1
PHYSICS
Q1
If frequency (F ), velocity (V ) and density (D) are considered as fundamental units, the dimensional formula for
momentum will be
(a.)
Sol.
(b.)
DVF 2
DV 2 F 1
(c.)
D 2V 2 F 2
(d.)
DV 4 F 3
(d)
P  mv
Assume [ p ]  [ F aV b D c ]  [ F ] a [V ]b [ D ]c
[ F ]  [T 1 ]
[V ]  [ LT 1 ]
[ D]  [ ML3 ]
[ MLT 1 ]  M c Lb 3c T  a b
C 1
b  3c  1  b  4
 a  b  1  a  b  1  a  3
Q2
Write the dimensions of a  b in the relation E 
b  x2
. where E is the energy, x is the displacement and t is
at
time.
(a.)
Sol.
(b.)
ML2T
M 1 L2T 1
(c.)
ML2T 2
(d.)
MLT 2
(b)
E
b  x2
at
[b]  [ x 2 ]  [ L2 ]
 (b  x 2 ) 
 ML2T  2

 at 

Also 

 L2 
[ at ]   2 2   [ M 1T 2 ]
 ML T 
[ a ]  [ M 1T 1 ]
Hence [b  a ]  [ M 1 L2T 1 ]
Q3
If the percentage errors of A, B and C are a, b and c, respectively, then the total percentage error in the product
ABC is
101, Blue Diamond Complex, Fatehgunj, Vadodara-02 .Tel. No. 2787268 / 2750018 / 2782232
2
(a.)
Sol.
(b.)
abc
abc
(c.)
1 1 1
 
a b c
(d.) ab  bc  ca
(b)
Let Z  ABC
Z
A
B
C
%
%
%
%
Z
A
B
C
Z
 100%  a  b  c
Z
Q4
Fig. shows a particle starting from point A, travelling upto B with a speed s, then upto point C with a speed 2 s
and finally upto A with a speed of 3s. Determine its average speed
(a.) 1.2
Sol.
s
(c.) 1.8
s
(d.) 1.5
s
(c)
Vavg 

Total Dis tan ce
Time
2R
2R

R 2R 5R
t1  t 2  t 3


2s
6 s 18s
2s
 1.8 s
1 1 5
 
2 3 18
Vavg 
Q5
(b.) 1.6
s
A particle starts from the origin with a velocity of 10 m / s and moves with a constant acceleration along a
straight line till the velocity increases to 50 m / s. At that instant, the acceleration is suddenly reversed keeping
the magnitude constant. What will be the speed of the particle when it returns to the starting point?
(a.) Zero
Sol.
(b.) 10
m/s
(c.)
50 m / s
(d.) 70
(d)
S AB 
2500  100 1200

2a
a
V A2  VC2  2 aS CA
V A2  2500  2a
1200
 4900
a
V A  70 m / s
101, Blue Diamond Complex, Fatehgunj, Vadodara-02 .Tel. No. 2787268 / 2750018 / 2782232
m/ s
3
Q6
A body starts from rest and travels a distance S with uniform acceleration, then moves a distance 2 S uniformly
and finally comes to rest after moving further 5S under uniform retardation (assume the motion to be on straight
line). The ratio of average velocity to maximum velocity is
(a.) 2/5
Sol.
(b.) 3/5
(c.) 4/7
(d.) 5/7
(c)
Area of I  S
1
t1Vm  S
2
t1 
2s
Vm
Area of II  2 S
Vm t 2  2 s
t2 
2s
Vm
Area of III  5S
1
Vm t 3  5 s
2
t3 
10s
Vm
v
Vm
I
II
III
t
t1
Vav 
t2
t3
8s
8s
8

