SM 17 18 XII XII Physics Unit-4 Section-A

Optics and Modern Physics

4

UNIT

Section A : Straight Objective Type 1. Answer (2) Light ray will fall normally on PR.

P

Using Snell's law

30º

1 × sin45º =  × sin30º

60º

45º

1 2



 



1a.

90º

30º

1 2



Q

2

Answer (1) sin C 

1 n



1 2

, C = =

 = 30°



 = 15°

[JEE(Advanced)-2016]

45° 

Also, 1 × sin45 = 

R



1 2

 sin 



C

2. Answer (3) Spherical aberration is inability of mirror to focus all the rays coming parallel to principal axis at a single point. So it exists in concave mirror. Chromatic aberration is due to behavior of light. It does not exist. 3.

Answer (2) For refraction at surface OX 1.sin60° =

60°

3 . sin r





sin r



r = = 30°

1

sinC





c

2

Y

R

sin(90  r )

OP 

X

r

In OP OPY Y , OP

P

O

sin C cos r



 .R

R

. cos r

Critical angle

.R



[sine law ]

cos r 1

C = =

.R



1 3 . cos 30

2R 3

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366 Optics and Modern Physics

4.

Success Magnet (Solutions)

Answer (4) By Snell’s law for refraction = 30° r =

3

In  ABC

2 60°

.tan30° BC = AC .tan30°  

A

r

2 3

30° 30 °

OB = OC OC – BC 3 r



r



2

60° 60 °

O

2 3

B

r C 2 3

r



5.

.

3

Answer (2) = –(2f 0 – 4), v = = –(2f 0 + 6), where f 0 is focal length of lens or f = = –f 0 u = using

6.

1

v

1



u



1

f

 f 0 = 12 cm

Answer (1) Image should be real and virtual for the same magnification for the two given position of the real object. Therefore 

x 2 – f x 1 – f

–1

⇒

– x 1 

f



x 2

– f

 2 f = = x 1 + x 2 

7.

x 1

| f | 

 x 2 2

Answer (3)

E

4E Radius of

Radius of

Radius of

Radius of

beam =

beam = 2r

beam = 2r

beam =

r

Arrangement Arrangem ent (1)

r

Arrangement Arrangem ent (2)

E represents represents energy incident (per unit time) which gets focussed on screen for arrangement (1).

I 1

I 1 I 2





E 4r

2

I 2





4E

 r    2 

4 

2

   4r 

 16

E

2

1 16




367

8. Answer (4) For no average deviation ( = ( – 1) A) (1.48 – 1)1 – (1.64 – 1)3º + (1.48 – 1)2 = 0  1 + 2 = 4º

9.

Answer (1) For no emergence, 

 min  cosec



 min  cosec

A

2 90 2

= cosec 45° 

2

= 1.414 10. Answer (3) v

fish (as seen by bird)

 v fish 

= =

5

55

1 4

 observer  v bird  object

 3  10

m/s

4

11. Answer (3) 2 >  (as it is converging the rays)

and 1 =  (No deviation) 12. Answer (2) Use the relation u 2

f



v

2

D – d

d

4D

d = v – u = 50 – 30 = 20 D

D = v + u = 50 + 30 = 80 2

f



f



(80) – (20) 4  80 75 4

2



v

(80  20) (80 – 20) 4  80



100  60 4  80

u

30 cm 50 cm

cm.

13. Answer (3) Deviation = 360 – 2 i or i =

360 – D 2



360 – 210 2

 75

Where ever the object is placed in between the mirror it's all the images and object will lie on a circle. Number of images = 4 Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456



14. Answer (1) After refraction through water, light is incident normally on the mirror and returns back from the same path and images formed at the radius of curvature of the mirror. 1







=



15



20

O

3 4

15 cm

4 5 cm

3

15. Answer (2) Ray is incident normally. It will not bend and move along x -axis i.e., y = 0

16. Answer (1) In new convention 2

1

–

v

u



1

 2 – 1

3

R

2

2 v 1

2



v 1

v 1

1  2 – 1   R





v 1

f v 1

 2 – 1 R

 2 R ( 2 – 1 )

The first image will behave like a virtual object for the second surface. Now the rays are refracted from second surface and will converge at focus 3 f

3 f

3 f

3 f

3 f

f



–

–







2 v 1



3 –  2 – R

 2 ( 2 – 1 )  3 –  2   2R – R  2 – 1 R



2 – 3 R

( 2 – 1   2 –  3 ) R

(2 2 – 1 –  3 ) R

 3R (2 2 – 1 –  3 )

Put  3 

4 3

, 2 

3 2

, 1 = 1 and R = 30 cm, f = 60 cm.



