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Narayana IIT Academy INDIA
Sec: Sr. IIT_IZ Jee-Advanced Date: 11-11-17 Time: 09:00 AM to 12:00 Noon 2011_P1 Model Max.Marks: 240
KEY SHEET
CHEMISTRY 1 C 2 D 3 D 4 B 5 A 6 D
7 B 8 AC 9 BC 10 ABD 11 ABD 12 A
13 C 14 D 15 B 16 B 17 9 18 2
19 4 20 7 21 8 22 3 23 6
PHYSICS 24 B 25 D 26 B 27 A 28 B 29 C
30 A 31 BCD 32 ABCD 33 ABC 34 ABC 35 B
36 C 37 C 38 D 39 C 40 7 41 8
42 2 43 9 44 2 45 4 46 2
MATHS 47 A 48 C 49 D 50 A 51 D 52 C
53 C 54 ABCD 55 ABD 56 AC 57 ABCD 58 B
59 B 60 C 61 A 62 B 63 0 64 6
65 5 66 6 67 6 68 7 69 9
Narayana IIT Academy 11-11-17_Sr.IIT_IZ_JEE-Adv_(2011_P1)_CTA-4_Key & Sol’s
Sec: Sr.IIT_IZ Page 2
CHEMISTRY 1. 2. (a) Acid strength H2SO4 > H2CO3.
Basic strength HCO3– > HSO4–. (b) Acid strength HCl > HF. Basic strength F– > Cl–. (d) Acid strength HSO4– > HCN. 3. Sol. Initial formula º A1B3 later the formula becomes A1–1/8 B3 i.e. A7/8 B3 i.e. A7B24 4. 5. 6. 7. HCOOH H+ + HCOO–
C – x x + 0.01
a
x 0.01 xK
0.01 x
neglect x w.r.t. 0.01 due to common ion effect
Ka = x
∴ X= [HCOO–] = 2 × 10–4
8. A) As ([H+] = [OH–]) , Hence (pH = pOH) for pure water
(B) pH = 1 pH = 3
[H+]1 = 10–1 [H+]2 = 10–3
= 100
(D) Acording to Lewis concept water acts only base
9. KSP for AgCl = 10–9 × (0.1) = 10–10
0/ /Cl AgCl AgE = +0.80 +
0.0591
log10–10 = 0.21 V.
10.
11. Water softened by ion exchange resin is completely free form minerals and is not
useful for drinking purpose.
12. 222Mg OH Mg OH
Narayana IIT Academy 11-11-17_Sr.IIT_IZ_JEE-Adv_(2011_P1)_CTA-4_Key & Sol’s
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0.2 0.2
0.1 x
2 120.1 1.6 10x
64 10x
6 log 4 5.48.6
pOHpH
13.
2 12
2 12
6
0.08 1.6 1020 1020 10
16 1.32
6 0.655.35
8.65
yyypOH
pH
14. moles of NaOH left
222Mg OH Mg OH
0.2 0.16
0.12 z
2 12
2 6
0.12 1.6 10160 1012
16 1.6 0.482
6 0.56 5.44
z
z
pOH
pH=8.56 15. Ans: B
(A) 24 4/ /
oPb PbSO SOE = 2
0.062
oPb PbE
log Ksp
= + 0.12 – 0.03 × (–16) = + 0.12 + 0.48 = 0.60V (B) E = 2 2
4 4/ /Pb PbSO SO Pb PbE E
= 2
160.06 10log2 0.1
oPb PbE
+ 2 20.06 1log2 [ ]
oPb PbE
Pb
Narayana IIT Academy 11-11-17_Sr.IIT_IZ_JEE-Adv_(2011_P1)_CTA-4_Key & Sol’s
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= – 0.03(–15) – 0.03 1
0.01 log
= 0.45 – 0.06 = 0.39 Ans.]
