On the surface of earth acceleration due gravity is `g` and gravitational potential is V. Match of following columns
`{:(,"Column-I",,"Column-II"),("(A)","At height h=R, value of g","(p)","Decreases by a factor" (1)/(4)),("(B)","At depth " h =(R)/(2)"," "value of g","(q)","Decrease by a factor" (1)/(2)),("(C)","At height "h=R",""value of V","(r)","Increase by a factor" (11)/(8)),("(D)","At height "h=(R)/(2)"," "value of V","(s)","Increase by a factor 2"),(,,"(t)","None"):}`

Answer

`(A rarr q,B rarr q,C rarr t,D rarr t)`

Explanation

`g=(GM)/(R^(2)),V=-(GM)/(R)`
At height `h=R,g'=(g)/(1+(h)/(R))=(g)/(2)rArrV'=-(GM)/(2R)`
i.e., decrease by a factor 1/2 and V' increase by a factor of 1/2.
At depth `h=(R)/(2)`
`g'=g(1-(h)/(R))=g(1-(1)/(2))=(g)/(2)`
`V'=-(GM)/(R^(3))[15R^(2)-0.5r^(2)]=-(GM)/(R^(3))[15 R^(2)-(1)/(8)R^(2)]=-(11)/(8)(GM)/(R)`
i.e., g' decrease by a factor `(1)/(2) and V'` also decreases by a factor `(11)/(8)`.