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HC Verma Solutions: Chapter 16 - Sound Waves | Physics Class 11 - NEET PDF Download

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 Page 1


16.1
SOLUTIONS TO CONCEPTS 
CHAPTER – 16
1. V
air
= 230 m/s. V
s
= 5200 m/s. Here S = 7 m
So, t = t
1
– t
2
= ?
?
?
?
?
?
?
5200
1
330
1
= 2.75 × 10
–3
sec = 2.75 ms.
2. Here given S = 80 m × 2 = 160 m.
v = 320 m/s
So the maximum time interval will be 
t = 5/v = 160/320 = 0.5 seconds.
3. He has to clap 10 times in 3 seconds.
So time interval between two clap = (3/10 second).
So the time taken go the wall = (3/2 × 10) = 3/20 seconds.
= 333 m/s.
4. a) for maximum wavelength n = 20 Hz.
as ?
?
?
?
?
?
?
? ?
1
b) for minimum wavelength, n = 20 kHz
? ? = 360/ (20 × 10
3
) = 18 × 10
–3
m = 18 mm
? x = (v/n) = 360/20 = 18 m. ?
5. a) for minimum wavelength n = 20 KHz
? v = n ? ? ? = ?
?
?
?
?
?
?
3
10 20
1450
= 7.25 cm.
b) for maximum wavelength n should be minium
? v = n ? ? ? = v/n ? 1450 / 20 = 72.5 m. ?
6. According to the question,
a) ? = 20 cm × 10 = 200 cm = 2 m
? v = 340 m/s
so, n = v/ ? = 340/2 = 170 Hz.
N = v/ ? ?
2
10 2
340
?
?
= 17.000 Hz = 17 KH
2
(because ? = 2 cm = 2 × 10
–2
m) ?
7. a) Given V
air
= 340 m/s , n = 4.5 ×10
6
Hz
? ?
air
= (340 / 4.5) × 10
–6
= 7.36 × 10
–5 
m.
b) V
tissue 
= 1500 m/s ? ?
t
= (1500 / 4.5) × 10
–6
= 3.3 × 10
–4
m. ?
8. Here given r
y
= 6.0 × 10
–5
m
a) Given 2 ?/ ? = 1.8 ? ? = (2 ?/1.8)
? So, 
?
? ?
?
?
?
2
s / m 10 ) 8 . 1 ( 0 . 6
r
5
y
= 1.7 × 10
–5
m
b) Let, velocity amplitude = V
y
V = dy/dt = 3600 cos (600 t – 1.8) × 10
–5
m/s
Here V
y
= 3600 × 10
–5
  m/s
Again, ? = 2 ?/1.8 and T = 2 ?/600 ? wave speed = v = ?/T = 600/1.8 = 1000 / 3 m/s.
So the ratio of (V
y
/v) = 
1000
10 3 3600
5 ?
? ?
. ?
9. a) Here given n = 100, v = 350 m/s
? ? = 
100
350
n
v
? = 3.5 m.
In 2.5 ms, the distance travelled by the particle is given by
?x = 350 × 2.5 × 10
–3
Page 2


16.1
SOLUTIONS TO CONCEPTS 
CHAPTER – 16
1. V
air
= 230 m/s. V
s
= 5200 m/s. Here S = 7 m
So, t = t
1
– t
2
= ?
?
?
?
?
?
?
5200
1
330
1
= 2.75 × 10
–3
sec = 2.75 ms.
2. Here given S = 80 m × 2 = 160 m.
v = 320 m/s
So the maximum time interval will be 
t = 5/v = 160/320 = 0.5 seconds.
3. He has to clap 10 times in 3 seconds.
So time interval between two clap = (3/10 second).
So the time taken go the wall = (3/2 × 10) = 3/20 seconds.
= 333 m/s.
4. a) for maximum wavelength n = 20 Hz.
as ?
?
?
?
?
?
?
? ?
1
b) for minimum wavelength, n = 20 kHz
? ? = 360/ (20 × 10
3
) = 18 × 10
–3
m = 18 mm
? x = (v/n) = 360/20 = 18 m. ?
5. a) for minimum wavelength n = 20 KHz
? v = n ? ? ? = ?
?
?
?
?
?
?
3
10 20
1450
= 7.25 cm.
b) for maximum wavelength n should be minium
? v = n ? ? ? = v/n ? 1450 / 20 = 72.5 m. ?
6. According to the question,
a) ? = 20 cm × 10 = 200 cm = 2 m
? v = 340 m/s
so, n = v/ ? = 340/2 = 170 Hz.
N = v/ ? ?
2
10 2
340
?
?
= 17.000 Hz = 17 KH
2
(because ? = 2 cm = 2 × 10
–2
m) ?
7. a) Given V
air
= 340 m/s , n = 4.5 ×10
6
Hz
? ?
air
= (340 / 4.5) × 10
–6
= 7.36 × 10
–5 
m.
b) V
tissue 
= 1500 m/s ? ?
t
= (1500 / 4.5) × 10
–6
= 3.3 × 10
–4
m. ?
8. Here given r
y
= 6.0 × 10
–5
m
a) Given 2 ?/ ? = 1.8 ? ? = (2 ?/1.8)
? So, 
?
? ?
?
?
?
2
s / m 10 ) 8 . 1 ( 0 . 6
r
5
y
= 1.7 × 10
–5
m
b) Let, velocity amplitude = V
y
V = dy/dt = 3600 cos (600 t – 1.8) × 10
–5
m/s
Here V
y
= 3600 × 10
–5
  m/s
Again, ? = 2 ?/1.8 and T = 2 ?/600 ? wave speed = v = ?/T = 600/1.8 = 1000 / 3 m/s.
