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Page 1 of 12 MOMENTUM : JBP. 1525, Wright Town, Ph. (0761) 4005358, 2400022,
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JAN ISSUE - PHYSICS FOR YOU - IIT JEE PAPERP H Y S I C S
1.(a) KEmax = (5 – φ) eV
when these electrons are accelerated through 5V,
they will reach the anode with maximum energy
= (5 – φ + 5)eV
∴ 10 – φ = 8
φφφφ = 2eV Ans.
Current is less than saturation current because
if slowest electron also reached the plate it would
have 5eV energy at the anode, but there it is
given that the minimum energy is 6eV.
2.(b) Maximum energy of emitted photon
= 100
Rch49
4800
= 49
48Rch
Energy released if electron jumps from level
n′ to level 1 = Rch
′−
2n
1
1
12
∴ Rch
′−
2n
1
1
12 =
49
48Rch
∴ n′ = 7∴ n = 6Each atom can emit a maximum of 6 photonsQ there are 100 atoms, maximum number ofphotons that can be emitted = 600.
3.(d) λ = p
h =
mE2
h∴ E = 2
2
m2
h
λ
∆E = m2
h2
λ−
λ 22
21
11
Put λ1 = 0.5 × 10–9 m
& λ2 = 2 × 10–9 m and solve.
4.(a) Frequency of sound heard directly
01 0
330250 252.2
330 5s
V Vf f Hz
V V
− = = = − + Frequency received by wall
1 0
256 330 256 33
330 325s
Vf f
V V S
× ×= = =
− − The same frequency is recived by observer throughreffected wave
259.9 252.2 7.7 .f Hz∴ ∆ = − =
5.(d) 0eHg = in a satellite
∴ pendulum does not oscillate
∴ time period is infinite.
6.(a)
21
02 2
eVGMmm
R h
− + = +
1 20
8
GMm GMm
R h R⇒ − + =
+
1 13
4h R
R h R⇒ = ⇒ =
+
7.(a) Energy of oscillation = 9 - 5 = 4 J
( )2 2 2
2
1 84
2 2 0.01mA ω ω= ⇒ =
×
20.01
0.01 100T
πω π⇒ = ⇒ = × =
8.(a) Open pipe 1 0
0 1
3300.55
2 2 2 300
v vf l m
l f= ⇒ = =
× .
1st overtone of closed pipe = first overtone of openpipe
0
3 24 2c
v v
l l
=
⇒ 0
3 30.55 0.4125
4 4c
l l m= = × =
9.(a) Distance between two succenive maxima =2
λ
14 13 3 2 1
∴ 13 0.13 0.022
mλ
λ= ⇒ =
∴
8103 10
1.5 100.02
Vf Hz
λ×
= = = × .
10.(d)
11.(c)
12.(a)
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JAN ISSUE - PHYSICS FOR YOU - IIT JEE PAPER
13.(a) 32
)1n(n=
−∴ n = 3
i.e., after excitation atom jumps to second excitedstate.Hence n
f = 3. So n
i can be 1 or 2
If ni = 1 then energy emitted is either equal to,
greater than or less than the energy absorbed.Hence the emitted wavelength is either equal to,less than or greater than the absorbedwavelength.Hence n
i ≠ 1.
If ni = 2, then E
e ≥ E
a.
Hence λe ≤ λ
0
14.(c) E3 – E
2 = 68 eV
∴ (13.6) (Z2)
−
9
1
4
1 = 68 ∴ Z = 6
15.(a) λmin
= 13 EE
12400
−=
−
9
11)6()6.13(
12400
2
= 2.435
12400 = 28.49 Ans.
