Answer
(a) more than
(b) $7.4fm$
(c) $5.9fm$
Work Step by Step
(a) We know that
$V\propto r^3$
As the volume is reduced to one half, the radius becomes $(\frac{1}{2})^{\frac{1}{3}}$ times its initial value. Thus, the radius of each of the two equal fragments of $U^{236}$ is $(\frac{1}{2})^{\frac{1}{3}}=0.8$ times that of the initial value. This shows that the new radius is more than half of the original radius.
(b) We can calculate the required radius as
$r=(1.2\times 10^{-15}m)(236)^{\frac{1}{3}}$
$r=7.4fm$
(c) The required radius can be calculated as
$r=(0.8)(7.4\times 10^{-15}m)$
$r=5.92\times 10^{-15}m=5.9fm$
