Question

Two cubes of masses 3 g and 6 g and sides of length $2\text{cm}$ and $3\text{cm}$ respectively are placed on a smooth table one over other. Calculate the pressure experienced by the table due to the system of boxes. [Take $g=10m/s^2]$

• A. 80 Pa
• B. 100 Pa
• C. 120 Pa
• D. 60 Pa

Answer 1

Answer: (B)

Step 1: Given data

Mass of the cubes(M₁)= 3 g

Mass of the cubes(M₂)= 6 g

Sides of cube(S₁)= 2 cm = $2\times {10}^{-2}m$

Sides of cube(S₂)= 3 cm = $3\times {10}^{-2}m$

Step 2: Formula used

Pressure $=\frac{\text{Force}}{\text{Area}}$

Step 3: Finding the pressure

Mass of the two boxes = 6 g + 3 g = 9 g $=0.009\text{kg}$

$\therefore$ Weight of the two boxes exerted on the table(W) $=mg=0.009\times g$

$W=0.09N$

Pressure $=\frac{\text{Force}}{\text{Area}}$

Let 3g cube put over 6g cube and cube of 6g with 3cm side is in contact with table.

Then area which is in contact with table is A = side*side = ${(3\times {10}^{-2})}^2{m}^2$

Putting all the values
P $=\frac{0.09}{(3\times {10}^{-2}{)}^2}$

P $=\frac{0.09\times {10}^4}{9}$

P $=100Pa$

Hence, the pressure is 100 Pa.