The acceleration due to gravity on the moon is only one sixth that of earth. If the earth and moon are assumed to have the same density, the ratio of the radii of moon and earth will be

The correct option is 1/6.

To find the ratio of the radii of the Moon and the Earth, we can establish the relationship between the acceleration due to gravity, the radius, and the density of a spherical celestial body.

The acceleration due to gravity $g$ on the surface of a spherical body of mass $M$ and radius $R$ is given by the formula:
$g=\frac{GM}{R^2}$
where $G$ is the universal gravitational constant.

Assuming the body is a uniform sphere, its mass can be expressed in terms of its average density $\rho$ and volume as:
$M=\text{Volume}\times \text{Density}=\frac{4}{3}\pi R^3\rho$

Substituting this expression for mass $M$ into the formula for $g$ gives:
$g=\frac{G\left(\frac{4}{3}\pi R^3\rho \right)}{R^2}$
Simplifying the expression, we get:
$g=\frac{4}{3}\pi G\rho R$

Since $G$ is a constant, and the Earth and the Moon are assumed to have the same average density ($\rho$ is constant), the acceleration due to gravity is directly proportional to the radius:
$g\propto R$

Therefore, we can write the ratio of the acceleration due to gravity on the Moon ($g_m$) to that on the Earth ($g_e$) as:
$\frac{g_m}{g_e}=\frac{R_m}{R_e}$
where $R_m$ is the radius of the Moon and $R_e$ is the radius of the Earth.

We are given that the acceleration due to gravity on the Moon is one-sixth that of the Earth:
$\frac{g_m}{g_e}=\frac{1}{6}$

Substituting this value into the ratio of the radii, we obtain:
$\frac{R_m}{R_e}=\frac{1}{6}$

Thus, the ratio of the radii of the Moon and the Earth is 1/6.