Answer
(a) $0.00131m^3$
(b) decrease
(c) $0.00130m^3$
Work Step by Step
(a) We know that
$V=\frac{m+\rho^{\prime}(\frac{4}{3}\pi r^3)}{\rho}$
We plug in the known values to obtain:
$V=\frac{1.2Kg+(7.86\times 10^3Kg/m^3)(\frac{4}{3}\pi (1.22\times 10^{-2}m))^{3}}{1000Kg/m^3}=0.00131m^3$
(b) We know that in order to support the same total weight of the wood and iron ball system, the same amount of the volume of water should be displaced. The volume of wood that is submerged will decrease because the ball is now submerged.
(c) We know that
$V=V_{wood}-\frac{4}{3}\pi r^3$
We plug in the known values to obtain:
$V=0.00131m^3-\frac{4}{3}\pi(1.22\times 10^{-2}m)^3$
$V=0.00130m^3$