Aiits HCT-7 Jee Adv Paper 1 Solution

FIITJEE
ALL INDIA INTEGRATED TEST SERIES

HALF COURSE TEST – VII
JEE (Advanced)-2022
PAPER –1
TEST DATE: AUGUST-2021
ANSWERS, HINTS & SOLUTIONS

Physics PART – I
SECTION – A
1. D
7
Sol. As air comprises of O2, N2 & H2 all have  =
5
7
  of air can be taken to be
5
For pv = constant
p v

p v
7
=  5% = 7%.
5
2. A
K
Sol. Maximum Acc = 2A =   A = 4m/s2
m
24
A  4  A = 1m.
6
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2
AIITS-HCT-VII (Paper-1)-PCM(Sol.)-JEE(Advanced)/2022
3. C
VA V
Sol. (a) 2 3 B
2L A 4LB
VA 3
   L A  LB 
VB 4
 ART BRT
VA  , VB 
MA MB A B
A MB 3 5 7
  , A  , B 
B MA 4 3 5
MA 400
 
MB 189
4. B
1 1 1
Sol.    r = 4 mm.
r r1 r2
5. C
GM v r
Sol. v0   1  2  r1  r2  v 2  v1
r v2 r1
6. C
Sol. Internal energy does not change in free expansion.
7. BCD
Sol. Magnitude and direction of angular momentum about C is constant. About point O,
magnitude of angular momentum is constant but direction is not constant.
8. AB
dv v 2 dv v 2
Sol.   & v 
dt R ds R
9. AD
GM GM
Sol. Vout   and Vin   3 (3R2  r 2 )
r 2R
10. BC
Sol. F.B.D. of ball in frame of container 4Vg
4Vg  Vg Y
ay   3g
V Va 4Va
L 1 L X
 3gt 2  t 
2 2 3g Vg
4Va  Va
ax =  3a
V
3
1 2L
L 3at 2  t
2 3a
2L L
Ball to collide at point Q, t    a = 2g
3a 3g
11. ABC
Sol. At steady state, the rate of heat gain and the rate of heat loss are equal.
12. AD
Sol. Property of friction force.
SECTION – C
13. 00004.00
Sol. Td = r d2r
T= r22
T r 2 2
Stress =  4
A A
14. 00001.80
Range: 1.90 to 2.00
Sol. (T + T cos 60) vw – T vb = 0
15. 00136.00
Range: 130.00 to 140.00
Sol. Let the bus be at O when it sends a signal that is detected B D C
by the detector as of frequency = 1500 Hz O 
 v 
 f=    1000  1500 l
 v  v B cos  
3
 cos =   = 30°
2 S
By the time signal reaches at S the bus reaches at D. Let
this time be t0
OS lcosec 
 t0 =  … (1)
v v
Now man fires and the bullet reaches C in time t1 (say). In the same time bus moves
from
D to C
l
 t1 = ; where u = speed of bullet
u
Also, OD + DC = lcot
vB t0 + vBt1 = lcot
 lcosec  
vB   + vB (l/u) = lcot 
 v 
2 2 v
2   3
3 3 3 3u
4
v 5
 
u 2
2 2
 u = v   340 = 00136.00 m/s
5 5
16. 00041.40
Range: 40.00 to 42.00
2GM GM
Sol. v esc  ; v orb 
R R
17. 00007.50
Range: 7.00 to 8.00
Sol. Initially M.R. 5M, 2R
Mx = 5 M(12-x)  x = 10R C.M.
x 12x
12R
18. 00020.00
Sol. According to the question
3 h 
 0  hg    0  g 
2 2 
h '  g h g

