FTS-2A Paper-1 Sol
FTS-2A Paper-1 Sol
FTS-2A Paper-1 Sol
A
CODE
Time: 3 hrs
FINAL TEST SERIES MM : 180
for
JEE (Advanced)-2023_Test-2A_(Paper-I)
ANSWERS
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Test-2A_Paper-1_(Code-A)_(Answers and Solutions) FINAL TEST SERIES for JEE (Advanced)-2023_PHASE-III
PART – I : PHYSICS 40
v cm
3
1. Answer (D)
x 3 –1 3
tan tan 40
1 VSD
p – 1
pq
40
3 40
4. Answer (A)
1
1 MSD
q
2. Answer (A)
1 1 1
v 10 40
1 1– 4 –3
v 40 40
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FINAL TEST SERIES for JEE (Advanced)-2023_PHASE-III Test-2A_Paper-1_(Code-A)_( Answers and Solutions)
4 3 1 t t t
2 30 4u dq q RC A RC B 2 RC
5 5 e RC e te – t e
dt RC R R
u = 20 m/s
t A t
B
t
Time of collision =
d
d
d q e RC
t e RC – t 2 e RC dt
R
R
30cos37 20cos53 24 12
d 72 q et t e dt – t
t 2
e t dt
= 2 second
36 36
v1 y 30
3
5
– 10 2 18 – 20 = – 2 m/s
q et t et – et – t 2et – 2 t et dt
4 3 t et – et – t 2et k
v 2 y 20 – 10 2 16 – 20 = –4 m/s
5
at t = 0 , q = 0
0=0–3+kk=3
q = 3 [t – 1] – t2 + 3e–t
9. Answer (00.00)
10. Answer (02.25)
1 So work done = 0
S2 16 2 – 10 4 32 – 20 12 m
2 Kmin. 2.25 J
h2 – h1 = 4 m
11. Answer (A, C, D)
For 1st
v2
aAC
5R
aBA aB /G aA /G
v2
aBA
R
10 1
h1 16 3 10 9
3 2 aB /G aBA aAG
= 10 + 45 = 55 m v2 v2 4v 2 v2 1
aB /G 1
h1 = 55 – 16 = 39 m R 5R 5R R 5
h2 = 39 + 4 = 43 m
v2 v2 6v 2
7. Answer (00.01) aD / G
R 5R 5R
8. Answer (00.55)
12. Answer (B, D)
Solution for Q. Nos. 7 and 8
q
iR v t
C
q dq
R At – Bt 2
C dt
dq q A B
t – t2
dt RC R R
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Test-2A_Paper-1_(Code-A)_(Answers and Solutions) FINAL TEST SERIES for JEE (Advanced)-2023_PHASE-III
dy 0 il l x
1 × sin60 = sin tan = ln
dx 2 x
d 0 i 0 l l x
dx
2
1 ind = cos t ln
1 dt 2 x
dy
sin2
0 i0 l l x
I cos t ln
3 1 1 8R x
sin2 2
2 2
dx
dx 1 Fnet = Il(BA – BB)
1
dy dy
0 i 0 l 2 l x 0 i 0 sin t l
= cos t ln
3 dx
2
dy
8R x 2 x l x
1 3 – 2x
2 dy dx 20 i 02 l 2 l x l
= sin 2t ln
322R x x l x
3 1
1
2 2 at x = 2l
3 – 2x
02 i02 l 3
13. Answer (C, D) Fnet sin t ln
2
6 32 R 2
Torque about OO of mg, normal reaction and
16. Answer (B, C)
spring force at any instant is zero so angular
momentum is conserved about OO
14. Answer (A, C, D)
16R = Rz2
z=4
For transition from 2nd excited state to 1st exited
state. v 2ax
1 1 v 8 2ax
E 13.6 16 –
4 9
64 32
a m/s2
5 2x x
13.6 16
36
volume l 3
= 30.2 eV Time to empty the chamber =
flow rate 8a
From 4th to 2nd excited state,
17. Answer (06)
1 1
E 13.6 16 –
9 25
13.6 16 16
225
For minimum, wavelength
n = to n = 1
E = 13.6 × 16(1) mV
R
12400
qB
A 57 Å
217.6
l
15. Answer (B) sin
R
0 i
ldx P P
2x R l
2eB 2eB
l x
0il dx
=
2
x
x
P
RP
eB
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FINAL TEST SERIES for JEE (Advanced)-2023_PHASE-III Test-2A_Paper-1_(Code-A)_( Answers and Solutions)
180
6
30
18. Answer (04)
3a
rA2 rB–
4
Q
Q Q dx hA T – TS 4
x 4 ((150)3 2 (45)3 )
3
dQ 64
·dx hA T – TS (150 45)3
dx 3 3
d 2T = 0.649
KS 2
dx hA T – TS S = R2
dx 23. Answer (D)
A = (2Rdx)
(I) CrCl2 NO 2 2 NH3 2
2
d T 2h
2
T – TS
dx KR
2 25 500
5
K 100
h 0.1
PART – II : CHEMISTRY
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Test-2A_Paper-1_(Code-A)_(Answers and Solutions) FINAL TEST SERIES for JEE (Advanced)-2023_PHASE-III
Amount of R = x =
2 80 75 50
2 10 330
100 100 100
= 1.98 g
Amount of w = y
80 60 50
= 2 102 1 1 122
100 100 100
= 0.58 g
26. Answer (83.00)
27. Answer (17.00)
Solution for Q. Nos. 26 and 27
A (S)
B(S) 2C(g)
Hence (1) and (2) are anomers.
2
P Hemiacetals are reducing in nature.
