The walls of a closed cubical box of edge 60 cm are made of material of thickness 1 mm and thermal conductivity,  `4 xx 10^(-4)" cal " s^(-1) cm ^(-1 @) C^(-1)` The interior of the box is maintained `1000^(@)C` above the outside temperature by a heater placed inside the box and connected across 400 V DC supply. The resistance of the heater is

Options

A. `4.41 Omega`
B. `44.1Omega`
C. `0.441Omega`
D. `441Omega`

Answer

C

Explanation

Key Ide Heat transfer through conduction wall is given by mathematical expression,
`(dQ)/(dt) = kA((T_(1) = T_(0)))/x`
where, `(dQ)/dt=` power transferred through the wall and
`x = ` thickness of the wall
Here, side of cube, `a = 60 ` cm. Hence, area ` A = 6a,`
`because` Total area = 6 ( area of one side of the cube )
thickness `x = 0.1` cm thermal conductivity,
` k = 4 xx 10 ^(-4)" cal " s^(-1) cm ^(-1) ""^(@)C^(-1)` temperature
difference, `(T_(1) - T_(0)) = 1000^(@) C` and potential of DC
source `V = 400 V`
Hence, the power
`P= (kA xx 4.184xx (T_(1) - T_(0)))/x`
`P= (k6a^(2) xx 4.184xx 10^(3))/x`
`P = (4xx 10^(-4) xx 6xx (60)^(2) xx 10^(3) xx 4.184)/0.1 J`
`P = 361.49 W`
Given, voltage supply of `DC = 400 V`
Hence, power generated in a resistance,
`P = V^(2) / R = 361. 49 rArr R = V^(2) /P`
`R = (400xx 400)/(361.49xx 10^(3) ) = 0.44 26 Omega`
Hence, the correct option is(c ) .