16-18 amplitude A = 4 cm. wavelength λ = 2 cm.
let the wave equation be
2 24siny t x
T
π πα
λ
= + +
[Q the wave is travelling in negative x direction]
at 0t = ; ( )4siny xπ α= +
at 0x = ; 2 2y =
2 2 4sinα∴ =1
sin2
α⇒ =
4πα⇒ =
16.(a) P
dyV
dxν = −
20 2 2 2π π ν⇒ =
10 / .v cm s⇒ =
17.(c) f vλ = 2 10f⇒ × =
5f Hz⇒ ×
2 2 4 5 40Vmax A Afω π π π∴ = = = × × = cm/s
18.(b) Wave equation 4sin 20.2 2 4
t xy
ππ = + +
19.(a) The plate is initially at extreme position. In
equilibrium, if extension in the spring is 0
Y in
equilibrium
0 0
MgkY Mg Y
k= ⇒ =
This is also the amplitude.Angular frequency for spring-mass system is
k
Mω =
20.(a) Time period of oscillation 2M
Tk
π=
given time is 4t T= .
∴ At this time the plate is back to its originalposition
21.(b) ( )nY t = Y-Co-ordinate of palete
Wrt the equilibrium position, the equation ofmotion is
cosMg
y tk
ω=
∴ Equilibrium position has Y Co-ordinate
= Mg
Dk
− +
∴ Position of plate =
( ) cosn
Mg MgY t D t
k kω = − + +
( )1 cosMg
D tk
ω = − + −
22. (A) → q ; (B) → r ; (C) → r ; (D)→(s)Activity of the sample II becomes half in minimumtime. Hence it has maximum disintegration
constant.(b) Activity of the sample III takes maximum lifeto become half therefore it has maximum half -life.(c) Parent nuclei will be left maximum in thesample, for which half life is maximum i.e.minimum decay.(d) It can not be compared without informationabout atomic weight as energy radiated willdepend upon no. of atoms, not upon amount ofsubstance.
23. (A) →s, (B) →q, (C) →s, (D) →s
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JAN ISSUE - PHYSICS FOR YOU - IIT JEE PAPERC H E M I S T R Y
24. (a)
|| || ||O O O
H
H
H
H
H
H
H H H
H
H
H
H
H
HH H
H H
H H H
28 C
22 H
3 O
Hence formula28 22 3C H O
25.(c)
*
*
**
**
**
Total 8 chiral carbons
26.(d) Protonation occurs on that OH group which can
produce more stable carbocation and whilerearrangement phenyl has higher migratoryaptitude over pro-sustituted nitrophenyl.
27.(a) Moving Reverse
2 3CH CH−
2 3CH CH−
Cl
4H−
2 3CH C CH CH CH≡ − − −
2 3CH CH−
Alkyne
3 2 2 2 3CH CH CH CH CH− − − −
3 2 2 3
H
ClCH CH C CH CH +
−− − − − →
28(a) According to formula, it should be either alkyne
or cycloalkene → but in hydrogenation process
only two hydrogens are adding
∴It is a cycloalkene.
As it gives only one monochloro isomer
∴ It should be cyclohexene.
2 /H Ni→ 2 4/Cl h→Cl
6 10C H 6 12C H
29.(c) According to given formula, it should be either
alkyne or cycloalkene → but in hydrogenation
process only two hydrogens are adding
∴ It is a cyclo alkene.
As it gives only one monochloro isomer
∴ It should be cyclooctene.
3O
2 /H Ni 2 4/Cl h Cl
2 2 2 2 2 2OCH CH CH CH CH CH CH CHO− −( )N octane : 1, 8 - dial
[O] [P]
[M]
30.(a) General mechanism will be
(CH3)
3C – X → (CH ) C3 3
+ + X
- , X
- + CH
3Y
→ X – CH3 + Y
-
Since I- is a good leaving group and a good
nucleophile hence reaction ‘a’ will be the fastest.
31.(c)
1
3
SN
AgNO → + AgCl ↓
)(
ClC)CH( 33
II
− 1
3
SN
AgNO →
ncarbocatio3
AgClC)CH( 33°
↓++
1
3
SN
AgNO → + AgCl ↓
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JAN ISSUE - PHYSICS FOR YOU - IIT JEE PAPER
→ 3AgNO N.R
32.(c)
(a) +HBr
(b) +2AgBF4
(c) + HBr
(d) + HBr
33.(c) Assertion is correct Reason is wrong.
