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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Thermal Properties of Matter

NTA Abhyas 2020 · Physics · Thermal Properties of Matter

A rod of length l with thermally insulated lateral surface consists...

A rod of length ll with thermally insulated lateral surface consists of material whose heat conductivity coefficient varies with temperature as K(x)=αTK\left(x\right) = \dfrac{\alpha}{T} (here α\alpha is a positive constant). The ends of the rod are kept at temperature T1{T}_{1} and T2{T}_{2} (T1>T2{T}_{1} > {T}_{2}).

A rod of length l with thermally insulated lateral surface consists of material whose heat conductivity coefficient

Find the heat current per unit cross-sectional area in the rod

Options

  1. Aαlln⁡(T1+T2T1)\dfrac{\alpha}{l} \ln \left(\dfrac{{T}_{1} + {T}_{2}}{{T}_{1}}\right)
  2. Bαlln⁡(T1T2)\dfrac{\alpha}{l} \ln \left(\dfrac{{T}_{1}}{{T}_{2}}\right)
  3. Cαlln⁡(2T1T1+T2)\dfrac{\alpha}{l} \ln \left(\dfrac{2{T}_{1}}{{T}_{1} + {T}_{2}}\right)
  4. Dαlln⁡(T1+T2T2)\dfrac{\alpha}{l} \ln \left(\dfrac{{T}_{1} + {T}_{2}}{{T}_{2}}\right)

Correct answer

B. αlln⁡(T1T2)\dfrac{\alpha}{l} \ln \left(\dfrac{{T}_{1}}{{T}_{2}}\right)

Step-by-step solution

A rod of length l with thermally insulated lateral surface consists of material whose heat conductivity coefficient

dQdt=−KA(T(x+dx)−T(x))dx⇒I0=−αTAdTdx\dfrac{dQ}{dt} = - KA\left(T\left(x + dx\right) - T\left(x\right)\right)dx ⇒ {I}_{0} = - \dfrac{α}{T}A\dfrac{dT}{dx}
⇒∫0lI0Aαdx=−∫T1T2dTT⇒ \mathop{∫}_{0}^{l}\dfrac{{I}_{0}}{A\alpha}dx = - \mathop{∫}_{{T}_{1}}^{{T}_{2}}\dfrac{dT}{T}
∴∣I0lAα∣=lnT1T2∴ \left|\dfrac{{I}_{0}l}{A\alpha}\right| = ln\dfrac{{T}_{1}}{{T}_{2}}
⇒∣I0A∣=αllnT1T2⇒ \left|\dfrac{{I}_{0}}{A}\right| = \dfrac{\alpha}{l}ln\dfrac{{T}_{1}}{{T}_{2}}
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