 Vm
2 s 2 s 10s 14
t1  t 2  t 3


Vm Vm Vm
Vav 4

Vm 7
Q7
A ball is thrown from the top of a tower in vertically upward direction. Velocity at a point h m below the point
of projection is twice of the velocity at a point h m above the point of projection. Find the maximum height
reached by the ball above the top of the tower
101, Blue Diamond Complex, Fatehgunj, Vadodara-02 .Tel. No. 2787268 / 2750018 / 2782232
4
(a.)
Sol.
(b.) 3h
2h
(c.)
(5 / 3)h
(d.) ( 4 / 3) h
(c)
v B2  v 2  2 gh
v 2A  v 2  2 gh
v B2  4v A2
v 2  2 gh  4v 2  8 gh
3v 2  10 gh
v2 
10
gh
3
For max height
Hm 
Q8
v2
10 gh
5

 h
2g
3 2g
3
For motion of an object along x  axis, the velocity v depends on the displacement x as v  3 x 2  2 x. What is
the acceleration at x  2m.
(a.)
Sol.
(b.) 80 ms 2
48 ms 2
(c.) 18 ms 2
(d.) 10 ms 2
(b)
v  3x 2  2 x
a
d
v  6 xv  2v
dt
at x  2
v  12  4  8
a  12(8)  16  96  16  80m / s 2
Q9
The acceleration versus time graph of a particle is shown in Fig. The corresponding v  t graph of the particle is
(a.)
Sol.
(b.)
(c.)
(d.)
(a)
For t1 a is linearly increasing
Hence slope of v  t graph should be increasing and for next t 2 ,slope of v  t graph should be decreasing
101, Blue Diamond Complex, Fatehgunj, Vadodara-02 .Tel. No. 2787268 / 2750018 / 2782232
5
Q10 A boy is cycling with a speed of 20 km/h in a direction making an angle 30  north of east (see Fig.)
Find the speed of the second boy moving towards north so that to him the first boy appears to be moving
towards east.
(a.) 10
Sol.
(b.) 20
km/h
km/h
(c.) 30 km/h
(d.)
10 3 km/h
(a)
N
E
V B1 / B2 should be along east
V B1 / B2 along north = 0
VB1  20 sin 30   0
VB1  10
Q11 A projectile is thrown in the upward direction making an angle of 60° with the horizontal direction with a
velocity of 147 ms 1 . Then the time after which its inclination with the horizontal is 45°, is
(a.)
Sol.
15( 3  1)s
(b.)
 ( 3  1) s
(c.)
7.5( 3  1) s
(d.)
7.5( 3  1) s
(c)
After time t
V
2

147
2
V y  147
3
 gt
2
101, Blue Diamond Complex, Fatehgunj, Vadodara-02 .Tel. No. 2787268 / 2750018 / 2782232
6
147
3
 147
 gt
2
2
gt 
147
[ 3  1]
2
t  7.5 [ 3  1]
Q12 The angular velocity of a particle moving in a circle of radius 50 cm is increased in 5 min from 100 revolutions
per minute to 400 revolutions per minute. Find the tangential acceleration of the particle.
(a.)
Sol.
60 m / s 2
(b.) 
(c.)
/ 30 m / s 2
 / 15 m / s 2
(d.) 
/ 60 m / s 2
(d)

300 sec . min
sec
 60
5 min
(min) 2
at  50  10  2 
at 
60  2m
3600s 2

m / s2
60
Q13 Two particles A and B are placed as shown in Fig. The particle A, on the top of a tower, is projected
horizontally with a velocity u and the particle B is projected along the surface towards the tower,
simultaneously. If both the particles meet each other. Then the speed of projection of particle B is [Ignore any
friction]
(a.)
d
Sol.
g
u
2H
(b.)
d
(c.)
g
2H
d
g
u
2H
(d.) u
(a)
Observing motion of B w.r.t A
A
H
g(m/s2)
d(V+v)
B (m/s)
For B to meet A
d  (V  v)t
101, Blue Diamond Complex, Fatehgunj, Vadodara-02 .Tel. No. 2787268 / 2750018 / 2782232
7
H
1 2
gt
2

