1.

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

Answer»

`19.6" m"//s^(2)`
`9.8" m"//s^(2)`
`4.9" m"//s^(2)`
`2.45" m"//s^(2)`

SOLUTION :`g=(GM)/(R^(2))=G.(4)/(3)(PIR^(3)xxrho)/(R^(2))=(4pi"G"rho"R")/(3)`
`rho_(2)=2rho_(1)` (given)
`:. (g_(1))/(g_(2))=(4piG)/(3). Rho_(1)R_(1)xx(3)/(4pi" G "rho_(2)R_(2))=(rho_(1))/(rho_(2)).(R_(1))/(R_(2))=(Rho_(1))/(rho_(2))=(rho_(1))/(2rho_(1)).(R )/(R )=(1)/(2)`
`RARR g_(2)=2g_(1)=2xx9.8=19.6" m"//s^(2)`


Discussion

No Comment Found

Related InterviewSolutions