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The the resonance tube experiment first resonant length is `l_(1)` and the second resonant length is `l_(2)`, then the third resonant length will be ?A. `5l`B. `2(l_(2)-l_(1))`C. `2l_(2)-l_(1)`D. `3l_(2)-2l_(1)` |
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Answer» Correct Answer - C For first resonance `l_(1) + epsilon = (V)/(4 f_(0))` for second resonance `l_(2) = epsilon = (3v)/(4f_(0))` for the third resonance `l_(3) + epsilon = (5v)/(4f_(0))` Solving get `l_(3) = 2l_(2) - l_(1)`. |
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