FS_G2
Test Date: 20/08/2023
Code-A
Corporate Office : Aakash Tower, 8, Pusa Road, New Delhi-110005, Phone : 011-47623456
Time : 3 Hrs
TERM EXAM for First Step JEE (Advanced)-2023-25
MM : 180
Test – 1A (Paper - 1)_Actual Pattern-2021
ANSWERS
PHYSICS
CHEMISTRY
MATHEMATICS
1.
(B)
20.
(B)
39.
(D)
2.
(A)
21.
(D)
40.
(B)
3.
(B)
22.
(B)
41.
(C)
4.
(D)
23.
(A)
42.
(D)
5.
(46.00)
24.
(03.00)
43.
(00.00)
6.
(15.00)
25.
(01.00)
44.
(04.00)
7.
(08.00)
26.
(03.00)
45.
(06.00)
8.
(59.00)
27.
(01.00)
46.
(01.00)
9.
(05.00)
28.
(85.00)
47.
(23.00)
10.
(01.00)
29.
(02.00)
48.
(15.00)
11.
(A)
30.
(A, B, D)
49.
(A, B)
12.
(B, C, D)
31.
(B, C, D)
50.
(B, D)
13.
(C, D)
32.
(A, B, C, D)
51.
(B, C)
14.
(A, C)
33.
(A, D)
52.
(A, C)
15.
(B, C, D)
34.
(A, B, D)
53.
(A, D)
16.
(A, B, D)
35.
(A, C)
54.
(B, D)
17.
(04)
36.
(02)
55.
(71)
18.
(02)
37.
(17)
56.
(36)
19.
(01)
38.
(04)
57.
(04)
Corporate Office : Aakash Tower, 8, Pusa Road, New Delhi-110005, Phone : 011-47623456
[1]
Test-1A_Paper-1 (Code-A)_G2_Answers & Solutions
Term Exam for First Step JEE (Advanced)-2023-25
Test Date: 20/08/2023
FS_G2
Code-A
Corporate Office : Aakash Tower, 8, Pusa Road, New Delhi-110005, Phone : 011-47623456
Time : 3 Hrs
TERM EXAM for First Step JEE (Advanced)-2023-25
MM : 180
Test – 1A (Paper - 1)_Actual Pattern-2021
ANSWERS & SOLUTIONS
PART –I : PHYSICS
1.
Answer (B)
5.
Answer (46.00)
6.
Answer (15.00)
Solution for Q. Nos. 5 & 6 :
∆x ∆a 2 ∆b 3∆c
= +
+
x
a
b
c
2.
Answer (A)
Number of significant figures in sum is 2.
3.
Answer (B)
 
1
=
S U AB t + aAB t 2
2
1
1
v 0 t − gt 2 − gt 2 =
0
2
2
t=
=
t
v0
g
dt
= 0 ⇒ x = 60 m
dx
v 02 1 v 02 H
−
=
g 2 g
2
=
t
∴
v 0 = gH
4.
Answer (D)
302 + (100 − x )2
802 + x 2
+
3
4
100 50
+
 46 s
3
4
∴ dtotal = 802 + 602 + 302 + 402 =
150 m
7.
Answer (08.00)
ui = 10 m/s, ai = 6 m/s2
u1 = ( 10 – 6) = 4 m/s, a1 =
6
= 3 m/s2
2
u2 = ( 4 – 3) = 1 m/s, a2 = 1.5 m/s2
T =
u2
1
2
∆t3 = =
=
s
a2 1.5 3
2v 0
g cos θ
1
=
Y v 0 cos θ.T − gT 2
2
2v 2
2v 02
2v 02
=0 −
=
tan2 θ
2
g
g
g cos θ
∴t=1+1+
8.
2
8
= s
3
3
Answer (59.00)
=
S1
102 − 42 100 − 16
= = 7m
2× 6
12
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[2]
Term Exam for First Step JEE (Advanced)-2023-25
S2 =
42 − 12
15
= = 2.5 m
2×3
2×3
S3 =
12 − 02
=
2 × 1.5
1
1
=
m
3 3
2×
2
1 59
m
∴ Stotal = 7 + 2.5 + =
3 6
9.
Answer (05.00)
Test-1A_Paper-1 (Code-A)_G2_Answers & Solutions
13. Answer (C, D)
While both are in air
areal = 0, vreal = constant
When, one is on ground
areal = g, vreal will be increasing
14. Answer (A, C)
[η] =ML−1T −1
Block will loose the contact when normal
reaction vanishes.
[η] =[m]x [D]y [v]z
⇒ x = 1, y = – 2, z = 1
10. Answer (01.00)
Draw F.B.D. at t = 3 s
∴ η =k
mv
D2
15. Answer (B, C, D)
11. Answer (A)
S = t3 + 5
am = µg = (0.1)10 = 1 m/s2
dS
= 3t 2
dt
a2m =
v = 3t2
1
m/s2
2
1 3

