The walls of a closed cubical box of edge 50cm are made of a material of thickness 1 mm and coefficient of thermal conductivity $4\times {10}^{-4}cal/s/cm/{}^{0}C.$ The interior of the box is maintained at ${100}^{0}C$ above the outside temperature by a heater placed inside the box and connected across 400V d.c. The resistance of the heater is nearly:

The rate of heat transmitted through the walls of the closed cubical box, 

$\text{ Area, }A=6\times \text{ area of each face }$

$H=\frac{q}{t}=\frac{KA({\theta }_2-{\theta }_1)}{d}=\frac{4\times {10}^{-4}\times 6\times 50\times 50\times 100}{0.1}=6000cal/s$

To maintain constant temperature difference between and inside the box, this heat escaped must be produced by the electric current in the heater. Let R be the resistance of the coil.  The heat produced per second is 

$H=\frac{Q}{t}=\frac{V^2}{R}$

$H=\frac{V^2}{JR}cal=\frac{V^2}{4.2R}cal$

$\therefore \frac{V^2}{4.2R}=6000$

$\Rightarrow R=\frac{400\times 400}{4.2\times 6000}=6.35\Omega$