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Corporate Office: CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005

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DATE : 09-11-2014

PART TEST-3 (PT-3)(JEE ADVANCED PATTERN)

TARGET : JEE(MAIN +ADVANCED)2016

COURSE : VIKAAS (JA)

HINTS & SOLUTIONS ladsr ,oa gy

Part-IMathematics

1. (1 + x) (1 + x + x2) .

Sol. Highest exponent in the product of first two is 3 = 1 + 2

Highest exponent in the product of first three is 6 = 1 + 2

+ 3

Similarly, Highest exponent in the product of first hundred

= 1 + 2 + ...... + 100 = 5050]

izFke nks xq.kuQy esa x dh mPpre ?kkr 3 = 1 + 2

izFke rhu ds xq.kuQy esa x dh mPpre ?kkr 6 = 1 + 2 + 3

blh izdkj izFke lkS ds xq.kuQy esa x dh mPpre ?kkr

= 1 + 2 + ...... + 100 = 5050]

2. If a,b,c are sides

;fn f=kHkqt a,b,c dh

Sol. R(b + c) = a bc

R(b + c) = 2RsinA bc

sin A = b c2 bc

now applying AMGM for b and c

vc b,oa c ij AMGM yxkusij

b c

2

bc b + c 2 bc

hence sin A1 which is not possible.

vr% sin A1 tks lEHko ugh gS

hence vr% sin A = 1 A = 90

A = 90 and b = c (3)]

3. A triangle has sides .

fdlh f=kHkqt dh Hkqtkvksa

Sol.

=

6 h

2= 3h ;

(where h is the altitude from A)

(tgk h, kh"kZ A ls Mkysx;s yEc dh yEckbZ gSA )

Also iqu%= 21 r 2

(using= rs) ( = r.s ds mi;ksx ls)

21 r 2

=6 h

2= 3h

r

h=

2

7

now APQ and ABC are similar

vcAPQ rFkkABC le:i gS

h r

h

=

PQ

6

r

1 h =PQ

6

2

1 7 =PQ

6

5

7 =

PQ

6 PQ =

30

7Ans.

4. If (1 + x 3x2)2145

.

;fn (1 + x 3x2)2145.

Sol. Put j[kus ij x = 1 ; ( 3)2145= a0 a1+ a2 a3+ ...... ;

(3)2145

= (34)536

3 ends in 3 ]

j[kus ij x = 1 ; ( 3)2145= a0 a1+ a2 a3+ ...... ;

(3)2145

= (34)536

3 vfUre vad 3 gS

5. Given the family of .

nh xbZ js[kkvksa dk .

Sol. point of intersection is A ( 2, 0) . The required line will be

one which passes through ( 2, 0) and is perpendicular to

the line joining ( 2, 0) and (2, 3) or taking (2, 3) as centreand radius equal to PA draw a circle, the required line will

be a tangent to the circle at ( 2, 0) ]

HindiizfrPNsn fcUnq A( 2, 0)gSA vHkh"V js[kk ,d gksxh tks ( 2, 0)

ls xqtjrh gS vkSj js[kk ( 2, 0) vkSj(2, 3) dks feykus okyh js[kk

ds yEcor~ gSA (2, 3) dks dsUnz vkSj PA dks f=kT;k ekudj ,d

ok [khpk tkrk gSA vHkh"V js[kk] ok ds fcUnq ( 2, 0) ij LikZ

js[kk gksxhA]

6. A, B and C are points .

A, B, C, xy lery esa fcUnq

Sol. Locus of C is a circle 8(x2+ y

2) 88x 104y + 544 = 0

(4)]

C dk fcUnqiFk ,d ok 8(x2+ y2) 88x 104y + 544 = 0

(4)gSA

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7. Number of positive .

lehdj.k (x1+ x2+ x3)

Sol. (x1+ x2+ x3) (y1+ y2) = 11 7 or 7 11

In the first case (x1+ x2+ x3) = 11 and (y1+ y2) = 7, which

have10

C26C1solutions (using beggar)

In the second case (x1+ x2+ x3) = 7 and (y1+ y2) = 11,

which have6C2

10C1solutions (using beggar)

total number of solutions = 10C2 6C1+ 6C2 10C1= 270+ 150 = 420 Ans. ]

(x1+ x2+ x3) (y1+ y2) = 11 7 or 7 11

(x1+ x2+ x3) = 11 ,oa (y1+ y2) = 7 ftlls 10C2 6C1 gy

izkIr gksrs gSA (csxj fof/k ls)

(x1+ x2+ x3) = 7 ,oa (y1+ y2) = 11 ftlls 6C2 10C1 gy

izkIr gksrs gSA (csxj fof/k ls)

gyksa dh dqy la[;k = 10C2 6C1+ 6C2 10C1= 270 + 150= 420 Ans. ]

8. The number of ten .

vadksa 1,2,3 dh lgk;rk

Sol. Case-I

Total number in this case

bl fLFkfr esass dqy la[;k, gSA1.2.1.2.1.2.1.2.1.2 = 2

5

CasefLFkfr-IITotal number in this case

bl fLFkfr esass dqy la[;k, gSA2.1.2.1.2.1.2.1.2.1 = 2

5

Total digits dqy la[;k,a = 25+ 25= 64

9. Number of divisors

25. 3

2.5

3 ds 2 ls .Sol. Let 2 3 5 be a divisor, then

can take 1 or 2, can take the value 0 and can take the values 0, 1, 2, 3

Number of ways = 2 1 4 = 8ekukfd 23 5 ,d Hk ktd g S ] rk s

dk eku 1 ;k 2 gk s ldrk g S ] dk eku 0 gk sldrk g S vk Sj ds eku 0, 1, 2, 3 gk s ldrs g SA dqy rjhd s= 2 1 4 = 8

10. Number of zeros .2002

C1001 ds vUr esa .

Sol.2002

C1001=(2002)!

