EM Waves Practice Questions - DPP for NEET

 Page 1


1. (d) m
0
 = 4p × 10
–7
, e
0
 = 8.85 × 10
–12
 
2
2
N – m
C
so c = 
00
1
me
 = 
8
meter
3 10
sec
´
.
2. (b) Wavelength of visible spectrum is 3900 Å
 
– 7800 Å.
3. (a) l
g–rays
 < l
x-rays
 < l
a–rays 
< l
b–rays
.
4. (c) Electric field between the plates of the capacitor is given
by
0
E
s
=
Î
   or  
0
q
A Î
Flux through the area considered
f = 
00
qAq
A 44
´=
ÎÎ
Displacement corrent i
d 
= Î
0
E
d
dt
f
            = Î
0 
× 
0
d qi
dt44
æö
=
ç÷
Î èø
5. (a) Electric field between the plates is
E = 
0
Q
A Î
 \ f
E 
= E. A  or  
0
Q
A Î
 × AA
\ i
d 
= Î
0
E
d
dt
f
   or  Î
0
0
dQ
dt
æö
ç÷
Î èø
\ i
d 
= 
dQ
dt
 = i (charging current)
Hence i
d 
= 1A
6. (c) E × B
7. (b) Since the direction of propogation of EM wave is given
by E × B \ ( )
^ ^ ^
j
i k ´ =-
8. (d) Speed of E.M. wave = 
0 0 rr
1
µµ ÎÎ
 in medium hence
it will travel with different speed in different medium.
9. (b) I = 
1
2
 Î
0
cE
0
2
E
0 
= 
0
2I
c Î
 or  
9
8
2 500 10 36
3 10
´ ´ ´p
p´´
E
0 
= 2
3
 × 10
2 
N/C
10. (b) I = 
P
A
   or   
21
4 48
=
p´p
 W/m
2
I = 
1
2
 Î
0 
E
0
2
c
E
0 
=
0
2I
c Î
 or 
9
8
2 1 36 10
8 3 10
´ ´ p´
p´´
= 30 N/C
11. (a)U
B  
= 
2
0
0
B
4µ
Also     
0
0
E
c
B
=
  \ B
0 
= 
0
E
c
Hence B
0 
= 
0
00
E
1
µ Î
\     U
B 
= 
2
0 00
0
Eµ
4µ
Î
  or   
12
100 8.84 10
4
-
´´
\  U
B 
= 2.21 × 10
–10 
 J/m
3
12. (a) The speed of electromagnetic waves and in a medium
is given by
n = 
1
(µ) e
Where µ and e are absolute permeability and absolute
permittivity of the medium.
We know that, µ = µ
0
µ
r 
and e = e
0
e
r
. Hence
( ) ( ) ( )
rr 0 r 0 r 00
1 11
v
µ µµ.µ
= =´
e eee
or v = 
( )
rr
c
µ e
  or  e
r 
= 
2
2
r
c
(µ) n
\ e
r 
= 
82
82
(3 10)
(2 10)1
´
´´
 = 2.25
13. (a) Given B
y 
= 2 × 10
–7 
sin (0.5 × 10
3
x + 1.5 × 10
11
t)
Comparing it with a standard equation for a progresive
wave travelling along the negative direction of x-axis
is
y = r sin
2p
l
 (x + vt) = r sin
2x 2vt pp æö
+
ç÷
èø ll
Page 2


1. (d) m
0
 = 4p × 10
–7
, e
0
 = 8.85 × 10
–12
 
2
2
N – m
C
so c = 
00
1
me
 = 
8
meter
3 10
sec
´
.
2. (b) Wavelength of visible spectrum is 3900 Å
 
