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(D) For an infinitesimally thin disc of uniform surface mass density $\sigma$,the gravitational field $g$ at a small distance $z$ from the center alon

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$2^{1/2}$

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(D) For an infinitesimally thin disc of uniform surface mass density $\sigma$,the gravitational field $g$ at a small distance $z$ from the center along the axis is given by $g = 2\pi G \sigma \left(1 - \frac{z}{\sqrt{z^2 + R^2}}ight)$.Since $z \ll R$,we use the binomial expansion: $\frac{z}{\sqrt{z^2 + R^2}} = \frac{z}{R(1 + z^2/R^2)^{1/2}} \approx \frac{z}{R}(1 - \frac{z^2}{2R^2}) \approx \frac{z}{R}$.Thus,$g \approx 2\pi G \sigma (1 - \frac{z}{R}) \approx 2\pi G \sigma$. This implies that for small $z$,the gravitational field is approximately constant,$g = g_0 = 2\pi G \sigma$.The equation of motion for a star at height $z$ is $\ddot{z} = -g_0 	ext{sgn}(z)$.This represents a motion with constant acceleration towards the disc. The time taken to travel from $z_0$ to $0$ is $t = \sqrt{2z_0/g_0}$.The total time period $T$ for one complete oscillation (from $z_0$ to $-z_0$ and back) is $T = 4t = 4\sqrt{\frac{2z_0}{g_0}}$.Therefore,$T \propto \sqrt{z_0}$.The ratio of the time periods is $\frac{T_A}{T_B} = \sqrt{\frac{2z_0}{z_0}} = \sqrt{2} = 2^{1/2}$.

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