16. 17. Ans. 9
Pressure at the surface of lake in terms of H2O = 1 atm = 10 m of water P at the bottom = 10 + 80 = 90 m of H2O Now applying boyle's law 10 V1 = 90 V2
2
1
VV
= 1/9 V1 - initial volume
18. Ans 2 Sol. CuSO4 . 5H2O(s) CuSO4(s) + 5H2O(g) Kp = 10–10 (atm).
10–10 cm =2
5H OP ⇒
2H OP = 10–2 atm. n = PVRT
= 210 2.5
1 30012
= 10–3
19. 20. Ans. 13 Sol. When NaCl is electrolysed, Q = it = » 0.1 At anode : 2Cl– Cl2 + 2e–
At cathode : 2H2O + 2e– H2 (g) + 2OH– 0.1 F 0.1 Mole So, [OH–] = 0.1 pOH = 1. 21. 22. 3
Sol. (D) t20 = 2.303 100 2.303 10log log
100 20 8
2.303[1 3 log 2]
t50 = 2.303 100 2.303log log 2
100 50
50
20
t log2 3t 1 3log2
or t50 = 3 t20 = 3t
23.
Narayana IIT Academy 11-11-17_Sr.IIT_IZ_JEE-Adv_(2011_P1)_CTA-4_Key & Sol’s
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PHYSICS 24. At the initial moment, angular velocity of rod is zero.
Acceleration of end B of rod with respect to end A is shown in figure.
Centripetal acceleration of point B with respect to A is zero 2 0 l
So at the initial moment, acceleration of end B with respect to end A is perpendicular
to the rod which is equal to 2 2a b
rela l
2 2a bl
where is angular acceleration
25. 2 2T.4a cos120 l a g a h g
2T.2a a g l h
2Tl ha g
26. Friction force between wedge and block is internal i.e. will not change motion of
COM. Friction force on the wedge by ground is external and causes COM to move
towards right. Gravitational force (mg) on block brings it downward hence COM
comes down.
27 When 4 coaches (m each) are attached with engine (2m) according to question
P = K 6mgv (1)
(constant power), (K being proportionality constant)
Since resistive force is proportional to weight
Now if 12 coaches are attached
1P K.14mg.v (2)
Narayana IIT Academy 11-11-17_Sr.IIT_IZ_JEE-Adv_(2011_P1)_CTA-4_Key & Sol’s
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Since engine power is constant
So by equation (1) and (2)
16Kmgv 14Kmgv 16v v
14
6 2014
6 10 607 7
1v 8.5m / sec
Similar for 6 coaches 2K6mgv K8mgv
26v 208
3 20 15 m / sec4
28. Force on table due to collision of balls:
3dynamic
dpF 2 20 20 10 5 0.5 2 Ndt
Net force on one leg 1 2 0.2 10 1N4
29. During 1st collision perpendicular component of v, v becomes e times, while 2nd
component IIv remain unchanged and similarly for second collision. The end result is
that both IIv and v becomes e times their initial value and hence v" ev (the (-) sign
indicates the reversal of direction).
30. Consider a parallel axis passing through center of base. The two axes are equidistant
from the C.M.
Narayana IIT Academy 11-11-17_Sr.IIT_IZ_JEE-Adv_(2011_P1)_CTA-4_Key & Sol’s
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31. At maximum extension both should move with equal velocity.
By momentum conservation,
5 3 2 10 5 2 V
V = 5 m/sec
Now, by energy conservation
2 2 2 21 1 1 15 3 2 10 5 2 V kx2 2 2 2
Put V and k
max1x m 25cm4
.
Also first maximum compression occurs at;
3T 3t 24 4 k
3 10 32 sec
4 7 1120 56
.
(where reduced mass, 1 2
1 2
m mm m
).
32. The ball has v ' , component of its velocity perpendicular to the length of rod
immediately after the collision. u is velocity of COM of the rod and is angular
velocity of the rod, just after collision. The ball strikes the rod with speed v cos53 in
perpendicular direction and its component along the length of the rod after the
collision is unchanged.
Using for the point of collision.
Narayana IIT Academy 11-11-17_Sr.IIT_IZ_JEE-Adv_(2011_P1)_CTA-4_Key & Sol’s
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Velocity of separation = Velocity of approach
3v l u v '5 4
(1)
Conserving linear momentum (or rod + particle), in the direction of the rod.
