JPT 3 SolutionBooklet English

8/12/2019 JPT 3 SolutionBooklet English

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RJPT3SOL030411-1

Hints & Solution

DATE : 03-04-2011 CLASS-XII/XIII

PAPER-1

PART-I (Chemistry)1. From law of equivalence,

Eq of KMnO 4 (v.f. = 5) = Eq. of SO 2 (v.f. = 2)

15858.1

5 2 = 2SOn 2

! 2SOn = 0.05 molesWhole of S from FeS 2 is converted into SO 2! Applying POAC on S :

2 2FeSn = 1 2SOn ! 2 120m

2FeS = 1 0.05

! 2FeSm = 3 g = x ! x = 3

2. A corner Si-atom and a Si-atom in tetrahedral void touch eachother.

!4

a3 = 2r.. ! a =3

r8

Also, number of atoms per unit cell = 8.

! % of occupied space = 33

a

r34

8 "# 100

= 3

3

3r8

r348

$$ % &

''( )

"# 100 = 34%.

! % of unoccupied space = 100 34 = 66%

3. Pb 3O4 + 8HCl *+ 3PbCl 2 + 4H 2O + Cl 2,2Pb 3O4 + 6H 2SO 4 *+ 6PbSO 4 + 6H 2O + O 2,

4. (A) Sodium metal can be produced by the electrolysis ofmolten NaCl.(B) CsOH has the maximum basicity and maximum solubilityamong all alkali metal hydroxides.(C) Gypsum when heated above 393 K forms Dead Burnt Plaster.

5. For a second order reaction,

Rate = dtdC

= K C 2.

Also, t 1/2 - C 01 2. So, t 1/2 C 0 = constant. Therefore, correct

graphs are :

JEE PREPARATORY TEST-3 JPT-3)TARGET IIT-JEE 2011

(i) (iv)

6. . m = S10K 3/#

2(1.25 10 5) =S

10104 34 // ##

! S = 1.6 10 2 mol/L. ! Ksp = 4S 3 0 1.64 10 5

7. K2S is water soluble and MnS (IVth group) is more soluble thanHgS (IInd group) due to higher K sp of MnS than HgS.

11. (A) For all elements, IE 1 < IE 2 < IE 3 .........(B) The minimum value of IE for an element in periodic table (Cs)

is also greater than the maxiumum value of 1Heg for an elementin periodic table (Cl). (w.r.t. magnitude only)(C) Due to poor shielding effect of 4f-electrons, IE for Au > IE forAg.

(D) 1Heg for S > 1Heg for O because of small size of O. (w.r.t.magnitude only).

12. P 0 = XAP A + XBP B = 82

300 +86

500 = 450 mm of Hg.

Now, RLVP =0

S0P

P P =

450420 450

=151

Also, RLVP = $$ % &

''( )

2Ninin

;151

=8

70i32

70i32

2

! i = 1.25 = (1 + (2 1) 3). So, 3 = 0.25 (or 25%)

! Cln produced = 10025

7032

=354

! 2PbCln precipitated = 21

354

=352

13. (C) The oxidising power of halogens follow the order : F 2 > Cl 2 >

Br2 > I 2.(D) On reaction of XeF 2 with water, redox change takes place.

2XeF 2 + 2H 2O *+ * 2Xe + 4HF + O 2 (slow)


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RJPT3SOL030411-2

14. (A)

(C)

It gives positive tollen s test.

15.

meso

16. The unknown compounds are as follows :(A) (NH4)2Cr2O7 (B) Cr 2O 3 (C) N 2(M) Ca(N) CaO (O) CaC 2 (P) NH 4NO2 (Q) NH 3Now, the reaction involved is :

CaC 2 + 2H 2O *+ * Ca(OH) 2 + C 2H2

17. The Cr metal present in compound (B) contains 6 unpairedelectrons in elemental state (3d 5 4s 1).

18. (A) Upon addition of Zn dust to a solution of compound NH 4NO 2,gas NH 3 is evolved and it is a redox change (NO 2

reduced toNH3).(C) A filter paper moistened with copper sulphate solution turnsblue on coming in contact with NH 3 gas.

