phase 2 pat 1dczdad

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liT -2014-PAT -A-LOT - TEST-I-CPM(I)-13

F I I T J E E PAT TEST(CTY 1214 A LOT Batch)

liT - JEE 2014

ANSWERS

SECTION - I (Chemistr y)PART-A

1. 8 2. 0 3. A 4. 85. 8 6. A 7. A 8. 89. A 10. 0 11. A, C 12. A, C13. A, C 14. A, C, O 15. A, 8

PART-C

1. 4 2. 4 3. 1 4. 25. 6

SECTION - II (Physics)

PART-A1. 8 2. A 3. A 4. C5. C 6. C 7. C 8. C9. 8 10. C 11. A, 8, C, O 12. A, C, O13. 8, C 14. A, C 15. 8, C

PART-C

1. 3 2. 4 3. 2 4. 45. 3

SECTION - III (Mathematics)

PART-A

1. A 2. 8 3. A 4. C5. A 6. 8 7. A 8. C9. C 10. 8 11. C, O 12. C, O13. A, 8, C 14. A, 8 15. A, 8, O

PART-C

1. 6 2. 6 3. 6 4. 25. 4

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liT -2014-PAT -A-LOT -TEST -1-CP M(I)-14

HINTS & SOLUTIONSSection - I (Chemistry)

PART-A1. B

Sol. Kp =Kc(RTtn ~n=2-3=-1

:.1.3 x 10-2

= Kc

(0.0821 x 1000t

Kc

=1.067

2. 0

2HI(g)~H2 (g)+I 2 (g)

Sol.

I-aa a

3.

Sol.

Sol.

5.

2 2

(~ Pr ) 2K =---p (l_ a )2 P :

=>~=2 fj(I-a '\j"'"p

=>a= 2 J K ;1 +2 J K ;

A

(N H4\ C0 3 (s)~2NH3 (g) +CO 2 (g) +H20(g)

2P P P

Total Pressure = 4P = 0.42

P =0.105

Kp

= (2 P) 2 x Px P

Kp

= 4.86 x 10-4 atm44. B

2HI ~ H 2 +12

I 0 0

I - .22 .11 .11

K =[HJ[IJ=0.IIXO.II=0.OI99

c [HIt (0.78)2

B

~ 4S2 (g )o

4x0.30

S8 (g )Initially I

At equilibrium (1- .30)

= 0.70 atm =1.2 atm

( p ; , r (1.2f 3Kp = -( ,) = -( -) = 2.9 6 atm

P 0.70s,

A

Those reaction which have more value of K proceeds towards completion.A

For the given equilibrium

Sol.

6.

Sol.

7.

Sol.

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l iT -2014-PAT -A-LOT -TEST -1-CPM(I)-15

2A B2 (g )

Initially 1

At equilibrium (I-a)

~2AB(g)+B2(g)

o 0

a al2

a

L :n = 1+-2

aa

1(I-a)

aSince, a is small compared to unity, so 1- a =:: 1 and 1+- =:: 1

2

Pa3

. K =_T_ p 2

B

PCI ---"5 .. --

8.

Sol.

Use the relation n . =1+aeqw

ninitial

9.

(vapour density) initial nequilibrium=---

(vapour density) equilibrium

104.16 1+a--=--

71 1

a =46.7%A

Sol. K = ( p so,r

=__(0_.3_3~1)_2_=2. 5

p ( )2 ( ) (0 .6 62 )2 (0 .1 01 )p.~o, Po,

( p so, rNow, Kp =( p ) ( p )

so , a,

1 1Then P o =:: - =- =0.4 atm

, Kp

2.5

10. D

A + B ~C+DSol. Initially 1 1

At equilibrium (1- .8) (1- .8) .8 .8

'K _[C][D]=0.8xO.8=16.. c [ A ] [ B ] 0.2xO.2

11. A, C

Sol. The degree of dissociation cannot be calculated from the vapour density data when the number of

moles remain unchanged before and after reaching equilibrium.12. A, C

Sol. Pressure does not disturb equilibrium state because b.n = and addition of catalyst increases ratesforward and backward reaction equally.

13. A,C

Sol. ':Kp =Kc(RTt'

14. A, C, D

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IIT-2014-PAT -A-LOT- TES T -1-CP M(I)-16

Sol. K p =(P H ,O r

':Kp=KC(RT)Z ':~n=2

KC = [ H20 J15. A, B

Sol. Doubling the volume would reduce the pressure and shift the equilibrium in the direction where thereis more number of moles of gases.

PART-C1. 4

Sol. e(s)+e02(g)~2eO(g)

K - ( P;o r - ~- 4p - - ( p ' ) -I-

co ,

2. 4

Sol. According to Le-chatlier's principle, when the temperature is increased, the equilibrium shifts in the

direction of the endothermic reaction. When the pressure is increased, the equilibrium shifts in thedirection of less number of moles. The removal of a reactant and addition of product shift theequlibbrium in the backward direction.

3. 1

Sol.

