1. Ans : b
Suppose they cross each other after a time t second
then,
1 2
1
gt + (50 t – gt2)
2
2
= 100
50 t = 100
t=2
2. Ans : c
u = 0. Let v be the maximum speed attained.
For accelerated motion,
v = u + at
v = a1 t1 …. (1)
For retarded motion, initial velocity is equal to the final velocity of accelerated motion
O = u – a2 t2 = v – a2 t2
 v = a2 t2……... (2)
From (1) and (2)
a1 t1 = a2 t2
t1 =
a 2t 2 4t 2
=
= 2t2
2
a1
Now t1 + t2 = 3
ie. 2t2 + t2 = 3
3t2 = 3  t2 = 1 second
 t1 = 2 second
(1)  v = 2  2 = 4 m/s
3. Ans: c
In train – bridge problems total distance to be traveled = length of train + length of bridge
Total distance = 300 + 200 = 500 m
v 2  u 2 5 2  32 16
=
=
=.016 m/s2
2s
2  500 1000
vu 53
t=
=
= 125 second
.016
a
a=
4. Ans : c, Knowledge based question
5. Ans : c
s = ut +
1 2
at
2
here u = 0
x1
=
x2 + x 1 =
x2
1 2
at
2
1
1 2
a (t + t)2 = 4 
at
2
2
1
= 4  at2 – x1
2
1
1
= 4  at2 – at2
2
2
1
= 3  at2 = 3x1
2
6. Ans : e
S = ut +
1 2
at
2
Suppose they cross each other after time t.
Then
x = 50 t +  1  18 t2
2
….. (1)
1
2
100 – x = 50 t –   18 t2 ….. (2)
(1) + (2)  100 = 100 t
= 1h
t
(1) 
x
= 50  1 +
= 50 + 9
= 59 km
7. Ans : b
1
2
18  12
Let the projectile is projected with a velocity u
Vertical component = u sin 
u sin 30
= 80
U

1
2
= 80
u
At t =
= 160 m/s
T
, projectile is a maximum height
2
 its vertical velocity
=0
= u cos 
= 160 cos 30
Its horizontal velocity
= 160
3
2
= 80 3 m/s
8. Ans : a
The inclination of resultant to horizontal vector is given by :
Tan  =
Q sin 
P  Q cos
………… (1)
According to questions,
 = 90o; Q = 2P
Hence from eqn. (1), we have
Tan 90 =
Q sin 
P  2 p cos
 =
Q sin 
P  2 p cos
Or
 P + 2P cos  = 0  cos  = –
1
2
  = cos–1 (–1/2)
  = 120o
Hint : cos 120 = –
9. b Exp.:
1
2
The reading of the spring balance
will be less because the weight of the bird will now be excluded
10. b
Exp.: Acceleration (retardation)
produced = 25 ms–1/0.1 s=250ms–2
Force applied
= 0.100 kg×250 ms–2 = 25 N
11. b
⃗⃗⃗ =
Exp.: 𝐹
⃗⃗⃗⃗⃗⃗
𝑑𝑝
𝑑𝑡
cos𝜃 =
12. Exp.:
⃗⃗⃗
𝑎 =
= −𝑖̂ 2 sin t + 𝑗̂ 2 cos t
⃗⃗⃗ ∙𝑃
⃗⃗⃗
𝐹
𝐹𝑝
= 0. Hence 𝜃 = 90°
𝑀1 −𝑀2
𝑀1 +𝑀2
( ⃗⃗⃗⃗
𝑔 − ⃗⃗⃗
𝑓)
2 𝑀 −𝑀
⃗⃗⃗
𝐹 = 𝑀 1+𝑀 2 ( ⃗⃗⃗⃗
𝑔 − ⃗⃗⃗
𝑓)
1
13. Exp.:
2
Here 4×a = 4×g sin 30° − 2×g = 0
Hence a = 0
14. a
Exp.: Friction on B due to floor
= 0.5× (8 + 2) × 10 = 50 N
Applied force = 25 N. So the block B is not moving.
15. d
Exp.: Man exerts force F = 144 N on
the gun while firing.
Same force would be exerted by the gun on the hand of man.
