eng model test 2

This paper contains 80 questions from Mathematics, 40 questions each fromPhysics and Chemistry.

Each of the 160 questions carries one mark. No negative marking for wronganswers.

Maximum trying allowed is 3 hours (180 minuites) Use Ball - Point Pen while entering the Hall - Ticket Number and filling in Part - A

of the First page.

Use H.B. Pencil only to darken the circle on OMR Answer Sheet. Over - writing or blackening of more than one circle will not count for marks. If you wish to change your answer, erase the already darkened circle completely

and then darken the appropriate circle.

Candidates are prohibited from carrying any sheet of paper to the ExaminationHall except the Hall - Ticket.

Do any rough / scratch work on the Test Paper itself. Calculators, watches with calculators, pagers & cellular phones will not be allowed

into the Examination Hall.

Candidates have to write suitable answers on the Answer Sheet only. Candidates have to return the Answer Sheets and the Question Papers at the time

of leaving the Examination Hall.

Candidates can leave the Examination Hall only in the last half - an hour beforethe close of the Test.

Total number of pages of the Test is 48.

INSTRUCTIONS TO CANDIDATES

EENADU - PRATIBHA

EAMCET GRAND TEST (ENGINEERING)

The questions of this paper were set by Senior Faculty Members ofSri Chaitanya EAMCET Coaching Centre, Vijayawada.

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MATHEMATICS

1. The domain of the function 2

2

log x 3f x

x 3x 2

is

22log x 3

f xx 3x 2

"xH ^O

1) R {1, 2} 2) 2, 3) R {1, 2, 3} 4) 3, 1, 2

2. If f(x) : R R satisfies the condition f(x + y) = f(x) + f(y) for all x, y R then f(x) isx, y R H~, f(x) : R R "O, f(x + y) = f(x) + f(y) x=x \, f(x) J##k1) zero every where ZC_ #O 2) an even function i"O3) an odd function "O 4) cannot be determine H#Q#=

3. If the sum of n terms of an A.P. is 24n 3n4

then nth term of the A.P. is____

'n ^ JOH} " `= 24n 3n4

J~ ` JOH}x n= ^=

1) 5n 14

2) 8n 74

3) 23n 24

4) 4n 85

4. If i 2 j 3k,3i 2 j k are sides of a parallelogram then a unit vector parallel to one of the diagonalsof the parallelogram is

i 2 j 3k,3i 2 j k k *QQ =O `~ K` ~[O U ^

7. If a 3i 5 j and b 6i 3j are two vectors and c is a vector such that c a b then a : b : c

a 3i 5 j , b 6i 3j ~O_ k =i c a b J~ ` a : b : c 1) 34 : 45 : 39 2) 34 : 45 : 39 3) 34 : 39 : 45 4) 39 : 35 : 34

8. Let ai b j ck, bi c j ak and ci a j bk be three coplanar vectors with a b ,and V i j k then V is perpendicular to

ai b j ck, bi c j ak =i ci a j bk `b k a b ,=iV i j k ,J~# V k O|OQ LO_ k1) 2) 3) 4) All the above

9. If the roots of the equation ax2 + ax + c = 0 are in the ratio p : q then p qq p

ax2 + ax + c = 0 gH~} = x+u p : q J~ ` p qq p

1) ac

2) ca

3) ac 4) c

a

10. If the roots of the equation x2 2ax + a2 + a 3 = 0 are real and less than 3, then x2 2ax + a2 + a 3 = 0 gH~} = "", =i 3H# `= J~ `.1) a < 2 2) 2 a 3 3) 3 a 4 4) a > 4

11. If , , are the roots of the equation x3 + 4x + 1 = 0 then 1 1 1 =

x3 + 4x + 1 = 0 gH~}xH = , , ~# 1 1 1 =1) 2 2) 2 3) 4 4) 1

12. If 2

A2

and 3A 125 , then the value of is

2A

2

=i

3A 125J~ ` q=

1) 1 2) 2 3) 3 4) 5

13. If

2x 1 2A 1 3x 3

2 3 6 2x

is a matrix such that trace (A) = 0 then values of x are

2x 1 2A 1 3x 3

2 3 6 2x

=i trace (A) (*_) = 0 J~# x q=

1) 2, 3 2) 2, 3 3) 5, 1 4) 6, 1

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14. Inverse of a skew symmetric matrix of odd order is `~Qu, J+= =uH q=O

1) a symmetric matrix XH += =uH2) a skew symmetric matrix XH J+= =uH3) diagonal matrix qH~ =uH 4) does not exist ==` O H ^

15. If the system of equations x+2y+3z= x , 3x+y+2z= y , 2x+3y+z= z has non trivial solution then=x + 2y + 3z = x , 3x + y + 2z = y , 2x + 3y + z = z =1) 6 2) 12 3) 18 4) 16

16. The rank of

1 1 11 1 11 1 1

is

1 1 11 1 11 1 1

=uH H\

1) 0 2) 1 3) 2 4) 317. A polygon has 44 diagonals, the number of its sides areXH H= lx H~ 44, JO ^ * OY

1) 13 2) 12 3) 11 4) 1018. How many different nine digit numbers can be formed from the number 223355888 by rearranging its

digits so that the odd digits occupy even positions223355888 x JOH# uiy J=i#, J=iH JOH, i

22.

2

22

x 1 A Bx Cx x 1x x 1

1 Acos

C

=

1) 6

2) 4

3) 3

4) 2

23.

n 1C n,0 C n,1 ..... C n, n

P n, n

d

=

1) e2 2) e2 + 1 3) e2 1 4) 1e

24. If A is a square matrix such that A(adj A) = 4(I3). Then adj(adj A) = ____A XH K` ~ =uH =i A(adj A) = 4(I3). J~ ` adj(adj A) = ____1) 4A 2) A 3) 16A 4) 64A

25. If n2 1

, then (J~#)

2 n 1cos cos 2 .cos 2 ........cos 2 =

1) n12 2) cos 3) 2 4) 2

n

26. If , , , are solutions of the equation tan 3 tan 34

, no two of which have equal tan-

gents then the value of tan tan tan tan =

tan 3 tan 34

gH~}xH , , , ^# J~ ` tan tan tan tan =

1) 1 2) 1 3) 2 4) 0

27. The most general value of satisfying 3

tan tan 24

is

3tan tan 2

4

J~# ^~} ^#

1) n 3

2) 2n 3

3) 2n 3

4) n2n 1 3

28. In a ABC if A = tan12, B = tan13 then C = ABC A = tan12, B = tan13 J~# C =

1) 3

2) 4

3) 6

4) 34

29. If x is an acute angle and xy log tan

4 2

then cosx.coshy =

x H}= =ixy log tan

4 2

J~ `, cosx.coshy =

1) 0 2) 1 3) 1 4) 2

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30. A tree is broken by wind, its upper part touches the ground at a point 10 meters from the foot of thetree and makes an angle 450 with the ground the entire length of the tree isXH K@ QeH qiy H# K@ ^O#O_ 10gII ^ ~O

