Tyre of a bicycle has volume $2\times {10}^{-3}{\mathrm{m}}^3$ . Initially the tube is filled to 75% of its volume by air at atmospheric pressure of $p_0={10}^5\mathrm{N}/{\mathrm{m}}^2$. When a rider rides the bicycle the area of contact of tyre with road is $A=24\times {10}^{-5}{\mathrm{m}}^2$ . The mass of rider with bicycle is 120 kg. The number of strokes which delivers,$V=500{\mathrm{cm}}^3$ volume of air in each stroke required to inflate the tyres is $\left[\text{ Take }g=m/s^2\right]$

The pressure of air in the tube is

$\begin{array}{r}p=\frac{mg}{A}+p_0\\ p=6\times {10}^5\mathrm{N}/{\mathrm{m}}^2\end{array}$

Final volume of air in the tube is,

$V=2\times {10}^{-3} {\mathrm{m}}^3$

The number of moles of air in the tube after pumping is, Initially volume of air in the tube is $n=\frac{pV}{RT}$where T is temperature. Then moles of air has volume V₁ at atmosphere pressure given by $V_1=\frac{nRT}{p_0}=\frac{6\times {10}^5\times 2\times {10}^{-3}}{{10}^5}{\mathrm{m}}^3$ $=12\times {10}^{-3}{\mathrm{m}}^3$

Initially volume of air in the tube is $V_0=0.75\times 2\times {10}^{-3}=1.5\times {10}^{-3} {\mathrm{m}}^3$ So, volume of air to be pumped is, $V=V_1-V_0=10.5\times {10}^{-5}{\mathrm{m}}^3$ So, number of strokes of pump required is, $n_0=\frac{V}{V_0}=\frac{10.5\times {10}^{-3}}{0.5\times {10}^{-3}}=21$