Suppose the gravitational force varies inversely as the nth power of distance. Then, the time period of a planet in circular orbit of radius R around the sun will be proportional to

The correct option is R⁽ⁿ⁺¹⁾/².

Let us analyze the relationship step-by-step:

According to the problem, the gravitational force $F$ varies inversely as the $n^{\text{th}}$ power of the distance $R$:
$F\propto \frac{1}{R^n}$

For a planet of mass $m$ moving in a circular orbit of radius $R$ around the sun, the gravitational force provides the necessary centripetal force $F_c$:

$F_c=m{\omega }^{2}R$

Here, $\omega$ is the angular velocity of the planet, which can be expressed in terms of its time period $T$ as:
$\omega =\frac{2\pi }{T}$

Substituting this value of $\omega$ into the centripetal force equation gives:
$F_c=m{\left(\frac{2\pi }{T}\right)}^{2}R=\frac{4{\pi }^{2}mR}{T^2}$

Since the centripetal force is provided by the gravitational force, we have:
$F_c\propto F$
$\frac{4{\pi }^{2}mR}{T^2}\propto \frac{1}{R^n}$

Since $m$ and $4{\pi }^2$ are constants, this simplifies to:
$\frac{R}{T^2}\propto \frac{1}{R^n}$

Rearranging the relation to solve for $T^2$:
$T^2\propto R\cdot R^n$
$T^2\propto R^{n+1}$

Taking the square root on both sides:
$T\propto R^{\frac{n+1}{2}}$

Thus, the time period $T$ of the planet is proportional to $R^{(n+1)/2}$.