KP C
P0 31. Answer (A, B, C, D)
2 In associated colloids, the dispersed phase is
P
Since Gº = –RT In KP = – RT In C made up of particles that behave as regular
P0 electrolytes at lower concentration but
aggregate at higher concentration to particles of
P
= – 2RT In C colloidal dimension. The aggregated molecules
P0 thus formed are called micelles.
Now Gº = Hº – TSº In colloidal solutions, smaller particles
aggregate to form bigger particles of colloidal
P dimension.
2RT In C H TSº
P0 In colloidal solution, the value of colligative
P H Sº properties is of lower order as compared to true
In C solution of same concentration because
P0 2RT 2R colloidal solutions contain less number of
particle as compared to true solution.
P H 10 4 Sº
In C Lyophobic colloids have higher refractive index,
P0 2R 10
4
T 2R
thus tyndall effect is more prominent in
H 2 5 3 1 lyophobic colloids as compared to lyophilic
Slope = colloids.
2R 10 4
6 12 6 2
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FINAL TEST SERIES for JEE (Advanced)-2023_PHASE-III Test-2A_Paper-1_(Code-A)_( Answers and Solutions)
12KCI
FeCI3 + 3KCNS Fe(CNS)3 + 3KCI
Red colour.
34. Answer (B, C)
Total isomes = 20
1 3 1 3 8 1
+ …. + (n – 1) term
4 4 9 4 9 16
1 1 1 1 1
....
2 1.2 2.3 3.4 (n 1)n
36. Answer (04)
1 1 1 1 1
1 ....
2 2 2 3 ( n 1)n
1 1 n 1
1
2 n 2n
n 1
Odd against :
are more basic than aniline n 1
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Test-2A_Paper-1_(Code-A)_(Answers and Solutions) FINAL TEST SERIES for JEE (Advanced)-2023_PHASE-III
1
2 5
3
46. Answer (01.00)
1
sin
x 2n x x 2 2x 1
41. Answer (B) 1
lim lim x 1
|A| = (2k + 1)3, |B| = 0 (since B is skew –
x n 2n 1
symmetric of order 3) x 3 1 x cos 2
x
det(adj A) = |A|n – 1 = ((2k + 1)3)2 = 106
47. Answer (01.00)
2k + 1 = 10 or 2k = 9
48. Answer (02.00)
[k] = 4
Solutions for Q. Nos. 47 and 48
42. Answer (C)
3
Clearly magnitude of faces F1, F2, F3 and F4 are Let tan–1x = t, 0 t ; A t3 t
2 2
same as it is a regular tetrahedron and for a
closed figure which is regular: 3 3 t
= t
V 1 V 2 V 3 V 4 0 8 2 2
Also g(0) = 0 2 2
B ,
8 4
g(x) > 0 x 0 f(x) < 0
49. Answer (A, B, C)
f(x) < 0 and g(x) > 0 x 0
3 1 2
45. Answer (05.00) P ( A) , P (B ) , P (C )
10 2 5
1
x 2n 1 sin ( x 2 2 x ( 1)) P(exactly one to pass) = P ABC +
lim x
n 1 1 44
P ABC P ABC
2n
x 3 x cos 2
x x 100
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FINAL TEST SERIES for JEE (Advanced)-2023_PHASE-III Test-2A_Paper-1_(Code-A)_( Answers and Solutions)
Split inequality into two weaker inequalities: P = 120°, Q = 150° (not valid); P = 120°; Q = R
= 30° q = r = 1
|z – 5| |z – 2|
|z + 1| |z – 2|
The first requires Re(z) 3.5 while the second
requires Re(z) 0.5
So obviously no z fulfils both inequalities
simultaneously.
1/ 4 1
51. Answer (A, B, C, D) Slope of RS =
3 3/4 3 3
log(n 1) log n
An +
3 1 1
log n log( n 1) Equation x 3 3 y ; k OE
2 6 6
log( n 1)
Let x 1
log n 7
RS
2
1
= x 0 0 1
x
1 1 3
>2 Area of SOE 0 1
2 6 48
1 3 1
If An 3 then x 1 1
x 4 4
Let y x 1 3 2 3
r
s 2
2
1 1 5
If y 1 then y2 – y 1 or y or
y 2 4
1 5
y
2
So x = y2 > 2.618
log(n 1)
But, using natural logs, x
log(n )
1 1 1 1 Let x2 – 5x + 4 = a
Then ... 2 2
12 22 k k a(y2 + y + 1) < 2y ay2 + y(a – 2) + a < 0
1 Vy R
On adding on both side
(1 k )2
a < 0 and D < 0 a < 0 and
1 1 1 1 1 1 (a – 2)2 – 4a2 < 0
2
2 .... 2 2
2
1 2 k (k 1) k (k 1)2
3a2 + 4a – 4 > 0
1 1 1
Now 2 2 will be true if a < 0 and a < –2 a < –2
k (k 1)2 k 1
x2 – 5x + 4 < –2 x2 – 5x + 6 < 0
1 1 1
x (2, 3)
k (k 1)2 k 1
57. Answer (03)
Sk + 1 < T k + 1
Let x = b + c – a, y = a + c – b, z = a + b – c
55. Answer (06)
y z zx xy
Then a , b , c
1 1 2 2 2
sin3x sin3x = (4sin3x)sin3x = (3sinx –
4 4
y z z x x y
sin3x)sin3x
x x y y z z
Now 1
1 1 1 6
= (2 × 3sin3x sinx – 2sin23x) =
4 2 8
a b c
[3(cos2x – cos4x) – (1 – cos6x)] 3
bc a ac b abc
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