So the ratio of (V
y
/v) = 
1000
10 3 3600
5 ?
? ?
. ?
9. a) Here given n = 100, v = 350 m/s
? ? = 
100
350
n
v
? = 3.5 m.
In 2.5 ms, the distance travelled by the particle is given by
?x = 350 × 2.5 × 10
–3
Chapter 16
16.2
So, phase difference ? = x
2
? ?
?
?
? ) 2 / ( 10 5 . 2 350
) 100 / 350 (
2
3
? ? ? ? ?
?
?
.
b) In the second case, Given ?? = 10 cm = 10
–1
m
? So, ? = 35 / 2
) 100 / 350 (
10 2
x
x
2
1
? ?
? ?
? ?
?
?
. ?
10. a) Given ?x = 10 cm, ? = 5.0 cm
? ? = ? ? ?
?
? 2
= ? ? ?
?
4 10
5
2
.
So phase difference is zero.
b) Zero, as the particle is in same phase because of having same path.
11. Given that p = 1.0 × 10
5
N/m
2
, T = 273 K, M = 32 g = 32 × 10
–3
kg
V = 22.4 litre = 22.4 × 10
–3
m
3
C/C
v
= r = 3.5 R / 2.5 R = 1.4
? V = 
4 . 22 / 32
10 0 . 1 4 . 1
f
rp
5 ?
? ?
? = 310 m/s (because ? = m/v) ?
12. V
1
= 330 m/s, V
2
= ?
T
1
= 273 + 17 = 290 K, T
2
= 272 + 32 = 305 K
We know v ? T
1
2 1
2
2
1
2
1
T
T V
V
T
T
V
V ?
? ? ?
= 
290
305
340 ? = 349 m/s.
13. T
1
= 273 V
2
= 2V
1
V
1
= v T
2
= ?
We know that V ? T ?
2
1
2
2
1
2
V
V
T
T
? ? T
2
= 273 × 2
2
= 4 × 273 K
So temperature will be (4 × 273) – 273 = 819°c.
14. The variation of temperature is given by
T = T
1
+ 
d
) T T (
2 2
?
x …(1)
We know that V ? T ?
273
T
V
V
T
? ? VT = 
273
T
v
? dt = 
T
273
V
du
V
dx
T
? ?
? t = 
?
? ?
d
0
2 / 1
1 2 1
] x ) d / ) T T ( T [
dx
V
273
= 
d
0
1 2
1
1 2
] x
d
T T
T [
T T
d 2
V
273 ?
?
?
? = 
1 2
1 2
T T
T T
273
V
d 2
? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= T = 
1 2
T T
273
V
d 2
?
Putting the given value we get
= 
310 280
273
330
33 2
?
?
?
= 96 ms.
??
?x
cm
20 cm
10cm
A
B
d
x
T 1
T 2
Page 3


16.1
SOLUTIONS TO CONCEPTS 
CHAPTER – 16
1. V
air
= 230 m/s. V
s
= 5200 m/s. Here S = 7 m
So, t = t
1
– t
2
= ?
?
?
?
?
?
?
5200
1
330
1
= 2.75 × 10
–3
sec = 2.75 ms.
2. Here given S = 80 m × 2 = 160 m.
v = 320 m/s
So the maximum time interval will be 
t = 5/v = 160/320 = 0.5 seconds.
3. He has to clap 10 times in 3 seconds.
So time interval between two clap = (3/10 second).
So the time taken go the wall = (3/2 × 10) = 3/20 seconds.
= 333 m/s.
4. a) for maximum wavelength n = 20 Hz.
as ?
?
?
?
?
?
?
? ?
1
b) for minimum wavelength, n = 20 kHz
? ? = 360/ (20 × 10
3
) = 18 × 10
–3
m = 18 mm
? x = (v/n) = 360/20 = 18 m. ?
5. a) for minimum wavelength n = 20 KHz
? v = n ? ? ? = ?
?
?
?
?
?
?
3
10 20
1450
= 7.25 cm.
b) for maximum wavelength n should be minium
? v = n ? ? ? = v/n ? 1450 / 20 = 72.5 m. ?
6. According to the question,
a) ? = 20 cm × 10 = 200 cm = 2 m
? v = 340 m/s
so, n = v/ ? = 340/2 = 170 Hz.
N = v/ ? ?
2
10 2
340
?
?
= 17.000 Hz = 17 KH
2
(because ? = 2 cm = 2 × 10
–2
m) ?
7. a) Given V
air
= 340 m/s , n = 4.5 ×10
6
Hz
? ?
air
= (340 / 4.5) × 10
–6
= 7.36 × 10
–5 
m.
b) V
tissue 
= 1500 m/s ? ?
t
= (1500 / 4.5) × 10
–6
= 3.3 × 10
–4
m. ?
8. Here given r
y
= 6.0 × 10
–5
m
a) Given 2 ?/ ? = 1.8 ? ? = (2 ?/1.8)
? So, 
?
? ?
?
?
?
2
s / m 10 ) 8 . 1 ( 0 . 6
r
5
y
= 1.7 × 10
–5
m
b) Let, velocity amplitude = V
y
V = dy/dt = 3600 cos (600 t – 1.8) × 10
–5
m/s
Here V
y
= 3600 × 10
–5
  m/s
Again, ? = 2 ?/1.8 and T = 2 ?/600 ? wave speed = v = ?/T = 600/1.8 = 1000 / 3 m/s.
