Find the stress developed inside a tooth cavity filled with copper when hot tea at temperature of…
Find the stress developed inside a tooth cavity filled with copper when hot tea at temperature of \(57^{\circ} \mathrm{C}\) is drunk. You can take body (tooth) temperature to be \(37^{\circ} \mathrm{C}\) and \(\alpha=1.7 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\), bulk modulus for copper \(B=140 \times 10^9 \mathrm{Nm}^{-2}\).
\(1.4 \times 10^8 \mathrm{Nm}^{-2}\)
\(1.9 \times 10^8 \mathrm{Nm}^{-2}\)
\(2.0 \times 10^8 \mathrm{Nm}^{-2}\)
\(3.4 \times 10^7 \mathrm{Nm}^{-2}\)
Solution
Temperature of hot tea, \(t_2=57^{\circ} \mathrm{C}\)
Normal temperature of tooth, \(t_1=37^{\circ} \mathrm{C}\)
\(\alpha=1.7 \times 10^{-5} \mathrm{C}^{-1}\)
Bulk modulus, \(B=140 \times 10^9 \mathrm{Nm}^{-2}\)
Thermal stress in tooth cavity
\(\begin{aligned}
& =\text {Bulk modulus } \times \text { Volumetric strain } \\
& =B \times \frac{\Delta V}{V}=B \times \gamma \cdot \Delta t \\
& {[\because \Delta V=V \gamma \Delta t]} \\
& =B \times 3 \alpha \times \Delta t \\
& {[\because \gamma=3 \alpha]} \\
& =3 B \alpha \Delta t=3 B \alpha\left(t_2-t_1\right) \\
& =3 \times 140 \times 10^9 \times 1.7 \times 10^{-5}(57-37) \\
& =1.4 \times 10^8 \mathrm{Nm}^{-2} \\
\end{aligned}\)