GT4
)4f(32
RESONANCE Page # 35
93. Three equal charges +q are placed at the three vertices of an equilateral triangle centred at the origin. Theyare held in equilibrium by a restoring force of magnitude F(r) = kr directed towards the origin, where k is aconstant. What is the distance of the three charges from the origin ?
(A)
3/12
0 kq
61
(B)
3/12
0 kq
123
(C)
3/22
0 kq
61
(D)
3/22
0 kq
43
Ans. (B)
Sol. a = 2r × 23
2
2
q
Kq2
2
q
Kq2cos30º
= r 3
kr = 2
2
q
kq2 cos30º
k.r.3r2 = 04
2
.q2 × 23
r =
3/1
0
2
k123q
94. Consider the infinite ladder circuit shown below :
For which angular frequency will the circuit behave like a pure inductance ?
(A) 2
LC(B)
LC
2(C)
LC
1(D)
C
L2
Ans. (B)
Sol.
Let the equivalent impedence of circuit be ZSo, Z = wL + Z
Z = L + C
C
XZ
ZX
= L +
C1
Z
C1
Z
On solving we get,
RESONANCE Page # 36
Z = C2
LC4CLLC 222
For Z to be purely inductive 0LC4CL 222 = 0
= LC
2Ans.
95. A narrow parallel beam of light falls on a glass sphere of radius R and refractive index at normal incidence.The distance of the image from the outer edge is given by :
(A) )1(2)2(R
(B) )1(2
)2(R
(C) )1(2
)2(R
(D) )1(2
)2(R
Ans. (A)
Sol.
1V1
= R
1 R
1v1
1
Rv1
)VR2(V1
1f
=
R1
1R
R2R
1V1
f
)RR2R2()1(
R1
=
21
R1
)2(
)2(
R
)1( =
2
2
R
1 v
f = )1(2
)2(R
96. A particle of mass m undergoes oscillations about x = 0 in a potential given by V(x) = 21
kx2 � V0cos
a
x,
where V0,k, a are constants. If the amplitude of oscillation is much smaller than a, the time period is given by
(A) 0
2
2
Vka
ma2
(B)
km
2 (C) 0
2
Vma
2 (D) 0
2
2
Vka
ma2
Ans. (A)
Sol. V(x)
= 21
kx2 � v0cos
ax
G = dxdv
= kx + v0 sin
a
x.a1
Since, x <<<< a So, sinax
�~ax
RESONANCE Page # 37
EQ = kx +
ax
.a
V0 = xa
VK
20
=
20
a
VK = 2
02
a
VKa K = m2
T =
2 =
02
2
vka
ma2
.
97. An ideal gas with heat capacity at constant volume CV undergoes a quasistatic process described by PV in
a P�V diagram, where is a constant. The heat capacity of the gas during this process is given by :
(A) CV
(B) CV +
1nR
(C) CV + nR (D) C
V + 21
nR
Ans. (B)
Sol. CV
PV
Direct formulae C = Cv+
1nR
.
98. An ideal gas with constant heat capacity CV =
23
nR is made to carry out
a cycle that is depicted by a triangle in the figure given below : The following statement is true about the cycle :
(A) The efficiency is given by 22
11
VPVP
1
(B) The efficiency is given by 22
11
VPVP
21
1
(C) Net heat absorbed in the cycle is (P2 � P
1) (V
2 � V
1)
(D) Heat absorbed in part AC is given by
2(P2V
2 � P
1V
1) +
21
(P1V
2 � P
2V
1)
Ans. (D)
Sol. CV =
23
nR, CP = C
v + R =
2nR5
f = 3
w = 21
(V2 � V
1) (P
2 � P
1)
A � B Q = nCpT = n
25
RT = 25
(P1V
1 � P
2V
2)
A � C, QAC
= 21
(P1 + P
2) (V
2� V
1) + nC
V T
QAC
= 21
(P1 � P
2) (V
2 � 1) +
23
(P2V
2 � P
1V
1)
= 23
nR (T2 � T
1) +
2VPVP)TT(nR 122112
= 2nR (T2 � T
1) +
2VPVP 1221
RESONANCE Page # 38
= )VPVP(
23
)VV)(PP(21
)PP)(VV(21
11221221
1212
= 112222221121
1212
VP3VP3)VPVPVPVP()PP)(VV(
= 12211122
11212122
VPVPVP4VP3PVPVVPVP
99. Two identical particles of mass m and charge q are shot at each other from a very great distance with aninitial speed v. The distance of closest approach of these charges is :
(A) 20
2
mv8
q
(B) 2
0
2
mv4
q
(C) 2
0
2
mv2
q
(D) 0
Ans. (B)
Sol. V V
0mv2
12 2
= 0 + 0
d
Kqq
d =
2
2
0 mv
q.
41
= 20
2
mv4
q
100. At time t = 0, a container has N0 radioactive atoms with a decay constant . In addition, c numbers of atoms
of the same type are being added to the container per unit time. How many atoms of this type are there att = T ?
