Let g be the acceleration due to gravity at the earth's surface and K the rotational kinetic energy of the earth. Suppose the earth's radius decreases by 2%. Keeping all other quantities constant, then

Answer

`4`

Explanation

Acceleration due to gravity is `g=(GM)/(R^(2))`
and if L be the angular momentum of the earth, then rotational `KE=(L^(2))/(2I)`
Where I be the moment of inertia
so `I=2/5MR^(2)`(for sphere)
`:.` Rotational `KE=(5L^(2))/(4MR^(2))`
sicne angular momentum remains conserved.
So rotational `KE(K)prop1/(R^(2))`
`:.` Both g and K are `prop R^(-2)`
`:.(Deltag)/g=(DeltaK)/K=-2xx(DeltaR)/R`
`:.` Both g and K would decrease by
`2xx2%=4%`