A thin rod having length L₀ at 0° C and coefficient of linear expansion α has its two ends maintained at temperatures θ₁ and θ₂, respectively. Find its new length.

This is a long answer type question as classified in NCERT Exemplar

Consider the diagram

Θ= θ₁+ θ₂/2

Let temperature varies linearly in the rod from its one end to other end, let θ be the temperature of the midpoint of the rod. At steady state

Rate of flow of heat,

dQ/dt = $\frac{KA({\theta }_1-\mathrm{\theta })}{\frac{L_o}{2}}=\frac{KA\left(\mathrm{\theta }-{\mathrm{\theta }\mathrm{}}_2\right)}{\frac{L_o}{2}}$

where k is the coefficient of thermal conductivity of the rod

so θ₁- θ= θ- θ₂

θ= θ₁+ θ₂/2

L=L₀ (1+ $\alpha \mathrm{\theta }$ )

L= L_o (1+ $\frac{\alpha \left({\mathrm{\theta }\mathrm{}}_1+{\mathrm{\theta }}_2\right)}{2}$ )

<p><span data-teams="true">This is a long answer type question as classified in NCERT Exemplar</span></p><p>Consider the diagram</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745477526phpPikL2m_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745477526phpPikL2m.jpeg" alt="" width="156" height="149"></picture></div></div><p>Θ= θ₁+ θ₂/2</p><p>Let temperature varies linearly in the rod from its one end to other end, let θ be the temperature of the midpoint of the rod. At steady state</p><p>Rate of flow of heat, </p><p>dQ/dt = <span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>K</mi> <mi>A</mi> <mo> (</mo> <msub> <mrow> <mrow> <mi>θ</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> <mo>-</mo> <mi mathvariant="normal">θ</mi> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>L</mi> </mrow> </mrow> <mrow> <mrow> <mi>o</mi> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <mi>K</mi> <mi>A</mi> <mfenced separators="|"> <mrow> <mrow> <mi mathvariant="normal">θ</mi> <mo>-</mo> <msub> <mrow> <mrow> <mi mathvariant="normal">θ</mi> <mi mathvariant="normal"> </mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfenced> </mrow> </mrow> <mrow> <mrow> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>L</mi> </mrow> </mrow> <mrow> <mrow> <mi>o</mi> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </mrow> </mrow> </mfrac> </math> </span></p><p>where k is the coefficient of thermal conductivity of the rod</p><p>so θ₁- θ= θ- θ₂</p><p>θ= θ₁+ θ₂/2</p><p>L=L₀ (1+<span contenteditable="false"> <math> <mi>α</mi> <mi mathvariant="normal">θ</mi> </math> </span>)</p><p>L= L_o (1+<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>α</mi> <mfenced separators="|"> <mrow> <mrow> <msub> <mrow> <mrow> <mi mathvariant="normal">θ</mi> <mi mathvariant="normal"> </mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mrow> <mi mathvariant="normal">θ</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfenced> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </math> </span>)</p>

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