H
1
d2
g
2 (V  v) 2
gd 2
2H
(V  v) 2 
gd 2
v
2H
V
g
v
2H
V d
Q14 A monkey A (mass = 6 kg) is climbing up a rope tied to a rigid support. The monkey B (mass = 2 kg) is holding
on the tail of monkey A (see Fig.). If the tail can tolerate a maximum tension of 30 N what maximum force can
monkey A apply on the rope in order to carry monkey B with it? ( g  10 m / s 2 )
(a.) 120
Sol.
(b.) 30
N
N
(c.) 60 N
(d.) 90
N
(a)
TBmax  30 N
a Bmax 
10
 5m / s2
2
Hence max acceleration of system  5m / s 2
T A  80  40  120 N
Q15 Two wooden blocks are moving on a smooth horizontal surface such that the mass m remains stationary with
respect to block of mass M as shown in the Fig. The magnitude of force P is
(a.) ( M
Sol.
 m) g tan 
(b.)
g tan 
(c.)
mg cos 
(d.) ( M
(a)
For system to move together
a
P
M m
101, Blue Diamond Complex, Fatehgunj, Vadodara-02 .Tel. No. 2787268 / 2750018 / 2782232
 m) g cos ec
8
N sin   ma
N cos   mg
a  g tan 
P  ( M  m) g tan 
Q16 In the Fig. if f , f , and T be the frictional forces on 2 kg block, 3 kg block & tension in string, respectively,
1
2
then their values respectively are
Sol.
(a.) 2
N, 6 N, 3.2 N
(b.) 2
(c.) 1
N, 6 N, 2 N
(d.) Data
N, 6 N, 0 N
insufficient to calculate the required value
(c)
f1 max  2 N
f 2max  6 N
8N
1N
1
2
Assume system accelerated to right
a
8 1  6  2
 ve value
5
Which is not possible
Hence a  0
1N
f1  1N
f1  1N
2
T
6N
T
3
8N
T  2N
f2  6N
T  2N
Q17 On a stationary block of mass 2 kg, a horizontal force F starts acting at t  0 whose variation with time is shown
in Fig. The coefficient of friction between the block and ground is 0.5. Then the velocity of the block
101, Blue Diamond Complex, Fatehgunj, Vadodara-02 .Tel. No. 2787268 / 2750018 / 2782232
9
at t  12 s
(a.)
Sol.
(b.)  12
20 m / s
m/ s
(c.)
 6 m/ s
(d.) Zero
(d)
f e max  10 N

 f
f
Fnet  F  f x  20  2t  10
Fnet  10  2t
anet 5  t
dv
5t
dt
t2
V  5t 
2
Partgick stops at v = 0
t  10 s
Also f from 10 to 12 s is less than f max
Hence V at t  12s  0
Q18 Two particles A and B are kept on a smooth horizontal surface. Particle A is fixed at origin. At the instant when
particle B was spotted at  4 ˆj 
16 ˆ
k moving with a constant velocity11 ˆj  2kˆ, particle A is projected with a
3
speed of 10 m / s. Both the particles always remain in the same horizontal plane. What should be the velocity of
particle A to hit the particle B.
(a.)
Sol.
(b.)
8 ˆj  6kˆ
8 ˆj  6kˆ
(c.)
6 ˆj  8kˆ
(d.) x
(a)
V B  11 ˆj  2 kˆ
16
r B  4 ˆj  kˆ
3
32
3
 132  32
100


3
3
V B . r B  44 
101, Blue Diamond Complex, Fatehgunj, Vadodara-02 .Tel. No. 2787268 / 2750018 / 2782232
10
Z
B
Y
A
cos 
100 / 3
5 5  20 / 3
1

5
sin  
2
5
20/3
VA=10
A
VB sin   V B ar
5 5
2
5
 10
ar
Hence for particles to collide the velocity of A should be  to line joining them
Let V A  ( yˆj  zkˆ )
16 

( yˆj  zkˆ) .   4 ˆj  kˆ   0
3 

16
4
4y  z
y z
3
3
2
2
Also y  z  100
16 2
z  z 2  100
9
25 z 2
 100
9
z  6 y 8
V  8 ˆj  6kˆ
A
Q19 A particle is moving in a circle of radius R with constant speed. The time period of particle is T  1s. In a
101, Blue Diamond Complex, Fatehgunj, Vadodara-02 .Tel. No. 2787268 / 2750018 / 2782232
11
time t 
T
, if the difference between average speed and average velocity of the particle is 2 m / s. Then the
6
radius R of the circle is.
(a.) 7 m
Sol.
(b.) 14 m
(c.)
(d.) 28 m
21m
(a)
B
A
T  1s
1
ln t 
6
2


