xis-ic-iit-spark (18.12.17) narayana - wordpress.com...2017/12/18 · xis-ic-iit-spark (18.12.17)...

XIS-IC-IIT-SPARK (18.12.17)

NARAYANA IIT/NEET ACADEMY (1)

NARAYANA I I T / N E E T A C A D E M Y

C o m m o n Pr a c t i c e T e s t – 1 3 XI-IC SPARK Date: 18.12.17

AANNSSWWEERR

PHYSICS CHEMISTRY MATHEMATICS 1. (C) 2. (C) 3. (A) 4. (B) 5. (C) 6. (A) 7. (B) 8. (A) 9. (B) 10. (B) 11. (D) 12. (D) 13. (A) 14. (C) 15. (D)

16. (A) 17. (A) 18. (D) 19. (A) 20. (D) 21. (B) 22. (A) 23. (D) 24. (C) 25. (C) 26. (B) 27. (D) 28. (D) 29. (A) 30. (D)

31. (C) 32. (B) 33. (A) 34. (A) 35. (C) 36. (B) 37. (C) 38. (B) 39. (B) 40. (B) 41. (C) 42. (B) 43. (C) 44. (A) 45. (A)

46. (A) 47. (B) 48. (D) 49. (B) 50. (A) 51. (B) 52. (C) 53. (B) 54. (B) 55. (B) 56. (A) 57. (C) 58. (B) 59. (A) 60. (C)

61. (C) 62. (A) 63. (D) 64. (D) 65. (A) 66. (D) 67. (D) 68. (B) 69. (A) 70. (D) 71. (B) 72. (A) 73. (C) 74. (A) 75. (C)

76. (A) 77. (C) 78. (B) 79. (A) 80. (A) 81. (A) 82. (A) 83. (C) 84. (A) 85. (C) 86. (A) 87. (B) 88. (A) 89. (A) 90. (A)

(Hint & SolutionA PART A : PHYSICS

1. (C) In this question, we will have to assume that temperature of enclosed air about water is

constant (or pV=constant)

0p p gh

0p A 500 H p A 300

Solving these two equations, we get H=206 mm Level fall=(206–200) mm=6 mm 4. (B) From equation of continuity (Av=constant)

2 28 0.25 2 v4 4

…... (i)

Here, v is the velocity of water with which water comes out of the syringe (Horizontally). Solving Eq. (i), we get

V=4 m/s The path of water after leaving the syringe will be a parabola. Substituting proper values in equation of trajectory.



2

2 2

gxy x tan2u cos

According to question, we have,

2o

2 2 o

10 R1.25 R tan 0

2 4 cos 0

(R=horizontal range) Solving this equation, we get R=2m. 7. (B) Stress = F/A , F =Mg 8. (A) Strain = /l l

9. (B) Net pressure =pressure due to atmosphere

10. (B) FY

lAl

11. (D) 20 0 1

h 1P g P v2 2

20 0 2

h 1P gh 2 g P 2 v2 2

1

2

ghv 1v 22gh

.

12. (D) Applying Bernoulli's equation at C and D, we have

20 0

1P 0 g 3.6 P v 02

v 6 2m / s

Volume blown per unit time 2av r v 96 2 Similarly, at A and C,

2A

1P v g 3.6 1.82

20

1P v 02

5 2AP 0.46 10 N / m

13. (A) The meniscus between two plates is cylinidrical in shape. Pressure at A(the lowest point of the meniscus )

A 0p p T / r Pressure at B=Pressure at C=p0=Pressure at A gh

B 0TP p gh,r

T 2Thgr gd

Alternative method : Force upward 2 T cos 2 T o0

Gravitational pull = (Volume×Density) g=lhdg

2lT=lhdg 2Thd g

.

14. (C) Let P0 be atmospheric pressure and P1 the pressure of air within the sealed tube.



Therefore equilibrium pressure just below the meniscus should be equal to atmospheric pressure because levels of water inside and outside the tube is same.

i.e., 1 02TP pr

or 1 02TP pr

If L=0.11m is the length of tube and x the

Length of immersed part, then from Boyle's law

1 1 2 2p V p V ; 0 02TP La Pr

L x a

Where a is the cross-sectional area of tube,

i.e., 0 02TP L P L xr

0 02TP L P L x L xr

o

4TLxP d 4T

15. (D) Energy released 2 2n 4 a 4 b

Now, 3 34 4n ra b3 3

or 3

3

bna

, Therefore, energy released is

3

2 23

b 4 a 4 ba

2 b4 b 1a

Now, 21 4 b2 3

2 2v 4 b b 1a

or1/2

6 1 1va b

.