3mv. mu mv '5 (2)
Conserving angular moment about point ‘D’ as shown in the figure
2l ml l0 0 mu u
4 12 3
(3)
By solving 24v 72vu , w55 55 l
Time taken to rotate by angle t
In the same time, distance travelled 2lu .t
3
Using angular impulse-angular momentum equation.
2l 1 ml 72vN.dt. .
4 3 4 55 l 24mvN.dt
55 or using impulse – momentum equation on
Rod 24mvNdt mu55
33. As V v
V 340 1mv 340
first Resonance depth (from upper end)
11R m 25cm
4 4
Narayana IIT Academy 11-11-17_Sr.IIT_IZ_JEE-Adv_(2011_P1)_CTA-4_Key & Sol’s
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34. U 3x 4y
xx
U / xFa 3m m
yy
F U / ya 4
m m
2a 5m / s
Let at time ‘t’ particle crosses y-axis
Then 216 3 t2
t 2sec
Along y-direction :
21y 4 2 82
particle crosses y-axis at y = -4
At (6, 4) : U = 34 & KE = 0
At (0, -4) : U = -16 KE = 50
or 21 mv 502
v = 10m/s while crossing y-axis
PASSAGE-1_(35, 36, 37)
Upon placing on the step due to symmetry B, D should be overall equally compressed.
Net torque must be zero about any axis, including diagonals, so changes in tension at
the opposite ends of a diagonal must be equal. This is why springs A, C should be
equally further compressed by an amount x and B, D should equally lengthen by same
amount x. This equality of changes in length ensures that net force provided by springs
to support weight of the table doesn’t change. Denoting upwards as positive. Vertical
displacement at A, B, C and D will be 8 – x, x, - x and x respectively. Displacement of
midpoint of AC = displacement of midpoint of 82
2 2x x x xBD x cm
.
PASSAGE-2_(38, 39)
cosla sl saS S S 40. 1 2x &x be the displacement from equilibrium position
Narayana IIT Academy 11-11-17_Sr.IIT_IZ_JEE-Adv_(2011_P1)_CTA-4_Key & Sol’s
Sec: Sr.IIT_IZ Page 10
Now for hollow sphere, applying A I
2 11 2
a5k x x r m r3 r
(1)
By angular momentum conservation (about A) of the system, 1 25 7m v r m v r3 5
1 225v 21v (2) 1 225x 21x (3)
Using (1) and (3) we get, 1 1 125 5k x x m a21 3
1 146 ka x35 m
1 46kf2 35m
41. 1v 2g h H / 2 2v 2g h
By continuity equation
1 2dhA a v vdt
dhA a 2g h H / 2 2g hdt
or H/2 t
H 0
A dh dta 2g h h H / 2
2A Ht 2 1
3a g
42. Work done by force F
ˆ ˆ ˆ ˆW F.dx yi xj . dxi dyj
(1)
2 2 2x y a xdx y dy 0
2 2x yydyW y xdy dyx x
a 2 2
2 20
a ady2a y
Narayana IIT Academy 11-11-17_Sr.IIT_IZ_JEE-Adv_(2011_P1)_CTA-4_Key & Sol’s
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It can be observed that the force is tangential to the curve at each point and the
magnitude is constant. The direction of force is opposite to the direction of motion of
the particle.
work done = (force) (distance) 2
2 2 a a ax y a J2 2 2
43. Maximum kinetic energy is gained when the semicircular disc has minimum potential
energy. By energy conservation,
20
4 r 1mg I3 2
; here 0I M.I about point of contact
2 2 22 2 2
20 cm cm 0
4r mr 4r mr 4r 4r 8rI I m r I m ; I m mr m m3 2 3 2 3 3 3
20
3 8I mr2 3
g4
9 16 r
rad/sec.
44. Due to error in max
l 100l lll 100 l l 100 l
When l 100 l is maximum then max
will be minimum, that means l 50 cm
45. 2
mg xYd / 4 l
2
mglY/ 4 d x
(1)
max
dY m l d x2Y m l d x
m 20.0 kg m 0.1kg l = 125m l 1 cm
d = 0.050 cm d 0.001cm x = 0.100 cm x 0.001cm
max
dY 0.1kg 1cm 0.001cm 0.001cm 100% 4.3%Y 20.0kg 125cm 0.05cm 0.100cm
46. Effective area should be taken.
Narayana IIT Academy 11-11-17_Sr.IIT_IZ_JEE-Adv_(2011_P1)_CTA-4_Key & Sol’s
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MATHS
47.
