Question

The walls of a closed cubical box of edge $5cm$ are of a material of thermal conductivity $4\times 10^{4}cal{s}^{-1}c{m}^{-1}{C}^{-1}$ and thickness $1mm$. The interior is maintained at 100${}^{\circ }$C above the outside temperature by a heater placed inside the box and connected across a $400V$ d.c. source. The resistance of the heater is

• A. 3.56 $\Omega$
• B. 6.35 $\Omega$
• C. 5.36 $\Omega$
• D. 36.5 $\Omega$

Answer 1

Answer: (B)

Using Q = $\frac{\lambda A({\theta }_2-{\theta }_1)t}{x}$ we get $\frac{Q}{t}=\frac{\lambda A({\theta }_2-{\theta }_1)}{x}$
= $\frac{(4\times 10^{-4})6\times 50\times 50\times 100}{0.1}$ ( $\because$ number of faces = 6 )
= 600 cal
Heat produced, H = $\frac{V^2}{R}i.e.6000\times 4.2=\frac{400\times 400}{R}$ or R = $\frac{400\times 400}{4.2\times 6000}$ = 6.35 $\Omega$