#### Question

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

#### Solution

(1) Weight of brick = Thrust = F = 2 kgf

Area of base = `20/100xx10/100=1/50m^2`

Pressure exerted P =`F/A =(2kgf)/(1/50)m^2 = 100kgf m^-2`

Area of base = area of top L × B

20 Cm × 10 = 200 cm^{2}

^{ `P= (Thrust)/A= (2kgf)/(200cm^2)`=`1/100 `= `0.01kgf cm^-2`}

^{(2)}

^{ }

^{Area of base = area of top }

^{= `5cm xx 10 cm = 50 cm^2`}

^{Pressure exerted = `P = F/A =( 2kgf)/ (50 cm^2)`= 0.04}

^{ P = `0.04 kgf cm^-2`}

^{(3) Weight of brickF = 2 kgfArea of base = L × B= 20 cm × 5 cm = 100 cm2}

^{}

^{Pressure exerted =`F/A = 2/100=0.02 kgf cm^-2`}

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Solution Figure Shows a Brick of Weight 2 Kgf and Dimensions 20 Cm X 10 Cm × 5 Cm Placed in Three Different Positions on the Ground. Find the Pressure Exerted by the Brick in Each Case. Concept: Pressure - Calculation of Pressure in Simple Cases.