34.(c) Assertion is correct Reason is wrong.
35.(b)
36. (d)
37.(c)
38.(b)
39.(a) In metamerism migration of 2CH takes place
to form metamer but here 2 5C H group is
migrating hence incorrect.
40.(b) Aldehydes give positive test with Tollen’s Reagent
but in case of formaldehyde.
H
H2C O H N= + →
OH
H
HC N=
OH
only one oxime is possible (No Geometrical
isomerism)
Anti & syn in case of Acetaldehyde oximes will
be formed.
3CH
H2C O H N= + →
OH
3CH
HC N=
OH
Anti
or 3CH
HC N=
OH
Syn
41.(d)
Sol. 42.(d), 43.(a), 44.(b)
(X)
H SO2 4
(1)
CH MgBr (1eq.)3
(2) (3)
(4)
LiAlH4
H SO / 2 4 ∆(80°)
– H O2
Me
Me
Me
O
Me
O
O
O
OH
HO
Me
COOH
(Y)
(Z)
Me
Me MeOH
(W)
OH
..
..
45. (A) →s, (B) → r,, (C) →q, (D) →p
46. (A) →q, (B) →s, (C) →p, (D) → r
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JAN ISSUE - PHYSICS FOR YOU - IIT JEE PAPERM A T H E M A T I C S
47.(b) en ∫ θθ−θθ−n
0
2 d)tan(sece = 1
put – θ = t
– en 2
0
(sec tan )
n
te t t dt
−
+∫ = 1
– en [ ] n
0
t ttane−
= 1 ⇒ – en [– e–n tan n] = 1
⇒ tan n = + 15
4n
π⇒ =
48.(d) Let I = ∫3
0
2 dx)x(·x f ; put x2 = t
⇒ 2x · dx = dt
= ∫9
0
dt)t(2
1f = 4·
2
1 = 2
49.(c) Put x = t4 ⇒ dx = 4t3dt
I = ∫ +
2
12
3
dttt
t4 = ∫ +
2
1
2
dtt1
t4
=4
+
+−∫2
1
2
dtt1
1)1t( = 4
++− ∫∫
2
1
2
1t1
dtdt)1t(
=4
++
−
2
1
2
1
2
)t1(nt2
tl
= 41
(0) ln 3 ln 22
− − + −
= 4
+
2
3n
2
1l = 4 ln
2
3 + 2
50.(b) Let f (x) = y
dx
dy – y = 2ex; I.F. = e–x
y · e–x = ∫− dxee2 xx
= 2x + C
y (0) = 0 ⇒ C = 0f (x) = y = 2xex
dx
dy = 2[xex + ex] = 0
x = – 1
A = 2 ∫∞−
0x dxxe = ( )
00
2 x xxe e dx−∞
−∞
−
∫
( ) ( )02 0 lim
2
x
xxe e e
−∞
→ −∞
= − − −
= −
Hence Area is 2
51.(b) 3 2 0dx dx dy
x y xdy x y
− = ⇒ + =
( ) cxykxyln22 =⇒=⇒
52.(a) Equation of BC = y – 0 = –4
6(x – 10)
∴ 2x + 3y = 20. Note that (2a, a) lies on ithence 4a + 3a = 20
⇒ a = 20
7
B(4, 4)
(a, a)
(0, 0)A
(2a, a)
(a, 0) (2a, 0) C(10, 0)x
y
∴ Area =400
49
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53.(c)
A
y
B
C
Ox
y=x
3
3
2 2 4x y+ =
(3, 3)
Radius of the smallest circle =2
AB
now AB = 3 + 3 2 + 2 = 5 + 3 2
∴ r = 5 3 2
2
+
54.(a)
M(h, k)
1
1/2
C
M is the mid point of chord . The given circle is(x-1)2+(y-1)2 =1.
