Question

What will be the acceleration due to gravity on the surface of the moon, if its radius is
1/4th the radius of the earth and its mass is 1/80 times the mass of the earth.

Answer 1

Info is quite incorrect, as instead of 80, it is 96.
Let radius of moon= $r_m$
radius of earth= $r_e$
mass of moon= $m_m$
mass of earth= $m_e$
Let acc. due to gravity on earth= $g_e$
acc. due to gravity on moon= $g_m$

Given, $r_e=4r_m$ and  $m_e=96m_m$
We know, 
$g_e= \frac{Gm_e}{r_e^2}$ ............(i)
So,
$g_m= \frac{Gm_m}{r_m^2}$ ...........(ii)
Where G is the Universal gravitational constant

Divide (i) by (ii)
$\frac{g_e}{g_m} = \frac{Gm_e}{r_e^2} * \frac{r_m^2}{Gm_m}$

$\frac{g_e}{g_m} = \frac{m_e}{r_e^2} * \frac{r_m^2}{m_m}$
Substitute the values-:
$\frac{g_e}{g_m} = \frac{96m_m}{(4r_m)^2} * \frac{r_m^2}{m_m}$
$\frac{g_e}{g_m} = 6$
$g_m= \frac{g_e}{6}$
We know, $g_e=9.8m/s^2$
So, $g_m= \frac{9.8}{6} = 1.633 m/s^2$