Model Question Paper (P+C+M)

Subject Topic Test # Date

P + C + M Term Examination IIT – XI - TE

JUL 2013

Model Question Paper (P+C+M)

1. Calculate the mass of a thin disc of radius and of uniform surface density 0 .

(a) 2

0

2

R (b)

20

3

R (c) 2

0 R (d) 2

0

4

R

2. Calculate the potential energy of a thin rod of length L with constant linear mass density

held vertically on ground

(a) 2

2

gL (b)

2

4

gL (c) 2gL (d)

2

3

gL

3. A projectile is moving at 120ms at its highest point, where it breaks into two equal parts

due to an internal explosion. One part moves vertically up at 130ms with respect to the

ground. Then the other part will move at

(a) 110 13ms (b) 120ms (c) 130ms (d) 150ms

4. A body is projected vertically upwards. The times corresponding to height h while

ascending and while descending are 1t and 2t respectively. Then the velocity of

projection is (g is acceleration due to gravity)

(a) 1 2

2

g t t (b) 1 2

2

g t t (c) 1 2

1 2

g t t

t t (d) 1 2g t t

5. The angle between two vectors of magnitudes 12N and 3N on a body is 60 the resultant

w.r.t. 12 N force is about

(a) 30 (b) 60 (c) 10 (d) 75

6. A cube has a side of length 21.2 10 m . Calculate its volume.

(a) 6 31.7 10 m (b) 6 31.73 10 m (c) 6 31.70 10 m (d) 6 31.732 10 m

7. A wire has a mass 0.3 0.003 ,g radius 0.5 0.005 mm and length 6 0.06 cm . The

maximum percentage error in the measurement of its density is

(a) 1 (b) 2 (c) 3 (d) 4



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8. The dimensions of a

b in the equation

2a tP

bx

where P is pressure, x is distance and t is

time are

(a) 2 3M LT

(b) 2MT

(c) 3 1ML T

(d) 3LT

9. In the relation; aZ

kp e

p is pressure, Z is distance, k is Boltzmann co9nstant and is

the temperature. The dimensional formula of will be

(a) 0 2 0M L T

(b) 1 2 1M L T

(c) 1 0 1M L T

(d) 0 2 1M L T

10. Square of the resultant of two equal vectors (forces) is equal to three times of its product.

The angle between them is

(a) 0 (b) 45 (c) 60 (d) 90

11. Given that P Q R and that R is perpendicular to P . If P R , then the angle between

P and Q is

(a) 4

rad

(b) 2

rad

(c) 34

rad

(d) rad

12. The horizontal range and maximum height attained by a projectile are R and H

respectively. If a constant horizontal acceleration 4

ga is imparted to the projectile due

to wind, then its horizontal range and maximum height will be

(a) ,2

HR H (b) ,2

2

HR H

(c) 2 ,R H H (d) ,R H H

13. The speed of a projectile when it is at its greatest height is 2

5times its speed at half the

maximum height. The angle of projection is;

(a) 30 (b) 60 (c) 45 (d) 1tan 3 /4

14. A projectile is thrown with a velocity of 110 2ms at an angle of 045 with horizontal. The

time interval between the moments when its speed is 1125ms is: 2(g 10m / s )

(a) 1.0 s (b) 1.5 s (c) 2.0 s (d) 0.5 s



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15. A projectile is fired at an angle of 030 to the horizontal such that the vertical component

of its initial velocity is 180 ms . Its time of flight is T . Its velocity at t T / 4 has a magnitude

nearly equal to:

(a) 200 m / s (b) 300 m / s (c) 140 m / s (d) 100 m / s

16. After one second the velocity of a projectile makes an angle of 045 with the horizontal.

After another one second it is travelling horizontally. The magnitude of its initial

velocity and the angle of projection are:

2(g 10m / s )

(a) 014.62m / s,60 (b) 114.62m / s, tan (2) (c) 122.36m / s, tan (2) (d) 022.36m / s,60

17. A relief aeroplane is flying at a constant height of 1960 m with speed 600 km/hr above

the ground towards a point directly over a person struggling in flood water. At what

angle of sight i.e. angle made by the displacement vector of the kit with the vertical,

should the pilot release a survival kit if it is to reach the person in water? 2(g 9.8m / s ) .