2 4
P0 h g

2 4
h = 20 m
5
Chemistry PART – II
SECTION – A
19. D
Sol. If the difference between hydration energy and lattice energy is maximum. The
compound will be most soluble in water.
20. C
Sol. Xenon undergoes sp3d2 hybridisation and holds two lone pairs in XeF4.
21. C
Sol. Al(OH)3 is not soluble in NH4OH as it can’t form complex with NH3.
22. A
Sol. Primary amines boil at higher temperature as they can form more number of hydrogen
bonds than 2o and 3o isomers.
23. A
Sol. BF3 is not hydrolysed in water or any polar solvent.
24. D
Sol. BN  NaOH  3H2O  Na B  OH4   NH3
25. ABD
Sol. Cyclobutanol contains one OH group. So it can’t form position isomers.
26. BD
Sol. Sulphur uses d-orbital in forming pi-bonds in SO2 and SO3.
27. ABD
Sol. Ag  Cl  AgCl 
H  CN  HCN  weak acid 
Ag  OH  AgOH 
KCN exerts common ion effect.
28. CD
Sol. Na2SO4 and Na2CO3 do not decompose as Na2O is very unstable to be formed as a
product.
29. AB
Sol. 2-butanol and 2-pentanol contain asymmetric carbon atom.
30. BD
Sol. This is due to keto-enol tautomerism.
6
SECTION – C
31. 00025.20
1
Sol. For 2nd order reaction, t1/2 
 A 0
32. 00011.40
K
Sol. KH  w
Ka
K w 10 14
or, K a    1011.4
K H 10 2.6
pK a   logK a   log1011.4  11.4
33. 00400.00
3RT 2RT
Sol. 
30 20
3R400 2RT
or  , T  400K
30 20
34. 00012.00
Sol. x = 2, y = 1, z = 9
35. 00019.00
Sol. The products
, and
Can form respectively 3, 10 and 6 hyperconjugation structures.
36. 00002.00
Sol. t½  [A]0
 It is a zero order reaction
For zero order reaction
A
t1/2  0
2k
A0 0.4
or k    0.01 or 10 2
2 t1/2 2  20
x=2
7
Mathematics PART – III
SECTION – A
37. D
Sol. Equation of pair of tangents by SS1 = T2 is (a2 – 1)y2 – x2 + 2ax – a2 = 0
If  be the angle between the tangents, then
2 h 2
 ab   
2  a 2  1  1
tan    2
ab a 2
2 a2  1

a2  2
  lies in II quadrant, than tan  < 0
2 a2  1
 0
a2  2
 a2 – 1 > 0 and a2 – 2 < 0
 1 < a2 < 2
 
a   2,  1  1, 2   
38. D
Sol. A = (a cos , b sin )
B = (a cos ( + ), b sin ( +))
C = (a cos ( + 2), b sin ( + 2))
 = Area of  ABC
1 a cos  b sin 
1
 1 a cos  a    b sin     
2
1 a cos    2  b sin    2 

 2ab sin2   sin 
2
     ab sin  1  cos  
ab
  2 sin   sin2 
2
 '   0
 cos   1
1
or cos   
2
cos   1 gives  = 0
1 3 3
cos    gives maximum value of   ab
2 4
8
39. C
8.9.17
Sol. 12 + 22 + 32 +-----------------+82 =
6
= 12.17 = 204
40. D
Sol. Maximum number of points = 8 P2  56
41. A
2
Sol. Equation of hyperbola x 2  y 2   x  y  1
2xy  2x  2y  1  0
 Equation of Asymptotes 2xy  2x  2y  1    0 . For a pair of straight lines   1
 Asymptoes :  y  1 x  1  0 or y  1 and x  1
42. D
Sol. If (x, y) be to incentre then P  2 cos , sin  
x2 y2
 2
1
1
 2 1  2  cos 
2  cos 
S  1, 0  2 S 1, 0 
43. AC
Sol. (4a – 5b)2 – c2 = 0
 (4a – 5b +c) (4a –5b –c) = 0
either 4a –5b + c = 0, or, 4a – 5b – c = 0  – 4a + 5b + c = 0
44. ABC
Sol. The given inequality can be written as
2 2
2cos ec x
 y  1 1  2
2
Since cos ec x  1 for all real x, we have …(i)
cos ec 2 x
2 2
2 2
Also  y  1  1  1   y  1 11 …(ii)
From Eqs. (i) and (ii), we get
2 2
2cosec x
 y  1 1 2
2
Therefore, from Equation (i) and (iv) , equality holds only when 2cos ec x  2 and
2
 y  1 11
2
 cos ec 2 x  1 and  y  1  1  1  sin x  1 and y  1
9
 3
 x , and y = 1
2 2
 3
Hence, the solution of the given inequality is x  , and y = 1
2 2
45. CD
Sol. Abscissa corresponding to the vertex is given by
1
x  1 is the vertex
sin 
The graph of f  x    sin   x 2  2x  b is shown
as  x  1
Therefore, the minimum of vertex
f  x    sin   x 2  2x  b  2 must be greater than
zero but minimum is at x = 1, i.e.
sin   2  b  2  0
0
 b  4  sin ,    0,   1 x = cosec 
 b  4 as sin   0 in (0, )
46. AC
Sol. Sides are in A.P. and a < min{b, c}
Therefore, order of A.P. can be b, c, a or c, b, a.
Case I :
If 2c = a + b
2 2 2
b2  c 2  a 2 b  c   2c  b  4b  3c
cos A   
2bc 2bc 2b
Case II
2 2 2
b2  c 2  a 2 b  c   2b  c  4c  3b
cos A   
2bc 2bc 2b
47. BD
Sol. Since, sin   2 cos   6x  x 2  11
   6x  x 2  11
 sin      ………..(i)
 3 2
 