 rad .  60 
6
3
(Vel ) avg
R

T /6
( Speed ) avg

R
 3
T /6
 6R
 R
2R  6 R  2
2
2
R

 7  7m
2  6
44  42
Q20 Two blocks of mass M and m are connected with either end of a massless string which passes over a massless
smooth pulley ( M  m). The pulley is fixed to the roof of an elevator moving with an acceleration ‘ a ’
upwards. The acceleration of M relative to m is’ 2 a ’. Then ratio of
(a.)
Sol.
1
(b.)
a
g
1
a
g
(c.)
1
M
is equal to
m
2a
g
(d.)
1
(c)
M m
Relative to elevator
T
M
M(g+a)
T
M
m(g+a)
101, Blue Diamond Complex, Fatehgunj, Vadodara-02 .Tel. No. 2787268 / 2750018 / 2782232
2a
g
12
M
m
a' 
(M  m)( g  a)
( M  m)
a M / m  2a'  2a (given)
a'  a
(M  m)
( g  a)  a
( M  m)
M
2a
 1
m
g
CHEMISTRY
Q1
100 ml of 0.1 M solution of AB(d  1.5 gm / ml) is mixed with 100 ml of 0.2 M solution of CB2 (d  2.5 gm/ ml).
Calculate
the
molarity
of B  in
final
solution
if
the
density
of
final
solution
is 4 gm / ml.
Assuming AB and CB2 are non-reacting & dissociate completely into A  , B  , C 2 .
(a.) 0.5 M
Sol.
(b.) 1.0
M
(c.) 0.25
M
(d.) 0.75
M
(a)
Volume of final solution
100  1.5  100  2.5
 100 ml
4
100
100
50
Moles of B  
 0.1 
 0 .2  2 
1000
1000
1000
50
/
1000
Molarity of B  
 1000  0.5 M
100
Q2 One gram of a sample of CaCO was strongly heated and the CO liberated absorbed in 100 ml of
3
2

0.5 M NaOH . Assuming 90% purity for the sample, how much ml of 0.5 M HCl would be required to react
with the solution of the alkali for the phenolphthalein end point?
(a.) 73
Sol.
ml
(b.) 41
ml
(c.) 82
ml
(d.) 97
ml
(c)
101, Blue Diamond Complex, Fatehgunj, Vadodara-02 .Tel. No. 2787268 / 2750018 / 2782232
13
0.9
 9  10 3 mole
100
For phenolphthalein end point 9  10 3 mole of CaCO3 is equivalent to the formation of 9 10 3 mole of
Assuming 90% purity, 1 gm of CaCO3 is
Na 2 CO3 and finally forming 9  10 3 mole of NaHCO3
[50  9]  10 3 mole  41 10 3 mole of NaOH must have reacted with HCl.
V

 0.5  41 10 3
1000
V  82 ml
Q3
2.36 g of sample of dolomite containing only CaCO3 and MgCO3 were dissolved in 700 ml of 0.1 M HCl . The
solution was diluted to 2.50 L. 25 ml of this solution required 20 ml of 0.01 N NaOH solution for complete
neutralization. Calculate % of CaCO3 .
(a.) 44.2%
Sol.
(b.) 15.2%
(c.) 68.8%
(d.) 26.3%
(c)
Let wt. of CaCO3  x, wt. of MgCO3  y
Meq. of HCl added to dolomite = 700  0.1 = 70
Meq. of HCl left in 25 ml = 20  0.01 = 0.2
Meq. of HCl left in 2.5 L =
0.2  2500
 20
25
Meq. Of HCl used for dolomite = 70 – 20 = 50
Meq. of CaCO3 + Meq. of MgCO3 = Meq. Of HCl used
x
y
 2  1000   2  1000  50
100
84
( x  y)  2.36 g
On solving eq (i) & (ii); x  1.625 g ., y  0.735 g
1.625
 % of CaCO3 
 100  68.9%,
2.36
%of MgCO3
 100  68.9  31.1%
Q4
(ii)
The mass of an electron is 9.1  10 31 kg . If its K.E. is 3.0  10 25 J , calculate its wavelength.
(a.)
(b.)