(a) 030 (b) 045 (c) 060 (d) not possible

18. There are two values of time for which a projectile is at the same height. The sum of

these two times is equal to:

(a) 3T / 2 (b) 4T / 3 (c) 3T / 4 (d) T

19. A ball is projected upwards from the foot of a tower. The ball crosses the top of the tower

twice after an interval of 6s and the ball reaches the ground after 12s. The height of the

tower is 2(g 10m / s ) :

(a) 120 m (b) 135 m (c) 175 m (d) 80 m

20. A v t graph is shown in the figure. We can draw the

following conclusions:

1. Between t 1 s to t 2 s speed of particle is

decreasing

2. Between t 2 s to t 3s speed of particle is increasing

3. Between t 5 s to t 6s acceleration of particle is negative

Between t 0 s tot 4s particle changes, its direction of motion twice

(a) only 1 and 2 are true (b) only 3 and 4 are true

O 1 2

3 4 5 6 t

v



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(c) only 1 and 4 are true (d) only 2 and 3 are true

21. Trajectories of two projectiles are shown in figure. Let 1T and 2T be the time periods and

1u and 2u their speeds of projection. Then :

(a) 2 1T T (b) 1 2T T

(c) 1 2u u (d) 1 2u u

22. From an elevated point a body is dropped and another body is thrown horizontally at

the same instant. Choose the incorrect option:

(a) Time of descent of projected body is greater than the time of descent dropped

body

(b) Time of descent of projected body is equal to that of the time of descent dropped

body

(c) The ratio of vertical displacement in successive equal intervals of time is 1:3:5 for

both the bodies

(d) The velocity of striking of projected body is more than the dropped body

23. A particle is moving along a straight line with uniform acceleration and has velocities

7 /m s at P and 17 m/s at Q. R is the mid-point of PQ . Then

(a) the average velocity between R and Q is 12 /m s

(b) the ratio of time to go from P to R and that from R to Q is 3 : 5

(c) the velocity at R is 10 /m s

(d) the average velocity between P and R is 10m/s

24. A projectile projected with the same speed has the same range R for two angles of

projections. If 1T and 2T be the times fight in the two cases, then

(a) 1 2 T T R (b) 21 2 T T R (c) 1

2

sinT

T (d) 1

2

cotT

T

25. From the given system of equations, solve for T in terms of 1 2 3m ,m ,m andg.

1 1 1

2 2 2

3 3 3

1 2 3

T m g m ( a )......(i)

T m g m ( a )......(ii)

2T m g m ( a )......(iii)

a a 2a 0........(iv)

1

2

x



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(a) 1 2 3

1 2 2 3 1 3

4m m mT g

4m m m m m m

(b) 1 2 3

1 2 2 3 1 3

m m mT g

4m m m m m m

(c) 1 2 3

1 2 2 3 1 3

4m m mT g

m m m m m m

(d) 1 2 3

1 2 2 3 1 3

m m mT g

m m m m m m

26. Choose the incorrect option in the following. The vectors A and B are drawn from a

common point and C A B , then angle between A and B is

(a) 90 if 2 2 2C A B (b) greater than 90 if 2 2 2C A B

(c) greater than 90 if 2 2 2C A B (d) less than 90 if 2 2 2C A B

27. If 2 2

3 2%, 4 0.12 and 5 4%, then the quanitity is given by,2

x yx y z

z

(a) 5 6.64% (b) 0.5 6.64% (c) 0.5 13.28% (d) 5 3.32%

28. Which of the following sets have different dimensions?

(a) Pressure, Young’s modulus, Stress

(b) Emf, Potential difference, Electric potential

(c) Heat, Work done, Energy

(d) Dipole moment, Electric flux, Electric field

29. Figure shows ABCDEF as a regular hexagon. The value of AB AC AD AE AF will be

(a) AO (b) 2AO (c) 4AO (d) 6AO

30. Many bullets with same speed u cover some area on ground.

The maximum area covered on ground is

(a) (b) 2 4

2

u

g

(c)

2u

g

(d)

4

2

u

g

Chemistry:

31. 41 gCH and 4g of compound X have equal number of moles. Thus, molar mass of X is:

(a) 116gmol (b) 132gmol (c) 14gmol (d) 164gmol

A B

C

DE

FO



JUL 2013

32. In the following reaction, 3 2 3 4 3 4 2 3Ba(NO ) Na PO Ba (PO ) NaNO The number of moles of

3 4 2Ba (PO ) formed from 2 moles each of 3 2Ba(NO ) and 3 4Na PO is

(a) 1 mol (b) 1/3 mol (c) 2/3 mol (d) 4 mol

33. Oxidation numbers of Cl atoms in 2CaOCl (bleaching powder) are

Ca

Cl

ClO

*

* *

(a) Zero on each (b) 1 on *Cl and +1 on **Cl

(c) *1 onCl and 1 on **Cl (d) 1 on each

34. Which of the following changes requires a reducing agent?

(a) 2 24 2 7CrO Cr O (b) 3BrO BrO

(c) 23 3 4H AsO HAsO (d) 3 4Al(OH) Al(OH)

35. For the reaction between 22 2 4MnO (s) C O in basic solution, the unbalanced equation is:

2 24 2 4 2 3MnO C O MnO (s) CO in a balanced equation, the number of OH ions is:

