Answer :
The ratio of solid cylinder to hollow cylinder travel time is 0.0648
How to compare the travel time?
If the hollow and solid cylinders are dropped simultaneously, they will hit the ground at the same time (ignoring air resistance). For an object rolling down a slope (rather than sliding) is actually turning its potential energy into two forms of kinetic energy viz. translational and rotational. For a given mass, a hollow cylinder has more material away from the axis than a solid cylinder, so its moment of inertia will be higher.
Now as the question mentions,
[tex]\frac{t_{bb} }{t_{hc}}[/tex]= [tex]\frac{1.52}{1.72}[/tex]
[tex]\frac{\Delta t_{bb} }{\Delta t_{hc}} = \sqrt{(\frac{\Delta t_{hc} }{t_{hc}})^{2}+(\frac{\Delta t_{bb} }{t_{bb}})^{2}}[/tex]
= [tex]\sqrt{(\frac{0.0987}{1.72}) ^{2}+{(\frac{0.0457}{1.5}) ^{2}}[/tex]
= [tex]\sqrt{0.0033 + 0.0009}[/tex]
= 0.0648
∴ [tex]\frac{t_{x} }{t_{hc} }[/tex] = 0.8721 ± 0.0648
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