InterviewSolution
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InterviewSolution
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In an organ pipe ( may be closed or open of `99 cm` length standing wave is setup , whose equation is given by longitudinal displacement `xi = (0.1 mm) cos ( 2pi)/( 0.8) ( y + 1 cm) cos 2 pi (400) t` where `y` is measured from the top of the tube in metres and `t` in second. Here `1 cm` is th end correction. Assume end correction approximately equals to `(0.3) xx`(diameter of tube) , estimate the moles of air pressure inside the tube (Assume tube is at `NTP` , and at `NTP , 22.4 litre` contain `1 mole`)A. `(10 pi)/(36 xx 22.4)`B. `(10 pi)/(18 xx 22.4)`C. `(10 pi)/(72 xx 22.4)`D. `(10 pi)/(60 xx22.4)` |
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Answer» Correct Answer - A End correction ` = (0.3)d = 1cm rArr d = (10)/(3) cm` Vol. of tube `= (pi (d^(2))/(4))l = (pi)/(4) ((10)/(3))^(2) xx 100 cm^(3)` (take `1 = 0.99 m ~~ 1m) = (10 pi)/(36) L` Moles `= ( 10 pi)/( 36 xx 22.4)` moles ( `22.4` 1t contains `1` mole `( 10 pi)/(36)` lt contains `( 10 pi)/(36 xx 22.4)` mole) |
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