arel =  1 +  =
m/s2

2 2
dv
= 6t
dt
At t = 2 s
(1) = (2)t –
v = 12 m/s
(12)2 144
⇒ =
ac
= = 7.2 m/s
20
20
dv
= 12 m/s2
dt
a=
ac2 + at2 =
≈ 14 m/s2
12. Answer (B, C, D)
dv
= 4 m/s2
at =
dt
aR =
v2
= 16 m/s2
R
t=
1 3  2
 t
22
2
s
3
16. Answer (A, B, D)
max
( 0.5 ) × 30 = 15 N
f=
s
122 + (7.2)2
3m
amax
= 0.5 × 10 = 5 m/s2
a=
10 5
= m/s2
6
3
17. Answer (04)
4F2 = 4F2 + F2 + 4F2cosθ
cosθ = −
1
4
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[3]
Test-1A_Paper-1 (Code-A)_G2_Answers & Solutions
18. Answer (02)
Term Exam for First Step JEE (Advanced)-2023-25
25. Answer (01.00)
Retarding force F = ma = µR = µmg
rmix
4/5
=
= 2=
rx
4 / 10
∴ a = µg
Now from equation of motion v2 = u2 – 2as
⇒ 0 = u 2 – 2as ⇒ s =
2
2
2
=
1
v 02
u
u
=
=
2a 2µg
2µg
36
4
36
= =
M(avg) 1 M(avg)
M(avg) = 9
19. Answer (01)
Suppose mole of H2 and CH4 are X and Y
mg
3
g
=
⇒ a=
m (g – a) 2
3
respectively.
9=
PART –II : CHEMISTRY
20. Answer (B)
2 × X + 16 × Y
(X + Y)
9X + 9Y = 2X + 16Y
Initially electrons are present in 6th state.
∴ Transition belonging to infrared region :
6 → 5, 6 → 4
6 → 3, 5 → 4, 5 → 3
4→3
X 1
=
Y 1
26. Answer (03.00)
N + e− → N− (Endothermic process)
21. Answer (D)
Number of radial nodes for given orbital
27. Answer (01.00)
=n––1=1
E.A. of Be is zero or negative.
28. Answer (85.00)
22. Answer (B)
Na+
Mg+
Al+
Si+
s0
s1
s2
s 3 p1
+
For the first excited state n = 2
Energy = −13.6 ×
+
Na has noble gas configuration while Al has
fully filled s-orbital.
23. Answer (A)
=
−13.6 ×
+
24. Answer (03.00)
Mavg 4
=
=
2
1
Mavg 16 M(avg)
=
=
2
1
2
Suppose mole of C2H4 and CO2 is x and y
32 =
28 × x + 44 × y
x+y
n2
25
22
=
−85 eV
rH = 0.529
n2
1
= 0.529 × 1 Å
M(avg) = 32
respectively
Z2
29. Answer (02.00)
O= N= O has bond angle equal to 180°.
rH2
=
rmix
Mx
M(mix)
= 0.529 Å
rBe3 + 0.529 ×
=
22
= 0.529 Å
4
30. Answer (A, B, D)
3dxy > 4dxy (penetration power)
31. Answer (B, C, D)
There are three possible values of spin quantum
4x = 12y
number it means an orbital can accommodate 3
x 12 3
= =
y
4
1
electrons. So 1s1 – 1s3, first period would have 3
vertical columns.
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[4]
Term Exam for First Step JEE (Advanced)-2023-25
Test-1A_Paper-1 (Code-A)_G2_Answers & Solutions
37. Answer (17)
32. Answer (A, B, C, D)
For mth line n2 = (m + 1)
Let the volume of each vessel be V L
1
1
1 
= Rz2  2 −
2
λm
1 (m + 1) 
V ×1
300 R
moles of Cl2 =
for nth line n2 = (n +1)
moles of Cl2O =
1
1
1 
= Rz2  2 −
2
λn
1 (n + 1) 
Cl2O → Cl2 +
λm (m + 1)2  (n + 1)2 − 1 
=


