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Tyre of a bicycle has volume 2 xx 10^(-3) m^(3). Initially the tube is filled 75% of its volume by air at atmospheric pressure 10^(5)Nm^(-2). When a rider is on the bicycle, the area of contact of tyre with road is 24 xx 10^(-4) m^(2). The mass of rider with bicycle is 120kg. If a pump delivers a volume 500 cm^(3)of air in each stroke, then the number of strokes required to inflate the tyre is (g = 10 ms^(-2)) |
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Answer» 10 `=(F)/(A) + p_("atm") = (120 xx 10)/(24 xx 10^(-4)) + 1 xx 10^(5) = 6 xx 10^(5) (N)/(m^(2))` Number of moles of air in the tube `n_(1) + (pV)/(RT) = (6 xx 10^(5) xx 2 xx 10^(-3))/(RT)` Volume of these moles at atmospheric pressure is `V_(1) = (nRT)/(p_("atm")) = (6 xx 10^(5) xx 2 xx 10^(-3))/(1 xx 10^(5))` `=12 xx 10^(-3)m^(3)` Initial volume of air inside tyre `V_(0) = (75)/(100) xx 2 xx 10^(-3) = 1.5 xx 10^(-3) m^(3)` So, to inflate the tube volume to be pumped in is `V_(2) - V_(1) - V_(0) = (12 - 1.5) xx 10^(-3)` `=10.5 xx 10^(-3) m^(3)` HENCE, number of strokes of pump required `=N = (V_(2))/(V_("pump")) = (10.5 xx 10^(-3))/(500 xx 10^(-6)) = 21` |
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