(1001)! (1001)!

no. of zeros in (2002)! are

400 + 80 + 16 + 3 = 499

no. of zeroes in (1001 !)2= 2(200 + 40 + 8 + 1) = 498

Hence no. of zeroes is2

(2002)!

(1001!)

= 1]

Hindi2002

C1001=(2002)!

(1001)! (1001)!

(2002)! esa kwU;ksa dh la[;k

400 + 80 + 16 + 3 = 499

(1001 !)2 esa kwU;ksa dh la[;k = 2(200 + 40 + 8 + 1) = 498

vr% kwU;ksa dh la[;k2

(2002)!

(1001!)= 1

11. If the circles x2+ y

2+ .

;fn ok x2+ y2+ 2ax +.Sol. S1 S2= 5ax + (c d)y + a + 1 = 0 and 5x + by a = 0

must represent the same line

S1 S2= 5ax + (c d)y + a + 1 = 0 vkSj 5x + by a = 0

leku js[kk dks O;Dr djrh gS

a c d a 1

1 b a

ab = c d and vkSj a2+ a + 1 = 0

a is imaginary only

a dsoy dkYifud gSso no value of a (2)]

blfy, a dk dksbZ eku ugha (2) ]

12. Let C be a circle .

ekuk fd C dksbZ ok

Sol. Equation of the line is

js[kk dk lehdj.k gS

y 0 = m(x + 1) ....(1)

solving it withgy djus ij x2+ y2= 1

x2+ m

2(x + 1)

2= 1

(m2+ 1)x

2+ 2m

2x + (m

2 1) = 0, mQ

x1x2=

2

2

m 1

m 1

but ijUrq x1= 1 x2=

2

2

1 m

1 m

= x coordinate of P

ijUrq x1= 1 x2=2

2

1 m

1 m

= P dk xfunsZkkad

since mQ hence x will be rational.

pwfd mQ vr% x ifjes; gksxkIf x is rational then y is also rational from (1) ]

;fn x ifjes; gS rc y Hkh ifjes; gksxk (1) ]

13. Consider the triangle .

fp=k esa ,d f=kHkqt .

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Sol. cos C =2 2 2(19) (17) c

2 19 17

> 0

c2< (19)2+ (17)2

c2< (18 + 1)

2+ (18 1)

2

c2< 2(18

2+ 1)

c2< 650 .... (1)

c > 2 and c < 26

2 < c < 26....(2)

c is 3, 4, 5, ....... , 25 number of integral values of c is 23 Ans. c ds iw.kkZad ekuksa dh la[;k 23 gksxh

14. Sides ofABC are .AB C dh Hk qtk, a lekUrj .

Sol. Sides are in A.P. and a < min{b, c} pa w fd

lekUrj J s

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x cos (+ 30) = d(1) and vkSj x sin= 1d (2)

dividing 3 cot =1 d

1 d

, squaring equation (2) and

putting

3 cot=1 d

1 d

dk s foHkkftr djds lehdj.k (2) dk oxZ

djds tksM+us ij

the value of cot, x2=1

3(4d

2 4d+ 4)

x = 22d d 1

3

]

cot dk eku gS x2=1

3(4d

2 4d + 4)

x = 22d d 1

3

]

20. A triangle ABC has

f=kHkqt ABC, Sol. 2[a cos A + b cos B + c cos C] = a + b + c

= [sin 2A + sin 2B + sin 2C] = sin A + sin B + sin C

(using a = 2R sin A etc.dk mi;ksx )

4 sin A sin B sin C = sin A + sin B + sin C

3

4abc

8R

=a b c

2R

2

abc

R

= (a + b + c) = 2s

4

R

= 2s

abcR

4

2

s

= R

2r = R Also 2r = R triangle is equilateral 2r = R rFkk f=kHkqt leckgq gSA

21. If 4a2 5b

2+ 6a + 1 = 0.

;fn 4a 2 5b 2+ 6a + 1 = 0, Sol. Let the ci rc le be (x h)

2+ (y k)

2= r

2

22 2 2

2 2

ah bk 1r r a b ah bk 1

a b

2 2 2 2 2 2

a h r b k r 2abhk 2ah 2bk 1 0 Now on comparing from the given result

4a2 5b

2+ 6a + 1 = 0

h2 r

2= 4, k

2 r

2= 5, hk = 0, 2h = 6, 2k = 02 2h 3, k 0 & r h 4 5

r 5

centre of circle 3, 0 , radius 5

Hindi ekuk fd ok (x h)2+ (y k)2 = r2

22 2 2

2 2

ah bk 1r r a b ah bk 1

a b

2 2 2 2 2 2a h r b k r 2abhk 2ah 2bk 1 0

fn, x, izfrcU/k ls rqyuk djus ij4a

2 5b

2+ 6a + 1 = 0

h2 r

2= 4, k

2 r

2= 5, hk = 0, 2h = 6, 2k = 02 2h 3, k 0 & r h 4 5

r 5 ok dk dsUnz (3, 0), rFkk f=kT;k = 5

22. The Number of ways in .

ikap fofHkUu iqLrdksa dks .Sol. Given answer is 150 which comes in BCD in A, it is

5C3 3! = 60]

izkIr mkj 150 gS tks BCD esa Hkh vkrk gS rFkk A esa ;g5C3 3! = 60 gSA

23. A family of linear functions

,d js[kh; Qyuksa dk .Sol. Line : y = 1 + c(x + 3) is tangent to circle

x

2

+ y

2

= 1js[kk : y = 1 + c(x + 3) ok x2+ y2= 1 dh LikZ js[kk gSA

p = r

2

1 3c

1 c

= 1

(1 + 3c)2= 1 + c

2

9c2+ 6c = c

2

8c2= 6c

c = 0 or c = 3/4 (1), (2) ]

25. For any positi ve .

m, n fdlh /kukRed .