38. The vertices of a triangle are (0, 0), 3,3 , 3,3 then the incentre is(0, 0), 3,3 , 3,3 j~Q Q u[ JO`~=$`HO^O1) (0, 2) 2) (2, 0) 3) (1, 1) 4) (1, 2)

39. If ( 4, 5) is one vertex and 7x y + 8 = 0 is one diagonal of a square then equation of the seconddiagonal isXH K` ~O XH j~O ( 4, 5) , XH H~O 7x y + 8 = 0 J~ ` ~O_= H~O gH~}O1) x + 7y 31 = 0 2) x + 7y 15 = 0 3) x + 7y + 8 = 0 4) 7x y 31 = 0

40. The distance of the point (2, 3) from the line 2x 3y + 9 = 0 measured along a line x y + 1 = 0x y + 1 = 0 ~Y "O|_, 2x 3y + 9 = 0 ~Y #O_ (2, 3) aO ^= Q ^~O.1) 2 2) 2 3) 2 2 4) 4 2

41. The distance from a point , to a pair of lines passing through the origin is d then equation to thepair of lines is

=aO ^=#O_ ` , #O_ d ^ ~O LO_ ~~Y QgH~}O

1) 2 2 2 2x y d x y 2) 2 2 2 2x y d x y

3) 2 2 2 2x y d x y 4) 2 2 2 2x y d x y 42. The circum radius of the triangle ABC with vertices A(2, 1, 1), B(1, 3, 5), C(3, 4,4) is

A(2, 1, 1), B(1, 3, 5), C(3, 4,4) j~Q Q u*O i=$ ` "~O

1) 62

2) 352

3) 412

4) 4143. If the line joining the points (2, 3, 4), (0, 1, 2) is perpendicular to the line joining the points (x, 0, 4),

(7, 4, 3) then x =(2, 3, 4), (0, 1, 2) aO^=# H ~Y (x, 0, 4), (7, 4, 3) aO ^=# H ~Y O|OQ LO> xq=

1) 1 2) 2 3) 3 4) 444. The equation of the perpendicular bisecting plane of the line segment joining (3, 1, 2), (7, 5, 4) is

(3, 1, 2), (7, 5, 4) aO ^=# H ~M YO_xH O|=k YO_# ` gH~}O1) 5x + 2y + z 19 = 0 2) 5x 2y z + 17 = 03) 5x + 2y z + 17 = 0 4) 5x 2y + z 7 = 0

45. If the lengths of the tangents from two points A, B to a circle are 6,7 respectively. If A, B are conjugatepoints then AB =A, B J

46. If the circle x2 + y2 + 4x + 22y + c = 0 bisects the circumference of the circle x2 + y2 2x + 8yd= 0then c + d =x2 + y2 + 4x + 22y + c = 0 =$ `O x2 + y2 2x + 8y d = 0 J# =$ `O ikx =kYO_# K c + d =1) 60 2) 50 3) 40 4) 56

47. If (1, 2), (4, 3) are the limiting points of a coaxal system then the equation of the circle in its conjugatesystem having minimum area is(1, 2), (4, 3) aO^= J=k aO^=QQ =$` ~} H OQ ~}H KOk, Hx+"OQ=$ ` gH~}O

1) x2 + y2 2x 4y + 5 = 0 2) x2 + y2 8x 6y + 25 = 03) x2 + y2 5x 5y + 10 = 0 4) x2 + y2 + 5x + 5y 10 = 0

48. If (2, 3, 5) is the one end of a diameter of a sphere x2 + y2 + z2 6x 12y 2z + 20 = 0 then the otherend of diameter isx2 + y2 + z2 6x 12y 2z + 20 = 0 QO (2, 3, 5) aO ^= XH "QO J~ ` ~O_= z=i aO ^=1) ( 4, 9, 3) 2) (4, 9, 3) 3) (4, 9, 3) 4) (4, 9, 3)

49. If PSP| is a focal chord of a parabola y2 = 4ax and SL is its semi latusrectum then SP, SL, SP| are iny2 = 4ax ~=xH PSP| 5

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54. The angle between the circles r a cos( ), f b sin( ) R B R B isr a cos( ), f b sin( ) R B R B =$` = ^ H}O

1) 2 2) 3) 2

4) 00

55. The quadratic equation whose roots are ,m where 3

0

3sin 4sinLtR

l

and

202 tan

m Lt1 tanR

l

is

3

0

3sin 4sinLtR

l

=i 20

2 tanm Lt

1 tanR

l

,m =Q Q =~ gH~}O

1) x2 + 5x + 6 = 0 2) x2 5x + 6 = 0 3) x2 5x 6 = 0 4) x2 + 5x 6 = 0

56. The points of discontinuity of f(x) =1

log x are

f(x) =1

log x "O qz#= aO^=

1) 0 , 2 2) 1, 2 3) 1,0 4) 0 , 3

57. I : If f(x) = 2a

axax

, then (J~#) f | (a) = 0.

II : If f(x) = x x2 + x3 x4 + ....... , x 1 , then (J~#) f | (x) = 21

1 x .

1) Only I is true I = `" x[O 2) Only II is true II = `" x[O3) Both I and II are true I, II x[O 4) Neither I nor II true I, II ~O_ x[O H ^

58. If y = sin x sin x sin x ..... then (J~#) dydx =

1) sin x2y 1 2)

cos x

2y 1 3) cos x

1 2y 4) sin x

1 2y

59. The function f(x) 1 sin x isf(x) 1 sin x J

60. If 2 2

12 2

x yu sin

x y

then (J~#) xux + yuy =

1) sinu 2) 0 3) cosu 4) cotu61. The radius of a circular plate is increasing at the rate of 0.01 cm/sec when radius is 12 cm, then the rate

of increase in its area isXH =$ ` "~O ~Q ^ 0.01 cm/sec J~ ` "~O 12 cm L#_ ^ x "O ~Q^ ~@1) 0.24 sq. cm/sec 2) 6.0 sq. cm/sec 3) 24 sq. cm/sec 4) 1.2 sq. cm/sec

62. If there is an error of 0.05 cm while measuring the side of an equilateral triangle as 'a' cm. Then thepercentage error in area isXH = u[O [O 'a' Q HK@ ^+O 0.05 cm J~ ` ^x "O ^+ `O

1) 103a 2) 53a 3)

10a

4) 5a

63. The sub tangent, ordinate and sub normal to the parabola y2 = 4ax at a point different from origin areiny2 = 4ax ~=O (0, 0) ^ Q~ ` , U ^

69.

