At what temperature a gold ring of diameter 6.230 cm be heated so that it can be fitted on a wooden bangle of diameter 6.241 cm ? Both the diameters have been measured at room temperature (27$^\circ$C).
(Given : coefficient of linear thermal expansion of gold $\alpha$L = 1.4 $\times$ 10$-$5 K$-$1)
Solution
<p>$\Delta D = D\alpha \Delta T$</p>
<p>$\Delta T = {{0.011} \over {6.230 \times 1.4 \times {{10}^{ - 5}}}}$</p>
<p>= 126.11$^\circ$C</p>
<p>$\Rightarrow$ T<sub>f</sub> = T + $\Delta$T</p>
<p>= (27 + 126.11)$^\circ$C</p>
<p>= 153.11$^\circ$C</p>
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
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