So the ratio of (V
y
/v) = 
1000
10 3 3600
5 ?
? ?
. ?
9. a) Here given n = 100, v = 350 m/s
? ? = 
100
350
n
v
? = 3.5 m.
In 2.5 ms, the distance travelled by the particle is given by
?x = 350 × 2.5 × 10
–3
Chapter 16
16.2
So, phase difference ? = x
2
? ?
?
?
? ) 2 / ( 10 5 . 2 350
) 100 / 350 (
2
3
? ? ? ? ?
?
?
.
b) In the second case, Given ?? = 10 cm = 10
–1
m
? So, ? = 35 / 2
) 100 / 350 (
10 2
x
x
2
1
? ?
? ?
? ?
?
?
. ?
10. a) Given ?x = 10 cm, ? = 5.0 cm
? ? = ? ? ?
?
? 2
= ? ? ?
?
4 10
5
2
.
So phase difference is zero.
b) Zero, as the particle is in same phase because of having same path.
11. Given that p = 1.0 × 10
5
N/m
2
, T = 273 K, M = 32 g = 32 × 10
–3
kg
V = 22.4 litre = 22.4 × 10
–3
m
3
C/C
v
= r = 3.5 R / 2.5 R = 1.4
? V = 
4 . 22 / 32
10 0 . 1 4 . 1
f
rp
5 ?
? ?
? = 310 m/s (because ? = m/v) ?
12. V
1
= 330 m/s, V
2
= ?
T
1
= 273 + 17 = 290 K, T
2
= 272 + 32 = 305 K
We know v ? T
1
2 1
2
2
1
2
1
T
T V
V
T
T
V
V ?
? ? ?
= 
290
305
340 ? = 349 m/s.
13. T
1
= 273 V
2
= 2V
1
V
1
= v T
2
= ?
We know that V ? T ?
2
1
2
2
1
2
V
V
T
T
? ? T
2
= 273 × 2
2
= 4 × 273 K
So temperature will be (4 × 273) – 273 = 819°c.
14. The variation of temperature is given by
T = T
1
+ 
d
) T T (
2 2
?
x …(1)
We know that V ? T ?
273
T
V
V
T
? ? VT = 
273
T
v
? dt = 
T
273
V
du
V
dx
T
? ?
? t = 
?
? ?
d
0
2 / 1
1 2 1
] x ) d / ) T T ( T [
dx
V
273
= 
d
0
1 2
1
1 2
] x
d
T T
T [
T T
d 2
V
273 ?
?
?
? = 
1 2
1 2
T T
T T
273
V
d 2
? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= T = 
1 2
T T
273
V
d 2
?
Putting the given value we get
= 
310 280
273
330
33 2
?
?
?
= 96 ms.
??
?x
cm
20 cm
10cm
A
B
d
x
T 1
T 2
Chapter 16
16.3
15. We know that v = ? / K
Where K = bulk modulus of elasticity
? K = v
2
? = (1330)
2
× 800 N/m
2
We know K = ?
?
?
?
?
?
? V / V
A / F
? ?V = 
800 1330 1330
10 2
K
essures Pr
5
? ?
?
?
So, ?V = 0.15 cm
3 ?
16. We know that,
Bulk modulus B = 
0
0
S 2
P
) V / V (
p
?
?
?
?
?
Where P
0
= pressure amplitude ? P
0
= 1.0 × 10
5
S
0
= displacement amplitude ? S
0
= 5.5 × 10
–6
m
? B = 
m 10 ) 5 . 5 ( 2
m 10 35 14
6
2
?
?
? ?
? ?
= 1.4 × 10
5
N/m
2
.
17. a) Here given V
air
= 340 m/s., Power = E/t = 20 W
f = 2,000 Hz, ? = 1.2 kg/m
3
So, intensity I = E/t.A
= 44
6 4
20
r 4
20
2 2
?
? ? ?
?
?
mw/m
2
(because r = 6m)
b) We know that I = 
air
2
0
V 2
P
?
?
air 0
V 2 1 P ? ? ?
? = 
3
10 44 340 2 . 1 2
?
? ? ? ? = 6.0 N/m
2
.
c) We know that I = V v S 2
2 2
0
2
? ? where S
0
= displacement amplitude
? S
0
= 
air
2 2
V
I
? ? ?
Putting the value we get S
g
= 1.2 × 10
–6
m.
18. Here I
1
= 1.0 × 10
–8
W
1
/m
2
; I
2
= ?
r
1
= 5.0 m, r
2
= 25 m.
We know that I ?
2
r
1
? I
1
r
1
2
= I
2
r
2
2
  ? I
2
= 
2
2
2
1 1
r
r I
= 
625
25 10 0 . 1
8
? ?
?
= 4.0 × 10
–10
W/m
2
.
19. We know that ? = 10 log
10
?
?
?
?
?
?
?
?
0
I
I
?
A
= 
o
A
I
I
log 10 , ?
B
= 
o
B
I
I
log 10
? I
A
/ I
0
= 
) 10 / (
A
10
?
? I
B
/I
o
= 
) 10 / (
B
10
?
?
2
2
A
2
B
B
A
5
50
r
r
I
I
?
?
?
?
?
?
? ? ?
2 ) (
10 10
B A
?
? ?
? 2
10
B A
?
? ? ?
? 20
B A
? ? ? ?
? ?
B
= 40 – 20 = 20 d ?. ?
