Question

AN IRON CUBE OF EDGE LENGTH OF 10 CM IS KEPT ON A HORIZONTAL SURFACE .IF THE DENSITY OF IRON IS 7800 KG M^-3 , FIND THE PRESSURE THAT THE CUBE APPLIES ON THE SURFACE

Answer 1

Dear Student ,
The volume of the iron cube = V = ${10}^3=1000c{m}^3=\frac{1000}{{10}^6}{m}^3={10}^{-3}{m}^3$
Now the density of the iron is , d = 7800 kg/m³
So the mass of the cube = m = Vd = $7800\times {10}^{-3}=7·8Kg$
So the pressure applied by the cube on the surface is ,
$P=\frac{F}{A}=\frac{Mg}{A}=\frac{7·8\times 9·8}{10\times 10\times {10}^{-4}}=7644N/m^2$
Regards