     • date post

02-Jan-2020
• Category

## Documents

• view

8

0

Embed Size (px)

### Transcript of JAN ISSUE-PHYSICS FOR YOU-IIT JEE ISSUE... JAN ISSUE - PHYSICS FOR YOU - IIT JEE PAPER P H Y S I C S

• Page 1 of 12 MOMENTUM : JBP. 1525, Wright Town, Ph. (0761) 4005358, 2400022,

NGP. : 24 , Pragati colony, Opp. sai mandir , Ph. (0712) 9371945613, 2222911 Website: www.momentumacademy.com

JAN ISSUE - PHYSICS FOR YOU - IIT JEE PAPER P 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

Rch 49

4800

= 49

48 Rch

Energy released if electron jumps from level

n′ to level 1 = Rch  

  

′ −

2n

1

1

1 2

∴ Rch  

  

′ −

2n

1

1

1 2 = 49

48 Rch

∴ n′ = 7 ∴ n = 6 Each atom can emit a maximum of 6 photons Q there are 100 atoms, maximum number of photons that can be emitted = 600.

3.(d) λ = p

h =

mE2

h ∴ E = 2

2

m2

h

λ

∆E = m2

h2 

 

λ −

λ 22 2 1

11

Put λ 1 = 0.5 × 10–9 m

& λ 2 = 2 × 10–9 m and solve.

4.(a) Frequency of sound heard directly

0 1 0

330 250 252.2

330 5 s

V V f f Hz

V V

 −  = = =   − +   Frequency received by wall

1 0

256 330 256 33

330 325s

V f f

V V S

  × × = = = 

− −  The same frequency is recived by observer through reffected 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)

2 1

0 2 2

eVGMm m R h

  − + = +  

1 2 0

8

GMm GMm

R h R ⇒ − + =

+

1 1 3

4 h R

R h R ⇒ = ⇒ =

+

7.(a) Energy of oscillation = 9 - 5 = 4 J

( ) 2 2 2

2

1 8 4

2 2 0.01 mA ω ω= ⇒ =

×

2 0.01

0.01 100 T

π ω π⇒ = ⇒ = × =

8.(a) Open pipe 1 0 0 1

330 0.55

2 2 2 300

v v f l m

l f = ⇒ = =

× .

1st overtone of closed pipe = first overtone of open pipe

0

3 2 4 2c

v v

l l

    =   

   

⇒ 0 3 3

0.55 0.4125 4 4

c l l m= = × =

9.(a) Distance between two succenive maxima = 2

λ

14 13 3 2 1

∴ 13 0.13 0.02 2

m λ

λ= ⇒ =

∴ 8

103 10 1.5 10

0.02

V f Hz

λ ×

= = = × .

10.(d)

11.(c)

12.(a)

• Page 2 of 12 MOMENTUM : JBP. 1525, Wright Town, Ph. (0761) 4005358, 2400022,

NGP. : 24 , Pragati colony, Opp. sai mandir , Ph. (0712) 9371945613, 2222911 Website: www.momentumacademy.com

JAN ISSUE - PHYSICS FOR YOU - IIT JEE PAPER

13.(a) 3 2

)1n(n =

− ∴ n = 3

i.e., after excitation atom jumps to second excited state. Hence n

f = 3. So n

i can be 1 or 2

If n i = 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 absorbed wavelength. Hence n

i ≠ 1.

If n i = 2, then E

e ≥ E

a .

Hence λ e ≤ λ

0

14.(c) E 3 – E

2 = 68 eV

∴ (13.6) (Z2)  

  

 −

9

1

4

1 = 68 ∴ Z = 6

15.(a) λ min

= 13 EE

12400

− =

 

  

 −

9

1 1)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 2 4siny t x

T

π π α

λ  

= + +   

[Q the wave is travelling in negative x direction]

at 0t = ; ( )4siny xπ α= +

at 0x = ; 2 2y =

2 2 4sinα∴ = 1

sin 2

α⇒ =

4 πα⇒ =

16.(a) P dy

V dx

ν = −   

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 2 0.2 2 4

t x y

π π  = + +   

19.(a) The plate is initially at extreme position. In

equilibrium, if extension in the spring is 0

Y in

equilibrium

0 0

Mg kY Mg Y

k = ⇒ =

This is also the amplitude. Angular frequency for spring-mass system is

k

M ω =

20.(a) Time period of oscillation 2 M

T k

π=

given time is 4t T= .

∴ At this time the plate is back to its original position

21.(b) ( )nY t = Y-Co-ordinate of palete Wrt the equilibrium position, the equation of motion is

cos Mg

y t k

ω=

∴ Equilibrium position has Y Co-ordinate

= Mg

D k

  − +   

∴ Position of plate =

( ) cosn Mg Mg

Y t D t k k

ω = − + +   

( )1 cosMgD t k

ω = − + −   22. (A) → q ; (B) → r ; (C) → r ; (D)→(s)

Activity of the sample II becomes half in minimum time. Hence it has maximum disintegration

constant. (b) Activity of the sample III takes maximum life to become half therefore it has maximum half - life. (c) Parent nuclei will be left maximum in the sample, for which half life is maximum i.e. minimum decay. (d) It can not be compared without information about atomic weight as energy radiated will depend upon no. of atoms, not upon amount of substance.

23. (A) →s, (B) →q, (C) →s, (D) →s

• Page 3 of 12 MOMENTUM : JBP. 1525, Wright Town, Ph. (0761) 4005358, 2400022,

NGP. : 24 , Pragati colony, Opp. sai mandir , Ph. (0712) 9371945613, 2222911 Website: www.momentumacademy.com

JAN ISSUE - PHYSICS FOR YOU - IIT JEE PAPER C 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 formula 28 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 while rearrangement phenyl has higher migratory aptitude 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

Cl CH 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

(CH 3 )

3 C – X → (CH ) C3 3

+ + X

- , X

- + CH

3 Y

→ X – CH 3 + 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 ↓

• Page 4 of 12 MOMENTUM : JBP. 1525, Wright Town, Ph. (0761) 4005358, 2400022,

NGP. : 24 , Pragati colony, Opp. sai mandir , Ph. (0712) 9371945613, 2222911 Website: www.momentumacademy.com

JAN ISSUE - PHYSICS FOR YOU - IIT JEE PAPER

 → 3 AgNO

N.R

32.(c)

(a) +HBr

(b) +2AgBF 4

(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