Sri Chaitanya IIT Academy, India: A Right Choice For The Real Aspirant Icon Central Office, Madhapur-Hyd
Sri Chaitanya IIT Academy, India: A Right Choice For The Real Aspirant Icon Central Office, Madhapur-Hyd
Sri Chaitanya IIT Academy, India: A Right Choice For The Real Aspirant Icon Central Office, Madhapur-Hyd
MATHS
1 2 3 4 5 6 7 8 9 10
ABCD AC ABD ABD BC ABCD ACD ABCD 5 1
11 12 13 14 15 16 17
2 2 1 1 7 2 A-Q; B-R; C-P; D-S
18 19 20
A-P; B-S; C-P; D-Q a-r, b-r ,c-s ,d-r a-r ,b-q, c-p, d-s
PHYSICS
21 22 23 24 25 26 27 28 29 30
ACD ABC D AD ACD BCD AB ACD 4 2
31 32 33 34 35 36 37 38 39 40
A-qr; B-q; A-q; B-q A-r; B-s A-r; B-s
4 4 5 2 1 6
C-pr; D-pr C-q; D-s C-p; D-q C-p; D-q
CHEMISTRY
41 42 43 44 45 46 47 48 49 50
B AC B BC ABCD AD AC ABCD 2 2
51 52 53 54 55 56
4 2 5 0 8 7
57 A -P Q S, B -R S, C -Q, D – RS 58 A - P; B - P; C - QS ; D – PRS
59 A-S; B-R; C-Q; D-P 60 A-RS; B-S; C-S; D-PQ
2 x 3 px 2 qx r 0
x a1 x a3 x a5 x a2 x a4 x a6 0 f x
f a1 a1 a2 a1 a4 a1 a6 0
f a2 a2 a1 a2 a3 a2 a5 0
f x 0 has a root as a1 , a2
& f a3 0 & f a4 0
f x 0 has a root is a3 , a4
3. a) Let h x f x g x
h x f x g x 0
h x e
b) Consider a continuous function defined on [0, 2], g(x)=f(x+1)-f(x)
g(0)=f(1)-f(0)= f(1)-f(2)
g(1)=f(2)-f(1)
g 0 g 1 0
g c 0; c 0,1
f c 1 f c ; c 0,1
f x1 f x2 ; x1 x2 1
ln 2 x , in 0,1
c) f x sin x, in 1,
2
ln 3 sin1
, x 1
2
b) 7 C 5! 2520
2
5! 5!
d) 3 150
2!2!1! 3!1!1!
5. Conceptual
7. a) ax 2 2hxy by 2 2 gx 2 fy 1 0
& h 2 ab 4 order
d) ax by 1 bx ay c 0
2 2
ax by 1 bx ay c
a 2 b 2 a 2 b2
c) 1 & relation between p & q is given.
p2 q2
2 2
b)
x a
y b 1 (a, b are given)
2
p q2
8. an21 10an an 1 an2 1 0
an2 10an 1an an21 1 0
an1 , an 1 are the roots of t 2 10tan an2 1 0
an 1 an1 10an , n 2, an 1.an1 an2 1 n 2
1
dx 2 x 2 dx
9. a = 4
1
2
2 0 x x 2 a 2 2 1
0
a2 x
x
1
4
x2 1 x2 1 dx
2 0 x x 2 a 2 2 1 0 x 4 x 2 a 2 2 1
2
x 1 1
I1 4 divide by x2 and put x t
0 x x a 2 1
2 2
x
x2 1 1
I2 0 x4 x2 a 2 2 1 dx divide by x2 and put x y
x
and evaluate
x
10. (1) Let x f t dt , ' x f x x 0,
0
x x 2
1 2 1
On differentiating f t dt f t dt
20 x 0
2
x ' x x ' x
We get 4 2 0
x x
' x 2 2
i.e., and on integrating we have
x x
x c.x 2 2
' x 2 2 c.x 1 2 , f x 2 2 c.x 1 2
Since f 1 2
2 c 1
1 1
0 f x dx 0 2 2 x
1 2
dx 1
11. n S 53 125
Number of favorable ways.
2, 1, 1
3, 1, 2 (or) 3, 2, 1
4, 2, 2 (or) 4, 1, 3
(or) 4, 3, 1
5, 2, 3 (or) 5, 3, 2 or 5, 4, 1
Or 5, 1, 4
n E 10
P E 10 2
125 25
12.
T
1
P
A 1 1 B C
O
1 BP
sin 1 BP
OC BC
BP 1 sin
ABP
2
2
AP 4 1 sin 4 1 sin sin
2
5 3sin 2 2sin
2
1 1 16
5 3 sin .
3 3 3
4
AP
3
13. r 1
p q
So, the equation z 3 p z 2 q z r 0
z 1 z 2 p 1 z 1 0
2 sin 2 x sin x 2 0
1
sin x , 2
2
5
x 2n or 2n
4 4
From equation (2) then only possible solution is x 2n
4
Number of solutions is 2
9 92 9999
1 2 ......... 999
17. a) p 10 10 999
10
9
10 9.
