Question

A cubical block of dimension 50cm×30cm×10cm is kept on a floor. the weight of block is 50N.
(i)find the pressure exerted by the block On the floor when the area of contact is smallest?​

Answer 1

Explanation:

Area = 3cm×3cm=9cm

2

=9×10

−4

m

2

& Pressure exterted = 5Pa

Since Pressure =

Area

Force

Force = Pressure × Area

Force = 9×10

−4

×5N

= 45×10

−4

N.

Answer By

gentryamansharma

Answer 2

Answer:

Smallest contact area=30(10)=300cm^2÷10000

=3/100m^2

Pressure=F/A

=50/(3/100)

=5000/3 Pa

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