Brilliant Tut Test2

8/8/2019 Brilliant Tut Test2

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SOLUTIONSTO

TEST- II

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(Solutions)PHYSICS

SECTION ISINGLE CORRECT CHOICE TYPE

1. VOLUME OF BALL

=m

V

ACCELERATION OF BALL INSIDE THE LIQUID

mm

Fa

weightupthrustnet ==

g

v=0

a

H

h

ORm

mggm

a

=))(3(

= 2G(UPWARDS)

VELOCITY OF BALL WHILE REACHING AT SURFACE

ghahv 42 ==

THE BALL WILL JUMP TO A HEIGHT

hg

gh

g

vH 2

2

4

2

2

===

(C)

2. Maximum heat is released when block reaches till pointA . glv = 220 , glv = 20 (b)

3. Loss in potential energy = gain in kinetic energy

22

22

123

= MLQL ;

ML

Q212=

( ) QLjkQLkji

Ep 3

330

00

===rrr

unit vector along the direction of

is

2

jk

( )jkML

Q 26 =r

(b)

4. BY CONSERVATION OF ENERGY

FOR 0 0 ga =




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2

max

2

2

mg

E

g

vt ==

TIME PERIOD IS 21 2tT + (C)

5. gRMMRTdt 2=

( )2

22

=MR

RdtT R

g

2=

(b)

6. mumu

dtN +=2

, mvmvdtN = ' ,

ghm

2

2

3 = )'( vvm , ghvv 23.0'=

(c)

SECTION II

One or more than one type

7. LET AT ANY INSTANT, THE VELOCITY IS vr

.

jvivv yx +=

rAND ( )BvqF

rrr

=

ixvjxva yx =

rALSO 2

22vvv yx =+

xvdy

dvv

dt

dva x

y

y

y

y ===

yyxx dvvdvv = (I)

xvdy

dvv xyy = (II)

xvdx

dvv

yx

x= (III)

FROM (I), (II) & (III)2

2x

vy = AND4

422 x

vvx =

ALSO,2

vvx =

2

3va = AND B= 0

WHEN VX= 0, X= (2V)

1/2

(B, C)

8. K.E. 2222

8

3

422

1mv

mRmR =

+=

ANGULAR MOMENTUM =

+= 2

22

4

3

42mR

mRmR

(A, C)9. (A, B, C, D)

10. NET = 0

CROSSING = E2RL




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R

lE

++= sin2

cossin2

32

1 AA

AEcrossing

++= sin

2cos2sin

2

22R

RlR

E

Er

A2

A1

A3

( )+= cos2sin2 RlREcrossing

(A3/2)sin

A2

A1

A2cos

(A1/2)sin

A3

(A, B, C)SECTION III

Reasoning type

11. (D)12. (A)13. (A)

14. (A)

SECTION IV

Comprehension Type

PASSAGE I

AA

Ndt

dN)3( = ;30

t

A eNN=

AB

BNN

dt

dN +=1

12

2

11 =A

B

N

N

ALSO,

AB

BNN

dt

dN+= 2

2

2 22 =A

B

N

N

15. (B)

16. (C)17. (D)

PASSAGE-II

18. As the process in cylinderA is adiabatic

PV

= constant

Initial pressure inA, P0

= 105

N/m2

Initial volume inA , V0 = 10-2

m2

Final volume inA = V0/4 = 25 104

Let final pressure isA = P1




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5.1

01

5.1

004

=

VPVP ;

018PP =

For equilibrium of pistons (let pressure in cylinderB finally = P2)

P1SP2(2S)

212PP =

024PP =

For gas in cylinderA, 000 RTM

mVP = (i)

For gas in cylinderB, 002'

2

5RT

M

mVP = (ii)

kg110' == mm (d)

19.15.1

48

1

020

01122

=

=VP

VP

r

VPVPu

==00

2 VPu 2000 J

(a)

20. N800080

== SPF (C)

PASSAGE-III

APPLYING KIRCHOFFS LAW IN A LOOP Qxyppp'

1243 1 =++ rIRI (1)

APPLYING KIRCHOFFS LAW IN A LOOP QQxypQ'

04

)( 11 =+R

IIrI (2)

SOLVING (1) AND (2) WE GET 1244

311 =+ rIRI

SINCE Rr>> 1241

=rI ; VrI 31

=

V

Q

P

yx

Q

IP

II1I1

I1

I

SIMILARLY IN THE SECOND CASE WHEN VOLTMETER IS CONNECTED AS IN FIGURE (2)READING OF VOLTMETER IS 9V.

