Question

a brick of weight 2 kgf and
dimensions 20 cm x 10 cm x 5 cm placed in
three different positions on the ground. Find the
pressure exerted by the brick in each case.​

Answer 1

Answer:  the pressure exerted when it is placed with dimension 5 cm will be max

Explanation:

P = Mass/ area as the area in the third case is the minimum the pressure will be more the minimum pressure will be exerted when placed with dim/. 20 cm

Answer 2

Answer:

We know that P =  F/A, here A=effective area  

In first case , P =  mg/A = 2 × 10 = 1000 Pa

In second case , P =  2 × 10/0.05 × 0.1 = 4000 Pa

In third case , P = 2 × 10/0.05 × 0.2 = 2000 Pa

Credit: Toppr - https://bit.ly/2X1Jabe