1.

The explosion of a fire cracker in the air at the a heigth of `40 m` produced a `100 dB` sound level at ground below. What is the instantaneous total radiated power? Assuming that it radiates as a point source.

Answer» Correct Answer - A::B
`l_(1) = 10 log_(10) ((I)/(I_(o)))`
`100 = 10 log_(10) ((I)/(10^(-12)))`
Solving we get, `I = 10^(-2) W//m^(2)`
Now, `I = (P)/(4pi r ^(2))`
:. `P = I (4 pi r^(2))`
= `(10^(-2)) (4 pi) (40)^(2)`
= `201 W`


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