(D) A water soluble salt of metal Ca gives white precipitate of amixed salt on adding Potassium ferrocyanide solution.

22. At closest distance of approach :

qV =rqKq 21

! 2e V = 4 5xe79)e2(K

(for 3-particle) ....... (i)

and e V =4 5y

e79)e(K(for proton) ... .. .. (ii)

For (i) and (ii), x = yx : y = 1 : 1

23. Limiting reagent may neither have the least mass nor the leastmoles among all the reactants available.

24. Elemental Boron can be obtained from Van Arkel method.

2Bl3 Tantalumor

Whotred * * * * + * 2B, + 3I 2, (Van Arkel method).

25. A chemical reaction having larger value of activation energy (E a)

is more sensitive to changes in temperature.

PART-II (Mathematics)

28. f(x) =67

68

9

:2; {0} > (1, =).

29. Required Area = ]223a3)a2[(32

dx]2)x(f[ 2 / 3a

1

///;/? .Differentiate w.r.t. a ,

@ f(a) 2 =32

ABC

DEF /# 32a2

23

@ f(a) = 2 a2 ; a G 1 or f(x) = 2 x2 ; x G 1

30. f(x) = f(6 x) ....(i)on diffn both side f H(x) = f H (6 x) ....(ii)

putting x = 0, 2, 3, 5 in (ii) we get

fH(0) = f H(6) = 0fH(2) = f H(4) = 0fH(3) = 0fH(5) = f H(1) = 0

fH(0) = 0 = f H(2) = f H(3) = f H(5) = f H(1) = f H(4) = f H(6)! fH(x) has minimum 7 roots in [0, 6]


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RJPT3SOL030411-3

now, consider a function y = f H(x)As fH(x) satisfy Roll's theorem in intervals [0, 1], [1, 2], [2, 3],[3, 4], [4, 5] and [5, 6] respectively.So by Rolle's theorem, the equation f HH(x) = 0 has minimum 6roots.

Nowg(x) = (f HH(x))2 + fH(x) fHHH(x) =dxd

(fH(x) (fHH(x)) = h H(x)where h(x) = f H(x) fHH(x)clearly h(x) = 0 has minimum 13 roots in [0, 6]hence again by Rolle's theorem, g(x) = h H(x) has minimum 12zeroes in [0, 6]

31. x2 + y 2 2x = 0 and x2 + y 2 + 2x = 0x > 0 x < 0centre of third circle will be on radical axis4x = 0 let the centre be (0, k)so distance from centre of one of the circle = 2

2k1 2 ;2k2 = 3

k = 3

x2 + y 2 32 y + 2 = 0, x 2 + y 2 + 32 y + 2 = 0

32. Equation of tangent at P (a cos I , b sin I ) is

acosx I

+b

siny I = 1

! Equation of the lines OA and OB is

x2 + y 2 a 2 2

bsiny

acosx $

% &'

( ) I2I = 0

since the lines OA and OB are perpendicular

! 1 cos 2I + 1 22

b

a sin 2I = 0

i.e. sin 2I + 1 = 22

b

a sin 2I

i.e. 22

a

b = I2

I2

2

sin1sin

! e 2 = 1 22

ab

= 1 I2I2

2

sin1sin

= I2 2sin11

! e is least if I = 2"

! the point P is (0, b)

33. Take 3, K, L, k are roots of equation P(x) = 0

P(x) = (x

3) (x

K) (x

L) (x

k)! n P(x) = ! n(x 3) + ! n(x K) + ! n(x L) + ! n(x k)take differentiation,

)x(P)x(PH

= )x(1

3/ + )x(1

K/ + )x(1

L/ + )kx(1/

x = 7 3 = 1, K = 2, L = 3

0 =17

1/ + 27

1/ + 37

1/ + k7

1/ @ k = 37

319

34. 1 =18606206

062

// = 0

11 =18616203

0611

////

/ M 0 : 12 =

18 1 06 3 6

011 2M 0

and 13 =1 603 206

11 62

M 0.