( ) ( )2 221 1CM h k= = − + −
2
2 11
2
= −
11
4= −
( ) ( )2 2 31 1
4h k= − + − =
so locus of M is ( ) ( )2 2 31 1
4x y− + − =
55.(c) Since reflection of (0,0) in 2 0x y− + = is
(-2, 2). Also since line of symmetry inclined at
an angle of 45° therefore reflection of 24y x=
will be an upward parabola with same Latus
rectum.
(-2, 2)
(-1,1)
(0, 0)
2 0x y− + =
so the required equation is 24( 2) ( 2)y x− = +
56.(c) { }[ ]( )
[ ]
0
0
0 0
0
0
sin
sin sin
sin sin
sin 0
sin
x dx
x x dx
x dx x dx
x dx
x dx
π
π
π π
π
π
=
−
= −
= −
=
∫∫∫ ∫∫∫
57.(b)
58.(a) Let us consider normal at
2
,b
aea
i.e. 2 2ax
ay a be
− = −
If it passes through the end of minor axis (0, )b−then
2 2ab a b= − 2 2
ab a e⇒ = 2b ae⇒ =
24 4 2 4 2
21 1 0
be e e e e
a⇒ = ⇒ = − ⇒ + − =
2 1 5
2e
− +⇒ =
59.(c) [ ]( )f x 2= − 1 3/ 2x≤ <
1= −3
22
x≤ <
0= 522
x≤ <
=1 5 32
x≤ <
2= 3x =
[ ]3 3 / 2 2 3
1 1 3 / 2 5 / 2( ) 2 1 1f x dx dx dx dx= + +∫ ∫ ∫ ∫
1 11 22 2
= + + =
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60.(c) Required area =261.(d) From the graph it is clear that
2 2
1 1( ) 0 let'say ( )f x dx A f x dx< =∫ ∫
3
2( ) 0f x∴ >∫ & is equal to A−
3
1( ) 0f x dx∴ =∫
[ ]2
1
3( )
2f x dx
−∴ =∫ where as [ ]
3
2
1( )
2f x dx =∫
but 2 3
1 2( ) ( )f x dx f x dx=∫ ∫
62.(d) [ ]( )dy
k y M tdx
= − −
( )
dyk dt
y M t= −
−∫ ∫
10
dyk dt C
y= − +
−∫ ∫
log( 10)y kt c⇒ − = − +
( 10) kty ce−⇒ − =
When 0, 200t y= =
190c⇒ = ∴ 10 190 kty e−− =
63.(c)100 40
200 010
dyk dt
y= −
−∫ ∫
[ ]100
200log ( 10) 40
ey k− = − ×
log 90 log190
40k
−=
−
log 9 log 19
40
e ek−
⇒ =−
log19 log 9
40
−=
64.(d)200
100 010
tdyk dt
y= −
−∫ ∫
[ ]200
400log 10
ey kt⇒ − = −
log190 log 390t
k
−⇒ =
−
log 39 log19t
k
−⇒ =
(log 39 log19)40
(log19 log 9)t
−⇒ =
−
Passage :
Let a possible figure be as
B
R
D
S
A
P (4,3)
C
Q
let the ellipse be
2 2
2 21
x y
a b+ =
Point (4, 3 ) lie on the ellipse
2 2
16 91
a b∴ + = ...(1)
65.(d) Infinitely many ellipses are possible and a cantake any value more than 4. But when a = 5, weget b = 5 which gives us a circle. For vertices ofan ellipse to exist, it cannot be a circle.
66.(c) Clearly, due to symmetry, normals at any twopoints out of P, Q, R, S, will meet on the principalaxes of the ellipse. Hence, the possible set of 4co-normal points are PQCD, RSCD, PSAB,QRAB and ABCD.
