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Four sources of sound each of sound level `10 dB` are sounded together in phase , the resultant intensity level will be `(log _(10) 2 = 0.3)`A. `40 dB`B. `26 dB`C. `22 dB`D. `13 dB` |
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Answer» Correct Answer - C Resulant ampulitude will become `4` times. Therefore , resultant intensity is `16` times `L_(2) - L_(1) = log_(10) (I_(2))/(I_(1))` or `I_(2) - 10 = 10 log _(10) (16)` or `L_(2) = 22 dB` |
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