Waves- 2 Practice Questions - DPP for NEET

 Page 1


1. (b)
1
n = Frequency of the police car horn observer heard
by motorcyclist
1
n = Frequency of the siren heard by motorcyclist.
v
2
= Speed of motor cyclist
12
330 330
176; 165
330 22 330
vv
nn
-+
= ´ =´
-
12
0 22/ n n v ms \ - =Þ=
2. (c) Frequency of first over tone of closed pipe = Frequency
of first over tone of open pipe
1 2 1 1 22
3 31
44
vv P PP
v
LL LL
éù
g gg
Þ = Þ = \=
êú
r rr
ëû
111
2
22
4 4
33
L L
L
rr
Þ==
rr
3. (b) For observer note of B will not change due to zero
relative motion.
Observed frequency of sound produced by 
A
(330 30)
660 600
330
Hz
-
==
\ No of beats = 600 - 596 = 4
4. (c) Open pipe resonanace frequency 
1
2
2
v
f
L
=
Closed pipe resonance frequency 
2
4
nv
f
L
=
21
4
n
ff = (where n is odd and 
21
) ff >
\ 5 n =
5. (b)
0
()
s
nvv
vv
-
-
6. (a) Wave number 
1
=
l
but 
11
'
s
v
vv
æö
= ==
ç÷
l l- èø
 and
3
s
v
v =
\
,
( . .) ( . .) 256
/3 2 /3
vv
WN WN
vv
æö
= =´
ç÷
èø-u
3
256 384
2
=´=
7. (d) Since there is no relative motion between observer and
source, therefore there is no apparent change in
frequency.
8. (c)
350
' 1200 1400
350 50
s
v
n n cps
vv
æö
æö
= =´=
ç÷
ç÷
èø
-- èø
9. (d) By using 
1
'
s
n vV
nn
v v n VS
æö
= Þ=
ç÷
èø --
10. (a)
0
330 33
' 100 90
330
vv
n n Hz
v
- - æö
= = ´=
ç÷
èø
11. (d) The apparent frequency heard by the observer is given
by
330 330
' 450 450 500
330 33 297
s
v
n x Hz
vv
= = ´= ´=
--
12. (b) Observer is moving ayay form siren 1 and towards the
siren 2.
v
Stationay Moving Stationay
siren 1 observer siren 2
Hearing frequency of sound emitted by siren 1
0
1
330 2
330 328
330
vv
n n Hz
v
- - æö æö
= ==
ç÷ ç÷
èø èø
Hearing frequency of sound emitted by siren 2
0
2
330 2
330 332
330
vv
n n Hz
v
- + æö æö
= ==
ç÷ ç÷
èø èø
Hence, beat frequency 
21
332 328 4 nn = - = -=
13. (b) At point 
A
, source is moving away from observer so
apparent frequency 
1
nn < (actual frequency). At point
B
source is coming towards observer so apparent
frequency 
2
nn > and point 
C
source is moving
perpendicular to observer so 
3
nn =
Hence 
2 31
n nn >>
14. (c) According the concept of sound image
3455
' .272 272 280
3455
person
person
vv
n Hz
vv
+
+
= = ´=
--
nD = Number of beats = 280 – 272 = 8 Hz
15. (a) The observer will hear two sound, one directly from
source and other from reflected image of sound
Page 2


