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`)

Options

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)`

Answer

A

Explanation

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)