16a.


Answer (1, 4)

369


As m = –2  v = 60 cm 1

v



1

1



u

1 ⇒

f

60



1 30



1 ⇒

f

f

 20 cm

…(i)

O 30 cm

For reflection from convex surface, 1

1



v

u

1



10

1



f

1

30



2

1 ⇒

10

R



1 30



31

2 ⇒

30

R



2 30

 R = 30 cm



2 R

...(ii)

By lens maker's formula, n

1

30



1 ⇒

20

n

1 3



1 ⇒

2

n

1

3 ⇒

2

n

 2.5

17. Answer (1)  sin = constant 3 sin 30 

sin =

 =

sin

17a.

3 2 2

–1

   

2 sin 

3



8

3  8

  

Answer (2)

(IIT-JEE 2008)

For total internal reflection, at IV,  depends only on I and IV 

17b.

sin 



 IV   I

1 8

Answer (1, 3, 4)


Since refraction occurs at a set of parallel surfaces, n1 sini = n2 sin f i

l

f

The lateral displacement will depend on how the n(z ) varies in the medium. Clearly, l does not depend on n2. Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456



18. Answer (3) Using Snell’s Law, sin45° =

T.I.R.

air

2 sin r

60°

1 2



sin r =

1



2

45°

sin r

30°

60°

30°

45°

1 2

 r = 30°

Now at point R , total internal reflection will take place. Clearly, incident and emergent ray are antiparallel. T = 180°

18a.

Answer (1, 2, 3)

IIT-JEE 2010

At P 1  sin 60



3 sin r

r = 30°



60°

From geometry, angle of incidence at Q is 45°

60° 30°

P

At Q

45°

3 sin(45) 

3 2

45°

1

90°

TIR takes place



Q

30° 75°

R

D

60°

At R angle of incidence is 30° By symmetry R = 60° From second diagram, angle of deviation is 90° 19. Answer (4) min =

2 sin

–1

A     sin  – A 2  

min = 2 sin–1(1.5 sin30°) – 60° = 37°

The maximum value of deviation is given by min = 90 + i min – A i min =

sin

–1

  2 – 1 sin A – cos A  

–1 2 = sin  (15) – 1 sin 60 – cos 60  

  using r 2  C    r 1  A – C    sin i min     sin r  

= sin–1(0.4682) = 28° max = 90° + 28° – 60° = 58° Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456


19a.


Answer (1)

371

(IIT-JEE 2008)

Angle of refraction at minimum deviation is

A 2

.

20. Answer (3) Using

1



v

1



u

1

for both Q  and P  (end points of the image of PQ in plane mirror) we get that image in

f

convex mirror is virtual, erect and 3 cm long.

Q

P 5 cm

Q 5 cm

P 

5 cm

15 cm

15 cm

30 cm

45 cm

21. Answer (4) The ray will be paraxial ray and does not converge at the focus Use the relation PQ = x =

x =

R –

x=

R –

x



R –

R

2 cos 

R

2 cos 30

30° 30°

R 2

2

  

R  1–

C

f

Q

x

P

3

   3 

1

22. Answer (4) Y tan 30

1 3

y







y 20

30°



X

y 20

30°

20 3



20 



3 

Coordinates of image are  20, 

 




23. Answer (3) The rays after refraction through lens incidence normally on the curved mirror and after reflection rays retraces their path and finally form the image at the object. The focal length of the combination 1

f 1

f

1



f 

2



1

–

f m

1

f 

1

–

f 



⇒

f m

1



F

2

f 

–

R   ∵ f   2  

2

O

m

m

R m

Consider u = object distance for the combined system v = image distance for the combined system

