If the density of the earth is doubled keeping its radius constant then acceleration due to gravity will be `(g=9.8 m//s^(2))`

Options

A. `4.9 m//s^(2)`
B. `9.8 m//s^(2)`
C. `19.6 m//s^(2)`
D. `15 m//s^(2)`

Answer

C

Explanation

Mass of the earth `= v xx rho`
`= (4)/(3)pi R^(3)rho`
`therefore g=(GM)/(R^(2))=(G)/(R^(2))[(4)/(3)pi R^(3)rho]=(4)/(3)pi RG rho`
Thus, `g prop rho` since R and G are constant.
`therefore` If `rho` is doubled, then,
`g'=2g=2xx9.8=19.6m//s^(2)`