The figure shows a system of two concentric spheres of radii r₁ and r₂ kept at temperature T₁ and T₂ respectively. The radial flow of heat in a substance between the two concentric spheres is proportional to

Correct Answer is : r1r2(r2−r1)\; r₁ r₂/(r₂ - r₁)(r2−r1)r1r2

Correct Answer is : r1r2(r2−r1)\; r₁ r₂/(r₂ - r₁)(r2−r1)r1r2

Test Index

CGPET 2013 Solved Paper

Section: Physics

Question : 6 of 150

Marks: +1, -0

r_1

r_2

T_1

T_2

Solution:

Consider a shell of thickness

(d r)

and of radii (

r

) and the temperature of inner and outer surfaces of this shell be

T

(T-dT)

\; \frac{dQ}{dt} =

rate of flow of heat through it

\; = \; \frac{kA[(T-dT)-T]}{dr} = \; \frac{-KAdT}{dr}

\; = -4\pi (kr)^2 \; \frac{dT}{dT} \;\; [ \because A = 4\pi r^2 ]

To measure the radial rate of heat flow, integration technique is used, since the area of the surface through which heat will flow is not constant.

Then,

\; \left( \; \frac{dQ}{dt} \right) \int\limits_{r_1}^{r_2} \; \frac{1}{r^2} dr = -4\pi K \int\limits_{T_1}^{T_2} dT

\;\; \frac{dQ}{dt} \left( \; \frac{1}{r_1} - \; \frac{1}{r_2} \right) = -4\pi K (T_2 - T_1)

\;\; \frac{dQ}{dt} = \; \frac{-4\pi k_1 r_2 (T_2 - T_1)}{(r_2 - r_1)}

\;\Rightarrow \;\; \; \frac{dQ}{dt} \propto \; \frac{(r_1 r_2)}{(r_2 - r_1)}

\;

Go to Question:

Prev Question

Next Question