x

3

04x 0

sin t dtLt

xl

=

1) 2.5 2) 5.2 3) 0.5 4) 0.25

70. If 3x sin xf x

1 2

then (J~#) a

a

f x dx

1) 0 2) 12 3) 3 4) 12

71. The area enclosed between the curves y2 = x and y = x isy2 = x, y = x =H = ^ "O

1) 23 2) 1 3) 16 4)

13

72. Solution of differential equation 2 2 2 2y xy 2x y dx x xy x y dy 0 is

2 2 2 2y xy 2x y dx x xy x y dy 0 J=H#gH~} ^#

1) 12 log x log y Cxy

2) 12 log y log x Cxy

3) 12log x log y Cxy

4) 12 log y log x Cxy

73. The solution of 22 2

dy 2x 1ydx 1 x 1 x

given y = 0, x = 1 is

y = 0, x = 1 J~# 22 2

dy 2x 1ydx 1 x 1 x

^#

1) 2 1y 1 x tan x 4 2) 2 1y 1 x tan x 4

3) 2 1y 1 x tan x 4 4) 2 1x 1 x tan x 4

74. A fair die is rolled, the probability that the 1st time 1 occurs at even throws isXH zH# iOY~ ZQ~"#_ " ^\i 1 =K@# O= `

1) 16 2) 5

11 3) 6

11 4) 5

3675. A bag contains 4 balls, two balls are drawn and found to be white, the probability that all the balls are

white isXH Oz 4 |O `

76. If the range of the random variable X is {0, 1, 2, 3 ........} with P(X = K) = KK 1

a3

, for K 0then a =

XH K~t X H " {0, 1, 2, 3 ........} =i P(X = K) = KK 1

a3 J~ ` a =

1) 23 2) 49 3)

827 4)

1681

77. Out of 10,000 families with 4 children each the probable number of families all of whose children aredaughters is10,000 @O u @O|O 4 Q~ LO>, "i JO ^~ P_ Hey=O_ @O

OY

1) 625 2) 1250 3) 2500 4) 9375

78. 1f x

1 x

for 0 x 1 and the interval (0, 1) is divided into 2 equal subintervals using trapezoi-

dal rule, the value of 1

0

dx1 x is

1f x1 x

, 0 x 1 =i (0, 1) # ~O_ = JO `~Q qlOz#_,

1

0

dx1 x q=,

>*~_ x=x J#iOz

1) 1716 2) 1712 3)

1724 4)

1748

79. 5 5 56n1Lt n 1 n 2 ....... 2nnld

=

1) 0 2) 212 3) 312 4)

323

80. Assertion (A) : The unknown coefficient of the equation x2 + bx + 3 = 0 is determined by throwingan ordinary six faced die, then the probability that the equation has real root is 1/2.^$_"Y (A) : x2 + bx + 3 = 0 gH~}O b q=# ^~} zH# ^ iOz H#Q#K b2 4ac 0. J~#The the correct answer is i# = ^#=1) A and R are true and R is the correct explanation of A. A =i R i#q. =i R J##k A i# q=~}.2) A and R are true and R is not the correct explanation of A. A =i R i#q. =i R J##k A i# q=~}H ^.3) A is true, R is false. A i#k, R i#k H^.4) A is false, R is true. A i#k H^, R i#k.

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PHYSICS81. Which logic gate is represented by the following combination of logic gatesHOk =O U lH ^~x zOk.

1) OR 2) NAND 3) AND 4) NOR82. Assertion (A): When two spheres carrying same charge but of different radii are connected by a conducting wire, the charge flows from smaller sphere to larger sphere."Y (A): qa# "~ Q ~O_ Q XH P"x He L

86. A locomotive of mass m starts moving so that its velocity varies as V K S , where K isa constant and S, is the distance traversed. The total work done by all the forces acting onthe locomotive during the first t seconds after the start of motion ism ^=~t Q "#O, ^ x "QO V K S gH~}x J#iOz KeOk. K ~~t =i S"#O }Oz# ^~O. "#O | ^i# =O #O_ " ^\ t H# HO "#OxK Jx | ^$ [iy# x

1) 4 212 m K t 2) 4 21

4m K t 3) 4 218 m K t 4)

4 2116

m K t

87. Particles of masses m,2m,3m,......nm grams are placed on the same line at distances,2 ,3 ,...l l l nl cm from a fixed point. The distance of centre of mass of the particles from the

fixed point in centimetre ism,2m,3m,......nm Q= ^=~t Q H} XH ~ aO ^= #O_ =~Q ,2 ,3 ,...l l l nl OIIg^~ L

92. The amount of work done in lifting a body of mass m from the surface of the earth to aheight equal to twice the radius of the earth isq Li `O #O_, "~xH ~O_ ~@ Z ` m ^=~t Q ==# fHx"O> [iy#

x

1) 3 2GMm

R2) 23

GMmR

3) 53GMm

R4) 35

GMmR

93. Two simple pendulums of lengths 100m and 121m start swinging together in the samedirection with same phase. They will swing together again in same phase after100 gII =i 121 gII _= Q ~O_ H XH k XH ^ ` , XH i HO#O KO ^ `

97. Two litre glass flask contains some mercury.It is found that at all temperatures the volume ofthe air inside the flask remains the same. The volume of mercury inside the flask is

06 1 4 0 19 10 1.8 10g HgC CB H

s s

~O_ b@~ #i=}O Q QA HO ` ^~O Q ^. Jx LQ` = ^ Q "

#i=}O ~O J~# Q ^~ #i=}O

06 1 4 0 19 10 1.8 10g HgC CB H

s s

1) 1500cc 2) 150 cc 3) 3000 cc 4) 300 cc

98. If an air bubble rises from the bottom of a mercury tank to the top its volume becomes 112

times. When normal pressure is 76cm of Hg then the depth of the Hg tank is ^~ ` > J_Q QO #O_ Qe |_Q H K~@\H ^ x #i=}O ` e #i=}xH

112 ~@J~Ok. ^~} _#O 76 OII ^~ _ ` ZO `?

