d
(d) Initially \(F = k.\frac{{{Q^2}}}{{{r^2}}}\) (fig. \(A\)). Finally when a third spherical conductor comes in contact alternately with \(B\) and \(C\) then removed, so charges on \(B\) and \(C\) are \(Q / 2\) and \(3Q / 4\) respectively (fig. \(B\))

Now force \(F' = k.\frac{{\left( {\frac{Q}{2}} \right)\,\left( {\frac{{3Q}}{4}} \right)}}{{{r^2}}} = \frac{3}{8}F\)