Jee-main-2019-modelpaper-4.pdf

JEE-MAIN MODEL GRAND TEST IMPORTANT INSTRUCTIONS: *

The test is of 3 hours duration.

*

The Test Booklet consists of 90 questions in three parts, having 30 questions in each part of equal weightage. Each question is allotted four marks for correct response. Maximum marks is 360.

*

One mark will be deducted for indicating incorrect response of each question.

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*

PART A — CHEMISTRY

1.

The degree of dissociation (α ) of a weak electrolyte, Ax B y is related to van’t Hoff

2.

i −1 ( x + y − 1)

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factor (i) by the expression x + y −1 x + y +1 1) α = 2) α = i −1 i −1

3) α =

4) α =

i −1 ( x + y + 1)

Which of the following compounds will not undergo Friedal–craft’s reaction easily 1) Nitro benzene

2) Toluene

3) Cumene

4) Xylene

OCH 3

Methoxy benzene is called anisole

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3.

.

1) 5

2) 4

3) 3

4) 2

In the following compounds

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4.

w

How many more structures can be drawn for the same forumula?

I)

O II)

N

III)

N

H

IV)

N

N

H

H

The order of basic strength is

1) IV>I>III>II

2) III>I>IV>II

3) II>I>III>IV

4) I>III>II>IV 1

5.

Find the product of the given reaction? CH3

H⊕ ∆

CH3 OH

?

CH3

CH3

1)

2) CH3

CH3 CH3

6.

The reagent(s) for the following conversion ?

Br

H

Br

H

is/are

2) Alcoholic KOH followed by NaNH2

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1) Alcoholic KOH

3) Aqueous KOH followed by NaNH2 7.

CH3

4)

CH3

CH3

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3)

4) Zn/CH3OH

A,B and C are the elements of the same period in the long form of the periodic table .Oxides of A is amphoteric, B is acidic and C is basic. The increasing order of atomic numbers of the elements is

4) B
1) CO3−2 < CO2 < CO

2) CO2 < CO32 − < CO

3) CO < CO32 − < CO2

4) CO < CO2 < CO32 −

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9.

3) A
The correct order of increasing C-O bond length of CO, CO32 − ,CO2 is

Which one of the following alkaline earth metal sulphates has its

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8.

2) C
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1) C
hydration

enthalpy greater than its lattice enthalpy? 1) CaSO4

2) BeSO4

3) BaSO4

4) SrSO4

10. Aluminium chloride exists as dimer, Al2Cl6 in solid state as well as in solution of nonpolar solvents such as benzene. When dissolved in water, it gives 1) Al 3+ + 3Cl − 3)  Al ( OH )6 





3−

2) Al2O3 + 6 HCl 4)  Al ( H 2O )6 





3+

+ 3Cl −

11. Which one of the following statements is wrong ? 2

1) Ionic carbides are formed by highly electropositive metals 2) Carborundum and boron carbide are true covalent carbides 3) Carborundum and boron carbide are used as abrasive 4) Mg 2C3 like CaC2 on hydrolysis evolves C2H2 12.

H 2O2 exists as ______ in alkaline medium 1) HO2-

2) HO2+

3) O22-

4) Both 1 & 3

13. The common components of photo chemical smog are

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1) O3, NO, acrolein , HCHO, Peroxyacetyl Nitrate 2) O3,O2,H2 3) NO + O3

14. Dacron is a polymer of

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4) only Peroxyacetyl Nitrate , acrolein

1) Ethylene glycol +Phenol

2) Ethylene glycol+ Vinyl acetate

3) Glycerol +phthalic anhydride

4) Ethylene glycol+ Terephthalic acid H

i) pentose

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iv) Ketose

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15. Which set of terms correctly identifies the carbohydrate shown? iii) Aldose

v) pyranose

vi) Furanose

2) i,iii and vi

3) ii,iii and v

OH

OH

H

H

OH

HOH2C

H

4) i,iv and vi

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1) i,iii and v

ii) Hexose

O

16. The correct statements among the following is/are a) phosphorus is an essential constituent of animal and plant matter b) phosphoproteins are present in milk and eggs c) selenium can form dichlorides and dibromides d) In the oxidising properties of sulphuric acid, with metals, non metals it is reduced to SO3 1) All are correct

2) a,b only

3) a,b,d only

4) a,b,c only

17. Which of the following can exist as dimer 3

1) BrCl

2) ClF3

3) ICl3

4) IF3

18. With which of the following Xenon fluorides can form cationic species 1) PF5 19.

2) NaF

a) MnO4− + e− → MnO42 − ;

3) SbF5

4) Both 1,3

E0 = x

b) MnO4− + 4 H ⊕ + 3e− → MnO2 + 2 H 2O ; E 0 = y c) MnO4− + 8 H ⊕ + 5e− → Mn +2 + 4 H 2O ; E 0 = z

The correct relation between x,y and z is 2) x rel="nofollow">y
20. Match the following Table-2

Ore

composition

A) Fools gold

i) CuFeS2

B) Chalcopyrites

ii) ZnCO3

C) Malachite

iii) FeS2

D) Calamine

iv) Cu2S

4) x
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Table-1

3) y>z>x

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1) x>y>z

v) CO3.Cu(OH)2

2) A-iii,B-v,C-i,D-ii

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1) A-iii,B-i,C-v,D-ii 3) A-i,B-iii,C-v,D-ii

4) A-iii,B-i,C-ii,D-v

21. Energy required to stop the ejection of electron from Cu plate is 0.24ev. Calculate

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1) 4.89ev

w

the work function when radiation of λ = 253.7 nm strikes the plate 2) 4.65 ev

3) 4.89 J

4) 7.835x10-19 ergs

22. One litre flask contains air, water vapour and a small amount of liquid water at a pressure of 200mmHg. If this is connected to another one litre evacuated flask, what will be the final pressure of the gas mixture at equilibrium? Assume the temperature to be 500C. Aqueous tension at 500C = 93 mmHg. 1) 53.5mm

