Question

A cubical box is to be constructed with iron sheets 1mm in thickness

Answer 1

here the total volume of iron used will be equal to difference of the total volume and the inside volume of block.

let each side be equal to x mm.

So, as the thickness is 1mm we have (subtract 1mm both sides)

inside volume = (x-2)3 mm3

total volume = x3 mm3

V = x3 - (x-2)3 mm3

now,

the weight exerted in the block will be

F = mg = ρVg = 8000 *g*[x3 - (x-2)3]........................(1)

now the buoyant force due to water will be

F' = ρ'V'g = 1000 *g*x3..............................(2)

so, the necessary condition for the block to not sink is

F = F'

or

8000 *g*[x3 - (x-2)3] = 1000 *g*x3

or

8[x3 - (x-2)3] = x3

or

x3 = 8x3 - 8(x-2)3

or

(x-2)3 = (7/8)x3

x-2 = (7/8)1/3x

x - 0.956x = 2

0.965x = 2

thus, the minimum value of external edge will be

x = 45.45 mm ~ 4.54 cm

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