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A bimetallic strip is made of aluminium and steel $\left(\alpha_{\mathrm{AL}}>\alpha_{\text {steel }}\right) .$ On heating, the strip will
(A) remain straight
(B) get twisted
(C) will bend with aluminium on the concave side.
(D) will bend with steel on the concave side.

Last updated date: 25th Jun 2024
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Hint: The thermal conductivity of a material can be defined as a measure of its ability to conduct heat. It is commonly denoted by the symbols $k, \lambda,$ or $\kappa$. Heat transfer occurs at a lower rate in materials with low thermal conductivity than in materials with a high thermal conductivity. For instance, metals typically have high thermal conductivity and are very efficient at conducting heat, while the insulating materials like Styrofoam have a very low thermal conductivity. Correspondingly, materials of high thermal conductivity are widely used in heat sink applications, and materials of low thermal conductivity are generally used as thermal insulation. The reciprocal of thermal conductivity is known as thermal resistivity. The defining equation for thermal conductivity is $\mathbf{q}=-k \nabla T$, where $\mathbf{q}$ is the heat flux, k is the thermal conductivity, and $\nabla T$ is the temperature gradient. This is known as the Fourier's Law for heat conduction. Although commonly expressed as a scalar, the most general form of thermal conductivity is a second-rank tensor.

Complete step by step answer
We know that the thermal conductivity of aluminium is greater than that of steel. Both strips of Aluminium and steel are fixed together initially in a bimetallic strip. When both are heated then expansion in steel will be smaller than aluminium.

So, Al strip will be convex side and steel on concave side verifies the option D as the correct answer.

Note: We must be thorough and clear with the basic concepts of thermal conduction and thermodynamics in case of different materials. Keeping this in mind we can solve most similar problems.