2) 93 mm

3) 146.5 mm

4) 107 mm

4

23. What will be the PH of an aqueous solution of 1.0M ammonium formate ? Given

= P K a 3.8, = P Kb 4.8 1) 7.5

2) 3.4

3) 6.5

4) 10.2

24. 2 moles of FeSO4 in acid medium are oxidized by ‘x’ moles of KMnO4 where as 2 moles of FeC2O4 in acid medium are oxidized by ‘y’ moles of KMnO4. The ratio of ‘x’ and ‘y’ is 1) 1/3

2) 1/2

3) 1/4

4) 1/5

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25. The equilibrium constant for a reaction is 1x1020 at 300K . The standard free energy change for this reaction is 1) -115 kJ

2) 115 kJ

3) 166 kJ

4) -166 kJ

26. For a gas having molar mass M, specific heat at constant pressure can be given as:

gR M (g- 1)

g RM

M R (g- 1)

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1)

2)

3)

4)

gRM g+ 1

27. In a solid ‘AB’ having the NaCl structure ‘A’ atoms occupy the corners of the cubic unit cell. If all the Face centred atoms along one of the axes are removed , then the resultant stoichiometry of the solid is 1) AB2

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2) A2B

28. For which of the following reactions

w

1) A + B → C Ea = 50kJ

4) A3B4

K310 would be maximum K300 2) x + y → z Ea = 40kJ 4) E + F → G Ea = 100kJ

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3) P + Q → R Ea = 60kJ

3) A4B3

29. Cyclohexene on ozonolysis followed by reaction with zinc dust and water gives compound E. Compound E on further treatment with aqueous KOH yields compound F. Compound F is 1)

CHO

CHO

2)

CO2H 3)

COOH

4)

CO2H 5

30. Efficiency of the following cell is 84% 2+

2+

A+ B (s)

A+ B

(aq)

(aq)

∆H = −285kJ

(s)

.

Then the standard electrode potential of the cell will be 1) 1.20v

2) 2.40v

3) 1.10v

4) 1.24v

PART B — MATHEMATICS

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1 1 1 31. If= D 1 1+ x 1 for x ≠ 0, y ≠ 0 then D is 1 1 1+ y

2) divisible by y but not x

3) divisible by neither x nor y

4) divisible by both x and y

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1) divisible by x but not y

 2x  32. Let f : (−1,1) → B , be a function defined by f ( x) = tan −1   , then f is both one 1 − x2  one and onto when B is the interval

 π 2)  0,   2

 π 1) 0,   2

 π π 3)  − ,   2 2

 π π 4)  − ,   2 2

2

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33. If z 2 + z + 1 = 0 , where z is a complex number, then the value of 2

2

2

2) 54

3) 6

4) 12

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1) 18

w

1  3 1  6 1  2 1   z +  +  z + 2  +  z + 3  + ... +  z + 6  is z z z z

34. The system of equations

α x + y + z = α − 1, x + α y + z = α − 1, x + y + α z = α − 1 hs no solutions, if α is 1) either -2 or 1

2) -2

3) 1

4) not -2

35. Let R = {(1,3),(4, 2),(2, 4),(2,3),(3,1)} be a relation on the set A = {1, 2,3, 4} . The relation R is 1) not symmetric

2) transitive

3) a function

4) reflexive

6

36. If the roots of the equation x 2 − bx + c = 0 be two consecutive integers, then

b 2 − 4c equals 1) 3

2) -2

3) 1

4) 2

37. How many ways are there to arrange the letters in the word SANDESH with the vowels in alphabetical order? 1) 360

2) 2520

3) 120

4) 480

38. If the sum of the coefficients in the expansion of ( a + b ) is 4096, then the greatest coefficient in the expansion is 1) 1594

 x cos θ tan −1   1 − x sin θ

3) 924

 −1  cos θ  − cot    x − sin θ

1) θ

2)

  is equals to 

4) 2924

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39.

2) 792

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n

π

2

−θ

3)

θ

2

4)

π 4

−

θ 2

40. In a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of this progression is equals 2)

5

1 2

(

)

5 −1

3)

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1)

(

1 1− 5 2

)

4)

1 5 2

    41. The vectors AB= 3i + 4k and AC =5i − 2 j + 4k are the sides of a triangle ABC. The

72

2)

33

3)

288

4) 18

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1)

w

length of the median through A is

42. = If cos α 1 1) 3 4

α 2cos β − 1 π ) , α + β π then tan is equal to (0 < α, β < = 2 2 − cos β 1 2) 3 2

3) 3

4) 32

43. Suppose a population A has 100 observation 101,102,…,200, and another population B has 100 observations 151,152,…,250.If V A and VB represent the variances of the two populations, respectively, then V A / VB is 1) 1

2) 9/4

3) 4/9

4) 2/3 7

44. A random variable X has poisson distribution with mean 2. The P ( X > 1.5 ) equals 1) 0

3) 3 / e2

4) 1−

3 e2

n.3n 1 then the range of x is (where n ∈ N ) lim = n n + 1 n 3 n → ∞ n ( x − 2 ) + n.3 −3

1) [2,5)

2) (1,5)

3) (-1,5)

47. Let ‘f’ be a function satisfy

4) ( −∞, ∞ )

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46. If

2) 2 / e 2

f ( x) f ( = y ) f ( x) + f ( y ) + f ( xy ) − 2∀x, y ∈ R and f(2) = 5 then lim f ( x) = x→4