1) 38cm 2) 132cm 3) 76cm 4) 49cm99. A gas is compressed at a constant pressure of 250 /N m , from a volume 310m to a volume

of 34m . 100J of heat is added to the gas then its internal energy50 #@

102. Two trains move towards each other with the same speed. The speed of sound is 1340ms .

If the height of the tone of the whistle of one of them heard on the other is 98 times theactual frequency, then the speed of each train should be~O_ ~ |O_ XH "QO` =u~Y k XH ^ x

106. A bar magnet of moment M is bent into a shape as shown below. If the length of each partis same, its new magnetic moment will beM =HO Q JO`x fQO =# _=Q

Q HO ^ K|_# PH~O =Oz# H ` JO ` =HO

1) 3M

2) 5M

3) 2M

4) 23 M

107. The period of oscillation of a suspended thin cylindrical magnet is 2 seconds. It isbroken into exactly three equal parts pependicular to its length and three parts are thenplaced on each other with their like poles together the time period of combination isKx H~ JO`x HO# JO ` =HO HOOK# _ " "QO 3.5 g/ J~#_ _= Z^\ X_# K~ Jk Ok# _ ZO`

( 0tan37 3/ 4 )1) 7.5 m 2) 22.5 m 3) 16.5 m 4) 9m

110. The value of current I, in the figure shown will be

HOk =O H~O@ I q=

1) 11A 2) 19 A 3) 13 A 4) 9 A

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111. The balancing lengths of potentiometer wire are 800 cm and 600 cm when two cells ofemfs 1E and 2E are connected in the secondary circuit first in series and then terminals of

one cell is reversed ; 12

EE is equal to

1E =i 2E q.K.| Q ~O_ q ^ `\# >xg@~ QO Q}=O " ^\

}#, ^ XH @O H H## u J#O ^#O K#

1

2

EE q= ZO `? ~O_ +~`

O `# _= =~Q 800 cm =i 600 cm

1) 111 2) 1411 3)

71 4)

43

112. The minimum force required to move a body up an inclined plane is two times the minimumforce required to prevent it from sliding down the plane. If coefficient of friction between the

body and inclined plane is 1/ 3 the angle of inclined plane is" Ò T~ =YOQ ==# KeOK_xH H=# |O, J ^ " Ò *~ `# ==#

x~kOK_xH H=# |xH ~\O. ~} Q}HO

1/ 3 J~# " Ò K H}O

1) 060 2) 045 3) 030 4) 015113. A circular coil of radius 2R is carrying current i . The ratio of magnetic fields at the centre of

the coil and at a point at a distance 6R from the centre of the coil on the axis of the coil is2R "~O Q =$`H~ fQK@ QO_ i H~O@ =Ok. HO ^ aO ^= = ^ =i JH ~Y

HO âO ^= #O_ 6R ^~O Q aO ^= = ^ ~} H ` f= ` = ^ x+u

1) 10 2) 10 10 3) 20 5 4) 20 10114. An ammeter whose resistance is 1808 gives full scale deflection when current is 2mA.

The shunt required to convert into an ammeter of reading 20 mA (in ohm)1808 x~ Ô Q Jg@~ 2mA H~O@ i J=~#O WOk. nxx 20 mA Jg@~Q=~O> H=# +O@ q= F"

1) 18 2) 20 3) 0.1 4) 10115. The wing span of an aeroplane is 20m. It is flying in a field, where the vertical component

of magnetic field of earth is 55 10s tesla, with velocity 360 km/h. The potential differenceproduced between the blades will beq=# ~H _= 20m. =O ` Hu[ OOO 55 10s > Q H Ò q=#O 360 km/h "QO` }Oz#, ~H ~O_ z=~ >x ^ O

1) 0.10 V 2) 0.15 V 3) 0.20 V 4) 0.30 V

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116. An emf 4cos 1000E t volt is applied to an L-R circuit of inductance 3mH and resistance48 . The amplitude of the current in the circuit is3mH ~H`, 48 x~ ^O Q L-R =xH 4cos 1000E t F q.K.| # J#=iO K#H~O@ HO# iqu

1) 1.0A 2) 0.8A 3) 47

A 4) 57 A

117. Two wires of same material have masses in the ratio 3:4 the ratio of their extensions underthe same load if their lengths are in the ratio 9:10 isXH ^ ~O` K|_# ~O_ fQ ^=~ x+u 3:4 =i XH ~x "_n# "\Q ^ x+u ZO `? "\ `e_= x+u 9:101) 5:3 2) 27:40 3) 6:5 4) 27:25

118. A cannon ball is fired with a velocity of 1200ms at an angle of 060 with the horizontal. Atthe highest point of its flight it explodes into 3 equal fragments. One fragment is going ver-tically upwards with a velocity of 100m/s and second going vertically downwards with avelocity of 1100ms The third fragment will be moving with a velocity ofXH |Oux Hu[ =O `~O` 60

0

H}O K

1200ms "QO` HO K~. Jk Qi+ Z ` = ^ 3=QQ q@#O KOk, JO ^ " ^\k x@x=Q H 100m/s "QO`, ~O_=k x@x=QHO^ 100m/s "QO` Ke =_=QO O ^"QO1) 1600 ms in the horizontal direction Hu[ =O `~OQ 1600 ms2) 1300 ms in the horizontal direction Hu[ =O `~OQ 1300 ms3) 1300 ms in the direction making 060 with the horizontal Hu[ =O `~O` 60

0

H}O K

1300 ms4) 1200 ms in the direction making 060 with the horizontal Hu[ =O `~O` 60

0

H}O K

1200 ms119. Match the following HOk "xx [ `~=

List - I List - IIa) pressure _#= e) 2 2 1ML T I b) Latent heat Q+= f) 0 0 1M L T c) Velocity gradient "Q=}` g) 1 2ML T d) Magnetic flux JO ` Ja"O h) 0 2 2M L T 1) a-h b-f c-g d-e 2) a-g b-h c-e d-f3) a-g b-h c-f d-e 4) a-f b-g c-e d-h

120. If the equation of motion of a projectile is 213 ,8y x x the range and maximum heightare respectively ( y and x are in metres)H= ^ gH~}O

213 ,8

y x x (x , y g@~) J~# " =i Qi+ Z ` =~Q

1) 18m and 24m 2) 24m and 18m 3) 24m and 6m 4) 12m and 9m

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CHEMISTRY121. The orbital in which the electron can only absorb but cannot emit radiation isU Pi\x ZH

129. The correct order of oxidation number of nitrogen in its oxidesI) N2O3 II) NO III) N2O5 IV) N2O I > II > IV 3) I > II > III > IV 4) III > II > I > IV130. Assertion (A): Equivalent weight of 4KMnO in acidic medium in 31.6"Y (A): P= #HO 4KMnO ` ~= 31.6Reason (R): In acidic medium the half reaction is 4 2MnO MnOmH~}= (R): P= #HOx J~ K~ 4 2MnO MnOm1) Both A and R are true R explains A A =i R i#q. R, A i# q=~}

2) Both A and R are true R do not explains A A =i R i#q. R , A i# q=~} H ^

3) A is true R is false 4) A is false R is true A i#k Hx R i#k H^ A i#k H^ R i#k

131. 1.36 g of calcium sulphate is present in 10 lit of water. The hardness of water in ppm10 bII Hi# [O 1.36 QII HeO \ H ^. P [ Hi# ` ppm 1) 200 ppm 2) 100 ppm 3) 25 ppm 4) 4.50 ppm

132. Which of the following is not a true peroxide ?HOk "x Uk x["# ~H _ H ^

1) 5CrO 2) 2 2Na O 3) 2PbO 4) 2BaO133. Not possible among the followingD HOk "x Uk [~Q^ ?