(a) 0 (b) 4 on the right (c) 4 on the left (d) 2 on

the left

36. Which has maximum number of milliequivalents?

(a) 2 4100mLof 0.01NH SO (b) 100mLof 0.01NHCl

(c) 3 4100mLof 0.01N H PO (d) Equal

37. Sucrose solution is 1 molal. Mole fraction of sucrose in the aqueous solution is:

(a) 0.018 (b) 0.015 (c) 0.036 (d) 0.009



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38. 1.06 g 2 3Na CO is dissolved in 100 mL solution. 10 mL of this solution can be neutralized

by:

(a) 10 mL of 0.1 N HCl (b) 10 mL of 0.1 M 3 4H PO

(c) 20 mL of 2 40.1MH SO (d) 20 mL of 0.1MHCl

39. 5.3 g of 2 3M CO is dissolved in 150 mL of 1 HCl. Unused acid required 100 mL of 0.5

N NaOH . Hence, equivalent weight of M is:

(a) 23 (b) 12 (c) 24 (d) 13

40. If the radius of first Bohr orbit is x , then de-Broglie wavelength of electron in 3rd orbit is

nearly:

(a) 2 x (b) 6 x (c) 9x (d) 3

x

41. 3 moles of mixture of CO and 2CO requires 80 g NaOH to convert all 2CO into 2 3Na CO .

How many additional grams of NaOH will be required for complete conversion into

2 3Na CO if CO is completely oxidized into 2CO

(a) 100 g (b) 120 g (c) 90 g (d) 160 g

42. If ,1 2 3

E E and E represent respectively the energies of an electron, an alpha particle and a

proton each having same de-broglie wavelength, then

(a) 1 3 2

E E E (b) 2 3 1

E E E (c) 1 2 3

E E E (d)1 2 3

E E E

43. A 3p orbital has:

(a) Two non-spherical nodes

(b) two spherical nodes

(c) One spherical and one non-spherical node

(d) one spherical and two non-spherical nodes



JUL 2013

44. The radii of two of the first four Bohr’s orbits of the hydrogen atom are in the ratio 1:4.

The energy difference between them may be:

(a) Either 12.09eVor3.4eV (b) either 2.55eVor10.2eV

(c) either 13.6eV or3.4eV (d) either 3.4eV or0.85eV

45. An ion aMn has the magnetic moment equal to 4.9 BM the value of a is:

(a) 3 (b) 4 (c) 2 (d) 5

46. For which of the following, the radius will be same as for hydrogen atom having n 1?

(a) He ,n 2 (b) 2Li ,n 2 (c) 3Be ,n 2 (d) 2Li ,n 3

47. The circumference of thn orbit in H-atom can be expressed in terms of de Broglie

wavelength as:

(a) (0.529)n (b) n (c) 13.6 (d) n

48. The wave number of first line of Balmer series of hydrogen is 115200cm . The wave

number of first line of Balmer series of 2Li ion is:

(a) 115200cm (b) 160800cm (c) 176000cm (d) 1136800cm

49. An electron of H-atom from an excited state returns to ground state in one or more steps

shows emission of ten lines. How many of these belongs to visible spectrum?

(a) 6 (b) 5 (c) 4 (d) 3

50. The orbital represented by 4,2,0 is:

(a) 2z4d (b) x4P (c) z4P (d) xy4d

51. Which is incorrect about 34.25g NH ?

(a) it contains 0.25 mol of 3NH

(b) it contains 0.75 mol of H atoms

(c) it contains total of 1.0 mol of N and H atoms

(d) it contains 236 10 molecules of 3NH



JUL 2013

52. Mole of fraction of ethanol 2 5(C H OH) in ethanol-water system is 0.25 . Thus, it has:

(a) 30% ethanol by weight of solution (b) 75% water by weight of

solution

(c) 46% ethanol by weight of solution (d) 60% water by weight of

solution

53. Which of the following represent redox reactions?

(a) 2 22 7 4 2Cr O 2OH 2CrO H O (b) 2 2

4 2 7 22CrO 2H Cr O H O

(c) 24 2 22MnO 3Mn 4OH 5MnO 2H O (d) 2 2

3 3 4Cu 4NH [Cu(NH ) ]

54. In the following cases, equivalent weight of the underlined may be equal to molecular

weight:

(a) 4 2 2 2 4 22CuSO 4KI Cu I 2K SO I (b) 2 2 3 2 4 62Na S O I 2NaI Na S O

(c) 4 2 2 2 3P 3NaOH 3H O 3NaH PO PH (d) 2 4 2 4 2H SO 2NaOH Na SO 2H O

55. If Aufbau rule and Hund’s rule are not considered, which of the following statements is

incorrect?