According to the question u = v

f  = 30 cm

R m = –30

Thus, u

f = 2 u

1 u

1

2 2

=

f  1

=



u

–

30

2

R m



1 30

2 30

u = 15 cm

24. Answer (2) The vessel should be filled to such a height that a ray diverging from O proceeds along the dotted line. Let h be the required height. Since the vessel is cubical in shape,  ACB = Angle of incidence = 45° Now NO = h – b = h tan r or

h

=



E

b

A

1– tan r

sin 45 sin r

or sin r



1

1–

h

1 2

B

N

O

C

2

b

Hence h =

D



b 2 – 1 2

2 – 1 – 1

2

2 – 1

10 2 

or h =

   

2

16 9

16

–1

9

 – 1 – 1  



10 23 23 – 3



10  4.8 4.8  3



26.7 cm




373

25. Answer (4) y O (0, a, 0)

I

y (0, b, 0)

O

(0, a, 0)

x

x

Case (1)

Case (1) I (0, –b, 0)

If in case (1), image is as shown (which is given) we can conclude that in case (2) image is as shown 26. Answer (2) 10 cm

20 cm

After refraction through the concave lens, rays become parallel. 27. Answer (2) Direction perpendicular to mirror is y axis. Shown are the components along y -axis, as only these components change. y

2 m/s Object x

3 m/s z

 Component of velocity of the image in y -direction

= 2 × 3 – 2 = 4 m/s ˆ  V image  i ˆ  4 j ˆ  5k 

m/s

28. Answer (2) At y =  y = 0

 x = 0  x = 2. Some where in between,  x =  .

So only three well defined maximas are formed. Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456



29. Answer (4) 

D d

As water < air So, if d is decreased, D is increased or  is increased original  can be restored. 30. Answer (1) I 1 I 2

2







a1 a2



2

a1

9I

2



a2

4I

3 2

2  a  a2  5   1       25  1   a1  a2 

I max I min

31. Answer (3) As slits are producing same intensities ( I ), 

I net = 4I cos2

2



I max = 4I



I at point =

=

3 4

3 4

th of maximum

 4I  3I



I net = 3I = 4I cos2

cos



  

2

2

2

= 2n ±

 = 4n ± 2



2

3

=





 6

 3

 x = 4n ±

 x =

 3

     4n   2  3 

 x = 2 n ±

 6



dy  = 2n + 6 D



y =

  D   2n   d  6 

For n = 0 y =

3  10 6

4

= 0.05 mm.



31a.


Answer (2) I

375

JEE (Advanced)-2013

  I max cos 2   2

1 2

  cos2   2



cos  = 0







 x 



 x  (2n  1)

 3 5  7 

, , , 2 2 2 2

 3 5  , , 4 4 4

 4

32. Answer (4) y

v

v

s i n



v cos

 

v cos





v

O

x

v sin

(see the figure) So, v  v cos 2iˆ  v sin 2jˆ 

I

v I  10 cos 2iˆ  10 sin 2j ˆ 

33. Answer (1) S 1

Path difference = 600 nm

2400

P

For maxima 1800

Path difference = n  

600 × 10–9 = n. 

600 n

0 0 3 0

S 2

× 10–9 m

 (in nm) = 600, 300, 200, 150

{for n = 1, 2, 3 & 4 }

only  = 600 nm falls in the visible range. 34. Answer (1) Optical path = n. x

(n = refractive index)

Optical path through glass = n × 4 Optical path through water = 

n × 4 =



n = 2.

4 3

4 3

6

6




35. Answer (2) Extra optical path introduced by upper glass = ( n – 1)t = (1.5 – 1) t =

t 2

= 0.5 t

Extra optical path introduced by lower slab  4  =  3 – 1 t (2)  

=

2t 3

 0.67 t



lower ray travels extra path at centre



hence central maxima will shift in downward direction.

36. Answer (3) Two independent sources cannot be coherent sources. 37. Answer (1) For constructive interference in refraction path difference = n 

2 t = n



t



n

=

2

1 300 2  125



300 2 .5

 120 nm

38. Answer (3) Two independent sources can not be coherent, which is a must to observe interference. 39. Answer (1) After reflection from mirror, there is a phase change of , which is equivalent to path difference of

 2

 After reflection from mirror, for destructive interference,

Geometrical path difference =  x = n  BC  CD 





3 x n

⇒

  3 x ,

3 x 2

,

3 x 3

3 x

etc.

