Question

Find the stress developed inside a tooth cavity filled with copper when hot tea at temperature of $57^{\circ} C$ is drunk. You can take body (tooth) temperature to be $37^{\circ} C$ and $a=1.7 \times 10^{-5}{ }^{\circ} C ^{-1}$, Bulk modulus for copper $B=140 \times 10^{9} N \cdot M^{-2}$

• A. $1.4 \times 10^{8} N \cdot m ^{-2}$
• B. $1.9 \times 10^{8} N \cdot m^{-2}$
• C. $2.0 \times 10^{8} N \cdot m ^{-2}$
• D. $3.4 \times 10^{7} N \cdot m ^{-2}$

Answer 1

Answer: (A)

Temperature of hot tea, $t_{2}=57^{\circ} C$
Normal temperature of tooth, $t_{1}=37^{\circ} C$
$\alpha=1.7 \times 10^{-5} C ^{-1}$
Bulk modulus, $B=140 \times 10^{9} Nm ^{-2}$
Thermal stress in tooth cavity
$=$ Bulk modulus $\times$ Volumetric strain
$=B \times \frac{\Delta V}{V}=B \times \gamma \cdot \Delta t$
$[\because \Delta V=V \gamma \Delta t]$
$=B \times 3 \alpha \times \Delta t\,\, [\because \gamma=3 \alpha]$
$=3 B \alpha \Delta t=3 B \alpha\left(t_{2}-t_{1}\right)$
$=3 \times 140 \times 10^{9} \times 1.7 \times 10^{-5}(57-37)$
$=1.4 \times 10^{8} Nm ^{-2}$