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An open pipe of sufficient length is dipping in water with a speed V vertically if at any instant l is length of tube above water then the rate at which fundamental frequency of pipe changes is (C is the speed of sound in air ) A. `(CV)/2l^(2)`B. `(CV)/4l^(2)`C. `(CV)/(2v^(2)t^(2))`D. `(CV)/(4v^(2)t^(2))` |
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Answer» Correct Answer - B At time t `l=(l-vt)=lambda//4` fundamental frequency `f_(0)=C/(4l)` `f_(0)=C/(4(l-vt))` `:. (df)/(dt)=-C/(4(l-vt)^(2))(-v)=(CV)/(4l^(2))` |
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