21. (c)

22. (b)

23. (b)




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(Solutions)CHEMISTRY


24. (B) INITIALLY THE PH WILL INCREASE SLOWLY DUE TO NEUTRALISATION OF WEAK ACID

BY NAOH. AFTER THE END POINT IS REACHED, PH WILL INCREASE RAPIDLY DUE TOADDITION OF NAOH.

25. (b)

26. (c)

27. (c) AH3

12

Ka;AHH + + AH2 2

2 Ka;HAH + + 2HA 3

3 Ka;AH + +

At second equivalence point, the only species present in appreciable concentration is2HA .

102

128

2

pKpKpH 32

aa=

+=

+=

1010H

+ =

and[ ]

[ ][ ] 7

321

3

3

310

KaKaKa

H

A

AH +

=

=

28. (b)1

1d

P3U =

( )12RMS PPd

3U =

( )20030075.03 =

2001004 == m/s

29. (c) (A) is Lunar caustic i.e. AgNO3

2AgNO3(A)

2NO2(B) + O2(C) + 2Ag(D)

AgNO3(A) + KCN AgCN + KNO3(white ppt.)

AgCN + KCN K[Ag(CN)2](Soluble complex

Potassium argentocyanide)

SECTION II

More than one type

30. (a, c, d)




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O

S

O

HO OS

O

OH,

OO

O

S

O

HO OS

O

OH,

O

O

S

O

HO O OH, CrO

OO O

O

31. (a, c, d)

C + O2 CO + CO2t = 0 0.1

t = t 0 x (0.1 x)

5CO + I2O5 5CO2 + I2t = 0 x

t = t 0 x x/5

Moles of I2 liberated =2

1moles of hypo consumed =

2

1 120 103 0.1 = 60 104

So,x = 5 60 104 = 0.03

Percentage of C converted to CO is 30% .

Moles of CO2 left = [0.1 x]1st reaction + [x]IInd reaction

= [0.1 0.03] + [0.03] = 0.1

CO2 + H2O H2CO30.1 0.1

H2CO3 + Ca(OH)2 CaCO3 + 2H2O0.1 0.1

32. (a, b, c)

Since, product is an alcohol, so compound (A) is a 10 aliphatic amine.

33. (c, d)

Le Chateliers principle is not quantitative. If both the stresses are caused in the same direction, the shift is

determinable. If both the stresses are caused in the opposite direction, the shift is undeterminable.

SECTION III

Reasoning type

34. (c)

35. (a) The conductivity due to free electrons is more than due to charged ions.

36. (b)

37. (a)0

11

1

10 epn + (-particle). -particle emission occurs when n/p ratio is higher than 1.6.

SECTION IV

Comprehension Type

PASSAGE I

THE REQUIRED REACTIONS OCCURING AT THE ELECTRODES ARE

CATHODE: 2H+

+ 2E

H2

CU2+

+ 2E

CU

ANODE: 2OH

O2 + H2O + 2E

38. (B) THE MOLAR VOLUME OF A GAS AT 1 ATM AND 273 K IS 22.4 L/MOLE. SINCE, THE

VOLUME OF H2 COLLECTED AT 0.1 ATM AND 273 K IS 2.024 CM3, SO THE VOLUME OF H2

COLLECTED AT 1 ATM AND 273 K (STP CONDITIONS) WOULD HAVE BEEN 0.2024 CM3.




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45. (c)

( )[ ] ( )pptYellowcolourBlue

NH4K2CNCuKCN2NHCu 322

43+++ ++

( ) ( )pptWhite

CNCuCN2CNCu2 2decomposes

2 +

( )[ ]Colourless

K3CNCuKCN3CuCN 34+ ++

1Cu + has a d10 configuration, so it is tetrahedral in shape with sp3 hybridization.

46. (d)

Gravimetric estimation of Cu 2+ would be possible only when the given procedure gives an insoluble compound

on treatment with a given reagent.