Hence, the system is inconsistent

35 . x = 2 + 3i @ x2 4x + 13 = 0! x4 x 3 + 10x 2 + 3x 5= (x 2 4x + 13) (x 2 + 3x + 9) 122 = 122

36. N NO 1


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RJPT3SOL030411-4

39. =+nlim

n1 N

2;$ % &'

( ) n2

1nr nr

f = ?2

1

dx)x(f

=+nlim

n1 N

;$ % &'

( ) 2n

1r nnr

f = ? 21

0

dx)x1(f = ?2

1

dt)t(f = ?2

1

dx)x(f

=+nlim

n1 N

;$ % &'

( ) n

1r nr

f = ?1

0

dx)x(f

=+nlim

n1 N

;$ % &'

( ) n2

1r nr

f = ?2

0

dx)x(f

40. Given function is discontinuous when a + sin "x = 1Now if a = 1 @ sin "x = 0 @ x = 1, 2, 3, 4, 5If a = 3 @ sin "x = 2 not possibleIf a = 0.5 @ sin "x = 0.5@ x has 6 values, 2 each for one cycle of period 2.

If a = 0 @ sin "x = 1 @ x =21

,25

,29

41. Since Q PRQ = " /2 and points P, Q, R lie on the circle with PQ asdiameter

Also PQ = 5Now, the maximum area of the triangle is 1max =

21

(5) $ % &'

( )

25

= 6.25

For area = 5, we have four symmetrical positions of pointP (P 1, P 2, P 3, P 4 )For area = 6.25 we have exactly two points.For area = 7, no such points exist.

42. Image of A in the plane 2x +2y + z = 1 is given by

21x /

=2

0y / =

11z /

= 2222 122

1102

22/22

= 94/

! Coordinates of the image are $ % &'

( ) /

95

,98

,91

! Equation of the reflected ray is9 / 10x /

=9 / 80y

//

=9 / 41z

//

i.e. 8x = y = 2z 2

or9 / 191x /

=9 / 898y

/

2 = 9 / 4

95z

/

/i.e. 8 72x = 9y + 8 = 18z 10

43. Equation of PQ is2

2 z3

3y1

1 x ;2; any point on PQ has

co-ordinate (1 +r, 3 + 3r, 2 + 2r)it is Q if 1 + r + ( 3 + 3r) 3 (2 + 2r) = 0

@ r = 4

! Q = ( 3, 15, 6) ~ k6 j15 i344. Let P H = (3, K, L) be the image of P in plane. then the middle point

of PP H $ % &'

( ) 2LK232

22

,2

3 ,

21

lies on plane

so,2

2.3

23

21 2LK2232 = 0

@ 3 + K 3L = 8also PP H is perpendicular to the plane, so,

3 2

13

11 L;2K;3 = )3( 3 11

)2 (3 31

2L2K23

=1116

! 3 =1127

, K=1117

,L=11

26

Clearly P H, Q, R are on the same line so QR has the equation

1010 6z

3737 15y

1515 3x 2;2;2

@ r"

= )k

10 j

37i

15()k

4 j

22i

12( 22.222

45. Three points P(1, 3, 2), Q( 3, 15, 6)

& P H $ % &'

( )

1126 ,

1117 ,

1127

lies on the plane

so equation of plane passing P,Q,P H is 11x 5y + 2z = 30

Sol.(46,47,48)

Sol. x2 3a

a2/ x + 3a

2a//

= 0

P : 3a2a

//

> 0 ( = , 2) > (3, =)

Q : 3a2a

//

< 0 (2 ,3)

R : D G 0 4a 2 4(a 2 5a + 6) G 0 5a 6 G 0

@ a G 56

set R is $ % &

DEF

2,56

> (3, =)

S : $ % &

DEF =,

56

T : $ % &'

( ) =/

56

,

46. Clearly, all option (A), (B), (C) are true.47. Clearly, T is a subset of P.48. Least positive integer for S is 2.

Least positive integer for R is 4.