40. Answer (2) At P ,  x = d sin ~d sin For maxima,  x = n . 41. Answer (3) As maximum path difference is 5000 Å, three maximas will be formed at 

 d

, 0.



41a.


Answer (1, 2)

377

(IIT-JEE 2008)

At centre,  x = 0 for which maxima is obtained. The path difference at a large distance from the screen  x  d . When d = , the path difference is between 0 to . Only central maxima exists in that case when  < d < 2,  x lies between 0 to 2 . So more than one maxima will be obtained. When both the slits give same intensity, dark fringes are perfectly dark. 42. Answer (4) As minimas of both the wavelengths are formed in the region of complete darkness. The region of completed darkness will satisfy the condition y  (2n – 1)

400  D 560 D  (2m – 1) 2d 2d

n = 4, m = 3 and n = 11, m = 8 can satisfy the above equation.

42a.

Answer (1, 2, 3) 

D

Also,

2  1

⇒

d

y



m

[JEE (Advanced)-2014]



⇒

m1

∵

2 > 1

 m2

3rd maxima of 2 lies at 3(600 nm) 

D d

 (1800 nm)

5th minima of 1 lies at (2  5  1), 400  Angular separation is

 D





⇒

d

D

2d

D d

 (1800 nm)

D d

It is more for 2.

43. Answer (4) 1.sin1 = sin2 sin 1





1 p

sin  2

q r

p



r s



p

2

s

r

44. Answer (3) tan 



dy dx



2 x

2

y



2 h

 cos = 1 

h



sec 



1  4h 1

2  1 4




45. Answer (4)  A  A    2   A  sin   2

sin 



n

2cos

A



2

n

1  n 

A = m =

2cos

  = 2

2sin

1

1 n

2

4

46. Answer (3) (

I

I)

I 0

I 



 I 0 (Initially)

2



4

When one slit is covered   

I



2

    I  4  I

I  

9 4

I 

9 I 0 16

47. Answer (2) t

N

5





 1  t ½    2 

N 0

N 0

 1     2 

7 2

N 0 = 60 grams

48. Answer (1) Here  > critical angle 1

    sin1  2   1 

Now, 1sin(90 – ) = 1 × sin  sin 

  

90 –



2

 1 cos   12  22

Since,  > critical angle sin 

 12   22




379

49. Answer (4) PQ = 2r cos V rel = 2 (PQ) sin

= 2 2r cos sin = 2r sin2 50. Answer (1) f 1



f 2



e –1 e 1 2

f 1 > f 2

51. Answer (1) tan 



tan 



t 

v sin   gt v cos  tan 



gt v cos 

(tan   tan )v cos  g

52. Answer (3) Path difference created by medium = ∫ ( ) �dx – dx  ∫ (1  ax )dx – dx  axdx = x

2

For minima at O,

al

2





⇒a

2



l

∫

0

2

axdx 

al

2

 2

l

53. Answer (3) ·sin45° = 1 × sin  sin 



1  0.4t 2

45° cos  

d  dt



0.4 2



d  dt



0.4 (1.4)2 2 1– 2



 2 rad/s

 = 2 rad/s

54. Answer (1)  2

Speed of man in medium 2 is

V

Speed of man in medium 2 is

V

1 2 1



3 2



V

3 1




55. Answer (3) R



R1 R 2

a = R 2 =  R 1R 2 a1 = R 12

a2 = R 22 a1 a2 = 2 (R 1R 2)2 a1a2



a

 R1 R2  a

a1a2

56. Answer (2)  R(t 1/2  1 min)

P t 0

( 4 N 0 )

Q t 0 ( N 0 )

 R (t 1/2  2 min)

After four minutes, the number of nuclei of P and Q are same and equal to

N 0 4

.