1) 2KCl + F2 m 2KF + Cl2 2) 2KI + F2 m 2KF + I23) 2KI + Br2 m 2KBr + I2 4) 2KCl + Br2 m 2KBr + Cl2

134. The largest hydrated cation of the following isHOk "x Ju K sH~} H@

136. List I (\H I) List II (\H II)(Noble gas, L ` $+ "=) (use, LQ=)A) Helium eO 1) Treatment of cancer H#~# #O K\xHB) Neon x

143. A false statement among the following isHOk "x Uk x[O Hx "Y

1) Faraday is the charge carried by one mole of electronsXH " ZH

147. The amount of energy released when 20ml of 0.5M NaOH mixed with 100ml of 0.1MHCl is x KJ, then heat of neutralisation is (KJ / mole)20ml, 0.5M NaOH =i 100ml, 0.1M HCl HQ x KJ L+= "=_#k. J~ ``@H~}+= q= (KJ / mole)

1) +100x 2) -50x 3) +50x 4) -100x148. Which among the following statements is correct

HOk "x Uk i# "Y

1) All VA group elements are tetratomic except bismuth a ` HO_ qye# VA =Hx K` ~ ~=}H J}=

2) Formation of ozone from oxygen is endothermic PH[

151. The couplings between base units of DNA through

DNA x H~ = ^ LO_ |

1) Vanderwaals forces 2) Electrostatic attractions "O_~" | q ^^H~} |

3) Covalent bonding 4) Hydrogen bonding O[h |O ^ _[ water W^ P> h~3) ethyl alcohol > water > phenol W^ P > h~ > phenol > ethyl alcohol h~ > W^ P

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155. Which of following responds to Aldol condensationHOk U "#O P O##O ##

1) HCHO 2) 3CH CHO 3) 6 5C H CHO 4) 3CCl CHO

156. 42

)6 5 2 5 ).

i KMnO OHii H H OC H C H X

}}}}}m . Here, compound X is

4

2

)6 5 2 5 ).

i KMnO OHii H H OC H C H X

}}}}}m . WK@ "#O X J##k

1) 6 5 2.C H CH COOH 2) 6 5C H COOH3) 6 5C H CHO 4) 6 5 2C H CH CHO

157. Nitro benzene undergoes reduction with Zn and .alc KOH to form a compound A. Thenumber of 'sigma' and 'pi' bonds in a molecule of A respectively are

1) 4 2) 3 3) 2 4) 1 5) 3 6) 3 7) 2 8) 4 9) 1 10) 1

11) 3 12) 3 13) 1 14) 4 15) 1 16) 4 17) 3 18) 3 19) 3 20) 1

21) 2 22) 3 23) 3 24) 1 25) 1 26) 4 27) 1 28) 2 29) 3 30) 3

31) 1 32) 3 33) 4 34) 2 35) 4 36) 3 37) 1 38) 1 39) 1 40) 4

41) 3 42) 3 43) 2 44) 1 45) 2 46) 2 47) 3 48) 2 49) 3 50) 1

51) 4 52) 1 53) 3 54) 3 55) 2 56) 3 57) 3 58) 2 59) 3 60) 2

61) 1 62) 3 63) 2 64) 2 65) 3 66) 2 67) 1 68) 1 69) 4 70) 1

71) 3 72) 1 73) 1 74) 2 75) 2 76) 2 77) 1 78) 3 79) 2 80) 1

81) 3 82) 1 83) 2 84) 2 85) 3 86) 3 87) 1 88) 3 89) 1 90) 1

91) 3 92) 2 93) 1 94) 4 95) 3 96) 1 97) 4 98) 1 99) 1 100) 2

101) 3 102) 1 103) 3 104) 1 105) 1 106) 2 107) 2 108) 3 109) 1 110) 1

111) 3 112) 1 113) 2 114) 2 115) 1 116) 2 117) 4 118) 2 119) 3 120) 2

121) 2 122) 3 123) 4 124) 2 125) 4 126) 4 127) 3 128) 2 129) 2 130) 3

131) 2 132) 3 133) 4 134) 1 135) 3 136) 2 137) 3 138) 4 139) 1 140) 3

141) 2 142) 4 143) 2 144) 1 145) 3 146) 1 147) 4 148) 2 149) 4 150) 1

151) 4 152) 4 153) 4 154) 1 155) 2 156) 2 157) 1 158) 1 159) 3 160) 2

EENADU - PRATIBHA


KEY SHEET

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Page No : 29

EENADU - PRATIBHA


HINTS & SOLUTIONSMaths

1. x + 3 > 0 and x2 + 3x + 2 x 0x > 3 x 1, 2x =

domain = ( 3, ) { 1, 2} d 2. f(x+y) = f(x) + f(y)

f(x) = kx, k Rf(x) = kx = f(x)

f(x) is odd function

3. Given 2

n

4n 3nS4

2 2

n n n 14n 3n 4(n 1) 3(n 1)

t S S4 4

4(2n 1) 3 8n 74 4

4.OA i 2J 3K

OB OA AB 4i 4J 4KAB 3i 2J K

=

Has direction unit vector = OB i J K

3OB

p

B>

A

C

O

5.

A

B CDAB ACAD 4i J 4K 33

2

6. BD AD AB (4a 5b)

Now

2 22BD 16 a 25 b 40 a b cos a, b B

>

A

CD

= 128 + 225 + 243 = 593

7. a 3i 5J a 9 25 34

b 6i 3J b 36 9 45

i J Kc a b 3 5 0 K

6 3 0 s

( 9+ 30) c 39

a : b : c 34 : 45 : 39

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Page No : 30

8. [ , , ] 0 B C H a b cb c a 0c a b

a b c 0

v. v. v. a b c 0= B C H

9. Ratio of the roots 2 2p : q pqa ac(p q) 2(p q) a q p a

pq c p q c

10. roots real 0% r a 3b

3B a 3 a 3 p 3 a (3 a) p 1 3 ap

3a > 1 3 a > 0

a < 2 a < 3

11.

0

4

1

B

BC

BCH

Now 1 1 1( ) ( ) ( ) B C C H H B

=

1 1 1

H B C

= 4BC

BCH

12. 3A 125 A 5 2 4 5B 3B p

13. x2 3x + 6 2x = 0

x2 5x + 6 = 0

x = 2 or 3

14. If A is a skew symetric odd order matrix A 0 =

Inverse of A does not exists

15. Non-trivial solution 1 2 3

3 1 2 02 3 1

M

M

M

, C1 mC1 + C2 + C3

1 2 3(6 ) 1 1 2 0

1 3 1 M M

M

6M

16.

1 1 11 1 1 11 1 1

(2) + 1 (0) + 1(2) = 4x

0

Rank = 3

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Page No : 31

17.n(n 3) 44

2

n(n3) = 88 = 11s

8

n = 11

18. ___ , ___ ,___ ,___ ,___ ,___ ,___ ,___ ,___ , = 4! 5!

