Answer
$\Delta V=29 cm^3$
Work Step by Step
As given that $r_{\circ}=10 cm= 0.1 m$ and $\Delta{T}=100 C^{\circ}$
we know that
$V_{\circ}=\frac{4}{3}{\pi}r_{\circ}^3=\frac{4}{3}{(3.14)}(0.1)^3=4.19\times10^{-3}m^3$
We also know that
$\Delta V=\gamma V_{\circ}\Delta{T}=(3\alpha)V_{\circ}\Delta{T}$
putting the values, we get
$\Delta V=(3)(23\times10^{-6})(4.19\times10^{-3})(100)$
$\Delta V= 2.89\times10^{-5}m^3$
$\Delta V= 28.9 cm^3=29 cm^3$
$\Delta V=29 cm^3$
