1. Ans : c
V2 – o2 = 2as
16
V2
=
= 2 m/s2
2S 2  4
a=
2. Ans : c
2h
 t1 =
g
t=
t1
=
t2
2h1
, t2 =
g
2 h2
g
h1
h2
3. Ans : c
Let h be the height
u = 0, take g = 10 m/s2
h=
1 2 1
gt =  10  42 = 80 m
2
2
4. Ans : a
The horizontal range is the same whether the projectile is thrown at an angle  with the
horizontal or at an angle  with the vertical.  a is the answer
5. Ans : b
Let the stone remain in air for n seconds.
Then :
a
(2n – 1)
2
10
=0+
(2n –1) = 10 n –5)
2
xn
=u+
Also xn is distance covered in first three seconds.
That is
xn = ut +
=0+
a
2
 t2
10
(3)2 = 45 m
2
Hence 10 n – 5 = 45
Which gives n = 5 seconds
6. Ans : c
Potential energies at the highest point are equal to the loss in kinetic energies. In the first case KE
=
1
mu2
2
= PE at the highest point
In the second case KE
=
1
m (u cos 60)2
2
=
1
1

m u  
2
2

=
1 1
 mu2
4 2
2
1
mu 2
2
The ratio is
=4:1
1 1
 mu 2
4 2
Note : If the projectile makes an angle 60o with vertical, it will make an angle 30o with the horizontal
 initial vertical velocity is
u sin  = u sin 30 = u cos 60.
7. a Exp.:
Action is directed opposite to the
reaction
8. b Exp.: The tension between P and Q
acceleration double the mass as compared to that between Q and R
9. d Exp.: Here mg sin 𝑎 = 8. To make it
move up with the same acceleration, we need to apply force F such that F – mg sin 𝑎.
Because acceleration down the inclined plane is g sin 𝑎.
10. c Exp.: Let the acceleration if system be 𝑎.
Then 𝑎 = 5 N/( 6 + 4) kg
= 0.5 ms –2
Force on the block of mass 4 kg
= 4× 0.5 N = 2 N
11. a
43 – 18 = 25 (Mn)
12.
c
Bond orders : H2+ : 21 , H2 : 1, H2 − :
H2 − has an anti bonding electron
13.
1
2
d
Look for sp carbon. As p orbital participation decreases, bonded pair of electrons is closer to
carbon.
14. c
Ti2+ : 3d2 and Ni2+ : 3d 8
15.
b
By removal of electron B.O. changes as
NO : 1 21 to 2
O2 : 2 to 2 21
F2 : 1 to 1 21
16. d
KO2 : K + and O2−
NO : . N̈ = O
17. N2 : 3, N2+ and N2− : 2 12 , N22−
18.
∶ 2
a
Bond order and related properties of diatomic molecules may be remembered thus.
Molecule B2 C2 N2
O2
F2
B. O
1
2
3
2
1
Count the total number of electrons in the diatomic molecule
Molecule
B2
C2
N2
O2
F2
–
No of e
10
12
14
16
18
B. O
1
2
3
2
1
19.
e
In tetro compounds (complexes in which O2– is the ligand ) the central atom transfers its electrons
in its outer shell to each O atoms and receives back O2 – in sp3 vacant orbitals
20.
e
Anhydrous AlCl3 covalent with Al in sp2 - vacant orbital on
Al is the reason for hydrolysis.
AlCl3 + 3 H2O ⟶ Al (OH)3 + 3 HCl
21.
e
1s, 2p , 3d and 4f have no node : Number of nodes = n −(ℓ + 1)
22.
b
1
1
1
𝑛22
1
𝜈 = 𝑅𝐻 ( 2 −
𝑛
) ; 𝑛1 = 1 and 𝑛2 = ∞
𝑐
Then 𝜈 = 𝑅𝐻 ; 𝜈 = 𝜆 and 𝜈 = 𝜆 = 𝑐 𝜈
E = h𝑣 = hc 𝜈 = hc RH
23.
a
E = 12 mv2
3
−19
E = 10× 10 × 1.6 × 10
2𝐸
1
2
J
2×10×103 ×1.6×10−19
ν = (𝑚) = (
9.1×10−31
1
2
7
) = 5.9× 10
24. c
1
1
1
𝑛22
RH (𝑛 2 −
𝜈=
) ; n2 = ∞
n1 = 1, 2 and 3 respectively
𝑐
𝜈=𝜆=𝑐𝜈
1.
2.
3.
25.
a
109677 ×102 ×3×108
12
109677 ×102 ×3×108
22
109677 ×102 ×3×108
32
1 eV = 1.6× 10−19 J
𝜆𝑂 =
ℎ𝑐
𝐸𝑂
=
6.63 ×10−36 ×3×108
2.3×1.6×10−19
= 5.4 × 10−7 m
(5.4 × 10−7 × 109 = 540 𝑛𝑚
26. a
The pyrenoids are small spherical protein bodies surrounded by starch deposition. They
are found singly or in numbers embedded in the chloroplast of many algae and
bryophytes.
27. b
The chief source of carragenin is red algae, Chondrus crispus.
28. a
Life cycle in Spirogyra is haplontic as dominant phase in life is haploid and diploid phase
is represented only by zygote or zygospore
29. c
Bryophytes are simple chlorophyll containing non vascular plants. They show
heteromorphic alternation of generation in which the gametophyte is the dominant phase.
30. C
In funaria, antheridium represents male sex organ and archegonium shows female sex
organ. It is situated on the apex of stem.
31. a
In Dahlia, roots do not originate from radicles and are therefore,adventitious. Roots of
radish, carrot and beet originate from radicle are the examples of modified tap root.
32. b
The main purpose of sub-aerial modification of stem is vegetative propagation, as found
in runner, offset, stolon and sucker. The examples of sucker are Chrysanthemum, Mint
etc
33. d
Bulbil is a structure rich in food materials and serves the function of vegetative
propogation.
34. a
The mustard (Brassica campestris) belongs to cruciferae (Brassicaeae). The stamens of
this family is characteristically tetradynamous.
35. B
When the stamens are united to petals the stamens are called epipetalous. Such condition
is found in solaaceae, malvaceae, compositae etc.
36. b
37. a
38. b
39. d
40. b
41. b
42. b
43. b
44. c
45. a