2!2! 2! 3!s

s

odd digits

3, 3, 5, 5 = 6 s

10even digits

2, 2, 8, 8, 8 = 60

19.r100 r

100 54r 1 rT C 5 4

100 r r,

4 5

are integers if r = 0, 20, 40, 60, 80, 100

=

number of rational terms = 6

20. (1 + x)6 [1 + (1 + x) + (1 + x)2 + ............. + (1 + x)9 ]10

6 (1 x) 1(1 x)1 x 1

610(1 x) [(1 x) 1]

x

16 6(1 x) (1 x)x x

Req coet = 16C7 = 16C9

21. Let 3 3.5 3.5.7S .......

4.8 4.8.12 4.8.12.16 d

2 3 41.3 1 1.3.5 1 1.3.5.7 1S . .......1.2 4 1.2.3 4 1.2.3.4 4

d

1/ 2 1/ 23 1 3 2S 14 2 2 3

2 3S3 4

22.2

2 2 2x 1 2x 1 2 A Bx c

x xx(x 1) x 1 x 1

A = 1, C = 2 1 1A 1

cos cosC 2 3

Q

23.n

2 3 n

n 1

2 1 1 1 1.2 (2) (2) ......... .2

n! 1! 2! 3! n!

d

= e2 1

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Page No : 32

24. A(adj A) = |A|.I |A| = 4Now adj(adj A) = (|A|)n2 .A = 432 .A = 4A

25. 2 n 1cos .cos2 .cos2 .........cos2 R R R R

n

nn n

n n4 n

n n

2sin sin2 1sin(2 ) 12 1

2 .sin 22 .sin 2 sin2 1 2 1

Q

Q

R

Q Q

R

26.3

21 tan 3 tan tan31 tan 1 3tan

R R R

R

R

1 + tanR

3tan2R

3tan3R

= (9tanR

3tan3R

) (1tanR

)

1 + tanR

3tan2R

3tan3R

= 9tanR

3tan3R

9tan2R

3tan4R

3 tan4R

6tan2R

+8tanR

1 = 0

tan 0 B

27.1 tan

tan 21 tan R

R

R

tanR

tan2R

1 + tanR

= 2 + 2tanR

2 2tan 3 ( 3) R n 3Q

R Q p

28. A + B + C = Q

A + B = Q

Ctan A tan B

tan C1 tan A tan B

tanC = 1 C 4Q

29.y

x1 tan2ex1 tan2

y

x1 tan2ex1 tan2

2y y

2

x1 tan2e e 2x1 tan2

1cosh y

cos x cos x.cosh y 1

30. h1

h2

h2

45010

h1 = 10, h22 = 200 2h 2 10 s

=

Req height = h1 + h2 = 10 ( 2 1)

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Page No : 33

31. 2R. tanA = 2R tanB = 2R tanc e%

l is equilateral

32.

A

B Ca

P1 11

a.p2%

1

1 aP 2

%

Similarly 2 3

1 b 1 C.

P 2 P 2

% %

But 2 1 32P P P 2 1 1 H.P.b a c

33. 2x or X X

x X 21

x 1x XX

2 22

1x 1

x X X

33

1x 1 1 2

x

s

2 4 16 1 31 i 3 2 i

2 2 2

p

B p p

2(cos isin ) 2(cos isin )3 3 3 3Q Q Q Q

B C

4 4 4 n 1n n2 2cos 2 C os3 3

Q Q

B C

n n 1 n n 3 n n 5 51 3 5sin n C cos sin C cos sin C cos sin ...................

R R R R R R R

5 3 3 5sin 6 6 cos sin 20 cos sin 6 cos sinR R R R R R R

y2 = 4x37. Length of AD = 2

=

Length of the side = 2 2 43 3s

B D C

A(1,2)

600 3x-4y+15=02

38.

(0, 0)

I12 12

( 3,3)( 3,3) 12Inentre = centroid (G) = (0, 2)

39.

7x -y+8= 0(-4, 5)

x +7y+K= 0

K = 31

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Page No : 34

40. x y + 1 = 0

m = 1

R

= 450P(x1 y1) = (2, 3) Let the Req distance be r

1 12 r. . 3 r.2 2

Lie on 2x 3y + 9 = 0

3r4 r 2 9 9 02 9

r4 02 r 4 2

41. Let P(x1 y1) be any point on the Req locus, OA PQ?

Area of e1 1OPQ x y OP d2 2% C B s s

l

O(0, 0)

P(x, y)A

B

Qd ( , )B C

2 2 2 2( x y) (x y )d C B

42. 2AB 1 4 36 41

BC 4 1 1 6 AB2 = BC2 = AC2

AC 1 9 25 35 Req circumradius = 412

43. AB dr's (l1 m1 n1) = (2, 2, 2)

CD dr's (l2 m2 n2) = (x 7, 4, 1)lrAB CD? l1 l2 + m1 m2 + n1 n2 = 0

2 (x 7) + 8 + 2 = 0

x = 2

44. AB dr's

(a, b, c) = (10, 4, 2)Mid point of AB = ( 2, 3, 3)Req plane eq is 10(x 2) + 4(y 3) + 2(z 3) = 0

10 x + 4y + 2z = 20 + 12 + 6

A(-3,1,2)

B(7, 5, 4) 5x + 24 + z 19 = 045. AB are conjugate point w.r.to the given circle

2 2 2AB 6 7 AB 36 49 85

46. R.A.

6x + 14y + c+ d = 0 (1, 4) lie on R.A.

6 56 + c + d = 0

c + d = 50

47. (1, 2) (4, 3) as diameter end points of a circle

(x 1) ( x 4) + ( y 2) (y 3) = 0

x2 + y2 5x 5y + 10 = 0

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Page No : 35

48. (2, 3, 5)

(3, 6, 11)

A C B

B = 2C A = (4, 9,3)

49.

PL

S

L1 P1y2=4ax

We know by polar coordinates 11 1 2

SP SP

l

11 1 2

SP SLSP

H.P

50.. Put X = x + a

y2 = 4aXTangent becomes

y = m(X a) + C = mX + C am

Now Req condition is aC amm

1C a(m )m

51.O 450

L

L1S

20 b / atan 45

ae

22

2b

e 1 ea

e2 + e 1 = 01 1 4 5 1

e2 2 p p

=

e < 15 1

e2

52. Req r?

l distinces product =

1 1

2 2

x ya b1 1a b

.