67.(a) If 1
2e= then by
22
21
be
a= − , we get
22 2
2
1 31 4 3
4 4
bb a
a= − = ⇒ = ....(2)
Solving (1) & (2), we get
2 7, 21a b= =Area of ellipse
= ( )2 2 2 1 1 4 3a bπ π π= =
68. (A) →s, (B) → r, (C) →q, (D) →p
(A) Equation of Normal at point θ is
sec cosec 2 2ax - by = a bθ θ −
and equation of tangent cos sinx y
+ = 1a b
θ θ
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∴ coordinates of &K L will be
2 2
,0 & ,0cos sec
a a b
aθ θ −
2 22 2 2 2
cos sec
a a bKL b a a e
aθ θ−
∴ = − = −
2 2
cos sec
a a ba
aθ θ−
⇒ − =2
11
cos sec
e
θ θ⇒ − =
21cos 1
cose θ
θ⇒ − = ⇒
2 21 cos1
cos
e θθ
−=
2 21 cos1,
cos
e θθ
−⇒ = because
2 21 cos 0e θ− >
⇒ 2 21 cos cose θ θ− = ⇒ 2 21 cos cose θ θ− = ±
⇒ 2 21 cos cose θ θ= ± 2 2
cos cos 1e θ θ⇒ + =
(B) 2 2 2 2
1 1 1 1
CP CQ a b+ = − 5
36=
(C) Equation of chord of contact is 2 3 9x y+ =
length of 4 9 9 2AD = + − =
∴ 2 2
2 12
13
CA ADAB
CA AD
⋅= =
+
∴ 1 1 12 4 24
.2 2 1313 13
ABD AB DE∆ = ⋅ = =
A
C
D
B
(2, 3)
E
2 3 9x y+ =
(D) 3 4 3 0x y K− − =
3 4 3 (1)x y K− = − − − − − −
4 33 (2)x y
K
+ = − − − − − −
locus of intersection of (1) & (2) is
( 3 )( 3 ) 48x y x y− + =
2 23 48x y− =
2 2
116 48
x y− = is a hyperbola with eccentricity e then
2 2 2( 1)b a e= −
248 16( 1)e⇒ = −
⇒ 24 e=
⇒ 2e =
69. (A) → r, (B) →q, (C) →s, (D) → r
(A)
2
2
-
( 1)
x x
x
e edx
e +∫
2
2 2
1 2-1
2 1 ( 1)
x x
x x
e dx e dx
e e+ +∫ ∫
2 -11 1log( 1) - tan , , 1
2 2
x xe e C A B= + + = = −
(B)( )
13 3
4
x xdx
x
−∫
( )1
2 3
3
1xdx
x
− −= ∫
Let 2 1x t− − =
3
2dx dt
x
−⇒ =
41 4
33 3
2
3 3 11
2 8 8
dtt t c
x
− − = = = − + − ∫
3, 0
8A B
−∴ = =
(C)( )cos8 cos 7
1 2cos5
x xdx
x
−
+∫
152sin sin
2 2
1 2cos5
x x
dxx
− =
+∫
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3
2
5 52 3sin 4sin sin
2 2 2
51 2 1 2sin
2
x x x
dxx
− = −
+ −
∫
2
2
5 5sin 3 4sin sin
2 2 22
(5 )3 4sin
2
x x x
dxx
− = − −
∫
52sin sin
2 2
x xdx= −∫
( )cos2 cos3xdx xdx= − −∫ ∫sin 2 sin 3 1 1
, ,2 3 2 3
x xc A B
− −= + + = = ,
(D) 3
lo g xd x
x∫log
1
ex t
dx dtx
=
=
2tt e dt
−= ∫2 2
1.2 2
t tt e edt c
− − = − + − −
∫
2 21
2 2
t tte e dt c
− −−= + +∫
22 1
2 2 2
ttt e
e c−
−= − + × +−
2 21
2 4
t tte e c
− −−= − +
( )2 2
log 1 1, , 1
2 4 2
e xc A B
x x
− −= − + = = −
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