Two identical sound `S_1` and `S_2` reach at a point P is phase. The resultant loudness at point P is `n` dB higher than the loudness of `S_1` the value of n is :A. `2`B. `4`C. `5`D. `6`

Two identical sound `S_1` and `S_2` reach at a point P is phase. The r

1.	Two identical sound `S_1` and `S_2` reach at a point P is phase. The resultant loudness at point P is `n` dB higher than the loudness of `S_1` the value of n is :A. `2`B. `4`C. `5`D. `6`
Answer» Correct Answer - D Let `a` be the amplitude due to `S_1` and `S_2` individually. Loudness due to `S_1=I_1=Ka^2` Loudness due to `S_(1)+S_(2)=I=K(2a)^(2)=4I_(I)` `:. n=10log_(10)((4l_1)/(l_1))` `=10log_(10)(4)=6`

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Two identical sound `S_1` and `S_2` reach at a point P is phase. The resultant loudness at point P is `n` dB higher than the loudness of `S_1` the value of n is :A. `2`B. `4`C. `5`D. `6`

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