8321 A
Sol.
(i)
(c.)

6500 A

8067 A
(d.)

8967 A
(d)
1 2
mv  3.0 10 25 J
2
Mass of the electron, m  9.1  10 31 kg
K.E. of the electron 
v2 

2  3.0 10 25 J 2  3.0 10 25 J

m
9.1 10 31 kg
2  3.0  10 25 kg.m 2 s 2
9.1  10 31 kg
1
 2  3.0  10 25 2 1  2
This gives, v  
m s   812 ms 1
31
9
.
1

10


101, Blue Diamond Complex, Fatehgunj, Vadodara-02 .Tel. No. 2787268 / 2750018 / 2782232
14

h
6.626  10 34 Js

 8.967 10 7 m
31
1
mv 9.1 10 kg  812ms
or   8.967  10 7 m  10 2 cm / m

 8.967  10 5 cm  8967 A
Q5
The wave number of the spectral line of shortest wavelength of Lyman series of He  ion is ( R  Rydberg’s
constant)
(a.)
Sol.
(b.) 3R
R
(c.)
4R
(d.)
R/3
(c)
Z  2, n1  1, n2  
1 
1
v  R (2) 2  2  2   4 R
1  
Q6
An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons
than the electrons find the symbol of the ion.
(a.)
Sol.
37
13
(b.)
Cl 
37
18
Cl 
(c.)
37
17
Cl 
(d.) 37 Cl 
12
(c)
Mass number of the ion = 37
Charge = 1 unit negative charge
Total number of electrons in the ion, = No. of protons + 1 = p + 1
Total number of the nucleons in the ion, n  p  37
So, number of neutrons in the ion, n  37  p
No. of electrons 100

No. of neutrons 111.1
No. of electrons
or
No. of electrons  No. of neutrons
100
100


 0.474
100  111.1 211.1
p 1
Then
 0.474
( p  1)  (37  p)
p 1
This given
 0.474
38
p  0.4784  38  1  17
From the given data.
37
So, n  37  17  20 and no. of symbol of the ion is 17
X
37
The element having atomic number 17, is chlorine, so the ion is 17
Cl 
Q7
What in the minimum error in position of an electron moving with a speed of 500 ms 1 measured to an accuracy
of 0.006 % (mass of the electron  9.1 10 31 kg )?
(a.) 1.932  10 3 m
Sol.
(b.) 0.122  10 1 m
(c.)
3.122 10 3 m
(d.) 2.236  10 2 m
(a)
p  mv  9.1 10 31  500 
0.006
100
101, Blue Diamond Complex, Fatehgunj, Vadodara-02 .Tel. No. 2787268 / 2750018 / 2782232
15
 2.73  10 31 kgms 1
h
6.626  10 34