Page 4


16.1
SOLUTIONS TO CONCEPTS 
CHAPTER – 16
1. V
air
= 230 m/s. V
s
= 5200 m/s. Here S = 7 m
So, t = t
1
– t
2
= ?
?
?
?
?
?
?
5200
1
330
1
= 2.75 × 10
–3
sec = 2.75 ms.
2. Here given S = 80 m × 2 = 160 m.
v = 320 m/s
So the maximum time interval will be 
t = 5/v = 160/320 = 0.5 seconds.
3. He has to clap 10 times in 3 seconds.
So time interval between two clap = (3/10 second).
So the time taken go the wall = (3/2 × 10) = 3/20 seconds.
= 333 m/s.
4. a) for maximum wavelength n = 20 Hz.
as ?
?
?
?
?
?
?
? ?
1
b) for minimum wavelength, n = 20 kHz
? ? = 360/ (20 × 10
3
) = 18 × 10
–3
m = 18 mm
? x = (v/n) = 360/20 = 18 m. ?
5. a) for minimum wavelength n = 20 KHz
? v = n ? ? ? = ?
?
?
?
?
?
?
3
10 20
1450
= 7.25 cm.
b) for maximum wavelength n should be minium
? v = n ? ? ? = v/n ? 1450 / 20 = 72.5 m. ?
6. According to the question,
a) ? = 20 cm × 10 = 200 cm = 2 m
? v = 340 m/s
so, n = v/ ? = 340/2 = 170 Hz.
N = v/ ? ?
2
10 2
340
?
?
= 17.000 Hz = 17 KH
2
(because ? = 2 cm = 2 × 10
–2
m) ?
7. a) Given V
air
= 340 m/s , n = 4.5 ×10
6
Hz
? ?
air
= (340 / 4.5) × 10
–6
= 7.36 × 10
–5 
m.
b) V
tissue 
= 1500 m/s ? ?
t
= (1500 / 4.5) × 10
–6
= 3.3 × 10
–4
m. ?
8. Here given r
y
= 6.0 × 10
–5
m
a) Given 2 ?/ ? = 1.8 ? ? = (2 ?/1.8)
? So, 
?
? ?
?
?
?
2
s / m 10 ) 8 . 1 ( 0 . 6
r
5
y
= 1.7 × 10
–5
m
b) Let, velocity amplitude = V
y
V = dy/dt = 3600 cos (600 t – 1.8) × 10
–5
m/s
Here V
y
= 3600 × 10
–5
  m/s
Again, ? = 2 ?/1.8 and T = 2 ?/600 ? wave speed = v = ?/T = 600/1.8 = 1000 / 3 m/s.
So the ratio of (V
y
/v) = 
1000
10 3 3600
5 ?
? ?
. ?
9. a) Here given n = 100, v = 350 m/s
? ? = 
100
350
n
v
? = 3.5 m.
In 2.5 ms, the distance travelled by the particle is given by
?x = 350 × 2.5 × 10
–3
Chapter 16
16.2
So, phase difference ? = x
2
? ?
?
?
? ) 2 / ( 10 5 . 2 350
) 100 / 350 (
2
3
? ? ? ? ?
?
?
.
b) In the second case, Given ?? = 10 cm = 10
–1
m
? So, ? = 35 / 2
) 100 / 350 (
10 2
x
x
2
1
? ?
? ?
? ?
?
?
. ?
10. a) Given ?x = 10 cm, ? = 5.0 cm
? ? = ? ? ?
?
? 2
= ? ? ?
?
4 10
5
2
.
So phase difference is zero.
b) Zero, as the particle is in same phase because of having same path.
11. Given that p = 1.0 × 10
5
N/m
2
, T = 273 K, M = 32 g = 32 × 10
–3
kg
V = 22.4 litre = 22.4 × 10
–3
m
3
C/C
v
= r = 3.5 R / 2.5 R = 1.4
? V = 
4 . 22 / 32
10 0 . 1 4 . 1
f
rp
5 ?
? ?
? = 310 m/s (because ? = m/v) ?
12. V
1
= 330 m/s, V
2
= ?
T
1
= 273 + 17 = 290 K, T
2
= 272 + 32 = 305 K
We know v ? T
1
2 1
2
2
1
2
1
T
T V
V
T
T
V
V ?
? ? ?
= 
290
305
340 ? = 349 m/s.
13. T
1
= 273 V
2
= 2V
1
V
1
= v T
2
= ?
We know that V ? T ?
2
1
2
2
1
2
V
V
T
T
? ? T
2
= 273 × 2
2
= 4 × 273 K
So temperature will be (4 × 273) – 273 = 819°c.
14. The variation of temperature is given by
T = T
1
+ 
d
) T T (
2 2
?
x …(1)
We know that V ? T ?
273
T
V
V
T
? ? VT = 
273
T
v
? dt = 
T
273
V
du
V
dx
T
? ?
? t = 
?
? ?
d
0
2 / 1
1 2 1
] x ) d / ) T T ( T [
dx
V
273
= 
d
0
1 2
1
1 2
] x
d
T T
T [
T T
d 2
V
273 ?
?
?
? = 
1 2
1 2
T T
T T
273
V
d 2
? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= T = 
1 2
T T
273
V
d 2
?
Putting the given value we get
= 
310 280
273
330
33 2
?
?
?
= 96 ms.
??
?x
cm
20 cm
10cm
A
B
d
x
T 1
T 2
Chapter 16
16.3
15. We know that v = ? / K
Where K = bulk modulus of elasticity
? K = v
2
? = (1330)
2
× 800 N/m
2
We know K = ?
