Question

A house has one external window that directly faces a nearby tall building across a narrow alley; it can withstand gauge pressures less than 60 Pascal. Assuming the house is completely closed, what is the minimum speed a wind has to be coming through the alley to break the window? Explain how you came to this conclusion.

Answer 1

Let's consider window breaks at 60 Pa. i.e. 60 N/m²

As velocity & pressure have inverse relation i.e. PV=Const.

Also A₁V₁=A₂V₂ . . But as Neither the Surface area of the window nor the alley is given lets consider it as same.

Now coming to PV= Const.

Now window can withstand 60Pa external pressure. As room is closed so lets consider the internal Pressure as 1 atm. = 101325Pa.

As house is closedvelocity will be almost negligible. So lets consider it as 1 unit.

Now, P₁V₁ = P₂V₂.

Here, P₁ = 101325 Pa, V₁ = 1 unit

  P₂ = 60Pa , V₂ = ?

Hence, V₂ = (101325 x 1)/ 60 = 1688.75m/s