JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Heat & Thermodynamics

Two thin metallic spherical shells of radii

r_{1}

and

r_{2}

(

r_{1} < r_{2}

) are placed with their centres coinciding. A material of thermal conductivity K is filled in the space between the shells. The inner shell is maintained at temperature

θ_{1}

and the outer shell at temperature

θ_{2}

(

θ_{1} < θ_{2}

). The rate at which heat flows radially through the material is:

(A) $\frac{π r _{1} r _{2} ( θ _{2} - θ _{1} )}{r _{2} - r _{1}}$
(B) $\frac{K ( θ _{2} - θ _{1} ) ( r _{2} - r _{1} )}{4 π r _{1} r _{2}}$
(C) $\frac{K ( θ _{2} - θ _{1} )}{r _{2} - r _{1}}$
(D) $\frac{4 π K r _{1} r _{2} ( θ _{2} - θ _{1} )}{r _{2} - r _{1}}$ ✓ Correct

Answer: (D)

Solution

In steady state the heat current

\frac{d Q}{d t}

is the same through every spherical layer. Fourier's law for a layer of radius

r

and area

4 π r^{2}

gives

\frac{d Q}{d t} = - K (4 π r^{2}) \frac{d θ}{d r} .

Separating variables and integrating from

r_{1}

r_{2}

as the temperature rises from

θ_{1}

θ_{2}

\frac{d Q}{d t} \int_{r_{1}}^{r_{2}} \frac{d r}{r ^{2}} = - 4 π K \int_{θ_{1}}^{θ_{2}} d θ \Rightarrow \frac{d Q}{d t} (\frac{1}{r _{1}} - \frac{1}{r _{2}}) = 4 π K (θ_{2} - θ_{1}) .

Hence

\frac{d Q}{d t} = \frac{4 π K r _{1} r _{2} ( θ _{2} - θ _{1} )}{r _{2} - r _{1}} .

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Single correct option

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JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Heat & Thermodynamics

Solution

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