Acceleration due to gravity at earth's surface is `10ms^(-2)`. The value of acceleration due to gravity at the surface of a planet of mass `(1/5)^(th)` and radius `1/2` of the earth is

Options

A. `4m//s^(2)`
B. `6m//s^(2)`
C. `8m//s^(2)`
D. `12m//s^(2)`

Answer

c

Explanation

`(g_(2))/(g_(1))=(m_(2))/(m_(1))xx((R_(1))/(R_(2)))^(2)=(1)/(5)xx(4)/(1)`
`g_(2)=(4)/(5)xx g_(1)`
`=(4)/(5)xx10=8m//s^(2).`