1. (b)
1
n = Frequency of the police car horn observer heard
by motorcyclist
1
n = Frequency of the siren heard by motorcyclist.
v
2
= Speed of motor cyclist
12
330 330
176; 165
330 22 330
vv
nn
-+
= ´ =´
-
12
0 22/ n n v ms \ - =Þ=
2. (c) Frequency of first over tone of closed pipe = Frequency
of first over tone of open pipe
1 2 1 1 22
3 31
44
vv P PP
v
LL LL
éù
g gg
Þ = Þ = \=
êú
r rr
ëû
111
2
22
4 4
33
L L
L
rr
Þ==
rr
3. (b) For observer note of B will not change due to zero
relative motion.
Observed frequency of sound produced by 
A
(330 30)
660 600
330
Hz
-
==
\ No of beats = 600 - 596 = 4
4. (c) Open pipe resonanace frequency 
1
2
2
v
f
L
=
Closed pipe resonance frequency 
2
4
nv
f
L
=
21
4
n
ff = (where n is odd and 
21
) ff >
\ 5 n =
5. (b)
0
()
s
nvv
vv
-
-
6. (a) Wave number 
1
=
l
but 
11
'
s
v
vv
æö
= ==
ç÷
l l- èø
 and
3
s
v
v =
\
,
( . .) ( . .) 256
/3 2 /3
vv
WN WN
vv
æö
= =´
ç÷
èø-u
3
256 384
2
=´=
7. (d) Since there is no relative motion between observer and
source, therefore there is no apparent change in
frequency.
8. (c)
350
' 1200 1400
350 50
s
v
n n cps
vv
æö
æö
= =´=
ç÷
ç÷
èø
-- èø
9. (d) By using 
1
'
s
n vV
nn
v v n VS
æö
= Þ=
ç÷
èø --
10. (a)
0
330 33
' 100 90
330
vv
n n Hz
v
- - æö
= = ´=
ç÷
èø
11. (d) The apparent frequency heard by the observer is given
by
330 330
' 450 450 500
330 33 297
s
v
n x Hz
vv
= = ´= ´=
--
12. (b) Observer is moving ayay form siren 1 and towards the
siren 2.
v
Stationay Moving Stationay
siren 1 observer siren 2
Hearing frequency of sound emitted by siren 1
0
1
330 2
330 328
330
vv
n n Hz
v
- - æö æö
= ==
ç÷ ç÷
èø èø
Hearing frequency of sound emitted by siren 2
0
2
330 2
330 332
330
vv
n n Hz
v
- + æö æö
= ==
ç÷ ç÷
èø èø
Hence, beat frequency 
21
332 328 4 nn = - = -=
13. (b) At point 
A
, source is moving away from observer so
apparent frequency 
1
nn < (actual frequency). At point
B
source is coming towards observer so apparent
frequency 
2
nn > and point 
C
source is moving
perpendicular to observer so 
3
nn =
Hence 
2 31
n nn >>
14. (c) According the concept of sound image
3455
' .272 272 280
3455
person
person
vv
n Hz
vv
+
+
= = ´=
--
nD = Number of beats = 280 – 272 = 8 Hz
15. (a) The observer will hear two sound, one directly from
source and other from reflected image of sound
84
DPP/ P 30
Hence number of beats heard per second
ss
vv
nn
vv vv
æ ö æö
=-
ç ÷ ç÷
-+ è ø èø
22
2 2 2563305
7.8
335 325
s
s
n
Hz
uu ´ ´´
= ==
´
u -u
16. (a) In closed pipe only odd harmonics are present.
17. (a) Maximum pressure at closed end will be atmospheric
pressure adding with acoustic wave pressure
So 
max0 A
P PP =+ and 
min0 A
P PP =-
Thus 
max0
min0
A
A
P PP
P PP
+
=
-
18. (b)
11
( ) 320 160
22
= =´=
closed open
n n Hz
19. (a)
1.21Å l=
N
N
N
A A
1.21 Å
20. (a)
211
21
122
1 270
50 13.5
1000
lnn
n l l cm
llnn
æö
µÞ = Þ= = ´=
ç÷
èø
21. (c) Loudness depends upon intensity while pitch depends
upon frequency .
22. (d) Using ( ) ( )
21 21
2ll v 2nll l= -Þ=-
2 215(63.2 30.7) 33280cm / s Þ´ -=
Actual speed of sound v
0
 = 332m / s = 33200cm / s
Hence error = 33280 – 33200 = 80cm / s
23. (c)
3 7
1
2
5
300
301 303 308
8
3
7
5
2
1
300
305 307
308
8
24. (a) Doppler shift doesn't depend upon the distance of listner
from the source.
25. (b) Since the edges are clamped, displacement of the edges
(, )0 uxy = for line -
y
C B
(L,L)
A
x
O
(0, 0)
(0,L)
OA i.e y = 0,
0 xL ££
AB
 i.e. ,0 y L yL = ££
BC
i.e. ,0 y L xL = ££
OC
i.e. 0,0 x yL = ££
The above conditions are satisfied only in alternatives
(b) and (c).
Note that (, )0 uxy = for all four values e.g. in
alternative (), (, )0 d uxy = for 0, y yL == But it is not
zero for 
0 x =
or 
xL =
. Similarly in option (a)
(,)0 uxy = at , x LyL == but it is not zero for
0 x =
or 0 y = , while in options (b) and (c),
(,)0 uxy = for 0, 0, x y xL = == and yL =
26. (b),   27. (d).
For fundamental force, s 2s
2
l
= Þl=
V elocity of waves is,  v = 
Y
r
 where Y is Y oung's modu-
lus of quartz and r is its density .
From f
0
 = 
4
v 2.87 10
s
´
=
l
  Þ 
4
Y 1 2.87 10
2ss
´
´=
r
Þ Y = 8.76 × 10
12
 N/m
2
For 3rd harmonic, f = 3f
0
 = 1.2 × 10
6
 Hz Þ 
4
3 2.87 10
s
´´
= 1.2 × 10
6
 Þ s = 0.07175 cm.
28. (d) As emission of light from atom is a random and rapid
phenomenon. The phase at a point due to two
independent light source will change rapidly and
randomly. Therefore, instead of beats, we shall get
uniform intensity . However, if light sources are LASER
beams of nearly equal frequencies, it may possible to
observe the phenomenon of beats in light.
29. (d) The person will hear the loud sound at nodes than at
antinodes. We know that at anti-nodes the displacement
is maximum and pressure change is minimum while at
nodes the displacement is zero and pressure change is
maximum. The sound is heared due to variation of
pressure.
Also in stationary waves particles in two different
segment vibrates in opposite phase.
30. (a) Stationary wave
N
N
N
A
A
A node is a place of zero amplitude and an antinode is
a place of maximum amplitude.