10-1.Thermometry, Thermal Expansion and Calorimetry
hard

A bimetallic strip is formed out of two identical strips, one of copper and other of brass. The coefficients of linear expansion of the two metals are ${\alpha _C}$ and ${\alpha _{B}}.$ On heating, the temperature of the strip goes up by $\Delta T$ and the strip bends to form an arc of radius of curvature $R.$ Then $R$ is

A

Proportional to $\Delta T$

B

Inversely proportional to $\Delta T$

C

Inversely proportional to $|{\alpha _B} - {\alpha _C}|$

D

Both $(B)$ and $(C)$

(IIT-1999)

Solution

(d) Let $L_0$ be the initial length of each strip before heating.

Length after heating will be

${L_B} = {L_0}(1 + {\alpha _B}\Delta T) = (R + d)\theta $

${L_C} = {L_0}(1 + {\alpha _C}\Delta T) = R\theta $

==> $\frac{{R + d}}{R} = \frac{{1 + {\alpha _B}\Delta T}}{{1 + {\alpha _C}\Delta T}}$

==> $1 + \frac{d}{R} = 1 + ({\alpha _B} – {\alpha _C})\Delta T$

==> $R = \frac{d}{{({\alpha _B} – {\alpha _C})\Delta T}}$ ==> $R \propto \frac{1}{{\Delta T}}$ and $R \propto \frac{1}{{({\alpha _B} – {\alpha _C})}}$

Standard 11
Physics

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