A star suddenly shrinks and its density becomes 10⁹ times its original value. The value of acceleration due to gravity on its surface will increase by a factor of

10⁶

$g=\frac{G M}{R^2}=\frac{G V \rho}{R^2}=G \frac{4}{3} \pi \frac{R^3 \rho}{R^2}=G \frac{4}{3} \pi R \rho$

$M=\frac{4}{3} \pi R^3 \rho=\frac{4}{3} \pi R^{\prime 3} \rho^{\prime}=\frac{4}{3} \pi R^{\prime 3} \times 10^9 \rho$

$\left(R^{\prime} \times 10^3\right)^3=R^3 ; R^{\prime}=\frac{R}{1000}=R \times 10^{-3}$

∴ $\frac{g^{\prime}}{g}=\frac{G \frac{4}{3} \pi R^{\prime} \rho^{\prime}}{G \frac{4}{3} \pi R \rho}=\frac{R \times 10^{-3}}{R} \times \frac{10^9 \rho}{1(\rho)}=10^6$