Greatest positive integer for T is 1.Hence option (B) is correct.


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RJPT3SOL030411-5

49. arg(z 1) arg(z 4) + arg(z 2)

arg(z 3) = 0

50. Two real roots ; b 2 4ac > 03 < 2 ; K > 23 < 0 < K @ f(0) = c < 0f( 1) = a b + c < 0f(1) = a + b + c < 0@ a + |b| + c < 0

f(

2) = 4a

2b + c < 0f(2) = 4a + 2b + c < 0@ 4a + 2|b| + c < 0.

51. f(x) = sin 1 [2 4x 2] 1 O [2 4x 2] O 1 1 O 2 4x 2 < 20 < 4x 2 O 3

0 < x 2 O 43

x R A

A

B

C

D

D

E

F/

23

,23

{0}

52. Let m 2 = 2 8 + 2 11 + 2 n

2n = m 2 28 9 = m 2 (24 . 3) 2

= (m 48) (m + 48)

Let m 48 = 2 s & m + 48 = 2 t

where s + t = n! m = 2 s + 48 = 2 t 48

= 2 t 2 s = 96 = 2 s (2 t s 1) = 96# 2 t s 1 is odd@ 2s (2 t s 1) = 2 5 3 @ s = 5, t = 7

53. F(x) = ? 2x

0dt)1t( @ F (x) = x + 1 > 0, S x R [2, 3]

Hence F(x) is increasing function in this interval.

Therefore greatest value is F(3) = ? 23

0

dt)1t(

and least value F(2) = ? 22

0

dt)1t(

hence required difference,

h = F(3) F(2) = ? 23

0dt)1t( ? 2

2

0dt)1t(

= ? 23

2

dt)1t( =27

54. f(1) = 7 f(2) = 16# fH(x) = 3x 2 " sin "x > 0 S x R [1, 2]@ f(x) is an strictly increasing function@ f(x) = 13 has excatly one solution in [1, 2]

PART-III (Physics)

55. # 2iaB = mg @ B = ia2mg

56. E"

= i

xV

TT

= $ % &'

( )

mkV

3000 x i

UE = ? sd.E ""

= 3000 0.1 0.1 2 10 3 Vm

@ q encl = UE .R0 = 2.67 10 8 C

57. U = 21

CV2

F = dxdU

=2V2/

dxdC

c =

d

)x(b 0R/! +d

Kbx 0R @ F =2

V2/(K 1)

d

b 0R

58. 2 capacitors in || el

@ Ceq = 2C = dKA

2 0R# =

F105.0

109.81.01.0523

12

/

/

######

= 1780 pF

59. At equilibrium

T = 2 " g!

= 2 sec

T = 2 " 22 ag 2

!

= 2 "

2

2

ga

1g 2

!

= 2 " gcos I!

= 2 Icos = 2 Vcos36 = 1.8 sec.

60. N.S.L.

Vdg V gW 6"XrY = Vd 2g

@ Y = $ % &'

( ) W/

X 2dgr

92 2

= Xgdr

272 2

61. 1p

T/L

L = const. @ T

dT 100 = p

dp1$$ % &

''( )

L/L

100

@ TdT

100 = 5.3

57

157

#

$$$$$

%

&

'''''

(

) $ % &'

( ) /

= 1%


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RJPT3SOL030411-6

62.

!

!

6 / v

2 / v

f

f

2

1

; = 3

@ f1 = 3f 2 = 3 2 = 6 Hz ( # f2 = 2Hz)@ beat frequency = f 1 f 2 = 6 2 = 4 Hz

63. h = gr45cosT2

WV

.............. (i)

2

h= gr

cosT2W

I ............. (ii)

(i) & (ii) @ cos I =2

1 @ I = 60

64. ms dtdT

= ZA $$ %

&''(

) $ % &'

( ) /

44

2T

T

@ dtdT

A1

= CZ

(T4 16T4

) =

4

2T

C15 $

% &'

( ) Z

(# ms = C)

65. # relative acceleration is zero so path will be straight line &since they collide so line must pass through origin.