So number of nuclei of R present at this time is N0  3N0   4 N 0  4   4   

9 N 0 2 i 4

57. Answer (4)

75



25



25



25



Given circuit i 3 = 0

Can be looked upon as i 1 = i 2 + i 4

i 2

i 1 = 0.2 A

75

i 1

i 2 = i 1 = 0.1 A



15 V

58. Answer (1) 238

92

90

U

X

234



90



91

234



U

Y

234



He

4

2

0

1

59. Answer (2) 2

13.6

Z n

2

 3.4

Z = 1,

n

2

13.6



3.4



4

n = 2 M

q 2m

L

e nh 2m 2



eh 2m




381

60. Answer (3)  m1 + m2

M



Rest



P 1

P 2

Using conservation of linear momentum 



P1  P 2

0





P1









P2

 P

h P

1

So,   1 2

61. Answer (3) k



=

hc



 W

1242eV 400nm

 2 eV  1.105 eV

Linear momentum = =

2  9.1 10

31

mv



2 mk

 1.105  10 19  1.6

= 5.67 × 10–25 kg m/s Now r 

mV qB

 B = 35.8 × 10–6 T

62. Answer (1) So, if A = 0, B = 1

A

A Y = A.B

Y = 0

(AND gate)

and if A = 1, B = 1 Y = 1

B

B

63. Answer (3) When A is at positive potential and B is at negative potential. Diodes are in forward bias. So, R eq = 8  When A is at negative potential and B is at positive potential  Reverse bias R eq = 19  Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456



64. Answer (4) At same temperature 1



v

m

v v

1



m 2

mv 1 1

m 1

m v 2 2

2

 H2  O2

m O2





mH 2

32 2





m 1 m 2



2 1

4

65. Answer (3) Using equation KE max



hc



– & 

hc

 cut off

Put values KE max = 1.5 eV & cutoff = 230 nm We get,  = 180 nm 66. Answer (1) Population covered = P × d 2 (P = population density) d



2Rh

On solving  Population covered = 76.8 lakhs 67. Answer (1) For perfectly absorbing body radiation, force is F

I  Area normal



to Intensity

Speed of light 2



F



I R c

68. Answer (1) Using photoelectric equation eV s

eV s





hc



–

1242

– 2

200

V s = 4.2 V Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456


68a.


Answer (7) hc







v

0



(IIT-JEE 2011)

ev 0

ne

4

0

1240



eV

200

r



xne



 4.7eV    eV  40 r 



6.2 – 4.7



n



383



9

 199  n  1.6  10 –19 10 –2

1.5  102 9  1.6  10 10

 1.04  107 

z

7

69. Answer (3) E = a(cos0t + cos t cos0t )

 Maximum frequency of incident photon is f max =

  0 2

 KE max = hf max – 

70. Answer (1) K max = hf –  KE max = hf – hf 0

Now KE 1 = h × 3 × 1014 – hf 0

...(i)

KE 2 = h × 2 × 10 14 – hf 0

...(ii)

Divide both (1) & (2) we get 2

3  10

14

– f 0

2  10

14

– f 0

 f 0 = 1014 Hz

70a.

Answer (1) 1 2 1 2

2

mu1



2



mu2

hc

1 hc

2 hc

2

 u 1  1  u   hc 2

2 4hc

2


 W

 W

 W  W

 4W 

hc

1

 W

 ∵ 

u 1 u 2



 2




384 Optics and Modern Physics 4hc

hc



2

1

 3 W

4  1240

⇒


310



1240 248



11 = 3 W



W = 3.7 eV

 3W  16 – 5 = 3 W

71. Answer (3) E



2hc

2E 

3hc

...(i)



–



–



...(ii)



Solve & then  

hc



72. Answer (2) E = h –  = 4.9 – 4.5 = 0.4 eV p 

2 mE



31

2  9.1 10

–25  0.4  1.6  1019 = 3.45 × 10 kg m/s

73. Answer (1) Applying conservation of linear momentum h



mH V H



mH = 1.67 × 10–27 kg

 = 122 nm

We get V H = 3.25 m/s 74. Answer (2) KE max = hf – 

 KE max is not solely dependent on f .

