Deduce The Weight Of A Truck From Its Flat Tires

First of all, let us calculate the Initial Volume.

Note: I shall leave the geometrical calculations to the reader.

$V_1 = \pi(0.5^2-0.1^2)\times 0.2 = 0.15079m^3$

The change in volume is clearly the volume of the squisshed part (volume of the partial cylinder)

$\Delta V = Area\times 0.2 = (A_{sector}-A_{triangle})\times 0.2 = 2.9363\times 10^{-3}m^3$

$\Rightarrow V_2 = V_1 - \Delta V = 0.14786m^3$

Now comes the key part.
Note that the problem mentions that the process is very fast.

Immediately, Adiabatic process comes into our mind.

$PV^\gamma = Constant$

Also,
$\displaystyle\gamma = \frac{f+2}{f}$ , where $f$ is the degree of freedom

Since the gas is diatomic ( $O_2$ and $N_2$ ),
$f = 5$

Thus,
$\gamma = \frac{7}{5}$

We also know,
Total Pressure = Gauge Pressure + Atmospheric Pressure

$P_1 = 101.325 + 200 kPa = 301325Pa$

Using all these values,
$P_1V_1^\gamma = P_2V_2^\gamma$

$\Rightarrow P_2 = 309717.0563Pa$

Now,
The area on which this final pressure acts is,

$A = 0.2\times \sqrt{0.5^2-0.45^2} = 0.0871779m^2$

Thus,
$F = P_2A = 27000.50362N$

Since there are 4 tires,
$F_{net} = 4F = 108002.0145N$

Hence,
$M_{total} = \frac{F}{g} = 11020.613\approx\boxed{11021Kg}$