66. Initially

[H = 3mg 2!

\H = 3m2 2!

+12

m 2!+ m

2

23

$$ %

&''(

) !=

23

m ! 2

[H = \H 3 @ 3 = !g

Hing Force

N1 = 3mg cos 60 = 23

mg

3mgsin60 N2 = 3m !g 3

! @ N2 = 2

3 mg

Hinge force = 222

1 NN 2 = 3 mg

67. j

xi

yV /;"

@ dtdx

= y & dtdy

= x @ dxdy

= yx

= x 2 + y 2 = c (circle)

speed = 22 yx 2 = c = constant

a"

=dtdy

i

dtdx

j

a"

= x i y j

@ j

myi

mxF //;"

(conservative)

68. from work energy theoremv1 = v 2 = v 3I 1 + I 2 = 90 & I 2 = 45

@ 231

2I;I2I @ I1, I 2 &I 3 are in AP

T1 = gsinu2 1I

@ T2 = g45sinu2 V

@ T3 = gcosu2 1I

@ 2

TT 232

1 2 = T22 @ T12 , T 22 & T32 are in AP

69. Just before entering the region,

tan I =x

y

u

u

as u x will increase u y remains constantso I will decrease

Sol.(70 to 72)B.T. at (1) & (2)

pa = p 2 + Wgh 2 + 21

Wv22

setting v 2 = 0

h2 = gpp 2a

W/

(h2)max = gpaW 4 50p2 ;# = 10 m Ans.

B.T. at (1) & (3)

pa = p a + 21

Wv2 Wgh 1

@ v = 1gh2 Ans.

From B.T. at (1) & (2)

@ v2 = )ghpp(2

22a W//W

@ (v2)max = )ghp(2

2a W/W (setting p 2 = 0) = 10 m/s Ans.


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RJPT3SOL030411-7

73.

u = 20 v =3

20/ cm m =uv/ =

31/

d = 161

2 2$$ % &

''( )

d =34

mm @ D =3

340cm

K =d

D. =

4

3400)(. = ( . )850

K = 2000 850 10 10 m @ K = 1.7 10 4 m.

74. As half reflecting surface of one part is painted black thenintensity of one source will decrease so intensity at maxima willdecrease and at minima it will increase (when intensity of bothsources is same then there is zero intesntiy at minima).

75. As source is moved away from the mirror both images will movetowards the mirrors 'D' will increase and 'd' will decrease.

76. \ 3P02

77. MRT

VsL

;

0

so if Domain is (0, =) Range will be 4 54log,1 3

32. Upon simplifying we getf(x) = 5 + 4 cos 2x 3 sin 2x= 5 + 5 cos (2x + 3).

33. log x +21

log x +41

log x + ....... = y

@ $ % &'

( ) 222 .....

41

211 log x = y

@

21

1

1

/ log x = y @ 2 log x = y ...(1)

Now )1y3(.....1074)1y2(.....531

22222/2222

= xlog720

@

N

N

;

;

2

/

y

1r

y

1r

)1r3(

)1r2(

=y2

)1y(y.3

y2

)1y(y.2

22

/2

=

ABCD

EF 22

/2

y2y3

2y3

)11y(y2

=

ABCDEF 2 25

2y3y

y 2

= 5y3y22 = xlog7

20

using equation (1)

5y3y22 = y7

40 @ 7y 2 = 60y + 100

@ 7y 2 60y 100 = 0

7 / 55 10,10x 7

10,10y

/;/;

34. y = f(x) & y = f 1(x) symmetric about y = x

# 4 5?"

0

dxxf + 4 5?"

/

0

1 dxxf = 2 1OAB

@ 4 5?"

/

0

1 dxxf = 2 1OAB 4 5?"

0

dxxf

@ 4 5?"

/

0

1 dxxf = 2. 21

(")(") ?"