?
?
?
?
?
? V / V
A / F
? ?V = 
800 1330 1330
10 2
K
essures Pr
5
? ?
?
?
So, ?V = 0.15 cm
3 ?
16. We know that,
Bulk modulus B = 
0
0
S 2
P
) V / V (
p
?
?
?
?
?
Where P
0
= pressure amplitude ? P
0
= 1.0 × 10
5
S
0
= displacement amplitude ? S
0
= 5.5 × 10
–6
m
? B = 
m 10 ) 5 . 5 ( 2
m 10 35 14
6
2
?
?
? ?
? ?
= 1.4 × 10
5
N/m
2
.
17. a) Here given V
air
= 340 m/s., Power = E/t = 20 W
f = 2,000 Hz, ? = 1.2 kg/m
3
So, intensity I = E/t.A
= 44
6 4
20
r 4
20
2 2
?
? ? ?
?
?
mw/m
2
(because r = 6m)
b) We know that I = 
air
2
0
V 2
P
?
?
air 0
V 2 1 P ? ? ?
? = 
3
10 44 340 2 . 1 2
?
? ? ? ? = 6.0 N/m
2
.
c) We know that I = V v S 2
2 2
0
2
? ? where S
0
= displacement amplitude
? S
0
= 
air
2 2
V
I
? ? ?
Putting the value we get S
g
= 1.2 × 10
–6
m.
18. Here I
1
= 1.0 × 10
–8
W
1
/m
2
; I
2
= ?
r
1
= 5.0 m, r
2
= 25 m.
We know that I ?
2
r
1
? I
1
r
1
2
= I
2
r
2
2
  ? I
2
= 
2
2
2
1 1
r
r I
= 
625
25 10 0 . 1
8
? ?
?
= 4.0 × 10
–10
W/m
2
.
19. We know that ? = 10 log
10
?
?
?
?
?
?
?
?
0
I
I
?
A
= 
o
A
I
I
log 10 , ?
B
= 
o
B
I
I
log 10
? I
A
/ I
0
= 
) 10 / (
A
10
?
? I
B
/I
o
= 
) 10 / (
B
10
?
?
2
2
A
2
B
B
A
5
50
r
r
I
I
?
?
?
?
?
?
? ? ?
2 ) (
10 10
B A
?
? ?
? 2
10
B A
?
? ? ?
? 20
B A
? ? ? ?
? ?
B
= 40 – 20 = 20 d ?. ?
Chapter 16
16.4
20. We know that, ? = 10 log
10
J/I
0
According to the questions
?
A
= 10 log
10
(2I/I
0
)
? ?
B
– ?
A
= 10 log (2I/I) = 10 × 0.3010 = 3 dB. ?
21. If sound level = 120 dB, then I = intensity = 1 W/m
2
Given that, audio output = 2W
Let the closest distance be x.
So, intensity = (2 / 4 ?x
2
) = 1 ? x
2
= (2/2 ?) ? x = 0.4 m = 40 cm. ?
22. ?
1
= 50 dB, ?
2
= 60 dB
? I
1
= 10
–7
W/m
2
, I
2
= 10
–6
W/m
2
(because ? = 10 log
10
(I/I
0
), where I
0
= 10
–12
W/m
2
)
Again, I
2
/I
1
= (p
2
/p
1
)
2
=(10
–6
/10
–7
) = 10 (where p = pressure amplitude).
? (p
2
/ p
1
) = 10 . ?
23. Let the intensity of each student be I.
According to the question
?
A
= 
0
10
I
I 50
log 10 ; ?
B
= 
?
?
?
?
?
?
?
?
0
10
I
I 100
log 10
? ?
B
– ?
A
= 
0
10
I
I 50
log 10 –
?
?
?
?
?
?
?
?
0
10
I
I 100
log 10
= 3 2 log 10
I 50
I 100
log 10
10
? ?
?
?
?
?
?
?
?
?
So, ?
A
= 50 + 3 = 53 dB. ?
24. Distance between tow maximum to a minimum is given by, ?/4 = 2.50 cm
? ? = 10 cm = 10
–1
m
We know, V = nx
? n = 
1
10
340 V
?
?
?
= 3400 Hz = 3.4 kHz. ?
25. a) According to the data
?/4 = 16.5 mm ? ? = 66 mm = 66 × 10
–6=3
m
? n = 
3
10 66
330 V
?
?
?
?
= 5 kHz.
b) I
minimum 
= K(A
1
– A
2
)
2
= I ? A
1
– A
2
= 11
I
maximum 
= K(A
1
+ A
2
)
2
= 9 ? A
1
+ A
2
= 31
So, 
4
3
A A
A A
2 1
2 1
?
?
?
? A
1
/A
2
= 2/1
So, the ratio amplitudes is 2. ?
26. The path difference of the two sound waves is given by
?L = 6.4 – 6.0 = 0.4 m
The wavelength of either wave = ? = 
?
?
?
320 V
(m/s)
For destructive interference ?L = 
2
) 1 n 2 ( ? ?
where n is an integers.
or 0.4 m = 
?
?
? 320
2
1 n 2
? ? = n = Hz
2
1 n 2
800
4 . 0
320 ?
? = (2n + 1) 400 Hz
Thus the frequency within the specified range which cause destructive interference are 1200 Hz, 
2000 Hz, 2800 Hz, 3600 Hz and 4400 Hz. ?
Page 5


16.1
SOLUTIONS TO CONCEPTS 
CHAPTER – 16
1. V
air
= 230 m/s. V
s
= 5200 m/s. Here S = 7 m
So, t = t
1
– t
2
= ?