Sol.n

Cm+n1

Cm+n2

Cm+ .............+m

Cm= Co-efficient of x

min (1+x)

n+ (1+x)

n1+ .......+ (1+x)

m

= (1+x)n+ (1+x)

n1+ .......+ (1+x)

m esa xm dk xq.kkad

= Co-efficient of xmin (1+x)

m

n m 1(1 x) 1

x

= (1+x)m

n m 1(1 x) 1

x

m esa xm dk xq.kkad

=n+1

Cm+1S =

nCm+ 2.

n1Cm+ 3.

n2Cm+ ............

S = Co-efficient of xmin(1+x)

n+ 2.(1+x)

n1+ 3(1+x)

n2+.......

S = (1+x)n + 2.(1+x)n1 + 3(1+x)n2 +.... m esa xm dkxq.kkad

Let ekuk S = (1+x)n + 2.(1+x)n1 + 3(1+x)n2 +..........+(nm+1)(1+x)

m......(i)

S

(1 x)= (1+x)

n1+ 2.(1+x)

n2+....+ (nm+1)(1+x)

m1

......(ii)

from (i) (ii) ls

xS

1 x= (1+x)

n+ (1+x)

n1+...+(1+x)

m(nm+1)(1+x)

m1

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xS

1 x= (1+x)

m

n m 1(1 x) 1

x

(nm+1)(1+x)m1

S =n 2 m 1 m

2

(1 x) (1 x) (n m 1)(1 x)

xx

S = Co-efficient of xmin S = n+2Cm+2 S = S esa xm dk xq.kkad = n+2Cm+2

26. In an examination, .

,d ijh{kk esa] ,d ijh{kkFkhZ Sol. Number of ways he can fai l is either one or two,

three or four subject then total of ways.4C1 +

4C2+

4C3+

4C4 = 2

4 1

Sol. mldsvlQy gksus ds rjhdksa esa og ;k rks ,d fo"k; esa ;k nks

fo"k; esa ;k rhu fo"k; esa ;k pkj fo"k;ksa esa vlQy gks ldrk

gks] vr% dqy rjhdksa dh la[;k = 4C1 +4C2 + 4C3 +4C4 =

24 1

27. IfnCr1= 36 ,

nCr= 84 .

;fn nC r 1= 36 , nC r .

Sol.

nr1

nr

C

C=

36

84and vk Sj

nr

nr 1

C

C =

84

126

3n 1 0r = 3 a ndvk Sj4n 10r = 6

n = 9 & r = 3 | n r| = 6

28. The coefficient of .

(1 + x + y z)9 ds .

Sol. (1 + x + y z)9= {(1 z) + (x + y)}

9

=9C0(1 z)

9+

9C1(1 z)

8(x + y) +

9C2(1 z)

7(x + y)

2

+....+9C7(1 z)

2(x + y)

7+.......

x3y4appears in (x + y)7only

(x + y)7 esa dsoy x3y4 gksxkA

so, the required coefficientblfy, vHkh"V xq.kkad

=9C7(2) .

7C4

29. Let AB C be an .

ekuk ABC U;wudks.k f=kHkqt Sol.

AE F : AF = bc osA , AE = cco sA

cosA =2 2 2 2 2b cos A c cos A EF

2bcosA.ccosA

(EF)2= (b 2 + c2 2b cco sA ) co s2A(EF)

2= a

2co s

2A

EF = a.cosA

30. One side of a rectangle.

,d vk;r dh ,d Hkqtk .

Sol.(1, 1)

( 3, 1) 4x + 7y + 5 = 0

Line perpendicular to 4x + 7y + 5 = 0 is 7x 4y += 0

4x + 7y + 5 = 0 ds yEcor~ js[kk7x 4y += 0It passes through ( 3, 1) and (1, 1)

;g ( 3, 1) rFkk (1, 1) ls xqtjrk gSA

11 4 += 0 = 257 4 += 0 = 3Hence lines are 7x 4y + 25 = 0, 7x 4y 3 = 0

vr% js[kk,a 7x 4y + 25 = 0, 7x 4y 3 = 0 gSAline parallel to 4x + 7y + 5 = 0 passing through (1, 1) is 4x

+ 7y += 04x + 7y + 5 = 0 ds lekUrj ljy js[kk 4x + 7y += 0 tks(1, 1) lsxqtjrh gSA

= 11 4x + 7y 11 = 0

32. The circle with equation .

ok x2+ y2= 1, js[kk .Sol.

Solving y = 7x + 5 and the circle x2+ y

2= 1

js[kk y = 7x + 5 vkSj ok x2+ y2= 1 dks gy djus ij

A3 4

,5 5

and vkSj B4 3

,5 5

mOA=4

3 ; mOB=

3

4

Hence vr% AOD = 90 ACB =4

= tan

1(1)

33. Statement 1:The sum

oDrO;-1 : ;ksxQy 40C0.

oDrO;-2 : 40 yM+ds vkSj 60 yM+fd;ksa esa ls 10 fo|kfFkZ;ksa ds

pquus ds ep; 100C10gSA

Sol. Obviously Li"Vr;rk%

34. Passing through a

Statement-1: Locus of

,d fcUnq A(6, 8) ls tkus okyh Sol. Let P (x1, y1) then w.r.t. the circle i.e. x

2+ y

2 6x 8y + 5

= 0

T = 0 is xx1+ yy1+ g(x + x1) + f (y + y1) + c = 0 it passes

through A(6, 8)

ekuk P (x1, y1) gS] rc ok x2+ y2 6x 8y + 5 = 0 ds lkis{k

T = 0, xx1+ yy1+ g(x + x1) + f (y + y1) + c = 0 gS tks A(6, 8)

ls xqtjrk gSA

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6x + 8y 3(x + 6) 4(y + 8) + 5 = 0

3x + 4y 18 32 + 5 = 0

3x + 4y 45 = 0 S-1 is incorrect vlR; gSAAlso the point A (6, 8) lies outside S.

fcUnq A(6, 8) Hkh S ds ckgj fLFkr gSAHence S-1 is false and S-2 is true.

vr% oDrO;-1 vlR; gS] oDrO;-2 lR; gSA

35. The value of

x =1

2ds fy;s

36. If ithterms is the

x =1

2ds fy;s i ok

37. If kthterms is having .

;fn k ok in vf/kdre.Sol. R = (1 + 2x)

n

put x = 1 to get sum of all the coefficients

x = 1j[kus ij lHkh xq.kkdksa dk ;ksxQy gS

3n= 6561 = 38 n = 8

(35) for x =

1

2; R =

8

2 1 ds fy,

consider

8 8 8

80

I f f

2 1 2 1 2 C 2 ........ even integer

ekuk fd 8 8 8

80

I f f

2 1 2 1 2 C 2 ........