 1.932 10 3 m
4mv 4    2.73  10 32
Q8 An element having electronic configuration [ Ar ]3d 2 ,4 s 2 belongs to
x 
(a.)
(b.)
d  block
f  block
(c.)
s  block
(d.)
p  block
Sol.
(a)
Fact
Q9 How many of the following molecules are planar.
CH 2  CH  CHO, C 2 H 4 , H 2C  C  CH 2 , XeF4 , CO 2 , SF4
(a.) 4
Sol.
(b.) 3
(c.) 6
(d.) 5
(a)
C 2 H 4 , H 2 C  C  CH 2 , XeF4 and CO2 are planar
Q10 Which one of the following ions is paramagnetic
(a.)
Sol.
(b.)
Ag 
Fe 2
(c.)
(d.)
K
Mg 2 
(b)
Fe 2 is [ Ar ]3d 6
i.e. 4 unpaired electrons, hence paramagnetic
Q11 An element (X) which occurs in the second period has an outer electronic configuration s 2 p 1  what is the
formula and nature of its oxide?
(a.)
Sol.
(b.)
XO3 , basic
X 2 O3 , acidic
(c.)
(d.)
X 2 O3 , basic
XO2 , acidic
(b)
Configuration - s 2 p 1
Hence valency = 3
Oxide = X 2 O3
X is from p block
Hence X 2 O 3 is acidic
Q12 Select correct order of electron affinities of following O, F, S, Cl
(a.) Cl
>F>O>S
(b.) F
> Cl > O > S
(c.) Cl
(d.) S
>F>S>O
> O > Cl > F
Sol.
(c)
Fact
Q13 Valency expresses
(a.) Total
electrons in an atom
(c.) Oxidation
number of an element
(b.) Atomicity of an
(d.) Combining
element
capacity of an element
Sol.
(d)
Fact
Q14 The crystal lattice of electrovalent compound is composed of:
Sol.
(a.) Atoms
(b.) Molecules
(c.) Oppositely charged ions
(d.) Both
molecules and ions
(c)
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Fact
Q15 Formation of   bond:
(a.) Increases
(b.) Distorts
bond length
(c.) Decreases
(d.) Makes
bond length
the geometry of the molecule
homoatomic molecule more reactive
Sol.
(c)
Fact
Q16 Select correct option about Cl O
2 7
I. All Cl  0 bonds are equivalent
II. 6 Cl  0 bonds are equivalent
III. Cl  atom is sp 3 hybridized
IV. There is no Cl  Cl bonds
(a.) I & IV
Sol.
(b.) II,
are correct
III & IV are correct
(c.) II
& III are correct
(d.) I,
& III are correct
(b)
O
O
O
Cl
Cl
O
O
O
O
Q17 The bonds present in N O are
2 5
Sol.
(a.) Only ionic
(b.) Covalent
and coordinate
(c.) Only covalent
(d.) Covalent
and ionic
(b)
O
O
N
N
O
O
O
Q18 10 g of CaCO contains
3
(a.) 10
(c.)
Sol.
(b.) 0.1 g atom of Ca
moles of CaCO3
(d.) 0.1 of equivalent
6 10 23 atoms of Ca
of Ca
(b)
Mw2 of CaCO3  40  12  48  100
10
Moles of CaCO3 in 10 g 
100
 0.1 mol  0.1 g atom
Q19 Two glucose solutions are mixed. One has a volume of 480 mL and a concentration of 1.50 M and the second
has a volume of 520 mL and concentration 1.20 M . The molarity of final solution is
(a.) 1.20
Sol.
(b.) 1.50
M
M
(c.) 1.344
M
(d.) 2.70
(c)
M 1V1  M 2V2  M 3V3
101, Blue Diamond Complex, Fatehgunj, Vadodara-02 .Tel. No. 2787268 / 2750018 / 2782232
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17
(V3  V1  V2 )
1.5  480  1.2  520  M 3  1000
M 3  1.344 M
Q20 Which of the following sets of quantum number is allowable:
Sol.
(a.)
n  2, l  1, m  0, s  1 / 2
(b.)
n  2, l  2, m  1, s  1 / 2
(c.)
n  2, l  2, m  1, s  1 / 2
(d.)
n  2, l  1, m  0, s  0
(a)
l cannot be more than or equal to n and it cannot be  ve
s cannot be 0
BIOLOGY
Q1
Glycoproteins are known to play an important role in cell recognition. The specificity of this recognition is
provided largely by
(a.)
Carbohydrate portion of these glycoproteins
(b.)
Protein portion of these glycoproteins
(c.)
Both carbohydrate and protein component of these glycoproteins
(d.)
Lipid portion of glycoproteins
Sol.
Q2
(a)
If living cells similar to those found on earth were found on another planet where there was no oxygen, which
cell organelle would most probably be absent?
(a.) Ribosomes
Sol.
Q3
Q4
(c.) Chromosomes
(d.) Cell
membrane
(b)
How many times mitotic division must occur in a cell to form 1024 cells?
(a.) 10
Sol.
(b.) Mitochondria
(b.) 20
(c.) 40
(d.) 64
(a)
Radioactive thymidine when added to the medium surrounding living mammalian cell gets incorporated into a
newly synthesized DNA. Which of the following type of chromatin is expected to be radioactive if cells are
exposed to radioactive thymidine as soon as they enter the S-phase?