2) 16

3) 21

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1) 4

4) 17

48. If y(x) is the solution of the a differential equation

( x + 2)

dy = x 2 + 4 x − 9, x ≠ −2 and y(0) = 0 then y(-4) is dx

1) 2

2) 0

3) 1

4) -1

49. If at each point on the curve y = x3 − ax 2 + x + 1a the tangent is inclined at an acute

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angle with the +ve direction of the x-axis, then 1) a > 0

2) − 3 ≤ a ≤ 3

3) a ≤ 3

4) ( −∞, ∞ )

w

π  π π 50. Let g : ( −∞, ∞ ) →  − ,  be a function given = by g ( x) 2 tan −1 e x − then g is 2  2 2

w w

1) Strictly decreasing in ( −∞, ∞ ) & odd function 2) Strictly increasing only in ( 0,∞ ) 3) Strictly increasing in ( −∞, ∞ ) , but neither even nor odd 4) Strictly increasing in ( −∞, ∞ ) & odd function 51. A curve g= ( x)

1)

37 7

∫x

27

(1 + x + x2 ) ( 6 x2 + 5x + 4) dx is passing through (0,0) then g(1) = 6

2)

36 6

3)

27 7

4)

26 6 8

1

b

0

b −1

et 52. If ∫ dt = a then t +1

∫

e −t dt = t − b −1 3) − a.e−b

2) − a.eb

1) a.eb

53. The area bounded by the curves= y

4) a.eb

x , 2 y − x= + 3 0 , x-axis and lying in the first

quadrant is 1) 3

2) 19

3) 6

4) 9

possible values of a is the interval 1) ( 0,∞ ) 55.

2) (1,∞ )

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54. The lines x + y = a and ax − y = 1 intersect each other in Q1 . Then the set of all 3) ( −1, ∞ )

4) (−1,1]

The radices of the circle of least size that passes through (-2,1) and touching both axes is 2) 2

3) 5

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1) 1

4) 6

56. If ( x, y ) lies on the ellipse 9 ( x − 2 ) + 16 ( y − 1) = 144 then the maximum value of 2

x + y is 1) 4

2

2) 8

3) 1

4) 16

57. The product of the perpendiculars from a point on 9 ( x + y ) − 16 ( x − y ) = 144 to its

144 9

2)

144 50

3)

144 25

2

4)

144 16

w

1)

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asymptotes is

2

w w

t 58. If the straight lines x = 1 + s, y = −3 − λ s, z = 1 + λ s and x = , y =+ 1 t , z =− 2 t with 2 parameters s and t respectively are coplanar then λ = 1) 1

2) 3

3) 2

4) -2

59. A line makes an angle θ with x-axis and y-axis, a possible θ is in

π π  1)  ,  4 2

π π  2)  ,  3 6

 π 3) 0,   2

 π 4) 0,   4

60. Let f ( x)= x + cot x and f is the inverse of g then g1 ( x ) = 1)

−1

[ x − g ( x)]

2

2)

x

[ x − g ( x)]

2

3)

x x − g ( x)

4)

1

[ x − g ( x)]2 9

PART C — PHYSICS

61. A coil in the shape of an equilateral triangle of side ‘l’ is suspended between the  pole pieces of a permanent magnet such that B is in plane of the coil. If due to a current ‘i’ in the triangle, a torque ‘T’ acts on it, side ‘l’ of the triangle is 1/2

 T  1) 2    3Bi 

2)

2 T    3  Bi 

3)

1/2

2 T    3  Bi 

4) None

62. Three persons P, Q and R of same mass travel with same speed u along an equilateral triangle of side ‘d’ such that each one faces the other always. After how

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much time will they meet each other Q u

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u P

1) d/u seconds 3) 2d /

( 3) u

R

u

2) 2 d/3 u seconds

seconds

4) d /

( 3) u

seconds

1) 30o

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   63. The resultant of two vectors A and B is perpendicular to the vector A and its   magnitude is equal to half of the magnitude of vector B . Then the angle between A  and B is 2) 45o

3) 150o

4) 120o

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64. Three blocks of masses m1, m2 and m3 are connected by massless string as shown in

w w

fig. on a frictionless table. They are pulled with a force T3 = 40 N. If m1 = 10 kg, m2 = 6 kg and m3 = 4 kg, the tension T2 will be

1) 20 N

T1 m1

2) 40 N

T3

T2 m3

m2

3) 10 N

4) 32 N

65. The kinetic energy K of a particle moving in a straight line depends on the distance s as K = as2. The force acting on the particle is 1) 2 as

2) 2 m as

3) 2 a

4)

as 2 10

66. A rigid spherical body is spinning around an axis without any external torque. Due to temperature its volume increases by 3%. Then percentage change in its angular speed is 1) -1%

2) 1%

3) -2%

4) -3%

67. Infinite number of masses each of 3 kg are placed along a straight line at the distances of 1 m, 2m, 4m, 8m, …. From a point O on the same line. If G is the universal gravitational constant, thent he magnitude to gravitational field intensity at O is : 2) 2.0 G

3) 3.0 G

4) 4.0 G

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1) 1.0 G

68. Two wires of the same length and radius are joined end to end and loaded. If the Young’s modulus of the materials of the wire are Y1 and Y2, the combination behaves as a single wire of Young’s modulus

(Y1 + Y2 )

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1) Y1 + Y2

2)

Y1Y2

3)

2

4)

2Y1Y2 (Y1 + Y2 )

69. What change in surface energy will be noticed when a drop of radius R splits up into 1000 droplets of radius r, surface tension T ? 1) 4π R 2T

2) 7π R 2T

3) 16π R 2T

4) 36π R 2T

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70. A flat plate of area 10 cm2 is separated from a large plate by a layer of glycerine 1 mm thick. If the coefficient of viscosity of glycerine is 20 dyne cm2 s, the force required to keep the plate moving with a velocity of 1 cm per sec is : 1) 80 dyne 2) 200 dyne 3) 800 dyne 4) 2000 dyne

w

71. A metal sphere of radius r and specific heat S is rotated about an axis passing

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through its centre at a speed of n rotations per second. It is suddenly stopped and 50% of its energy is used in increasing its temperature. Then the rise in temperature of the sphere is