=le iw.kkZad gS

since I is integer f + f must be an integerpwfd I iw.kkZad gS f + f iw.kkZad gksxkbut ijUrq 0 < f + f < 2 f + f = 1

f = 1 fnow vc n + R Rf

n + R(1 f) = n n

8 2 1 2 1

= 8 + 1 = 9 Ans.

(36) Tr + 1in (1 + 2x)8=

8Cr(2x)

r=

8Cr when tcx =

1

2

now vc Tr+1Tr

r 1

r

T1

T

8

r

8r 1

C1

C

r 1 rT T

8! (r 1)! (9 r)!

1r!(8 r)! 8!

(9 r)r 92r

for r = 1, 2, 3, 4 this is true ds fy, ;g lR; gSi.e. T5> T4

but for r = 5, T6< T5

T5is the greatest term (2) T5 vf/kdre in gS (2)

(37) again iqu% Tk +1= 8Ck 2k xk; Tk= 8Ck 1 2k 1 xk 1

Tk 1=8Ck 2 2

k 2 x

k 2

we want to find the term having the greatest coefficient

ge Kkr dj ldrs gS ftlesa vf/kdre xq.kkad j[krk gS

2k 1 8Ck 12k 8Ck .....(1)and vkSj 2k 1 8Ck 12k 2 8Ck 2 .....(2)

from (1)

k 1 k8!2 2 8!

(k 1)! (9 k)! k! (8 k)!

1 2

(9 k) k

k18 3k k6

again iqu% 2k 1 8Ck 12k 2 8Ck2k 1 k 28!2 2 8!

(k 1)! (9 k)! (k 2)! (10 k)!

2 1

k 1 10 k

20 2kk 1 213k k7 6k7 T6and T7term has the greatest coefficient

T6 vkSjT7 vf/kdre xq.kkad j[krk gS

k = 6 or;k 7 sum;ksxQy= 6 + 7 = 13

Ans.]

38. Number of ways .

16 f[kykfM;ks dks ..

39. Number of ways in ..

bu f[kykfM;ksa dks pkj ..

40. Number of ways in ..

bu 16 f[kykfM;ksa dks ..

Sol.(38) Number of ways rjhdks la[;k

=4

(16)!

(4!) 4!=

88

r 12 8! (2r 1)

4!4!4!4!4!

=

82 8765

(24)(24)(24)(24)

=

92 35

(24)(24)(24)

=

8

r 1

35(2r 1)

27

Ans.

(39) 16

P ,P ,P ,P1 2 3 4

12 others

12 others can be divided into 4 equal groups in 4(12)!(3!)

=(12) (11)!

6 6 6 6 =

(11)!

108Ans.

vU; 12 dks 4 leku lewgks esa foHkkftr djus ds rjhds4

(12)!

(3!)

=(12) (11)!

6 6 6 6 =

(11)!

108Ans.

(40)16

P ,P ,P ,P ,P1 2 3 4 5P6P P (10)7 16

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; 10

3

7

=10!

3! 7!

0 0 0+P6

4 players

x x x x x x xP to P1 5

12 players

Now , 12

4

4

4

=3

12!

(4!) 3!

Total ways dqy rjhds =(10)!

3! 7!

3

(12)!

(4!) 3!

=3

12!

(4!)

20 k = 20 Ans.]

41. The radius of the circle

okks S1= 0 ; S2= 0 ,oa..

42. If the circle S = 0 is ..

;fn ok S = 0, okksa S1= 0

43. The radius of the circle

okks S1= 0 ,oaS2= 0 dks.Sol.

(1,0)O(1,0)

S1

S2

(41)

(1,0)

2

r

(a,b)(0,1)

2

(0,0) 1

r

S1 S = 03

S3

r2= a

2+ b

2+ 1 = (a + 1)

2+ b

2+ 4 and

(a + 1)2+ b

2+ 4 = a

2+ (b + 1)

2+ 4

2a + 4 = 02a = 2b

a = 2, b = 2

r2= 9 r = 3Ans.

(42) S1 S2= 0 x = 1S2 S3 y = 1 Radical centre ewyk{k dsUnz = (1, 1)

radiusf=kT;k LT= 1S = 1

equation of circle is ok dh lehdj.k gS(x 1)

2+ (y 1)

2= 1

radius f=kT;k= 1 and ,oa a = 1 ; b = 1 a + b + r = 3Ans.

(43)

(3,2)

(1,0)

x = 1

family of circles touches the line x 1 = 0 at (1, 0) is

js[kk x 1 = 0 dks fcUnq (1, 0) ij LikZ djus okys ok

fudk; dk lehdj.k gS(x 1)

2+ (y 0)

2+(x 1) = 0

passing through (3, 2) 4 + 4 + 2= 0 = 4vHkh ok (3, 2) xqtjrk gS vr% 4 + 4 + 2= 0 = 4 x2+ y2 6x + 5 = 0

radius f=kT;k= 9 5 = 2Ans.]

Part-II

Physics

1. A uniform solid cuboid of mass m..................

mnzO;eku dk ,d Bksl ?kukHk fpduh ..................

Sol. F0(a) = mgb

2

t =ak2

mgb

2. Equal volumes of two immiscible..................

leku vk;ru ds rFkk 3 ?kuRo..................