16. (A) Let the mass of the needle be m. As the liquid surface is distorted, the surface tension forces acing on both sides of the needle make an angle , say, with vertical. Since the forces acting on the needle are F, F and mg, resolving the forces vertically for its equilibrium, we have



yF Fcos Fcos mg 0

This gives 2Fcosmg

where F=Tl

Then 2Tlcosmg

For m to be maximum, cos 1

Hence, max2Tlmg

.

17. (A) 1t H/2 1/2

0 H

Adt y dya 2g

or H

1 H/2

2At ya 2g

or 12A Ht H

2a 2g

or 1A Ht 2 1a g

…. (ii)

Similarly, 2t 0 1/2

0 H/ 2

Adt y dya 2g

or 2A Hta g

….. (iii)

From Eqs. (ii) and (iii), we get 1

2

t 2 1t

or 1

2

t 0.414t .

18. (D) (a) Applying Bernoulli's equation between points (1) and (2) 2 2

1 1 1 2 2 21 1P v gh P v gh2 2

Since, area of reservoir >> area of pipe 1v 0,

also 1 2P P atmospheric pressure

So, 2 1 2v 2g h h 2 9.8 7 =11.7 m/s

(b) The minimum, pressure in the bend will be at A. Therefore, applying Bernoulli's equation between (1) and (A)

2 21 1 1 A A A

1 1P v gh P v gh2 2

Again, 1v 0 and from conservation of mass A 2v v

or 2A 1 1 A 2

1P P g h h v2

Therefore, substituting the values, we have



25A

1P 1.01 10 1000 9.8 1 1000 11.72

4 22.27 10 N / m 19. (A) As mass of the air is conserved, 1 2n n n (as PV=nRT)

1 1 2 2

1 2

P V P V PVRT RT RT

As temperature is constant,

1 2T T T

1 1 2 2P V P V PV

3 30 1 0 2

1 2

4S 4 4S 4P r P rr 3 r 3

30

4S 4P rr 3

Solving, this we get

3 3 30 1 2

2 2 21 2

P r r rS

4 r r r

.

20. (D) v 2g 10 h ….. (i)

Component of its velocity parallel to the plane is v cos 30°. Let the stream strikes the plane after time t. Then

o o0 vcos 30 g sin 30 T

ovcot 30t

g

Further 2 ov cot 30x vt

g 3y

or 2 o

2v cot 30 13 h gtg 2

2 2 2 o

2

3v g v cot 303 hg 2 g

or 2 2v 3 vh

g 2 g

25 v h

2 g

or 5 10 h h

h = 8.33 m.

21. (B) 2F 1 vA 2



or 21F Av2

…. (i)

Here, v is the velocity of liquid, with which it comes out of the hole.

Further V= Ax ….. (ii)

Vtsv

….. (iii)

and W = F.x ….(iv) From the above four equations,

21 VW A v2 A

2 3

2 2 2 2

1 V 1 V.V2 s t 2 s t

.

22. (A) From continuity equation, A B 0v v v

A Bp gH p 0

B Ap p gH …(i) Now let us make pressure equation from manometer.

gA Hp g h H Bgh p

Putting B Ap p gh we get h=0.

23. (D) Radius of meniscus R rcos

P due to spherical surface 2r

2 cos

R

B atmP p

24. (C) Viscous force F area

So let F kA 0 1 2F k A A ……(i)

and 1T k A ……(ii) Dividing Eq.(ii) by Eq. (i) we get,

1

0 1 2

ATF A A

0 1

1 2

F ATA A

25. (C) Top2H Hh 1.5H

2

Since this point lies in the tank. So hole should be made at this point. 26. (B) Force from right hand side liquid on left hand side liquid. (i) Due to surface tension force = 2RT(towards right) (ii) Due to liquid pressure force



x h

0x 0

p gh 2R.x dx

202p Rh R gh (towards left)

Net force is 202p Rh R gh 2RT .

