The walls of a closed cubical box of edge` 40 cm` are made of a material of thickness `1 mm` and thermal conductivity `4 xx 10^(-4) cals^(-1@) C^(-1)`. The interior of the box is maintained at `100 ^(@) C` above the outside temperature by a heater placed inside the box and connected across `400 V dc` . Calculate the resistance of the heater.

Explanation

To solve the problem step by step, we will follow the concepts of thermal conductivity and electrical resistance.

### Step 1: Understand the Problem
We have a closed cubical box with an edge of 40 cm, made of a material with a thickness of 1 mm and thermal conductivity of \(4 \times 10^{-4} \, \text{cal/s/°C}\). The interior of the box is maintained at \(100 \, \text{°C}\) above the outside temperature by a heater connected to a \(400 \, \text{V}\) DC source. We need to calculate the resistance of the heater.

### Step 2: Convert Units
1. Convert the edge length of the cube from centimeters to meters:
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