First an object is slowly lifted from the bottom(point -`A`) of a shaft of depth `h_(1)=R/2` to earth's surface (point `-B`) and then it slowly lifted still higher to attain an altitude `h_(2)=R/2` above the earth's surface (Point `C`).
`W_(1)` and `W_(2)` are the work performed i two cases respectively. Choose the correct option(s)

Options

A. `W_(1)gtW_(2)`
B. `W_(1)ltW_(2)`
C. `|(W_(1)+W_(2))/(W_(1)+W_(2))|=17`
D. `|(W_(1)+W_(2))/(W_(1)-W_(2))|=9`

Answer

A, C

Explanation

`F_("ex")={((GMm)/(R^(3))vecr, if, 0le r le R),((GMm)/(r^(3)) vecr, if , R le r lt oo):}`
`impliesW_(1)gtW_(2)`
`W_(1)=int_(R//2)^(R) 4/3 piGrhomdx dx=(3GMm)/(8R)`
`W_(2)=int_(R)^(3R//2) (GMm)/(x^(2)) dx=(GMm)/(3R)`
`implies(W_(1))/(W_(2))=9/8 implies|(W_(1)+W_(2))/(W_(1)-W_(2))|=17`