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(B) The gravitational acceleration is given by $g = \frac{4}{3} \pi G \rho R$. Thus,$\frac{g'}{g} = \frac{\rho' R'}{\rho R}$. Given $\frac{g'}{g} = \f

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(B) The gravitational acceleration is given by $g = \frac{4}{3} \pi G \rho R$. Thus,$\frac{g'}{g} = \frac{\rho' R'}{\rho R}$. Given $\frac{g'}{g} = \frac{\sqrt{6}}{11}$ and $\frac{\rho'}{\rho} = \frac{2}{3}$,we have $\frac{\sqrt{6}}{11} = \frac{2}{3} \cdot \frac{R'}{R}$,which implies $\frac{R'}{R} = \frac{3\sqrt{6}}{22}$. The escape velocity is $v_e = \sqrt{2gR} = \sqrt{2 (\frac{4}{3} \pi G \rho R) R} = R \sqrt{\frac{8}{3} \pi G \rho}$. Therefore,$\frac{v_e'}{v_e} = \frac{R'}{R} \sqrt{\frac{\rho'}{\rho}} = \left( \frac{3\sqrt{6}}{22} \right) \sqrt{\frac{2}{3}} = \frac{3\sqrt{6}}{22} \cdot \frac{\sqrt{2}}{\sqrt{3}} = \frac{3\sqrt{2}\sqrt{3}}{22} \cdot \frac{\sqrt{2}}{\sqrt{3}} = \frac{3 \cdot 2}{22} = \frac{6}{22} = \frac{3}{11}$. Given $v_e = 11 \ km/s$,we get $v_e' = 11 \cdot \frac{3}{11} = 3 \ km/s$.

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