A perpendicular force of 50 N acting on a surface generates a pressure of 250 pa. calculate the area of cross section of the surface on which pressure is acting .

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Solution

Given,
Pressure $P = 250 P a = 250 N m^{- 2}$
Force applied $F = 50 N$
As pressure $= \frac{F o r c e}{A r e a}$

$\Rightarrow$ Area $= \frac{F o r c e}{P r e s s u r e}$

$\Rightarrow$ Area $= \frac{50}{250} = \frac{1}{5} m^{2} = 0.2 m^{2}$ .

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A perpendicular force of 50 N acting on a surface generates a pressure of 250 pa. calculate the area of cross section of the surface on which pressure is acting .

Given,Pressure P=250 Pa =250 Nm−2Force applied F=50 NAs pressure =ForceArea ⇒ Area =ForcePressure ⇒ Area =50250=15m2=0.2m2.

Given,
Pressure $P = 250 P a = 250 N m^{- 2}$
Force applied $F = 50 N$
As pressure $= \frac{F o r c e}{A r e a}$

$\Rightarrow$ Area $= \frac{F o r c e}{P r e s s u r e}$

$\Rightarrow$ Area $= \frac{50}{250} = \frac{1}{5} m^{2} = 0.2 m^{2}$ .