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