Sol.

P

P0+

gh

3 + 1

2v2= P

0+ 0 +

1

2v

12

v1=

2gh

3

P0+

3 gh

3

+gh+

1

2v2= P

0+ 0 +

1

2 3v

22

v2=

4gh

3

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3. A sphere made of material of specific..................

8 fofk"V xq:Ro ds inkFkZ ds xksys esa..................

Sol.3 31 2

4 4R R 8 g

3 3

=

31

4R g

3

4. A uniform rod of length L and weight..................

L yEckbZ rFkkWR Hkkj dh ,dleku NM+..................Sol. The forces on rod hinged at O, which cause torque about

O are as shown.

Oij fuyfEcr dh xbZ NM+ ij cy tks O ds ifjr% cyk?kw.kZ

mRiUu djrs gS] fn[kk;s x;s gSA

Since net torque on rod about hinge at O is zero

pwafd NM+ ij fuyEcu ds ifjr% usV vk?kw.kZ kwU; gSA

WRW

T sin /2

T sin = WR 2

+ W

or;k T =

RW

W2

sin

6. A circular wooden loop of mass m..................

m nzO;eku o R f=kT;k dk ,d okkdkj..................

Sol. Conservation of angular momentum about C.O.M. of m of

loop of mass m gives

ywi dh nzO;eku dsUnz ds ifjr% dks.kh; laosx laj{k.k }kjk

mVR

2=

2 2

2 R R

m R m m w2 2

V = 3R

=V

3RAns. (B)

7. A rod of mass M and length is..................

M nzO;eku rFkk yEckbZ dh ,d NM+..................

Sol. Initial momentum is along negative x-axis. Hence final

total momentum of system will also be along negative

x-direction.

izkjfEHkd laosx _.kkRed x-v{k ds vuqfnk gSA vr% fudk; dk

vfUre laosx Hkh _.kkRed x-v{k ds vuqfnk gksxkA

8. A cylinder A rolls without slipping..................

,d csyu A tks fd r[rs B ij fcuk..................Sol. For rolling, vA= 2 m/s

yksVuh (rolling) xfr ds fy,vA= 2 m/s

or 4 1.= 2

or = 2 rad/sec. (clockwise) (nf{k.kkorZ)

9. A thin uniform rod of mass M and..................

nzO;ekuM rFkk yEckbZ L dh ,d iryh..................

Sol. By conservationof energy tkZ laj{k.k ls

MgL

2

+ 0 = 0 +1

2I2

Mg

L

2 =

2

21 ML

2 3

=3g

L

Speed of end point of the rod =L = 3gL

NM+ ds vfUre fljs dh pky =L = 3gL

10. A smooth disc is rotating with uniform..................

,d fpduh pdrh ,d leku dks.kh;..................Sol. The only force block exerts on disc is parallel to axis of

rotation of disc. This additional force does not cause any

torque on disc,

Hence angular momentum of disc remains same.Since there is no friction between block and disc, the block

remains in its position.

og cy tks CykWd pdrh ij vkjksfir djrk gS pdrh ds ?kw.kZu

v{k ds lekUrj gSA ;g vfrfjDr cy pdrh ij dksbZ vk?kw.kZ

mRiUu ugh djrk gSA blfy;spdrh dk dks.kh; laosx leku

jgrk gSA pwafd CykWd rFkk pdrh ds chp dksbZ ?k"kZ.k ugh gS]

blfy;s CykWd viuh fLFkfr esa gh jgsxkA

11. Two vessels A and B of different shapes..................

fHkUu fHkUu vkdkj ds nks crZu A vkSjB..................

Sol. WA> WBas mass of water in A is more than in B

WA> WB pwafd A dk nzO;eku B ls T;knk gSPA= PBArea of A = Area of B

A dk {ks=kQy = B dk {ks=kQy

or;k PAAreaA= PBAreaBPA {ks=kQyA= PB {ks=kQyB

or;k FA = FB.12. An equilateral prism of mass m rests..................

m nzO;eku dk ,d leckgq fizTe ?k"kZ.k..................Sol. The tendency of rotating will be about the pont C. For

minimum force, the torque of F about C has to be equal to

the torque of mg about C.

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fizTe dh fcUnq C ds izfr ?kw.kZu xfr djus dh izofr gSA U;wure

cy ds fy;s F dk fcUnqC ds ifjr% cy vk?kw.kZmg dk Cds

ifjr% cy vk?kw.kZ ds cjkcj gksuk pkfg,

F3

a2

= mga

2

F =mg

3Ans.

13. In the shown figure a force F

is..................

m nzO;eku rFkk yEckbZ dh le:i..................

Sol. The force applied by the hinge may act in the same

direction as force F.

The force applied by the hinge may act in the opposite

direction as force F.

The force applied by the hinge may be zero.

vkyEcu }kjk NM+ ij vkjksfir fd;k x;k cy] cy F ds leku

fnkk esa gks ldrk gSA

vkyEcu }kjk NM+ ij vkjksfir fd;k x;k cy] cy F ds

foifjr fnkk esa gks ldrk gSA

vkyEcu }kjk NM+ ij vkjksfir fd;k x;k cy kwU; gks ldrk

gSA14. A uniform rigid body is given initial..................

,d le:i n r2, then the customer will get lesser amount ofvegetables, as compared to the measured weight on that

balance.

;fnr1= r2 gks, rks rqyk] lCth dh lgh ek=kk iznku djrh gSA

;fn r1 > r2 gks, rks rqyk }kjk xzkgd dks iznku dh lCth dh

ek=kk] ekfir Hkkj dh rqyuk esa de gksxhA

16. Two discs A and B of radius R and 2R................

nks pdrh;k A rFkk B ftudh f=kT;k R o 2R................Sol. If B is rotated one complete revolution, then A has rotated

one complete revolution in that time.

If A is rotated one complete revolution, then B has rotated

two complete revolution in that time.

