Physics - Heat & Thermo Question with Solution | TestHub

Thermal resistance $$\begin{array}{rl}& R=\frac{l}{KA}\\ & \therefore R_A=\frac{L}{(2K)(4Lw)}=\frac{1}{8Kw}\\ & \text{ (Here, }w=\text{ width) }\\ & R_B=\frac{4L}{3K(Lw)}=\frac{4}{3Kw}\\ & R_C=\frac{4L}{(4K)(2Lw)}=\frac{1}{2Kw}\\ & R_D=\frac{4L}{(5K)(Lw)}=\frac{4}{5Kw}\\ & R_E=\frac{L}{(6K)(Lw)}=\frac{1}{6Kw}\\ & R_A:R_B:R_C:R_D:R_E\\ & =15:160:60:96:12\end{array}$$ So, let us write, $R_A=15R,R_B=160R$ etc and draw a simple electrical circuit as shown in figure.
$H=$ Heat current $=$ Rate of heat flow. $$H_A=H_E=H$$ [let] $\therefore$ Option (a) is correct. In parallel, current distributes in inverse ratio of resistance. $$\begin{array}{rl}& \therefore H_B:H_C:H_D=\frac{1}{R_B}:\frac{1}{R_C}:\frac{1}{R_D}\\ & =\frac{1}{160}:\frac{1}{60}:\frac{1}{96}\\ & =9:24:15\\ & \therefore H_B=\left(\frac{9}{9+24+15}\right)H=\frac{3}{16}H\\ & H_C=\left(\frac{24}{9+24+15}\right)H=\frac{1}{2}H\\ & \text{ and }{H}_D=\left(\frac{15}{9+24+15}\right)H=\frac{5}{16}H\\ & \end{array}$$ $$H_C=H_B+H_D$$ $\therefore$ Option (d) is correct. Temperature difference (let us call it $T$ ) $=($ Heat current $)\times ($ Thermal resistance $)$ $T_A=H_A{R}_A=(H)(15R)=15\mathrm{HR}$ $T_B=H_B{R}_B=\left(\frac{3}{16}H\right)(160R)=30HR$ $T_C=H_C{R}_C=\left(\frac{1}{2}H\right)(60R)=30HR$ $T_D=H_D{R}_D=\left(\frac{5}{16}H\right)(96R)=30HR$ $T_E=H_E{R}_E=(H)(12R)=12\mathrm{HR}$ Here, $T_E$ is minimum. Therefore, option (c) is also correct. $\therefore$ Correct options are (a), (c) and (d). Analysis of Question (i) From calculation point of view, question is difficult otherwise question is simple. (ii) In heat transfer, questions are mainly asked from conduction and radiation topic.