The area of the triangle \(\mathrm{ABC}\) is \(10 \sqrt{3} \mathrm{~cm}^{2}\), angle \(\mathrm{B}\) is \(60^{\circ}\) and its perimeter is \(20 \mathrm{~cm}\),
then \(\ell(\mathrm{AC})=\)

(D)
\(\begin{array}{l}
\text { Area }=\frac{\operatorname{acsin} B}{2}=10 \sqrt{3}=\frac{\operatorname{acsin} 60}{2} \\
10 \sqrt{3}=\frac{\operatorname{ac} \sqrt{3}}{4} a c=40
\end{array}\)
Now
\(\begin{aligned} & \cos \mathrm{B}=\frac{\mathrm{a}^{2}+\mathrm{c}^{2}-\mathrm{b}^{2}}{2 \mathrm{ac}} \\ & \cos 60 \times 2 \mathrm{ac}=\mathrm{a}^{2}+\mathrm{c}^{2}-\mathrm{b}^{2} \\ \therefore & \frac{1}{2} \times 2 \times 40=(\mathrm{a}+\mathrm{c})^{2}-2 \mathrm{ac}-\mathrm{b}^{2} \end{aligned}\)
\(\therefore 40=(20-b)^{2}-(2 \times 40)-b^{2} \quad \ldots[a+b+c=20\), given \(]\)
\(\begin{aligned} &=400+b^{2}-40 b-80-b^{2} \\ \therefore \quad 40 b &=280 \Rightarrow b=7 \end{aligned}\)