10
p=1
b) cos cos x sin x cos sin x cos x
2
c)
d)
19. a) If line cuts the graph at 6 points then using LMVT there exists atleast 5 points
where f ' x p (slope of the line )
using Rolle’s theorem there exists atleast 4 points where f '' x 0
b)
(0, 1)
y = -c
b2 b3 .............b6
b1
6
ai2 6 ai2
f x c 0 , f ' x 2
i 1 x bi i 1 x bi
2
d) B A
1
20. a) ee , eh 2
2
x2 2 y 2 p 2
2 p22
x y
2
2
9x
36
y2
b) x1 2 R cos A
a 2 R sin A
dx1
tan A
da
tan A tan B tan C 3 3
dx dy dz
3 1 1 1 15 6
da db dc
c)
F
(0, x)
A B (1, 0)
E
(0, 0) (x, 0)
x 2 1 x
Area CDFE 1 A
2 2
1 x x 2
2
1 5
“A” is maximum when x and Amax
2 8
d) i)
A /6
Q
P
A
4 2
A cos ecA
2 2
iii) slope OQ > slope OP
A cot A
3 3
Q
P
3 10 3
Buoyant force B = 3 × 10–4 × 103 × g = 3 N
TB = 10 – 3 = 7 N
Reading of B = 14 + 3 = 17 N
Reading of C = {14 + (10 – 3) + 3}N = 24 N
27. capacitance of system is increased. Charge supplied by battery
increases. Potential difference across A is increased.
28. CONCEPTUAL
29. Geometrical path ‘d’ is equivalent to d / 2 is medium. At high temp
TH 4TLow .
d V
So .
2 2 f
f min 4 Hz
where A 2 xdy 2 y dy
4
y 3/2
0 k 0 3 k
1
Since y at 2
2
B 2 3/2 3
a t cos
3 K
d
By Faraday’s law, Eind
dt
Ba 3/ 2 2 3 d
or Eind t sin 2
3t cos
3 K dt
When the frame turns through /4,
t
4
y 2 a 2a
R 2 x .2 t
k k 2 k 4
Eind 12 Ba k 4 Ba 12
2 3/ 2
I 2
R 192 k 2a 48 2
1
32. let time taken to hit OB are t1 & t2 respectively a g cos t12
2
1 t
b g tan t12 1 4 .
2 t2
33. Relative to ‘A’ , ‘C’ goes down with speed 4 m/s. ground frame speed of
c 4 2 32 5 m / s .
34. = r2 [where is a constant]
Charge in the shell (element)
dq = (4r2dr) = (4)r4dr
r 5
Charge enclosed in sphere of radius r, q = 4 r 4 dr 4 r
0
5
By Gauss's theorem,
at r = R/2
v0 vb
e= vb = v0 (1 e) = 2 gh
(1 e)
=1
v0 vb 1 e 1 e
36. = I
m 2
mg sin 45° × =
2 3
N
N
45°
mg
3g
=
2 2
17
net hing force N 2 N||2 mg 34 = 5.83 = 6 (rounded of to nearest integer)
4 2
42. AC
Sol: gold sol is obtained from AuCl3 by reduction while FeCl3 is hydrolysed to obtain
ferric hydroxide sol
Spondumene is simple chain silicate
Na2 S 2 O3 .5 H 2 O Na2 S 5 Na2 SO4 H 2O
43. B
44. BC
Br
Br Br Br
alc KOH
+
Br Br
45. ABCD
It is an example of Sulphonation of aromatic ring; B is perchlorate and C is
hydronium ion
46. AD
Diazotization of aniline happens at a pH of around 6
B and C have pH above 7
47. AC
Eq of ferrous oxalate used for excess acid=2X1.5X3=9meq
Eq of excess H2SO4 (nf =6)=9
Eq of H2SO4 added for ammonia released from salt=3X2X2=12 (nf=2)
Eq of excess H2SO4 with nf 2 = 9X2/6=3meq
Eq of H2SO4 used up for NH3 = 12-3 = 9
Meq of ammonia in 1L=9X4=36
So W of ammonium chloride = 0.036X53.5=1.926g
%purity=19.26
48. ABCD
Integer:
49. 2
Sol: = CRT × i
hence i = 1.1; so = 0.1
Sec: Sr.IPLCO/IC/ISB/LIIT _P2_Solutions Page 12
Sri Chaitanya IIT Academy 21-04-16_ Sr.IPLCO/IC/ISB/LIIT _JEE-ADV_(New Model-VII_P2)_GTA-5_Q.P
and [H+] = c = 0.1 x 0.1 = 10-2 M
pH = 2
50. 2
Sol: BaCrO4 – Yellow, soluble in dil HNO3
Hg2CrO4 – Red, soluble in conc. HNO3
ZnS – White, soluble in Conc. HNO3
BaSO4 – White, insoluble in all mineral acids
CH3COOAg and AgNO2 all are white solid and are soluble in dilute HNO3
solution
51. 4
sol: Let x% be the decrease in density of Ge crystals. Let ‘a’ be the total no. of Ge
atoms missing and ‘b’ be the no. of boron atoms replacing Ge atoms.
b
100 2.376 ..........(i )
a
72.6a 11b
1 1 0.01x
N AV N AV
72.6a 11b
0.01x ……….(ii)
N AV N AV
150 N
Also, b 6 AV 1 0.01x ……………(iii)
10 11
Solving (i), (ii) and (iii)
X = 4%
52. 2
53: 5
Sol:
PCl3 3H 2O H 3 PO3 3HCl
54. 0
55. 8
Sol: 1st titration
1 1
Meq. Of NaOH Meq. of Na2CO3 Meq. of HCl 50 5
2 10
2nd titration
56. 7
Matrix matching:
57. (A p, q, s), (B r, s), (C q), (D r, s)
58. A p; B p; C q,s ; D pr