2 π 2n2r 2 1) 5 S

2 π 2n2 2) 10 r 2 S

7 3) π r 2 n 2 S 8

 π rn  4) 5    145 

2

72. In the P-V diagram shown in Fig. ABC is a semicircle. The work done in the process ABC is

11

y C

P (Nm-2)

3 B

2

A

1

0

1

2 V (m3)

2) 4 J

3)

π 2

4) −

J

π 2

J

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1) zero

73. An engine has an efficiency of 1/3. The amount of work this engine can perform per kilocalorie of heat input is 1) 1400 cal

2) 700 cal

3) 700 J

4) 1400 J

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74. Two plates of same area are placed in contact. Their thickness as well as thermal conductivities are in the ratio 2 : 3. The outer surface of one plate is maintained at 10oC and that of other at 0oC. What is the temp. of the common surface surface ? 1) 0oC

2) 2.5oC

3) 5oC

4) 6.5oC

75. A spring mass system oscillates with a time period T1. when a certain mass is

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attached to the spring. When a different mass is attached after removing the first mass the time period becomes T2. When both the mass are attached to the free end

2) T1 + T2

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1) T1T2

w

of the spring with the other end fixed, the time period of oscillation is : 3)

T1 + T2 2

4)

T12 + T22

76. Mono chromatic radiation of wave length ' λ ' is incident on a hydrogen sample in ground state. Hydrogen atoms absorb a fraction of light and subsequently emits six difference wave length. The value of ' λ ' is o

1) 970.6 A

o

2) 1970 A

o

3) 1276 A

o

4) 1658 A

77. In the circuit shown, the potential difference across the capacitor is 12V. The cell is ideal, the emf of the cell is 12

R

R

C

R

R

R

E

1) 9V

2) 12V

3) 15V

4) 18V

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78. The radius of second orbit of an electron in hydrogen atom is 2.116Ao. The deBroglie wavelength associated with the electron in this orbit would be o

o

1) 1.058 A

2) 3.32 A

o

3) 6.64 A

o

4) 13.28 A

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79. Statement-1 : If the frequency of the incident light on a photosensitive surface is doubled, the maximum K.E of emitted photo electrons becomes double. Statement-2 : The intensity of the light will set doubled by doubling the frequency of light falling on surface, keeping the distance between the source of the light and the surface same

1) Statement-1 and Statement-2 are true.

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2) Statement-1 and Statement-2 are false

3) Statement-1 is true and Statement-2 is false. 4) Statement-1 is false and Statement-2 is true.

w

80. With a concave mirror, an object is placed at a distance ‘X1’ from the principal

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focus, on the principal axis and the image is formed at a distance X2 from the principal focus. Then the focal length of the mirror is 1)

x1x2

2) x1x2

3)

x1 + x2 2

4)

x1 x2

81. If 10% of a radio active material decays in 5 days then the amount of the original material left after 20 days is approximately 1) 60%

2) 65%

3) 70%

4) 75%

13

82. In Young’s double slit experiment, the intensity at a point where path difference is

λ 6 1)

is I. If Io denotes the maximum intensity, then

1 2

2)

3 4

3)

I is equal to I0

1 2

4)

3 2

83. The transfer ratio ' β ' of a transistor is 50. The input resistance of the transistor when used in the common emitter configuration is 500Ω . The peak value of the collector ac current for a peak value of ac input voltage of 0.02 V is 2) 2 mA

3) 4 mA

4) 20 mA

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1) 1 mA

84. An electric bulb has a rated power of 50w at 100V. Calculate the inductance of a choke coil to be used is series with this bulb operating on 220V, 50 Hz ac source is 2) 1 H

3) 0.1 H

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1) 1.25 H

4) 1.25 mH

85. A T.V transmission tower at a particular station has a height of 160m. By how much the height of tower be increased to double its coverage range (Given radius of earth is 6400 km) 1) 640 m

2) 160 m

3) 320 m

4) 480 m

86. Consider the charge configuration and a spherical Gaussian surface as shown in

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figure. When calculating the flux of electric field over the spherical surface, the electric field will be due to

w w

w

+ q1

+ q2

−q1

1) q2

2) only positive charges

3) + q1 Q − q1

4) all the charges

87. A magnetizing field of 1600 Am-1 produces a magnetic flux of 2.4 × 10−5 wb in an iron bar of cross sectional area of 0.2 cm2. The susceptibility of the iron bar is 1) 298

2) 596

3) 1192

4) 1788

14

88. Five identical plates are connected across a battery as shown in figure. If charge on plate ‘I’ be +q, then the charges on plates 2, 3, 4 and 5 are : 1 2 3 4 5 +

-

2) − q, + 2q, − 2q, + q

3) −2q, + 2q, − 2q, + q

4) none of these

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1) − q, + q, − q, + q

89. The pressure P exerted by a liquid is given by P = ax +

b

c + t2

, where x is distance, t

1) ML−2T −2 ; ML−1; T 2

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is time; a, b, c are arbitrary constants. The dimensions of a, b, c are respectively 2) ML−1T −2 ; T 2 ; L2

