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(D) The rate of heat flow $H$ through a cylindrical rod is given by $H = \frac{kA \Delta 	heta}{\ell}$, where $k$ is thermal conductivity, $A = \pi r

Vedclass · Accepted Answer

$3: 10$

Vedclass · Answer

(D) The rate of heat flow $H$ through a cylindrical rod is given by $H = \frac{kA \Delta 	heta}{\ell}$, where $k$ is thermal conductivity, $A = \pi r^2$ is the cross-sectional area, $\Delta 	heta$ is the temperature difference, and $\ell$ is the length.Since the material is the same, $k_A = k_B$.Given: $\frac{\Delta 	heta_A}{\Delta 	heta_B} = \frac{2}{3}$, $\frac{H_A}{H_B} = \frac{5}{9}$, and $\frac{r_A}{r_B} = \frac{1}{2}$.Using the formula $\frac{H_A}{H_B} = \left( \frac{r_A}{r_B} ight)^2 \left( \frac{\Delta 	heta_A}{\Delta 	heta_B} ight) \left( \frac{\ell_B}{\ell_A} ight)$:$\frac{5}{9} = \left( \frac{1}{2} ight)^2 \left( \frac{2}{3} ight) \left( \frac{\ell_B}{\ell_A} ight)$$\frac{5}{9} = \left( \frac{1}{4} ight) \left( \frac{2}{3} ight) \left( \frac{\ell_B}{\ell_A} ight)$$\frac{5}{9} = \frac{2}{12} \left( \frac{\ell_B}{\ell_A} ight) = \frac{1}{6} \left( \frac{\ell_B}{\ell_A} ight)$$\frac{\ell_B}{\ell_A} = \frac{5}{9} 	imes 6 = \frac{30}{9} = \frac{10}{3}$Therefore, $\frac{\ell_A}{\ell_B} = \frac{3}{10}$.

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