We know, when force is constant, pressure ∞ 1/Area.Therefore, pressure is maximum when area is minimum and vice versa.
By calculating the area of three faces, we get,
⇒A₁ = (20×10)= 200 sq m.
⇒A₂ = (10×5) = 50 sq m.
⇒A₃ = (20×5) = 100 sq m.
∴ A₁ is maximum and A₂ is minimum.
∴ Minimum pressure P₁ = F/A₁ = 30/200 = 0.15 Nm⁻².
∴ Maximum pressure P₂ = F/A₂ = 30/50 = 0.6 Nm⁻². (Ans)
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⇒A₂ = (10×5) = 50 sq m.
⇒A₃ = (20×5) = 100 sq m.
∴A₁ is maximum and A₂ is minimum.
∴Minimum pressure P₁ = F/A₁ = 30/200 = 0.15 Nm⁻².
∴Maximum pressure P₂ = F/A₂ = 30/50 = 0.6 Nm⁻
We know, when force is constant, pressure ∞ 1/Area.Therefore, pressure is maximum when area is minimum and vice versa.
By calculating the area of three faces, we get,
⇒A₁ = (20×10)= 200 sq m.
⇒A₂ = (10×5) = 50 sq m.
⇒A₃ = (20×5) = 100 sq m.
∴ A₁ is maximum and A₂ is minimum.
∴ Minimum pressure P₁ = F/A₁ = 30/200 = 0.15 Nm⁻².
∴ Maximum pressure P₂ = F/A₂ = 30/50 = 0.6 Nm⁻². (Ans)