75. Answer (2) Photon emitted from coolidge tube have wavelength greater than cut off wavelength & till infinity. 76. Answer (2) As voltage across is increased all wavelength will decrease i.e. in graph shifted towards left. 77. Answer (4) f k



f 1 f 2

or



3 4

Rc ( z – 1)

2

2

 Z – 1   31– 1        1  Z 2 – 1   51– 1  f 1 f 2



9 25

so

f 2



25 9

2

f 1




385

78. Answer (1)  a(Z  b )

f

Where a is slope of

and

a



Rc

 1   n2  1

f

Z graph

,

  2  n2  1

–

for k  n2 = 3 & n1 = 1 k , n2 = 2 & n1 = 1 ak







ak 

32 27

79. Answer (2) 99.8% of energy is lost as heat in the target Heat in target =



VI 

99.8 100

J/s =

238.75 cal/s

80. Answer (4) Value of a is different for different members of a series. 81. Answer (4) mvr





v 1

i

nh 2



 ev 1 

h 2 mr 1

e v 1 2r 1



(for n = 1) eh 2

2

4 mr 1

Magnetic dipole moment, M1



 i  A1 

  

eh 2

2

4  mr 1

 nr12 

 

  M  B  M B sin30º 

eh 4 m

ehB

8m

82. Answer (2) r



0.53 n

2

n

Z

2

4



1 1

 n = 2

Z is Atomic Number r is orbital radius

83. Answer (1) No spectral line is emitted as energy required is at least 10.2 eV. Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456



84. Answer (2) K.E. = |E | = 3.4 eV h



2 m(K.E.)

85. Answer (4) h

 

p

, for e– in uniform magnetic field p is constant   for this e– is constant

But for e– in uniform electric field  may increase or decrease as p is variable. 86. Answer (1) Binding energy

2

E

13.6Z



n

2

eV

i.e. E = 13.6 Z2 eV

For third Balmer series hc



 1  2  2

2



13.6Z

–

1 5

2

  

Put values of h, c,  and calculating 13.6Z2 = 54.43 eV as binding energy. 87. Answer (4) All these series are found in emission spectrum. 88. Answer (1) T n  n3

 n    1   n2 

8

3

⇒

n1



2

n2

 n1 = 4 & n2 = 2

89. Answer (1) 1   1  2 2   2

1

 R 

 

R 

4 3644

R 4

1

,

C

1  1  2 2 1  

 R( z  1) 

z – 1 = 30.2



z



31

90. Answer (4) E 1



12375 1085

E 1 + E 2 = 1 n

2



,

E 2

12375



12375 304

1 1   1 2  1085  304  = 13.6  (2) 1  n 2 

0.04




387

91. Answer (2) Shortest wavelength of Lyman series is x 1 x

 R

Now wavelength of first member of Balmer series 1



 1

 R 

1 



–

 4

9 

36 x

 

5

91a.

Answer (1) f =

f 0

(IIT-JEE 2011)

 v  v  u   v  u   v    

=

8

=

8

 32 0  1 0   320 –10    33 21

= 8.5 kHz 92. Answer (3) 1







R 

1

 2

For n = 3,

–

2

1

 1

1 n



For n = 4,  

2

  

from n = 3, 4, 5 .......

5R 36

3R 16

93. Answer (1) Lyman series is in UV range & when system in question will radiate it will radiate in UV region. 94. Answer (1) Since no energy is absorbed by atom in ground state 

collision must be elastic

95. Answer (2) 1





1 



9 

 R 1 –



96. Answer (2) As energy is shared by H atom and e – so electron is emitted with lesser energy and greater wavelength. Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456



97. Answer (1) f

 R 

R 2

cos 

A 

 R > 2f AB = R – f

C

BC = (R – f )sin30º BC 

BC 

R

º 0 3

f B

C

R–F

 f 2

f 2

98. Answer (1) N 1  1 = N 2  2 N A N B

=

 B x   A 1

99. Answer (3) Activity,



A

– dN dt

A = + N 0 e–t A at t = 0 is

 m N A

A = N 0 =

99a.