λn
(n + 1)2  (m + 1)2 − 1
V × 1.5
300 R
1
O2
2
After reaction
total moles of Cl2
33. Answer (A, D)
=
V
1.5 V
2.5 V
5V
+
=
=
300 R 300 R 300 R 600 R
moles of O2 =
1.5 V
1.5 V
=
2 × 300 R 600 R
Total moles of gases in two vessel after the
34. Answer (A, B, D)
The dipole moment of CH2Cl2 is greater than
reaction =
5V
1.5 V
6.5 V
+
=
600 R 600 R 600 R
CHCl3.
When the two vessel are kept in water bath at
35. Answer (A, C)
different temperatures diffusion will take place till
En = –13.6 ×
Z2
n
2
the pressure of the gases in two vessel becomes
eV
∴ 3.4 eV = –13.6 ×
equal. Let the moles of gases left in one of vessel
22
2
n
⇒n=4
n = 4, l = 2, m = 0 belongs to 4d orbital
Number of angular nodes in an orbital = value
of l
Number of radial node in 4d = 1
Since it is a unielectronic species, there will be
no shielding and the nuclear charge felt will
maintained at 27°C be x. Thus moles of gases in
another vessel maintained at 52°C would be
 6.5 V

− x .

600
R


 R × 325
x × R × 300  6.5 V
= 
− x
V
 600 R
 V
x = 5.65 × 10−3
V
R
P = 5.65 × 10 −3
V R × 300
×
=
1.695 ~ 1.7 atm .
R
V
be 2e.
36. Answer (02)
1
1
1
= RH z2  2 − 2 
λ
 n1 n2 
38. Answer (04)
Energy of incident photon should be higher than
for n1 = 1, n2 = ?
work function to slow photoelectric effect
1 1
= 1.097 × 107 × 4  2 − 2 
30.4 × 10
1 n2 
Energy of photon
1
−9
n2 = 2
=
6.62 × 10−34 × 3 × 108
300 × 10−9 × 1.6 × 10−19
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= 4.14 eV
[5]
Test-1A_Paper-1 (Code-A)_G2_Answers & Solutions
Term Exam for First Step JEE (Advanced)-2023-25
PART –III : MATHEMATICS
39. Answer (D)
| 2x − 5 | < 1
−1 < 2 x − 5 < 1
4 < 2x < 6
2<x<3
x ∈ (2, 3)
40. Answer (B)
D≥0
 −2 2 
,
⇒ 3a2 – 4 ≤ 0 ⇒ a ∈ 

