Answer
(a) $17\mu m$
(b) Yes
(c) $21Kg$
Work Step by Step
(a) We know that
$V=\frac{m}{\rho}$
$\implies V=\frac{11Kg}{11300Kg/m^3}=9.73\times 10^{-4}m^3$
Now the thickness of the foil can be determined as follows:
$t=\frac{V}{A}$
$\implies t=\frac{9.73\times 10^{-4}Kg/m^3}{55.7m^2}=1.75\times 10^{-5}m^2=17\mu m$
(b) We can calculate the required mass as follows:
$M=(Volume \space of\space balloon)(Density \space of\space air-Density \space of helium)$
$M=(3.049m)^3(1.29Kg/m^3-0.179Kg/m^3)$
$M=31.5Kg$
As this mass is larger than the mass of the balloon (which is $11Kg$), hence the balloon will float.
(c) We know that the mass that the balloon could lift in addition to its own mass can be determined as:
$m=31.5Kg-11Kg$
$m=21Kg$