(A)
cexp(�T) � N
0exp(�T) (B)
cexp(�T) + N
0exp(�T)
(C)
c(1 � exp(�T)) + N
0exp(�T) (D)
c(1 + exp(�T) + N
0exp(�T)
Ans. (C)
Sol. N0 Initial nucleon
at t = 0 N
0N
Add. at a constan rate (N � C) = dtdN
k
0
dt =
N
N0
CNdN
t = NN0
)CN(n1
RESONANCE Page # 39
t =
CNCN
n1
0
e�t = CNCN
0
N � C = e�lt (N0 � C)
N =
teC(N
0 � C)
=
C + N
0e�t �
C.e�(t)
=
C[1 � e�(t)] + N
0 e�dt .
CHEMISTRY
101. 2.52 g of oxalic acid dihydrate was dissolved in 100 mL of water. 10 mL of this solution was diluted is 500 mL.
The normality of the final solution and the amount of oxalic acid (mg/mL) in the solution are respectively.
(A) 0.16 N, 5.04 (B) 0.08 N, 3.60 (C) 0.04 N, 3.60 (D) 0.02 N, 10.08
Ans. (.....)
Sol. H2C
2O
4.2H
2O (GMM = 126)
Molarity (M) = 100126100052.2
= 0.2M
M1V
1 = M
2V
2
0.2 × 10 = M2 × 500
M2 = 4 × 10�3.
valence factor of H2C
2O
4.2H
2O = 2.
Normality = [M] × V.F. = 2 × [4 × 10�3] = 8 × 10�3 N
Amount of oxalic acid (mg/mL) = 3
33
10
10126104
mLmg
= 0.504.
Note : Options are not match with solution.
102. Two isomeric compounds I and II are heated with HBr.
I II
The products obtained are
RESONANCE Page # 40
(A) (B)
(C) (D)
Ans. (A)
Sol. H
Br
HBr + CH
3OH
103. The number of possible enantiomeric pair(s) produced from the bromination of I and II, respectively, are
(A) 0, 1 (B) 1 , 0 (C) 0, 2 (D) 1, 1
Ans. (A)
Sol.
RESONANCE Page # 41
104. For the reaction A B, H° = 7.5 kJ mol�1 and S° = 25 J mol�1, the value of G° and the temperature, at
which the reaction reaches equilibrium are, respectively,
(A) 0 kJ mol�1 and 400 K (B) � 2.5 kJ mol�1 and 400 K
(C) 2.5 kJ mol�1 and 200 K (D) 0 kJ mol�1 and 300 K
Ans. (D)
Sol. A B
Hº = 7.5 kJ/mole ; Sº = 25 J/mole.
Gº = Hº � TSº at equilibrium Gº = 0
0 = 7.5 × 103 � T(25)
T = 25
105.7 3
= 300 K
105. The solubility product of Mg(OH)2 is 1.0 10�12. Concentrated aqueous NaOH solution is added to a 0.01
M aqueous solution of MgCl2. The pH at which precipitation occurs is
(A) 7.2 (B) 7.8 (C) 8.0 (D) 9.0
Ans. (D)
Sol. [Mg2+] = 10�2 M
)2)OH(Mg(SPK = [Mg2+] [OH�]2 = 1.0 × 10�12
= [OH�]2 = 2
12
10
100.1
= 10�10
[OH�] = 1 × 10�5
pOH = 5 ; pH = 9
106. A metal with an atomic radius of 141.4 pm crystallizes in the face centred cubic structure. The volume of
the unit cell in pm3 is
(A) 2.74 107 (B) 6.40 107 (C) 2.19 107 (D) 9.20 107
Ans. (C)
Sol. For FCC unit cell
4r = a2
a =
2
4.1414 = 400 pm
Volume of unit cell a3 = (400)3 = 6.4 × 107 pm3.
RESONANCE Page # 42
107. Identify the cyclic silicate ion given in the figure below
(A) [Si6O
24]24� (B) [Si
6O
18]18� (C) [Si
6O
18]12� (D) [Si
6O
24]12�
Ans. (B)
Sol. Formula of cyclic silicate : [SinO
3n]2n�
Cyclic silicates : (SiO3
2�)n or (SiO
3)
n2n�
108. Diborane is formed the elements as shown in equation (1)2B (s) + 3H
2(g) B
2H
6(g) ... (1)
Given thatH
2O () H
2O (g) H
10 = 44 kJ
2B + 3/2 O2 (g) B
2O
3(s) H
20 = � 1273 kJ
B2H
6(g) + 3 O
2 (g) B
2O
3 (s) + 3 H
2O (g) H
30 = � 2035 kJ
H2(g) + 1/2 O
2 (g) H
2O () H
40 = 286 kJ
the H0 for the reaction (1) is :(A) 36 kJ (B) 520 kJ (C) 509 kJ (D) � 3550 kJ
Ans. (A)
Sol. Target eq.2B (s) + 3H
2(g) B
2H
6(g)
Given thatH
2O () H
2O (g) H
10 = 44 kJ .....(i)
2B + 3/2 O2 (g) B
2O
3(s) H
20 = � 1273 kJ .....(ii)
B2H
6(g) + 3 O
2 (g) B
2O
3 (s) + 3 H
2O (g) H
30 = � 2035 kJ .....(iii)
H2(g) + 1/2 O
2 (g) H
2O () H
40 = 286 kJ .....(iv)
Target eq. : Eq.(ii) � Eq.(iii) + 3[Eq.(iv)] + 3 [Eq.(i)]
= � 1273 � [� 2035] + 3[�286] + 3[44]
= 36 KJ/mole.