At any instant, the angular velocity of A is equal to angular

velocity of B.

At any instant, the angular acceleration of A is equal to

angular acceleration of B.

;fn B ,d pDdj iw.kZ djrh gS rksbl le; vUrjky esa A Hkh

,d pDdj iwjk djrh gSA

;fn A ,d pDdj iw.kZ djrh gS rksbl le; vUrjky esa B nks

pDdj iwjs djrh gSA

fdlh Hkh {k.k ij A dk dks.kh; osx B ds dks.kh; osx dsleku

gksxk

fdlh Hkh {k.k ij A dk dks.kh; Roj.k B ds dks.kh; Roj.k ds

leku gksxk

17. A uniform rigid body, (mass M, Radius R)....................

,d le:i n

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backward direction or can be zero. Consequently friction

can act in backward direction or forward or can be zero.

C nzO;eku dsUnz gS

VP= Vcmrcm, ijUrq nzO;eku dsUnz C fcUnq P ls fdlh nwjh

rCM ij gSA vr% Vp dk eku kwU; ;k vxz fnkk esa ;k ip fnkk

esa dqN Hkh gks ldrk gS vkSj blh izdkj ?k"kZ.k cy Hkh kwU; ;k

vxz fnkk esa;k ip fnkk esa dqN Hkh gks ldrk gSA

20. A uniform rod AB of mass m and.................

m nzO;eku o yEckbZdh ,d .................

Sol. Impulse = change in momentum

vkosx= laosx ifjorZu

P.2

=

2m

12

.

(about centre of AB) (ABds dsUnz ds ifjr% )

=6P

m

For= 2

ds fy, ; 2

=t t = 2

=m

2 6p

t =m

12p

Ans.

21. A uniform body of circular cross-section.................

okkdkj dkV {ks=k (f=kT;k R) dh ,d.................Sol. Body comes to rest and applying angular momentum

conservation about any point on the ground.

oLrq fLFkj voLFkk esa vk tk;sxh rFkk tehu ij fLFkr fdlh

fcUnq ds lkis{k dks.kh; laosx laj{k.k ds }kjkmv0R mK

20= 0

22. Two men of equal masses stand.................

fu;r dks.kh; osx ls ?kweus okyh ,d.................Sol. 11=22

Since, men move towards middle of turn table 2decreases hence2increases.

pwafd, O;fDr dsUnz dh vksj xfr djrs gS blfy, 2 ?kVsxk

QyLo:i 2 c

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Let y be the height of water surface above hole

dy

dt=

av

A=

a 2gy

A

0 t

3h / 4 0

dy adt

A2gy t =

A 3h

a 2g

fNnz ds ckgj vkus ij ty dk izkjfEHkd osx {kSfrt gS rFkk fNnz

{kSfrt lrg lsh

4pkbZ ij gSA vr% ty }kjk {kSfrt lrg

rd igqpus es yxk; le; t =2(h/ 4)

ggksxk tks fu;r

jgrk gSA

x = vt tgk v ckgj vkus dk osx ....

pawfd v le; ds lkFk ?kV jgk gSa

v

v

y

vr% x ?kVsxkA ekuk fNnz ls ty lrg dh apkbZ y gS

dy

dt=

av

A=

a 2gy

A

0 t

3h / 4 0

dy adt

A2gy

t =A 3h

a 2g

39. The friction force acting on.............

NM+ ij vjksfir ?k"kZ.k cy.............Sol. N + T = mg sin30

Net= 0, mg sin30L

4=

3L

4T

T =mg

6 f = mg cos30 =

3mg

2

40. Just after the string is cut, the normal.............

jLlh dks dkVus ds rqjUr ipkr~.............Sol. Torque about the point at which normal acts.

tgkW vfHkyEc cy dk;Zjr gSA ml fcUnq ds lkis{k cyk?kw.kZ

=mg

2

L

422mL L

m12 4

=

mgL

8

=6g

7L

maCM=mg

2 N N =

mg

2 m

L

4 N =

2mg

7

41. The moment of inertia of system.............

oy; ds v{k ds lkis{k fudk;.............

42. The acceleration of centre of mass.............

fudk; ds nzO;eku dsUnz.............

43. The minimum value of coefficient.............

fQlyu jksdus dsfy, U;wure.............

Sol. (41 to 43)

= 2

2

2M R 2 R

I M 4 mR12 2

= 20 kgm2.

(4M + m)g sin f = (4M + m)a.

f.R. =a

R

F

Solving gy djusij

7ga

24

f = 20a (4M + m)g cos 30

5

12 3

min=5

12 3

5

12 3

Part-III

Chemistry1. An aqueous solution having pH = 7 .............

80C ij pH = 7 okyk tyh; foy;u ------------------

Sol. As temp, Kwso at 80C, neutral point will at pH < 7

solution of pH = 7 will be basic.

Sol. rki, Kw vr% 80C ij mnklhu fcUnqpH < 7 ij gksxkA

pH = 7 dk foy;u {kkjh; gksxkA

2. Following five solutions of Ca(OH)2................

dejs ds rki ij Ca(OH)2 dsfuEu ikp foy;u --------------------

Sol. All solutions have [Ca(OH)2] = 0.01M

lHkh foy;u esa [Ca(OH)2] = 0.01M gSA

[OH ] = 2102M

pOH = 1.7

pH = 12.3

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3. Initially for the equilibrium x2+ y2 2xy ...................

lkE;x2+ y2 2xy dsfy, izkjEHk esa .................

Sol. x2 + y2 2xy

Initial conc.2

5M

2

5M 0

at equilibrium 0.4a 0.4a 2a = 0.7

a = 0.35M

[x2]eq= [y2]eq= 0.05MSol. x2 + y2 2xy

izkjfEHkd2

5M

2

5M 0

lkanzrk

lkE; ij 0.4a 0.4a 2a = 0.7

a = 0.35M

[x2]eq= [y2]eq= 0.05M

4. For a real gas (vander Waals gas) behaving ...............

,d okLrfod xSl okMajoky xSl tks vknkZ -------------------------

Sol. A real gas behaves ideally at Boyles temperature.

