The height a which the weight of a body becomes `1//16th` its weight on the surface of earth (radius `R`) is

Options

A. 5R
B. 15R
C. 3R
D. 4R

Answer

C

Explanation

Acceleration due to gravity at a height h from the surface of earth is g.= `(g)/((1 + h/R)^2)` ...(i)
where g is the acceleration due to gravity at the surface of earth and R is the radius of earth. Multiplying (i) by m (mass of the body) on both sides we get
`mg. = (mg)/((1 + h/R)^2)`
`therefore ` Weight of body at height h, W. = mg.
Weight of body at surface of earth, W= mg
Given `W. = 1/16 W`
` therefore 1/16 = (1)/(( 1+ h/R)^2) " or " (1 + h/R)^2 = 16 " or " 1 + h/R = 4`
or `h/R = 3 "or " h = 3R`