A body rolled on ice with a velocity of $8{\mathrm{ }\mathrm{m}\mathrm{s}}^{-1}$ comes to rest after travelling $4\mathrm{ }\mathrm{m}$. The coefficient of friction of ice is ........... .

Given,

$u=8m/s,v=0,S=4m$

$v^2=u^2\text{ + 2as}$ 

$a=\frac{64}{2\times 4}=8m{s}^{-2}$

$\mu =\frac{F}{N}=\frac{ma}{mg}=\frac{a}{g}=\frac{8}{9.8}=0.8164$

So, the coefficient of friction of ice is $0.8164$.

Hence, the correct answer is $0.8164$.