Given density distribution

The mass density varies as $\rho(r)=\frac{K}{r}$ for large distance $r$ from the centre.

Mass enclosed within radius $R$

For a spherically symmetric galaxy, the mass enclosed inside radius $R$ is $M(R)=\int_0^R \rho(r)\, 4\pi r^2\,dr$

Substituting $\rho(r)=\dfrac{K}{r}$, $M(R)=\int_0^R \frac{K}{r}\,4\pi r^2\,dr = 4\pi K \int_0^R r\,dr$

$M(R)=4\pi K \left[\frac{r^2}{2}\right]_0^R = 2\pi K R^2$

So, $M(R)\propto R^2$

Gravitational force provides centripetal force

For a star of mass $m$ in circular orbit of radius $R$, $\frac{G M(R)m}{R^2}=\frac{mv^2}{R}$

Since $M(R)\propto R^2$, $\frac{G M(R)}{R^2}=\text{constant}$

Therefore, $\frac{v^2}{R}=\text{constant}$ which gives $v^2\propto R$ and hence $v\propto \sqrt{R}$

Relation for time period

The time period of revolution is $T=\frac{2\pi R}{v}$

Since $v\propto \sqrt{R}$, $T\propto \frac{R}{\sqrt{R}}=\sqrt{R}$

Squaring both sides, $T^2\propto R$

Checking options

• A: $T^2 \propto R$ ✅
• B: $T^2 \propto R^3$ ❌
• C: $T \propto R$ ❌ since $T\propto \sqrt{R}$
• D: $T^2 \propto \dfrac{1}{R^3}$ ❌

Final answer

The correct option is: $\boxed{A}$