3) ML−2T 2 ; MT −2 ; ML−1

4) MT −2 ; L2 ; T 2

90. In the equation for a stationary wave given by y = 5cos

πx 25

sin100π t. Where ‘X’ is in

cm and ‘t’ in second, a node will not occur at

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2) 25 cm

3) 37.5 cm

4) 62.5

w w

w

1) 12.5 cm

15

JEE-MAIN MODEL TEST KEY SHEET

2) 3

3) 4

4) 2

5) 4

6) 3

11) 2

12) 1

13) 1

14) 4

15) 1

21) 1

22) 3

23) 4

24) 2

MATHEMATICS

7) 2

8) 3

9) 1

10) 2

16) 3

17) 4

18) 1

19) 2

20) 4

25) 1

26) 2

27) 2

28) 4

29) 1

30) 2

ad up ra tib

1) 1

ha .n et

CHESMISTRY

32) 2

33) 2

34) 4

35) 1

36) 1

37) 1

38) 4

39) 2

40) 4

41) 4

42) 4

43) 1

44) 4

45) 1

46) 2

47) 3

48) 4

49) 3

50) 1

51) 2

52) 3

53) 3

54) 1

55) 1

56) 1

57) 1

58) 4

59) 3

60) 4

w

w w

PHYSICS

.e en

31) 1

61) 1

62) 2

63) 3

64) 4

65) 1

66) 3

67) 4

68) 4

69) 4

70) 4

71) 1

72) 3

73) 4

74) 3

75) 4

76) 1

77) 3

78) 3

79) 4

80) 1

81) 2

82) 2

83) 2

84) 1

85) 4

86) 4

87) 2

88) 3

89) 1

90) 2

16

JEE-MAIN MODEL TEST CHEMISTRY SOLUTIONS

9.

Due to small size of Be+2

10.

Al 2 Cl6 + 3H 2 O → Al 2 O3 + 3HCl

11.

Mg 2C3 + 4 H 2O → 2 Mg ( OH )2 + CH 3 − C ≡ CH (propyne)

12.

H 2 O2 + OH − → H 2 O + HO2−

13.

Photochemical smog contains PAN, NO,

1

0

0

1−α

xα

yα

i −1 ( x + y − 1)

-NO2 is electron withdrawing group. CH2OH

O and

3.

O3 and acrolein

CH3

14.

4.

I is the most basic as the lonepair belongs IV is the least basic as the lonepair of

15.

electrons on N forms a part of aromatic sextet H

OH CH3

⊕

CH3 CH3 -H2O O+ H 2

CH3 ⊕

w

⊕ CH3 − H

w w

(30 carbocation)

6.

Simple

CH3

.e en

CH3 CH3

alkyl

CH3

⊕

CH3

(20 carbocation)

Methyl

Shift

halides

are

require strongest base like

NH 2

for

elimination

7.

Basic nature of Oxides decreases and acidic nature of oxides increases from left to right in a period

Carbohydrate contains 5 carbons, H atoms and aldehyde. It was in pyranose structure.

16.

Selenium usually forms tetrahalides

17.

As monomer IF3 is less stable

18.

XeF2 + PF3 → [XeF]+ [PF6 ]−

19.

E 0 value for MnO −4 to MnO2 is less,

compared to other two conversions

dehydrohalogenated by using strong base like alc.KOH. Whereas vinyl halides

Polymer of terephthalic acid and ethylene glycol is commercially called dacron

sp3 orbital therefore available for donation.

5.

ha .n et

Ax By → xA+ y + y.B − x

α= 2.

Bond length ∝

ad up ra tib

1.

1 Bond order

8.

20.

Composition of minerals

21.

Ephoton= work function + KE absorbed (1.602×10-19J=1ev) = work function + ev0 Where e = electronic charge ; v0 = stopping potential

17

Þ CP =

22. 1lit

Aqueous tension remain same in both

27.

8 = 1 8 4 No.of ‘B’ atoms in unit cell= = 2 2 The formula is AB2 No.of ‘A’ atoms in unitcell=

plasks p1v1

=

(intial)

p2v2

p1v1

=

(final)

107x1

= p2x2

Intial

pressure p1 = 107 = p2 = 53.5mm 2 v1 =1 pressure of gaseous mixture

p2v2

200-93=107

28.

energy 29.

p2=? (since aqueous tension in plask =93)

CHO

CHO

23.

1 kw  p + p ka − p kb  p H= 2

24.

Fe +2 → Fe +3 + e −

.e en

Mn +7 + 5e − → Mn +2

C2O → 2CO2 + 2e

30.

−

31.

5mole FeSO4 = 1mole Mn+7

CHO

CHO 0

∆G −nFE 0 Efficiency = = ∆H 0 ∆H 0

w

w w

2 6 x 2 5 1 = x= = y ; = × = 5 5 y 5 6 3

25.

∆G 0 = −2.303RTLogK

26.

R CP - CV = M CP R C = (Q P = n) n M CV æ 1ö R Þ CP ç = 1- ÷ ÷ ç ÷ ç è nø M

1 1 1 = D 1 1+ x 1 1 1 1+ y

1 0 0 = 0 x 0= 1( xy − 0)= xy 1 0 y

2 mole of FeSO4 = ?

CP -

∆

(Apply C2 → C2 − C1 , C3 → C3 − C1 )

5mole of FeC2O4 = 3moles of Mn+7

=

OH

OH ( − )

MATHEMATICS SOLUTIONS

Fe +2 → Fe +3 + e − −2 4

i)O3 ii)Zn/H2O

ad up ra tib

(1+1)

Increase in rate constant is maximum for the reaction having maximum activation

=53.5+93=146.5mm

v2=2

Along one axis two Faces are present so

ha .n et

v=1lit

nR M (n- 1)

Hence, D is divisible by both x and y 32.

For x ∈( −1,1)  2x  we have f ( x ) = tan −1   1 − x 2 

 2 tan θ  tan −1  ∴ f ( tan θ ) =  1 − tan 2 θ  (By x = tan θ )

18

36.

−1 = tan = tan 2θ 2 tan −1 x

33.