?
?
?
?
?
?
5200
1
330
1
= 2.75 × 10
–3
sec = 2.75 ms.
2. Here given S = 80 m × 2 = 160 m.
v = 320 m/s
So the maximum time interval will be 
t = 5/v = 160/320 = 0.5 seconds.
3. He has to clap 10 times in 3 seconds.
So time interval between two clap = (3/10 second).
So the time taken go the wall = (3/2 × 10) = 3/20 seconds.
= 333 m/s.
4. a) for maximum wavelength n = 20 Hz.
as ?
?
?
?
?
?
?
? ?
1
b) for minimum wavelength, n = 20 kHz
? ? = 360/ (20 × 10
3
) = 18 × 10
–3
m = 18 mm
? x = (v/n) = 360/20 = 18 m. ?
5. a) for minimum wavelength n = 20 KHz
? v = n ? ? ? = ?
?
?
?
?
?
?
3
10 20
1450
= 7.25 cm.
b) for maximum wavelength n should be minium
? v = n ? ? ? = v/n ? 1450 / 20 = 72.5 m. ?
6. According to the question,
a) ? = 20 cm × 10 = 200 cm = 2 m
? v = 340 m/s
so, n = v/ ? = 340/2 = 170 Hz.
N = v/ ? ?
2
10 2
340
?
?
= 17.000 Hz = 17 KH
2
(because ? = 2 cm = 2 × 10
–2
m) ?
7. a) Given V
air
= 340 m/s , n = 4.5 ×10
6
Hz
? ?
air
= (340 / 4.5) × 10
–6
= 7.36 × 10
–5 
m.
b) V
tissue 
= 1500 m/s ? ?
t
= (1500 / 4.5) × 10
–6
= 3.3 × 10
–4
m. ?
8. Here given r
y
= 6.0 × 10
–5
m
a) Given 2 ?/ ? = 1.8 ? ? = (2 ?/1.8)
? So, 
?
? ?
?
?
?
2
s / m 10 ) 8 . 1 ( 0 . 6
r
5
y
= 1.7 × 10
–5
m
b) Let, velocity amplitude = V
y
V = dy/dt = 3600 cos (600 t – 1.8) × 10
–5
m/s
Here V
y
= 3600 × 10
–5
  m/s
Again, ? = 2 ?/1.8 and T = 2 ?/600 ? wave speed = v = ?/T = 600/1.8 = 1000 / 3 m/s.
So the ratio of (V
y
/v) = 
1000
10 3 3600
5 ?
? ?
. ?
9. a) Here given n = 100, v = 350 m/s
? ? = 
100
350
n
v
? = 3.5 m.
In 2.5 ms, the distance travelled by the particle is given by
?x = 350 × 2.5 × 10
–3
Chapter 16
16.2
So, phase difference ? = x
2
? ?
?
?
? ) 2 / ( 10 5 . 2 350
) 100 / 350 (
2
3
? ? ? ? ?
?
?
.
b) In the second case, Given ?? = 10 cm = 10
–1
m
? So, ? = 35 / 2
) 100 / 350 (
10 2
x
x
2
1
? ?
? ?
? ?
?
?
. ?
10. a) Given ?x = 10 cm, ? = 5.0 cm
? ? = ? ? ?
?
? 2
= ? ? ?
?
4 10
5
2
.
So phase difference is zero.
b) Zero, as the particle is in same phase because of having same path.
11. Given that p = 1.0 × 10
5
N/m
2
, T = 273 K, M = 32 g = 32 × 10
–3
kg
V = 22.4 litre = 22.4 × 10
–3
m
3
C/C
v
= r = 3.5 R / 2.5 R = 1.4
? V = 
4 . 22 / 32
10 0 . 1 4 . 1
f
rp
5 ?
? ?
? = 310 m/s (because ? = m/v) ?
12. V
1
= 330 m/s, V
2
= ?
T
1
= 273 + 17 = 290 K, T
2
= 272 + 32 = 305 K
We know v ? T
1
2 1
2
2
1
2
1
T
T V
V
T
T
V
V ?
? ? ?
= 
290
305
340 ? = 349 m/s.
13. T
1
= 273 V
2
= 2V
1
V
1
= v T
2
= ?
We know that V ? T ?
2
1
2
2
1
2
V
V
T
T
? ? T
2
= 273 × 2
2
= 4 × 273 K
So temperature will be (4 × 273) – 273 = 819°c.
14. The variation of temperature is given by
T = T
1
+ 
d
) T T (
2 2
?
x …(1)
We know that V ? T ?
273
T
V
V
T
? ? VT = 
273
T
v
? dt = 
T
273
V
du
V
dx
T
? ?
? t = 
?
? ?
d
0
2 / 1
1 2 1
] x ) d / ) T T ( T [
dx
V
273
= 
d
0
1 2
1
1 2
] x
d
T T
T [
T T
d 2
V
273 ?
?
?
? = 
1 2
1 2
T T
T T
273
V
d 2
? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= T = 
1 2
T T
273
V
d 2
?
Putting the given value we get
= 
310 280
273
330
33 2
?
?
?
= 96 ms.
??
?x
cm
20 cm
10cm
A
B
d
x
T 1
T 2
Chapter 16
16.3
15. We know that v = ? / K
Where K = bulk modulus of elasticity
? K = v
2
? = (1330)
2
× 800 N/m
2
We know K = ?
?
?
?
?
?
? V / V
A / F
? ?V = 
800 1330 1330
10 2
K
essures Pr
5
? ?
