Answer:
Suppose \[\upsilon \] is speed of the car at the
topmost point of the loop.
\[\therefore \] \[mgh=\frac{1}{2}m{{\upsilon }^{2}}\]
or \[m{{\upsilon }^{2}}=2mgh\] ...(i)
At the top of the loop,
\[mg+R=\frac{m{{\upsilon }^{2}}}{r}=\frac{2mgh}{r}\]
where R is the normal reaction.
For h to be minimum, R = minimum = 0 ;
\[\therefore \] \[\frac{2mg{{h}_{\min .}}}{r}=mg\] ; \[{{h}_{\min
}}=r/2\]
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