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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)



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



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 ↓




→ 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



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




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

= + + =




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




∴ 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

− =

+∫




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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