Hence, from Newton’s third law, we have :
→
𝐹𝑚𝑎𝑛 = 𝐹𝑔𝑢𝑛
That is, 𝐹𝑚𝑎𝑛
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑜𝑓 𝑔𝑢𝑛
𝑡𝑖𝑚𝑒
or 𝐹𝑚𝑎𝑛 =
𝑃𝑓−𝑃𝑖
𝑡
=
Since, initial momentum, Pi = 0
There fore , 𝐹𝑚𝑎𝑛 =
𝑃𝑓
𝑡
=
𝑚𝑣
𝑡
If n bullets are fired by the gun. So we have:
𝐹𝑚𝑎𝑛 = 𝑛
𝑚𝑣
𝑡
⇒𝑛=𝑛=
𝐹𝑚𝑎𝑛 ×t
𝑚𝑣
Substituting 𝐹𝑚𝑎𝑛 = 144 N;
t = 1s ; m = 40 g = 40× 10−3kg;
v = 1200 ms –1
We get : n = 3
16.
d
dx2-y2 and dz2 are along the coordinate axes which also are the directions of approach of ligands in
octahedral complexes
17.
b
NaCl (crystal →
UO
Na+ (g) + Cl– (g)
– IE + EA
− ∆ Hf
Na(s) + 21 Cl2 (g)
−∆𝐻𝑠𝑢𝑏𝑙𝑖
− 21 ∆𝐻𝑑𝑖𝑠𝑠
Na (g) + Cl (g)
Uo = ∆ Hf + ∆ Hsubl + 21 ∆ Hdiss + IE − EA
= 410 + 108.7 + 120.9 + 493.8 – 368.2 =766
(Look at the direction of the change to assign positive or negative sign to
various energy changes)
18.
a
1. NF3 : Pyramidal
2. NO3− : Trigonal planar
3. BF3 : Trigonal planar
4. H3O+ : pyramidal
5. HN3 : H – N = N ⃗⃗⃗⃗− N
19.
c
Cation small, anion large : covalent
Cation large, anion small : ionic
To get idea about lattice energy look at the charge density of cation and anion. High charge
density implies high lattice energy.
20. a
O = C = O linear and HgCl2 (g) covalent and linear Cl-Hg-Cl
21. c
For quick calculation of B.O. count the total number of electrons in the diatomic
molecule. Refer to answer 22.Knowledge of M.O configuration need to know whether
the molecule is paramagnetic or diamagnetic
22.
e
E1 for H = −1.312 × 106 J
1.312 ×106
1.458×105
23.
= 9 ie 32
b
1
1
𝜈 = =RH
𝜈=
𝑐
𝜆
𝜆
= c 𝜈 = c RH ×
1
1
(22 − ∞2 )
4
= 3× 108 × 109677 × 102 × 0.25 = 8.22 × 1014 𝑠 −1
(RH = 109677𝑐𝑚−1 )
24.
d
1
E0 ∝
4000
KE1 ∝
KE2 ∝
𝜆2 ∝
25.
d
1
1
−
2000
4000
2
4000
1
4000
+
2
4000
∝
=
1
4000
3
4000
=
1
1333
The orbital is dx2-y2
26. c
1
Exp.: y = 𝑥 ⟹ 𝑥𝑦 = 1
∴ A is the set of all points on the rectangular hyperbola 𝑥𝑦 = 1 with branches in I and III
quadrants. 𝑦 = −𝑥 represents a line with slope ′ − 1′ and 𝑐 equal to ‘O’.
∴ B is the set of all points on this line. Since the graphs of 𝑥𝑦 = 1 and 𝑦 = −𝑥 are noninteresting, we have A ∩ B = 𝜙.
27. d
Exp.: A = {3,6,9,12, . . . .) and
B = {6,12,18,24, . . . .)
∴
A – B = {3,9,15,21, . . . .)
= {𝑥 𝜖 𝑁 ∶ 𝑥 is a multiple of 3 but not
of 6.}
28. c
Exp.: 𝑓(−𝑥) = (−𝑥)5 + cos(−𝑥)
= −𝑥 5 + cos 𝑥 ≠ 𝑓(𝑥).
Also 𝑓(−𝑥) = −𝑥 5 + cos 𝑥
= −(𝑥 5 − cos 𝑥) = −𝑓(𝑥)
∴ 𝑓(𝑥) is neither even nor odd.