27. (D) The bubble will detach if, Buoyant force Surface tension force

sin T dl T 2 r sin

3w

4 R g T dlsin3

3w

4 R g T 2 r sin3

rsinR

Solving, 4

2w w2 R g 2 gr R3T 3T

28. (D) Using geometry

b bcos RR 2 cos

2

Using pressure equation along the path

0 02Sp h g pR

Substituting the value of R, we get

2S 2Sh cosR g b g 2

.

29. (A) Terminal velocity 2

T s L2r gv9

and viscous force F= T6 rv Rate of production of heat (power): as viscous force is the only dissipative force. Hence.

T TdQ Fv 6 rvdt

2T Tv 6 rv

22 2

2 5s L s L

2 r g 8 g6 r r9 27

or 5dQ rdt

.

30. (D) Decrease in surface energy=heat required in vaporization.



T dS L dm

2T 2 4 r dr L 4 r dr 2TrL

.

PART B : CHEMISTRY 33. (A) (i) OH show +R effect (ii) OH show –i effect (iii) NO2 show –R effect (iv) NO2 show –i effect 37. (C) It has more no. of bonds than other. 39. (B) All are acceptor –i power - 2NO CN F 40. (B) in (ii) second is aromatic while first is antiaromatic 41. (C) electron participated in conjugation (resonance) are counted. 42. (B) Nitrogen with double bond is SP2 hybridized. 43. (C) SP3 nitrogen is more basic than SP2 nitrogen. 50. (A)

12

3

4

56 7

8

60. (C)

12

34

56

CH3C H2 5

PART C : MATHEMATICS

61. (C) Let P x, y be the point of contact dy dy2y 4a and 2x 4adx dx

. For the tangency of curves, 24a 2x xy 4a ,2y 4a

which is the

required locus. 62. (A) Locus of (h, k) is 22y a b 2x a b Which is equation of parabola.

63. (D) Solve 22 2yx , x y 1 1a

We get 1y 2a

Put in 2y ax

22

2a 1 1x 0 aa 2

64. (D) 1 2 1 2m m 1 t t 4 Equation of PQ is 1 2 1 2y t t 2x 2at t 0 y = 0 x= 4a

65. (A)

2 11 1

1 2

1 2

1 1t tt ttan 1 t t 11

t t

, 2 221 2 1 2 1 2tan 1 t t t t 0 4t t or

2 2

22

h k 4htan 1 ora a a

, 22 2tan h a k 4ah

66. (D)



67. (D)

68. (B)

69. (A) Circumcircle of ABC passes through focus 1 ,04k 4

of the given parabola.

Putting y = 0, in equation of circumcircle, we get either 14k 4

= 2 or 1 34k 4



70. (D) Slope of AB × slope of BC = -1 1 12 2 2

1 1

k t t 2 1k 4 t 4 t 4

71. (B) Slopes 2 1m,3m 4m ;3m

, Eliminates ‘m’ we have 23

16

72. (A) All the circles described on focal chords of a parabola contains same constant term hence the common chords will pass through origin.

73. (C) Family of lines passes thorough focus hence latus rectum will makes shortest intercept.

74. (A) Solving the equation 2

2 2 2px y p and y 2px2

we get

2

2 3p p 3px px 0 x or x4 2 2

But 3px2

is not acceptable hence the required points are p p, p or , p2 2

75. (C)

76. (A) r 2 or r 3 and r 1 or r 5 r 2 or r 5

Also 2 2 11r r 6 r 6r 5 r5

77. (C) Given ellipse is 2 2x y 1 ... i

16 4

Equation of a circle centered at (1,0) can be written as 2 2 2x 1 y r ... ii The abscissa of the intersection points of the circle and the ellipse is given by the equation

2

2 216 xx 1 r4

i.e. 2 2 24 x 2x 1 16 x 4r , i.e. 2 23x 3y 6x 8 0

78. (B)

2x 2y 3 2x y 1

5 5 14 1/ 4

PA PB 2a 2 2 4



79. (A)

80. (A)



81. (A)

82. (A)

83. (C)



84. (A)

85. (C)

86. (A) 2 2

5 61 15ab a a 5 3,e , a 10, b2 2



87. (B)

88. (A)



89. (A)

90. (A) The given ratio is b 1

a 2

Now 2

22

b 1 3 3e 1 1 ea 4 4 2

xis-ic-iit-spark (18.12.17) narayana - wordpress.com...2017/12/18 · xis-ic-iit-spark (18.12.17)...

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