iron cube of side 10 cm is kept on a horizontal table . If the density of iron cube is 8000 kg /m cube , find the pressure on the portion of the table where the cube is kept?

weight= mg=8000kg/m^3 * (.1m)^3

pressure= force/area=mg/(.1)^2=
= 8000kg*9.8N/kg*.1 N/m^2

To find the pressure on the portion of the table where the iron cube is kept, we need to use the formula for pressure:

Pressure = Force / Area

First, let's calculate the force exerted by the iron cube. The force is equal to the weight of the cube, and we can find the weight using the formula:

Weight = Mass * Gravity

Since density is given in kg/m^3, and we know the volume of the cube (side^3), we can calculate its mass:

Mass = Density * Volume

Volume = Side^3

Now, let's plug in the values:

Side = 10 cm = 0.1 m
Density = 8000 kg/m^3

Volume = (0.1)^3 = 0.001 m^3
Mass = 8000 * 0.001 = 8 kg

Next, we need to calculate the force using the formula for weight:

Weight = Mass * Gravity

Gravity is approximately 9.8 m/s^2 (acceleration due to gravity):

Weight = 8 kg * 9.8 m/s^2 = 78.4 N

Now, we have the force exerted by the cube (78.4 N).

To find the pressure, we need to know the area on which the force is acting. Since the cube is kept on a horizontal table, the force is distributed over the entire bottom surface area of the cube, which is a square.

The area of the square can be calculated using the formula:

Area = Length * Width

Since the cube has equal sides, the length and width are both equal to the side length:

Area = Side * Side = 0.1 m * 0.1 m = 0.01 m^2

Finally, we can calculate the pressure using the formula:

Pressure = Force / Area

Pressure = 78.4 N / 0.01 m^2 = 7840 N/m^2

Therefore, the pressure on the portion of the table where the iron cube is kept is 7840 N/m^2 or 7840 Pa (Pascals).