==>$\frac{{\Delta T}}{T} = \frac{1}{2}\frac{{\Delta l}}{l} = \frac{1}{2}\alpha \Delta \theta $
Also according to thermal expansion $l' = (1 + \alpha \Delta \theta )$
$\frac{{\Delta l}}{l} = \alpha + \theta $. Hence $\frac{{\Delta T}}{T} = \frac{1}{2}\frac{{\Delta l}}{l} = \frac{1}{2}\alpha \Delta \theta $
$ = \frac{1}{2} \times 12 \times {10^{ - 6}} \times (40 - 20) = 12 \times {10^{ - 5}}$
$ \Rightarrow \Delta T = 12 \times {10^{ - 5}} \times 86400\, seconds / day$
$\Delta T \approx 10.3\, seconds/day$
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$[1]$ The magnetic field strength may have been increased while the particle was travelling in air
$[2]$ The particle lost energy by ionising the air
$[3]$ The particle lost charge by ionising the air