20

dx)xsinx(

=AAB

CDDE

F2"/" 2

2

22 =

242 /"

a = 4, b = 2 @ a + b = 6.

35. f(x) = max {(1 x), (1 + x), 2}, S x R ( = , =)shaded graph is the graph of f(x).

y = 1 +

x y = 1 x

1

2

1

1

! f(x) =67

689

O2OO

O

x1,x11x1 ,2

1 xx 1

We can see f(x) is continuous everywhere and non-differen-tiable at x = 1.

36. Let the roots of z 3 + az 2 + bz + c = 0 be 3,K,L then

|c| = | 3KL| = 1 & | 3K + KL + L3| = L2

K2

3111

= |3 + K + L| = |a|@ |a| = |b| & |c| = 1Also , |a| = | 3 + K + L| O |3| + | K| + | L| O 3

Put, |a| = 2k + 1 , when 1k21 OO/

Thus , z 3 + |a|z 2 + |b|z + |c|= z 3 + (2k + 1)z 2 + (2k + 1)z + 1= (z + 1) (z 2 + 2kz + 1)

It has roots 1 & 1kk 2 /^/ since, k 2 O 1, we

we have 1k2 / = 2k1i /

So,2

2 1kk /^/ = k2 + 1 k2 = 1

So, all the roots have modulus 1.


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RJPT3SOL030411-12

37. (A )

?

?

ABC

DEF 2

=+

=+

y

1y

y

1

1

y

dxx1

1lim

dx]x[tanlim

=1 y

1tan ylimy =+

= 1

(B) Let x = tan I , then tan 1 IItan 1

sec : I

= a : 1

tan 1 II

tan1 sec

= tan 1 II

sincos 1

=2I

! a : 1 =2I

: I @ a =21

(C) =+nlim

$ % &'

( ) 2

n11n

)!nsin(.n 3 / 2 = =+nlim

3 / 1n1 .

n11

)!n(sin

2

= =1

.1

1 &1betweennumbera = 0.

(D) |cos x| = sin 1 sinx

(i)2

x2

"


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RJPT3SOL030411-13

PART-III (Physics)

43.

Beat = f f ABC

DEF I

Csinv C

=Csinvf I

f'

as u I , beats also I

44. T = mgsin37 +f mgcos37 Fmin = T + f (2+4) g

= 20 53

+ 0.2 20 54

+0.2 60

= 12 + 3.2 + 12 = 27.2 N

45. Ncos I = mg .......... (i)Nsin I = m(R sin I5 g2 ............. (ii)

(i) & (ii) @ g = IcosRg

46. Normal stress

2a2

2F7

2F

$$

%

&''

(

) 2

=4 5

2a2

F17 2

47. Torque balance for 2 nd brick from the top

$ % &'

( ) / y

2mg

! = mgy

@ y =4!

Torque balance for 3 rd brick from the top

$ % &'

( ) / x

2mg

! = 2mgx

@ x =6!

48.

Rab = 30 h @ Vcd = 3900

= 300V

Vef = 3300

= 100 V

& Vgh

=3

100V = reading of voltmeter

Vcd = 300 V

from KCL at junction d, reading of Ammeter = 10 A

49. e + e+ + L + L2E L = 2 0.00055 930 MeV @ EL = 0.512 MeV

50. Energy radiated outside

= ? 2""ctanh

0

22s

)hr(4

Prdr2

= ]c[secn2

P 2s ! .

51. # qE = qvB

@ v = BE

= 3 km/s (only one possible speed)

52.

Ey = E z = 0

yE

TT

"

is not zero, E x at A and B may be different.

53.A

|J| \;

" (where A is cross sectional area and \ is current),

JE""

W; (W is resistivity)

Current is uniform in the conductor.