(a) 2Fe will have configuration as Ar

(b) 2Cu is colourless ion

(c) Mn has magnetic moment of 3 BM

(d) K belongs to d block



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56. Of the following quantum state designations, which is/are not allowed state for an

electron in an atom:

n l m

(a) 3 2 -2

(b) 3 1 0

(c) 3 0 -1

(d) 3 2 0

57. For the reaction 3 4 2 2 4 2H PO 2Ca(OH) Ca HPO 2H O . Select the incorrect statement.

1mole 1mole

(a) the equivalent weight of 3 4H PO is 98

(b) the resulting solution is neutralized by 1 mole of KOH

(c) 1 mole of 3 4H PO is completely neutralized by 1.5 mole of 2Ca(OH)

(d) None of these

58. The set of quantum numbers not possible for an electron is

(a) 1,1,1, 1 2 (b) 1,0,0, 1 2 (c) 1,0,0, 1 2 (d) 2,0,0, 1 2

59. The number of d-electrons in 2 26Fe Z is not equal to that of the

(a) p electrons in . 10Ne At no

(b) s electrons in . 12Mg At no

(c) d -electrons in . 26Fe At no

(d) p electrons in . . 17Cl At no of Cl

60. Without looking at the periodic table, select the elements of III A group of the periodic

table.

(Atomic numbers are given)

(a) 3, 11, 19, 37 (b) 5, 13, 21, 39 (c) 7, 15, 31, 49 (d) 5,13, 31, 49



JUL 2013

Maths:

61. If 24 36 48log 12 , log 24 , log 36a b c then 1 abc

(a) 2ab (b) 2bc (c) 2ca (d) 0

62. If 2.3 0.23 1000x y then

1 1

x y

(a) 3 (b) 2 (c) 1

3 (d)

1

2

63. If coslog sin 0ecx x then

(a) 0x (b) 0x (c) 1 1x (d) Not possible

64. If sin sin2 , cos cos2x y then 2 2 2 2 3x y x y

(a) 2x (b) 2y (c) x (d) y

65. If 21 sin sin .....to 4 2 3 thenx x x

(a) 6

(b)

4

(c)

3

(d)

3

or

2

3

66. For 133 ,2 cos2

AA

(a) 1 sin 1 sinA A (c) 1 sin 1 sinA A

(c) 1 sin 1 sinA A (d) 1 sin 1 sinA A

67. If

3

4then

2

12 cot

sin

(a) 1 cot (b) 1 cot (c) 1 cot (d) 1 cot

68. cot16 cot 44 cot 44 cot76 cot76 cot16

(a) 0 (b) 1 (c) 2 (d) 3

69. If 1 tan 2 cot 62

3tan 152 cot 88

k

then the value of k is

(a) 1 (b) -1 (c) 1

2 (d)

1

2



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70. Which of the following is true

(a) If 3 3 3 30 then a b c a b c abc

(b) If 4 4 4 0 then 2a b c a b c ab bc ca

(c) If 0a b then a b ab

(d) None of these

71. In a right angled triangle, if the hypotenuse is 4 times the length of the perpendicular

drawn to it from its opposite vertex, then one of the angles of the triangle is

(a) 30 (b) 45 (c) 60 (d) 75

72. If , ,A B C are angles of a triangle such that tan tan tan 6A B C , tan tan 2A B then which

of the following is true.

(a) tan 1A

(b) tan 2B

(c) tan 3C

(d) None of these

73. If

22 2

3 5log log

4 4 2x x

x

then the value of x can be

(a) 1

32 (b) 1

32

(c) 2

32 (d) 1

42

74. If tan , tan are roots of the equation 2 0 0x px q p then

(a) 2 2sin sin cos cosp q q

(b) tan1

p

q

(c) cos 1 q

(d) sin p

75. If 2log.log5 xa a then x

(a) 125 (b) 2a (c) 25 (d) 5

76. If 222 cba then cc baba )()( log

1

log

1

(a) 2

1 (b) 1 (c) 2 (d) 3

77. If zyx 7553 then zyxy

(a) 2xz (b) xz (c) 3xz (d) 4xz

78. If 3 2 3 0a x b x c then 3323 cxbxa

(a) 0 (b) 3abcx (c) abc (d) 1



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79. If 13

5)sin(,

5

4)cos( and , lies between 0 and

4

then 2tan

(a) 33

56 (b) 56

33 (c) 65

16 (d) 61

60

80. If qec cotcos , then the value of eccos

(a) q

q1

(b) q

q1

(c)

qq

1

2

1 (d) none of these

81. If 41

41

33

x , 41

41

33

y then 2223 yx

(a) 1 (b) 64 (c) 49 (d) 81

82. If 001.0log 1.0x , 81log9y , then yx 2

(a) 223 (b) 23 (c) 12 (d) 22

83. 40tan20tan340tan20tan is equal to

(a) 2

3 (b) 3 (c)

4

3 (d) 1

84. 22

tantantantan1

(a) 22 tantan (b) 22 secsec (c) 22 cottan (d)

22 cossec

85. If is any angle then 24 secsec is equal to

(a) 24 tantan (b) 24 tantan (c) 42 tantan (d) 2tan2

86. The value of 2266 sincos3cossin is

(a) 0 (b) 1 (c) 2 (d) 3

87.

ac

ac

cacb

bc

cbba

ba

ab

x

x

x

x

x

x

22

22

22

(a) 32

1

ax

(b) 1 (c) x (d) none of these

88.