29. b
Exp.: Let R1 = {(𝑥, 𝑦): 𝑦 2 = 𝑥; 𝑥, 𝑦 𝜖 𝑅}
We have (4,2), (4, - 2 ) 𝜖 𝑅1
∴ 4 has two images 2, - 2
∴ R1 is not a function
Let R2 = {(x,y) : y = |x| ; 𝑥, 𝑦 𝜖 𝑅}
R2 is a function because for each
x 𝜖 𝑅, |x| is uniquely determined
Let R3 = {(x,y) : 𝑥 2 + 𝑦 2 = 1, 𝑥, 𝑦 𝜖 𝑅}
We have (0, - 1 ) , (0,1) 𝜖 𝑅3
∴ 0 has two images - 1, 1
∴ R3 is not a function
Let R4 = {(x,y) : 𝑥 2 − 𝑦 2 = 1, 𝑥, 𝑦 𝜖 𝑅}
𝑥2 − 𝑦2 = 1 ⟹ 𝑦2 = 𝑥2 − 1
⟹ y = ± √𝑥 2 − 1
𝑥 = √2 implies y = ±√2 − 1 = ±1
∴ √2 has two images – 1, 1
∴ R4 is not a function
∴ the correct answer is b
30. a
Exp.: sec𝜃 tan𝜃 are roots of
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0
∴ a sec2𝜃 + b sec𝜃 + c = 0
(1)
a tan2𝜃 + b tan𝜃 +c = 0
(2)
(1) – (2) gives
a(sec2𝜃 −tan2𝜃) + b(sec𝜃 −tan𝜃) = 0
a + b (sec𝜃 − tan𝜃) = 0
sec𝜃 −tan 𝜃 =
31. e
Exp.:
−𝑎
𝑏
𝜋
A+B=2
𝜋
cot (A + B) = cot 4 = 1
cot A cotB−1
cotA + cotB
=1
cot A cotB – 1 = cotA + cotB
cotA cotB – cotA – cotB – 1 = 0
cotA (cotB – 1 ) – (cotB – 1 ) = 2
(cotA – 1 ) (cotB – 1 ) = 2
32. b
Exp.: sinA −√6 cosA = √7 cosA
sinA = (√7 + √6) cosA
sinA =
(√7+ √6)(√7 −√6)
(√7 − √6)
sinA =
7−6
√7−√6
cosA
sinA =
1
7−
√ √6
cosA
cosA
√7 sinA − √6 sinA = cosA
√7 sinA = √6 sinA + cosA
33. d
Exp.: cos20° + cos100° + cos140°
= cos20° + cos(180 – 80)°
+ cos(180 – 40 )°
cos20° −cos80° −cos40°
= ( cos20° −cos40°) −cos80°
= 2 sin30° sin10° −cos80°
1
= 2 × 2 × sin10° −sin10°
(cos80° = sin10°) = 0
34. b
Exp.:
Distance between
(a cos𝛼, a sin𝛼) and
(a cos𝛽, a sin𝛽) is
√𝑎2 (𝑐𝑜𝑠𝛼 − 𝑐𝑜𝑠𝛽)2 + 𝑎2 (𝑠𝑖𝑛𝛼 − 𝑠𝑖𝑛𝛽)2
= |a|√1 + 1 − 2𝑐𝑜𝑠𝛼 𝑐𝑜𝑠𝛽 − 2𝑠𝑖𝑛𝛼 𝑠𝑖𝑛𝛽
= |a| √2 − 2 cos(𝛼 − 𝛽)
= |a| √2(1 − 𝑐𝑜𝑠(𝛼 − 𝛽)
= |a| √2 × 2
𝑠𝑖𝑛2 (𝛼−𝛽)
2
(𝛼−𝛽)
|
2
= 2 |a| |sin
35. b
Exp.:
𝑠𝑖𝑛𝑥
𝑐𝑜𝑠𝑥
𝑠𝑒𝑐𝑥
𝑡𝑎𝑛𝑥
× 𝑐𝑜𝑠𝑒𝑐𝑥 × 𝑐𝑜𝑡𝑥 = 9
tanx × tanx× tanx× tanx = 9
tan4 x = 9 . so tan2 x = 3
tanx= √3 and x 𝜖 [0, 𝜋⁄2]
x = 𝜋⁄3
36. b
4
Exp.: cos (𝛼 + 𝛽) = 5
16
3
sin(𝛼 + 𝛽) = √1 − 25 = 5
tan (𝛼 + 𝛽) =
3
4
sin(𝛼 − 𝛽) =
5
13
cos(𝛼 − 𝛽) =
12
13
tan (𝛼 − 𝛽) =
5
12
tan (2𝛼) = tan [𝛼 + 𝛽 + 𝛼 − 𝛽]
𝑡𝑎𝑛(𝛼+𝛽)+𝑡𝑎𝑛(𝛼−𝛽)
= 1−𝑡𝑎𝑛(𝛼+𝛽)𝑡𝑎𝑛(𝛼−𝛽)
=
3 5
+
4 12
3 5
1− ×
4 12
56
= 33
37. e
Exp.:
2𝛼 = −1 − 𝑖√3 and
2𝛽 = −1 + 𝑖√3
So 𝛼 =
−1−𝑖√3
2
𝛽=
−1+𝑖√3
2
and
i.e., 𝛼 = 𝑤 and 𝛽 = 𝑤 2
where w is a non real cube root of unity.