∴ ( x + α ) ( x + α + 1) = x 2 − bx + c

π

 2x  π < tan −1  < .  1 − x 2  2 2

z + z + 1 =0 ⇒ z = ω,ω 2

2

Comparing

2

(

2

)

2

(

+ 2 ω3 + ω3

)

we

get

= c α2 + α

2

( 2α + 1) 2 − 4 (α 2 + α )=

∴ b 2 − 4c =

1

2

37.

( )

= 4 ( −1) + 2 22 = 4 + 8 = 12 2

sides

b 2α + 1 ⇒ −=

1  1 1   ∴  z +  +  z 2 + 2  + ...  z 6 + 6    z  z  z 

=4 ω + ω2

both

Number of letters = 7 Number of vowels = 2 namely A.E these

ha .n et

⇒−

Let α , α + 1 are consecutive integer

alphabets can be arrange themselves by 2! 34.

For no solution A = 0 and (adj A) (B) ≠ 0

α

1

Ways

1 1 = 0

38.

⇒ α 3 − 3α + 2 = 0 ⇒ (α − 1) 2 ( α + 2) = 0 ⇒ α =1, −2

∴ 2n = C0 + C1 + C2 + .... + Cn

= 2n 4096 = 212

.e en

solution .Also the each equation becomes

⇒n= 12 ( even)

x+ y+ z = 0 again for α = −2

Now (a + b) n =( a + b )

12

A = 0 but (adj A)(B) ≠ 0

As n = 12 is even so coefficient of greatest

w

⇒ ∃ no solution.

w w

term is n = Cn

R is a function as A = {1, 2,3, 4} and

R is not reflexive as (1,1) ∉ R

R is not transitive as

( 3,1) ∈ R

but (1,1) ∉ R

(1,3) ∈ R

=

but

( 3, 2) ∉ R

39. and

= C12

12

C6

12 11 10 9 8 7 × . . . 6 5 4 3 2 1

=

( 2,3) ∈ R

12

2

and ( 2,3) ∈ R

R is not symmetric as

C0 a n + C1a n −1b + C2 a n − 2b 2 + ... + Cnb n

Putting a= b= 1

⇒ For α = 1 there exist infinitely many

( 2, 4) ∈ R

Consider

( a + b)=n

But for α = 1 , A = 0 and (adj (A) (B) = 0

35.

7! = 2520 2!

ad up ra tib

Now A = 0⇒ 1 α 1 1 α

Therefore, number of words =

11 × 9.8.7 =× 11 3.4.7 = 924 3.2.1

 x cos θ tan −1   1 − x sin θ

 −1  x − sin θ   − tan     cos θ 

−1 = tan = ( tan θ ) θ

40.

Given, a = ar + ar 2 ⇒ r 2 + r − 1 = 0 19

45.

−1 + 5 ⇒r= 2

Median through any vertex divide the opposite side into    AB + AC = 2 AD

two

equal

parts 46.

 1   1  AB + AC = ⇒ AD= [8i − 2 j + 8k ]   2 2  ∴ AD = 33

42.

−1 <

43.

=⇒ 3 cos α =3 − 2 1 1 − cos α = 34 1 + cos α

48.

⇒y=

x2 + 2 x − 13log x + 2 + c 2

y ( 0) = 0 ⇒ c = 13log 2

Series a

= 101,102,103,………………,200

.e en

Series B

= 151,152,153,………………,250

Therefore, y ( −4) = 0

49.

Series B is obtained by adding a fixed

dy ≥0 dx

⇒ 3 x 2 − 2ax + 1 ≥ 0∀x

quantity to each item of series A, we know

∆ ≤ 0 ⇒ 4a 2 − 12 ≤ 0

origin both series have the same variance

⇒− 3≤a≤ 3

w w

w

that variance is independent of change of so ratio of their variances is 1. 44.

x2 + 4x − 9 ∫ dy = ∫ x + 2 dx

ad up ra tib

α  ∴ tan  =  2

25 = 5 + 5 + f ( 4 ) − 2

f ( 4 ) = 17

cos 2 α + 4 cos α + 1 = 0 2

Put

x =2 ,y =2 , f ( 2 ) f ( 2 ) =f ( 2 ) + f ( 2 ) + f ( 4 ) − 2

−2 cos α − 1 cos α = 2 + cos α

( cos α + 2 )

x−2 <1 3

⇒ −1 < x < 5 ⇒ x ∈( −1,5) 47.

− cos α cos β = cos (π − α ) =

n

Divide Nr & Dr by n.3

ha .n et

41.

P ( X= r= )

e− λ λ r ( λ= mean) r!

∴ P ( X =r > 1.5) =P ( 2) + P ( 3) + .....∞

= 1 −  P ( 0) + P (1)   3 e −2 × 22  = = 1 − e −2 + 1− 2  2  e 

50.

2e x = g ( x) > 0∀x ∈( −∞, ∞ ) 1 + e2 x 1

( )

& = g ( − x ) 2 tan −1 e − x −  1 π = 2 tan −1  x  − e  2

= 2cot −1 e x −

π 2

π

2

π  π = 2  − tan −1 e x  − 2  2 20

π

= − 2 tan −1 e x = − g ( x) 2

x1 − x2

y1 − y2

z1 − z2

a1

b1

c1

a2

b2

c2

58.

Therefore, g is odd function

∫ x (1 + x + x ) ( 6 x

g= ( x)

51.

24

∫(x

=

2 6

4

+ x5 + x 6

) (6x 6

5

5

)

+ 5 x 4 + 4 x3 dx

⇒ g ( x) + cot g ( x) = x

6

(

)

7

3 ∴ g (1) = 7

Take t − b = − y and proceed

53.

R.A =

54.

a>0 2

∫ ( 2 y + 3 − y )dy 3

2

0

2

1

⇒x= 1

⇒ y = a −1> 0

a ∈(1, ∞ )

.e en

a>1 55.