,d okLrfod xSl ckW;y rki ij vknkZ O;ogkj djrh gSA

Z =MPV

RT= 1

P =

M

RT

V=

M

aR

Rb

V=

M

a

V .b

5. At critical temperature, which of the .................

kfUrd rki ij dkSulk xzkQ --------------------

Sol. TC< TB(Boyles temperature)

So PV

P

Sol. TC< TB(ckW;y rki)

blfy, PV

P

6. If K1and K2are the equilibrium ...............

;fn ,d mRe.kh; vfHkf;k ds fy, T1K -------------------

Sol. IfH = 0

K1= K2at all temperature.

Sol. ;fnH = 0

rks lHkh rki ijK1= K2

9. In which case the-bond pair ..............

fdl fLFkfr esa -ca/k ;qXe rFkk cU/k ------------------

Sol. (I, m)

12. A simplified application of MO theory ...............

,d dkYifud v.kq OF ds fy, MO fl)kUr ---------------------

Sol. OF is derivative of O2 and isoelectronic with O2.

So (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x =

2p2y) (*2px2=*2p1y)

The bond order of OF 1/2(10 7) = 1.5.

gy OF v.kq O2 dk O;qRiUu gSvkS j O2 ls lebysDVkWfud gSA

vr% (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x=

2p2y) (*2px2=*2p1y)

OF dk ca/k e 1/2(10 7) = 1.5.

13. Select correct statement(s) .................

lgh dFku@dFkuksa dks pqfu, ---------------------------

Sol. KP of a particular reaction depends only on

temperature.

On increasing pressure reaction NH4HS(s)

NH3(g) + H2S(g) moves in backward direction, so

dissociation is suppressed.

High pressure is favourable for metting of ice.

During dissociation of PCl5(g), number of moles

increases, so Mavgdecreases.

Sol. ,d fufpr vfHkf;k dk KP dsoy rki ij fuHkZj djrk

gSA nkc c

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15. For the reaction

H2(g) + I2(g) 2HI(g) ...............

vfHkf;k ds fy,

H2(g) + I2(g) 2HI(g) .................

Sol. Changing volume will change partial pressure of all

species.

Changing temperature will change KP, so partial

pressure of all species will change

Adding inert gas at constant volume & temperature

will not change the partial pressure of any species.

Adding inert gas at constant pressure & temperature

will increase volume, so partial pressure of all

species will decrease.

Sol. vk;ru cnyus ij lHkh fLikht dk vkafkd nkc cny

tkrk gSA

lHkh Lihkht dk vkafkd nkc cny tkrk gSA

fLFkj vk;ru o rkieku ij vf; xSl feykus ij fdlh

Hkh Lihkht dk vkafkd nkc ifjofrZr ugha gksrk gSA

fLFkj nkc o rki ij vf; xSl feykus ij lHkh Lihkht

dk vkafkd nkc ?kV tkrk gSA

16. Consider the reaction ...............

vfHkf;k P(g) + 2Q(g)

K =10C12

R(g) + S(g) ..........

Sol. P(g) + 2Q(g) R(g) + S(g)

KC= 1012

Initial conc. 2M 4M 0 0

Since KC it very large, reaction will almost take place to

completion.at equilibrium conc. x 2x 2M 2M

(x0)

KC= 1012

=2

(2)(2)

x(2x) x = 104M

[P(g)]eq= 104

M

[Q(g)eq= 2104

M

[R(g) = [S(g) = 2M

Sol. P(g) + 2Q(g) R(g) + S(g)

KC= 1012

izkjfEHkd 2M 4M 0 0lkUnzrk

D;ksafdKC cgqr vf/kd gS] vfHkf;k yxHkx iw.kZ gks tk;sxhA

lkE; lkUnzrk ij x 2x 2M 2M

(x0)

KC= 1012

=2

(2)(2)

x(2x) x = 104M

[P(g)]eq= 104

M

[Q(g)eq= 2104

M

[R(g) = [S(g) = 2M

17. The reaction pKa= 14 pKbis ...............

fuEu esa ls fdl ;qXe@;qXeksa ds fy, pKa.............

Sol. pKa= 14 pKbis applicable for weak conjugate acid base

pair.

Sol. nqcZy la;qXeh vEy {kkj ;qXe ds fy, pKa= 14 pKb ykxw gksrk

gSA

18. Select correct statement .................

lgh dFku@dFkuksa dk p;u --------------------

Sol. Vc= 3b = 3(4VNA) = 12(VNA)

= 12(Volume of 1 mole of gaseous molecules)

Tb> Tc, so gas cant be liquefied above Tb.

At high pressure (P) (V-nb) = nRT

V =nRT

P+ nb

At Boyles temperature, second virial coefficient

becomes zero, on solving the given equation

second virial coefficient comes out to be

2 RT

.

2B

RT

= 0

TB=R

Sol. Vc= 3b = 3(4VNA) = 12(VNA)

= 12(1 eksy xSlh; v.kqvksadk vk;ru )

Tb> Tc, vr% xSlTb ds ij nzohr ugha gksxhA

mPp nkc ij(P) (V-nb) = nRT

V =nRT

P+ nb

ckW;y rki ij] f}rh;d fofj;y xq.kkad kwU; gksxk nh xbZ

lehdj.k dks gy djus ij f}rh;d fofj;y xq.kkad

2

RT

gksxkA

2B

RT

= 0

TB=R

19. A vessel contains a mixture of H2...............

,d ik=k esaH2 oD2 xSlksa dk feJ.k ----------------------

Sol. D2 is heavier than H2hence rate of effusion of D2 is less

than that of H2.

Mole fraction of D2 in the gas remaining in the vessel

will increase with time.