3

1

3

432

23

31525

15

9863

(a) 228 (b) 325 (c) 229 (d) 320



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89. The moon’s distance from the earth is 0000,35 km and its distance subtends an angle of 31

’ at the eye of the observer. The diameter of the moon is:

(a) 27

113157 km (b)

27

11315 km (c)

27

113050 km (d) none of these

90.

3

10

5log

1

1.0log

15 7

(a) 0 (b) 1 (c) 2 (d) 3



JUL 2013

Answers:

Physics:

1. c 2. a 3. a 4. a 5. c 6. a 7. d 8. b 9. a 10. c

11. c 12. d 13. b 14. a 15. c 16. c 17. c 18. d 19. b 20. b

21. d 22. a 23. d 24. a 25. a 26. c 27. b 28. d 29. d 30. c

Chemistry:

31. d 32. c 33. b 34. b 35. c 36. d 37. a 38. d 39. a 40. b

41. d 42. a 43. c 44. b 45. a 46. c 47. d 48. d 49. d 50. a

51. d 52. c 53. c 54. b 55. b 56. c 57. a 58. a 59. d 60. d

Maths:

61. B 62. c 63. d 64. b 65. d 66. b 67. b 68. d 69. b 70. b

71. d 72. c 73. b 74. a 75. c 76. c 77. a 78. b 79. a 80. c

81. b 82. c 83. b 84. b 85. b 86. b 87. a 88. a 89. a 90. c

Solutions: (P+C+M)

Physics:

1. 02dm rdr

Integrate to get M

20 M R

2. Conceptual

3. Here 2 2 220 30 v i.e., 11300 10 13v ms



JUL 2013

4. Time of ascent = time of descent i.e., 1 2t t

Using, v u at , we get

Velocity of projection 1 10 . .u gt i e u gt

1t can be replaced with 1 21 2

2 2

g t tt tu

5. Using 2

1 2

sintan

cos

F

F F

we get

3sin 60 3 3 / 2 3 3tan

12 3 1/ 2 2712 3cos60

5.19620.19245 10.9

2

6. 3

3 21.2 10V l m

6 31.728 10 m

Length l has two significant figures; the volume V will also have two significant figures.

Therefore, the correct answer is 6 31.7 10V m .

7. Density 2

m

r L

100 2 100m r L

m r L

8. 2a t

Pbx

2Pbx a t

2Pbx a T

2 2

1 4

1 2or

T Tb M T

P x ML T L

2

2

1 4

TaMT

b M T

9. 0 0 0ZM L T

k



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k

Z

Further P

k

P ZP

Dimensions of k are that to energy.

Hence, 2 2

0 2 0

1 2

ML TM L T

LML T

10. Using 2 2 2 2 cosR A B AB we get

2 2 22 cosR A A A A B is given

i.e., 2 22 2 cosR A A (given)

but 23R A

2 22 2 2

2

3 2 13 2 2 cos cos . . 60

22

A AA A A or i e

A

11. Here P Q R

and Q R P

Since R P , therefore 2 2 2Q R P

Also, R P , therefore 2 22Q P or 2Q P

Now, 2 2 22 cosP Q PQ R i.e., 2 2 2 22 2 2 cosP P P P

Which gives, 1

cos2

i.e., 3

4

12. 22

, and2

y yx

u uT H R u T

g g Where T time of flight of the projectile and xu and yu are

respectively the horizontal and vertical components o initial velocity.

When a horizontal acceleration is also given to the projectile yu and hence T and H will remain

unchanged while the range will become

21

2xR u T aT

2

2

41

2 4

yugR R H

g

13. Let be the angle of projection and u its initial speed. The maximum height will be:

2 2 2 2sin sin

2 2

u uH gH

y



JUL 2013

Now, cosHv u

or 2 2 2cosHv u …(1)

2 2 2/2 2

2H

Hv u g u gH

2 2

2 sin

2

uu

… (2)

Now it is given that

or 2 2/2 /2

2 2or

5 5H H H Hv v v v

Substituting the values from Equations (1) and (2), we get

2 2

2 2 22 sincos

5 2

uu u

or 2

2 sin5cos 2 1

2

or 2 2 2 35 1 sin 2 sin or sin

4

or 3

sin or 602

14. 0 1xu ucos45 10ms

0 1yu u sin45 10ms

2 2 2y xV v v

125= 2 1y yv 100 v 5ms

y yv u gt

5 10 10t

1

2

1t sec

t 1S2

t 3 / 2sec



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15. 0xx

y

cot30 3 u 80 3m / su

u

x2uT T 16s

g

At x

Tt 4s,v 80 3m / s

4

yv 80 10 4 40m / s 2 2v (80 3) (40) 140m / s

16. Total time of flight is T 4s and if u its initial speed and the angle of projection. Then

2usinT 4

g

Or usin 2g

After 1 s velocity vector makes an angle of 045 with horizontal i.e,

Or ucos (usin ) (gt)

