An iron cube of side 10 cm is kept on a horizontal table. If the density of iron is 8000 kg m~3. Find the pressure on the portion of the table where the cube is kept. (g = 10 ms^-2)


#1

Side of the cube = 10 cm
Its volume = 10 x 10 x 10 = 1000 cm^3 = 10^-3 m^3
Area of the face of the cube = 10 x 10 = 100 cm^2
= 10^-2 m^2
Density of iron = 8000 kg m^-3 Therefore, mass ‘m’ of the iron cube
= 8000 x 10^-3 kg = 8 kg
Pressure on the portion of the table in contact with the table

Pressure = Weight / Area
= 8 x 10 / 10^-2
= 8 x 10^3 Pa