( )pptBlack

K2SCNCuKSCN2Cu 22 ++ ++

( ) ( )pptWhite

SCNCuSCN2SCNCu2 2decomposes

2 +

With other reagents, either there is no reaction or even if it reacts, then the product is a soluble compound.




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(Solutions)MATHEMATICS


47. (C)

48. (B)

49. (A)

50. (A) LET THREE NUMBERS ARE 2,2 AND 2

( )5

37

3

.......321=

++++++

n

n

( )

( )35

37

2

1

+

=++ nnn

(NMUST BE IN THE FORM OF ( )35 +N ) AND ++ MUST BE NATURAL NUMBER LESSTHAN 33 n THUS POSSIBLE VALUES OF NARE 8 AND 13.

51. (D) ( ) dxkkxx

xkkkI

++=

0

22

ln LET ktx =

( )

++=

0

21

lndt

tt

ktkkI

( ) ( ) ( )ktt

dt

IkkI ln11 20

++=

( )22

0

2

3

2

1

ln

+

+

=

t

dtk

( )6

2

3

2ln

3

=

k

23ek=

52. (A)1

2. =n

n nS NOW FOR ANY VALUE OF N,

nS WILL END IN TWO ZEROS IF THE PRODUCT.

( )( )times1.....2.2.22. 1 = nnn n CONTAINS EXACTLY TWO 5 AS FACTORS.

THIS WILL HAPPEN FOR N= 25, 50, 75, 100

So required probability = 2511004 = SECTION II

One or more than one type

53. (B), (C)

Z= 0 IS ONE OF THE PLANE PERPENDICULAR TO 1,0,0 ===+ xyxyx

NOW THE TRIANGLE FORMED IS TAKE ONE LET ON Z-AXIS I.E., ( )k,0,0 , THEN2222 211 =++ k 2=k




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NOW THE DIRECTION RATIO OF PLANE HAVING POINT ( ( )0,1,1,2,0,0 AND ( )0,1,1 IS( )1,0,2 AND THE DIRECTION RATIO OF PLANE HAVING POINT ( ) ( )0,1,1,2,0,0 AND ( )0,1,1

IS ( )1,0,2 .

(1, 1, 0)

(1, 1, 0)

(0, 0, 0)

B

A

C

54. (A), (B), (C)

( ) ( ) ( ) ( ) ( )( )yxyxyyxfyxyxfyx +=++ 2 LET vyxuyx =+= ;

( ) ( ) ( )uvuvuvfvuf = 2 ( ) ( )uv

u

uf

v

vf=

( ) ( )

=

u

u

ufv

v

vf=CONSTANT

LET( )

= xx

xf

( ) ( 2xxxf += ( ) 21 =f

21 =+ 1= ( ) xxxf += 2

55. (B) LET X [0, 2]. SINCE ( )xf SATISFIES ALL THE CONDITIONS OF L.M.V. THEOREM ON [0, 2].

IT ALSO SATISFIES ON [0, X] [0, 2]

( ) ( )

( )10

0xf

x

fxf =

WHERE 20 1


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SECTION III

Reasoning type

57. (A) STATEMENT-1: ( ) ( ) ( ) abcbacacbcba


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WHEN GAME ENDS WITH TWO HEADS THE PROBABILITY IS .......9

4

9

4

9

432

+

+

+

THEN FOR MINIMUM NUMBER OF THROWS, PROBABILITY IS

( )

9

5

941

1

9

4

94

=

63. (D) GAME END WITH EITHER 2 HEADS OR 2 TAILS THEN IT IS 1.

PASSAGE-II

E D

C

B

A

O

SOL.:Q bOEaODbOCbaOBaOArrrrrr

===+== ,,,,

bABr

= AND aBCr

=

|||| BCAB = |||| barr

=

NOW, OCAB = AND CBOA =

AND CBOAOCAB ===

OABCIS A RHOMBUS

AND ODOA, AND OCEO, ARE COLLINEAR

OAOD =[ AND ]OCOE =

53== ABCAOC AND 52== OCBOAB

I.E., ANGLE BETWEEN ar

AND br

IS5

3

NOW bbaABDCrrr

==

( ) ( ) ( )2222 .2 bbabarrrrr

=+ 5

2cos2

=

64. (B)( )

5

2cos211

1 +==

=

a

a

BC

ADr

r

65. (A)

66. (D)

PASSAGE-III

67. (A)68. (A)

69. (C)


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