?
?
So, ?V = 0.15 cm
3 ?
16. We know that,
Bulk modulus B = 
0
0
S 2
P
) V / V (
p
?
?
?
?
?
Where P
0
= pressure amplitude ? P
0
= 1.0 × 10
5
S
0
= displacement amplitude ? S
0
= 5.5 × 10
–6
m
? B = 
m 10 ) 5 . 5 ( 2
m 10 35 14
6
2
?
?
? ?
? ?
= 1.4 × 10
5
N/m
2
.
17. a) Here given V
air
= 340 m/s., Power = E/t = 20 W
f = 2,000 Hz, ? = 1.2 kg/m
3
So, intensity I = E/t.A
= 44
6 4
20
r 4
20
2 2
?
? ? ?
?
?
mw/m
2
(because r = 6m)
b) We know that I = 
air
2
0
V 2
P
?
?
air 0
V 2 1 P ? ? ?
? = 
3
10 44 340 2 . 1 2
?
? ? ? ? = 6.0 N/m
2
.
c) We know that I = V v S 2
2 2
0
2
? ? where S
0
= displacement amplitude
? S
0
= 
air
2 2
V
I
? ? ?
Putting the value we get S
g
= 1.2 × 10
–6
m.
18. Here I
1
= 1.0 × 10
–8
W
1
/m
2
; I
2
= ?
r
1
= 5.0 m, r
2
= 25 m.
We know that I ?
2
r
1
? I
1
r
1
2
= I
2
r
2
2
  ? I
2
= 
2
2
2
1 1
r
r I
= 
625
25 10 0 . 1
8
? ?
?
= 4.0 × 10
–10
W/m
2
.
19. We know that ? = 10 log
10
?
?
?
?
?
?
?
?
0
I
I
?
A
= 
o
A
I
I
log 10 , ?
B
= 
o
B
I
I
log 10
? I
A
/ I
0
= 
) 10 / (
A
10
?
? I
B
/I
o
= 
) 10 / (
B
10
?
?
2
2
A
2
B
B
A
5
50
r
r
I
I
?
?
?
?
?
?
? ? ?
2 ) (
10 10
B A
?
? ?
? 2
10
B A
?
? ? ?
? 20
B A
? ? ? ?
? ?
B
= 40 – 20 = 20 d ?. ?
Chapter 16
16.4
20. We know that, ? = 10 log
10
J/I
0
According to the questions
?
A
= 10 log
10
(2I/I
0
)
? ?
B
– ?
A
= 10 log (2I/I) = 10 × 0.3010 = 3 dB. ?
21. If sound level = 120 dB, then I = intensity = 1 W/m
2
Given that, audio output = 2W
Let the closest distance be x.
So, intensity = (2 / 4 ?x
2
) = 1 ? x
2
= (2/2 ?) ? x = 0.4 m = 40 cm. ?
22. ?
1
= 50 dB, ?
2
= 60 dB
? I
1
= 10
–7
W/m
2
, I
2
= 10
–6
W/m
2
(because ? = 10 log
10
(I/I
0
), where I
0
= 10
–12
W/m
2
)
Again, I
2
/I
1
= (p
2
/p
1
)
2
=(10
–6
/10
–7
) = 10 (where p = pressure amplitude).
? (p
2
/ p
1
) = 10 . ?
23. Let the intensity of each student be I.
According to the question
?
A
= 
0
10
I
I 50
log 10 ; ?
B
= 
?
?
?
?
?
?
?
?
0
10
I
I 100
log 10
? ?
B
– ?
A
= 
0
10
I
I 50
log 10 –
?
?
?
?
?
?
?
?
0
10
I
I 100
log 10
= 3 2 log 10
I 50
I 100
log 10
10
? ?
?
?
?
?
?
?
?
?
So, ?
A
= 50 + 3 = 53 dB. ?
24. Distance between tow maximum to a minimum is given by, ?/4 = 2.50 cm
? ? = 10 cm = 10
–1
m
We know, V = nx
? n = 
1
10
340 V
?
?
?
= 3400 Hz = 3.4 kHz. ?
25. a) According to the data
?/4 = 16.5 mm ? ? = 66 mm = 66 × 10
–6=3
m
? n = 
3
10 66
330 V
?
?
?
?
= 5 kHz.
b) I
minimum 
= K(A
1
– A
2
)
2
= I ? A
1
– A
2
= 11
I
maximum 
= K(A
1
+ A
2
)
2
= 9 ? A
1
+ A
2
= 31
So, 
4
3
A A
A A
2 1
2 1
?
?
?
? A
1
/A
2
= 2/1
So, the ratio amplitudes is 2. ?
26. The path difference of the two sound waves is given by
?L = 6.4 – 6.0 = 0.4 m
The wavelength of either wave = ? = 
?
?
?
320 V
(m/s)
For destructive interference ?L = 
2
) 1 n 2 ( ? ?
where n is an integers.
or 0.4 m = 
?
?
? 320
2
1 n 2
? ? = n = Hz
2
1 n 2
800
4 . 0
320 ?
? = (2n + 1) 400 Hz
Thus the frequency within the specified range which cause destructive interference are 1200 Hz, 
2000 Hz, 2800 Hz, 3600 Hz and 4400 Hz. ?
Chapter 16
16.5
27. According to the given data
V = 336 m/s, 
?/4 = distance between maximum and minimum intensity
= (20 cm) ? ? = 80 cm
? n = frequency = 
2
10 80
336 V
?
?
?
?
