Answer
0.32 m
Work Step by Step
Here we use Boyle's law $P_{1}V_{1}=P_{2}V_{2}$ to find the distance 'H' that the biker pushes down on the handle before air begins to flow from the pump to the inner tube.
$P_{1}V_{1}=P_{2}V_{2}=>P_{1}Ah_{1}=P_{2}Ah_{2}=>P_{1}h_{1}=P_{2}h_{2}$
$h_{2}=\frac{P_{1}h_{1}}{P_{2}}-(1)$
We can write,
$H=h_{1}-h_{2}-(2)$
(1)=>(2),
$H=h_{1}-\frac{P_{1}h_{1}}{P_{2}}=h_{1}(1-\frac{P_{1}}{P_{2}})$
Let's plug known values into this equation.
$H=(0.55\space m)(1-\frac{1\times10^{5}Pa}{2.4\times10^{5}Pa})=0.32\space m$
