The walls of a closed cubical box of edge $60 \mathrm{~cm}$ are made of material of thickness $1…
The walls of a closed cubical box of edge $60 \mathrm{~cm}$ are made of material of thickness $1 \mathrm{~mm}$ and thermal conductivity, $4 \times 10^{-4} \mathrm{cal} \mathrm{s}^{-1} \mathrm{~cm}^{-1}{ }^{\circ} \mathrm{C}^{-1}$. The interior of the box is maintained $1000^{\circ} \mathrm{C}$ above the outside temperature by a heater placed inside the box and connected across $400 \mathrm{~V}$ DC supply. The resistance of the heater is
$4.41 \Omega$
$44.1 \Omega$
$0.441 \Omega$
$441 \Omega$
Solution
Key Idea Heat transfer through conduction wall is given by mathematical expression,
$
\frac{d Q}{d t}=k A \frac{\left(T_1-T_0\right)}{x}
$
where, $\frac{d Q}{d t}=$ power transferred through the wall and $x=$ thickness of the wall
Here, side of cube, $a=60 \mathrm{~cm}$. Hence, area $A=6 a$,
$\because$ Total area $=6$ (area of one side of the cube)
thickness $x=0.1 \mathrm{~cm}$, thermal conductivity, $k=4 \times 10^{-4} \mathrm{cal} \mathrm{s}^{-1} \mathrm{~cm}^{-1}{ }^{\circ} \mathrm{C}^{-1}$, temperature difference, $\left(T_1-T_0\right)=1000^{\circ} \mathrm{C}$ and potential of DC source $V=400 \mathrm{~V}$
Hence, the power
$
\begin{aligned}
P & =\frac{k A \times 4.184 \times\left(T_1-T_0\right)}{x} \\
P & =\frac{k 6 a^2 \times 4.184 \times 10^3}{x} \\
P & =\frac{4 \times 10^{-4} \times 6 \times(60)^2 \times 10^3 \times 4.184}{0.1} \mathrm{~J} \\
P & =361.49 \mathrm{~W}
\end{aligned}
$
Given, voltage supply of $\mathrm{DC}=400 \mathrm{~V}$ Hence, power generated in a resistance,
$
\begin{aligned}
& P=\frac{V^2}{R}=361.49 \Rightarrow R=\frac{V^2}{P} \\
& R=\frac{400 \times 400}{361.49 \times 10^3}=0.4426 \Omega
\end{aligned}
$
Hence, the correct option is (c)