Center C = (-r,r) ,P(-2,1)

⇒ r=1

P

61.

B

S

l

l

60o

Q

O

o sin 60 =

PO =

N

R

PO PO = = PQ l

3 2

3l Area of ∆PQR is 2

+ 4 cos θ + 3sin θ = 3= +34cos θ + 3sin θ

A=

w

x + y = 2 + 4cos θ + 1 + 3sin θ Maximum = 3 + 16 + 9 = 8

57.

PHYSICS SOLUTIONS

w w

CP = r 56.

Diff w.r.t x

+c 7

( 0,0) ⇒ c =0 52.

⇒ cot g ( x ) = x − g ( x)

7

ad up ra tib

= g ( x)

ha .n et

Put x + x + x = t

x 6 + x5 + x 4

f  g ( x )  = x

f −1 ( x) = g ( x)

60.

5

∴ 0 ≤ 2cos 2 θ ≤ 1

π π  ⇒ θ ∈  ,  ( cos θ is decreasing) 4 2

)

+ 5 x 4 + 4 x3 dx

= ∫ t 6 dt 4

2cos 2 θ = sin 2 α

59.

= a 2 8,= b2 2

=0

9 2

QR × PO l 3l = × 2 2 2

= T MB sin θ = ⇒T Torque, 2

x + y x − y  2   2  − = 1 9 8 2

i 3l 2 = T = B ×1 4

( iA) B sin 90o

3 Bi l 2 4

 T  Therefore, l = 2   3Bi 

1/ 2

a 2b 2 144 Product = = 2 2 a +b 50 21

62.

The persons P, Q, R at the vertices of an

Proceeding as in the above equation,

equilateral triangle will meet at O, i.e., the

F= F= 2 as t

centre of median of triangle PQR, Fig. So the person at P will travel a distance PO,

66.

with velocity along PO = u cos30o Here,

Or

d 2 d PO = PB sec 30o = × = 2 3 3 ∴ time of meeting,

∆R ∆V / V 3 = == 1% R 3 3 2 Moment of = intertia, I MR 2 or I ∝ R 2 5 = 2% ∴

dis tan ce d/ 3 d / 3 2d = = = o velocity u cos 30 u 3 / 2 3u Q

As no external torque is being applied, = L I= ω cons tan t

u

30o

u

O

30o u

B

63.

R

.e en

B2 = A2 + B 2 + 2 AB ( − A / B ) = B 2 − A2 4 3 2

w

A2 3 A Or = = or B2 4 B

w w

3 From i), cos θ = cos150o − = 2

Or

θ = 150

by

2%,

angular

by -2%.

67.

o

F 64. Common acceleration, a = m1 + m2 + m3 40 = = 2m / s 2 10 + 6 + 4 Equation of motion of m3 is T3 – T2 = m3 a

I=

GM GM GM + 2 + 2 + ..... 12 2 4

 1 1  = GM 1 + + + ......  4 16 

   1  4 4 = GM  = GM = × G × 3 = 4G  1 3  1−  3  4

1/2 B R = =  A2 + B 2 + 2 AB cos θ  2

Or

increases

velocity ω will decrease by 2% or increase

B sin θ tan 90o or A + = B cos θ 0 = A + B cos θ A or cos θ = − ….i) B

I

Since,

A

ad up ra tib

C

∆V ∆R × 100 = 3 × 100 V R

ha .n et

= t

4 As V = π R 3 , we find V ∝ R 3 3

68.

Since the two wires are joined end to end and loaded, each wire will have the same extension force say F. If ∆l1 and ∆l2 are the extensions in the two wires and k1 and k2 are their force constants, then Y1 = ∴ k1 =

F l A ∆l1

and Y2 =

F l A ∆l2

F YA F Y1 A and = k2 = 2 = ∆l2 l ∆l1 l

40 − T2 =4 × 2, T2 =32 N 65.

When the particle is moving in a straight line, Fc = 0 22

If k is the restoring force per unit extension of the combined wires in series, then k1k2 YA and k = , where Y is the 2l k1 + k2

total Young’s modulus of the combination of two wires.

 ∆Q   ∆Q    =   ∆t 1  ∆t  2 K1 A1

Increase in surface energy =

S.T ×

Or

increase in surface area

= 20 ×10 ×

Here,

1 = 2000 dyne 0.1

∴

75.

50 (KE of rotation) = s m θ 100

2 2 (10 − θ ) =θ 3 3

T1 = 2π

or θ = 5o C

m1 kT 2 or m1 = 12 k 4π kT22 4π 2

w

w w

m + m2 = π 1 2π T 2= k

Refer to fig. In the part AB, volume of the gas is decreasing, and in the part BC, volume of the gas increasing. Therefore,

76.

From nth energy state, total number of wavelengths emitted n ( n − 1) = 6∴ n 2 − n − 12 = 0 2

WAB is negative and WBC is positive. As

=

WBC > WAB ,

Solving, n = 4

work

is

kT12 kT22 + 4π 2 4π 2 k

T12 + T22

=

net

x2

then

2 π 2n2r 2 5 S

therefore,

(θ − 0 )

end of string, then at one end of string,

12 2 2  m r  ( 2π n ) = S m θ 45 

72.

A2

When two masses are kept together at one

11 2  Iω  = S mθ 22 

θ=

2

K1 x (10 − θ ) =1 θ K2 x2

Similarly, m2 =

.e en

71.

dυ dx

x1

ad up ra tib

Viscous force, F = η A

(10 − θ ) = K

x1 K1 2 Now = A1 A2 , = = x2 K 2 3

  R2 =× − R2  = T 4π 1000 × 36 π R 2T 100   70.