The average molecular mass of gas remaining in the

vessel will increase with time.

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Sol. D2,H2 ls Hkkjh gksrh gSA vr%D2 dh folj.k dh njH2 dh rqyuk

esa de gksrh gSA

ik=k esa ks"k cph xSl esa D2 dk eksy fHkUu le; ds lkFk

c

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35. The initial partial pressure of C2H4.............

C2H4 dk izkjfEHkd vkafkd nkc .................

Sol.2 4C H

P + COP + 2NP = 250 mm of Hg

2O added

P = 250 mm of Hg

(PTotal)after reaction= 370 mm of Hg

2CO formedP = 200 mm of Hg

(PTotal)after reaction=2CO

(P ) +2N

P + V.P. of H2O +2O

P left

370 = 200 + 20 +2N

P +2O

P left

2NP +

2O leftP = 150 mm of Hg .......(i)

let2 4C H

P = a C2H4+ 3O2 2CO2+ 2H2O()

PCO= b CO +1

2O2 CO2

2NP = 250ab

2CO formedP = 200 = 2a + b ....(ii)

2O used

P = 3a +b

2

2O left

P = 250 3a b

2

from equation (i)

lehdj.k (i) ls

(250 a b) + (250 3a

b

2 ) = 150

8a + 3b = 700 ....(iii)

Solving equation (ii) & (iii)

lehdj.k (ii) ls (iii) dks gy djus ij

a = 50 =2 4C H

P

b = 100 = PCO

36. Total O2consumed in mm ................

dqy mi;ksx gksus okyh O2 mm ...............

Sol. O2used = 3a +b

2= 150 + 50 = 200 mm

37. The sum of initial pressure of CO .................

LikfdZx (sparking)ls igys CO ------------------------

Sol.2CO O

P P = 100 + 250 = 350 MM

38. What is the equilibrium constant ..............

vfHkf;k ds fy, lkE; fLFkjkad --------------------

Sol. 2HI(g) H2(g) + I2(g)

Initial mole 1 0 0

at equilibrium mole 1 2

2

KC=2

.2 2

(1 )

=

2

2

(0.25)

4 (0.75)=

1

36= 0.027

Sol. 2HI(g) H2(g) + I2(g)

izkjfEHkd eksy 1 0 0

lkE; eksy 1 2

2

KC=2

.2 2

(1 )

=

2

2

(0.25)

4 (0.75)=

1

36= 0.027

39. What is the degree of dissociation ...............

;fn iz;ksx dks ,d eksyHI ----------------------

Sol. IfI2(g) is also taken initially, then degree of dissociation of

HIwill be less than 25%.

Sol. ;fn izkjEHk esa I2(g) dksfy;k tkrk gS rks HI dk fo;kstu25% ls

de gksrk gSA

40. What is the degree of dissociation ...............

;fn vfHkf;k dks,d eksy HI -------------------------

Sol. If one mole He is also taken initially then degree of

dissociation will remain same at same temperature.

;fn izkjEHk esa ,d eksy Hedks Hkh fy;k tkrk gS rks leku rki

ij fo;kstu dh ek=kk leku jgrh gSA

43. Which of the following hydrogen ...........

fuEu esa ls dkSu gkbMkstu cU/k --------------------

Sol. Because of highest electronegativity of F, hydrogen

bonding in F H - - - - F is strongest.

Sol. F dh fo|qr_.krk vf/kd gksus ls F H - - - - F esagkbM kstu

ca/kqrk izcy gksrh gSA

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DATE : 09-11-2014

PART TEST-3 (PT-3)(JEE ADVANCED PATTERN)

TARGET : JEE(MAIN +ADVANCED)2016

COURSE : VIKAAS (JA)

ANSWER KEY

CODE-0

MATHEMATICS

1. (B) 2. (C) 3. (C) 4. (B) 5. (A) 6. (D) 7. (C)

8. (B) 9. (A) 10. (B) 11. (B) 12. (D) 13. (BC) 14. (BD)

15. (ABCD) 16. (ABC) 17. (BD) 18. (BCD) 19. (AC) 20. (BC) 21. (BD)

22. (BCD) 23. (AB) 24. (ABCD) 25. (ABC) 26. (CD) 27. (ABCD) 28. (BD)

29. (BC) 30. (ACD) 31. (BCD) 32. (BD) 33. (A) 34. (D) 35. (C)

36. (B) 37. (D) 38. (D) 39. (B) 40. (D) 41. (C) 42. (D)

43. (C)

PHYSICS

1. (C) 2. (B) 3. (D) 4. (D) 5. (B) 6. (B) 7. (B)

8. (A) 9. (C) 10. (C) 11. (B) 12. (A) 13. (BCD) 14. (CD)

15. (AB) 16. (ABCD) 17. (AC) 18. (BD) 19. (ABC) 20. (AC) 21. (ABC)

22. (BCD) 23. (AB) 24. (BC) 25. (ABD) 26. (BC) 27. (BCD) 28. (ABCD)

29. (ABD) 30. (AB) 31. (ABCD) 32. (BC) 33. (D) 34. (D) 35. (B)

36. (C) 37. (A) 38. (B) 39. (C) 40. (C) 41. (A) 42. (C)

43. (B)

CHEMISTRY

1. (B) 2. (C) 3. (D) 4. (C) 5. (A) 6. (D) 7. (D)

8. (D) 9. (C) 10. (B) 11. (A) 12. (B) 13. (BD) 14. (ACD)

15. (B) 16. (ACD) 17. (AB) 18. (ACD) 19. (BC) 20. (ABC) 21. (AB)

22. (AB) 23. (ACD) 24. (ACD) 25. (ABD) 26. (CD) 27. (ABC) 28. (BCD)

29. (AB) 30. (AB) 31. (ACD) 32. (BC) 33. (D) 34. (B) 35. (B)

36. (C) 37. (D) 38. (B) 39. (A) 40. (C) 41. (D) 42. (D)

43. (D)

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