1 1

2 2

x ya b1 1a b

2 2

2 2

2 2

1 a b1 1 a ba b

53.2 2

x y 1r 2 5 r

represents an ellepse

r 2 0, 5 r 0

2 r 5

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Page No : 36

54. r = a cos ( )R B 2r ar[cos cos sin sin ] R B R B2 2x y a cos x a sin y 0 B B

r = b sin ( )R B 2r br[sin cos cos sin ] R B R B2 2x y bsin x b cos y 0 B B

Now 2g1g2 + 2f1f2 = ab2

abcos sin cos sin 0

2B B B B

55.0 0

0 0

2 tan tan 2m Lt Lt 2

2 tan tan 2m Lt Lt 2

R R

R R

R R

R R

R R

R R

Req Quadratic equation is

x2 (l+m) x + lm = 0

x2 5 x +6 = 0

56. f(x) discontinuous log x 0 x or undefine

x 1 x or x x

0 x 1.0 p

57. I) 2ax af(x)

ax

2a . x.f(x) ax a

11a. .f(x) a. x.f (x) a2 x

at x = a 11

.f(a) a.f (a) a2

a + a. f1(a) = a

f1(a) = 0

II) 1 f(x) = (1 + x)1 1 21f (x) (1 x)

12

1f (x) (1 x)

58. y2 = sin x + ydy dy2y. cosxdx dx

dy cos xdx 2y 1

59.1 sin x if x 0

f(x)1 sin x if x 0 r

cosx if x 0f (x)

cos x if x 0r

Also x 10 x 10Lt f(x) Lt f(x) f(0)

But f (0 ) f (0 ) x

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Page No : 37

60. Sin u= 2 2

2 2x yx y

sin u is a homogenious function of degree n = 0

By Eular's formula d dn. (sin u) 4. (sin u) 0dx dy

. sin u

x yx.U y.U 0

61. rE = 0.01 sec. r = 12 cm

Area A = 2rQ

AE = 2Q r r 24 0.01E Q s = 0.24 Q

sq cmn/sec

62. aE = 0.05 cm a m side

Equilateral e% l Area 23A a

4

log A log 3 / 4 2 log a dA da2.A a

dA da 10100 2. 100A a a s s

63. y2 = 4axdy2y 4adx

1

2am

y Let P(x1, y1)

(y1.m). y1. 1y

m

are in G.P.

64. 2r + rR

= constant

Kr

2

R

21A r2 R

2

21 KA .2 ( 2)R

R

r

rR

R

r

dA 0d

R

2 2

4K ( 2) 2( 2) 02 ( 2)

R R R

R

2R 65. y = xn1.log x

n 1 n 21

1y x . (n 1).x . log xx

xy1 = xn1

+ (n1)yDiff. (n1) times

n n 1x.y (n 1).y n 1(n 1)! (n 1)y

n

(n 1)!yx

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Page No : 38

66.3 2

2cos x (1 sin x)dx cosxdx

sin x(1 sin x)sin x sin x

(1 sin x)cos xdx

sin x

log sin x sin x c

67. Put Tan1x = t 2dx dt

1 x

R.I. = t 2e (Tant Sec t)dt

te Tan t C

1Tan xxe C

68. Period of x [x] is 1100 1

0 0(x [x])dx 100. (x [x])dx

1

0100. x.dx

1100 0 502

69.

x3

04x 0

sin t dtLt

xm

3

3x 0

1 sin x. Lt

4 xm

1 0.254

70. f(x) = 2x3 sinx

f(x) is odd function a

a

f(x) 0

71. Req Area = 1

0

2 1 1( x x)dx3 2 6

72. xy(ydx + xdy) + x2y2 (2ydx xdy) = 0dividing with x3y3

2d(xy) 1 12. dx dy 0

x y(xy)

integrating 1 2 log | x | log | y | C

xy

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Page No : 39

73. 2 2 2dy 2x 1ydx 1 x (1 x )

I.F =. 2

2x dx 21 xe 1 x

Solution y(1+x2) = 22 21 (1 x )dx

(1 x )

y(1 + x2) = 21 dx

1 x

y(1 + x2) = tan1 x + Cat x =1, y = 0 0 / 4 C Q C / 4 Qy(1+x2) = tan1x / 4Q

74. R.P. 55 1 5 5 5 1 5 1

. .......

6 6 6 6 6 6 6 6

d

2 45 5 51 ...........36 6 6

d

5 1 5. 2536 111

36

75. R.P.

4 42 2

2 3 42 2 2

4 4 42 2 2

1. C / C

31 C 1 C 1 C

. .

3 3 3C C C

4 balls

2w, 2 other balls

3w, 1 other balls

4w

= 3/5

76.k 0

P(x k) 1d

2 32 3 41.a .a a a ....... 13 3 3 d

2 31 1 1

a. 1 2. 3. 4. ...... 13 3 3

d

1 1.1/ 3a. 1

1 1/ 3 4 / 9

3a. 3 / 2 1

4

4a

9

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Page No : 40

77.1 1p ,q2 2 n = 4

44

4 41 1 1P(x 4) C2 162

No. of families = 110,000 625

16s

78.1 0 1h

2 2

0, 1/2, 1

1

0

1 1 1 2 17dx 1 2.1 x 4 2 3 24

79. G.E. 6n1Lt .nmd

n5

r 1(n r)

15 1n 6 65

nr 1 0 0

1 r (1 x) 2 1 21Lt 1 (1 x) dxn n 6 6 2md

80. x2 + bx + 3 = 0 0% r2b 12 0 r

b = 4, 5, 6

=

R.P. = 3 1/ 26

PHYSICS81. A B A.B A.B

82. for a given charge 1VRt

83. Conceptual

84. When B

particle is emitted Z n by 2 & A nby 4

When C

particle is emitted Z n by 2 & A is unchanged.When H particle is emitted Z & A are unchanged.

85. According to L.C.L.M

1 1 2 2 1 1 2 2m u m u m v m v+ = +

( )1 21 1 2 22 1

0 0 1m vm u m vm u

+ = + =

Given 12f i

k k =

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Page No : 41

( )2

2 2 2 12 2 1 1 2

1 2

1 1 1 1. 2

2 2 2 2v m

m v m uu m

= =

from (1) & (2) 21

12

ve

u= =

86. W Fs mas

( )2 2 21 1 12 2

ma at ma t = = Given v k s=

( ) ( )1 1 22 21 22 2 2dv ds k ka k s s k sdt dt

= = = =

From (1) & (2)4 21

8W mk t

87. 1 2 2 21 2

.....