A +B ~e +DInitial moles 4 4 0 0

At equilibrium (4 -2) (4 -2) 2 2

K =[e ] [D ]c [ A ] [ B ]

K =(2/V)2 =1

C (2/V)2

4. 2

Sol. .,' The unit of Kc

is molz [2,

Hence ~n=2

5. 6

N20

4(g) ~ 2NO

z(g)

Sol. Initially I 0

At equilibrium (1- 0.2) 2 x 0.2

Total moles = 1.2

Now PV=nRT

P x I x R x 300 = 1.2 x R x 1500I

P =6 atm

Section - III (Mathematics)1.

Sol.

A

a+bIf a, b>O, then x = A.M. of a and b =>x = - - . . . . (1)

2

(b ) 1 / 3

a,y,z,b in G.P. => b = ~ = ar3

=>r= - ; ;

(b ) 1 / 3 1 1 3 ( b ) Z 1 3

~ = Y = ar = a -;; = ( a2

b) , ~ = z = a/ = a -;;

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2.

Sol.

3.

Sol.

4.

Sol.

5.

Sol.

6.

Sol.

liT -2014-PAT -A-LOT -TEST -1-CPM(I)-17

(2)113 3 3 2 2

= ab . This => y +z =a b+ab =ab(a+b) ...(2)

And XYz=( a;b)(a2bt3.(ab2t3

=(a+b)ab=!(/+/), by (2) =>/+/ =22 . 2 x y z

S

Given x,y,z in G.P. and so / =xz

1 , 1 , 1 in H.P. => 1+ logz.1 + logy, 1+ logz in AP. => logx, logy, logz in1+ logx 1+ logy 1+ logz

AP.

=> 210gy = logx +logz => / = xz, which is trueA

2221.3 +2 .5 +3.7 + ...

Series formed by first factors is

1+ 2 + 3 + ...1

=n.

Series formed by second factors isn

32

+52

+i +...1 " =[3 +(n-I)2J =(2n +lr,

Sn = I,ln = I,n.(2n + 1)2 = I,n( 4n2

+ 4n + 1)= 4I,n3 + 4I,n2 + I,n

=4[~(n + 1)]2 + in(n + 1)(2n + 1)+ ~(n+ 1) =262

This => S = (20x 21)2 +3.x 20x 21 x 41 + lOx 21 = 18809020 3

C

a,a ,a ,...,a are in H.P.I 2 3 n

This => ~,~,~, .... ,_1 are in AP. with common difference = da a a a

J 2 3 n

1 1 1 1 1 1 a-a a -a a -I-a:. d=---=---= ... or d=_I __ 2 = 2 3 - = n-I n. This =>

a a a a a a aa a a a a2 I 3 2 n n-I I 2 2 3 n n-I=>daa =a -a2,daa =a2-a, ....,daa =a -aI 2 1 2 3 3 n n-I n-I n

Adding, d(aa +a a +.... +a a )=a -a ....(1)1 2 2 3 " n-I 1 n

1 1 a-aAlso T =-=-+(n-l)d=> _I_n =(n-I)d

n a a aan I n I

Put this in (1), a1a2 +a2a3 +...+anan_1=(n-l)anajA

x +y +z=15...(1),9,x,y,z,a in AP. with common dif. =d

S" =~(a+l) gives

59+(x+ y +z) +a =-(9+a) => 2(9 +15 +a) =5(9+a)

2

=> a=1. same result is obtained with the second result.

S

13

13

+i 13

+i+33

-+--+----+.....1 1+3 1+3+5

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liT -2014-PAT -A-LOT -TEST -1-CP M(I)-18

7.

Sol.

8.

Sol.

9.

Sol.

10.

Sol.

T = I ' +2' + 3' + ...+n' _ [i(n+l)r = (n+I)'n 1+3+S+ .....nterms ~[2+2(n-I)J 4

8 ='l.:.T=> 48 = 'l.:.(n+1 ) 2 ='l.:.n2+2'l.:.n+nn n n

n n=>48n =6'(n+l)(2n+l)+2'2(n+l)+n

= ~(n+ 1)[2n +7 ] +n6

If n = 168 = 16x17x39 +.!i = 446, 16 6x 4 4

A2

Let the numbers be a,ar,ar

a+ar+a/ =14 ...(1)

(a +1),( ar +1),( ar2 -I) are in A.P.

2(ar+I)=a+ a/ .... (2 )

Solving (1) and (2)

We get the answer 8.

e

8 = 1+ 2 +3 + ... . +n =~(n+ 1)2

By assumption, 8+669=(n-8r or ~(n+I)+669=n2+64-I6n or, n2

-33n-I2IO=0 or2

( n - SS)(n + 22) = 0 .'. n = S5 .

n 55(56)Total number of balls = -( n + 1) = -15402 2

e150,146, I42, ....,a = I50,d = -4.

Suppose 150 workers complete work in n days

. 11 worker does In one day --thpart of the work.