Or ucos 2g g(t 1)

Or ucos g

Squaring and adding Eqs. (1) and (2), we get

2 2 2 2 2u 5g 5(10) m / s u 22.36m / s

Dividing eq. (1) by (2), we get, tan 2 or 1tan (2)

17. As the plane is flying at a speed of 600 (5 /18), i.e.,

(500 / 3)m / s horizontally (at a height of 1960 m above

the ground), the taken by the kit to reach the ground

2h 2 1960t 20s

g 9.8

And in this time the kit will move horizontal by

x ut (500 / 3) 20 (10,000 / 3)m

So the angle of sight x 10000 10

tan 1.7 3h 3 1960 5.88

1 0tan ( 3) 60

x

h

A

P



JUL 2013

C

B

u A

h

18.

OA OB BC OBt t t t T

19. BCt 6 / 2 3s

AC a AC

ut 12 / 2 6s t t

g

160u ms

6 3 3AB AC BCt t t S

Further 2AB AB

1h ut gt

2

21(60)(3) (10)(3)

2 135m

20. Acceleration is slope of v t graph is negative between 5 to 6t t . Between t 0 to t 4s particle

changes its direction of motion at t 1s and at t 3s

21. Maximum height and time of flight depends on the vertical component of initial velocity

1 2 y1 y2H H u u

Hence, 1 2T T

Range 2u sin 2 2(usin )(ucos )

Rg g

x y2u u

g

2 1 xz x1R R u u or 2 1u u

22. (b) for dropped body d

2Ht ;

g for projected body d

2Ht

g they are suffering same vertical

displacements

(c) Ratio of displacements 1:3:5 from nS u a(n 1/ 2)

(d) 1V 2gh dropped body 22V u 2gh for projected body

So (a) is incorrect.



JUL 2013

23. Let the distance between P and Q is 2s .

Then 2 2 2 2v u a s

or 2 2

17 7 4as

2 24 240 /as m s

or 2 22 120 /as m s

velocity of particle at R is

2,2 2 2 7 120 169v u as

, 13 /v m s

Now, let 1 2andPR RQt t t t

Then ,1v u at

or 113 7 at

or 1 6 /at m s …(1)

Similarly, ,2v v at

or 217 13 at

or 2 4 /at m s …(2)

From equations (1) and (2) 1

2

6 3

4 2

t

t

Average velocity between andR Q is

, 22 2

,2

2 2

1122

RQ

v t ats

v v att t

1

13 4 15 /2

m s

Similarly, average velocity between p and R is

P R Q

s s

7 /u m s

,v 17 /v m s



JUL 2013

21 1

11 1

1122

PR

ut ats

v u att t

1

7 6 10 /2

m s

24. We know that 2 2sin 2 2 sin cosu u

Rg g

…(1)

Also 90

R R

Now, 12 sinu

Tg

and

2

2 sin 90 2 cosu uT

g g

2 2

1 2 2

4 sin cos 2 sin cos 2 2u u RT T

g g gg

or 1 2T T R

25. Substituting 1 2a ,a and 3a from eqn. (i), (ii) and (iii) respectively in eqn. (iv) we have

1 2 3

T T 2Tg g 2 g 0

m m m

This gives 1 2 3

1 2 2 3 1 3

4m m mT g

4m m m m m m

26. From the addition of two vectors, we know that 2 2 2 2 cosC A B AB

From this expression it is clear that

(a) 2 2 2 when 90C A B

(b) 2 2 2 when 90C A B

(d) and 2 2 2 when 90C A B ; Hence C is incorrect

27. 2 2 9 4% 16 6% 9 0.36 16 0.96 25 1.32 25 5.28% 5 2.64% x y

2 10 4% z . Therefore, 2 2

5 2.64%0.5 6.64%

2 10 4%

x y

z Answer is b.

28. Dipole Moment = (Charge) × (Distance)

Electric Flux = (Electric Field) × (Area)

29. It is equal to 5 4 3 2AB BC CD DE EF . But AB DE and BC EF

So 3 3 6AB BC CD AD AO .

30. Area

22 4

2 u uR

g g



JUL 2013

Chemistry:

31. Equating No. of moles 1 4

x 6416 x

32. 3 2Ba(NO ) is limiting reagent 3moles 3 2Ba(NO ) 1mole 3 4 2Ba (PO )

2 12moles ___________?

3

33.

