1. Find the positive root $\alpha$ of $p(x)=x^2-x-2=0$

   Factorizing: $x^2-x-2=(x-2)(x+1)=0$ So the roots are $2$ and $-1$. The positive root is $\alpha=2.$

2. Substitute $\alpha=2$ into the limit We need to evaluate $\lim_{x\to \alpha^+}\frac{\sqrt{1-\cos(p(x))}}{x+\alpha-4}$ Since $\alpha=2$, $x+\alpha-4=x+2-4=x-2.$ Also, $p(x)=x^2-x-2=(x-2)(x+1).$ Hence the limit becomes $\lim_{x\to 2^+}\frac{\sqrt{1-\cos((x-2)(x+1))}}{x-2}.$

3. Use the standard small-angle approximation As $x\to 2^+$, we have $p(x)=(x-2)(x+1)\to 0^+.$ For small $t$, $1-\cos t \sim \frac{t^2}{2}.$ Therefore, $\sqrt{1-\cos t}\sim \sqrt{\frac{t^2}{2}}=\frac{|t|}{\sqrt{2}}.$

   Here $t=p(x)=(x-2)(x+1)$. Since $x\to 2^+$, we have $x-2>0$ and $x+1>0$, so $p(x)>0 \implies |p(x)|=p(x).$ Thus, \sqrt{1-\cos(p(x))}\sim \frac{p(x)}{\sqrt{2}}=rac{(x-2)(x+1)}{\sqrt{2}}.

4. Substitute into the limit

   =\lim_{x\to 2^+}\frac{\frac{(x-2)(x+1)}{\sqrt{2}}}{x-2}$$ $$=\lim_{x\to 2^+}\frac{x+1}{\sqrt{2}}=rac{3}{\sqrt{2}}.$$

5. Check options $\frac{3}{\sqrt{2}}$ matches Option C.

Final Answer: $\boxed{\frac{3}{\sqrt{2}}}$