 1  sin      1
 3
2
6x  x  11
 1  1
2
 2  6x  x 2  11  2
 x=3
 
From Equation (i), sin      1
 3 
 
   2n 
3 2
10
5
   2n 
6
7  19
For  ,  0    4 
6 6
7 19 
or  x,     3,  ,  3, 
 6   6 
48. ABC
Sol. 9 cos12 x  cos2 2x  1  6cos6 x cos 2x  6 cos 6 x  2cos 2x  0
2
  3 cos 6
x  1  cos 2x  0
 
cos2 x 3 cos4 x  2  0 
 cos x  0

 x  n  , n  I
2
2
and cos4 x 
3
2
 cos2 x  
3
2
 cos x   4  
3
2
 x  n  cos1 4  , n  I
3
SECTION – C
49. 00026.40
Sol. Sum of digits in the tens place
  4  1 !  4  5  6  7 
 6  22
= 132
50. 00001.20
Sol.  sin 3x  sin x   sin 2x
  cos 3x  cos x   cos 2x
or  2 cos x  1 sin 2x  cos 2x   0
1
or cos x   , tan 2x  1
2
2 4  5 9 13
 x , , , , ,
3 3 8 8 8 8
11
51. 00000.75
Sol. For point p, y coordinate = 4 y
 Given ellipse is 16x 2  25y 2  400

2
16x 2  25  4   400 (0, 4)
Coordinate of P is (0, 4)
16 9
e2  1   ‘x x
25 25 (–3, 0)S’ C S (3, 0)
foci   ae, 0  i.e.  3, 0 

x y
Equation of reflected ray (i.e. PS) is  1
3 4
or 4x  3y  12
   12, y'
52. 00003.75
Sol. Let (hk) be the middle point the xh  yk  h2  k 2 …(1)
Also equation of normal is x cos   y cot   2a …(2)
(1) and (2) are identical  k  8
53. 00007.50
Sol. Let AB  n, AC  n  1, BC  n  2 . A
sin A sinB sinC
Now,  
n  2 n 1 n
 A  2C
 B  180o  3C n
n+1
sin2C sin3C sin C
  
n2 n 1 n
2 cos C 3  4 sin2 C 1
  
n2 n 1 n B C
n2 2n  1 n+2
 cos C  and sin2 C 
2n 4n
2
n 2 2n  1
    1
 2n  4n
 n4
54. 00033.75
Sol. Let the 3 elements in the union be a, b, c .
Now, a can be in S or T or both i.e. a can be placed in 3 ways.
Similarly b and c can be placed in 3 ways.
 27 ways to pick S ant T.
 6 C3  27  20  27  540 ways

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FIITJEE

ALL INDIA INTEGRATED TEST SERIES

ANSWERS, HINTS & SOLUTIONS

Can form respectively 3, 10 and 6 hyperconjugation structures.

Mathematics PART – III

 Given ellipse is 16x 2  25y 2  400

foci   ae, 0  i.e.  3, 0 

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