34. That which undergoes reduction, requires reducing agent for that process

3BrO BrO under goes reduction

35. 7 3 4 4

2 24 4 2(s) 3MnO C O MnO C O

2 2

4 34 2 2 24OH 2MnO 3C O 2MnO 6CO 2H O

36. meq N V(ml)

37. Sucrose 21molin1000gH O

moles of 1000

water18

moles of sucrose =1

total moles 1000

118

mole fraction of sucrose 1 18

0.0181000 1018

118

38. 100 ml soln has-1.06 gm

10ml_______?0.106gm

2 3meq of of Na CO meq of acid

Ca

+2 Cl-1

O-Cl-2 +1



JUL 2013

0.106

1000 2 meq53

acid (NVml)

39. 2 3 2 2M CO 2HCl 2MCl H O CO

M is a monovalent cation.

Thus, equivalent weight of 2 3

molar massM CO M 30

2

Let unused HCl be VmL

V mL of 1N HCl 100mL of 0.5 N NaOH

V 50mL

1 N HCl used 150 50 100mL 0.1 equiv.

Equiv. of 2 3

5.3M CO 0.1

M 30

M=23

40. n

nhmvr

2

20n a nh

mv :Z 2

0a x Bohr’s radius of H-atom n 3,z 1

9x 3hmv

1 2

hmv

6 x

h6 x

mv

41. 2 2 3 2CO 2NaOH Na CO H O

1mole 2 40gm The mixture has 2moles of co So it requires 4 moles of

NaOH

42. As mass decreases, kinetic energy increases to have same wave length.

43. Radial node n 1 angular node .

44. The two orbits are either I and II or II and IV

n2

n1

r 4

r 1 and 2

nr n



JUL 2013

Thus, 2 1

13.6E E 13.6 10.2ev

4

and 4 2

13.6 13.6E E 2.55eV

16 4

45. Magnetic moment n(n 2) : where n is no. of unpaired electron n(n 2) or n 4

Thus, electronic configuration of aMn having 4 unpaired electron is

3 2 2 6 2 6 425Mn : ls ,2s ,2p ,3s 3p 3d

46. 3 212

r Hr Be 2

4

2 3 n2 1 n

r H( r H r H 2 andr Be )

n

47. h

mu and

nhmur

2

2 r n i.e., circumference of an orbit is integer multiple of

48. H 2H 12 2

1 1 1R

n n

2H 2

12 2Li

1 1 1R 9

n n

2

H

Li9

49. 10 lines means electron must be in 5th orbit

(5 1) 4 3 2 1 3 lines in balmer senes

50. General representation is n,l,m

51. Moles of 3

wt 4.25NH 0.25

mol.wt 17

moles of H-atoms = 0.25 3 0.75

total moles of atoms= 0.75 0.25 1

No. of molecules 230.25 6.023 10 231.5 10

52. Let 2 5C H OH in the mixture 1w g

and 2H O 2w g

moles of 2 5C H OH ,



JUL 2013

11

22 2

wn

46

wMole of H O,n

18

Total moles 1 21 2

w wn n

46 18

Mole fraction of 12 5

1 2

n 1C H OH 0.25

n n 4

1

2

n 1mol 46g

n 3mol 54g

Total mass of solution =46+54=100g

Thus, 100 g contains 46 g 2 5C H OH and 254gH O Thus, (c) and (d) are correct.

53. There should be increase and decrease in oxidation state

54. Eq . wt=mol .wt

Change in oxidation. State per molecule Change in oxidation state must be one

2 2.5

2 2 3 2 2 4 6Na S O I NaI Na S O

55. Conceptual

56. In ‘C’ m-value cannot be -1

57. 3 4 2 4 2H PO Ca(OH) CaHPO 2H O

1 mole 1 mole

(a) Eq. wt of 3 4

98H PO 49

2

(b) One mole of 2Ca(OH) can neutralize only two gm equivalent of base is verified

(c) No. of gm equivalents are equal.

58. ‘ l ’ value never greater then ‘n'

59. In 2 6Fe d electrons are present

60. Conceptual



JUL 2013

Maths:

61. Ans: (b) 2

48 48 48 481 log 48 log 12 log 24 2log 24 2abc bc

62. Ans: (c) 310

10

32.3 10 , .log 2.3 3

log 2.3

xx x

310

10

30.23 10 , log 0.23 3

log 0.23

yy y

63. Ans: (d) Since cos 1 sin 1ecx x which is impossible.

64. Ans: (b) Use verification by taking 0 .

65. Ans: (d) Sum of infinite terms of G.P1

a

r

1 1 4 2 3 4 2 3 34 2 3 1 sin 1

1 sin 4 24 2 3 4 2 3x

x

3 2sin (or) .

2 3 3x x

66. Ans: (b) 3

, 1 sin cos sin ,4 2 4 2 2

A A AA

cos sin 0,cos sin 0

2 2 2 2

A A A A .

2cos cos sin cos sin cos sin cos sin 1 sin 1 sin2 2 2 2 2 2 2 2 2

A A A A A A A A AA A

67. Ans: (b) Put 5

6

and verify.

