Answer
$V = 2720~cm^3$
Work Step by Step
In part (a), we found that the density of the ball is $~~1.5~g/cm^3$
When the density of the liquid is zero, the kinetic energy after falling $4.0~cm$ is $1.6~J$
We can find the speed after falling $4.0~cm$:
$v_f^2 = v_0^2+2ad$
$v_f^2 = 0+2ad$
$v_f = \sqrt{2ad}$
$v_f = \sqrt{(2)(9.8~m/s^2)(0.040~m)}$
$v_f = 0.8854~m/s$
We can find the mass of the ball:
$K = \frac{1}{2}mv^2 = 1.60~J$
$m = \frac{(2)(1.60~J)}{v^2}$
$m = \frac{(2)(1.60~J)}{(0.8854~m/s)^2}$
$m = 4.082~kg$
We can find the volume of the ball:
$V = \frac{m}{\rho}$
$V = \frac{4.082~kg}{1500~kg/m^3}$
$V = 0.00272~m^3$
$V = 2720~cm^3$