– 7800 Å.
3. (a) l
g–rays
 < l
x-rays
 < l
a–rays 
< l
b–rays
.
4. (c) Electric field between the plates of the capacitor is given
by
0
E
s
=
Î
   or  
0
q
A Î
Flux through the area considered
f = 
00
qAq
A 44
´=
ÎÎ
Displacement corrent i
d 
= Î
0
E
d
dt
f
            = Î
0 
× 
0
d qi
dt44
æö
=
ç÷
Î èø
5. (a) Electric field between the plates is
E = 
0
Q
A Î
 \ f
E 
= E. A  or  
0
Q
A Î
 × AA
\ i
d 
= Î
0
E
d
dt
f
   or  Î
0
0
dQ
dt
æö
ç÷
Î èø
\ i
d 
= 
dQ
dt
 = i (charging current)
Hence i
d 
= 1A
6. (c) E × B
7. (b) Since the direction of propogation of EM wave is given
by E × B \ ( )
^ ^ ^
j
i k ´ =-
8. (d) Speed of E.M. wave = 
0 0 rr
1
µµ ÎÎ
 in medium hence
it will travel with different speed in different medium.
9. (b) I = 
1
2
 Î
0
cE
0
2
E
0 
= 
0
2I
c Î
 or  
9
8
2 500 10 36
3 10
´ ´ ´p
p´´
E
0 
= 2
3
 × 10
2 
N/C
10. (b) I = 
P
A
   or   
21
4 48
=
p´p
 W/m
2
I = 
1
2
 Î
0 
E
0
2
c
E
0 
=
0
2I
c Î
 or 
9
8
2 1 36 10
8 3 10
´ ´ p´
p´´
= 30 N/C
11. (a)U
B  
= 
2
0
0
B
4µ
Also     
0
0
E
c
B
=
  \ B
0 
= 
0
E
c
Hence B
0 
= 
0
00
E
1
µ Î
\     U
B 
= 
2
0 00
0
Eµ
4µ
Î
  or   
12
100 8.84 10
4
-
´´
\  U
B 
= 2.21 × 10
–10 
 J/m
3
12. (a) The speed of electromagnetic waves and in a medium
is given by
n = 
1
(µ) e
Where µ and e are absolute permeability and absolute
permittivity of the medium.
We know that, µ = µ
0
µ
r 
and e = e
0
e
r
. Hence
( ) ( ) ( )
rr 0 r 0 r 00
1 11
v
µ µµ.µ
= =´
e eee
or v = 
( )
rr
c
µ e
  or  e
r 
= 
2
2
r
c
(µ) n
\ e
r 
= 
82
82
(3 10)
(2 10)1
´
´´
 = 2.25
13. (a) Given B
y 
= 2 × 10
–7 
sin (0.5 × 10
3
x + 1.5 × 10
11
t)
Comparing it with a standard equation for a progresive
wave travelling along the negative direction of x-axis
is
y = r sin
2p
l
 (x + vt) = r sin
2x 2vt pp æö
+
ç÷
èø ll
DPP/ P 48
132
   = r sin
2x
2t
pæö
+ pn
ç÷
l
èø
2pn = 1.5 × 10
11
n = 
11
1.5 10
2
´
p
= 23.9 × 10
9 
Hz = 23.9 Hz
14. (c) The given equation
E
y
 = 0.5 cos[2p × 10
8 
(t – x/c)] ..... (1)
indicates that the electromagnetic waves are
propagating along the positive direction of X-axis.
The standard equation of electromagnetic wave is given
by  E
y 
= E
0 
cosw(t – x/c) ..... (2)
Comparing the given eq. (1) with the standard eq. (2),
we get w = 2p × 10
8
or 2pn = 2p × 10
8
\ n = 10
8 
per second
Now, l = 
c
n
 = 
8
8
3 10
10
´
 = 3 m
15. (a) The maximum value of magnetic field (B
0
) is given by
B
0 
= 
0
E
c
 = 
0
E
c
 = 10
–6 
tesla
The magnetic field will be along Z-axis
The maximum magnetic force on the electron is
F
b
= |q (v × B)| = q n B
0
= (1.6 × 10
–19
) × (2.0 × 10
7
) × (10
–6
)
= 3.2 × 10
–18 
N
16. (c) b-rays are beams of fast electrons.
17. (b)
18. (d) v = 
rr
c
me
 = 
8
3 10
1.3 2.14
´
´
 = 1.8 ×10
8
m/sec
19. (c) l
m
 > l
v
 > l
x
20. (d) Direction of wave propagation is given by
E×B
rr
.
21. (c) Wave impedance 
0
0
m m
=´
ee
r
r
Z
   = 
50
2
×376.6=1883W
22. (a)
23. (a) b- rays are beams of fast electrons.
24. (a) Refractive index = 
00
me
me
Þ Then refractive index = 
0
2
e
=
e
\ Speed and wavelength of wave becomes half and
frequency remain unchanged.
For 25 - 27
h = 150 km = 150 × 10
3
 m
N
m
 = 9 × 10
10
 per m
3
D = 250 km = 250 × 10
3
 m
25. (a) Critical frequency of layer
106
cm
f 9 N 9 910 2.710 Hz. = =´ ´ =´
26. (b) Maximum usuable frequency
2
23
6
c
23
D 250 10
f f1 2.7101
4h 4 150 10
æö
´
= + = ´ ´+
ç÷
´´ èø
                             = 3.17 × 10
6
 Hz.
27. (c) If angle of incidence at this layer
is f
i
, from second law of f = f
c
 sec
f
i
.
6
i
6
c
f 3.17 10
sec 1.174
f
2.7 10
´
f===
´
1
i
sec (1.174) 31.6
-
f= =°
28. (d) The electromagnetic waves of shorter wavelength do
not suffer much diffraction from the obstacles of earth's
atmosphere so they can travel long distance.
29. (b) The wavelength of these waves ranges between 300 Å
and 4000 Å that is smaller wavelength and higher
frequency. They are absorbed by atmosphere and
convert oxygen into ozone. They cause skin diseases
and they are harmful to eye and cause permanent
blindness.
30. (b) Radio waves can be polarised because they are
transverse in nature. Sound waves in air are longitudinal
in nature.