5 𝛼 4 + 5 𝛽 4 + 7𝛼−1 𝛽−1
= 5𝑤 4+ 5𝑤 8 + 7𝑤 −1 × 𝑤 −2
= 5w + 5 𝑤 2 + 7
[𝑤 3 = 1. So 𝑤 4 = 𝑤 3 × 𝑤
=1×𝑤 = 𝑤
𝑤 8 = (𝑤 6 )𝑤 2 = (𝑤 3 ) 2 𝑤 2
= 1 × 𝑤2
1
and 𝑤 −3 = 𝑤 3 = 1]
But 1 + w + 𝑤 2 = 0
So w + 𝑤 2 = −1
Hence 5 w + 5 𝑤 2 + 7
= 5(𝑤 + 𝑤 2 ) + 7
= 5 (−1) + 7 = 2
38. b
Exp.: 249 | 𝑍 |=|√5 + √3 𝑖 |
33
= (√5 + 3)
33
33
= (√8)
= (2
3⁄ 33
2)
|Z|=2
99
= 22
99⁄ −49
2
1
= 22 = √2
39. a
Exp.: Take the roots as 𝛼 and 𝛽
𝛼 + 𝛽 = 2𝑎 and 𝛼𝛽 = 𝑎2
A.M. =
𝛼+𝛽
2
= 𝑎 and
G.M. =√𝛼𝛽 = 𝑎
Hence A = G
40. e
Exp.: Z = √3 + 𝑖
𝑍 2 = 3 + 2√3 𝑖 − 1 = 2 + 2√3 𝑖
𝑍 2 𝑒 𝑍−𝑖 = (2 + 2√3𝑖) 𝑒 √3
𝑒 √3 is a real number
Hence real part of 𝑍 2 𝑒 𝑍−𝑖 = 2 𝑒 √3
and imaginary part = 2√3 𝑒 √3
Argument = 𝑡𝑎𝑛−1 [
2√3 𝑒 √3
2 𝑒 √3
]
= 𝑡𝑎𝑛−1 [√3]
𝜋
=3
41. b
1
1
1
Exp.: 𝑥+𝑝 + 𝑥+𝑞 = 𝑟
(𝑥 + q) r + (𝑥 + p) r = (𝑥 + p ) (𝑥+q)
(2𝑥 + p + q) r = 𝑥 2 + (p + q) 𝑥 + pq
𝑥 2 + [p + q – 2 r] 𝑥 + pq – pr –qr = 0
Given that roots are equal in magnitude and opposite in sign. Hence sum of roots = 0
P + q – 2 r = 0. So p + q = 2 r
42. a
Exp.: Complex roots occurs in pairs. 2 – i is a root.
So 2 + i is also a root. Sum of roots = 2 – i + 2 + i = 4.
Product of roots = (2 – i) (2 + i) = 5
−12
𝑎
Hence
= 4 and
𝑏
𝑎
=5
a = −3 and b = −15
ab = 45
43. c
Exp.: 𝑥 + 𝑖𝑦 = (2 + 3 𝑖)3
= 8 + 9 𝑖 × 4 + 27 𝑖 2 × 2 + 27 𝑖 3
= −46 + 9 𝑖
𝑥 = −46 and y = 9
3 𝑥 + 2 𝑦 = −138 + 18 = −120
44. e
3
7
3
7
Exp.: The roots are 2 + 2 𝑖 and 2 − 2 𝑖
9
Sum = 3 and product = 4 +
Sum =
3=
+6
𝑎
+6
𝑎
and product =
and
29
2
49
4
=
29
2
𝑏
𝑎
𝑏
=𝑎
a = 2 and b = 29
a + b = 31
45. d
Exp.: To get integral roots for a quadratic equation of the form 𝑥 2 + 𝑏𝑥 + 𝑐 = 0, we
have to split the coefficient of 𝑥 as sum of integral factors of the constant term.
Constant Term = −16
Possible factors are
16 × −1, −16 × 1,8 × −2, −8 × 2
and 4 × −4.
Sums are 15, −15, 6, −6, 0.
Hence there are 5 possible values for b.
15, −15, 6, −6 and zero.