1000 × 4.2 joule = 1400 J 3

Let θ be the temp. of common surface. As

4 3 4 3 1000 × π= r π R or= r R /10 3 3

= T 1000 × 4π r 2 − 4π R 2 

W 1 W= η Q1= ×1000 cals. Q1 3

η=

=

74.

 Y1 A   Y2 A  YA  l   l  2Y1Y2 = ∴ = or Y Y A Y A 2l Y1 + Y2 1 + 2 l l 69.

73.

ha .n et

k=

1 2 1 π 2 πr π= joule = (1) 2 2 2

=

positive. Wnet = area of semicircle ABC 23

E −13.6 As Em =21 ⇒ E4 = 2 = −0.85 ev . 4 n

Energy required to take hydrogen atom

E=

78.

According to Bohr’s postulate of stationary

from ground state to n = 4 is

orbits

∆E =E4 − E1 =−0.85 − ( −13.6) eV

nh 2h h mvR = ⇒ mvR = ( or ) mv = 2π 2π πR

= 12.75eV

∴ deBroglie wavelength,

hc = 12.75 ×1.6 ×10−19

λ

−34

= λ

8

6.6 × 10 × 3 × 10 12.75 × 1.6 × 10−19

= 6.64 Ao

970.6 Ao ⇒λ = In steady state, there will be no current

79.

through the branch containing capacitor.

ad up ra tib

circuit. The current through different arms

will be as shown in figure. Potential I difference across C and D is VC − VD =R 3

= hv + ( K .Emax )1

=

VD − VF = IR.

R

D

I

E

Statement – 2 → Incident light intensity is

R

R

I = nhv.

F

If ‘ υ ’ becomes = 2v,

I

= I 1 nh = then ( 2v ) 2 I

∴VC − VF = (VC − VD ) + (VD − VF ) IR 4 = + I R = IR 3 3 As VF = VG . since no current in arm FG VC − VG=

Thus Statement – 1 : is false

G

.e en I 3

w w

I

2I 3

C

w

A

i.e., ( K .Emax )2 > 2 ( K .Emax )1 ,

R

R

Thus statement-2 is true. 80.

we have

4 I R= 12 ( or ) IR= 9 3

2I 5 R + IR= IR 3 3

Object distance, U= x1 + f

and Image

distance, V= x2 + f , using mirror formula, 1 1 1 1 1 = + = + t u v x1 + f x2 + f

on simplification, we get f = x1 x2

In the closed circuit EADFE we have E=

( K .Emax )1 + φ0 + ( K .Emax )1

= 2 ( K .Emax )1 + φ0

potential difference across D and F is

I 3

Statement-1 : ( K .Emax )= hv − φ0 1

h ( 2v) − φ0 ( K .E= max ) 2 = h ( 2v) −  hv − ( k .Emax )1 

Let ‘I’ be the current supplied to the

B

h R 3.14 × 2.116 = π= mv

ha .n et

λ= 77.

5 × 9 = 15V (∴ IR = 9 ) 3

81.

Percentage of material left after 5 days = 100 – 10 = 90% 24

Fraction of material left after 5 days= f

As X = ,L WL= L

90 = 0.9 . 100

L Number of fractions involved in 20 days = 20 = 4 5

for material left behind, = n N As = NO

N ⇒ = NO

( 0.9 )

4

Coverage range, d = 2 Rh (or) d ∝ h . d1 ∴ = d

= 0.656

h1 ⇒ 2= h

h1 ( or ) h=1 4h h

∴ h1 =4 ×160 =640m

N × 100 = 65.6% NO

Increase in height of tower = h1 − h

If I 1 is intensity of each of each source of

= 640 − 160 = 480m

light, then

I max= I= 4 I 1 . When path 0

86.

difference, ∆X = then phase difference, 6 2π

λ

∆X ⇒ ∆φ=

π

electric flux and algebraic sum of charges

3

enclosed is zero.

Therefore, resultant intensisty, I = I 1 + I 1 + 2 I 1I 1

presence of ' q2 ' does not a effect the

ad up ra tib

∆φ=

87.

π  cos   3

1 I = 2 I 1 + 2 I 1 × ⇒ I = 3I 1 2 1 I 3I 3 ∴ = = 1 I0 4I 4

.e en

B=

w

∆Vi 0.02 ∆I in = ∆I b = = =× 4 10−5 A Ri 500

w w

∆V ∆I C = β ∆I B = β i = 50 × 4 × 10−5 Ri

µ= r

P 50 V 100 84. = = 0.5 A &= R = = 200Ω V 100 I 0.5 when operated with a c source, EV EV 220 = = = 440Ω 0.5 IV I Z 2 − R2

φ

B 1.2 = 1.2T & µ= = A H 1600

µ 1.2 = = 597 µ0 1600 × 4π ×10−7

χ = µr − 1= 597 − 1= 596 88.

Plates connected across a battery

89.

ax= P, a=

P ML−1T −2 = = ML−2T −2 x L

b 2 c= t = T 2 ; 2= P T

= 2 × 10−3 A = 2mA

∴ R 2 + X L2 = Z 2 ⇒ X L =

Hence, H = 1600 Am −1 ,= φ 2.4 ×10−5 wb A 0.2 ×10−4 m 2 =

= β 50, R = 500Ω I

= Z

The electric field will be there due to all the charges. It is a separate matter that

λ

83.

4402 − 2002 = 1.25 H . 2π × 50

ha .n et

82.

(f)

n

85.

Z 2 − R2 2π v

XL = w

or b= PT 2 = ML−1T −2 × T 2 = ML−1 90.

Node will occur where amplitude of stationary wave is zero. It will be so if cos

πx 25

= 0 (or)

πx 25

=

π 3π 5π

, , ,...... (or) 2 2 2

x = 12.5 cm, 37.5 cm, 62.5 cm……etc. Thus node will not occur at x = 25 cm. 25