2019-03-01_Physics_Times

The intensity in decibles is

0

10log

I

I



 

  

 

 

1 2

1 0

0

20 10log 10

I

I I

I

 

   

 

(^220)   4
0
40 10log 10
I
I I
I
 
   
 
2
1
100
I
I

Let intensity of sound is I and increased by
100 I
Loudness of sound in decibel
1 10
0
10log
I
I

 
  
 
When intensity of sound becomes 100I then new
decibel level 2 10
0
100
10log
I
I

 
  
 
  2   1 10log 100 20 10  dB
 descibel rise by 20 dB
option (d) is correct.
Given that f540 , 15Hz rad/s
v and v 1 2 are speed of whistle
v v 1       2 r 15 2 30 /m s
Maximum frequency heard is
max
s
v
f f
v v
 
  
  
Minimum frequency heard is
min
s
v
f f
v v
 
  
  
max
min
330 30 360
1.2
330 30 300
s
s
f v v
f v v
 
   
 
Frequency received by the cliff is
1 0
s
v
f f
v v
 
  
  
Frequency received by the source after reflection
is
2 1 0
2
s s 1 s
s
v v v v v
f f f f
v v v v
       
      
      
0.990f
0
1
f v
f
v u


 
2 0
v u
f f
v


2 1 0
u v v
f f f
v v u
  
    
  
  
2
0
2 1
u f
f f ve
v v u

   

 f f 1 2.
Given that velocity of source is
vs39.6kmph m s11 /
The frequency of the horn is
1 1
30
f
T
 
As the source is approaching the stationary ob-
server.
1 330
'
s 30 330 11
v
f f
v v
   
    
     
2.Sol:
3.Sol:
4.Sol:
5.Sol:
6.Sol:
7.Sol:
1
' 29
'
T
f
  seconds
Velocity of sound in a gas medium is
RT
v
M


2 2
2 2
2 1
32 4
o H
H o
v M
v M
  