3
 3
If p > q, the two curves intersect at 8 points; 2 in
Also f(2) > 0
each quadrant
⇒ a2 – 2a + 3 > 0 ⇒ a ∈ R
 x.y < 0 represents second and fourth
−b
<2
2a
quadrant. So, required number of ordered pairs
⇒
−( −a )
<2
2 ×1
⇒ a<4
41. Answer (C)
If p < q, the two curves do not intersect.
will be 4.
45. Answer (06.00)
46. Answer (01.00)
Solution of Q. Nos. 45 & 46 :
( A ∪ (B ∪ C )) =
{1, 2, 3, 4, 5, 6}
∴
n ( A ∪ (B ∪ C )) =
6
A × B = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}
B × C = {(3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}
( A × B ) ∩ (B × C ) =
{(3, 4)}
42. Answer (D)
n (( A × B ) ∩ (B × C )) =
1
A = {2, 3, 5, 7}, B = {1, 2, 3, 4, 5}
47. Answer (23.00)
∴ A ∩ B = {2, 3, 5}
48. Answer (15.00)
∴ n[(A × B) ∩ (B × A)] = [n(A ∩ B)]2 =
32 = 9 ordered pairs
43. Answer (00.00)
44. Answer (04.00)
Solution for Q. Nos. 43 & 44 :
Solution for Q. Nos. 47 & 48 :
sin x cos y =
1
4
3 tan x = 4 tan y
3 sin x 4 sin y
=
cos x
cos y
3 sin x cos y = 4 sin y cos x
3×
1
3
=
4 sin y cos x ⇒ sin y cos x =
4
16
∴
sin(=
x + y ) sin x cos y + cos x sin y
=
4+3
1 3
7
+
=
=
16
4 16
16
∴ a + b = 7 + 16 = 23
Corporate Office : Aakash Tower, 8, Pusa Road, New Delhi-110005, Phone : 011-47623456
[6]
Term Exam for First Step JEE (Advanced)-2023-25
Test-1A_Paper-1 (Code-A)_G2_Answers & Solutions
53. Answer (A, D)
sin(=
x − y ) sin x cos y − cos x sin y
=
 α + β = –k and αβ = 9
1 3
1
−
=
4 16
16
α + 2β = 2αβ2
∴ a – b = 16 – 1 = 15
⇒ –k + β = 18β ⇒ β = −
49. Answer (A, B)
x + 2y = 5
2
 k 
 k 
So,  −  + k  −  + 9 =
0
17


 17 
For x = 1, y = 2
For x = 3, y = 1
If x ≥ 5 , then y is negative which does not
belong to natural number.
If x = 2, 4, then y become rational number.
51
⇒ k=
±
4
54. Answer (B, D)
Coffee is greater than choco.
∴ Domain = {1, 3}
∴ x>y
Range = {2, 1}
Each pack is at least 10 g
50. Answer (B, D)
where x ≠
Let z = x + iy
(x + iy)2 – (x – iy) – (x2 + y2) +
55. Answer (71)
64
(
64
| x |5
x + y ≥ 10
∴
1
2
)
Z = (6 + 5i)2
=0
5
= 36 +25i2 + 60i
x2 + y 2 2
= 11 + 60i
⇒ 2xy – y = 0 ⇒ y = 0
Also, − x +
∴
= 0 ⇒ x= 2
Z= 11 − 60i
Re(Z ) − Im(Z ) = 11 – (–60) = 71
56. Answer (36)
So, z = 2
If x and y are integers then sinx = siny
51. Answer (B, C)
⇒ x=y
 1 ∉ A ∪ (B ∩ {1, 2, 3})
 n(A × A) = 81
⇒ 1 ∉ A and 1 ∉ B ∩ {1, 2, 3}
Out of these 81 ordered pairs (1, 1), (2, 2) ...,
(9, 9)
⇒ 1 ∉ A and 1 ∉ B
⇒ 1∉A∪B
Cannot satisfy the given inequality.
⇒ 1 ∈ (A ∪ B)′
Required number of ordered pairs
 4 ∉ B ∩ {1, 2, 3} and 5 ∉ B ∩ {1, 2, 3}
=
So, the smallest possible set A = {4, 5}
Also, smallest possible set B = φ (when A = {2,
3, 4, 5})
81 − 9
= 36
2
57. Answer (04)
52. Answer (A, C)
tan2 θ
sec 2 θ − 1 (sec θ + 1)(sec θ − 1)
=
=
(1 + sec θ)2 (1 + sec θ)2
(1 + sec θ)2
1
−1
sec θ − 1 cos θ
=
=
1
sec θ + 1
+1
cos θ
=
k
17
tan10º tan80º + tan20º tan70º + tan30º tan60º +
tan40º tan50º
= tan10º cot10º + tan20º cot20º + tan30º cot30º
+ tan40º cot40º
tan θ cot(90º − θ)
 =




1
θ
=
tan


cot θ
=1+1+1+1
1 − cos θ
1 + cos θ
=4
  
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[7]