Question

Sumit has a cube of side 0.2m of mass 50 kg. The pressure exerted by the cube on the table is

(Take g = $10m/s^2$)

• A. $25N{m}^{-2}$
• B. $1200N{m}^{-2}$
• C. $1250N{m}^{-2}$
• D. $12500N{m}^{-2}$

Answer 1

The correct option is (D)

Given,

Mass = 50 kg

Side of the cube = 0.2 m

Now we can write

Force of the table = 500 N

Area of the cube in contact with table = $0.2\times 0.2=0.04m^2$

We know,

$Pressure=\frac{Force}{Area}=\frac{500}{0.04}=12500N{m}^{-2}$

Hence the pressure of the table is 12500 $N{m}^{-2}$