Question

Two rods of same length but made of different material are taken for a study as shown in the figure. Area of cross-section of rod $1$ is $12{\text{cm}}^2$ and that of rod $2$ is $16{\text{cm}}^2$. Thermal conductivity of rod $1$ is $0.0008{\text{cal/cm-s-}}^{\circ }\text{C}$. If rate of loss of heat due to conduction is equal, find the thermal conductivity of rod $2$.

• A. $0.0006{\text{cal/cm-s-}}^{\circ }\text{C}$
• B. $0.0008{\text{cal/cm-s-}}^{\circ }\text{C}$
• C. $0.0010{\text{cal/cm-s-}}^{\circ }\text{C}$
• D. $0.0004{\text{cal/cm-s-}}^{\circ }\text{C}$

Answer 1

The correct option is A $0.0006{\text{cal/cm-s-}}^{\circ }\text{C}$

Given, that lengths of two rods are equal.
Cross-sectional area of rod $1$ $\left(A_1\right)=12{\text{cm}}^2$
Cross-sectional area of rod $2$$\left(A_2\right)=16{\text{cm}}^2$
Thermal conductivity of rod $1$ $\left(k_1\right)=0.0008{\text{cal/cm-s-}}^{\circ }\text{C}$
Rate of loss of heat in rod $1$ and $2$
$Q_1=\frac{k_1{A}_1(T_1-T_2)}{l_1}$
$Q_2=\frac{k_2{A}_2(T_1-T_2)}{l_2}$
$\because Q_1=Q_2$ and $l_1=l_2;\Delta T_1=\Delta T_2$
We get, $k_1{A}_1=k_2{A}_2$
$\Rightarrow k_2=\frac{k_1{A}_1}{A_2}=\frac{0.0008\times 12}{16}$
$=0.0006{\text{cal/cm-s-}}^{\circ }\text{C}$