Answer to Question #207893 in Mechanics | Relativity for karen

Question

An aluminum tube is observed to be 60.4 cm long at 300 C. The tube is then set up in a linear expansion apparatus adjusted so that the initial gauge reading is zero. Steam at 1000 C is then allowed to enter the tube. After thermal equilibrium is attained at 100o C, it is observed that the change in pointer position of the apparatus is 100 divisions, where 1 pointer division change is equivalent to an expansion of 0.01mm. From this data, compute the coefficient of linear expansion of the rod and the percentage error. The accepted value of this coefficient for the rod is 2.4 x 10-5 / C0 . Express your answer to three significant figures.

Answer 1

Explanations & Calculations

• We are given the initial length $\small l_0 =0.604m \,@\,30 \,^0C$ and the final length at the equilibrium at $\small 100\,^0C$ is unknown.
• But it can be assessed with help of given data in terms of change of the length.

• The length change is

$\qquad\qquad \begin{aligned} \small \Delta l&=\small 0.01mm\,/div\times100\,div\\ &=\small 1\,mm \end{aligned}$

• The relationship for the length change in terms of the expansion of the tube is,

$\qquad\qquad \begin{aligned} \small \Delta l&=\small l_0.\alpha.\Delta\theta\\ \end{aligned}$

• Then the coefficient of linear expansion can be estimated as follows,

$\qquad\qquad \begin{aligned} \small \alpha &=\small \frac{\Delta l}{l_0.\Delta \theta}=\frac{1\times 10^{-3}m}{0.604m\times(100-30)\,^0C}\\ &=\small \bold{2.37\times10^{-5}\,/^0C} \end{aligned}$

• Then the percentage error is

$\qquad\qquad \begin{aligned} \small \%\,E &=\small |\frac{\alpha_{cal}-\alpha_{acc}}{\alpha_{acc}}|\times100\%\\ &=\small |\frac{2.37\times10^{-5}-2.4\times10^{-5}}{2.4\times10^{-5}}|\times100\%\\ &=\small \bold{1.25\%} \end{aligned}$