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(D) The change in volume of the mercury is given by $\Delta V = V_0 \gamma \Delta 	heta$. This change in volume causes the mercury to rise in the ste

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$4.5 \, cm$

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(D) The change in volume of the mercury is given by $\Delta V = V_0 \gamma \Delta 	heta$. This change in volume causes the mercury to rise in the stem,so $\Delta V = A \Delta l$,where $A$ is the cross-sectional area and $\Delta l$ is the change in length. Equating the two,we get $A \Delta l = V_0 \gamma \Delta 	heta$. Given: $V_0 = 10^{-6} \, m^3$,$\gamma = 18 	imes 10^{-5} \, ^\circ C^{-1}$,$\Delta 	heta = 100 \, ^\circ C$,and $A = 0.004 \, cm^2 = 0.004 	imes 10^{-4} \, m^2 = 4 	imes 10^{-7} \, m^2$. Substituting the values: $\Delta l = \frac{V_0 \gamma \Delta 	heta}{A} = \frac{10^{-6} 	imes 18 	imes 10^{-5} 	imes 100}{4 	imes 10^{-7}}$. $\Delta l = \frac{18 	imes 10^{-9}}{4 	imes 10^{-7}} = 4.5 	imes 10^{-2} \, m = 4.5 \, cm$.

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