68. Ans: (d) cot cot 60 cot 60 cot 60 cot cot 60 3 ,R n n z .

69. Ans: (b)

1 cot 90 2 cot 62 1 cot88 cot 62 cot88 cot 62 13

cot 62 cot88 cot 62 cot88tan 90 62 cot88k

cot 88 62 cot150 3

1k

70. Ans: (b) 3 3 30 3x y z x y z xyz

(a) Put 3 3 3 3 3 3 3, , 0 3x a y b z c a b c a b c abc .

(b) 4 4 4 4 4 40a b c a b c .

Squaring on both sides 4 42. 2.a b ab c a b c ab

Squaring on both sides 2 2 2 4a b c ab bc ca ab

2 2 2a b c ab bc ca



JUL 2013

71. Ans: (d) sin ,sin 904 4

r r

y x

2 22 2

2 2sin cos 1 1

16 16

r r

y x

2 2

2 2 2 2

2 2

1 11 16

16

rx y x y

y x

22 2 4 4 1 0

y yx y xy

x x

tan 2 3 15 (or) 75

y

x

72. Ans: (c) In 1, tan tan tan tan tan tan 6 2tan tan 3 tan 3ABC A B C A B C C C C

Now tan tan 3, tan tan 2A B A B tan , tan 1,2A B

73. Ans: (b)Put

23 5 24 4

23 5 1

log . Now 2 log 24 4 2

tt

xt

x t x tt

.

23 23 4 5 1

3 4 5 2 04 2

t tt t t

t

1

2 32

1log 1 (or) 2 (or) - 2 (or) 2 (or) 2

3t x x

74. Ans: (a) tan tan

tan tan , tan tan tan1 tan tan 1 1

p pp q

q q

2 2sin sin cos cosp q

22 22

2 22 2

1 1 1. 1 . 1

1 1

p q q qp p p q q qq

p q p q

31. Ans: (c) 25 5log log 2 log 2 5 25aa x x x

32. c 2 2 2 2 2 2a b c a b c

1 1

log loglog log

c ca b a b

a b a bc c

2 2 2log log 2c ca b c

33. Ans: (a) 3 5 75x y z k 11 1

3 ,5 ,75yx zk k k

2 1125 5 3 y xzk k k

1 2 12

xyx y

z y x z 2 2xy xz zy xy zy xz

34. Ans: (b) 3 2 3 0a x b x c

33 2 3 3 2 33 . .a x b x c a x b x c 3 2 3 3 3a x b x c abcx

90

4

r

r

y

x



JUL 2013

35. a 4 5

cos ,sin5 13

3 5

tan , tan4 12

tan tantan 2 tan

1 tan tan

3 5

4 12tan 23 5

14 12

36 20 56

48 15 33

36. Ans: (c) qec cotcos , q

ec1

cotcos

qqec

qqecec

1

2

1cos

1cotcoscotcos

37. Ans (b) 233 21

21

2

x , 233 21

21

2

y ,

LHS2

21

21

2

21

21

21

21

33122332333

3

15123323312 2

12

11

641643

11512

38. Ans: (c) 31.0log31.0log 1.03

1.0 , 29log29log 92

9 y

Then 122232 yx

ba,xyyxyxba thatsuch 2

39. Ans: (b) 360tan 0 0tan 40 20 3 0 0tan 40 tan 20

31 tan 40 tan 20

0 0 0 0tan 20 tan 40 3 tan 20 tan 40 3 .

40. Ans: (b) LHS 2 2

1 tan tan tan tan

tantan2tantantantan2tantan1 2222

22 tan1tan1 22 secsec

85. Ans: (b) 4 2 2 2 2 2sec sec sec sec 1 sec tan 2 2 2 41 tan tan tan tan .

86. Ans: (b) We have, 6 6 2 2sin cos 3sin cos (1)

222266 cossincossin3cossin 1cossin322

87. Ans: (a) 2 2 2 2 2 2

ab a b a b b c bc b c ca c a c ax x x



JUL 2013

3 3 3 33 3 3 3 3 3 3 3 3

3

2

2

1 a b c ab c a b b c c a a

ax x x x x

x

88. Ans: (a)

1

3 2 2 2

4 4 112 2 3 3 33

3 3 2 2 7

5 5 3 5 3

4 1 3 4 11 53 2 2

3 3 32 2

2 4 6 2 42

3 3 3

3 2 7 3 2 7

5 5

228

5

2283

0

0

.

89. Ans: (a) Let d be the diameter of moon and t its distance from the earth. If the moon

subtend an angle at the eye of the observer, then 27

113157

18060

31350000

ld cm.

90. Ans: (c) 5 5log 7 log 7 33

310

10

1 15 5 7 1

log 101log

10

283

Model Question Paper (P+C+M)

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g t t c

g t t b

g t tt t d

date p c

r b203r c

hr h c

r h h d

gl b24gl c