M

Answer (9) I AB =

2



(IIT-JEE 2011)

5 4

2



2 md

 2 md 2 ma2    = 9 × 10–4 kg-m2 + 2  2  5 4

100. Answer (2)  = i + e – A

(for same deviation there are two values of i ) the second value of i for same deviation is the angle of emergence for first value of i So, 23º = 15º + 35º – A  A = 27º

101. Answer (3) A1 =  N 0

e

A2 =  N 0 A2 A1

A2



e

 t 1

e

 t 2

 ( t 1 – t 2 )

 A1 e

 t 1 – t 2     T 




389

102. Answer (3) During  and  decay either atomic number or atomic mass changes. But during  -decay neither the atomic number nor the atomic mass changes. 103. Answer (1) We know, N = N 0 e–t N 1 = N 0 e–11t N 2 = N 0 e–t N 1



N 2

1



5

1

e



2



e

–10t

t

104. Answer (3) 2 1H



2 1H



1 1H

2 1H



3 1H



4 2 He



3 1H







1 0n

4.03 MeV

 17.59

MeV

10 40 ( 4.03  17.59)  10 –13   3 1016

t

t = 1012 approx

105. Answer (2) It is chargeless but has spin. 106. Answer (3) Let N 0 be R n concentration initially. Approximately 5 half lifes are therefore Rn to convert to Po and 20000 half lifes for Po to Pb and conversion of Pb to Bi is after 10.6 hrs. Pb has maxium number of atoms after five minutes, but Bismuth has least mass.



107. Answer (2) From Carbon Nitrogen cycle. 108. Answer (1)  a( z  b ) (Moseley's law)

f

for K  line b = 1 f1



f2



c

1 c

2

 a 2 (z 1  1)2  a 2 (z 2  1)2



1 ( z 2  1)2   2 ( z 1  1)2



( z  1)   2 4 (11  1)2

2

 z 2 = 6 Aakash Educational Services Pvt. Ltd. Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456



108a. Answer (2)  Cu

2

(ZMo  1)2



Mo


(ZCu

 41     2.14 2  28   1)

109. Answer (4) E = 13.6 (Z–1) 2

 Z = 1.7

110. Answer (3) mvr







v

h

r 

2

mv eB

2

or

2

m v eB



h 2

heB 2m

2

111. Answer (1)  hc

Hard X-ray means smaller wavelength 

 



 eV 



V is increased.



112. Answer (1) As V is increased then min decreases but k remains same  ( k – min) increases.

113. Answer (4) Both characteristic & continuous X-rays are possible. 114. Answer (3) Since collision is head on 

One photon hits electron first & no energy is gained by electron as E P <  & same phenomena with other photon



No emission of electron.

115. Answer (4) By, Einstein’s photoelectric equation, 1

mv

2



2

k 1 = k 2 = k 2 –

k 2 =

hc



hc

3

3

3



–

k 1 

4

–

–

4 hc

4



 3

k 1 

 3

k 2 is more than

4 3

1/ 2

k 1 so

 4  v 2 is more than    3 

v

.




391

116. Answer (2) 1   30 × 42.5    100   13.6 1 – 2     n  1 n

n

3  42.5

 1–

2

2

136



136



136 – 12 7.5

136 8.5

 16

n = 4

Number of lines in emission spectra =

4(4 – 1) 6 2

117. Answer (2) 9

 N0 e – t

N0

10

N = N0 (1 – e– 2 ) t

 81    100 

N

= 1– 

N0

N

 100 = 19%

N0

118. Answer (3) M = i A



eV

r 2 

2r

e Vr 2



e

mVr

2m



neh 4m

for first excited state n = 2 m

eh



2m

119. Answer (2) 1

e

2

�

40



r

1 8

 1 e2  �  8 r  0  



 r = 16 r 0  r = n2r 0  n = 4

 

E = E 2 – E 1 = 13.6 1 – E  1 2

   E = 12.75 eV 16  1

1  m1m2  1 m  2 3 2 2   v1 – v 2  (1– e )    v  2  m1  m2  2 2  4

2 mv





 34eV




120. Answer (4)  0      2 

Most energetic electrons will be obtained from radiations of frequency     0     0     KEmax    h    – KEmax  2   2 

h

 3.6  1015  6  1014  –19   0.592  10  2  22  

  6.6  10 –34 

15–34

 

7  0.3  4.2  10 2

– 0.592  10

–19

= (4.41 – 0.592) × 10 –19 J

2.39 eV








SM 17 18 XII XII Physics Unit-4 Section-A

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