(a.)
Both heterochromatin and euchromatin
(b.)
Heterochromatin
(c.)
Euchromatin
(d.)
Neither heterochromatin nor euchromatin but only the nucleolus
Sol.
Q5
(a)
Nitrifying bacteria are an example of __________
(a.) Photoautotroph
Sol.
(b.) Chemoheterotroph
(c.) Chemoautotroph
(d.) Photoheterotroph
(c)
101, Blue Diamond Complex, Fatehgunj, Vadodara-02 .Tel. No. 2787268 / 2750018 / 2782232
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Q6
Which of the following bacteria can oxidize hydrogen sulphide to sulphur?
(a.) Rhizobium
Sol.
Q7
(c.) Azotobacter
the time of germination of zygospore
(c.) During asexual
reproduction
(b.) During the
formation of germ sporangium
(d.) During the
formation of gametangium
Q8
(a)
Select the statement that doesn’t apply to kingdom fungi
(a.)
Fungi are eukaryotic, multicellular, ingestive heterotroph
(b.)
Some fungi form beneficial inter relationship with plants
(c.)
Fungal life cycle typically includes a spore stage
(d.)
Certain fungi are natural source of Antibiotics
Sol.
Q9
(a)
Among the following which is NOT a characteristic feature of Bryophytes?
(a.) Presence
(b.) Water
of Archegonia
(c.) Photosynthetically independent
Sol.
(d.) Beggiatoa
(d)
The reduction division in the life cycle of Rhizopus occurs
(a.) At
Sol.
(b.) Leptothrix
sporophyte
is essential for fertilization
(d.) Motile antherozoids
(c)
Q10 Formation of a sporophyte from a vegetative portion of prothallus without sexual fusion is called
(a.) Apocarpy
Sol.
(b.) Apogamy
(c.) Apospory
(d.) Adventive
embryony
(b)
Q11 If the haploid number of chromosomes in a gymnosperm is 12, the number of chromosomes in its endosperm
cells will be
(a.) 12
Sol.
(b.) 24
(c.) 36
(d.) 6
(a)
Q12 An Amoeba transferred from a container X to another container Y developed a new contractile vacuole, but the
vacuole disappeared again when the Amoeba was transferred back to container X. The containers X and Y
respectively contain
(a.) Fresh and
(c.) Both
Sol.
(b.) Marine and fresh
marine water
(d.) Both
contain fresh water
water
contain marine water
(b)
Q13 Triploblastic, unsegmented, acoelomate, exhibiting bilateral symmetry and reproducing both asexually and
sexually, with parasitic forms belong to phylum
(a.) Porifera
Sol.
(b.) Platyhelminthes
(c.) Annelida
(d.) Ctenophora
(b)
Q14 Which one of the following phyla is correctly matched with its two general characteristics i.e its Morphological
feature?
(a.)
Animal-Scorpion, spider, cockroach – Morphological features-Ventral solid central nervous system
(b.)
Animal-Cockroach, locust, Taenia – Morphological features-Metameric segmentation
101, Blue Diamond Complex, Fatehgunj, Vadodara-02 .Tel. No. 2787268 / 2750018 / 2782232
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(c.)
Animal-Liver fluke, sea anemone, sea cucumber – Morphological features-Bilateral symmetry
(d.)
Animal-Centipede, prawn, sea urchin – Morphological features-Jointed appendages
Sol.
(a)
Q15 Heart pumps only impure blood in case of
(a.) Shark
Sol.
(b.) Whale
(c.) Lizard
(d.) Frog
(c.) Fat
(d.) Starch
(a)
Q16 Fehling’s solution is used to detect
(a.) Sucrose
Sol.
(b.) Glucose
(b)
Q17 K value of Enzyme is substrate concentration at
m
(a.)
Sol.
1
V max
4
(b.) V
(c.) 2V
max
max
(d.)
1
V max
2
(d)
Q18 Bond present between Ribose sugar and Gaunine in Guanosine is ________
(a.) Ester bond
Sol.
(b.) Peptide
bond
(c.) Glycosidic bond
(d.) Phosphodiester bond
(c)
Q19 Which of these is not a function of capsid in virus ____________
(a.)
Protect genetic material from Nuclease attack
(b.)
Attachment and injection of viral genome into host
(c.)
Determines antigenic specificity of virus
(d.)
Replication of genetic material in host cell
Sol.
(d)
Q20 A kingdom common to unicellular plant-like and animal-like organism is
(a.) Monera
Sol.
(b.) Protista
(c.) Fungi
(d.) Plantae
(b)
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101, Blue Diamond Complex, Fatehgunj, Vadodara-02 .Tel. No. 2787268 / 2750018 / 2782232