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))`

Options

A. 10
B. 11
C. 21
D. 20

Answer

C

Explanation

Pressure against which pump has to deliver air = Pressure due to weight of rider and cycle + Atmospheric pressure initially inside tyre
`=(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`