The mass of a planet is $\frac{1}{10}$ th that of the earth and its diameter is half that of the earth. The acceleration due to gravity on that planet is:

$3.92\, m\, s^{-2}$

The correct answer is option (4) : $3.92\, m\, s^{-2}$

 The acceleration due to gravity (g) on a planet is given by the formula:

 $g=G\frac{M}{R^2}$

where:

$G$  is the universal gravitational constant,

$M$  is the mass of the planet, and

$R$  is the radius of the planet.

In this question, we know that:

The mass of the planet $M_p$  is $\frac{1}{10}$ of the mass of the earth $M_e$, so $M_p=\frac{M_e}{10}$.The diameter of the planet is half that of earth. Since the diameter is twice the radius, a diameter half that of earth implies the radius $R_p$  is also half the radius of the earth $R_e$. Therefore, $R_p=\frac{R_e}{2}$.

Using the formula for acceleration due to gravity and substituting the above information:

 $g_p=G\frac{M_p}{R^2_p}=G\frac{\frac{M_e}{10}}{(\frac{R_e}{2})^2}$

Simplify the expression:

 $g_p=G\frac{M_e}{10}.\frac{4}{R^2_e}=\frac{4}{10}G\frac{M_e}{R^2_e}$

Knowing that $G\frac{M_e}{R^2_e}$  is the acceleration due to gravity on Earth $g_e ≈ 9.8\, m/s^2$ , we substitute this value into the equation:

$g_p=\frac{4}{10}.9.8m/s^2=3.92m/s^2$

Thus, the acceleration due to gravity on the planet is $3.92\, m/s^2$, which corresponds to:

Option 4 : $3.92\,m/s^2$