Answer
$21 m$
Work Step by Step
We know that in a such situation, the force of friction $f$ and centripetal force will $F_c$ be equal. That is,
$F_c=f$
Also, as $F_c=\frac{mv^2}{r}$ and $f={\mu}N$
$ \frac{mv^2}{r}={\mu}N$
Since $N=mg$,
$\frac{mv^2}{r}={\mu}mg$
After solving the above equation, we get
$r=\frac{v^2}{{\mu}g}$
We know that $v=29\frac{Km}{h}=\frac{29\times 10^3}{3600}\frac{m}{s}=8.05\frac{m}{s}$. Putting the values in the formula, we get
$r=\frac{(8.05)^2}{(0.32)(9.8)}$
$r=20.66 m$
Rounding it off,
$r\approx21 m$
