Accomplished silver workers in india can pound silver into incredibly thin sheets, as thin as 3.00 10-7 m (about one-hundredth of the thickness of a sheet of paper). find the area of such a sheet that can be formed from 1.70 kg of silver.

The density of silver is ρ = 10500 kg/m³ approximately.

Given:
m = 1.70 kg, the mass of silver
t = 3.0 x 10⁻⁷ m, the thickness of the sheet

Let A be the area.
Then, by definition,
m = (t*A)*ρ

Therefore
A = m/(t*ρ)
    = (1.7 kg)/ [(3.0 x 10⁻⁷ m)*(10500 kg/m³)]
    = 539.7 m²

539.7 m²

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