= 420 Hz.
28. Here given ? = d/2
Initial path difference is given by = d d 2
2
d
2
2
2
? ? ?
?
?
?
?
?
If it is now shifted a distance x then path difference will be
= ?
?
?
?
?
?
? ? ? ? ? ?
?
?
?
?
?
4
d
d 2
4
d
d ) x d 2 (
2
d
2
2
2
?
64
d 169
) x d 2 (
2
d
2
2
2
? ? ? ?
?
?
?
?
?
?
2
d
64
153
? ? ? x d 2 1.54 d ? x = 1.54 d – 1.414 d = 0.13 d.
29. As shown in the figure the path differences 2.4 = ?x = 2 . 3 ) 4 . 2 ( ) 2 . 3 (
2 2
? ?
Again, the wavelength of the either sound waves = 
?
320
We know, destructive interference will be occur
If ?x = 
2
) 1 n 2 ( ? ?
?
?
?
? ? ?
320
2
) 1 n 2 (
) 2 . 3 ( ) 4 . 2 ( ) 2 . 3 (
2 2
Solving we get
? V = ) 1 n 2 ( 200
2
400 ) 1 n 2 (
? ?
?
where n = 1, 2, 3, …… 49. (audible region) ?
30. According to the data
? = 20 cm, S
1
S
2
= 20 cm, BD = 20 cm
Let the detector is shifted to left for a distance x for hearing the 
minimum sound.
So path difference AI = BC – AB 
=
2 2 2 2
) x 10 ( ) 20 ( ) x 10 ( ) 20 ( ? ? ? ? ?
So the minimum distances hearing for minimum 
= 
2
20
2 2
) 1 n 2 (
?
?
?
? ?
= 10 cm
??
2 2 2 2
) x 10 ( ) 20 ( ) x 10 ( ) 20 ( ? ? ? ? ? = 10 solving we get x = 12.0 cm. ?
31.
Given, F = 600 Hz, and v = 330 m/s ? ? = v/f = 330/600 = 0.55 mm
S
D
=x/4
20cm
2 d
D
S
2 2
d 2 ) 2 / d ( ?
x
d
2 2
) 4 . 2 ( ) 2 . 3 ( ?
A
A A
B
A
x
C
20cm
20cm
X
S
? ? P
R
1m
Q
y
S 1
S 2
O
1m
D
S 1
S 2
O ? ?
P
Q
X
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FAQs on HC Verma Solutions: Chapter 16 - Sound Waves - Physics Class 11 - NEET

1. What are sound waves and how are they produced?
Ans. Sound waves are longitudinal mechanical waves that propagate through a medium, such as air, water, or solids. They are produced by the vibration of a source, such as a speaker or a musical instrument. These vibrations create compressions and rarefactions in the medium, which then propagate as sound waves.
2. How does the frequency of a sound wave affect its pitch?
Ans. The frequency of a sound wave directly affects its pitch. Pitch refers to how high or low a sound is perceived to be. Higher frequency waves have a higher pitch, while lower frequency waves have a lower pitch. For example, a high-frequency sound wave will have a higher pitch, like that of a whistle, while a low-frequency sound wave will have a lower pitch, like that of a bass drum.
3. What is the speed of sound in different mediums?
Ans. The speed of sound varies depending on the medium through which it travels. In dry air at 20 degrees Celsius, the speed of sound is approximately 343 meters per second. However, the speed of sound is faster in solids and liquids compared to gases. For example, in water, sound travels at around 1482 meters per second, and in steel, it can reach speeds of up to 5960 meters per second.
4. How does the amplitude of a sound wave affect its loudness?
Ans. The amplitude of a sound wave directly affects its loudness. The amplitude refers to the maximum displacement of particles in the medium caused by the wave. A higher amplitude corresponds to a louder sound, while a lower amplitude corresponds to a softer sound. This is because a higher amplitude wave compresses the particles in the medium more, resulting in a stronger perception of sound.
5. Can sound waves travel in a vacuum?
Ans. No, sound waves cannot travel in a vacuum. Sound waves require a medium to propagate, such as air, water, or solids. In a vacuum, there are no particles to vibrate and transmit the sound waves. Therefore, sound cannot travel through empty space or vacuum. This is why sound cannot be heard in outer space, where there is no air or other medium for sound to travel through.
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Document Description: HC Verma Solutions: Chapter 16 - Sound Waves for NEET 2024 is part of Physics Class 11 preparation. The notes and questions for HC Verma Solutions: Chapter 16 - Sound Waves have been prepared according to the NEET exam syllabus. Information about HC Verma Solutions: Chapter 16 - Sound Waves covers topics like and HC Verma Solutions: Chapter 16 - Sound Waves Example, for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for HC Verma Solutions: Chapter 16 - Sound Waves.
Introduction of HC Verma Solutions: Chapter 16 - Sound Waves in English is available as part of our Physics Class 11 for NEET & HC Verma Solutions: Chapter 16 - Sound Waves in Hindi for Physics Class 11 course. Download more important topics related with notes, lectures and mock test series for NEET Exam by signing up for free. NEET: HC Verma Solutions: Chapter 16 - Sound Waves | Physics Class 11 - NEET
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Full syllabus notes, lecture & questions for HC Verma Solutions: Chapter 16 - Sound Waves | Physics Class 11 - NEET - NEET | Plus excerises question with solution to help you revise complete syllabus for Physics Class 11 | Best notes, free PDF download
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Additional Information about HC Verma Solutions: Chapter 16 - Sound Waves for NEET Preparation

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