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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
$$ %
&''(
) !=
23
m ! 2
[H = \H 3 @ 3 = !g
Hing Force
N1 = 3mg cos 60 = 23
mg
3mgsin60 N2 = 3m !g 3
! @ N2 = 2
3 mg
Hinge force = 222
1 NN 2 = 3 mg
67. j
xi
yV /;"
@ dtdx
= y & dtdy
= x @ dxdy
= yx
= x 2 + y 2 = c (circle)
speed = 22 yx 2 = c = constant
a"
=dtdy
i
dtdx
j
a"
= x i y j
@ j
myi
mxF //;"
(conservative)
68. from work energy theoremv1 = v 2 = v 3I 1 + I 2 = 90 & I 2 = 45
@ 231
2I;I2I @ I1, I 2 &I 3 are in AP
T1 = gsinu2 1I
@ T2 = g45sinu2 V
@ T3 = gcosu2 1I
@ 2
TT 232
1 2 = T22 @ T12 , T 22 & T32 are in AP
69. Just before entering the region,
tan I =x
y
u
u
as u x will increase u y remains constantso I will decrease
Sol.(70 to 72)B.T. at (1) & (2)
pa = p 2 + Wgh 2 + 21
Wv22
setting v 2 = 0
h2 = gpp 2a
W/
(h2)max = gpaW 4 50p2 ;# = 10 m Ans.
B.T. at (1) & (3)
pa = p a + 21
Wv2 Wgh 1
@ v = 1gh2 Ans.
From B.T. at (1) & (2)
@ v2 = )ghpp(2
22a W//W
@ (v2)max = )ghp(2
2a W/W (setting p 2 = 0) = 10 m/s Ans.
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73.
u = 20 v =3
20/ cm m =uv/ =
31/
d = 161
2 2$$ % &
''( )
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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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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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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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
8/12/2019 JPT 3 SolutionBooklet English
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