Two thin metallic spherical shells of radii r₁ and r₂(r₁ [31 Aug 2021 Shift 2]

Correct Answer is : 4πKr1r2(θ2−θ1)r2−r1{4 π K r₁ r₂ (θ₂ - θ₁)}{r₂ - r₁}r2−r14πKr1r2(θ2−θ1)

{4 π K r₁ r₂ (θ₂ - θ₁)}{r₂ - r₁}

{π r₁ r₂ (θ₂ - θ₁)}{r₂ - r₁}

{K(θ₂ - θ₁)}{r₂ - r₁}

{K(θ₂ - θ₁)(r₂ - r₁)}{4 π r₁ r₂}

Correct Answer is : 4πKr1r2(θ2−θ1)r2−r1{4 π K r₁ r₂ (θ₂ - θ₁)}{r₂ - r₁}r2−r14πKr1r2(θ2−θ1)

Test Index

Thermodynamics Part 3

Section: Physics

Question : 62 of 100

Marks: +1, -0

r_{1}

r_{2}(r_{1} < r_{2})

\theta_{1}

\theta_{2}(\theta+1 < \theta_{2})

Solution:

Given that,

r_{1}

and

r_{2}

be the radii of inner and outer shells.

A material of thermal conductivity (K) is filled between region.

Consider an elementary sphere of radius r and thickness dr.

Now, thermal resistance of elementary sphere

dR=\frac{dr}{K 4 \pi r^{2}}

[∵ inner surface area,

A=4 \pi r^{2}

]

So, total thermal resistance,

R=\int dR=\int\limits_{r_1}^{r_2} \frac{dr}{4 \pi K r^{2}} = -\frac{1}{4 \pi K} \left[\frac{1}{r}\right]_{r_1}^{r_2}

R=\frac{r_{2} - r_{1}}{4 \pi K r_{1} r_{2}}

Hence, heat current (or the rate of flow of heat)

\frac{dQ}{dt} = \frac{\Delta \theta}{R} = \frac{(\theta_{2} - \theta_{1})}{\frac{(r_{2} - r_{1})}{4 \pi K r_{1} r_{2}}} = \frac{4 \pi K r_{1} r_{2} (\theta_{2} - \theta_{1})}{r_{2} - r_{1}}

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