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A pop-gun consistsofa cylindrical barrel 3 cm^(3) in cross-section closed at one end by a cork and having a well fitting piston at the other. If the piston is pushed slowly in, the cork is finally ejected, giving a pop, the frequency of which is found to be 512 Hz. Assuming that the initial distance between the cork and the pistonwas25 cm and that there isno leakage of air,calculate the force required to eject the cork. Atmospheric pressure = 1kg wt//cm^(2), v = 340 m/s. |
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Answer» <P> Solution : When the cork is ejected,the situation is shown in figure-6.77. Let the position ofthe piston fromendy4 belcm. So, this forms a closedpipeof length1. It produces a note of frequency512 Hz. Now `N = (v)/(4L) ` or `512 = (340)/(4L)` `L = (340)/(512 xx 4) = 0.166 m` = 16.6 CM Before the piston is moved, `P = 1 kg wt//cm^(2)` and `V = 25 xx 3 cm^(3)`. When the cork is ejected, let pressure be p'. The volume of air inside `v' = 16.6 xx 3 cm^(3)` From BOYLE's law, `PV = P'V' ` ` (1)(25xx 3) = `P' xx (16.6 xx 3)` `P' ((25)/(16.6))kg wt//cm^(2)` = 1.5 kg wt// cm^(2)` Thus pressure inside the barrel = `1.5 kg wt//cm^(2)` Pressure difference = pressure inside - pressure outside = `(1.5 - I)kg wt//cm^(2)` = `0.5 kg wt//cm^(2)` Force on piston = pressure xx area = `0.5 xx3` 1.5 kg wt |
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