2
2 30
1 .cos1 cos
I d by partsa
22 2
3 400
1 3 sin. sin
1 cos 1 cosa d
a a
2 1 20 I I I
48. Let
24
22 4 2
2 11 cos sin1 1 1
x xx dx dx x x
But 22
2sin 1 cos1
xx
/4
0 2 4 2dGI
49.
50. 2 2 2 0 / 2y z z y
3 0 01 2 1
x y z
1 2 3 0 3x z z z x
3,0,0 1, 2,1 9 36 45 15.6 21, 2,1 1 4 1
R D
51. P.I =(4,3,5) equation of plane 4x+3y+5z=50
50 50 50, , . 12 16 10 384 3 5
G E
52. (1,1,1) 0
21 1 3 1:
1 2 2x yL
53.
1 1 2 1 1 1 2 1 1 1 2 12 3 2 0 1 4 0 0 1 4 03 1 3 0 3 7 0 0 0 7 3 4 0
Infinite solutions 4 3 7 rectangular hyper 2xy e
Rad= 2 2 7 14e area =14
54. Split areas into trapeziums approx value of1
klnx dx
Narayana IIT Academy 11-11-17_Sr.IIT_IZ_JEE-Adv_(2011_P1)_CTA-4_Key & Sol’s
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Is 1 2 2 3 1...........
3 2 2f f f f f x f x
f
1 1 !2
lnk ln k
1
1ln ln ln 1 !2
kx dx k k
1ln 1 ln ln 1 !2
k k k k k
re… we get !kkk e k
e
d) / / /2 2 2
11
k n k n k n
proceedn nn
k
55. 1 1sin .
1j x
jeS xe dx G E
e
56. 0,0,0 , 1,1,1 are vertices third vertex (l,2l,3l)
Area = 56 2l (2,4,6)
5 7 831, , 3
3 3 9G OG GE
57. , , 1 , 1 , 1r x y z z y c x z c y x c
Also x+y+z=2 solving 0
12 / 33
4 / 3
xy cz
2 40, ,
3 3r
58.
59. Solving 41 __ 1y x
Given 41 ___ 2y x
R.A= 40 1
4
1 0
21 15
x dx x dx
60.
61. 2
2 2 , 11 2 1, 1
p x xf x
q x x x
Narayana IIT Academy 11-11-17_Sr.IIT_IZ_JEE-Adv_(2011_P1)_CTA-4_Key & Sol’s
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R.A= 1/8 1
2
1 2
2 2 12 2x xx dx x x dx
y
x
2
1-1
-2
257192
1
3p q
p q
2
1 12 ,, 1 1,2 1
xxf x
xx x
p=2
q=-1
62. 2, 1 3f f x f x or f x
2 11/ 2x
x
No solutions
63. 22 2b b
a aGS k b a k f x dx f x dx k R
You get minimum looking it as quadratic in k 1, 1 0
64. 2 2 33 2
0 01 1 1 2x dx x x dx
2 22 23 30 0
. 2 2 2y d y y y ydy
232 2 2 4 6y y y
65. clearly 4
d and K=10 5GE
66. 2 16
3kf x x R f x fx x
Narayana IIT Academy 11-11-17_Sr.IIT_IZ_JEE-Adv_(2011_P1)_CTA-4_Key & Sol’s
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2
2 2 0 0x
xf x dx c dA f x f x x R
kf xx
67. 2 22 2r xa y a b r x a y a b
2
2 2 1 31 3 6 66
xx y r xa a b
68. Let 2 2, 4P x y x y
2 22 2 21 4 1 25 4PA PB x x y x
2cos3,5
2sinxy
2 23 5 34 7GE sum of digit
69. A= 2tan A tan B tanC (tanA+tanB+tanC)3
1458, 48.630
{49,50,51…}
49K