Two equal masses each in are hung from a balance whose scale pans differ in vertical height by `h`. The error in weighing in terms of density of the earth `rho` is

Explanation

`g_(h)=[1-(2h)/(R_(e))],W_(2),W_(1)=mg_(2)-mg_(1)=2mg[(h_(1))/(R_(e))-(h_(2))/(R_(e))]=2m(GM)/(R_(e)^(2))xx(h)/(R_(e))[becauseg=(GM)/(R_(e)^(2))&h_(1)-h_(2)=h]`
Error in weighing `=W_(2)-W_(1)=2mG(4)/(3)piR_(e)^(3)rho(h)/(R_(e)^(3))=(8pi)/(3)Gmrhoh`
Due to depth
Assuming density of earth remains same throughout
At earth surface, `g=(4)/(3)piGR_(e)rho` ...(i) at depth d inside the earth
For point P only mass of the inner sphere is effective `g_(d)=(GM')/(r^(2))`
`g_(d)=(G)/(r^(2))xx(Mr^(2))/(R_(e)^(2))=(GM)/(R_(e)^(2))xx(r)/(R_(e))=(GM)/(R_(e)^(2))xx(R_(e)-d)/(R_(e))`
Mass of sphere of radius `r=M'`
`M'=(4)/(3)pir^(3)rho=(4)/(3)pir^(3)xx(M)/(4//3piR^(3))=M'=(M)/(R^(3))r^(3)`
`g_(d)=g[1-(d)/(R_(e))]` valid for any depth.