A rope of length 30cm is on a horizontal table with maximum length hanging from edge A of the table the coefficient of friction between  the rope and table is 0.5. The distance of center of mass of the rope from A is $\frac{\text{b}\sqrt{\text{17}}}{\text{3}}\text{cm}$ , Then b= …….

$\text{Fractional length hanging is }$

$\begin{array}{l}\frac{l}{\text{L}}=\frac{\mu }{1+\mu }\Rightarrow \frac{l}{\text{30}}=\frac{0.5}{1+0.5}\Rightarrow l=10\text{ cm}\\ \text{Let }\rho \text{ be the linear density. The co-ordinates of 20}\rho \text{ and 10}\rho \text{ are (}\rho \text{,0)}\\ \text{and (0,5) respectively from A}\\ {\text{x}}_{\text{cm}}=\frac{20\rho \left(10\right)+10\rho \left(0\right)}{20\rho +10\rho }=\frac{\text{200}\rho }{\text{30}\rho }=\frac{\text{20}}{\text{3}}\\ y_{\text{cm}}=\frac{20\rho \left(0\right)+10\rho \left(5\right)}{20\rho +10\rho }=\frac{\text{50}\rho }{\text{30}\rho }=\frac{\text{5}}{\text{3}}\\ r_{\text{cm}}=\sqrt{x_{cm}^2+y_{cm}^2}=\sqrt{{\left(\frac{5\text{X4}}{\text{3}}\right)}^{\text{2}}+{\left(\frac{5}{3}\right)}^2}=\frac{5}{3}\sqrt{17}\text{ cm}\end{array}$