Tyre of a bicycle has volume $2 \times 10^{-3} \mathrm{~m}^3$. Initially, the tube is filled $75 \%$ of its…
Tyre of a bicycle has volume $2 \times 10^{-3} \mathrm{~m}^3$. Initially, the tube is filled $75 \%$ of its volume by air at atmospheric pressure $10^5 \mathrm{Nm}^{-2}$. When a rider is on the bicycle, the area of contact of tyre with road is $24 \times 10^{-4} \mathrm{~m}^2$. The mass of rider with bicycle is $120 \mathrm{~kg}$. If a pump delivers a volume $500 \mathrm{~cm}^3$ of air in each stroke, then the number of strokes required to inflate the tyre is $\left(g=10 \mathrm{~ms}^{-2}\right)$
10
11
21
20
Solution
Pressure against which pump has to deliver air
$=$ Pressure due to weight of rider and cycle
+ Atmospheric pressure initially inside tyre
$
=\frac{F}{A}+p_{\mathrm{atm}}=\frac{120 \times 10}{24 \times 10^{-4}}+1 \times 10^5=6 \times 10^5 \frac{\mathrm{N}}{\mathrm{m}^2}
$
Number of moles of air in the tube
$
n_1=\frac{\mathrm{pV}}{\mathrm{RT}}=\frac{6 \times 10^5 \times 2 \times 10^{-3}}{\mathrm{RT}}
$
Volume of these moles at atmospheric pressure is
$
\begin{aligned}
V_1 & =\frac{n R T}{p_{\text {atm }}}=\frac{6 \times 10^5 \times 2 \times 10^{-3}}{1 \times 10^5} \\
& =12 \times 10^{-3} \mathrm{~m}^3
\end{aligned}
$
Initial volume of air inside tyre
$
V_0=\frac{75}{100} \times 2 \times 10^{-3}=1.5 \times 10^{-3} \mathrm{~m}^3
$
So, to inflate the tube volume to be pumped in is
$
\begin{aligned}
V_2 & =V_1-V_0=(12-1.5) \times 10^{-3} \\
& =10.5 \times 10^{-3} \mathrm{~m}^3
\end{aligned}
$
Hence, number of strokes of pump required
$
=N=\frac{V_2}{V_{\text {pump }}}=\frac{10.5 \times 10^{-3}}{500 \times 10^{-6}}=21
$