Answer:
Correct option is
B
16dB
Sound level at A is ⊥o⊥B
If the intensity of sound at A 4 cm away from the source is I 
A
 
 
then. 
⊥o=10log 
I 
o
 
 
I 
A
 
 
 
 Where I 
o
 
 =10 
−12
 w m 
−2
 
⇒⊥=log 
10 
−12
 
I 
A
 
 
 
 
⇒ 
10 
−12
 
I 
A
 
 
 
 =10
⇒I 
A
 
 =10 
−11
 wm 
−2
 
If the source (s) is emmitting the sound energy of power P
I 
A
 
 = 
4×(4) 
2
 
P
 
 
⇒P=10 
−11
 ×64π w
⇒ Intensity of sound (I 
B
 
 ) ar point B,2 m
I 
B
 
 = 
4π(2) 
2
 
P
 
 
I 
B
 
 = 
16π
64π×10 
−11
 
 
 10 m 
−2
 
⇒I 
B
 
 =4×10 
−11
 wm 
−2
 
⇒ Sound level at 2m=10log( 
I
I 
B )
=10log(10 −12
 
4×10 −11)
=10log40
=16.02
Hence Sound level at 2m from sourceis 16 dB