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Question: Find the stress developed inside a tooth cavity filled with copper when hot tea at temperature of 57...

Find the stress developed inside a tooth cavity filled with copper when hot tea at temperature of 57°C is durnk. (Take temperature of tooth to be 37°C, α=1.7×10−5oC−1\alpha = 1.7 \times 1{0^{- 5}}^{o}C^{- 1} and bulk modulus for copper =140×109Nm−2= 140 \times 10^{9}Nm^{- 2}

A

1.43×108Nm−21.43 \times 10^{8}Nm^{- 2}

B

4.13×108Nm−24.13 \times 10^{8}Nm^{- 2}

C

2.12×104Nm−22.12 \times 10^{4}Nm^{- 2}

D

3.12×104Nm−23.12 \times 10^{4}Nm^{- 2}

Answer

1.43×108Nm−21.43 \times 10^{8}Nm^{- 2}

Explanation

Solution

Volumetric strain in tooth cavity=ΔVV= \frac{\Delta V}{V}

Let γ\gammabe the coefficient of volume expansion with the change in temperature of ΔT∘C.\Delta T{^\circ}C.

Change in volume is ΔV=γVΔT\Delta V = \gamma V\Delta TOr ΔVV=γΔT\frac{\Delta V}{V} = \gamma\Delta T

Thermal stress in tooth cavity=β×= \beta \timesVolumteric strain =β×γΔT= \beta \times \gamma\Delta T

=β×3αΔT= \beta \times 3\alpha\Delta T (∵γ=3α\because\gamma = 3\alpha)

=140×109×3×1.7×10−5×(57∘C−37∘C)= 140 \times 10^{9} \times 3 \times 1.7 \times 10^{- 5} \times (57{^\circ}C - 37{^\circ}C)

=1.73×108Nm−2= 1.73 \times 10^{8}Nm^{- 2}