Answer
\[\underline{\text{3}\text{.5 cm}}\]
Work Step by Step
(1) Density (d) is defined as mass per unit volume.
\[d=m/V\]
Here, m is the mass of mercury and V is the volume of mercury.
Rearrange this equation to calculate the volume of mercury.
\[V=m\text{ }/d\] (1)
(2) Calculate the volume of mercury at \[\text{0}\text{.0}{}^\circ \text{C}\] by substituting the values in equation (1) as follows:
\[\begin{align}
& V\ \text{(0}\text{.0}{}^\circ \text{C})=m\text{/}d \\
& =\frac{3.380\text{ }g}{13.596\ \text{g/c}{{\text{m}}^{\text{3}}}} \\
& \text{= 0}\text{.2486 }c{{m}^{3}}
\end{align}\]
(3) Calculate the volume of mercury at \[\text{25}\text{.0}{}^\circ \text{C}\] by substituting the values in equation (1) as follows:
\[\begin{align}
& V\text{ (25}\text{.0}{}^\circ \text{C})=m\text{/}d \\
& =\frac{3.380\text{ }g}{13.534\ \text{g/c}{{m}^{3}}} \\
& \text{= 0}\text{.2497 }c{{m}^{3}}
\end{align}\]
(4) Calculate the change in the volume of mercury as follows:
\[\begin{align}
& Change\text{ }in\text{ }volume=V\text{ (25}\text{.0}{}^\circ \text{C})-V\text{ (0}\text{.0}{}^\circ \text{C}) \\
& =\text{0}\text{.2497 }c{{m}^{3}}-\text{0}\text{.2486 }c{{m}^{3}} \\
& =\text{0}\text{.0011 }c{{m}^{3}}
\end{align}\]
(5) Capillary is cylindrical in shape.
Let the rise in the capillary be \[x\text{ cm}\].
Calculate the change in volume in the capillary with the temperature as follows:
\[\begin{align}
& \text{Change in volume = 3}\text{.14}\times {{\text{(Radius of capillary)}}^{2}}\times \text{Rise in the capillary} \\
& \text{=3}\text{.14}\times {{\left\{ \left( \frac{0.200\text{ mm}}{2} \right)\left( \frac{1\text{ cm}}{10\text{ mm}} \right) \right\}}^{2}}\times \left( x\text{ cm} \right) \\
& =\text{0}\text{.000314}x\text{ c}{{\text{m}}^{3}}
\end{align}\]
(6) Calculate the rise in the capillary by equating the two-volume changes as follows:
\[\begin{align}
& \text{0}\text{.000314}x\text{ c}{{\text{m}}^{3}}\text{ = 0}\text{.0011 c}{{\text{m}}^{\text{3}}} \\
& x=\text{ 3}\text{.5 cm}
\end{align}\]
Rise in the mercury level in thermometer is \[\underline{\text{3}\text{.5 cm}}\].
