Question

Find the rate of heat flow through a cross-section of the tapered conical rod shown in figure $($Temperature of the big end$\left({\theta }_2\right)>$ Temperature of the small end$\left({\theta }_1\right))$, with radius $r_1$ and $r_2$ at the small and big end respectively. Thermal conductivity of the material of the rod is K.

• A. $\frac{K\pi r_1^2{r}_2^2({\theta }_2-{\theta }_1)}{L^3}$
• B. $\frac{K\pi r_1{r}_2({\theta }_2-{\theta }_1)}{L}$
• C. $\frac{K\pi r_1^3{r}_2^3({\theta }_2-{\theta }_1)}{L^5}$
• D. $K\pi \sqrt{\frac{r_1{r}_2}{l}}({\theta }_2-{\theta }_1)$

Answer 1

The correct option is B

$\frac{K\pi r_1{r}_2({\theta }_2-{\theta }_1)}{L}$

As is evident the cross sectional area is not constant, it varies with x and the corresponding radius r

By the fact that the slope of the line is constant

$\frac{r_2-r_1}{l}=\frac{r-r_1}{x}$

$\Rightarrow r=\left(\frac{r_2-r_1}{l}\right)x+r_1$ ..........(i)

Now, consider a cylindrical shell with infinitesimal thickness dx, let d$\theta$ be the temperature difference

$\frac{\Delta Q}{\Delta t}=KA\frac{d\theta }{dx}=K\pi r^2\frac{d\theta }{dx}$ ......(ii)

From (i) and (ii)

$\frac{\Delta Q}{\Delta t}=K\pi {[\left(\frac{r_2-r_1}{l}\right)x+r_1]}^2\frac{d\theta }{dx}$

$\Rightarrow \frac{\Delta Q}{\Delta t}{\int }_0^L\frac{dx}{{(r_1+\left(\frac{r_2-r_1}{L}\right)x)}^2}=K\pi {\int }_{{\theta }_1}^{\theta }d\theta$ ..(iii)

Let , y=$r_1+\left(\frac{r_2-r_1}{l}\right)x$

$\Rightarrow {\int }_{\theta }^L\frac{dx}{{(r_1+\left(\frac{r_2-r_1}{L}\right)x)}^2}=\frac{l}{r_2-r_1}{\int }_{r_1}^{r_2}\frac{dy}{y^2}$

=$\frac{l}{r_2-r_1}\times (\frac{l}{r_1}-\frac{l}{r_2})$

=$\frac{l}{r_1{r}_2}$

From (iii) and (iv)

$\frac{\Delta Q}{\Delta t}=\frac{K\pi r_1{r}_2({\theta }_2-{\theta }_1)}{l}$