54. . cutoff = 8400012400

= 0.15 A

99% of power = 84000 10 10 3 0.99 = 300 0.04 4.2 dtdT

@ dtdT = 16.5 C/sec


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RJPT3SOL030411-14

55. Transition that should occur is n = 1 to n = 4

@ E0 = E 4 E 1 = ( 0.85 + 13.6) eV = 12.75 eV

maximum 3 different lines can be emitted in emission spectrumas shown

56. B = 3

20

x2

iRf (x > > R)

UB = 322

0

x2

riR "f

@ i = dtd BU/

=dtdx

x2

riR34

220 "f

= 24

20

RN

vr23 "f

58. (A) In p,q,r,s external forces are acting.

(B) In 'p' there is not any point about which j[ external = 0 ,in 'q'j[ external = 0 about any point on horizontal surface, in 'r' j[ = 0about sun, in 's' mg is non impulsive so torque of mg can beneglected about top most point of step. (for very small timeinterval).

(C) In 'q' kinetic friction is acting, in 's' energy will not conservedue to inelastic collision.

(D) In 't' there is no loss in kinetic energy due to perfectly elasticcollision.

59. Before S is closed

Steady current through inductor =200120

15050120 ;2 = 0.6 AA

After S is closed

current through it is 302n

5x150

e6.050

120 !#/

/

= 2.4 26.0

= 2.1 Amp (towards right)

60. WY

= fF

10 and W =Af

@ 200

Y100

YAF .;; @ . = 2

61. V0 = ef

eh U/

if f = 0

V0 = eU/ = 1V @ U = 1eV

62. O point will appear at its original

position, & for point A, refraction

from outer surface

52 / 3

v1

// =

1023

1

/

/

v1

= 205

103

201 /;/ = 4

1

@ v = 4 cm @ distance from O = 10 4 = 6 cm

63. k = A(n - 1)

O\ = k 9 "

= 4 V"

180 $$ %

&''( ) / 1

23

9 "

= 1 cm


15/15

RJPT3SOL030411-15

CODE 0

PAPER-1

ANSWER KEY

PAPER-2CODE 0

Q.No. 1 2 3 4 5 6 7 8 9 10

Ans. B C A D B B A D B B

Q. No. 11 12 13 14 15 16 17 18 19 20

Ans. CD ABD AB AC ABCD D D B B B

Q. No. 21 22 23 24 25 26 27 28 29 30

Ans. B T F F F F F D A C

Q. No. 31 32 33 34 35 36 37 38 39 40

Ans. B D B C A B C AB BC ABCD

Q. No. 41 42 43 44 45 46 47 48 49 50

Ans. ABCD AC D A B D C B T T

Q. No. 51 52 53 54 55 56 57 58 59 60

Ans. T T F T B A D A B C

Q. No. 61 62 63 64 65 66 67 68 69 70

Ans. A D C D CD ABD ACD BCD AC B

Q. No. 71 72 73 74 75 76 77 78 79 80

Ans. D C B B A T T F F F

Q. No. 81

Ans. T

Q.No. 1 2 3 4 5 6 7 8 9 10

Ans. A D D C A A B B AC BD

Q.No. 11 12 13 14 15 16 17 18 19 20

Ans. AD ACD BC ABCD (A p, q) ; (B q) ;

(C q, s, t) ; (D r, t).(A q,r) ; (B p,t) ;(C q,r) ; (D p,t)

6 5 3 2

Q.No. 21 22 23 24 25 26 27 28 29 30

Ans. 4 C C D A A A A A ABC

Q.No. 31 32 33 34 35 36 37 38 39 40

Ans. AD BD AB ABD AC (A - q), (B - r),(C -r), (D - r)(A p) ; (B q) ;(C r) ; (D s) 2 5 9

Q.No. 41 42 43 44 45 46 47 48 49 50

Ans. 5 5 A D B B D D A C

Q.No. 51 52 53 54 55 56 57 58 59 60

Ans. AD AC ABD AC BC BC (A) p, q,t, ; (B) p,r ;

(C) p, q,r,s ; (D) q(A) t ; (B) q, r, s, t ;

(C) p, r, t ; (D) t 3 2

Q.No. 61 62 63

Ans. 1 6 1

JPT 3 SolutionBooklet English

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