Find the stress developed inside a tooth cavity filled with copper when hot tea at temperature of $57^\circ C$ is drunk. (Take temperature of tooth to be $37^\circ C$, $\alpha = 1.7 \times {10^{ - 5}}{\text{ }}^\circ {C^{ - 1}}$ and bulk modulus for copper is \[140 \times {10^9}N{m^{ - 2}}\] )
A. $1.43 \times {10^8}N{m^{ - 2}}$
B. $4.13 \times {10^8}N{m^{ - 2}}$
C. $2.12 \times {10^4}N{m^{ - 2}}$
D. $3.12 \times {10^4}N{m^{ - 2}}$

Hint: Stress is the internal restoring force acting per unit area of a deformed body and strain is the fractional change in configuration (in case of bulk modulus volumetric strain is considered).
Bulk modulus is the ratio of normal stress to the volumetric strain of the deformed body within the elastic limit.
The thermal stress produced in a volume is given by $K\gamma \Delta \theta $ where $K$ is the bulk modulus of elasticity, $\gamma $ is the coefficient of cubical expansion and $\Delta \theta $ is the change in temperature.
The relation between the coefficient of cubical expansion $\gamma $ and the coefficient of linear expansion $\alpha $ is $\gamma = 3\alpha $

Complete step by step answer:
Let us first discuss stress, strain and bulk modulus.
Stress is the internal restoring force acting per unit area of a deformed body and strain is the fractional change in configuration (in case of bulk modulus volumetric strain is considered).
Bulk modulus is the ratio of normal stress to the volumetric strain of the deformed body within the elastic limit.
According to the question, the stress is developed due to change in temperature. This type of stress is called thermal stress. The thermal stress produced in a volume is given by $K\gamma \Delta \theta $ where $K$ is the bulk modulus of elasticity, $\gamma $ is the coefficient of cubical expansion and $\Delta \theta $ is the change in temperature.
This formula is derived as
The change in volume due to temperature change is given by $\Delta V = V\gamma \Delta \theta $ where $V$ is the original volume of the body.
So, the volumetric strain can be given by $\dfrac{{\Delta V}}{V} = \gamma \Delta \theta $
Therefore, ${\text{Thermal Stress }} = {\text{ }}K \times \dfrac{{\Delta V}}{V} = K\gamma \Delta \theta $
We know that the relation between the coefficient of cubical expansion $\gamma $ and the coefficient of linear expansion $\alpha $ is $\gamma = 3\alpha $ .
According to the question, the change in temperature will be $\Delta \theta = 57^\circ C - 37^\circ C = 20^\circ C$
Therefore, the thermal stress
$S = K \times 3\alpha \Delta \theta $
On substituting the values we have
$S = 140 \times {10^9} \times 3 \times 1.7 \times {10^{ - 5}} \times 20$
On simplifying we have
$S = 1.43 \times {10^8}N{m^{ - 2}}$
Hence, the stress developed inside the tooth cavity will be $1.43 \times {10^8}N{m^{ - 2}}$ .

Note: Here as we are given Bulk Modulus it is important to consider volumetric expansion. If we were given Young’s Modulus then we would consider linear expansion and solve the problem, also note the relationship between coefficient of linear expansion and coefficient of volumetric expansion.