.............

n ncm

n

m x m x m xx

m m m

+ + +=

+ + +

( ) ( )2 2 .........2 ...........

m m nm n

m m nm

+ + +=

+ + +

2 2 21 2 .............1 2 ...........

n

n

+ + + =

+ + +

1 2 1 2 11 36

2

n n n n

n n

88.

here N = FTo move the duster upwards F1 = wt + fF Wt N Wt FN N a

2 0.4 5 4F N = + =89. 2T

n = Ti + Tc Ti = 2 270 10 = 5300C

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Page No : 42

90. PE at the top = KE at the bottom

212

Mgh MV=

( )21 52Mgh M gr=52rh =

91. AC = Range = ut

22 hAC Vg

=

( )2 22 rVg

=

4 rVg

=

92. f iW PE PE=

1 1GMm GMm GMmR h R R R h

= = + +

1 1 22 3

GMmGMmR R R R

= = +

93. 1L

S

n

n=

121 111 100 10

n

n= =

i.e., after the shortest pendulum completes 11 oscillations(or)After the longer pendulum completes 10 oscillations

94.1

2

1 0 0 0

2 0 0 0

2 13 2

KE h hKE h h+ + + +

+ + + +

1 1

2 2

12

V kEV kE

95. 010 20

cos cos60h

mm

= = =

96. Amount of fluid flowing per sec = 2A r =2 2 2 2

1 1 2 2 3 3 4 4: : : 4 :98 : 45:15r r r r =Ans: a,d,c,b

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Page No : 43

97. L L C C L L C CV t V t V VH H H H% % 4 61.8 10 2000 27 10LV =

300LV cc=98. 1 1 2 2PV PV=

1 2H h V HV

1 1376 762

h V V s

h = 38 cm99. dQ du dW

dQ du P V % (work is done on the system dW ve= )( )100 50 10 4du=

400du J= Increases

100.

1 2Q Q Q ( ) ( ) ( )0 90 90KA KA KAL L L

= +

180 2 = 060 c =

101. ( )1nT a N d= + ( )2 64 1 4 252a a a Hz= + =( )15 16 1 252 15 4 312T a d Hz= + = + =

102.( )0

s

v vn n

v v

=

98

v xn n

v x

+ =

( )0 sv v x= =9 340 20 /8 340

xx m s

x

+= =

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Page No : 44

103. A : Image and scale magnified twiceB : Spherical abberation can be reduced by sharing deiviation at different surfaces.

104.

1 2 3

1 1 1 1F f f f= + +

2 2 21 1 1g w gR R RN N N

2 1 2 20.5 0.520 3 20 20

= +

1 1 1 1 1 15 ,20 30 20 15

F cm convexF= + = =

105. Dichroism = Double Refractin + selective absorption

106.

2 21 45 5

R = + =

25 5

m MM = =

107. 3M MMn

= = ------------ (1)

22 112 12 3 3

m mI

= =

27II =

------------- (2)

1 1 2

2 2 1

33

I T I M I MTM T I M I Ma

t

333

27

I MI M

22

2 23 sec3

TT= =

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Page No : 45

108. 2 2F to F 4 F= and is serious to 12 F4 12 3

16F= =

and is parallel to 2 F

3 2 5ABC F = + =

109.

drift (x) =

0sin37R bV V t

( )0 0sin37 cos37R b bdV V

V=

3 603.5 5 45 55

=

0.5 15 7.5m= =

110. Apply Kirchoffs first law at every junction

111.1 2 1 1 1 2

1 2 2 2 1 2

800 600 7800 600 1

E E EE E E

112. 2up downF F=

[ ] [ ]sin cos 2 sin cosMg Mg / + = / / /3 cos sin =tan 3 =

01tan 3 603

= =

113.

20 0

1 2 2 2

(2 )2(2 ) 2 (6 ) (2 )

i i RB BR R R

N N

32 2

13

2

40 80 10 10 10(2 ) 8 RB

B R

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Page No : 46

114. 1

GSn

20 102

G

i MAn

i MA 180 20

9= 8S

115.5 55 10 20 360 0.1

18

s s s s

ve B V V

116. 23 & 4 Z= R 5 = L LX WL L R L X Y r

00

4 0.85

Ei AZ

117.

m Vd A dF

e mAY Ad

= = = =

2

, &F Fe F Y d are samem mdYYd

221 1 2

2 2 1

e me

m e m

=

81 4 27100 3 25

= =

118.

According to L.C.L.M( )1 1 2 2 3 3cosmu m v m v m v = + +

31200 100 1002 3 3 3

m m mm v = +

V3 = 300 m/s in horizontal direction

119.2

1 22

F MLTP ML TA L

= = =

2 22 2Q ML TL L T

m M

= = =

Velocity gradient = 1

1dv LT Tdx L

= =

2 22 2 1F MLT LBA A ML T I

m ILG

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Page No : 47

120. Given equation is of the form 2y Ax Bx

Range = 3 2418

Am

B= =

2 9 1814 48

AH mB

= = =

CHEMISTRY121. 1s can absorb energy, but does not emit

122. M t 1m

. Electron has least mass

123. Configuration of Cr is 1s2 2s2 2p6 3s2 3p6 3d5 4s1

124. 22O has no unpaired electrons.125. Na has larger size and forms cation easily126. 99% Completion of first order reaction takes 6.7 half lives

99.9% completion requires 10 half lives.

127. B AA B

M M2r 1r M M 4

128. Hydrogen bonding between molecules.129. Oxidation state of N is +1 in N2O, +2 in NO, +3 in N2O3 and +5 in N2O5130. Change in oxidation state of Mn is 5.

131. s s 8 8

24

2

weightt of salt 10 1.36 10Hardness 10 ppmGMW H O sample in ml 136 10

132. PbO2 is dioxide. Oxidation state of oxygen is 2133. Br2 cannot displace chloride.134. Smallar the size of ion, more is the hydration ability.135. Si forms four single bonds with sp3 orbitals formed in excited state.136. Liquid He is cryogenic. Beacon lights of Ne for vison through fog. Kr in helmet caps and

Rn in cancer treatment ointments.137. NO + O3 m NO2 + O2138. CH3 has 3 bond pairs and 1 lone pair

139. 2 4x C H= and 2 5y C H OH=2 3

02 5 2 4350Al O

cC H OH C H

140.

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Page No : 48

141. % s 1b b1 2

w 1000T k .m w

s s

6 10000.52 0.57760 90

142. Only Po has simple cubic structure143. Oxidation occurs at anode

2 2H 2OH 2H O 2e

144. 3 2p NH H Sk P .P [ ][ ]3 2ck NH H S=

units = atm. atm = atm2 units =m m

l l 2

m

l

145. Physical adsorption is weak.146. Hot gases leave through chimnee.147. 0.01 moles gives ----- x kJ

1 mole gives --------- ?Then, heat of neutralisation = 100 x kj / mole

148. 3O2 + 69K cal 2O3149. Ligand is is different. 1, 3-prapanediamine and 1, 2-propanediamine150. CH2CHOH is glycolic acid

CH3CHOH COOH is lactic acid151. Hydrogen bonds152. Asprin153. A = C2H5Cl

Ethyl chloride is not used to detect primary amines (Chloroforms is used)

154. principle functional group is COOH

155. Compound which contains - hydrogens undergoes aldol condensation156. C6H5 COOH

157.

27, 6 = =158. C2H5OH oxidises to CH3CHO

CH3CHO is substituted to give CCl3CHO159. Cane sugar is not reducing as anomenic carbon atoms can not involve in isomerisation.

160. Buffer capacity = -number of moles of OH added per litre

change in pH

8/400.2=change in pH

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