I50 n

Under condition II, workers complete the work in (n +8) days. Total workers

=( n ;8)[ 2x I50+(n +8-1)( -4)J = (n+8)(136-2n)

1Total workers. -- = 1=> (n + 8)(136 - 2n) = I50n

ISO n

=> (n+ 8)( 68- n) = 7Sn => n2

+ ISn-544 = 0

( n + 32) ( n - 17) = 0 => n = 17 =>n + 8 = 17 + 8 = 25 days.

B

6.6.6 ....n terms = 6+6.10+6.102

+6.103

+ ...n terms = 6[1+10+102

+103

+ .....mtermsJ

6.(JOn - 1 ) =~(10n -1)JO-I 9

Similarly 8.8.8.8 ....n terms

= % (JO n -1 )

Given sum = [%(IOn - I ) r +~(IOn -1)

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Sol.

12.

Sol.

13.

Sol.

14.

Sol.

15.

liT -2014-PAT -A-LOT -TES T -1-CP M(I)-19

= ~[(lO n -If + 2( lO n _1) ]= ~(102 n -1)11. C,O

x,x2

+2, / +10 in G.P.

=:>( / +2 f= x.( / +10) =:>/ +4+4/ =/ + lO x2 I

= :> 4x -10 x+ 4= 0= :> x= 2,- 2

x ,x2

+2, / +10= 2,6,18 if x =2 ...(1) and .!.,2,!! if x =.!. ...(2)2 4 8 2

Next term of (1) is 18x 3 =54

. 81 9 729Next term of (2) IS -x- =-

8 2 16

C,O

. H 4By assumption, - =-

G 5

5 ( 2ab ) r;-=:>5H =4G =:> - 4"ab

a+ b

=:>5 J ; ; E = 2(a+b) =:>5 = 2( K + ~), where K =( ~r 2

2 1 (a ) I/2 a 4 .= :> 2K -5K+2=0=:>K=2,-.lfK=2,then - =2 or-=-. Th ls= :> a:b = 4:1

2 b b I

or 1 : 4

A, B,C

Last term of nth row

I=1+2 +3 +... +n = - n(n + I )

2

As terms in the nth row forms an A.P. with common difference 1.

First term = last term - (n- l) (1) = = ~ ( n2 - n+ 2 )

n1=lastterm = -(n+l)

2

a =first term =~ ( n 2 - n +2)

nSum = -(a+l)

2

=:>S um = = ~ n [~ (n 2 -n + 2 )+ ~ (n 2 + n )]

= ~ n ( n 2 + 1 )

A,B

1/3We haveS=--=

1-1/3 2

(A) Thus, (0.25yog,(S) =(0.5)210g,(1/2)=(0.5(IOg,(2) = (0.5f2 =4,

(B) (0.008)=(0.2)2 ,=C~J =53

(0.008yog

s(S)= (0.008tOgs2 =(5-3tOgS2 =(5)3108,2

=i =8A, B, 0

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liT -2014-PAT -A-LOT -TEST-I-CPM(I)-20

Sol.

1.

Sol.

2.

Sol.

3.

Sol.

4.

Sol.

5.

(a_b)2 +(b-C)2 +(c_a)2 +(a_l )2 +(b-ll +(c_l)2 =0

PART-C

6

Given S =3S in AP.Zn n

:::)2n[ 2a+ (2n -l)dJ = 3~[2a+(n-l)dJ2 2

:::)2a =(n +I)d

: : : ) S 3 n =3n[2a+(3n-l)dJ+~[2a+(n-l)dJS 2 2

n

=3.(4nd)=62nd

6

Given al,aZ,.....,aIO

in AP.

hi' hz, ...,hl o in H.P ... (1)

al =hl =2 ...(2)a

lo =hlo =3 ...(3) a4h7 =?

a = a +(10-1) d :::)3 = 2 +9d U d = !10 I 9

3 7 1 1 I .a =a +3d=2+-=-. Now(1) :::)-,-,- arelnAP .

4 I 93 hhhI Z 3

_I =~+(IO-l)d :::)~=!+9d :::)d =__1h h I 3 2 I I 54

10 I

I I I ( I) 7 18h= 2 " + (7 - I) dl = 2 " + 6 - 54 = 1 8 : : : ) \ = 7

7

h =~.~=64 7 3 7

6

Four numbers are a, b, c, d.

1According to problem, a, b, c in AP. with a =6 and b,c,d in G.P. with common ration = -

2b b b b

:. b,c,d = b,-,-z = b,-,-2 2 2 4

:. a,b,c,d =6,b,~,~ (1). Now 6, b, ~ in AP.242

:::)2b = 6 +~:::) b = 4. Again b,~,~ in G.P. :::) (~)Z = b.~2 24 2 4Which is true. Put b = 4 in (1), we find a, b, c, d = 6,4,2,1.2

. 2xzx,y,z are In H.P. :::) Y = - - , Now

x+z

[4 X Z ]log(x +z) +log(x+z -2y) = log(x+ z)+ log x+ z ---

x+z

= log(x +z) +logr - z ) ; = log(x +z) +2 lo g(x - z )-log(x +z)x+z

=2log(x -z)4

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