RESONANCE Page # 43
109. The Crystal Field stabilization Energy (CFSE) and the spin only magnetic moment in Bohr Magneton (BM)for the complex K
3[Fe(CN)
6] are, respectively.
(A) 0.0 and 35 BM (B) � 2.0
and 3 BM
(C) � 0.4 and 24 BM (D) � 2.4
and 0 BM
Ans. (B)
Sol. K3[Fe(CN)
6]
3 + x � 6 = 0 x = + 3
26Fe = 3d6 4s2
26Fe3+ d5 with S.L. t
2g2,2,1, eg0,0 CFSE = �2.0
0 + 2P ~
�2.0
0
Number of unpaired electron = 1.
So, = )2n(n = 3 B.M
110. A solution containing 8.0 g of nicotine in 92 g of water freezes 0.925 degrees below the normal freezing pointof water. If the molal freezing point depression constant, K
f = 1.85ºC mol�1 then the molar mass of nicotine is
:(A) 16 (B) 80 (C) 320 (D) 160
Ans. (D)
Sol. Tf = K
f × molality
= 0.925 = 1.85 ×
92M
10008
M = 173.9 ~ 174
Closest answer = 160.
BIOLOGY111. A hot cell has intracellular bacterial symbionts. If the growth rate of the bacterial symbiont is always 10%
higher than that of the host cell, after 10 generations of the host cell the density of bacteria in host cells will
increase
(A) by 10% (B) two-fold (C) ten-fold (D) hundred-fold
Ans. (B)
112. In a diploid organism, there are three different alleles for a particular gene. Of these three alleles one is
recessive and the other two alleles exhibit co-dominance. How many phenotypes are possible with this set
of alleles?
(A) 3 (B) 6 (C) 4 (D) 2
Ans. (C)
RESONANCE Page # 44
113. Two students are given two different double stranded DNA molecules of equal length. They are asked to
denature the DNA molecules by heating. The DNA given to student A has a following composition of base
(A:G:T:C::35:15:35:15). Which of the following statements is true?
(A) Both the DNA molecules would denature at the same rate
(B) The information given is insufficient to draw any conclusion
(C) DNA molecule given to student B would denature faster than that of student A
(D) DNA molecule given to student A would denature faster than that given to student B
Ans. (D)
114. The amino acid sequences of a bacterial protein and human protein carrying out similar function are found to
be 60% identical. However, the DNA sequences of the genes coding for these proteins are only 45% identi-
cal. This is possible because
(A) Protein sequence does not depend on DNA sequence
(B) DNA codons having different nucleotides in the third position can code for the same amino acids
(C) DNA codons having different nucleotides in the second position can code for the same amino acids
(D) Same DNA codons can code for multiple amino acids.
Ans. (B)
115. The following DNA sequence (5´ 3´) specifies part of a protein coding sequence, starting form position 1.
Which of the following mutations will give rise to a protein that is shorter than the full-length protein?
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
A T G C A A G A T A T A G C T
(A) Deletion of nucleotide 13
(B) Deletion of nucleotide 8
(C) Insertion of a single nucleotide between 3 and 4
(D) Insertion of a single nucleotide between 10 and 11
Ans. (D)
116. Which of the following correctly represents the results of an enzymatic reaction? Enzyme is E, Substrate is
S and Products are P1 and P2
(A) P1 + S P2 + E (B) E + S P1 + P2
(C) P1 + P2 + E S (D) E + S P1 + P2 + E
Ans. (D)
117. Four species of birds have different egg colors: [1] while with no markings, [2] pale brown with ano markings,
[3] gray-brown with dark streaks and spots, [4] pale blue with dark blue-green spots. Based on egg color,
which species is most likely to nest in a deep tree hole?
(A) 1 (B) 2 (C) 3 (D) 4
Ans. (C)
118. Consider a locus with two alleles, A and a, If the frequency of AA is 0.25, what is the frequency of A under
Hardy-Weinberg equilibrium?
(A) 1 (B) 0.25 (C) 0.5 (D) 0
Ans. (C)
RESONANCE Page # 45
119. Which of the following graphs accurately represents the insulin levels (Y-axis) in the body as a function of
time (X-axis) after eating sugar and bread/roti?
(A) (B)
(C) (D)
Ans. (A)
120. You marked two ink-spots along the height at the base of a coconut tree and also at the top of the tree. When
you examine the spots next year when the tree has grown taller, you will see
(A) the two spots at the to have grown more apart than the two sports at the bottom
(B) the top two spots have grown less apart than the bottom two spots
(C) both sets of spots have grown apart to the same extent
(D) both sets of spots remain un-altered.
Ans. (C)
