narayana iit academy file11.11.2017 · narayana iit academy india sec: sr. iit_iz jee-advanced...

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

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


Sec: Sr.IIT_IZ

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


Sec: Sr.IIT_IZ

= – 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.


Sec: Sr.IIT_IZ

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)


Sec: Sr.IIT_IZ

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.


Sec: Sr.IIT_IZ

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.


Sec: Sr.IIT_IZ

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


Sec: Sr.IIT_IZ

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


Sec: Sr.IIT_IZ

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


Sec: Sr.IIT_IZ

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.


Sec: Sr.IIT_IZ

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


Sec: Sr.IIT_IZ

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


Sec: Sr.IIT_IZ

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


Sec: Sr.IIT_IZ

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

narayana iit academy file11.11.2017 · narayana iit academy india sec: sr. iit_iz jee-advanced...

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