Question

A cubical box filled with water has a small hole located in one of the bottom corner. When the box is full and placed on a horizontal surface, complete opening of the hole results in a flow of water with a speed v₀, as shown in figure (1). When the box becomes half empty, it is tilted by 45° so that the hole is at the lowest point. Now the water will flow out with a speed of

• A. V₀
• B. V₀/2
• C. V₀/√2
• D. V₀/√[4]{2}

Answer 1

Answer: (D)

V₀/√[4]{2}

Answer 2

V_o = $\sqrt{2gh}$

V₂ = $\sqrt{2g\frac{h}{\sqrt{2}}} = \frac{V_{0}}{\sqrt[4]{2}}$