A rectangular block has amass of 60 kg and sides 2mx1mx3mx ij g=10m/s then what is the minimum pressure that is it can exert as it rest on a surface A.100pa
B.200pa
C.50pa
D.300pa

Explanation:

The largest face of this rectangular block is 2*3 = 6 square meters. You simply pick the two largest sides and multiply them to get the largest area.

Pressure is defined as the force divided by area. The larger the area gets, the smaller the pressure will be. This is because the pressure is spread out over more area to reduce the per square unit pressure amount.

The block is 60 kg and the acceleration of gravity is roughly 10 m/s^2

So the force pulling down on the block is roughly 60*10 = 600 newtons.

Now we can find the pressure

P = pressure = unknown

F = force = 600 N

A = area = 6 square m

P = F/A

P = (600 N)/(6 m^2)

P = 100 (N/m^2)

P = 100 pa

Where "pa" stands for the unit "pascals", which is one unit for pressure.

Note what happens when we make A as small as possible. That happens with the 1 by 2 side, so the area is A = 1*2 = 2, and we get

P = F/A = 600/2 = 300 pascals

Reducing A enlarges P. Consequently, minimizing A leads to maximizing P. So the max pressure occurs on the face that is 1 m by 2 m.